Introduction
The setting
Let n ∈ { 2 , 3 , … } n \in\{2,3,\ldots\} n ∈ { 2 , 3 , … } a i ∈ N 0 = { 0 , 1 , 2 , … } a_i \in\mathbb{N}_0 = \{0,1,2,\ldots\} a i ∈ N 0 = { 0 , 1 , 2 , … } i ∈ { 0 , 1 , … , n − 1 } i \in\{0,1,\ldots,n-1\} i ∈ { 0 , 1 , … , n − 1 } a 0 a n − 1 ≥ 1 a_0a_{n-1} \ge1 a 0 a n − 1 ≥ 1
f ( x ) = 1 − a 0 x − a 1 x 2 − ⋯ − a n − 1 x n . (1) f(x) = 1 - \frac{a_0}{x} - \frac{a_1}{x^2} - \cdots- \frac{a_{n-1}}{x^n}. \tag*{(1)} f ( x ) = 1 − x a 0 − x 2 a 1 − ⋯ − x n a n − 1 . (1) We have f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 x > 0 x>0 x > 0 f ( a 0 ) < 0 f(a_0)<0 f ( a 0 ) < 0 lim x → ∞ f ( x ) = 1 \lim_{x\to\infty} f(x)=1 lim x → ∞ f ( x ) = 1
1 = a 0 β + a 1 β 2 + ⋯ + a n − 1 β n , β > 0 , (2) 1=\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\cdots+\frac{a_{n-1}}{\beta^n},\qquad\beta>0, \tag*{(2)} 1 = β a 0 + β 2 a 1 + ⋯ + β n a n − 1 , β > 0 , (2) has a unique solution. In this context, we see that β > 1 \beta>1 β > 1 n n n
p ( z ) = z n − a 0 z n − 1 − ⋯ − a n − 1 . (3) p(z)=z^n-a_0z^{n-1}-\cdots-a_{n-1}. \tag*{(3)} p ( z ) = z n − a 0 z n − 1 − ⋯ − a n − 1 . (3) Let T β ( x ) = x β − ⌊ x β ⌋ T_\beta(x)=x\beta-\lfloor x\beta\rfloor T β ( x ) = x β − ⌊ x β ⌋ β \beta β { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1
a 0 = ⌊ β ⌋ , a 1 = ⌊ β T β ( 1 ) ⌋ , … , a n − 1 = ⌊ β T β n − 1 ( 1 ) ⌋ , T β n ( 1 ) = 0. (4) a_0=\lfloor\beta\rfloor,\qquad a_1=\lfloor\beta T_\beta(1)\rfloor,\qquad\ldots,\qquad a_{n-1}=\lfloor\beta T_\beta^{n-1}(1)\rfloor,\qquad T_\beta^n(1)=0. \tag*{(4)} a 0 = ⌊ β ⌋ , a 1 = ⌊ β T β ( 1 )⌋ , … , a n − 1 = ⌊ β T β n − 1 ( 1 )⌋ , T β n ( 1 ) = 0. (4) We call such a j a_j a j [10 ] ) found the necessary and sufficient conditions which the coefficients must satisfy in order to also obey (2) and (4) : denoting by ≻ \succ ≻
[ a 0 , a 1 , … , a n − 1 , 0 , … ] ≻ [ a j , a j + 1 , … , a n − 1 , 0 , … ] for all 1 ≤ j ≤ n − 1. (5) [a_0,a_1,\ldots,a_{n-1},0,\ldots]\succ[a_j,a_{j+1},\ldots,a_{n-1},0,\ldots]\quad\text{for all }1\leq j\leq n-1. \tag*{(5)} [ a 0 , a 1 , … , a n − 1 , 0 , … ] ≻ [ a j , a j + 1 , … , a n − 1 , 0 , … ] for all 1 ≤ j ≤ n − 1. (5) Definition 1.1. We say that a polynomial p ( z ) p(z) p ( z ) (3) is a Parry polynomial if the coefficients { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1 p ( z ) p(z) p ( z ) n − 1 n-1 n − 1 p ( z ) p(z) p ( z ) β \beta β
Definition 1.2. We say that β > 1 \beta>1 β > 1 p ( z ) = z n + ∑ k = 1 n c k z n − k p(z)=z^n+\sum_{k=1}^{n}c_kz^{n-k} p ( z ) = z n + ∑ k = 1 n c k z n − k c k ∈ Z c_k\in\mathbb{Z} c k ∈ Z p ( β ) = 0 p(\beta)=0 p ( β ) = 0 p p p
Remark 1.1. Note that a Parry polynomial may fail to be a Parry-Pisot polynomial although its corresponding β \beta β
z 5 − z 4 − 1 = ( z 2 − z + 1 ) ( z 3 − z − 1 ) (6) z^5-z^4-1=(z^2-z+1)(z^3-z-1) \tag*{(6)} z 5 − z 4 − 1 = ( z 2 − z + 1 ) ( z 3 − z − 1 ) (6) where n = 5 n=5 n = 5 a 0 = a 4 = 1 a_0=a_4=1 a 0 = a 4 = 1 a 1 = a 2 = a 3 = 0 a_1=a_2=a_3=0 a 1 = a 2 = a 3 = 0 (13) . Here β \beta β z 3 − z − 1 z^3-z-1 z 3 − z − 1 p ( z ) p(z) p ( z ) (6) is not Parry-Pisot because two of its roots are e ± i π / 3 e^{\pm i\pi/3} e ± iπ /3 z 3 − z − 1 z^3-z-1 z 3 − z − 1
For any β > 1 \beta>1 β > 1 P : L p ( [ 0 , 1 ] ) → L p ( [ 0 , 1 ] ) \mathcal{P}:L^p([0,1])\to L^p([0,1]) P : L p ([ 0 , 1 ]) → L p ([ 0 , 1 ]) 1 ≤ p ≤ ∞ 1\leq p\leq\infty 1 ≤ p ≤ ∞
( P f ) ( x ) = β − 1 ∑ k = 0 a 0 f ( x + k β ) , 0 ≤ x ≤ 1. (7) (\mathcal{P}f)(x)=\beta^{-1}\sum_{k=0}^{a_0}f\left(\frac{x+k}{\beta}\right),\qquad0\leq x\leq1. \tag*{(7)} ( P f ) ( x ) = β − 1 k = 0 ∑ a 0 f ( β x + k ) , 0 ≤ x ≤ 1. (7) Here we always assume f ( x ) = 0 f(x)=0 f ( x ) = 0 x ∉ [ 0 , 1 ] x\notin[0,1] x ∈ / [ 0 , 1 ]
Let us also introduce the composition (Koopman) operator K : L p ( [ 0 , 1 ] ) → L p ( [ 0 , 1 ] ) K : L^p([0,1]) \to L^p([0,1]) K : L p ([ 0 , 1 ]) → L p ([ 0 , 1 ])
( K f ) ( x ) = f ( T β ( x ) ) = f ( β x − ⌊ β x ⌋ ) . (8) (Kf)(x)=f(T_\beta(x))=f(\beta x-\lfloor\beta x\rfloor). \tag*{(8)} ( K f ) ( x ) = f ( T β ( x )) = f ( β x − ⌊ β x ⌋) . (8) There is an important duality relation between these two operators. For all g ∈ L 1 ( [ 0 , 1 ] ) g \in L^1([0,1]) g ∈ L 1 ([ 0 , 1 ]) f ∈ L ∞ ( [ 0 , 1 ] ) f \in L^\infty([0,1]) f ∈ L ∞ ([ 0 , 1 ])
∫ [ 0 , 1 ] g ( x ) ( K f ) ( x ) d x = ∫ [ 0 , 1 ] ( P g ) ( x ) f ( x ) d x . (9) \int_{[0,1]} g(x)(Kf)(x)\,\mathrm{d}x=\int_{[0,1]}(\mathcal{P}g)(x)f(x)\,\mathrm{d}x. \tag*{(9)} ∫ [ 0 , 1 ] g ( x ) ( K f ) ( x ) d x = ∫ [ 0 , 1 ] ( P g ) ( x ) f ( x ) d x . (9) More on the necessary and sufficient conditions for having a finite Parry sequence
Following [10 ] we will give here a few more details on the necessary and sufficient conditions which the coefficients must obey when they satisfy (4) and (5) .
The first equality in (4) demands a 0 < β < a 0 + 1 a_0 < \beta< a_0+1 a 0 < β < a 0 + 1 β > a 0 \beta> a_0 β > a 0 (2) ). Furthermore, by T β : [ 0 , 1 ] → [ 0 , 1 ) T_\beta: [0,1] \to[0,1) T β : [ 0 , 1 ] → [ 0 , 1 ) (2) and (4) , we have for all 1 ≤ k ≤ n − 1 1 \leq k \leq n-1 1 ≤ k ≤ n − 1
b 0 : = 1 , b 1 : = a 1 β + ⋯ + a n − 1 β n − 1 = T β ( 1 ) = β − a 0 ∈ ( 0 , 1 ) , b 2 : = a 2 β + ⋯ + a n − 1 β n − 2 = T β 2 ( 1 ) = β 2 − β a 0 − a 1 ∈ ( 0 , 1 ) , ⋮ b k : = a k β + ⋯ + a n − 1 β n − k = T β k ( 1 ) = β k − β k − 1 a 0 − β k − 2 a 1 − ⋯ − a k − 1 ∈ ( 0 , 1 ) . (10) \begin{aligned}
b_0&:=1,\\
b_1&:=\frac{a_1}{\beta}+\cdots+\frac{a_{n-1}}{\beta^{n-1}}=T_\beta(1)=\beta-a_0\in(0,1),\\
b_2&:=\frac{a_2}{\beta}+\cdots+\frac{a_{n-1}}{\beta^{n-2}}=T_\beta^2(1)=\beta^2-\beta a_0-a_1\in(0,1),\\
&\vdots\\
b_k&:=\frac{a_k}{\beta}+\cdots+\frac{a_{n-1}}{\beta^{n-k}}=T_\beta^k(1)=\beta^k-\beta^{k-1}a_0-\beta^{k-2}a_1-\cdots-a_{k-1}\in(0,1). \tag*{(10)}
\end{aligned} b 0 b 1 b 2 b k := 1 , := β a 1 + ⋯ + β n − 1 a n − 1 = T β ( 1 ) = β − a 0 ∈ ( 0 , 1 ) , := β a 2 + ⋯ + β n − 2 a n − 1 = T β 2 ( 1 ) = β 2 − β a 0 − a 1 ∈ ( 0 , 1 ) , ⋮ := β a k + ⋯ + β n − k a n − 1 = T β k ( 1 ) = β k − β k − 1 a 0 − β k − 2 a 1 − ⋯ − a k − 1 ∈ ( 0 , 1 ) . (10) We gather a few important facts implied by (4) .
Lemma 1.2. We have b j ≠ b k b_j \ne b_k b j = b k j ≠ k j \ne k j = k
Proof. If j > k j>k j > k T j ( 1 ) = T k ( 1 ) T^j(1)=T^k(1) T j ( 1 ) = T k ( 1 ) T n − j T^{n-j} T n − j T j ( 1 ) T^j(1) T j ( 1 ) T k ( 1 ) T^k(1) T k ( 1 ) □ \square □
Lemma 1.3. Let a 0 ≥ max { a 1 , … , a n − 1 } a_0 \geq\max\{a_1,\ldots,a_{n-1}\} a 0 ≥ max { a 1 , … , a n − 1 } a 0 a n − 1 ≥ 1 a_0a_{n-1} \geq1 a 0 a n − 1 ≥ 1 (5) . Let β \beta β (2) . Then a 0 < β < a 0 + 1 a_0 < \beta< a_0+1 a 0 < β < a 0 + 1 2 ≤ n ≤ 3 2 \leq n \leq3 2 ≤ n ≤ 3 a 0 ≥ max { a 1 , … , a n − 1 } a_0 \geq\max\{a_1,\ldots,a_{n-1}\} a 0 ≥ max { a 1 , … , a n − 1 }
Proof. Indeed, from (1) we get
f ( a 0 ) < 0 , f ( a 0 + 1 ) > 1 − ∑ j ≥ 1 a 0 ( a 0 + 1 ) j = 0 f(a_0)<0,\qquad f(a_0+1)>1-\sum_{j\geq1}\frac{a_0}{(a_0+1)^j}=0 f ( a 0 ) < 0 , f ( a 0 + 1 ) > 1 − j ≥ 1 ∑ ( a 0 + 1 ) j a 0 = 0 which proves that a 0 < β < a 0 + 1 a_0 < \beta< a_0+1 a 0 < β < a 0 + 1 2 ≤ n ≤ 3 2 \leq n \leq3 2 ≤ n ≤ 3 (5) is always satisfied. □ \square □
Three important examples of finite Parry sequences
The first example we consider is
a 0 ≥ a 1 ≥ ⋯ ≥ a n − 1 ≥ 1. (11) a_0 \ge a_1 \ge\cdots\ge a_{n-1} \ge1. \tag*{(11)} a 0 ≥ a 1 ≥ ⋯ ≥ a n − 1 ≥ 1. (11) Brauer [1 ] proved that (11) is also sufficient for p ( z ) p(z) p ( z ) Theorem 2.1 . For completeness, we give a short alternative proof of Brauer’s result in Lemma A.1.
Another example where (4) and (5) are satisfied is when there exists 0 ≤ j ≤ n − 2 0 \le j \le n-2 0 ≤ j ≤ n − 2
a 0 ≥ a 1 ≥ ⋯ ≥ a j > a j + k ; 1 ≤ j + 1 ≤ j + k ≤ n − 1 ; a 0 a n − 1 ≥ 1. (12) a_0 \ge a_1 \ge\cdots\ge a_j > a_{j+k}; \qquad1 \le j+1 \le j+k \le n-1; \qquad a_0a_{n-1} \ge1. \tag*{(12)} a 0 ≥ a 1 ≥ ⋯ ≥ a j > a j + k ; 1 ≤ j + 1 ≤ j + k ≤ n − 1 ; a 0 a n − 1 ≥ 1. (12) Here p ( z ) p(z) p ( z ) (12) with the more restrictive condition a 0 ≥ 2 + a 1 + ⋯ + a n − 1 a_0 \ge2+a_1+\cdots+a_{n-1} a 0 ≥ 2 + a 1 + ⋯ + a n − 1 (12) with j = 0 j=0 j = 0 p ( z ) p(z) p ( z )
A third example where (4) and (5) hold is
n ≥ 4 , a 0 ≥ a n − 1 ≥ 1 , a 1 = ⋯ = a n − 2 = 0. (13) n \ge4,\qquad a_0 \ge a_{n-1} \ge1,\qquad a_1=\cdots=a_{n-2}=0. \tag*{(13)} n ≥ 4 , a 0 ≥ a n − 1 ≥ 1 , a 1 = ⋯ = a n − 2 = 0. (13) We also prove in Appendix E.§ 2 that all the roots of the polynomial p ( z ) = z n − z n − 1 − 1 p(z)=z^n-z^{n-1}-1 p ( z ) = z n − z n − 1 − 1 n → ∞ n \to\infty n → ∞ n = 6 j + 5 n=6j+5 n = 6 j + 5 j ≥ 0 j \ge0 j ≥ 0 e ± i π / 3 e^{\pm i\pi/3} e ± iπ /3 (6) when j = 0 j=0 j = 0
The Main Results
Asymptotics of P r ( F ) \mathcal{P}^r(F) P r ( F ) F F F
Let us introduce the following linearly independent (due to Lemma 1.2) characteristic functions
ψ 1 ( x ) : = χ [ 0 , 1 ] ( x ) , ψ 2 ( x ) : = χ [ 0 , b 1 ] ( x ) , … , ψ n ( x ) : = χ [ 0 , b n − 1 ] ( x ) , \psi_1(x):=\chi_{[0,1]}(x),\qquad\psi_2(x):=\chi_{[0,b_1]}(x),\qquad\ldots,\qquad\psi_n(x):=\chi_{[0,b_{n-1}]}(x), ψ 1 ( x ) := χ [ 0 , 1 ] ( x ) , ψ 2 ( x ) := χ [ 0 , b 1 ] ( x ) , … , ψ n ( x ) := χ [ 0 , b n − 1 ] ( x ) , and
u 1 ( x ) = C 0 ( ψ 1 ( x ) + 1 β ψ 2 ( x ) + ⋯ + 1 β n − 1 ψ n ( x ) ) > 0 , (14) u_1(x)=C_0\left(\psi_1(x)+\frac{1}{\beta}\psi_2(x)+\cdots+\frac{1}{\beta^{n-1}}\psi_n(x)\right)>0, \tag*{(14)} u 1 ( x ) = C 0 ( ψ 1 ( x ) + β 1 ψ 2 ( x ) + ⋯ + β n − 1 1 ψ n ( x ) ) > 0 , (14) where C 0 > 0 C_0>0 C 0 > 0 ∫ [ 0 , 1 ] u 1 ( x ) d x = 1 \int_{[0,1]}u_1(x)\,\mathrm{d}x=1 ∫ [ 0 , 1 ] u 1 ( x ) d x = 1 [10 ] (see also (A.9)) that u 1 u_1 u 1 P f = f \mathcal{P}f=f P f = f L 1 ( [ 0 , 1 ] ) L^1([0,1]) L 1 ([ 0 , 1 ])
We now formulate the first result of our paper:
Theorem 2.1. Let n ≥ 2 n\ge2 n ≥ 2 { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1 p ( z ) p(z) p ( z ) P \mathcal{P} P (7) . There exists a function v 1 ≠ 0 v_1\ne0 v 1 = 0
v 1 v_1 v 1
P v 1 = β − 1 v 1 \mathcal{P}v_1=\beta^{-1}v_1 P v 1 = β − 1 v 1 ∫ [ 0 , 1 ] v 1 ( x ) d x = 0 \int_{[0,1]}v_1(x)\,\mathrm{d}x=0 ∫ [ 0 , 1 ] v 1 ( x ) d x = 0
and there exist N 0 ≥ 1 N_0\ge1 N 0 ≥ 1 ε > 0 \varepsilon>0 ε > 0 N ≥ N 0 N\ge N_0 N ≥ N 0 C < ∞ C<\infty C < ∞ F ∈ C N ( [ 0 , 1 ] ) F\in C^N([0,1]) F ∈ C N ([ 0 , 1 ]) r ≥ 1 r\ge1 r ≥ 1
∥ ( P r F ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 F d x − β − r 4 ( F ( 1 ) − F ( 0 ) ) v 1 ( ⋅ ) ∥ L ∞ ( [ 0 , 1 ] ) ≤ C ∥ F ∥ C N ( [ 0 , 1 ] ) β − ( 1 + ε ) r . \left\|(\mathcal{P}^rF)(\cdot)-u_1(\cdot)\int_0^1F\,\mathrm{d}x-\frac{\beta^{-r}}{4}\bigl(F(1)-F(0)\bigr)v_1(\cdot)\right\|_{L^\infty([0,1])}
\le C\|F\|_{C^N([0,1])}\beta^{-(1+\varepsilon)r}. ( P r F ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 F d x − 4 β − r ( F ( 1 ) − F ( 0 ) ) v 1 ( ⋅ ) L ∞ ([ 0 , 1 ]) ≤ C ∥ F ∥ C N ([ 0 , 1 ]) β − ( 1 + ε ) r . The proof of this theorem is in § 4.1 .
Remark 2.2. Theorem 2.1 generalizes the results of [3 ] where only the case n = 2 n=2 n = 2 Lemma A.3 ) that p ( z ) p(z) p ( z ) n ≤ 3 n \le3 n ≤ 3 n ≥ 4 n \ge4 n ≥ 4
Nevertheless, other conditions like in (11) are sufficient for both having a Parry sequence and for p ( z ) p(z) p ( z ) a 0 ≥ 2 + a 1 + ⋯ + a n − 1 a_0 \ge2+a_1+\cdots+a_{n-1} a 0 ≥ 2 + a 1 + ⋯ + a n − 1 (12) and Lemma A.2).
