Introduction
Main result
A Rademacher random completely multiplicative function f : N → { − 1 , 1 } f : \mathbb{N} \to\{-1,1\} f : N → { − 1 , 1 } is obtained by taking ( f ( p ) ) p (f(p))_p ( f ( p ) ) p to be independent Rademacher random variables and extending f f f completely multiplicatively. Throughout the paper, we refer to this model as the extended Rademacher RMF. The more familiar Rademacher RMF is supported only on squarefree integers, and it goes back to Wintner [32 ] , who introduced it as a probabilistic model for the Möbius function. Although it is clear nowadays that the Rademacher RMF does not capture much behavior of Möbius function, we believe that the extended Rademacher RMF is a good model to predict the typical behavior of Dirichlet quadratic characters. Write f ( p ) = ϵ p f(p)=\epsilon_p f ( p ) = ϵ p and
S f ( x ) = ∑ n ≤ x f ( n ) . S_f(x)=\sum_{n \le x} f(n). S f ( x ) = n ≤ x ∑ f ( n ) .
In this paper, our purpose is to prove a quantitative two-sided fluctuation result for S f ( x ) S_f(x) S f ( x ) . One can distinguish this model from the Rademacher RMF g ( n ) g(n) g ( n ) supported on squarefree integers, by noting that g ( n ) = μ 2 ( n ) f ( n ) g(n)=\mu^2(n)f(n) g ( n ) = μ 2 ( n ) f ( n ) , where μ \mu μ is the Möbius function, so that g ( n ) = 0 g(n)=0 g ( n ) = 0 unless n n n is squarefree. The distinction is substantial. The variables g ( n ) g(n) g ( n ) , indexed by squarefree n n n , has perfect orthogonality, whereas E f ( n ) \mathbb{E}f(n) E f ( n ) equals 1 1 1 when n n n is a square and equals 0 0 0 otherwise; in particular, E S f ( x ) = ⌊ x ⌋ \mathbb{E}S_f(x)=\lfloor\sqrt{x}\rfloor E S f ( x ) = ⌊ x ⌋ . The two Dirichlet series are related by
∑ n ≥ 1 f ( n ) n s = ∏ p ( 1 − ϵ p p s ) − 1 = ζ ( 2 s ) ∏ p ( 1 + ϵ p p s ) , Re s > 1. (1) \sum_{n \ge1}\frac{f(n)}{n^s}=\prod_p\left(1-\frac{\epsilon_p}{p^s}\right)^{-1}=\zeta(2s)\prod_p\left(1+\frac{\epsilon_p}{p^s}\right),\qquad\operatorname{Re}s>1. \tag*{(1)} n ≥ 1 ∑ n s f ( n ) = p ∏ ( 1 − p s ϵ p ) − 1 = ζ ( 2 s ) p ∏ ( 1 + p s ϵ p ) , Re s > 1. (1)
Thus the completely multiplicative property introduces a zeta factor with a pole at s = 1 / 2 s=1/2 s = 1/2 . This pole is the analytic expression of the contribution from squares and can be a principal obstruction to transferring some results for g g g directly to f f f .
The study of the large fluctuations of the partial sums of RMFs has been very active. In the Rademacher case, Wintner [32 ] and Erdős [12 ] obtained some of the earliest results in
2020 Mathematics Subject Classification. Primary 11K65; Secondary 11N37, 60F15.
Key words and phrases: extended Rademacher random multiplicative functions, partial sums, multiplicative chaos, large fluctuations. this direction, and Halász [15 ] later obtained an important almost sure Ω \Omega Ω -result. Harper [17 ] subsequently improved Halász’s result by means of Gaussian comparison. His later work proved that, for every deterministic function V ( x ) V(x) V ( x ) tending to infinity, almost surely there are arbitrarily large x x x for which
∣ ∑ n ≤ x g ( n ) ∣ ≥ x ( log log x ) 1 / 4 V ( x ) . (2) \left|\sum_{n\le x} g(n)\right| \ge\frac{\sqrt{x}(\log\log x)^{1/4}}{V(x)}. \tag*{(2)} n ≤ x ∑ g ( n ) ≥ V ( x ) x ( log log x ) 1/4 . (2)
He also proved the corresponding result for the Steinhaus RMF; see [20 ] . This was the first almost sure lower bound for the partial sums which exceeds x \sqrt{x} x by a factor tending to infinity. The proof conditions on the values at the smaller primes and applies a multivariate Gaussian approximation to the contribution from the remaining large primes. Its main difficulty is to show that sufficiently many of the resulting conditional covariances are small. Although the theorem is stated in the absolute value form (1.2), in the Rademacher case the conditional symmetry of the large prime vector gives the corresponding large fluctuations of either sign.
Harper further suggested that the exponent 1 / 4 1/4 1/4 in (1.2) should be optimal. More precisely, the conjectured upper bound is that, for every ε > 0 \varepsilon> 0 ε > 0 , almost surely
∑ n ≤ x g ( n ) ≪ ε x ( log log x ) 1 / 4 + ε . (3) \sum_{n\le x} g(n) \ll_{\varepsilon} \sqrt{x}(\log\log x)^{1/4+\varepsilon}. \tag*{(3)} n ≤ x ∑ g ( n ) ≪ ε x ( log log x ) 1/4 + ε . (3)
One heuristic for this exponent comes from the low moments of the partial sums. The results of Harper [19 ] show that their typical size is smaller than x \sqrt{x} x by a factor of order ( log log x ) 1 / 4 (\log\log x)^{1/4} ( log log x ) 1/4 . Combining this with the usual log log x \sqrt{\log\log x} log log x correction suggested by the law of the iterated logarithm leads precisely to the scale x ( log log x ) 1 / 4 \sqrt{x}(\log\log x)^{1/4} x ( log log x ) 1/4 . The relevant low and high moment estimates for the underlying random Euler products are developed in [18 , 19 ] .
We give a quick summary of the substantial progress on the upper bounds: see, for example, [7 , 9 , 15 , 26 , 27 ] and the references therein. Very recently, Durkan and Pearce-Crump [10 ] obtained the matching upper bound
∑ n ≤ x g ( n ) ≪ ε , g x ( log log x ) 1 / 4 + ε (4) \sum_{n\le x} g(n) \ll_{\varepsilon,g} \sqrt{x}(\log\log x)^{1/4+\varepsilon} \tag*{(4)} n ≤ x ∑ g ( n ) ≪ ε , g x ( log log x ) 1/4 + ε (4)
almost surely.
They proved the same result for the Steinhaus RMF. Together with Harper’s lower bound, this determines the optimal iterated logarithmic exponent in both classical RMFs and settles Harper’s conjecture. In an independent work, Verreault [31 ] obtained the sharper estimate
∑ n ≤ x g ( n ) ≪ ε , g x ( log 2 x ) 1 / 4 log 3 x log 4 x ( log 5 x ) 1 / 2 + ε (5) \sum_{n\le x} g(n) \ll_{\varepsilon,g} \sqrt{x}(\log_2 x)^{1/4}\log_3 x\log_4 x(\log_5 x)^{1/2+\varepsilon} \tag*{(5)} n ≤ x ∑ g ( n ) ≪ ε , g x ( log 2 x ) 1/4 log 3 x log 4 x ( log 5 x ) 1/2 + ε (5)
almost surely,
for both the Rademacher and Steinhaus RMFs, where log k \log_k log k denotes the k k k -fold iterated logarithm. We also note that in the weighted case, Atherfold [2 ] proved that almost surely
∑ n ≤ x g ( n ) n ≪ ( log log x ) 3 / 4 + ε , \sum_{n\le x}\frac{g(n)}{\sqrt{n}} \ll(\log\log x)^{3/4+\varepsilon}, n ≤ x ∑ n g ( n ) ≪ ( log log x ) 3/4 + ε ,
and obtained the sharper exponent 1 / 4 + ε 1/4+\varepsilon 1/4 + ε under the restriction P + ( n ) > x P^{+}(n)>\sqrt{x} P + ( n ) > x . He conjectured that this exponent is optimal. He also showed that almost surely
S f ( x ) ≪ x ( log x ) 1 + ε S_f(x)\ll\sqrt{x}(\log x)^{1+\varepsilon} S f ( x ) ≪ x ( log x ) 1 + ε
for extended Rademacher RMFs. See also the work by Hardy on the Steinhaus case [16 ] .
It is in general the case that the three standard RMFs may exhibit different levels of difficulty and the Steinhaus case is usually the least technical among the three. So far, we in general know less in the Rademacher case, while the extended Rademacher RMF is arguably even more challenging. This is indeed the case in the fluctuation problem considered here. As we described before, large fluctuations in both the Rademacher and Steinhaus RMFs have been known since the work of Harper [20 ] , whereas no corresponding result was previously available for the extended Rademacher RMF. In fact, even the weaker original question of Erdős remained open1 . The formulation in the 1999 problem collection Some of Paul’s favorite problems asks whether lim sup x → ∞ ∣ S f ( x ) ∣ / x = + ∞ \limsup_{x\to\infty} |S_f(x)|/\sqrt{x}=+\infty lim sup x → ∞ ∣ S f ( x ) ∣/ x = + ∞ almost surely, while the current online formulation, recorded as Erdős Problem #1144, asks the stronger one-sided question whether lim sup x → ∞ S f ( x ) / x = + ∞ \limsup_{x\to\infty} S_f(x)/\sqrt{x}=+\infty lim sup x → ∞ S f ( x ) / x = + ∞ almost surely [8 , 30 ] . Our main theorem gives a much stronger quantitative two-sided strengthening of both formulations.
Theorem 1.1. Let f f f be an extended Rademacher RMF. Then for any fixed deterministic function V ( x ) → ∞ V(x)\to\infty V ( x ) → ∞ , we have, almost surely
S f ( x ) = Ω ± ( x ( log log x ) 1 / 4 V ( x ) ) . (6) S_f(x)=\Omega_{\pm}\left(\frac{\sqrt{x}(\log\log x)^{1/4}}{V(x)}\right). \tag*{(6)} S f ( x ) = Ω ± ( V ( x ) x ( log log x ) 1/4 ) . (6)
Here A ( x ) = Ω ± ( B ( x ) ) A(x)=\Omega_{\pm}(B(x)) A ( x ) = Ω ± ( B ( x )) means that lim sup x → ∞ A ( x ) / B ( x ) > 0 \limsup_{x\to\infty} A(x)/B(x)>0 lim sup x → ∞ A ( x ) / B ( x ) > 0 and lim inf x → ∞ A ( x ) / B ( x ) < 0 \liminf_{x\to\infty} A(x)/B(x)<0 lim inf x → ∞ A ( x ) / B ( x ) < 0 for an eventually positive function B ( x ) B(x) B ( x ) .
In particular, the positive limsup assertion implies lim sup x → ∞ S f ( x ) / x = + ∞ \limsup_{x\to\infty} S_f(x)/\sqrt{x}=+\infty lim sup x → ∞ S f ( x ) / x = + ∞ , answering Erdős Problem #1144 affirmatively in the stronger form.
Remark 1.2. It is worth pointing out that our argument may work for a more general model than the extended Rademacher RMF considered here. What is essentially used is the Euler product factorisation
∑ n ≥ 1 P + ( n ) ≤ X f ( n ) n s = F X ( s ) Z X ( 2 s ) R X ( s ) , \sum_{\substack{n\geq1\\P^{+}(n)\leq X}}\frac{f(n)}{n^s}=F_X(s)Z_X(2s)R_X(s), n ≥ 1 P + ( n ) ≤ X ∑ n s f ( n ) = F X ( s ) Z X ( 2 s ) R X ( s ) ,
where F X ( s ) = ∏ p ≤ X ( 1 + ε p p − s ) F_X(s)=\prod_{p\leq X}(1+\varepsilon_p p^{-s}) F X ( s ) = ∏ p ≤ X ( 1 + ε p p − s ) is the squarefree Rademacher Euler product, Z X ( s ) = ∏ p ≤ X ( 1 − p − s ) − 1 Z_X(s)=\prod_{p\leq X}(1-p^{-s})^{-1} Z X ( s ) = ∏ p ≤ X ( 1 − p − s ) − 1 and R X R_X R X is a “well-behaved” factor (in our case R X ( s ) ≡ 1 R_X(s)\equiv1 R X ( s ) ≡ 1 ). Under some suitable moment and regularity assumptions on R X R_X R X , the same argument applies more generally. For instance, the m m m -free extended Rademacher RMF has factorisation
F X ( s ) Z X ( 2 s ) ∏ p ≤ X ( 1 − ε p m p − m s ) . F_X(s)Z_X(2s)\prod_{p\leq X}(1-\varepsilon_p^m p^{-ms}). F X ( s ) Z X ( 2 s ) p ≤ X ∏ ( 1 − ε p m p − m s ) .
For m ≥ 3 m\geq3 m ≥ 3 the last product is regular on ℜ s = 1 / 2 \Re s=1/2 ℜ s = 1/2 , whereas for m = 2 m=2 m = 2 it cancels Z X ( 2 s ) Z_X(2s) Z X ( 2 s ) completely, giving the usual Rademacher RMF. Actually, by combining Harper [20 ] ’s original argument with our conditional symmetry observation used in Section 8, one can obtain the Rademacher RMF analogues of Theorems 1.1 and 1.4.
Another related family is obtained by taking f ( p ) f(p) f ( p ) uniformly among the k k k -th roots of unity and extending completely multiplicatively. Its Euler product decomposition contains the deterministic factor Z X ( k s ) Z_X(ks) Z X ( k s ) . Thus k = 2 k=2 k = 2 is the critical case in which the associated zeta singularity occurs at s = 1 / 2 s=1/2 s = 1/2 , while for k ≥ 3 k\geq3 k ≥ 3 it lies to the left of the critical line; as k → ∞ k\to\infty k → ∞ , the uniform distribution on the k k k -th roots of unity converges weakly to the uniform distribution on the unit circle, corresponding to the Steinhaus RMF. We do not pursue the details of these extensions here.
Remark 1.3. We also remark that our bound should be sharp up to ( log log x ) o ( 1 ) (\log\log x)^{o(1)} ( log log x ) o ( 1 ) . One might be able to prove the corresponding upper bound by combining some techniques here and arguments of Caich [9 ] with the recent developments [10 , 31 ] .
It is worth noting that, very recently, Atherfold, Gerspach and Hamdan [3 ] announced their results on low moments of extended Rademacher RMFs. They show that the typical size can behave very differently with the additional completely multiplicative property, compared with the usual Rademacher case. While our result shows that its influence on large fluctuations is more limited.
Following the general strategy of Harper [20 ] , we first prove a conditional fluctuation estimate for X 8 / 7 ≤ x ≤ X 4 / 3 X^{8/7} \le x \le X^{4/3} X 8/7 ≤ x ≤ X 4/3 . For X ≥ 3 X \ge3 X ≥ 3 let
F X = σ ( f ( p ) : p ≤ X ) , P X ( ⋅ ) = P ( ⋅ ∣ F X ) . \mathcal{F}_X = \sigma(f(p) : p \le X), \qquad\mathbb{P}_X(\cdot) = \mathbb{P}(\cdot\mid\mathcal{F}_X). F X = σ ( f ( p ) : p ≤ X ) , P X ( ⋅ ) = P ( ⋅ ∣ F X ) .
Thus F X \mathcal{F}_X F X records the small prime variables, while conditional on F X \mathcal{F}_X F X the variables ( f ( p ) ) p > X (f(p))_{p>X} ( f ( p ) ) p > X remain independent Rademacher variables. Using the conditional Borel–Cantelli lemma (see Lemma 2.2 below), Theorem 1.1 follows from the following result.
Theorem 1.4. Let f f f be an extended Rademacher RMF. There is an absolute constant W 0 > 0 W_0 > 0 W 0 > 0 such that for all sufficiently large X X X and all W 0 ≤ W ≤ 1 30 log log log X W_0 \le W \le\frac{1}{30}\log\log\log X W 0 ≤ W ≤ 30 1 log log log X , we have
P X ( max X 8 / 7 ≤ x ≤ X 4 / 3 + 1 x ∑ n ≤ x f ( n ) ≥ ( log log X ) 1 / 4 e 1.2 W ) ≥ 1 3 . \mathbb{P}_X\left(\max_{X^{8/7}\le x\le X^{4/3}} \frac{+1}{\sqrt{x}}\sum_{n\le x} f(n) \ge\frac{(\log\log X)^{1/4}}{e^{1.2W}}\right) \ge\frac{1}{3}. P X ( X 8/7 ≤ x ≤ X 4/3 max x + 1 n ≤ x ∑ f ( n ) ≥ e 1.2 W ( log log X ) 1/4 ) ≥ 3 1 .
with probability ≥ 1 − O ( e − W / 10 ) \ge1-O(e^{-W/10}) ≥ 1 − O ( e − W /10 ) . The same result with + 1 +1 + 1 replaced by − 1 -1 − 1 holds.
Proof of Theorem 1.1 assuming Theorem 1.4. We prove the positive assertion; the negative assertion follows in exactly the same way.
Since V ( x ) → ∞ V(x)\to\infty V ( x ) → ∞ , we may choose a deterministic increasing sequence X ℓ → ∞ X_\ell\to\infty X ℓ → ∞ such that, for every sufficiently large ℓ \ell ℓ ,
X ℓ 4 / 3 < X ℓ + 1 , W 0 ≤ W = 20 log ℓ ≤ 1 30 log log log X ℓ , X_\ell^{4/3}<X_{\ell+1}, \qquad W_0\le W=20\log\ell\le\frac{1}{30}\log\log\log X_\ell, X ℓ 4/3 < X ℓ + 1 , W 0 ≤ W = 20 log ℓ ≤ 30 1 log log log X ℓ ,
and V ( x ) ≥ ℓ 24 V(x)\ge\ell^{24} V ( x ) ≥ ℓ 24 for every x ≥ X ℓ 8 / 7 x\ge X_\ell^{8/7} x ≥ X ℓ 8/7 .
Let
E ℓ = { max X ℓ 8 / 7 ≤ x ≤ X ℓ 4 / 3 1 x ∑ n ≤ x f ( n ) ≥ ( log log X ℓ ) 1 / 4 ℓ 24 } . \mathcal{E}_\ell=\left\{\max_{X_\ell^{8/7}\le x\le X_\ell^{4/3}}\frac{1}{\sqrt{x}}\sum_{n\le x}f(n)\ge\frac{(\log\log X_\ell)^{1/4}}{\ell^{24}}\right\}. E ℓ = { X ℓ 8/7 ≤ x ≤ X ℓ 4/3 max x 1 n ≤ x ∑ f ( n ) ≥ ℓ 24 ( log log X ℓ ) 1/4 } .
By Theorem 1.4, with W = 20 log ℓ W=20\log\ell W = 20 log ℓ , we have
P ( P ( E ℓ ∣ F X ℓ ) < 1 / 3 ) ≪ e − 2 log ℓ = ℓ − 2 , \mathbb{P}\left(\mathbb{P}(\mathcal{E}_\ell\mid\mathcal{F}_{X_\ell})<1/3\right)\ll e^{-2\log\ell}=\ell^{-2}, P ( P ( E ℓ ∣ F X ℓ ) < 1/3 ) ≪ e − 2 l o g ℓ = ℓ − 2 ,
where the outer probability is taken over all realizations of the small primes ( f ( p ) ) p ≤ X ℓ (f(p))_{p\le X_\ell} ( f ( p ) ) p ≤ X ℓ . Since ∑ ℓ ≥ 2 ℓ − 2 < ∞ \sum_{\ell\ge2}\ell^{-2}<\infty ∑ ℓ ≥ 2 ℓ − 2 < ∞ , the first Borel–Cantelli lemma shows that almost surely P ( E ℓ ∣ F X ℓ ) ≥ 1 / 3 \mathbb{P}(\mathcal{E}_\ell\mid\mathcal{F}_{X_\ell})\ge1/3 P ( E ℓ ∣ F X ℓ ) ≥ 1/3 for every sufficiently large ℓ \ell ℓ . The condition X ℓ 4 / 3 < X ℓ + 1 X_\ell^{4/3}<X_{\ell+1} X ℓ 4/3 < X ℓ + 1 implies that E ℓ ∈ F X ℓ + 1 \mathcal{E}_\ell\in\mathcal{F}_{X_{\ell+1}} E ℓ ∈ F X ℓ + 1 . Thus,
∑ ℓ ≥ 2 P ( E ℓ ∣ F X ℓ ) = ∞ almost surely . \sum_{\ell\ge2}\mathbb{P}(\mathcal{E}_\ell\mid\mathcal{F}_{X_\ell})=\infty\quad\text{almost surely}. ℓ ≥ 2 ∑ P ( E ℓ ∣ F X ℓ ) = ∞ almost surely .
Applying Lemma 2.2 to the filtration ( F X ℓ + 1 ) ℓ ≥ 2 (\mathcal{F}_{X_{\ell+1}})_{\ell\ge2} ( F X ℓ + 1 ) ℓ ≥ 2 shows that E ℓ \mathcal{E}_\ell E ℓ occurs infinitely often almost surely. Every occurrence of E ℓ \mathcal{E}_\ell E ℓ therefore provides a point satisfying (1.6), this completes the proof. □
Note that here we use both the first and the conditional Borel–Cantelli lemmas. This differs from Harper’s deduction [20 ] , which uses only the first Borel–Cantelli lemma. In both arguments, however, no independence assumption is required.
An application to sign changes
We recall some known results on sign changes in the two Rademacher RMFs. For the Rademacher RMF, Aymone et al. [6 ] proved almost surely infinitely many sign changes for the unweighted partial sums, and Aymone [4 ] later proved the corresponding result for weighted sums ∑ n ≤ x g ( n ) / n \sum_{n \le x} g(n)/\sqrt{n} ∑ n ≤ x g ( n ) / n . Quantitative almost sure lower bounds for the number of sign changes were subsequently obtained by Geis and Hiary [13 ] and improved by Klurman et al. [24 ] ; see also Aymone [5 ] for an average lower bound. For the extended Rademacher RMF, Aymone [4 ] proved almost surely infinitely many sign changes for the unweighted partial sums, while Angelo and Xu [1 ] settled the weighted case ∑ n ≤ x f ( n ) / n \sum_{n \le x} f(n)/\sqrt{n} ∑ n ≤ x f ( n ) / n , and also studied the probability that the partial sums remain nonnegative after conditioning on f ( p ) = 1 f(p)=1 f ( p ) = 1 for all sufficiently small primes, answering a question of Kucheriaviy [25 ] . Note that, however, no quantitative almost sure lower bounds on the number of sign changes are currently known for the extended Rademacher RMF. Our Corollary 1.5 provides the first such quantitative lower bound.
For h ∈ { f , g } h \in\{f,g\} h ∈ { f , g } , let N h ( x ) N_h(x) N h ( x ) denote the number of sign changes of the partial sums ∑ n ≤ y h ( n ) \sum_{n \le y} h(n) ∑ n ≤ y h ( n ) when y ∈ [ 1 , x ] y \in[1,x] y ∈ [ 1 , x ] . The conditional form of Theorem 1.4 has a simple consequence for sign changes. The point is that, on each of a sequence of well-separated logarithmic scales, the theorem gives a fixed positive conditional probability of producing a value with either choice of sign. Choosing the sign opposite to the last nonzero partial sum on the current scale then forces a new sign change.
Corollary 1.5. Let h ∈ { f , g } h \in\{f,g\} h ∈ { f , g } . We have, almost surely for all sufficiently large x x x ,
N h ( x ) ≫ log log x . N_h(x) \gg\log\log x. N h ( x ) ≫ log log x .
For the Rademacher RMF this improves the previous almost sure lower bound ≫ log log x / log log log log x \gg\log\log x/\log\log\log\log x ≫ log log x / log log log log x of Klurman et al. [24 ] 1 .
Proof of Corollary 1.5 assuming Theorem 1.4. Let X ℓ = exp ( 2 ℓ ) X_\ell=\exp(2^\ell) X ℓ = exp ( 2 ℓ ) and W ℓ = ( log ℓ ) / 60 W_\ell=(\log\ell)/60 W ℓ = ( log ℓ ) /60 . For all sufficiently large ℓ \ell ℓ , these parameters lie in the range of Theorem 1.4, and X ℓ 4 / 3 < X ℓ + 1 = X ℓ 2 X_\ell^{4/3}<X_{\ell+1}=X_\ell^2 X ℓ 4/3 < X ℓ + 1 = X ℓ 2 . Let Ω ℓ \Omega_\ell Ω ℓ be the F X ℓ \mathcal{F}_{X_\ell} F X ℓ -measurable event on which the two conditional estimates in Theorem 1.4 hold simultaneously. Then P ( Ω ℓ c ) ≪ ℓ − 1 / 600 \mathbb{P}(\Omega_\ell^c)\ll\ell^{-1/600} P ( Ω ℓ c ) ≪ ℓ − 1/600 and, on Ω ℓ \Omega_\ell Ω ℓ , either sign occurs somewhere in [ X ℓ 8 / 7 , X ℓ 4 / 3 ] [X_\ell^{8/7},X_\ell^{4/3}] [ X ℓ 8/7 , X ℓ 4/3 ] with conditional probability at least 1 / 3 1/3 1/3 .
Let E ℓ : = { N f ( X ℓ + 1 ) − N f ( X ℓ ) ≥ 1 } \mathcal{E}_\ell:=\{N_f(X_{\ell+1})-N_f(X_\ell)\ge1\} E ℓ := { N f ( X ℓ + 1 ) − N f ( X ℓ ) ≥ 1 } be the event that S f ( y ) S_f(y) S f ( y ) has a new sign change between X ℓ X_\ell X ℓ and X ℓ + 1 X_{\ell+1} X ℓ + 1 . The sign of the last nonzero value of S f ( y ) S_f(y) S f ( y ) with y ≤ X ℓ y\le X_\ell y ≤ X ℓ is F X ℓ \mathcal{F}_{X_\ell} F X ℓ -measurable. We may therefore apply Theorem 1.4 with the opposite sign, and hence, if q ℓ = P ( E ℓ ∣ F X ℓ ) q_\ell=\mathbb{P}(\mathcal{E}_\ell\mid\mathcal{F}_{X_\ell}) q ℓ = P ( E ℓ ∣ F X ℓ ) , then q ℓ ≥ 1 3 1 Ω ℓ q_\ell\ge\frac{1}{3}\mathbf{1}_{\Omega_\ell} q ℓ ≥ 3 1 1 Ω ℓ .
Since E ℓ ∈ F X ℓ + 1 \mathcal{E}_\ell\in\mathcal{F}_{X_{\ell+1}} E ℓ ∈ F X ℓ + 1 , the variables D ℓ : = 1 E ℓ − q ℓ = 1 E ℓ − E ( 1 E ℓ ∣ F X ℓ ) D_\ell:=\mathbf{1}_{\mathcal{E}_\ell}-q_\ell=\mathbf{1}_{\mathcal{E}_\ell}-\mathbb{E}(\mathbf{1}_{\mathcal{E}_\ell}\mid\mathcal{F}_{X_\ell}) D ℓ := 1 E ℓ − q ℓ = 1 E ℓ − E ( 1 E ℓ ∣ F X ℓ ) form a martingale-difference sequence with respect to the filtration ( F X ℓ + 1 ) ℓ (\mathcal{F}_{X_{\ell+1}})_\ell ( F X ℓ + 1 ) ℓ . Indeed, E ( D ℓ ∣ F X ℓ ) = 0 \mathbb{E}(D_\ell\mid\mathcal{F}_{X_\ell})=0 E ( D ℓ ∣ F X ℓ ) = 0 , and for j < ℓ j<\ell j < ℓ the variable D j D_j D j is F X ℓ \mathcal{F}_{X_\ell} F X ℓ -measurable, whence E ( D j D ℓ ) = 0 \mathbb{E}(D_jD_\ell)=0 E ( D j D ℓ ) = 0 . Thus for any fixed sufficiently large ℓ 0 \ell_0 ℓ 0 ,
E ∣ ∑ ℓ 0 ≤ ℓ ≤ L D ℓ ∣ 2 = ∑ ℓ 0 ≤ ℓ ≤ L E ∣ D ℓ ∣ 2 ≤ L . \mathbb{E}\left|\sum_{\ell_0\le\ell\le L}D_\ell\right|^2
=\sum_{\ell_0\le\ell\le L}\mathbb{E}|D_\ell|^2\le L. E ℓ 0 ≤ ℓ ≤ L ∑ D ℓ 2 = ℓ 0 ≤ ℓ ≤ L ∑ E ∣ D ℓ ∣ 2 ≤ L .
Moreover, E ∑ ℓ 0 ≤ ℓ ≤ L 1 Ω ℓ c ≪ L 1 − 1 / 600 \mathbb{E}\sum_{\ell_0\le\ell\le L}\mathbf{1}_{\Omega_\ell^c}\ll L^{1-1/600} E ∑ ℓ 0 ≤ ℓ ≤ L 1 Ω ℓ c ≪ L 1 − 1/600 . Since q ℓ ≥ 1 3 1 Ω ℓ q_\ell\ge\frac{1}{3}\mathbf{1}_{\Omega_\ell} q ℓ ≥ 3 1 1 Ω ℓ , outside the event that Ω ℓ \Omega_\ell Ω ℓ fails for a positive proportion of ℓ ∈ [ ℓ 0 , L ] \ell\in[\ell_0,L] ℓ ∈ [ ℓ 0 , L ] , we have ∑ ℓ 0 ≤ ℓ ≤ L q ℓ ≫ L \sum_{\ell_0\le\ell\le L}q_\ell\gg L ∑ ℓ 0 ≤ ℓ ≤ L q ℓ ≫ L . On the other hand, we have ∑ ℓ 0 ≤ ℓ ≤ L 1 E ℓ ≤ N f ( X L + 1 ) \sum_{\ell_0\le\ell\le L}\mathbf{1}_{\mathcal{E}_\ell}\le N_f(X_{L+1}) ∑ ℓ 0 ≤ ℓ ≤ L 1 E ℓ ≤ N f ( X L + 1 ) . Thus, if N f ( X L + 1 ) ≤ c L N_f(X_{L+1})\le cL N f ( X L + 1 ) ≤ c L for some sufficiently small absolute constant c > 0 c > 0 c > 0 , then this forces either Ω ℓ \Omega_\ell Ω ℓ fails for a positive proportion of ℓ ∈ [ ℓ 0 , L ] \ell\in[\ell_0,L] ℓ ∈ [ ℓ 0 , L ] or ∣ ∑ ℓ 0 ≤ ℓ ≤ L D ℓ ∣ ≫ L \left|\sum_{\ell_0\leq\ell\leq L}D_\ell\right| \gg L ∑ ℓ 0 ≤ ℓ ≤ L D ℓ ≫ L . Hence
P ( N f ( X L + 1 ) < c L ) ≤ P ( ∑ ℓ 0 ≤ ℓ ≤ L 1 Ω ℓ c ≫ L ) + P ( ∣ ∑ ℓ 0 ≤ ℓ ≤ L D ℓ ∣ ≫ L ) ≪ 1 L E ∑ ℓ 0 ≤ ℓ ≤ L 1 Ω ℓ c + 1 L 2 E ∣ ∑ ℓ 0 ≤ ℓ ≤ L D ℓ ∣ 2 ≪ L − 1 / 600 + L − 1 . \begin{aligned}
\mathbb{P}\left(N_f(X_{L+1}) < cL\right) \leq\mathbb{P}\left(\sum_{\ell_0\leq\ell\leq L}\mathbf{1}_{\Omega_\ell^c} \gg L\right) + \mathbb{P}\left(\left|\sum_{\ell_0\leq\ell\leq L}D_\ell\right| \gg L\right) \\
&\ll\frac{1}{L}\mathbb{E}\sum_{\ell_0\leq\ell\leq L}\mathbf{1}_{\Omega_\ell^c}+\frac{1}{L^2}\mathbb{E}\left|\sum_{\ell_0\leq\ell\leq L}D_\ell\right|^2 \\
&\ll L^{-1/600}+L^{-1}.
\end{aligned} P ( N f ( X L + 1 ) < c L ) ≤ P ( ℓ 0 ≤ ℓ ≤ L ∑ 1 Ω ℓ c ≫ L ) + P ( ℓ 0 ≤ ℓ ≤ L ∑ D ℓ ≫ L ) ≪ L 1 E ℓ 0 ≤ ℓ ≤ L ∑ 1 Ω ℓ c + L 2 1 E ℓ 0 ≤ ℓ ≤ L ∑ D ℓ 2 ≪ L − 1/600 + L − 1 .
Taking L = 2 m L=2^m L = 2 m and applying the first Borel–Cantelli lemma, we obtain almost surely N f ( X 2 m + 1 ) ≫ 2 m N_f(X_{2^m+1})\gg2^m N f ( X 2 m + 1 ) ≫ 2 m for all sufficiently large m m m . If X 2 m + 1 ≤ x < X 2 m + 1 + 1 X_{2^m+1}\leq x<X_{2^{m+1}+1} X 2 m + 1 ≤ x < X 2 m + 1 + 1 , then N f ( x ) ≫ 2 m ≫ log log x N_f(x)\gg2^m\gg\log\log x N f ( x ) ≫ 2 m ≫ log log x for all sufficiently large x x x .
Finally the Rademacher RMF case follows in exactly the same way from the squarefree analogue of Theorem 1.4 noted above. □
Remark 1.6. The proof of Corollary 1.5 uses only the sign information in Theorem 1.4; the size of the fluctuation plays no role. Moreover, even if two large values of opposite signs occur, the partial sums between them may change sign many times, while our argument counts only once. We believe there is a great room to gain more information by modifying our argument in the proof of Theorem 1.4, rather than using it only as a black box.
Outline of the proof of Theorem 1.4
Fix a large integer X X X . Suppose that x < X 2 x<X^2 x < X 2 . Every integer n ≤ x n\leq x n ≤ x has at most one prime factor exceeding X X X , and using multiplicativity we can break up the sums as
1 x ∑ n ≤ x f ( n ) = 1 x ∑ n ≤ x P + ( n ) ≤ X f ( n ) + 1 x ∑ X < p ≤ x f ( p ) ∑ m ≤ x / p P + ( m ) ≤ X f ( m ) . \frac{1}{\sqrt{x}}\sum_{n\leq x}f(n)=\frac{1}{\sqrt{x}}\sum_{\substack{n\leq x\\P^+(n)\leq X}}f(n)+\frac{1}{\sqrt{x}}\sum_{X<p\leq x}f(p)\sum_{\substack{m\leq x/p\\P^+(m)\leq X}}f(m). x 1 n ≤ x ∑ f ( n ) = x 1 n ≤ x P + ( n ) ≤ X ∑ f ( n ) + x 1 X < p ≤ x ∑ f ( p ) m ≤ x / p P + ( m ) ≤ X ∑ f ( m ) .
After conditioning on ( f ( p ) ) p ≤ X (f(p))_{p\leq X} ( f ( p ) ) p ≤ X , the first term is fixed, whereas the second is a centred linear form in the independent Rademacher variables ( f ( p ) ) X < p ≤ x (f(p))_{X<p\leq x} ( f ( p ) ) X < p ≤ x .
