Introduction
In this paper, we study small values of the Riemann zeta–function and its negative moments. The questions we are interested in are how frequently small values occur, and how small they can be on a typical short interval, as the distance to the critical line tends to zero. Throughout the paper, t t t [ T , 2 T ] [T,2T] [ T , 2 T ] P \mathbb{P} P
Negative moments near the critical line
We first state our sharp bounds for negative moments. For k > 0 k>0 k > 0 α > 0 \alpha>0 α > 0
I − k ( α , T ) : = 1 T ∫ T 2 T ∣ ζ ( 1 2 + α + i t ) ∣ − 2 k d t . I_{-k}(\alpha,T):=\frac{1}{T}\int_T^{2T}\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-2k}\,dt. I − k ( α , T ) := T 1 ∫ T 2 T ζ ( 2 1 + α + i t ) − 2 k d t . Gonek [25 ] conjectured that for every fixed k > 0 k>0 k > 0 1 / log T ≤ α ≤ 1 1/\log T\leq\alpha\leq1 1/ log T ≤ α ≤ 1
I − k ( α , T ) ≍ k α − k 2 . I_{-k}(\alpha,T)\asymp_k\alpha^{-k^2}. I − k ( α , T ) ≍ k α − k 2 . For smaller shifts, his conjecture predicts that for 0 < α ≤ 1 / log T 0<\alpha\leq1/\log T 0 < α ≤ 1/ log T
I − k ( α , T ) ≍ k { ( log T ) k 2 0 < k < 1 / 2 , ( log T ) 1 / 4 log e α log T k = 1 / 2. I_{-k}(\alpha,T)\asymp_k
\begin{cases}
(\log T)^{k^2} & 0<k<1/2,\\
(\log T)^{1/4}\log\frac{e}{\alpha\log T} & k=1/2.
\end{cases} I − k ( α , T ) ≍ k { ( log T ) k 2 ( log T ) 1/4 log α l o g T e 0 < k < 1/2 , k = 1/2. Assuming the Riemann Hypothesis (RH), Gonek proved lower bounds of the conjectured order for all k > 0 k>0 k > 0 1 / log T ≤ α ≤ 1 1/\log T\leq\alpha\leq1 1/ log T ≤ α ≤ 1 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 0 < α ≤ 1 / log T 0<\alpha\leq1/\log T 0 < α ≤ 1/ log T [14 ] proved non-trivial upper bounds for certain ranges of shifts, and asymptotic formulas when the shift is sufficiently large. More precisely, they obtained asymptotic formulas under RH for every fixed k , ε , C > 0 k,\varepsilon,C>0 k , ε , C > 0
α ≥ max { C ( log log T ) 4 / k + ε ( log T ) 1 / ( 2 k ) , ( 1 + ε ) log log T 2 log T } . \alpha\geq\max\left\{\frac{C(\log\log T)^{4/k+\varepsilon}}{(\log T)^{1/(2k)}},\frac{(1+\varepsilon)\log\log T}{2\log T}\right\}. α ≥ max { ( log T ) 1/ ( 2 k ) C ( log log T ) 4/ k + ε , 2 log T ( 1 + ε ) log log T } . Their results also give upper bounds for smaller shifts, but these do not have the conjectured order throughout that range.
In this paper, we improve the upper bounds for small shifts. In particular, for 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 1 / log T 1/\log T 1/ log T
Theorem 1.1. Assume RH. For every 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 0 < c < 1 0<c<1 0 < c < 1
I − k ( α , T ) ≪ { α − k 2 if 1 ≪ α log T ≪ log log T , ( log T ) O ( 1 ) if ( log log log T ) − c ≪ α log T ≪ 1. I_{-k}(\alpha,T)\ll
\begin{cases}
\alpha^{-k^2} & \text{if }1\ll\alpha\log T\ll\log\log T,\\
(\log T)^{O(1)} & \text{if }(\log\log\log T)^{-c}\ll\alpha\log T\ll1.
\end{cases} I − k ( α , T ) ≪ { α − k 2 ( log T ) O ( 1 ) if 1 ≪ α log T ≪ log log T , if ( log log log T ) − c ≪ α log T ≪ 1. We note that the improvement includes the range 1 ≪ α log T = o ( log log T ) 1\ll\alpha\log T=o(\log\log T) 1 ≪ α log T = o ( log log T ) [14 ] . Combining our result with the larger-shift results in [14 ] gives the conjectured upper bound for all α = o ( 1 ) \alpha=o(1) α = o ( 1 ) α ≫ 1 / log T \alpha\gg1/\log T α ≫ 1/ log T 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 [25 ] shows that when 1 / log T ≪ α = o ( 1 ) 1/\log T\ll\alpha=o(1) 1/ log T ≪ α = o ( 1 ) 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2
I − k ( α , T ) ≍ α − k 2 . I_{-k}(\alpha,T)\asymp\alpha^{-k^2}. I − k ( α , T ) ≍ α − k 2 . To the left of 1 / log T 1/\log T 1/ log T log T \log T log T exp ( O ( log log T ( log log log T ) 2 ) ) \exp(O(\log\log T(\log\log\log T)^2)) exp ( O ( log log T ( log log log T ) 2 )) [14 ] (see [14 (1.8)] ).
The next theorem bounds the moment I − 1 / 2 ( α , T ) I_{-1/2}(\alpha,T) I − 1/2 ( α , T ) [14 ] .
Theorem 1.2. Assume RH. For every 0 < ε < 1 0<\varepsilon<1 0 < ε < 1 0 < c < 1 0<c<1 0 < c < 1 C = C ( c ) > 0 C=C(c)>0 C = C ( c ) > 0 C 1 = C 1 ( ε ) > 0 C_1=C_1(\varepsilon)>0 C 1 = C 1 ( ε ) > 0
I − 1 / 2 ( α , T ) ≪ { exp ( C log log T ( log log log T log log log log T ) 2 ) if ( log log log T ) − c ≪ α log T ≪ 1 , exp ( ( log T ) ε ) if C 1 ( log log T ) − 1 ≤ α log T ≪ ( log log log T ) − c . (1) I_{-1/2}(\alpha,T)\ll
\begin{cases}
\exp\left(C\log\log T\left(\frac{\log\log\log T}{\log\log\log\log T}\right)^2\right) & \text{if }(\log\log\log T)^{-c}\ll\alpha\log T\ll1,\\
\exp\left((\log T)^\varepsilon\right) & \text{if }C_1(\log\log T)^{-1}\leq\alpha\log T\ll(\log\log\log T)^{-c}.
\end{cases} \tag*{(1)} I − 1/2 ( α , T ) ≪ ⎩ ⎨ ⎧ exp ( C log log T ( l o g l o g l o g l o g T l o g l o g l o g T ) 2 ) exp ( ( log T ) ε ) if ( log log log T ) − c ≪ α log T ≪ 1 , if C 1 ( log log T ) − 1 ≤ α log T ≪ ( log log log T ) − c . (1) Note that by comparison, Theorem 1.2 in [14 ] gives in the range (1.3)
I − 1 / 2 ( α , T ) ≪ exp ( ( 4 + ε ) log T log log log T log log T ) , I_{-1/2}(\alpha,T)\ll\exp\left(\frac{(4+\varepsilon)\log T\log\log\log T}{\log\log T}\right), I − 1/2 ( α , T ) ≪ exp ( log log T ( 4 + ε ) log T log log log T ) , and in the range (1.4) it gives exp ( ε log T ) \exp(\varepsilon\log T) exp ( ε log T )
Finally, when k > 1 / 2 k>1/2 k > 1/2
Theorem 1.3. Assume RH. For every k > 1 / 2 k>1/2 k > 1/2 C = C ( k ) > 0 C=C(k)>0 C = C ( k ) > 0
α = o ( 1 ) , α ≥ C ( log T ) 1 / ( 2 k ) log log T , \alpha=o(1),\qquad\alpha\geq\frac{C}{(\log T)^{1/(2k)}\log\log T}, α = o ( 1 ) , α ≥ ( log T ) 1/ ( 2 k ) log log T C , we have
I − k ( α , T ) ≪ α − k 2 . I_{-k}(\alpha,T)\ll\alpha^{-k^2}. I − k ( α , T ) ≪ α − k 2 . We note that one can study the corresponding negative moment problem in random matrix theory as well. The moment calculations of Keating and Snaith [31 ] give order N k 2 N^{k^2} N k 2 2 k 2k 2 k 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 [21 ] and Strahov and Fyodorov [41 ] . Berry and Keating [13 ] predicted that high negative moments are governed by rare clusters of eigenvalues. Related calculations were carried out by Fyodorov and Keating [23 ] and Forrester and Keating [20 ] . For unitary families, these predictions lead to transitions in the asymptotic formulas at k = 1 / 2 , 3 / 2 , … k=1/2,3/2,\ldots k = 1/2 , 3/2 , … α ≤ 1 / log T \alpha\leq1/\log T α ≤ 1/ log T k ≥ 3 / 2 k\geq3/2 k ≥ 3/2 [14 Section 1] ). For integral k k k [19 ] , Conrey, Farmer, and Zirnbauer [16 ] , and Conrey and Snaith [17 ] for conjectures and applications in this direction.
Averages of the Möbius function
The main application of our negative moment bounds lies in obtaining cancellation in averages of the Möbius function. Assuming RH, Soundararajan [40 ] proved that
∑ n ≤ x μ ( n ) ≪ x exp ( log x ( log log x ) 14 ) . (2) \sum_{n\leq x}\mu(n)\ll\sqrt{x}\exp\left(\sqrt{\log x}(\log\log x)^{14}\right). \tag*{(2)} n ≤ x ∑ μ ( n ) ≪ x exp ( log x ( log log x ) 14 ) . (2) Balazard and de Roton [12 ] optimized his argument, replacing the exponent 14 14 14 5 / 2 + ε 5/2+\varepsilon 5/2 + ε [12 ] as the limit of this method. For earlier bounds on ∑ n ≤ x μ ( n ) \sum_{n\leq x}\mu(n) ∑ n ≤ x μ ( n ) [33 ] , Landau [32 ] , Titchmarsh [42 ] , and Maier and Montgomery [34 ] . Bui and Florea [14 ] used negative-moment estimates for ζ ( s ) \zeta(s) ζ ( s ) μ k \mu_k μ k ζ ( s ) − k \zeta(s)^{-k} ζ ( s ) − k k k k k k k
Following the method in [14 ] , we use Theorems 1.2 and 1.3 to control the mean of ∣ 1 / ζ ( s ) k ∣ |1/\zeta(s)^k| ∣1/ ζ ( s ) k ∣ log log x \log\log x log log x
Theorem 1.4. Assume RH. For every k ∈ N k \in\mathbb{N} k ∈ N C = C ( k ) > 0 C=C(k)>0 C = C ( k ) > 0
∑ n ≤ x μ k ( n ) ≪ x exp ( C ( log x log log x ) k / ( k + 1 ) ) . \sum_{n\le x}\mu_k(n)\ll\sqrt{x}\exp\left(C\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right). n ≤ x ∑ μ k ( n ) ≪ x exp ( C ( log log x log x ) k / ( k + 1 ) ) . By comparison, Bui and Florea [14 Theorem 1.5] established the bound
x exp ( ( log x ) k / ( k + 1 ) ( log log x ) 7 / ( k + 1 ) + ε ) \sqrt{x}\exp\left((\log x)^{k/(k+1)}(\log\log x)^{7/(k+1)+\varepsilon}\right) x exp ( ( log x ) k / ( k + 1 ) ( log log x ) 7/ ( k + 1 ) + ε ) for every k ∈ N k \in\mathbb{N} k ∈ N k k k Theorem 8.1 for a more general statement).
The minimum on short intervals
We next consider the minimum on a typical interval of bounded length. On the critical line, any interval containing a zero has minimum zero. We therefore shift a small distance α > 0 \alpha>0 α > 0
M α ( t ) : = min ∣ h ∣ ≤ 1 ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ . M_\alpha(t):=\min_{|h|\le1}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|. M α ( t ) := ∣ h ∣ ≤ 1 min ζ ( 2 1 + α + i ( t + h ) ) . We are interested in small shifts α → 0 \alpha\to0 α → 0 M α ( t ) M_\alpha(t) M α ( t ) α \alpha α
We determine the leading logarithmic order of this minimum in the following range.
Theorem 1.5. Assume RH. Fix 0 < c < 1 0<c<1 0 < c < 1
0 < α ≪ ( log T ) − c , α log T log log T ⟶ ∞ . 0<\alpha\ll(\log T)^{-c},\qquad\frac{\alpha\log T}{\log\log T}\longrightarrow\infty. 0 < α ≪ ( log T ) − c , log log T α log T ⟶ ∞. For every ε , δ > 0 \varepsilon,\delta>0 ε , δ > 0 T T T
P ( α 1 + δ ≤ M α ( t ) ≤ α 1 − δ ) ≥ 1 − ε . \mathbb{P}\left(\alpha^{1+\delta}\le M_\alpha(t)\le\alpha^{1-\delta}\right)\ge1-\varepsilon. P ( α 1 + δ ≤ M α ( t ) ≤ α 1 − δ ) ≥ 1 − ε . Note that in Theorem 9.2 we prove a lower bound result with the expected leading order term for the minimum in intervals of length ( log T ) γ (\log T)^\gamma ( log T ) γ y y y Theorem 1.5 follows by combining Theorem 9.5 and Theorem 9.7, and corresponds to the case γ = 0 \gamma=0 γ = 0 Theorem 9.2 .
Theorem 1.5 gives the leading order predicted by Conjecture 1.6 below. We describe the connection with the maximum problem and the heuristic behind this prediction.
The extreme values of the Riemann zeta-function in short intervals have attracted considerable attention, in part because of their connection with random matrix theory and logarithmically correlated random fields. Fyodorov, Hiary, and Keating [22 ] , and Fyodorov and Keating [24 ] predicted the size and fluctuations of the maximum on a typical interval of bounded length on the critical line. Their prediction arose from modelling the zeta-function by characteristic polynomials of random unitary matrices and applying the freezing-transition heuristic from statistical mechanics [22 , 24 ] . They predicted that for every ε > 0 \varepsilon>0 ε > 0 C = C ( ε ) > 0 C=C(\varepsilon)>0 C = C ( ε ) > 0 T T T
P ( ∣ max ∣ h ∣ ≤ 1 log ∣ ζ ( 1 2 + i ( t + h ) ) ∣ − log log T + 3 4 log log log T ∣ ≤ C ) ≥ 1 − ε . \mathbb{P}\left(\left|\max_{|h|\le1}\log\left|\zeta\left(\frac{1}{2}+i(t+h)\right)\right|-\log\log T+\frac{3}{4}\log\log\log T\right|\le C\right)\ge1-\varepsilon. P ( ∣ h ∣ ≤ 1 max log ζ ( 2 1 + i ( t + h ) ) − log log T + 4 3 log log log T ≤ C ) ≥ 1 − ε . Najnudel [36 ] proved the first-order asymptotic under RH, and Arguin, Belius, Bourgade, Radziwiłł, and Soundararajan [5 ] proved it unconditionally. Namely, in probability, they showed that
max ∣ h ∣ ≤ 1 log ∣ ζ ( 1 2 + i ( t + h ) ) ∣ = ( 1 + o ( 1 ) ) log log T . \max_{|h|\le1}\log\left|\zeta\left(\frac{1}{2}+i(t+h)\right)\right|=(1+o(1))\log\log T. ∣ h ∣ ≤ 1 max log ζ ( 2 1 + i ( t + h ) ) = ( 1 + o ( 1 )) log log T . Najnudel’s first-order upper bound is itself unconditional; RH is used for the lower bound. Harper [28 ] obtained an upper bound with the correct second-order term; subsequent work of Arguin, Bourgade, and Radziwiłł [6 ] improved the error term and established the predicted upper-tail bound. A matching lower bound was subsequently established by the same authors [7 ] . For an account of the earlier developments, see Harper’s Bourbaki survey [29].
The Fyodorov–Hiary–Keating (FHK) Conjecture has extensions to intervals whose length is a power of log T \log T log T − 1 < γ < 0 -1 < \gamma< 0 − 1 < γ < 0 [24 ] formulated a mesoscopic version of the conjecture. Arguin and Hamdan [10 ] proved the sharp upper bound, including the second-order coefficient 3 / 4 3/4 3/4 [11 ] proposed the following extension: for every fixed γ > 0 \gamma> 0 γ > 0 t t t [ T , 2 T ] [T,2T] [ T , 2 T ]
max ∣ h ∣ ≤ ( log T ) γ log ∣ ζ ( 1 2 + i ( t + h ) ) ∣ = 1 + γ log log T − 1 4 1 + γ log log log T + O P ( 1 ) , \max_{|h|\leq(\log T)^\gamma}\log\left|\zeta\left(\frac{1}{2}+i(t+h)\right)\right|=\sqrt{1+\gamma}\log\log T-\frac{1}{4\sqrt{1+\gamma}}\log\log\log T+O_{\mathbb{P}}(1), ∣ h ∣ ≤ ( l o g T ) γ max log ζ ( 2 1 + i ( t + h ) ) = 1 + γ log log T − 4 1 + γ 1 log log log T + O P ( 1 ) , where the error is bounded in probability. They established the leading-order asymptotic unconditionally for 0 < γ ≤ 3 0 < \gamma\leq3 0 < γ ≤ 3 γ > 3 \gamma> 3 γ > 3 [1 ] subsequently proved the corresponding upper bound with the conjectured second-order term and an error bounded in probability for 0 < γ < 3 0 < \gamma< 3 0 < γ < 3 1 / 4 1/4 1/4 γ ↓ 0 \gamma\downarrow0 γ ↓ 0 3 / 4 3/4 3/4
The random-matrix analogue of the FHK Conjecture concerns the characteristic polynomial of an N × N N \times N N × N U N U_N U N [22 ] predicted that
max 0 ≤ θ < 2 π log ∣ det ( I − e − i θ U N ) ∣ = log N − 3 4 log log N + O P ( 1 ) , \max_{0\leq\theta<2\pi}\log\left|\det(I-e^{-i\theta}U_N)\right|=\log N-\frac{3}{4}\log\log N+O_{\mathbb{P}}(1), 0 ≤ θ < 2 π max log det ( I − e − i θ U N ) = log N − 4 3 log log N + O P ( 1 ) , corresponding to the zeta-function prediction under the identification N ≍ log T N \asymp\log T N ≍ log T [4 ] established the leading order, and Paquette and Zeitouni [37 ] obtained the second-order term.
To formulate an FHK-type conjecture for the minimum, we revisit the original conjecture of Fyodorov, Hiary and Keating. Their conjecture can be motivated by approximating log ∣ ζ ∣ \log|\zeta| log ∣ ζ ∣ [26 ] . For t t t [ T , 2 T ] [T,2T] [ T , 2 T ] p − i t p^{-it} p − i t X ( p ) X(p) X ( p )
log ∣ ζ ( 1 2 + i ( t + h ) ) ∣ ≈ X 0 ( h ) : = ℜ ∑ p ≤ T X ( p ) p 1 / 2 + i h . \log\left|\zeta\left(\frac{1}{2}+i(t+h)\right)\right|\approx X_0(h):=\Re\sum_{p\leq T}\frac{X(p)}{p^{1/2+ih}}. log ζ ( 2 1 + i ( t + h ) ) ≈ X 0 ( h ) := ℜ p ≤ T ∑ p 1/2 + ih X ( p ) . Its covariance satisfies
Cov ( X 0 ( h ) , X 0 ( h ′ ) ) = 1 2 ∑ p ≤ T cos ( ( h − h ′ ) log p ) p = 1 2 log 1 ∣ h − h ′ ∣ + 1 / log T + O ( 1 ) . \operatorname{Cov}(X_0(h),X_0(h'))=\frac{1}{2}\sum_{p\leq T}\frac{\cos((h-h')\log p)}{p}=\frac{1}{2}\log\frac{1}{|h-h'|+1/\log T}+O(1). Cov ( X 0 ( h ) , X 0 ( h ′ )) = 2 1 p ≤ T ∑ p cos (( h − h ′ ) log p ) = 2 1 log ∣ h − h ′ ∣ + 1/ log T 1 + O ( 1 ) . In particular, the variance is 1 2 log log T + O ( 1 ) \frac{1}{2}\log\log T+O(1) 2 1 log log T + O ( 1 ) 1 / log T 1/\log T 1/ log T 1 1 1 log log p \log\log p log log p
A similar heuristic applies to the minimum of zeta to the right of the critical line. We write
W α ( t ) : = log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 . W_\alpha(t):=\log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}. W α ( t ) := log ζ ( 2 1 + α + i t ) − 1 . Suppose that α = o ( 1 ) \alpha=o(1) α = o ( 1 ) α log T ≫ 1 \alpha\log T\gg1 α log T ≫ 1
W α ( t + h ) = log ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 1 ≈ − X α ( h ) : = − ℜ ∑ p ≤ T X ( p ) p 1 / 2 + α + i h . W_\alpha(t+h)=\log\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-1}\approx-X_\alpha(h):=-\Re\sum_{p\leq T}\frac{X(p)}{p^{1/2+\alpha+ih}}. W α ( t + h ) = log ζ ( 2 1 + α + i ( t + h ) ) − 1 ≈ − X α ( h ) := − ℜ p ≤ T ∑ p 1/2 + α + ih X ( p ) . Hence
Var ( X α ( h ) ) = 1 2 ∑ p ≤ T 1 p 1 + 2 α = 1 2 log 1 α + O ( 1 ) . \operatorname{Var}(X_\alpha(h))=\frac{1}{2}\sum_{p\leq T}\frac{1}{p^{1+2\alpha}}=\frac{1}{2}\log\frac{1}{\alpha}+O(1). Var ( X α ( h )) = 2 1 p ≤ T ∑ p 1 + 2 α 1 = 2 1 log α 1 + O ( 1 ) . and
Cov ( X α ( h ) , X α ( h ′ ) ) = 1 2 log 1 α + ∣ h − h ′ ∣ + O ( 1 ) . \operatorname{Cov}(X_{\alpha}(h),X_{\alpha}(h'))=\frac{1}{2}\log\frac{1}{\alpha+\lvert h-h'\rvert}+O(1). Cov ( X α ( h ) , X α ( h ′ )) = 2 1 log α + ∣ h − h ′ ∣ 1 + O ( 1 ) . Thus the model retains the same logarithmic correlation structure, with α \alpha α 1 / log T 1/\log T 1/ log T
Since the joint distribution of the X ( p ) X(p) X ( p ) X ( p ) X(p) X ( p ) − X ( p ) -X(p) − X ( p ) X α X_{\alpha} X α − X α -X_{\alpha} − X α log T \log T log T 1 / α 1/\alpha 1/ α
Conjecture 1.6. Suppose that
α = o ( 1 ) , α log T ≫ 1. \alpha=o(1),\qquad\alpha\log T\gg1. α = o ( 1 ) , α log T ≫ 1. For every ε > 0 \varepsilon>0 ε > 0 C = C ( ε ) > 0 C=C(\varepsilon)>0 C = C ( ε ) > 0 T T T
P ( ∣ log M α ( t ) + log 1 α − 3 4 log log 1 α ∣ ≤ C ) ≥ 1 − ε . \mathbb{P}\left(\left\lvert\log M_{\alpha}(t)+\log\frac{1}{\alpha}-\frac{3}{4}\log\log\frac{1}{\alpha}\right\rvert\leq C\right)\geq1-\varepsilon. P ( log M α ( t ) + log α 1 − 4 3 log log α 1 ≤ C ) ≥ 1 − ε . In particular, the conjectured scale of the minimum is α ( log 1 α ) 3 / 4 \alpha(\log\frac{1}{\alpha})^{3/4} α ( log α 1 ) 3/4
Large deviations
The main analytic input for the negative-moment estimates is a precise upper bound for the probability that − log ∣ ζ ∣ -\log\lvert\zeta\rvert − log ∣ ζ ∣ [38 ] describes log ∣ ζ ( 1 / 2 + i t ) ∣ \log\lvert\zeta(1/2+it)\rvert log ∣ ζ ( 1/2 + i t )∣ log log T \sqrt{\log\log T} log log T [43 ] .) Moments, however, depend on much rarer values of size comparable to log log T \log\log T log log T [39 ] used large-deviation estimates to obtain, under RH, positive-moment upper bounds with exponent k 2 / 2 + ε k^2/2+\varepsilon k 2 /2 + ε log T \log T log T [27 ] removed this ε \varepsilon ε [1 ] proved, unconditionally, that for every fixed 0 < κ < 2 0<\kappa<2 0 < κ < 2 V = κ log log T V=\kappa\log\log T V = κ log log T
P ( log ∣ ζ ( 1 2 + i t ) ∣ > V ) ≪ 1 log log T exp ( − V 2 log log T ) . \mathbb{P}\left(\log\left\lvert\zeta\left(\frac{1}{2}+it\right)\right\rvert>V\right)\ll\frac{1}{\sqrt{\log\log T}}\exp\left(-\frac{V^2}{\log\log T}\right). P ( log ζ ( 2 1 + i t ) > V ) ≪ log log T 1 exp ( − log log T V 2 ) . Their estimate recovers, for 0 < κ < 2 0<\kappa<2 0 < κ < 2 [30 ] . More recently, Arguin, Bailey, and Roberts [3 ] established the same large-deviation upper bound under RH for every fixed κ > 0 \kappa>0 κ > 0 [6 ] with the methods of Soundararajan [39 ] and Harper [27 ] .
Lower bounds for large deviations have also been studied. Arguin and Bailey [2 ] obtained an unconditional lower bound of the conjectured Gaussian order at 1 / 2 + c / log T 1/2+c/\log T 1/2 + c / log T c c c κ \kappa κ [9 ] proved the corresponding unconditional lower bound on the critical line, for every fixed κ > 0 \kappa>0 κ > 0 V ∼ κ log log T V\sim\kappa\log\log T V ∼ κ log log T L L L [8 ] for Dirichlet L L L [18 ] for quadratic twists of elliptic curves.
Here we study the corresponding large deviations of − log ∣ ζ ( 1 / 2 + α + i t ) ∣ -\log\lvert\zeta(1/2+\alpha+it)\rvert − log ∣ ζ ( 1/2 + α + i t )∣ 1 2 log 1 α \frac{1}{2}\log\frac{1}{\alpha} 2 1 log α 1 V = κ log 1 α V=\kappa\log\frac{1}{\alpha} V = κ log α 1 α = c / log T \alpha=c/\log T α = c / log T c > 0 c>0 c > 0 0 < κ < 1 / 2 0<\kappa<1/2 0 < κ < 1/2 κ > 0 \kappa>0 κ > 0
Theorem 1.7. Assume RH. Fix 0 < κ 0 < κ 1 < 1 / 2 0<\kappa_0<\kappa_1<1/2 0 < κ 0 < κ 1 < 1/2
α = o ( 1 ) , α log T ≫ 1. \alpha=o(1),\qquad\alpha\log T\gg1. α = o ( 1 ) , α log T ≫ 1. Then, uniformly for κ 0 ≤ κ ≤ κ 1 \kappa_0 \le\kappa\le\kappa_1 κ 0 ≤ κ ≤ κ 1 V = κ log 1 α V = \kappa\log\frac{1}{\alpha} V = κ log α 1
P ( W α ( t ) > V ) ≪ 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) . \mathbb{P}(W_\alpha(t)>V)\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right). P ( W α ( t ) > V ) ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) V 2 ) . Moreover, for every ε > 0 \varepsilon>0 ε > 0 0 < κ 0 < κ 1 < ∞ 0<\kappa_0<\kappa_1<\infty 0 < κ 0 < κ 1 < ∞ κ 0 ≤ κ ≤ κ 1 \kappa_0\le\kappa\le\kappa_1 κ 0 ≤ κ ≤ κ 1
α log T ≫ ( log 1 α ) 12 + ε . \alpha\log T\gg\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. α log T ≫ ( log α 1 ) 12 + ε . The proof separates larger shifts from smaller shifts, which require a recursive argument. For larger deviation levels at small shifts, we obtain the following weaker estimate.
Theorem 1.8. Assume RH. Fix ε , ε 1 > 0 \varepsilon,\varepsilon_1>0 ε , ε 1 > 0 κ 1 ≥ 1 / 2 \kappa_1\ge1/2 κ 1 ≥ 1/2
1 ≪ α log T ≪ ( log 1 α ) 12 + ε . 1\ll\alpha\log T\ll\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. 1 ≪ α log T ≪ ( log α 1 ) 12 + ε . Then, uniformly for 1 / 2 ≤ κ ≤ κ 1 1/2\le\kappa\le\kappa_1 1/2 ≤ κ ≤ κ 1 V = κ log 1 α V=\kappa\log\frac{1}{\alpha} V = κ log α 1
P ( W α ( t ) > V ) ≪ exp ( − V + ( 1 4 + ε 1 ) log 1 α ) = α κ − 1 / 4 − ε 1 . \mathbb{P}(W_\alpha(t)>V)\ll\exp\left(-V+\left(\frac{1}{4}+\varepsilon_1\right)\log\frac{1}{\alpha}\right)=\alpha^{\kappa-1/4-\varepsilon_1}. P ( W α ( t ) > V ) ≪ exp ( − V + ( 4 1 + ε 1 ) log α 1 ) = α κ − 1/4 − ε 1 . Further applications
The lower bound for the minimum in Theorem 1.5 (or, to be more precise, Theorem 9.2) also gives bounds for negative moments on almost all short intervals. The following result is the analogue of Corollary 1.4 in [1 ] and Corollary 3.2 in [3 ] . Here k γ k_\gamma k γ
Theorem 1.9. Assume RH. Fix k > 0 k>0 k > 0 ε > 0 \varepsilon>0 ε > 0 0 < c < 1 0<c<1 0 < c < 1 M > 0 M>0 M > 0 γ ≥ 0 \gamma\ge0 γ ≥ 0 H = ( log T ) γ H=(\log T)^\gamma H = ( log T ) γ
0 < α ≪ ( log T ) − c and α log T ≫ ( log 1 α ) 12 + ε . 0<\alpha\ll(\log T)^{-c}\qquad\text{and}\qquad\alpha\log T\gg\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. 0 < α ≪ ( log T ) − c and α log T ≫ ( log α 1 ) 12 + ε . Then, uniformly for
1 ≤ A ≪ α − M , 1\le A\ll\alpha^{-M}, 1 ≤ A ≪ α − M , we have
1 2 H ∫ ∣ h ∣ ≤ H ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 2 k d h ≪ A α − k 2 , \frac{1}{2H}\int_{|h|\le H}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-2k}\,dh\ll A\alpha^{-k^2}, 2 H 1 ∫ ∣ h ∣ ≤ H ζ ( 2 1 + α + i ( t + h ) ) − 2 k d h ≪ A α − k 2 , for all t ∈ [ T , 2 T ] t\in[T,2T] t ∈ [ T , 2 T ] ≪ T / A \ll T/A ≪ T / A
For k > k γ k>k_\gamma k > k γ
1 2 H ∫ ∣ h ∣ ≤ H ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 2 k d h ≪ A k / k γ α − ( 2 k k γ − k γ 2 ) ( k − k γ ) log ( 1 / α ) ( log 1 α ) C ( k − k γ ) , (3) \frac{1}{2H}\int_{|h|\le H}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-2k}\,dh\ll\frac{A^{k/k_\gamma}\alpha^{-(2kk_\gamma-k_\gamma^2)}}{(k-k_\gamma)\sqrt{\log(1/\alpha)}}\left(\log\frac{1}{\alpha}\right)^{C(k-k_\gamma)}, \tag*{(3)} 2 H 1 ∫ ∣ h ∣ ≤ H ζ ( 2 1 + α + i ( t + h ) ) − 2 k d h ≪ ( k − k γ ) log ( 1/ α ) A k / k γ α − ( 2 k k γ − k γ 2 ) ( log α 1 ) C ( k − k γ ) , (3) for all t ∈ [ T , 2 T ] t\in[T,2T] t ∈ [ T , 2 T ] ≪ T / A \ll T/A ≪ T / A
For positive moments on the critical line in typical intervals, Arguin, Ouimet, and Radziwiłł [11 Theorem 1.1] determined the leading powers of log T \log T log T k = 1 + γ k=\sqrt{1+\gamma} k = 1 + γ γ ≥ 0 \gamma\ge0 γ ≥ 0 0 ≤ γ ≤ 3 0\le\gamma\le3 0 ≤ γ ≤ 3 γ > 3 \gamma>3 γ > 3 [1 Corollary 1.4] obtained sharper upper bounds for 0 ≤ γ < 3 0\le\gamma<3 0 ≤ γ < 3 [3 Corollary 3.2] extended these bounds to every fixed γ ≥ 0 \gamma\ge0 γ ≥ 0 k > 1 + γ k>\sqrt{1+\gamma} k > 1 + γ A > 1 A>1 A > 1
1 ( log T ) γ ∫ ∣ h ∣ ≤ ( log T ) γ ∣ ζ ( 1 2 + i ( t + h ) ) ∣ 2 k d h ≪ ( log T ) 2 k 1 + γ − ( 1 + γ ) ( log log T ) k / ( 2 1 + γ ) , \frac{1}{(\log T)^\gamma}\int_{|h|\le(\log T)^\gamma}\left|\zeta\left(\frac{1}{2}+i(t+h)\right)\right|^{2k}\,dh\ll\frac{(\log T)^{2k\sqrt{1+\gamma}-(1+\gamma)}}{(\log\log T)^{k/(2\sqrt{1+\gamma})}}, ( log T ) γ 1 ∫ ∣ h ∣ ≤ ( l o g T ) γ ζ ( 2 1 + i ( t + h ) ) 2 k d h ≪ ( log log T ) k / ( 2 1 + γ ) ( log T ) 2 k 1 + γ − ( 1 + γ ) , outside a set of measure ≪ T / A \ll T/A ≪ T / A k = 1 k=1 k = 1 [28 Corollary 1] proved an upper bound of order log T / log log T \log T/\sqrt{\log\log T} log T / log log T
We note that for our result, when k > k γ k > k_{\gamma} k > k γ 2 k k γ − k γ 2 2k k_{\gamma} - k_{\gamma}^{2} 2 k k γ − k γ 2 k k k k 2 k^{2} k 2 Theorem 9.2 . The extra exponential factor in the bound (1.6) comes from the error of size ( log 1 α ) C (\log\frac{1}{\alpha})^{C} ( log α 1 ) C Theorem 9.2 ). In the positive-moment problem, Arguin and Bailey [1, Corollary 1.3] have a more precise bound for the maximum, which allows them to obtain the logarithmic factor in their moment estimate. Our bound therefore has the analogous leading power, but is less precise in its lower-order terms.
Ideas of proof
The main analytic issue is to estimate small values close enough to the critical line that the contribution of zeros begins to matter. Large deviations give the sharp negative-moment bounds, while separate estimates for more extreme values provide the bound needed for the Möbius application.
To obtain the large-deviation estimates, we build on the method of Arguin, Bailey, and Roberts [3 ] . The main additional difficulty is that the contribution of zeros becomes increasingly important as the shift approaches 1 / log T 1/\log T 1/ log T [15 ] , in the form used by Bui and Florea [14 ] , bound − log ∣ ζ ∣ -\log|\zeta| − log ∣ ζ ∣ α \alpha α
For larger shifts (i.e., larger than ( log log T ) O ( 1 ) / log T (\log\log T)^{O(1)}/\log T ( log log T ) O ( 1 ) / log T 1 / log T ≪ α ≪ ( log log T ) O ( 1 ) / log T 1/\log T \ll\alpha\ll(\log\log T)^{O(1)}/\log T 1/ log T ≪ α ≪ ( log log T ) O ( 1 ) / log T [3 ] . We estimate separately the contribution from points where all partial sums remain within the prescribed barriers and from points where one of these bounds first fails. We keep the dependence on the shift precise enough to recover the full Gaussian upper bound, including the factor 1 / log ( 1 / α ) 1/\sqrt{\log(1/\alpha)} 1/ log ( 1/ α ) Theorem 1.7 . In the smaller-shift regime, the same argument gives a weaker large-deviation bound when κ ≥ 1 / 2 \kappa\ge1/2 κ ≥ 1/2 Theorem 1.8 .
We then bound negative moments by integrating the deviation probabilities against 2 k e 2 k V 2k e^{2kV} 2 k e 2 k V k > 0 k > 0 k > 0 O ( α − k 2 ) O(\alpha^{-k^{2}}) O ( α − k 2 ) k k k V = k log 1 α V = k\log\frac{1}{\alpha} V = k log α 1 Theorem 1.7 controls a neighbourhood of this level for 0 < k < 1 / 2 0 < k < 1/2 0 < k < 1/2 1 / log T 1/\log T 1/ log T k > 1 / 2 k > 1/2 k > 1/2 Theorem 1.3 . In either case, the Gaussian estimate alone does not control the full moment; we also need separate (non-Gaussian) bounds for larger deviation levels.
For 0 < k < 1 / 2 0 < k < 1/2 0 < k < 1/2 Theorem 1.8 beyond the Gaussian range. For all V ≥ B log log T V \ge B\log\log T V ≥ B log log T Lemma 7.1 bounds the deviation probability by O ( e − ρ V ) O(e^{-\rho V}) O ( e − ρ V ) 2 k < ρ < 1 2k < \rho< 1 2 k < ρ < 1 B B B 2 k e 2 k V 2k e^{2kV} 2 k e 2 k V o ( 1 ) o(1) o ( 1 )
For k > 1 / 2 k > 1/2 k > 1/2 e 2 k V e^{2kV} e 2 k V [39 ] and keeping track of the shift-dependent error in the explicit formula. In the range of Theorem 1.3 , these probabilities decay fast enough to make their integrals against e 2 k V e^{2kV} e 2 k V
The Möbius application requires a bound on the moment I − 1 / 2 ( α , T ) I_{-1/2}(\alpha,T) I − 1/2 ( α , T ) I − 1 / 2 I_{-1/2} I − 1/2 Theorem 1.2 is not of the conjectured order, it is sufficient for the arithmetic application. Following the contour argument of Bui and Florea [14 ] , we use Perron’s formula to express the average of the Möbius function in terms of 1 / ζ ( s ) 1/\zeta(s) 1/ ζ ( s ) I − 1 / 2 I_{-1/2} I − 1/2 Theorem 1.4 in the case k = 1 k = 1 k = 1 Theorem 1.3 , yields the estimates for the generalized Möbius function in the case k > 1 k > 1 k > 1
Finally, we turn to the minimum in short intervals, for which the two directions require different arguments. A lower bound for the minimum must rule out values that are too small anywhere in the interval. We therefore combine the explicit inequality for log ∣ ζ ∣ − 1 \log|\zeta|^{-1} log ∣ ζ ∣ − 1
An upper bound for the minimum instead requires the existence of at least one sufficiently small value. For the prime sums, we adapt the argument of Arguin, Belius, Bourgade, Radziwiłł, and Soundararajan [5 ] to produce a large negative value. We then show that the large negative value of the prime sum gives a correspondingly small value of ∣ ζ ∣ |\zeta| ∣ ζ ∣ α \alpha α [15 ] bound log ∣ ζ ∣ \log|\zeta| log ∣ ζ ∣ 1 / log T 1/\log T 1/ log T ℜ s = 1 / 2 + α \Re s = 1/2 + \alpha ℜ s = 1/2 + α
Finally, the upper bounds for the negative moments in almost all short intervals in Theorem 1.9 are an application of Theorem 9.2 .
Organization of the paper
In Section 2 we first gather a few preliminary results we will use throughout the paper. We prove Theorem 1.7 when the shift α \alpha α Theorem 1.7 in the case of small α \alpha α Theorems 1.7 and 1.8 . We finish the proofs of Theorems 1.7 and 1.8 in Section 6. In Section 7 we prove the upper bounds in Theorems 1.1 , 1.2 and 1.3 , and we prove Theorem 1.4 in Section 8. Theorem 1.5 is proved in Section 9, and finally, we prove Theorem 1.9 in Section 10.
Notation
We write P \mathbb{P} P E \mathbb{E} E t t t [ T , 2 T ] [T,2T] [ T , 2 T ]
Let ( X ( p ) ) p (X(p))_p ( X ( p ) ) p X X X
X ( 1 ) = 1 , X ( n ) = ∏ p r ∥ n X ( p ) r . X(1)=1,\qquad X(n)=\prod_{p^r\parallel n}X(p)^r. X ( 1 ) = 1 , X ( n ) = p r ∥ n ∏ X ( p ) r . We denote probability and expectation for these variables by P r a n d \mathbb{P}_{\mathrm{rand}} P rand E r a n d \mathbb{E}_{\mathrm{rand}} E rand P ( t ) = ∑ n ≤ N a n n − i t P(t)=\sum_{n\le N}a_n n^{-it} P ( t ) = ∑ n ≤ N a n n − i t
P ~ = ∑ n ≤ N a n X ( n ) . \widetilde{P}=\sum_{n\le N}a_nX(n). P = n ≤ N ∑ a n X ( n ) . Thus a tilde means that each p − i t p^{-it} p − i t X ( p ) X(p) X ( p ) p − 2 i t p^{-2it} p − 2 i t X ( p ) 2 X(p)^2 X ( p ) 2
The notation f ≪ g f\ll g f ≪ g ∣ f ∣ ≤ C g |f|\le Cg ∣ f ∣ ≤ C g C > 0 C>0 C > 0 f ≍ g f\asymp g f ≍ g f ≪ g f\ll g f ≪ g g ≪ f g\ll f g ≪ f f , g f,g f , g ε \varepsilon ε C C C
Throughout, ∑ u ∈ [ L , U ] \sum_{u\in[L,U]} ∑ u ∈ [ L , U ] u ∈ Z ∩ [ L , U + 1 ] u\in\mathbb{Z}\cap[L,U+1] u ∈ Z ∩ [ L , U + 1 ]
For ease of notation, we will write M α ( t ) = M α , 0 ( t ) M_{\alpha}(t)=M_{\alpha,0}(t) M α ( t ) = M α , 0 ( t )
M α , γ ( t ) : = min ∣ h ∣ ≤ ( log T ) γ ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ . M_{\alpha,\gamma}(t):=\min_{|h|\le(\log T)^\gamma}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|. M α , γ ( t ) := ∣ h ∣ ≤ ( l o g T ) γ min ζ ( 2 1 + α + i ( t + h ) ) . We also write
W α ( t ) = log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 . W_{\alpha}(t)=\log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}. W α ( t ) = log ζ ( 2 1 + α + i t ) − 1 . AI Disclosure: The authors used ChatGPT to assist with the proofs of Propositions 5.1 and 5.2. They also used ChatGPT to test the choice of parameters, and with checking and proofreading the mathematical arguments. Additionally, after finishing a draft of this paper, the authors asked Claude for a report on the paper; the report suggested a possible improvement in the log log x \log\log x log log x exp ( C ( log x ) k / ( k + 1 ) ) \exp\left(C(\log x)^{k/(k+1)}\right) exp ( C ( log x ) k / ( k + 1 ) ) exp ( C ( log x log log x ) k / ( k + 1 ) ) \exp\left(C\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right) exp ( C ( l o g l o g x l o g x ) k / ( k + 1 ) )
Acknowledgments: The authors would like to thank Louis-Pierre Arguin and Adam Harper for helpful comments. AF was supported by the National Science Foundation under NSF CAREER grant DMS-2339274.
Preliminary results
We gather a few results we will need in the proof of the main theorems. First we need the following lemma from [14 ] .
Lemma 2.1. Assume RH. For α > 0 \alpha> 0 α > 0
log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 ≤ ℜ ( ∑ n ≤ x b α ( n ; x ) n 1 / 2 + α + i t ) + log t 2 π log x log 1 1 − x − α + O ( ( log x ) 2 x 1 + ( log x ) t + ( log x ) log ( 1 + ( log x ) t ) t + 1 ) . \log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1} \leq\Re\left(\sum_{n\leq x}\frac{b_\alpha(n;x)}{n^{1/2+\alpha+it}}\right)+\frac{\log\frac{t}{2\pi}}{\log x}\log\frac{1}{1-x^{-\alpha}}+O\left(\frac{(\log x)^2\sqrt{x}}{1+(\log x)t}+\frac{(\log x)\log(1+(\log x)\sqrt{t})}{\sqrt{t}}+1\right). log ζ ( 2 1 + α + i t ) − 1 ≤ ℜ ( n ≤ x ∑ n 1/2 + α + i t b α ( n ; x ) ) + log x log 2 π t log 1 − x − α 1 + O ( 1 + ( log x ) t ( log x ) 2 x + t ( log x ) log ( 1 + ( log x ) t ) + 1 ) . where
b α ( n ; x ) = − Λ ( n ) a α ( n ; x ) , b_\alpha(n;x)=-\Lambda(n)a_\alpha(n;x), b α ( n ; x ) = − Λ ( n ) a α ( n ; x ) , and
a α ( n ; x ) = ∑ j = 0 ∞ ( j + 1 log n + j log x x − α j − ( j + 1 ) n 2 α ( j + 2 ) log x − log n x − α ( j + 2 ) ) . a_\alpha(n;x)=\sum_{j=0}^{\infty}\left(\frac{j+1}{\log n+j\log x}x^{-\alpha j}-\frac{(j+1)n^{2\alpha}}{(j+2)\log x-\log n}x^{-\alpha(j+2)}\right). a α ( n ; x ) = j = 0 ∑ ∞ ( log n + j log x j + 1 x − α j − ( j + 2 ) log x − log n ( j + 1 ) n 2 α x − α ( j + 2 ) ) . We will also prove the following refinement of the inequality (2.6) in [14 ] .
Lemma 2.2. For n ≤ e 2 π Δ n\leq e^{2\pi\Delta} n ≤ e 2 π Δ
a α ( n ; e 2 π Δ ) = { 1 log n + O ( 1 Δ ) if 2 π Δ α ≥ 1 , 1 log n + 2 ( 2 π Δ − log n ) ( 2 π Δ ) 2 log 1 2 π Δ α + O ( 1 Δ ) if 2 π Δ α < 1. a_\alpha(n;e^{2\pi\Delta})=
\begin{cases}
\frac{1}{\log n}+O\left(\frac{1}{\Delta}\right) & \text{if }2\pi\Delta\alpha\geq1,\\[6pt]
\frac{1}{\log n}+\frac{2(2\pi\Delta-\log n)}{(2\pi\Delta)^2}\log\frac{1}{2\pi\Delta\alpha}+O\left(\frac{1}{\Delta}\right) & \text{if }2\pi\Delta\alpha<1.