Remark 2.3. The estimate we obtain in Theorem 2.1 does not hold when F F F β − r \beta^{-r} β − r
Remark 2.4. The case in which the coefficients { a j } j ≥ 0 \{a_j\}_{j\geq0} { a j } j ≥ 0
When p ( z ) p(z) p ( z ) P \mathcal{P} P β − α r \beta^{-\alpha r} β − α r 0 < α < 1 0<\alpha<1 0 < α < 1 F F F p ( z ) p(z) p ( z )
Theorem 2.5. Let n = 4 n=4 n = 4 a 0 = 3 a_0=3 a 0 = 3 a 1 = 2 a_1=2 a 1 = 2 a 2 = 0 a_2=0 a 2 = 0 a 3 = 2 a_3=2 a 3 = 2 3 < β < 4 3<\beta<4 3 < β < 4
(1) There exist two functions u 2 u_2 u 2 v 1 v_1 v 1
u 2 u_2 u 2 v 1 v_1 v 1
P u 2 = − 1 β u 2 \mathcal{P}u_2=-\frac{1}{\beta}u_2 P u 2 = − β 1 u 2 P v 1 = 1 β v 1 \mathcal{P}v_1=\frac{1}{\beta}v_1 P v 1 = β 1 v 1 ∫ [ 0 , 1 ] u 2 ( x ) d x = ∫ [ 0 , 1 ] v 1 ( x ) d x = 0 \int_{[0,1]}u_2(x)\,\mathrm{d}x=\int_{[0,1]}v_1(x)\,\mathrm{d}x=0 ∫ [ 0 , 1 ] u 2 ( x ) d x = ∫ [ 0 , 1 ] v 1 ( x ) d x = 0
(2) If q ( x ) q(x) q ( x ) M ≥ 1 M\geq1 M ≥ 1 [ 0 , 1 ] [0,1] [ 0 , 1 ] ε > 0 \varepsilon>0 ε > 0 C < ∞ C<\infty C < ∞ C 2 , q C_{2,q} C 2 , q C 3 , q C_{3,q} C 3 , q q q q r ≥ 1 r\geq1 r ≥ 1
∥ ( P r q ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 q ( x ) d x − ( − 1 ) r β r C 2 , q u 2 ( ⋅ ) − 1 β r C 3 , q v 1 ( ⋅ ) ∥ L ∞ ≤ C ∥ q ∥ C M β − ( 1 + ε ) r . \left\|(\mathcal{P}^{r}q)(\cdot)-u_1(\cdot)\int_0^1q(x)\,\mathrm{d}x-\frac{(-1)^r}{\beta^r}C_{2,q}u_2(\cdot)-\frac{1}{\beta^r}C_{3,q}v_1(\cdot)\right\|_{L^\infty}\leq C\|q\|_{C^M}\beta^{-(1+\varepsilon)r}. ( P r q ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 q ( x ) d x − β r ( − 1 ) r C 2 , q u 2 ( ⋅ ) − β r 1 C 3 , q v 1 ( ⋅ ) L ∞ ≤ C ∥ q ∥ C M β − ( 1 + ε ) r . This theorem is proved in § 4.2 .
The spectrum of the transfer operator when restricted to L p ( [ 0 , 1 ] ) L^p([0,1]) L p ([ 0 , 1 ])
Let { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1
b 1 = T β ( 1 ) = β − a 0 = a 1 β + ⋯ + a n − 1 β n − 1 ∈ ( 0 , 1 ) . b_1=T_\beta(1)=\beta-a_0=\frac{a_1}{\beta}+\cdots+\frac{a_{n-1}}{\beta^{n-1}}\in(0,1). b 1 = T β ( 1 ) = β − a 0 = β a 1 + ⋯ + β n − 1 a n − 1 ∈ ( 0 , 1 ) . Theorem 2.6. The following statements hold true:
(1) For any g ∈ C 0 ∞ ( ( 0 , 1 ) ) g\in C^\infty_0((0,1)) g ∈ C 0 ∞ (( 0 , 1 )) ψ 0 ( x ) = g ( β x ) χ [ 0 , b 1 / β ] ( x ) − g ( β x − a 0 ) χ [ a 0 / β , 1 ] ( x ) \psi_0(x)=g(\beta x)\chi_{[0,b_1/\beta]}(x)-g(\beta x-a_0)\chi_{[a_0/\beta,1]}(x) ψ 0 ( x ) = g ( β x ) χ [ 0 , b 1 / β ] ( x ) − g ( β x − a 0 ) χ [ a 0 / β , 1 ] ( x )
( P ψ 0 ) ( x ) = 0 for a.e. x ∈ [ 0 , 1 ] . (15) (\mathcal{P}\psi_0)(x)=0\quad\text{for a.e. }x\in[0,1]. \tag*{(15)} ( P ψ 0 ) ( x ) = 0 for a.e. x ∈ [ 0 , 1 ] . (15) (2) Let u 1 u_1 u 1 (14) . Then K ~ p : = u 1 1 / p K u 1 − 1 / p \widetilde{K}_p := u_1^{1/p} K u_1^{-1/p} K p := u 1 1/ p K u 1 − 1/ p L p ( [ 0 , 1 ] ) L^p([0,1]) L p ([ 0 , 1 ]) 1 ≤ p ≤ ∞ 1 \le p \le\infty 1 ≤ p ≤ ∞ K ~ p \widetilde{K}_p K p K K K
(3) Let 1 ≤ p , p ′ ≤ ∞ 1 \le p,p' \le\infty 1 ≤ p , p ′ ≤ ∞ 1 p + 1 p ′ = 1 \frac{1}{p}+\frac{1}{p'}=1 p 1 + p ′ 1 = 1 ∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1
ψ z : = u 1 1 / p ′ ( Id − z K ~ p ) − 1 ( ψ 0 u 1 1 / p ′ ) ∈ L p ( [ 0 , 1 ] ) . \psi_z := u_1^{1/p'}\left(\operatorname{Id}-z\widetilde{K}_p\right)^{-1}\left(\frac{\psi_0}{u_1^{1/p'}}\right) \in L^p([0,1]). ψ z := u 1 1/ p ′ ( Id − z K p ) − 1 ( u 1 1/ p ′ ψ 0 ) ∈ L p ([ 0 , 1 ]) . Then P ψ z = z ψ z \mathcal{P}\psi_z=z\psi_z P ψ z = z ψ z
The proof of this theorem is given in § 4.3 .
The resolvent of the transfer operator when restricted to functions of bounded variation
The next result is about the resolvent and the spectrum of P \mathcal{P} P B V ( [ 0 , 1 ] ) ⊂ L ∞ ( [ 0 , 1 ] ) BV([0,1])\subset L^\infty([0,1]) B V ([ 0 , 1 ]) ⊂ L ∞ ([ 0 , 1 ])
∥ f ∥ B V : = inf { Var 0 1 ( g ) + ∥ g ∥ L ∞ : f = g Lebesgue almost everywhere on [ 0 , 1 ] } . (16) \|f\|_{BV}:=\inf\left\{\operatorname{Var}_0^1(g)+\|g\|_{L^\infty}: f=g\ \text{Lebesgue almost everywhere on }[0,1]\right\}. \tag*{(16)} ∥ f ∥ B V := inf { Var 0 1 ( g ) + ∥ g ∥ L ∞ : f = g Lebesgue almost everywhere on [ 0 , 1 ] } . (16) where
Var 0 1 ( g ) : = sup m ≥ 1 sup { 0 = x 1 < x 2 < ⋯ < x m = 1 } ∑ j = 1 m − 1 ∣ g ( x j + 1 ) − g ( x j ) ∣ . (17) \operatorname{Var}_0^1(g):=\sup_{m\ge1}\sup_{\{0=x_1<x_2<\cdots<x_m=1\}}\sum_{j=1}^{m-1}|g(x_{j+1})-g(x_j)|. \tag*{(17)} Var 0 1 ( g ) := m ≥ 1 sup { 0 = x 1 < x 2 < ⋯ < x m = 1 } sup j = 1 ∑ m − 1 ∣ g ( x j + 1 ) − g ( x j ) ∣. (17) One may also use ∥ f ∥ L 1 \|f\|_{L^1} ∥ f ∥ L 1 ∥ f ∥ L ∞ \|f\|_{L^\infty} ∥ f ∥ L ∞ B V BV B V [ 0 , 1 ] [0,1] [ 0 , 1 ]
∥ f ∥ L 1 ≤ ∥ f ∥ L ∞ , ∥ f ∥ L ∞ ≤ Var 0 1 ( f ) + ∥ f ∥ L 1 . \|f\|_{L^1}\le\|f\|_{L^\infty},\qquad\|f\|_{L^\infty}\le\operatorname{Var}_0^1(f)+\|f\|_{L^1}. ∥ f ∥ L 1 ≤ ∥ f ∥ L ∞ , ∥ f ∥ L ∞ ≤ Var 0 1 ( f ) + ∥ f ∥ L 1 . Let Π \Pi Π n n n
Span { χ [ 0 , b 0 ] , … , χ [ 0 , b n − 1 ] } ⊂ L 2 ( [ 0 , 1 ] ) \operatorname{Span}\{\chi_{[0,b_0]},\ldots,\chi_{[0,b_{n-1}]}\}\subset L^2([0,1]) Span { χ [ 0 , b 0 ] , … , χ [ 0 , b n − 1 ] } ⊂ L 2 ([ 0 , 1 ]) where the b j b_j b j (10) . The matrix A = { ⟨ χ [ 0 , b k ] , χ [ 0 , b j ] ⟩ L 2 } k , j = 0 n − 1 A=\{\langle\chi_{[0,b_k]},\chi_{[0,b_j]}\rangle_{L^2}\}_{k,j=0}^{n-1} A = {⟨ χ [ 0 , b k ] , χ [ 0 , b j ] ⟩ L 2 } k , j = 0 n − 1 u ~ j ∈ Range ( Π ) \widetilde{u}_j\in\operatorname{Range}(\Pi) u j ∈ Range ( Π ) u ~ j = ∑ k = 0 n − 1 ( A − 1 ) j k χ [ 0 , b k ] \widetilde{u}_j=\sum_{k=0}^{n-1}(A^{-1})_{jk}\chi_{[0,b_k]} u j = ∑ k = 0 n − 1 ( A − 1 ) j k χ [ 0 , b k ]
∫ 0 1 u ~ j ( t ) χ [ 0 , b m ] ( t ) d t = δ j m \int_0^1\widetilde{u}_j(t)\chi_{[0,b_m]}(t)\,\mathrm{d}t=\delta_{jm} ∫ 0 1 u j ( t ) χ [ 0 , b m ] ( t ) d t = δ j m and
( Π f ) ( x ) = ∑ j = 0 n − 1 χ [ 0 , b j ] ( x ) ∫ 0 1 u ~ j ( t ) f ( t ) d t . (18) (\Pi f)(x)=\sum_{j=0}^{n-1}\chi_{[0,b_j]}(x)\int_0^1\widetilde{u}_j(t)f(t)\,\mathrm{d}t. \tag*{(18)} ( Π f ) ( x ) = j = 0 ∑ n − 1 χ [ 0 , b j ] ( x ) ∫ 0 1 u j ( t ) f ( t ) d t . (18) Theorem 2.7. Both P \mathcal{P} P (7) and Π \Pi Π (18) leave B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) P Π = Π P Π \mathcal{P}\Pi=\Pi\mathcal{P}\Pi P Π = Π P Π P n P_n P n β Π P Π \beta\Pi\mathcal{P}\Pi β Π P Π B V BV B V
( P − z ) − 1 − Π ( β − 1 P n − z ) − 1 Π (\mathcal{P}-z)^{-1}-\Pi(\beta^{-1}P_n-z)^{-1}\Pi ( P − z ) − 1 − Π ( β − 1 P n − z ) − 1 Π has an analytic extension from ∣ z ∣ > ∥ P ∥ B V ≥ 1 |z|>\|\mathcal{P}\|_{BV}\ge1 ∣ z ∣ > ∥ P ∥ B V ≥ 1 ∣ z ∣ > β − 1 |z|>\beta^{-1} ∣ z ∣ > β − 1 ( P − z ) − 1 (\mathcal{P}-z)^{-1} ( P − z ) − 1 ∣ z ∣ > β − 1 |z|>\beta^{-1} ∣ z ∣ > β − 1 ζ \zeta ζ λ = ζ β \lambda=\zeta\beta λ = ζ β ∣ λ ∣ > 1 |\lambda|>1 ∣ λ ∣ > 1 p ( z ) p(z) p ( z ) (3) .
Moreover, the essential spectrum of P \mathcal{P} P ∣ z ∣ ≤ β − 1 |z|\le\beta^{-1} ∣ z ∣ ≤ β − 1 z z z ∣ z ∣ < β − 1 |z|<\beta^{-1} ∣ z ∣ < β − 1
The proof of this theorem is given in § 4.4 .
Asymptotics of P r ( f ) \mathcal{P}^{r}(f) P r ( f ) f ∈ B V ( [ 0 , 1 ] ) f \in BV([0,1]) f ∈ B V ([ 0 , 1 ])
Here P \mathcal{P} P B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) Π \Pi Π (18) , together with the matrix P n \mathcal{P}_{n} P n β Π P Π \beta\Pi\mathcal{P}\Pi β Π P Π { β ≡ λ 1 , λ 2 , … , λ n } \{\beta\equiv\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\} { β ≡ λ 1 , λ 2 , … , λ n } p ( z ) p(z) p ( z ) (3) . From Theorem 2.7 we know that the essential spectrum of P \mathcal{P} P B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) ∣ z ∣ ≤ β − 1 |z| \leq\beta^{-1} ∣ z ∣ ≤ β − 1 ∣ z ∣ < β − 1 |z| < \beta^{-1} ∣ z ∣ < β − 1 Theorem 2.7 we know that all the numbers ζ j : = λ j β − 1 \zeta_{j}:=\lambda_{j}\beta^{-1} ζ j := λ j β − 1 ∣ ζ j ∣ > β − 1 |\zeta_{j}| > \beta^{-1} ∣ ζ j ∣ > β − 1 P \mathcal{P} P ζ 1 = 1 \zeta_{1}=1 ζ 1 = 1 p ( z ) p(z) p ( z )
Corollary 2.8. Let Q j Q_{j} Q j ζ j \zeta_{j} ζ j ∣ ζ j ∣ > β − 1 |\zeta_{j}|>\beta^{-1} ∣ ζ j ∣ > β − 1 Q = ∑ ∣ ζ j ∣ > β − 1 Q j Q=\sum_{|\zeta_{j}|>\beta^{-1} }Q_{j} Q = ∑ ∣ ζ j ∣ > β − 1 Q j 1 ≤ m j < n 1\leq m_{j}<n 1 ≤ m j < n Q j Q_{j} Q j
(1) We have
P r Q = Q 1 + ∑ 1 > ∣ ζ j ∣ > β − 1 ζ j r ( 1 + ( N j / ζ j ) ) r Q j \mathcal{P}^{r}Q=Q_{1}+\sum_{1>|\zeta_{j}|>\beta^{-1}}\zeta_{j}^{r}\left(1+\left(N_{j}/\zeta_{j}\right)\right)^{r}Q_{j} P r Q = Q 1 + 1 > ∣ ζ j ∣ > β − 1 ∑ ζ j r ( 1 + ( N j / ζ j ) ) r Q j where N j : = ( P − ζ j ) Q j N_{j}:=(\mathcal{P}-\zeta_{j})Q_{j} N j := ( P − ζ j ) Q j N j m j = 0 N_{j}^{m_{j}}=0 N j m j = 0
( 1 + N j / ζ j ) r = ∑ l = 0 m j − 1 ( N j / ζ j ) l ( r l ) . \left(1+N_{j}/\zeta_{j}\right)^{r}=\sum_{l=0}^{m_{j}-1}\left(N_{j}/\zeta_{j}\right)^{l}\binom{r}{l}. ( 1 + N j / ζ j ) r = l = 0 ∑ m j − 1 ( N j / ζ j ) l ( l r ) . (2) Let 0 < α < 1 0<\alpha<1 0 < α < 1 r ≥ 1 r\geq1 r ≥ 1
P r = Q 1 + ∑ 1 > ∣ ζ j ∣ > β − 1 ζ j r ( 1 + ( N j / ζ j ) ) r Q j + O ( β − α r ) , (19) \mathcal{P}^{r}=Q_{1}+\sum_{1>|\zeta_{j}|>\beta^{-1}}\zeta_{j}^{r}\left(1+\left(N_{j}/\zeta_{j}\right)\right)^{r}Q_{j}+O(\beta^{-\alpha r}), \tag*{(19)} P r = Q 1 + 1 > ∣ ζ j ∣ > β − 1 ∑ ζ j r ( 1 + ( N j / ζ j ) ) r Q j + O ( β − α r ) , (19) where O ( β − α r ) O(\beta^{-\alpha r}) O ( β − α r ) B V BV B V
Q 1 f = u 1 ∫ 0 1 f ( x ) d x . (20) Q_{1}f=u_{1}\int_{0}^{1}f(x)\,\mathrm{d}x. \tag*{(20)} Q 1 f = u 1 ∫ 0 1 f ( x ) d x . (20) (3) Assume that all eigenvalues obeying 1 > ∣ ζ j ∣ > β − 1 1>|\zeta_{j}|>\beta^{-1} 1 > ∣ ζ j ∣ > β − 1 m j = 1 m_{j}=1 m j = 1 u j ∈ Range ( Π ) u_{j}\in\operatorname{Range}(\Pi) u j ∈ Range ( Π ) U ^ ( j ) ∈ B V ( [ 0 , 1 ] ) ∗ \widehat{U}^{(j)}\in BV([0,1])^{*} U ( j ) ∈ B V ([ 0 , 1 ] ) ∗ ( Q j f ) ( x ) = u j ( x ) U ^ ( j ) ( f ) (Q_{j}f)(x)=u_{j}(x)\widehat{U}^{(j)}(f) ( Q j f ) ( x ) = u j ( x ) U ( j ) ( f ) U ^ ( j ) ( u k ) = δ j k \widehat{U}^{(j)}(u_{k})=\delta_{jk} U ( j ) ( u k ) = δ j k 0 < α < 1 0<\alpha<1 0 < α < 1 C α > 0 C_{\alpha}>0 C α > 0 f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ]) r ≥ 1 r\geq1 r ≥ 1
∥ ( P r f ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 f ( t ) d t − ∑ 1 > ∣ ζ j ∣ > β − 1 ζ j r u j ( ⋅ ) U ^ ( j ) ( f ) ∥ B V ≤ C α ∥ f ∥ B V β − α r . (21) \left\|\left(\mathcal{P}^{r}f\right)(\cdot)-u_{1}(\cdot)\int_{0}^{1}f(t)\,\mathrm{d}t-\sum_{1>|\zeta_{j}|>\beta^{-1}}\zeta_{j}^{r}u_{j}(\cdot)\widehat{U}^{(j)}(f)\right\|_{BV}\leq C_{\alpha}\|f\|_{BV}\beta^{-\alpha r}. \tag*{(21)} ( P r f ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 f ( t ) d t − 1 > ∣ ζ j ∣ > β − 1 ∑ ζ j r u j ( ⋅ ) U ( j ) ( f ) B V ≤ C α ∥ f ∥ B V β − α r . (21) (4) If p ( z ) p(z) p ( z ) C > 0 C>0 C > 0 f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ]) r ≥ 1 r\geq1 r ≥ 1
∥ ( P r f ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 f ( t ) d t ∥ B V ≤ C ∥ f ∥ B V β − r . \left\|\left(\mathcal{P}^{r}f\right)(\cdot)-u_{1}(\cdot)\int_{0}^{1}f(t)\,\mathrm{d}t\right\|_{BV}\leq C\|f\|_{BV}\beta^{-r}. ( P r f ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 f ( t ) d t B V ≤ C ∥ f ∥ B V β − r . The proof of this corollary is given in § 4.5 .
The eigenvectors of P ∗ \mathcal{P}^{*} P ∗ BV ( [ 0 , 1 ] ) ∗ \operatorname{BV}([0,1])^{*} BV ([ 0 , 1 ] ) ∗
Here we want to give an “explicit” formula for the functionals U ^ ( j ) \widehat{U}^{(j)} U ( j ) (21) . Before we can do that we need to introduce some notation, which will also play an important role in several other proofs.