The first term is one of the points at which the extended Rademacher RMF differs from the Rademacher RMF. The exact convolution identity f = ( μ 2 f ) ∗ 1 □ f=(\mu^2f)*1_{\square} f = ( μ 2 f ) ∗ 1 □ gives, for x < X 2 x<X^2 x < X 2 ,
1 x ∑ n ≤ x P + ( n ) ≤ X f ( n ) = ∑ a ≤ x P + ( a ) ≤ X g ( a ) a − 1 x ∑ a ≤ x P + ( a ) ≤ X g ( a ) { x a } . (7) \frac{1}{\sqrt{x}}\sum_{\substack{n\leq x\\P^+(n)\leq X}}f(n)=\sum_{\substack{a\leq x\\P^+(a)\leq X}}\frac{g(a)}{\sqrt{a}}-\frac{1}{\sqrt{x}}\sum_{\substack{a\leq x\\P^+(a)\leq X}}g(a)\left\{\sqrt{\frac{x}{a}}\right\}. \tag*{(7)} x 1 n ≤ x P + ( n ) ≤ X ∑ f ( n ) = a ≤ x P + ( a ) ≤ X ∑ a g ( a ) − x 1 a ≤ x P + ( a ) ≤ X ∑ g ( a ) { a x } . (7)
The variables g ( a ) g(a) g ( a ) , with a a a squarefree, form an orthonormal system, so the second term on the right has second moment at most 1. The first term is handled by Parseval’s identity (2.10). With a horizontal shift σ = 1 / log X \sigma=1/\log X σ = 1/ log X , the relevant identity is
∫ 1 ∞ ∣ ∑ a ≤ x P + ( a ) ≤ X g ( a ) a ∣ 2 d x x 1 + 2 σ = 1 2 π ∫ R 1 σ 2 + t 2 ∣ ∏ p ≤ X ( 1 + f ( p ) p 1 / 2 + σ + i t ) ∣ 2 d t . \int_1^\infty\left|\sum_{\substack{a\leq x\\P^+(a)\leq X}}\frac{g(a)}{\sqrt{a}}\right|^2\frac{\mathrm{d}x}{x^{1+2\sigma}}=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{1}{\sigma^2+t^2}\left|\prod_{p\leq X}\left(1+\frac{f(p)}{p^{1/2+\sigma+\mathrm{i}t}}\right)\right|^2\mathrm{d}t. ∫ 1 ∞ a ≤ x P + ( a ) ≤ X ∑ a g ( a ) 2 x 1 + 2 σ d x = 2 π 1 ∫ R σ 2 + t 2 1 p ≤ X ∏ ( 1 + p 1/2 + σ + i t f ( p ) ) 2 d t .
The multiplicative chaos estimates from [19 , 20 ] control the Euler product on intervals of length 1 in the frequency variable t t t . The kernel is singular near t = 0 t=0 t = 0 , and this range requires a separate argument. We factor the Euler product at primes up to exp ( 1 / ∣ t ∣ ) \exp(1/|t|) exp ( 1/∣ t ∣ ) , compare its value at t t t with its value at t = 0 t=0 t = 0 , and control the latter using a standard large-deviation estimate for weighted Rademacher sums. This gives
P ( 1 log X ∫ X 8 / 7 X 4 / 3 ∣ 1 x ∑ n ≤ x P + ( n ) ≤ X f ( n ) ∣ 2 d x x > ( log log X ) 1 / 3 ) ≪ ( log log X ) − 1 / 200 . \mathbb{P}\left(\frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}}\left|\frac{1}{\sqrt{x}}\sum_{\substack{n\le x\\P^{+}(n)\le X}}f(n)\right|^2\frac{\mathrm{d}x}{x}>(\log\log X)^{1/3}\right)\ll(\log\log X)^{-1/200}. P log X 1 ∫ X 8/7 X 4/3 x 1 n ≤ x P + ( n ) ≤ X ∑ f ( n ) 2 x d x > ( log log X ) 1/3 ≪ ( log log X ) − 1/200 .
Thus no pointwise estimate for either term in (1.7) is needed.
This is another point where the extended Rademacher RMF differs from the Rademacher RMF. In the latter, orthogonality gives an easy second moment bound, showing that the X X X -smooth contribution is small at most points of the fixed logarithmic grid X 8 / 7 e 2 π r X^{8/7}e^{2\pi r} X 8/7 e 2 π r , as in [20, Section 3.1]. Here the square contributions destroy this orthogonality, the corresponding easy second moment estimate is no longer available. We therefore turn to prove a logarithmic mean square estimate over X 8 / 7 ≤ x ≤ X 4 / 3 X^{8/7}\le x\le X^{4/3} X 8/7 ≤ x ≤ X 4/3 . To transfer this continuous integral estimate to a logarithmically spaced grid, we introduce an additional parameter θ ∈ [ 0 , 2 π ) \theta\in[0,2\pi) θ ∈ [ 0 , 2 π ) .
We sample at logarithmically spaced points
x = X 8 / 7 e θ + 2 π r , 0 ≤ θ < 2 π , x=X^{8/7}e^{\theta+2\pi r},\qquad0\le\theta<2\pi, x = X 8/7 e θ + 2 π r , 0 ≤ θ < 2 π ,
where r r r ranges over the integers for which 4 x ≤ X 4 / 3 4x\le X^{4/3} 4 x ≤ X 4/3 . There are ≍ log X \asymp\log X ≍ log X such points. Averaging over the common shift θ \theta θ , the preceding logarithmic mean square estimate shows that, for all but a proportion O ( e 1.2 W ( log log X ) − 1 / 12 ) O(e^{1.2W}(\log\log X)^{-1/12}) O ( e 1.2 W ( log log X ) − 1/12 ) of the shifts, all but the same proportion of the indices r r r satisfy
∣ 1 y ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∣ ≤ A : = e − 1.2 W ( log log X ) 1 / 4 \left|\frac{1}{\sqrt{y}}\sum_{\substack{n\le y\\P^{+}(n)\le X}}f(n)\right|\le A:=e^{-1.2W(\log\log X)^{1/4}} y 1 n ≤ y P + ( n ) ≤ X ∑ f ( n ) ≤ A := e − 1.2 W ( l o g l o g X ) 1/4
both at y ∈ { x , 4 x } y\in\{x,4x\} y ∈ { x , 4 x } . It is important here that the same shift θ \theta θ is used for every r r r : this preserves the logarithmic spacing log x r , θ − log x s , θ = 2 π ( r − s ) \log x_{r,\theta}-\log x_{s,\theta}=2\pi(r-s) log x r , θ − log x s , θ = 2 π ( r − s ) needed in the covariance argument.
We next consider the contribution from primes exceeding X X X . Rather than study its value at the point x x x , we take the difference between the normalised values at x x x and 4 x 4x 4 x , which we also call the large prime increment at x x x , namely,
1 x ∑ X < p ≤ 4 x f ( p ) ( 1 2 ∑ m ≤ 4 x / p P + ( m ) ≤ X f ( m ) − ∑ m ≤ x / p P + ( m ) ≤ X f ( m ) ) . (8) \frac{1}{\sqrt{x}}\sum_{X<p\le4x}f(p)\left(\frac{1}{2}\sum_{\substack{m\le4x/p\\P^{+}(m)\le X}}f(m)-\sum_{\substack{m\le x/p\\P^{+}(m)\le X}}f(m)\right). \tag*{(8)} x 1 X < p ≤ 4 x ∑ f ( p ) 2 1 m ≤ 4 x / p P + ( m ) ≤ X ∑ f ( m ) − m ≤ x / p P + ( m ) ≤ X ∑ f ( m ) . (8)
Viewed as a function of x / p x/p x / p , the expression in parentheses has Mellin transform
4 s − 1 / 2 − 1 s ∏ p ≤ X ( 1 + f ( p ) p s ) ∏ p ≤ X ( 1 − 1 p 2 s ) − 1 . (9) \frac{4^{s-1/2}-1}{s}\prod_{p\le X}\left(1+\frac{f(p)}{p^s}\right)\prod_{p\le X}\left(1-\frac{1}{p^{2s}}\right)^{-1}. \tag*{(9)} s 4 s − 1/2 − 1 p ≤ X ∏ ( 1 + p s f ( p ) ) p ≤ X ∏ ( 1 − p 2 s 1 ) − 1 . (9)
The second product is the finite analogue of the factor ζ ( 2 s ) \zeta(2s) ζ ( 2 s ) in (1.1). Near s = 1 / 2 s=1/2 s = 1/2 it has singular growth as X → ∞ X\to\infty X → ∞ , but 4 s − 1 / 2 − 1 4^{s-1/2}-1 4 s − 1/2 − 1 vanishes to first order there. The product ( 4 s − 1 / 2 − 1 ) ∏ p ≤ X ( 1 − p − 2 s ) − 1 (4^{s-1/2}-1)\prod_{p\le X}(1-p^{-2s})^{-1} ( 4 s − 1/2 − 1 ) ∏ p ≤ X ( 1 − p − 2 s ) − 1 is therefore regular (uniformly bounded) near s = 1 / 2 s=1/2 s = 1/2 . On a fixed frequency interval, for example 1 / 3 ≤ ∣ t ∣ ≤ 1 / 2 1/3\le|t|\le1/2 1/3 ≤ ∣ t ∣ ≤ 1/2 , it is also bounded away from zero. The factor 4 s − 1 / 2 − 1 4^{s-1/2}-1 4 s − 1/2 − 1 thus removes the low frequency singularity while keeping the fixed frequency contribution needed for the lower bound.
Remark 1.7. The use of an auxiliary factor vanishing at a zeta type singularity is classical in spirit. A closely related cancellation occurs in the work of Soundararajan [29 ] , Section 5.3], where the factor ( 1 − 2 1 − 2 s ) ζ ( 2 s ) (1-2^{1-2s})\zeta(2s) ( 1 − 2 1 − 2 s ) ζ ( 2 s ) appears, with the zero at s = 1 / 2 s=1/2 s = 1/2 canceling the pole of ζ ( 2 s ) \zeta(2s) ζ ( 2 s ) . This is essentially the same as ours. In the present setting, our canceling factor is not inserted ad hoc: it arises naturally in (1.9) from (1.8).
Conditioning on the values f ( p ) f(p) f ( p ) , p ≤ X p \le X p ≤ X , the large prime part (1.8) has mean zero, and its conditional variance is given by the corresponding square sum over X < p ≤ 4 x X < p \le4x X < p ≤ 4 x . A short interval prime estimate bounds this square sum from below by a weighted integral. Parseval’s identity (2.10), together with (1.9), then turns this integral into an Euler product integral on the Mellin line. Restricting the t t t -integral to 1 / 3 ≤ ∣ t ∣ ≤ 1 / 2 1/3 \le|t| \le1/2 1/3 ≤ ∣ t ∣ ≤ 1/2 , the multiplicative chaos lower bound from [20 ] gives, with probability 1 − O ( e − 0.1 W ) 1-O(e^{-0.1W}) 1 − O ( e − 0.1 W ) ,
1 x ∑ X < p ≤ 4 x ( 1 2 ∑ m ≤ 4 x / p P + ( m ) ≤ X f ( m ) − ∑ m ≤ x / p P + ( m ) ≤ X f ( m ) ) 2 ≫ e − 2.2 W log log X , \frac{1}{x}\sum_{X<p\le4x}\left(\frac{1}{2}\sum_{\substack{m\le4x/p\\ P^{+}(m)\le X}}f(m)-\sum_{\substack{m\le x/p\\ P^{+}(m)\le X}}f(m)\right)^2 \gg\frac{e^{-2.2W}}{\sqrt{\log\log X}}, x 1 X < p ≤ 4 x ∑ 2 1 m ≤ 4 x / p P + ( m ) ≤ X ∑ f ( m ) − m ≤ x / p P + ( m ) ≤ X ∑ f ( m ) 2 ≫ log log X e − 2.2 W ,
simultaneously for all x r , θ x_{r,\theta} x r , θ in the shifted grid.
This variance estimate also explains the exponent 1 / 4 1/4 1/4 . If one had ( log X ) 0.19 (\log X)^{0.19} ( log X ) 0.19 independent Gaussian variables with variances of the preceding order, then their maximum would naturally have size
e − 2.2 W log log X log ( ( log X ) 0.19 ) ≍ e − 1.1 W ( log log X ) 1 / 4 . \sqrt{\frac{e^{-2.2W}}{\sqrt{\log\log X}}}\sqrt{\log\left((\log X)^{0.19}\right)}\asymp e^{-1.1W}(\log\log X)^{1/4}. log log X e − 2.2 W log ( ( log X ) 0.19 ) ≍ e − 1.1 W ( log log X ) 1/4 .
This is larger than the target A = e − 1.2 W ( log log X ) 1 / 4 A=e^{-1.2W}(\log\log X)^{1/4} A = e − 1.2 W ( log log X ) 1/4 , since W ≥ W 0 W\ge W_0 W ≥ W 0 large enough. The main remaining problem is therefore to find sufficiently many points whose conditional covariances are small.
For this we prove a weighted version of Harper’s high mixed moment estimate. Outside an F X \mathcal{F}_X F X -measurable event of small probability, uniformly for every common shift θ \theta θ , each increment has conditional covariance exceeding ( log log X ) − 0.59 (\log\log X)^{-0.59} ( log log X ) − 0.59 with at most ( log X ) 0.8 (\log X)^{0.8} ( log X ) 0.8 of the other increments. Since all but a small proportion of the original ≍ log X \asymp\log X ≍ log X grid points have small X X X -smooth contribution, a greedy selection among these points leaves at least ( log X ) 0.19 (\log X)^{0.19} ( log X ) 0.19 points with all pairwise covariances below this threshold. Dividing by the variance lower bound, the corresponding correlations are at most
( log log X ) − 0.59 e − 2.2 W ( log log X ) − 1 / 2 = e 2.2 W ( log log X ) − 0.09 = o ( 1 ) . \frac{(\log\log X)^{-0.59}}{e^{-2.2W}(\log\log X)^{-1/2}}=e^{2.2W}(\log\log X)^{-0.09}=o(1). e − 2.2 W ( log log X ) − 1/2 ( log log X ) − 0.59 = e 2.2 W ( log log X ) − 0.09 = o ( 1 ) .
We briefly indicate the high moment argument. After Perron inversion, the conditional covariance is represented by a double integral involving the Rademacher Euler product in (1.9), a short prime sum, and the deterministic multiplier 4 s − 1 / 2 − 1 4^{s-1/2}-1 4 s − 1/2 − 1 . Taking a 2 k 2k 2 k -th moment, where k = ⌊ log log X 5000 log log log X ⌋ k=\left\lfloor\frac{\log\log X}{5000\log\log\log X}\right\rfloor k = ⌊ 5000 l o g l o g l o g X l o g l o g X ⌋ , and expanding introduces frequencies t 1 , … , t 2 k t_1,\ldots,t_{2k} t 1 , … , t 2 k . Summation over the logarithmic grid produces the geometric sum
∑ r e − 2 π i r Ξ ≪ min ( log X , 1 ∥ Ξ ∥ R / Z ) , \sum_r e^{-2\pi i r\Xi}\ll\min\left(\log X,\frac{1}{\|\Xi\|_{\mathbb{R}/\mathbb{Z}}}\right), r ∑ e − 2 π i r Ξ ≪ min ( log X , ∥Ξ ∥ R / Z 1 ) ,
where Ξ : = t 1 + ⋯ + t k − t k + 1 − ⋯ − t 2 k \Xi:=t_1+\cdots+t_k-t_{k+1}-\cdots-t_{2k} Ξ := t 1 + ⋯ + t k − t k + 1 − ⋯ − t 2 k . Away from the integers this already gives the required saving. On the near-integer range, we use the strong barrier condition and, as in [20 ] , Sections 3.3–3.4, order the frequencies and decompose according to the sizes of the gaps between consecutive frequencies. Roughly speaking, when many consecutive frequencies are close together, the barrier gives a large saving, while when many of the gaps are fairly large, the near-integer constraint gives the required volume saving.
The main point to keep in mind when adapting Harper’s argument is that every Euler product integral carries the deterministic multiplier from (1.9). On the relevant frequency range, this multiplier is O ( ( log log log X ) 2 ) O((\log\log\log X)^2) O (( log log log X ) 2 ) , so it contributes only a harmless additional loss in the 2 k 2k 2 k -th moment estimate. The resulting weighted moment estimate gives the required sparsity of the large conditional covariances.
Conditional on the primes up to X X X , the selected increments are linear forms in independent Rademacher variables. We use Harper’s multivariate normal approximation to compare their maximum with that of a centred Gaussian vector having the same covariance matrix; the required error terms are controlled by the fixed third- and fourth-moment estimates from Lemma 2.1. After standardisation, the Gaussian variables have uniformly small pairwise correlations. Harper’s normal comparison result, followed by the normal approximation, then shows that the maximum of the selected increments exceeds 4 A = 4 e − 1.2 W ( log log X ) 1 / 4 4A = 4e^{-1.2W}(\log\log X)^{1/4} 4 A = 4 e − 1.2 W ( log log X ) 1/4 with conditional probability 1 − o ( 1 ) 1-o(1) 1 − o ( 1 ) .
Such an increment implies that the large prime contributions at the selected endpoints have a range of at least 4 A 4A 4 A . Conditional on the primes up to X X X , these contributions form a centrally symmetric vector: simultaneously replacing f ( p ) f(p) f ( p ) by − f ( p ) -f(p) − f ( p ) for X < p ≤ X 4 / 3 X < p \le X^{4/3} X < p ≤ X 4/3 changes the vector to its negative. Hence, with conditional probability 1 − o ( 1 ) 1-o(1) 1 − o ( 1 ) , either the maximum is at least 2 A 2A 2 A or the minimum is at most − 2 A -2A − 2 A . By symmetry, each of these two events has conditional probability at least 1 / 2 − o ( 1 ) 1/2-o(1) 1/2 − o ( 1 ) . By our choice of the selected points, the X X X -smooth contribution at both corresponding endpoints has absolute value at most A A A . Adding this contribution back therefore gives a large value of either sign and completes the proof of Theorem 1.4.
Organisation
The paper is organised as follows. Section 2 records the large prime decomposition, the square divisor identity, and the probabilistic inputs. Section 3 proves the logarithmic mean square estimate for the X X X -smooth contribution. Section 4 introduces the multiplier and establishes the lower bound for the conditional variance. Section 5 proves the weighted high moment estimate and the sparsity of large conditional covariances. Section 6 establishes the conditional Gaussian approximation, and Section 7 uses a common logarithmic shift to transfer the logarithmic mean square estimate to most points of the grid. Finally, Section 8 combines these ingredients and completes the proof of Theorem 1.4.
Acknowledgement
The work was initiated during the Universality in Number Theory Program at CRM Montreal 2026. The authors thank the hospitality of CRM and the organizers. We thank Adam Harper for helpful discussions on this problem. H.G. would like to express sincere gratitude to Professors Andrew Granville and Jianya Liu for their guidance and support during his Ph.D. studies. He also thanks financial support from the China Scholarship Council (CSC).
Preliminaries
Throughout the paper p p p denotes a prime, P + ( n ) P^{+}(n) P + ( n ) is the largest prime factor of n n n , and P + ( 1 ) = 1 P^{+}(1)=1 P + ( 1 ) = 1 . Constants implicit in O O O - and ≪ \ll ≪ -notation are absolute unless a dependence is displayed. We use the same probability space for
f ( n ) = ∏ p ν p ∥ n ϵ p ν p and g ( n ) = μ 2 ( n ) f ( n ) . f(n)=\prod_{p^{\nu_p}\Vert n}\epsilon_p^{\nu_p}
\qquad\text{and}\qquad
g(n)=\mu^2(n)f(n). f ( n ) = p ν p ∥ n ∏ ϵ p ν p and g ( n ) = μ 2 ( n ) f ( n ) .
The function g ( n ) g(n) g ( n ) is the usual Rademacher RMF supported on squarefree integers. Since ϵ p 2 = 1 \epsilon_p^2=1 ϵ p 2 = 1 , one has the exact identity
f ( n ) = ∑ d 2 ∣ n g ( n / d 2 ) . (10) f(n)=\sum_{d^2\mid n}g(n/d^2). \tag*{(10)} f ( n ) = d 2 ∣ n ∑ g ( n / d 2 ) . (10)
Exactly one summand on the right is non-zero.
For X ≥ 3 X \ge3 X ≥ 3 , let F X = σ ( ϵ p : p ≤ X ) \mathcal{F}_X = \sigma(\epsilon_p : p \le X) F X = σ ( ϵ p : p ≤ X ) . We write P X ( ⋅ ) = P ( ⋅ ∣ F X ) \mathbb{P}_X(\cdot) = \mathbb{P}(\cdot\mid\mathcal{F}_X) P X ( ⋅ ) = P ( ⋅ ∣ F X ) and E X ( ⋅ ) = E ( ⋅ ∣ F X ) \mathbb{E}_X(\cdot) = \mathbb{E}(\cdot\mid\mathcal{F}_X) E X ( ⋅ ) = E ( ⋅ ∣ F X ) , and denote the corresponding conditional variance and covariance by Var X ( ⋅ ) \operatorname{Var}_X(\cdot) Var X ( ⋅ ) and Cov X ( ⋅ ) \operatorname{Cov}_X(\cdot) Cov X ( ⋅ ) , respectively. We also write
C X ( y ) = 1 y ∑ n ≤ y P + ( n ) ≤ X f ( n ) . (11) C_X(y) = \frac{1}{\sqrt{y}} \sum_{\substack{n \le y \\ P^{+}(n) \le X}} f(n). \tag*{(11)} C X ( y ) = y 1 n ≤ y P + ( n ) ≤ X ∑ f ( n ) . (11)
An empty sum is understood when y < 1 y < 1 y < 1 . The Rademacher Euler product will be denoted by
F X ( s ) = ∏ p ≤ X ( 1 + ϵ p p − s ) . (12) F_X(s) = \prod_{p \le X}(1+\epsilon_p p^{-s}). \tag*{(12)} F X ( s ) = p ≤ X ∏ ( 1 + ϵ p p − s ) . (12)
We also introduce the deterministic Euler product
Z X ( s ) = ∏ p ≤ X ( 1 − p − s ) − 1 . (13) Z_X(s) = \prod_{p \le X}(1-p^{-s})^{-1}. \tag*{(13)} Z X ( s ) = p ≤ X ∏ ( 1 − p − s ) − 1 . (13)
The Dirichlet series of f f f restricted to the X X X -smooth integers factors as
∑ n ≥ 1 P + ( n ) ≤ X f ( n ) n s = ∏ p ≤ X ( 1 − ϵ p p − s ) − 1 = F X ( s ) Z X ( 2 s ) , Re s > 0. (14) \sum_{\substack{n \ge1 \\ P^{+}(n) \le X}} \frac{f(n)}{n^s} = \prod_{p \le X}(1-\epsilon_p p^{-s})^{-1} = F_X(s)Z_X(2s), \qquad\operatorname{Re} s > 0. \tag*{(14)} n ≥ 1 P + ( n ) ≤ X ∑ n s f ( n ) = p ≤ X ∏ ( 1 − ϵ p p − s ) − 1 = F X ( s ) Z X ( 2 s ) , Re s > 0. (14)
For 1 ≤ x < X 2 1 \le x < X^2 1 ≤ x < X 2 , since every n < X 2 n < X^2 n < X 2 has at most one prime factor exceeding X X X , we have
S f ( x ) x = C X ( x ) + U X ( x ) , U X ( x ) = ∑ X < p ≤ x ϵ p p C X ( x / p ) . (15) \frac{S_f(x)}{\sqrt{x}} = C_X(x) + U_X(x), \qquad U_X(x) = \sum_{X<p \le x} \frac{\epsilon_p}{\sqrt{p}} C_X(x/p). \tag*{(15)} x S f ( x ) = C X ( x ) + U X ( x ) , U X ( x ) = X < p ≤ x ∑ p ϵ p C X ( x / p ) . (15)
Conditional on F X \mathcal{F}_X F X , the coefficients in U X ( x ) U_X(x) U X ( x ) are fixed, whereas ( ϵ p ) p > X (\epsilon_p)_{p>X} ( ϵ p ) p > X are independent Rademacher variables.
We shall compare the endpoints x x x and 4 x 4x 4 x . Accordingly, set
B X ( x ) = 1 2 ∑ n ≤ 4 x P + ( n ) ≤ X f ( n ) − ∑ n ≤ x P + ( n ) ≤ X f ( n ) . (16) B_X(x) = \frac{1}{2}\sum_{\substack{n \le4x \\ P^{+}(n) \le X}} f(n) - \sum_{\substack{n \le x \\ P^{+}(n) \le X}} f(n). \tag*{(16)} B X ( x ) = 2 1 n ≤ 4 x P + ( n ) ≤ X ∑ f ( n ) − n ≤ x P + ( n ) ≤ X ∑ f ( n ) . (16)
If 4 x < X 2 4x < X^2 4 x < X 2 , we define the large prime increment by
Y x : = U X ( 4 x ) − U X ( x ) = 1 x ∑ X < p ≤ 4 x ϵ p B X ( x / p ) . (17) Y_x := U_X(4x)-U_X(x) = \frac{1}{\sqrt{x}}\sum_{X<p \le4x} \epsilon_p B_X(x/p). \tag*{(17)} Y x := U X ( 4 x ) − U X ( x ) = x 1 X < p ≤ 4 x ∑ ϵ p B X ( x / p ) . (17)
Only fixed moments are needed for the normal approximation. The following crude bounds are sufficient for our purpose.
Lemma 2.1 (Moments of the X X X -smooth partial sums). Let $0 \le y < X. ∗ F o r e a c h ∗ . *For each* . ∗ F or e a c h ∗ r \in {2,3,4}∗ t h e r e i s ∗ *there is* ∗ t h er e i s ∗ A_r > 0 such that
E ∣ ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∣ r ≪ r ( 1 + y ) r / 2 log A r ( 2 + y ) . (18) \mathbb{E}\left|\sum_{\substack{n \le y \\ P^{+}(n) \le X}} f(n)\right|^r \ll_r (1+y)^{r/2}\log^{A_r}(2+y). \tag*{(18)} E n ≤ y P + ( n ) ≤ X ∑ f ( n ) r ≪ r ( 1 + y ) r /2 log A r ( 2 + y ) . (18)
The same conclusion holds for B X ( y ) B_X(y) B X ( y ) whenever 4 y < X 4y < X 4 y < X .
Proof. In the stated range every integer not exceeding y y y is automatically X X X -smooth. Using (2.1), we therefore have
∑ n ≤ y P + ( n ) ≤ X f ( n ) = ∑ d ≤ y ∑ m ≤ y / d 2 g ( m ) . \sum_{\substack{n \le y \\ P^{+}(n) \le X}} f(n) = \sum_{d \le\sqrt{y}} \sum_{m \le y/d^2} g(m). n ≤ y P + ( n ) ≤ X ∑ f ( n ) = d ≤ y ∑ m ≤ y / d 2 ∑ g ( m ) .
For a random variable Y Y Y and q ≥ 1 q \ge1 q ≥ 1 , write ∥ Y ∥ q = ( E ∣ Y ∣ q ) 1 / q \lVert Y\rVert_q = \left(\mathbb{E}|Y|^q\right)^{1/q} ∥ Y ∥ q = ( E ∣ Y ∣ q ) 1/ q . By the hypercontractive inequality for Rademacher RMFs (see, for example, [18 ] , Probability Result 2.3), for k ∈ { 1 , 2 } k \in\{1,2\} k ∈ { 1 , 2 } ,
E ∣ ∑ m ≤ z g ( m ) ∣ 2 k ≤ ( ∑ m ≤ z μ 2 ( m ) τ 2 k − 1 ( m ) ) k . \mathbb{E}\left|\sum_{m \le z} g(m)\right|^{2k} \le\left(\sum_{m \le z} \mu^2(m)\tau_{2k-1}(m)\right)^k. E m ≤ z ∑ g ( m ) 2 k ≤ ( m ≤ z ∑ μ 2 ( m ) τ 2 k − 1 ( m ) ) k .
where τ j \tau_j τ j is the j j j -fold divisor function. It follows that, by the standard upper bound
∑ m ≤ z τ k ( m ) ≪ k z log k − 1 ( 2 + z ) , \sum_{m \le z} \tau_k(m) \ll_k z\log^{k-1}(2+z), m ≤ z ∑ τ k ( m ) ≪ k z log k − 1 ( 2 + z ) ,
∥ ∑ m ≤ z g ( m ) ∥ 2 k ≪ k z log k − 1 ( 2 + z ) ( k = 1 , 2 ) , \left\lVert\sum_{m \le z}g(m)\right\rVert_{2k}\ll_k \sqrt{z}\log^{k-1}(2+z)\qquad(k=1,2), m ≤ z ∑ g ( m ) 2 k ≪ k z log k − 1 ( 2 + z ) ( k = 1 , 2 ) ,
Minkowski’s inequality now gives us
∥ ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∥ 2 k ≤ ∑ d ≤ y ∥ ∑ m ≤ y / d 2 g ( m ) ∥ 2 k ≪ k y log k − 1 ( 2 + y ) ∑ d ≤ y 1 d ≪ k y log k ( 2 + y ) . \left\lVert\sum_{\substack{n\le y\\P^+(n)\le X}}f(n)\right\rVert_{2k}
\le\sum_{d\le\sqrt{y}}\left\lVert\sum_{m\le y/d^2}g(m)\right\rVert_{2k}
\ll_k \sqrt{y}\log^{k-1}(2+y)\sum_{d\le\sqrt{y}}\frac{1}{d}
\ll_k \sqrt{y}\log^k(2+y). n ≤ y P + ( n ) ≤ X ∑ f ( n ) 2 k ≤ d ≤ y ∑ m ≤ y / d 2 ∑ g ( m ) 2 k ≪ k y log k − 1 ( 2 + y ) d ≤ y ∑ d 1 ≪ k y log k ( 2 + y ) .
This proves the cases r = 2 r=2 r = 2 and r = 4 r=4 r = 4 , while the case r = 3 r=3 r = 3 follows from Cauchy–Schwarz,
E ∣ ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∣ 3 ≤ ( E ∣ ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∣ 2 ) 1 / 2 ( E ∣ ∑ n ≤ y P + ( n ) ≤ X f ( n ) ∣ 4 ) 1 / 2 . \mathbb{E}\left|\sum_{\substack{n\le y\\P^+(n)\le X}}f(n)\right|^3
\le\left(\mathbb{E}\left|\sum_{\substack{n\le y\\P^+(n)\le X}}f(n)\right|^2\right)^{1/2}
\left(\mathbb{E}\left|\sum_{\substack{n\le y\\P^+(n)\le X}}f(n)\right|^4\right)^{1/2}. E n ≤ y P + ( n ) ≤ X ∑ f ( n ) 3 ≤ E n ≤ y P + ( n ) ≤ X ∑ f ( n ) 2 1/2 E n ≤ y P + ( n ) ≤ X ∑ f ( n ) 4 1/2 .
If 4 y < X 4y<X 4 y < X , the same bounds apply to the two partial sums in (2.7); the assertion for B X ( y ) B_X(y) B X ( y ) follows from the triangle inequality. □ \square □
We shall repeatedly use the following form of Parseval’s identity. If A ( s ) = ∑ n ≥ 1 a n n − s A(s)=\sum_{n\ge1}a_n n^{-s} A ( s ) = ∑ n ≥ 1 a n n − s converges absolutely for Re s > c \operatorname{Re}s>c Re s > c , then, whenever both sides converge,
∫ 0 ∞ ∣ ∑ n ≤ x a n ∣ 2 x − 1 − 2 σ d x = 1 2 π ∫ R ∣ A ( σ + i t ) σ + i t ∣ 2 d t . (19) \int_0^\infty\left|\sum_{n\le x}a_n\right|^2x^{-1-2\sigma}\,dx
=\frac{1}{2\pi}\int_{\mathbb{R}}\left|\frac{A(\sigma+it)}{\sigma+it}\right|^2\,dt. \tag*{(19)} ∫ 0 ∞ n ≤ x ∑ a n 2 x − 1 − 2 σ d x = 2 π 1 ∫ R σ + i t A ( σ + i t ) 2 d t . (19)
This is Parseval’s identity for Dirichlet series; see [28 ] , Section 5.1, especially (5.26). All applications below are made after a positive horizontal shift, where the required convergence is immediate.
We shall also use the following Lévy’s extension of the second Borel–Cantelli lemma; see [11 ] , Theorem 4.3.4.
Lemma 2.2 (Conditional Borel–Cantelli lemma). Let ( F ℓ ) ℓ ≥ 0 (\mathcal{F}_\ell)_{\ell\ge0} ( F ℓ ) ℓ ≥ 0 be a filtration, and let E ℓ ∈ F ℓ E_\ell\in\mathcal{F}_\ell E ℓ ∈ F ℓ for every ℓ ≥ 1 \ell\ge1 ℓ ≥ 1 . Then, almost surely the following events are the same
{ E ℓ infinitely often } = { ∑ ℓ ≥ 1 P ( E ℓ ∣ F ℓ − 1 ) = ∞ } . \{E_\ell\ \text{infinitely often}\}
=
\left\{\sum_{\ell\ge1}\mathbb{P}(E_\ell\mid\mathcal{F}_{\ell-1})=\infty\right\}. { E ℓ infinitely often } = { ℓ ≥ 1 ∑ P ( E ℓ ∣ F ℓ − 1 ) = ∞ } .
The multiplicative chaos estimates used below are those proved in [20 ] , Sections 2.3–2.4. We use in particular Multiplicative Chaos Result 3 for low-moment upper bounds, Multiplicative Chaos Result 4 for a lower bound on a fixed frequency interval, and the barrier estimates from Sections 3.3–3.4. All applications are made to the squarefree Euler product F X F_X F X or to its prime truncations; we do not apply these results directly to the Euler product F X ( s ) Z X ( 2 s ) F_X(s)Z_X(2s) F X ( s ) Z X ( 2 s ) of the extended Rademacher RMF. The relevant parameters are specified at each application.
We also use Normal Approximation Result 1 and Normal Comparison Result 1 of [20 ] . The former is used in Section 6 to replace the large prime increments by a Gaussian vector with the same covariance matrix, while the latter is used in the proof of Theorem 1.4 to control the maximum of the resulting weakly correlated Gaussian variables.
The X-smooth contribution
The first issue peculiar to the extended Rademacher RMF is that, after conditioning on ( f ( p ) ) p ≤ X (f(p))_{p \le X} ( f ( p ) ) p ≤ X , there remains a non-centred, F X \mathcal{F}_X F X -measurable sum supported on X X X -smooth integers. The identity f = ( μ 2 f ) ∗ 1 □ f=(\mu^2 f) \ast\mathbf{1}_{\square} f = ( μ 2 f ) ∗ 1 □ isolates this term in a useful form. Throughout this section, g g g and C X C_X C X have the meanings fixed in (10) and (11) . Only the values X 8 / 7 ≤ x ≤ X 4 / 3 X^{8/7} \le x \le X^{4/3} X 8/7 ≤ x ≤ X 4/3 will be needed.
For 1 ≤ x < X 2 1 \le x < X^2 1 ≤ x < X 2 , every d ≤ x d \le\sqrt{x} d ≤ x is automatically X X X -smooth. Hence (10) gives
C X ( x ) = 1 x ∑ a ≤ x P + ( a ) ≤ X g ( a ) ⌊ x a ⌋ . C_X(x)=\frac{1}{\sqrt{x}}\sum_{\substack{a \le x\\ P^+(a)\le X}}g(a)\left\lfloor\sqrt{\frac{x}{a}}\right\rfloor. C X ( x ) = x 1 a ≤ x P + ( a ) ≤ X ∑ g ( a ) ⌊ a x ⌋ .