\end{cases} a α ( n ; e 2 π Δ ) = ⎩ ⎨ ⎧ l o g n 1 + O ( Δ 1 ) l o g n 1 + ( 2 π Δ ) 2 2 ( 2 π Δ − l o g n ) log 2 π Δ α 1 + O ( Δ 1 ) if 2 π Δ α ≥ 1 , if 2 π Δ α < 1. Proof. We first consider the regime 2 π Δ α ≥ 1 2\pi\Delta\alpha\geq1 2 π Δ α ≥ 1
a α ( n ; e 2 π Δ ) = 1 log n − n 2 α 4 π Δ − log n e − 4 π Δ α + ∑ j ≥ 1 ( j + 1 log n + 2 π j Δ e − 2 π j Δ α − ( j + 1 ) n 2 α 2 π Δ ( j + 2 ) − log n e − 2 π ( j + 2 ) Δ α ) . a_\alpha(n;e^{2\pi\Delta})=\frac{1}{\log n}-\frac{n^{2\alpha}}{4\pi\Delta-\log n}e^{-4\pi\Delta\alpha}+\sum_{j\geq1}\left(\frac{j+1}{\log n+2\pi j\Delta}e^{-2\pi j\Delta\alpha}-\frac{(j+1)n^{2\alpha}}{2\pi\Delta(j+2)-\log n}e^{-2\pi(j+2)\Delta\alpha}\right). a α ( n ; e 2 π Δ ) = log n 1 − 4 π Δ − log n n 2 α e − 4 π Δ α + j ≥ 1 ∑ ( log n + 2 π j Δ j + 1 e − 2 π j Δ α − 2 π Δ ( j + 2 ) − log n ( j + 1 ) n 2 α e − 2 π ( j + 2 ) Δ α ) . Let E E E a α ( n ; e 2 π Δ ) − 1 / log n a_\alpha(n;e^{2\pi\Delta})-1/\log n a α ( n ; e 2 π Δ ) − 1/ log n
∑ j ≥ 1 j + 1 log n + 2 π j Δ e − 2 π j Δ α ≤ 1 π Δ ∑ j ≥ 1 e − 2 π j Δ α ≤ 1 π Δ ( e − 1 ) . \sum_{j\geq1}\frac{j+1}{\log n+2\pi j\Delta}e^{-2\pi j\Delta\alpha}\leq\frac{1}{\pi\Delta}\sum_{j\geq1}e^{-2\pi j\Delta\alpha}\leq\frac{1}{\pi\Delta(e-1)}. j ≥ 1 ∑ log n + 2 π j Δ j + 1 e − 2 π j Δ α ≤ π Δ 1 j ≥ 1 ∑ e − 2 π j Δ α ≤ π Δ ( e − 1 ) 1 . We also have that
∑ j ≥ 1 ( j + 1 ) n 2 α 2 π Δ ( j + 2 ) − log n e − 2 π ( j + 2 ) Δ α ≤ 1 2 π Δ ( e − 1 ) . \sum_{j\geq1}\frac{(j+1)n^{2\alpha}}{2\pi\Delta(j+2)-\log n}e^{-2\pi(j+2)\Delta\alpha}\leq\frac{1}{2\pi\Delta(e-1)}. j ≥ 1 ∑ 2 π Δ ( j + 2 ) − log n ( j + 1 ) n 2 α e − 2 π ( j + 2 ) Δ α ≤ 2 π Δ ( e − 1 ) 1 . It follows that
∣ E ∣ ≤ 1 2 π Δ ( 1 + 3 e − 1 ) . |E| \le\frac{1}{2\pi\Delta}\left(1+\frac{3}{e-1}\right). ∣ E ∣ ≤ 2 π Δ 1 ( 1 + e − 1 3 ) . Now suppose that 2 π Δ α < 1 2\pi\Delta\alpha<1 2 π Δ α < 1
a α ( n ; e 2 π Δ ) = 1 log n + ∑ j = 1 ∞ j + 1 log n + 2 π Δ j e − 2 π Δ α j − ∑ j = 0 ∞ ( j + 1 ) n 2 α 2 π Δ ( j + 2 ) − log n e − 2 π Δ α ( j + 2 ) . a_{\alpha}(n;e^{2\pi\Delta})=\frac{1}{\log n}+\sum_{j=1}^{\infty}\frac{j+1}{\log n+2\pi\Delta j}e^{-2\pi\Delta\alpha j}-\sum_{j=0}^{\infty}\frac{(j+1)n^{2\alpha}}{2\pi\Delta(j+2)-\log n}e^{-2\pi\Delta\alpha(j+2)}. a α ( n ; e 2 π Δ ) = log n 1 + j = 1 ∑ ∞ log n + 2 π Δ j j + 1 e − 2 π Δ α j − j = 0 ∑ ∞ 2 π Δ ( j + 2 ) − log n ( j + 1 ) n 2 α e − 2 π Δ α ( j + 2 ) . Note that we can replace n 2 α n^{2\alpha} n 2 α 1 1 1 log n ≤ 2 π Δ \log n\le2\pi\Delta log n ≤ 2 π Δ 0 ≤ n 2 α − 1 ≤ e 4 π Δ α − 1 0\le n^{2\alpha}-1\le e^{4\pi\Delta\alpha}-1 0 ≤ n 2 α − 1 ≤ e 4 π Δ α − 1 j ≥ 0 j\ge0 j ≥ 0
j + 1 2 π Δ ( j + 2 ) − log n ≤ 1 2 π Δ . \frac{j+1}{2\pi\Delta(j+2)-\log n}\le\frac{1}{2\pi\Delta}. 2 π Δ ( j + 2 ) − log n j + 1 ≤ 2 π Δ 1 . Summing the geometric series,
∑ j ≥ 0 ( j + 1 ) e − 2 π Δ α ( j + 2 ) 2 π Δ ( j + 2 ) − log n ≤ 1 2 π Δ ∑ j ≥ 0 e − 2 π Δ α ( j + 2 ) = 1 2 π Δ e − 4 π Δ α 1 − e − 2 π Δ α , \sum_{j\ge0}\frac{(j+1)e^{-2\pi\Delta\alpha(j+2)}}{2\pi\Delta(j+2)-\log n}\le\frac{1}{2\pi\Delta}\sum_{j\ge0}e^{-2\pi\Delta\alpha(j+2)}=\frac{1}{2\pi\Delta}\frac{e^{-4\pi\Delta\alpha}}{1-e^{-2\pi\Delta\alpha}}, j ≥ 0 ∑ 2 π Δ ( j + 2 ) − log n ( j + 1 ) e − 2 π Δ α ( j + 2 ) ≤ 2 π Δ 1 j ≥ 0 ∑ e − 2 π Δ α ( j + 2 ) = 2 π Δ 1 1 − e − 2 π Δ α e − 4 π Δ α , and hence,
( n 2 α − 1 ) ∑ j ≥ 0 ( j + 1 ) e − 2 π Δ α ( j + 2 ) 2 π Δ ( j + 2 ) − log n ≤ ( e 4 π Δ α − 1 ) e − 4 π Δ α 2 π Δ ( 1 − e − 2 π Δ α ) = 1 + e − 2 π Δ α 2 π Δ ≤ 1 π Δ . (n^{2\alpha}-1)\sum_{j\ge0}\frac{(j+1)e^{-2\pi\Delta\alpha(j+2)}}{2\pi\Delta(j+2)-\log n}\le\frac{(e^{4\pi\Delta\alpha}-1)e^{-4\pi\Delta\alpha}}{2\pi\Delta(1-e^{-2\pi\Delta\alpha})}=\frac{1+e^{-2\pi\Delta\alpha}}{2\pi\Delta}\le\frac{1}{\pi\Delta}. ( n 2 α − 1 ) j ≥ 0 ∑ 2 π Δ ( j + 2 ) − log n ( j + 1 ) e − 2 π Δ α ( j + 2 ) ≤ 2 π Δ ( 1 − e − 2 π Δ α ) ( e 4 π Δ α − 1 ) e − 4 π Δ α = 2 π Δ 1 + e − 2 π Δ α ≤ π Δ 1 . Then we write
a α ( n ; e 2 π Δ ) = 1 log n + ∑ j = 1 ∞ j + 1 log n + 2 π Δ j e − 2 π Δ α j − ∑ j ≥ 2 ( j − 1 ) e − 2 π Δ α j 2 π Δ j − log n + O ( 1 Δ ) . a_{\alpha}(n;e^{2\pi\Delta})=\frac{1}{\log n}+\sum_{j=1}^{\infty}\frac{j+1}{\log n+2\pi\Delta j}e^{-2\pi\Delta\alpha j}-\sum_{j\ge2}\frac{(j-1)e^{-2\pi\Delta\alpha j}}{2\pi\Delta j-\log n}+O\left(\frac{1}{\Delta}\right). a α ( n ; e 2 π Δ ) = log n 1 + j = 1 ∑ ∞ log n + 2 π Δ j j + 1 e − 2 π Δ α j − j ≥ 2 ∑ 2 π Δ j − log n ( j − 1 ) e − 2 π Δ α j + O ( Δ 1 ) . The j = 1 j=1 j = 1 2 e − 2 π Δ α / ( 2 π Δ + log n ) ≤ 1 / ( π Δ ) 2e^{-2\pi\Delta\alpha}/(2\pi\Delta+\log n)\le1/(\pi\Delta) 2 e − 2 π Δ α / ( 2 π Δ + log n ) ≤ 1/ ( π Δ ) j ≥ 2 j\ge2 j ≥ 2
j + 1 log n + 2 π Δ j − j − 1 2 π Δ j − log n = 2 j ( 2 π Δ − log n ) ( 2 π Δ j ) 2 − ( log n ) 2 . \frac{j+1}{\log n+2\pi\Delta j}-\frac{j-1}{2\pi\Delta j-\log n}=\frac{2j(2\pi\Delta-\log n)}{(2\pi\Delta j)^2-(\log n)^2}. log n + 2 π Δ j j + 1 − 2 π Δ j − log n j − 1 = ( 2 π Δ j ) 2 − ( log n ) 2 2 j ( 2 π Δ − log n ) . Therefore
a α ( n ; e 2 π Δ ) = 1 log n + 2 ( 2 π Δ − log n ) ∑ j ≥ 2 j e − 2 π Δ α j ( 2 π Δ j ) 2 − ( log n ) 2 + O ( 1 Δ ) . a_{\alpha}(n;e^{2\pi\Delta})=\frac{1}{\log n}+2(2\pi\Delta-\log n)\sum_{j\ge2}\frac{je^{-2\pi\Delta\alpha j}}{(2\pi\Delta j)^2-(\log n)^2}+O\left(\frac{1}{\Delta}\right). a α ( n ; e 2 π Δ ) = log n 1 + 2 ( 2 π Δ − log n ) j ≥ 2 ∑ ( 2 π Δ j ) 2 − ( log n ) 2 j e − 2 π Δ α j + O ( Δ 1 ) . Since log n ≤ 2 π Δ \log n\le2\pi\Delta log n ≤ 2 π Δ j ≥ 2 j\ge2 j ≥ 2
0 ≤ j ( 2 π Δ j ) 2 − ( log n ) 2 − 1 ( 2 π Δ ) 2 j = ( log n ) 2 ( 2 π Δ ) 2 j ( ( 2 π Δ j ) 2 − ( log n ) 2 ) ≪ 1 Δ 2 j 3 . 0\le\frac{j}{(2\pi\Delta j)^2-(\log n)^2}-\frac{1}{(2\pi\Delta)^2j}=\frac{(\log n)^2}{(2\pi\Delta)^2j\left((2\pi\Delta j)^2-(\log n)^2\right)}\ll\frac{1}{\Delta^2j^3}. 0 ≤ ( 2 π Δ j ) 2 − ( log n ) 2 j − ( 2 π Δ ) 2 j 1 = ( 2 π Δ ) 2 j ( ( 2 π Δ j ) 2 − ( log n ) 2 ) ( log n ) 2 ≪ Δ 2 j 3 1 . Hence
a α ( n ; e 2 π Δ ) = 1 log n + 2 ( 2 π Δ − log n ) ( 2 π Δ ) 2 ∑ j ≥ 2 e − 2 π Δ α j j + O ( 1 Δ ) . a_{\alpha}(n;e^{2\pi\Delta})=\frac{1}{\log n}+\frac{2(2\pi\Delta-\log n)}{(2\pi\Delta)^2}\sum_{j\ge2}\frac{e^{-2\pi\Delta\alpha j}}{j}+O\left(\frac{1}{\Delta}\right). a α ( n ; e 2 π Δ ) = log n 1 + ( 2 π Δ ) 2 2 ( 2 π Δ − log n ) j ≥ 2 ∑ j e − 2 π Δ α j + O ( Δ 1 ) . Finally,
∑ j ≥ 2 e − 2 π Δ α j j = log 1 1 − e − 2 π Δ α − e − 2 π Δ α = log 1 2 π Δ α + O ( 1 ) . \sum_{j\ge2}\frac{e^{-2\pi\Delta\alpha j}}{j}=\log\frac{1}{1-e^{-2\pi\Delta\alpha}}-e^{-2\pi\Delta\alpha}=\log\frac{1}{2\pi\Delta\alpha}+O(1). j ≥ 2 ∑ j e − 2 π Δ α j = log 1 − e − 2 π Δ α 1 − e − 2 π Δ α = log 2 π Δ α 1 + O ( 1 ) . This finishes the proof.
Let
S x ( t ) = ℜ ( ∑ p ≤ x b α ( p ; x ) p 1 / 2 + α + i t + ∑ p 2 ≤ x b α ( p 2 ; x ) p 1 + 2 α + 2 i t ) . (4) S_x(t)=\Re\left(\sum_{p\le x}\frac{b_{\alpha}(p;x)}{p^{1/2+\alpha+it}}+\sum_{p^2\le x}\frac{b_{\alpha}(p^2;x)}{p^{1+2\alpha+2it}}\right). \tag*{(4)} S x ( t ) = ℜ p ≤ x ∑ p 1/2 + α + i t b α ( p ; x ) + p 2 ≤ x ∑ p 1 + 2 α + 2 i t b α ( p 2 ; x ) . (4) Using Lemma 2.2 , if α log x ≫ 1 \alpha\log x \gg1 α log x ≫ 1
b α ( p ; x ) = − 1 + O ( log p log x ) and b α ( p 2 ; x ) = − 1 2 + O ( log p log x ) . (5) b_{\alpha}(p;x)=-1+O\left(\frac{\log p}{\log x}\right)\qquad\text{and}\qquad b_{\alpha}(p^{2};x)=-\frac{1}{2}+O\left(\frac{\log p}{\log x}\right). \tag*{(5)} b α ( p ; x ) = − 1 + O ( log x log p ) and b α ( p 2 ; x ) = − 2 1 + O ( log x log p ) . (5) For simplicity of notation, for a sequence β 1 < … < β K \beta_{1}<\ldots<\beta_{K} β 1 < … < β K
b α ( n ; j ) = b α ( n ; T β j ) . b_{\alpha}(n;j)=b_{\alpha}(n;T^{\beta_{j}}). b α ( n ; j ) = b α ( n ; T β j ) . Let
S ℓ ( j ) ( t ) = ℜ ( ∑ p ≤ T ℓ b α ( p ; j ) p 1 / 2 + α + i t + ∑ p ≤ T ℓ p 2 ≤ T j b α ( p 2 ; j ) p 1 + 2 α + 2 i t ) . S_{\ell}^{(j)}(t)=\Re\left(\sum_{p\leq T_{\ell}}\frac{b_{\alpha}(p;j)}{p^{1/2+\alpha+it}}+\sum_{\substack{p\leq T_{\ell}\\p^{2}\leq T_{j}}}\frac{b_{\alpha}(p^{2};j)}{p^{1+2\alpha+2it}}\right). S ℓ ( j ) ( t ) = ℜ p ≤ T ℓ ∑ p 1/2 + α + i t b α ( p ; j ) + p ≤ T ℓ p 2 ≤ T j ∑ p 1 + 2 α + 2 i t b α ( p 2 ; j ) . Note that the definition of S ℓ ( j ) S_{\ell}^{(j)} S ℓ ( j ) [3 ] ; unlike in [3 ] , we do not consider the prime squares for which p 2 ≤ T ℓ p^{2}\leq T_{\ell} p 2 ≤ T ℓ
We have the following lemma which computes the variance of S ℓ ( j ) S_{\ell}^{(j)} S ℓ ( j )
Lemma 2.3. Let j ≥ ℓ j\geq\ell j ≥ ℓ 2 ≤ T ℓ ≤ T 1 / 4 − ε 2\leq T_{\ell}\leq T^{1/4-\varepsilon} 2 ≤ T ℓ ≤ T 1/4 − ε α ≥ T j − 1 \alpha\geq T_{j}^{-1} α ≥ T j − 1 α < 1 \alpha<1 α < 1 α log T j ≥ 1 \alpha\log T_{j}\geq1 α log T j ≥ 1
1 T ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = 1 2 min { log 1 α , log log T ℓ } + O ( 1 ) . \frac{1}{T}\int_{T}^{2T}S_{\ell}^{(j)}(t)^{2}\,\mathrm{d}t=\frac{1}{2}\min\left\{\log\frac{1}{\alpha},\log\log T_{\ell}\right\}+O(1). T 1 ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = 2 1 min { log α 1 , log log T ℓ } + O ( 1 ) . If α log T j < 1 \alpha\log T_{j}<1 α log T j < 1
1 T ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = 1 2 log log T ℓ + O ( 1 + log 2 1 α log T j ) . \frac{1}{T}\int_{T}^{2T}S_{\ell}^{(j)}(t)^{2}\,\mathrm{d}t=\frac{1}{2}\log\log T_{\ell}+O\left(1+\log^{2}\frac{1}{\alpha\log T_{j}}\right). T 1 ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = 2 1 log log T ℓ + O ( 1 + log 2 α log T j 1 ) . Proof. The Dirichlet polynomial whose real part is S ℓ ( j ) S_{\ell}^{(j)} S ℓ ( j ) T ℓ 2 ≤ T 1 / 2 − ε T_{\ell}^{2}\leq T^{1/2-\varepsilon} T ℓ 2 ≤ T 1/2 − ε
1 T ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = ( 1 + O ( T − ε ) ) ( 1 2 ∑ p ≤ T ℓ b α ( p ; j ) 2 p 1 + 2 α + 1 2 ∑ p ≤ T ℓ p 2 ≤ T j b α ( p 2 ; j ) 2 p 2 + 4 α ) . (6) \frac{1}{T}\int_{T}^{2T}S_{\ell}^{(j)}(t)^{2}\,\mathrm{d}t=\left(1+O(T^{-\varepsilon})\right)\left(\frac{1}{2}\sum_{p\leq T_{\ell}}\frac{b_{\alpha}(p;j)^{2}}{p^{1+2\alpha}}+\frac{1}{2}\sum_{\substack{p\leq T_{\ell}\\p^{2}\leq T_{j}}}\frac{b_{\alpha}(p^{2};j)^{2}}{p^{2+4\alpha}}\right). \tag*{(6)} T 1 ∫ T 2 T S ℓ ( j ) ( t ) 2 d t = ( 1 + O ( T − ε ) ) 2 1 p ≤ T ℓ ∑ p 1 + 2 α b α ( p ; j ) 2 + 2 1 p ≤ T ℓ p 2 ≤ T j ∑ p 2 + 4 α b α ( p 2 ; j ) 2 . (6) Put
Z = max { 0 , log 1 α log T j } . Z=\max\left\{0,\log\frac{1}{\alpha\log T_{j}}\right\}. Z = max { 0 , log α log T j 1 } . By Lemma 2.2 , for r = 1 , 2 r=1,2 r = 1 , 2 p r ≤ T j p^{r}\leq T_{j} p r ≤ T j
b α ( p r ; j ) = − 1 r + O ( ( 1 + Z ) log p log T j ) . b_{\alpha}(p^{r};j)=-\frac{1}{r}+O\left(\frac{(1+Z)\log p}{\log T_{j}}\right). b α ( p r ; j ) = − r 1 + O ( log T j ( 1 + Z ) log p ) . Since α ≥ T j − 1 \alpha\geq T_{j}^{-1} α ≥ T j − 1 Z ≪ log T j Z\ll\log T_{j} Z ≪ log T j
∑ p ≤ T ℓ p 2 ≤ T j b α ( p 2 ; j ) 2 p 2 + 4 α ≪ ∑ p 1 + ( log p ) 2 p 2 ≪ 1. \sum_{\substack{p\leq T_{\ell}\\p^{2}\leq T_{j}}}\frac{b_{\alpha}(p^{2};j)^{2}}{p^{2+4\alpha}}\ll\sum_{p}\frac{1+(\log p)^{2}}{p^{2}}\ll1. p ≤ T ℓ p 2 ≤ T j ∑ p 2 + 4 α b α ( p 2 ; j ) 2 ≪ p ∑ p 2 1 + ( log p ) 2 ≪ 1. Using the coefficient estimate above and
∑ p ≤ T ℓ ( log p ) n p ≪ ( log T ℓ ) n , \sum_{p\leq T_{\ell}}\frac{(\log p)^{n}}{p}\ll(\log T_{\ell})^{n}, p ≤ T ℓ ∑ p ( log p ) n ≪ ( log T ℓ ) n , for n = 1 , 2 n=1,2 n = 1 , 2
∑ p ≤ T ℓ ∣ b α ( p ; j ) 2 − 1 ∣ p 1 + 2 α ≪ ( 1 + Z ) log T ℓ log T j + ( 1 + Z ) 2 ( log T ℓ ) 2 ( log T j ) 2 ≪ 1 + Z 2 . \sum_{p\leq T_{\ell}}\frac{\left|b_{\alpha}(p;j)^{2}-1\right|}{p^{1+2\alpha}}\ll(1+Z)\frac{\log T_{\ell}}{\log T_{j}}+(1+Z)^{2}\frac{(\log T_{\ell})^{2}}{(\log T_{j})^{2}}\ll1+Z^{2}. p ≤ T ℓ ∑ p 1 + 2 α b α ( p ; j ) 2 − 1 ≪ ( 1 + Z ) log T j log T ℓ + ( 1 + Z ) 2 ( log T j ) 2 ( log T ℓ ) 2 ≪ 1 + Z 2 . Finally, Mertens’ theorem and partial summation give
∑ p ≤ T ℓ 1 p 1 + 2 α = min { log 1 α , log log T ℓ } + O ( 1 ) . \sum_{p\le T_\ell}\frac{1}{p^{1+2\alpha}}=\min\left\{\log\frac{1}{\alpha},\log\log T_\ell\right\}+O(1). p ≤ T ℓ ∑ p 1 + 2 α 1 = min { log α 1 , log log T ℓ } + O ( 1 ) . If α log T j ≥ 1 \alpha\log T_j\ge1 α log T j ≥ 1 Z = 0 Z=0 Z = 0 α log T j < 1 \alpha\log T_j<1 α log T j < 1 α log T ℓ < 1 \alpha\log T_\ell<1 α log T ℓ < 1 log log T ℓ \log\log T_\ell log log T ℓ
We recall that ( X ( p ) ) p (X(p))_p ( X ( p ) ) p
Lemma 2.4. For any q ∈ N q\in\mathbb{N} q ∈ N γ p , δ p \gamma_p,\delta_p γ p , δ p
E r a n d ( ∣ ∑ p γ p X ( p ) + ∑ p δ p X ( p ) 2 ∣ 2 q ) ≤ 2 2 q q ! ( ∑ p ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . \mathbb{E}_{\mathrm{rand}}\left(\left|\sum_p\gamma_pX(p)+\sum_p\delta_pX(p)^2\right|^{2q}\right)\le2^{2q}q!\left(\sum_p\left(|\gamma_p|^2+|\delta_p|^2\right)\right)^q. E rand p ∑ γ p X ( p ) + p ∑ δ p X ( p ) 2 2 q ≤ 2 2 q q ! ( p ∑ ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . Proof. Let
ξ ~ 1 : = ∑ p γ p X ( p ) and ξ ~ 2 : = ∑ p δ p X ( p ) 2 . \widetilde{\xi}_1:=\sum_p\gamma_pX(p)\qquad\text{and}\qquad\widetilde{\xi}_2:=\sum_p\delta_pX(p)^2. ξ 1 := p ∑ γ p X ( p ) and ξ 2 := p ∑ δ p X ( p ) 2 . By Minkowski’s inequality we have
E r a n d ( ( ∣ ξ ~ 1 + ξ ~ 2 ∣ 2 q ) 1 / ( 2 q ) ) ≤ E r a n d ( ( ∣ ξ ~ 1 ∣ 2 q ) 1 / ( 2 q ) ) + E r a n d ( ( ∣ ξ ~ 2 ∣ 2 q ) 1 / ( 2 q ) ) ≤ 2 E r a n d ( ∣ ξ ~ 1 ∣ 2 q ) 1 / q + E r a n d ( ∣ ξ ~ 2 ∣ 2 q ) 1 / q . (7) \begin{aligned}
\mathbb{E}_{\mathrm{rand}}\left((|\widetilde{\xi}_1+\widetilde{\xi}_2|^{2q})^{1/(2q)}\right)\le\mathbb{E}_{\mathrm{rand}}\left((|\widetilde{\xi}_1|^{2q})^{1/(2q)}\right)+\mathbb{E}_{\mathrm{rand}}\left((|\widetilde{\xi}_2|^{2q})^{1/(2q)}\right) \\
&\le\sqrt{2}\sqrt{\mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_1|^{2q}\right)^{1/q}+\mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_2|^{2q}\right)^{1/q}}. \tag*{(7)}
\end{aligned} E rand ( ( ∣ ξ 1 + ξ 2 ∣ 2 q ) 1/ ( 2 q ) ) ≤ E rand ( ( ∣ ξ 1 ∣ 2 q ) 1/ ( 2 q ) ) + E rand ( ( ∣ ξ 2 ∣ 2 q ) 1/ ( 2 q ) ) ≤ 2 E rand ( ∣ ξ 1 ∣ 2 q ) 1/ q + E rand ( ∣ ξ 2 ∣ 2 q ) 1/ q . (7) Now
E r a n d ( ∣ ξ ~ 1 ∣ 2 q ) = E r a n d ( ∣ ∑ p 1 , … , p r p i ≠ p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ( q α 1 , … , α r ) ∏ i = 1 r γ p i α i X ( p i ) α i ∣ 2 ) = ∑ p 1 , … , p r p i ≠ p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ( q α 1 , … , α r ) 2 ∏ i = 1 r ∣ γ p i ∣ 2 α i ≤ q ! ∑ p 1 , … , p r p i ≠ p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ( q α 1 , … , α r ) ∏ i = 1 r ∣ γ p i ∣ 2 α i = q ! ( ∑ p ∣ γ p ∣ 2 ) q . \begin{aligned}
\mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_1|^{2q}\right)=\mathbb{E}_{\mathrm{rand}}\left(\left|\sum_{\substack{p_1,\ldots,p_r\\p_i\ne p_j}}\sum_{\substack{\alpha_i\ge1\\\sum_{i=1}^r\alpha_i=q}}\binom{q}{\alpha_1,\ldots,\alpha_r}\prod_{i=1}^r\gamma_{p_i}^{\alpha_i}X(p_i)^{\alpha_i}\right|^2\right) \\
&=\sum_{\substack{p_1,\ldots,p_r\\p_i\ne p_j}}\sum_{\substack{\alpha_i\ge1\\\sum_{i=1}^r\alpha_i=q}}\binom{q}{\alpha_1,\ldots,\alpha_r}^{2}\prod_{i=1}^r|\gamma_{p_i}|^{2\alpha_i} \\
&\le q!\sum_{\substack{p_1,\ldots,p_r\\p_i\ne p_j}}\sum_{\substack{\alpha_i\ge1\\\sum_{i=1}^r\alpha_i=q}}\binom{q}{\alpha_1,\ldots,\alpha_r}\prod_{i=1}^r|\gamma_{p_i}|^{2\alpha_i}
=q!\left(\sum_p|\gamma_p|^2\right)^q.
\end{aligned} E rand ( ∣ ξ 1 ∣ 2 q ) = E rand p 1 , … , p r p i = p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ∑ ( α 1 , … , α r q ) i = 1 ∏ r γ p i α i X ( p i ) α i 2 = p 1 , … , p r p i = p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ∑ ( α 1 , … , α r q ) 2 i = 1 ∏ r ∣ γ p i ∣ 2 α i ≤ q ! p 1 , … , p r p i = p j ∑ α i ≥ 1 ∑ i = 1 r α i = q ∑ ( α 1 , … , α r q ) i = 1 ∏ r ∣ γ p i ∣ 2 α i = q ! ( p ∑ ∣ γ p ∣ 2 ) q . A similar expression holds for E r a n d ( ∣ ξ ~ 2 ∣ 2 q ) \mathbb{E}_{\mathrm{rand}}(|\widetilde{\xi}_2|^{2q}) E rand ( ∣ ξ 2 ∣ 2 q )
E r a n d ( ∣ ξ ~ 1 + ξ ~ 2 ∣ 2 q ) ≤ 2 2 q q ! ( ∑ p ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . \mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_1+\widetilde{\xi}_2|^{2q}\right)\le2^{2q}q!\left(\sum_p\left(|\gamma_p|^2+|\delta_p|^2\right)\right)^q. E rand ( ∣ ξ 1 + ξ 2 ∣ 2 q ) ≤ 2 2 q q ! ( p ∑ ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . This proves the lemma.
We also have the following. (See also Lemma A.1 in [3 ] .)
Lemma 2.5. For j ≥ 1 j\ge1 j ≥ 1 q ∈ N q\in\mathbb{N} q ∈ N 2 q ≤ log T / log T 1 2q\le\log T/\log T_1 2 q ≤ log T / log T 1
1 T ∫ T 2 T ( S 1 ( j ) ( t ) ) 2 q d t ≪ 4 q q ! Var ( S 1 ( j ) ) q . \frac{1}{T}\int_T^{2T}\left(S_1^{(j)}(t)\right)^{2q}\,dt\ll4^q q!\operatorname{Var}\left(S_1^{(j)}\right)^q. T 1 ∫ T 2 T ( S 1 ( j ) ( t ) ) 2 q d t ≪ 4 q q ! Var ( S 1 ( j ) ) q . Moreover, for 1 ≤ k < ℓ ≤ j 1\le k<\ell\le j 1 ≤ k < ℓ ≤ j q ∈ N q\in\mathbb{N} q ∈ N 2 q ≤ log T / log T ℓ 2q\le\log T/\log T_\ell 2 q ≤ log T / log T ℓ
1 T ∫ T 2 T ( S ℓ ( j ) ( t ) − S k ( j ) ( t ) ) 2 q d t ≪ 4 q q ! ( Var ( S ℓ ( j ) ) − Var ( S k ( j ) ) ) q . \frac{1}{T}\int_T^{2T}\left(S_\ell^{(j)}(t)-S_k^{(j)}(t)\right)^{2q}\,dt\ll4^q q!\left(\operatorname{Var}\left(S_\ell^{(j)}\right)-\operatorname{Var}\left(S_k^{(j)}\right)\right)^q. T 1 ∫ T 2 T ( S ℓ ( j ) ( t ) − S k ( j ) ( t ) ) 2 q d t ≪ 4 q q ! ( Var ( S ℓ ( j ) ) − Var ( S k ( j ) ) ) q . Proof. We prove the second statement. The treatment of the first one is similar. We denote
ξ 1 ( t ) : = ∑ p γ p p − i t , ξ 2 ( t ) : = ∑ p δ p p − 2 i t . \xi_1(t):=\sum_p \gamma_p p^{-it}, \qquad\xi_2(t):=\sum_p \delta_p p^{-2it}. ξ 1 ( t ) := p ∑ γ p p − i t , ξ 2 ( t ) := p ∑ δ p p − 2 i t . and
ξ ~ 1 : = ∑ p γ p X ( p ) , ξ ~ 2 : = ∑ p δ p X ( p ) 2 . \widetilde{\xi}_1:=\sum_p \gamma_p X(p), \qquad\widetilde{\xi}_2:=\sum_p \delta_p X(p)^2. ξ 1 := p ∑ γ p X ( p ) , ξ 2 := p ∑ δ p X ( p ) 2 . where
γ p = b α ( p ; j ) p 1 / 2 + α 1 T k < p ≤ T ℓ and δ p = b α ( p 2 ; j ) p 1 + 2 α 1 T k < p ≤ T ℓ . \gamma_p=\frac{b_\alpha(p;j)}{p^{1/2+\alpha}}\mathbf{1}_{T_k<p\le T_\ell}
\qquad\text{and}\qquad
\delta_p=\frac{b_\alpha(p^2;j)}{p^{1+2\alpha}}\mathbf{1}_{T_k<p\le T_\ell}. γ p = p 1/2 + α b α ( p ; j ) 1 T k < p ≤ T ℓ and δ p = p 1 + 2 α b α ( p 2 ; j ) 1 T k < p ≤ T ℓ . Then
S ℓ ( j ) − S k ( j ) = ℜ ( ξ 1 ( t ) + ξ 2 ( t ) ) . (8) S_\ell^{(j)}-S_k^{(j)}=\Re\bigl(\xi_1(t)+\xi_2(t)\bigr). \tag*{(8)} S ℓ ( j ) − S k ( j ) = ℜ ( ξ 1 ( t ) + ξ 2 ( t ) ) . (8) and, by (6) ,
Var ( S ℓ ( j ) ) − Var ( S k ( j ) ) ≪ 1 2 ( ∑ p ∣ γ p ∣ 2 + ∑ p ∣ δ p ∣ 2 ) . (9) \operatorname{Var}(S_\ell^{(j)})-\operatorname{Var}(S_k^{(j)})
\ll\frac{1}{2}\left(\sum_p|\gamma_p|^2+\sum_p|\delta_p|^2\right). \tag*{(9)} Var ( S ℓ ( j ) ) − Var ( S k ( j ) ) ≪ 2 1 ( p ∑ ∣ γ p ∣ 2 + p ∑ ∣ δ p ∣ 2 ) . (9) By expanding through the binomial formula and using the mean-value theorem for Dirichlet polynomials, we have
1 T ∫ T 2 T ∣ ξ 1 ( t ) + ξ 2 ( t ) ∣ 2 q d t ≪ E r a n d ( ∣ ξ ~ 1 + ξ ~ 2 ∣ 2 q ) . (10) \frac{1}{T}\int_T^{2T}|\xi_1(t)+\xi_2(t)|^{2q}\,dt
\ll\mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_1+\widetilde{\xi}_2|^{2q}\right). \tag*{(10)} T 1 ∫ T 2 T ∣ ξ 1 ( t ) + ξ 2 ( t ) ∣ 2 q d t ≪ E rand ( ∣ ξ 1 + ξ 2 ∣ 2 q ) . (10) Also, from Lemma 2.4 we obtain that
E r a n d ( ∣ ξ ~ 1 + ξ ~ 2 ∣ 2 q ) ≤ 2 2 q q ! ( ∑ p ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . (11) \mathbb{E}_{\mathrm{rand}}\left(|\widetilde{\xi}_1+\widetilde{\xi}_2|^{2q}\right)
\le2^{2q}q!\left(\sum_p\left(|\gamma_p|^2+|\delta_p|^2\right)\right)^q. \tag*{(11)} E rand ( ∣ ξ 1 + ξ 2 ∣ 2 q ) ≤ 2 2 q q ! ( p ∑ ( ∣ γ p ∣ 2 + ∣ δ p ∣ 2 ) ) q . (11) Observe that by (8) ,
1 T ∫ T 2 T ( S ℓ ( j ) − S k ( j ) ) 2 q d t = 1 T ∫ T 2 T ℜ ( ξ 1 ( t ) + ξ 2 ( t ) ) 2 q d t ≤ 1 T ∫ T 2 T ∣ ξ 1 ( t ) + ξ 2 ( t ) ∣ 2 q d t . \frac{1}{T}\int_T^{2T}\left(S_\ell^{(j)}-S_k^{(j)}\right)^{2q}\,dt
=\frac{1}{T}\int_T^{2T}\Re\left(\xi_1(t)+\xi_2(t)\right)^{2q}\,dt
\le\frac{1}{T}\int_T^{2T}|\xi_1(t)+\xi_2(t)|^{2q}\,dt. T 1 ∫ T 2 T ( S ℓ ( j ) − S k ( j ) ) 2 q d t = T 1 ∫ T 2 T ℜ ( ξ 1 ( t ) + ξ 2 ( t ) ) 2 q d t ≤ T 1 ∫ T 2 T ∣ ξ 1 ( t ) + ξ 2 ( t ) ∣ 2 q d t . The second statement of the lemma now follows by combining with (10) , (2.8) and (2.6)`. □
Proof of Theorem 1.7 in the large α \alpha α
We will split the proof of Theorem 1.7 into two parts, depending on the size of α \alpha α α \alpha α α \alpha α
We denote by H H H
H = { log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 > V } . (12) H=\left\{\log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}>V\right\}. \tag*{(12)} H = { log ζ ( 2 1 + α + i t ) − 1 > V } . (12) We first assume that
α log T ≫ ( log 1 α ) 12 + ε . (13) \alpha\log T\gg\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. \tag*{(13)} α log T ≫ ( log α 1 ) 12 + ε . (13) Let
x = exp ( 2 log ( α log T ) α ) , (14) x=\exp\left(\frac{2\log(\alpha\log T)}{\alpha}\right), \tag*{(14)} x = exp ( α 2 log ( α log T ) ) , (14) and let
σ x 2 : = E ( S x 2 ) = 1 2 log 1 α + O ( 1 ) , (15) \sigma_x^2:=\mathbb{E}(S_x^2)=\frac{1}{2}\log\frac{1}{\alpha}+O(1), \tag*{(15)} σ x 2 := E ( S x 2 ) = 2 1 log α 1 + O ( 1 ) , (15) which follows from Lemma 2.3. From Lemma 2.1, since the choice of x x x (14) gives
log t 2 π log x log 1 1 − x − α = o ( 1 ) , \frac{\log\frac{t}{2\pi}}{\log x}\log\frac{1}{1-x^{-\alpha}}=o(1), log x log 2 π t log 1 − x − α 1 = o ( 1 ) , we have that
∣ ζ ( 1 2 + α + i t ) ∣ − 1 ≤ S x ( t ) + C , \left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}\le S_x(t)+C, ζ ( 2 1 + α + i t ) − 1 ≤ S x ( t ) + C , for some absolute constant C C C A > 1 A>1 A > 1
P ( H ) ≤ P ( S x ( t ) ≥ V − C ) ≤ P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) + ∑ V − C ≤ u ≤ A σ x 2 P ( S x ( t ) ∈ [ u − 1 , u ] ) . (16) \mathbb{P}(H)\le\mathbb{P}(S_x(t)\ge V-C)\le\mathbb{P}(S_x(t)\ge(A-1)\sigma_x^2)+\sum_{V-C\le u\le A\sigma_x^2}\mathbb{P}(S_x(t)\in[u-1,u]). \tag*{(16)} P ( H ) ≤ P ( S x ( t ) ≥ V − C ) ≤ P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) + V − C ≤ u ≤ A σ x 2 ∑ P ( S x ( t ) ∈ [ u − 1 , u ]) . (16) We first evaluate the first term in (16) . By Markov’s inequality, we have that
P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) ≤ 1 ( ( A − 1 ) σ x 2 ) 2 q E ( S x ( t ) 2 q ) , \mathbb{P}(S_x(t)\ge(A-1)\sigma_x^2)\le\frac{1}{((A-1)\sigma_x^2)^{2q}}\mathbb{E}\left(S_x(t)^{2q}\right), P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) ≤ (( A − 1 ) σ x 2 ) 2 q 1 E ( S x ( t ) 2 q ) , where q q q q ≤ α log T 4 log ( α log T ) q\le\frac{\alpha\log T}{4\log(\alpha\log T)} q ≤ 4 l o g ( α l o g T ) α l o g T
P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) ≪ exp ( q log q + q log 4 e ( A − 1 ) 2 + log q 2 − 2 q log σ x ) . (17) \mathbb{P}(S_x(t)\ge(A-1)\sigma_x^2)\ll\exp\left(q\log q+q\log\frac{4}{e(A-1)^2}+\frac{\log q}{2}-2q\log\sigma_x\right). \tag*{(17)} P ( S x ( t ) ≥ ( A − 1 ) σ x 2 ) ≪ exp ( q log q + q log e ( A − 1 ) 2 4 + 2 log q − 2 q log σ x ) . (17) Choosing q = ⌊ σ x 2 ⌋ q=\lfloor\sigma_x^2\rfloor q = ⌊ σ x 2 ⌋ q ≤ α log T 4 log ( α log T ) q\le\frac{\alpha\log T}{4\log(\alpha\log T)} q ≤ 4 l o g ( α l o g T ) α l o g T α log T ≫ ( log 1 α ) 12 + ε \alpha\log T\gg\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon} α log T ≫ ( log α 1 ) 12 + ε A A A κ \kappa κ
( 3.6 ) ≤ 1 σ x exp ( − V 2 2 σ x 2 ) . (18) (3.6)\le\frac{1}{\sigma_x}\exp\left(-\frac{V^2}{2\sigma_x^2}\right). \tag*{(18)} ( 3.6 ) ≤ σ x 1 exp ( − 2 σ x 2 V 2 ) . (18) We are left with bounding the second term in inequality (16) . Using Proposition 5.3 , we have that
∑ V − C ≤ u ≤ A σ x 2 P ( S x ( t ) ∈ [ u − 1 , u ] ) ≪ ∑ V − C ≤ u ≤ A σ x 2 1 σ x exp ( − u 2 2 σ x 2 ) ≪ σ x V − C exp ( − ( V − C ) 2 2 σ x 2 ) ≪ 1 σ x exp ( − V 2 2 σ x 2 ) . \begin{aligned}
\sum_{V-C\le u\le A\sigma_x^2}\mathbb{P}(S_x(t)\in[u-1,u])\ll\sum_{V-C\le u\le A\sigma_x^2}\frac{1}{\sigma_x}\exp\left(-\frac{u^2}{2\sigma_x^2}\right) \\
&\ll\frac{\sigma_x}{V-C}\exp\left(-\frac{(V-C)^2}{2\sigma_x^2}\right)\ll\frac{1}{\sigma_x}\exp\left(-\frac{V^2}{2\sigma_x^2}\right).
\end{aligned} V − C ≤ u ≤ A σ x 2 ∑ P ( S x ( t ) ∈ [ u − 1 , u ]) ≪ V − C ≤ u ≤ A σ x 2 ∑ σ x 1 exp ( − 2 σ x 2 u 2 ) ≪ V − C σ x exp ( − 2 σ x 2 ( V − C ) 2 ) ≪ σ x 1 exp ( − 2 σ x 2 V 2 ) . Combining this and the bound (18) finishes the proof of Theorem 1.7 in the regime (13) .
Preliminary steps – small α \alpha α
In this section, we consider the more difficult regime when
1 ≪ α log T ≪ ( log 1 α ) 12 + ε . (19) 1\ll\alpha\log T\ll\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. \tag*{(19)} 1 ≪ α log T ≪ ( log α 1 ) 12 + ε . (19) More specifically, assume
M 0 ≤ α log T ≤ M 1 ( log 1 α ) 12 + ε , M_0\le\alpha\log T\le M_1\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}, M 0 ≤ α log T ≤ M 1 ( log α 1 ) 12 + ε , for some M 0 , M 1 > 0 M_0,M_1>0 M 0 , M 1 > 0 T j = T β j T_j=T^{\beta_j} T j = T β j
β 1 = 1 ( log log T ) 30 , β j = r j − 1 β 1 , (20) \beta_1=\frac{1}{(\log\log T)^{30}},\qquad\beta_j=r^{j-1}\beta_1, \tag*{(20)} β 1 = ( log log T ) 30 1 , β j = r j − 1 β 1 , (20) for some r > 1 r>1 r > 1 r r r § 7 for more details). We stop when θ / r < β K ≤ θ \theta/r<\beta_K\le\theta θ / r < β K ≤ θ θ \theta θ θ \theta θ Propositions 5.1 and 5.2 and Lemmas 6.1 , 6.2 , 6.3 , 6.4. We do not record them all here, but choose θ \theta θ
The “good events” G ℓ G_\ell G ℓ 1 ≤ ℓ ≤ K 1\le\ell\le K 1 ≤ ℓ ≤ K
G ℓ = { S ℓ ( j ) ∈ [ L ℓ , U ℓ ] , ∀ j ≥ ℓ } ∩ G ℓ − 1 , G_\ell=\left\{S_\ell^{(j)}\in[L_\ell,U_\ell],\ \forall j\ge\ell\right\}\cap G_{\ell-1}, G ℓ = { S ℓ ( j ) ∈ [ L ℓ , U ℓ ] , ∀ j ≥ ℓ } ∩ G ℓ − 1 , where
L ℓ = V log log T log log T ℓ − c ℓ , U ℓ = V log log T log log T ℓ + c ℓ . (21) L_\ell=\frac{V}{\log\log T}\log\log T_\ell-c_\ell,\qquad U_\ell=\frac{V}{\log\log T}\log\log T_\ell+c_\ell. \tag*{(21)} L ℓ = log log T V log log T ℓ − c ℓ , U ℓ = log log T V log log T ℓ + c ℓ . (21) with
c ℓ = ( 1 β ℓ ) 1 / 90 . c_\ell= \left(\frac{1}{\beta_\ell}\right)^{1/90}. c ℓ = ( β ℓ 1 ) 1/90 . We use the conventions
S 0 ( j ) = 0 , G 0 = [ T , 2 T ] . S_0^{(j)}=0,\qquad G_0=[T,2T]. S 0 ( j ) = 0 , G 0 = [ T , 2 T ] . For simplicity of notation, let
z ℓ = log 1 β ℓ . z_\ell=\log\frac{1}{\beta_\ell}. z ℓ = log β ℓ 1 . Let
σ ℓ , j 2 = Var ( S ℓ ( j ) ) . \sigma_{\ell,j}^2=\operatorname{Var}(S_\ell^{(j)}). σ ℓ , j 2 = Var ( S ℓ ( j ) ) . Using Lemma 2.3 , we have that
σ ℓ , j 2 = t ℓ 2 + { O ( 1 ) if α log T j ≥ 1 , O ( z ℓ 2 ) if α log T j < 1. (22) \sigma_{\ell,j}^2=\frac{t_\ell}{2}+
\begin{cases}
O(1) & \text{if }\alpha\log T_j\ge1,\\
O(z_\ell^2) & \text{if }\alpha\log T_j<1.
\end{cases} \tag*{(22)} σ ℓ , j 2 = 2 t ℓ + { O ( 1 ) O ( z ℓ 2 ) if α log T j ≥ 1 , if α log T j < 1. (22) where
t ℓ : = min { log 1 α , log log T ℓ } . t_\ell:=\min\left\{\log\frac{1}{\alpha},\log\log T_\ell\right\}. t ℓ := min { log α 1 , log log T ℓ } . We will frequently use the following corollary to Lemma 2.2 .
Corollary 4.1. We have
b α ( p ; j ) = − 1 + O ( z j log p log T j ) , p ≤ T j , b_\alpha(p;j)=-1+O\left(\frac{z_j\log p}{\log T_j}\right),\qquad p\le T_j, b α ( p ; j ) = − 1 + O ( log T j z j log p ) , p ≤ T j , b α ( p 2 ; j ) = − 1 2 + O ( z j log p log T j ) , p 2 ≤ T j . b_\alpha(p^2;j)=-\frac{1}{2}+O\left(\frac{z_j\log p}{\log T_j}\right),\qquad p^2\le T_j. b α ( p 2 ; j ) = − 2 1 + O ( log T j z j log p ) , p 2 ≤ T j . For 2 ≤ ℓ ≤ K 2\le\ell\le K 2 ≤ ℓ ≤ K
A ℓ = { ∣ S ℓ ( j ) − S ℓ − 1 ( j ) ∣ ≤ 2 c ℓ , ∀ ℓ ≤ j ≤ K } . A_\ell=\left\{\lvert S_\ell^{(j)}-S_{\ell-1}^{(j)}\rvert\le2c_\ell,\ \forall\ell\le j\le K\right\}. A ℓ = { ∣ S ℓ ( j ) − S ℓ − 1 ( j ) ∣ ≤ 2 c ℓ , ∀ ℓ ≤ j ≤ K } . We have that
P ( H ) ≤ P ( H ∩ G 1 c ) + ∑ ℓ = 2 K ( P ( G ℓ − 1 ∩ A ℓ c ) + P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) \mathbb{P}(H)\le\mathbb{P}(H\cap G_1^c)+\sum_{\ell=2}^{K}\left(\mathbb{P}(G_{\ell-1}\cap A_\ell^c)+\mathbb{P}\left(G_{\ell-1}\cap A_\ell\cap\{\exists j\ge\ell:S_\ell^{(j)}>U_\ell\}\right)\right. P ( H ) ≤ P ( H ∩ G 1 c ) + ℓ = 2 ∑ K ( P ( G ℓ − 1 ∩ A ℓ c ) + P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) + P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ) + P ( H ∩ G K ) . (23) \left.+\mathbb{P}\left(H\cap G_{\ell-1}\cap A_\ell\cap\{\exists j\ge\ell:S_\ell^{(j)}<L_\ell\}\right)\right)+\mathbb{P}(H\cap G_K). \tag*{(23)} + P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ) + P ( H ∩ G K ) . (23) We will bound the probabilities above in § 7 . Before doing that, in the next section, we prove some key estimates which are used to bound (23) .
Some localization results
In this section, we will prove three key localization bounds, which are analogues of Lemmas B.1 and B.2 in [3 ] . These will be used in the proof of Theorem 1.7 .