We split the interval [ 0 , 1 ] [0,1] [ 0 , 1 ]
t k 1 j 1 : = { j 1 β , 0 ≤ j 1 ≤ a 0 − 1 when k 1 = 1 , a 0 β + j 1 β 2 , 0 ≤ j 1 ≤ a 1 − 1 when k 1 = 2 , a 0 β + a 1 β 2 + j 1 β 3 , 0 ≤ j 1 ≤ a 2 − 1 when k 1 = 3 , ⋮ a 0 β + a 1 β 2 + ⋯ + a n − 2 β n − 1 + j 1 β n , 0 ≤ j 1 ≤ a n − 1 − 1 when k 1 = n . t_{k_1}^{j_1}:=
\begin{cases}
\dfrac{j_1}{\beta}, & 0\leq j_1\leq a_0-1 \quad\text{when } k_1=1,\\
\dfrac{a_0}{\beta}+\dfrac{j_1}{\beta^2}, & 0\leq j_1\leq a_1-1 \quad\text{when } k_1=2,\\
\dfrac{a_0}{\beta}+\dfrac{a_1}{\beta^2}+\dfrac{j_1}{\beta^3}, & 0\leq j_1\leq a_2-1 \quad\text{when } k_1=3,\\
\vdots& \\
\dfrac{a_0}{\beta}+\dfrac{a_1}{\beta^2}+\cdots+\dfrac{a_{n-2}}{\beta^{n-1}}+\dfrac{j_1}{\beta^n}, & 0\leq j_1\leq a_{n-1}-1 \quad\text{when } k_1=n.
\end{cases} t k 1 j 1 := ⎩ ⎨ ⎧ β j 1 , β a 0 + β 2 j 1 , β a 0 + β 2 a 1 + β 3 j 1 , ⋮ β a 0 + β 2 a 1 + ⋯ + β n − 1 a n − 2 + β n j 1 , 0 ≤ j 1 ≤ a 0 − 1 when k 1 = 1 , 0 ≤ j 1 ≤ a 1 − 1 when k 1 = 2 , 0 ≤ j 1 ≤ a 2 − 1 when k 1 = 3 , 0 ≤ j 1 ≤ a n − 1 − 1 when k 1 = n . where we use the convention that if a i − 1 = 0 a_{i-1}=0 a i − 1 = 0 i ∈ { 2 , … , n − 1 } i\in\{2,\ldots,n-1\} i ∈ { 2 , … , n − 1 } k 1 = i k_1=i k 1 = i β − i \beta^{-i} β − i
t k a k − 1 − 1 + β − k = t k + 1 0 , a k − 1 a k > 0 , 1 ≤ k ≤ n − 1 , t_k^{a_{k-1}-1}+\beta^{-k}=t_{k+1}^{0},\qquad a_{k-1}a_k>0,\qquad1\leq k\leq n-1, t k a k − 1 − 1 + β − k = t k + 1 0 , a k − 1 a k > 0 , 1 ≤ k ≤ n − 1 , t n a n − 1 − 1 + β − n = 1. t_n^{a_{n-1}-1}+\beta^{-n}=1. t n a n − 1 − 1 + β − n = 1. Figure 1. Plot of the first splitting of the interval $[0,1]..
A second splitting of each subinterval is given by
t k 1 k 2 j 1 j 2 : = t k 1 j 1 + β − k 1 t k 2 j 2 t_{k_1k_2}^{j_1j_2}:=t_{k_1}^{j_1}+\beta^{-k_1}t_{k_2}^{j_2} t k 1 k 2 j 1 j 2 := t k 1 j 1 + β − k 1 t k 2 j 2 and after m m m
t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m : = t k 1 k 2 ⋯ k m − 1 j 1 j 2 ⋯ j m − 1 + β − ( k 1 + k 2 + ⋯ + k m − 1 ) t k m j m . t_{k_1k_2\cdots k_m}^{j_1j_2\cdots j_m}:=t_{k_1k_2\cdots k_{m-1}}^{j_1j_2\cdots j_{m-1}}+\beta^{-(k_1+k_2+\cdots+k_{m-1})}t_{k_m}^{j_m}. t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m := t k 1 k 2 ⋯ k m − 1 j 1 j 2 ⋯ j m − 1 + β − ( k 1 + k 2 + ⋯ + k m − 1 ) t k m j m . To simplify notation, we write
k m : = ( k 1 , k 2 , … , k m ) , \mathbf{k}_m:=(k_1,k_2,\ldots,k_m), k m := ( k 1 , k 2 , … , k m ) , j m : = ( j 1 , j 2 , … , j m ) , \mathbf{j}_m:=(j_1,j_2,\ldots,j_m), j m := ( j 1 , j 2 , … , j m ) , t k m j m : = t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m , t_{\mathbf{k}_m}^{\mathbf{j}_m}:=t_{k_1k_2\cdots k_m}^{j_1j_2\cdots j_m}, t k m j m := t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m , ∣ k m ∣ : = k 1 + k 2 + ⋯ + k m , \lvert\mathbf{k}_m\rvert:=k_1+k_2+\cdots+k_m, ∣ k m ∣ := k 1 + k 2 + ⋯ + k m , m ≤ ∣ k m ∣ ≤ m n . (22) m\leq\lvert\mathbf{k}_m\rvert\leq mn. \tag*{(22)} m ≤ ∣ k m ∣ ≤ mn . (22) We will use the notation
∑ t k m j m : = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 − 1 ⋯ ∑ k m = 1 n ∑ j m = 0 a k m − 1 − 1 \sum_{t_{k_m}^{j_m}}:=\sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1-1}-1}\cdots\sum_{k_m=1}^{n}\sum_{j_m=0}^{a_{k_m-1}-1} t k m j m ∑ := k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 − 1 ⋯ k m = 1 ∑ n j m = 0 ∑ a k m − 1 − 1 with the same convention that if a k − 1 = 0 a_{k-1}=0 a k − 1 = 0 k k k
Theorem 2.9. Let ζ = λ / β \zeta=\lambda/\beta ζ = λ / β 1 < ∣ λ ∣ ≤ β 1<|\lambda|\leq\beta 1 < ∣ λ ∣ ≤ β λ \lambda λ
1 = ∑ j = 1 n a j − 1 λ j = ∑ k = 1 n λ − k ∑ j k = 0 a k − 1 − 1 1. (23) 1=\sum_{j=1}^{n}\frac{a_{j-1}}{\lambda^{j}}=\sum_{k=1}^{n}\lambda^{-k}\sum_{j_k=0}^{a_{k-1}-1}1. \tag*{(23)} 1 = j = 1 ∑ n λ j a j − 1 = k = 1 ∑ n λ − k j k = 0 ∑ a k − 1 − 1 1. (23) Let f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ]) m ≥ 1 m\geq1 m ≥ 1
U ^ m ( f ) : = ∑ t k m j m λ − ∣ k m ∣ f ( t k m j m ) . \widehat{U}_m(f):=\sum_{t_{k_m}^{j_m}}\lambda^{-|k_m|}f\left(t_{k_m}^{j_m}\right). U m ( f ) := t k m j m ∑ λ − ∣ k m ∣ f ( t k m j m ) . Then:
(i) U ^ ( f ) : = lim m → ∞ U ^ m ( f ) \widehat{U}(f):=\lim_{m\to\infty}\widehat{U}_m(f) U ( f ) := lim m → ∞ U m ( f ) B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ])
U ^ ( P f ) = ζ U ^ ( f ) . (24) \widehat{U}(\mathcal{P}f)=\zeta\widehat{U}(f). \tag*{(24)} U ( P f ) = ζ U ( f ) . (24) When ζ = 1 \zeta=1 ζ = 1 U ^ ( f ) = ∫ 0 1 f ( x ) d x \widehat{U}(f)=\int_0^1f(x)\,\mathrm{d}x U ( f ) = ∫ 0 1 f ( x ) d x
(ii) Take any y ∈ ( 0 , 1 ) y\in(0,1) y ∈ ( 0 , 1 ) β \beta β y = ∑ j ≥ 1 y j − 1 β j y=\sum_{j\geq1}\frac{y_{j-1}}{\beta^j} y = ∑ j ≥ 1 β j y j − 1 y j − 1 = ⌊ β T j − 1 ( y ) ⌋ y_{j-1}=\lfloor\beta T^{j-1}(y)\rfloor y j − 1 = ⌊ β T j − 1 ( y )⌋ j ≥ 1 j\geq1 j ≥ 1
U ^ ( χ [ 0 , y ] ) = ∑ j ≥ 1 y j − 1 λ j . (25) \widehat{U}(\chi_{[0,y]})=\sum_{j\geq1}\frac{y_{j-1}}{\lambda^j}. \tag*{(25)} U ( χ [ 0 , y ] ) = j ≥ 1 ∑ λ j y j − 1 . (25) Formula (25) coincides with the one obtained in [12 Proposition 4.2] . If h h h [ 0 , 1 ] [0,1] [ 0 , 1 ]
U ^ ( h ) = h ( 1 ) − ∑ j = 1 ∞ λ − j ∫ 0 1 y j − 1 h ′ ( y ) d y . (26) \widehat{U}(h)=h(1)-\sum_{j=1}^{\infty}\lambda^{-j}\int_0^1y_{j-1}h'(y)\,\mathrm{d}y. \tag*{(26)} U ( h ) = h ( 1 ) − j = 1 ∑ ∞ λ − j ∫ 0 1 y j − 1 h ′ ( y ) d y . (26) Also, by fixing u = ∑ j = 0 n − 1 λ − j − 1 χ [ 0 , b j ] u=\sum_{j=0}^{n-1}\lambda^{-j-1}\chi_{[0,b_j]} u = ∑ j = 0 n − 1 λ − j − 1 χ [ 0 , b j ] P u = ζ u \mathcal{P}u=\zeta u P u = ζ u
U ^ ( u ) = h n ′ ( λ ) , h n ( z ) : = 1 − ∑ k = 0 n − 1 a k z k + 1 = z − n p ( z ) . (27) \widehat{U}(u)=h_n'(\lambda),\qquad h_n(z):=1-\sum_{k=0}^{n-1}\frac{a_k}{z^{k+1}}=z^{-n}p(z). \tag*{(27)} U ( u ) = h n ′ ( λ ) , h n ( z ) := 1 − k = 0 ∑ n − 1 z k + 1 a k = z − n p ( z ) . (27) Thus U ^ ( u ) ≠ 0 \widehat{U}(u)\ne0 U ( u ) = 0 λ \lambda λ
(iii) If β − 1 < ∣ ζ ∣ < 1 \beta^{-1}<|\zeta|<1 β − 1 < ∣ ζ ∣ < 1
( 0 , 1 ) ∋ y ↦ U ^ ( χ [ 0 , y ] ) ∈ C (0,1)\ni y\mapsto\widehat{U}(\chi_{[0,y]})\in\mathbb{C} ( 0 , 1 ) ∋ y ↦ U ( χ [ 0 , y ] ) ∈ C is nowhere differentiable. See also [12 Theorem 4.3(v)] .
(iv) If β − 1 < ∣ ζ ∣ < 1 \beta^{-1}<|\zeta|<1 β − 1 < ∣ ζ ∣ < 1 μ \mu μ U ^ ( f ) = ∫ 0 1 f ( x ) d μ ( x ) \widehat{U}(f)=\int_0^1f(x)\,\mathrm{d}\mu(x) U ( f ) = ∫ 0 1 f ( x ) d μ ( x ) [12 Theorem 4.5] .
(v) Let β − 1 < ∣ ζ ∣ < 1 \beta^{-1}<|\zeta|<1 β − 1 < ∣ ζ ∣ < 1 c 1 = 1 / λ c_1=1/\lambda c 1 = 1/ λ U ^ ( u ζ ) = h n ′ ( λ ) \widehat{U}(u_\zeta)=h'_n(\lambda) U ( u ζ ) = h n ′ ( λ ) (27) . There exists a sequence { C s } s ≥ 1 ⊂ C \{C_s\}_{s\geq1}\subset\mathbb{C} { C s } s ≥ 1 ⊂ C C s C_s C s ( s + 1 ) n × ( s + 1 ) n (s+1)n\times(s+1)n ( s + 1 ) n × ( s + 1 ) n q q q M ≥ 0 M\geq0 M ≥ 0 h n h_n h n (27) we have
U ^ ( q ) = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 − 1 ζ − k 1 ( ∫ t k 1 j 1 t k 1 j 1 + β − k 1 q ( u ) d u + h n ′ ( λ ) ∑ s = 1 M + 1 C s β − k 1 s ( q ( s − 1 ) ( t k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( t k 1 j 1 ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) ) . (28) \widehat{U}(q)=\sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1-1}-1}\zeta^{-k_1}\left(\int_{t_{k_1}^{j_1}}^{t_{k_1}^{j_1}+\beta^{-k_1}}q(u)\,\mathrm{d}u
+h'_n(\lambda)\sum_{s=1}^{M+1}C_s\beta^{-k_1s}\left(q^{(s-1)}\left(t_{k_1}^{j_1}+\beta^{-k_1}\right)-q^{(s-1)}\left(t_{k_1}^{j_1}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}\right). \tag*{(28)} U ( q ) = k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 − 1 ζ − k 1 ( ∫ t k 1 j 1 t k 1 j 1 + β − k 1 q ( u ) d u + h n ′ ( λ ) s = 1 ∑ M + 1 C s β − k 1 s ( q ( s − 1 ) ( t k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( t k 1 j 1 ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) ) . (28) The proof of this theorem is given in Section 4.6.
Remark 2.10. Using (26) in order to compute U ^ ( q ) \widehat{U}(q) U ( q ) (28) would need to compute infinitely many y j y_j y j y y y (28) only requires a finite number of computations.
Acknowledgements
This work was supported by The Danish Council for Independent Research – Natural Sciences, grant DFF–10.46540/2032-00005B.
Proofs of the Main Results
Proof of Theorem 2.1
From (22) we see that t k m j m t_{k_m}^{j_m} t k m j m t k m j m + β − ∣ k m ∣ t_{k_m}^{j_m}+\beta^{-|k_m|} t k m j m + β − ∣ k m ∣ b = t k m j m + β − ∣ k m ∣ b=t_{k_m}^{j_m}+\beta^{-|k_m|} b = t k m j m + β − ∣ k m ∣ a = t k m j m a=t_{k_m}^{j_m} a = t k m j m
F ( y ) = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + ∑ s = 1 N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) B ~ s ( β ∣ k m ∣ ( y − t k m j m ) ) s ! + O ( β − ∣ k m ∣ N ∥ F ( N ) ∥ C ( [ 0 , 1 ] ) ) , ∀ y ∈ [ t k m j m , t k m j m + β − ∣ k m ∣ ] . (29) \begin{aligned}
F(y)={}&\beta^{|k_m|}\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-|k_m|}}F(u)\,\mathrm{d}u\\
&+\sum_{s=1}^{N}\beta^{-(s-1)|k_m|}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-|k_m|}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{\widetilde{B}_s\left(\beta^{|k_m|}(y-t_{k_m}^{j_m})\right)}{s!}\\
&+O\left(\beta^{-|k_m|N}\lVert F^{(N)}\rVert_{C([0,1])}\right),\qquad\forall y\in\left[t_{k_m}^{j_m},t_{k_m}^{j_m}+\beta^{-|k_m|}\right]. \tag*{(29)}
\end{aligned} F ( y ) = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + s = 1 ∑ N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! B s ( β ∣ k m ∣ ( y − t k m j m ) ) + O ( β − ∣ k m ∣ N ∥ F ( N ) ∥ C ([ 0 , 1 ]) ) , ∀ y ∈ [ t k m j m , t k m j m + β − ∣ k m ∣ ] . (29) Definition 4.1. Let B s B_s B s s ∈ N 0 s\in\mathbb{N}_0 s ∈ N 0
G s , k m j m ( x ) : = β ∣ k m ∣ ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ∣ k m ∣ ( x − t k m j m ) ) χ [ t k m j m , t k m j m + β − ∣ k m ∣ ] ( x ) . G_{s,k_m}^{j_m}(x):=\frac{\beta^{|k_m|}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{|k_m|}(x-t_{k_m}^{j_m})\right)\chi_{[t_{k_m}^{j_m},t_{k_m}^{j_m}+\beta^{-|k_m|}]}(x). G s , k m j m ( x ) := ∥ B s ∥ L 1 ([ 0 , 1 ]) β ∣ k m ∣ B s ( β ∣ k m ∣ ( x − t k m j m ) ) χ [ t k m j m , t k m j m + β − ∣ k m ∣ ] ( x ) . Moreover, for j ∈ { 1 , … , n } j\in\{1,\ldots,n\} j ∈ { 1 , … , n } b j b_j b j (10) we define
H s n + j ( x ) : = 1 b j − 1 B s ( 1 b j − 1 x ) ∥ B s ∥ L 1 ( [ 0 , 1 ] ) χ [ 0 , b j − 1 ] ( x ) , H 1 = ψ 1 = χ [ 0 , 1 ] . H_{sn+j}(x):=\frac{1}{b_{j-1}}\frac{B_s\left(\frac{1}{b_{j-1}}x\right)}{\lVert B_s\rVert_{L^1([0,1])}}\chi_{[0,b_{j-1}]}(x),\qquad H_1=\psi_1=\chi_{[0,1]}. H s n + j ( x ) := b j − 1 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) B s ( b j − 1 1 x ) χ [ 0 , b j − 1 ] ( x ) , H 1 = ψ 1 = χ [ 0 , 1 ] . We now let y y y (29) to belong to [ 0 , 1 ] [0,1] [ 0 , 1 ]