Writing the integer part as its argument minus its fractional part, we obtain
C X ( x ) = M X ( x ) − D X ( x ) , C_X(x)=M_X(x)-D_X(x), C X ( x ) = M X ( x ) − D X ( x ) ,
where
M X ( x ) = ∑ a ≤ x P + ( a ) ≤ X g ( a ) a , D X ( x ) = 1 x ∑ a ≤ x P + ( a ) ≤ X g ( a ) { x a } . M_X(x)=\sum_{\substack{a \le x\\ P^+(a)\le X}}\frac{g(a)}{\sqrt{a}},\qquad D_X(x)=\frac{1}{\sqrt{x}}\sum_{\substack{a \le x\\ P^+(a)\le X}}g(a)\left\{\sqrt{\frac{x}{a}}\right\}. M X ( x ) = a ≤ x P + ( a ) ≤ X ∑ a g ( a ) , D X ( x ) = x 1 a ≤ x P + ( a ) ≤ X ∑ g ( a ) { a x } .
The random variables g ( a ) g(a) g ( a ) , with a a a squarefree, form an orthonormal system. Hence
E D X ( x ) 2 = 1 x ∑ a ≤ x P + ( a ) ≤ X { x a } 2 ≤ 1 x ∑ a ≤ x 1 ≤ 1. \mathbb{E}D_X(x)^2=\frac{1}{x}\sum_{\substack{a \le x\\ P^+(a)\le X}}\left\{\sqrt{\frac{x}{a}}\right\}^2\le\frac{1}{x}\sum_{a\le x}1\le1. E D X ( x ) 2 = x 1 a ≤ x P + ( a ) ≤ X ∑ { a x } 2 ≤ x 1 a ≤ x ∑ 1 ≤ 1.
We next prove the estimate used to control the X X X -smooth contribution at most endpoints. Recall the Rademacher Euler product F X F_X F X from (12) ; equivalently, F X ( s ) = ∑ P + ( n ) ≤ X g ( n ) n − s F_X(s)=\sum_{P^+(n)\le X}g(n)n^{-s} F X ( s ) = ∑ P + ( n ) ≤ X g ( n ) n − s . The multiplicative chaos input is the following low-moment bound. It is Multiplicative Chaos Result 3 of [20 ] ; see also [19 ] , Key Proposition 3 and the discussion following it, and the formulation in [2 ] , Proposition 2.9. For q = 2 / 3 q=2/3 q = 2/3 , σ = 1 / log X \sigma=1/\log X σ = 1/ log X , and ∣ N ∣ ≤ ( log X ) 1000 |N|\le(\log X)^{1000} ∣ N ∣ ≤ ( log X ) 1000 , it gives
E ( ∫ N − 1 / 2 N + 1 / 2 ∣ F X ( 1 2 + σ + i t ) ∣ 2 d t ) q ≪ ( log log ( ∣ N ∣ + 10 ) ) q ( log X log log X ) q . (20) \mathbb{E}\left(\int_{N-1/2}^{N+1/2}\left|F_X\left(\frac{1}{2}+\sigma+it\right)\right|^2\,dt\right)^q\ll(\log\log(|N|+10))^q\left(\frac{\log X}{\sqrt{\log\log X}}\right)^q. \tag*{(20)} E ( ∫ N − 1/2 N + 1/2 F X ( 2 1 + σ + i t ) 2 d t ) q ≪ ( log log ( ∣ N ∣ + 10 ) ) q ( log log X log X ) q . (20)
The weighted integral below is closely related to the methods used by Gerspach [14 ] in his study of low pseudomoments, and by Hardy [16 ] for weighted Steinhaus RMFs, and, more directly in the Rademacher setting, by Atherfold [2 ] . A related weighted Euler product integral also appears in the recent work of Harper–Soundararajan–Xu [21 ] , Section 2, where a short interval Fourier transform also produces a decay factor there. In the present setting, however, the Parseval kernel ( σ 2 + t 2 ) − 1 (\sigma^2+t^2)^{-1} ( σ 2 + t 2 ) − 1 becomes singular at t = 0 t=0 t = 0 as σ = 1 / log X → 0 \sigma=1/\log X\to0 σ = 1/ log X → 0 , so the main additional point is to control the contribution of the very small frequencies.
Lemma 3.1 (A weighted mean square bound for the Rademacher Euler product). Let σ = 1 / log X \sigma=1/\log X σ = 1/ log X . Then, for all sufficiently large X X X ,
P ( 1 log X ∫ − ∞ ∞ ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ≫ ( log log X ) − 1 / 200 ) ≪ ( log log X ) − 1 / 200 . (21) \mathbb{P}\left(\frac{1}{\log X}\int_{-\infty}^{\infty}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\gg(\log\log X)^{-1/200}\right)\ll(\log\log X)^{-1/200}. \tag*{(21)} P ( log X 1 ∫ − ∞ ∞ σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ≫ ( log log X ) − 1/200 ) ≪ ( log log X ) − 1/200 . (21)
Proof. Set T 0 = ( log log X ) − 1 / 10 T_0=(\log\log X)^{-1/10} T 0 = ( log log X ) − 1/10 . We divide the t t t -integral in (21) into ∣ t ∣ ≤ T 0 |t|\le T_0 ∣ t ∣ ≤ T 0 , T 0 < ∣ t ∣ ≤ 1 T_0<|t|\le1 T 0 < ∣ t ∣ ≤ 1 , 1 < ∣ t ∣ ≤ ( log X ) 900 1<|t|\le(\log X)^{900} 1 < ∣ t ∣ ≤ ( log X ) 900 , and the remaining tail.
We begin with the range containing the singularity. If σ ≤ T ≤ T 0 \sigma\le T \le T_0 σ ≤ T ≤ T 0 is a dyadic multiple of σ \sigma σ , put Y = exp ( 1 / T ) Y=\exp(1/T) Y = exp ( 1/ T ) . For r = p − 1 / 2 − σ r=p^{-1/2-\sigma} r = p − 1/2 − σ and real ϕ \phi ϕ , direct averaging over a Rademacher variable ϵ \epsilon ϵ gives the exact identity
E ϵ ∣ 1 + ϵ r e − i ϕ 1 + ϵ r ∣ 2 = 1 + 4 r 2 ( 1 − cos ϕ ) ( 1 − r 2 ) 2 . (22) \mathbb{E}_{\epsilon}\left|\frac{1+\epsilon r e^{-i\phi}}{1+\epsilon r}\right|^2=1+\frac{4r^2(1-\cos\phi)}{(1-r^2)^2}. \tag*{(22)} E ϵ 1 + ϵr 1 + ϵr e − i ϕ 2 = 1 + ( 1 − r 2 ) 2 4 r 2 ( 1 − cos ϕ ) . (22)
Taking the product of (3.3) over p ≤ Y p\le Y p ≤ Y and using independence, we obtain
E ∏ p ≤ Y ∣ 1 + f ( p ) p − 1 / 2 − σ − i t 1 + f ( p ) p − 1 / 2 − σ ∣ 2 = ∏ p ≤ Y ( 1 + 4 p − 1 − 2 σ ( 1 − cos ( t log p ) ) ( 1 − p − 1 − 2 σ ) 2 ) . \mathbb{E}\prod_{p\le Y}\left|\frac{1+f(p)p^{-1/2-\sigma-it}}{1+f(p)p^{-1/2-\sigma}}\right|^2=\prod_{p\le Y}\left(1+\frac{4p^{-1-2\sigma}(1-\cos(t\log p))}{(1-p^{-1-2\sigma})^2}\right). E p ≤ Y ∏ 1 + f ( p ) p − 1/2 − σ 1 + f ( p ) p − 1/2 − σ − i t 2 = p ≤ Y ∏ ( 1 + ( 1 − p − 1 − 2 σ ) 2 4 p − 1 − 2 σ ( 1 − cos ( t log p )) ) .
Since ( 1 − p − 1 − 2 σ ) − 2 ≪ 1 (1-p^{-1-2\sigma})^{-2}\ll1 ( 1 − p − 1 − 2 σ ) − 2 ≪ 1 uniformly in p p p , taking logarithms and using log ( 1 + v ) ≤ v \log(1+v)\le v log ( 1 + v ) ≤ v and 1 − cos v ≪ v 2 1-\cos v\ll v^2 1 − cos v ≪ v 2 , we have
log E ∏ p ≤ Y ∣ 1 + f ( p ) p − 1 / 2 − σ − i t 1 + f ( p ) p − 1 / 2 − σ ∣ 2 ≪ t 2 ∑ p ≤ Y ( log p ) 2 p ≪ t 2 ( log Y ) 2 . \log\mathbb{E}\prod_{p\le Y}\left|\frac{1+f(p)p^{-1/2-\sigma-it}}{1+f(p)p^{-1/2-\sigma}}\right|^2\ll t^2\sum_{p\le Y}\frac{(\log p)^2}{p}\ll t^2(\log Y)^2. log E p ≤ Y ∏ 1 + f ( p ) p − 1/2 − σ 1 + f ( p ) p − 1/2 − σ − i t 2 ≪ t 2 p ≤ Y ∑ p ( log p ) 2 ≪ t 2 ( log Y ) 2 .
For ∣ t ∣ ≤ 2 T |t|\le2T ∣ t ∣ ≤ 2 T and log Y = 1 / T \log Y=1/T log Y = 1/ T , the last expression is O ( 1 ) O(1) O ( 1 ) . Consequently,
E ∏ p ≤ Y ∣ 1 + f ( p ) p − 1 / 2 − σ − i t 1 + f ( p ) p − 1 / 2 − σ ∣ 2 ≪ 1 (23) \mathbb{E}\prod_{p\le Y}\left|\frac{1+f(p)p^{-1/2-\sigma-it}}{1+f(p)p^{-1/2-\sigma}}\right|^2\ll1 \tag*{(23)} E p ≤ Y ∏ 1 + f ( p ) p − 1/2 − σ 1 + f ( p ) p − 1/2 − σ − i t 2 ≪ 1 (23)
uniformly for ∣ t ∣ ≤ 2 T |t|\le2T ∣ t ∣ ≤ 2 T . Independence and Mertens’ estimate also give
E ∏ Y < p ≤ X ∣ 1 + f ( p ) p − 1 / 2 − σ − i t ∣ 2 = ∏ Y < p ≤ X ( 1 + p − 1 − 2 σ ) ≪ log X log Y = T log X . (24) \mathbb{E}\prod_{Y<p\le X}|1+f(p)p^{-1/2-\sigma-it}|^2=\prod_{Y<p\le X}(1+p^{-1-2\sigma})\ll\frac{\log X}{\log Y}=T\log X. \tag*{(24)} E Y < p ≤ X ∏ ∣1 + f ( p ) p − 1/2 − σ − i t ∣ 2 = Y < p ≤ X ∏ ( 1 + p − 1 − 2 σ ) ≪ log Y log X = T log X . (24)
It remains to control the real-axis factor that was divided out in (3.4). Set u = log ( 1 / T ) u=\log(1/T) u = log ( 1/ T ) . Since σ log Y ≤ 1 \sigma\log Y\le1 σ log Y ≤ 1 , Mertens’ estimate and partial summation show uniformly that
∑ p ≤ Y p − 1 − 2 σ = u + O ( 1 ) . (25) \sum_{p\le Y}p^{-1-2\sigma}=u+O(1). \tag*{(25)} p ≤ Y ∑ p − 1 − 2 σ = u + O ( 1 ) . (25)
Taylor expansion, with an absolutely summable remainder, yields
log ∣ F Y ( 1 / 2 + σ ) ∣ 2 = 2 ∑ p ≤ Y f ( p ) p 1 / 2 + σ − ∑ p ≤ Y 1 p 1 + 2 σ + O ( 1 ) . (26) \log|F_Y(1/2+\sigma)|^2=2\sum_{p\le Y}\frac{f(p)}{p^{1/2+\sigma}}-\sum_{p\le Y}\frac{1}{p^{1+2\sigma}}+O(1). \tag*{(26)} log ∣ F Y ( 1/2 + σ ) ∣ 2 = 2 p ≤ Y ∑ p 1/2 + σ f ( p ) − p ≤ Y ∑ p 1 + 2 σ 1 + O ( 1 ) . (26)
The event ∣ F Y ( 1 / 2 + σ ) ∣ 2 > T 1 / 3 = e − u / 3 |F_Y(1/2+\sigma)|^2>T^{1/3}=e^{-u/3} ∣ F Y ( 1/2 + σ ) ∣ 2 > T 1/3 = e − u /3 therefore forces
∑ p ≤ Y f ( p ) p 1 / 2 + σ > u 3 − O ( 1 ) . \sum_{p\le Y}\frac{f(p)}{p^{1/2+\sigma}}>\frac{u}{3}-O(1). p ≤ Y ∑ p 1/2 + σ f ( p ) > 3 u − O ( 1 ) .
Hoeffding’s inequality consequently gives
P ( ∣ F Y ( 1 / 2 + σ ) ∣ 2 > T 1 / 3 ) ≤ exp ( − ( u / 3 − O ( 1 ) ) 2 2 ( u + O ( 1 ) ) ) ≪ T 1 / 19 . (27) \mathbb{P}\left(|F_Y(1/2+\sigma)|^2>T^{1/3}\right)\le\exp\left(-\frac{(u/3-O(1))^2}{2(u+O(1))}\right)\ll T^{1/19}. \tag*{(27)} P ( ∣ F Y ( 1/2 + σ ) ∣ 2 > T 1/3 ) ≤ exp ( − 2 ( u + O ( 1 )) ( u /3 − O ( 1 ) ) 2 ) ≪ T 1/19 . (27)
Let Ω X ( 0 ) \Omega_X^{(0)} Ω X ( 0 ) be the barrier event that ∣ F exp ( 1 / T ) ( 1 / 2 + σ ) ∣ 2 ≤ T 1 / 3 |F_{\exp(1/T)}(1/2+\sigma)|^2\le T^{1/3} ∣ F e x p ( 1/ T ) ( 1/2 + σ ) ∣ 2 ≤ T 1/3 for every T = 2 j σ ≤ T 0 T=2^j\sigma\le T_0 T = 2 j σ ≤ T 0 , where j ≥ 0 j\ge0 j ≥ 0 . These dyadic intervals, together with ∣ t ∣ ≤ σ |t|\le\sigma ∣ t ∣ ≤ σ , cover ∣ t ∣ ≤ T 0 |t|\le T_0 ∣ t ∣ ≤ T 0 . Summing (3.8) over the geometric sequence gives
P ( ( Ω X ( 0 ) ) c ) ≪ ∑ j ≥ 0 2 j σ ≤ T 0 ( 2 j σ ) 1 / 19 ≪ T 0 1 / 19 ≪ ( log log X ) − 1 / 200 . (28) \mathbb{P}\left((\Omega_X^{(0)})^c\right)\ll\sum_{\substack{j\ge0\\2^j\sigma\le T_0}}(2^j\sigma)^{1/19}\ll T_0^{1/19}\ll(\log\log X)^{-1/200}. \tag*{(28)} P ( ( Ω X ( 0 ) ) c ) ≪ j ≥ 0 2 j σ ≤ T 0 ∑ ( 2 j σ ) 1/19 ≪ T 0 1/19 ≪ ( log log X ) − 1/200 . (28)
We now explain why no independence between Ω X ( 0 ) \Omega_X^{(0)} Ω X ( 0 ) and the Euler product ratio is required. On this event, and for T ≤ ∣ t ∣ ≤ 2 T T \le|t| \le2T T ≤ ∣ t ∣ ≤ 2 T , one has pointwise
1 Ω X ( 0 ) ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 ≤ T 1 / 3 ∣ F Y ( 1 / 2 + σ + i t ) F Y ( 1 / 2 + σ ) ∣ 2 × ∏ Y < p ≤ X ∣ 1 + f ( p ) p − 1 / 2 − σ − i t ∣ 2 . \mathbf{1}_{\Omega_X^{(0)}} |F_X(1/2+\sigma+it)|^2 \le T^{1/3}\left|\frac{F_Y(1/2+\sigma+it)}{F_Y(1/2+\sigma)}\right|^2 \times\prod_{Y<p\le X}|1+f(p)p^{-1/2-\sigma-it}|^2. 1 Ω X ( 0 ) ∣ F X ( 1/2 + σ + i t ) ∣ 2 ≤ T 1/3 F Y ( 1/2 + σ ) F Y ( 1/2 + σ + i t ) 2 × Y < p ≤ X ∏ ∣1 + f ( p ) p − 1/2 − σ − i t ∣ 2 .
The ratio uses only primes at most Y Y Y , whereas the last product uses only primes exceeding Y Y Y . Independence and (3.4)–(3.5) therefore give
E [ 1 Ω X ( 0 ) ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 ] ≪ T 4 / 3 log X . \mathbb{E}\left[\mathbf{1}_{\Omega_X^{(0)}}|F_X(1/2+\sigma+it)|^2\right] \ll T^{4/3}\log X. E [ 1 Ω X ( 0 ) ∣ F X ( 1/2 + σ + i t ) ∣ 2 ] ≪ T 4/3 log X .
Since the dyadic interval has length O ( T ) O(T) O ( T ) and its kernel is O ( T − 2 ) O(T^{-2}) O ( T − 2 ) , it follows that
E [ 1 Ω X ( 0 ) 1 log X ∫ T ≤ ∣ t ∣ ≤ 2 T ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ] ≪ T 1 / 3 . \mathbb{E}\left[\mathbf{1}_{\Omega_X^{(0)}}\frac{1}{\log X}\int_{T\le|t|\le2T}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right]\ll T^{1/3}. E [ 1 Ω X ( 0 ) log X 1 ∫ T ≤ ∣ t ∣ ≤ 2 T σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ] ≪ T 1/3 .
For ∣ t ∣ ≤ σ |t|\le\sigma ∣ t ∣ ≤ σ , take Y = X Y=X Y = X . On Ω X ( 0 ) \Omega_X^{(0)} Ω X ( 0 ) we have ∣ F X ( 1 / 2 + σ ) ∣ 2 ≤ σ 1 / 3 |F_X(1/2+\sigma)|^2\le\sigma^{1/3} ∣ F X ( 1/2 + σ ) ∣ 2 ≤ σ 1/3 , while (3.4) remains valid. Since 1 log X ∫ ∣ t ∣ ≤ σ d t σ 2 + t 2 ≪ 1 \frac{1}{\log X}\int_{|t|\le\sigma}\frac{dt}{\sigma^2+t^2}\ll1 l o g X 1 ∫ ∣ t ∣ ≤ σ σ 2 + t 2 d t ≪ 1 , the same argument gives
E [ 1 Ω X ( 0 ) 1 log X ∫ ∣ t ∣ ≤ σ ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ] ≪ σ 1 / 3 . \mathbb{E}\left[\mathbf{1}_{\Omega_X^{(0)}}\frac{1}{\log X}\int_{|t|\le\sigma}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right]\ll\sigma^{1/3}. E [ 1 Ω X ( 0 ) log X 1 ∫ ∣ t ∣ ≤ σ σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ] ≪ σ 1/3 .
Summing over the dyadic values of T T T therefore yields
E [ 1 Ω X ( 0 ) 1 log X ∫ ∣ t ∣ ≤ T 0 ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ] ≪ σ 1 / 3 + ∑ 2 j σ ≤ T 0 ( 2 j σ ) 1 / 3 ≪ T 0 1 / 3 . \mathbb{E}\left[\mathbf{1}_{\Omega_X^{(0)}}\frac{1}{\log X}\int_{|t|\le T_0}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right]\ll\sigma^{1/3}+\sum_{2^j\sigma\le T_0}(2^j\sigma)^{1/3}\ll T_0^{1/3}. E [ 1 Ω X ( 0 ) log X 1 ∫ ∣ t ∣ ≤ T 0 σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ] ≪ σ 1/3 + 2 j σ ≤ T 0 ∑ ( 2 j σ ) 1/3 ≪ T 0 1/3 .
Markov’s inequality now gives
P ( Ω X ( 0 ) and 1 log X ∫ ∣ t ∣ ≤ T 0 ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t > ( log log X ) − 1 / 200 ) ≪ ( log log X ) 1 / 200 T 0 1 / 3 ≪ ( log log X ) − 1 / 40 . (29) \mathbb{P}\left(\Omega_X^{(0)}\text{ and }\frac{1}{\log X}\int_{|t|\le T_0}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt>(\log\log X)^{-1/200}\right)
\ll(\log\log X)^{1/200}T_0^{1/3}\ll(\log\log X)^{-1/40}. \tag*{(29)} P ( Ω X ( 0 ) and log X 1 ∫ ∣ t ∣ ≤ T 0 σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t > ( log log X ) − 1/200 ) ≪ ( log log X ) 1/200 T 0 1/3 ≪ ( log log X ) − 1/40 . (29)
The other three ranges follow from (3.1). Since q = 2 / 3 q=2/3 q = 2/3 and ( a + b ) q ≤ a q + b q (a+b)^q\le a^q+b^q ( a + b ) q ≤ a q + b q , decomposition into unit intervals gives
E ( 1 log X ∫ T 0 < ∣ t ∣ ≤ 1 ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ) q ≪ T 0 − 2 q ( log log X ) − q / 2 = ( log log X ) − 1 / 5 . (30) \mathbb{E}\left(\frac{1}{\log X}\int_{T_0<|t|\le1}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right)^q\ll T_0^{-2q}(\log\log X)^{-q/2}=(\log\log X)^{-1/5}. \tag*{(30)} E ( log X 1 ∫ T 0 < ∣ t ∣ ≤ 1 σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ) q ≪ T 0 − 2 q ( log log X ) − q /2 = ( log log X ) − 1/5 . (30)
For 1 < ∣ t ∣ ≤ ( log X ) 900 1<|t|\le(\log X)^{900} 1 < ∣ t ∣ ≤ ( log X ) 900 , cover the range by intervals [ N − 1 / 2 , N + 1 / 2 ] [N-1/2,N+1/2] [ N − 1/2 , N + 1/2 ] . The factor ( 1 + N 2 ) − 1 (1+N^2)^{-1} ( 1 + N 2 ) − 1 from the kernel similarly yields
E ( 1 log X ∫ 1 < ∣ t ∣ ≤ ( log X ) 900 ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ) q ≪ ( log log X ) − q / 2 ∑ N ∈ Z ( log log ( ∣ N ∣ + 10 ) ) q ( 1 + N 2 ) q ≪ ( log log X ) − 1 / 3 . \mathbb{E}\left(\frac{1}{\log X}\int_{1<|t|\le(\log X)^{900}}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right)^q
\ll(\log\log X)^{-q/2}\sum_{N\in\mathbb{Z}}\frac{(\log\log(|N|+10))^q}{(1+N^2)^q}\ll(\log\log X)^{-1/3}. E ( log X 1 ∫ 1 < ∣ t ∣ ≤ ( l o g X ) 900 σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ) q ≪ ( log log X ) − q /2 N ∈ Z ∑ ( 1 + N 2 ) q ( log log ( ∣ N ∣ + 10 ) ) q ≪ ( log log X ) − 1/3 .
The series converges because 2 q = 4 / 3 > 1 2q=4/3>1 2 q = 4/3 > 1 . Finally, E ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 = ∏ p ≤ X ( 1 + p − 1 − 2 σ ) ≪ log X \mathbb{E}|F_X(1/2+\sigma+it)|^2=\prod_{p\le X}(1+p^{-1-2\sigma})\ll\log X E ∣ F X ( 1/2 + σ + i t ) ∣ 2 = ∏ p ≤ X ( 1 + p − 1 − 2 σ ) ≪ log X , so the first moment of the tail is
E ( 1 log X ∫ ∣ t ∣ > ( log X ) 900 ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t ) ≪ ∫ ∣ t ∣ > ( log X ) 900 d t 1 + t 2 ≪ ( log X ) − 900 . \mathbb{E}\left(\frac{1}{\log X}\int_{|t|>(\log X)^{900}}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,dt\right)\ll\int_{|t|>(\log X)^{900}}\frac{dt}{1+t^2}\ll(\log X)^{-900}. E ( log X 1 ∫ ∣ t ∣ > ( l o g X ) 900 σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t ) ≪ ∫ ∣ t ∣ > ( l o g X ) 900 1 + t 2 d t ≪ ( log X ) − 900 .
Applying Markov’s inequality with threshold ( log log X ) − 1 / 200 (\log\log X)^{-1/200} ( log log X ) − 1/200 multiplies the first two q q q -th-moment estimates by ( log log X ) q / 200 = ( log log X ) 1 / 300 (\log\log X)^{q/200}=(\log\log X)^{1/300} ( log log X ) q /200 = ( log log X ) 1/300 . The resulting probabilities, as well as the first-moment tail probability, are o ( ( log log X ) − 1 / 200 ) o((\log\log X)^{-1/200}) o (( log log X ) − 1/200 ) . Combining these estimates with (3.9) and (3.10) proves the lemma. □
We can now state the estimate for the X X X -smooth contribution in the form needed later.
Proposition 3.2 (Mean square of the X X X -smooth contribution).
P ( 1 log X ∫ X 8 / 7 X 4 / 3 ∣ C X ( x ) ∣ 2 d x x > ( log log X ) 1 / 3 ) ≪ ( log log X ) − 1 / 200 . (31) \mathbb{P}\left(\frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}} |C_X(x)|^2\frac{\mathrm{d}x}{x}>(\log\log X)^{1/3}\right)\ll(\log\log X)^{-1/200}. \tag*{(31)} P ( log X 1 ∫ X 8/7 X 4/3 ∣ C X ( x ) ∣ 2 x d x > ( log log X ) 1/3 ) ≪ ( log log X ) − 1/200 . (31)
Proof. Set σ = 1 / log X \sigma=1/\log X σ = 1/ log X . Applying Parseval’s identity (2.10) to the Dirichlet polynomial F X ( 1 / 2 + s ) F_X(1/2+s) F X ( 1/2 + s ) gives
∫ 1 ∞ ∣ M X ( x ) ∣ 2 x − 1 − 2 σ d x = 1 2 π ∫ − ∞ ∞ ∣ F X ( 1 / 2 + σ + i t ) ∣ 2 σ 2 + t 2 d t . \int_1^\infty|M_X(x)|^2x^{-1-2\sigma}\,\mathrm{d}x=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{|F_X(1/2+\sigma+it)|^2}{\sigma^2+t^2}\,\mathrm{d}t. ∫ 1 ∞ ∣ M X ( x ) ∣ 2 x − 1 − 2 σ d x = 2 π 1 ∫ − ∞ ∞ σ 2 + t 2 ∣ F X ( 1/2 + σ + i t ) ∣ 2 d t .
Since x 2 σ ≤ exp ( 8 / 3 ) x^{2\sigma}\le\exp(8/3) x 2 σ ≤ exp ( 8/3 ) for x ≤ X 4 / 3 x\le X^{4/3} x ≤ X 4/3 , Lemma 3.1 and the preceding identity imply, outside an event of probability O ( ( log log X ) − 1 / 200 ) O((\log\log X)^{-1/200}) O (( log log X ) − 1/200 ) , that
1 log X ∫ X 8 / 7 X 4 / 3 ∣ M X ( x ) ∣ 2 d x x ≪ ( log log X ) − 1 / 200 . \frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}} |M_X(x)|^2\frac{\mathrm{d}x}{x}\ll(\log\log X)^{-1/200}. log X 1 ∫ X 8/7 X 4/3 ∣ M X ( x ) ∣ 2 x d x ≪ ( log log X ) − 1/200 .
On the other hand, Tonelli’s theorem and the estimate E ∣ D X ( x ) ∣ 2 ≤ 1 \mathbb{E}|D_X(x)|^2\le1 E ∣ D X ( x ) ∣ 2 ≤ 1 give
E ( 1 log X ∫ X 8 / 7 X 4 / 3 ∣ D X ( x ) ∣ 2 d x x ) = 1 log X ∫ X 8 / 7 X 4 / 3 E ∣ D X ( x ) ∣ 2 d x x ≤ 1 log X log ( X 4 / 3 − 8 / 7 ) = 4 21 . \mathbb{E}\left(\frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}} |D_X(x)|^2\frac{\mathrm{d}x}{x}\right)=\frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}}\mathbb{E}|D_X(x)|^2\frac{\mathrm{d}x}{x}
\le\frac{1}{\log X}\log\left(X^{4/3-8/7}\right)=\frac{4}{21}. E ( log X 1 ∫ X 8/7 X 4/3 ∣ D X ( x ) ∣ 2 x d x ) = log X 1 ∫ X 8/7 X 4/3 E ∣ D X ( x ) ∣ 2 x d x ≤ log X 1 log ( X 4/3 − 8/7 ) = 21 4 .
Thus Markov’s inequality shows that, outside an event of probability O ( ( log log X ) − 1 / 3 ) O((\log\log X)^{-1/3}) O (( log log X ) − 1/3 ) , the last integral is at most ( log log X ) 1 / 3 / 4 (\log\log X)^{1/3}/4 ( log log X ) 1/3 /4 . For sufficiently large X X X , the corresponding integral for M X M_X M X is also at most ( log log X ) 1 / 3 / 4 (\log\log X)^{1/3}/4 ( log log X ) 1/3 /4 . Since ∣ C X ( x ) ∣ 2 ≤ 2 ∣ M X ( x ) ∣ 2 + 2 ∣ D X ( x ) ∣ 2 |C_X(x)|^2\le2|M_X(x)|^2+2|D_X(x)|^2 ∣ C X ( x ) ∣ 2 ≤ 2∣ M X ( x ) ∣ 2 + 2∣ D X ( x ) ∣ 2 , the required bound follows by taking the union of the two exceptional events and using ( log log X ) − 1 / 3 ≤ ( log log X ) − 1 / 200 (\log\log X)^{-1/3}\le(\log\log X)^{-1/200} ( log log X ) − 1/3 ≤ ( log log X ) − 1/200 . □
An immediate consequence, useful when the endpoints are logarithmically shifted, is that whenever the integral in (3.12) is at most ( log log X ) 1 / 3 (\log\log X)^{1/3} ( log log X ) 1/3 , one has, for every A > 0 A>0 A > 0 ,
1 log X ∫ X 8 / 7 X 4 / 3 1 { ∣ C X ( x ) ∣ > A } d x x ≤ ( log log X ) 1 / 3 A 2 . \frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}}\mathbf{1}_{\{|C_X(x)|>A\}}\frac{\mathrm{d}x}{x}\le\frac{(\log\log X)^{1/3}}{A^2}. log X 1 ∫ X 8/7 X 4/3 1 { ∣ C X ( x ) ∣ > A } x d x ≤ A 2 ( log log X ) 1/3 .
The multiplier and conditional variance
The multiplier
Recall F X F_X F X , B X B_X B X and Z X Z_X Z X from (2.3), (2.7) and (2.4). The Dirichlet series of the extended Rademacher RMF restricted to X X X -smooth integers is
∏ p ≤ X ( 1 − f ( p ) p − s ) − 1 = F X ( s ) Z X ( 2 s ) . \prod_{p\le X}(1-f(p)p^{-s})^{-1}=F_X(s)Z_X(2s). p ≤ X ∏ ( 1 − f ( p ) p − s ) − 1 = F X ( s ) Z X ( 2 s ) .
This elementary factorisation is the point at which the extended Rademacher RMF separates from the Rademacher RMF.
For σ ≥ 0 \sigma\ge0 σ ≥ 0 and t ∈ R t\in\mathbb{R} t ∈ R , define
m X , σ ( t ) : = ( 4 σ + i t − 1 ) Z X ( 1 + 2 σ + 2 i t ) . (32) m_{X,\sigma}(t):=(4^{\sigma+it}-1)Z_X(1+2\sigma+2it). \tag*{(32)} m X , σ ( t ) := ( 4 σ + i t − 1 ) Z X ( 1 + 2 σ + 2 i t ) . (32)
The factor 4 σ + i t − 1 4^{\sigma+it}-1 4 σ + i t − 1 removes the singular growth of Z X Z_X Z X near σ = t = 0 \sigma=t=0 σ = t = 0 . We shall repeatedly use the following bounds.
Lemma 4.1. For sufficiently large X X X , suppose that 0 ≤ σ ≤ 100 ( log log X ) 1 / 100 / log X 0 \le\sigma\le100(\log\log X)^{1/100}/\log X 0 ≤ σ ≤ 100 ( log log X ) 1/100 / log X . Then the following hold.
(1) If ∣ t ∣ ≤ 1 |t| \le1 ∣ t ∣ ≤ 1 , then ∣ m X , σ ( t ) ∣ ≪ 1 |m_{X,\sigma}(t)| \ll1 ∣ m X , σ ( t ) ∣ ≪ 1 .
(2) For every t ∈ R t \in\mathbb{R} t ∈ R ,
∣ m X , σ ( t ) ∣ ≪ min ( log X , ( 1 + log ( 2 + ∣ t ∣ ) ) 2 ) . |m_{X,\sigma}(t)| \ll\min\left(\log X,\left(1+\log(2+|t|)\right)^2\right). ∣ m X , σ ( t ) ∣ ≪ min ( log X , ( 1 + log ( 2 + ∣ t ∣ ) ) 2 ) .
(3) If 1 / 3 ≤ ∣ t ∣ ≤ 1 / 2 1/3 \le|t| \le1/2 1/3 ≤ ∣ t ∣ ≤ 1/2 , then ∣ m X , σ ( t ) ∣ ≍ 1 |m_{X,\sigma}(t)| \asymp1 ∣ m X , σ ( t ) ∣ ≍ 1 .
All implied constants are absolute.
Proof. Expanding the logarithm of the Euler product, the terms coming from prime powers p k p^k p k , k ≥ 2 k \ge2 k ≥ 2 , contribute O ( 1 ) O(1) O ( 1 ) uniformly in σ \sigma σ . Hence
log ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ = ℜ ∑ p ≤ X 1 p 1 + 2 σ + 2 i t + O ( 1 ) . (33) \log|Z_X(1+2\sigma+2it)|=\Re\sum_{p\le X}\frac{1}{p^{1+2\sigma+2it}}+O(1). \tag*{(33)} log ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ = ℜ p ≤ X ∑ p 1 + 2 σ + 2 i t 1 + O ( 1 ) . (33)
Suppose first that ∣ t ∣ ≤ 1 |t| \le1 ∣ t ∣ ≤ 1 . By the prime number theorem and partial summation, we have
ℜ ∑ p ≤ X 1 p 1 + 2 σ + 2 i t = ∫ log 2 log X e − 2 σ v cos ( 2 t v ) d v v + O ( 1 ) = log ( 1 + min ( log X , 1 σ + ∣ t ∣ ) ) + O ( 1 ) . \begin{aligned}
\Re\sum_{p\le X}\frac{1}{p^{1+2\sigma+2it}}=\int_{\log2}^{\log X}e^{-2\sigma v}\cos(2tv)\frac{\mathrm{d}v}{v}+O(1) \\
&=\log\left(1+\min\left(\log X,\frac{1}{\sigma+|t|}\right)\right)+O(1).
\end{aligned} ℜ p ≤ X ∑ p 1 + 2 σ + 2 i t 1 = ∫ l o g 2 l o g X e − 2 σ v cos ( 2 t v ) v d v + O ( 1 ) = log ( 1 + min ( log X , σ + ∣ t ∣ 1 ) ) + O ( 1 ) .