We first assume that
1 ≪ α log T ≪ ( log 1 α ) 12 + ε . 1\ll\alpha\log T\ll\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}. 1 ≪ α log T ≪ ( log α 1 ) 12 + ε . We denote by
G ℓ ( j ) ( u ) = G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } . G_\ell^{(j)}(u)=G_{\ell-1}\cap\{S_\ell^{(j)}\in[u-1,u]\}. G ℓ ( j ) ( u ) = G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} . For a sufficiently large constant C C C
δ l o c = C ( β 1 ( log log T ) 11 + ∑ i = 2 K β i ( 1 + z i 2 ) c i 45 / 2 ) . (24) \delta_{\mathrm{loc}}=C\left(\beta_1(\log\log T)^{11}+\sum_{i=2}^{K}\beta_i(1+z_i^2)c_i^{45/2}\right). \tag*{(24)} δ loc = C ( β 1 ( log log T ) 11 + i = 2 ∑ K β i ( 1 + z i 2 ) c i 45/2 ) . (24) Fix η < 1 \eta< 1 η < 1 θ \theta θ T T T
δ l o c ≤ 1 − η 2 . \delta_{\mathrm{loc}} \le\frac{1-\eta}{2}. δ loc ≤ 2 1 − η . Proposition 5.1. Let 1 ≤ ℓ ≤ j ≤ K 1 \le\ell\le j \le K 1 ≤ ℓ ≤ j ≤ K Q Q Q T η T^\eta T η T ℓ T_\ell T ℓ η < 1 \eta< 1 η < 1
η + δ l o c < 1. (25) \eta+\delta_{\mathrm{loc}}<1. \tag*{(25)} η + δ loc < 1. (25) Put
L ℓ ′ : = L ℓ − 3 c ℓ = V log log T log log T ℓ − 4 c ℓ , U ℓ ′ : = U ℓ + 3 c ℓ = V log log T log log T ℓ + 4 c ℓ . L'_\ell:= L_\ell-3c_\ell=\frac{V}{\log\log T}\log\log T_\ell-4c_\ell,\qquad U'_\ell:= U_\ell+3c_\ell=\frac{V}{\log\log T}\log\log T_\ell+4c_\ell. L ℓ ′ := L ℓ − 3 c ℓ = log log T V log log T ℓ − 4 c ℓ , U ℓ ′ := U ℓ + 3 c ℓ = log log T V log log T ℓ + 4 c ℓ . Assume ∣ u ∣ ≤ A σ ℓ , j 2 |u|\le A\sigma_{\ell,j}^{2} ∣ u ∣ ≤ A σ ℓ , j 2 A > 0 A>0 A > 0 ℓ ≥ 2 \ell\ge2 ℓ ≥ 2 u ∈ [ L ℓ ′ , U ℓ ′ ] u\in[L'_\ell,U'_\ell] u ∈ [ L ℓ ′ , U ℓ ′ ]
E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) ) ≪ E r a n d ( ∣ Q ~ ∣ 2 ) e − u 2 / ( 2 σ ℓ , j 2 ) σ ℓ , j . \mathbb{E}\left(|Q|^2\mathbf{1}\left(G_\ell^{(j)}(u)\right)\right)\ll\mathbb{E}_{\mathrm{rand}}\left(|\widetilde{Q}|^2\right)\frac{e^{-u^2/(2\sigma_{\ell,j}^2)}}{\sigma_{\ell,j}}. E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) ) ≪ E rand ( ∣ Q ∣ 2 ) σ ℓ , j e − u 2 / ( 2 σ ℓ , j 2 ) . Proof. Let
Y i = S i ( j ) − S i − 1 ( j ) Y_i=S_i^{(j)}-S_{i-1}^{(j)} Y i = S i ( j ) − S i − 1 ( j ) for 1 ≤ i ≤ ℓ 1\le i\le\ell 1 ≤ i ≤ ℓ j j j Y i Y_i Y i j j j
Δ 1 = C 1 σ 1 , j , Δ i = c i ( 2 ≤ i ≤ ℓ ) , \Delta_1=C_1\sigma_{1,j},\qquad\Delta_i=c_i\qquad(2\le i\le\ell), Δ 1 = C 1 σ 1 , j , Δ i = c i ( 2 ≤ i ≤ ℓ ) , where C 1 C_1 C 1 θ \theta θ
∑ i ≤ K Δ i − 1 ≤ 1 2 and 2 ∑ i ≤ K Δ i − 3 ≤ 1 2 . \sum_{i\le K}\Delta_i^{-1}\le\frac12\qquad\text{and}\qquad2\sum_{i\le K}\Delta_i^{-3}\le\frac12. i ≤ K ∑ Δ i − 1 ≤ 2 1 and 2 i ≤ K ∑ Δ i − 3 ≤ 2 1 . Fix τ ∈ [ 0 , 1 ) \tau\in[0,1) τ ∈ [ 0 , 1 ) [ u i , u i + Δ i − 1 ) [u_i,u_i+\Delta_i^{-1}) [ u i , u i + Δ i − 1 )
u i : = ( n i + τ ) ( Δ i − 1 + 2 Δ i − 4 ) , u_i:=(n_i+\tau)(\Delta_i^{-1}+2\Delta_i^{-4}), u i := ( n i + τ ) ( Δ i − 1 + 2 Δ i − 4 ) , with n i ∈ Z n_i\in\mathbb{Z} n i ∈ Z i i i n i n_i n i u i u_i u i 2 Δ i − 4 2\Delta_i^{-4} 2 Δ i − 4
Suppose that each Y i Y_i Y i 1 ≤ i ≤ ℓ 1\le i\le\ell 1 ≤ i ≤ ℓ u i u_i u i
0 ≤ Y i − u i < Δ i − 1 . 0\le Y_i-u_i<\Delta_i^{-1}. 0 ≤ Y i − u i < Δ i − 1 . Summing these inequalities, we obtain that
0 ≤ S q ( j ) − ∑ i = 1 q u i = ∑ i = 1 q ( Y i − u i ) < ∑ i = 1 q Δ i − 1 ≤ 1 2 , 0\le S_q^{(j)}-\sum_{i=1}^{q}u_i=\sum_{i=1}^{q}(Y_i-u_i)<\sum_{i=1}^{q}\Delta_i^{-1}\le\frac12, 0 ≤ S q ( j ) − i = 1 ∑ q u i = i = 1 ∑ q ( Y i − u i ) < i = 1 ∑ q Δ i − 1 ≤ 2 1 , for every 1 ≤ q ≤ ℓ 1\le q\le\ell 1 ≤ q ≤ ℓ S q ( j ) ∈ [ L q , U q ] S_q^{(j)}\in[L_q,U_q] S q ( j ) ∈ [ L q , U q ] ∑ i ≤ q u i ∈ [ L q − 1 , U q ] \sum_{i\le q}u_i\in[L_q-1,U_q] ∑ i ≤ q u i ∈ [ L q − 1 , U q ] S ℓ ( j ) ∈ [ u − 1 , u ] S_\ell^{(j)}\in[u-1,u] S ℓ ( j ) ∈ [ u − 1 , u ] ∑ i ≤ ℓ u i ∈ [ u − 3 2 , u ] \sum_{i\le\ell}u_i\in[u-\frac32,u] ∑ i ≤ ℓ u i ∈ [ u − 2 3 , u ]
Now let I ( u ) \mathcal{I}(u) I ( u ) u = ( u 1 , … , u ℓ ) \mathbf{u}=(u_1,\ldots,u_\ell) u = ( u 1 , … , u ℓ )
u i = ( n i + τ ) ( Δ i − 1 + 2 Δ i − 4 ) , n i ∈ Z ( 1 ≤ i ≤ ℓ ) , u_i=(n_i+\tau)(\Delta_i^{-1}+2\Delta_i^{-4}),\qquad n_i\in\mathbb{Z}\qquad(1\le i\le\ell), u i = ( n i + τ ) ( Δ i − 1 + 2 Δ i − 4 ) , n i ∈ Z ( 1 ≤ i ≤ ℓ ) , ∑ i ≤ q u i ∈ [ L q − 1 , U q ] ( q < ℓ ) , \sum_{i\le q}u_i\in[L_q-1,U_q]\qquad(q<\ell), i ≤ q ∑ u i ∈ [ L q − 1 , U q ] ( q < ℓ ) , ∑ i ≤ ℓ u i ∈ [ u − 3 2 , u ] . \sum_{i\le\ell}u_i\in\left[u-\frac32,u\right]. i ≤ ℓ ∑ u i ∈ [ u − 2 3 , u ] . If t ∈ G ℓ − 1 t \in G_{\ell-1} t ∈ G ℓ − 1 S ℓ ( j ) ( t ) ∈ [ u − 1 , u ] S_{\ell}^{(j)}(t) \in[u-1,u] S ℓ ( j ) ( t ) ∈ [ u − 1 , u ] Y i ( t ) Y_i(t) Y i ( t ) ( u 1 , … , u ℓ ) ∈ I ( u ) (u_1,\ldots,u_\ell) \in\mathcal{I}(u) ( u 1 , … , u ℓ ) ∈ I ( u )
For each i i i τ \tau τ Y i Y_i Y i 2 Δ i − 4 2\Delta_i^{-4} 2 Δ i − 4
2 Δ i − 4 Δ i − 1 + 2 Δ i − 4 ≤ 2 Δ i − 3 . \frac{2\Delta_i^{-4}}{\Delta_i^{-1}+2\Delta_i^{-4}} \le2\Delta_i^{-3}. Δ i − 1 + 2 Δ i − 4 2 Δ i − 4 ≤ 2 Δ i − 3 . So we have
meas { τ ∈ [ 0 , 1 ) : Y i lies in a gap for some i ≤ ℓ } ≤ 2 ∑ i ≤ ℓ Δ i − 3 ≤ 1 2 , \operatorname{meas}\{\tau\in[0,1):Y_i\text{ lies in a gap for some }i\le\ell\}\le2\sum_{i\le\ell}\Delta_i^{-3}\le\frac{1}{2}, meas { τ ∈ [ 0 , 1 ) : Y i lies in a gap for some i ≤ ℓ } ≤ 2 i ≤ ℓ ∑ Δ i − 3 ≤ 2 1 , and, hence, for a set of phases of measure at least 1 / 2 1/2 1/2 Y i Y_i Y i
Multiplying by ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) |\mathcal{Q}|^2\mathbf{1}(G_{\ell}^{(j)}(u)) ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) τ \tau τ
∫ 0 1 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) 1 { every Y i lies in one of the displayed intervals at phase τ } ) d τ ≥ 1 2 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) ) . \int_0^1 \mathbb{E}\left(|\mathcal{Q}|^2\mathbf{1}(G_{\ell}^{(j)}(u))\mathbf{1}_{\{\text{every }Y_i\text{ lies in one of the displayed intervals at phase }\tau\}}\right)\,d\tau
\ge\frac{1}{2}\mathbb{E}\left(|\mathcal{Q}|^2\mathbf{1}(G_{\ell}^{(j)}(u))\right). ∫ 0 1 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) 1 { every Y i lies in one of the displayed intervals at phase τ } ) d τ ≥ 2 1 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) ) . Hence, for at least one phase τ \tau τ 1 2 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) ) \frac{1}{2}\mathbb{E}(|\mathcal{Q}|^2\mathbf{1}(G_{\ell}^{(j)}(u))) 2 1 E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )))
∑ u ∈ I ( u ) E ( ∣ Q ∣ 2 ∏ i ≤ ℓ 1 ( Y i ∈ [ u i , u i + Δ i − 1 ] ) ) . \sum_{u\in\mathcal{I}(u)}\mathbb{E}\left(|\mathcal{Q}|^2\prod_{i\le\ell}\mathbf{1}(Y_i\in[u_i,u_i+\Delta_i^{-1}])\right). u ∈ I ( u ) ∑ E ( ∣ Q ∣ 2 i ≤ ℓ ∏ 1 ( Y i ∈ [ u i , u i + Δ i − 1 ]) ) . Note that we must have
∣ u 1 ∣ ≪ V and ∣ u i ∣ ≪ c i ( i ≥ 2 ) . (26) |u_1|\ll V \qquad\text{and}\qquad|u_i|\ll c_i\quad(i\ge2). \tag*{(26)} ∣ u 1 ∣ ≪ V and ∣ u i ∣ ≪ c i ( i ≥ 2 ) . (26) We use the polynomials in equation (B.7) of [3 ] with the parameter a = 4 a=4 a = 4 Δ \Delta Δ X ≥ 1 X\ge1 X ≥ 1 D Δ , X D_{\Delta,X} D Δ , X 100 X Δ 12 100X\Delta^{12} 100 X Δ 12 n n n ( 2 π ) n n ! 2 Δ 8 ( n + 2 ) \frac{(2\pi)^n}{n!}2\Delta^{8(n+2)} n ! ( 2 π ) n 2 Δ 8 ( n + 2 ) ∣ x ∣ ≤ X |x|\le X ∣ x ∣ ≤ X
1 ( x ∈ [ 0 , Δ − 1 ] ) ( 1 − e − Δ 2 ) ≤ ∣ D Δ , X ( x ) ∣ 2 ≤ 1 ( x ∈ [ − Δ − 4 , Δ − 1 + Δ − 4 ] ) + e − Δ 2 . (27) \mathbf{1}(x\in[0,\Delta^{-1}])(1-e^{-\Delta^2})\le|D_{\Delta,X}(x)|^2\le\mathbf{1}(x\in[-\Delta^{-4},\Delta^{-1}+\Delta^{-4}])+e^{-\Delta^2}. \tag*{(27)} 1 ( x ∈ [ 0 , Δ − 1 ]) ( 1 − e − Δ 2 ) ≤ ∣ D Δ , X ( x ) ∣ 2 ≤ 1 ( x ∈ [ − Δ − 4 , Δ − 1 + Δ − 4 ]) + e − Δ 2 . (27) Take
X 1 = C 2 Δ 10 , X i = C 2 ( 1 + z i 2 ) c i 8 ( i ≥ 2 ) X_1=C_2\Delta^{10},\qquad X_i=C_2(1+z_i^2)c_i^8\quad(i\ge2) X 1 = C 2 Δ 10 , X i = C 2 ( 1 + z i 2 ) c i 8 ( i ≥ 2 ) with C 2 C_2 C 2 D i = D Δ i , X i D_i=D_{\Delta_i,X_i} D i = D Δ i , X i
d 1 ≤ 100 C 2 Δ 1 22 ≪ ( log log T ) 11 and d i ≤ 100 C 2 ( 1 + z i 2 ) c i 20 ( i ≥ 2 ) . d_1\le100C_2\Delta_1^{22}\ll(\log\log T)^{11}\qquad\text{and}\qquad d_i\le100C_2(1+z_i^2)c_i^{20}\quad(i\ge2). d 1 ≤ 100 C 2 Δ 1 22 ≪ ( log log T ) 11 and d i ≤ 100 C 2 ( 1 + z i 2 ) c i 20 ( i ≥ 2 ) . Expanding the product ∣ Q ∣ 2 ∏ i ∣ D i ( Y i − u i ) ∣ 2 |\mathcal{Q}|^2\prod_i|D_i(Y_i-u_i)|^2 ∣ Q ∣ 2 ∏ i ∣ D i ( Y i − u i ) ∣ 2
( Q ∏ i ≤ ℓ ( S i ( j ) − S i − 1 ( j ) ) r i ) ( Q ∏ i ≤ ℓ ( S i ( j ) − S i − 1 ( j ) ) s i ) ‾ , r i + s i ≤ 2 d i . \left(\mathcal{Q}\prod_{i\le\ell}(S_i^{(j)}-S_{i-1}^{(j)})^{r_i}\right)\overline{\left(\mathcal{Q}\prod_{i\le\ell}(S_i^{(j)}-S_{i-1}^{(j)})^{s_i}\right)},\qquad r_i+s_i\le2d_i. ( Q i ≤ ℓ ∏ ( S i ( j ) − S i − 1 ( j ) ) r i ) ( Q i ≤ ℓ ∏ ( S i ( j ) − S i − 1 ( j ) ) s i ) , r i + s i ≤ 2 d i . Each of the two Dirichlet polynomials in such a term has length at most T η ∏ i ≤ ℓ T i 4 d i T^\eta\prod_{i\le\ell}T_i^{4d_i} T η ∏ i ≤ ℓ T i 4 d i
T η ∏ i ≤ ℓ T i 4 d i ≤ T η + C β 1 ( log log T ) 11 + C ∑ i = 2 ℓ β i ( 1 + z i 2 ) c i 20 ≤ T η + δ l o c < T , T^\eta\prod_{i\le\ell}T_i^{4d_i}\le T^{\eta+C\beta_1(\log\log T)^{11}+C\sum_{i=2}^{\ell}\beta_i(1+z_i^2)c_i^{20}}\le T^{\eta+\delta_{\mathrm{loc}}}<T, T η i ≤ ℓ ∏ T i 4 d i ≤ T η + C β 1 ( l o g l o g T ) 11 + C ∑ i = 2 ℓ β i ( 1 + z i 2 ) c i 20 ≤ T η + δ loc < T , by (5.2). Applying Montgomery–Vaughan [35 Corollary 2, (1.8)] gives an error of size O ( T η + δ l o c − 1 ) O(T^{\eta+\delta_{\mathrm{loc}}-1}) O ( T η + δ loc − 1 )
E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u ) ) ) ≪ E r a n d ( ∣ Q ~ ∣ 2 ) ∑ u ∈ I ( u ) ∏ i ≤ ℓ E r a n d ( ∣ D i ( Y ~ i − u i ) ∣ 2 ) . (28) \mathbb{E}\left(|\mathcal{Q}|^2\mathbf{1}(G_{\ell}^{(j)}(u))\right)\ll\mathbb{E}_{\mathrm{rand}}(|\widetilde{\mathcal{Q}}|^2)\sum_{u\in\mathcal{I}(u)}\prod_{i\le\ell}\mathbb{E}_{\mathrm{rand}}\left(|D_i(\widetilde{Y}_i-u_i)|^2\right). \tag*{(28)} E ( ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) ) ≪ E rand ( ∣ Q ∣ 2 ) u ∈ I ( u ) ∑ i ≤ ℓ ∏ E rand ( ∣ D i ( Y i − u i ) ∣ 2 ) . (28) Note that
Var ( Y ~ 1 ) = σ 1 , j 2 + o ( 1 ) ≍ log log T , Var ( Y ~ i ) ≪ z i 2 ( i ≥ 2 ) . \operatorname{Var}(\widetilde{Y}_1)=\sigma_{1,j}^2+o(1)\asymp\log\log T,\qquad\operatorname{Var}(\widetilde{Y}_i)\ll z_i^2\quad(i\ge2). Var ( Y 1 ) = σ 1 , j 2 + o ( 1 ) ≍ log log T , Var ( Y i ) ≪ z i 2 ( i ≥ 2 ) . Let
σ ~ 2 = Var ( S ~ ℓ ( j ) ) = σ ℓ , j 2 + o ( 1 ) . (29) \widetilde{\sigma}^{2}=\operatorname{Var}\left(\widetilde{S}_{\ell}^{(j)}\right)=\sigma_{\ell,j}^{2}+o(1). \tag*{(29)} σ 2 = Var ( S ℓ ( j ) ) = σ ℓ , j 2 + o ( 1 ) . (29) We write
Y ~ i = ∑ T i − 1 < p ≤ T i C j ( p ) , \widetilde{Y}_{i}=\sum_{T_{i-1}<p\leq T_i} C_j(p), Y i = T i − 1 < p ≤ T i ∑ C j ( p ) , where T 0 = 1 T_0=1 T 0 = 1
C j ( p ) = R ( b α ( p ; j ) X ( p ) p 1 / 2 + α + b α ( p 2 ; j ) X ( p ) p 1 + 2 α 1 { p 2 ≤ T j } ) . (30) C_j(p)=\mathfrak{R}\left(\frac{b_\alpha(p;j)X(p)}{p^{1/2+\alpha}}+\frac{b_\alpha(p^2;j)X(p)}{p^{1+2\alpha}}\mathbf{1}_{\{p^2\leq T_j\}}\right). \tag*{(30)} C j ( p ) = R ( p 1/2 + α b α ( p ; j ) X ( p ) + p 1 + 2 α b α ( p 2 ; j ) X ( p ) 1 { p 2 ≤ T j } ) . (30) Using Corollary 4.1, we have
m p : = ∥ C j ( p ) ∥ ∞ ≤ C p 1 / 2 + α ( 1 + z j log p log T j ) . (31) m_p:=\lVert C_j(p)\rVert_\infty\leq\frac{C}{p^{1/2+\alpha}}\left(1+\frac{z_j\log p}{\log T_j}\right). \tag*{(31)} m p := ∥ C j ( p ) ∥ ∞ ≤ p 1/2 + α C ( 1 + log T j z j log p ) . (31) Moreover,
∑ T i − 1 < p ≤ T i E r a n d ( C j ( p ) 2 ) = Var ( Y ~ i ) . \sum_{T_{i-1}<p\leq T_i}\mathbb{E}_{\mathrm{rand}}\left(C_j(p)^2\right)=\operatorname{Var}(\widetilde{Y}_i). T i − 1 < p ≤ T i ∑ E rand ( C j ( p ) 2 ) = Var ( Y i ) . Now Hoeffding’s lemma gives
E r a n d ( e λ Y ~ i ) ≤ exp ( λ 2 w i 2 ) (32) \mathbb{E}_{\mathrm{rand}}\left(e^{\lambda\widetilde{Y}_i}\right)\leq\exp\left(\frac{\lambda^2w_i}{2}\right) \tag*{(32)} E rand ( e λ Y i ) ≤ exp ( 2 λ 2 w i ) (32) for every λ ∈ R \lambda\in\mathbb{R} λ ∈ R
w i : = ∑ T i − 1 < p ≤ T i m p 2 . w_i:=\sum_{T_{i-1}<p\leq T_i}m_p^2. w i := T i − 1 < p ≤ T i ∑ m p 2 . We have
w 1 ≪ log log T and w i ≪ z i 2 ( i ≥ 2 ) . w_1\ll\log\log T\qquad\text{and}\qquad w_i\ll z_i^2\quad(i\geq2). w 1 ≪ log log T and w i ≪ z i 2 ( i ≥ 2 ) . For ∣ λ ∣ ≤ 3 A |\lambda|\leq3A ∣ λ ∣ ≤ 3 A E r a n d ( C j ( p ) ) = 0 \mathbb{E}_{\mathrm{rand}}(C_j(p))=0 E rand ( C j ( p )) = 0
log ( E r a n d ( e λ C j ( p ) ) ) = λ 2 2 E r a n d ( C j ( p ) 2 ) + O ( m p 3 ) . \log\left(\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda C_j(p)}\right)\right)=\frac{\lambda^2}{2}\mathbb{E}_{\mathrm{rand}}\left(C_j(p)^2\right)+O(m_p^3). log ( E rand ( e λ C j ( p ) ) ) = 2 λ 2 E rand ( C j ( p ) 2 ) + O ( m p 3 ) . By independence,
E r a n d ( e λ Y ~ i ) = exp ( λ 2 2 Var ( Y ~ i ) + O ( ∑ T i − 1 < p ≤ T i m p 3 ) ) . (33) \mathbb{E}_{\mathrm{rand}}\left(e^{\lambda\widetilde{Y}_i}\right)=\exp\left(\frac{\lambda^2}{2}\operatorname{Var}(\widetilde{Y}_i)+O\left(\sum_{T_{i-1}<p\leq T_i}m_p^3\right)\right). \tag*{(33)} E rand ( e λ Y i ) = exp 2 λ 2 Var ( Y i ) + O T i − 1 < p ≤ T i ∑ m p 3 . (33) From the coefficient bound (5.8), we get that
∑ T i − 1 < p ≤ T i m p 3 ≪ { 1 i = 1 , z i 3 T i − 1 − 1 / 2 i ≥ 2. \sum_{T_{i-1}<p\leq T_i}m_p^3\ll
\begin{cases}
1 & i=1,\\
z_i^3T_{i-1}^{-1/2} & i\geq2.
\end{cases} T i − 1 < p ≤ T i ∑ m p 3 ≪ { 1 z i 3 T i − 1 − 1/2 i = 1 , i ≥ 2. Now note that we have ∣ D i ( x ) ∣ ≤ 2 Δ i 16 e 2 π Δ i 8 ∣ x ∣ |\mathcal{D}_i(x)|\leq2\Delta_i^{16}e^{2\pi\Delta_i^8|x|} ∣ D i ( x ) ∣ ≤ 2 Δ i 16 e 2 π Δ i 8 ∣ x ∣
1 ( ∣ x ∣ > X i ) ≤ e 4 π Δ i 8 ( ∣ x ∣ − X i ) . (34) \mathbf{1}(|x|>X_i)\leq e^{4\pi\Delta_i^8(|x|-X_i)}. \tag*{(34)} 1 ( ∣ x ∣ > X i ) ≤ e 4 π Δ i 8 ( ∣ x ∣ − X i ) . (34) Hence
∣ D i ( Y ~ i − u i ) ∣ 2 1 ( ∣ Y ~ i − u i ∣ > X i ) ≤ 4 Δ i 32 e − 4 π Δ i 8 X i e 8 π Δ i 8 ∣ Y ~ i − u i ∣ . |\mathcal{D}_i(\widetilde{Y}_i-u_i)|^2\mathbf{1}(|\widetilde{Y}_i-u_i|>X_i)\leq4\Delta_i^{32}e^{-4\pi\Delta_i^8X_i}e^{8\pi\Delta_i^8|\widetilde{Y}_i-u_i|}. ∣ D i ( Y i − u i ) ∣ 2 1 ( ∣ Y i − u i ∣ > X i ) ≤ 4 Δ i 32 e − 4 π Δ i 8 X i e 8 π Δ i 8 ∣ Y i − u i ∣ . Using the triangle inequality,
E r a n d ( ∣ D i ( Y ~ i − u i ) ∣ 2 1 ( ∣ Y ~ i − u i ∣ > X i ) ) ≤ 4 Δ i 32 e − 4 π Δ i 8 X i e 8 π Δ i 8 ∣ u i ∣ E r a n d ( e 8 π Δ i 8 ∣ Y ~ i ∣ ) . (35) \mathbb{E}_{\mathrm{rand}}\left(|\mathcal{D}_i(\widetilde{Y}_i-u_i)|^2\mathbf{1}(|\widetilde{Y}_i-u_i|>X_i)\right)\leq4\Delta_i^{32}e^{-4\pi\Delta_i^8X_i}e^{8\pi\Delta_i^8|u_i|}\mathbb{E}_{\mathrm{rand}}\left(e^{8\pi\Delta_i^8|\widetilde{Y}_i|}\right). \tag*{(35)} E rand ( ∣ D i ( Y i − u i ) ∣ 2 1 ( ∣ Y i − u i ∣ > X i ) ) ≤ 4 Δ i 32 e − 4 π Δ i 8 X i e 8 π Δ i 8 ∣ u i ∣ E rand ( e 8 π Δ i 8 ∣ Y i ∣ ) . (35) Since
E r a n d ( e 8 π Δ i 8 ∣ Y ~ i ∣ ) ≤ E r a n d ( e 8 π Δ i 8 Y ~ i ) + E r a n d ( e − 8 π Δ i 8 Y ~ i ) , \mathbb{E}_{\mathrm{rand}}\left(e^{8\pi\Delta_i^8|\widetilde{Y}_i|}\right)\leq\mathbb{E}_{\mathrm{rand}}\left(e^{8\pi\Delta_i^8\widetilde{Y}_i}\right)+\mathbb{E}_{\mathrm{rand}}\left(e^{-8\pi\Delta_i^8\widetilde{Y}_i}\right), E rand ( e 8 π Δ i 8 ∣ Y i ∣ ) ≤ E rand ( e 8 π Δ i 8 Y i ) + E rand ( e − 8 π Δ i 8 Y i ) , and, by Hoeffding’s bound (5.9),
E r a n d ( e ± 8 π Δ i 8 Y ~ i ) ≤ e 32 π 2 Δ i 16 w i , \mathbb{E}_{\mathrm{rand}}\left(e^{\pm8\pi\Delta_i^8\widetilde{Y}_i}\right)\leq e^{32\pi^2\Delta_i^{16}w_i}, E rand ( e ± 8 π Δ i 8 Y i ) ≤ e 32 π 2 Δ i 16 w i , it follows that
E r a n d ( ∣ D i ( Y ~ i − u i ) ∣ 2 1 ( ∣ Y ~ i − u i ∣ > X i ) ) ≤ 8 Δ i 32 exp ( − 4 π Δ i 8 X i + 8 π Δ i 8 ∣ u i ∣ + 32 π 2 Δ i 16 w i ) ≤ e − Δ i 2 . (36) \mathbb{E}_{\mathrm{rand}}\left(\left|\mathcal{D}_i(\widetilde{Y}_i-u_i)\right|^2\mathbf{1}\left(\left|\widetilde{Y}_i-u_i\right|>X_i\right)\right)\leq8\Delta_i^{32}\exp\left(-4\pi\Delta_i^8X_i+8\pi\Delta_i^8|u_i|+32\pi^2\Delta_i^{16}w_i\right)\leq e^{-\Delta_i^2}. \tag*{(36)} E rand ( D i ( Y i − u i ) 2 1 ( Y i − u i > X i ) ) ≤ 8 Δ i 32 exp ( − 4 π Δ i 8 X i + 8 π Δ i 8 ∣ u i ∣ + 32 π 2 Δ i 16 w i ) ≤ e − Δ i 2 . (36) Combining (36) with (27) on the range ∣ Y ~ i − u i ∣ ≤ X i \left|\widetilde{Y}_i-u_i\right|\leq X_i Y i − u i ≤ X i
E r a n d ( ∣ D i ( Y ~ i − u i ) ∣ 2 ) ≤ P r a n d ( Y ~ i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) + 2 e − Δ i 2 . (37) \mathbb{E}_{\mathrm{rand}}\left(\left|\mathcal{D}_i(\widetilde{Y}_i-u_i)\right|^2\right)\leq\mathbb{P}_{\mathrm{rand}}\left(\widetilde{Y}_i\in\left[u_i-\Delta_i^{-4},u_i+\Delta_i^{-1}+\Delta_i^{-4}\right]\right)+2e^{-\Delta_i^2}. \tag*{(37)} E rand ( D i ( Y i − u i ) 2 ) ≤ P rand ( Y i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) + 2 e − Δ i 2 . (37) For simplicity of notation, let
P i ( u i ) = P r a n d ( Y ~ i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) . (38) P_i(u_i)=\mathbb{P}_{\mathrm{rand}}\left(\widetilde{Y}_i\in\left[u_i-\Delta_i^{-4},u_i+\Delta_i^{-1}+\Delta_i^{-4}\right]\right). \tag*{(38)} P i ( u i ) = P rand ( Y i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) . (38) Take
λ = u σ ~ 2 . \lambda=\frac{u}{\widetilde{\sigma}^2}. λ = σ 2 u . Since ∣ u ∣ ≤ A σ ℓ , j 2 |u|\leq A\sigma_{\ell,j}^2 ∣ u ∣ ≤ A σ ℓ , j 2 σ ~ 2 = σ ℓ , j 2 + o ( 1 ) \widetilde{\sigma}^2=\sigma_{\ell,j}^2+o(1) σ 2 = σ ℓ , j 2 + o ( 1 ) (29) , we have ∣ λ ∣ ≤ 3 A |\lambda|\leq3A ∣ λ ∣ ≤ 3 A T T T (33) with this choice.
For a tuple u ∈ I ( u ) \mathbf{u}\in I(u) u ∈ I ( u ) ∑ i u i ∈ [ u − 2 , u ] \sum_i u_i\in[u-2,u] ∑ i u i ∈ [ u − 2 , u ] 1 ≤ e λ ( ∑ i u i − u ) + 2 ∣ λ ∣ 1\leq e^{\lambda(\sum_i u_i-u)+2|\lambda|} 1 ≤ e λ ( ∑ i u i − u ) + 2∣ λ ∣ (28) –(37) give
∑ u ∈ I ( u ) ∏ i ≤ ℓ ( P i ( u i ) + 2 e − Δ i 2 ) ≤ e 2 ∣ λ ∣ − λ u ∑ u ∈ I ( u ) ∏ i ≤ ℓ e λ u i ( P i ( u i ) + 2 e − Δ i 2 ) . \sum_{\mathbf{u}\in I(u)}\prod_{i\leq\ell}\left(P_i(u_i)+2e^{-\Delta_i^2}\right)\leq e^{2|\lambda|-\lambda u}\sum_{\mathbf{u}\in I(u)}\prod_{i\leq\ell}e^{\lambda u_i}\left(P_i(u_i)+2e^{-\Delta_i^2}\right). u ∈ I ( u ) ∑ i ≤ ℓ ∏ ( P i ( u i ) + 2 e − Δ i 2 ) ≤ e 2∣ λ ∣ − λ u u ∈ I ( u ) ∑ i ≤ ℓ ∏ e λ u i ( P i ( u i ) + 2 e − Δ i 2 ) . Let B i ( u i ) B_i(u_i) B i ( u i ) i t h i^{\mathrm{th}} i th
∑ u ∈ I ( u ) ∏ i ≤ ℓ B i ( u i ) = ∑ ∣ u i ∣ ≤ C c i 2 ≤ i ≤ ℓ ( ∏ i = 2 ℓ B i ( u i ) ) ∑ u 1 : ( u 1 , … , u ℓ ) ∈ I ( u ) B 1 ( u 1 ) . \sum_{\mathbf{u}\in I(u)}\prod_{i\leq\ell}B_i(u_i)=\sum_{\substack{|u_i|\leq Cc_i\\2\leq i\leq\ell}}\left(\prod_{i=2}^{\ell}B_i(u_i)\right)\sum_{\substack{u_1:(u_1,\ldots,u_\ell)\in I(u)}}B_1(u_1). u ∈ I ( u ) ∑ i ≤ ℓ ∏ B i ( u i ) = ∣ u i ∣ ≤ C c i 2 ≤ i ≤ ℓ ∑ ( i = 2 ∏ ℓ B i ( u i ) ) u 1 : ( u 1 , … , u ℓ ) ∈ I ( u ) ∑ B 1 ( u 1 ) . Fixing ( u i ) i ≥ 2 (u_i)_{i\geq2} ( u i ) i ≥ 2 ( u 1 , … , u ℓ ) ∈ I ( u ) (u_1,\ldots,u_\ell)\in I(u) ( u 1 , … , u ℓ ) ∈ I ( u ) u 1 u_1 u 1 I I I O ( 1 ) O(1) O ( 1 ) I ^ \widehat I I I I I ∣ I ^ ∣ = ∣ I ∣ + 4 = O ( 1 ) |\widehat I|=|I|+4=O(1) ∣ I ∣ = ∣ I ∣ + 4 = O ( 1 )
If u 1 ∈ I u_1\in I u 1 ∈ I Y ~ 1 ∈ [ u 1 − Δ 1 − 4 , u 1 + Δ 1 − 1 + Δ 1 − 4 ] \widetilde{Y}_1\in[u_1-\Delta_1^{-4},u_1+\Delta_1^{-1}+\Delta_1^{-4}] Y 1 ∈ [ u 1 − Δ 1 − 4 , u 1 + Δ 1 − 1 + Δ 1 − 4 ] ∣ Y ~ 1 − u 1 ∣ ≤ 2 |\widetilde{Y}_1-u_1|\leq2 ∣ Y 1 − u 1 ∣ ≤ 2 Y ~ 1 ∈ I ^ \widetilde{Y}_1\in\widehat I Y 1 ∈ I ∣ λ ∣ ≤ 3 A |\lambda|\leq3A ∣ λ ∣ ≤ 3 A e λ u 1 ≪ e λ Y ~ 1 e^{\lambda u_1}\ll e^{\lambda\widetilde{Y}_1} e λ u 1 ≪ e λ Y 1 Y ~ 1 \widetilde{Y}_1 Y 1
∑ u 1 ∈ I e λ u 1 P r a n d ( Y ~ 1 ∈ [ u 1 − Δ 1 − 4 , u 1 + Δ 1 − 1 + Δ 1 − 4 ] ) ≪ E r a n d ( e λ Y ~ 1 1 ( Y ~ 1 ∈ I ^ ) ) = E r a n d ( e λ Y ~ 1 ) P λ ( Y ~ 1 ∈ I ^ ) . (39) \begin{aligned}
\sum_{u_1\in I}e^{\lambda u_1}\mathbb{P}_{\mathrm{rand}}\left(\widetilde{Y}_1\in\left[u_1-\Delta_1^{-4},u_1+\Delta_1^{-1}+\Delta_1^{-4}\right]\right)\ll\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda\widetilde{Y}_1}\mathbf{1}(\widetilde{Y}_1\in\widehat I)\right) \\
&=\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda\widetilde{Y}_1}\right)\mathbb{P}_{\lambda}(\widetilde{Y}_1\in\widehat I). \tag*{(39)}
\end{aligned} u 1 ∈ I ∑ e λ u 1 P rand ( Y 1 ∈ [ u 1 − Δ 1 − 4 , u 1 + Δ 1 − 1 + Δ 1 − 4 ] ) ≪ E rand ( e λ Y 1 1 ( Y 1 ∈ I ) ) = E rand ( e λ Y 1 ) P λ ( Y 1 ∈ I ) . (39) where P λ \mathbb{P}_{\lambda} P λ
P λ ( E ) : = E r a n d [ e λ Y ~ 1 1 E ] E r a n d ( e λ Y ~ 1 ) . \mathbb{P}_{\lambda}(E):=\frac{\mathbb{E}_{\mathrm{rand}}\left[e^{\lambda\widetilde{Y}_1}\mathbf{1}_E\right]}{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda\widetilde{Y}_1}\right)}. P λ ( E ) := E rand ( e λ Y 1 ) E rand [ e λ Y 1 1 E ] . We denote E P λ \mathbb{E}_{\mathbb{P}_{\lambda}} E P λ Var P λ \operatorname{Var}_{\mathbb{P}_{\lambda}} Var P λ P λ \mathbb{P}_{\lambda} P λ
sup x P λ ( Y ~ 1 ∈ [ x , x + O ( 1 ) ] ) ≪ 1 σ ℓ , j . (40) \sup_x\mathbb{P}_{\lambda}\left(\widetilde{Y}_1\in[x,x+O(1)]\right)\ll\frac{1}{\sigma_{\ell,j}}. \tag*{(40)} x sup P λ ( Y 1 ∈ [ x , x + O ( 1 )] ) ≪ σ ℓ , j 1 . (40) where we used the fact that Var P λ ( Y ~ 1 ) = σ 1 , j 2 + O ( 1 ) ≍ σ ℓ , j 2 \operatorname{Var}_{\mathbb{P}_{\lambda}}(\widetilde{Y}_1)=\sigma_{1,j}^2+O(1)\asymp\sigma_{\ell,j}^2 Var P λ ( Y 1 ) = σ 1 , j 2 + O ( 1 ) ≍ σ ℓ , j 2 (26) , and we have
∑ ∣ u 1 ∣ ≪ log log T e λ u 1 2 e − Δ 1 2 ≪ Δ 1 log log T e C log log T − Δ 1 2 ≤ e − Δ 1 2 / 2 ≤ 1 σ ℓ , j . \sum_{|u_1|\ll\log\log T}e^{\lambda u_1}2e^{-\Delta_1^2}\ll\Delta_1\log\log T\,e^{C\log\log T-\Delta_1^2}\leq e^{-\Delta_1^2/2}\leq\frac{1}{\sigma_{\ell,j}}. ∣ u 1 ∣ ≪ l o g l o g T ∑ e λ u 1 2 e − Δ 1 2 ≪ Δ 1 log log T e C l o g l o g T − Δ 1 2 ≤ e − Δ 1 2 /2 ≤ σ ℓ , j 1 . Since E r a n d ( e λ Y ~ 1 ) ≥ 1 \mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_1}) \ge1 E rand ( e λ Y 1 ) ≥ 1
∑ u 1 : ( u 1 , … , u ℓ ) ∈ I ( u ) B 1 ( u 1 ) ≪ E r a n d ( e λ Y ~ 1 ) σ ℓ , j . (41) \sum_{u_1:(u_1,\ldots,u_\ell)\in I(u)} B_1(u_1) \ll\frac{\mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_1})}{\sigma_{\ell,j}}. \tag*{(41)} u 1 : ( u 1 , … , u ℓ ) ∈ I ( u ) ∑ B 1 ( u 1 ) ≪ σ ℓ , j E rand ( e λ Y 1 ) . (41) Now for i ≥ 2 i \ge2 i ≥ 2 ∣ u i ∣ ≪ c i |u_i| \ll c_i ∣ u i ∣ ≪ c i
∑ ∣ u i ∣ ≪ c i e λ u i 2 e − Δ i 2 ≪ c i 2 e − c i 2 + O ( c i ) . \sum_{|u_i|\ll c_i} e^{\lambda u_i}2e^{-\Delta_i^2} \ll c_i^2e^{-c_i^2+O(c_i)}. ∣ u i ∣ ≪ c i ∑ e λ u i 2 e − Δ i 2 ≪ c i 2 e − c i 2 + O ( c i ) . For the term involving P i ( u i ) P_i(u_i) P i ( u i ) i = 1 i=1 i = 1
∑ ∣ u i ∣ ≪ c i e λ u i P i ( u i ) ≤ e O ( Δ i − 1 + Δ i − 4 ) E r a n d ( e λ Y ~ i ) . \sum_{|u_i|\ll c_i} e^{\lambda u_i}P_i(u_i) \le e^{O(\Delta_i^{-1}+\Delta_i^{-4})}\mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_i}). ∣ u i ∣ ≪ c i ∑ e λ u i P i ( u i ) ≤ e O ( Δ i − 1 + Δ i − 4 ) E rand ( e λ Y i ) . Since E r a n d ( e λ Y ~ i ) ≥ 1 \mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_i}) \ge1 E rand ( e λ Y i ) ≥ 1
∑ u i B i ( u i ) ≤ e O ( Δ i − 1 + Δ i − 4 ) E r a n d ( e λ Y ~ i ) ( 1 + O ( c i 2 e − c i 2 + O ( c i ) ) ) . \sum_{u_i} B_i(u_i) \le e^{O(\Delta_i^{-1}+\Delta_i^{-4})}\mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_i})\left(1+O(c_i^2e^{-c_i^2+O(c_i)})\right). u i ∑ B i ( u i ) ≤ e O ( Δ i − 1 + Δ i − 4 ) E rand ( e λ Y i ) ( 1 + O ( c i 2 e − c i 2 + O ( c i ) ) ) . For θ \theta θ
exp ( O ( ∑ i ≥ 2 ( Δ i − 1 + Δ i − 4 ) ) ) ∏ i ≥ 2 ( 1 + O ( c i 2 e − c i 2 + O ( c i ) ) ) ≪ 1. \exp\left(O\left(\sum_{i\ge2}(\Delta_i^{-1}+\Delta_i^{-4})\right)\right)\prod_{i\ge2}\left(1+O(c_i^2e^{-c_i^2+O(c_i)})\right)\ll1. exp ( O ( i ≥ 2 ∑ ( Δ i − 1 + Δ i − 4 ) ) ) i ≥ 2 ∏ ( 1 + O ( c i 2 e − c i 2 + O ( c i ) ) ) ≪ 1. Hence, combining the above with (41) and (33) , we get that
∑ u ∈ I ( u ) ∏ i ≤ ℓ ( P r a n d ( Y ~ i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) + 2 e − Δ i 2 ) ≪ e − λ u σ ℓ , j ∏ i ≤ ℓ E r a n d ( e λ Y ~ i ) e O ( 1 ) ≪ 1 σ ℓ , j exp ( − λ u + λ 2 σ ~ 2 2 ) , \begin{aligned}
\sum_{u\in I(u)}\prod_{i\le\ell}\left(\mathbb{P}_{\mathrm{rand}}\left(\widetilde{Y}_i\in[u_i-\Delta_i^{-4},u_i+\Delta_i^{-1}+\Delta_i^{-4}]\right)+2e^{-\Delta_i^2}\right)
\ll\frac{e^{-\lambda u}}{\sigma_{\ell,j}}\prod_{i\le\ell}\mathbb{E}_{\mathrm{rand}}(e^{\lambda\widetilde{Y}_i})e^{O(1)} \\
&\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\lambda u+\frac{\lambda^2\widetilde{\sigma}^2}{2}\right),
\end{aligned} u ∈ I ( u ) ∑ i ≤ ℓ ∏ ( P rand ( Y i ∈ [ u i − Δ i − 4 , u i + Δ i − 1 + Δ i − 4 ] ) + 2 e − Δ i 2 ) ≪ σ ℓ , j e − λ u i ≤ ℓ ∏ E rand ( e λ Y i ) e O ( 1 ) ≪ σ ℓ , j 1 exp ( − λ u + 2 λ 2 σ 2 ) , using (5.6). Since
− λ u + λ 2 σ ~ 2 2 = − u 2 2 σ ~ 2 , -\lambda u+\frac{\lambda^2\widetilde{\sigma}^2}{2}=-\frac{u^2}{2\widetilde{\sigma}^2}, − λ u + 2 λ 2 σ 2 = − 2 σ 2 u 2 , by the choice for λ \lambda λ
We also need the following proposition.
Proposition 5.2. Let 2 ≤ ℓ ≤ K 2\le\ell\le K 2 ≤ ℓ ≤ K ℓ ≤ j , m ≤ K \ell\le j,m\le K ℓ ≤ j , m ≤ K
L ℓ ′ : = L ℓ − 3 c ℓ = V log log T log log T ℓ − 4 c ℓ , U ℓ ′ : = U ℓ + 3 c ℓ = V log log T log log T ℓ + 4 c ℓ . L'_\ell:=L_\ell-3c_\ell=\frac{V}{\log\log T}\log\log T_\ell-4c_\ell,\qquad U'_\ell:=U_\ell+3c_\ell=\frac{V}{\log\log T}\log\log T_\ell+4c_\ell. L ℓ ′ := L ℓ − 3 c ℓ = log log T V log log T ℓ − 4 c ℓ , U ℓ ′ := U ℓ + 3 c ℓ = log log T V log log T ℓ + 4 c ℓ . Assume
u , w ∈ [ L ℓ ′ , U ℓ ′ ] , ∣ u ∣ ≤ A σ ℓ , j 2 and ∣ w ∣ ≤ A σ ℓ , m 2 . u,w\in[L'_\ell,U'_\ell],\qquad|u|\le A\sigma_{\ell,j}^2\qquad\text{and}\qquad|w|\le A\sigma_{\ell,m}^2. u , w ∈ [ L ℓ ′ , U ℓ ′ ] , ∣ u ∣ ≤ A σ ℓ , j 2 and ∣ w ∣ ≤ A σ ℓ , m 2 . Then there exists C > 0 C>0 C > 0
P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } ∩ { S ℓ ( m ) ∈ [ w − 1 , w ] } ) ≪ 1 σ ℓ , j exp ( − ( u + w ) 2 8 σ ℓ , j 2 − C ( w − u ) 2 z ℓ 2 + O ( z ℓ 2 ) ) . (42) \mathbb{P}\left(G_{\ell-1}\cap\{S_\ell^{(j)}\in[u-1,u]\}\cap\{S_\ell^{(m)}\in[w-1,w]\}\right)
\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{(u+w)^2}{8\sigma_{\ell,j}^2}-C\frac{(w-u)^2}{z_\ell^2}+O(z_\ell^2)\right). \tag*{(42)} P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} ∩ { S ℓ ( m ) ∈ [ w − 1 , w ]} ) ≪ σ ℓ , j 1 exp ( − 8 σ ℓ , j 2 ( u + w ) 2 − C z ℓ 2 ( w − u ) 2 + O ( z ℓ 2 ) ) . (42) Also, for ℓ = 1 \ell=1 ℓ = 1 1 ≤ j , m ≤ K 1\le j,m\le K 1 ≤ j , m ≤ K H > 0 H>0 H > 0 ∣ u ∣ ≤ A σ 1 , j 2 |u|\le A\sigma_{1,j}^2 ∣ u ∣ ≤ A σ 1 , j 2 ∣ w ∣ ≤ A σ 1 , m 2 |w|\le A\sigma_{1,m}^2 ∣ w ∣ ≤ A σ 1 , m 2
P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S 1 ( m ) ∈ [ w − 1 , w ] } ) ≪ 1 σ 1 , j exp ( − ( u + w ) 2 8 σ 1 , j 2 − H ∣ w − u ∣ + O ( z 1 2 ) ) . (43) \mathbb{P}\left(\{S_1^{(j)}\in[u-1,u]\}\cap\{S_1^{(m)}\in[w-1,w]\}\right)
\ll\frac{1}{\sigma_{1,j}}\exp\left(-\frac{(u+w)^2}{8\sigma_{1,j}^2}-H|w-u|+O(z_1^2)\right). \tag*{(43)} P ( { S 1 ( j ) ∈ [ u − 1 , u ]} ∩ { S 1 ( m ) ∈ [ w − 1 , w ]} ) ≪ σ 1 , j 1 exp ( − 8 σ 1 , j 2 ( u + w ) 2 − H ∣ w − u ∣ + O ( z 1 2 ) ) . (43) Proof. The proof is similar to the proof of Lemma B.2 in [3 ] and follows along similar lines as the proof of Proposition 5.1.