∑ : = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 − 1 ⋯ ∑ k m = 1 n ∑ j m = 0 a k m − 1 − 1 \sum:= \sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1-1}-1}\cdots\sum_{k_m=1}^{n}\sum_{j_m=0}^{a_{k_m-1}-1} ∑ := k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 − 1 ⋯ k m = 1 ∑ n j m = 0 ∑ a k m − 1 − 1 with the same convention that if a k − 1 = 0 a_{k-1}=0 a k − 1 = 0 k k k m ≥ 1 m\ge1 m ≥ 1 ∣ k m ∣ ≥ m \lvert k_m\rvert\ge m ∣ k m ∣ ≥ m
F ( y ) = ∑ { G 0 , k m j m ( y ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + ∑ s = 1 N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) G s , k m j m ( y ) } + O ( β − m N ∥ F ( N ) ∥ C ( [ 0 , 1 ] ) ) , ∀ y ∈ [ 0 , 1 ] . (30) \begin{aligned}
F(y)={}&\sum\left\{G_{0,k_m}^{j_m}(y)\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-\lvert k_m\rvert}}F(u)\,\mathrm{d}u\right.\\
&\quad+\sum_{s=1}^{N}\beta^{-\lvert k_m\rvert s}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-\lvert k_m\rvert}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}G_{s,k_m}^{j_m}(y)\bigg\}\\
&\quad+O\left(\beta^{-mN}\lVert F^{(N)}\rVert_{C([0,1])}\right),\qquad\forall y\in[0,1]. \tag*{(30)}
\end{aligned} F ( y ) = ∑ { G 0 , k m j m ( y ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + s = 1 ∑ N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) G s , k m j m ( y ) } + O ( β − m N ∥ F ( N ) ∥ C ([ 0 , 1 ]) ) , ∀ y ∈ [ 0 , 1 ] . (30) By applying the transfer operator a sufficient number of times we can bring the Bernoulli polynomials from the small intervals in the splitting back to the full interval [ 0 , 1 ] [0,1] [ 0 , 1 ]
Lemma 4.1. Let s ∈ N 0 s\in\mathbb{N}_0 s ∈ N 0 x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ] independently of the choice of j m j_m j m
( P ∣ k m ∣ G s , k m j m ) ( x ) = H s n + 1 ( x ) . \left(\mathcal{P}^{\lvert k_m\rvert}G_{s,k_m}^{j_m}\right)(x)=H_{s n+1}(x). ( P ∣ k m ∣ G s , k m j m ) ( x ) = H s n + 1 ( x ) . The next step is to apply the above identity to (30) while controlling the L ∞ L^\infty L ∞ ( a 0 + 1 ) r / β r (a_0+1)^r/\beta^r ( a 0 + 1 ) r / β r P r \mathcal{P}^r P r L ∞ L^\infty L ∞ 1 ≤ a 0 < β < a 0 + 1 1\le a_0<\beta<a_0+1 1 ≤ a 0 < β < a 0 + 1 ( a 0 + 1 ) / β < 2 (a_0+1)/\beta<2 ( a 0 + 1 ) / β < 2
( P r F ) ( y ) = ∑ { ( P r − ∣ k m ∣ H 1 ) ( y ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + ∑ s = 1 N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) ( P r − ∣ k m ∣ H s n + 1 ) ( y ) } + O ( 2 r β − m N ∥ F ( N ) ∥ C ( [ 0 , 1 ] ) ) . (31) \begin{aligned}
(\mathcal{P}^rF)(y)={}&\sum\left\{(\mathcal{P}^{r-\lvert k_m\rvert}H_1)(y)\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-\lvert k_m\rvert}}F(u)\,\mathrm{d}u\right.\\
&\quad+\sum_{s=1}^{N}\beta^{-\lvert k_m\rvert s}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-\lvert k_m\rvert}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}(\mathcal{P}^{r-\lvert k_m\rvert}H_{s n+1})(y)\bigg\}\\
&\quad+O\left(2^r\beta^{-mN}\lVert F^{(N)}\rVert_{C([0,1])}\right). \tag*{(31)}
\end{aligned} ( P r F ) ( y ) = ∑ { ( P r − ∣ k m ∣ H 1 ) ( y ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + s = 1 ∑ N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) ( P r − ∣ k m ∣ H s n + 1 ) ( y ) } + O ( 2 r β − m N ∥ F ( N ) ∥ C ([ 0 , 1 ]) ) . (31) The first term is treated with Lemma A.4 (iv), and also using that ∣ k m ∣ ≤ n m \lvert k_m\rvert\le nm ∣ k m ∣ ≤ nm
∑ ( P r − ∣ k m ∣ H 1 ) ( x ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u = u 1 ( x ) ∫ 0 1 F ( u ) d u + O ( ( ∣ λ ∗ ∣ β ) r − m n ∥ F ∥ C ( [ 0 , 1 ] ) ) . \sum(\mathcal{P}^{r-\lvert k_m\rvert}H_1)(x)\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-\lvert k_m\rvert}}F(u)\,\mathrm{d}u=u_1(x)\int_0^1F(u)\,\mathrm{d}u+O\left(\left(\frac{\lvert\lambda_*\rvert}{\beta}\right)^{r-mn}\lVert F\rVert_{C([0,1])}\right). ∑ ( P r − ∣ k m ∣ H 1 ) ( x ) ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u = u 1 ( x ) ∫ 0 1 F ( u ) d u + O ( ( β ∣ λ ∗ ∣ ) r − mn ∥ F ∥ C ([ 0 , 1 ]) ) . We split the second term from (31) in s = 1 s=1 s = 1 s ≥ 2 s\ge2 s ≥ 2 s = 1 s=1 s = 1 Lemma A.5 along with the Mean Value Theorem to get
∑ j m k m β − ∣ k m ∣ ( F ( t k m j m + β − ∣ k m ∣ ) − F ( t k m j m ) ) ∥ B 1 ∥ L 1 ( [ 0 , 1 ] ) ( G r − ∣ k m ∣ H n + 1 ) ( x ) = 4 − 1 β − r ( F ( 1 ) − F ( 0 ) ) v 1 ( x ) + β − m ∥ F ′ ∥ C ( [ 0 , 1 ] ) O ( ( ∣ λ ∗ ∣ β ) r − n m ) . \sum_{\substack{j_m\\ k_m}}\beta^{-|k_m|}\left(F\left(t_{k_m}^{j_m}+\beta^{-|k_m|}\right)-F\left(t_{k_m}^{j_m}\right)\right)\lVert B_1\rVert_{L^1([0,1])}\left(\mathcal{G}^{r-|k_m|}H_{n+1}\right)(x)
=4^{-1}\beta^{-r}\left(F(1)-F(0)\right)v_1(x)+\beta^{-m}\lVert F'\rVert_{C([0,1])}O\left(\left(\frac{|\lambda_*|}{\beta}\right)^{r-nm}\right). j m k m ∑ β − ∣ k m ∣ ( F ( t k m j m + β − ∣ k m ∣ ) − F ( t k m j m ) ) ∥ B 1 ∥ L 1 ([ 0 , 1 ]) ( G r − ∣ k m ∣ H n + 1 ) ( x ) = 4 − 1 β − r ( F ( 1 ) − F ( 0 ) ) v 1 ( x ) + β − m ∥ F ′ ∥ C ([ 0 , 1 ]) O ( ( β ∣ λ ∗ ∣ ) r − nm ) . For s ≥ 2 s\ge2 s ≥ 2 ν \nu ν ∣ k m ∣ ≥ m \lvert k_m\rvert\ge m ∣ k m ∣ ≥ m
∑ j m k m ∑ s = 2 N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) ( G r − ∣ k m ∣ H s n + 1 ) ( y ) = max k m ∑ s = 2 N β − s ∣ k m ∣ ∥ F ( s ) ∥ C ( [ 0 , 1 ] ) O ( ( 1 β ) r − ∣ k m ∣ ) = O ( ( 1 β ) r + m ∥ F ∥ C N ( [ 0 , 1 ] ) ) . \sum_{\substack{j_m\\ k_m}}\sum_{s=2}^{N}\beta^{-|k_m|s}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-|k_m|}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}\left(\mathcal{G}^{r-|k_m|}H_{sn+1}\right)(y)
=\max_{k_m}\sum_{s=2}^{N}\beta^{-s|k_m|}\lVert F^{(s)}\rVert_{C([0,1])}O\left(\left(\frac{1}{\beta}\right)^{r-|k_m|}\right)
=O\left(\left(\frac{1}{\beta}\right)^{r+m}\lVert F\rVert_{C^N([0,1])}\right). j m k m ∑ s = 2 ∑ N β − ∣ k m ∣ s ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) ( G r − ∣ k m ∣ H s n + 1 ) ( y ) = k m max s = 2 ∑ N β − s ∣ k m ∣ ∥ F ( s ) ∥ C ([ 0 , 1 ]) O ( ( β 1 ) r − ∣ k m ∣ ) = O ( ( β 1 ) r + m ∥ F ∥ C N ([ 0 , 1 ]) ) . Combining all of this leads to
( G r F ) ( x ) = u 1 ( x ) ∫ 0 1 F ( u ) d u + F ( 1 ) − F ( 0 ) 4 β r v 1 ( x ) + O ( ( ∣ λ ∗ ∣ β ) r − n m ∥ F ∥ C N ( [ 0 , 1 ] ) ) + β − m ∥ F ′ ∥ C ( [ 0 , 1 ] ) O ( ( ∣ λ ∗ ∣ β ) r − n m ) + O ( ( 1 β ) r + m ∥ F ∥ C N ( [ 0 , 1 ] ) ) + O ( 2 r β − m N ∥ F ( N ) ∥ C ( [ 0 , 1 ] ) ) . \begin{aligned}
(\mathcal{G}^rF)(x)={}&u_1(x)\int_0^1F(u)\,du+\frac{F(1)-F(0)}{4\beta^r}v_1(x)+O\left(\left(\frac{|\lambda_*|}{\beta}\right)^{r-nm}\lVert F\rVert_{C^N([0,1])}\right)\\
&+\beta^{-m}\lVert F'\rVert_{C([0,1])}O\left(\left(\frac{|\lambda_*|}{\beta}\right)^{r-nm}\right)+O\left(\left(\frac{1}{\beta}\right)^{r+m}\lVert F\rVert_{C^N([0,1])}\right)\\
&+O\left(2^r\beta^{-mN}\lVert F^{(N)}\rVert_{C([0,1])}\right).
\end{aligned} ( G r F ) ( x ) = u 1 ( x ) ∫ 0 1 F ( u ) d u + 4 β r F ( 1 ) − F ( 0 ) v 1 ( x ) + O ( ( β ∣ λ ∗ ∣ ) r − nm ∥ F ∥ C N ([ 0 , 1 ]) ) + β − m ∥ F ′ ∥ C ([ 0 , 1 ]) O ( ( β ∣ λ ∗ ∣ ) r − nm ) + O ( ( β 1 ) r + m ∥ F ∥ C N ([ 0 , 1 ]) ) + O ( 2 r β − m N ∥ F ( N ) ∥ C ([ 0 , 1 ]) ) . We will now show that there exists an ε > 0 \varepsilon>0 ε > 0 N N N δ ∈ ( 0 , 1 ) \delta\in(0,1) δ ∈ ( 0 , 1 ) 2 r β − m N = β − r ( 1 + δ ) 2^r\beta^{-mN}=\beta^{-r(1+\delta)} 2 r β − m N = β − r ( 1 + δ )
m δ , r , N = r ln ( 2 ) + ( 1 + δ ) ln ( β ) N ln ( β ) m_{\delta,r,N}=r\frac{\ln(2)+(1+\delta)\ln(\beta)}{N\ln(\beta)} m δ , r , N = r N ln ( β ) ln ( 2 ) + ( 1 + δ ) ln ( β ) which is less than r / n r/n r / n N N N r > n r>n r > n m δ , r , N m_{\delta,r,N} m δ , r , N
∣ λ ∗ ∣ r − n m δ , r , N β − r + n m δ , r , N ≤ β − r ( 1 + δ ) . |\lambda_*|^{r-nm_{\delta,r,N}}\beta^{-r+nm_{\delta,r,N}}\le\beta^{-r(1+\delta)}. ∣ λ ∗ ∣ r − n m δ , r , N β − r + n m δ , r , N ≤ β − r ( 1 + δ ) . This holds for all r r r
δ ≤ ln ( ∣ λ ∗ ∣ − 1 ) ( 1 − n ln ( 2 β ) N ln ( β ) ) − n ln ( 2 β ) N ( n ln ( ∣ λ ∗ ∣ − 1 ) N + n N ln ( β ) + ln ( β ) ) , \delta\le\frac{\ln\left(|\lambda_*|^{-1}\right)\left(1-\frac{n\ln(2\beta)}{N\ln(\beta)}\right)-\frac{n\ln(2\beta)}{N}}{\left(\frac{n\ln(|\lambda_*|^{-1})}{N}+\frac{n}{N}\ln(\beta)+\ln(\beta)\right)}, δ ≤ ( N n l n ( ∣ λ ∗ ∣ − 1 ) + N n ln ( β ) + ln ( β ) ) ln ( ∣ λ ∗ ∣ − 1 ) ( 1 − N l n ( β ) n l n ( 2 β ) ) − N n l n ( 2 β ) , a condition which can be satisfied if N ≥ N 0 N\ge N_0 N ≥ N 0 N 0 N_0 N 0 δ < ln ( ∣ λ ∗ ∣ − 1 ) ln ( β ) \delta<\frac{\ln(|\lambda_*|^{-1})}{\ln(\beta)} δ < l n ( β ) l n ( ∣ λ ∗ ∣ − 1 ) β − m \beta^{-m} β − m β − r − m δ , r , N \beta^{-r-m_{\delta,r,N}} β − r − m δ , r , N
ε = min { δ , ln ( 2 ) + ( 1 + δ ) ln ( β ) N 0 ln ( β ) } . \varepsilon=\min\left\{\delta,\frac{\ln(2)+(1+\delta)\ln(\beta)}{N_0\ln(\beta)}\right\}. ε = min { δ , N 0 ln ( β ) ln ( 2 ) + ( 1 + δ ) ln ( β ) } . and we are done.
Proof of Theorem 2.5
Here n = 4 n=4 n = 4 a 0 = 3 a_0=3 a 0 = 3 a 1 = 2 a_1=2 a 1 = 2 a 2 = 0 a_2=0 a 2 = 0 a 3 = 2 a_3=2 a 3 = 2 a 0 a_0 a 0 p ( z ) = z 4 − ( 3 z 3 + 2 z 2 + 2 ) p(z)=z^4-(3z^3+2z^2+2) p ( z ) = z 4 − ( 3 z 3 + 2 z 2 + 2 ) 3 < β < 4 3<\beta<4 3 < β < 4 p ( − 1 ) = 0 p(-1)=0 p ( − 1 ) = 0 p ′ ( − 1 ) ≠ 0 p'(-1)\ne0 p ′ ( − 1 ) = 0 − 1 -1 − 1 p ( z ) p(z) p ( z ) f ( x ) = p ( x ) / x 3 = x − 3 − 2 / x − 2 / x 3 f(x)=p(x)/x^3=x-3-2/x-2/x^3 f ( x ) = p ( x ) / x 3 = x − 3 − 2/ x − 2/ x 3 p p p f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ( − ∞ , 0 ) (-\infty,0) ( − ∞ , 0 ) f f f p p p − 1 -1 − 1 β \beta β p p p z 3 = z 4 ‾ z_3=\overline{z_4} z 3 = z 4 − 2 -2 − 2
∣ z 3 ∣ 2 = ∣ z 4 ∣ 2 = 2 / β |z_3|^2=|z_4|^2=2/\beta ∣ z 3 ∣ 2 = ∣ z 4 ∣ 2 = 2/ β hence ∣ z 3 ∣ = ∣ z 4 ∣ < 1 |z_3|=|z_4|<1 ∣ z 3 ∣ = ∣ z 4 ∣ < 1 β \beta β
p ( z ) z + 1 = z 3 − 4 z 2 + 2 z − 2 , \frac{p(z)}{z+1}=z^3-4z^2+2z-2, z + 1 p ( z ) = z 3 − 4 z 2 + 2 z − 2 , whose roots are β \beta β z 3 z_3 z 3 z 4 z_4 z 4
This choice of coefficients corresponds to a restriction matrix P 4 P_4 P 4 λ 1 = β \lambda_1=\beta λ 1 = β λ 2 = − 1 \lambda_2=-1 λ 2 = − 1 ∣ λ 3 ∣ = ∣ λ 4 ∣ < 1 |\lambda_3|=|\lambda_4|<1 ∣ λ 3 ∣ = ∣ λ 4 ∣ < 1
Here we have ∣ λ ∗ ∣ = 1 |\lambda_*|=1 ∣ λ ∗ ∣ = 1 A 4 , 1 A_{4,1} A 4 , 1 1 1 1 − 1 / β -1/\beta − 1/ β ∣ λ 3 / β ∣ = ∣ λ 4 / β ∣ < β − 1 |\lambda_3/\beta|=|\lambda_4/\beta|<\beta^{-1} ∣ λ 3 / β ∣ = ∣ λ 4 / β ∣ < β − 1 4 × 4 4\times4 4 × 4 − 1 / β -1/\beta − 1/ β p j 1 p_{j1} p j 1 1 ≤ j ≤ 4 1\leq j\leq4 1 ≤ j ≤ 4 u 2 ( x ) = ∑ j = 1 4 p j 1 H j ( x ) u_2(x)=\sum_{j=1}^{4}p_{j1}H_j(x) u 2 ( x ) = ∑ j = 1 4 p j 1 H j ( x ) − 1 / β -1/\beta − 1/ β
( P r H 1 ) ( x ) = u 1 ( x ) + ( − 1 ) r β − r u 2 ( x ) + O ( ( ∣ z 3 ∣ β ) r ) . (\mathcal{P}^rH_1)(x)=u_1(x)+(-1)^r\beta^{-r}u_2(x)+O\left(\left(\frac{|z_3|}{\beta}\right)^r\right). ( P r H 1 ) ( x ) = u 1 ( x ) + ( − 1 ) r β − r u 2 ( x ) + O ( ( β ∣ z 3 ∣ ) r ) . The function v 1 v_1 v 1 P \mathcal{P} P { H j } j = 1 8 \{H_j\}_{j=1}^{8} { H j } j = 1 8 + 1 / β +1/\beta + 1/ β