To get the last line, we split the integral at min ( log X , 1 / ( σ + ∣ t ∣ ) ) \min(\log X,1/(\sigma+|t|)) min ( log X , 1/ ( σ + ∣ t ∣ )) and use exponential decay when σ ≥ ∣ t ∣ \sigma\ge|t| σ ≥ ∣ t ∣ and integration by parts when ∣ t ∣ > σ |t|>\sigma ∣ t ∣ > σ . Since ∣ 4 σ + i t − 1 ∣ ≪ σ + ∣ t ∣ |4^{\sigma+it}-1| \ll\sigma+|t| ∣ 4 σ + i t − 1∣ ≪ σ + ∣ t ∣ , we obtain
∣ m X , σ ( t ) ∣ ≪ ( σ + ∣ t ∣ ) ( 1 + min ( log X , 1 σ + ∣ t ∣ ) ) ≪ 1 , |m_{X,\sigma}(t)|\ll(\sigma+|t|)\left(1+\min\left(\log X,\frac{1}{\sigma+|t|}\right)\right)\ll1, ∣ m X , σ ( t ) ∣ ≪ ( σ + ∣ t ∣ ) ( 1 + min ( log X , σ + ∣ t ∣ 1 ) ) ≪ 1 ,
which proves the first assertion.
For ∣ t ∣ ≥ 1 |t|\ge1 ∣ t ∣ ≥ 1 , we have ∣ 4 σ + i t − 1 ∣ ≪ 1 |4^{\sigma+it}-1|\ll1 ∣ 4 σ + i t − 1∣ ≪ 1 , while trivially ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ log X |Z_X(1+2\sigma+2it)|\ll\log X ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ log X . If log X ≥ C log 2 ( 2 + ∣ t ∣ ) \log X\ge C\log^2(2+|t|) log X ≥ C log 2 ( 2 + ∣ t ∣ ) for some sufficiently large constant C C C , then by Mertens’ theorem and the prime number theorem, we have
∑ p ≤ exp ( C log 2 ( 2 + ∣ t ∣ ) ) 1 p 1 + 2 σ + 2 i t ≤ 2 log log ( 2 + ∣ t ∣ ) + O ( 1 ) , \sum_{p\le\exp(C\log^2(2+|t|))}\frac{1}{p^{1+2\sigma+2it}}\le2\log\log(2+|t|)+O(1), p ≤ e x p ( C l o g 2 ( 2 + ∣ t ∣ )) ∑ p 1 + 2 σ + 2 i t 1 ≤ 2 log log ( 2 + ∣ t ∣ ) + O ( 1 ) ,
and
∑ exp ( C log 2 ( 2 + ∣ t ∣ ) ) < p ≤ X 1 p 1 + 2 σ + 2 i t ≪ 1 ∣ t ∣ log 2 ( 2 + ∣ t ∣ ) + ( 1 + ∣ t ∣ ) e − c C log ( 2 + ∣ t ∣ ) ≪ 1. \sum_{\exp(C\log^2(2+|t|))<p\le X}\frac{1}{p^{1+2\sigma+2it}}\ll\frac{1}{|t|\log^2(2+|t|)}+(1+|t|)e^{-c\sqrt{C}\log(2+|t|)}\ll1. e x p ( C l o g 2 ( 2 + ∣ t ∣ )) < p ≤ X ∑ p 1 + 2 σ + 2 i t 1 ≪ ∣ t ∣ log 2 ( 2 + ∣ t ∣ ) 1 + ( 1 + ∣ t ∣ ) e − c C l o g ( 2 + ∣ t ∣ ) ≪ 1.
It follows from (4.2) that
∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ ( 1 + log ( 2 + ∣ t ∣ ) ) 2 . |Z_X(1+2\sigma+2it)|\ll(1+\log(2+|t|))^2. ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ ( 1 + log ( 2 + ∣ t ∣ ) ) 2 .
If instead log X ≤ C log 2 ( 2 + ∣ t ∣ ) \log X\le C\log^2(2+|t|) log X ≤ C log 2 ( 2 + ∣ t ∣ ) , the same bound follows clearly from
∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ log X ≪ log 2 ( 2 + ∣ t ∣ ) . |Z_X(1+2\sigma+2it)|\ll\log X\ll\log^2(2+|t|). ∣ Z X ( 1 + 2 σ + 2 i t ) ∣ ≪ log X ≪ log 2 ( 2 + ∣ t ∣ ) .
Together with the trivial bound, this proves
∣ m X , σ ( t ) ∣ ≪ min ( log X , ( 1 + log ( 2 + ∣ t ∣ ) ) 2 ) . |m_{X,\sigma}(t)|\ll\min\left(\log X,(1+\log(2+|t|))^2\right). ∣ m X , σ ( t ) ∣ ≪ min ( log X , ( 1 + log ( 2 + ∣ t ∣ ) ) 2 ) .
For 1 / 3 ≤ ∣ t ∣ ≤ 1 / 2 1/3\le|t|\le1/2 1/3 ≤ ∣ t ∣ ≤ 1/2 , the prime number theorem and partial summation give, uniformly in σ \sigma σ ,
∑ p > X 1 p 1 + 2 σ + 2 i t ≪ 1 ∣ t ∣ log X + e − c log X = o ( 1 ) , \sum_{p>X}\frac{1}{p^{1+2\sigma+2it}}\ll\frac{1}{|t|\log X}+e^{-c\sqrt{\log X}}=o(1), p > X ∑ p 1 + 2 σ + 2 i t 1 ≪ ∣ t ∣ log X 1 + e − c l o g X = o ( 1 ) ,
while the terms with prime powers p k p^k p k , k ≥ 2 k \ge2 k ≥ 2 , contribute O ( X − 1 ) O(X^{-1}) O ( X − 1 ) . Comparing the logarithms of the corresponding Euler products, we obtain
Z X ( 1 + 2 σ + 2 i t ) = ζ ( 1 + 2 σ + 2 i t ) ( 1 + o ( 1 ) ) . Z_X(1+2\sigma+2it)=\zeta(1+2\sigma+2it)(1+o(1)). Z X ( 1 + 2 σ + 2 i t ) = ζ ( 1 + 2 σ + 2 i t ) ( 1 + o ( 1 )) .
Since ζ ( 1 + 2 σ + 2 i t ) \zeta(1+2\sigma+2it) ζ ( 1 + 2 σ + 2 i t ) and 4 σ + i t − 1 4^{\sigma+it}-1 4 σ + i t − 1 are both bounded above and away from zero uniformly on this range, it follows that ∣ m X , σ ( t ) ∣ ≍ 1 \lvert m_{X,\sigma}(t)\rvert\asymp1 ∣ m X , σ ( t )∣ ≍ 1 , which proves the third assertion. □
For Re s > 1 \operatorname{Re}s>1 Re s > 1 , by termwise integration we obtain the exact identity
∫ 0 ∞ B X ( z ) z − s − 1 d z = ( 4 s − 1 / 2 − 1 ) F X ( s ) Z X ( 2 s ) s ( Re s > 1 ) . (34) \int_0^\infty B_X(z)z^{-s-1}\,dz=\left(4^{s-1/2}-1\right)\frac{F_X(s)Z_X(2s)}{s}
\qquad(\operatorname{Re}s>1). \tag*{(34)} ∫ 0 ∞ B X ( z ) z − s − 1 d z = ( 4 s − 1/2 − 1 ) s F X ( s ) Z X ( 2 s ) ( Re s > 1 ) . (34)
For fixed X X X , the number of X X X -smooth integers at most z z z is at most
∏ p ≤ X ( 1 + log z log p ) ≪ X ( 1 + log z ) π ( X ) . \prod_{p\le X}\left(1+\frac{\log z}{\log p}\right)\ll_X(1+\log z)^{\pi(X)}. p ≤ X ∏ ( 1 + log p log z ) ≪ X ( 1 + log z ) π ( X ) .
Thus the Mellin transform on the left of (34) converges absolutely and is holomorphic for Re s > 0 \operatorname{Re}s>0 Re s > 0 ; there is no contribution near zero because B X ( z ) = 0 B_X(z)=0 B X ( z ) = 0 for z < 1 / 4 z<1/4 z < 1/4 . The right-hand side is also holomorphic in this half-plane. Since both sides are holomorphic for Re s > 0 \operatorname{Re}s>0 Re s > 0 and agree for Re s > 1 \operatorname{Re}s>1 Re s > 1 , the identity theorem extends (34) to Re s > 0 \operatorname{Re}s>0 Re s > 0 . In particular, on writing s = 1 / 2 + σ + i t s=1/2+\sigma+it s = 1/2 + σ + i t , the right-hand side of (34) is
m X , σ ( t ) F X ( 1 / 2 + σ + i t ) 1 / 2 + σ + i t . \frac{m_{X,\sigma}(t)F_X(1/2+\sigma+it)}{1/2+\sigma+it}. 1/2 + σ + i t m X , σ ( t ) F X ( 1/2 + σ + i t ) .
Thus the ζ \zeta ζ -type growth of Z X ( 2 s ) Z_X(2s) Z X ( 2 s ) near s = 1 / 2 s=1/2 s = 1/2 is cancelled uniformly in X X X by the multiplier 4 s − 1 / 2 − 1 4^{s-1/2}-1 4 s − 1/2 − 1 .
The conditional variance
Conditioning on the variables ( f ( p ) ) p ≤ X (f(p))_{p\le X} ( f ( p ) ) p ≤ X , the quantities B X ( x / p ) B_X(x/p) B X ( x / p ) are fixed, while the variables f ( p ) f(p) f ( p ) , p > X p>X p > X , remain independent Rademacher random variables. Recall the definition of Y x Y_x Y x , we have
Var X ( Y x ) = Var ( Y x ∣ ( f ( p ) ) p ≤ X ) = 1 x ∑ X < p ≤ 4 x B X ( x / p ) 2 . \operatorname{Var}_X(Y_x)=\operatorname{Var}\left(Y_x\mid(f(p))_{p\le X}\right)=\frac{1}{x}\sum_{X<p\le4x}B_X(x/p)^2. Var X ( Y x ) = Var ( Y x ∣ ( f ( p ) ) p ≤ X ) = x 1 X < p ≤ 4 x ∑ B X ( x / p ) 2 .
The following proposition shows that this conditional variance is typically not too small. The proof is an adaptation of the corresponding variance estimate in [17 ] , with the multiplier accounting for the additional Euler factor arising in the extended Rademacher RMF.
Proposition 4.2 (Conditional variance of the large-prime increment). Let W 0 ≤ W ≤ ( log log X ) 1 / 100 W_0\le W\le(\log\log X)^{1/100} W 0 ≤ W ≤ ( log log X ) 1/100 , where W 0 W_0 W 0 is a sufficiently large absolute constant, and suppose that X 8 / 7 ≤ x ≤ X 4 / 3 / 4 X^{8/7}\le x\le X^{4/3}/4 X 8/7 ≤ x ≤ X 4/3 /4 . Then we have
Var X ( Y x ) ≫ e − 2.2 W log log X (35) \operatorname{Var}_X(Y_x)\gg\frac{e^{-2.2W}}{\sqrt{\log\log X}} \tag*{(35)} Var X ( Y x ) ≫ log log X e − 2.2 W (35)
with probability ≥ 1 − O ( e − 0.1 W ) \ge1-O(e^{-0.1W}) ≥ 1 − O ( e − 0.1 W ) .
Proof. We first use the constancy of B X B_X B X on intervals of length 1 / 4 1/4 1/4 , together with Huxley’s theorem, to bound the prime sum from below by a weighted integral of B X 2 B_X^2 B X 2 . We then apply Parseval’s identity (19) and multiplicative-chaos estimates to obtain the required lower bound.
For each integer 1 ≤ r ≤ ⌈ 2 X 1 / 7 ⌉ − 1 1\le r\le\lceil2X^{1/7}\rceil-1 1 ≤ r ≤ ⌈ 2 X 1/7 ⌉ − 1 , consider the interval 4 x / ( r + 1 ) < p ≤ 4 x / r 4x/(r+1)<p\le4x/r 4 x / ( r + 1 ) < p ≤ 4 x / r . Its left endpoint exceeds X X X for all sufficiently large X X X , since x ≥ X 8 / 7 x\ge X^{8/7} x ≥ X 8/7 , while its right endpoint is at most 4 x 4x 4 x . Its length is ≍ x / r 2 \asymp x/r^2 ≍ x / r 2 and its left endpoint is ≍ x / r \asymp x/r ≍ x / r . Moreover, r 8 ≪ x r^8 \ll x r 8 ≪ x , and hence x / r 2 ≫ ( x / r ) 6 / 7 x/r^2 \gg(x/r)^{6/7} x / r 2 ≫ ( x / r ) 6/7 . Since 6 / 7 > 7 / 12 6/7 > 7/12 6/7 > 7/12 , Huxley’s theorem on primes in short intervals [23 ] is available. Hence
# { p : 4 x r + 1 < p ≤ 4 x r } ≫ x r ( r + 1 ) log X . \#\left\{p:\frac{4x}{r+1}<p\le\frac{4x}{r}\right\}\gg\frac{x}{r(r+1)\log X}. # { p : r + 1 4 x < p ≤ r 4 x } ≫ r ( r + 1 ) log X x .
For these primes, x / p ∈ [ r / 4 , ( r + 1 ) / 4 ) x/p\in[r/4,(r+1)/4) x / p ∈ [ r /4 , ( r + 1 ) /4 ) , so B X ( x / p ) = B X ( r / 4 ) B_X(x/p)=B_X(r/4) B X ( x / p ) = B X ( r /4 ) . Since ∫ r / 4 ( r + 1 ) / 4 z − 2 d z = 4 / ( r ( r + 1 ) ) \int_{r/4}^{(r+1)/4}z^{-2}\,\mathrm{d}z=4/(r(r+1)) ∫ r /4 ( r + 1 ) /4 z − 2 d z = 4/ ( r ( r + 1 )) , summing over r r r we get
1 x ∑ X < p ≤ 4 x B X ( x / p ) 2 ≫ 1 log X ∑ 1 ≤ r ≤ ⌈ 2 X 1 / 7 ⌉ − 1 B X ( r / 4 ) 2 r ( r + 1 ) ≫ 1 log X ∫ 1 / 4 X 1 / 7 / 2 B X ( z ) 2 d z z 2 . \frac{1}{x}\sum_{X<p\le4x}B_X(x/p)^2\gg\frac{1}{\log X}\sum_{1\le r\le\lceil2X^{1/7}\rceil-1}\frac{B_X(r/4)^2}{r(r+1)}\gg\frac{1}{\log X}\int_{1/4}^{X^{1/7}/2}B_X(z)^2\frac{\mathrm{d}z}{z^2}. x 1 X < p ≤ 4 x ∑ B X ( x / p ) 2 ≫ log X 1 1 ≤ r ≤ ⌈ 2 X 1/7 ⌉ − 1 ∑ r ( r + 1 ) B X ( r /4 ) 2 ≫ log X 1 ∫ 1/4 X 1/7 /2 B X ( z ) 2 z 2 d z .
It remains to estimate the integral. Let η = 42 W / log X \eta=42W/\log X η = 42 W / log X . Since B X ( z ) = 0 B_X(z)=0 B X ( z ) = 0 for z < 1 / 4 z<1/4 z < 1/4 and ( X 1 / 7 / 2 ) − η ≤ 2 e − 6 W (X^{1/7}/2)^{-\eta}\le2e^{-6W} ( X 1/7 /2 ) − η ≤ 2 e − 6 W . By comparing z − 2 z^{-2} z − 2 with z − 2 − 2 η z^{-2-2\eta} z − 2 − 2 η on [ 1 / 4 , X 1 / 7 / 2 ] [1/4,X^{1/7}/2] [ 1/4 , X 1/7 /2 ] and controlling the tail z > X 1 / 7 / 2 z>X^{1/7}/2 z > X 1/7 /2 by z − 2 − η z^{-2-\eta} z − 2 − η , we obtain
∫ 1 / 4 X 1 / 7 / 2 B X ( z ) 2 d z z 2 ≥ 4 − 2 η ( ∫ 0 ∞ B X ( z ) 2 d z z 2 + 2 η − 2 e − 6 W ∫ 0 ∞ B X ( z ) 2 d z z 2 + η ) . \int_{1/4}^{X^{1/7}/2}B_X(z)^2\frac{\mathrm{d}z}{z^2}\ge4^{-2\eta}\left(\int_0^\infty B_X(z)^2\frac{\mathrm{d}z}{z^{2+2\eta}}-2e^{-6W}\int_0^\infty B_X(z)^2\frac{\mathrm{d}z}{z^{2+\eta}}\right). ∫ 1/4 X 1/7 /2 B X ( z ) 2 z 2 d z ≥ 4 − 2 η ( ∫ 0 ∞ B X ( z ) 2 z 2 + 2 η d z − 2 e − 6 W ∫ 0 ∞ B X ( z ) 2 z 2 + η d z ) .
Parseval’s identity (2.10) and (4.3) give
∫ 0 ∞ B X ( z ) 2 d z z 2 + 2 η = 1 2 π ∫ R ∣ m X , η ( t ) F X ( 1 / 2 + η + i t ) 1 / 2 + η + i t ∣ 2 d t . (36) \int_0^\infty B_X(z)^2\frac{\mathrm{d}z}{z^{2+2\eta}}=\frac{1}{2\pi}\int_{\mathbb{R}}\left|\frac{m_{X,\eta}(t)F_X(1/2+\eta+it)}{1/2+\eta+it}\right|^2\mathrm{d}t. \tag*{(36)} ∫ 0 ∞ B X ( z ) 2 z 2 + 2 η d z = 2 π 1 ∫ R 1/2 + η + i t m X , η ( t ) F X ( 1/2 + η + i t ) 2 d t . (36)
The third assertion of Lemma 4.1, followed by Multiplicative Chaos Result 4 of [20 ] , Section 2.4, shows that, outside an event of probability O ( e − 0.1 W ) O(e^{-0.1W}) O ( e − 0.1 W ) , this is
≫ ∫ 1 / 3 1 / 2 ∣ F X ( 1 / 2 + η + i t ) ∣ 2 d t ≫ log X 42 W e 2.1 W log log X . \gg\int_{1/3}^{1/2}|F_X(1/2+\eta+it)|^2\mathrm{d}t\gg\frac{\log X}{42We^{2.1W}\sqrt{\log\log X}}. ≫ ∫ 1/3 1/2 ∣ F X ( 1/2 + η + i t ) ∣ 2 d t ≫ 42 W e 2.1 W log log X log X .
We next show that the second integral in the truncation error is negligible. We claim that
E ( ∫ R ∣ m X , η / 2 ( t ) F X ( 1 / 2 + η / 2 + i t ) 1 / 2 + η / 2 + i t ∣ 2 d t ) 2 / 3 ≪ ( log X W log log X ) 2 / 3 . (37) \mathbb{E}\left(\int_{\mathbb{R}}\left|\frac{m_{X,\eta/2}(t)F_X(1/2+\eta/2+it)}{1/2+\eta/2+it}\right|^2\mathrm{d}t\right)^{2/3}\ll\left(\frac{\log X}{W\sqrt{\log\log X}}\right)^{2/3}. \tag*{(37)} E ( ∫ R 1/2 + η /2 + i t m X , η /2 ( t ) F X ( 1/2 + η /2 + i t ) 2 d t ) 2/3 ≪ ( W log log X log X ) 2/3 . (37)
For each integer N N N with ∣ N ∣ ≤ log 1000 X |N|\le\log^{1000}X ∣ N ∣ ≤ log 1000 X , apply Multiplicative Chaos Result 3 of [20 ] , Section 2.4, on [ N − 1 / 2 , N + 1 / 2 ] [N-1/2,N+1/2] [ N − 1/2 , N + 1/2 ] . Lemma 4.1 contributes at most O ( ( 1 + log ( 2 + ∣ N ∣ ) ) 4 ) O((1+\log(2+|N|))^4) O (( 1 + log ( 2 + ∣ N ∣ ) ) 4 ) before taking the 2 / 3 2/3 2/3 power, and the resulting bounds are summable because
∑ N ∈ Z ( 1 + log ( 2 + ∣ N ∣ ) ) 8 / 3 ( log log ( ∣ N ∣ + 10 ) ) 2 / 3 ( 1 + N 2 ) 2 / 3 < ∞ . \sum_{N\in\mathbb{Z}}\frac{(1+\log(2+|N|))^{8/3}(\log\log(|N|+10))^{2/3}}{(1+N^2)^{2/3}}<\infty. N ∈ Z ∑ ( 1 + N 2 ) 2/3 ( 1 + log ( 2 + ∣ N ∣ ) ) 8/3 ( log log ( ∣ N ∣ + 10 ) ) 2/3 < ∞.
The subadditivity of u 2 / 3 u^{2/3} u 2/3 allows us to sum the contributions of these intervals. On the complementary range, ∣ m X , η / 2 ( t ) ∣ ≪ log X |m_{X,\eta/2}(t)|\ll\log X ∣ m X , η /2 ( t ) ∣ ≪ log X and E ∣ F X ( 1 / 2 + η / 2 + i t ) ∣ 2 ≪ log X \mathbb{E}|F_X(1/2+\eta/2+it)|^2\ll\log X E ∣ F X ( 1/2 + η /2 + i t ) ∣ 2 ≪ log X , so
E ∫ ∣ t ∣ > log 1000 X ∣ m X , η / 2 ( t ) F X ( 1 / 2 + η / 2 + i t ) 1 / 2 + η / 2 + i t ∣ 2 d t ≪ log 3 X ∫ ∣ t ∣ > log 1000 X d t 1 + t 2 ≪ log − 900 X . \mathbb{E}\int_{|t|>\log^{1000}X}\left|\frac{m_{X,\eta/2}(t)F_X(1/2+\eta/2+it)}{1/2+\eta/2+it}\right|^2\mathrm{d}t\ll\log^3X\int_{|t|>\log^{1000}X}\frac{\mathrm{d}t}{1+t^2}\ll\log^{-900}X. E ∫ ∣ t ∣ > l o g 1000 X 1/2 + η /2 + i t m X , η /2 ( t ) F X ( 1/2 + η /2 + i t ) 2 d t ≪ log 3 X ∫ ∣ t ∣ > l o g 1000 X 1 + t 2 d t ≪ log − 900 X .
Together with concavity, this proves (4.6).
Parseval’s identity (2.10) with shift η / 2 \eta/2 η /2 and Markov’s inequality now show that, outside an event of probability O ( e − 2 W ) O(e^{-2W}) O ( e − 2 W ) ,
∫ 0 ∞ B X ( z ) 2 d z z 2 + η = 1 2 π ∫ R ∣ m X , η / 2 ( t ) F X ( 1 / 2 + η / 2 + i t ) 1 / 2 + η / 2 + i t ∣ 2 d t ≪ e 3 W log X W log log X . \int_0^\infty B_X(z)^2\frac{\mathrm{d}z}{z^{2+\eta}}=\frac{1}{2\pi}\int_{\mathbb{R}}\left|\frac{m_{X,\eta/2}(t)F_X(1/2+\eta/2+it)}{1/2+\eta/2+it}\right|^2\mathrm{d}t\ll\frac{e^{3W}\log X}{W\sqrt{\log\log X}}. ∫ 0 ∞ B X ( z ) 2 z 2 + η d z = 2 π 1 ∫ R 1/2 + η /2 + i t m X , η /2 ( t ) F X ( 1/2 + η /2 + i t ) 2 d t ≪ W log log X e 3 W log X .
After multiplication by 2 e − 6 W 2e^{-6W} 2 e − 6 W , this is smaller than the main term once W 0 W_0 W 0 is sufficiently large. Since 4 − 2 η ≍ 1 4^{-2\eta}\asymp1 4 − 2 η ≍ 1 , we conclude, outside an event of probability O ( e − 0.1 W ) O(e^{-0.1W}) O ( e − 0.1 W ) , that
Var X ( Y x ) = 1 x ∑ X < p ≤ 4 x B X ( x / p ) 2 ≫ e − 2.1 W W log log X ≫ e − 2.2 W log log X . \operatorname{Var}_{X}(Y_x)=\frac{1}{x}\sum_{X<p\le4x}B_X(x/p)^2\gg\frac{e^{-2.1W}}{W\sqrt{\log\log X}}\gg\frac{e^{-2.2W}}{\sqrt{\log\log X}}. Var X ( Y x ) = x 1 X < p ≤ 4 x ∑ B X ( x / p ) 2 ≫ W log log X e − 2.1 W ≫ log log X e − 2.2 W .
The probabilistic estimates depend only on F X F_X F X , so the same exceptional event works simultaneously for all X 8 / 7 ≤ x ≤ X 4 / 3 / 4 X^{8/7}\le x\le X^{4/3}/4 X 8/7 ≤ x ≤ X 4/3 /4 . This proves (35) . □
Large conditional covariance estimates
We next prove a weighted version of Harper’s covariance estimate. For 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π let
G X ( θ ) : = { X 8 / 7 e θ + 2 π r : r ∈ Z ≥ 0 , X 8 / 7 e θ + 2 π r ≤ X 4 / 3 / 4 } . \mathcal{G}_X(\theta):=\left\{X^{8/7}e^{\theta+2\pi r}:r\in\mathbb{Z}_{\ge0},\ X^{8/7}e^{\theta+2\pi r}\le X^{4/3}/4\right\}. G X ( θ ) := { X 8/7 e θ + 2 π r : r ∈ Z ≥ 0 , X 8/7 e θ + 2 π r ≤ X 4/3 /4 } .
For x , y ∈ G X ( θ ) x,y\in\mathcal{G}_X(\theta) x , y ∈ G X ( θ ) we have
Cov X ( Y x , Y y ) = 1 x y ∑ X < p ≤ X 4 / 3 B X ( x / p ) B X ( y / p ) . \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{\sqrt{xy}}\sum_{X<p\le X^{4/3}}B_X(x/p)B_X(y/p). Cov X ( Y x , Y y ) = x y 1 X < p ≤ X 4/3 ∑ B X ( x / p ) B X ( y / p ) .
Terms with p > 4 x p>4x p > 4 x or p > 4 y p>4y p > 4 y vanish. Throughout this section, θ \theta θ is regarded as a fixed common shift. All estimates below are uniform in θ \theta θ , and the exceptional events may be chosen independently of θ \theta θ .
Proposition 5.1 (Sparsity of large conditional covariances). Outside an event of probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) , we have, simultaneously for every 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π ,
max x ∈ G X ( θ ) # { y ∈ G X ( θ ) ∖ { x } : ∣ Cov X ( Y x , Y y ) ∣ > ( log log X ) − 0.59 } ≤ ( log X ) 0.8 . \max_{x\in\mathcal{G}_X(\theta)}\#\left\{y\in\mathcal{G}_X(\theta)\setminus\{x\}:|\operatorname{Cov}_X(Y_x,Y_y)|>(\log\log X)^{-0.59}\right\}\le(\log X)^{0.8}. x ∈ G X ( θ ) max # { y ∈ G X ( θ ) ∖ { x } : ∣ Cov X ( Y x , Y y ) ∣ > ( log log X ) − 0.59 } ≤ ( log X ) 0.8 .
We organise the proof in two stages. The first three lemmas form the analytic reduction of the conditional covariance. We first use Perron’s formula to represent the covariance by a double integral, then restrict this integral to the near-diagonal region, and finally insert the barrier condition. The outcome is the approximation
Cov X ( Y x , Y y ) = 1 4 π 2 I ( x , y ) + O ( ( log log X ) − 0.6 ) \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{4\pi^2}\mathcal{I}(x,y)+O((\log\log X)^{-0.6}) Cov X ( Y x , Y y ) = 4 π 2 1 I ( x , y ) + O (( log log X ) − 0.6 )
outside a suitably small exceptional set. The last two lemmas estimate I ( x , y ) \mathcal{I}(x,y) I ( x , y ) . The proposition then follows from the covariance representation under the barrier condition and the uniform high moment bound by a counting argument.
Recall the definition of m X , σ ( t ) m_{X,\sigma}(t) m X , σ ( t ) in Lemma 4.1 and the definition of F X ( s ) F_X(s) F X ( s ) in (12) . For convenience, we also introduce the following shorthand throughout the proof.
T = X 3 / 4 , K = ( log X ) 20 , d = ( log log X ) 100 log X , P X ( u ) : = ∑ X < p ≤ X 4 / 3 1 p 1 + i u . T=X^{3/4},\qquad K=(\log X)^{20},\qquad d=\frac{(\log\log X)^{100}}{\log X},\qquad P_X(u):=\sum_{X<p\le X^{4/3}}\frac{1}{p^{1+iu}}. T = X 3/4 , K = ( log X ) 20 , d = log X ( log log X ) 100 , P X ( u ) := X < p ≤ X 4/3 ∑ p 1 + i u 1 .
Lemma 5.2 (Perron formula and frequency truncation). Outside an event of probability O ( log − 1 / 2 X ) O(\log^{-1/2}X) O ( log − 1/2 X ) , we have, simultaneously for every 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π and all x , y ∈ G X ( θ ) x,y\in\mathcal{G}_X(\theta) x , y ∈ G X ( θ ) ,
Cov X ( Y x , Y y ) = 1 4 π 2 ∫ − K K ∫ − K K m X , 0 ( v ) F X ( 1 / 2 + i v ) x i v 1 / 2 + i v m X , 0 ( t ) F X ( 1 / 2 + i t ) y − i t 1 / 2 − i t P X ( v − t ) d v d t + O ( log − 3 X ) . (38) \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{4\pi^2}\int_{-K}^{K}\int_{-K}^{K}\frac{m_{X,0}(v)F_X(1/2+iv)x^{iv}}{1/2+iv}\frac{m_{X,0}(t)F_X(1/2+it)y^{-it}}{1/2-it}P_X(v-t)\,\mathrm{d}v\,\mathrm{d}t+O(\log^{-3}X). \tag*{(38)} Cov X ( Y x , Y y ) = 4 π 2 1 ∫ − K K ∫ − K K 1/2 + i v m X , 0 ( v ) F X ( 1/2 + i v ) x i v 1/2 − i t m X , 0 ( t ) F X ( 1/2 + i t ) y − i t P X ( v − t ) d v d t + O ( log − 3 X ) . (38)
Proof. Applying the truncated Perron formula [28 ] , Theorem 5.2 and Corollary 5.3 to the two partial sums in (2.7), we obtain
B X ( x / p ) x = 1 2 π i x ∫ 1 − i T 1 + i T ( 4 s − 1 / 2 − 1 ) F X ( s ) Z X ( 2 s ) ( x / p ) s s d s + E ( x , p ) , \frac{B_X(x/p)}{\sqrt{x}}=\frac{1}{2\pi i\sqrt{x}}\int_{1-iT}^{1+iT}\left(4^{s-1/2}-1\right)F_X(s)Z_X(2s)\frac{(x/p)^s}{s}\,\mathrm{d}s+E(x,p), x B X ( x / p ) = 2 π i x 1 ∫ 1 − i T 1 + i T ( 4 s − 1/2 − 1 ) F X ( s ) Z X ( 2 s ) s ( x / p ) s d s + E ( x , p ) ,
where by the Corollary 5.3 of [28 ] with σ 0 = 1 \sigma_0=1 σ 0 = 1 , we get the error term is
E ( x , p ) ≪ 1 x ∑ a ∈ { 1 , 4 } ( 1 a x / p ∈ Z + ∑ a x / ( 2 p ) < n < 2 a x / p n ≠ a x / p min ( 1 , a x T ∣ a x − p n ∣ ) + ( 1 + a x / p ) log X T ) . (39) E(x,p)\ll\frac{1}{\sqrt{x}}\sum_{a\in\{1,4\}}\left(\mathbf{1}_{ax/p\in\mathbb{Z}}+\sum_{\substack{ax/(2p)<n<2ax/p\\n\ne ax/p}}\min\left(1,\frac{ax}{T\lvert ax-pn\rvert}\right)+\frac{(1+ax/p)\log X}{T}\right). \tag*{(39)} E ( x , p ) ≪ x 1 a ∈ { 1 , 4 } ∑ 1 a x / p ∈ Z + a x / ( 2 p ) < n < 2 a x / p n = a x / p ∑ min ( 1 , T ∣ a x − p n ∣ a x ) + T ( 1 + a x / p ) log X . (39)
The indicator accounts for the possible endpoint contribution.
Since the covariance is a quadratic sum over p p p , we do not have to need a pointwise estimate for E ( x , p ) E(x,p) E ( x , p ) . It is enough to obtain a power-saving bound for its ℓ 2 \ell^2 ℓ 2 -norm over the primes,
∑ X < p ≤ X 4 / 3 ∣ E ( x , p ) ∣ 2 ≪ X − c (40) \sum_{X<p\le X^{4/3}}\lvert E(x,p)\rvert^2\ll X^{-c} \tag*{(40)} X < p ≤ X 4/3 ∑ ∣ E ( x , p ) ∣ 2 ≪ X − c (40)
for some absolute c > 0 c>0 c > 0 . After the contour shift, we may combine this with the corresponding ℓ 2 \ell^2 ℓ 2 -bounds for the vertical and horizontal contributions by Cauchy–Schwarz. We therefore estimate the Perron remainders collectively over p p p .
We fix a ∈ { 1 , 4 } a\in\{1,4\} a ∈ { 1 , 4 } and make the change of variables m = p n m=pn m = p n in the double sum above. Since m < 2 a x ≤ 2 X 4 / 3 < X 2 m<2ax\le2X^{4/3}<X^2 m < 2 a x ≤ 2 X 4/3 < X 2 , each such m m m has at most one prime divisor p > X p>X p > X . Hence
∑ X < p ≤ X 4 / 3 ( 1 a x / p ∈ Z + ∑ a x / ( 2 p ) < n < 2 a x / p n ≠ a x / p min ( 1 , a x T ∣ a x − p n ∣ ) ) ≪ 1 + ∑ a x / 2 < m < 2 a x m ≠ a x min ( 1 , a x T ∣ a x − m ∣ ) ≪ 1 + x log X T . \sum_{X<p\le X^{4/3}}\left(\mathbf{1}_{ax/p\in\mathbb{Z}}+\sum_{\substack{ax/(2p)<n<2ax/p\\n\ne ax/p}}\min\left(1,\frac{ax}{T\lvert ax-pn\rvert}\right)\right)
\ll1+\sum_{\substack{ax/2<m<2ax\\m\ne ax}}\min\left(1,\frac{ax}{T\lvert ax-m\rvert}\right)\ll1+\frac{x\log X}{T}. X < p ≤ X 4/3 ∑ 1 a x / p ∈ Z + a x / ( 2 p ) < n < 2 a x / p n = a x / p ∑ min ( 1 , T ∣ a x − p n ∣ a x ) ≪ 1 + a x /2 < m < 2 a x m = a x ∑ min ( 1 , T ∣ a x − m ∣ a x ) ≪ 1 + T x log X .
The last inequality follows immediately by summing according to the distance of m m m from a x ax a x . Using also the fact that # { X < p ≤ X 4 / 3 } ≪ X 4 / 3 / log X \#\{X<p\le X^{4/3}\}\ll X^{4/3}/\log X # { X < p ≤ X 4/3 } ≪ X 4/3 / log X , and ∑ X < p ≤ X 4 / 3 1 / p ≪ 1 \sum_{X<p\le X^{4/3}}1/p\ll1 ∑ X < p ≤ X 4/3 1/ p ≪ 1 , we obtain
∑ X < p ≤ X 4 / 3 ∣ E ( x , p ) ∣ ≪ 1 x ( 1 + x log X T + X 4 / 3 T ) ≪ X 1 / 20 . \sum_{X<p\le X^{4/3}}\lvert E(x,p)\rvert\ll\frac{1}{\sqrt{x}}\left(1+\frac{x\log X}{T}+\frac{X^{4/3}}{T}\right)\ll X^{1/20}. X < p ≤ X 4/3 ∑ ∣ E ( x , p )∣ ≪ x 1 ( 1 + T x log X + T X 4/3 ) ≪ X 1/20 .