Put
Y i = S i ( j ) − S i − 1 ( j ) and Y i ′ = S i ( m ) − S i − 1 ( m ) . Y_i=S_i^{(j)}-S_{i-1}^{(j)} \qquad\text{and}\qquad Y_i'=S_i^{(m)}-S_{i-1}^{(m)}. Y i = S i ( j ) − S i − 1 ( j ) and Y i ′ = S i ( m ) − S i − 1 ( m ) . Take
Δ 1 = C 3 σ 1 , j , a 1 = 4 , X 1 = C 3 Δ 1 10 . (44) \Delta_1=C_3\sigma_{1,j},\qquad a_1=4,\qquad X_1=C_3\Delta_1^{10}. \tag*{(44)} Δ 1 = C 3 σ 1 , j , a 1 = 4 , X 1 = C 3 Δ 1 10 . (44) and, for i ≥ 2 i\ge2 i ≥ 2
Δ i = c i , a i = 9 2 , X i = C 2 z i 2 c i 9 . (45) \Delta_i=c_i,\qquad a_i=\frac{9}{2},\qquad X_i=C_2z_i^2c_i^9. \tag*{(45)} Δ i = c i , a i = 2 9 , X i = C 2 z i 2 c i 9 . (45) where C 3 C_3 C 3 θ \theta θ
∑ i ≤ K Δ i − 1 ≤ 1 2 , 4 ∑ i ≤ K Δ i 1 − a i ≤ 1 2 . \sum_{i\le K}\Delta_i^{-1}\le\frac{1}{2},\qquad4\sum_{i\le K}\Delta_i^{1-a_i}\le\frac{1}{2}. i ≤ K ∑ Δ i − 1 ≤ 2 1 , 4 i ≤ K ∑ Δ i 1 − a i ≤ 2 1 . For τ ∈ [ 0 , 1 ) \tau\in[0,1) τ ∈ [ 0 , 1 ) [ u i , u i + Δ i − 1 ) [u_i,u_i+\Delta_i^{-1}) [ u i , u i + Δ i − 1 ) [ w i , w i + Δ i − 1 ) [w_i,w_i+\Delta_i^{-1}) [ w i , w i + Δ i − 1 )
u i = ( n i + τ ) ( Δ i − 1 + 2 Δ i − a i ) , w i = ( m i + τ ) ( Δ i − 1 + 2 Δ i − a i ) u_i=(n_i+\tau)(\Delta_i^{-1}+2\Delta_i^{-a_i}),\qquad w_i=(m_i+\tau)(\Delta_i^{-1}+2\Delta_i^{-a_i}) u i = ( n i + τ ) ( Δ i − 1 + 2 Δ i − a i ) , w i = ( m i + τ ) ( Δ i − 1 + 2 Δ i − a i ) with n i , m i ∈ Z n_i,m_i\in\mathbb{Z} n i , m i ∈ Z I ( u ) \mathcal{I}(u) I ( u ) I ( w ) \mathcal{I}(w) I ( w ) u ∈ I ( u ) \mathbf{u}\in\mathcal{I}(u) u ∈ I ( u ) w ∈ I ( w ) \mathbf{w}\in\mathcal{I}(w) w ∈ I ( w ) ℓ ≥ 2 \ell\ge2 ℓ ≥ 2
∣ u i ∣ + ∣ w i ∣ ≪ c i ( 2 ≤ i ≤ ℓ ) , |u_i|+|w_i|\ll c_i\qquad(2\le i\le\ell), ∣ u i ∣ + ∣ w i ∣ ≪ c i ( 2 ≤ i ≤ ℓ ) , and
∣ u 1 ∣ + ∣ w 1 ∣ ≪ V ≪ log log T , ∣ u 1 − w 1 ∣ ≪ c 1 . (46) |u_1|+|w_1|\ll V\ll\log\log T,\qquad|u_1-w_1|\ll c_1. \tag*{(46)} ∣ u 1 ∣ + ∣ w 1 ∣ ≪ V ≪ log log T , ∣ u 1 − w 1 ∣ ≪ c 1 . (46) By the averaging argument of Proposition 5.1, we may fix a phase τ \tau τ
P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } ∩ { S ℓ ( m ) ∈ [ w − 1 , w ] } ) ≪ ∑ u ∈ I ( u ) w ∈ I ( w ) P ( Y i ∈ [ u i , u i + Δ i − 1 ) , Y i ′ ∈ [ w i , w i + Δ i − 1 ) for all i ≤ ℓ ) . \mathbb{P}\left(G_{\ell-1}\cap\{S_\ell^{(j)}\in[u-1,u]\}\cap\{S_\ell^{(m)}\in[w-1,w]\}\right)
\ll\sum_{\substack{\mathbf{u}\in\mathcal{I}(u)\\\mathbf{w}\in\mathcal{I}(w)}}\mathbb{P}\left(Y_i\in[u_i,u_i+\Delta_i^{-1}),\ Y_i'\in[w_i,w_i+\Delta_i^{-1})\text{ for all }i\le\ell\right). P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} ∩ { S ℓ ( m ) ∈ [ w − 1 , w ]} ) ≪ u ∈ I ( u ) w ∈ I ( w ) ∑ P ( Y i ∈ [ u i , u i + Δ i − 1 ) , Y i ′ ∈ [ w i , w i + Δ i − 1 ) for all i ≤ ℓ ) . We choose the polynomial D i D_i D i C j ( p ) C_j(p) C j ( p )
Y ~ i = ∑ T i − 1 < p ≤ T i C j ( p ) and Y ~ i ′ = ∑ T i − 1 < p ≤ T i C m ( p ) . \widetilde{Y}_i=\sum_{T_{i-1}<p\le T_i}C_j(p)\qquad\text{and}\qquad\widetilde{Y}_i'=\sum_{T_{i-1}<p\le T_i}C_m(p). Y i = T i − 1 < p ≤ T i ∑ C j ( p ) and Y i ′ = T i − 1 < p ≤ T i ∑ C m ( p ) . Before passing to the random model, we check the lengths required by the mean-value theorem. As in Proposition 5.1, expanding ∏ i ≤ ℓ ∣ D i ( Y i − u i ) ∣ 2 ∣ D i ( Y i ′ − w i ) ∣ 2 \prod_{i\le\ell}|D_i(Y_i-u_i)|^2|D_i(Y_i'-w_i)|^2 ∏ i ≤ ℓ ∣ D i ( Y i − u i ) ∣ 2 ∣ D i ( Y i ′ − w i ) ∣ 2
∏ i ≤ ℓ T i 8 d i ≤ T C β 1 ( log log T ) 11 + C ∑ i = 2 ℓ β i ( 1 + z i 2 ) c i 45 / 2 ≤ T δ l o c < T . \prod_{i\le\ell}T_i^{8d_i}\le T^{C\beta_1(\log\log T)^{11}+C\sum_{i=2}^{\ell}\beta_i(1+z_i^2)c_i^{45/2}}\le T^{\delta_{\mathrm{loc}}}<T. i ≤ ℓ ∏ T i 8 d i ≤ T C β 1 ( l o g l o g T ) 11 + C ∑ i = 2 ℓ β i ( 1 + z i 2 ) c i 45/2 ≤ T δ loc < T . The same mean-value argument as in the proof of Proposition 5.1 shows that
P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } ∩ { S ℓ ( m ) ∈ [ w − 1 , w ] } ) ≪ ∑ u ∈ I ( u ) w ∈ I ( w ) E r a n d ( ∏ i = 1 ℓ ∣ D i ( Y ~ i − u i ) ∣ 2 ∣ D i ( Y ~ i ′ − w i ) ∣ 2 ) . \mathbb{P}\left(G_{\ell-1}\cap\{S_\ell^{(j)}\in[u-1,u]\}\cap\{S_\ell^{(m)}\in[w-1,w]\}\right)
\ll\sum_{\substack{\mathbf{u}\in\mathcal{I}(u)\\\mathbf{w}\in\mathcal{I}(w)}}\mathbb{E}_{\mathrm{rand}}\left(\prod_{i=1}^{\ell}|D_i(\widetilde{Y}_i-u_i)|^2|D_i(\widetilde{Y}_i'-w_i)|^2\right). P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} ∩ { S ℓ ( m ) ∈ [ w − 1 , w ]} ) ≪ u ∈ I ( u ) w ∈ I ( w ) ∑ E rand ( i = 1 ∏ ℓ ∣ D i ( Y i − u i ) ∣ 2 ∣ D i ( Y i ′ − w i ) ∣ 2 ) . We first remove the ranges on which either ∣ Y ~ i − u i ∣ > X i |\widetilde{Y}_i-u_i|>X_i ∣ Y i − u i ∣ > X i ∣ Y ~ i ′ − w i ∣ > X i |\widetilde{Y}'_i-w_i|>X_i ∣ Y i ′ − w i ∣ > X i ≪ e − Δ i a i − 2 \ll e^{-\Delta_i^{a_i-2}} ≪ e − Δ i a i − 2
≪ ∑ u ∈ I ( u ) , w ∈ I ( w ) ∏ i = 1 ℓ ( P i ( u i , w i ) + ε i ) . \ll\sum_{\substack{\mathbf{u}\in\mathcal{I}(\mathbf{u}),\mathbf{w}\in\mathcal{I}(\mathbf{w})}} \prod_{i=1}^{\ell}\left(P_i(u_i,w_i)+\varepsilon_i\right). ≪ u ∈ I ( u ) , w ∈ I ( w ) ∑ i = 1 ∏ ℓ ( P i ( u i , w i ) + ε i ) . where, for simplicity of notation,
P i ( u i , w i ) = P r a n d ( Y ~ i ∈ [ u i − Δ i − a i , u i + Δ i − 1 + Δ i − a i ] , Y ~ i ′ ∈ [ w i − Δ i − a i , w i + Δ i − 1 + Δ i − a i ] ) P_i(u_i,w_i)=\mathbb{P}_{\mathrm{rand}}\left(\widetilde{Y}_i\in\left[u_i-\Delta_i^{-a_i},u_i+\Delta_i^{-1}+\Delta_i^{-a_i}\right],\widetilde{Y}'_i\in\left[w_i-\Delta_i^{-a_i},w_i+\Delta_i^{-1}+\Delta_i^{-a_i}\right]\right) P i ( u i , w i ) = P rand ( Y i ∈ [ u i − Δ i − a i , u i + Δ i − 1 + Δ i − a i ] , Y i ′ ∈ [ w i − Δ i − a i , w i + Δ i − 1 + Δ i − a i ] ) and
ε i = C e − Δ i a i − 2 . \varepsilon_i=Ce^{-\Delta_i^{a_i-2}}. ε i = C e − Δ i a i − 2 . Define
Z p = C j ( p ) + C m ( p ) 2 , F p = C m ( p ) − C j ( p ) , Z_p=\frac{C_j(p)+C_m(p)}{2},\qquad F_p=C_m(p)-C_j(p), Z p = 2 C j ( p ) + C m ( p ) , F p = C m ( p ) − C j ( p ) , where C j ( p ) C_j(p) C j ( p ) Z i , F i Z_i,F_i Z i , F i T i − 1 < p ≤ T i T_{i-1}<p\leq T_i T i − 1 < p ≤ T i
Z : = ∑ i ≤ ℓ Z i = 1 2 ( S ~ ℓ ( j ) + S ~ ℓ ( m ) ) , Z:=\sum_{i\leq\ell}Z_i=\frac{1}{2}\left(\widetilde{S}^{(j)}_\ell+\widetilde{S}^{(m)}_\ell\right), Z := i ≤ ℓ ∑ Z i = 2 1 ( S ℓ ( j ) + S ℓ ( m ) ) , and
F : = ∑ i ≤ ℓ F i = S ~ ℓ ( m ) − S ~ ℓ ( j ) . F:=\sum_{i\leq\ell}F_i=\widetilde{S}^{(m)}_\ell-\widetilde{S}^{(j)}_\ell. F := i ≤ ℓ ∑ F i = S ℓ ( m ) − S ℓ ( j ) . Corollary 4.1 gives
∥ F p ∥ ∞ ≪ z ℓ log p log T ℓ p − 1 / 2 − α , \|F_p\|_\infty\ll z_\ell\frac{\log p}{\log T_\ell}p^{-1/2-\alpha}, ∥ F p ∥ ∞ ≪ z ℓ log T ℓ log p p − 1/2 − α , and using (5.8) as well gives
∑ p ∥ F p ∥ ∞ 2 ≪ z ℓ 2 , ∑ p ∥ Z p ∥ ∞ ∥ F p ∥ ∞ ≪ z ℓ 2 , ∑ p ∥ Z p ∥ ∞ 3 ≪ 1. (47) \sum_p\|F_p\|_\infty^2\ll z_\ell^2,\qquad\sum_p\|Z_p\|_\infty\|F_p\|_\infty\ll z_\ell^2,\qquad\sum_p\|Z_p\|_\infty^3\ll1. \tag*{(47)} p ∑ ∥ F p ∥ ∞ 2 ≪ z ℓ 2 , p ∑ ∥ Z p ∥ ∞ ∥ F p ∥ ∞ ≪ z ℓ 2 , p ∑ ∥ Z p ∥ ∞ 3 ≪ 1. (47) and
∑ p ( ( ∥ Z p ∥ ∞ 2 + ∥ C j ( p ) ∥ ∞ 2 ) ∥ F p ∥ ∞ + max p ∥ F p ∥ ∞ ) ≪ z ℓ log T ℓ . (48) \sum_p\left(\left(\|Z_p\|_\infty^2+\|C_j(p)\|_\infty^2\right)\|F_p\|_\infty+\max_p\|F_p\|_\infty\right)\ll\frac{z_\ell}{\log T_\ell}. \tag*{(48)} p ∑ ( ( ∥ Z p ∥ ∞ 2 + ∥ C j ( p ) ∥ ∞ 2 ) ∥ F p ∥ ∞ + p max ∥ F p ∥ ∞ ) ≪ log T ℓ z ℓ . (48) Hence
Var ( F ) ≪ z ℓ 2 and σ + 2 : = Var ( Z ) = σ ℓ , j 2 + O ( z ℓ 2 ) . \operatorname{Var}(F)\ll z_\ell^2\qquad\text{and}\qquad\sigma_+^2:=\operatorname{Var}(Z)=\sigma_{\ell,j}^2+O(z_\ell^2). Var ( F ) ≪ z ℓ 2 and σ + 2 := Var ( Z ) = σ ℓ , j 2 + O ( z ℓ 2 ) . Let
K ( λ , μ ) = log E r a n d ( e λ Z + μ F ) . K(\lambda,\mu)=\log\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z+\mu F}\right). K ( λ , μ ) = log E rand ( e λ Z + μ F ) . For bounded ∣ λ ∣ \lvert\lambda\rvert ∣ λ ∣
K ( λ , 0 ) = ∑ p log E r a n d ( e λ Z p ) = λ 2 2 ∑ p E r a n d ( Z p 2 ) + O ( ∑ p ∥ Z p ∥ ∞ 3 ) = λ 2 σ + 2 2 + O ( 1 ) . K(\lambda,0)=\sum_p\log\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p}\right)=\frac{\lambda^2}{2}\sum_p\mathbb{E}_{\mathrm{rand}}\left(Z_p^2\right)+O\left(\sum_p\|Z_p\|_\infty^3\right)=\frac{\lambda^2\sigma_+^2}{2}+O(1). K ( λ , 0 ) = p ∑ log E rand ( e λ Z p ) = 2 λ 2 p ∑ E rand ( Z p 2 ) + O ( p ∑ ∥ Z p ∥ ∞ 3 ) = 2 λ 2 σ + 2 + O ( 1 ) . Moreover, we have
K ( λ , μ ) − K ( λ , 0 ) = ∑ p log E r a n d ( e λ Z p + μ F p ) E r a n d ( e λ Z p ) . K(\lambda,\mu)-K(\lambda,0)=\sum_p\log\frac{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p+\mu F_p}\right)}{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p}\right)}. K ( λ , μ ) − K ( λ , 0 ) = p ∑ log E rand ( e λ Z p ) E rand ( e λ Z p + μ F p ) . For each p p p
log E r a n d ( e λ Z p + μ F p ) E r a n d ( e λ Z p ) ≤ μ E r a n d ( F p e λ Z p ) E r a n d ( e λ Z p ) + μ 2 2 ∥ F p ∥ ∞ 2 . \log\frac{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p+\mu F_p}\right)}{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p}\right)}
\leq\mu\frac{\mathbb{E}_{\mathrm{rand}}\left(F_pe^{\lambda Z_p}\right)}{\mathbb{E}_{\mathrm{rand}}\left(e^{\lambda Z_p}\right)}+\frac{\mu^2}{2}\|F_p\|_\infty^2. log E rand ( e λ Z p ) E rand ( e λ Z p + μ F p ) ≤ μ E rand ( e λ Z p ) E rand ( F p e λ Z p ) + 2 μ 2 ∥ F p ∥ ∞ 2 . Since E r a n d ( F p ) = 0 \mathbb{E}_{\rm rand}(F_p)=0 E rand ( F p ) = 0 E r a n d ( e λ Z p ) ≥ 1 \mathbb{E}_{\rm rand}(e^{\lambda Z_p})\ge1 E rand ( e λ Z p ) ≥ 1 ∣ e λ Z p − 1 ∣ ≪ ∣ Z p ∣ |e^{\lambda Z_p}-1|\ll|Z_p| ∣ e λ Z p − 1∣ ≪ ∣ Z p ∣
∣ E r a n d ( F p e λ Z p ) E r a n d ( e λ Z p ) ∣ = ∣ E r a n d ( F p ( e λ Z p − 1 ) ) E r a n d ( e λ Z p ) ∣ ≪ ∥ Z p ∥ ∞ ∥ F p ∥ ∞ . \left|\frac{\mathbb{E}_{\rm rand}(F_p e^{\lambda Z_p})}{\mathbb{E}_{\rm rand}(e^{\lambda Z_p})}\right|=\left|\frac{\mathbb{E}_{\rm rand}(F_p(e^{\lambda Z_p}-1))}{\mathbb{E}_{\rm rand}(e^{\lambda Z_p})}\right|\ll\|Z_p\|_\infty\|F_p\|_\infty. E rand ( e λ Z p ) E rand ( F p e λ Z p ) = E rand ( e λ Z p ) E rand ( F p ( e λ Z p − 1 )) ≪ ∥ Z p ∥ ∞ ∥ F p ∥ ∞ . Therefore, by (47) we have
K ( λ , μ ) − K ( λ , 0 ) ≪ z ℓ 2 ( ∣ μ ∣ + μ 2 ) . K(\lambda,\mu)-K(\lambda,0)\ll z_\ell^2(|\mu|+\mu^2). K ( λ , μ ) − K ( λ , 0 ) ≪ z ℓ 2 ( ∣ μ ∣ + μ 2 ) . Hence, for some constants C 1 , C 2 > 0 C_1,C_2>0 C 1 , C 2 > 0
K ( λ , μ ) ≤ λ 2 σ + 2 2 + C 1 z ℓ 2 ( ∣ μ ∣ + μ 2 ) + C 2 . (49) K(\lambda,\mu)\le\frac{\lambda^2\sigma_+^2}{2}+C_1z_\ell^2(|\mu|+\mu^2)+C_2. \tag*{(49)} K ( λ , μ ) ≤ 2 λ 2 σ + 2 + C 1 z ℓ 2 ( ∣ μ ∣ + μ 2 ) + C 2 . (49) For ℓ ≥ 2 \ell\ge2 ℓ ≥ 2
λ = u + w 2 σ + 2 , μ = η w − u z ℓ 2 , (50) \lambda=\frac{u+w}{2\sigma_+^2},\qquad\mu=\eta\frac{w-u}{z_\ell^2}, \tag*{(50)} λ = 2 σ + 2 u + w , μ = η z ℓ 2 w − u , (50) where η > 0 \eta>0 η > 0
∣ λ ∣ ≪ 1 , ∣ μ ∣ ≪ c ℓ z ℓ 2 , |\lambda|\ll1,\qquad|\mu|\ll\frac{c_\ell}{z_\ell^2}, ∣ λ ∣ ≪ 1 , ∣ μ ∣ ≪ z ℓ 2 c ℓ , and, by (48) ,
∣ μ ∣ ( max p ∥ F p ∥ ∞ + ∑ p ∥ C j ( p ) ∥ ∞ 2 ∥ F p ∥ ∞ ) = o ( 1 ) . (51) |\mu|\left(\max_p\|F_p\|_\infty+\sum_p\|C_j(p)\|_\infty^2\|F_p\|_\infty\right)=o(1). \tag*{(51)} ∣ μ ∣ ( p max ∥ F p ∥ ∞ + p ∑ ∥ C j ( p ) ∥ ∞ 2 ∥ F p ∥ ∞ ) = o ( 1 ) . (51) If ℓ = 1 \ell=1 ℓ = 1
λ = u + w 2 σ + 2 , μ = ( H + 1 ) sgn ( w − u ) . \lambda=\frac{u+w}{2\sigma_+^2},\qquad\mu=(H+1)\operatorname{sgn}(w-u). λ = 2 σ + 2 u + w , μ = ( H + 1 ) sgn ( w − u ) . Here H > 0 H>0 H > 0 (51) also holds.
For an admissible pair of tuples ( u ) (\mathbf{u}) ( u ) ( w ) (\mathbf{w}) ( w ) I ( u ) \mathcal{I}(\mathbf{u}) I ( u ) I ( w ) \mathcal{I}(\mathbf{w}) I ( w )
1 ≤ exp ( − λ w + u 2 − μ ( w − u ) + C ( ∣ λ ∣ + ∣ μ ∣ ) + ∑ i ≤ ℓ ( λ u i + w i 2 + μ ( w i − u i ) ) ) . (52) 1\le\exp\left(-\lambda\frac{w+u}{2}-\mu(w-u)+C(|\lambda|+|\mu|)+\sum_{i\le\ell}\left(\lambda\frac{u_i+w_i}{2}+\mu(w_i-u_i)\right)\right). \tag*{(52)} 1 ≤ exp ( − λ 2 w + u − μ ( w − u ) + C ( ∣ λ ∣ + ∣ μ ∣ ) + i ≤ ℓ ∑ ( λ 2 u i + w i + μ ( w i − u i ) ) ) . (52) On the event in P i ( u i , w i ) P_i(u_i,w_i) P i ( u i , w i )
∣ u i + w i 2 − Z i ∣ ≤ Δ i − 1 + Δ i − a i and ∣ ( w i − u i ) − F i ∣ ≤ 2 ( Δ i − 1 + Δ i − a i ) . \left|\frac{u_i+w_i}{2}-Z_i\right|\le\Delta_i^{-1}+\Delta_i^{-a_i}
\qquad\text{and}\qquad
\left|(w_i-u_i)-\mathcal{F}_i\right|\le2(\Delta_i^{-1}+\Delta_i^{-a_i}). 2 u i + w i − Z i ≤ Δ i − 1 + Δ i − a i and ∣ ( w i − u i ) − F i ∣ ≤ 2 ( Δ i − 1 + Δ i − a i ) . Now we deal with the first block. Fix all the ( u i ) i ≥ 2 (u_i)_{i\ge2} ( u i ) i ≥ 2 ( w i ) i ≥ 2 (w_i)_{i\ge2} ( w i ) i ≥ 2 I ( u ) \mathcal{I}(\mathbf{u}) I ( u ) I ( w ) \mathcal{I}(\mathbf{w}) I ( w ) u 1 u_1 u 1 w 1 w_1 w 1 O ( 1 ) O(1) O ( 1 ) Y ~ 1 \widetilde{Y}_1 Y 1 I ^ \widehat{I} I O ( 1 ) O(1) O ( 1 )
∑ u 1 , w 1 e λ ( u 1 + w 1 ) / 2 + μ ( w 1 − u 1 ) P 1 ( u 1 , w 1 ) ≪ E r a n d ( e λ Z 1 + μ F 1 1 ( Y ~ 1 ∈ I ^ ) ) . \sum_{u_1,w_1}e^{\lambda(u_1+w_1)/2+\mu(w_1-u_1)}P_1(u_1,w_1)\ll\mathbb{E}_{\rm rand}\left(e^{\lambda Z_1+\mu\mathcal{F}_1}\mathbf{1}(\widetilde{Y}_1\in\widehat{I})\right). u 1 , w 1 ∑ e λ ( u 1 + w 1 ) /2 + μ ( w 1 − u 1 ) P 1 ( u 1 , w 1 ) ≪ E rand ( e λ Z 1 + μ F 1 1 ( Y 1 ∈ I ) ) . Define the tilted probability P λ , μ \mathbb{P}_{\lambda,\mu} P λ , μ
P λ , μ ( E ) : = E r a n d ( e λ Z 1 + μ F 1 1 E ) E r a n d ( e λ Z 1 + μ F 1 ) . \mathbb{P}_{\lambda,\mu}(E):=\frac{\mathbb{E}_{\rm rand}(e^{\lambda Z_1+\mu\mathcal{F}_1}\mathbf{1}_E)}{\mathbb{E}_{\rm rand}(e^{\lambda Z_1+\mu\mathcal{F}_1})}. P λ , μ ( E ) := E rand ( e λ Z 1 + μ F 1 ) E rand ( e λ Z 1 + μ F 1 1 E ) . Then
E r a n d ( e λ Z 1 + μ F 1 1 ( Y ~ 1 ∈ I ^ ) ) = E r a n d ( e λ Z 1 + μ F 1 ) P λ , μ ( Y ~ 1 ∈ I ^ ) . \mathbb{E}_{\rm rand}\left(e^{\lambda Z_1+\mu\mathcal{F}_1}\mathbf{1}(\widetilde{Y}_1\in\widehat{I})\right)=\mathbb{E}_{\rm rand}(e^{\lambda Z_1+\mu\mathcal{F}_1})\mathbb{P}_{\lambda,\mu}(\widetilde{Y}_1\in\widehat{I}). E rand ( e λ Z 1 + μ F 1 1 ( Y 1 ∈ I ) ) = E rand ( e λ Z 1 + μ F 1 ) P λ , μ ( Y 1 ∈ I ) . We write E λ , μ \mathbb{E}_{\lambda,\mu} E λ , μ Var λ , μ \operatorname{Var}_{\lambda,\mu} Var λ , μ
∑ p ≤ T 1 E λ , μ ∣ C j ( p ) − E λ , μ C j ( p ) ∣ 3 ≪ 1 and Var λ , μ ( Y ~ 1 ) = σ 1 , j 2 + O ( 1 ) ≍ σ ℓ , j 2 . \sum_{p\le T_1}\mathbb{E}_{\lambda,\mu}|C_j(p)-\mathbb{E}_{\lambda,\mu}C_j(p)|^3\ll1
\qquad\text{and}\qquad
\operatorname{Var}_{\lambda,\mu}(\widetilde{Y}_1)=\sigma_{1,j}^2+O(1)\asymp\sigma_{\ell,j}^2. p ≤ T 1 ∑ E λ , μ ∣ C j ( p ) − E λ , μ C j ( p ) ∣ 3 ≪ 1 and Var λ , μ ( Y 1 ) = σ 1 , j 2 + O ( 1 ) ≍ σ ℓ , j 2 . The Berry–Esseen theorem therefore gives
P λ , μ ( Y ~ 1 ∈ I ) ≪ 1 σ ℓ , j . \mathbb{P}_{\lambda,\mu}(\widetilde{Y}_1 \in I) \ll\frac{1}{\sigma_{\ell,j}}. P λ , μ ( Y 1 ∈ I ) ≪ σ ℓ , j 1 . We now bound the error from the block i = 1 i=1 i = 1 u i , w i u_i,w_i u i , w i i ≥ 2 i\geq2 i ≥ 2 ℓ ≥ 2 \ell\geq2 ℓ ≥ 2 (46) we have that
∑ u 1 , w 1 e λ ( u 1 + w 1 ) / 2 + μ ( w 1 − u 1 ) ε 1 ≪ Δ 1 2 e − Δ 1 2 + O ( V + C c 1 2 ) ≤ E r a n d ( e λ Z 1 + μ F 1 ) σ ℓ , j . \sum_{u_1,w_1}e^{\lambda(u_1+w_1)/2+\mu(w_1-u_1)}\varepsilon_1 \ll\Delta_1^2e^{-\Delta_1^2+O(V+C c_1^2)}\leq\frac{\mathbb{E}_{\mathrm{rand}}(e^{\lambda Z_1+\mu F_1})}{\sigma_{\ell,j}}. u 1 , w 1 ∑ e λ ( u 1 + w 1 ) /2 + μ ( w 1 − u 1 ) ε 1 ≪ Δ 1 2 e − Δ 1 2 + O ( V + C c 1 2 ) ≤ σ ℓ , j E rand ( e λ Z 1 + μ F 1 ) . If ℓ = 1 \ell=1 ℓ = 1 ∣ u 1 ∣ + ∣ w 1 ∣ ≪ log log T |u_1|+|w_1|\ll\log\log T ∣ u 1 ∣ + ∣ w 1 ∣ ≪ log log T ∣ μ ∣ ≤ H + 1 |\mu|\leq H+1 ∣ μ ∣ ≤ H + 1
≪ Δ 1 2 exp ( − Δ 1 2 + O ( log log T ) ) ≤ E r a n d ( e λ Z 1 + μ F 1 ) σ ℓ , j , \ll\Delta_1^2\exp\left(-\Delta_1^2+O(\log\log T)\right)\leq\frac{\mathbb{E}_{\mathrm{rand}}(e^{\lambda Z_1+\mu F_1})}{\sigma_{\ell,j}}, ≪ Δ 1 2 exp ( − Δ 1 2 + O ( log log T ) ) ≤ σ ℓ , j E rand ( e λ Z 1 + μ F 1 ) , where we used Jensen’s inequality in the last step. Hence
∑ u 1 , w 1 e λ ( u 1 + w 1 ) / 2 + μ ( w 1 − u 1 ) ( P 1 ( u 1 , w 1 ) + ε 1 ) ≪ E r a n d ( e λ Z 1 + μ F 1 ) σ ℓ , j . (53) \sum_{u_1,w_1}e^{\lambda(u_1+w_1)/2+\mu(w_1-u_1)}(P_1(u_1,w_1)+\varepsilon_1)\ll\frac{\mathbb{E}_{\mathrm{rand}}(e^{\lambda Z_1+\mu F_1})}{\sigma_{\ell,j}}. \tag*{(53)} u 1 , w 1 ∑ e λ ( u 1 + w 1 ) /2 + μ ( w 1 − u 1 ) ( P 1 ( u 1 , w 1 ) + ε 1 ) ≪ σ ℓ , j E rand ( e λ Z 1 + μ F 1 ) . (53) We now treat the subsequent blocks. Recall that ∣ u i ∣ + ∣ w i ∣ ≪ c i |u_i|+|w_i|\ll c_i ∣ u i ∣ + ∣ w i ∣ ≪ c i i ≥ 2 i\geq2 i ≥ 2 i ≥ 2 i\geq2 i ≥ 2 Proposition 5.1 ,
∑ u i , w i e λ ( u i + w i ) / 2 + μ ( w i − u i ) P i ( u i , w i ) ≤ e O ( ( ∣ λ ∣ + ∣ μ ∣ ) ( Δ i − 1 + Δ i − a i ) ) E r a n d ( e λ Z i + μ F i ) , \sum_{u_i,w_i}e^{\lambda(u_i+w_i)/2+\mu(w_i-u_i)}P_i(u_i,w_i)\leq e^{O((|\lambda|+|\mu|)(\Delta_i^{-1}+\Delta_i^{-a_i}))}\mathbb{E}_{\mathrm{rand}}(e^{\lambda Z_i+\mu F_i}), u i , w i ∑ e λ ( u i + w i ) /2 + μ ( w i − u i ) P i ( u i , w i ) ≤ e O (( ∣ λ ∣ + ∣ μ ∣ ) ( Δ i − 1 + Δ i − a i )) E rand ( e λ Z i + μ F i ) , and again as in Proposition 5.1 , we get that
∏ i = 2 ℓ ∑ u i , w i e λ ( u i + w i ) / 2 + μ ( w i − u i ) ( P i ( u i , w i ) + C ε i ) ≪ ∏ i = 2 ℓ E r a n d ( e λ Z i + μ F i ) . \prod_{i=2}^{\ell}\sum_{u_i,w_i}e^{\lambda(u_i+w_i)/2+\mu(w_i-u_i)}(P_i(u_i,w_i)+C\varepsilon_i)\ll\prod_{i=2}^{\ell}\mathbb{E}_{\mathrm{rand}}(e^{\lambda Z_i+\mu F_i}). i = 2 ∏ ℓ u i , w i ∑ e λ ( u i + w i ) /2 + μ ( w i − u i ) ( P i ( u i , w i ) + C ε i ) ≪ i = 2 ∏ ℓ E rand ( e λ Z i + μ F i ) . Combining this with (53) and (52) shows that
P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } ∩ { S ℓ ( m ) ∈ [ w − 1 , w ] } ) ≪ 1 σ ℓ , j exp ( − λ u + w 2 − μ ( w − u ) + C ( ∣ λ ∣ + ∣ μ ∣ ) + K ( λ , μ ) ) . \mathbb{P}\left(G_{\ell-1}\cap\{S_{\ell}^{(j)}\in[u-1,u]\}\cap\{S_{\ell}^{(m)}\in[w-1,w]\}\right)
\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\lambda\frac{u+w}{2}-\mu(w-u)+C(|\lambda|+|\mu|)+K(\lambda,\mu)\right). P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} ∩ { S ℓ ( m ) ∈ [ w − 1 , w ]} ) ≪ σ ℓ , j 1 exp ( − λ 2 u + w − μ ( w − u ) + C ( ∣ λ ∣ + ∣ μ ∣ ) + K ( λ , μ ) ) . For ℓ ≥ 2 \ell\geq2 ℓ ≥ 2 (50) and using (49) , the exponent above is at most
− ( u + w ) 2 8 σ + 2 − η ( w − u ) 2 z ℓ 2 + C 1 η 2 ( w − u ) 2 z ℓ 2 + C η ∣ w − u ∣ + O ( 1 ) . -\frac{(u+w)^2}{8\sigma_+^2}-\frac{\eta(w-u)^2}{z_\ell^2}+\frac{C_1\eta^2(w-u)^2}{z_\ell^2}+C\eta|w-u|+O(1). − 8 σ + 2 ( u + w ) 2 − z ℓ 2 η ( w − u ) 2 + z ℓ 2 C 1 η 2 ( w − u ) 2 + C η ∣ w − u ∣ + O ( 1 ) . We choose η \eta η C 1 η 2 ≤ η / 4 C_1\eta^2\leq\eta/4 C 1 η 2 ≤ η /4
C η ∣ w − u ∣ ≤ η ( w − u ) 2 4 z ℓ 2 + C 2 η z ℓ 2 , C\eta|w-u|\leq\frac{\eta(w-u)^2}{4z_\ell^2}+C^2\eta z_\ell^2, C η ∣ w − u ∣ ≤ 4 z ℓ 2 η ( w − u ) 2 + C 2 η z ℓ 2 , the exponent is at most
− ( u + w ) 2 8 σ + 2 − η ( w − u ) 2 2 z ℓ 2 + O ( z ℓ 2 ) . -\frac{(u+w)^2}{8\sigma_+^2}-\frac{\eta(w-u)^2}{2z_\ell^2}+O(z_\ell^2). − 8 σ + 2 ( u + w ) 2 − 2 z ℓ 2 η ( w − u ) 2 + O ( z ℓ 2 ) . Finally, σ + 2 = σ ℓ , j 2 + O ( z ℓ 2 ) \sigma_+^2=\sigma_{\ell,j}^2+O(z_\ell^2) σ + 2 = σ ℓ , j 2 + O ( z ℓ 2 ) ( u + w ) 2 ≪ σ ℓ , j 4 (u+w)^2\ll\sigma_{\ell,j}^4 ( u + w ) 2 ≪ σ ℓ , j 4
− ( u + w ) 2 8 σ ℓ , j 2 − C ( w − u ) 2 z ℓ 2 + O ( z ℓ 2 ) , -\frac{(u+w)^2}{8\sigma_{\ell,j}^2}-\frac{C(w-u)^2}{z_\ell^2}+O(z_\ell^2), − 8 σ ℓ , j 2 ( u + w ) 2 − z ℓ 2 C ( w − u ) 2 + O ( z ℓ 2 ) , which proves (42) . For ℓ = 1 \ell=1 ℓ = 1 μ = ( H + 1 ) sgn ( w − u ) \mu=(H+1)\operatorname{sgn}(w-u) μ = ( H + 1 ) sgn ( w − u ) − μ ( w − u ) = − ( H + 1 ) ∣ w − u ∣ -\mu(w-u)=-(H+1)|w-u| − μ ( w − u ) = − ( H + 1 ) ∣ w − u ∣ (49) bounds the remaining terms involving μ \mu μ O ( z 1 2 ) O(z_1^2) O ( z 1 2 ) (43) . □
We now prove a simplified version of Proposition 5.1, in the easier regime (13) .
Proposition 5.3. Assume (13) and α = o ( 1 ) \alpha=o(1) α = o ( 1 ) x x x S x S_x S x σ x \sigma_x σ x (14) , (4) , (15) , respectively. For every A ≥ 1 A\ge1 A ≥ 1 ∣ u ∣ ≤ A σ x 2 |u|\le A\sigma_x^2 ∣ u ∣ ≤ A σ x 2
P ( S x ∈ [ u − 1 , u ] ) ≪ 1 σ x exp ( − u 2 2 σ x 2 ) . (54) \mathbb{P}(S_x\in[u-1,u])\ll\frac{1}{\sigma_x}\exp\left(-\frac{u^2}{2\sigma_x^2}\right). \tag*{(54)} P ( S x ∈ [ u − 1 , u ]) ≪ σ x 1 exp ( − 2 σ x 2 u 2 ) . (54) Proof. We repeat the proof of Proposition 5.1 for the single sum S x ( t ) S_x(t) S x ( t ) Q = 1 Q=1 Q = 1
Take
Δ = C σ x , X = C Δ 10 , D = D Δ , X \Delta=C\sigma_x,\qquad X=C\Delta^{10},\qquad\mathcal{D}=\mathcal{D}_{\Delta,X} Δ = C σ x , X = C Δ 10 , D = D Δ , X from (27) with a = 4 a=4 a = 4 [ u − 1 , u ] [u-1,u] [ u − 1 , u ] Δ − 1 \Delta^{-1} Δ − 1 u u u [ v , v + Δ − 1 ] [v,v+\Delta^{-1}] [ v , v + Δ − 1 ] [ v − Δ − 4 , v + Δ − 1 + Δ − 4 ] [v-\Delta^{-4},v+\Delta^{-1}+\Delta^{-4}] [ v − Δ − 4 , v + Δ − 1 + Δ − 4 ]
The degree of D \mathcal{D} D d ≪ σ x 2 ≪ ( log 1 α ) 11 d\ll\sigma_x^2\ll\left(\log\frac{1}{\alpha}\right)^{11} d ≪ σ x 2 ≪ ( log α 1 ) 11 ∣ D ( S x − v ) ∣ 2 |\mathcal{D}(S_x-v)|^2 ∣ D ( S x − v ) ∣ 2
x 2 d = T O ( ( log ( 1 / α ) ) 11 log ( α log T ) α log T ) = T o ( 1 ) . x^{2d}=T^{O\left((\log(1/\alpha))^{11}\frac{\log(\alpha\log T)}{\alpha\log T}\right)}=T^{o(1)}. x 2 d = T O ( ( l o g ( 1/ α ) ) 11 α l o g T l o g ( α l o g T ) ) = T o ( 1 ) . Since α log x = 2 log ( α log T ) → ∞ \alpha\log x=2\log(\alpha\log T)\to\infty α log x = 2 log ( α log T ) → ∞ (5) holds. The coefficient bounds therefore give
∑ p ≤ x m p 2 ≪ log 1 α , ∑ p ≤ x m p 3 ≪ 1 , \sum_{p\le x}m_p^2\ll\log\frac{1}{\alpha},\qquad\sum_{p\le x}m_p^3\ll1, p ≤ x ∑ m p 2 ≪ log α 1 , p ≤ x ∑ m p 3 ≪ 1 , and the variance is σ x 2 + o ( 1 ) \sigma_x^2+o(1) σ x 2 + o ( 1 )
The truncation estimate (36) applies with ∣ v ∣ ≪ log 1 α |v|\ll\log\frac{1}{\alpha} ∣ v ∣ ≪ log α 1 (39) , (40) then give
P ( S x ∈ [ u − 1 , u ] ) ≪ 1 σ x exp ( − λ u + λ 2 2 Var r a n d ( S ~ x ) ) + O ( Δ e − Δ 2 ) \mathbb{P}(S_x\in[u-1,u])\ll\frac{1}{\sigma_x}\exp\left(-\lambda u+\frac{\lambda^2}{2}\operatorname{Var}_{\mathrm{rand}}(\widetilde{S}_x)\right)+O(\Delta e^{-\Delta^2}) P ( S x ∈ [ u − 1 , u ]) ≪ σ x 1 exp ( − λ u + 2 λ 2 Var rand ( S x ) ) + O ( Δ e − Δ 2 ) for bounded λ \lambda λ λ = u / Var r a n d ( S ~ x ) \lambda=u/\operatorname{Var}_{\mathrm{rand}}(\widetilde{S}_x) λ = u / Var rand ( S x ) σ x 2 \sigma_x^2 σ x 2 (54) . □
Proof of Theorems 1.7 and 1.8 in the small α \alpha α
Throughout this section we assume (19) , and let V = κ log 1 α V=\kappa\log\frac{1}{\alpha} V = κ log α 1 0 < κ 0 ≤ κ ≤ κ 1 < ∞ 0<\kappa_0\le\kappa\le\kappa_1<\infty 0 < κ 0 ≤ κ ≤ κ 1 < ∞ (19) . In the range (19) , we have
log 1 α = log log T + O ( log log log T ) . \log\frac{1}{\alpha}=\log\log T+O(\log\log\log T). log α 1 = log log T + O ( log log log T ) . For Theorem 1.7, where κ 1 < 1 / 2 \kappa_1<1/2 κ 1 < 1/2 2 κ 1 < ρ < 1 2\kappa_1<\rho<1 2 κ 1 < ρ < 1 0 < ρ < 1 0<\rho<1 0 < ρ < 1 1 1 1
( 1 − ρ ) ( κ 1 − 1 2 ) + ( 1 − ρ ) 2 4 < ϵ 1 2 . (55) (1-\rho)\left(\kappa_1-\frac{1}{2}\right)+\frac{(1-\rho)^2}{4}<\frac{\epsilon_1}{2}. \tag*{(55)} ( 1 − ρ ) ( κ 1 − 2 1 ) + 4 ( 1 − ρ ) 2 < 2 ϵ 1 . (55) Recall the definition of r > 1 r>1 r > 1 (20) in the definition of T β j T^{\beta_j} T β j r > 1 r>1 r > 1 D > 1 D>1 D > 1 a < 1 a<1 a < 1
ρ r D < a < 1. (56) \rho rD<a<1. \tag*{(56)} ρ r D < a < 1. (56) Also recall that θ \theta θ
δ l o c ≤ 1 − a 2 , (57) \delta_{\mathrm{loc}}\le\frac{1-a}{2}, \tag*{(57)} δ loc ≤ 2 1 − a , (57) with δ l o c \delta_{\mathrm{loc}} δ loc (24) . Thus
q β m + δ l o c ≤ 1 + a 2 < 1 q\beta_m+\delta_{\mathrm{loc}}\le\frac{1+a}{2}<1 q β m + δ loc ≤ 2 1 + a < 1 whenever q = ⌊ a / β m ⌋ q=\lfloor a/\beta_m\rfloor q = ⌊ a / β m ⌋
Lemma 6.1. For 2 ≤ ℓ ≤ K 2\le\ell\le K 2 ≤ ℓ ≤ K
P ( G ℓ − 1 ∩ A ℓ c ) ≪ e − C c ℓ z ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \mathbb{P}(G_{\ell-1}\cap A_\ell^c)\ll\frac{e^{-Cc_\ell z_\ell}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), P ( G ℓ − 1 ∩ A ℓ c ) ≪ log ( 1/ α ) e − C c ℓ z ℓ exp ( − log ( 1/ α ) V 2 ) , for some C > 0 C>0 C > 0
Proof. As in [3 ] , we have that
P ( G ℓ − 1 ∩ A ℓ c ) ≪ ∑ j = ℓ K ∑ u ∈ [ L ℓ − 1 , U ℓ − 1 ] E ( ( S ℓ ( j ) − S ℓ − 1 ( j ) ) 2 q ( 2 c ℓ ) 2 q 1 ( G ℓ − 1 ( j ) ( u ) ) ) . \mathbb{P}(G_{\ell-1}\cap A_\ell^c)\ll\sum_{j=\ell}^{K}\sum_{u\in[L_{\ell-1},U_{\ell-1}]} \mathbb{E}\left(\frac{(S_\ell^{(j)}-S_{\ell-1}^{(j)})^{2q}}{(2c_\ell)^{2q}}\mathbf{1}(G_{\ell-1}^{(j)}(u))\right). P ( G ℓ − 1 ∩ A ℓ c ) ≪ j = ℓ ∑ K u ∈ [ L ℓ − 1 , U ℓ − 1 ] ∑ E ( ( 2 c ℓ ) 2 q ( S ℓ ( j ) − S ℓ − 1 ( j ) ) 2 q 1 ( G ℓ − 1 ( j ) ( u )) ) . We choose q = ⌊ δ c ℓ ⌋ q=\lfloor\delta c_\ell\rfloor q = ⌊ δ c ℓ ⌋ δ > 0 \delta>0 δ > 0 S ℓ ( j ) − S ℓ − 1 ( j ) = ℜ Q ( t ) S_\ell^{(j)}-S_{\ell-1}^{(j)}=\Re Q(t) S ℓ ( j ) − S ℓ − 1 ( j ) = ℜ Q ( t ) ∣ ℜ Q ( t ) ∣ 2 q ≤ ∣ Q ( t ) ∣ 2 q |\Re Q(t)|^{2q}\leq|Q(t)|^{2q} ∣ℜ Q ( t ) ∣ 2 q ≤ ∣ Q ( t ) ∣ 2 q
E ( ( S ℓ ( j ) − S ℓ − 1 ( j ) ) 2 q ( 2 c ℓ ) 2 q 1 ( G ℓ − 1 ( j ) ( u ) ) ) ≪ E r a n d ( ∣ Q ~ ∣ 2 q ) ( 2 c ℓ ) 2 q σ ℓ − 1 , j 2 q exp ( − u 2 2 σ ℓ − 1 , j 2 + O ( z ℓ 2 ) ) . \mathbb{E}\left(\frac{(S_\ell^{(j)}-S_{\ell-1}^{(j)})^{2q}}{(2c_\ell)^{2q}}\mathbf{1}(G_{\ell-1}^{(j)}(u))\right)\ll\frac{\mathbb{E}_{\mathrm{rand}}(|\widetilde{Q}|^{2q})}{(2c_\ell)^{2q}\sigma_{\ell-1,j}^{2q}}\exp\left(-\frac{u^2}{2\sigma_{\ell-1,j}^2}+O(z_\ell^2)\right). E ( ( 2 c ℓ ) 2 q ( S ℓ ( j ) − S ℓ − 1 ( j ) ) 2 q 1 ( G ℓ − 1 ( j ) ( u )) ) ≪ ( 2 c ℓ ) 2 q σ ℓ − 1 , j 2 q E rand ( ∣ Q ∣ 2 q ) exp ( − 2 σ ℓ − 1 , j 2 u 2 + O ( z ℓ 2 ) ) . Summing over u ∈ [ L ℓ − 1 , U ℓ − 1 ] u\in[L_{\ell-1},U_{\ell-1}] u ∈ [ L ℓ − 1 , U ℓ − 1 ] j ≥ ℓ j\geq\ell j ≥ ℓ
P ( G ℓ − 1 ∩ A ℓ c ) ≪ ( K − ℓ + 1 ) c ℓ − 1 σ ℓ − 1 , j ( 2 q ) ! ( C z ℓ 2 ) q 2 q q ! ( 2 c ℓ ) 2 q exp ( − L ℓ − 1 2 2 σ ℓ − 1 , j 2 + O ( z ℓ 2 ) ) . \mathbb{P}(G_{\ell-1}\cap A_\ell^c)\ll\frac{(K-\ell+1)c_{\ell-1}}{\sigma_{\ell-1,j}}\frac{(2q)!(Cz_\ell^2)^q}{2^q q!(2c_\ell)^{2q}}\exp\left(-\frac{L_{\ell-1}^2}{2\sigma_{\ell-1,j}^2}+O(z_\ell^2)\right). P ( G ℓ − 1 ∩ A ℓ c ) ≪ σ ℓ − 1 , j ( K − ℓ + 1 ) c ℓ − 1 2 q q ! ( 2 c ℓ ) 2 q ( 2 q )! ( C z ℓ 2 ) q exp ( − 2 σ ℓ − 1 , j 2 L ℓ − 1 2 + O ( z ℓ 2 ) ) . for some C > 0 C>0 C > 0
≪ e − C c ℓ z ℓ σ ℓ − 1 , j exp ( − V 2 2 σ ℓ − 1 , j 2 ) ≪ e − C c ℓ z ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \ll\frac{e^{-Cc_\ell z_\ell}}{\sigma_{\ell-1,j}}\exp\left(-\frac{V^2}{2\sigma_{\ell-1,j}^2}\right)\ll\frac{e^{-Cc_\ell z_\ell}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), ≪ σ ℓ − 1 , j e − C c ℓ z ℓ exp ( − 2 σ ℓ − 1 , j 2 V 2 ) ≪ log ( 1/ α ) e − C c ℓ z ℓ exp ( − log ( 1/ α ) V 2 ) , where we used the fact that in the regime (4.1), we have
σ ℓ , j 2 = 1 2 log 1 α + O ( 1 ) , (58) \sigma_{\ell,j}^2=\frac{1}{2}\log\frac{1}{\alpha}+O(1), \tag*{(58)} σ ℓ , j 2 = 2 1 log α 1 + O ( 1 ) , (58) by the variance estimate (4.4). This proves the lemma. □
Lemma 6.2. For 2 ≤ ℓ ≤ K 2\leq\ell\leq K 2 ≤ ℓ ≤ K
P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) ≪ e − C c ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \mathbb{P}\left(G_{\ell-1}\cap A_\ell\cap\{\exists j\geq\ell:S_\ell^{(j)}>U_\ell\}\right)\ll\frac{e^{-Cc_\ell}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) ≪ log ( 1/ α ) e − C c ℓ exp ( − log ( 1/ α ) V 2 ) , for some C > 0 C>0 C > 0
Proof. Note that on the event G ℓ − 1 ∩ A ℓ G_{\ell-1}\cap A_\ell G ℓ − 1 ∩ A ℓ S ℓ ( j ) ≤ U ℓ − 1 + 2 c ℓ ≤ U ℓ + 3 c ℓ = : U ℓ ′ S_\ell^{(j)}\leq U_{\ell-1}+2c_\ell\leq U_\ell+3c_\ell=:U_\ell' S ℓ ( j ) ≤ U ℓ − 1 + 2 c ℓ ≤ U ℓ + 3 c ℓ =: U ℓ ′ S ℓ ( j ) ∈ [ U ℓ , U ℓ ′ ] S_\ell^{(j)}\in[U_\ell,U_\ell'] S ℓ ( j ) ∈ [ U ℓ , U ℓ ′ ]
P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) ≤ ∑ j = ℓ K ∑ u ∈ [ U ℓ , U ℓ ′ ] P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ] } ) . (59) \mathbb{P}\left(G_{\ell-1}\cap A_\ell\cap\{\exists j\geq\ell:S_\ell^{(j)}>U_\ell\}\right)\leq\sum_{j=\ell}^{K}\sum_{u\in[U_\ell,U_\ell']} \mathbb{P}\left(G_{\ell-1}\cap\{S_\ell^{(j)}\in[u-1,u]\}\right). \tag*{(59)} P ( G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) > U ℓ } ) ≤ j = ℓ ∑ K u ∈ [ U ℓ , U ℓ ′ ] ∑ P ( G ℓ − 1 ∩ { S ℓ ( j ) ∈ [ u − 1 , u ]} ) . (59) Note that because we are in the regime (4.1), we have that ∣ u ∣ ≤ A σ ℓ , j 2 |u|\leq A\sigma_{\ell,j}^2 ∣ u ∣ ≤ A σ ℓ , j 2 A > 0 A>0 A > 0
( 6.5 ) ≪ ∑ j = ℓ K ∑ u ∈ [ U ℓ , U ℓ ′ ] 1 σ ℓ , j exp ( − u 2 2 σ ℓ , j 2 ) ≪ ( K − ℓ + 1 ) c ℓ σ ℓ , j exp ( − U ℓ 2 2 σ ℓ , j 2 ) . ≪ e − C c ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \begin{aligned}
(6.5)\ll\sum_{j=\ell}^{K}\sum_{u\in[U_\ell,U_\ell']} \frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}\right)\ll\frac{(K-\ell+1)c_\ell}{\sigma_{\ell,j}}\exp\left(-\frac{U_\ell^2}{2\sigma_{\ell,j}^2}\right). \\
&\ll\frac{e^{-Cc_{\ell}}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right),
\end{aligned} ( 6.5 ) ≪ j = ℓ ∑ K u ∈ [ U ℓ , U ℓ ′ ] ∑ σ ℓ , j 1 exp ( − 2 σ ℓ , j 2 u 2 ) ≪ σ ℓ , j ( K − ℓ + 1 ) c ℓ exp ( − 2 σ ℓ , j 2 U ℓ 2 ) . ≪ log ( 1/ α ) e − C c ℓ exp ( − log ( 1/ α ) V 2 ) , for some C > 0 C>0 C > 0 (58) again.