If q q q M M M n = 4 n=4 n = 4 m = 1 m=1 m = 1 N = M + 1 N=M+1 N = M + 1 (31) we see that the error term is identically zero and
( P r q ) ( y ) = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 { ( P r − k 1 H 1 ) ( y ) ∫ t k 1 j 1 t k 1 j 1 + β − k 1 q ( u ) d u + ∑ s = 1 M + 1 β − k 1 s ( q ( s − 1 ) ( t k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( t k 1 j 1 ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) ( P r − k 1 H 4 s + 1 ) ( y ) } . (32) (\mathcal{P}^r q)(y)=\sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1}-1}\left\{(\mathcal{P}^{r-k_1}H_1)(y)\int_{t_{k_1}^{j_1}}^{t_{k_1}^{j_1}+\beta^{-k_1}}q(u)\,\mathrm{d}u
+\sum_{s=1}^{M+1}\beta^{-k_1s}\left(q^{(s-1)}\left(t_{k_1}^{j_1}+\beta^{-k_1}\right)-q^{(s-1)}\left(t_{k_1}^{j_1}\right)\right)\frac{1}{s!}\|B_s\|_{L^1([0,1])}(\mathcal{P}^{r-k_1}H_{4s+1})(y)\right\}. \tag*{(32)} ( P r q ) ( y ) = k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 { ( P r − k 1 H 1 ) ( y ) ∫ t k 1 j 1 t k 1 j 1 + β − k 1 q ( u ) d u + s = 1 ∑ M + 1 β − k 1 s ( q ( s − 1 ) ( t k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( t k 1 j 1 ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) ( P r − k 1 H 4 s + 1 ) ( y ) } . (32) Note that the term with s = M + 1 s=M+1 s = M + 1 q ( M ) q^{(M)} q ( M ) P r − k 1 H 4 s + 1 \mathcal{P}^{r-k_1}H_{4s+1} P r − k 1 H 4 s + 1 0 ≤ s ≤ M 0\leq s\leq M 0 ≤ s ≤ M k 1 ∈ { 1 , 2 , 4 } k_1\in\{1,2,4\} k 1 ∈ { 1 , 2 , 4 } k 1 = 3 k_1=3 k 1 = 3 a 2 = 0 a_2=0 a 2 = 0
Thus P r q \mathcal{P}^r q P r q ν = M + 1 \nu=M+1 ν = M + 1 n = 4 n=4 n = 4 1 1 1 P r − k 1 H 1 \mathcal{P}^{r-k_1}H_1 P r − k 1 H 1 P r − k 1 H 4 s + 1 \mathcal{P}^{r-k_1}H_{4s+1} P r − k 1 H 4 s + 1 s ≥ 1 s\geq1 s ≥ 1 u 1 u_1 u 1 − 1 / β -1/\beta − 1/ β C 2 , q C_{2,q} C 2 , q 1 / β 1/\beta 1/ β C 3 , q C_{3,q} C 3 , q 1 / β 1/\beta 1/ β
( P r q ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 q ( u ) d u − C 2 , q ( − β ) − r u 2 ( ⋅ ) − C 3 , q β − r v 1 ( ⋅ ) (\mathcal{P}^{r}q)(\cdot)-u_{1}(\cdot)\int_{0}^{1}q(u)\,\mathrm{d}u-C_{2,q}(-\beta)^{-r}u_{2}(\cdot)-C_{3,q}\beta^{-r}v_{1}(\cdot) ( P r q ) ( ⋅ ) − u 1 ( ⋅ ) ∫ 0 1 q ( u ) d u − C 2 , q ( − β ) − r u 2 ( ⋅ ) − C 3 , q β − r v 1 ( ⋅ ) will decay like β − r ( 1 + ε ) \beta^{-r(1+\varepsilon)} β − r ( 1 + ε )
Proof of Theorem 2.6
(i). Let us first prove (2.2). Using that β − a 0 = b 1 \beta-a_{0}=b_{1} β − a 0 = b 1
β ( P ψ 0 ) ( x ) = g ( x ) χ [ 0 , b 1 ] ( x ) − g ( x ) χ [ 0 , b 1 ] ( x ) = 0. \beta(\mathcal{P}\psi_{0})(x)=g(x)\chi_{[0,b_{1}]}(x)-g(x)\chi_{[0,b_{1}]}(x)=0. β ( P ψ 0 ) ( x ) = g ( x ) χ [ 0 , b 1 ] ( x ) − g ( x ) χ [ 0 , b 1 ] ( x ) = 0. (ii). Now let us show that K ~ p \widetilde{K}_{p} K p K K K P \mathcal{P} P K ~ p \widetilde{K}_{p} K p
∫ 0 1 ∣ K ~ p f ∣ p ( x ) d x = ∫ 0 1 u 1 ( x ) ( ∣ f ∣ u 1 1 / p ) p ( T β ( x ) ) d x = ∫ 0 1 u 1 ( x ) K ( ( ∣ f ∣ u 1 1 / p ) p ) ( x ) d x = ∫ 0 1 ∣ f ∣ p ( x ) d x , \begin{aligned}
\int_{0}^{1}\left|\widetilde{K}_{p}f\right|^{p}(x)\,\mathrm{d}x
=\int_{0}^{1}u_{1}(x)\left(\frac{|f|}{u_{1}^{1/p}}\right)^{p}(T_{\beta}(x))\,\mathrm{d}x
=\int_{0}^{1}u_{1}(x)K\left(\left(\frac{|f|}{u_{1}^{1/p}}\right)^{p}\right)(x)\,\mathrm{d}x \\
&=\int_{0}^{1}|f|^{p}(x)\,\mathrm{d}x,
\end{aligned} ∫ 0 1 K p f p ( x ) d x = ∫ 0 1 u 1 ( x ) ( u 1 1/ p ∣ f ∣ ) p ( T β ( x )) d x = ∫ 0 1 u 1 ( x ) K ( ( u 1 1/ p ∣ f ∣ ) p ) ( x ) d x = ∫ 0 1 ∣ f ∣ p ( x ) d x , where in the second equality we have used that K K K (9) between K K K P \mathcal{P} P P u 1 = u 1 \mathcal{P}u_{1}=u_{1} P u 1 = u 1
Now let us prove that K ~ p \widetilde{K}_{p} K p f ∈ L p f\in L^{p} f ∈ L p K ~ p f = ψ 0 \widetilde{K}_{p}f=\psi_{0} K p f = ψ 0 ψ 0 \psi_{0} ψ 0
0 < ∫ 0 1 u 1 ( x ) − 1 / p ψ 0 2 ( x ) d x = ∫ 0 1 ψ 0 ( x ) ( K ( u 1 − 1 / p f ) ) ( x ) d x = 0 , 0<\int_{0}^{1}u_{1}(x)^{-1/p}\psi_{0}^{2}(x)\,\mathrm{d}x
=\int_{0}^{1}\psi_{0}(x)\left(K(u_{1}^{-1/p}f)\right)(x)\,\mathrm{d}x=0, 0 < ∫ 0 1 u 1 ( x ) − 1/ p ψ 0 2 ( x ) d x = ∫ 0 1 ψ 0 ( x ) ( K ( u 1 − 1/ p f ) ) ( x ) d x = 0 , where in the last equality we again used duality (9) between K K K P \mathcal{P} P P ψ 0 = 0 \mathcal{P}\psi_{0}=0 P ψ 0 = 0 K ~ p \widetilde{K}_{p} K p [2 Lemma 3.1] we get that the spectrum of K ~ p \widetilde{K}_{p} K p K K K K ~ p \widetilde{K}_{p} K p
(iii). If p = 2 p=2 p = 2 K ∗ = P K^{*}=\mathcal{P} K ∗ = P
K ~ 2 ∗ = 1 u 1 P u 1 , K ~ 2 ∗ K ~ 2 = Id . \widetilde{K}_{2}^{*}=\frac{1}{\sqrt{u_{1}}}\mathcal{P}\sqrt{u_{1}},\qquad\widetilde{K}_{2}^{*}\widetilde{K}_{2}=\operatorname{Id}. K 2 ∗ = u 1 1 P u 1 , K 2 ∗ K 2 = Id . The last identity can be written as u 1 − 1 / 2 P u 1 K u 1 − 1 / 2 = Id u_{1}^{-1/2}\mathcal{P}u_{1}Ku_{1}^{-1/2}=\operatorname{Id} u 1 − 1/2 P u 1 K u 1 − 1/2 = Id L 2 L^{2} L 2 P u 1 K f = u 1 f \mathcal{P}u_{1}Kf=u_{1}f P u 1 K f = u 1 f f f f f ∈ L p f\in L^{p} f ∈ L p 1 ≤ p < ∞ 1\leq p<\infty 1 ≤ p < ∞ L ∞ ( [ 0 , 1 ] ) L^{\infty}([0,1]) L ∞ ([ 0 , 1 ]) L 2 ( [ 0 , 1 ] ) L^{2}([0,1]) L 2 ([ 0 , 1 ]) 1 p + 1 p ′ = 1 \frac{1}{p}+\frac{1}{p'}=1 p 1 + p ′ 1 = 1
u 1 − 1 / p ′ P u 1 1 / p ′ u 1 1 / p K u 1 − 1 / p = u 1 − 1 / p ′ P u 1 1 / p ′ K ~ p = Id on L p , 1 ≤ p ≤ ∞ . (33) u_{1}^{-1/p'}\mathcal{P}u_{1}^{1/p'}u_{1}^{1/p}Ku_{1}^{-1/p}=u_{1}^{-1/p'}\mathcal{P}u_{1}^{1/p'}\widetilde{K}_{p}=\operatorname{Id}\quad\text{on }L^{p},\quad1\leq p\leq\infty. \tag*{(33)} u 1 − 1/ p ′ P u 1 1/ p ′ u 1 1/ p K u 1 − 1/ p = u 1 − 1/ p ′ P u 1 1/ p ′ K p = Id on L p , 1 ≤ p ≤ ∞. (33) If ∣ z ∣ < 1 |z| < 1 ∣ z ∣ < 1 z K ~ p z\widetilde{K}_p z K p ( Id − z K ~ p ) − 1 (\operatorname{Id}-z\widetilde{K}_p)^{-1} ( Id − z K p ) − 1 L p L^p L p
ψ z = ∑ j = 0 ∞ z j u 1 1 / p ′ K ~ p j ( 1 u 1 1 / p ′ ψ 0 ) = ψ 0 + ∑ j = 1 ∞ z j u 1 1 / p ′ K ~ p j ( 1 u 1 1 / p ′ ψ 0 ) . \psi_z=\sum_{j=0}^{\infty}z^j u_1^{1/p'}\widetilde{K}_p^j\left(\frac{1}{u_1^{1/p'}}\psi_0\right)=\psi_0+\sum_{j=1}^{\infty}z^j u_1^{1/p'}\widetilde{K}_p^j\left(\frac{1}{u_1^{1/p'}}\psi_0\right). ψ z = j = 0 ∑ ∞ z j u 1 1/ p ′ K p j ( u 1 1/ p ′ 1 ψ 0 ) = ψ 0 + j = 1 ∑ ∞ z j u 1 1/ p ′ K p j ( u 1 1/ p ′ 1 ψ 0 ) . Moreover:
P ψ z = P ψ 0 + z ∑ j = 1 ∞ z j − 1 u 1 1 / p ′ ( u 1 − 1 / p ′ P u 1 1 / p ′ ) ( K ~ p j ( 1 u 1 1 / p ′ ψ 0 ) ) = 0 + z ∑ j = 1 ∞ z j − 1 u 1 1 / p ′ K ~ p j − 1 ( 1 u 1 1 / p ′ ψ 0 ) = z ψ z , \begin{aligned}
\mathcal{P}\psi_z=\mathcal{P}\psi_0+z\sum_{j=1}^{\infty}z^{j-1}u_1^{1/p'}\left(u_1^{-1/p'}\mathcal{P}u_1^{1/p'}\right)\left(\widetilde{K}_p^j\left(\frac{1}{u_1^{1/p'}}\psi_0\right)\right) \\
&=0+z\sum_{j=1}^{\infty}z^{j-1}u_1^{1/p'}\widetilde{K}_p^{j-1}\left(\frac{1}{u_1^{1/p'}}\psi_0\right)=z\psi_z,
\end{aligned} P ψ z = P ψ 0 + z j = 1 ∑ ∞ z j − 1 u 1 1/ p ′ ( u 1 − 1/ p ′ P u 1 1/ p ′ ) ( K p j ( u 1 1/ p ′ 1 ψ 0 ) ) = 0 + z j = 1 ∑ ∞ z j − 1 u 1 1/ p ′ K p j − 1 ( u 1 1/ p ′ 1 ψ 0 ) = z ψ z , where in the second equality we have used (33) . This shows that every z ∈ C z\in\mathbb{C} z ∈ C ∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1 P \mathcal{P} P L p L^p L p
Proof of Theorem 2.7
Let us first deal with the spectrum in ∣ z ∣ < β − 1 |z|<\beta^{-1} ∣ z ∣ < β − 1 z z z B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) K ~ 1 \widetilde{K}_1 K 1 p = 1 p=1 p = 1 j ≥ 1 j\geq1 j ≥ 1 f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ])
( K ~ 1 j f ) ( x ) = u 1 ( x ) u 1 ( T β ( x ) ) ( K ~ 1 j − 1 f ) ( T β ( x ) ) = ⋯ = u 1 ( x ) u 1 ( T β j ( x ) ) f ( T β j ( x ) ) . (\widetilde{K}_1^j f)(x)=\frac{u_1(x)}{u_1(T_\beta(x))}(\widetilde{K}_1^{j-1}f)(T_\beta(x))=\cdots=\frac{u_1(x)}{u_1(T_\beta^j(x))}f(T_\beta^j(x)). ( K 1 j f ) ( x ) = u 1 ( T β ( x )) u 1 ( x ) ( K 1 j − 1 f ) ( T β ( x )) = ⋯ = u 1 ( T β j ( x )) u 1 ( x ) f ( T β j ( x )) . Let ψ 0 \psi_0 ψ 0
( K ~ 1 j ψ 0 ) ( x ) = u 1 ( x ) u 1 ( T β j ( x ) ) ( g ( β T β j ( x ) ) χ [ 0 , b 1 / β ] ( T β j ( x ) ) − g ( β T β j ( x ) − a 0 ) χ [ a 0 / β , 1 ] ( T β j ( x ) ) ) . (\widetilde{K}_1^j\psi_0)(x)=\frac{u_1(x)}{u_1(T_\beta^j(x))}\left(g(\beta T_\beta^j(x))\chi_{[0,b_1/\beta]}(T_\beta^j(x))-g(\beta T_\beta^j(x)-a_0)\chi_{[a_0/\beta,1]}(T_\beta^j(x))\right). ( K 1 j ψ 0 ) ( x ) = u 1 ( T β j ( x )) u 1 ( x ) ( g ( β T β j ( x )) χ [ 0 , b 1 / β ] ( T β j ( x )) − g ( β T β j ( x ) − a 0 ) χ [ a 0 / β , 1 ] ( T β j ( x )) ) . We now show that the total variation of this function is of order β j \beta^j β j L ∞ L^\infty L ∞
When x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ] T β T_\beta T β [ 0 , 1 ) [0,1) [ 0 , 1 ) [ 0 , b 1 ] [0,b_1] [ 0 , b 1 ] T β 2 T_\beta^2 T β 2 [ 0 , 1 ) [0,1) [ 0 , 1 ) [ 0 , b 1 ] [0,b_1] [ 0 , b 1 ] [ 0 , b 2 ] [0,b_2] [ 0 , b 2 ] T β j T_\beta^j T β j [ 0 , 1 ) [0,1) [ 0 , 1 ) [ 0 , b k ] [0,b_k] [ 0 , b k ] 1 ≤ k ≤ n − 1 1\leq k\leq n-1 1 ≤ k ≤ n − 1 T β j T_\beta^j T β j j ≥ 1 j\geq1 j ≥ 1 c = min { b 1 , … , b n − 1 } ∈ ( 0 , 1 ) c=\min\{b_1,\ldots,b_{n-1}\}\in(0,1) c = min { b 1 , … , b n − 1 } ∈ ( 0 , 1 ) T β j T_\beta^j T β j β j \beta^j β j T β j T_\beta^j T β j c / β j c/\beta^j c / β j N j N_j N j T β j T_\beta^j T β j [ 0 , 1 ] [0,1] [ 0 , 1 ] N j c / β j ≤ 1 N_jc/\beta^j\leq1 N j c / β j ≤ 1
N j ≤ c − 1 β j . N_j\leq c^{-1}\beta^j. N j ≤ c − 1 β j . Thus the total variation of each factor grows, up to a constant, as β j \beta^j β j ∣ z ∣ < β − 1 |z|<\beta^{-1} ∣ z ∣ < β − 1 ψ z = ∑ j = 0 ∞ z j K ~ 1 j ( ψ 0 ) \psi_z=\sum_{j=0}^{\infty}z^j\widetilde{K}_1^j(\psi_0) ψ z = ∑ j = 0 ∞ z j K 1 j ( ψ 0 ) B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) P ψ z = z ψ z \mathcal{P}\psi_z=z\psi_z P ψ z = z ψ z g g g
We now investigate the domain ∣ z ∣ > β − 1 |z|>\beta^{-1} ∣ z ∣ > β − 1
Lemma 4.2. Let f ∈ B V ( [ 0 , 1 ] ) f \in BV([0,1]) f ∈ B V ([ 0 , 1 ]) g 1 = f ( 1 ) β − 1 χ [ 0 , b 1 ] g_1=f(1)\beta^{-1}\chi_{[0,b_1]} g 1 = f ( 1 ) β − 1 χ [ 0 , b 1 ] b 1 b_1 b 1 (10) . Then
Var 0 1 ( P f − g 1 ) ≤ β − 1 Var 0 1 ( f ) . (34) \operatorname{Var}_0^1(\mathcal{P}f-g_1)\leq\beta^{-1}\operatorname{Var}_0^1(f). \tag*{(34)} Var 0 1 ( P f − g 1 ) ≤ β − 1 Var 0 1 ( f ) . (34) The proof of this lemma is in Appendix § 8 .§ 1 .
Lemma 4.3. Let ∣ λ ∗ ∣ < β |\lambda_*|<\beta ∣ λ ∗ ∣ < β p ( z ) p(z) p ( z ) (3) . There exists C > 0 C>0 C > 0 ∣ λ ∗ ∣ < β |\lambda_*|<\beta ∣ λ ∗ ∣ < β r ≥ 1 r\geq1 r ≥ 1 f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ]) ∫ 0 1 f ( x ) d x = 0 \int_0^1 f(x)\,\mathrm{d}x=0 ∫ 0 1 f ( x ) d x = 0
∥ P r f ∥ B V ≤ C Var 0 1 ( f ) ⋅ { β − r if ∣ λ ∗ ∣ < 1 , r n β − r if ∣ λ ∗ ∣ = 1 , r n − 1 ( ∣ λ ∗ ∣ / β ) r if 1 < ∣ λ ∗ ∣ < β . (35) \|\mathcal{P}^r f\|_{BV}\leq C\operatorname{Var}_0^1(f)\cdot
\begin{cases}
\beta^{-r} & \text{if }|\lambda_*|<1,\\
r^n\beta^{-r} & \text{if }|\lambda_*|=1,\\
r^{n-1}(|\lambda_*|/\beta)^r & \text{if }1<|\lambda_*|<\beta.