On the other hand, from the same harmonic-sum estimate applied pointwise, together with a x / p ≪ X 1 / 3 ax/p\ll X^{1/3} a x / p ≪ X 1/3 , we obtain max X < p ≤ X 4 / 3 ∣ E ( x , p ) ∣ ≪ x − 1 / 2 ≪ X − 1 / 2 \max_{X<p\le X^{4/3}}\lvert E(x,p)\rvert\ll x^{-1/2}\ll X^{-1/2} max X < p ≤ X 4/3 ∣ E ( x , p )∣ ≪ x − 1/2 ≪ X − 1/2 . Thus (5.3) follows.
We now move the contour to Re s = 1 / 2 \operatorname{Re}s=1/2 Re s = 1/2 . No pole is crossed. We first show that the two horizontal segments make a negligible contribution. Since Z X Z_X Z X is deterministic, by Cauchy–Schwarz and independence we obtain, uniformly for 1 / 2 ≤ σ ≤ 1 1/2\le\sigma\le1 1/2 ≤ σ ≤ 1 ,
E ∣ F X ( σ + i T ) Z X ( 2 σ + 2 i T ) ∣ ≤ ∣ Z X ( 2 σ + 2 i T ) ∣ E ∣ F X ( σ + i T ) ∣ 2 ≪ ∏ p ≤ X ( 1 − p − 2 σ ) − 1 ∏ p ≤ X ( 1 + p − 2 σ ) 1 / 2 ≪ log 3 / 2 X . \begin{aligned}
\mathbb{E}\lvert F_X(\sigma+iT)Z_X(2\sigma+2iT)\rvert\le\lvert Z_X(2\sigma+2iT)\rvert\sqrt{\mathbb{E}\lvert F_X(\sigma+iT)\rvert^2} \\
&\ll\prod_{p\le X}(1-p^{-2\sigma})^{-1}\prod_{p\le X}(1+p^{-2\sigma})^{1/2}\ll\log^{3/2}X.
\end{aligned} E ∣ F X ( σ + i T ) Z X ( 2 σ + 2 i T )∣ ≤ ∣ Z X ( 2 σ + 2 i T )∣ E ∣ F X ( σ + i T ) ∣ 2 ≪ p ≤ X ∏ ( 1 − p − 2 σ ) − 1 p ≤ X ∏ ( 1 + p − 2 σ ) 1/2 ≪ log 3/2 X .
Hence, by Markov’s inequality,
∫ 1 / 2 1 ∣ F X ( σ + i T ) Z X ( 2 σ + 2 i T ) ∣ d σ ≤ log 2 X \int_{1/2}^{1}\lvert F_X(\sigma+iT)Z_X(2\sigma+2iT)\rvert\,\mathrm{d}\sigma\le\log^2X ∫ 1/2 1 ∣ F X ( σ + i T ) Z X ( 2 σ + 2 i T )∣ d σ ≤ log 2 X
outside an event of probability O ( log − 1 / 2 X ) O(\log^{-1/2} X) O ( log − 1/2 X ) ; the same conclusion holds for the lower horizontal segment. On this event, the absolute value of either horizontal contribution corresponding to ( x , p ) (x,p) ( x , p ) is
≪ 1 x ∫ 1 / 2 1 ∣ F X ( σ ± i T ) Z X ( 2 σ ± 2 i T ) ∣ ( x / p ) σ ∣ σ ± i T ∣ d σ ≪ log 2 X T ( 1 p + x p ) , \ll\frac{1}{\sqrt{x}}\int_{1/2}^{1} \left|F_X(\sigma\pm iT)Z_X(2\sigma\pm2iT)\right|\frac{(x/p)^\sigma}{|\sigma\pm iT|}\,\mathrm{d}\sigma\ll\frac{\log^2 X}{T}\left(\frac{1}{\sqrt{p}}+\frac{\sqrt{x}}{p}\right), ≪ x 1 ∫ 1/2 1 ∣ F X ( σ ± i T ) Z X ( 2 σ ± 2 i T ) ∣ ∣ σ ± i T ∣ ( x / p ) σ d σ ≪ T log 2 X ( p 1 + p x ) ,
where we used ∣ σ ± i T ∣ ≍ T |\sigma\pm iT| \asymp T ∣ σ ± i T ∣ ≍ T and ( x / p ) σ ≪ x / p + x / p (x/p)^\sigma\ll\sqrt{x/p}+x/p ( x / p ) σ ≪ x / p + x / p if 1 2 ≤ σ ≤ 1 \frac{1}{2}\leq\sigma\leq1 2 1 ≤ σ ≤ 1 . Consequently,
∑ X < p ≤ X 4 / 3 [ log 2 X T ( 1 p + x p ) ] 2 ≪ log 4 X T 2 ( 1 + x X + x X ) ≪ X − c \sum_{X<p\leq X^{4/3}}\left[\frac{\log^2 X}{T}\left(\frac{1}{\sqrt{p}}+\frac{\sqrt{x}}{p}\right)\right]^2 \ll\frac{\log^4 X}{T^2}\left(1+\frac{x}{X}+\sqrt{\frac{x}{X}}\right)\ll X^{-c} X < p ≤ X 4/3 ∑ [ T log 2 X ( p 1 + p x ) ] 2 ≪ T 2 log 4 X ( 1 + X x + X x ) ≪ X − c
for some absolute c > 0 c>0 c > 0 , since T = X 3 / 4 T=X^{3/4} T = X 3/4 and x ≤ X 4 / 3 / 4 x\leq X^{4/3}/4 x ≤ X 4/3 /4 . Thus the horizontal contributions have power-saving ℓ 2 \ell^2 ℓ 2 -norm over the primes and may later be absorbed by Cauchy–Schwarz in the covariance calculation.
On the vertical segment, writing s = 1 / 2 + i v s=1/2+iv s = 1/2 + i v , the contribution to B X ( x / p ) / x B_X(x/p)/\sqrt{x} B X ( x / p ) / x is
1 2 π p ∫ − T T m X , 0 ( v ) F X ( 1 / 2 + i v ) x i v p − i v 1 / 2 + i v d v . \frac{1}{2\pi\sqrt{p}}\int_{-T}^{T}\frac{m_{X,0}(v)F_X(1/2+iv)x^{iv}p^{-iv}}{1/2+iv}\,\mathrm{d}v. 2 π p 1 ∫ − T T 1/2 + i v m X , 0 ( v ) F X ( 1/2 + i v ) x i v p − i v d v .
Moreover, from the trivial multiplier bound, Cauchy–Schwarz, and E ∣ F X ( 1 / 2 + i v ) ∣ 2 ≪ log X \mathbb{E}|F_X(1/2+iv)|^2\ll\log X E ∣ F X ( 1/2 + i v ) ∣ 2 ≪ log X , we obtain
E ∫ − T T ∣ m X , 0 ( v ) F X ( 1 / 2 + i v ) ∣ 1 + ∣ v ∣ d v ≪ log 5 / 2 X . \mathbb{E}\int_{-T}^{T}\frac{|m_{X,0}(v)F_X(1/2+iv)|}{1+|v|}\,\mathrm{d}v\ll\log^{5/2}X. E ∫ − T T 1 + ∣ v ∣ ∣ m X , 0 ( v ) F X ( 1/2 + i v ) ∣ d v ≪ log 5/2 X .
Thus, outside an event of probability O ( log − 3 / 2 X ) O(\log^{-3/2}X) O ( log − 3/2 X ) , the last integral is at most log 4 X \log^4 X log 4 X . It follows that the squared sum over p p p of the vertical contributions is O ( log 8 X ) O(\log^8 X) O ( log 8 X ) .
Combining this bound with (40) and the preceding bound for the horizontal contributions, Cauchy–Schwarz shows that every covariance term containing either a Perron remainder or a horizontal contribution is O ( X − c ) O(X^{-c}) O ( X − c ) for some c > 0 c>0 c > 0 . Consequently, outside an event of probability O ( log − 1 / 2 X ) O(\log^{-1/2}X) O ( log − 1/2 X ) , the covariance is given, up to O ( X − c ) O(X^{-c}) O ( X − c ) , by the product of the two vertical integrals.
Since F X ( 1 / 2 − i t ) = F X ( 1 / 2 + i t ) ‾ F_X(1/2-it)=\overline{F_X(1/2+it)} F X ( 1/2 − i t ) = F X ( 1/2 + i t ) and m X , 0 ( − t ) = m X , 0 ( t ) m_{X,0}(-t)=m_{X,0}(t) m X , 0 ( − t ) = m X , 0 ( t ) , by changing t t t to − t -t − t in the second vertical integral and summing over p p p , we obtain
Cov X ( Y x , Y y ) = 1 4 π 2 ∫ − T T ∫ − T T m X , 0 ( v ) F X ( 1 / 2 + i v ) x i v 1 / 2 + i v m X , 0 ( t ) F X ( 1 / 2 + i t ) y − i t 1 / 2 − i t P X ( v − t ) d v d t + O ( X − c ) . (41) \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{4\pi^2}\int_{-T}^{T}\int_{-T}^{T}\frac{m_{X,0}(v)F_X(1/2+iv)x^{iv}}{1/2+iv}\frac{m_{X,0}(t)F_X(1/2+it)y^{-it}}{1/2-it}P_X(v-t)\,\mathrm{d}v\,\mathrm{d}t+O(X^{-c}). \tag*{(41)} Cov X ( Y x , Y y ) = 4 π 2 1 ∫ − T T ∫ − T T 1/2 + i v m X , 0 ( v ) F X ( 1/2 + i v ) x i v 1/2 − i t m X , 0 ( t ) F X ( 1/2 + i t ) y − i t P X ( v − t ) d v d t + O ( X − c ) . (41)
It remains to truncate the frequencies. Applying Number Theory Result 1 of [20 ] , Section 2.2 with a p = p i t / log p a_p=p^{it}/\log p a p = p i t / log p for X < p ≤ X 4 / 3 X<p\leq X^{4/3} X < p ≤ X 4/3 and a n = 0 a_n=0 a n = 0 otherwise, we obtain uniformly in t t t ,
∫ − T T ∣ P X ( v − t ) ∣ 2 d v ≪ ∑ X < p ≤ X 4 / 3 1 p log p ≪ 1 log X . \int_{-T}^{T}|P_X(v-t)|^2\,\mathrm{d}v\ll\sum_{X<p\leq X^{4/3}}\frac{1}{p\log p}\ll\frac{1}{\log X}. ∫ − T T ∣ P X ( v − t ) ∣ 2 d v ≪ X < p ≤ X 4/3 ∑ p log p 1 ≪ log X 1 .
Here the condition p ≥ T 1.01 p\geq T^{1.01} p ≥ T 1.01 is satisfied since T = X 3 / 4 T=X^{3/4} T = X 3/4 . Moreover, by Cauchy–Schwarz and independence, we have E ∣ F X ( 1 / 2 + i v ) F X ( 1 / 2 + i t ) ∣ ≪ log X \mathbb{E}|F_X(1/2+iv)F_X(1/2+it)|\ll\log X E ∣ F X ( 1/2 + i v ) F X ( 1/2 + i t ) ∣ ≪ log X , while Lemma 4.1 gives ∣ m X , 0 ( v ) m X , 0 ( t ) ∣ ≪ log 2 X |m_{X,0}(v)m_{X,0}(t)|\ll\log^2 X ∣ m X , 0 ( v ) m X , 0 ( t ) ∣ ≪ log 2 X .
By symmetry, it is enough to consider the part of (5.4) where ∣ v ∣ ≥ max ( K , ∣ t ∣ ) |v|\geq\max(K,|t|) ∣ v ∣ ≥ max ( K , ∣ t ∣ ) . Using the preceding bounds for the Euler products and the multiplier, the expected absolute value of this contribution is
≪ log 3 X ∫ ∣ t ∣ ≤ T d t 1 + ∣ t ∣ ∫ max ( K , ∣ t ∣ ) ≤ ∣ v ∣ ≤ T ∣ P X ( v − t ) ∣ 1 + ∣ v ∣ d v . \ll\log^{3} X \int_{|t|\le T} \frac{dt}{1+|t|} \int_{\max(K,|t|)\le|v|\le T} \frac{|P_X(v-t)|}{1+|v|}\,dv. ≪ log 3 X ∫ ∣ t ∣ ≤ T 1 + ∣ t ∣ d t ∫ m a x ( K , ∣ t ∣ ) ≤ ∣ v ∣ ≤ T 1 + ∣ v ∣ ∣ P X ( v − t ) ∣ d v .
For fixed t t t , Cauchy–Schwarz and the mean value estimate for P X P_X P X give
∫ max ( K , ∣ t ∣ ) ≤ ∣ v ∣ ≤ T ∣ P X ( v − t ) ∣ 1 + ∣ v ∣ d v ≤ ( ∫ − T T ∣ P X ( v − t ) ∣ 2 d v ) 1 / 2 ( ∫ ∣ v ∣ ≥ max ( K , ∣ t ∣ ) d v ( 1 + ∣ v ∣ ) 2 ) 1 / 2 ≪ 1 log X max ( K , ∣ t ∣ ) . \begin{aligned}
\int_{\max(K,|t|)\le|v|\le T} \frac{|P_X(v-t)|}{1+|v|}\,dv \le\left(\int_{-T}^{T}|P_X(v-t)|^2\,dv\right)^{1/2}\left(\int_{|v|\ge\max(K,|t|)}\frac{dv}{(1+|v|)^2}\right)^{1/2} \\
&\ll\frac{1}{\sqrt{\log X}\sqrt{\max(K,|t|)}}.
\end{aligned} ∫ m a x ( K , ∣ t ∣ ) ≤ ∣ v ∣ ≤ T 1 + ∣ v ∣ ∣ P X ( v − t ) ∣ d v ≤ ( ∫ − T T ∣ P X ( v − t ) ∣ 2 d v ) 1/2 ( ∫ ∣ v ∣ ≥ m a x ( K , ∣ t ∣ ) ( 1 + ∣ v ∣ ) 2 d v ) 1/2 ≪ log X max ( K , ∣ t ∣ ) 1 .
Consequently the expected high frequency contribution is
≪ log 5 / 2 X ∫ R d t ( 1 + ∣ t ∣ ) max ( K , ∣ t ∣ ) ≪ log 5 / 2 X log K K . \ll\log^{5/2}X\int_{\mathbb{R}}\frac{dt}{(1+|t|)\sqrt{\max(K,|t|)}}\ll\frac{\log^{5/2}X\log K}{\sqrt K}. ≪ log 5/2 X ∫ R ( 1 + ∣ t ∣ ) max ( K , ∣ t ∣ ) d t ≪ K log 5/2 X log K .
Here the last estimate follows by splitting the t t t -integral at ∣ t ∣ = K |t|=K ∣ t ∣ = K . Since K = ( log X ) 20 K=(\log X)^{20} K = ( log X ) 20 , this is O ( log − 7 X ) O(\log^{-7}X) O ( log − 7 X ) . Markov’s inequality therefore shows that the range max ( ∣ v ∣ , ∣ t ∣ ) ≥ K \max(|v|,|t|)\ge K max ( ∣ v ∣ , ∣ t ∣ ) ≥ K contributes O ( log − 3 X ) O(\log^{-3}X) O ( log − 3 X ) outside an event of probability O ( log − 4 X ) O(\log^{-4}X) O ( log − 4 X ) . Combining this with (41) proves (38) , and hence the lemma. □
The next lemma isolates the narrow near-diagonal frequency range on which the barrier argument will be applied.
Lemma 5.3 (Near-diagonal reduction). Let
Δ = { ( v , t ) : ( log log X ) − 2 ≤ ∣ v ∣ , ∣ t ∣ ≤ ( log log X ) 2 , ∣ v − t ∣ ≤ d } . \Delta=\{(v,t):(\log\log X)^{-2}\le|v|,|t|\le(\log\log X)^2,\ |v-t|\le d\}. Δ = {( v , t ) : ( log log X ) − 2 ≤ ∣ v ∣ , ∣ t ∣ ≤ ( log log X ) 2 , ∣ v − t ∣ ≤ d } .
Outside an event of probability O ( ( log log X ) − 1.2 ) O((\log\log X)^{-1.2}) O (( log log X ) − 1.2 ) , uniformly for every 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π and all x , y ∈ G X ( θ ) x,y\in G_X(\theta) x , y ∈ G X ( θ ) , the contribution from [ − K , K ] 2 ∖ Δ [-K,K]^2\setminus\Delta [ − K , K ] 2 ∖ Δ to the double integral in (38) is O ( ( log log X ) − 0.7 ) O((\log\log X)^{-0.7}) O (( log log X ) − 0.7 ) .
Proof. We first discard the off-diagonal range ∣ v − t ∣ > d |v-t|>d ∣ v − t ∣ > d . After removing the finitely many primes below 200, Euler Product Result 2 of [20 ] shows that, uniformly for ∣ v ∣ , ∣ t ∣ ≤ K |v|,|t|\le K ∣ v ∣ , ∣ t ∣ ≤ K ,
E ∣ F X ( 1 / 2 + i v ) F X ( 1 / 2 + i t ) ∣ ≪ log X ( 1 + min ( log X , ∣ v − t ∣ − 1 / 2 ) + min ( log X , ∣ v + t ∣ − 1 / 2 ) + log ( 2 + ∣ v ∣ + ∣ t ∣ ) ) . (42) \mathbb{E}\left|F_X(1/2+iv)F_X(1/2+it)\right|\ll\sqrt{\log X}\left(1+\min\left(\sqrt{\log X},|v-t|^{-1/2}\right)+\min\left(\sqrt{\log X},|v+t|^{-1/2}\right)+\log(2+|v|+|t|)\right). \tag*{(42)} E ∣ F X ( 1/2 + i v ) F X ( 1/2 + i t ) ∣ ≪ log X ( 1 + min ( log X , ∣ v − t ∣ − 1/2 ) + min ( log X , ∣ v + t ∣ − 1/2 ) + log ( 2 + ∣ v ∣ + ∣ t ∣ ) ) . (42)
At v = t v=t v = t or v = − t v=-t v = − t , the second entry in the corresponding minimum is interpreted as + ∞ +\infty + ∞ .
On the other hand, Number Theory Result 2 of [20 ] , combined with the trivial bound ∣ P X ( u ) ∣ ≪ 1 |P_X(u)|\ll1 ∣ P X ( u ) ∣ ≪ 1 , gives
∣ P X ( u ) ∣ ≪ min ( 1 , 1 ∣ u ∣ log X ) + O ( ( 1 + K ) e − c log X ) ( ∣ u ∣ ≤ 2 K ) . (43) |P_X(u)|\ll\min\left(1,\frac{1}{|u|\log X}\right)+O\left((1+K)e^{-c\sqrt{\log X}}\right)\qquad(|u|\le2K). \tag*{(43)} ∣ P X ( u ) ∣ ≪ min ( 1 , ∣ u ∣ log X 1 ) + O ( ( 1 + K ) e − c l o g X ) ( ∣ u ∣ ≤ 2 K ) . (43)
Since d > 1 / log X d>1/\log X d > 1/ log X , on the range ∣ v − t ∣ > d |v-t|>d ∣ v − t ∣ > d we have ∣ P X ( v − t ) ∣ ≪ ( ∣ v − t ∣ log X ) − 1 |P_X(v-t)|\ll(|v-t|\log X)^{-1} ∣ P X ( v − t ) ∣ ≪ ( ∣ v − t ∣ log X ) − 1 , up to the exponentially small error. Lemma 4.1, (42) , and (43) show that the expected absolute value of the part with ∣ v − t ∣ > d |v-t|>d ∣ v − t ∣ > d is
≪ ( log log X ) 4 log X ∬ ∣ v ∣ , ∣ t ∣ ≤ K ∣ v − t ∣ > d 1 ( 1 + ∣ v ∣ ) ( 1 + ∣ t ∣ ) ∣ v − t ∣ ( 1 + min ( log X , ∣ v − t ∣ − 1 / 2 ) + min ( log X , ∣ v + t ∣ − 1 / 2 ) + log ( 2 + ∣ v ∣ + ∣ t ∣ ) ) d v d t . (44) \ll\frac{(\log\log X)^4}{\sqrt{\log X}}\iint_{\substack{|v|,|t|\le K\\|v-t|>d}}\frac{1}{(1+|v|)(1+|t|)|v-t|}\left(1+\min\left(\sqrt{\log X},|v-t|^{-1/2}\right)+\min\left(\sqrt{\log X},|v+t|^{-1/2}\right)+\log(2+|v|+|t|)\right)\,dv\,dt. \tag*{(44)} ≪ log X ( log log X ) 4 ∬ ∣ v ∣ , ∣ t ∣ ≤ K ∣ v − t ∣ > d ( 1 + ∣ v ∣ ) ( 1 + ∣ t ∣ ) ∣ v − t ∣ 1 ( 1 + min ( log X , ∣ v − t ∣ − 1/2 ) + min ( log X , ∣ v + t ∣ − 1/2 ) + log ( 2 + ∣ v ∣ + ∣ t ∣ ) ) d v d t . (44)
Here ( log log X ) 4 (\log\log X)^4 ( log log X ) 4 comes from the two multiplier bounds ∣ m X , 0 ( v ) ∣ |m_{X,0}(v)| ∣ m X , 0 ( v ) ∣ , ∣ m X , 0 ( t ) ∣ ≪ ( log log X ) 2 |m_{X,0}(t)| \ll(\log\log X)^2 ∣ m X , 0 ( t ) ∣ ≪ ( log log X ) 2 on this range. To estimate the regular terms, change variables u = v − t u=v-t u = v − t and use
∫ R d r ( 1 + ∣ r ∣ ) ( 1 + ∣ r − u ∣ ) ≪ log ( 2 + ∣ u ∣ ) 1 + ∣ u ∣ . \int_{\mathbb{R}} \frac{dr}{(1+|r|)(1+|r-u|)} \ll\frac{\log(2+|u|)}{1+|u|}. ∫ R ( 1 + ∣ r ∣ ) ( 1 + ∣ r − u ∣ ) d r ≪ 1 + ∣ u ∣ log ( 2 + ∣ u ∣ ) .
It follows that the terms 1 + log ( 2 + ∣ v ∣ + ∣ t ∣ ) 1+\log(2+|v|+|t|) 1 + log ( 2 + ∣ v ∣ + ∣ t ∣ ) in (5.7) contribute O ( log log X ) O(\log\log X) O ( log log X ) to the double integral, while the singularity at v − t = 0 v-t=0 v − t = 0 contributes
≪ ∫ d 1 d u u 3 / 2 + log log X ≪ d − 1 / 2 + log log X . (45) \ll\int_d^1 \frac{du}{u^{3/2}}+\log\log X \ll d^{-1/2}+\log\log X. \tag*{(45)} ≪ ∫ d 1 u 3/2 d u + log log X ≪ d − 1/2 + log log X . (45)
For the term singular at v + t = 0 v+t=0 v + t = 0 , we make the change of variables u = v − t u=v-t u = v − t and s = v + t s=v+t s = v + t . Up to absolute constants arising from the Jacobian and the rescaling of the two denominator factors, the relevant contribution is bounded by
∫ d < ∣ u ∣ ≤ 2 K d u ∣ u ∣ ∫ ∣ s ∣ ≤ 2 K min ( log X , ∣ s ∣ − 1 / 2 ) ( 1 + ∣ s + u ∣ ) ( 1 + ∣ s − u ∣ ) d s ≪ log ( 2 / d ) . (46) \int_{d<|u|\le2K}\frac{du}{|u|}\int_{|s|\le2K}\frac{\min\left(\sqrt{\log X},|s|^{-1/2}\right)}{(1+|s+u|)(1+|s-u|)}\,ds \ll\log(2/d). \tag*{(46)} ∫ d < ∣ u ∣ ≤ 2 K ∣ u ∣ d u ∫ ∣ s ∣ ≤ 2 K ( 1 + ∣ s + u ∣ ) ( 1 + ∣ s − u ∣ ) min ( log X , ∣ s ∣ − 1/2 ) d s ≪ log ( 2/ d ) . (46)
The inner integral is O ( 1 ) O(1) O ( 1 ) for ∣ u ∣ ≤ 2 |u|\le2 ∣ u ∣ ≤ 2 , while for ∣ u ∣ > 2 |u|>2 ∣ u ∣ > 2 it is O ( ( 1 + log ∣ u ∣ ) ∣ u ∣ − 3 / 2 ) O((1+\log|u|)|u|^{-3/2}) O (( 1 + log ∣ u ∣ ) ∣ u ∣ − 3/2 ) after splitting near s = ± u s=\pm u s = ± u and on the complement.
Combining the three estimates, we have
( 5.7 ) ≪ ( log log X ) 4 log X ( d − 1 / 2 + log log X + log ( 2 / d ) ) ≪ ( log log X ) − 46 . (5.7)\ll\frac{(\log\log X)^4}{\sqrt{\log X}}\left(d^{-1/2}+\log\log X+\log(2/d)\right)\ll(\log\log X)^{-46}. ( 5.7 ) ≪ log X ( log log X ) 4 ( d − 1/2 + log log X + log ( 2/ d ) ) ≪ ( log log X ) − 46 .
Markov’s inequality therefore shows that the region ∣ v − t ∣ > d |v-t|>d ∣ v − t ∣ > d may be discarded with error O ( ( log log X ) − 20 ) O((\log\log X)^{-20}) O (( log log X ) − 20 ) outside an event of probability O ( ( log log X ) − 26 ) O((\log\log X)^{-26}) O (( log log X ) − 26 ) .
We are now left with ∣ v ∣ , ∣ t ∣ ≤ K |v|,|t|\le K ∣ v ∣ , ∣ t ∣ ≤ K and ∣ v − t ∣ ≤ d |v-t|\le d ∣ v − t ∣ ≤ d . We next discard the ranges where max ( ∣ v ∣ , ∣ t ∣ ) ≥ ( log log X ) 2 \max(|v|,|t|)\ge(\log\log X)^2 max ( ∣ v ∣ , ∣ t ∣ ) ≥ ( log log X ) 2 or min ( ∣ v ∣ , ∣ t ∣ ) ≤ ( log log X ) − 2 \min(|v|,|t|)\le(\log\log X)^{-2} min ( ∣ v ∣ , ∣ t ∣ ) ≤ ( log log X ) − 2 , leaving precisely the region Δ \Delta Δ . We shall use the L 1 L^1 L 1 -norm of P X ( u ) P_X(u) P X ( u ) on the near-diagonal range. From (5.6),
∫ ∣ u ∣ ≤ d ∣ P X ( u ) ∣ d u ≪ ∫ ∣ u ∣ ≤ 1 / log X d u + 1 log X ∫ 1 / log X < ∣ u ∣ ≤ d d u ∣ u ∣ ≪ 1 + log ( d log X ) log X ≪ log log log X log X . (47) \begin{aligned}
\int_{|u|\le d}|P_X(u)|\,du\ll\int_{|u|\le1/\log X}du+\frac{1}{\log X}\int_{1/\log X<|u|\le d}\frac{du}{|u|} \\
&\ll\frac{1+\log(d\log X)}{\log X}\ll\frac{\log\log\log X}{\log X}. \tag*{(47)}
\end{aligned} ∫ ∣ u ∣ ≤ d ∣ P X ( u ) ∣ d u ≪ ∫ ∣ u ∣ ≤ 1/ l o g X d u + log X 1 ∫ 1/ l o g X < ∣ u ∣ ≤ d ∣ u ∣ d u ≪ log X 1 + log ( d log X ) ≪ log X log log log X . (47)
Moreover, Cauchy–Schwarz in probability gives, uniformly in v v v and t t t ,
E ∣ F X ( 1 / 2 + i v ) F X ( 1 / 2 + i t ) ∣ ≪ log X . \mathbb{E}|F_X(1/2+iv)F_X(1/2+it)|\ll\log X. E ∣ F X ( 1/2 + i v ) F X ( 1/2 + i t ) ∣ ≪ log X .
Thus the factor log X \log X log X from the Euler product moment is cancelled by the factor 1 / log X 1/\log X 1/ log X in (5.10).
Consider first the range max ( ∣ v ∣ , ∣ t ∣ ) ≥ ( log log X ) 2 \max(|v|,|t|)\ge(\log\log X)^2 max ( ∣ v ∣ , ∣ t ∣ ) ≥ ( log log X ) 2 . Since d = o ( ( log log X ) 2 ) d=o((\log\log X)^2) d = o (( log log X ) 2 ) , one has ∣ v ∣ ≍ ∣ t ∣ |v|\asymp|t| ∣ v ∣ ≍ ∣ t ∣ there. Using Lemma 4.1 and integrating first in u = v − t u=v-t u = v − t , the expected absolute value of the contribution of this range to (5.1) is
≪ ( log log log X ) ∫ ∣ r ∣ ≥ ( log log X ) 2 / 2 ( 1 + log ( 2 + ∣ r ∣ ) ) 4 ( 1 + ∣ r ∣ ) 2 d r ≪ ( log log log X ) 5 ( log log X ) 2 . \ll(\log\log\log X)\int_{|r|\ge(\log\log X)^2/2}\frac{(1+\log(2+|r|))^4}{(1+|r|)^2}\,dr\ll\frac{(\log\log\log X)^5}{(\log\log X)^2}. ≪ ( log log log X ) ∫ ∣ r ∣ ≥ ( l o g l o g X ) 2 /2 ( 1 + ∣ r ∣ ) 2 ( 1 + log ( 2 + ∣ r ∣ ) ) 4 d r ≪ ( log log X ) 2 ( log log log X ) 5 .
Here the factor log log log X \log\log\log X log log log X comes from (5.10), while the remaining integral accounts for the two multiplier factors and the two Mellin denominators.
Next consider min ( ∣ v ∣ , ∣ t ∣ ) ≤ ( log log X ) − 2 \min(|v|,|t|)\le(\log\log X)^{-2} min ( ∣ v ∣ , ∣ t ∣ ) ≤ ( log log X ) − 2 . Since d = o ( ( log log X ) − 2 ) d=o((\log\log X)^{-2}) d = o (( log log X ) − 2 ) , both v v v and t t t are O ( ( log log X ) − 2 ) O((\log\log X)^{-2}) O (( log log X ) − 2 ) . Lemma 4.1 then gives bounded multipliers, and the remaining frequency interval has length O ( ( log log X ) − 2 ) O((\log\log X)^{-2}) O (( log log X ) − 2 ) . Using (5.10) again, the expected absolute value of this contribution is therefore ≪ log log log X / ( log log X ) 2 \ll\log\log\log X/(\log\log X)^2 ≪ log log log X / ( log log X ) 2 . Thus the total expected contribution from these two frequency ranges is ≪ ( log log log X ) 5 / ( log log X ) 2 \ll(\log\log\log X)^5/(\log\log X)^2 ≪ ( log log log X ) 5 / ( log log X ) 2 .
Markov’s inequality shows that they may be discarded with total error O ( ( log log X ) − 0.7 ) O((\log\log X)^{-0.7}) O (( log log X ) − 0.7 ) outside an event of probability O ( ( log log X ) − 1.2 ) O((\log\log X)^{-1.2}) O (( log log X ) − 1.2 ) .
Together with the preceding removal of the region ∣ v − t ∣ > d |v-t|>d ∣ v − t ∣ > d , this restricts the integral in (5.1) to Δ \Delta Δ with the claimed error and exceptional probability. After taking absolute values, the phase factors x i v x^{iv} x i v and y − i t y^{-it} y − i t disappear, so all the majorants above are independent of x x x , y y y , and the common shift θ \theta θ . Hence the same exceptional event works uniformly for all the grids under consideration. □
We now insert the same strong barrier condition as in [20, Section 3.3]. For integers 0 ≤ j ≤ ⌊ log log X ⌋ 0\le j\le\lfloor\log\log X\rfloor 0 ≤ j ≤ ⌊ log log X ⌋ , write
F X , j ( s ) : = ∏ p ≤ X e − j ( 1 + f ( p ) p − s ) . F_{X,j}(s):=\prod_{p\le Xe^{-j}}(1+f(p)p^{-s}). F X , j ( s ) := p ≤ X e − j ∏ ( 1 + f ( p ) p − s ) .
Following [20, Section 2.4], for each t ∈ R t\in\mathbb{R} t ∈ R we define a sequence of discretisations t ( j ) t(j) t ( j ) by setting t ( − 1 ) = t t(-1)=t t ( − 1 ) = t and, for 0 ≤ j ≤ ⌊ log log X ⌋ − 1 0\le j\le\lfloor\log\log X\rfloor-1 0 ≤ j ≤ ⌊ log log X ⌋ − 1 ,
t ( j ) : = max { u ≤ t ( j − 1 ) : u = n ( ( log X ) / e j ) log ( ( log X ) / e j ) for some n ∈ Z } . t(j):=\max\left\{u\le t(j-1):u=\frac{n}{((\log X)/e^j)\log((\log X)/e^j)}\text{ for some }n\in\mathbb{Z}\right\}. t ( j ) := max { u ≤ t ( j − 1 ) : u = (( log X ) / e j ) log (( log X ) / e j ) n for some n ∈ Z } .
For t ∈ R t\in\mathbb{R} t ∈ R , let A ( t ) A(t) A ( t ) be the event that, for every integer 0 ≤ j ≤ ⌊ log log X ⌋ − 1 0\le j\le\lfloor\log\log X\rfloor-1 0 ≤ j ≤ ⌊ log log X ⌋ − 1 ,
∣ F X , j ( 1 / 2 + i t ) ∣ ≤ { log X e j ( log log X ) 1000 , 0 ≤ j ≤ ⌊ 0.99 log log X ⌋ , log X e j ( log log X ) 6 , ⌊ 0.99 log log X ⌋ < j ≤ ⌊ log log X ⌋ − 1. (48) |F_{X,j}(1/2+it)|\le
\begin{cases}
\dfrac{\log X}{e^j(\log\log X)^{1000}}, & 0\le j\le\lfloor0.99\log\log X\rfloor,\\
\dfrac{\log X}{e^j}(\log\log X)^6, & \lfloor0.99\log\log X\rfloor<j\le\lfloor\log\log X\rfloor-1.
\end{cases} \tag*{(48)} ∣ F X , j ( 1/2 + i t ) ∣ ≤ ⎩ ⎨ ⎧ e j ( log log X ) 1000 log X , e j log X ( log log X ) 6 , 0 ≤ j ≤ ⌊ 0.99 log log X ⌋ , ⌊ 0.99 log log X ⌋ < j ≤ ⌊ log log X ⌋ − 1. (48)
Define
I ( x , y ) : = ∬ Δ 1 A ( v ) m X , 0 ( v ) F X ( 1 / 2 + i v ) x i v 1 / 2 + i v 1 A ( t ) m X , 0 ( t ) F X ( 1 / 2 + i t ) y − i t 1 / 2 − i t P X ( v − t ) d v d t . (49) \mathcal{I}(x,y):=\iint_{\Delta}\frac{\mathbf{1}_{A(v)}m_{X,0}(v)F_X(1/2+iv)x^{iv}}{1/2+iv}\frac{\mathbf{1}_{A(t)}m_{X,0}(t)F_X(1/2+it)y^{-it}}{1/2-it}P_X(v-t)\,dv\,dt. \tag*{(49)} I ( x , y ) := ∬ Δ 1/2 + i v 1 A ( v ) m X , 0 ( v ) F X ( 1/2 + i v ) x i v 1/2 − i t 1 A ( t ) m X , 0 ( t ) F X ( 1/2 + i t ) y − i t P X ( v − t ) d v d t . (49)
Lemma 5.4 (Inserting the barrier condition). Outside an event of probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) , we have, simultaneously for every 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π and all x , y ∈ G X ( θ ) x,y\in\mathcal{G}_X(\theta) x , y ∈ G X ( θ ) ,
Cov X ( Y x , Y y ) = 1 4 π 2 I ( x , y ) + O ( ( log log X ) − 0.6 ) . \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{4\pi^2}\mathcal{I}(x,y)+O((\log\log X)^{-0.6}). Cov X ( Y x , Y y ) = 4 π 2 1 I ( x , y ) + O (( log log X ) − 0.6 ) .