Lemma 6.3. Suppose 0 < ρ < 1 0<\rho<1 0 < ρ < 1 2 ≤ ℓ ≤ K 2\leq\ell\leq K 2 ≤ ℓ ≤ K
P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ e − C c ℓ log ( 1 / α ) exp ( − ρ V + ρ 2 4 log 1 α ) , (60) \mathbb{P}\left(H\cap G_{\ell-1}\cap A_{\ell}\cap\left\{\exists j\geq\ell:S_{\ell}^{(j)}<L_{\ell}\right\}\right)\ll\frac{e^{-Cc_{\ell}}}{\sqrt{\log(1/\alpha)}}\exp\left(-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}\right), \tag*{(60)} P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ log ( 1/ α ) e − C c ℓ exp ( − ρ V + 4 ρ 2 log α 1 ) , (60) for some C > 0 C>0 C > 0 2 κ 1 < ρ 2\kappa_1<\rho 2 κ 1 < ρ
P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ e − C c ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , (61) \mathbb{P}\left(H\cap G_{\ell-1}\cap A_{\ell}\cap\left\{\exists j\geq\ell:S_{\ell}^{(j)}<L_{\ell}\right\}\right)\ll\frac{e^{-Cc_{\ell}}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), \tag*{(61)} P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ log ( 1/ α ) e − C c ℓ exp ( − log ( 1/ α ) V 2 ) , (61) for some $C>0.
Proof. Note that on the event G ℓ − 1 ∩ A ℓ G_{\ell-1}\cap A_{\ell} G ℓ − 1 ∩ A ℓ S ℓ ( j ) > L ℓ ′ : = L ℓ − 3 c ℓ S_{\ell}^{(j)}>L_{\ell}' := L_{\ell}-3c_{\ell} S ℓ ( j ) > L ℓ ′ := L ℓ − 3 c ℓ
P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≤ ∑ j = ℓ K ∑ u ∈ [ L ℓ ′ , L ℓ ] P ( { log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 − S ℓ ( j ) > V − u } ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) . (62) \mathbb{P}\left(H\cap G_{\ell-1}\cap A_{\ell}\cap\left\{\exists j\geq\ell:S_{\ell}^{(j)}<L_{\ell}\right\}\right)
\leq\sum_{j=\ell}^{K}\sum_{u\in[L_{\ell}',L_{\ell}]}\mathbb{P}\left(\left\{\log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}-S_{\ell}^{(j)}>V-u\right\}\cap G_{\ell}^{(j)}(u)\cap A_{\ell}\right). \tag*{(62)} P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≤ j = ℓ ∑ K u ∈ [ L ℓ ′ , L ℓ ] ∑ P ( { log ζ ( 2 1 + α + i t ) − 1 − S ℓ ( j ) > V − u } ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) . (62) Recall that D > 1 D>1 D > 1 1 1 1 Lemma 2.1 with x = T m x=T_m x = T m
log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 ≤ S m ( m ) + R m + C , \log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}\leq S_m^{(m)}+R_m+C, log ζ ( 2 1 + α + i t ) − 1 ≤ S m ( m ) + R m + C , for some constant C > 0 C>0 C > 0
R m = 1 β m log 1 1 − T m − α . R_m=\frac{1}{\beta_m}\log\frac{1}{1-T_m^{-\alpha}}. R m = β m 1 log 1 − T m − α 1 . Hence if
log ∣ ζ ( 1 2 + α + i t ) ∣ − 1 − S ℓ ( j ) > V − u , \log\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-1}-S_{\ell}^{(j)}>V-u, log ζ ( 2 1 + α + i t ) − 1 − S ℓ ( j ) > V − u , then
V − u < S m ( m ) − S ℓ ( j ) + R m + C . (63) V-u<S_m^{(m)}-S_{\ell}^{(j)}+R_m+C. \tag*{(63)} V − u < S m ( m ) − S ℓ ( j ) + R m + C . (63) Note that we have
c ℓ < V − u < C 1 c ℓ , c_{\ell}<V-u<C_1c_{\ell}, c ℓ < V − u < C 1 c ℓ , for some constant C 1 > 0 C_1>0 C 1 > 0
D log ( V − u ) V − u > β K . D\frac{\log(V-u)}{V-u}>\beta_K. D V − u log ( V − u ) > β K . Note that this implies that V − u = O ( 1 ) V-u=O(1) V − u = O ( 1 ) c ℓ = O ( 1 ) c_{\ell}=O(1) c ℓ = O ( 1 ) z ℓ = O ( 1 ) z_{\ell}=O(1) z ℓ = O ( 1 ) Proposition 5.1 with Q = 1 Q=1 Q = 1
P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≤ P ( G ℓ ( j ) ( u ) ) ≪ 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \mathbb{P}\left(H\cap G_{\ell}^{(j)}(u)\cap A_{\ell}\right)\leq\mathbb{P}\left(G_{\ell}^{(j)}(u)\right)\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≤ P ( G ℓ ( j ) ( u ) ) ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) V 2 ) , where we used equation (58) again. There are O ( 1 ) O(1) O ( 1 ) u u u j j j c ℓ = O ( 1 ) c_{\ell}=O(1) c ℓ = O ( 1 ) e − C c ℓ e^{-Cc_{\ell}} e − C c ℓ
We now assume that
D log ( V − u ) V − u ≤ β K . D\frac{\log(V-u)}{V-u}\leq\beta_K. D V − u log ( V − u ) ≤ β K . Choose m m m
β m ≥ D log ( V − u ) V − u . \beta_m \ge D\frac{\log(V-u)}{V-u}. β m ≥ D V − u log ( V − u ) . Such an index now exists with m ≤ K m \le K m ≤ K V − u ≍ c ℓ V-u \asymp c_\ell V − u ≍ c ℓ β ℓ = c ℓ − 90 \beta_\ell=c_\ell^{-90} β ℓ = c ℓ − 90 θ \theta θ
β ℓ < D log ( V − u ) V − u . \beta_\ell< D\frac{\log(V-u)}{V-u}. β ℓ < D V − u log ( V − u ) . Thus m > ℓ m>\ell m > ℓ m m m
D log ( V − u ) V − u ≤ β m < r D log ( V − u ) V − u . (64) D\frac{\log(V-u)}{V-u}\le\beta_m<rD\frac{\log(V-u)}{V-u}. \tag*{(64)} D V − u log ( V − u ) ≤ β m < r D V − u log ( V − u ) . (64) From the choice of m m m
R m ≤ V − u D log ( V − u ) log 1 1 − e − M 0 D log ( V − u ) / ( V − u ) ≤ V − u D log ( V − u ) log ( 1 + V − u M 0 D log ( V − u ) ) . R_m\le\frac{V-u}{D\log(V-u)}\log\frac{1}{1-e^{-M_0D\log(V-u)/(V-u)}}\le\frac{V-u}{D\log(V-u)}\log\left(1+\frac{V-u}{M_0D\log(V-u)}\right). R m ≤ D log ( V − u ) V − u log 1 − e − M 0 D l o g ( V − u ) / ( V − u ) 1 ≤ D log ( V − u ) V − u log ( 1 + M 0 D log ( V − u ) V − u ) . Since V − u ≥ c ℓ ≥ c K ≥ θ − 1 / 90 V-u\ge c_\ell\ge c_K\ge\theta^{-1/90} V − u ≥ c ℓ ≥ c K ≥ θ − 1/90 θ \theta θ
R m + C ≤ V − u D . R_m+C\le\frac{V-u}{D}. R m + C ≤ D V − u . Combining this with (63) we get
S m ( m ) − S ℓ ( j ) > ( V − u ) ( 1 − 1 D ) . S_m^{(m)}-S_\ell^{(j)}>(V-u)\left(1-\frac{1}{D}\right). S m ( m ) − S ℓ ( j ) > ( V − u ) ( 1 − D 1 ) . Hence, by writing
S m ( m ) − S ℓ ( j ) = S m ( m ) − S ℓ ( m ) + S ℓ ( m ) − S ℓ ( j ) , S_m^{(m)}-S_\ell^{(j)}=S_m^{(m)}-S_\ell^{(m)}+S_\ell^{(m)}-S_\ell^{(j)}, S m ( m ) − S ℓ ( j ) = S m ( m ) − S ℓ ( m ) + S ℓ ( m ) − S ℓ ( j ) , we obtain that
P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≤ P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) + P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) , (65) \mathbb{P}\left(H\cap G_\ell^{(j)}(u)\cap A_\ell\right)\le\mathbb{P}\left(G_\ell^{(j)}(u)\cap\left\{S_m^{(m)}-S_\ell^{(m)}>W\right\}\right)
+\mathbb{P}\left(G_\ell^{(j)}(u)\cap A_\ell\cap\left\{S_\ell^{(m)}-S_\ell^{(j)}>W\right\}\right), \tag*{(65)} P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≤ P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) + P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) , (65) where W = D − 1 2 D ( V − u ) W=\frac{D-1}{2D}(V-u) W = 2 D D − 1 ( V − u )
We first bound the first probability on the right hand side of (65) . Write S m ( m ) − S ℓ ( m ) = ℜ A ( t ) S_m^{(m)}-S_\ell^{(m)}=\Re A(t) S m ( m ) − S ℓ ( m ) = ℜ A ( t ) q ∈ N q\in\mathbb{N} q ∈ N Q = A ( t ) q Q=A(t)^q Q = A ( t ) q Q Q Q T ℓ T_\ell T ℓ ∣ ℜ A ( t ) ∣ 2 q ≤ ∣ A ( t ) ∣ 2 q |\Re A(t)|^{2q}\le|A(t)|^{2q} ∣ℜ A ( t ) ∣ 2 q ≤ ∣ A ( t ) ∣ 2 q
P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≤ E ( ∣ Q ∣ 2 W 2 q 1 ( G ℓ ( j ) ( u ) ) ) . (66) \mathbb{P}\left(G_\ell^{(j)}(u)\cap\left\{S_m^{(m)}-S_\ell^{(m)}>W\right\}\right)\le\mathbb{E}\left(\frac{|Q|^2}{W^{2q}}1(G_\ell^{(j)}(u))\right). \tag*{(66)} P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≤ E ( W 2 q ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) ) . (66) Now choose
q = ⌊ a β m ⌋ . (67) q=\left\lfloor\frac{a}{\beta_m}\right\rfloor. \tag*{(67)} q = ⌊ β m a ⌋ . (67) Note that the length of Q Q Q
T m q ≤ T β m q < T , T_m^q\le T^{\beta_mq}<T, T m q ≤ T β m q < T , and the condition (5.2) in Proposition 5.1 follows from (6.3). Proposition 5.1 then gives
E ( ∣ Q ∣ 2 W 2 q 1 ( G ℓ ( j ) ( u ) ) ) ≪ 1 σ ℓ , j exp ( − u 2 2 σ ℓ , j 2 ) E r a n d ( ∣ Q ~ ∣ 2 ) W 2 q . (68) \mathbb{E}\left(\frac{|Q|^2}{W^{2q}}1(G_\ell^{(j)}(u))\right)\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}\right)\frac{\mathbb{E}_{\mathrm{rand}}(|\widetilde{Q}|^2)}{W^{2q}}. \tag*{(68)} E ( W 2 q ∣ Q ∣ 2 1 ( G ℓ ( j ) ( u )) ) ≪ σ ℓ , j 1 exp ( − 2 σ ℓ , j 2 u 2 ) W 2 q E rand ( ∣ Q ∣ 2 ) . (68) Using Lemma 2.4 and (4.4) gives that
E r a n d ( ∣ Q ~ ∣ 2 ) ≪ q ! ( C z ℓ 2 ) q . (69) \mathbb{E}_{\mathrm{rand}}(|\widetilde{Q}|^2)\ll q!(Cz_\ell^2)^q. \tag*{(69)} E rand ( ∣ Q ∣ 2 ) ≪ q ! ( C z ℓ 2 ) q . (69) for some C > 0 C > 0 C > 0 (66) , (68) and (69) , it follows that
P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≪ 1 σ ℓ , j exp ( − u 2 2 σ ℓ , j 2 ) exp ( q log q + 2 q log z ℓ − 2 q log W + O ( q ) ) . \mathbb{P}\left(G_{\ell}^{(j)}(u)\cap\left\{S_m^{(m)}-S_{\ell}^{(m)}>W\right\}\right)
\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}\right)\exp\left(q\log q+2q\log z_\ell-2q\log W+O(q)\right). P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≪ σ ℓ , j 1 exp ( − 2 σ ℓ , j 2 u 2 ) exp ( q log q + 2 q log z ℓ − 2 q log W + O ( q ) ) . Hence by (64) and (67) , we have
P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≪ 1 σ ℓ , j exp ( − u 2 2 σ ℓ , j 2 ) exp ( − a ( V − u ) r D + O ( ( V − u ) log log ( V − u ) log ( V − u ) ) ) . (70) \mathbb{P}\left(G_{\ell}^{(j)}(u)\cap\left\{S_m^{(m)}-S_{\ell}^{(m)}>W\right\}\right)
\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}\right)\exp\left(-\frac{a(V-u)}{rD}+O\left(\frac{(V-u)\log\log(V-u)}{\log(V-u)}\right)\right). \tag*{(70)} P ( G ℓ ( j ) ( u ) ∩ { S m ( m ) − S ℓ ( m ) > W } ) ≪ σ ℓ , j 1 exp ( − 2 σ ℓ , j 2 u 2 ) exp ( − r D a ( V − u ) + O ( log ( V − u ) ( V − u ) log log ( V − u ) ) ) . (70) Since ρ < a r D \rho<\frac{a}{rD} ρ < r D a
(6.16) ≪ 1 log ( 1 / α ) exp ( − u 2 2 σ ℓ , j 2 − ρ ( V − u ) ) . (71) \text{(6.16)}\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)\right). \tag*{(71)} (6.16) ≪ log ( 1/ α ) 1 exp ( − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) ) . (71) Now we bound the second term on the right hand side of (65) . Similarly as before, since we are on the event G ℓ − 1 ∩ A ℓ G_{\ell-1}\cap A_\ell G ℓ − 1 ∩ A ℓ S ℓ ( m ) ∈ [ L ℓ ′ , U ℓ ′ ] S_\ell^{(m)}\in[L'_\ell,U'_\ell] S ℓ ( m ) ∈ [ L ℓ ′ , U ℓ ′ ] S ℓ ( m ) ∈ [ w − 1 , w ] S_\ell^{(m)}\in[w-1,w] S ℓ ( m ) ∈ [ w − 1 , w ] w ∈ [ L ℓ ′ , U ℓ ′ ] w\in[L'_\ell,U'_\ell] w ∈ [ L ℓ ′ , U ℓ ′ ] S ℓ ( m ) − S ℓ ( j ) > W S_\ell^{(m)}-S_\ell^{(j)}>W S ℓ ( m ) − S ℓ ( j ) > W w − u > W − 1 w-u>W-1 w − u > W − 1 θ \theta θ W − 1 > W 2 W-1>\frac{W}{2} W − 1 > 2 W
P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) ≪ 1 log ( 1 / α ) ∑ w ∈ [ L ℓ ′ , U ℓ ′ ] w − u > W 2 exp ( − ( u + w ) 2 8 σ ℓ , j 2 − C ( w − u ) 2 z ℓ 2 + O ( z ℓ 2 ) ) . \mathbb{P}\left(G_{\ell}^{(j)}(u)\cap A_\ell\cap\left\{S_\ell^{(m)}-S_\ell^{(j)}>W\right\}\right)
\ll\frac{1}{\sqrt{\log(1/\alpha)}}\sum_{\substack{w\in[L'_\ell,U'_\ell]\\w-u>\frac{W}{2}}}\exp\left(-\frac{(u+w)^2}{8\sigma_{\ell,j}^2}-C\frac{(w-u)^2}{z_\ell^2}+O(z_\ell^2)\right). P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) ≪ log ( 1/ α ) 1 w ∈ [ L ℓ ′ , U ℓ ′ ] w − u > 2 W ∑ exp ( − 8 σ ℓ , j 2 ( u + w ) 2 − C z ℓ 2 ( w − u ) 2 + O ( z ℓ 2 ) ) . Note that w > u w>u w > u ( u + w ) 2 / ( 8 σ ℓ , j 2 ) > u 2 / ( 2 σ ℓ , j 2 ) (u+w)^2/(8\sigma_{\ell,j}^2)>u^2/(2\sigma_{\ell,j}^2) ( u + w ) 2 / ( 8 σ ℓ , j 2 ) > u 2 / ( 2 σ ℓ , j 2 )
( w − u ) 2 z ℓ 2 ≥ ( D − 1 ) 2 16 D 2 c ℓ ( V − u ) z ℓ 2 ≥ ( ρ + 1 ) ( V − u ) . \frac{(w-u)^2}{z_\ell^2}\geq\frac{(D-1)^2}{16D^2}\frac{c_\ell(V-u)}{z_\ell^2}\geq(\rho+1)(V-u). z ℓ 2 ( w − u ) 2 ≥ 16 D 2 ( D − 1 ) 2 z ℓ 2 c ℓ ( V − u ) ≥ ( ρ + 1 ) ( V − u ) . Summing over w w w
P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) ≪ 1 log ( 1 / α ) exp ( − u 2 2 σ ℓ , j 2 − ρ ( V − u ) ) . \mathbb{P}\left(G_{\ell}^{(j)}(u)\cap A_\ell\cap\left\{S_\ell^{(m)}-S_\ell^{(j)}>W\right\}\right)
\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)\right). P ( G ℓ ( j ) ( u ) ∩ A ℓ ∩ { S ℓ ( m ) − S ℓ ( j ) > W } ) ≪ log ( 1/ α ) 1 exp ( − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) ) . Combining this with (71) and (62) gives that
P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ 1 log ( 1 / α ) ∑ j = ℓ κ ∑ u ∈ [ L ℓ ′ , L ℓ ] exp ( − u 2 2 σ ℓ , j 2 − ρ ( V − u ) ) . \mathbb{P}\left(H\cap G_{\ell-1}\cap A_\ell\cap\left\{\exists j\geq\ell:S_\ell^{(j)}<L_\ell\right\}\right)
\ll\frac{1}{\sqrt{\log(1/\alpha)}}\sum_{j=\ell}^{\kappa}\sum_{u\in[L'_\ell,L_\ell]}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)\right). P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ log ( 1/ α ) 1 j = ℓ ∑ κ u ∈ [ L ℓ ′ , L ℓ ] ∑ exp ( − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) ) . The preceding argument used only ρ < a / ( r D ) \rho<a/(rD) ρ < a / ( r D ) ρ \rho ρ ( ρ + a / ( r D ) ) / 2 (\rho+a/(rD))/2 ( ρ + a / ( r D )) /2
P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≪ 1 σ ℓ , j exp ( − u 2 2 σ ℓ , j 2 − ρ ( V − u ) − 1 2 ( a r D − ρ ) ( V − u ) ) . \mathbb{P}\left(H\cap G_{\ell}^{(j)}(u)\cap A_\ell\right)\ll\frac{1}{\sigma_{\ell,j}}\exp\left(-\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)-\frac{1}{2}\left(\frac{a}{rD}-\rho\right)(V-u)\right). P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≪ σ ℓ , j 1 exp ( − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) − 2 1 ( r D a − ρ ) ( V − u ) ) . Using the fact that
− u 2 2 σ ℓ , j 2 ≤ − u 2 log ( 1 / α ) + C z ℓ 2 . (72) -\frac{u^2}{2\sigma_{\ell,j}^2}\leq-\frac{u^2}{\log(1/\alpha)}+Cz_\ell^2. \tag*{(72)} − 2 σ ℓ , j 2 u 2 ≤ − log ( 1/ α ) u 2 + C z ℓ 2 . (72) for some C > 0 C>0 C > 0
− u 2 2 σ ℓ , j 2 − ρ ( V − u ) ≤ − ρ V + ρ 2 4 log 1 α − ( u − ρ 2 log ( 1 / α ) ) 2 log ( 1 / α ) + C z ℓ 2 ≤ − ρ V + ρ 2 4 log 1 α + C z ℓ 2 . \begin{aligned}
-\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)\le-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}-\frac{\left(u-\frac{\rho}{2}\log(1/\alpha)\right)^2}{\log(1/\alpha)}+Cz_\ell^2 \\
&\le-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}+Cz_\ell^2.
\end{aligned} − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) ≤ − ρ V + 4 ρ 2 log α 1 − log ( 1/ α ) ( u − 2 ρ log ( 1/ α ) ) 2 + C z ℓ 2 ≤ − ρ V + 4 ρ 2 log α 1 + C z ℓ 2 . There are O ( c ℓ ) O(c_\ell) O ( c ℓ ) u u u O ( z ℓ ) O(z_\ell) O ( z ℓ ) j j j c K c_K c K θ \theta θ ℓ ≤ K \ell\le K ℓ ≤ K
C z ℓ 2 + log ( C c ℓ ( 1 + z ℓ ) ) ≤ 1 4 ( a τ D − ρ ) c ℓ . Cz_\ell^2+\log\bigl(Cc_\ell(1+z_\ell)\bigr)\le\frac{1}{4}\left(\frac{a}{\tau D}-\rho\right)c_\ell. C z ℓ 2 + log ( C c ℓ ( 1 + z ℓ ) ) ≤ 4 1 ( τ D a − ρ ) c ℓ . Hence, combining the equations above and using the fact that V − u ≥ c ℓ V-u\ge c_\ell V − u ≥ c ℓ
∑ ℓ = 1 K ∑ j = 1 ℓ ∑ u ∈ [ L ℓ ′ , L ℓ ] P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≪ 1 log ( 1 / α ) exp ( − ρ V + ρ 2 4 log 1 α − 1 4 ( a τ D − ρ ) c ℓ ) . \sum_{\ell=1}^{K}\sum_{j=1}^{\ell}\sum_{u\in[L'_\ell,L_\ell]}\mathbb{P}\left(H\cap G_\ell^{(j)}(u)\cap A_\ell\right)\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}-\frac{1}{4}\left(\frac{a}{\tau D}-\rho\right)c_\ell\right). ℓ = 1 ∑ K j = 1 ∑ ℓ u ∈ [ L ℓ ′ , L ℓ ] ∑ P ( H ∩ G ℓ ( j ) ( u ) ∩ A ℓ ) ≪ log ( 1/ α ) 1 exp ( − ρ V + 4 ρ 2 log α 1 − 4 1 ( τ D a − ρ ) c ℓ ) . which proves the bound (6.6).
Finally, assume 2 κ 1 < ρ 2\kappa_1<\rho 2 κ 1 < ρ u 2 = V 2 − 2 V ( V − u ) + ( V − u ) 2 u^2=V^2-2V(V-u)+(V-u)^2 u 2 = V 2 − 2 V ( V − u ) + ( V − u ) 2
− u 2 2 σ ℓ , j 2 − ρ ( V − u ) ≤ − V 2 log ( 1 / α ) − ( ρ − 2 κ 1 ) ( V − u ) + C z ℓ 2 . -\frac{u^2}{2\sigma_{\ell,j}^2}-\rho(V-u)\le-\frac{V^2}{\log(1/\alpha)}-(\rho-2\kappa_1)(V-u)+Cz_\ell^2. − 2 σ ℓ , j 2 u 2 − ρ ( V − u ) ≤ − log ( 1/ α ) V 2 − ( ρ − 2 κ 1 ) ( V − u ) + C z ℓ 2 . Since ρ − 2 κ 1 > 0 \rho-2\kappa_1>0 ρ − 2 κ 1 > 0 V − u ≥ c ℓ V-u\ge c_\ell V − u ≥ c ℓ j j j u u u
P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ e − C c ℓ log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \mathbb{P}\left(H\cap G_{\ell-1}\cap A_\ell\cap\{\exists j\ge\ell:S_\ell^{(j)}<L_\ell\}\right)\ll\frac{e^{-Cc_\ell}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), P ( H ∩ G ℓ − 1 ∩ A ℓ ∩ { ∃ j ≥ ℓ : S ℓ ( j ) < L ℓ } ) ≪ log ( 1/ α ) e − C c ℓ exp ( − log ( 1/ α ) V 2 ) , for some C > 0 C>0 C > 0
Lemma 6.4
Lemma 6.4. Suppose 0 < ρ < 1 0<\rho<1 0 < ρ < 1 We have
P ( H ∩ G 1 c ) ≪ e − C c 1 log ( 1 / α ) exp ( − ρ V + ρ 2 4 log 1 α ) , (73) \mathbb{P}(H\cap G_1^c)\ll\frac{e^{-Cc_1}}{\sqrt{\log(1/\alpha)}}\exp\left(-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}\right), \tag*{(73)} P ( H ∩ G 1 c ) ≪ log ( 1/ α ) e − C c 1 exp ( − ρ V + 4 ρ 2 log α 1 ) , (73) for some C > 0 C>0 C > 0 If, in addition, 2 κ 1 < ρ 2\kappa_1<\rho 2 κ 1 < ρ
P ( H ∩ G 1 c ) ≪ e − C c 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) (74) \mathbb{P}(H\cap G_1^c)\ll\frac{e^{-Cc_1}}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right) \tag*{(74)} P ( H ∩ G 1 c ) ≪ log ( 1/ α ) e − C c 1 exp ( − log ( 1/ α ) V 2 ) (74) for some $C>0.
Proof. The proof is similar to the proofs of Lemmas 6.3 and 6.2, except that we first restrict the values of S 1 ( j ) S_1^{(j)} S 1 ( j ) [ u − 1 , u ] [u-1,u] [ u − 1 , u ] ∣ u ∣ ≤ C log log T |u|\le C\log\log T ∣ u ∣ ≤ C log log T C > 0 C>0 C > 0
P ( max 1 ≤ m ≤ K ∣ S 1 ( m ) ∣ > C log log T ) ≪ exp ( − c log log T ) , \mathbb{P}\left(\max_{1\le m\le K}|S_1^{(m)}|>C\log\log T\right)\ll\exp(-c\log\log T), P ( 1 ≤ m ≤ K max ∣ S 1 ( m ) ∣ > C log log T ) ≪ exp ( − c log log T ) , for some c > 0 c>0 c > 0
P ( H ∩ G 1 c ) ≤ P ( max 1 ≤ m ≤ K ∣ S 1 ( m ) ∣ > C log log T ) + ∑ j = 1 K P ( H ∩ { max 1 ≤ m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ∩ { S 1 ( j ) > U 1 } ) + ∑ j = 1 K P ( H ∩ { max 1 ≤ m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ∩ { S 1 ( j ) < L 1 } ) , (75) \begin{aligned}
\mathbb{P}(H\cap G_1^c)\le{}&\mathbb{P}\left(\max_{1\le m\le K}|S_1^{(m)}|>C\log\log T\right)\\
&+\sum_{j=1}^{K}\mathbb{P}\left(H\cap\left\{\max_{1\le m\le K}|S_1^{(m)}|\le C\log\log T\right\}\cap\{S_1^{(j)}>U_1\}\right)\\
&+\sum_{j=1}^{K}\mathbb{P}\left(H\cap\left\{\max_{1\le m\le K}|S_1^{(m)}|\le C\log\log T\right\}\cap\{S_1^{(j)}<L_1\}\right), \tag*{(75)}
\end{aligned} P ( H ∩ G 1 c ) ≤ P ( 1 ≤ m ≤ K max ∣ S 1 ( m ) ∣ > C log log T ) + j = 1 ∑ K P ( H ∩ { 1 ≤ m ≤ K max ∣ S 1 ( m ) ∣ ≤ C log log T } ∩ { S 1 ( j ) > U 1 } ) + j = 1 ∑ K P ( H ∩ { 1 ≤ m ≤ K max ∣ S 1 ( m ) ∣ ≤ C log log T } ∩ { S 1 ( j ) < L 1 } ) , (75) we are left with the second and the third probabilities on the right hand side of (75) .
For the second term in (75) , the same argument as in Lemma 6.2 goes through yielding the bound (6.20), and also (6.19).
For the third term, proceeding as in Lemma 6.3 , on H ∩ { S 1 ( j ) ∈ [ u − 1 , u ] } H \cap\{S_1^{(j)} \in[u-1,u]\} H ∩ { S 1 ( j ) ∈ [ u − 1 , u ]}
S m ( m ) − S 1 ( j ) > D − 1 D ( V − u ) , S_m^{(m)}-S_1^{(j)}>\frac{D-1}{D}(V-u), S m ( m ) − S 1 ( j ) > D D − 1 ( V − u ) , where we similarly choose m m m
β m ≥ D log ( V − u ) V − u . \beta_m \ge D\frac{\log(V-u)}{V-u}. β m ≥ D V − u log ( V − u ) . Here,
− C log log T ≤ u ≤ L 1 + 1 , -C\log\log T \le u \le L_1+1, − C log log T ≤ u ≤ L 1 + 1 , and note that c 1 − 1 ≤ V − u ≪ log log T c_1-1 \le V-u \ll\log\log T c 1 − 1 ≤ V − u ≪ log log T
P ( H ∩ { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { max m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ) ≤ P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S m ( m ) − S 1 ( m ) > W } ) + P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S 1 ( m ) − S 1 ( j ) > W } ∩ { max m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ) , (76) \begin{aligned}
\mathbb{P}\left(H\cap\{S_1^{(j)}\in[u-1,u]\}\cap\left\{\max_{m\le K}|S_1^{(m)}|\le C\log\log T\right\}\right) \\
&\le\mathbb{P}\left(\{S_1^{(j)}\in[u-1,u]\}\cap\{S_m^{(m)}-S_1^{(m)}>W\}\right) \\
&+\mathbb{P}\left(\{S_1^{(j)}\in[u-1,u]\}\cap\{S_1^{(m)}-S_1^{(j)}>W\}\cap\left\{\max_{m\le K}|S_1^{(m)}|\le C\log\log T\right\}\right), \tag*{(76)}
\end{aligned} P ( H ∩ { S 1 ( j ) ∈ [ u − 1 , u ]} ∩ { m ≤ K max ∣ S 1 ( m ) ∣ ≤ C log log T } ) ≤ P ( { S 1 ( j ) ∈ [ u − 1 , u ]} ∩ { S m ( m ) − S 1 ( m ) > W } ) + P ( { S 1 ( j ) ∈ [ u − 1 , u ]} ∩ { S 1 ( m ) − S 1 ( j ) > W } ∩ { m ≤ K max ∣ S 1 ( m ) ∣ ≤ C log log T } ) , (76) with W = D − 1 2 D ( V − u ) W=\frac{D-1}{2D}(V-u) W = 2 D D − 1 ( V − u )
For the first probability in (76) , as in Lemma 6.3 , take q = ⌊ a / β m ⌋ q=\lfloor a/\beta_m\rfloor q = ⌊ a / β m ⌋
Q ( t ) = ( ∑ T 1 < p ≤ T m b α ( p ; m ) p 1 / 2 + α + i t + ∑ T 1 < p p 2 ≤ T m b α ( p 2 ; m ) p 1 + 2 α + 2 i t ) q . Q(t)=\left(\sum_{T_1<p\le T_m}\frac{b_\alpha(p;m)}{p^{1/2+\alpha+it}}+\sum_{\substack{T_1<p\\p^2\le T_m}}\frac{b_\alpha(p^2;m)}{p^{1+2\alpha+2it}}\right)^q. Q ( t ) = T 1 < p ≤ T m ∑ p 1/2 + α + i t b α ( p ; m ) + T 1 < p p 2 ≤ T m ∑ p 1 + 2 α + 2 i t b α ( p 2 ; m ) q . Proposition 5.1 then implies that
P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S m ( m ) − S 1 ( m ) > W } ) ≪ 1 σ 1 , j exp ( − u 2 2 σ 1 , j 2 ) E r a n d ∣ Q ~ ∣ 2 W 2 q . \mathbb{P}\left(\{S_1^{(j)}\in[u-1,u]\}\cap\{S_m^{(m)}-S_1^{(m)}>W\}\right)\ll\frac{1}{\sigma_{1,j}}\exp\left(-\frac{u^2}{2\sigma_{1,j}^2}\right)\frac{\mathbb{E}_{\mathrm{rand}}|\widetilde{Q}|^2}{W^{2q}}. P ( { S 1 ( j ) ∈ [ u − 1 , u ]} ∩ { S m ( m ) − S 1 ( m ) > W } ) ≪ σ 1 , j 1 exp ( − 2 σ 1 , j 2 u 2 ) W 2 q E rand ∣ Q ∣ 2 . By Corollary 4.1, we have
∑ T 1 < p ≤ T m ∣ b α ( p ; m ) ∣ 2 p 1 + 2 α + ∑ T 1 < p p 2 ≤ T m ∣ b α ( p 2 ; m ) ∣ 2 p 2 + 4 α ≪ z 1 2 . \sum_{T_1<p\le T_m}\frac{|b_\alpha(p;m)|^2}{p^{1+2\alpha}}+\sum_{\substack{T_1<p\\p^2\le T_m}}\frac{|b_\alpha(p^2;m)|^2}{p^{2+4\alpha}}\ll z_1^2. T 1 < p ≤ T m ∑ p 1 + 2 α ∣ b α ( p ; m ) ∣ 2 + T 1 < p p 2 ≤ T m ∑ p 2 + 4 α ∣ b α ( p 2 ; m ) ∣ 2 ≪ z 1 2 . Hence Lemma 2.4 yields
E r a n d ∣ Q ~ ∣ 2 ≪ q ! ( C z 1 2 ) q . \mathbb{E}_{\mathrm{rand}}|\widetilde{Q}|^2\ll q!(Cz_1^2)^q. E rand ∣ Q ∣ 2 ≪ q ! ( C z 1 2 ) q . Proceeding as in Lemma 6.3 we get that the first probability in (76) is
≪ 1 σ 1 , j exp ( − u 2 2 σ 1 , j 2 − ρ ( V − u ) − 1 2 ( a r D − ρ ) ( V − u ) ) . \ll\frac{1}{\sigma_{1,j}}\exp\left(-\frac{u^2}{2\sigma_{1,j}^2}-\rho(V-u)-\frac{1}{2}\left(\frac{a}{rD}-\rho\right)(V-u)\right). ≪ σ 1 , j 1 exp ( − 2 σ 1 , j 2 u 2 − ρ ( V − u ) − 2 1 ( r D a − ρ ) ( V − u ) ) . For the second probability in (76) , suppose S 1 ( m ) ∈ [ w − 1 , w ] S_1^{(m)}\in[w-1,w] S 1 ( m ) ∈ [ w − 1 , w ]
− C log log T ≤ w ≤ C log log T + 1. -C\log\log T\le w\le C\log\log T+1. − C log log T ≤ w ≤ C log log T + 1. We use Lemma 5.1 (in particular, the bound (5.20)). Note that we have
− ( u + w ) 2 8 σ 1 , j 2 = − u 2 2 σ 1 , j 2 − u ( w − u ) 2 σ 1 , j 2 − ( w − u ) 2 8 σ 1 , j 2 ≤ − u 2 2 σ 1 , j 2 + A 2 ( w − u ) . -\frac{(u+w)^2}{8\sigma_{1,j}^2}=-\frac{u^2}{2\sigma_{1,j}^2}-\frac{u(w-u)}{2\sigma_{1,j}^2}-\frac{(w-u)^2}{8\sigma_{1,j}^2}\le-\frac{u^2}{2\sigma_{1,j}^2}+\frac{A}{2}(w-u). − 8 σ 1 , j 2 ( u + w ) 2 = − 2 σ 1 , j 2 u 2 − 2 σ 1 , j 2 u ( w − u ) − 8 σ 1 , j 2 ( w − u ) 2 ≤ − 2 σ 1 , j 2 u 2 + 2 A ( w − u ) . Hence, for some H H H
P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S 1 ( m ) − S 1 ( j ) > W } ∩ { max m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ) ≪ exp ( − u 2 / ( 2 σ 1 , j 2 ) + O ( z 1 2 ) ) σ 1 , j ∑ w − u > W / 2 e − ( H − A / 2 ) ( w − u ) ≪ 1 σ 1 , j exp ( − u 2 2 σ 1 , j 2 − a r D ( V − u ) + O ( z 1 2 ) ) ≪ 1 σ 1 , j exp ( − u 2 2 σ 1 , j 2 − ρ ( V − u ) − 1 2 ( a r D − ρ ) ( V − u ) ) . \begin{aligned}
\mathbb{P}\left(\left\{S_1^{(j)}\in[u-1,u]\right\}\cap\left\{S_1^{(m)}-S_1^{(j)}>W\right\}\cap\left\{\max_{m\le K}\left|S_1^{(m)}\right|\le C\log\log T\right\}\right) \\
&\ll\frac{\exp\left(-u^2/(2\sigma_{1,j}^2)+O(z_1^2)\right)}{\sigma_{1,j}}\sum_{w-u>W/2}e^{-(H-A/2)(w-u)} \\
&\ll\frac{1}{\sigma_{1,j}}\exp\left(-\frac{u^2}{2\sigma_{1,j}^2}-\frac{a}{rD}(V-u)+O(z_1^2)\right) \\
&\ll\frac{1}{\sigma_{1,j}}\exp\left(-\frac{u^2}{2\sigma_{1,j}^2}-\rho(V-u)-\frac{1}{2}\left(\frac{a}{rD}-\rho\right)(V-u)\right).
\end{aligned} P ( { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { S 1 ( m ) − S 1 ( j ) > W } ∩ { m ≤ K max S 1 ( m ) ≤ C log log T } ) ≪ σ 1 , j exp ( − u 2 / ( 2 σ 1 , j 2 ) + O ( z 1 2 ) ) w − u > W /2 ∑ e − ( H − A /2 ) ( w − u ) ≪ σ 1 , j 1 exp ( − 2 σ 1 , j 2 u 2 − r D a ( V − u ) + O ( z 1 2 ) ) ≪ σ 1 , j 1 exp ( − 2 σ 1 , j 2 u 2 − ρ ( V − u ) − 2 1 ( r D a − ρ ) ( V − u ) ) . In the last step we used z 1 2 = o ( c 1 ) z_1^2=o(c_1) z 1 2 = o ( c 1 ) V − u ≥ c 1 − 1 V-u\ge c_1-1 V − u ≥ c 1 − 1 (76) gives that
P ( H ∩ { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { max m ≤ K ∣ S 1 ( m ) ∣ ≤ C log log T } ) ≪ 1 log ( 1 / α ) exp ( − u 2 log ( 1 / α ) − ρ ( V − u ) − 1 2 ( a r D − ρ ) ( V − u ) + O ( z 1 2 ) ) . \begin{aligned}
\mathbb{P}\left(H\cap\left\{S_1^{(j)}\in[u-1,u]\right\}\cap\left\{\max_{m\le K}\left|S_1^{(m)}\right|\le C\log\log T\right\}\right) \\
&\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{u^2}{\log(1/\alpha)}-\rho(V-u)-\frac{1}{2}\left(\frac{a}{rD}-\rho\right)(V-u)+O(z_1^2)\right).
\end{aligned} P ( H ∩ { S 1 ( j ) ∈ [ u − 1 , u ] } ∩ { m ≤ K max S 1 ( m ) ≤ C log log T } ) ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) u 2 − ρ ( V − u ) − 2 1 ( r D a − ρ ) ( V − u ) + O ( z 1 2 ) ) . Similar argument as in the end of Lemma 6.3 finishes the proof of the lemma. □
Finally, the following lemma holds.
Lemma 6.5. We have that
P ( H ∩ G K ) ≪ 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) . \mathbb{P}(H\cap G_K)\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right). P ( H ∩ G K ) ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) V 2 ) . Proof. Note that on G K G_K G K S K K ∈ [ L K , U K ] S_K^K\in[L_K,U_K] S K K ∈ [ L K , U K ] L K = V + O ( 1 ) L_K=V+O(1) L K = V + O ( 1 ) U K = V + O ( 1 ) U_K=V+O(1) U K = V + O ( 1 )
P ( H ∩ G K ) ≤ ∑ u ∈ [ L K , U K ] P ( G K ( u ) ) ≪ 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) , \mathbb{P}(H\cap G_K)\le\sum_{u\in[L_K,U_K]}\mathbb{P}(G_K^{(u)})\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right), P ( H ∩ G K ) ≤ u ∈ [ L K , U K ] ∑ P ( G K ( u ) ) ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) V 2 ) , where we used Proposition 5.1 and the fact that there are O ( 1 ) O(1) O ( 1 ) [ L K , U K ] [L_K,U_K] [ L K , U K ]
Proof of Theorems 1.7 and 1.8. For the proof of Theorem 1.7, we have that 2 κ 1 < ρ 2\kappa_1<\rho 2 κ 1 < ρ (61) ), 6.4 (namely equation (74) ) and 6.5.
For the proof of Theorem 1.8, we have ρ \rho ρ (60) ), 6.4 (namely equation (73) ) and 6.5, we have that
P ( H ) ≪ 1 log ( 1 / α ) exp ( − ρ V + ρ 2 4 log 1 α ) . (77) \mathbb{P}(H)\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\rho V+\frac{\rho^2}{4}\log\frac{1}{\alpha}\right). \tag*{(77)} P ( H ) ≪ log ( 1/ α ) 1 exp ( − ρ V + 4 ρ 2 log α 1 ) . (77) Finally, by the choice (6.1) and the identity
− ρ κ + ρ 2 4 = − κ + 1 4 + ( 1 − ρ ) ( κ − 1 2 ) + ( 1 − ρ ) 2 4 , -\rho\kappa+\frac{\rho^2}{4}=-\kappa+\frac{1}{4}+(1-\rho)\left(\kappa-\frac{1}{2}\right)+\frac{(1-\rho)^2}{4}, − ρ κ + 4 ρ 2 = − κ + 4 1 + ( 1 − ρ ) ( κ − 2 1 ) + 4 ( 1 − ρ ) 2 , we have − ρ κ + ρ 2 / 4 ≤ − κ + 1 / 4 + ϵ 1 / 2 -\rho\kappa+\rho^2/4\le-\kappa+1/4+\epsilon_1/2 − ρ κ + ρ 2 /4 ≤ − κ + 1/4 + ϵ 1 /2 (77) yields
P ( H ) ≪ exp ( − V + ( 1 4 + ϵ 1 ) log 1 α ) , \mathbb{P}(H)\ll\exp\left(-V+\left(\frac{1}{4}+\epsilon_1\right)\log\frac{1}{\alpha}\right), P ( H ) ≪ exp ( − V + ( 4 1 + ϵ 1 ) log α 1 ) , which finishes the proof. □
Negative moments of ζ ( s ) \zeta(s) ζ ( s )
Before proving Theorem 1.1 , we need to prove the following lemma.
Lemma 7.1. Assume RH. Fix 0 < c < 1 0<c<1 0 < c < 1 0 < ρ < 1 0<\rho<1 0 < ρ < 1
1 ( log log log T ) c ≪ α log T ≪ log log T . (78) \frac{1}{(\log\log\log T)^c}\ll\alpha\log T\ll\log\log T. \tag*{(78)} ( log log log T ) c 1 ≪ α log T ≪ log log T . (78) There exist C 0 , C 1 , a > 0 C_0,C_1,a>0 C 0 , C 1 , a > 0
V 0 = C 0 log log T ( log log log T log log log log T ) 2 , V_0=C_0\log\log T\left(\frac{\log\log\log T}{\log\log\log\log T}\right)^2, V 0 = C 0 log log T ( log log log log T log log log T ) 2 , we have
P ( W α > V ) ≪ { e − ρ V , C 1 log log T ≤ V < V 0 , exp ( − V − a V log V ) , V ≥ V 0 . (79) \mathbb{P}(W_\alpha>V)\ll
\begin{cases}
e^{-\rho V}, & C_1\log\log T\le V<V_0,\\
\exp\left(-V-a\frac{V}{\log V}\right), & V\ge V_0.