\end{cases} \tag*{(35)} ∥ P r f ∥ B V ≤ C Var 0 1 ( f ) ⋅ ⎩ ⎨ ⎧ β − r r n β − r r n − 1 ( ∣ λ ∗ ∣/ β ) r if ∣ λ ∗ ∣ < 1 , if ∣ λ ∗ ∣ = 1 , if 1 < ∣ λ ∗ ∣ < β . (35) This lemma is proved in Appendix § 8 .§ 2 . The last technical result we need is the following:
Lemma 4.4. There exists C > 0 C>0 C > 0 f ∈ B V ( [ 0 , 1 ] ) f\in BV([0,1]) f ∈ B V ([ 0 , 1 ]) r ≥ 1 r\geq1 r ≥ 1 h r ∈ Span { χ [ 0 , b j ] : 0 ≤ j ≤ n − 1 } h_r\in\operatorname{Span}\{\chi_{[0,b_j]}:0\leq j\leq n-1\} h r ∈ Span { χ [ 0 , b j ] : 0 ≤ j ≤ n − 1 } b j b_j b j (10) such that
Var 0 1 ( P r f − h r ) ≤ C ∥ f ∥ B V β − r . \operatorname{Var}_0^1(\mathcal{P}^r f-h_r)\leq C\|f\|_{BV}\beta^{-r}. Var 0 1 ( P r f − h r ) ≤ C ∥ f ∥ B V β − r . The proof of this lemma is given in Appendix § 8 .§ 3 . Now we are ready to prove the theorem. Let us fix ε > 0 \varepsilon>0 ε > 0 z z z ∣ z ∣ ≥ β − 1 + ε |z|\geq\beta^{-1}+\varepsilon ∣ z ∣ ≥ β − 1 + ε β − 1 P n \beta^{-1}P_n β − 1 P n P n P_n P n β z \beta z β z r 0 > 1 r_0>1 r 0 > 1 z z z z r 0 z^{r_0} z r 0 P r 0 \mathcal{P}^{r_0} P r 0 z z z P \mathcal{P} P
We remind that Π \Pi Π L 2 ( [ 0 , 1 ] ) L^2([0,1]) L 2 ([ 0 , 1 ]) χ [ 0 , b j ] \chi_{[0,b_j]} χ [ 0 , b j ] 0 ≤ j ≤ n − 1 0\leq j\leq n-1 0 ≤ j ≤ n − 1 (18) . Because these functions are all characteristic functions (thus in B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) Π \Pi Π B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) § 5 that P \mathcal{P} P Π \Pi Π
P r − z r = [ ( β − 1 P n ) r − z r Π P r ( 1 − Π ) 0 ( 1 − Π ) P r ( 1 − Π ) − z r ( 1 − Π ) ] , r ≥ 1. (36) \mathcal{P}^r-z^r=
\begin{bmatrix}
(\beta^{-1}P_n)^r-z^r & \Pi\mathcal{P}^r(1-\Pi)\\
0 & (1-\Pi)\mathcal{P}^r(1-\Pi)-z^r(1-\Pi)
\end{bmatrix},
\qquad r\geq1. \tag*{(36)} P r − z r = [ ( β − 1 P n ) r − z r 0 Π P r ( 1 − Π ) ( 1 − Π ) P r ( 1 − Π ) − z r ( 1 − Π ) ] , r ≥ 1. (36) On the range of 1 − Π 1-\Pi 1 − Π χ [ 0 , 1 ] \chi_{[0,1]} χ [ 0 , 1 ] Π \Pi Π P \mathcal{P} P P r ( 1 − Π ) \mathcal{P}^r(1-\Pi) P r ( 1 − Π ) ∫ 0 1 f ( x ) d x = 0 \int_0^1 f(x)\,\mathrm{d}x=0 ∫ 0 1 f ( x ) d x = 0 Var 0 1 ( f ) ≤ ∥ f ∥ B V ≤ 2 Var 0 1 ( f ) \operatorname{Var}_0^1(f)\leq\|f\|_{BV}\leq2\operatorname{Var}_0^1(f) Var 0 1 ( f ) ≤ ∥ f ∥ B V ≤ 2 Var 0 1 ( f ) Lemma 4.4 implies that
∥ ( 1 − Π ) P r ( 1 − Π ) ∥ B V ≤ C β − r , ∀ r ≥ 1. (37) \|(1-\Pi)\mathcal{P}^r(1-\Pi)\|_{BV}\leq C\beta^{-r},\qquad\forall r\geq1. \tag*{(37)} ∥ ( 1 − Π ) P r ( 1 − Π ) ∥ B V ≤ C β − r , ∀ r ≥ 1. (37) Now there exists some r 0 > 1 r_0>1 r 0 > 1
∣ z r 0 ∣ = β − r 0 ( 1 + ∣ β z ∣ − 1 ) r 0 ≥ β − r 0 ( 1 + β ε ) r 0 > 2 C β − r 0 |z^{r_0}|=\beta^{-r_0}(1+|\beta z|-1)^{r_0}\geq\beta^{-r_0}(1+\beta\varepsilon)^{r_0}>2C\beta^{-r_0} ∣ z r 0 ∣ = β − r 0 ( 1 + ∣ β z ∣ − 1 ) r 0 ≥ β − r 0 ( 1 + β ε ) r 0 > 2 C β − r 0 with the same C C C (37) and
A ( z ) : = ( 1 − Π ) P r 0 ( 1 − Π ) − z r 0 ( 1 − Π ) A(z):=(1-\Pi)\mathcal{P}^{r_0}(1-\Pi)-z^{r_0}(1-\Pi) A ( z ) := ( 1 − Π ) P r 0 ( 1 − Π ) − z r 0 ( 1 − Π ) is invertible with
A ( z ) − 1 = − ∑ k ≥ 0 z − r 0 ( k + 1 ) ( ( 1 − Π ) P r 0 ( 1 − Π ) ) k (38) A(z)^{-1}=-\sum_{k\geq0}z^{-r_{0}(k+1)}\left((1-\Pi)\mathcal{P}^{r_{0}}(1-\Pi)\right)^{k} \tag*{(38)} A ( z ) − 1 = − k ≥ 0 ∑ z − r 0 ( k + 1 ) ( ( 1 − Π ) P r 0 ( 1 − Π ) ) k (38) in the Banach space ( 1 − Π ) B V ( [ 0 , 1 ] ) (1-\Pi)BV([0,1]) ( 1 − Π ) B V ([ 0 , 1 ])
( P r 0 − z r 0 ) − 1 = [ ( β − 1 P n ) r 0 − z r 0 ) − 1 − ( ( β − 1 P n ) r 0 − z r 0 ) − 1 Π P r 0 ( 1 − Π ) A ( z ) − 1 0 A ( z ) − 1 ] (39) (\mathcal{P}^{r_{0}}-z^{r_{0}})^{-1}=
\begin{bmatrix}
(\beta^{-1}\mathcal{P}_{n})^{r_{0}}-z^{r_{0}})^{-1} & -((\beta^{-1}\mathcal{P}_{n})^{r_{0}}-z^{r_{0}})^{-1}\Pi\mathcal{P}^{r_{0}}(1-\Pi)A(z)^{-1}\\
0 & A(z)^{-1}
\end{bmatrix} \tag*{(39)} ( P r 0 − z r 0 ) − 1 = [ ( β − 1 P n ) r 0 − z r 0 ) − 1 0 − (( β − 1 P n ) r 0 − z r 0 ) − 1 Π P r 0 ( 1 − Π ) A ( z ) − 1 A ( z ) − 1 ] (39) The polynomial ξ r 0 − z r 0 \xi^{r_{0}}-z^{r_{0}} ξ r 0 − z r 0 ( ξ − z ) ∑ k = 0 r 0 − 1 ξ k z r 0 − 1 − k (\xi-z)\sum_{k=0}^{r_{0}-1}\xi^{k}z^{r_{0}-1-k} ( ξ − z ) ∑ k = 0 r 0 − 1 ξ k z r 0 − 1 − k
( P − z ) − 1 = ( ∑ k = 0 r 0 − 1 z r 0 − 1 − k P k ) ( P r 0 − z r 0 ) − 1 = [ ( ( β − 1 P n ) − z ) − 1 − ( ( β − 1 P n ) − z ) − 1 Π P r 0 ( 1 − Π ) A ( z ) − 1 0 ( 1 − Π ) ∑ k = 0 r 0 − 1 z r 0 − 1 − k P k ( 1 − Π ) A ( z ) − 1 ] (40) \begin{aligned}
(\mathcal{P}-z)^{-1}=\left(\sum_{k=0}^{r_{0}-1}z^{r_{0}-1-k}\mathcal{P}^{k}\right)(\mathcal{P}^{r_{0}}-z^{r_{0}})^{-1} \\
&=\begin{bmatrix}
((\beta^{-1}\mathcal{P}_{n})-z)^{-1} & -((\beta^{-1}\mathcal{P}_{n})-z)^{-1}\Pi\mathcal{P}^{r_{0}}(1-\Pi)A(z)^{-1}\\
0 & (1-\Pi)\sum_{k=0}^{r_{0}-1}z^{r_{0}-1-k}\mathcal{P}^{k}(1-\Pi)A(z)^{-1}
\end{bmatrix} \tag*{(40)}
\end{aligned} ( P − z ) − 1 = ( k = 0 ∑ r 0 − 1 z r 0 − 1 − k P k ) ( P r 0 − z r 0 ) − 1 = [ (( β − 1 P n ) − z ) − 1 0 − (( β − 1 P n ) − z ) − 1 Π P r 0 ( 1 − Π ) A ( z ) − 1 ( 1 − Π ) ∑ k = 0 r 0 − 1 z r 0 − 1 − k P k ( 1 − Π ) A ( z ) − 1 ] (40) which shows that z z z P \mathcal{P} P P \mathcal{P} P B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) ( P − z ) − 1 (\mathcal{P}-z)^{-1} ( P − z ) − 1 ∣ z ∣ > β − 1 |z|>\beta^{-1} ∣ z ∣ > β − 1 λ / β \lambda/\beta λ / β λ \lambda λ 1 < ∣ λ ∣ ≤ β 1<|\lambda|\leq\beta 1 < ∣ λ ∣ ≤ β
Proof of Corollary 2.8
The key starting point is formula (40) . Choose ε > 0 \varepsilon>0 ε > 0 β − 1 P n \beta^{-1}\mathcal{P}_{n} β − 1 P n 1 ≥ ∣ ζ j ∣ > β − 1 1\geq|\zeta_j|>\beta^{-1} 1 ≥ ∣ ζ j ∣ > β − 1 ∣ z ∣ > β − 1 + 2 ε |z|>\beta^{-1}+2\varepsilon ∣ z ∣ > β − 1 + 2 ε A ( z ) − 1 A(z)^{-1} A ( z ) − 1 (38) is analytic on ∣ z ∣ > β − 1 + ε |z|>\beta^{-1}+\varepsilon ∣ z ∣ > β − 1 + ε r 0 r_{0} r 0 ε \varepsilon ε q j q_j q j β − 1 P n \beta^{-1}\mathcal{P}_{n} β − 1 P n 1 ≥ ∣ ζ j ∣ > β − 1 1\geq|\zeta_j|>\beta^{-1} 1 ≥ ∣ ζ j ∣ > β − 1 (40) we obtain:
Q j = Π q j Π − Π q j Π P r 0 A ( ζ j ) − 1 ( 1 − Π ) . (41) Q_j=\Pi q_j\Pi-\Pi q_j\Pi\mathcal{P}^{r_{0}}A(\zeta_j)^{-1}(1-\Pi). \tag*{(41)} Q j = Π q j Π − Π q j Π P r 0 A ( ζ j ) − 1 ( 1 − Π ) . (41) The norm estimate in (19) which needs an 0 < α < 1 0<\alpha<1 0 < α < 1 lim r → ∞ ∥ ( P ( 1 − Q ) ) r ∥ 1 / r \lim_{r\to\infty}\|\left(\mathcal{P}(1-Q)\right)^r\|^{1/r} lim r → ∞ ∥ ( P ( 1 − Q ) ) r ∥ 1/ r P ( 1 − Q ) \mathcal{P}(1-Q) P ( 1 − Q ) β − 1 \beta^{-1} β − 1 (19) .
Now let us prove (20) . We must have Q 1 f = u 1 C ( f ) Q_1f=u_1C(f) Q 1 f = u 1 C ( f ) C ( f ) C(f) C ( f ) (19) we have that P r f − u 1 C ( f ) \mathcal{P}^rf-u_1C(f) P r f − u 1 C ( f ) r → ∞ r\to\infty r → ∞ P \mathcal{P} P f f f C ( f ) = ∫ 0 1 f ( x ) d x C(f)=\int_0^1 f(x)\,\mathrm{d}x C ( f ) = ∫ 0 1 f ( x ) d x
Point (iii) is just a reformulation of (19) in the case of simple eigenvalues, and we can identify each U ^ j \widehat{U}_j U j (41) .
Finally, point (iv) gives a sharper norm estimate of the error without the extra α \alpha α (35) where we choose the function as f − u 1 ∫ 0 1 f ( x ) d x f-u_1\int_0^1f(x)\,\mathrm{d}x f − u 1 ∫ 0 1 f ( x ) d x
Remark 4.5. For completeness, let us give an example of a Parry polynomial of degree 4 with roots outside the closed unit disk. Let L ≥ 3 L\geq3 L ≥ 3 p ( z ) = z 4 − ( ( L + 1 ) z 3 + L z 2 + L ) p(z)=z^4-((L+1)z^3+Lz^2+L) p ( z ) = z 4 − (( L + 1 ) z 3 + L z 2 + L ) n = 4 n=4 n = 4 a 0 = L + 1 a_0=L+1 a 0 = L + 1 a 2 = 0 a_2=0 a 2 = 0 a 1 = a 3 = L a_1=a_3=L a 1 = a 3 = L a 0 a_0 a 0 (12) ). Let
f ( z ) = − p ( z ) z 3 = L + 1 + L z + L z 3 − z , z ≠ 0. f(z)=-\frac{p(z)}{z^3}=L+1+\frac{L}{z}+\frac{L}{z^3}-z,\qquad z\ne0. f ( z ) = − z 3 p ( z ) = L + 1 + z L + z 3 L − z , z = 0. The zeros of f f f p p p x ∈ R ∖ { 0 } x\in\mathbb{R}\setminus\{0\} x ∈ R ∖ { 0 } f ′ ( x ) = − L / x 2 − 3 L / x 4 − 1 < 0 f'(x)=-L/x^2-3L/x^4-1<0 f ′ ( x ) = − L / x 2 − 3 L / x 4 − 1 < 0 lim x → − ∞ f ( x ) = + ∞ \lim_{x\to-\infty}f(x)=+\infty lim x → − ∞ f ( x ) = + ∞ lim x ↗ 0 f ( x ) = − ∞ \lim_{x\nearrow0}f(x)=-\infty lim x ↗ 0 f ( x ) = − ∞
We have f ( − 1 ) = − L + 2 < 0 f(-1)=-L+2<0 f ( − 1 ) = − L + 2 < 0 z 2 < − 1 z_2<-1 z 2 < − 1 f f f p p p [11 ] we also know that ∣ z 2 ∣ < φ = 5 + 1 2 |z_2|<\varphi=\frac{\sqrt{5}+1}{2} ∣ z 2 ∣ < φ = 2 5 + 1
Proof of Theorem 2.9
Proof of (i). When ζ = 1 \zeta=1 ζ = 1 U ^ m ( F ) \widehat{U}_m(F) U m ( F ) F F F 1 / β < ∣ ζ ∣ < 1 1/\beta<|\zeta|<1 1/ β < ∣ ζ ∣ < 1 1 < ∣ λ ∣ < β 1<|\lambda|<\beta 1 < ∣ λ ∣ < β j m j_m j m k m k_m k m k m ∈ { 1 , 2 , … , n } k_m\in\{1,2,\ldots,n\} k m ∈ { 1 , 2 , … , n } j m ∈ { 0 , … , a k m − 1 − 1 } j_m\in\{0,\ldots,a_{k_m-1}-1\} j m ∈ { 0 , … , a k m − 1 − 1 }
∑ j m − 1 k m − 1 λ − k 1 − k 2 − ⋯ − k m − 1 ∑ k m = 1 n ∑ j m = 0 a k m − 1 − 1 λ − k m f ( j m − 1 k m − 1 + β − ∣ k m − 1 ∣ j m k m ) . \sum_{j_{m-1}}^{k_{m-1}}\lambda^{-k_1-k_2-\cdots-k_{m-1}}\sum_{k_m=1}^{n}\sum_{j_m=0}^{a_{k_m-1}-1}\lambda^{-k_m}f\left(\frac{j_{m-1}}{k_{m-1}}+\beta^{-|k_{m-1}|}\frac{j_m}{k_m}\right). j m − 1 ∑ k m − 1 λ − k 1 − k 2 − ⋯ − k m − 1 k m = 1 ∑ n j m = 0 ∑ a k m − 1 − 1 λ − k m f ( k m − 1 j m − 1 + β − ∣ k m − 1 ∣ k m j m ) . Here (23) plays a crucial role: the above sum over k m k_m k m j m j_m j m
f ( j m − 1 k m − 1 ) + ∑ k m = 1 n ∑ j m = 0 a k m − 1 − 1 λ − k m ( f ( j m − 1 k m − 1 + β − ∣ k m − 1 ∣ j m k m ) − f ( j m − 1 k m − 1 ) ) . f\left(\frac{j_{m-1}}{k_{m-1}}\right)+\sum_{k_m=1}^{n}\sum_{j_m=0}^{a_{k_m-1}-1}\lambda^{-k_m}\left(f\left(\frac{j_{m-1}}{k_{m-1}}+\beta^{-|k_{m-1}|}\frac{j_m}{k_m}\right)-f\left(\frac{j_{m-1}}{k_{m-1}}\right)\right). f ( k m − 1 j m − 1 ) + k m = 1 ∑ n j m = 0 ∑ a k m − 1 − 1 λ − k m ( f ( k m − 1 j m − 1 + β − ∣ k m − 1 ∣ k m j m ) − f ( k m − 1 j m − 1 ) ) . The key estimate is as follows:
∑ k m ∑ j m ∣ λ ∣ − k m ∣ f ( j m − 1 k m − 1 + β − ∣ k m − 1 ∣ j m k m ) − f ( j m − 1 k m − 1 ) ∣ ≤ ∑ k = 1 n a k − 1 ∣ λ ∣ k Var 0 1 ( f ) . \sum_{k_m}\sum_{j_m}|\lambda|^{-k_m}\left|f\left(\frac{j_{m-1}}{k_{m-1}}+\beta^{-|k_{m-1}|}\frac{j_m}{k_m}\right)-f\left(\frac{j_{m-1}}{k_{m-1}}\right)\right|
\leq\sum_{k=1}^{n}\frac{a_{k-1}}{|\lambda|^k}\operatorname{Var}_{0}^{1}\left(f\right). k m ∑ j m ∑ ∣ λ ∣ − k m f ( k m − 1 j m − 1 + β − ∣ k m − 1 ∣ k m j m ) − f ( k m − 1 j m − 1 ) ≤ k = 1 ∑ n ∣ λ ∣ k a k − 1 Var 0 1 ( f ) . Then when we sum over the partition with m − 1 m-1 m − 1
∣ U ^ m ( f ) − U ^ m − 1 ( f ) ∣ ≤ ∣ λ ∣ − m + 1 ∑ k = 1 n a k − 1 ∣ λ ∣ k Var 0 1 ( f ) \left|\widehat{U}_m(f)-\widehat{U}_{m-1}(f)\right|\leq|\lambda|^{-m+1}\sum_{k=1}^{n}\frac{a_{k-1}}{|\lambda|^k}\operatorname{Var}_{0}^{1}(f) U m ( f ) − U m − 1 ( f ) ≤ ∣ λ ∣ − m + 1 k = 1 ∑ n ∣ λ ∣ k a k − 1 Var 0 1 ( f ) which defines a convergent series because ∣ λ ∣ > 1 |\lambda|>1 ∣ λ ∣ > 1 U ^ ( f ) \widehat{U}(f) U ( f )
Now let us prove (24) . We will show that for all m ≥ n + 1 m\geq n+1 m ≥ n + 1
U ^ m ( P f ) = ζ U ^ m − n ( f ) + O ( ∣ λ ∣ − m ) . (42) \widehat{U}_m(\mathcal{P}f)=\zeta\widehat{U}_{m-n}(f)+O(|\lambda|^{-m}). \tag*{(42)} U m ( P f ) = ζ U m − n ( f ) + O ( ∣ λ ∣ − m ) . (42) From the definition of P \mathcal{P} P
U ^ m ( P f ) = β − 1 ∑ s = 0 a 0 − 1 ∑ j m k m λ − ∣ k m ∣ f ( s / β + β − 1 j m k m ) + β − 1 ∑ j m k m λ − ∣ k m ∣ f ( a 0 / β + β − 1 j m k m ) . \widehat{U}_m(\mathcal{P}f)=\beta^{-1}\sum_{s=0}^{a_0-1}\sum_{j_m}^{k_m}\lambda^{-|k_m|}f\left(s/\beta+\beta^{-1}\frac{j_m}{k_m}\right)+\beta^{-1}\sum_{j_m}^{k_m}\lambda^{-|k_m|}f\left(a_0/\beta+\beta^{-1}\frac{j_m}{k_m}\right). U m ( P f ) = β − 1 s = 0 ∑ a 0 − 1 j m ∑ k m λ − ∣ k m ∣ f ( s / β + β − 1 k m j m ) + β − 1 j m ∑ k m λ − ∣ k m ∣ f ( a 0 / β + β − 1 k m j m ) . The first term contains
s / β + β − 1 j 1 ⋯ j m k 1 ⋯ k m = s j 1 ⋯ j m k 1 ⋯ k m . s/\beta+\beta^{-1}\frac{j_1\cdots j_m}{k_1\cdots k_m}=\frac{s j_1\cdots j_m}{k_1\cdots k_m}. s / β + β − 1 k 1 ⋯ k m j 1 ⋯ j m = k 1 ⋯ k m s j 1 ⋯ j m . which will eventually run over all partition elements with “m + 1 m+1 m + 1 k k k j j j a 0 − 1 a_0-1 a 0 − 1