Proof. We first introduce the weaker discretised barrier needed to apply the multiplicative-chaos estimates. For t ∈ R t\in\mathbb{R} t ∈ R , let D ( t ) D(t) D ( t ) be the event
∣ F X , j ( 1 / 2 + i t ( j ) ) ∣ ≤ ( log X ) e − j ( log log X ) 5 ( 0 ≤ j ≤ ⌊ log log X ⌋ − 1 ) , |F_{X,j}(1/2+it(j))|\le(\log X)e^{-j}(\log\log X)^5\qquad(0\le j\le\lfloor\log\log X\rfloor-1), ∣ F X , j ( 1/2 + i t ( j )) ∣ ≤ ( log X ) e − j ( log log X ) 5 ( 0 ≤ j ≤ ⌊ log log X ⌋ − 1 ) ,
where t ( j ) t(j) t ( j ) is the discretisation defined above. Let D D D be the event that D ( t ) D(t) D ( t ) holds for every ∣ t ∣ ≤ ( log log X ) 2 |t|\le(\log\log X)^2 ∣ t ∣ ≤ ( log log X ) 2 .
Applying Multiplicative Chaos Result 1 of [20, Section 2.4] with the choice e W = ( log log X ) 3 e^W=(\log\log X)^3 e W = ( log log X ) 3 , the upper barrier is at most ( log X ) e − j ( log log X ) 5 (\log X)e^{-j}(\log\log X)^5 ( log X ) e − j ( log log X ) 5 , and the exceptional probability is O ( ( log log X ) − 6 ) O((\log\log X)^{-6}) O (( log log X ) − 6 ) . Hence, for any fixed N ∈ R N\in\mathbb{R} N ∈ R , D ( t ) D(t) D ( t ) holds for every ∣ t − N ∣ ≤ 1 / 2 |t-N|\le1/2 ∣ t − N ∣ ≤ 1/2 outside such an exceptional event. A union bound over O ( ( log log X ) 2 ) O((\log\log X)^2) O (( log log X ) 2 ) intervals of length 1 covering [ − ( log log X ) 2 , ( log log X ) 2 ] [-(\log\log X)^2,(\log\log X)^2] [ − ( log log X ) 2 , ( log log X ) 2 ] therefore yields
P ( D c ) ≪ ( log log X ) − 4 . \mathbb{P}(D^c)\ll(\log\log X)^{-4}. P ( D c ) ≪ ( log log X ) − 4 .
On D D D , the events D ( v ) D(v) D ( v ) and D ( t ) D(t) D ( t ) hold throughout Δ \Delta Δ , so their indicators may be inserted without any error.
We now pass to the strong pointwise barrier A ( t ) \mathcal{A}(t) A ( t ) . The range ( log log X ) − 2 ≤ ∣ t ∣ ≤ ( log log X ) 2 (\log\log X)^{-2} \le|t| \le(\log\log X)^2 ( log log X ) − 2 ≤ ∣ t ∣ ≤ ( log log X ) 2 lies within the range of Multiplicative Chaos Result 2 of [20 ] , Section 2.4]. Hence, uniformly in this range,
E ( ∣ F X ( 1 / 2 + i t ) ∣ 2 1 D ( t ) 1 A ( t ) c ) ≪ log X ( log log log X ) 7 log log X . (50) \mathbb{E}\left(\left|F_X(1/2+it)\right|^2\mathbf{1}_{\mathcal{D}(t)}\mathbf{1}_{\mathcal{A}(t)^c}\right) \ll\frac{\log X(\log\log\log X)^7}{\log\log X}. \tag*{(50)} E ( ∣ F X ( 1/2 + i t ) ∣ 2 1 D ( t ) 1 A ( t ) c ) ≪ log log X log X ( log log log X ) 7 . (50)
Integrating against ∣ 1 / 2 + i t ∣ − 2 |1/2+it|^{-2} ∣1/2 + i t ∣ − 2 and applying Markov’s inequality, we obtain, outside an event of probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) ,
∫ ( log log X ) − 2 ≤ ∣ t ∣ ≤ ( log log X ) 2 1 D ( t ) 1 A ( t ) c ∣ F X ( 1 / 2 + i t ) ∣ 2 ∣ 1 / 2 + i t ∣ 2 d t ≪ log X ( log log log X ) 7 ( log log X ) 0.9 . (51) \int_{(\log\log X)^{-2}\leq|t|\leq(\log\log X)^2}\frac{\mathbf{1}_{\mathcal{D}(t)}\mathbf{1}_{\mathcal{A}(t)^c}|F_X(1/2+it)|^2}{|1/2+it|^2}\,\mathrm{d}t \ll\frac{\log X(\log\log\log X)^7}{(\log\log X)^{0.9}}. \tag*{(51)} ∫ ( l o g l o g X ) − 2 ≤ ∣ t ∣ ≤ ( l o g l o g X ) 2 ∣1/2 + i t ∣ 2 1 D ( t ) 1 A ( t ) c ∣ F X ( 1/2 + i t ) ∣ 2 d t ≪ ( log log X ) 0.9 log X ( log log log X ) 7 . (51)
We also need an unrestricted mean square bound for the Euler product. Take q = 9 / 10 q=9/10 q = 9/10 in Multiplicative Chaos Result 3 of [20 ] , Section 2.4]. Decomposing the t t t -range into intervals [ N − 1 / 2 , N + 1 / 2 ] [N-1/2,N+1/2] [ N − 1/2 , N + 1/2 ] and using ( a + b ) q ≤ a q + b q (a+b)^q\leq a^q+b^q ( a + b ) q ≤ a q + b q , we obtain
E ( ∫ − ( log log X ) 2 ( log log X ) 2 ∣ F X ( 1 / 2 + i t ) ∣ 2 ∣ 1 / 2 + i t ∣ 2 d t ) q ≪ ∑ ∣ N ∣ ≤ ( log log X ) 2 + 1 1 ( 1 + N 2 ) q E ( ∫ N − 1 / 2 N + 1 / 2 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ) q ≪ ( log X log log X ) q ∑ N ∈ Z ( log ( log ( ∣ N ∣ + 10 ) ) ) q ( 1 + N 2 ) q ≪ ( log X log log X ) q . \begin{aligned}
\mathbb{E}\left(\int_{-(\log\log X)^2}^{(\log\log X)^2}\frac{|F_X(1/2+it)|^2}{|1/2+it|^2}\,\mathrm{d}t\right)^q
\ll\sum_{|N|\leq(\log\log X)^2+1}\frac{1}{(1+N^2)^q}\mathbb{E}\left(\int_{N-1/2}^{N+1/2}|F_X(1/2+it)|^2\,\mathrm{d}t\right)^q \\
&\ll\left(\frac{\log X}{\sqrt{\log\log X}}\right)^q\sum_{N\in\mathbb{Z}}\frac{(\log(\log(|N|+10)))^q}{(1+N^2)^q}\ll\left(\frac{\log X}{\sqrt{\log\log X}}\right)^q.
\end{aligned} E ( ∫ − ( l o g l o g X ) 2 ( l o g l o g X ) 2 ∣1/2 + i t ∣ 2 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) q ≪ ∣ N ∣ ≤ ( l o g l o g X ) 2 + 1 ∑ ( 1 + N 2 ) q 1 E ( ∫ N − 1/2 N + 1/2 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) q ≪ ( log log X log X ) q N ∈ Z ∑ ( 1 + N 2 ) q ( log ( log ( ∣ N ∣ + 10 )) ) q ≪ ( log log X log X ) q .
Here the last series converges since 2 q = 9 / 5 > 1 2q=9/5>1 2 q = 9/5 > 1 . Therefore, by Markov’s inequality,
P ( ∫ − ( log log X ) 2 ( log log X ) 2 ∣ F X ( 1 / 2 + i t ) ∣ 2 ∣ 1 / 2 + i t ∣ 2 d t > log X ( log log X ) 0.35 ) ≪ ( log X log log X ) q ( ( log log X ) 0.35 log X ) q = ( log log X ) − 0.135 . \mathbb{P}\left(\int_{-(\log\log X)^2}^{(\log\log X)^2}\frac{|F_X(1/2+it)|^2}{|1/2+it|^2}\,\mathrm{d}t>\frac{\log X}{(\log\log X)^{0.35}}\right)
\ll\left(\frac{\log X}{\sqrt{\log\log X}}\right)^q\left(\frac{(\log\log X)^{0.35}}{\log X}\right)^q=(\log\log X)^{-0.135}. P ( ∫ − ( l o g l o g X ) 2 ( l o g l o g X ) 2 ∣1/2 + i t ∣ 2 ∣ F X ( 1/2 + i t ) ∣ 2 d t > ( log log X ) 0.35 log X ) ≪ ( log log X log X ) q ( log X ( log log X ) 0.35 ) q = ( log log X ) − 0.135 .
Thus, outside an event of probability O ( ( log log X ) − 0.135 ) O((\log\log X)^{-0.135}) O (( log log X ) − 0.135 ) ,
∫ − ( log log X ) 2 ( log log X ) 2 ∣ F X ( 1 / 2 + i t ) ∣ 2 ∣ 1 / 2 + i t ∣ 2 d t ≪ log X ( log log X ) 0.35 . (52) \int_{-(\log\log X)^2}^{(\log\log X)^2}\frac{|F_X(1/2+it)|^2}{|1/2+it|^2}\,\mathrm{d}t\ll\frac{\log X}{(\log\log X)^{0.35}}. \tag*{(52)} ∫ − ( l o g l o g X ) 2 ( l o g l o g X ) 2 ∣1/2 + i t ∣ 2 ∣ F X ( 1/2 + i t ) ∣ 2 d t ≪ ( log log X ) 0.35 log X . (52)
It remains to estimate the change in the near-diagonal integral caused by inserting the strong barrier. By (5.10),
sup v ∫ { t : ( v , t ) ∈ Δ } ∣ P X ( v − t ) ∣ d t + sup t ∫ { v : ( v , t ) ∈ Δ } ∣ P X ( v − t ) ∣ d v ≪ log log log X log X . \sup_v\int_{\{t:(v,t)\in\Delta\}}|P_X(v-t)|\,\mathrm{d}t+\sup_t\int_{\{v:(v,t)\in\Delta\}}|P_X(v-t)|\,\mathrm{d}v\ll\frac{\log\log\log X}{\log X}. v sup ∫ { t : ( v , t ) ∈ Δ } ∣ P X ( v − t ) ∣ d t + t sup ∫ { v : ( v , t ) ∈ Δ } ∣ P X ( v − t ) ∣ d v ≪ log X log log log X .
Then for any square-integrable functions F F F and G G G , Cauchy–Schwarz gives us
∬ Δ ∣ P X ( v − t ) ∣ ∣ F ( v ) G ( t ) ∣ d v d t ≤ ( ∬ Δ ∣ P X ( v − t ) ∣ ∣ F ( v ) ∣ 2 d v d t ) 1 / 2 ( ∬ Δ ∣ P X ( v − t ) ∣ ∣ G ( t ) ∣ 2 d v d t ) 1 / 2 ≪ log log log X log X ∥ F ∥ 2 ∥ G ∥ 2 . (53) \iint_{\Delta}|P_X(v-t)||F(v)G(t)|\,\mathrm{d}v\,\mathrm{d}t
\leq\left(\iint_{\Delta}|P_X(v-t)||F(v)|^2\,\mathrm{d}v\,\mathrm{d}t\right)^{1/2}
\left(\iint_{\Delta}|P_X(v-t)||G(t)|^2\,\mathrm{d}v\,\mathrm{d}t\right)^{1/2}
\ll\frac{\log\log\log X}{\log X}\|F\|_2\|G\|_2. \tag*{(53)} ∬ Δ ∣ P X ( v − t ) ∣∣ F ( v ) G ( t ) ∣ d v d t ≤ ( ∬ Δ ∣ P X ( v − t ) ∣∣ F ( v ) ∣ 2 d v d t ) 1/2 ( ∬ Δ ∣ P X ( v − t ) ∣∣ G ( t ) ∣ 2 d v d t ) 1/2 ≪ log X log log log X ∥ F ∥ 2 ∥ G ∥ 2 . (53)
For all v , t ∈ R v,t\in\mathbb{R} v , t ∈ R , we have the elementary inequality
1 − 1 A ( v ) 1 A ( t ) ≤ 1 A ( v ) c + 1 A ( t ) c . 1-\mathbf{1}_{\mathcal{A}(v)}\mathbf{1}_{\mathcal{A}(t)}\leq\mathbf{1}_{\mathcal{A}(v)^c}+\mathbf{1}_{\mathcal{A}(t)^c}. 1 − 1 A ( v ) 1 A ( t ) ≤ 1 A ( v ) c + 1 A ( t ) c .
On the event D \mathcal{D} D , we have throughout Δ \Delta Δ , 1 A ( v ) c = 1 D ( v ) 1 A ( v ) c \mathbf{1}_{\mathcal{A}(v)^{c}}=\mathbf{1}_{\mathcal{D}(v)}\mathbf{1}_{\mathcal{A}(v)^{c}} 1 A ( v ) c = 1 D ( v ) 1 A ( v ) c , 1 A ( t ) c = 1 D ( t ) 1 A ( t ) c \mathbf{1}_{\mathcal{A}(t)^{c}}=\mathbf{1}_{\mathcal{D}(t)}\mathbf{1}_{\mathcal{A}(t)^{c}} 1 A ( t ) c = 1 D ( t ) 1 A ( t ) c . Hence, after taking absolute values, the difference between near-diagonal integral (i.e., the double integral in (5.1) restricted to Δ \Delta Δ ) and I ( x , y ) \mathcal{I}(x,y) I ( x , y ) is bounded by the sum of two terms, the first of which is
∬ Δ ∣ P X ( v − t ) ∣ ∣ m X , 0 ( v ) m X , 0 ( t ) ∣ 1 D ( v ) 1 A ( v ) c ∣ F X ( 1 / 2 + i v ) ∣ ∣ 1 / 2 + i v ∣ ∣ F X ( 1 / 2 + i t ) ∣ ∣ 1 / 2 − i t ∣ d v d t . \iint_{\Delta}|P_X(v-t)||m_{X,0}(v)m_{X,0}(t)|\frac{\mathbf{1}_{\mathcal{D}(v)}\mathbf{1}_{\mathcal{A}(v)^{c}}|F_X(1/2+iv)|}{|1/2+iv|}\frac{|F_X(1/2+it)|}{|1/2-it|}\,dv\,dt. ∬ Δ ∣ P X ( v − t ) ∣∣ m X , 0 ( v ) m X , 0 ( t ) ∣ ∣1/2 + i v ∣ 1 D ( v ) 1 A ( v ) c ∣ F X ( 1/2 + i v ) ∣ ∣1/2 − i t ∣ ∣ F X ( 1/2 + i t ) ∣ d v d t .
By Lemma 4.1, throughout Δ \Delta Δ , we have ∣ m X , 0 ( v ) ∣ |m_{X,0}(v)| ∣ m X , 0 ( v ) ∣ , ∣ m X , 0 ( t ) ∣ ≪ ( log log log X ) 2 |m_{X,0}(t)|\ll(\log\log\log X)^2 ∣ m X , 0 ( t ) ∣ ≪ ( log log log X ) 2 . Together with (5.16), the displayed integral above is therefore
≪ ( log log log X ) 5 log X ( ∫ ( log log X ) − 2 ≤ ∣ v ∣ ≤ ( log log X ) 2 1 D ( v ) 1 A ( v ) c ∣ F X ( 1 / 2 + i v ) ∣ 2 ∣ 1 / 2 + i v ∣ 2 d v ) 1 / 2 \ll\frac{(\log\log\log X)^5}{\log X}\left(\int_{(\log\log X)^{-2}\leq|v|\leq(\log\log X)^2}\frac{\mathbf{1}_{\mathcal{D}(v)}\mathbf{1}_{\mathcal{A}(v)^{c}}|F_X(1/2+iv)|^2}{|1/2+iv|^2}\,dv\right)^{1/2} ≪ log X ( log log log X ) 5 ( ∫ ( l o g l o g X ) − 2 ≤ ∣ v ∣ ≤ ( l o g l o g X ) 2 ∣1/2 + i v ∣ 2 1 D ( v ) 1 A ( v ) c ∣ F X ( 1/2 + i v ) ∣ 2 d v ) 1/2
× ( ∫ − ( log log X ) 2 ( log log X ) 2 ∣ F X ( 1 / 2 + i t ) ∣ 2 ∣ 1 / 2 + i t ∣ 2 d t ) 1 / 2 ≪ ( log log log X ) 5 log X ( log X ( log log log X ) 7 ( log log X ) 0.9 ) 1 / 2 ( log X ( log log X ) 0.35 ) 1 / 2 ≪ ( log log X ) − 0.6 . \begin{aligned}
{}\times\left(\int_{-(\log\log X)^2}^{(\log\log X)^2}\frac{|F_X(1/2+it)|^2}{|1/2+it|^2}\,dt\right)^{1/2} \\
&\ll\frac{(\log\log\log X)^5}{\log X}\left(\frac{\log X(\log\log\log X)^7}{(\log\log X)^{0.9}}\right)^{1/2}\left(\frac{\log X}{(\log\log X)^{0.35}}\right)^{1/2}\ll(\log\log X)^{-0.6}.
\end{aligned} × ( ∫ − ( l o g l o g X ) 2 ( l o g l o g X ) 2 ∣1/2 + i t ∣ 2 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) 1/2 ≪ log X ( log log log X ) 5 ( ( log log X ) 0.9 log X ( log log log X ) 7 ) 1/2 ( ( log log X ) 0.35 log X ) 1/2 ≪ ( log log X ) − 0.6 .
Here in the last line we have used (5.14) and (5.15). The same bound holds for the second term in which A ( t ) \mathcal{A}(t) A ( t ) fails.
Thus, outside an event of probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) , inserting 1 A ( v ) 1 A ( t ) \mathbf{1}_{\mathcal{A}(v)}\mathbf{1}_{\mathcal{A}(t)} 1 A ( v ) 1 A ( t ) into the near-diagonal integral changes it by O ( ( log log X ) − 0.6 ) O((\log\log X)^{-0.6}) O (( log log X ) − 0.6 ) .
Combining this with Lemmas 5.2 and 5.3, we obtain
Cov X ( Y x , Y y ) = 1 4 π 2 I ( x , y ) + O ( ( log log X ) − 0.6 ) . \operatorname{Cov}_X(Y_x,Y_y)=\frac{1}{4\pi^2}\mathcal{I}(x,y)+O((\log\log X)^{-0.6}). Cov X ( Y x , Y y ) = 4 π 2 1 I ( x , y ) + O (( log log X ) − 0.6 ) .
The union of the exceptional events arising above and in Lemmas 5.2 and 5.3 has probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) .
Finally, all these exceptional events depend only on the variables ( f ( p ) ) p ≤ X (f(p))_{p\leq X} ( f ( p ) ) p ≤ X . The dependence on x x x and y y y in the integral occurs only through the factors x i v x^{iv} x i v and y − i t y^{-it} y − i t , both of modulus one, so all the estimates are uniform in x x x , y y y , and the common shift θ \theta θ . Hence the same exceptional event works simultaneously for every 0 ≤ θ ≤ 2 π 0\leq\theta\leq2\pi 0 ≤ θ ≤ 2 π and all x , y ∈ G X ( θ ) x,y\in\mathcal{G}_X(\theta) x , y ∈ G X ( θ ) . □
We next adapt Harper’s high moment argument [20 ] to the weighted integral I ( x , y ) \mathcal{I}(x,y) I ( x , y ) . We retain the local mean square integrals of the Euler product at this stage, while keeping track of the additional deterministic multiplier.
Lemma 5.5 (Weighted high moment estimate). For any large X X X and every integer 1 ≤ k ≤ log 1 / 4 X 1\leq k\leq\log^{1/4}X 1 ≤ k ≤ log 1/4 X , outside an event of probability at most ( log log X ) − 2 k (\log\log X)^{-2k} ( log log X ) − 2 k , we have, simultaneously for every 0 ≤ θ ≤ 2 π 0\leq\theta\leq2\pi 0 ≤ θ ≤ 2 π ,
max x ∈ G X ( θ ) ∑ y ∈ G X ( θ ) ∣ I ( x , y ) ∣ 2 k ≪ log X ( 1 log 1 / 3 X ∑ ∣ N ∣ ≤ ( log log X ) 2 + 1 1 1 + N 2 ( C ( log log log X ) 5 log X ∫ N N + 1 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ) 2 k \max_{x\in\mathcal{G}_X(\theta)}\sum_{y\in\mathcal{G}_X(\theta)}|\mathcal{I}(x,y)|^{2k}
\ll\log X\left(\frac{1}{\log^{1/3}X}\sum_{|N|\leq(\log\log X)^2+1}\frac{1}{1+N^2}\left(\frac{C(\log\log\log X)^5}{\log X}\int_N^{N+1}|F_X(1/2+it)|^2\,dt\right)^{2k}
\right. x ∈ G X ( θ ) max y ∈ G X ( θ ) ∑ ∣ I ( x , y ) ∣ 2 k ≪ log X log 1/3 X 1 ∣ N ∣ ≤ ( l o g l o g X ) 2 + 1 ∑ 1 + N 2 1 ( log X C ( log log log X ) 5 ∫ N N + 1 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) 2 k
+ 1 log 1 / 3 X ( k 29 ( log log X ) 17 ) 2 k + ( k 29 ( log log X ) 989 ) 2 k ) . (54) \left.
{}+\frac{1}{\log^{1/3}X}\left(k^{29}(\log\log X)^{17}\right)^{2k}+\left(\frac{k^{29}}{(\log\log X)^{989}}\right)^{2k}\right). \tag*{(54)} + log 1/3 X 1 ( k 29 ( log log X ) 17 ) 2 k + ( ( log log X ) 989 k 29 ) 2 k ) . (54)
Proof. We follow the proof of [20 ] , keeping track of the two modifications arising in the present setting. The common shift θ \theta θ contributes only a phase of modulus one. In addition, each copy of I ( x , y ) \mathcal{I}(x,y) I ( x , y ) contains the two deterministic factors m X , 0 ( v ) m_{X,0}(v) m X , 0 ( v ) and m X , 0 ( t ) m_{X,0}(t) m X , 0 ( t ) . On Δ \Delta Δ , by Lemma 4.1 we have ∣ m X , 0 ( v ) ∣ , ∣ m X , 0 ( t ) ∣ ≪ ( log log log X ) 2 |m_{X,0}(v)|, |m_{X,0}(t)| \ll(\log\log\log X)^2 ∣ m X , 0 ( v ) ∣ , ∣ m X , 0 ( t ) ∣ ≪ ( log log log X ) 2 , so the 2 k 2k 2 k -th moment expansion contains 4 k 4k 4 k multiplier factors and hence incurs the loss
( C log log log X ) 8 k . (55) (C\log\log\log X)^{8k}. \tag*{(55)} ( C log log log X ) 8 k . (55)
For each integer N N N , let Δ N \Delta_N Δ N be the part of Δ \Delta Δ on which N ≤ max ( v , t ) < N + 1 N \le\max(v,t) < N+1 N ≤ max ( v , t ) < N + 1 , and let I N ( x , y ) \mathcal{I}_N(x,y) I N ( x , y ) denote the corresponding part of I ( x , y ) \mathcal{I}(x,y) I ( x , y ) . By Hölder’s inequality we get
∣ I ( x , y ) ∣ 2 k ≪ C 2 k ∑ N 1 1 + N 2 ∣ ( 1 + N 2 ) I N ( x , y ) ∣ 2 k . (56) |\mathcal{I}(x,y)|^{2k} \ll C^{2k}\sum_N \frac{1}{1+N^2}\left|(1+N^2)\mathcal{I}_N(x,y)\right|^{2k}. \tag*{(56)} ∣ I ( x , y ) ∣ 2 k ≪ C 2 k N ∑ 1 + N 2 1 ( 1 + N 2 ) I N ( x , y ) 2 k . (56)
On Δ N \Delta_N Δ N , we have ∣ ( 1 / 2 + i v ) ( 1 / 2 − i t ) ∣ ≍ 1 + N 2 |(1/2+iv)(1/2-it)| \asymp1+N^2 ∣ ( 1/2 + i v ) ( 1/2 − i t ) ∣ ≍ 1 + N 2 , thus the factor 1 + N 2 1+N^2 1 + N 2 introduced in (56) absorbs the two denominators in I N ( x , y ) \mathcal{I}_N(x,y) I N ( x , y ) .
We now fix N N N and expand the 2 k 2k 2 k -th power. The first k k k copies of I N ( x , y ) \mathcal{I}_N(x,y) I N ( x , y ) and the remaining k k k copies of its conjugate introduce variables ( v 1 , t 1 ) , … , ( v 2 k , t 2 k ) (v_1,t_1),\ldots,(v_{2k},t_{2k}) ( v 1 , t 1 ) , … , ( v 2 k , t 2 k ) . Writing
Ξ = t 1 + ⋯ + t k − t k + 1 − ⋯ − t 2 k , \Xi=t_1+\cdots+t_k-t_{k+1}-\cdots-t_{2k}, Ξ = t 1 + ⋯ + t k − t k + 1 − ⋯ − t 2 k ,
the y y y -dependent phase is y − i Ξ y^{-i\Xi} y − i Ξ . For y = X 8 / 7 e θ + 2 π u y=X^{8/7}e^{\theta+2\pi u} y = X 8/7 e θ + 2 π u , this becomes X − 8 i Ξ / 7 e − i θ Ξ e − 2 π i u Ξ X^{-8i\Xi/7}e^{-i\theta\Xi}e^{-2\pi iu\Xi} X − 8 i Ξ/7 e − i θ Ξ e − 2 π i u Ξ . The first two factors are independent of u u u and have modulus one, summing over the admissible values of u u u gives, uniformly in θ \theta θ ,
∣ ∑ u e − 2 π i u Ξ ∣ ≪ min ( log X , 1 ∥ Ξ ∥ R / Z ) , (57) \left|\sum_u e^{-2\pi iu\Xi}\right| \ll\min\left(\log X,\frac{1}{\|\Xi\|_{\mathbb{R}/\mathbb{Z}}}\right), \tag*{(57)} u ∑ e − 2 π i u Ξ ≪ min ( log X , ∥Ξ ∥ R / Z 1 ) , (57)
where ∥ ⋅ ∥ R / Z \|\cdot\|_{\mathbb{R}/\mathbb{Z}} ∥ ⋅ ∥ R / Z denotes distance to the nearest integer.
By (43) and (47) , throughout ∣ u ∣ ≤ d |u|\le d ∣ u ∣ ≤ d we have
∣ P X ( u ) ∣ ≪ min ( 1 , 1 ∣ u ∣ log X ) , ∫ ∣ u ∣ ≤ d ∣ P X ( u ) ∣ d u ≪ log log log X log X . (58) |P_X(u)|\ll\min\left(1,\frac{1}{|u|\log X}\right),\qquad\int_{|u|\le d}|P_X(u)|\,du\ll\frac{\log\log\log X}{\log X}. \tag*{(58)} ∣ P X ( u ) ∣ ≪ min ( 1 , ∣ u ∣ log X 1 ) , ∫ ∣ u ∣ ≤ d ∣ P X ( u ) ∣ d u ≪ log X log log log X . (58)
We first consider ∥ Ξ ∥ R / Z ≥ log − 2 / 3 X \|\Xi\|_{\mathbb{R}/\mathbb{Z}}\ge\log^{-2/3}X ∥Ξ ∥ R / Z ≥ log − 2/3 X . Using (57) with the bound log 2 / 3 X \log^{2/3}X log 2/3 X , discarding the barrier indicators, and integrating the v j v_j v j variables by (58) , Harper’s argument gives, apart from the additional factor (55) ,
∑ y ∈ G X ( θ ) ∣ ( 1 + N 2 ) I N ( x , y ) ∣ 2 k ≪ ( C log log log X ) 8 k log 2 / 3 X ( C log log log X log X ∫ N N + 1 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ) 2 k ≪ log X log 1 / 3 X ( C ( log log log X ) 5 log X ∫ N N + 1 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ) 2 k . \begin{aligned}
\sum_{y\in G_X(\theta)}\left|(1+N^2)\mathcal{I}_N(x,y)\right|^{2k}
\ll(C\log\log\log X)^{8k}\log^{2/3}X\left(\frac{C\log\log\log X}{\log X}\int_N^{N+1}|F_X(1/2+it)|^2\,dt\right)^{2k} \\
&\ll\frac{\log X}{\log^{1/3}X}\left(\frac{C(\log\log\log X)^5}{\log X}\int_N^{N+1}|F_X(1/2+it)|^2\,dt\right)^{2k}.
\end{aligned} y ∈ G X ( θ ) ∑ ( 1 + N 2 ) I N ( x , y ) 2 k ≪ ( C log log log X ) 8 k log 2/3 X ( log X C log log log X ∫ N N + 1 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) 2 k ≪ log 1/3 X log X ( log X C ( log log log X ) 5 ∫ N N + 1 ∣ F X ( 1/2 + i t ) ∣ 2 d t ) 2 k .
This is the first term in (54) . In Harper’s Proposition 3 the corresponding power is only log log log X \log\log\log X log log log X ; the additional fourth power here is exactly the loss coming from the multiplier.
We next turn to ∥ Ξ ∥ R / Z < log − 2 / 3 X \|\Xi\|_{\mathbb{R}/\mathbb{Z}}<\log^{-2/3}X ∥Ξ ∥ R / Z < log − 2/3 X . Since t j = N + O ( 1 ) t_j=N+O(1) t j = N + O ( 1 ) for every j j j , we have
Ξ = ∑ j ≤ k ( t j − N ) − ∑ j > k ( t j − N ) = O ( k ) . \Xi=\sum_{j\le k}(t_j-N)-\sum_{j>k}(t_j-N)=O(k). Ξ = j ≤ k ∑ ( t j − N ) − j > k ∑ ( t j − N ) = O ( k ) .
Hence ∣ Ξ − m ∣ < log − 2 / 3 X |\Xi-m|<\log^{-2/3}X ∣Ξ − m ∣ < log − 2/3 X for some integer ∣ m ∣ ≤ C k |m|\le Ck ∣ m ∣ ≤ C k . Moreover,
∣ ∑ j ≤ k v j − ∑ j > k v j − m ∣ ≤ ∣ Ξ − m ∣ + ∑ j = 1 2 k ∣ v j − t j ∣ ≤ log − 2 / 3 X + 2 k d ≤ 2 log − 2 / 3 X \left|\sum_{j\le k}v_j-\sum_{j>k}v_j-m\right|
\le|\Xi-m|+\sum_{j=1}^{2k}|v_j-t_j|
\le\log^{-2/3}X+2kd\le2\log^{-2/3}X j ≤ k ∑ v j − j > k ∑ v j − m ≤ ∣Ξ − m ∣ + j = 1 ∑ 2 k ∣ v j − t j ∣ ≤ log − 2/3 X + 2 k d ≤ 2 log − 2/3 X
for sufficiently large X X X .
We bound the geometric sum in (57) trivially by log X \log X log X . Using
∏ j = 1 2 k 1 A ( t j ) 1 A ( v j ) ∣ F X ( 1 2 + i t j ) F X ( 1 2 + i v j ) ∣ ≤ 1 2 ∏ j = 1 2 k 1 A ( t j ) ∣ F X ( 1 2 + i t j ) ∣ 2 + 1 2 ∏ j = 1 2 k 1 A ( v j ) ∣ F X ( 1 2 + i v j ) ∣ 2 \begin{aligned}
\prod_{j=1}^{2k}1_{\mathcal{A}(t_j)}1_{\mathcal{A}(v_j)}\left|F_X\left(\frac{1}{2}+it_j\right)F_X\left(\frac{1}{2}+iv_j\right)\right| \\
&\le\frac{1}{2}\prod_{j=1}^{2k}1_{\mathcal{A}(t_j)}\left|F_X\left(\frac{1}{2}+it_j\right)\right|^2+\frac{1}{2}\prod_{j=1}^{2k}1_{\mathcal{A}(v_j)}\left|F_X\left(\frac{1}{2}+iv_j\right)\right|^2
\end{aligned} j = 1 ∏ 2 k 1 A ( t j ) 1 A ( v j ) F X ( 2 1 + i t j ) F X ( 2 1 + i v j ) ≤ 2 1 j = 1 ∏ 2 k 1 A ( t j ) F X ( 2 1 + i t j ) 2 + 2 1 j = 1 ∏ 2 k 1 A ( v j ) F X ( 2 1 + i v j ) 2
and the symmetry between v v v and t t t , we need only consider the first term. Integrating each v j v_j v j by (47) , this part is bounded by
( C log log log X ) 8 k log X ( C log log log X log X ) 2 k ∑ ∣ m ∣ ≤ C k ∫ I m ∏ j = 1 2 k 1 A ( t j ) ∣ F X ( 1 / 2 + i t j ) ∣ 2 d t , (59) (C\log\log\log X)^{8k}\log X\left(\frac{C\log\log\log X}{\log X}\right)^{2k}\sum_{|m|\le Ck}\int_{I_m}\prod_{j=1}^{2k}1_{\mathcal{A}(t_j)}\left|F_X(1/2+it_j)\right|^2\,dt, \tag*{(59)} ( C log log log X ) 8 k log X ( log X C log log log X ) 2 k ∣ m ∣ ≤ C k ∑ ∫ I m j = 1 ∏ 2 k 1 A ( t j ) ∣ F X ( 1/2 + i t j ) ∣ 2 d t , (59)
where I m I_m I m denotes the part of the relevant 2 k 2k 2 k -dimensional box on which
∣ ∑ j ≤ k t j − ∑ j > k t j − m ∣ ≤ 2 log − 2 / 3 X . \left|\sum_{j\le k}t_j-\sum_{j>k}t_j-m\right|\le2\log^{-2/3}X. j ≤ k ∑ t j − j > k ∑ t j − m ≤ 2 log − 2/3 X .