\end{cases} \tag*{(79)} P ( W α > V ) ≪ { e − ρ V , exp ( − V − a l o g V V ) , C 1 log log T ≤ V < V 0 , V ≥ V 0 . (79) Proof. We first prove the bound for V ≥ V 0 V\ge V_0 V ≥ V 0 [14 Lemma 2.2, equation (2.8)] gives
W α ( t ) ≤ V ∗ : = log T 2 log log T ( log 1 α − log log log T + C ) ≍ log T , (80) W_\alpha(t)\le V_*:=\frac{\log T}{2\log\log T}\left(\log\frac{1}{\alpha}-\log\log\log T+C\right)\asymp\log T, \tag*{(80)} W α ( t ) ≤ V ∗ := 2 log log T log T ( log α 1 − log log log T + C ) ≍ log T , (80) for some C > 0 C>0 C > 0 V 0 ≤ V ≤ V ∗ V_0\le V\le V_* V 0 ≤ V ≤ V ∗
J = ⌊ log V ⌋ , β = J V , x = T J / V . J=\lfloor\log V\rfloor,\qquad\beta=\frac{J}{V},\qquad x=T^{J/V}. J = ⌊ log V ⌋ , β = V J , x = T J / V . Using Lemma 2.1 in [14 ] , we have that
W α ( t ) ≤ ℜ Q x ( t ) + V J log 1 1 − T − α J / V + O ( log log log T + max { 0 , log 1 α β log T } ) , (81) W_\alpha(t)\le\Re Q_x(t)+\frac{V}{J}\log\frac{1}{1-T^{-\alpha J/V}}+O\left(\log\log\log T+\max\left\{0,\log\frac{1}{\alpha\beta\log T}\right\}\right), \tag*{(81)} W α ( t ) ≤ ℜ Q x ( t ) + J V log 1 − T − α J / V 1 + O ( log log log T + max { 0 , log α β log T 1 } ) , (81) where
Q x ( t ) = ∑ p ≤ x b α ( p ; x ) p 1 / 2 + α + i t . Q_x(t)=\sum_{p\le x}\frac{b_\alpha(p;x)}{p^{1/2+\alpha+it}}. Q x ( t ) = p ≤ x ∑ p 1/2 + α + i t b α ( p ; x ) . From (78) , we have that
V J log 1 1 − T − α J / V ≤ V J ( log V − log J + c log log log log T + O ( 1 ) ) , \frac{V}{J}\log\frac{1}{1-T^{-\alpha J/V}}\le\frac{V}{J}\left(\log V-\log J+c\log\log\log\log T+O(1)\right), J V log 1 − T − α J / V 1 ≤ J V ( log V − log J + c log log log log T + O ( 1 ) ) , and hence
V − V J log 1 1 − T − α J / V ≥ V J ( log J − c log log log log T − O ( 1 ) ) ≥ 3 ( 1 − c ) V log J 4 J . V-\frac{V}{J}\log\frac{1}{1-T^{-\alpha J/V}}\ge\frac{V}{J}\left(\log J-c\log\log\log\log T-O(1)\right)\ge\frac{3(1-c)V\log J}{4J}. V − J V log 1 − T − α J / V 1 ≥ J V ( log J − c log log log log T − O ( 1 ) ) ≥ 4 J 3 ( 1 − c ) V log J . So, if W α ( t ) > V W_\alpha(t)>V W α ( t ) > V
∣ Q x ( t ) ∣ > w : = ( 1 − c ) V log J 4 J . |Q_x(t)|>w:=\frac{(1-c)V\log J}{4J}. ∣ Q x ( t ) ∣ > w := 4 J ( 1 − c ) V log J . Using Lemma 2.2, we also have that
∣ b α ( p ; x ) ∣ ≪ 1 + J log p log x ( 1 − log p log x ) . (82) |b_\alpha(p;x)|\ll1+J\frac{\log p}{\log x}\left(1-\frac{\log p}{\log x}\right). \tag*{(82)} ∣ b α ( p ; x ) ∣ ≪ 1 + J log x log p ( 1 − log x log p ) . (82) Now we divide the primes in consecutive intervals 1 , T 1 / V , T 2 / V , … , T J / V = x 1,T^{1/V},T^{2/V},\ldots,T^{J/V}=x 1 , T 1/ V , T 2/ V , … , T J / V = x
Q x = ∑ j = 0 J − 1 Q j , Q j ( t ) = ∑ T j / V < p ≤ T ( j + 1 ) / V b α ( p ; x ) p 1 / 2 + α + i t . (83) Q_x=\sum_{j=0}^{J-1}Q_j,\qquad Q_j(t)=\sum_{T^{j/V}<p\le T^{(j+1)/V}}\frac{b_\alpha(p;x)}{p^{1/2+\alpha+it}}. \tag*{(83)} Q x = j = 0 ∑ J − 1 Q j , Q j ( t ) = T j / V < p ≤ T ( j + 1 ) / V ∑ p 1/2 + α + i t b α ( p ; x ) . (83) For the primes in Q 0 Q_0 Q 0 (82) we have that b α ( p ; x ) ≪ 1 b_\alpha(p;x) \ll1 b α ( p ; x ) ≪ 1 1 ≤ j < J 1 \le j < J 1 ≤ j < J Q j Q_j Q j (82) again) b α ( p ; x ) ≪ J − j b_\alpha(p;x) \ll J-j b α ( p ; x ) ≪ J − j
∑ p ≤ T 1 / V ∣ b α ( p ; x ) ∣ 2 p 1 + 2 α ≪ log log T , \sum_{p\le T^{1/V}} \frac{|b_\alpha(p;x)|^2}{p^{1+2\alpha}} \ll\log\log T, p ≤ T 1/ V ∑ p 1 + 2 α ∣ b α ( p ; x ) ∣ 2 ≪ log log T , ∑ T j / V < p ≤ T ( j + 1 ) / V ∣ b α ( p ; x ) ∣ 2 p 1 + 2 α ≪ ( J − j ) 2 j + 1 , (84) \sum_{T^{j/V}<p\le T^{(j+1)/V}} \frac{|b_\alpha(p;x)|^2}{p^{1+2\alpha}} \ll\frac{(J-j)^2}{j+1}, \tag*{(84)} T j / V < p ≤ T ( j + 1 ) / V ∑ p 1 + 2 α ∣ b α ( p ; x ) ∣ 2 ≪ j + 1 ( J − j ) 2 , (84) for 1 ≤ j < J 1 \le j < J 1 ≤ j < J
Now note that
P ( W α > V ) ≤ P ( ∣ Q 0 ∣ > w 4 ) + ∑ j = 1 J − 1 P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) . \mathbb{P}(W_\alpha>V)\le\mathbb{P}\left(|Q_0|>\frac{w}{4}\right)+\sum_{j=1}^{J-1}\mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right). P ( W α > V ) ≤ P ( ∣ Q 0 ∣ > 4 w ) + j = 1 ∑ J − 1 P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) . For 0 ≤ j < J 0\le j<J 0 ≤ j < J
q j = ⌊ V j + 1 ⌋ . (85) q_j=\left\lfloor\frac{V}{j+1}\right\rfloor. \tag*{(85)} q j = ⌊ j + 1 V ⌋ . (85) The polynomial Q j q j Q_j^{q_j} Q j q j T q j ( j + 1 ) / V ≤ T T^{q_j(j+1)/V}\le T T q j ( j + 1 ) / V ≤ T (84) , followed by Markov’s inequality and Stirling’s formula, we obtain that
P ( ∣ Q 0 ∣ > w 4 ) ≪ ( C V log log T w 2 ) [ V ] , (86) \mathbb{P}\left(|Q_0|>\frac{w}{4}\right)\ll\left(\frac{CV\log\log T}{w^2}\right)^{[V]}, \tag*{(86)} P ( ∣ Q 0 ∣ > 4 w ) ≪ ( w 2 C V log log T ) [ V ] , (86) P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) ≪ ( C V ( J − j ) 6 ( j + 1 ) 2 w 2 ) q j ≤ ( C V ( J − j ) 8 J 2 w 2 ) q j . (87) \mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right)\ll\left(\frac{CV(J-j)^6}{(j+1)^2w^2}\right)^{q_j}\le\left(\frac{CV(J-j)^8}{J^2w^2}\right)^{q_j}. \tag*{(87)} P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) ≪ ( ( j + 1 ) 2 w 2 C V ( J − j ) 6 ) q j ≤ ( J 2 w 2 C V ( J − j ) 8 ) q j . (87) Substituting w = ( 1 − c ) V log J / ( 4 J ) w=(1-c)V\log J/(4J) w = ( 1 − c ) V log J / ( 4 J )
P ( ∣ Q 0 ∣ > w 4 ) ≪ ( C J 2 log log T V ( log J ) 2 ) [ V ] ≪ e − 2 V . \mathbb{P}\left(|Q_0|>\frac{w}{4}\right)\ll\left(\frac{CJ^2\log\log T}{V(\log J)^2}\right)^{[V]}\ll e^{-2V}. P ( ∣ Q 0 ∣ > 4 w ) ≪ ( V ( log J ) 2 C J 2 log log T ) [ V ] ≪ e − 2 V . For j ≥ 1 j\ge1 j ≥ 1 (87) becomes
P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) ≪ ( C ( J − j ) 8 V ( log J ) 2 ) q j . \mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right)\ll\left(\frac{C(J-j)^8}{V(\log J)^2}\right)^{q_j}. P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) ≪ ( V ( log J ) 2 C ( J − j ) 8 ) q j . We have that
log V ( log J ) 2 C ( J − j ) 8 ≥ j + 2 log log J − log C − 9 ≥ j + 5 \log\frac{V(\log J)^2}{C(J-j)^8}\ge j+2\log\log J-\log C-9\ge j+5 log C ( J − j ) 8 V ( log J ) 2 ≥ j + 2 log log J − log C − 9 ≥ j + 5 for sufficiently large T T T
P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) ≪ e − q j ( j + 5 ) , \mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right)\ll e^{-q_j(j+5)}, P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) ≪ e − q j ( j + 5 ) , and since
q j ( j + 5 ) ≥ V + 3 V J , q_j(j+5)\ge V+\frac{3V}{J}, q j ( j + 5 ) ≥ V + J 3 V , it follows that
∑ j = 1 J − 1 P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) ≪ J e − V − 3 V / J ≤ e − V − 2 V / J . \sum_{j=1}^{J-1}\mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right)\ll Je^{-V-3V/J}\le e^{-V-2V/J}. j = 1 ∑ J − 1 P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) ≪ J e − V − 3 V / J ≤ e − V − 2 V / J . Together with the bound for Q 0 Q_0 Q 0 (79) .
For C 1 log log T ≤ V < V 0 C_1\log\log T\le V<V_0 C 1 log log T ≤ V < V 0 D D D 1 < D < 1 / ρ 1<D<1/\rho 1 < D < 1/ ρ
J = ⌊ D log V ⌋ , β = J V , x = T J / V . J=\lfloor D\log V\rfloor,\qquad\beta=\frac{J}{V},\qquad x=T^{J/V}. J = ⌊ D log V ⌋ , β = V J , x = T J / V . Using (7.1), we have that
V J log 1 1 − T − α J / V ≤ V J ( log V − log J + c log log log log T + O ( 1 ) ) = V D + o ( V ) . \frac{V}{J}\log\frac{1}{1-T^{-\alpha J/V}}\leq\frac{V}{J}\left(\log V-\log J+c\log\log\log\log T+O(1)\right)=\frac{V}{D}+o(V). J V log 1 − T − α J / V 1 ≤ J V ( log V − log J + c log log log log T + O ( 1 ) ) = D V + o ( V ) . The error in (81) is o ( V ) o(V) o ( V )
{ W α > V } ⊆ { ∣ Q x ∣ > w } , \{W_\alpha>V\}\subseteq\{|Q_x|>w\}, { W α > V } ⊆ { ∣ Q x ∣ > w } , for w = D − 1 2 D V w=\frac{D-1}{2D}V w = 2 D D − 1 V (82) , the division (83) , and the estimate (87) hold as before. Now we have
log J 2 w 2 C V ( J − j ) 8 ≥ j + 1 D + 2 log J − C ≥ j + 1 D + 4 , \log\frac{J^2w^2}{CV(J-j)^8}\geq\frac{j+1}{D}+2\log J-C\geq\frac{j+1}{D}+4, log C V ( J − j ) 8 J 2 w 2 ≥ D j + 1 + 2 log J − C ≥ D j + 1 + 4 , so
∑ j = 1 J − 1 P ( ∣ Q j ∣ > w 4 ( J − j ) 2 ) ≪ J exp ( − V − J D − 4 ( V J − 1 ) ) ≪ e − V / D . \sum_{j=1}^{J-1}\mathbb{P}\left(|Q_j|>\frac{w}{4(J-j)^2}\right)\ll J\exp\left(-\frac{V-J}{D}-4\left(\frac{V}{J}-1\right)\right)\ll e^{-V/D}. j = 1 ∑ J − 1 P ( ∣ Q j ∣ > 4 ( J − j ) 2 w ) ≪ J exp ( − D V − J − 4 ( J V − 1 ) ) ≪ e − V / D . Including the contribution from Q 0 Q_0 Q 0
P ( W α > V ) ≪ ( C log log T V ) ⌊ V ⌋ + e − V / D . \mathbb{P}(W_\alpha>V)\ll\left(\frac{C\log\log T}{V}\right)^{\lfloor V\rfloor}+e^{-V/D}. P ( W α > V ) ≪ ( V C log log T ) ⌊ V ⌋ + e − V / D . Since 1 / D > ρ 1/D>\rho 1/ D > ρ C 1 C_1 C 1 O ( e − ρ V ) O(e^{-\rho V}) O ( e − ρ V ) (79) . □
Using similar arguments, we can also prove the following variant. We will skip the proof since it is very similar to the proof of Lemma 7.1.
Lemma 7.2. Assume RH. Fix 0 < c < 1 0<c<1 0 < c < 1 0 < ε < 1 0<\varepsilon<1 0 < ε < 1 C > 0 C>0 C > 0
C log log T ≤ α log T ≪ 1 ( log log log T ) c \frac{C}{\log\log T}\leq\alpha\log T\ll\frac{1}{(\log\log\log T)^c} log log T C ≤ α log T ≪ ( log log log T ) c 1 and ( log T ) ε ≤ V ≤ V ∗ (\log T)^\varepsilon\leq V\leq V_\ast ( log T ) ε ≤ V ≤ V ∗ V ∗ V_\ast V ∗ (80) ), we have
P ( W α > V ) ≪ exp ( − V − 2 V log V ) . \mathbb{P}(W_\alpha>V)\ll\exp\left(-V-\frac{2V}{\log V}\right). P ( W α > V ) ≪ exp ( − V − log V 2 V ) . Now suppose that k > 1 / 2 k>1/2 k > 1/2
C k ( log T ) 1 / ( 2 k ) log log T ≤ α ≪ ( log log T ) 4 / k + ε ( log T ) 1 / ( 2 k ) . (88) \frac{C_k}{(\log T)^{1/(2k)}\log\log T}\leq\alpha\ll\frac{(\log\log T)^{4/k+\varepsilon}}{(\log T)^{1/(2k)}}. \tag*{(88)} ( log T ) 1/ ( 2 k ) log log T C k ≤ α ≪ ( log T ) 1/ ( 2 k ) ( log log T ) 4/ k + ε . (88) We will prove the following lemma which holds in this regime.
Lemma 7.3. Assume RH. Fix k > 1 / 2 k>1/2 k > 1/2 b b b B B B c k c_k c k C k > 0 C_k>0 C k > 0 (88) , we have
P ( W α > V ) ≪ { e − ( 2 k + 2 ) V , b log ( 1 / α ) ≤ V ≤ B α log T , exp ( − ( 2 k + c k log log T ) V ) , V > B α log T . (89) \mathbb{P}(W_\alpha>V)\ll
\begin{cases}
e^{-(2k+2)V}, & b\log(1/\alpha)\leq V\leq B\alpha\log T,\\
\exp\left(-\left(2k+\frac{c_k}{\log\log T}\right)V\right), & V>B\alpha\log T.
\end{cases} \tag*{(89)} P ( W α > V ) ≪ { e − ( 2 k + 2 ) V , exp ( − ( 2 k + l o g l o g T c k ) V ) , b log ( 1/ α ) ≤ V ≤ B α log T , V > B α log T . (89) Proof. The proof follows along similar lines as Lemma 7.1, so we will only sketch the details. We first prove the bound in the second case. The pointwise bound (80) still holds, so we restrict to V ≤ V ∗ V\leq V_\ast V ≤ V ∗
J = ⌊ log ( V α log T ) − log log ( V α log T ) + 2 ⌋ , β = J V , x = T J / V . J=\left\lfloor\log\left(\frac{V}{\alpha\log T}\right)-\log\log\left(\frac{V}{\alpha\log T}\right)+2\right\rfloor,\qquad\beta=\frac{J}{V},\qquad x=T^{J/V}. J = ⌊ log ( α log T V ) − log log ( α log T V ) + 2 ⌋ , β = V J , x = T J / V . As in Lemma 7.1, we have
V − V J log 1 1 − T − J α / V ≥ 3 V 4 J . V-\frac{V}{J}\log\frac{1}{1-T^{-J\alpha/V}}\geq\frac{3V}{4J}. V − J V log 1 − T − J α / V 1 ≥ 4 J 3 V . Then
{ W α > V } ⊆ { ∣ Q x ∣ > w } , \{W_\alpha> V\} \subseteq\{|Q_x| > w\}, { W α > V } ⊆ { ∣ Q x ∣ > w } , for
w = V 2 J . w = \frac{V}{2J}. w = 2 J V . The coefficient bound (7.5), the partition (7.6), and the q j q_j q j w w w q j ( j + 1 ) ≥ V − J q_j(j+1) \ge V-J q j ( j + 1 ) ≥ V − J q j ≥ V / J − 1 q_j \ge V/J-1 q j ≥ V / J − 1
q j log V C ( J − j ) δ ≥ ( V − J ) + ( V J − 1 ) ( log V − J − C ) = ( V J − 1 ) ( log V − C ) . q_j \log\frac{V}{C(J-j)^{\delta}} \ge(V-J)+\left(\frac{V}{J}-1\right)(\log V-J-C)=\left(\frac{V}{J}-1\right)(\log V-C). q j log C ( J − j ) δ V ≥ ( V − J ) + ( J V − 1 ) ( log V − J − C ) = ( J V − 1 ) ( log V − C ) . Thus the sum over j ≥ 1 j \ge1 j ≥ 1
J exp ( − ( V J − 1 ) ( log V − C ) ) ≪ exp ( − V J ( log V − C ) ) . J\exp\left(-\left(\frac{V}{J}-1\right)(\log V-C)\right)\ll\exp\left(-\frac{V}{J}(\log V-C)\right). J exp ( − ( J V − 1 ) ( log V − C ) ) ≪ exp ( − J V ( log V − C ) ) . For the contribution from the small primes, (7.9) still holds, and since V > B α log T V>B\alpha\log T V > B α log T
( C J 2 log log T V ) ⌊ V ⌋ ≤ exp ( − V 3 log V ) . \left(\frac{CJ^2\log\log T}{V}\right)^{\lfloor V\rfloor}\le\exp\left(-\frac{V}{3}\log V\right). ( V C J 2 log log T ) ⌊ V ⌋ ≤ exp ( − 3 V log V ) . Hence
P ( W α > V ) ≪ k exp ( − V J ( log V − C ) ) . (90) \mathbb{P}(W_\alpha>V)\ll_k\exp\left(-\frac{V}{J}(\log V-C)\right). \tag*{(90)} P ( W α > V ) ≪ k exp ( − J V ( log V − C ) ) . (90) We next show that
log V − 2 k J ≥ C + 1. (91) \log V-2kJ\ge C+1. \tag*{(91)} log V − 2 k J ≥ C + 1. (91) We have
J ≤ log V α log T and e J ≤ e 3 V α log T log ( V / α log T ) , J\le\log\frac{V}{\alpha\log T}
\qquad\text{and}\qquad
e^J\le\frac{e^3V}{\alpha\log T\log(V/\alpha\log T)}, J ≤ log α log T V and e J ≤ α log T log ( V / α log T ) e 3 V , and hence
( α log T ) J e J ≪ V . (92) (\alpha\log T)Je^J\ll V. \tag*{(92)} ( α log T ) J e J ≪ V . (92) Moreover, (7.11) and (7.3) give
log V ≍ log log T \log V\asymp\log\log T log V ≍ log log T uniformly for B α log T < V ≤ V ∗ B\alpha\log T<V\le V_* B α log T < V ≤ V ∗
J > log V − C − 1 2 k ≫ log log T . J>\frac{\log V-C-1}{2k}\gg\log\log T. J > 2 k log V − C − 1 ≫ log log T . Substituting this lower bound into (7.15) would give
α log T log log T ≪ V 1 − 1 / ( 2 k ) ≪ ( log T ) 1 − 1 / ( 2 k ) , \alpha\log T\log\log T\ll V^{1-1/(2k)}\ll(\log T)^{1-1/(2k)}, α log T log log T ≪ V 1 − 1/ ( 2 k ) ≪ ( log T ) 1 − 1/ ( 2 k ) , where the last inequality uses k > 1 / 2 k>1/2 k > 1/2 C k C_k C k
Finally,
log V − C J ≥ 2 k + 1 J ≥ 2 k + c k log log T . \frac{\log V-C}{J}\ge2k+\frac{1}{J}\ge2k+\frac{c_k}{\log\log T}. J log V − C ≥ 2 k + J 1 ≥ 2 k + log log T c k . Together with (7.13), this proves the second bound in (7.12).
For b log ( 1 / α ) ≤ V ≤ B α log T b\log(1/\alpha)\le V\le B\alpha\log T b log ( 1/ α ) ≤ V ≤ B α log T Q x Q_x Q x β = A 0 / V \beta=A_0/V β = A 0 / V x = T β x=T^\beta x = T β A 0 A_0 A 0 α log T β ≥ A 0 / B \alpha\log T\beta\ge A_0/B α log T β ≥ A 0 / B V / 8 V/8 V /8 O ( log log log T ) = o ( V ) O(\log\log\log T)=o(V) O ( log log log T ) = o ( V )
P ( W α > V ) ≤ P ( ∣ Q x ∣ > V 2 ) . \mathbb{P}(W_\alpha>V)\le\mathbb{P}\left(|Q_x|>\frac{V}{2}\right). P ( W α > V ) ≤ P ( ∣ Q x ∣ > 2 V ) . Moreover, Lemma 2.2 gives ∣ b α ( p ; x ) ∣ ≪ 1 |b_{\alpha}(p;x)| \ll1 ∣ b α ( p ; x ) ∣ ≪ 1
∑ p ≤ x ∣ b α ( p ; x ) ∣ 2 p 1 + 2 α ≪ log 1 α . \sum_{p\le x}\frac{|b_{\alpha}(p;x)|^2}{p^{1+2\alpha}}\ll\log\frac{1}{\alpha}. p ≤ x ∑ p 1 + 2 α ∣ b α ( p ; x ) ∣ 2 ≪ log α 1 . Take q = ⌊ V / A 0 ⌋ q=\lfloor V/A_0\rfloor q = ⌊ V / A 0 ⌋ Q x q Q_x^q Q x q x q ≤ T x^q\le T x q ≤ T
P ( W α > V ) ≪ ( C q log ( 1 / α ) V 2 ) q ≤ ( C log ( 1 / α ) A 0 V ) q . \mathbb{P}(W_{\alpha}>V)\ll\left(\frac{Cq\log(1/\alpha)}{V^2}\right)^q\le\left(\frac{C\log(1/\alpha)}{A_0V}\right)^q. P ( W α > V ) ≪ ( V 2 C q log ( 1/ α ) ) q ≤ ( A 0 V C log ( 1/ α ) ) q . Choose b b b C / ( A 0 b ) ≤ e − 4 A 0 ( k + 1 ) C/(A_0b)\le e^{-4A_0(k+1)} C / ( A 0 b ) ≤ e − 4 A 0 ( k + 1 ) V ≥ b log 1 α V\ge b\log\frac{1}{\alpha} V ≥ b log α 1 T T T q ≥ V / ( 2 A 0 ) q\ge V/(2A_0) q ≥ V / ( 2 A 0 ) O ( e − ( 2 k + 2 ) V ) O(e^{-(2k+2)V}) O ( e − ( 2 k + 2 ) V ) (89) . □
Proof of Theorem 1.1 . We have
I − k ( α , T ) = 1 T ∫ T 2 T ∣ ζ ( 1 2 + α + i t ) ∣ − 2 k d t = 2 k ∫ − ∞ ∞ e 2 k V P ( W α > V ) d V . (93) I_{-k}(\alpha,T)=\frac{1}{T}\int_T^{2T}\left|\zeta\left(\frac{1}{2}+\alpha+it\right)\right|^{-2k}\,dt=2k\int_{-\infty}^{\infty}e^{2kV}\mathbb{P}(W_{\alpha}>V)\,dV. \tag*{(93)} I − k ( α , T ) = T 1 ∫ T 2 T ζ ( 2 1 + α + i t ) − 2 k d t = 2 k ∫ − ∞ ∞ e 2 k V P ( W α > V ) d V . (93) First assume that we are in the range (1.1). Fix
k ( 1 − k ) < b < 1 2 , 2 k < ρ < 1 , \sqrt{k(1-k)}<b<\frac{1}{2},\qquad2k<\rho<1, k ( 1 − k ) < b < 2 1 , 2 k < ρ < 1 , and let C 1 > 1 C_1>1 C 1 > 1 Lemma 7.1 . We split the integral as follows:
I − k ( α , T ) = 2 k ( ∫ − ∞ ( k / 4 ) log ( 1 / α ) + ∫ ( k / 4 ) log ( 1 / α ) b log ( 1 / α ) + ∫ b log ( 1 / α ) 1 2 log ( 1 / α ) + ∫ 1 2 log ( 1 / α ) B log log T + ∫ B log log T ∞ ) e 2 k V P ( W α > V ) d V = ∑ j = 1 5 I j , \begin{aligned}
I_{-k}(\alpha,T)=2k\left(\int_{-\infty}^{(k/4)\log(1/\alpha)}+\int_{(k/4)\log(1/\alpha)}^{b\log(1/\alpha)}+\int_{b\log(1/\alpha)}^{\frac{1}{2}\log(1/\alpha)}+\int_{\frac{1}{2}\log(1/\alpha)}^{B\log\log T}+\int_{B\log\log T}^{\infty}\right)e^{2kV}\mathbb{P}(W_{\alpha}>V)\,dV \\
&=\sum_{j=1}^{5}I_j,
\end{aligned} I − k ( α , T ) = 2 k ( ∫ − ∞ ( k /4 ) l o g ( 1/ α ) + ∫ ( k /4 ) l o g ( 1/ α ) b l o g ( 1/ α ) + ∫ b l o g ( 1/ α ) 2 1 l o g ( 1/ α ) + ∫ 2 1 l o g ( 1/ α ) B l o g l o g T + ∫ B l o g l o g T ∞ ) e 2 k V P ( W α > V ) d V = j = 1 ∑ 5 I j , say. The endpoints are in the displayed order for sufficiently large T T T
log 1 α = log log T + O ( log log log T ) . \log\frac{1}{\alpha}=\log\log T+O(\log\log\log T). log α 1 = log log T + O ( log log log T ) . For the first integral, we use P ( W α > V ) ≤ 1 \mathbb{P}(W_{\alpha}>V)\le1 P ( W α > V ) ≤ 1
I 1 = 2 k ∫ − ∞ ( k / 4 ) log ( 1 / α ) e 2 k V P ( W α > V ) d V ≤ α − k 2 / 2 . I_1=2k\int_{-\infty}^{(k/4)\log(1/\alpha)}e^{2kV}\mathbb{P}(W_{\alpha}>V)\,dV\le\alpha^{-k^2/2}. I 1 = 2 k ∫ − ∞ ( k /4 ) l o g ( 1/ α ) e 2 k V P ( W α > V ) d V ≤ α − k 2 /2 . For the second integral, Theorem 1.7 gives
I 2 ≪ α − k 2 log ( 1 / α ) ∫ R exp ( − ( V − k log ( 1 / α ) ) 2 log ( 1 / α ) ) d V ≪ α − k 2 . I_2\ll\frac{\alpha^{-k^2}}{\sqrt{\log(1/\alpha)}}\int_{\mathbb{R}}\exp\left(-\frac{(V-k\log(1/\alpha))^2}{\log(1/\alpha)}\right)\,dV\ll\alpha^{-k^2}. I 2 ≪ log ( 1/ α ) α − k 2 ∫ R exp ( − log ( 1/ α ) ( V − k log ( 1/ α ) ) 2 ) d V ≪ α − k 2 . For the third integral, monotonicity and Theorem 1.7 give
P ( W α > V ) ≤ P ( W α > b log 1 α ) ≪ α b 2 log ( 1 / α ) \mathbb{P}(W_{\alpha}>V)\le\mathbb{P}\left(W_{\alpha}>b\log\frac{1}{\alpha}\right)\ll\frac{\alpha^{b^2}}{\sqrt{\log(1/\alpha)}} P ( W α > V ) ≤ P ( W α > b log α 1 ) ≪ log ( 1/ α ) α b 2 for V ≥ b log 1 α V\ge b\log\frac{1}{\alpha} V ≥ b log α 1
I 3 ≪ α − ( k − b 2 ) log ( 1 / α ) ≪ α − k 2 , I_3\ll\frac{\alpha^{-(k-b^2)}}{\sqrt{\log(1/\alpha)}}\ll\alpha^{-k^2}, I 3 ≪ log ( 1/ α ) α − ( k − b 2 ) ≪ α − k 2 , since k − b 2 < k 2 k-b^2<k^2 k − b 2 < k 2
For the fourth integral, we use Theorem 1.8 with ε 1 = ( 1 / 2 − k ) 2 / 2 \varepsilon_1=(1/2-k)^2/2 ε 1 = ( 1/2 − k ) 2 /2 κ 1 = 2 B \kappa_1=2B κ 1 = 2 B 2 k < 1 2k<1 2 k < 1
I 4 ≪ α − 1 / 4 − ε 1 ∫ 1 2 log ( 1 / α ) B log log T e − ( 1 − 2 k ) V d V ≪ α − ( k − 1 / 4 + ε 1 ) = α − k 2 + ( 1 / 2 − k ) 2 / 2 ≪ α − k 2 . I_4\ll\alpha^{-1/4-\varepsilon_1}\int_{\frac{1}{2}\log(1/\alpha)}^{B\log\log T}e^{-(1-2k)V}\,dV\ll\alpha^{-(k-1/4+\varepsilon_1)}=\alpha^{-k^2+(1/2-k)^2/2}\ll\alpha^{-k^2}. I 4 ≪ α − 1/4 − ε 1 ∫ 2 1 l o g ( 1/ α ) B l o g l o g T e − ( 1 − 2 k ) V d V ≪ α − ( k − 1/4 + ε 1 ) = α − k 2 + ( 1/2 − k ) 2 /2 ≪ α − k 2 . For the fifth integral, Lemma 7.1 and ρ > 2 k \rho> 2k ρ > 2 k
I 5 ≪ ∫ B log log T ∞ e − ( ρ − 2 k ) V d V ≪ e − ( ρ − 2 k ) B log log T = o ( 1 ) . (94) I_5 \ll\int_{B\log\log T}^{\infty} e^{-(\rho-2k)V}\,dV \ll e^{-(\rho-2k)B\log\log T}=o(1). \tag*{(94)} I 5 ≪ ∫ B l o g l o g T ∞ e − ( ρ − 2 k ) V d V ≪ e − ( ρ − 2 k ) B l o g l o g T = o ( 1 ) . (94) Combining all the five bounds finishes the proof for the first range.
In the second range (1.2), we instead split the integral into
I − k ( α , T ) = 2 k ( ∫ − ∞ B log log T + ∫ B log log T ∞ ) e 2 k V P ( W α > V ) d V . I_{-k}(\alpha,T)=2k\left(\int_{-\infty}^{B\log\log T}+\int_{B\log\log T}^{\infty}\right)e^{2kV}\mathbb{P}(W_\alpha>V)\,dV. I − k ( α , T ) = 2 k ( ∫ − ∞ B l o g l o g T + ∫ B l o g l o g T ∞ ) e 2 k V P ( W α > V ) d V . The first integral is bounded by e 2 k B log log T = ( log T ) O ( 1 ) e^{2kB\log\log T}=(\log T)^{O(1)} e 2 k B l o g l o g T = ( log T ) O ( 1 ) (94) holds. This finishes the proof in this case. □
We now prove Theorem 1.2.
Proof of Theorem 1.2. Let V 0 V_0 V 0 V ∗ V_* V ∗ (80) . Let α \alpha α (1) . We use Lemma 7.1 to obtain that
I − 1 / 2 ( α , T ) = ∫ − ∞ V 0 e V P ( W α > V ) d V + ∫ V 0 V ∗ e V P ( W α > V ) d V ≤ e V 0 + O ( ( log T ) exp ( − a V 0 log V 0 ) ) = e V 0 + o ( 1 ) . \begin{aligned}
I_{-1/2}(\alpha,T)=\int_{-\infty}^{V_0}e^V\mathbb{P}(W_\alpha>V)\,dV+\int_{V_0}^{V_*}e^V\mathbb{P}(W_\alpha>V)\,dV \\
&\le e^{V_0}+O\left((\log T)\exp\left(-a\frac{V_0}{\log V_0}\right)\right)=e^{V_0}+o(1).
\end{aligned} I − 1/2 ( α , T ) = ∫ − ∞ V 0 e V P ( W α > V ) d V + ∫ V 0 V ∗ e V P ( W α > V ) d V ≤ e V 0 + O ( ( log T ) exp ( − a log V 0 V 0 ) ) = e V 0 + o ( 1 ) . For α \alpha α
I − 1 / 2 ( α , T ) = ( ∫ − ∞ ( log T ) ε + ∫ ( log T ) ε V ∗ ) e V P ( W α > V ) d V , I_{-1/2}(\alpha,T)=\left(\int_{-\infty}^{(\log T)^\varepsilon}+\int_{(\log T)^\varepsilon}^{V_*}\right)e^V\mathbb{P}(W_\alpha>V)\,dV, I − 1/2 ( α , T ) = ( ∫ − ∞ ( l o g T ) ε + ∫ ( l o g T ) ε V ∗ ) e V P ( W α > V ) d V , where V ∗ V_* V ∗ (80) . The first integral is bounded by exp ( ( log T ) ε ) \exp((\log T)^\varepsilon) exp (( log T ) ε ) o ( 1 ) o(1) o ( 1 )
Finally, we prove Theorem 1.3.
Proof of Theorem 1.3. We note that this theorem is already proven in [14 ] in the range
α ≫ ( log log T ) 4 / k + ε ( log T ) 1 / ( 2 k ) . \alpha\gg\frac{(\log\log T)^{4/k+\varepsilon}}{(\log T)^{1/(2k)}}. α ≫ ( log T ) 1/ ( 2 k ) ( log log T ) 4/ k + ε . Hence, we assume that
C k ( log T ) 1 / ( 2 k ) log log T ≤ α ≪ ( log log T ) 4 / k + ε ( log T ) 1 / ( 2 k ) . (95) \frac{C_k}{(\log T)^{1/(2k)}\log\log T}\leq\alpha\ll\frac{(\log\log T)^{4/k+\varepsilon}}{(\log T)^{1/(2k)}}. \tag*{(95)} ( log T ) 1/ ( 2 k ) log log T C k ≤ α ≪ ( log T ) 1/ ( 2 k ) ( log log T ) 4/ k + ε . (95) Equation (93) still holds, and we write
I − k ( α , T ) = 2 k ( ∫ − ∞ ( k / 4 ) log 1 α + ∫ ( k / 4 ) log 1 α b log 1 α + ∫ b log 1 α B α log T + ∫ B α log T V ∗ ) e 2 k V P ( W α > V ) d V = I 1 + I 2 + I 3 + I 4 , \begin{aligned}
I_{-k}(\alpha,T)=2k\left(\int_{-\infty}^{(k/4)\log\frac{1}{\alpha}}+\int_{(k/4)\log\frac{1}{\alpha}}^{b\log\frac{1}{\alpha}}+\int_{b\log\frac{1}{\alpha}}^{B\alpha\log T}+\int_{B\alpha\log T}^{V_*}\right)e^{2kV}\mathbb{P}(W_\alpha>V)\,dV \\
&=I_1+I_2+I_3+I_4,
\end{aligned} I − k ( α , T ) = 2 k ( ∫ − ∞ ( k /4 ) l o g α 1 + ∫ ( k /4 ) l o g α 1 b l o g α 1 + ∫ b l o g α 1 B α l o g T + ∫ B α l o g T V ∗ ) e 2 k V P ( W α > V ) d V = I 1 + I 2 + I 3 + I 4 , say.
For I 1 I_1 I 1
I 1 ≪ ∫ − ∞ ( k / 4 ) log ( 1 / α ) e 2 k V d V ≪ α − k 2 / 2 . I_1\ll\int_{-\infty}^{(k/4)\log(1/\alpha)}e^{2kV}\,dV\ll\alpha^{-k^2/2}. I 1 ≪ ∫ − ∞ ( k /4 ) l o g ( 1/ α ) e 2 k V d V ≪ α − k 2 /2 . We use Theorem 1.7 for the second integral. Since from (95) it follows that
α log T ≫ ( log 1 α ) 12 + ε , \alpha\log T \gg\left(\log\frac{1}{\alpha}\right)^{12+\varepsilon}, α log T ≫ ( log α 1 ) 12 + ε , we then have that
I 2 ≪ 1 log ( 1 / α ) ∫ ( k / 4 ) log ( 1 / α ) b log ( 1 / α ) exp ( 2 k V − V 2 log ( 1 / α ) ) d V ≪ α − k 2 . I_2 \ll\frac{1}{\sqrt{\log(1/\alpha)}}\int_{(k/4)\log(1/\alpha)}^{b\log(1/\alpha)} \exp\left(2kV-\frac{V^2}{\log(1/\alpha)}\right)\,dV \ll\alpha^{-k^2}. I 2 ≪ log ( 1/ α ) 1 ∫ ( k /4 ) l o g ( 1/ α ) b l o g ( 1/ α ) exp ( 2 k V − log ( 1/ α ) V 2 ) d V ≪ α − k 2 . Finally, for the third and fourth integrals, we use Lemma 7.3 which gives that I 3 = o ( 1 ) I_3=o(1) I 3 = o ( 1 ) I 4 = o ( 1 ) I_4=o(1) I 4 = o ( 1 )
Partial sums of the generalized Möbius function
Our arguments work for general k > 0 k>0 k > 0 μ k ( n ) \mu_k(n) μ k ( n ) n th n^{\text{th}} n th ζ ( s ) − k \zeta(s)^{-k} ζ ( s ) − k
Z − k ( s ) = 1 s ( ( s − 1 ) ζ ( s ) ) k . Z_{-k}(s)=\frac{1}{s((s-1)\zeta(s))^k}. Z − k ( s ) = s (( s − 1 ) ζ ( s ) ) k 1 . The function Z − k ( s ) Z_{-k}(s) Z − k ( s ) ∣ s − 1 ∣ < 1 |s-1|<1 ∣ s − 1∣ < 1
Z − k ( s ) = ∑ j = 0 ∞ γ − k , j ( s − 1 ) j . Z_{-k}(s)=\sum_{j=0}^{\infty}\gamma_{-k,j}(s-1)^j. Z − k ( s ) = j = 0 ∑ ∞ γ − k , j ( s − 1 ) j . The following more general theorem holds.
Theorem 8.1. Assume RH. For every k > 0 k>0 k > 0 C = C ( k ) > 0 C=C(k)>0 C = C ( k ) > 0
∑ n ≤ x μ k ( n ) = − sin ( k π ) π x ( log x ) k + 1 ( ∑ j = 0 N ( − 1 ) j Γ ( k + j + 1 ) γ − k , j ( log x ) j + O ( ( N + 1 ) k + 1 / 2 ( N + 1 log x ) N + 1 ) ) + O ( x exp ( C ( log x log log x ) k / ( k + 1 ) ) ) , \sum_{n\le x}\mu_k(n)
=-\frac{\sin(k\pi)}{\pi}\frac{x}{(\log x)^{k+1}}\left(\sum_{j=0}^{N}\frac{(-1)^j\Gamma(k+j+1)\gamma_{-k,j}}{(\log x)^j}+O\left((N+1)^{k+1/2}\left(\frac{N+1}{\log x}\right)^{N+1}\right)\right)
+O\left(\sqrt{x}\exp\left(C\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right)\right), n ≤ x ∑ μ k ( n ) = − π sin ( k π ) ( log x ) k + 1 x ( j = 0 ∑ N ( log x ) j ( − 1 ) j Γ ( k + j + 1 ) γ − k , j + O ( ( N + 1 ) k + 1/2 ( log x N + 1 ) N + 1 ) ) + O ( x exp ( C ( log log x log x ) k / ( k + 1 ) ) ) , for any N ≥ 0 N\ge0 N ≥ 0
We remark that the main term above vanishes if k ∈ N k\in\mathbb{N} k ∈ N Theorem 1.4 .
We shall prove the case k ∈ N k\in\mathbb{N} k ∈ N k ∉ N k\notin\mathbb{N} k ∈ / N [14 Section 9.1] . As in [14 Section 9.1] , the main term in the theorem comes from the Hankel-type contour and only appears when k ∉ N k\notin\mathbb{N} k ∈ / N
The case k = 1 k=1 k = 1
From [40 (9)] we have
∑ n ≤ x μ ( n ) = 1 2 π i ∫ c − i ⌊ x ⌋ c + i ⌊ x ⌋ x s ζ ( s ) d s s + O ( log x ) \sum_{n\le x}\mu(n)=\frac{1}{2\pi i}\int_{c-i\lfloor x\rfloor}^{c+i\lfloor x\rfloor}\frac{x^s}{\zeta(s)}\frac{ds}{s}+O(\log x) n ≤ x ∑ μ ( n ) = 2 π i 1 ∫ c − i ⌊ x ⌋ c + i ⌊ x ⌋ ζ ( s ) x s s d s + O ( log x ) with c = 1 + 1 log x c=1+\frac{1}{\log x} c = 1 + l o g x 1
F 0 ( x ) = log x log log x F_0(x)=\log x\log\log x F 0 ( x ) = log x log log x and let
x 0 = exp ( log x log log x ) , α 0 = C F 0 ( x 0 ) , J = ⌊ log 2 ∣ x ∣ x 0 ⌋ ≪ log x , x_0=\exp\left(\sqrt{\frac{\log x}{\log\log x}}\right),\qquad\alpha_0=\frac{C}{F_0(x_0)},\qquad J=\left\lfloor\log_2\frac{|x|}{x_0}\right\rfloor\ll\log x, x 0 = exp ( log log x log x ) , α 0 = F 0 ( x 0 ) C , J = ⌊ log 2 x 0 ∣ x ∣ ⌋ ≪ log x , where C = C 1 ( 1 / 3 ) C=C_1(1/3) C = C 1 ( 1/3 ) Theorem 1.2 .
We now deform the contour as follows. We replace the line segment c + i t c+it c + i t ∣ t ∣ ≤ ⌊ x ⌋ |t|\le\lfloor x\rfloor ∣ t ∣ ≤ ⌊ x ⌋
⋃ j = 1 J ( V j ∪ H j ) ∪ V 0 . \bigcup_{j=1}^{J}(V_j\cup H_j)\cup V_0. j = 1 ⋃ J ( V j ∪ H j ) ∪ V 0 . where
V 0 : s = 1 2 + α 0 + i t , ∣ t ∣ ≤ x 0 , V j : s = 1 2 + C F 0 ( 2 j − 1 x 0 ) + i t , 2 j − 1 x 0 ≤ ∣ t ∣ ≤ min { 2 j x 0 , ⌊ x ⌋ } , 1 ≤ j ≤ J , H j : s = σ ± i 2 j x 0 , 1 2 + C F 0 ( 2 j x 0 ) ≤ σ ≤ 1 2 + C F 0 ( 2 j − 1 x 0 ) , 1 ≤ j ≤ J − 1 , H J : s = σ ± i ⌊ x ⌋ , 1 2 + C F 0 ( 2 J − 1 x 0 ) ≤ σ ≤ 1 + 1 log x . \begin{aligned}
V_0:\quad&s=\frac{1}{2}+\alpha_0+it,\qquad|t|\le x_0,\\
V_j:\quad&s=\frac{1}{2}+\frac{C}{F_0(2^{j-1}x_0)}+it,\qquad2^{j-1}x_0\le|t|\le\min\{2^jx_0,\lfloor x\rfloor\},\qquad1\le j\le J,\\
H_j:\quad&s=\sigma\pm i2^jx_0,\qquad\frac{1}{2}+\frac{C}{F_0(2^jx_0)}\le\sigma\le\frac{1}{2}+\frac{C}{F_0(2^{j-1}x_0)},\qquad1\le j\le J-1,\\
H_J:\quad&s=\sigma\pm i\lfloor x\rfloor,\qquad\frac{1}{2}+\frac{C}{F_0(2^{J-1}x_0)}\le\sigma\le1+\frac{1}{\log x}.
\end{aligned} V 0 : V j : H j : H J : s = 2 1 + α 0 + i t , ∣ t ∣ ≤ x 0 , s = 2 1 + F 0 ( 2 j − 1 x 0 ) C + i t , 2 j − 1 x 0 ≤ ∣ t ∣ ≤ min { 2 j x 0 , ⌊ x ⌋} , 1 ≤ j ≤ J , s = σ ± i 2 j x 0 , 2 1 + F 0 ( 2 j x 0 ) C ≤ σ ≤ 2 1 + F 0 ( 2 j − 1 x 0 ) C , 1 ≤ j ≤ J − 1 , s = σ ± i ⌊ x ⌋ , 2 1 + F 0 ( 2 J − 1 x 0 ) C ≤ σ ≤ 1 + log x 1 . We encounter no singularity in moving the contour. From [15 Theorem 2] we have
∣ ζ ( s ) ∣ − 1 ≪ ( ∣ t ∣ + 2 ) ( 1 / 2 + ε ) log ( 1 / α 0 ) log log ( ∣ t ∣ + 10 ) (96) |\zeta(s)|^{-1}\ll(|t|+2)^{(1/2+\varepsilon)\frac{\log(1/\alpha_0)}{\log\log(|t|+10)}} \tag*{(96)} ∣ ζ ( s ) ∣ − 1 ≪ ( ∣ t ∣ + 2 ) ( 1/2 + ε ) l o g l o g ( ∣ t ∣ + 10 ) l o g ( 1/ α 0 ) (96) for s ∈ V 0 s\in V_0 s ∈ V 0 V 0 V_0 V 0
≪ x exp ( α 0 log x + ( 1 2 + ε ) log x 0 log log x 0 log 1 α 0 ) ≪ x exp ( C 0 log x log log x ) , (97) \ll\sqrt{x}\exp\left(\alpha_0\log x+\left(\frac{1}{2}+\varepsilon\right)\frac{\log x_0}{\log\log x_0}\log\frac{1}{\alpha_0}\right)\ll\sqrt{x}\exp\left(C_0\sqrt{\frac{\log x}{\log\log x}}\right), \tag*{(97)} ≪ x exp ( α 0 log x + ( 2 1 + ε ) log log x 0 log x 0 log α 0 1 ) ≪ x exp ( C 0 log log x log x ) , (97) for some C 0 > 0 C_0>0 C 0 > 0
For the contribution from the vertical segments ⋃ j = 1 J V j \bigcup_{j=1}^{J}V_j ⋃ j = 1 J V j Theorem 1.2 1
≪ x ∑ j = 1 J exp ( log x F 0 ( 2 j − 1 x 0 ) + ( log ( 2 j x 0 ) ) 1 / 3 ) ≪ x J exp ( α 0 log x + ( log x ) 1 / 3 ) ≪ x exp ( C 0 log x log log x ) . \begin{aligned}
\ll\sqrt{x}\sum_{j=1}^{J}\exp\left(\frac{\log x}{F_0(2^{j-1}x_0)}+(\log(2^jx_0))^{1/3}\right)\ll\sqrt{x}J\exp\left(\alpha_0\log x+(\log x)^{1/3}\right) \\
&\ll\sqrt{x}\exp\left(C_0\sqrt{\frac{\log x}{\log\log x}}\right).