The second term contains
a 0 β + β − 1 t k m j m \frac{a_0}{\beta}+\beta^{-1}t^{j_m}_{k_m} β a 0 + β − 1 t k m j m and we will show that summing over t k m j m t^{j_m}_{k_m} t k m j m k 1 ≥ 2 k_1 \ge2 k 1 ≥ 2
In order to make the main idea of the proof more transparent, we first consider a “simpler case”:
Case 1: a 0 > a j a_0>a_j a 0 > a j j ≥ 1 j\ge1 j ≥ 1 We have
b 1 = a 1 β + a 2 β 2 + ⋯ + a n − 1 β n − 1 = t 1 × ( n − 1 ) a 1 ⋯ a n − 1 b_1=\frac{a_1}{\beta}+\frac{a_2}{\beta^2}+\cdots+\frac{a_{n-1}}{\beta^{n-1}}=t^{a_1\cdots a_{n-1}}_{1\times(n-1)} b 1 = β a 1 + β 2 a 2 + ⋯ + β n − 1 a n − 1 = t 1 × ( n − 1 ) a 1 ⋯ a n − 1 where the notation 1 × ( n − 1 ) 1\times(n-1) 1 × ( n − 1 ) k 1 = ⋯ = k n − 1 = 1 k_1=\cdots=k_{n-1}=1 k 1 = ⋯ = k n − 1 = 1 f ( a 0 / β + β − 1 t ) = 0 f(a_0/\beta+\beta^{-1}t)=0 f ( a 0 / β + β − 1 t ) = 0 t > b 1 t>b_1 t > b 1 t k m j m < b 1 t^{j_m}_{k_m}<b_1 t k m j m < b 1
k 1 = 1 , j 1 < a 1 k_1=1,\quad j_1<a_1 k 1 = 1 , j 1 < a 1
k 1 = 1 , j 1 = a 1 ; k 2 = 1 , j 2 < a 2 k_1=1,\quad j_1=a_1;\quad k_2=1,\quad j_2<a_2 k 1 = 1 , j 1 = a 1 ; k 2 = 1 , j 2 < a 2
k 1 = 1 , j 1 = a 1 ; k 2 = 1 , j 2 = a 2 ; k 3 = 1 , j 3 < a 3 k_1=1,\quad j_1=a_1;\quad k_2=1,\quad j_2=a_2;\quad k_3=1,\quad j_3<a_3 k 1 = 1 , j 1 = a 1 ; k 2 = 1 , j 2 = a 2 ; k 3 = 1 , j 3 < a 3
⋮ \vdots ⋮
In the first case where k 1 = 1 k_1=1 k 1 = 1 j 1 < a 1 j_1<a_1 j 1 < a 1
a 0 β + β − 1 ( t 1 j 1 + β − 1 t k 2 ⋯ k m j 2 ⋯ j m ) = t 2 j 1 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m = t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m . \frac{a_0}{\beta}+\beta^{-1}\left(t^{j_1}_{1}+\beta^{-1}t^{j_2\cdots j_m}_{k_2\cdots k_m}\right)=t^{j_1}_{2}+\beta^{-2}t^{j_2\cdots j_m}_{k_2\cdots k_m}=t^{j_1j_2\cdots j_m}_{2k_2\cdots k_m}. β a 0 + β − 1 ( t 1 j 1 + β − 1 t k 2 ⋯ k m j 2 ⋯ j m ) = t 2 j 1 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m = t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m . In the second case where k 1 = k 2 = 1 k_1=k_2=1 k 1 = k 2 = 1 j 1 = a 1 < a 0 j_1=a_1<a_0 j 1 = a 1 < a 0 j 2 < a 2 j_2<a_2 j 2 < a 2
a 0 β + β − 1 ( a 1 β + β − 1 t 1 k 3 ⋯ k m j 2 ⋯ j m ) = a 0 β + a 1 β 2 + j 2 β 3 + β − 3 t k 3 ⋯ k m j 2 ⋯ j m = t 3 k 2 ⋯ k m j 2 ⋯ j m . \frac{a_0}{\beta}+\beta^{-1}\left(\frac{a_1}{\beta}+\beta^{-1}t^{j_2\cdots j_m}_{1k_3\cdots k_m}\right)=\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\frac{j_2}{\beta^3}+\beta^{-3}t^{j_2\cdots j_m}_{k_3\cdots k_m}=t^{j_2\cdots j_m}_{3k_2\cdots k_m}. β a 0 + β − 1 ( β a 1 + β − 1 t 1 k 3 ⋯ k m j 2 ⋯ j m ) = β a 0 + β 2 a 1 + β 3 j 2 + β − 3 t k 3 ⋯ k m j 2 ⋯ j m = t 3 k 2 ⋯ k m j 2 ⋯ j m . Now we show that “no points are missing”. Let p > 2 p>2 p > 2
t 2 k 2 k 3 ⋯ k p j 1 j 2 ⋯ j p = a 0 β + j 1 β 2 + β − 2 t k 2 ⋯ k p j 2 ⋯ j p = a 0 β + β − 1 t 1 k 2 ⋯ k p j 1 j 2 ⋯ j p t^{j_1j_2\cdots j_p}_{2k_2k_3\cdots k_p}=\frac{a_0}{\beta}+\frac{j_1}{\beta^2}+\beta^{-2}t^{j_2\cdots j_p}_{k_2\cdots k_p}=\frac{a_0}{\beta}+\beta^{-1}t^{j_1j_2\cdots j_p}_{1k_2\cdots k_p} t 2 k 2 k 3 ⋯ k p j 1 j 2 ⋯ j p = β a 0 + β 2 j 1 + β − 2 t k 2 ⋯ k p j 2 ⋯ j p = β a 0 + β − 1 t 1 k 2 ⋯ k p j 1 j 2 ⋯ j p and if i ≥ 3 i\ge3 i ≥ 3
t i k 2 k 3 ⋯ k p j 1 j 2 ⋯ j p = a 0 β + a 1 β 2 + ⋯ + a i − 2 β i − 1 + j 1 β i + β − i t k 2 ⋯ k p j 2 ⋯ j p = a 0 β + β − 1 t 1 × ( i − 1 ) , k 2 ⋯ k p a 1 ⋯ a i − 2 j 1 ⋯ j p . (43) \begin{aligned}
t^{j_1j_2\cdots j_p}_{ik_2k_3\cdots k_p}
&=\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\frac{j_1}{\beta^i}+\beta^{-i}t^{j_2\cdots j_p}_{k_2\cdots k_p}\\
&=\frac{a_0}{\beta}+\beta^{-1}t^{a_1\cdots a_{i-2}j_1\cdots j_p}_{1\times(i-1),k_2\cdots k_p}. \tag*{(43)}
\end{aligned} t i k 2 k 3 ⋯ k p j 1 j 2 ⋯ j p = β a 0 + β 2 a 1 + ⋯ + β i − 1 a i − 2 + β i j 1 + β − i t k 2 ⋯ k p j 2 ⋯ j p = β a 0 + β − 1 t 1 × ( i − 1 ) , k 2 ⋯ k p a 1 ⋯ a i − 2 j 1 ⋯ j p . (43) where the notation 1 × ( i − 1 ) 1\times(i-1) 1 × ( i − 1 ) i − 1 i-1 i − 1 k k k (43) we see that given m m m i i i a 0 / β + β − 1 t 1 × ( i − 1 ) , k 2 ⋯ k m a 1 ⋯ a i − 2 j 1 ⋯ j m a_0/\beta+\beta^{-1}t^{a_1\cdots a_{i-2}j_1\cdots j_m}_{1\times(i-1),k_2\cdots k_m} a 0 / β + β − 1 t 1 × ( i − 1 ) , k 2 ⋯ k m a 1 ⋯ a i − 2 j 1 ⋯ j m t i k 2 k 3 ⋯ k m − i + 2 j 1 j 2 ⋯ j m − i + 2 t^{j_1j_2\cdots j_{m-i+2}}_{ik_2k_3\cdots k_{m-i+2}} t i k 2 k 3 ⋯ k m − i + 2 j 1 j 2 ⋯ j m − i + 2 k k k (43) is exactly one unit larger than the sum of k k k U ^ m ( P f ) \widehat{U}_m(\mathcal{P}f) U m ( P f ) U ^ m ( f ) \widehat{U}_m(f) U m ( f ) β / λ \beta/\lambda β / λ
Thus
β λ U ^ m ( P f ) = ∑ t 1 k 1 ⋯ k m j 1 ⋯ j m λ − 1 − ∣ k m ∣ f ( t 1 k 1 ⋯ k m j 1 ⋯ j m ) + ∑ t 2 k 2 ⋯ k m j 1 ⋯ j m λ − 2 − k 2 − ⋯ − k m f ( t 2 k 2 ⋯ k m j 1 ⋯ j m ) + ∑ t 3 k 2 ⋯ k m − 1 j 1 ⋯ j m − 1 λ − 3 − k 2 − ⋯ − k m − 1 f ( t 3 k 2 ⋯ k m − 1 j 1 ⋯ j m − 1 ) + ⋯ + ∑ t n k 2 ⋯ k m − n + 2 j 1 ⋯ j m − n + 2 λ − n − k 2 − ⋯ − k m − n + 2 f ( t n k 2 ⋯ k m − n + 2 j 1 ⋯ j m − n + 2 ) . (44) \frac{\beta}{\lambda}\widehat{U}_m(\mathcal{P}f)=\sum_{t^{j_1\cdots j_m}_{1k_1\cdots k_m}}\lambda^{-1-\lvert k_m\rvert}f\left(t^{j_1\cdots j_m}_{1k_1\cdots k_m}\right)+\sum_{t^{j_1\cdots j_m}_{2k_2\cdots k_m}}\lambda^{-2-k_2-\cdots-k_m}f\left(t^{j_1\cdots j_m}_{2k_2\cdots k_m}\right)
+\sum_{t^{j_1\cdots j_{m-1}}_{3k_2\cdots k_{m-1}}}\lambda^{-3-k_2-\cdots-k_{m-1}}f\left(t^{j_1\cdots j_{m-1}}_{3k_2\cdots k_{m-1}}\right)+\cdots+\sum_{t^{j_1\cdots j_{m-n+2}}_{nk_2\cdots k_{m-n+2}}}\lambda^{-n-k_2-\cdots-k_{m-n+2}}f\left(t^{j_1\cdots j_{m-n+2}}_{nk_2\cdots k_{m-n+2}}\right). \tag*{(44)} λ β U m ( P f ) = t 1 k 1 ⋯ k m j 1 ⋯ j m ∑ λ − 1 − ∣ k m ∣ f ( t 1 k 1 ⋯ k m j 1 ⋯ j m ) + t 2 k 2 ⋯ k m j 1 ⋯ j m ∑ λ − 2 − k 2 − ⋯ − k m f ( t 2 k 2 ⋯ k m j 1 ⋯ j m ) + t 3 k 2 ⋯ k m − 1 j 1 ⋯ j m − 1 ∑ λ − 3 − k 2 − ⋯ − k m − 1 f ( t 3 k 2 ⋯ k m − 1 j 1 ⋯ j m − 1 ) + ⋯ + t n k 2 ⋯ k m − n + 2 j 1 ⋯ j m − n + 2 ∑ λ − n − k 2 − ⋯ − k m − n + 2 f ( t n k 2 ⋯ k m − n + 2 j 1 ⋯ j m − n + 2 ) . (44) Each sum on the right hand side is performed over a different number of layers varying from m − n + 1 m-n+1 m − n + 1 m + 1 m+1 m + 1 f f f O ( ∣ λ ∣ − m ) O(\lvert\lambda\rvert^{-m}) O (∣ λ ∣ − m ) (42) . Then take m → ∞ m\to\infty m → ∞ (24) when a 0 > a j a_0>a_j a 0 > a j 1 ≤ j ≤ n − 1 1\leq j\leq n-1 1 ≤ j ≤ n − 1
Case 2: general Parry coefficients. In Appendix B.2 we will prove the following result:
Lemma 4.6. Any number t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m t^{j_1j_2\cdots j_m}_{k_1k_2\cdots k_m} t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m with k 1 ≥ 2 k_1\geq2 k 1 ≥ 2 can be written as
t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = a 0 β + β − 1 t ~ with t ~ = t k ~ 1 ⋯ k ~ p j ~ 1 ⋯ j ~ p , t^{j_1j_2\cdots j_m}_{k_1k_2\cdots k_m}=\frac{a_0}{\beta}+\beta^{-1}\widetilde{t}
\quad\text{with}\quad
\widetilde{t}=t^{\widetilde{j}_1\cdots\widetilde{j}_p}_{\widetilde{k}_1\cdots\widetilde{k}_p}, t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = β a 0 + β − 1 t with t = t k 1 ⋯ k p j 1 ⋯ j p , where
k 1 + k 2 + ⋯ + k m = 1 + k ~ 1 + ⋯ + k ~ p , m ≤ p ≤ n − 1 + m . k_1+k_2+\cdots+k_m=1+\widetilde{k}_1+\cdots+\widetilde{k}_p,\qquad m\leq p\leq n-1+m. k 1 + k 2 + ⋯ + k m = 1 + k 1 + ⋯ + k p , m ≤ p ≤ n − 1 + m . The other way around, let t k 1 ⋯ k m j 1 ⋯ j m < b 1 = a 1 / β + ⋯ + a n − 1 / β n − 1 t^{j_1\cdots j_m}_{k_1\cdots k_m}<b_1=a_1/\beta+\cdots+a_{n-1}/\beta^{n-1} t k 1 ⋯ k m j 1 ⋯ j m < b 1 = a 1 / β + ⋯ + a n − 1 / β n − 1 Then there exists p ≥ 1 p\geq1 p ≥ 1 and t k ~ p j ~ p < 1 t^{\widetilde{j}_p}_{\widetilde{k}_p}<1 t k p j p < 1 with k ~ 1 ≥ 2 \widetilde{k}_1\geq2 k 1 ≥ 2 such that
t k ~ p j ~ p = a 0 β + β − 1 t k m j m . t^{\widetilde{j}_p}_{\widetilde{k}_p}=\frac{a_0}{\beta}+\beta^{-1}t^{j_m}_{k_m}. t k p j p = β a 0 + β − 1 t k m j m . This technical result replaces the identity (43) , which was the crucial ingredient in proving (44) . This time we cannot exactly say how many layers we loose or gain by replacing the summation over the indices k ~ \widetilde{k} k j ~ \widetilde{j} j k k k 2 , 3 , … , n 2,3,\ldots,n 2 , 3 , … , n (42) still holds, and this finishes the proof of (i).
Proof of (ii). Let us introduce a modified version of U ^ m ( F ) \widehat{U}_m(F) U m ( F )
U ~ m ( f ) = ∑ t k m j m λ − ∣ k m ∣ f ( t k m j m + β − ∣ k m ∣ ) . \widetilde{U}_m(f)=\sum_{t^{j_m}_{k_m}}\lambda^{-\lvert k_m\rvert}f\left(t^{j_m}_{k_m}+\beta^{-\lvert k_m\rvert}\right). U m ( f ) = t k m j m ∑ λ − ∣ k m ∣ f ( t k m j m + β − ∣ k m ∣ ) . It turns out that one can explicitly compute U ~ m ( f ) \widetilde{U}_m(f) U m ( f ) f = χ [ 0 , y ] f=\chi_{[0,y]} f = χ [ 0 , y ] 0 < y < 1 0<y<1 0 < y < 1 t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p t^{w_1w_2\cdots w_p}_{u_1u_2\cdots u_p} t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p
t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p ≤ y = ∑ r ≥ 1 y r − 1 β r < t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p + β − u 1 − ⋯ − u p , p ≥ 1. t^{w_1w_2\cdots w_p}_{u_1u_2\cdots u_p}\leq y=\sum_{r\geq1}\frac{y_{r-1}}{\beta^r}<t^{w_1w_2\cdots w_p}_{u_1u_2\cdots u_p}+\beta^{-u_1-\cdots-u_p},\qquad p\geq1. t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p ≤ y = r ≥ 1 ∑ β r y r − 1 < t u 1 u 2 ⋯ u p w 1 w 2 ⋯ w p + β − u 1 − ⋯ − u p , p ≥ 1. For example
t u 1 w 1 = a 0 β + ⋯ + a u 1 − 2 β u 1 − 1 + w 1 β u 1 , w 1 = y u 1 − 1 , t^{w_1}_{u_1}=\frac{a_0}{\beta}+\cdots+\frac{a_{u_1-2}}{\beta^{u_1-1}}+\frac{w_1}{\beta^{u_1}},\qquad w_1=y_{u_1-1}, t u 1 w 1 = β a 0 + ⋯ + β u 1 − 1 a u 1 − 2 + β u 1 w 1 , w 1 = y u 1 − 1 , where by convention, if u 1 = 1 u_1=1 u 1 = 1 y 0 / β y_0/\beta y 0 / β u 1 > 1 u_1>1 u 1 > 1 y 0 = a 0 y_0=a_0 y 0 = a 0 y 1 = a 1 , … , y u 1 − 2 = a u 1 − 2 y_1=a_1,\ldots,y_{u_1-2}=a_{u_1-2} y 1 = a 1 , … , y u 1 − 2 = a u 1 − 2
For m = 1 m=1 m = 1
U ~ 1 ( χ [ 0 , y ] ) = ∑ k 1 = 1 n λ − k 1 ∑ j 1 = 0 a k 1 − 1 − 1 χ [ 0 , y ] ( t k 1 j 1 + β − k 1 ) = ∑ r = 1 u 1 y r − 1 β r . \widetilde{U}_1(\chi_{[0,y]})=\sum_{k_1=1}^{n}\lambda^{-k_1}\sum_{j_1=0}^{a_{k_1-1}-1}\chi_{[0,y]}\left(t_{k_1}^{j_1}+\beta^{-k_1}\right)=\sum_{r=1}^{u_1}\frac{y_{r-1}}{\beta^r}. U 1 ( χ [ 0 , y ] ) = k 1 = 1 ∑ n λ − k 1 j 1 = 0 ∑ a k 1 − 1 − 1 χ [ 0 , y ] ( t k 1 j 1 + β − k 1 ) = r = 1 ∑ u 1 β r y r − 1 . where we see that when k 1 = u 1 k_1=u_1 k 1 = u 1 j 1 = 0 , 1 , … , w 1 − 1 j_1=0,1,\ldots,w_1-1 j 1 = 0 , 1 , … , w 1 − 1 w 1 = y u 1 − 1 w_1=y_{u_1-1} w 1 = y u 1 − 1
For m = 2 m=2 m = 2
U ~ 2 ( χ [ 0 , y ] ) = ∑ k 2 = 1 n λ − k 2 ∑ j 2 = 0 a k 2 − 1 − 1 ∑ k 1 = 1 n λ − k 1 ∑ j 1 = 0 a k 1 − 1 − 1 χ [ 0 , y ] ( t k 1 j 1 + β − k 1 t k 2 j 2 + β − k 1 − k 2 ) . \widetilde{U}_2(\chi_{[0,y]})=\sum_{k_2=1}^{n}\lambda^{-k_2}\sum_{j_2=0}^{a_{k_2-1}-1}\sum_{k_1=1}^{n}\lambda^{-k_1}\sum_{j_1=0}^{a_{k_1-1}-1}\chi_{[0,y]}\left(t_{k_1}^{j_1}+\beta^{-k_1}t_{k_2}^{j_2}+\beta^{-k_1-k_2}\right). U 2 ( χ [ 0 , y ] ) = k 2 = 1 ∑ n λ − k 2 j 2 = 0 ∑ a k 2 − 1 − 1 k 1 = 1 ∑ n λ − k 1 j 1 = 0 ∑ a k 1 − 1 − 1 χ [ 0 , y ] ( t k 1 j 1 + β − k 1 t k 2 j 2 + β − k 1 − k 2 ) . If k 1 < u 1 k_1<u_1 k 1 < u 1 k 1 = u 1 k_1=u_1 k 1 = u 1 j 1 < w 1 j_1<w_1 j 1 < w 1 k 2 k_2 k 2 j 2 j_2 j 2 U ~ 1 \widetilde{U}_1 U 1 k 1 = u 1 k_1=u_1 k 1 = u 1 j 1 = w 1 j_1=w_1 j 1 = w 1 k 2 k_2 k 2 j 2 j_2 j 2
∑ k 2 = 1 n ∑ j 2 = 0 a k 2 − 1 − 1 λ − u 1 − k 2 χ [ 0 , y ] ( t u 1 w 1 + β − u 1 t k 2 j 2 + β − u 1 − k 2 ) = ∑ r = u 1 + 1 u 1 + u 2 y r − 1 λ r . \sum_{k_2=1}^{n}\sum_{j_2=0}^{a_{k_2-1}-1}\lambda^{-u_1-k_2}\chi_{[0,y]}\left(t_{u_1}^{w_1}+\beta^{-u_1}t_{k_2}^{j_2}+\beta^{-u_1-k_2}\right)=\sum_{r=u_1+1}^{u_1+u_2}\frac{y_{r-1}}{\lambda^r}. k 2 = 1 ∑ n j 2 = 0 ∑ a k 2 − 1 − 1 λ − u 1 − k 2 χ [ 0 , y ] ( t u 1 w 1 + β − u 1 t k 2 j 2 + β − u 1 − k 2 ) = r = u 1 + 1 ∑ u 1 + u 2 λ r y r − 1 . In the same way we may prove