For each permutation σ ∈ S 2 k \sigma\in S_{2k} σ ∈ S 2 k , we restrict to the region t σ ( 1 ) < ⋯ < t σ ( 2 k ) t_{\sigma(1)}<\cdots<t_{\sigma(2k)} t σ ( 1 ) < ⋯ < t σ ( 2 k ) and further decompose it according to the sizes of the successive differences. For 1 ≤ j < 2 k 1\le j<2k 1 ≤ j < 2 k , let h j = 0 h_j=0 h j = 0 when t σ ( j + 1 ) − t σ ( j ) ≤ e / log X t_{\sigma(j+1)}-t_{\sigma(j)}\le e/\log X t σ ( j + 1 ) − t σ ( j ) ≤ e / log X , and otherwise choose h j ≥ 1 h_j\ge1 h j ≥ 1 so that e h j / log X < t σ ( j + 1 ) − t σ ( j ) ≤ e h j + 1 / log X e^{h_j}/\log X<t_{\sigma(j+1)}-t_{\sigma(j)}\le e^{h_j+1}/\log X e h j / log X < t σ ( j + 1 ) − t σ ( j ) ≤ e h j + 1 / log X . As in Harper’s proof, we let h ~ j = min ( h j , ⌊ log log X − 2 log log log X ⌋ ) \widetilde h_j=\min(h_j,\lfloor\log\log X-2\log\log\log X\rfloor) h j = min ( h j , ⌊ log log X − 2 log log log X ⌋) . Then X e − h ~ j = exp ( e log log X − h ~ j ) ≥ exp ( ( log log X ) 2 ) ≥ 1000 k 6 Xe^{-\widetilde h_j}=\exp(e^{\log\log X-\widetilde h_j})\ge\exp((\log\log X)^2)\ge1000k^6 X e − h j = exp ( e l o g l o g X − h j ) ≥ exp (( log log X ) 2 ) ≥ 1000 k 6 for sufficiently large X X X .
For 1 ≤ j < 2 k 1\le j<2k 1 ≤ j < 2 k , we write
F X ( 1 / 2 + i t σ ( j ) ) = F X , h ~ j ( 1 / 2 + i t σ ( j ) ) F X ( 1 / 2 + i t σ ( j ) ) F X , h ~ j ( 1 / 2 + i t σ ( j ) ) . F_X(1/2+it_{\sigma(j)})=F_{X,\widetilde h_j}(1/2+it_{\sigma(j)})\frac{F_X(1/2+it_{\sigma(j)})}{F_{X,\widetilde h_j}(1/2+it_{\sigma(j)})}. F X ( 1/2 + i t σ ( j ) ) = F X , h j ( 1/2 + i t σ ( j ) ) F X , h j ( 1/2 + i t σ ( j ) ) F X ( 1/2 + i t σ ( j ) ) .
On A ( t σ ( j ) ) \mathcal{A}(t_{\sigma(j)}) A ( t σ ( j ) ) , by the definition of barrier event (48) , we have
∣ F X , h ~ j ( 1 / 2 + i t σ ( j ) ) ∣ ≤ { log X e h ~ j ( log log X ) 1000 , h ~ j ≤ 0.99 log log X , log X e h ~ j ( log log X ) 6 , h ~ j > 0.99 log log X . \left|F_{X,\widetilde h_j}(1/2+it_{\sigma(j)})\right|\le
\begin{cases}
\frac{\log X}{e^{\widetilde h_j}(\log\log X)^{1000}},&\widetilde h_j\le0.99\log\log X,\\[6pt]
\frac{\log X}{e^{\widetilde h_j}}(\log\log X)^6,&\widetilde h_j>0.99\log\log X.
\end{cases} F X , h j ( 1/2 + i t σ ( j ) ) ≤ ⎩ ⎨ ⎧ e h j ( l o g l o g X ) 1000 l o g X , e h j l o g X ( log log X ) 6 , h j ≤ 0.99 log log X , h j > 0.99 log log X .
Hence
∏ l = 1 2 k 1 A ( t l ) ∣ F X ( 1 2 + i t l ) ∣ 2 ≤ ∏ 1 ≤ j < 2 k h ~ j ≤ 0.99 log log X ( log X e h ~ j ) 2 ∏ 1 ≤ j < 2 k h ~ j > 0.99 log log X ( log log X ) − 2000 \prod_{l=1}^{2k}1_{\mathcal{A}(t_l)}\left|F_X\left(\frac{1}{2}+it_l\right)\right|^2
\le\prod_{\substack{1\le j<2k\\\widetilde h_j\le0.99\log\log X}}\left(\frac{\log X}{e^{\widetilde h_j}}\right)^2
\prod_{\substack{1\le j<2k\\\widetilde h_j>0.99\log\log X}}(\log\log X)^{-2000} l = 1 ∏ 2 k 1 A ( t l ) F X ( 2 1 + i t l ) 2 ≤ 1 ≤ j < 2 k h j ≤ 0.99 l o g l o g X ∏ ( e h j log X ) 2 1 ≤ j < 2 k h j > 0.99 l o g l o g X ∏ ( log log X ) − 2000
× ∏ 1 ≤ j < 2 k h ~ j > 0.99 log log X ( log log X ) 12 ∣ F X ( 1 2 + i t σ ( 2 k ) ) ∣ 2 ∏ j = 1 2 k − 1 ∣ F X ( 1 / 2 + i t σ ( j ) ) ∣ 2 ∣ F X , h ~ j ( 1 / 2 + i t σ ( j ) ) ∣ 2 . (60) {}\times\prod_{\substack{1\le j<2k\\\widetilde h_j>0.99\log\log X}}(\log\log X)^{12}\left|F_X\left(\frac{1}{2}+it_{\sigma(2k)}\right)\right|^2\prod_{j=1}^{2k-1}\frac{\left|F_X(1/2+it_{\sigma(j)})\right|^2}{\left|F_{X,\widetilde h_j}(1/2+it_{\sigma(j)})\right|^2}. \tag*{(60)} × 1 ≤ j < 2 k h j > 0.99 l o g l o g X ∏ ( log log X ) 12 F X ( 2 1 + i t σ ( 2 k ) ) 2 j = 1 ∏ 2 k − 1 F X , h j ( 1/2 + i t σ ( j ) ) 2 F X ( 1/2 + i t σ ( j ) ) 2 . (60)
Moreover,
F X ( 1 / 2 + i t σ ( j ) ) F X , h ~ j ( 1 / 2 + i t σ ( j ) ) = ∏ X e − h ~ j < p ≤ X ( 1 + f ( p ) p 1 / 2 + i t σ ( j ) ) , \frac{F_X(1/2+it_{\sigma(j)})}{F_{X,\widetilde h_j}(1/2+it_{\sigma(j)})}
=\prod_{Xe^{-\widetilde h_j}<p\le X}\left(1+\frac{f(p)}{p^{1/2+it_{\sigma(j)}}}\right), F X , h j ( 1/2 + i t σ ( j ) ) F X ( 1/2 + i t σ ( j ) ) = X e − h j < p ≤ X ∏ ( 1 + p 1/2 + i t σ ( j ) f ( p ) ) ,
so every Euler factor occurring in these quotients corresponds to a prime p > 1000 k 6 p>1000k^6 p > 1000 k 6 . For a fixed prime p > 1000 k 6 p > 1000k^{6} p > 1000 k 6 , let J p J_p J p denote the set of frequencies whose Euler factors at p p p occur in the last line of (60) . Thus ∣ J p ∣ ≤ 2 k |J_p| \le2k ∣ J p ∣ ≤ 2 k , and
∣ 1 + f ( p ) p − 1 / 2 − i t j ∣ 2 = 1 + 1 p + 2 f ( p ) p cos ( t j log p ) . |1+f(p)p^{-1/2-it_j}|^2=1+\frac{1}{p}+\frac{2f(p)}{\sqrt{p}}\cos(t_j\log p). ∣1 + f ( p ) p − 1/2 − i t j ∣ 2 = 1 + p 1 + p 2 f ( p ) cos ( t j log p ) .
Expanding the product over j ∈ J p j \in J_p j ∈ J p , averaging over f ( p ) f(p) f ( p ) , and then using log ( 1 + u ) ≤ u \log(1+u) \le u log ( 1 + u ) ≤ u , we obtain log E ∏ j ∈ J p ∣ 1 + f ( p ) p − 1 / 2 − i t j ∣ 2 \log\mathbb{E}\prod_{j\in J_p}|1+f(p)p^{-1/2-it_j}|^2 log E ∏ j ∈ J p ∣1 + f ( p ) p − 1/2 − i t j ∣ 2 is
≤ ∣ J p ∣ p + 2 p ∑ j < l j , l ∈ J p ( cos ( ( t l − t j ) log p ) + cos ( ( t l + t j ) log p ) ) + O ( k 3 p 3 / 2 ) . (61) \leq\frac{|J_p|}{p}+\frac{2}{p}\sum_{\substack{j<l\\j,l\in J_p}}\left(\cos((t_l-t_j)\log p)+\cos((t_l+t_j)\log p)\right)+O\left(\frac{k^3}{p^{3/2}}\right). \tag*{(61)} ≤ p ∣ J p ∣ + p 2 j < l j , l ∈ J p ∑ ( cos (( t l − t j ) log p ) + cos (( t l + t j ) log p ) ) + O ( p 3/2 k 3 ) . (61)
The primes p ≤ 1000 k 6 p \le1000k^6 p ≤ 1000 k 6 occur only in the full factor F X ( 1 / 2 + i t σ ( 2 k ) ) F_X(1/2+it_{\sigma(2k)}) F X ( 1/2 + i t σ ( 2 k ) ) , and their contribution to the expectation is ∏ p ≤ 1000 k 6 ( 1 + 1 / p ) ≪ log ( 1000 k 6 ) \prod_{p\leq1000k^6}(1+1/p)\ll\log(1000k^6) ∏ p ≤ 1000 k 6 ( 1 + 1/ p ) ≪ log ( 1000 k 6 ) .
The error terms in (61) are harmless, since ∑ p > 1000 k 6 k 3 / p 3 / 2 ≪ 1 \sum_{p>1000k^6} k^3/p^{3/2}\ll1 ∑ p > 1000 k 6 k 3 / p 3/2 ≪ 1 . The one-point terms are estimated by Mertens’ theorem. For a pair of Euler product factors whose common prime range is Q < p ≤ X Q<p\leq X Q < p ≤ X , Number Theory Result 2 of [20 ] , Section 2.2 gives
∣ ∑ Q < p ≤ X cos ( u log p ) p ∣ ≪ 1 ∣ u ∣ log Q , Q ≥ e ( log log X ) 2 . \left|\sum_{Q<p\leq X}\frac{\cos(u\log p)}{p}\right|\ll\frac{1}{|u|\log Q},\qquad Q\geq e^{(\log\log X)^2}. Q < p ≤ X ∑ p cos ( u log p ) ≪ ∣ u ∣ log Q 1 , Q ≥ e ( l o g l o g X ) 2 .
All the frequencies occurring in a fixed Δ N \Delta_N Δ N have the same sign, since d = o ( ( log log X ) − 2 ) d=o((\log\log X)^{-2}) d = o (( log log X ) − 2 ) . Hence, for j < l j<l j < l ,
∣ t σ ( l ) + t σ ( j ) ∣ ≥ ∣ t σ ( l ) − t σ ( j ) ∣ . |t_{\sigma(l)}+t_{\sigma(j)}|\geq|t_{\sigma(l)}-t_{\sigma(j)}|. ∣ t σ ( l ) + t σ ( j ) ∣ ≥ ∣ t σ ( l ) − t σ ( j ) ∣.
Thus the terms involving t σ ( l ) + t σ ( j ) t_{\sigma(l)}+t_{\sigma(j)} t σ ( l ) + t σ ( j ) , which also occur in Harper’s Rademacher case, are bounded by the same estimates as the terms involving t σ ( l ) − t σ ( j ) t_{\sigma(l)}-t_{\sigma(j)} t σ ( l ) − t σ ( j ) . The resulting pair correlations are therefore controlled by the successive differences t σ ( r + 1 ) − t σ ( r ) t_{\sigma(r+1)}-t_{\sigma(r)} t σ ( r + 1 ) − t σ ( r ) exactly as in [20 ] , Section 3.4.
From this point, the estimates for the ordered frequency regions and the summation over the successive gap sizes proceed as in [20 ] , Section 3.4. After summing over N N N with the weights ( 1 + N 2 ) − 1 (1+N^2)^{-1} ( 1 + N 2 ) − 1 , the expectation of the contribution under consideration, with the deterministic multiplier factor omitted, is
≪ log X ( 1 log 1 / 3 X ( C k 29 ( log log X ) 15 log log log X ) 2 k + ( C k 29 log log log X ( log log X ) 991 ) 2 k ) . \ll\log X\left(\frac{1}{\log^{1/3}X}(Ck^{29}(\log\log X)^{15}\log\log\log X)^{2k}+\left(\frac{Ck^{29}\log\log\log X}{(\log\log X)^{991}}\right)^{2k}\right). ≪ log X ( log 1/3 X 1 ( C k 29 ( log log X ) 15 log log log X ) 2 k + ( ( log log X ) 991 C k 29 log log log X ) 2 k ) .
Applying Markov’s inequality at ( log log X ) 2 k (\log\log X)^{2k} ( log log X ) 2 k times this expectation gives an exceptional probability at most ( log log X ) − 2 k (\log\log X)^{-2k} ( log log X ) − 2 k . Restoring the multiplier factor (5.18) amounts to inserting ( C log log log X ) 4 (C\log\log\log X)^4 ( C log log log X ) 4 inside each 2 k 2k 2 k -th power. Thus the two quantities inside the powers are bounded by C k 29 ( log log X ) 16 ( log log log X ) 5 ≤ k 29 ( log log X ) 17 Ck^{29}(\log\log X)^{16}(\log\log\log X)^5\leq k^{29}(\log\log X)^{17} C k 29 ( log log X ) 16 ( log log log X ) 5 ≤ k 29 ( log log X ) 17 and C k 29 ( log log log X ) 5 / ( log log X ) 990 ≤ k 29 / ( log log X ) 989 Ck^{29}(\log\log\log X)^5/(\log\log X)^{990}\leq k^{29}/(\log\log X)^{989} C k 29 ( log log log X ) 5 / ( log log X ) 990 ≤ k 29 / ( log log X ) 989 , respectively, for sufficiently large X X X . Hence the last two terms are unchanged from Harper’s final estimate. The only visible effect of the deterministic multiplier is the replacement log log log X \log\log\log X log log log X by ( log log log X ) 5 (\log\log\log X)^5 ( log log log X ) 5 in the coefficient of the unit-interval mean square term.
Combining the two ranges and then using (5.19) gives (5.17). The argument is uniform for 1 ≤ k ≤ log 1 / 4 X 1\leq k\leq\log^{1/4}X 1 ≤ k ≤ log 1/4 X , since, for sufficiently large X X X ,
2 k d ≤ 2 log − 2 / 3 X , X e − h ~ j ≥ e ( log log X ) 2 ≥ 1000 k 6 . 2kd\leq2\log^{-2/3}X,\qquad Xe^{-\tilde h_j}\geq e^{(\log\log X)^2}\geq1000k^6. 2 k d ≤ 2 log − 2/3 X , X e − h ~ j ≥ e ( l o g l o g X ) 2 ≥ 1000 k 6 .
Finally, the only dependence on the common shift is through e − i θ Ξ e^{-i\theta\Xi} e − i θ Ξ , which has modulus one. Hence the same exceptional event works simultaneously for every 0 ≤ θ ≤ 2 π 0\leq\theta\leq2\pi 0 ≤ θ ≤ 2 π .
Lemma 5.6 (Uniform high moment bound). For every integer 1 ≤ k ≤ log 1 / 4 X 1 \le k \le\log^{1/4} X 1 ≤ k ≤ log 1/4 X , outside an event of probability O ( ( log log X ) − 1 ) O((\log\log X)^{-1}) O (( log log X ) − 1 ) , we have, simultaneously for every 0 ≤ θ ≤ 2 π 0 \le\theta\le2\pi 0 ≤ θ ≤ 2 π ,
max x ∈ G X ( θ ) ∑ y ∈ G X ( θ ) ∣ I ( x , y ) ∣ 2 k ≪ log X ( ( k 29 ( log log X ) 17 ) 2 k log 1 / 3 X + ( k 29 ( log log X ) 989 ) 2 k ) . (62) \max_{x\in\mathcal{G}_X(\theta)}\sum_{y\in\mathcal{G}_X(\theta)}|\mathcal{I}(x,y)|^{2k}\ll\log X\left(\frac{(k^{29}(\log\log X)^{17})^{2k}}{\log^{1/3}X}+\left(\frac{k^{29}}{(\log\log X)^{989}}\right)^{2k}\right). \tag*{(62)} x ∈ G X ( θ ) max y ∈ G X ( θ ) ∑ ∣ I ( x , y ) ∣ 2 k ≪ log X ( log 1/3 X ( k 29 ( log log X ) 17 ) 2 k + ( ( log log X ) 989 k 29 ) 2 k ) . (62)
Proof. By Markov’s inequality and a union bound, we have
P ( max ∣ N ∣ ≤ ( log log X ) 2 + 1 ∫ N N + 1 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t > log X ( log log X ) 3 ) ≤ 1 log X ( log log X ) 3 ∑ ∣ N ∣ ≤ ( log log X ) 2 + 1 ∫ N N + 1 E ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ≪ ( log log X ) 2 ( log log X ) − 3 ≪ ( log log X ) − 1 , \begin{aligned}
\mathbb{P}\left(\max_{|N|\le(\log\log X)^2+1}\int_N^{N+1}|F_X(1/2+it)|^2\,dt>\log X(\log\log X)^3\right) \\
&\le\frac{1}{\log X(\log\log X)^3}\sum_{|N|\le(\log\log X)^2+1}\int_N^{N+1}\mathbb{E}|F_X(1/2+it)|^2\,dt \\
&\ll(\log\log X)^2(\log\log X)^{-3}\ll(\log\log X)^{-1},
\end{aligned} P ( ∣ N ∣ ≤ ( l o g l o g X ) 2 + 1 max ∫ N N + 1 ∣ F X ( 1/2 + i t ) ∣ 2 d t > log X ( log log X ) 3 ) ≤ log X ( log log X ) 3 1 ∣ N ∣ ≤ ( l o g l o g X ) 2 + 1 ∑ ∫ N N + 1 E ∣ F X ( 1/2 + i t ) ∣ 2 d t ≪ ( log log X ) 2 ( log log X ) − 3 ≪ ( log log X ) − 1 ,
where E ∣ F X ( 1 / 2 + i t ) ∣ 2 = ∏ p ≤ X ( 1 + 1 / p ) ≪ log X \mathbb{E}|F_X(1/2+it)|^2=\prod_{p\le X}(1+1/p)\ll\log X E ∣ F X ( 1/2 + i t ) ∣ 2 = ∏ p ≤ X ( 1 + 1/ p ) ≪ log X uniformly in t t t . Thus, outside an event of probability O ( ( log log X ) − 1 ) O((\log\log X)^{-1}) O (( log log X ) − 1 ) ,
∫ N N + 1 ∣ F X ( 1 / 2 + i t ) ∣ 2 d t ≤ log X ( log log X ) 3 ( ∣ N ∣ ≤ ( log log X ) 2 + 1 ) . \int_N^{N+1}|F_X(1/2+it)|^2\,dt\le\log X(\log\log X)^3\qquad(|N|\le(\log\log X)^2+1). ∫ N N + 1 ∣ F X ( 1/2 + i t ) ∣ 2 d t ≤ log X ( log log X ) 3 ( ∣ N ∣ ≤ ( log log X ) 2 + 1 ) .
Substituting this bound into the first term of (5.17), we obtain it is
≪ 1 log 1 / 3 X ( C ( log log log X ) 5 ( log log X ) 3 ) 2 k ≪ 1 log 1 / 3 X ( k 29 ( log log X ) 17 ) 2 k , \ll\frac{1}{\log^{1/3}X}\left(C(\log\log\log X)^5(\log\log X)^3\right)^{2k}\ll\frac{1}{\log^{1/3}X}\left(k^{29}(\log\log X)^{17}\right)^{2k}, ≪ log 1/3 X 1 ( C ( log log log X ) 5 ( log log X ) 3 ) 2 k ≪ log 1/3 X 1 ( k 29 ( log log X ) 17 ) 2 k ,
for sufficiently large X X X . Lemma 5.5 therefore gives (5.25).
The exceptional probability from Lemma 5.5 is at most ( log log X ) − 2 k ≪ ( log log X ) − 2 (\log\log X)^{-2k}\ll(\log\log X)^{-2} ( log log X ) − 2 k ≪ ( log log X ) − 2 , so together with the preceding exceptional event the total probability is O ( ( log log X ) − 1 ) O((\log\log X)^{-1}) O (( log log X ) − 1 ) . Both estimates hold simultaneously for every common shift θ \theta θ , which completes the proof. □
Proof of Proposition 5.1. We take k k k to be the largest integer ≤ log log X / ( 5000 log log log X ) \le\log\log X/(5000\log\log\log X) ≤ log log X / ( 5000 log log log X ) . By Lemma 5.4, there is an absolute constant c > 0 c>0 c > 0 such that
∣ Cov X ( Y x , Y y ) ∣ > ( log log X ) − 0.59 ⟹ ∣ I ( x , y ) ∣ ≥ c ( log log X ) − 0.59 . |\operatorname{Cov}_X(Y_x,Y_y)|>(\log\log X)^{-0.59}\quad\Longrightarrow\quad|\mathcal{I}(x,y)|\ge c(\log\log X)^{-0.59}. ∣ Cov X ( Y x , Y y ) ∣ > ( log log X ) − 0.59 ⟹ ∣ I ( x , y ) ∣ ≥ c ( log log X ) − 0.59 .
For fixed x ∈ G X ( θ ) x\in\mathcal{G}_X(\theta) x ∈ G X ( θ ) , let D x D_x D x denote the number of y ∈ G X ( θ ) y\in\mathcal{G}_X(\theta) y ∈ G X ( θ ) satisfying the latter inequality. Then
D x c 2 k ( log log X ) − 1.18 k ≤ ∑ y ∈ G X ( θ ) ∣ I ( x , y ) ∣ 2 k . D_xc^{2k}(\log\log X)^{-1.18k}\le\sum_{y\in\mathcal{G}_X(\theta)}|\mathcal{I}(x,y)|^{2k}. D x c 2 k ( log log X ) − 1.18 k ≤ y ∈ G X ( θ ) ∑ ∣ I ( x , y ) ∣ 2 k .
Lemma 5.6 therefore gives
D x ≪ c − 2 k ( log log X ) 1.18 k log X ( ( k 29 ( log log X ) 17 ) 2 k log 1 / 3 X + ( k 29 ( log log X ) 989 ) 2 k ) ≪ log 2 / 3 X ( ( log log X ) 50 ) 2 k + log X ( ( log log X ) − 950 ) 2 k ≪ ( log X ) 0.70 ≤ ( log X ) 0.8 \begin{aligned}
D_x\ll c^{-2k}(\log\log X)^{1.18k}\log X\left(\frac{(k^{29}(\log\log X)^{17})^{2k}}{\log^{1/3}X}+\left(\frac{k^{29}}{(\log\log X)^{989}}\right)^{2k}\right) \\
&\ll\log^{2/3}X\left((\log\log X)^{50}\right)^{2k}+\log X\left((\log\log X)^{-950}\right)^{2k} \\
&\ll(\log X)^{0.70}\le(\log X)^{0.8}
\end{aligned} D x ≪ c − 2 k ( log log X ) 1.18 k log X ( log 1/3 X ( k 29 ( log log X ) 17 ) 2 k + ( ( log log X ) 989 k 29 ) 2 k ) ≪ log 2/3 X ( ( log log X ) 50 ) 2 k + log X ( ( log log X ) − 950 ) 2 k ≪ ( log X ) 0.70 ≤ ( log X ) 0.8
for sufficiently large X X X .
The exceptional events in Lemmas 5.4 and 5.6 have probabilities O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) and O ( ( log log X ) − 1 ) O((\log\log X)^{-1}) O (( log log X ) − 1 ) , respectively. Both conclusions hold simultaneously for every common shift θ \theta θ , so their union has probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) . This proves the proposition. □
Conditional Gaussian approximation
We now compare the conditional law of the large prime increments Y x Y_x Y x in (2.8) with that of a centred Gaussian vector having the same conditional covariance matrix. Once the variables ( f ( p ) ) p ≤ X (f(p))_{p \le X} ( f ( p ) ) p ≤ X are fixed, this is a deterministic multivariate normal-approximation problem. The Gaussian comparison itself is supplied by Normal Approximation Result 1 of [20 ] . The only additional point needed here is to verify that its error terms remain negligible for the coefficients arising from the extended Rademacher RMF.
Conditional on F X \mathcal{F}_X F X , let ( Z x ) x ∈ T (Z_x)_{x \in\mathcal{T}} ( Z x ) x ∈ T be a centred Gaussian vector with the same covariance matrix as ( Y x ) x ∈ T (Y_x)_{x \in\mathcal{T}} ( Y x ) x ∈ T , and write P Z \mathbb{P}_Z P Z for probability with respect to this Gaussian vector.
Lemma 6.1 (Conditional Gaussian approximation). Let T 0 ⊂ [ X 8 / 7 , X 4 / 3 / 4 ] \mathcal{T}_0 \subset[X^{8/7},X^{4/3}/4] T 0 ⊂ [ X 8/7 , X 4/3 /4 ] be deterministic with # T 0 ≪ log X \#\mathcal{T}_0 \ll\log X # T 0 ≪ log X , and let ∅ ≠ T ⊆ T 0 \varnothing\ne\mathcal{T} \subseteq\mathcal{T}_0 ∅ = T ⊆ T 0 be F X \mathcal{F}_X F X -measurable. Conditional on F X \mathcal{F}_X F X , let ( Z x ) x ∈ T (Z_x)_{x \in\mathcal{T}} ( Z x ) x ∈ T be a centred Gaussian vector with the same covariance matrix as ( Y x ) x ∈ T (Y_x)_{x \in\mathcal{T}} ( Y x ) x ∈ T . Then, for every u ∈ R u \in\mathbb{R} u ∈ R ,
P X ( max x ∈ T Y x ≤ u ) ≤ P Z ( max x ∈ T Z x ≤ u + 1 ) + Err X ( T 0 ) , (63) \mathbb{P}_X\left(\max_{x \in\mathcal{T}} Y_x \le u\right)
\le
\mathbb{P}_Z\left(\max_{x \in\mathcal{T}} Z_x \le u+1\right)
+\operatorname{Err}_X(\mathcal{T}_0), \tag*{(63)} P X ( x ∈ T max Y x ≤ u ) ≤ P Z ( x ∈ T max Z x ≤ u + 1 ) + Err X ( T 0 ) , (63)
where Err X ( T 0 ) \operatorname{Err}_X(\mathcal{T}_0) Err X ( T 0 ) is F X \mathcal{F}_X F X -measurable and satisfies
E Err X ( T 0 ) ≪ X − 1 / 2 log C X (64) \mathbb{E}\operatorname{Err}_X(\mathcal{T}_0) \ll X^{-1/2}\log^C X \tag*{(64)} E Err X ( T 0 ) ≪ X − 1/2 log C X (64)
for some absolute constant C > 0 C>0 C > 0 .
Proof. Conditional on F X \mathcal{F}_X F X , we write Y x = ∑ X < p ≤ X 4 / 3 a p , x f ( p ) Y_x=\sum_{X<p\le X^{4/3}}a_{p,x}f(p) Y x = ∑ X < p ≤ X 4/3 a p , x f ( p ) , where a p , x : = B X ( x / p ) / x a_{p,x}:=B_X(x/p)/\sqrt{x} a p , x := B X ( x / p ) / x . The coefficients a p , x a_{p,x} a p , x are then deterministic, while the variables ( f ( p ) ) p > X (f(p))_{p>X} ( f ( p ) ) p > X remain independent Rademacher random variables. Normal Approximation Result 1 of [20 ] , applied conditionally on F X \mathcal{F}_X F X with smoothing parameter η = 1 \eta=1 η = 1 , therefore gives
P X ( max x ∈ T Y x ≤ u ) ≤ P Z ( max x ∈ T Z x ≤ u + 1 ) + O ( E 2 ( T ) + E 3 ( T ) ) , \mathbb{P}_X\left(\max_{x \in\mathcal{T}}Y_x\le u\right)
\le
\mathbb{P}_Z\left(\max_{x \in\mathcal{T}}Z_x\le u+1\right)
+O\left(E_2(\mathcal{T})+E_3(\mathcal{T})\right), P X ( x ∈ T max Y x ≤ u ) ≤ P Z ( x ∈ T max Z x ≤ u + 1 ) + O ( E 2 ( T ) + E 3 ( T ) ) ,
where
E 2 ( T ) : = ∑ x , y ∈ T ( ∑ X < p ≤ X 4 / 3 a p , x 2 a p , y 2 ) 1 / 2 and E 3 ( T ) : = ∑ X < p ≤ X 4 / 3 ( ∑ x ∈ T ∣ a p , x ∣ ) 3 . E_2(\mathcal{T})
:=
\sum_{x,y\in\mathcal{T}}
\left(
\sum_{X<p\le X^{4/3}}a_{p,x}^2a_{p,y}^2
\right)^{1/2}
\qquad\text{and}\qquad
E_3(\mathcal{T})
:=
\sum_{X<p\le X^{4/3}}
\left(\sum_{x\in\mathcal{T}}|a_{p,x}|\right)^3. E 2 ( T ) := x , y ∈ T ∑ X < p ≤ X 4/3 ∑ a p , x 2 a p , y 2 1/2 and E 3 ( T ) := X < p ≤ X 4/3 ∑ ( x ∈ T ∑ ∣ a p , x ∣ ) 3 .
Since every term in these expressions is non-negative, then E j ( T ) ≤ E j ( T 0 ) E_j(\mathcal{T})\le E_j(\mathcal{T}_0) E j ( T ) ≤ E j ( T 0 ) for j = 2 , 3 j=2,3 j = 2 , 3 . Thus it suffices to show that
E E 2 ( T 0 ) + E E 3 ( T 0 ) ≪ X − 1 / 2 log C X (65) \mathbb{E}E_2(\mathcal{T}_0)+\mathbb{E}E_3(\mathcal{T}_0)
\ll X^{-1/2}\log^C X \tag*{(65)} E E 2 ( T 0 ) + E E 3 ( T 0 ) ≪ X − 1/2 log C X (65)
for some absolute constant C > 0 C>0 C > 0 .
We first estimate E 2 ( T 0 ) E_2(\mathcal{T}_0) E 2 ( T 0 ) . If x ∈ T 0 x\in\mathcal{T}_0 x ∈ T 0 and p > X p>X p > X , then 4 x / p ≤ X 1 / 3 < X 4x/p\le X^{1/3}<X 4 x / p ≤ X 1/3 < X . Hence the moment estimates for B X B_X B X in Lemma 2.1 apply to B X ( x / p ) B_X(x/p) B X ( x / p ) . For x , y ∈ T 0 x,y\in\mathcal{T}_0 x , y ∈ T 0 , by Cauchy–Schwarz and Lemma 2.1 with r = 4 r=4 r = 4 we get
E ( a p , x 2 a p , y 2 ) ≪ 1 x y ( E ∣ B X ( x / p ) ∣ 4 ) 1 / 2 ( E ∣ B X ( y / p ) ∣ 4 ) 1 / 2 ≪ ( 1 + x / p ) ( 1 + y / p ) x y log C X . \mathbb{E}\left(a_{p,x}^2a_{p,y}^2\right)
\ll
\frac{1}{xy}
\left(\mathbb{E}|B_X(x/p)|^4\right)^{1/2}
\left(\mathbb{E}|B_X(y/p)|^4\right)^{1/2}
\ll
\frac{(1+x/p)(1+y/p)}{xy}\log^C X. E ( a p , x 2 a p , y 2 ) ≪ x y 1 ( E ∣ B X ( x / p ) ∣ 4 ) 1/2 ( E ∣ B X ( y / p ) ∣ 4 ) 1/2 ≪ x y ( 1 + x / p ) ( 1 + y / p ) log C X .
Moreover, we have a p , x a p , y = 0 a_{p,x}a_{p,y}=0 a p , x a p , y = 0 unless p ≤ 4 min ( x , y ) p\le4\min(x,y) p ≤ 4 min ( x , y ) . Suppose, without loss of generality, that x ≤ y x\le y x ≤ y . Then
∑ X < p ≤ X 4 / 3 E ( a p , x 2 a p , y 2 ) ≪ log C X x y ∑ X < p ≤ 4 x ( 1 + x p ) ( 1 + y p ) ≪ log C X ( 1 y + 1 x ∑ p > X 1 p + ∑ p > X 1 p 2 ) ≪ X − 1 log C X . \begin{aligned}
\sum_{X<p\le X^{4/3}}\mathbb{E}\left(a_{p,x}^2a_{p,y}^2\right)
\ll
\frac{\log^C X}{xy}
\sum_{X<p\le4x}
\left(1+\frac{x}{p}\right)\left(1+\frac{y}{p}\right) \\
&\ll
\log^C X
\left(
\frac{1}{y}
+\frac{1}{x}\sum_{p>X}\frac{1}{p}
+\sum_{p>X}\frac{1}{p^2}
\right)
\ll X^{-1}\log^C X.
\end{aligned} X < p ≤ X 4/3 ∑ E ( a p , x 2 a p , y 2 ) ≪ x y log C X X < p ≤ 4 x ∑ ( 1 + p x ) ( 1 + p y ) ≪ log C X y 1 + x 1 p > X ∑ p 1 + p > X ∑ p 2 1 ≪ X − 1 log C X .
Here we used x , y ≥ X 8 / 7 x,y \ge X^{8/7} x , y ≥ X 8/7 . By Jensen’s inequality,
E ( ∑ X < p ≤ X 4 / 3 a p , x 2 a p , y 2 ) 1 / 2 ≤ ( ∑ X < p ≤ X 4 / 3 E ( a p , x 2 a p , y 2 ) ) 1 / 2 ≪ X − 1 / 2 log C X . \mathbb{E}\left(\sum_{X<p\le X^{4/3}}a_{p,x}^{2}a_{p,y}^{2}\right)^{1/2}\le\left(\sum_{X<p\le X^{4/3}}\mathbb{E}\left(a_{p,x}^{2}a_{p,y}^{2}\right)\right)^{1/2}\ll X^{-1/2}\log^{C}X. E X < p ≤ X 4/3 ∑ a p , x 2 a p , y 2 1/2 ≤ X < p ≤ X 4/3 ∑ E ( a p , x 2 a p , y 2 ) 1/2 ≪ X − 1/2 log C X .