\end{aligned} ≪ x j = 1 ∑ J exp ( F 0 ( 2 j − 1 x 0 ) log x + ( log ( 2 j x 0 ) ) 1/3 ) ≪ x J exp ( α 0 log x + ( log x ) 1/3 ) ≪ x exp ( C 0 log log x log x ) . For s ∈ H j s\in H_j s ∈ H j 1 ≤ j ≤ J − 1 1\le j\le J-1 1 ≤ j ≤ J − 1
∣ ζ ( s ) ∣ − 1 ≪ ( 2 j x 0 ) 1 / 2 + ε . |\zeta(s)|^{-1}\ll(2^jx_0)^{1/2+\varepsilon}. ∣ ζ ( s ) ∣ − 1 ≪ ( 2 j x 0 ) 1/2 + ε . So the contribution from the horizontal segments ⋃ j = 1 J − 1 H j \bigcup_{j=1}^{J-1}H_j ⋃ j = 1 J − 1 H j
≪ x ∑ j = 1 J − 1 ( 2 j x 0 ) − 1 / 2 + ε exp ( log x F 0 ( 2 j − 1 x 0 ) ) ≪ x J x 0 − 1 / 2 + ε exp ( α 0 log x ) ≪ x exp ( C 0 log x log log x ) . \begin{aligned}
\ll\sqrt{x}\sum_{j=1}^{J-1}(2^jx_0)^{-1/2+\varepsilon}\exp\left(\frac{\log x}{F_0(2^{j-1}x_0)}\right)\ll\sqrt{x}Jx_0^{-1/2+\varepsilon}\exp(\alpha_0\log x) \\
&\ll\sqrt{x}\exp\left(C_0\sqrt{\frac{\log x}{\log\log x}}\right).
\end{aligned} ≪ x j = 1 ∑ J − 1 ( 2 j x 0 ) − 1/2 + ε exp ( F 0 ( 2 j − 1 x 0 ) log x ) ≪ x J x 0 − 1/2 + ε exp ( α 0 log x ) ≪ x exp ( C 0 log log x log x ) . We are left with the integral along H J H_J H J
max 1 2 + C F 0 ( x ) ≤ σ ≤ 1 + 1 log x exp ( ( σ − 1 ) log x − log ∣ ζ ( σ ± i ⌊ x ⌋ ) ∣ ) . \max_{\frac{1}{2}+\frac{C}{F_0(x)}\le\sigma\le1+\frac{1}{\log x}}\exp\left((\sigma-1)\log x-\log|\zeta(\sigma\pm i\lfloor x\rfloor)|\right). 2 1 + F 0 ( x ) C ≤ σ ≤ 1 + l o g x 1 max exp ( ( σ − 1 ) log x − log ∣ ζ ( σ ± i ⌊ x ⌋) ∣ ) . For 1 2 + C F 0 ( x ) ≤ σ ≤ 1 2 + O ( 1 log log x ) \frac{1}{2}+\frac{C}{F_0(x)}\leq\sigma\leq\frac{1}{2}+O\left(\frac{1}{\log\log x}\right) 2 1 + F 0 ( x ) C ≤ σ ≤ 2 1 + O ( l o g l o g x 1 ) [14 ] implies that this is
≪ exp ( ( σ − 1 ) log x + log x 2 log log x log 1 1 − ( log x ) 1 − 2 σ + O ( log x log log x ) + O ( log x ( log log x ) 2 log 1 1 − ( log x ) 1 − 2 σ ) ) ≪ exp ( ( σ − 1 ) log x + log x 2 log log x log 1 ( 2 σ − 1 ) log log x + O ( log x log log x ) + O ( log x ( log log x ) 2 log 1 2 σ − 1 ) ) , \begin{aligned}
\ll\exp\left((\sigma-1)\log x+\frac{\log x}{2\log\log x}\log\frac{1}{1-(\log x)^{1-2\sigma}}+O\left(\frac{\log x}{\log\log x}\right)+O\left(\frac{\log x}{(\log\log x)^2}\log\frac{1}{1-(\log x)^{1-2\sigma}}\right)\right) \\
&\ll\exp\left((\sigma-1)\log x+\frac{\log x}{2\log\log x}\log\frac{1}{(2\sigma-1)\log\log x}+O\left(\frac{\log x}{\log\log x}\right)+O\left(\frac{\log x}{(\log\log x)^2}\log\frac{1}{2\sigma-1}\right)\right),
\end{aligned} ≪ exp ( ( σ − 1 ) log x + 2 log log x log x log 1 − ( log x ) 1 − 2 σ 1 + O ( log log x log x ) + O ( ( log log x ) 2 log x log 1 − ( log x ) 1 − 2 σ 1 ) ) ≪ exp ( ( σ − 1 ) log x + 2 log log x log x log ( 2 σ − 1 ) log log x 1 + O ( log log x log x ) + O ( ( log log x ) 2 log x log 2 σ − 1 1 ) ) , which is decreasing with respect to σ \sigma σ
≪ exp ( − log x 2 + C log x F 0 ( x ) + log x 2 log log x log F 0 ( x ) 2 log log x + O ( log x log log x ) ) ≪ exp ( O ( log x log log x ) ) ≪ x ε . \begin{aligned}
\ll\exp\left(-\frac{\log x}{2}+\frac{C\log x}{F_0(x)}+\frac{\log x}{2\log\log x}\log\frac{F_0(x)}{2\log\log x}+O\left(\frac{\log x}{\log\log x}\right)\right) \\
&\ll\exp\left(O\left(\frac{\log x}{\log\log x}\right)\right)\ll x^\varepsilon.
\end{aligned} ≪ exp ( − 2 log x + F 0 ( x ) C log x + 2 log log x log x log 2 log log x F 0 ( x ) + O ( log log x log x ) ) ≪ exp ( O ( log log x log x ) ) ≪ x ε . For 1 − 1 log log x ≤ σ ≤ 1 + 1 log x 1-\frac{1}{\log\log x}\leq\sigma\leq1+\frac{1}{\log x} 1 − l o g l o g x 1 ≤ σ ≤ 1 + l o g x 1 [15 ] leads to a bound of size
≪ exp ( ( σ − 1 ) log x + log log log x + O ( 1 ) ) ≪ log log x . \ll\exp\left((\sigma-1)\log x+\log\log\log x+O(1)\right)\ll\log\log x. ≪ exp ( ( σ − 1 ) log x + log log log x + O ( 1 ) ) ≪ log log x . Finally, for 1 2 + O ( 1 log log x ) ≤ σ ≤ 1 − 1 log log x \frac{1}{2}+O\left(\frac{1}{\log\log x}\right)\leq\sigma\leq1-\frac{1}{\log\log x} 2 1 + O ( l o g l o g x 1 ) ≤ σ ≤ 1 − l o g l o g x 1 [15 ] to get the bound
≪ exp ( ( σ − 1 ) log x + ε log x + ( log x ) 2 − 2 σ ( 1 − σ ) log log x + O ( ( log x ) 2 − 2 σ ( 1 − σ ) 2 ( log log x ) 2 ) ) ≪ x ε . (98) \ll\exp\left((\sigma-1)\log x+\varepsilon\log x+\frac{(\log x)^{2-2\sigma}}{(1-\sigma)\log\log x}+O\left(\frac{(\log x)^{2-2\sigma}}{(1-\sigma)^2(\log\log x)^2}\right)\right)\ll x^\varepsilon. \tag*{(98)} ≪ exp ( ( σ − 1 ) log x + ε log x + ( 1 − σ ) log log x ( log x ) 2 − 2 σ + O ( ( 1 − σ ) 2 ( log log x ) 2 ( log x ) 2 − 2 σ ) ) ≪ x ε . (98) Combining the estimates (97) –(98) we obtain Theorem 1.4 in the case k = 1 k=1 k = 1
The case k ∈ N k\in\mathbb{N} k ∈ N k > 1 k>1 k > 1
The proof follows closely that of the case k = 1 k=1 k = 1
From [14 (9.1)] we have
∑ n ≤ x μ k ( n ) = 1 2 π i ∫ c − i ⌊ x ⌋ c + i ⌊ x ⌋ x s ζ ( s ) k d s s + O ( ( log x ) k ) \sum_{n\leq x}\mu_k(n)=\frac{1}{2\pi i}\int_{c-i\lfloor x\rfloor}^{c+i\lfloor x\rfloor}\frac{x^s}{\zeta(s)^k}\frac{ds}{s}+O((\log x)^k) n ≤ x ∑ μ k ( n ) = 2 π i 1 ∫ c − i ⌊ x ⌋ c + i ⌊ x ⌋ ζ ( s ) k x s s d s + O (( log x ) k ) with c = 1 + 1 log x c=1+\frac{1}{\log x} c = 1 + l o g x 1
F 1 ( x ) = ( log x ) 1 / k log log x F_1(x)=(\log x)^{1/k}\log\log x F 1 ( x ) = ( log x ) 1/ k log log x and let
x 1 = exp ( ( log x log log x ) k / ( k + 1 ) ) , α 1 = C F 1 ( x 1 ) , J = ⌈ log 2 ⌊ x ⌋ x 1 ⌉ ≪ log x , x_1=\exp\left(\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right),\qquad\alpha_1=\frac{C}{F_1(x_1)},\qquad J=\left\lceil\log_2\frac{\lfloor x\rfloor}{x_1}\right\rceil\ll\log x, x 1 = exp ( ( log log x log x ) k / ( k + 1 ) ) , α 1 = F 1 ( x 1 ) C , J = ⌈ log 2 x 1 ⌊ x ⌋ ⌉ ≪ log x , with C = C ( k / 2 ) C=C(k/2) C = C ( k /2 ) c + i t c+it c + i t ∣ t ∣ ≤ ⌊ x ⌋ |t|\leq\lfloor x\rfloor ∣ t ∣ ≤ ⌊ x ⌋
⋃ j = 1 J ( V j ′ ∪ H j ′ ) ∪ V 0 ′ , \bigcup_{j=1}^{J}(V'_j\cup H'_j)\cup V'_0, j = 1 ⋃ J ( V j ′ ∪ H j ′ ) ∪ V 0 ′ , where
V 0 ′ : s = 1 2 + α 1 + i t , ∣ t ∣ ≤ x 1 , V'_0:\qquad s=\frac{1}{2}+\alpha_1+it,\qquad|t|\leq x_1, V 0 ′ : s = 2 1 + α 1 + i t , ∣ t ∣ ≤ x 1 , V j ′ : s = 1 2 + C F 1 ( 2 j − 1 x 1 ) + i t , 2 j − 1 x 1 ≤ ∣ t ∣ ≤ min { 2 j x 1 , ⌊ x ⌋ } , 1 ≤ j ≤ J , V'_j:\qquad s=\frac{1}{2}+\frac{C}{F_1(2^{j-1}x_1)}+it,\qquad2^{j-1}x_1\leq|t|\leq\min\{2^jx_1,\lfloor x\rfloor\},\qquad1\leq j\leq J, V j ′ : s = 2 1 + F 1 ( 2 j − 1 x 1 ) C + i t , 2 j − 1 x 1 ≤ ∣ t ∣ ≤ min { 2 j x 1 , ⌊ x ⌋} , 1 ≤ j ≤ J , H j ′ : s = σ ± i 2 j x 1 , 1 2 + C F 1 ( 2 j x 1 ) ≤ σ ≤ 1 2 + C F 1 ( 2 j − 1 x 1 ) , 1 ≤ j ≤ J − 1 , H'_j:\qquad s=\sigma\pm i2^j x_1,\qquad\frac{1}{2}+\frac{C}{F_1(2^j x_1)}\leq\sigma\leq\frac{1}{2}+\frac{C}{F_1(2^{j-1}x_1)},\qquad1\leq j\leq J-1, H j ′ : s = σ ± i 2 j x 1 , 2 1 + F 1 ( 2 j x 1 ) C ≤ σ ≤ 2 1 + F 1 ( 2 j − 1 x 1 ) C , 1 ≤ j ≤ J − 1 , H J ′ : s = σ ± i [ x ] , 1 2 + C F 1 ( 2 J − 1 x 1 ) ≤ σ ≤ 1 + 1 log x . H'_J:\qquad s=\sigma\pm i[x],\qquad\frac{1}{2}+\frac{C}{F_1(2^{J-1}x_1)}\leq\sigma\leq1+\frac{1}{\log x}. H J ′ : s = σ ± i [ x ] , 2 1 + F 1 ( 2 J − 1 x 1 ) C ≤ σ ≤ 1 + log x 1 . We cross no singularity in moving the contour.
For the integral along V 0 ′ V'_0 V 0 ′
1 2 π i ∫ 1 / 2 + α 1 − i x 1 1 / 2 + α 1 + i x 1 x s ζ ( s ) k d s s ≪ x exp ( α 1 log x + ( k 2 + ε ) log x 1 log log x 1 log 1 α 1 ) ≪ x exp ( C 0 ( log x log log x ) k / ( k + 1 ) ) , (99) \frac{1}{2\pi i}\int_{1/2+\alpha_1-i x_1}^{1/2+\alpha_1+i x_1}\frac{x^s}{\zeta(s)^k}\frac{ds}{s}\ll\sqrt{x}\exp\left(\alpha_1\log x+\left(\frac{k}{2}+\varepsilon\right)\frac{\log x_1}{\log\log x_1}\log\frac{1}{\alpha_1}\right)
\ll\sqrt{x}\exp\left(C_0\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right), \tag*{(99)} 2 π i 1 ∫ 1/2 + α 1 − i x 1 1/2 + α 1 + i x 1 ζ ( s ) k x s s d s ≪ x exp ( α 1 log x + ( 2 k + ε ) log log x 1 log x 1 log α 1 1 ) ≪ x exp ( C 0 ( log log x log x ) k / ( k + 1 ) ) , (99) for some C 0 > 0 C_0>0 C 0 > 0
For the contribution from the vertical segments ⋃ j = 1 J V j ′ \bigcup_{j=1}^J V'_j ⋃ j = 1 J V j ′
≪ x ∑ j = 1 J exp ( C log x F 1 ( 2 j − 1 x 1 ) + O ( log log ( 2 j x 1 ) ) ) ≪ x J exp ( α 1 log x + O ( log log x ) ) ≪ x exp ( C 0 ( log x log log x ) k / ( k + 1 ) ) . (100) \begin{aligned}
\ll\sqrt{x}\sum_{j=1}^J\exp\left(\frac{C\log x}{F_1(2^{j-1}x_1)}+O\left(\log\log(2^j x_1)\right)\right)\ll\sqrt{x}J\exp\left(\alpha_1\log x+O(\log\log x)\right) \\
&\ll\sqrt{x}\exp\left(C_0\left(\frac{\log x}{\log\log x}\right)^{k/(k+1)}\right). \tag*{(100)}
\end{aligned} ≪ x j = 1 ∑ J exp ( F 1 ( 2 j − 1 x 1 ) C log x + O ( log log ( 2 j x 1 ) ) ) ≪ x J exp ( α 1 log x + O ( log log x ) ) ≪ x exp ( C 0 ( log log x log x ) k / ( k + 1 ) ) . (100) The same bound holds for ⋃ j = 1 J − 1 H j ′ \bigcup_{j=1}^{J-1}H'_j ⋃ j = 1 J − 1 H j ′ H J ′ H'_J H J ′ O ( x ε ) O(x^\varepsilon) O ( x ε ) k ∈ N k\in\mathbb{N} k ∈ N k > 1 k>1 k > 1
The minimum of ζ ( s ) \zeta(s) ζ ( s )
Recall that
M α , γ ( t ) : = min ∣ h ∣ ≤ ( log T ) γ ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ . M_{\alpha,\gamma}(t):=\min_{|h|\leq(\log T)^\gamma}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|. M α , γ ( t ) := ∣ h ∣ ≤ ( l o g T ) γ min ζ ( 2 1 + α + i ( t + h ) ) . The log-correlated prime-sum model for log ∣ ζ ∣ − 1 \log|\zeta|^{-1} log ∣ ζ ∣ − 1 [11 ] .
For γ ≥ 0 \gamma\geq0 γ ≥ 0
k γ : = 1 + γ log log T log ( 1 / α ) . (101) k_\gamma:=\sqrt{1+\frac{\gamma\log\log T}{\log(1/\alpha)}}. \tag*{(101)} k γ := 1 + log ( 1/ α ) γ log log T . (101) Conjecture 9.1. Suppose that
α = o ( 1 ) , α log T ≫ 1. \alpha=o(1),\qquad\alpha\log T\gg1. α = o ( 1 ) , α log T ≫ 1. Fix γ ≥ 0 \gamma\geq0 γ ≥ 0 γ > 0 \gamma>0 γ > 0 α ≤ ( log T ) − c \alpha\leq(\log T)^{-c} α ≤ ( log T ) − c 0 < c < 1 0<c<1 0 < c < 1 ε > 0 \varepsilon>0 ε > 0 C > 0 C>0 C > 0 T T T
(1) If γ = 0 \gamma=0 γ = 0
P ( ∣ log M α , 0 ( t ) + log 1 α − 3 4 log log 1 α ∣ ≤ C ) ≥ 1 − ε . \mathbb{P}\left(\left|\log M_{\alpha,0}(t)+\log\frac{1}{\alpha}-\frac{3}{4}\log\log\frac{1}{\alpha}\right|\leq C\right)\geq1-\varepsilon. P ( log M α , 0 ( t ) + log α 1 − 4 3 log log α 1 ≤ C ) ≥ 1 − ε . (2) If γ > 0 \gamma>0 γ > 0
P ( ∣ log M α , γ ( t ) + k γ log 1 α − 1 4 k γ log log 1 α ∣ ≤ C ) ≥ 1 − ε . \mathbb{P}\left(\left|\log M_{\alpha,\gamma}(t)+k_\gamma\log\frac{1}{\alpha}-\frac{1}{4k_\gamma}\log\log\frac{1}{\alpha}\right|\leq C\right)\geq1-\varepsilon. P ( log M α , γ ( t ) + k γ log α 1 − 4 k γ 1 log log α 1 ≤ C ) ≥ 1 − ε . Lower bounds for the minimum
We will prove the following analogue to Corollary 3.1 in [3 ] .
Theorem 9.2. Assume RH. Fix 0 < c < 1 0<c<1 0 < c < 1 γ ≥ 0 \gamma\ge0 γ ≥ 0 k γ k_\gamma k γ (101) . Suppose that
0 < α ≪ ( log T ) − c , α log T log log T log log log T → ∞ . (102) 0<\alpha\ll(\log T)^{-c},\qquad\frac{\alpha\log T}{\log\log T\log\log\log T}\to\infty. \tag*{(102)} 0 < α ≪ ( log T ) − c , log log T log log log T α log T → ∞. (102) There exists C > 0 C>0 C > 0 T T T
0 ≤ y ≪ log 1 α , 0\le y\ll\log\frac{1}{\alpha}, 0 ≤ y ≪ log α 1 , we have
P ( M α , γ ( t ) ≤ e − y α k γ ( log 1 α ) − C ) ≤ C exp ( − 2 k γ y − y 2 log ( 1 / α ) ) . (103) \mathbb{P}\left(M_{\alpha,\gamma}(t)\le e^{-y\alpha^{k_\gamma}}\left(\log\frac{1}{\alpha}\right)^{-C}\right)\le C\exp\left(-2k_\gamma y-\frac{y^2}{\log(1/\alpha)}\right). \tag*{(103)} P ( M α , γ ( t ) ≤ e − y α k γ ( log α 1 ) − C ) ≤ C exp ( − 2 k γ y − log ( 1/ α ) y 2 ) . (103) Remark 9.3. Note that the lower threshold in the theorem above is
e − y α k γ ( log 1 α ) − C = e − y α k γ + o ( 1 ) . e^{-y\alpha^{k_\gamma}}\left(\log\frac{1}{\alpha}\right)^{-C}=e^{-y\alpha^{k_\gamma+o(1)}}. e − y α k γ ( log α 1 ) − C = e − y α k γ + o ( 1 ) . Also note that when γ = 0 \gamma=0 γ = 0
M α , 0 ( t ) ≥ e − y α 1 + o ( 1 ) M_{\alpha,0}(t)\ge e^{-y\alpha^{1+o(1)}} M α , 0 ( t ) ≥ e − y α 1 + o ( 1 ) outside a set of measure
≪ T exp ( − 2 y − y 2 log ( 1 / α ) ) . \ll T\exp\left(-2y-\frac{y^2}{\log(1/\alpha)}\right). ≪ T exp ( − 2 y − log ( 1/ α ) y 2 ) . Before beginning the proof of Theorem 9.2, we first need the following lemma.
Lemma 9.4. Let 1 ≤ H ≤ T / 10 1\le H\le T/10 1 ≤ H ≤ T /10
P ( t ) = ∑ p ≤ x a p p − i t . P(t)=\sum_{p\le x}a_p p^{-it}. P ( t ) = p ≤ x ∑ a p p − i t . For any q ∈ N q\in\mathbb{N} q ∈ N 2 q ≤ log T / log x 2q\le\log T/\log x 2 q ≤ log T / log x
∫ T 2 T max ∣ h ∣ ≤ H ∣ P ( t + h ) ∣ 2 q d t ≪ T ( 1 + q H log x ) q ! ( ∑ p ≤ x ∣ a p ∣ 2 ) q . (104) \int_T^{2T}\max_{|h|\le H}|P(t+h)|^{2q}\,\mathrm{d}t\ll T(1+qH\log x)q!\left(\sum_{p\le x}|a_p|^2\right)^q. \tag*{(104)} ∫ T 2 T ∣ h ∣ ≤ H max ∣ P ( t + h ) ∣ 2 q d t ≪ T ( 1 + q H log x ) q ! ( p ≤ x ∑ ∣ a p ∣ 2 ) q . (104) The same estimate holds for a polynomial of the form ∑ p 2 ≤ x a p p − 2 i t \sum_{p^2\le x}a_p p^{-2it} ∑ p 2 ≤ x a p p − 2 i t
Proof. We shall prove (104) . The treatment of the other case is similar.
We use the Sobolev inequality [5 (6)] to get that for fixed t t t
max ∣ h ∣ ≤ H ∣ P ( t + h ) ∣ 2 q ≤ ∣ P ( t − H ) ∣ 2 q + ∣ P ( t + H ) ∣ 2 q 2 + q ∫ − H H ∣ P ( t + v ) ∣ 2 q − 1 ∣ P ′ ( t + v ) ∣ d v . \max_{|h|\le H}|P(t+h)|^{2q}\le\frac{|P(t-H)|^{2q}+|P(t+H)|^{2q}}{2}+q\int_{-H}^{H}|P(t+v)|^{2q-1}|P'(t+v)|\,\mathrm{d}v. ∣ h ∣ ≤ H max ∣ P ( t + h ) ∣ 2 q ≤ 2 ∣ P ( t − H ) ∣ 2 q + ∣ P ( t + H ) ∣ 2 q + q ∫ − H H ∣ P ( t + v ) ∣ 2 q − 1 ∣ P ′ ( t + v ) ∣ d v . Integrating over T ≤ t ≤ 2 T T\le t\le2T T ≤ t ≤ 2 T
∫ T 2 T max ∣ h ∣ ≤ H ∣ P ( t + h ) ∣ 2 q d t ≤ ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v + 2 q H ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q − 1 ∣ P ′ ( v ) ∣ d v ≤ ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v + 2 q H ( ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v ) 1 − 1 / ( 2 q ) ( ∫ T − H 2 T + H ∣ P ′ ( v ) ∣ 2 q d v ) 1 / ( 2 q ) . (105) \begin{aligned}
\int_T^{2T}\max_{|h|\le H}|P(t+h)|^{2q}\,\mathrm{d}t
&\le\int_{T-H}^{2T+H}|P(v)|^{2q}\,\mathrm{d}v
+2qH\int_{T-H}^{2T+H}|P(v)|^{2q-1}|P'(v)|\,\mathrm{d}v\\
&\le\int_{T-H}^{2T+H}|P(v)|^{2q}\,\mathrm{d}v
+2qH\left(\int_{T-H}^{2T+H}|P(v)|^{2q}\,\mathrm{d}v\right)^{1-1/(2q)}
\left(\int_{T-H}^{2T+H}|P'(v)|^{2q}\,\mathrm{d}v\right)^{1/(2q)}. \tag*{(105)}
\end{aligned} ∫ T 2 T ∣ h ∣ ≤ H max ∣ P ( t + h ) ∣ 2 q d t ≤ ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v + 2 q H ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q − 1 ∣ P ′ ( v ) ∣ d v ≤ ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v + 2 q H ( ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v ) 1 − 1/ ( 2 q ) ( ∫ T − H 2 T + H ∣ P ′ ( v ) ∣ 2 q d v ) 1/ ( 2 q ) . (105) Now we bound these two moments. By [39 Lemma 3] ,
∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v ≪ T q ! ( ∑ p ≤ x ∣ a p ∣ 2 ) q . \int_{T-H}^{2T+H}|P(v)|^{2q}\,\mathrm{d}v\ll Tq!\left(\sum_{p\le x}|a_p|^2\right)^q. ∫ T − H 2 T + H ∣ P ( v ) ∣ 2 q d v ≪ T q ! ( p ≤ x ∑ ∣ a p ∣ 2 ) q . Since
P ′ ( v ) = − i ∑ p ≤ x a p ( log p ) p − i v , P'(v)=-i\sum_{p\le x}a_p(\log p)p^{-iv}, P ′ ( v ) = − i p ≤ x ∑ a p ( log p ) p − i v , the same estimate applied to P ′ P' P ′
∫ T − H 2 T + H ∣ P ′ ( v ) ∣ 2 q d v ≪ T q ! ( ∑ p ≤ x ∣ a p ∣ 2 ( log p ) 2 ) q ≤ T q ! ( log x ) 2 q ( ∑ p ≤ x ∣ a p ∣ 2 ) q . \int_{T-H}^{2T+H}|P'(v)|^{2q}\,dv\ll Tq!\left(\sum_{p\le x}|a_p|^2(\log p)^2\right)^q\le Tq!(\log x)^{2q}\left(\sum_{p\le x}|a_p|^2\right)^q. ∫ T − H 2 T + H ∣ P ′ ( v ) ∣ 2 q d v ≪ T q ! ( p ≤ x ∑ ∣ a p ∣ 2 ( log p ) 2 ) q ≤ T q ! ( log x ) 2 q ( p ≤ x ∑ ∣ a p ∣ 2 ) q . The two bounds above and (105) yield (104) . □
Proof of Theorem 9.2 . Set
u = log ( α log T ) , x = exp ( u α ) . (106) u=\log(\alpha\log T),\qquad x=\exp\left(\frac{u}{\alpha}\right). \tag*{(106)} u = log ( α log T ) , x = exp ( α u ) . (106) By (102) we have
u → ∞ , u ≤ ( 1 − c ) log log T + O ( 1 ) , c log log T + O ( 1 ) ≤ log 1 α ≤ log log T . u\to\infty,\qquad u\le(1-c)\log\log T+O(1),\qquad c\log\log T+O(1)\le\log\frac{1}{\alpha}\le\log\log T. u → ∞ , u ≤ ( 1 − c ) log log T + O ( 1 ) , c log log T + O ( 1 ) ≤ log α 1 ≤ log log T . We also have that
log x = u α = o ( log T ) , \log x=\frac{u}{\alpha}=o(\log T), log x = α u = o ( log T ) , and hence x = T o ( 1 ) x=T^{o(1)} x = T o ( 1 )
Now Lemma 2.1 gives
W α ( t ) ≤ ℜ ( ∑ n ≤ x b α ( n ; x ) n 1 / 2 + α + i t ) + log t 2 π log x log 1 1 − x − α + O ( 1 ) . (107) W_\alpha(t)\le\Re\left(\sum_{n\le x}\frac{b_\alpha(n;x)}{n^{1/2+\alpha+it}}\right)+\frac{\log\frac{t}{2\pi}}{\log x}\log\frac{1}{1-x^{-\alpha}}+O(1). \tag*{(107)} W α ( t ) ≤ ℜ ( n ≤ x ∑ n 1/2 + α + i t b α ( n ; x ) ) + log x log 2 π t log 1 − x − α 1 + O ( 1 ) . (107) Since x − α = 1 α log T x^{-\alpha}=\frac{1}{\alpha\log T} x − α = α l o g T 1
log t 2 π log x log 1 1 − x − α ≪ α log T u ⋅ 1 α log T ≪ 1 u . \frac{\log\frac{t}{2\pi}}{\log x}\log\frac{1}{1-x^{-\alpha}}\ll\frac{\alpha\log T}{u}\cdot\frac{1}{\alpha\log T}\ll\frac{1}{u}. log x log 2 π t log 1 − x − α 1 ≪ u α log T ⋅ α log T 1 ≪ u 1 . We also separate the prime and prime-square contributions by setting
P 1 ( t ) = ∑ p ≤ x b α ( p ; x ) p 1 / 2 + α + i t and P 2 ( t ) = ∑ p 2 ≤ x b α ( p 2 ; x ) p 1 + 2 α + 2 i t . P_1(t)=\sum_{p\le x}\frac{b_\alpha(p;x)}{p^{1/2+\alpha+it}}
\qquad\text{and}\qquad
P_2(t)=\sum_{p^2\le x}\frac{b_\alpha(p^2;x)}{p^{1+2\alpha+2it}}. P 1 ( t ) = p ≤ x ∑ p 1/2 + α + i t b α ( p ; x ) and P 2 ( t ) = p 2 ≤ x ∑ p 1 + 2 α + 2 i t b α ( p 2 ; x ) . The terms supported on p r p^r p r r ≥ 3 r\ge3 r ≥ 3 O ( 1 ) O(1) O ( 1 ) t t t
W α ( t ) ≤ ℜ P 1 ( t ) + ℜ P 2 ( t ) + C 0 W_\alpha(t)\le\Re P_1(t)+\Re P_2(t)+C_0 W α ( t ) ≤ ℜ P 1 ( t ) + ℜ P 2 ( t ) + C 0 for some absolute constant C 0 > 0 C_0>0 C 0 > 0
Since α log x → ∞ \alpha\log x\to\infty α log x → ∞ x = T o ( 1 ) x=T^{o(1)} x = T o ( 1 ) Lemma 2.3 give
B 1 : = ∑ p ≤ x ∣ b α ( p ; x ) ∣ 2 p 1 + 2 α = log 1 α + O ( 1 ) . (108) B_1:=\sum_{p\le x}\frac{|b_\alpha(p;x)|^2}{p^{1+2\alpha}}=\log\frac{1}{\alpha}+O(1). \tag*{(108)} B 1 := p ≤ x ∑ p 1 + 2 α ∣ b α ( p ; x ) ∣ 2 = log α 1 + O ( 1 ) . (108) We now take H = ( log T ) γ H=(\log T)^\gamma H = ( log T ) γ
Δ = C 1 log log 1 α , \Delta=C_1\log\log\frac{1}{\alpha}, Δ = C 1 log log α 1 , for C 1 C_1 C 1
U = k γ log 1 α + 2 Δ + y . U=k_\gamma\log\frac{1}{\alpha}+2\Delta+y. U = k γ log α 1 + 2Δ + y . By Lemma 9.4 and Markov’s inequality, for every q ∈ N q\in\mathbb{N} q ∈ N 2 q ≤ log T / log x 2q\le\log T/\log x 2 q ≤ log T / log x
P ( max ∣ h ∣ ≤ H ∣ P 1 ( t + h ) ∣ > U ) ≪ ( 1 + q H log x ) q q ! B 1 q U 2 q . (109) \mathbb{P}\left(\max_{|h|\le H}|P_1(t+h)|>U\right)\ll(1+qH\log x)^q\frac{q!B_1^q}{U^{2q}}. \tag*{(109)} P ( ∣ h ∣ ≤ H max ∣ P 1 ( t + h ) ∣ > U ) ≪ ( 1 + q H log x ) q U 2 q q ! B 1 q . (109) We choose
q = ⌊ U 2 B 1 ⌋ . q=\left\lfloor\frac{U^2}{B_1}\right\rfloor. q = ⌊ B 1 U 2 ⌋ . By Stirling’s formula,
q ! B 1 q U 2 q ≪ q exp ( − U 2 B 1 ) . (110) \frac{q!B_1^q}{U^{2q}}\ll\sqrt{q}\exp\left(-\frac{U^2}{B_1}\right). \tag*{(110)} U 2 q q ! B 1 q ≪ q exp ( − B 1 U 2 ) . (110) We also have that
log ( 1 + q H log x ) = k γ 2 log 1 α + O ( log log 1 α ) , (111) \log(1+qH\log x)=k_\gamma^2\log\frac{1}{\alpha}+O\left(\log\log\frac{1}{\alpha}\right), \tag*{(111)} log ( 1 + q H log x ) = k γ 2 log α 1 + O ( log log α 1 ) , (111) where we used u ≪ log 1 α u\ll\log\frac{1}{\alpha} u ≪ log α 1 q ≍ log 1 α q\asymp\log\frac{1}{\alpha} q ≍ log α 1 B 1 = log 1 α + O ( 1 ) B_1=\log\frac{1}{\alpha}+O(1) B 1 = log α 1 + O ( 1 )
U 2 B 1 ≥ k γ 2 log 1 α + 2 k γ y + y 2 log ( 1 / α ) + 2 Δ . (112) \frac{U^2}{B_1}\geq k_\gamma^2\log\frac{1}{\alpha}+2k_\gamma y+\frac{y^2}{\log(1/\alpha)}+2\Delta. \tag*{(112)} B 1 U 2 ≥ k γ 2 log α 1 + 2 k γ y + log ( 1/ α ) y 2 + 2Δ. (112) Combining (109) –(112) , and using Δ ≫ log log 1 α \Delta\gg\log\log\frac{1}{\alpha} Δ ≫ log log α 1
P ( max ∣ h ∣ ≤ H ∣ P 1 ( t + h ) ∣ > U ) ≪ exp ( − 2 k γ y − y 2 log ( 1 / α ) − 2 Δ + O ( log log 1 α ) ) . (113) \mathbb{P}\left(\max_{|h|\leq H}|P_1(t+h)|>U\right)\ll\exp\left(-2k_\gamma y-\frac{y^2}{\log(1/\alpha)}-2\Delta+O\left(\log\log\frac{1}{\alpha}\right)\right). \tag*{(113)} P ( ∣ h ∣ ≤ H max ∣ P 1 ( t + h ) ∣ > U ) ≪ exp ( − 2 k γ y − log ( 1/ α ) y 2 − 2Δ + O ( log log α 1 ) ) . (113) Now using (5) we have that
max ∣ h ∣ ≤ H ℜ P 2 ( t + h ) ≤ log log log T 2 + O ( 1 ) ≤ Δ . (114) \max_{|h|\leq H}\Re P_2(t+h)\leq\frac{\log\log\log T}{2}+O(1)\leq\Delta. \tag*{(114)} ∣ h ∣ ≤ H max ℜ P 2 ( t + h ) ≤ 2 log log log T + O ( 1 ) ≤ Δ. (114) Let C ≥ 4 C\geq4 C ≥ 4
E y = { t ∈ [ T , 2 T ] : max ∣ h ∣ ≤ H W α ( t + h ) ≥ k γ log 1 α + C Δ + y } . (115) E_y=\left\{t\in[T,2T]:\max_{|h|\leq H}W_\alpha(t+h)\geq k_\gamma\log\frac{1}{\alpha}+C\Delta+y\right\}. \tag*{(115)} E y = { t ∈ [ T , 2 T ] : ∣ h ∣ ≤ H max W α ( t + h ) ≥ k γ log α 1 + C Δ + y } . (115) If t ∈ E y t\in E_y t ∈ E y ∣ h ∣ ≤ H |h|\leq H ∣ h ∣ ≤ H
ℜ P 1 ( t + h ) + ℜ P 2 ( t + h ) > k γ log 1 α + ( C − 1 ) Δ + y . \Re P_1(t+h)+\Re P_2(t+h)>k_\gamma\log\frac{1}{\alpha}+(C-1)\Delta+y. ℜ P 1 ( t + h ) + ℜ P 2 ( t + h ) > k γ log α 1 + ( C − 1 ) Δ + y . Thus, by (114) , we obtain that
∣ P 1 ( t + h ) ∣ > k γ log 1 α + 2 Δ + y . |P_1(t+h)|>k_\gamma\log\frac{1}{\alpha}+2\Delta+y. ∣ P 1 ( t + h ) ∣ > k γ log α 1 + 2Δ + y . We deduce from (113) that
meas ( E y ) ≪ T exp ( − 2 k γ y − y 2 log ( 1 / α ) − Δ ) ≪ T exp ( − 2 k γ y − y 2 log ( 1 / α ) ) . (116) \operatorname{meas}(E_y)\ll T\exp\left(-2k_\gamma y-\frac{y^2}{\log(1/\alpha)}-\Delta\right)\ll T\exp\left(-2k_\gamma y-\frac{y^2}{\log(1/\alpha)}\right). \tag*{(116)} meas ( E y ) ≪ T exp ( − 2 k γ y − log ( 1/ α ) y 2 − Δ ) ≪ T exp ( − 2 k γ y − log ( 1/ α ) y 2 ) . (116) This completes the proof of the theorem. □
We show below that the range (102) in Theorem 9.2 can be slightly enlarged at the expense of a weaker error term. We obtain the following leading order bound with probability 1 − o ( 1 ) 1-o(1) 1 − o ( 1 ) y y y
Theorem 9.5. Assume RH. Fix 0 < c < 1 0<c<1 0 < c < 1 γ ≥ 0 \gamma\geq0 γ ≥ 0 δ > 0 \delta>0 δ > 0 k γ k_\gamma k γ (101) . Uniformly for
0 < α ≤ ( log T ) − c , α log T log log T ⟶ ∞ , 0<\alpha\leq(\log T)^{-c},\qquad\frac{\alpha\log T}{\log\log T}\longrightarrow\infty, 0 < α ≤ ( log T ) − c , log log T α log T ⟶ ∞ , we have
P ( M α , γ ( t ) < α k γ + δ ) = o ( 1 ) . \mathbb{P}\left(M_{\alpha,\gamma}(t)<\alpha^{k_\gamma+\delta}\right)=o(1). P ( M α , γ ( t ) < α k γ + δ ) = o ( 1 ) . Proof. Let H = ( log T ) γ H=(\log T)^\gamma H = ( log T ) γ Theorem 9.2 but replace (106) by
u = 2 log ( α log T log log T ) , x = exp ( u α ) . u=2\log\left(\frac{\alpha\log T}{\log\log T}\right),\qquad x=\exp\left(\frac{u}{\alpha}\right). u = 2 log ( log log T α log T ) , x = exp ( α u ) . Then u → ∞ u\to\infty u → ∞
log ( 1 / α ) log x log T ≪ log log T α log T log ( α log T log log T ) = o ( 1 ) . (117) \frac{\log(1/\alpha)\log x}{\log T}\ll\frac{\log\log T}{\alpha\log T}\log\left(\frac{\alpha\log T}{\log\log T}\right)=o(1). \tag*{(117)} log T log ( 1/ α ) log x ≪ α log T log log T log ( log log T α log T ) = o ( 1 ) . (117) The remainder in (107) is now
log T log x log 1 1 − x − α + O ( 1 ) ≪ α log T u e − u + 1 = o ( log 1 α ) . \frac{\log T}{\log x}\log\frac{1}{1-x^{-\alpha}}+O(1)\ll\frac{\alpha\log T}{u}e^{-u}+1=o\left(\log\frac{1}{\alpha}\right). log x log T log 1 − x − α 1 + O ( 1 ) ≪ u α log T e − u + 1 = o ( log α 1 ) . Note that estimate (108) still holds. For P 1 P_1 P 1
U = ( k γ + δ 3 ) log 1 α and q = ⌊ U 2 B 1 ⌋ U=\left(k_\gamma+\frac{\delta}{3}\right)\log\frac{1}{\alpha}
\qquad\text{and}\qquad
q=\left\lfloor\frac{U^2}{B_1}\right\rfloor U = ( k γ + 3 δ ) log α 1 and q = ⌊ B 1 U 2 ⌋ in (9.9). Then $q\asymp\log\frac{1}{\alpha}$. Thus (9.17) implies that $P_1^q$ has length $T^{o(1)}$. Also, since $\log x=u/\alpha$ and $1\leq u\ll\log\frac{1}{\alpha}$, equation {xref}(9.11) still holds. Hence, {xref}(9.9) and {xref}(9.10)` give
P ( max ∣ h ∣ ≤ H ∣ P 1 ( t + h ) ∣ > U ) ≪ ( 1 + q H log x ) q exp ( − U 2 B 1 ) ≪ exp ( − ( ( k γ + δ 3 ) 2 − k γ 2 + o ( 1 ) ) log 1 α ) = o ( 1 ) . \begin{aligned}
\mathbb{P}\left(\max_{|h|\leq H}|P_1(t+h)|>U\right)\ll(1+qH\log x)\sqrt{q}\exp\left(-\frac{U^2}{B_1}\right) \\
&\ll\exp\left(-\left(\left(k_\gamma+\frac{\delta}{3}\right)^2-k_\gamma^2+o(1)\right)\log\frac{1}{\alpha}\right)=o(1).
\end{aligned} P ( ∣ h ∣ ≤ H max ∣ P 1 ( t + h ) ∣ > U ) ≪ ( 1 + q H log x ) q exp ( − B 1 U 2 ) ≪ exp ( − ( ( k γ + 3 δ ) 2 − k γ 2 + o ( 1 ) ) log α 1 ) = o ( 1 ) . For P 2 P_2 P 2 (114) is still true. So we get that outside the exceptional set for P 1 P_1 P 1
W α ( t + h ) ≤ ( k γ + δ 3 ) log 1 α + o ( log 1 α ) < ( k γ + δ ) log 1 α . W_\alpha(t+h)\leq\left(k_\gamma+\frac{\delta}{3}\right)\log\frac{1}{\alpha}+o\left(\log\frac{1}{\alpha}\right)<(k_\gamma+\delta)\log\frac{1}{\alpha}. W α ( t + h ) ≤ ( k γ + 3 δ ) log α 1 + o ( log α 1 ) < ( k γ + δ ) log α 1 . Taking exponentials proves the assertion.
Proposition 9.6. Assume RH. Fix δ > 0 \delta>0 δ > 0
1 ≪ α log T ≪ log log T , 1\ll\alpha\log T\ll\log\log T, 1 ≪ α log T ≪ log log T , we have
P ( M α , 0 ( t ) ≥ α 5 / 4 + δ ) ⟶ 1. \mathbb{P}(M_{\alpha,0}(t)\geq\alpha^{5/4+\delta})\longrightarrow1. P ( M α , 0 ( t ) ≥ α 5/4 + δ ) ⟶ 1. Proof. Fix 0 < k < 1 / 2 0<k<1/2 0 < k < 1/2 [14 ] and Theorem 1.1 to get that
1 T ∫ T / 2 3 T ∣ ζ ( 1 2 + β + i t ) ∣ − 2 k d t ≪ α − k 2 \frac{1}{T}\int_{T/2}^{3T}\left|\zeta\left(\frac{1}{2}+\beta+it\right)\right|^{-2k}\,dt\ll\alpha^{-k^2} T 1 ∫ T /2 3 T ζ ( 2 1 + β + i t ) − 2 k d t ≪ α − k 2 for α / 2 ≤ β ≤ 3 α / 2 \alpha/2\leq\beta\leq3\alpha/2 α /2 ≤ β ≤ 3 α /2 ∣ ζ ∣ − 2 k |\zeta|^{-2k} ∣ ζ ∣ − 2 k α / 2 \alpha/2 α /2 1 / 2 + α + i ( t + h ) 1/2+\alpha+i(t+h) 1/2 + α + i ( t + h ) ∣ h ∣ ≤ 1 |h|\leq1 ∣ h ∣ ≤ 1 α / 2 ≤ ℜ s − 1 / 2 ≤ 3 α / 2 \alpha/2\leq\Re s-1/2\leq3\alpha/2 α /2 ≤ ℜ s − 1/2 ≤ 3 α /2 t − 2 ≤ ℑ s ≤ t + 2 t-2\leq\Im s\leq t+2 t − 2 ≤ ℑ s ≤ t + 2
1 T ∫ T 2 T max ∣ h ∣ ≤ 1 ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 2 k d t ≪ 1 α 2 ∫ α / 2 3 α / 2 1 T ∫ T − 2 2 T + 2 ∣ ζ ( 1 2 + β + i v ) ∣ − 2 k d v d β \frac{1}{T}\int_T^{2T}\max_{|h|\leq1}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-2k}\,dt
\ll\frac{1}{\alpha^2}\int_{\alpha/2}^{3\alpha/2}\frac{1}{T}\int_{T-2}^{2T+2}\left|\zeta\left(\frac{1}{2}+\beta+iv\right)\right|^{-2k}\,dv\,d\beta T 1 ∫ T 2 T ∣ h ∣ ≤ 1 max ζ ( 2 1 + α + i ( t + h ) ) − 2 k d t ≪ α 2 1 ∫ α /2 3 α /2 T 1 ∫ T − 2 2 T + 2 ζ ( 2 1 + β + i v ) − 2 k d v d β ≪ α − 1 − k 2 . \ll\alpha^{-1-k^2}. ≪ α − 1 − k 2 . Thus,
P ( M α , 0 ( t ) < α 5 / 4 + δ ) ≪ α 2 k ( 5 / 4 + δ ) − 1 − k 2 = o ( 1 ) , \mathbb{P}\left(M_{\alpha,0}(t)<\alpha^{5/4+\delta}\right)\ll\alpha^{2k(5/4+\delta)-1-k^2}=o(1), P ( M α , 0 ( t ) < α 5/4 + δ ) ≪ α 2 k ( 5/4 + δ ) − 1 − k 2 = o ( 1 ) , on choosing k k k 1 / 2 1/2 1/2
The upper bound for the minimum
We now focus on the case γ = 0 \gamma=0 γ = 0 ∣ h ∣ ≤ 1 |h|\le1 ∣ h ∣ ≤ 1 M α M_\alpha M α
Theorem 9.7. Assume RH. Fix 0 < c , δ < 1 0<c,\delta<1 0 < c , δ < 1 M 0 > 0 M_0>0 M 0 > 0
M 0 log T ≤ α ≪ ( log T ) − c , \frac{M_0}{\log T}\le\alpha\ll(\log T)^{-c}, log T M 0 ≤ α ≪ ( log T ) − c , we have
P ( M α , 0 ( t ) > α 1 − δ ) = o ( 1 ) . \mathbb{P}\left(M_{\alpha,0}(t)>\alpha^{1-\delta}\right)=o(1). P ( M α , 0 ( t ) > α 1 − δ ) = o ( 1 ) . We first need the following lemma. We will only sketch the proof, since it is similar to the proof of Proposition 3.2 in [5 ] .