Since # T 0 ≪ log X \#\mathcal{T}_0\ll\log X # T 0 ≪ log X , it follows that
E E 2 ( T 0 ) ≪ X − 1 / 2 log C X . (66) \mathbb{E}E_2(\mathcal{T}_0)\ll X^{-1/2}\log^{C}X. \tag*{(66)} E E 2 ( T 0 ) ≪ X − 1/2 log C X . (66)
We next estimate E 3 ( T 0 ) E_3(\mathcal{T}_0) E 3 ( T 0 ) . Again by Lemma 2.1 with r = 3 r=3 r = 3 ,
E ∣ a p , x ∣ 3 ≪ x − 3 / 2 ( 1 + x p ) 3 / 2 log C X , \mathbb{E}|a_{p,x}|^3\ll x^{-3/2}\left(1+\frac{x}{p}\right)^{3/2}\log^{C}X, E ∣ a p , x ∣ 3 ≪ x − 3/2 ( 1 + p x ) 3/2 log C X ,
and a p , x = 0 a_{p,x}=0 a p , x = 0 unless p ≤ 4 x p\le4x p ≤ 4 x . Since ( 1 + u ) 3 / 2 ≪ 1 + u 3 / 2 (1+u)^{3/2}\ll1+u^{3/2} ( 1 + u ) 3/2 ≪ 1 + u 3/2 , we obtain
∑ X < p ≤ X 4 / 3 ∑ x ∈ T 0 E ∣ a p , x ∣ 3 ≪ # T 0 ∑ X < p ≤ X 4 / 3 E ∣ a p , x ∣ 3 ≪ log C X ( x − 3 / 2 # { X < p ≤ 4 x } + ∑ p > X 1 p 3 / 2 ) ≪ X − 1 / 2 log C X , (67) \begin{aligned}
\sum_{X<p\le X^{4/3}}\sum_{x\in\mathcal{T}_0}\mathbb{E}|a_{p,x}|^3\ll\#\mathcal{T}_0\sum_{X<p\le X^{4/3}}\mathbb{E}|a_{p,x}|^3 \\
&\ll\log^{C}X\left(x^{-3/2}\#\{X<p\le4x\}+\sum_{p>X}\frac{1}{p^{3/2}}\right)\ll X^{-1/2}\log^{C}X, \tag*{(67)}
\end{aligned} X < p ≤ X 4/3 ∑ x ∈ T 0 ∑ E ∣ a p , x ∣ 3 ≪ # T 0 X < p ≤ X 4/3 ∑ E ∣ a p , x ∣ 3 ≪ log C X x − 3/2 # { X < p ≤ 4 x } + p > X ∑ p 3/2 1 ≪ X − 1/2 log C X , (67)
where # T 0 ≪ log X \#\mathcal{T}_0\ll\log X # T 0 ≪ log X has again been absorbed into the power of log X \log X log X . Now since
( ∑ x ∈ T 0 ∣ a p , x ∣ ) 3 ≤ ( # T 0 ) 2 ∑ x ∈ T 0 ∣ a p , x ∣ 3 , \left(\sum_{x\in\mathcal{T}_0}|a_{p,x}|\right)^3\le(\#\mathcal{T}_0)^2\sum_{x\in\mathcal{T}_0}|a_{p,x}|^3, ( x ∈ T 0 ∑ ∣ a p , x ∣ ) 3 ≤ ( # T 0 ) 2 x ∈ T 0 ∑ ∣ a p , x ∣ 3 ,
taking expectations and summing over p p p , together with (6.5), we obtain
E E 3 ( T 0 ) ≪ X − 1 / 2 log C X . (68) \mathbb{E}E_3(\mathcal{T}_0)\ll X^{-1/2}\log^{C}X. \tag*{(68)} E E 3 ( T 0 ) ≪ X − 1/2 log C X . (68)
Combining (6.4) and (6.6) proves (6.3) and hence completes the proof. □
A common logarithmic shift
The estimate in Proposition 3.2 controls the X X X -smooth contribution only after averaging over the endpoint, whereas the high moment argument in Section 5 relies on an exactly logarithmically spaced grid. We therefore average over common logarithmic shifts of this grid. Outside a small exceptional event, this shows that, for all but a small proportion of the shifts, the X X X -smooth contribution is small at most relevant endpoints.
For sufficiently large X X X , define
J : = ⌊ ( 4 / 21 ) log X − log 4 − 2 π 2 π ⌋ ≍ log X J:=\left\lfloor\frac{(4/21)\log X-\log4-2\pi}{2\pi}\right\rfloor\asymp\log X J := ⌊ 2 π ( 4/21 ) log X − log 4 − 2 π ⌋ ≍ log X
so that e 2 π X 8 / 7 ≤ x r , θ , 4 x r , θ ≤ X 4 / 3 e^{2\pi}X^{8/7}\le x_{r,\theta},4x_{r,\theta}\le X^{4/3} e 2 π X 8/7 ≤ x r , θ , 4 x r , θ ≤ X 4/3 , where x r , θ : = X 8 / 7 e θ + 2 π r x_{r,\theta}:=X^{8/7}e^{\theta+2\pi r} x r , θ := X 8/7 e θ + 2 π r for 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π and 1 ≤ r ≤ J 1\le r\le J 1 ≤ r ≤ J . Moreover, we have log x r , θ − log x s , θ = 2 π ( r − s ) \log x_{r,\theta}-\log x_{s,\theta}=2\pi(r-s) log x r , θ − log x s , θ = 2 π ( r − s ) , which is independent of θ \theta θ .
Lemma 7.1 (Averaging over a common logarithmic shift). Let A > 0 A>0 A > 0 . For 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π , let R θ ( A ) R_\theta(A) R θ ( A ) be the set of 1 ≤ r ≤ J 1\le r\le J 1 ≤ r ≤ J such that ∣ C X ( x r , θ ) ∣ ≤ A |C_X(x_{r,\theta})|\le A ∣ C X ( x r , θ ) ∣ ≤ A and ∣ C X ( 4 x r , θ ) ∣ ≤ A |C_X(4x_{r,\theta})|\le A ∣ C X ( 4 x r , θ ) ∣ ≤ A . Then
1 2 π ∫ 0 2 π J − # R θ ( A ) J d θ ≪ 1 A 2 log X ∫ X 8 / 7 X 4 / 3 ∣ C X ( z ) ∣ 2 d z z . (69) \frac{1}{2\pi}\int_0^{2\pi}\frac{J-\#R_\theta(A)}{J}\,d\theta\ll\frac{1}{A^2\log X}\int_{X^{8/7}}^{X^{4/3}}|C_X(z)|^2\frac{dz}{z}. \tag*{(69)} 2 π 1 ∫ 0 2 π J J − # R θ ( A ) d θ ≪ A 2 log X 1 ∫ X 8/7 X 4/3 ∣ C X ( z ) ∣ 2 z d z . (69)
Proof. Starting with the elementary inequality
J − # R θ ( A ) ≤ ∑ r = 1 J ( 1 { ∣ C X ( x r , θ ) ∣ > A } + 1 { ∣ C X ( 4 x r , θ ) ∣ > A } ) , J-\#R_\theta(A)\le\sum_{r=1}^{J}\left(\mathbf{1}_{\{|C_X(x_{r,\theta})|>A\}}+\mathbf{1}_{\{|C_X(4x_{r,\theta})|>A\}}\right), J − # R θ ( A ) ≤ r = 1 ∑ J ( 1 { ∣ C X ( x r , θ ) ∣ > A } + 1 { ∣ C X ( 4 x r , θ ) ∣ > A } ) ,
we obtain
∫ 0 2 π ( J − # R θ ( A ) ) d θ ≤ 1 A 2 ∫ 0 2 π ∑ r = 1 J ( ∣ C X ( x r , θ ) ∣ 2 + ∣ C X ( 4 x r , θ ) ∣ 2 ) d θ . \int_0^{2\pi}(J-\#R_\theta(A))\,d\theta\le\frac{1}{A^2}\int_0^{2\pi}\sum_{r=1}^{J}\left(|C_X(x_{r,\theta})|^2+|C_X(4x_{r,\theta})|^2\right)\,d\theta. ∫ 0 2 π ( J − # R θ ( A )) d θ ≤ A 2 1 ∫ 0 2 π r = 1 ∑ J ( ∣ C X ( x r , θ ) ∣ 2 + ∣ C X ( 4 x r , θ ) ∣ 2 ) d θ .
For each of the two families x r , θ x_{r,\theta} x r , θ and 4 x r , θ 4x_{r,\theta} 4 x r , θ , the changes of variables z = x r , θ z=x_{r,\theta} z = x r , θ and z = 4 x r , θ z=4x_{r,\theta} z = 4 x r , θ give d z / z = d θ \mathrm{d}z/z=\mathrm{d}\theta d z / z = d θ . As r r r varies, the corresponding logarithmic intervals are disjoint apart from their endpoints and lie in [ X 8 / 7 , X 4 / 3 ] [X^{8/7},X^{4/3}] [ X 8/7 , X 4/3 ] . Hence
∫ 0 2 π ∑ r = 1 J ( ∣ C X ( x r , θ ) ∣ 2 + ∣ C X ( 4 x r , θ ) ∣ 2 ) d θ ≪ ∫ X 8 / 7 X 4 / 3 ∣ C X ( z ) ∣ 2 d z z . \int_{0}^{2\pi}\sum_{r=1}^{J}\left(\lvert C_X(x_{r,\theta})\rvert^2+\lvert C_X(4x_{r,\theta})\rvert^2\right)\,\mathrm{d}\theta\ll\int_{X^{8/7}}^{X^{4/3}}\lvert C_X(z)\rvert^2\frac{\mathrm{d}z}{z}. ∫ 0 2 π r = 1 ∑ J ( ∣ C X ( x r , θ ) ∣ 2 + ∣ C X ( 4 x r , θ ) ∣ 2 ) d θ ≪ ∫ X 8/7 X 4/3 ∣ C X ( z ) ∣ 2 z d z .
Since J ≍ log X J\asymp\log X J ≍ log X , this proves (7.1). □
Taking now A = e − 1.2 W ( log log X ) 1 / 4 A=e^{-1.2W}(\log\log X)^{1/4} A = e − 1.2 W ( log log X ) 1/4 and using Proposition 3.2, the right-hand side of (7.1) is ≪ e 2.4 W ( log log X ) − 1 / 6 \ll e^{2.4W}(\log\log X)^{-1/6} ≪ e 2.4 W ( log log X ) − 1/6 . By Markov’s inequality, we have
# R θ ( A ) ≥ ( 1 − O ( e 1.2 W ( log log X ) − 1 / 12 ) ) J \#\mathcal{R}_{\theta}(A)\geq\left(1-O\left(e^{1.2W}(\log\log X)^{-1/12}\right)\right)J # R θ ( A ) ≥ ( 1 − O ( e 1.2 W ( log log X ) − 1/12 ) ) J
for all but O ( e 1.2 W ( log log X ) − 1 / 12 ) O(e^{1.2W}(\log\log X)^{-1/12}) O ( e 1.2 W ( log log X ) − 1/12 ) of the shifts θ \theta θ .
From increments to two-sided fluctuations
We now assemble the four ingredients needed to prove Theorem 1.4: smallness of the X X X -smooth contribution on most shifted endpoints, a lower bound for the conditional variances of the increments, sparsity of their large conditional covariances, and the conditional Gaussian approximation. These are supplied, respectively, by Proposition 3.2 together with Lemma 7.1, Proposition 4.2, Proposition 5.1, and Lemma 6.1.
The exceptional events in these four inputs arise separately. Before carrying out the Gaussian comparison, we collect them into a single F X \mathcal{F}_X F X -measurable good event and restrict to a common set of shifts on which all four conclusions hold simultaneously. The next lemma records this reduction.
Throughout this section we write A : = e − 1.2 W ( log log X ) 1 / 4 A:=e^{-1.2W}(\log\log X)^{1/4} A := e − 1.2 W ( log log X ) 1/4 .
Lemma 8.1 (A simultaneous good event). Suppose that W 0 ≤ W ≤ ( log log X ) 1 / 100 W_0\leq W\leq(\log\log X)^{1/100} W 0 ≤ W ≤ ( log log X ) 1/100 . Then there is an F X \mathcal{F}_X F X -measurable event Ω X ( W ) \Omega_X(W) Ω X ( W ) satisfying
P ( Ω X ( W ) c ) ≪ ( log log X ) − 1 / 200 + e − W / 10 . (70) \mathbb{P}\left(\Omega_X(W)^{\mathrm{c}}\right)\ll(\log\log X)^{-1/200}+e^{-W/10}. \tag*{(70)} P ( Ω X ( W ) c ) ≪ ( log log X ) − 1/200 + e − W /10 . (70)
On Ω X ( W ) \Omega_X(W) Ω X ( W ) , there is a measurable set Θ X ⊂ [ 0 , 2 π ) \Theta_X\subset[0,2\pi) Θ X ⊂ [ 0 , 2 π ) of normalised measure
1 − O ( e 1.2 W ( log log X ) − 1 / 12 + X − 1 / 8 ) 1-O\left(e^{1.2W}(\log\log X)^{-1/12}+X^{-1/8}\right) 1 − O ( e 1.2 W ( log log X ) − 1/12 + X − 1/8 )
such that, for every θ ∈ Θ X \theta\in\Theta_X θ ∈ Θ X , the following hold:
(1)
# R θ ( A ) ≥ ( 1 − O ( e 1.2 W ( log log X ) − 1 / 12 ) ) J . \#\mathcal{R}_{\theta}(A)\geq\left(1-O\left(e^{1.2W}(\log\log X)^{-1/12}\right)\right)J. # R θ ( A ) ≥ ( 1 − O ( e 1.2 W ( log log X ) − 1/12 ) ) J .
(2) For every 1 ≤ r ≤ J 1\leq r\leq J 1 ≤ r ≤ J ,
Var X ( Y x r , θ ) ≫ e − 2.2 W ( log log X ) − 1 / 2 . (71) \operatorname{Var}_X(Y_{x_{r,\theta}})\gg e^{-2.2W}(\log\log X)^{-1/2}. \tag*{(71)} Var X ( Y x r , θ ) ≫ e − 2.2 W ( log log X ) − 1/2 . (71)
(3) The graph on { 1 , … , J } \{1,\ldots,J\} { 1 , … , J } in which r ≠ s r\neq s r = s are joined when
∣ Cov X ( Y x r , θ , Y x s , θ ) ∣ > ( log log X ) − 0.59 (72) \left\lvert\operatorname{Cov}_X(Y_{x_{r,\theta}},Y_{x_{s,\theta}})\right\rvert>(\log\log X)^{-0.59} \tag*{(72)} Cov X ( Y x r , θ , Y x s , θ ) > ( log log X ) − 0.59 (72)
has maximum degree at most ( log X ) 0.8 (\log X)^{0.8} ( log X ) 0.8 .
(4) For T 0 ( θ ) = { x r , θ : 1 ≤ r ≤ J } T_0(\theta)=\{x_{r,\theta}:1\leq r\leq J\} T 0 ( θ ) = { x r , θ : 1 ≤ r ≤ J } , we have
Err X ( T 0 ( θ ) ) ≤ X − 1 / 8 . \operatorname{Err}_X(T_0(\theta))\leq X^{-1/8}. Err X ( T 0 ( θ )) ≤ X − 1/8 .
Proof. By Proposition 3.2, outside an F X \mathcal{F}_X F X -measurable event of probability O ( ( log log X ) − 1 / 200 ) O((\log\log X)^{-1/200}) O (( log log X ) − 1/200 ) , we have
1 log X ∫ X 8 / 7 X 4 / 3 ∣ C X ( z ) ∣ 2 d z z ≤ ( log log X ) 1 / 3 . \frac{1}{\log X}\int_{X^{8/7}}^{X^{4/3}}\lvert C_X(z)\rvert^2\frac{\mathrm{d}z}{z}\leq(\log\log X)^{1/3}. log X 1 ∫ X 8/7 X 4/3 ∣ C X ( z ) ∣ 2 z d z ≤ ( log log X ) 1/3 .
Since A = e − 1.2 W ( log log X ) 1 / 4 A=e^{-1.2W}(\log\log X)^{1/4} A = e − 1.2 W ( log log X ) 1/4 , Lemma 7.1 then gives
1 2 π ∫ 0 2 π J − # R θ ( A ) J d θ ≪ A − 2 ( log log X ) 1 / 3 = e 2.4 W ( log log X ) − 1 / 6 . \frac{1}{2\pi}\int_{0}^{2\pi}\frac{J-\#\mathcal{R}_{\theta}(A)}{J}\,d\theta\ll A^{-2}(\log\log X)^{1/3}=e^{2.4W}(\log\log X)^{-1/6}. 2 π 1 ∫ 0 2 π J J − # R θ ( A ) d θ ≪ A − 2 ( log log X ) 1/3 = e 2.4 W ( log log X ) − 1/6 .
Applying Markov’s inequality in θ \theta θ with threshold e 1.2 W ( log log X ) − 1 / 12 e^{1.2W}(\log\log X)^{-1/12} e 1.2 W ( log log X ) − 1/12 , we see that
1 2 π meas { θ ∈ [ 0 , 2 π ) : J − # R θ ( A ) J ≫ e 1.2 W ( log log X ) − 1 / 12 } ≪ e − 1.2 W ( log log X ) 1 / 12 ⋅ 1 2 π ∫ 0 2 π J − # R θ ( A ) J d θ ≪ e 1.2 W ( log log X ) − 1 / 12 . \begin{aligned}
\frac{1}{2\pi}\operatorname{meas}\left\{\theta\in[0,2\pi):\frac{J-\#\mathcal{R}_{\theta}(A)}{J}\gg e^{1.2W}(\log\log X)^{-1/12}\right\} \\
&\ll e^{-1.2W}(\log\log X)^{1/12}\cdot\frac{1}{2\pi}\int_{0}^{2\pi}\frac{J-\#\mathcal{R}_{\theta}(A)}{J}\,d\theta\ll e^{1.2W}(\log\log X)^{-1/12}.
\end{aligned} 2 π 1 meas { θ ∈ [ 0 , 2 π ) : J J − # R θ ( A ) ≫ e 1.2 W ( log log X ) − 1/12 } ≪ e − 1.2 W ( log log X ) 1/12 ⋅ 2 π 1 ∫ 0 2 π J J − # R θ ( A ) d θ ≪ e 1.2 W ( log log X ) − 1/12 .
This proves the first assertion.
Proposition 4.2 gives (71) , simultaneously for all relevant endpoints and hence for every shifted grid, outside an event of probability O ( e − W / 10 ) O(e^{-W/10}) O ( e − W /10 ) . Likewise, Proposition 5.1 gives the third assertion simultaneously for every θ \theta θ , outside an event of probability O ( ( log log X ) − 0.1 ) O((\log\log X)^{-0.1}) O (( log log X ) − 0.1 ) .
It remains to control the Gaussian-approximation error for most shifts. For each fixed θ \theta θ , the set T 0 ( θ ) \mathcal{T}_0(\theta) T 0 ( θ ) is deterministic. Hence by Lemma 6.1, Fubini’s theorem, and Markov’s inequality we have
P ( 1 2 π ∫ 0 2 π Err X ( T 0 ( θ ) ) d θ > X − 1 / 4 ) ≪ X 1 / 4 E [ 1 2 π ∫ 0 2 π Err X ( T 0 ( θ ) ) d θ ] = X 1 / 4 1 2 π ∫ 0 2 π E Err X ( T 0 ( θ ) ) d θ ≪ X 1 / 4 ⋅ X − 1 / 2 log C X = X − 1 / 4 log C X . \begin{aligned}
\mathbb{P}\left(\frac{1}{2\pi}\int_{0}^{2\pi}\operatorname{Err}_{X}(\mathcal{T}_0(\theta))\,d\theta>X^{-1/4}\right)\ll X^{1/4}\mathbb{E}\left[\frac{1}{2\pi}\int_{0}^{2\pi}\operatorname{Err}_{X}(\mathcal{T}_0(\theta))\,d\theta\right] \\
&=X^{1/4}\frac{1}{2\pi}\int_{0}^{2\pi}\mathbb{E}\operatorname{Err}_{X}(\mathcal{T}_0(\theta))\,d\theta\ll X^{1/4}\cdot X^{-1/2}\log^{C}X=X^{-1/4}\log^{C}X.
\end{aligned} P ( 2 π 1 ∫ 0 2 π Err X ( T 0 ( θ )) d θ > X − 1/4 ) ≪ X 1/4 E [ 2 π 1 ∫ 0 2 π Err X ( T 0 ( θ )) d θ ] = X 1/4 2 π 1 ∫ 0 2 π E Err X ( T 0 ( θ )) d θ ≪ X 1/4 ⋅ X − 1/2 log C X = X − 1/4 log C X .
Outside this exceptional event, another application of Markov’s inequality, now in θ \theta θ , yields
1 2 π meas { θ : Err X ( T 0 ( θ ) ) > X − 1 / 8 } ≤ X 1 / 8 1 2 π ∫ 0 2 π Err X ( T 0 ( θ ) ) d θ ≤ X − 1 / 8 . \frac{1}{2\pi}\operatorname{meas}\{\theta:\operatorname{Err}_{X}(\mathcal{T}_0(\theta))>X^{-1/8}\}\leq X^{1/8}\frac{1}{2\pi}\int_{0}^{2\pi}\operatorname{Err}_{X}(\mathcal{T}_0(\theta))\,d\theta\leq X^{-1/8}. 2 π 1 meas { θ : Err X ( T 0 ( θ )) > X − 1/8 } ≤ X 1/8 2 π 1 ∫ 0 2 π Err X ( T 0 ( θ )) d θ ≤ X − 1/8 .
Thus the fourth assertion holds outside a set of shifts of normalised measure at most X − 1 / 8 X^{-1/8} X − 1/8 .
Finally, intersecting the four F X \mathcal{F}_{X} F X -measurable good events gives Ω X ( W ) \Omega_X(W) Ω X ( W ) , and the union bound gives (8.1). Intersecting the sets of shifts arising from the first and fourth assertions gives Θ X \Theta_X Θ X , whose normalised measure is at least
1 − O ( e 1.2 W ( log log X ) − 1 / 12 + X − 1 / 8 ) . 1-O\left(e^{1.2W}(\log\log X)^{-1/12}+X^{-1/8}\right). 1 − O ( e 1.2 W ( log log X ) − 1/12 + X − 1/8 ) .
This completes the proof. □
With Lemma 8.1 in place, we are now in a position to prove Theorem 1.4.
Proof of Theorem 1.4. Let W 0 W_0 W 0 sufficiently large, suppose that W 0 ≤ W ≤ ( 1 / 30 ) log log log X W_0\leq W\leq(1/30)\log\log\log X W 0 ≤ W ≤ ( 1/30 ) log log log X . By Lemma 8.1, we have
P ( Ω X ( W ) c ) ≪ e − W / 10 + ( log log X ) − 1 / 200 ≪ e − W / 10 , \mathbb{P}(\Omega_X(W)^c)\ll e^{-W/10}+(\log\log X)^{-1/200}\ll e^{-W/10}, P ( Ω X ( W ) c ) ≪ e − W /10 + ( log log X ) − 1/200 ≪ e − W /10 ,
and the complement of Θ X \Theta_X Θ X has normalised measure
≪ e 1.2 W ( log log X ) − 1 / 12 + X − 1 / 8 ≪ ( log log X ) − 13 / 300 . \ll e^{1.2W}(\log\log X)^{-1/12}+X^{-1/8}\ll(\log\log X)^{-13/300}. ≪ e 1.2 W ( log log X ) − 1/12 + X − 1/8 ≪ ( log log X ) − 13/300 .
Thus ∣ Θ X ∣ / ( 2 π ) = 1 − O ( ( log log X ) − 13 / 300 ) |\Theta_X|/(2\pi)=1-O((\log\log X)^{-13/300}) ∣ Θ X ∣/ ( 2 π ) = 1 − O (( log log X ) − 13/300 ) .
Work on Ω X ( W ) \Omega_X(W) Ω X ( W ) , and fix θ ∈ Θ X \theta\in\Theta_X θ ∈ Θ X . By parts (1) and (3) of Lemma 8.1, and by the greedy algorithm, there exists a measurable subset I θ ⊆ R θ ( A ) I_\theta\subseteq\mathcal{R}_\theta(A) I θ ⊆ R θ ( A ) such that
# I θ ≥ ( 1 − O ( e 1.2 W ( log log X ) − 1 / 12 ) ) J 1 + ( log X ) 0.8 ≥ ( log X ) 0.19 . (73) \#I_\theta\geq\frac{(1-O(e^{1.2W}(\log\log X)^{-1/12}))J}{1+(\log X)^{0.8}}\geq(\log X)^{0.19}. \tag*{(73)} # I θ ≥ 1 + ( log X ) 0.8 ( 1 − O ( e 1.2 W ( log log X ) − 1/12 )) J ≥ ( log X ) 0.19 . (73)
Let ( Z r ) r ∈ I θ (Z_r)_{r\in I_\theta} ( Z r ) r ∈ I θ be the centred Gaussian vector having the same covariance matrix as ( Y x r , θ ) r ∈ I θ (Y_{x_{r,\theta}})_{r\in I_\theta} ( Y x r , θ ) r ∈ I θ . For distinct r , s ∈ I θ r,s\in I_\theta r , s ∈ I θ , the part (2) and (3) of Lemma 8.1 give
∣ Cov ( Z r Var X ( Y x r , θ ) , Z s Var X ( Y x s , θ ) ) ∣ ≪ ( log log X ) − 0.59 e − 2.2 W ( log log X ) − 1 / 2 ≪ ( log log X ) − 1 / 60 . \left\lvert\operatorname{Cov}\left(\frac{Z_r}{\sqrt{\operatorname{Var}_X(Y_{x_{r,\theta}})}},\frac{Z_s}{\sqrt{\operatorname{Var}_X(Y_{x_{s,\theta}})}}\right)\right\rvert\ll\frac{(\log\log X)^{-0.59}}{e^{-2.2W}(\log\log X)^{-1/2}} \ll(\log\log X)^{-1/60}. Cov ( Var X ( Y x r , θ ) Z r , Var X ( Y x s , θ ) Z s ) ≪ e − 2.2 W ( log log X ) − 1/2 ( log log X ) − 0.59 ≪ ( log log X ) − 1/60 .
Thus, for sufficiently large X X X , Normal Comparison Result 1 of [20 ] applies to these standardised Gaussian variables with δ = 1 / 100 \delta=1/100 δ = 1/100 . Moreover, since log # I θ ≥ 0.19 log log X \log\# I_\theta\ge0.19\log\log X log # I θ ≥ 0.19 log log X , after choosing W 0 W_0 W 0 sufficiently large we have, uniformly for r ∈ I θ r\in I_\theta r ∈ I θ ,
5 A Var X ( Y x r , θ ) ≪ e − 0.1 W log log X ≤ ( 2 − 1 100 ) log # I θ . \frac{5A}{\sqrt{\operatorname{Var}_X(Y_{x_{r,\theta}})}}\ll e^{-0.1W}\sqrt{\log\log X}\le\sqrt{\left(2-\frac{1}{100}\right)\log\# I_\theta}. Var X ( Y x r , θ ) 5 A ≪ e − 0.1 W log log X ≤ ( 2 − 100 1 ) log # I θ .
It follows that
P Z ( max r ∈ I θ Z r ≤ 5 A ) ≤ P Z ( max r ∈ I θ Z r Var X ( Y x r , θ ) ≤ ( 2 − 1 100 ) log # I θ ) = o ( 1 ) , \mathbb{P}_{Z}\left(\max_{r\in I_\theta}Z_r\le5A\right)\le\mathbb{P}_{Z}\left(\max_{r\in I_\theta}\frac{Z_r}{\sqrt{\operatorname{Var}_X(Y_{x_{r,\theta}})}}\le\sqrt{\left(2-\frac{1}{100}\right)\log\# I_\theta}\right)=o(1), P Z ( r ∈ I θ max Z r ≤ 5 A ) ≤ P Z ( r ∈ I θ max Var X ( Y x r , θ ) Z r ≤ ( 2 − 100 1 ) log # I θ ) = o ( 1 ) ,
by Normal Comparison Result 1 of [20 ] .
Applying Lemma 6.1 with T = { x r , θ : r ∈ I θ } T=\{x_{r,\theta}:r\in I_\theta\} T = { x r , θ : r ∈ I θ } and T 0 = T 0 ( θ ) T_0=T_0(\theta) T 0 = T 0 ( θ ) , and using Err X ( T 0 ( θ ) ) ≤ X − 1 / 8 \operatorname{Err}_X(T_0(\theta))\le X^{-1/8} Err X ( T 0 ( θ )) ≤ X − 1/8 , we obtain
P X ( max r ∈ I θ ( U X ( 4 x r , θ ) − U X ( x r , θ ) ) ≤ 4 A ) ≤ P Z ( max r ∈ I θ Z r ≤ 5 A ) + X − 1 / 8 = o ( 1 ) , \mathbb{P}_X\left(\max_{r\in I_\theta}\left(U_X(4x_{r,\theta})-U_X(x_{r,\theta})\right)\le4A\right)\le\mathbb{P}_Z\left(\max_{r\in I_\theta}Z_r\le5A\right)+X^{-1/8}=o(1), P X ( r ∈ I θ max ( U X ( 4 x r , θ ) − U X ( x r , θ ) ) ≤ 4 A ) ≤ P Z ( r ∈ I θ max Z r ≤ 5 A ) + X − 1/8 = o ( 1 ) ,
since A = e − 1.2 W ( log log X ) 1 / 4 ≥ ( log log X ) 21 / 100 A=e^{-1.2W}(\log\log X)^{1/4}\ge(\log\log X)^{21/100} A = e − 1.2 W ( log log X ) 1/4 ≥ ( log log X ) 21/100 , we have 4 A + 1 ≤ 5 A 4A+1\le5A 4 A + 1 ≤ 5 A for sufficiently large X X X . Thus
P X ( max r ∈ I θ ( U X ( 4 x r , θ ) − U X ( x r , θ ) ) ≥ 4 A ) = 1 − o ( 1 ) . (74) \mathbb{P}_X\left(\max_{r\in I_\theta}\left(U_X(4x_{r,\theta})-U_X(x_{r,\theta})\right)\ge4A\right)=1-o(1). \tag*{(74)} P X ( r ∈ I θ max ( U X ( 4 x r , θ ) − U X ( x r , θ ) ) ≥ 4 A ) = 1 − o ( 1 ) . (74)
On the event in (74) , the range of the values U X ( x r , θ ) U_X(x_{r,\theta}) U X ( x r , θ ) and U X ( 4 x r , θ ) U_X(4x_{r,\theta}) U X ( 4 x r , θ ) , r ∈ I θ r\in I_\theta r ∈ I θ , is at least 4 A 4A 4 A . Conditional on F X \mathcal{F}_X F X , these values form a centrally symmetric vector, since simultaneously replacing f ( p ) f(p) f ( p ) by − f ( p ) -f(p) − f ( p ) for X < p ≤ X 4 / 3 X<p\le X^{4/3} X < p ≤ X 4/3 replaces every ϵ p \epsilon_p ϵ p in the definition (2.6) of U X U_X U X by its negative. Hence
P X ( max r ∈ I θ z ∈ { x r , θ , 4 x r , θ } U X ( z ) ≥ 2 A ) = P X ( min r ∈ I θ z ∈ { x r , θ , 4 x r , θ } U X ( z ) ≤ − 2 A ) ≥ 1 2 − o ( 1 ) . \mathbb{P}_X\left(\max_{\substack{r\in I_\theta\\ z\in\{x_{r,\theta},4x_{r,\theta}\}}}U_X(z)\ge2A\right)=\mathbb{P}_X\left(\min_{\substack{r\in I_\theta\\ z\in\{x_{r,\theta},4x_{r,\theta}\}}}U_X(z)\le-2A\right)\ge\frac{1}{2}-o(1). P X r ∈ I θ z ∈ { x r , θ , 4 x r , θ } max U X ( z ) ≥ 2 A = P X r ∈ I θ z ∈ { x r , θ , 4 x r , θ } min U X ( z ) ≤ − 2 A ≥ 2 1 − o ( 1 ) .
Since I θ ⊆ R θ ( A ) I_\theta\subseteq R_\theta(A) I θ ⊆ R θ ( A ) , every endpoint appearing above satisfies ∣ C X ( z ) ∣ ≤ A \lvert C_X(z)\rvert\le A ∣ C X ( z )∣ ≤ A . Recalling S f ( z ) / z = C X ( z ) + U X ( z ) S_f(z)/\sqrt{z}=C_X(z)+U_X(z) S f ( z ) / z = C X ( z ) + U X ( z ) , we obtain, uniformly for θ ∈ Θ X \theta\in\Theta_X θ ∈ Θ X ,
P X ( max r ∈ I θ z ∈ { x r , θ , 4 x r , θ } S f ( z ) z ≥ A ) , P X ( min r ∈ I θ z ∈ { x r , θ , 4 x r , θ } S f ( z ) z ≤ − A ) ≥ 1 2 − o ( 1 ) . \mathbb{P}_X\left(\max_{\substack{r\in I_\theta\\ z\in\{x_{r,\theta},4x_{r,\theta}\}}}\frac{S_f(z)}{\sqrt{z}}\ge A\right),\quad
\mathbb{P}_X\left(\min_{\substack{r\in I_\theta\\ z\in\{x_{r,\theta},4x_{r,\theta}\}}}\frac{S_f(z)}{\sqrt{z}}\le-A\right)\ge\frac{1}{2}-o(1). P X r ∈ I θ z ∈ { x r , θ , 4 x r , θ } max z S f ( z ) ≥ A , P X r ∈ I θ z ∈ { x r , θ , 4 x r , θ } min z S f ( z ) ≤ − A ≥ 2 1 − o ( 1 ) .
All these endpoints lie in [ X 8 / 7 , X 4 / 3 ] [X^{8/7},X^{4/3}] [ X 8/7 , X 4/3 ] . Averaging over the common shift and using ∣ Θ X ∣ / ( 2 π ) = 1 − o ( 1 ) \lvert\Theta_X\rvert/(2\pi)=1-o(1) ∣ Θ X ∣ / ( 2 π ) = 1 − o ( 1 ) , we therefore have
P X ( max X 8 / 7 ≤ z ≤ X 4 / 3 S f ( z ) z ≥ A ) ≥ 1 2 π ∫ Θ X P X ( max r ∈ I θ z ∈ { x r , θ , 4 x r , θ } S f ( z ) z ≥ A ) d θ ≥ ∣ Θ X ∣ 2 π ( 1 2 − o ( 1 ) ) = 1 2 − o ( 1 ) ≥ 1 3 . \begin{aligned}
\mathbb{P}_X\left(\max_{X^{8/7}\le z\le X^{4/3}}\frac{S_f(z)}{\sqrt{z}}\ge A\right)\ge\frac{1}{2\pi}\int_{\Theta_X}\mathbb{P}_X\left(\max_{\substack{r\in I_\theta\\ z\in\{x_{r,\theta},4x_{r,\theta}\}}}\frac{S_f(z)}{\sqrt{z}}\ge A\right)\,d\theta\\
&\ge\frac{\lvert\Theta_X\rvert}{2\pi}\left(\frac{1}{2}-o(1)\right)=\frac{1}{2}-o(1)\ge\frac{1}{3}.
\end{aligned} P X ( X 8/7 ≤ z ≤ X 4/3 max z S f ( z ) ≥ A ) ≥ 2 π 1 ∫ Θ X P X r ∈ I θ z ∈ { x r , θ , 4 x r , θ } max z S f ( z ) ≥ A d θ ≥ 2 π ∣ Θ X ∣ ( 2 1 − o ( 1 ) ) = 2 1 − o ( 1 ) ≥ 3 1 .
and, in the same way,
P X ( min X 8 / 7 ≤ z ≤ X 4 / 3 S f ( z ) z ≤ − A ) ≥ 1 3 . \mathbb{P}_X\left(\min_{X^{8/7}\le z\le X^{4/3}} \frac{S_f(z)}{\sqrt{z}} \le-A\right) \ge\frac{1}{3}. P X ( X 8/7 ≤ z ≤ X 4/3 min z S f ( z ) ≤ − A ) ≥ 3 1 .
Therefore we finish the proof of Theorem 1.4. □
References
School of Mathematics, Shandong University, Jinan 250100, China
Département de mathématiques et de statistique, Université de Montréal, C.P. 6128, succ. Centre-ville, Montréal, QC H3C 3J7, Canada. Email address: hao zheogu@gmail.com
Yau Mathematical Sciences Center, Tsinghua University, Beijing, 100084, China
Beijing Institute of Mathematical Sciences and Applications, Beijing, 101408, China
Email address: maxxu1729@gmail.com, maxxu@tsinghua.edu.cn