Lemma 9.8. Fix 0 < c 1 < c 2 < 1 0<c_1<c_2<1 0 < c 1 < c 2 < 1 0 < ν < 1 0<\nu<1 0 < ν < 1
( log T ) c 1 ≤ Y ≤ ( log T ) c 2 , (\log T)^{c_1}\le Y\le(\log T)^{c_2}, ( log T ) c 1 ≤ Y ≤ ( log T ) c 2 , we have
P ( min ∣ h ∣ ≤ 1 / 2 ( ℜ ∑ p ≤ e Y p − 1 / 2 − i ( t + h ) ) < − ( 1 − ν ) log Y ) = 1 − o ( 1 ) . \mathbb{P}\left(\min_{|h|\le1/2}\left(\Re\sum_{p\le e^Y}p^{-1/2-i(t+h)}\right)<-(1-\nu)\log Y\right)=1-o(1). P ∣ h ∣ ≤ 1/2 min ℜ p ≤ e Y ∑ p − 1/2 − i ( t + h ) < − ( 1 − ν ) log Y = 1 − o ( 1 ) . Proof. Let K K K [5 (10)] and the corresponding prime sums with
J j = ( exp ( ( log T ) j / K ) , exp ( ( log T ) ( j + 1 ) / K ) ] , P j ( v ) = − ℜ ∑ p ∈ J j p − 1 / 2 − i v , J_j=\left(\exp\left((\log T)^{j/K}\right),\exp\left((\log T)^{(j+1)/K}\right)\right],\qquad P_j(v)=-\Re\sum_{p\in J_j}p^{-1/2-iv}, J j = ( exp ( ( log T ) j / K ) , exp ( ( log T ) ( j + 1 ) / K ) ] , P j ( v ) = − ℜ p ∈ J j ∑ p − 1/2 − i v , for 1 ≤ j ≤ K − 3 1\le j\le K-3 1 ≤ j ≤ K − 3 [5 ] states that
P ( max ∣ h ∣ ≤ 1 / 4 min 1 ≤ j ≤ K − 3 ( ℜ ∑ p ∈ J j p − σ 0 − i ( t + h ) ) > 1 − ν / 4 K log log T ) = 1 − o ( 1 ) , \mathbb{P}\left(\max_{|h|\le1/4}\min_{1\le j\le K-3}\left(\Re\sum_{p\in J_j}p^{-\sigma_0-i(t+h)}\right)>\frac{1-\nu/4}{K}\log\log T\right)=1-o(1), P ∣ h ∣ ≤ 1/4 max 1 ≤ j ≤ K − 3 min ℜ p ∈ J j ∑ p − σ 0 − i ( t + h ) > K 1 − ν /4 log log T = 1 − o ( 1 ) , where σ 0 = 1 / 2 + ( log T ) 3 / ( 2 K ) − 1 \sigma_0=1/2+(\log T)^{3/(2K)-1} σ 0 = 1/2 + ( log T ) 3/ ( 2 K ) − 1 [5 ] leads to the same conclusion with coefficients − p − 1 / 2 -p^{-1/2} − p − 1/2 p − σ 0 p^{-\sigma_0} p − σ 0
P ( max ∣ h ∣ ≤ 1 / 4 min 1 ≤ j ≤ K − 3 P j ( t + h ) > 1 − ν / 4 K log log T ) = 1 − o ( 1 ) . (118) \mathbb{P}\left(\max_{|h|\le1/4}\min_{1\le j\le K-3}P_j(t+h)>\frac{1-\nu/4}{K}\log\log T\right)=1-o(1). \tag*{(118)} P ( ∣ h ∣ ≤ 1/4 max 1 ≤ j ≤ K − 3 min P j ( t + h ) > K 1 − ν /4 log log T ) = 1 − o ( 1 ) . (118) Now note that e Y e^Y e Y e Y e^Y e Y m m m
( log T ) ( m + 1 ) / K ≤ Y < ( log T ) ( m + 2 ) / K . (\log T)^{(m+1)/K}\le Y<(\log T)^{(m+2)/K}. ( log T ) ( m + 1 ) / K ≤ Y < ( log T ) ( m + 2 ) / K . Note that
m = ⌊ K log Y log log T ⌋ − 1. m=\left\lfloor\frac{K\log Y}{\log\log T}\right\rfloor-1. m = ⌊ log log T K log Y ⌋ − 1. and
m K log log T ≥ log Y − 2 K log log T . \frac{m}{K}\log\log T\ge\log Y-\frac{2}{K}\log\log T. K m log log T ≥ log Y − K 2 log log T . Thus, at the point provided by (118) , we have
∑ j = 1 m P j ( t + h ) > ( 1 − ν 4 ) ( log Y − 2 K log log T ) ≥ ( 1 − ν 2 ) log Y , \sum_{j=1}^{m}P_j(t+h)>\left(1-\frac{\nu}{4}\right)\left(\log Y-\frac{2}{K}\log\log T\right)\ge\left(1-\frac{\nu}{2}\right)\log Y, j = 1 ∑ m P j ( t + h ) > ( 1 − 4 ν ) ( log Y − K 2 log log T ) ≥ ( 1 − 2 ν ) log Y , provided K c 1 ≥ 8 / ν Kc_1\ge8/\nu K c 1 ≥ 8/ ν log Y ≥ c 1 log log T \log Y\ge c_1\log\log T log Y ≥ c 1 log log T e Y e^Y e Y
R ( v ) = ∑ p ≤ exp ( ( log T ) 1 / K ) p − 1 / 2 − i v + ∑ exp ( ( log T ) ( m + 1 ) / K ) < p ≤ e Y p − 1 / 2 − i v . R(v)=\sum_{p\leq\exp((\log T)^{1/K})}p^{-1/2-iv}+\sum_{\exp((\log T)^{(m+1)/K})<p\leq e^Y}p^{-1/2-iv}. R ( v ) = p ≤ e x p (( l o g T ) 1/ K ) ∑ p − 1/2 − i v + e x p (( l o g T ) ( m + 1 ) / K ) < p ≤ e Y ∑ p − 1/2 − i v . Mertens’ estimate and the definition of m m m
∑ p ≤ exp ( ( log T ) 1 / K ) 1 p + ∑ exp ( ( log T ) ( m + 1 ) / K ) < p ≤ e Y 1 p = log log T K + log Y − ( m + 1 ) log log T K + O ( 1 ) ≤ 2 log log T K + O ( 1 ) ≤ 2 log Y K ε + O ( 1 ) . \sum_{p\leq\exp((\log T)^{1/K})}\frac{1}{p}+\sum_{\exp((\log T)^{(m+1)/K})<p\leq e^Y}\frac{1}{p}
=\frac{\log\log T}{K}+\log Y-\frac{(m+1)\log\log T}{K}+O(1)\leq\frac{2\log\log T}{K}+O(1)\leq\frac{2\log Y}{K\varepsilon}+O(1). p ≤ e x p (( l o g T ) 1/ K ) ∑ p 1 + e x p (( l o g T ) ( m + 1 ) / K ) < p ≤ e Y ∑ p 1 = K log log T + log Y − K ( m + 1 ) log log T + O ( 1 ) ≤ K 2 log log T + O ( 1 ) ≤ K ε 2 log Y + O ( 1 ) . Now we apply Lemma 9.4 with H = 1 H=1 H = 1 q = ⌊ log Y ⌋ q=\lfloor\log Y\rfloor q = ⌊ log Y ⌋ q Y = o ( log T ) qY=o(\log T) q Y = o ( log T )
P ( max ∣ h ∣ ≤ 1 ∣ R ( t + h ) ∣ > ν 2 log Y ) ≪ ( 1 + q Y ) ( C K + C log Y ) q = o ( 1 ) . \mathbb{P}\left(\max_{|h|\leq1}|R(t+h)|>\frac{\nu}{2}\log Y\right)\ll(1+qY)\left(\frac{C}{K}+\frac{C}{\log Y}\right)^q=o(1). P ( ∣ h ∣ ≤ 1 max ∣ R ( t + h ) ∣ > 2 ν log Y ) ≪ ( 1 + q Y ) ( K C + log Y C ) q = o ( 1 ) . Outside the two exceptional sets (the one obtained above and the one in (118) ), at the same point ∣ h ∣ ≤ 1 / 4 |h|\leq1/4 ∣ h ∣ ≤ 1/4
− ℜ ∑ p ≤ e Y p − 1 / 2 − i ( t + h ) = ∑ j = 1 m P j ( t + h ) − ℜ R ( t + h ) > ( 1 − ν ) log Y . -\Re\sum_{p\leq e^Y}p^{-1/2-i(t+h)}=\sum_{j=1}^{m}P_j(t+h)-\Re R(t+h)>(1-\nu)\log Y. − ℜ p ≤ e Y ∑ p − 1/2 − i ( t + h ) = j = 1 ∑ m P j ( t + h ) − ℜ R ( t + h ) > ( 1 − ν ) log Y . This finishes the proof. □
We are now ready to begin the proof of Theorem 9.7.
Proof of Theorem 9.7. We first prove the assertion for ∣ h ∣ ≤ 1 / 2 |h|\leq1/2 ∣ h ∣ ≤ 1/2
( log T ) − 1 + η ≤ α ≤ ( log T ) − c , (119) (\log T)^{-1+\eta}\leq\alpha\leq(\log T)^{-c}, \tag*{(119)} ( log T ) − 1 + η ≤ α ≤ ( log T ) − c , (119) where η > 0 \eta>0 η > 0
η < 1 − c and ( 1 − η ) 2 > 1 − δ . \eta<1-c\qquad\text{and}\qquad(1-\eta)^2>1-\delta. η < 1 − c and ( 1 − η ) 2 > 1 − δ . Let u , x u,x u , x (106) . The extremal minorant of Carneiro and Chandee [15 Lemma 5 and Section 2.1] , inserted in the explicit formula, gives the inequality
W α ( v ) ≥ ℜ ∑ n ≤ x b α − ( n ; x ) n 1 / 2 + α + i v − log T log x log ( 1 + x − α ) − O ( 1 ) , (120) W_\alpha(v)\geq\Re\sum_{n\leq x}\frac{b^-_\alpha(n;x)}{n^{1/2+\alpha+iv}}-\frac{\log T}{\log x}\log(1+x^{-\alpha})-O(1), \tag*{(120)} W α ( v ) ≥ ℜ n ≤ x ∑ n 1/2 + α + i v b α − ( n ; x ) − log x log T log ( 1 + x − α ) − O ( 1 ) , (120) with v ∈ [ T − 1 , 2 T + 1 ] v\in[T-1,2T+1] v ∈ [ T − 1 , 2 T + 1 ]
b α − ( n ; x ) = − Λ ( n ) ∑ j ≥ 0 ( − 1 ) j ( j + 1 log n + j log x x − j α − ( j + 1 ) n 2 α ( j + 2 ) log x − log n x − ( j + 2 ) α ) . b^-_\alpha(n;x)=-\Lambda(n)\sum_{j\geq0}(-1)^j\left(\frac{j+1}{\log n+j\log x}x^{-j\alpha}-\frac{(j+1)n^{2\alpha}}{(j+2)\log x-\log n}x^{-(j+2)\alpha}\right). b α − ( n ; x ) = − Λ ( n ) j ≥ 0 ∑ ( − 1 ) j ( log n + j log x j + 1 x − j α − ( j + 2 ) log x − log n ( j + 1 ) n 2 α x − ( j + 2 ) α ) . As in Lemma 2.2 , we have
b α − ( p r ; x ) = − 1 r + O ( log p log x ) , ∣ b α − ( p r ; x ) ∣ ≪ 1 r . (121) b^-_\alpha(p^r;x)=-\frac{1}{r}+O\left(\frac{\log p}{\log x}\right),\qquad|b^-_\alpha(p^r;x)|\ll\frac{1}{r}. \tag*{(121)} b α − ( p r ; x ) = − r 1 + O ( log x log p ) , ∣ b α − ( p r ; x ) ∣ ≪ r 1 . (121) Fix 0 < ν < 1 / 2 0<\nu<1/2 0 < ν < 1/2 δ \delta δ x 0 = exp ( α − 1 + ν ) x_0=\exp(\alpha^{-1+\nu}) x 0 = exp ( α − 1 + ν ) log x 0 / log x = α ν / u = o ( 1 ) \log x_0/\log x=\alpha^\nu/u=o(1) log x 0 / log x = α ν / u = o ( 1 ) x 0 < x x_0<x x 0 < x
R 1 ( v ) : = ∑ p ≤ x 0 b α − ( p ; x ) p − α + 1 p 1 / 2 + i v + ∑ x 0 < p ≤ x b α − ( p ; x ) p 1 / 2 + α + i v , (122) R_1(v):=\sum_{p\leq x_0}\frac{b^-_\alpha(p;x)p^{-\alpha}+1}{p^{1/2+iv}}+\sum_{x_0<p\leq x}\frac{b^-_\alpha(p;x)}{p^{1/2+\alpha+iv}}, \tag*{(122)} R 1 ( v ) := p ≤ x 0 ∑ p 1/2 + i v b α − ( p ; x ) p − α + 1 + x 0 < p ≤ x ∑ p 1/2 + α + i v b α − ( p ; x ) , (122) R 2 ( v ) : = ∑ p 2 ≤ x b α − ( p 2 ; x ) p 1 + 2 α + 2 i v . (123) R_2(v) := \sum_{p^2\le x} \frac{b_\alpha^-(p^2;x)}{p^{1+2\alpha+2iv}}. \tag*{(123)} R 2 ( v ) := p 2 ≤ x ∑ p 1 + 2 α + 2 i v b α − ( p 2 ; x ) . (123) Then
∑ p ≤ x b α − ( p ; x ) p 1 / 2 + α + i v = − ∑ p ≤ x 0 p − 1 / 2 − i v + R 1 ( v ) . \sum_{p\le x}\frac{b_\alpha^-(p;x)}{p^{1/2+\alpha+iv}}=-\sum_{p\le x_0}p^{-1/2-iv}+R_1(v). p ≤ x ∑ p 1/2 + α + i v b α − ( p ; x ) = − p ≤ x 0 ∑ p − 1/2 − i v + R 1 ( v ) . The higher prime powers are bounded uniformly in v v v (120) becomes
W α ( v ) ≥ − ℜ ∑ p ≤ x 0 p − 1 / 2 − i v + ℜ R 1 ( v ) + ℜ R 2 ( v ) − C , W_\alpha(v)\ge-\Re\sum_{p\le x_0}p^{-1/2-iv}+\Re R_1(v)+\Re R_2(v)-C, W α ( v ) ≥ − ℜ p ≤ x 0 ∑ p − 1/2 − i v + ℜ R 1 ( v ) + ℜ R 2 ( v ) − C , for some C > 0 C>0 C > 0
We next bound R 1 R_1 R 1 R 2 R_2 R 2 v = t + h v=t+h v = t + h ∣ h ∣ ≤ 1 |h|\le1 ∣ h ∣ ≤ 1 p ≤ x 0 p\le x_0 p ≤ x 0 (121) and the fact that 1 − e − s ≤ s 1-e^{-s}\le s 1 − e − s ≤ s s ≥ 0 s\ge0 s ≥ 0
∣ b α − ( p ; x ) ‾ p − α + 1 ∣ ≤ ∣ b α − ( p ; x ) + 1 ∣ p − α + 1 − p − α ≪ ( α + 1 log x ) log p . \left|\overline{b_\alpha^-(p;x)}p^{-\alpha}+1\right|\le\left|b_\alpha^-(p;x)+1\right|p^{-\alpha}+1-p^{-\alpha}\ll\left(\alpha+\frac{1}{\log x}\right)\log p. b α − ( p ; x ) p − α + 1 ≤ b α − ( p ; x ) + 1 p − α + 1 − p − α ≪ ( α + log x 1 ) log p . It follows that
∑ p ≤ x 0 ∣ b α − ( p ; x ) ‾ p − α + 1 ∣ 2 p ≪ ( α + 1 log x ) 2 ∑ p ≤ x 0 ( log p ) 2 p ≪ ( α + 1 log x ) 2 ( log x 0 ) 2 ≪ α 2 ν . \sum_{p\le x_0}\frac{\left|\overline{b_\alpha^-(p;x)}p^{-\alpha}+1\right|^2}{p}\ll\left(\alpha+\frac{1}{\log x}\right)^2\sum_{p\le x_0}\frac{(\log p)^2}{p}\ll\left(\alpha+\frac{1}{\log x}\right)^2(\log x_0)^2\ll\alpha^{2\nu}. p ≤ x 0 ∑ p b α − ( p ; x ) p − α + 1 2 ≪ ( α + log x 1 ) 2 p ≤ x 0 ∑ p ( log p ) 2 ≪ ( α + log x 1 ) 2 ( log x 0 ) 2 ≪ α 2 ν . For the remaining primes, we use x 0 < e 1 / α < x x_0<e^{1/\alpha}<x x 0 < e 1/ α < x (122) and (123) satisfy
∑ p ≤ x 0 ∣ b α − ( p ; x ) ‾ p − α + 1 ∣ 2 p + ∑ x 0 < p ≤ x ∣ b α − ( p ; x ) ∣ 2 p 1 + 2 α ≪ ν log 1 α , \sum_{p\le x_0}\frac{\left|\overline{b_\alpha^-(p;x)}p^{-\alpha}+1\right|^2}{p}+\sum_{x_0<p\le x}\frac{|b_\alpha^-(p;x)|^2}{p^{1+2\alpha}}\ll\nu\log\frac{1}{\alpha}, p ≤ x 0 ∑ p b α − ( p ; x ) p − α + 1 2 + x 0 < p ≤ x ∑ p 1 + 2 α ∣ b α − ( p ; x ) ∣ 2 ≪ ν log α 1 , ∑ p 2 ≤ x ∣ b α − ( p 2 ; x ) ∣ 2 p 2 + 4 α ≪ 1. \sum_{p^2\le x}\frac{|b_\alpha^-(p^2;x)|^2}{p^{2+4\alpha}}\ll1. p 2 ≤ x ∑ p 2 + 4 α ∣ b α − ( p 2 ; x ) ∣ 2 ≪ 1. Let q = ⌊ log 1 α ⌋ q=\left\lfloor\log\frac{1}{\alpha}\right\rfloor q = ⌊ log α 1 ⌋
q log x log T = q log ( α log T ) α log T ≪ ( log log T ) 2 ( log T ) η = o ( 1 ) . \frac{q\log x}{\log T}=\frac{q\log(\alpha\log T)}{\alpha\log T}\ll\frac{(\log\log T)^2}{(\log T)^\eta}=o(1). log T q log x = α log T q log ( α log T ) ≪ ( log T ) η ( log log T ) 2 = o ( 1 ) . Applying Lemma 9.4 with H = 1 H=1 H = 1 R 1 R_1 R 1 R 2 R_2 R 2
E ( max ∣ h ∣ ≤ 1 ∣ R 1 ( t + h ) ∣ 2 q ) ≪ ( 1 + q log x ) q ! ( C ( ν log 1 α ) ) q , \mathbb{E}\left(\max_{|h|\le1}|R_1(t+h)|^{2q}\right)\ll(1+q\log x)q!\left(C\left(\nu\log\frac{1}{\alpha}\right)\right)^q, E ( ∣ h ∣ ≤ 1 max ∣ R 1 ( t + h ) ∣ 2 q ) ≪ ( 1 + q log x ) q ! ( C ( ν log α 1 ) ) q , E ( max ∣ h ∣ ≤ 1 ∣ R 2 ( t + h ) ∣ 2 q ) ≪ ( 1 + q log x ) q ! C q . \mathbb{E}\left(\max_{|h|\le1}|R_2(t+h)|^{2q}\right)\ll(1+q\log x)q!C^q. E ( ∣ h ∣ ≤ 1 max ∣ R 2 ( t + h ) ∣ 2 q ) ≪ ( 1 + q log x ) q ! C q . Markov’s inequality and Stirling again give
P ( max ∣ h ∣ ≤ 1 ∣ R 1 ( t + h ) ∣ > δ 8 log 1 α ) ≪ ( 1 + q log x ) ( C δ 2 ( ν + 1 log ( 1 / α ) ) ) q . (124) \mathbb{P}\left(\max_{|h|\le1}|R_1(t+h)|>\frac{\delta}{8}\log\frac{1}{\alpha}\right)\ll(1+q\log x)\left(\frac{C}{\delta^2}\left(\nu+\frac{1}{\log(1/\alpha)}\right)\right)^q. \tag*{(124)} P ( ∣ h ∣ ≤ 1 max ∣ R 1 ( t + h ) ∣ > 8 δ log α 1 ) ≪ ( 1 + q log x ) ( δ 2 C ( ν + log ( 1/ α ) 1 ) ) q . (124) P ( max ∣ h ∣ ≤ 1 ∣ R 2 ( t + h ) ∣ > δ 8 log 1 α ) ≪ ( 1 + q log x ) ( C δ 2 log ( 1 / α ) ) q . (125) \mathbb{P}\left(\max_{|h|\le1}|R_2(t+h)|>\frac{\delta}{8}\log\frac{1}{\alpha}\right)\ll(1+q\log x)\left(\frac{C}{\delta^2\log(1/\alpha)}\right)^q. \tag*{(125)} P ( ∣ h ∣ ≤ 1 max ∣ R 2 ( t + h ) ∣ > 8 δ log α 1 ) ≪ ( 1 + q log x ) ( δ 2 log ( 1/ α ) C ) q . (125) Now note that
log ( 1 + q log x ) = log 1 α + O ( log log 1 α ) . \log(1+q\log x)=\log\frac{1}{\alpha}+O\left(\log\log\frac{1}{\alpha}\right). log ( 1 + q log x ) = log α 1 + O ( log log α 1 ) . We choose ν \nu ν ( 1 − ν ) 2 ≥ 1 − δ / 4 (1-\nu)^2 \ge1-\delta/4 ( 1 − ν ) 2 ≥ 1 − δ /4 C ν / δ 2 ≤ e − 4 C\nu/\delta^2 \le e^{-4} C ν / δ 2 ≤ e − 4 e − 3 e^{-3} e − 3 T T T
≪ exp ( − 2 log 1 α + O ( log log 1 α ) ) = o ( 1 ) . \ll\exp\left(-2\log\frac{1}{\alpha}+O\left(\log\log\frac{1}{\alpha}\right)\right)=o(1). ≪ exp ( − 2 log α 1 + O ( log log α 1 ) ) = o ( 1 ) . We also have the right-hand side of (9.25) is
≪ exp ( − log 1 α log log 1 α + O ( log 1 α ) ) = o ( 1 ) . \ll\exp\left(-\log\frac{1}{\alpha}\log\log\frac{1}{\alpha}+O\left(\log\frac{1}{\alpha}\right)\right)=o(1). ≪ exp ( − log α 1 log log α 1 + O ( log α 1 ) ) = o ( 1 ) . We next use Lemma 9.8 with Y = log x 0 Y=\log x_0 Y = log x 0 Y log Y = o ( log T ) Y\log Y=o(\log T) Y log Y = o ( log T ) o ( T ) o(T) o ( T ) ∣ h ∣ ≤ 1 / 2 |h|\le1/2 ∣ h ∣ ≤ 1/2
ℜ ∑ p ≤ x 0 p − 1 / 2 − i ( t + h ) < − ( 1 − ν ) log log x 0 = − ( 1 − ν ) 2 log 1 α ≤ − ( 1 − δ 4 ) log 1 α . \Re\sum_{p\le x_0}p^{-1/2-i(t+h)}<-(1-\nu)\log\log x_0=-(1-\nu)^2\log\frac{1}{\alpha}\le-\left(1-\frac{\delta}{4}\right)\log\frac{1}{\alpha}. ℜ p ≤ x 0 ∑ p − 1/2 − i ( t + h ) < − ( 1 − ν ) log log x 0 = − ( 1 − ν ) 2 log α 1 ≤ − ( 1 − 4 δ ) log α 1 . Outside the union of this exceptional set and the two sets in (9.24)–(9.25), we have
W α ( t + h ) > ( 1 − δ / 2 ) log 1 α − C > ( 1 − δ ) log 1 α , W_\alpha(t+h)>(1-\delta/2)\log\frac{1}{\alpha}-C>(1-\delta)\log\frac{1}{\alpha}, W α ( t + h ) > ( 1 − δ /2 ) log α 1 − C > ( 1 − δ ) log α 1 , for sufficiently large T T T
min ∣ h ∣ ≤ 1 / 2 ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ = exp ( − max ∣ h ∣ ≤ 1 / 2 W α ( t + h ) ) < α 1 − δ . \min_{|h|\le1/2}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|=\exp\left(-\max_{|h|\le1/2}W_\alpha(t+h)\right)<\alpha^{1-\delta}. ∣ h ∣ ≤ 1/2 min ζ ( 2 1 + α + i ( t + h ) ) = exp ( − ∣ h ∣ ≤ 1/2 max W α ( t + h ) ) < α 1 − δ . This finishes the proof in the range (9.19).
Now we assume that
M 0 log T ≤ α ≤ ( log T ) − 1 + η . \frac{M_0}{\log T}\le\alpha\le(\log T)^{-1+\eta}. log T M 0 ≤ α ≤ ( log T ) − 1 + η . We begin with the Sobolev inequality used in the proof of [5 Proposition 2.1, eq. (6)] : for every continuously differentiable complex-valued function g g g [ a , b ] [a,b] [ a , b ]
max a ≤ v ≤ b ∣ g ( v ) ∣ 2 ≤ ∣ g ( a ) ∣ 2 + ∣ g ( b ) ∣ 2 2 + ∫ a b ∣ g ( v ) g ′ ( v ) ∣ d v . (126) \max_{a\le v\le b}|g(v)|^2\le\frac{|g(a)|^2+|g(b)|^2}{2}+\int_a^b|g(v)g'(v)|\,dv. \tag*{(126)} a ≤ v ≤ b max ∣ g ( v ) ∣ 2 ≤ 2 ∣ g ( a ) ∣ 2 + ∣ g ( b ) ∣ 2 + ∫ a b ∣ g ( v ) g ′ ( v ) ∣ d v . (126) We apply this to f ( s ) = ζ ( s ) − 1 / 4 f(s)=\zeta(s)^{-1/4} f ( s ) = ζ ( s ) − 1/4
R t = { s : 1 2 + α ≤ ℜ s ≤ 2 , ∣ ℑ s − t ∣ ≤ 1 } . R_t=\left\{s:\frac{1}{2}+\alpha\le\Re s\le2,\ |\Im s-t|\le1\right\}. R t = { s : 2 1 + α ≤ ℜ s ≤ 2 , ∣ℑ s − t ∣ ≤ 1 } . On either vertical side, take g ( v ) = f ( σ + i v ) g(v)=f(\sigma+iv) g ( v ) = f ( σ + i v ) [ t − 1 , t + 1 ] [t-1,t+1] [ t − 1 , t + 1 ] t t t
1 T ∫ T 2 T max ∣ h ∣ ≤ 1 ∣ f ( σ + i ( t + h ) ) ∣ 2 d t ≪ 1 T ∫ T − 2 2 T + 2 ( ∣ f ( σ + i v ) ∣ 2 + ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ ) d v . (127) \frac{1}{T}\int_T^{2T}\max_{|h|\le1}|f(\sigma+i(t+h))|^2\,dt\ll\frac{1}{T}\int_{T-2}^{2T+2}\left(|f(\sigma+iv)|^2+|f(\sigma+iv)f'(\sigma+iv)|\right)\,dv. \tag*{(127)} T 1 ∫ T 2 T ∣ h ∣ ≤ 1 max ∣ f ( σ + i ( t + h )) ∣ 2 d t ≪ T 1 ∫ T − 2 2 T + 2 ( ∣ f ( σ + i v ) ∣ 2 + ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ ) d v . (127) On the two horizontal sides, we apply (9.26) to g ( σ ) = f ( σ + i ( t ± 1 ) ) g(\sigma)=f(\sigma+i(t\pm1)) g ( σ ) = f ( σ + i ( t ± 1 )) [ 1 / 2 + α , 2 ] [1/2+\alpha,2] [ 1/2 + α , 2 ]
1 T ∫ T 2 T max 1 / 2 + α ≤ σ ≤ 2 ∣ f ( σ + i ( t ± 1 ) ) ∣ 2 d t ≤ 1 2 ∑ σ ∈ { 1 / 2 + α , 2 } 1 T ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) ∣ 2 d v + ∫ 1 / 2 + α 2 ( 1 T ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ d v ) d σ . \frac{1}{T}\int_T^{2T}\max_{1/2+\alpha\le\sigma\le2}|f(\sigma+i(t\pm1))|^2\,dt\le\frac{1}{2}\sum_{\sigma\in\{1/2+\alpha,2\}}\frac{1}{T}\int_{T-2}^{2T+2}|f(\sigma+iv)|^2\,dv
+\int_{1/2+\alpha}^{2}\left(\frac{1}{T}\int_{T-2}^{2T+2}|f(\sigma+iv)f'(\sigma+iv)|\,dv\right)d\sigma. T 1 ∫ T 2 T 1/2 + α ≤ σ ≤ 2 max ∣ f ( σ + i ( t ± 1 )) ∣ 2 d t ≤ 2 1 σ ∈ { 1/2 + α , 2 } ∑ T 1 ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) ∣ 2 d v + ∫ 1/2 + α 2 ( T 1 ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ d v ) d σ . Thus, after averaging over t t t ∣ f ∣ 2 |f|^2 ∣ f ∣ 2 ∣ f f ′ ∣ |ff'| ∣ f f ′ ∣ ∣ f ∣ 2 |f|^2 ∣ f ∣ 2 k = 1 / 4 k=1/4 k = 1/4 ( 1 / 2 + ε ) log log T / log T (1/2+\varepsilon)\log\log T/\log T ( 1/2 + ε ) log log T / log T [14 Theorem 1.4] applies to the larger shifts. We obtain that
1 T ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) ∣ 2 d v ≪ ( σ − 1 2 ) − 1 / 16 ≪ α − 1 / 16 . (128) \frac{1}{T}\int_{T-2}^{2T+2}|f(\sigma+iv)|^2\,dv\ll\left(\sigma-\frac{1}{2}\right)^{-1/16}\ll\alpha^{-1/16}. \tag*{(128)} T 1 ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) ∣ 2 d v ≪ ( σ − 2 1 ) − 1/16 ≪ α − 1/16 . (128) Next, we bound the mean of ∣ f ′ ∣ 2 |f'|^2 ∣ f ′ ∣ 2 1 / 2 + α ≤ σ ≤ 2 1/2+\alpha\leq\sigma\leq2 1/2 + α ≤ σ ≤ 2 r = α / 4 r=\alpha/4 r = α /4
f ′ ( σ + i v ) = 1 2 π r ∫ 0 2 π f ( σ + i v + r e i θ ) e − i θ d θ , f'(\sigma+iv)=\frac{1}{2\pi r}\int_0^{2\pi}f(\sigma+iv+re^{i\theta})e^{-i\theta}\,d\theta, f ′ ( σ + i v ) = 2 π r 1 ∫ 0 2 π f ( σ + i v + r e i θ ) e − i θ d θ , so after using Cauchy–Schwarz, we get
∣ f ′ ( σ + i v ) ∣ 2 ≤ 1 2 π r 2 ∫ 0 2 π ∣ f ( σ + i v + r e i θ ) ∣ 2 d θ . |f'(\sigma+iv)|^2\leq\frac{1}{2\pi r^2}\int_0^{2\pi}|f(\sigma+iv+re^{i\theta})|^2\,d\theta. ∣ f ′ ( σ + i v ) ∣ 2 ≤ 2 π r 2 1 ∫ 0 2 π ∣ f ( σ + i v + r e i θ ) ∣ 2 d θ . Integrating over v v v
1 T ∫ T − 2 2 T + 2 ∣ f ′ ( σ + i v ) ∣ 2 d v ≪ α − 2 − 1 / 16 . \frac{1}{T}\int_{T-2}^{2T+2}|f'(\sigma+iv)|^2\,dv\ll\alpha^{-2-1/16}. T 1 ∫ T − 2 2 T + 2 ∣ f ′ ( σ + i v ) ∣ 2 d v ≪ α − 2 − 1/16 . The Cauchy–Schwarz inequality then gives
1 T ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ d v ≪ α − 17 / 16 . (129) \frac{1}{T}\int_{T-2}^{2T+2}|f(\sigma+iv)f'(\sigma+iv)|\,dv\ll\alpha^{-17/16}. \tag*{(129)} T 1 ∫ T − 2 2 T + 2 ∣ f ( σ + i v ) f ′ ( σ + i v ) ∣ d v ≪ α − 17/16 . (129) Substituting (9.28) and (9.29) into (9.27) bounds the average of each vertical-side maximum by O ( α − 17 / 16 ) O(\alpha^{-17/16}) O ( α − 17/16 ) O ( α − 17 / 16 ) O(\alpha^{-17/16}) O ( α − 17/16 )
1 T ∫ T 2 T max s ∈ R t ∣ ζ ( s ) ∣ − 1 / 2 d t ≪ α − 17 / 16 . \frac{1}{T}\int_T^{2T}\max_{s\in\mathcal{R}_t}|\zeta(s)|^{-1/2}\,dt\ll\alpha^{-17/16}. T 1 ∫ T 2 T s ∈ R t max ∣ ζ ( s ) ∣ − 1/2 d t ≪ α − 17/16 . Since α ≥ M 0 / log T \alpha\geq M_0/\log T α ≥ M 0 / log T
P ( max s ∈ R t ∣ ζ ( s ) ∣ − 1 / 2 > ( log T ) 3 ) ≪ ( log T ) − 31 / 16 = o ( 1 ) . (130) \mathbb{P}\left(\max_{s\in\mathcal{R}_t}|\zeta(s)|^{-1/2}>(\log T)^3\right)\ll(\log T)^{-31/16}=o(1). \tag*{(130)} P ( s ∈ R t max ∣ ζ ( s ) ∣ − 1/2 > ( log T ) 3 ) ≪ ( log T ) − 31/16 = o ( 1 ) . (130) Now let β = ( log T ) − 1 + η \beta=(\log T)^{-1+\eta} β = ( log T ) − 1 + η
max ∣ h ∣ ≤ 1 / 2 ∣ ζ ( 1 2 + β + i ( t + h ) ) ∣ − 1 / 2 ≥ β − ( 1 − η ) / 2 (131) \max_{|h|\leq1/2}\left|\zeta\left(\frac{1}{2}+\beta+i(t+h)\right)\right|^{-1/2}\geq\beta^{-(1-\eta)/2} \tag*{(131)} ∣ h ∣ ≤ 1/2 max ζ ( 2 1 + β + i ( t + h ) ) − 1/2 ≥ β − ( 1 − η ) /2 (131) with probability 1 − o ( 1 ) 1-o(1) 1 − o ( 1 ) ∣ h ∣ ≤ 1 / 2 |h|\leq1/2 ∣ h ∣ ≤ 1/2
∣ ζ ( 1 2 + β + i ( t + h ) ) ∣ − 1 ≥ β − ( 1 − η ) , \left|\zeta\left(\frac{1}{2}+\beta+i(t+h)\right)\right|^{-1}\geq\beta^{-(1-\eta)}, ζ ( 2 1 + β + i ( t + h ) ) − 1 ≥ β − ( 1 − η ) , and put
z = s − ( 1 2 + α + i ( t + h ) ) , g ( s ) = exp ( 32 log log T z ( z − ( 3 / 2 − α ) ) ) ζ ( s ) . z=s-\left(\frac{1}{2}+\alpha+i(t+h)\right),\qquad g(s)=\frac{\exp(32\log\log T\,z(z-(3/2-\alpha)))}{\zeta(s)}. z = s − ( 2 1 + α + i ( t + h ) ) , g ( s ) = ζ ( s ) exp ( 32 log log T z ( z − ( 3/2 − α ))) . This function is holomorphic on a neighbourhood of R t \mathcal{R}_t R t z = x + i y z=x+iy z = x + i y
ℜ ( z ( z − ( 3 / 2 − α ) ) ) = x ( x − ( 3 / 2 − α ) ) − y 2 . \Re\left(z(z-(3/2-\alpha))\right)=x(x-(3/2-\alpha))-y^2. ℜ ( z ( z − ( 3/2 − α )) ) = x ( x − ( 3/2 − α )) − y 2 . On the left and right sides, x = 0 x=0 x = 0 x = 3 / 2 − α x=3/2-\alpha x = 3/2 − α ∣ g ( s ) ∣ ≤ ζ ( 2 ) |g(s)|\leq\zeta(2) ∣ g ( s ) ∣ ≤ ζ ( 2 ) ∣ y ∣ ≥ 1 / 2 |y|\geq1/2 ∣ y ∣ ≥ 1/2 0 ≤ x ≤ 3 / 2 − α 0\leq x\leq3/2-\alpha 0 ≤ x ≤ 3/2 − α ( log T ) − 8 (\log T)^{-8} ( log T ) − 8 1 / ∣ ζ ( s ) ∣ ≤ ( log T ) 6 1/|\zeta(s)|\leq(\log T)^6 1/∣ ζ ( s ) ∣ ≤ ( log T ) 6 ∣ g ( s ) ∣ ≤ ( log T ) − 2 |g(s)|\leq(\log T)^{-2} ∣ g ( s ) ∣ ≤ ( log T ) − 2 s = 1 / 2 + β + i ( t + h ) s=1/2+\beta+i(t+h) s = 1/2 + β + i ( t + h )
∣ g ( s ) ∣ = e − 32 log log T ( β − α ) ( 3 / 2 − β ) ∣ ζ ( s ) ∣ − 1 ≥ e − 48 β log log T β − ( 1 − η ) ⟶ ∞ . |g(s)|=e^{-32\log\log T(\beta-\alpha)(3/2-\beta)}|\zeta(s)|^{-1}\ge e^{-48\beta\log\log T}\beta^{-(1-\eta)}\longrightarrow\infty. ∣ g ( s ) ∣ = e − 32 l o g l o g T ( β − α ) ( 3/2 − β ) ∣ ζ ( s ) ∣ − 1 ≥ e − 48 β l o g l o g T β − ( 1 − η ) ⟶ ∞. The maximum modulus principle therefore forces a value at least this large on the left side. Since the exponential factor has modulus at most 1 there, we conclude that
max ∣ h ∣ ≤ 1 ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 1 ≥ e − 48 β log log T β − ( 1 − η ) . \max_{|h|\le1}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-1}\ge e^{-48\beta\log\log T}\beta^{-(1-\eta)}. ∣ h ∣ ≤ 1 max ζ ( 2 1 + α + i ( t + h ) ) − 1 ≥ e − 48 β l o g l o g T β − ( 1 − η ) . Here β log log T = o ( 1 ) \beta\log\log T=o(1) β log log T = o ( 1 )
M α , 0 ( t ) ≤ e 48 β log log T β 1 − η ≤ 2 ( log T ) − ( 1 − η ) 2 ≤ α 1 − δ . M_{\alpha,0}(t)\le e^{48\beta\log\log T}\beta^{1-\eta}\le2(\log T)^{-(1-\eta)^2}\le\alpha^{1-\delta}. M α , 0 ( t ) ≤ e 48 β l o g l o g T β 1 − η ≤ 2 ( log T ) − ( 1 − η ) 2 ≤ α 1 − δ . for sufficiently large T T T ( 1 − η ) 2 > 1 − δ (1-\eta)^2>1-\delta ( 1 − η ) 2 > 1 − δ α ≥ M 0 / log T \alpha\ge M_0/\log T α ≥ M 0 / log T
Moments in short intervals
Proof of Theorem 1.9 . For simplicity of notation, let
Z k ( t ) = 1 2 H ∫ ∣ h ∣ ≤ H ∣ ζ ( 1 2 + α + i ( t + h ) ) ∣ − 2 k d h . Z_k(t)=\frac{1}{2H}\int_{|h|\le H}\left|\zeta\left(\frac{1}{2}+\alpha+i(t+h)\right)\right|^{-2k}\,dh. Z k ( t ) = 2 H 1 ∫ ∣ h ∣ ≤ H ζ ( 2 1 + α + i ( t + h ) ) − 2 k d h . Let
S t ( V ) = 1 2 H meas { ∣ h ∣ ≤ H : W α ( t + h ) > V } , S_t(V)=\frac{1}{2H}\operatorname{meas}\left\{|h|\le H:W_\alpha(t+h)>V\right\}, S t ( V ) = 2 H 1 meas { ∣ h ∣ ≤ H : W α ( t + h ) > V } , and note that we have
Z k ( t ) = 2 k ∫ − ∞ ∞ e 2 k V S t ( V ) d V . Z_k(t)=2k\int_{-\infty}^{\infty}e^{2kV}S_t(V)\,dV. Z k ( t ) = 2 k ∫ − ∞ ∞ e 2 k V S t ( V ) d V . Since
2 k ∫ − ∞ 0 e 2 k V S t ( V ) d V ≤ 1 , 2k\int_{-\infty}^{0}e^{2kV}S_t(V)\,dV\le1, 2 k ∫ − ∞ 0 e 2 k V S t ( V ) d V ≤ 1 , we get that
Z k ( t ) ≤ 1 + 2 k ∫ 0 ∞ e 2 k V S t ( V ) d V . Z_k(t)\le1+2k\int_{0}^{\infty}e^{2kV}S_t(V)\,dV. Z k ( t ) ≤ 1 + 2 k ∫ 0 ∞ e 2 k V S t ( V ) d V . Now pick any 0 < a < min { k / 2 , 1 / ( 4 k ) } 0<a<\min\{k/2,1/(4k)\} 0 < a < min { k /2 , 1/ ( 4 k )} b > a b>a b > a
b > max { k , 1 + 2 γ c 0 + M 2 } + 2. b>\max\left\{k,\sqrt{1+\frac{2\gamma}{c_0}+\frac{M}{2}}\right\}+2. b > max { k , 1 + c 0 2 γ + 2 M } + 2. We write
2 k ∫ 0 ∞ e 2 k V S t ( V ) d V = 2 k ∫ 0 a log 1 α e 2 k V S t ( V ) d V + 2 k ∫ a log 1 α b log 1 α e 2 k V S t ( V ) d V + 2 k ∫ b log 1 α ∞ e 2 k V S t ( V ) d V . (132) 2k\int_{0}^{\infty}e^{2kV}S_t(V)\,dV
=2k\int_{0}^{a\log\frac{1}{\alpha}}e^{2kV}S_t(V)\,dV
+2k\int_{a\log\frac{1}{\alpha}}^{b\log\frac{1}{\alpha}}e^{2kV}S_t(V)\,dV
+2k\int_{b\log\frac{1}{\alpha}}^{\infty}e^{2kV}S_t(V)\,dV. \tag*{(132)} 2 k ∫ 0 ∞ e 2 k V S t ( V ) d V = 2 k ∫ 0 a l o g α 1 e 2 k V S t ( V ) d V + 2 k ∫ a l o g α 1 b l o g α 1 e 2 k V S t ( V ) d V + 2 k ∫ b l o g α 1 ∞ e 2 k V S t ( V ) d V . (132) Note that
2 k ∫ 0 a log 1 α e 2 k V S t ( V ) d V ≤ e 2 k a log ( 1 / α ) ≤ α − k 2 , (133) 2k\int_{0}^{a\log\frac{1}{\alpha}}e^{2kV}S_t(V)\,dV
\le e^{2ka\log(1/\alpha)}
\le\alpha^{-k^2}, \tag*{(133)} 2 k ∫ 0 a l o g α 1 e 2 k V S t ( V ) d V ≤ e 2 k a l o g ( 1/ α ) ≤ α − k 2 , (133) where we bounded S t ( V ) ≤ 1 S_t(V)\le1 S t ( V ) ≤ 1 a < k / 2 a<k/2 a < k /2
Now using Theorem 9.2 (see equations (115) and (116) }), outside on a set of measure ≪ T / A \ll T/A ≪ T / A
max ∣ h ∣ ≤ ( log T ) γ W α ( t + h ) < k γ log 1 α + C log log 1 α + log A 2 k γ . \max_{|h|\le(\log T)^\gamma}W_\alpha(t+h)
<k_\gamma\log\frac{1}{\alpha}
+C\log\log\frac{1}{\alpha}
+\frac{\log A}{2k_\gamma}. ∣ h ∣ ≤ ( l o g T ) γ max W α ( t + h ) < k γ log α 1 + C log log α 1 + 2 k γ log A . for C > 0 C > 0 C > 0 b b b
max ∣ h ∣ ≤ ( log T ) γ W α ( t + h ) < b log 1 α , \max_{|h|\le(\log T)^\gamma} W_\alpha(t+h) < b\log\frac{1}{\alpha}, ∣ h ∣ ≤ ( l o g T ) γ max W α ( t + h ) < b log α 1 , for all t t t ≪ T / A \ll T/A ≪ T / A t t t V > b log 1 α V>b\log\frac{1}{\alpha} V > b log α 1
S t ( V ) = 0 , S_t(V)=0, S t ( V ) = 0 , and hence
2 k ∫ b log 1 α ∞ e 2 k V S t ( V ) d V = 0. 2k\int_{b\log\frac{1}{\alpha}}^\infty e^{2kV}S_t(V)\,dV=0. 2 k ∫ b l o g α 1 ∞ e 2 k V S t ( V ) d V = 0. We now focus on the middle term in (10.1). From Theorem 1.7 we have that
1 T ∫ T 2 T S t ( V ) d t = 1 2 T H ∫ T 2 T ∫ − H H 1 ( W α ( t + h ) > V ) d h d t ≤ 1 T meas { u ∈ [ T − H , 2 T + H ] : W α ( u ) > V } ≪ 1 log ( 1 / α ) exp ( − V 2 log ( 1 / α ) ) . (134) \begin{aligned}
\frac{1}{T}\int_T^{2T}S_t(V)\,dt=\frac{1}{2TH}\int_T^{2T}\int_{-H}^{H}\mathbf{1}(W_\alpha(t+h)>V)\,dh\,dt \\
&\le\frac{1}{T}\operatorname{meas}\{u\in[T-H,2T+H]:W_\alpha(u)>V\} \\
&\ll\frac{1}{\sqrt{\log(1/\alpha)}}\exp\left(-\frac{V^2}{\log(1/\alpha)}\right). \tag*{(134)}
\end{aligned} T 1 ∫ T 2 T S t ( V ) d t = 2 T H 1 ∫ T 2 T ∫ − H H 1 ( W α ( t + h ) > V ) d h d t ≤ T 1 meas { u ∈ [ T − H , 2 T + H ] : W α ( u ) > V } ≪ log ( 1/ α ) 1 exp ( − log ( 1/ α ) V 2 ) . (134) By Fubini’s theorem and (10.3), we obtain that
1 T ∫ T 2 T ( 2 k ∫ a log 1 α b log 1 α e 2 k V S t ( V ) d V ) d t ≪ 1 log ( 1 / α ) ∫ a log 1 α b log 1 α exp ( 2 k V − V 2 log ( 1 / α ) ) d V ≤ α − k 2 log ( 1 / α ) ∫ R exp ( − ( V − k log ( 1 / α ) ) 2 log ( 1 / α ) ) d V ≪ α − k 2 . \begin{aligned}
\frac{1}{T}\int_T^{2T}\left(2k\int_{a\log\frac{1}{\alpha}}^{b\log\frac{1}{\alpha}}e^{2kV}S_t(V)\,dV\right)\,dt
\ll\frac{1}{\sqrt{\log(1/\alpha)}}\int_{a\log\frac{1}{\alpha}}^{b\log\frac{1}{\alpha}}\exp\left(2kV-\frac{V^2}{\log(1/\alpha)}\right)\,dV \\
&\le\frac{\alpha^{-k^2}}{\sqrt{\log(1/\alpha)}}\int_{\mathbb{R}}\exp\left(-\frac{(V-k\log(1/\alpha))^2}{\log(1/\alpha)}\right)\,dV\ll\alpha^{-k^2}.
\end{aligned} T 1 ∫ T 2 T ( 2 k ∫ a l o g α 1 b l o g α 1 e 2 k V S t ( V ) d V ) d t ≪ log ( 1/ α ) 1 ∫ a l o g α 1 b l o g α 1 exp ( 2 k V − log ( 1/ α ) V 2 ) d V ≤ log ( 1/ α ) α − k 2 ∫ R exp ( − log ( 1/ α ) ( V − k log ( 1/ α ) ) 2 ) d V ≪ α − k 2 . By Markov’s inequality we get that, outside a set of measure ≪ T / A \ll T/A ≪ T / A
2 k ∫ a log 1 α b log 1 α e 2 k V S t ( V ) d V ≪ A α − k 2 . 2k\int_{a\log\frac{1}{\alpha}}^{b\log\frac{1}{\alpha}}e^{2kV}S_t(V)\,dV\ll A\alpha^{-k^2}. 2 k ∫ a l o g α 1 b l o g α 1 e 2 k V S t ( V ) d V ≪ A α − k 2 . Combining with (10.2) we have
Z k ( t ) ≪ A α − k 2 , Z_k(t)\ll A\alpha^{-k^2}, Z k ( t ) ≪ A α − k 2 , except for on a set of measure ≪ T / A \ll T/A ≪ T / A
Now suppose that k > k γ k>k_\gamma k > k γ
U : = k γ log 1 α + C log log 1 α + log A 2 k γ . U:=k_\gamma\log\frac{1}{\alpha}+C\log\log\frac{1}{\alpha}+\frac{\log A}{2k_\gamma}. U := k γ log α 1 + C log log α 1 + 2 k γ log A . Since
exp ( 2 k V − V 2 log ( 1 / α ) ) ≤ α − k γ 2 e 2 ( k − k γ ) V , \exp\left(2kV-\frac{V^2}{\log(1/\alpha)}\right)\le\alpha^{-k_\gamma^2}e^{2(k-k_\gamma)V}, exp ( 2 k V − log ( 1/ α ) V 2 ) ≤ α − k γ 2 e 2 ( k − k γ ) V , equation (10.3), Fubini’s theorem, and Markov’s inequality, together with the fact that 2 a k < 1 / 2 2ak<1/2 2 ak < 1/2
Z k ( t ) ≪ α − 1 / 2 + A α − k γ 2 log ( 1 / α ) ∫ a log 1 α U e 2 ( k − k γ ) V d V ≪ α − 1 / 2 + A α − k γ 2 e 2 ( k − k γ ) U ( k − k γ ) log ( 1 / α ) , Z_k(t)\ll\alpha^{-1/2}+\frac{A\alpha^{-k_\gamma^2}}{\sqrt{\log(1/\alpha)}}\int_{a\log\frac{1}{\alpha}}^Ue^{2(k-k_\gamma)V}\,dV
\ll\alpha^{-1/2}+\frac{A\alpha^{-k_\gamma^2}e^{2(k-k_\gamma)U}}{(k-k_\gamma)\sqrt{\log(1/\alpha)}}, Z k ( t ) ≪ α − 1/2 + log ( 1/ α ) A α − k γ 2 ∫ a l o g α 1 U e 2 ( k − k γ ) V d V ≪ α − 1/2 + ( k − k γ ) log ( 1/ α ) A α − k γ 2 e 2 ( k − k γ ) U , outside a set of measure ≪ T / A \ll T/A ≪ T / A U U U
Department of Mathematics, University of Manchester, Manchester M13 9PL, UK Email address: hung.bui@manchester.ac.uk
UC Irvine, Mathematics Department, Rowland Hall, Irvine 92697, USA Email address: floreaa@uci.edu
Department of Mathematics, University of Manchester, Manchester M13 9PL, UK Email address: andreheyсock@gmail.com