U ~ m ( χ [ 0 , y ] ) = ∑ r = 1 u 1 + ⋯ + u m y r − 1 λ r , m ≥ 1. \widetilde{U}_m(\chi_{[0,y]})=\sum_{r=1}^{u_1+\cdots+u_m}\frac{y_{r-1}}{\lambda^r},\qquad m\ge1. U m ( χ [ 0 , y ] ) = r = 1 ∑ u 1 + ⋯ + u m λ r y r − 1 , m ≥ 1. The final step is to take m → ∞ m\to\infty m → ∞
In order to prove (26) we use
h ( 1 ) − h ( t ) = ∫ t 1 h ′ ( y ) d y = ∫ 0 1 χ [ 0 , y ] ( t ) h ′ ( y ) d y h(1)-h(t)=\int_t^1 h'(y)\,dy=\int_0^1\chi_{[0,y]}(t)h'(y)\,dy h ( 1 ) − h ( t ) = ∫ t 1 h ′ ( y ) d y = ∫ 0 1 χ [ 0 , y ] ( t ) h ′ ( y ) d y and apply the formula for U ^ ( χ [ 0 , y ] \widehat{U}(\chi_{[0,y]} U ( χ [ 0 , y ]
Now let us prove (27) . Using (25) we have
U ^ ( u ) = ∑ k = 0 n − 1 λ − k − 1 U ^ ( χ [ 0 , b k ] ) = ∑ k = 0 n − 1 ∑ j ≥ 1 λ − k − j − 1 a k + j − 1 = ∑ p ≥ 1 p a p − 1 λ p + 1 = h n ′ ( λ ) . \widehat{U}(u)=\sum_{k=0}^{n-1}\lambda^{-k-1}\widehat{U}(\chi_{[0,b_k]})=\sum_{k=0}^{n-1}\sum_{j\ge1}\lambda^{-k-j-1}a_{k+j-1}=\sum_{p\ge1}p\frac{a_{p-1}}{\lambda^{p+1}}=h'_n(\lambda). U ( u ) = k = 0 ∑ n − 1 λ − k − 1 U ( χ [ 0 , b k ] ) = k = 0 ∑ n − 1 j ≥ 1 ∑ λ − k − j − 1 a k + j − 1 = p ≥ 1 ∑ p λ p + 1 a p − 1 = h n ′ ( λ ) . Proof of (iii). Denote by f ( y ) = U ^ ( χ [ 0 , y ] ) f(y)=\widehat{U}(\chi_{[0,y]}) f ( y ) = U ( χ [ 0 , y ] )
y = ∑ m ≥ 1 y k m β k m , y k m > 0. y=\sum_{m\ge1}\frac{y_{k_m}}{\beta^{k_m}},\qquad y_{k_m}>0. y = m ≥ 1 ∑ β k m y k m , y k m > 0. Let Y M = y k M − 1 β k M + ∑ m ≠ M y k m β k m Y_M=\frac{y_{k_M-1}}{\beta^{k_M}}+\sum_{m\ne M}\frac{y_{k_m}}{\beta^{k_m}} Y M = β k M y k M − 1 + ∑ m = M β k m y k m f ( y ) − f ( Y M ) = λ − k M f(y)-f(Y_M)=\lambda^{-k_M} f ( y ) − f ( Y M ) = λ − k M y − Y M = β − k M y-Y_M=\beta^{-k_M} y − Y M = β − k M M M M f ′ ( y ) f'(y) f ′ ( y )
Now assume that y = ∑ k = 1 K y k β k y=\sum_{k=1}^{K}\frac{y_k}{\beta^k} y = ∑ k = 1 K β k y k y K > 0 y_K>0 y K > 0 m > K + 2 m>K+2 m > K + 2 Y m = y + β − m < 1 Y_m=y+\beta^{-m}<1 Y m = y + β − m < 1 f ( Y m ) − f ( y ) = λ − m f(Y_m)-f(y)=\lambda^{-m} f ( Y m ) − f ( y ) = λ − m Y m − y = β − m Y_m-y=\beta^{-m} Y m − y = β − m m m m f ′ ( y ) f'(y) f ′ ( y )
Proof of (iv). The proof is identical with the one in [12 Theorem 4.5] . Assume that such a μ \mu μ μ ( [ 0 , y ] ) \mu([0,y]) μ ([ 0 , y ])
Proof of (v). If q q q M M M m = 1 m=1 m = 1 N = M + 1 N=M+1 N = M + 1 (31) we see that the error term is identically zero and for r ≥ n r\ge n r ≥ n
( P r q ) ( y ) = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 − 1 { ( P r − k 1 H 1 ) ( y ) ∫ k 1 j 1 k 1 j 1 + β − k 1 q ( u ) d u + ∑ s = 1 M + 1 β − k 1 s ( q ( s − 1 ) ( k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( k 1 j 1 ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) ( P r − k 1 H n s + 1 ) ( y ) } . (45) (\mathcal{P}^{r}q)(y)=\sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1-1}-1}\left\{(\mathcal{P}^{r-k_1}H_1)(y)\int_{k_1^{j_1}}^{k_1^{j_1}+\beta^{-k_1}}q(u)\,\mathrm{d}u
+\sum_{s=1}^{M+1}\beta^{-k_1s}\left(q^{(s-1)}\left(k_1^{j_1}+\beta^{-k_1}\right)-q^{(s-1)}\left(k_1^{j_1}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}(\mathcal{P}^{r-k_1}H_{ns+1})(y)\right\}. \tag*{(45)} ( P r q ) ( y ) = k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 − 1 { ( P r − k 1 H 1 ) ( y ) ∫ k 1 j 1 k 1 j 1 + β − k 1 q ( u ) d u + s = 1 ∑ M + 1 β − k 1 s ( q ( s − 1 ) ( k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( k 1 j 1 ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) ( P r − k 1 H n s + 1 ) ( y ) } . (45) Note that the term with s = M + 1 s=M+1 s = M + 1 q ( M ) q^{(M)} q ( M )
Let 1 / β < ∣ ζ ∣ < 1 1/\beta<|\zeta|<1 1/ β < ∣ ζ ∣ < 1 P ~ ν , n \widetilde{\mathcal{P}}_{\nu,n} P ν , n π ζ \pi_\zeta π ζ
π ζ ( H n s + 1 ) = C s u ζ , 0 ≤ s ≤ M , \pi_\zeta(H_{ns+1})=C_su_\zeta,\qquad0\le s\le M, π ζ ( H n s + 1 ) = C s u ζ , 0 ≤ s ≤ M , where u ζ u_\zeta u ζ c 1 = 1 / λ c_1=1/\lambda c 1 = 1/ λ U ^ ( u ζ ) \widehat{U}(u_\zeta) U ( u ζ ) (27) . We also have C 0 = 1 / U ^ ( u ζ ) C_0=1/\widehat{U}(u_\zeta) C 0 = 1/ U ( u ζ ) π ζ ( χ [ 0 , 1 ] ) = u ζ 1 U ^ ( u ζ ) U ^ ( χ [ 0 , 1 ] ) \pi_\zeta(\chi_{[0,1]})=u_\zeta\frac{1}{\widehat{U}(u_\zeta)}\widehat{U}(\chi_{[0,1]}) π ζ ( χ [ 0 , 1 ] ) = u ζ U ( u ζ ) 1 U ( χ [ 0 , 1 ] ) U ^ ( χ [ 0 , 1 ] ) = 1 \widehat{U}(\chi_{[0,1]})=1 U ( χ [ 0 , 1 ] ) = 1
U ^ \widehat{U} U P \mathcal{P} P B V ( [ 0 , 1 ] ) BV([0,1]) B V ([ 0 , 1 ]) P \mathcal{P} P U ^ \widehat{U} U (45) we obtain
ζ r U ^ ( q ) = ∑ k 1 = 1 n ∑ j 1 = 0 a k 1 − 1 − 1 ζ r − k 1 ∫ k 1 j 1 k 1 j 1 + β − k 1 q ( u ) d u + U ^ ( u ζ ) ∑ s = 1 M + 1 C s β − k 1 s ( q ( s − 1 ) ( k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( k 1 j 1 ) ) 1 s ! ∥ B s ∥ L 1 ( [ 0 , 1 ] ) . \zeta^r\widehat{U}(q)=\sum_{k_1=1}^{n}\sum_{j_1=0}^{a_{k_1-1}-1}\zeta^{r-k_1}\int_{k_1^{j_1}}^{k_1^{j_1}+\beta^{-k_1}}q(u)\,\mathrm{d}u
+\widehat{U}(u_\zeta)\sum_{s=1}^{M+1}C_s\beta^{-k_1s}\left(q^{(s-1)}\left(k_1^{j_1}+\beta^{-k_1}\right)-q^{(s-1)}\left(k_1^{j_1}\right)\right)\frac{1}{s!}\lVert B_s\rVert_{L^1([0,1])}. ζ r U ( q ) = k 1 = 1 ∑ n j 1 = 0 ∑ a k 1 − 1 − 1 ζ r − k 1 ∫ k 1 j 1 k 1 j 1 + β − k 1 q ( u ) d u + U ( u ζ ) s = 1 ∑ M + 1 C s β − k 1 s ( q ( s − 1 ) ( k 1 j 1 + β − k 1 ) − q ( s − 1 ) ( k 1 j 1 ) ) s ! 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) . Dividing by ζ r \zeta^r ζ r (28) . When q ( x ) = 1 q(x)=1 q ( x ) = 1 U ^ ( χ [ 0 , 1 ] ) = 1 \widehat{U}(\chi_{[0,1]})=1 U ( χ [ 0 , 1 ] ) = 1
The invariant subspace
We recall the definition of the following linearly independent (due to Lemma 1.2) characteristic functions
ψ 1 ( x ) = χ [ 0 , 1 ] ( x ) , ψ 2 ( x ) = χ [ 0 , b 1 ] ( x ) , … , ψ n ( x ) = χ [ 0 , b n − 1 ] ( x ) . \psi_1(x)=\chi_{[0,1]}(x),\qquad\psi_2(x)=\chi_{[0,b_1]}(x),\qquad\ldots,\qquad\psi_n(x)=\chi_{[0,b_{n-1}]}(x). ψ 1 ( x ) = χ [ 0 , 1 ] ( x ) , ψ 2 ( x ) = χ [ 0 , b 1 ] ( x ) , … , ψ n ( x ) = χ [ 0 , b n − 1 ] ( x ) . Applying the transfer operator P \mathcal{P} P 1 ≤ j ≤ n − 1 1\le j\le n-1 1 ≤ j ≤ n − 1 x x x [3 ] ):
β ( P ψ j ) ( x ) = a j − 1 ψ 1 ( x ) + ψ j + 1 ( x ) , \beta(\mathcal{P}\psi_j)(x)=a_{j-1}\psi_1(x)+\psi_{j+1}(x), β ( P ψ j ) ( x ) = a j − 1 ψ 1 ( x ) + ψ j + 1 ( x ) , and
β ( P ψ n ) ( x ) = a n − 1 ψ 1 ( x ) . \beta(\mathcal{P}\psi_n)(x)=a_{n-1}\psi_1(x). β ( P ψ n ) ( x ) = a n − 1 ψ 1 ( x ) . This shows that the functions ψ 1 , … , ψ n \psi_1,\ldots,\psi_n ψ 1 , … , ψ n β P \beta\mathcal{P} β P
P n = ( a 0 a 1 ⋯ ⋯ a j − 1 ⋯ a n − 2 a n − 1 1 0 ⋯ 0 ⋯ 0 0 0 0 1 0 0 0 0 0 0 ⋮ ⋮ ⋱ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 0 0 0 0 0 1 0 0 0 0 ⋮ ⋮ 0 ⋱ ⋮ ⋮ ⋮ 0 0 0 0 0 ⋯ 1 0 ) . (46) \mathcal{P}_n =
\begin{pmatrix}
a_0 & a_1 & \cdots& \cdots& a_{j-1} & \cdots& a_{n-2} & a_{n-1}\\
1 & 0 & \cdots& 0 & \cdots& 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
\vdots& \vdots& \ddots& 0 & \vdots& \vdots& \vdots& \vdots\\
0 & 0 & 0 & \cdots& 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
\vdots& \vdots& & 0 & \ddots& \vdots& \vdots& \vdots\\
0 & 0 & 0 & 0 & 0 & \cdots& 1 & 0
\end{pmatrix}. \tag*{(46)} P n = a 0 1 0 ⋮ 0 0 ⋮ 0 a 1 0 1 ⋮ 0 0 ⋮ 0 ⋯ ⋯ 0 ⋱ 0 0 0 ⋯ 0 0 0 ⋯ 1 0 0 a j − 1 ⋯ 0 ⋮ 0 0 ⋱ 0 ⋯ 0 0 ⋮ 0 0 ⋮ ⋯ a n − 2 0 0 ⋮ 0 0 ⋮ 1 a n − 1 0 0 ⋮ 0 0 ⋮ 0 . (46) Let ψ ℓ , j ( x ) = x ℓ ψ j ( x ) \psi_{\ell,j}(x)=x^\ell\psi_j(x) ψ ℓ , j ( x ) = x ℓ ψ j ( x ) 1 ≤ j ≤ n 1\leq j\leq n 1 ≤ j ≤ n ℓ ≥ 1 \ell\geq1 ℓ ≥ 1 j ≤ n − 1 j\leq n-1 j ≤ n − 1
β ( P ψ ℓ , j ) ( x ) = a j − 1 x ℓ β ℓ ψ 1 ( x ) + x ℓ β ℓ ψ j + 1 ( x ) + R ℓ − 1 , j ( x ) = a j − 1 β ℓ ψ ℓ , 1 ( x ) + 1 β ℓ ψ ℓ , j + 1 ( x ) + ∑ k = 0 ℓ − 1 ∑ m = 1 n C ℓ , j , k , m ψ k , m . \begin{aligned}
\beta(\mathcal{P}\psi_{\ell,j})(x)
=
\frac{a_{j-1}x^\ell}{\beta^\ell}\psi_1(x)
+
\frac{x^\ell}{\beta^\ell}\psi_{j+1}(x)
+
R_{\ell-1,j}(x) \\
&=
\frac{a_{j-1}}{\beta^\ell}\psi_{\ell,1}(x)
+
\frac{1}{\beta^\ell}\psi_{\ell,j+1}(x)
+
\sum_{k=0}^{\ell-1}\sum_{m=1}^{n}C_{\ell,j,k,m}\psi_{k,m}.
\end{aligned} β ( P ψ ℓ , j ) ( x ) = β ℓ a j − 1 x ℓ ψ 1 ( x ) + β ℓ x ℓ ψ j + 1 ( x ) + R ℓ − 1 , j ( x ) = β ℓ a j − 1 ψ ℓ , 1 ( x ) + β ℓ 1 ψ ℓ , j + 1 ( x ) + k = 0 ∑ ℓ − 1 m = 1 ∑ n C ℓ , j , k , m ψ k , m . where R ℓ − 1 , j R_{\ell-1,j} R ℓ − 1 , j ℓ − 1 \ell-1 ℓ − 1 x x x
β ( P ψ ℓ , n ) ( x ) = a n − 1 β ℓ ψ ℓ , 1 ( x ) + ∑ k = 0 ℓ − 1 ∑ m = 1 n C ℓ , k , m ψ k , m . (47) \beta(\mathcal{P}\psi_{\ell,n})(x)
=
\frac{a_{n-1}}{\beta^\ell}\psi_{\ell,1}(x)
+
\sum_{k=0}^{\ell-1}\sum_{m=1}^{n}C_{\ell,k,m}\psi_{k,m}. \tag*{(47)} β ( P ψ ℓ , n ) ( x ) = β ℓ a n − 1 ψ ℓ , 1 ( x ) + k = 0 ∑ ℓ − 1 m = 1 ∑ n C ℓ , k , m ψ k , m . (47) Now let us renumber the functions ψ ℓ , j ( x ) = x ℓ ψ j ( x ) \psi_{\ell,j}(x)=x^\ell\psi_j(x) ψ ℓ , j ( x ) = x ℓ ψ j ( x ) 0 ≤ ℓ ≤ ν − 1 0\leq\ell\leq\nu-1 0 ≤ ℓ ≤ ν − 1 1 ≤ j ≤ n 1\leq j\leq n 1 ≤ j ≤ n
Ψ 1 = ψ 0 , 1 , … , Ψ n = ψ 0 , n , Ψ n + 1 = ψ 1 , 1 , … , Ψ 2 n = ψ 1 , n , … , Ψ ν n = ψ ν , n . (48) \Psi_1=\psi_{0,1},\ldots,\Psi_n=\psi_{0,n},\ \Psi_{n+1}=\psi_{1,1},\ldots,\Psi_{2n}=\psi_{1,n},\ldots,\Psi_{\nu n}=\psi_{\nu,n}. \tag*{(48)} Ψ 1 = ψ 0 , 1 , … , Ψ n = ψ 0 , n , Ψ n + 1 = ψ 1 , 1 , … , Ψ 2 n = ψ 1 , n , … , Ψ ν n = ψ ν , n . (48) We see that the restriction matrix P ν , n \mathcal{P}_{\nu,n} P ν , n ν n \nu n ν n Ψ k \Psi_k Ψ k 1 ≤ k ≤ ν n 1\leq k\leq\nu n 1 ≤ k ≤ ν n ν \nu ν n × n n\times n n × n β − k P n \beta^{-k}\mathcal{P}_n β − k P n 0 ≤ k ≤ ν − 1 0\leq k\leq\nu-1 0 ≤ k ≤ ν − 1
P ν , n = ( P n ⋆ ⋆ ⋯ ⋆ 0 n β − 1 P n ⋆ ⋯ ⋆ 0 n 0 n β − 2 P n ⋯ ⋆ ⋮ ⋮ ⋱ ⋱ ⋆ 0 n 0 n 0 n ⋱ ⋆ 0 n 0 n 0 n ⋯ β − ( ν − 1 ) P n ) . (49) \mathcal{P}_{\nu,n}=
\begin{pmatrix}
\mathcal{P}_n & \star& \star& \cdots& \star\\
\mathbf{0}_n & \beta^{-1}\mathcal{P}_n & \star& \cdots& \star\\
\mathbf{0}_n & \mathbf{0}_n & \beta^{-2}\mathcal{P}_n & \cdots& \star\\
\vdots& \vdots& \ddots& \ddots& \star\\
\mathbf{0}_n & \mathbf{0}_n & \mathbf{0}_n & \ddots& \star\\
\mathbf{0}_n & \mathbf{0}_n & \mathbf{0}_n & \cdots& \beta^{-(\nu-1)}\mathcal{P}_n
\end{pmatrix}. \tag*{(49)} P ν , n = P n 0 n 0 n ⋮ 0 n 0 n ⋆ β − 1 P n 0 n ⋮ 0 n 0 n ⋆ ⋆ β − 2 P n ⋱ 0 n 0 n ⋯ ⋯ ⋯ ⋱ ⋱ ⋯ ⋆ ⋆ ⋆ ⋆ ⋆ β − ( ν − 1 ) P n . (49) This implies that the spectrum of P ν , n \mathcal{P}_{\nu,n} P ν , n
σ ( P ν , n ) = ⋃ ℓ = 0 ν − 1 β − ℓ σ ( P n ) . (50) \sigma(\mathcal{P}_{\nu,n})=\bigcup_{\ell=0}^{\nu-1}\beta^{-\ell}\sigma(\mathcal{P}_n). \tag*{(50)} σ ( P ν , n ) = ℓ = 0 ⋃ ν − 1 β − ℓ σ ( P n ) . (50) Lemma A.1. We have that σ ( P n ) = { β , λ 2 , … , λ n } \sigma(\mathcal{P}_n)=\{\beta,\lambda_2,\ldots,\lambda_n\} σ ( P n ) = { β , λ 2 , … , λ n } β > 1 \beta>1 β > 1 0 < ∣ λ j ∣ < β 0<|\lambda_j|<\beta 0 < ∣ λ j ∣ < β 2 ≤ j ≤ n 2\leq j\leq n 2 ≤ j ≤ n
Proof. Another way of formulating the lemma is that β \beta β Det ( z 1 n − P n ) \operatorname{Det}(z\mathbf{1}_n-\mathcal{P}_n) Det ( z 1 n − P n ) β \beta β
Det ( z 1 n − P n ) = ( z − a 0 ) z n − 1 − a 1 z n − 2 − ⋯ − a n − 1 = z n − ( a 0 z n − 1 + a 1 z n − 2 + ⋯ + a n − 1 ) . \operatorname{Det}(z\mathbf{1}_n-\mathcal{P}_n)=(z-a_0)z^{n-1}-a_1z^{n-2}-\cdots-a_{n-1}=z^n-(a_0z^{n-1}+a_1z^{n-2}+\cdots+a_{n-1}). Det ( z 1 n − P n ) = ( z − a 0 ) z n − 1 − a 1 z n − 2 − ⋯ − a n − 1 = z n − ( a 0 z n − 1 + a 1 z n − 2 + ⋯ + a n − 1 ) . Then z z z z z z
f ( z ) = 1 − ( a 0 z − 1 + a 1 z − 2 + ⋯ + a n − 1 z − n ) . (51) f(z)=1-(a_0z^{-1}+a_1z^{-2}+\cdots+a_{n-1}z^{-n}). \tag*{(51)} f ( z ) = 1 − ( a 0 z − 1 + a 1 z − 2 + ⋯ + a n − 1 z − n ) . (51) We already know that β > a 0 \beta>a_0 β > a 0 f ( x ) = 0 f(x)=0 f ( x ) = 0 x > 0 x>0 x > 0 Det ( z 1 n − P n ) = z n f ( z ) \operatorname{Det}(z\mathbf{1}_n-\mathcal{P}_n)=z^nf(z) Det ( z 1 n − P n ) = z n f ( z ) f ′ ( β ) ≠ 0 f'(\beta)\ne0 f ′ ( β ) = 0 β \beta β
Next, let us assume that ∣ z ∣ > β |z|>\beta ∣ z ∣ > β
∣ f ( z ) ∣ ≥ 1 − ∑ j = 0 n − 1 a j ∣ z ∣ − j − 1 > f ( β ) = 0. |f(z)|\geq1-\sum_{j=0}^{n-1}a_j|z|^{-j-1}>f(\beta)=0. ∣ f ( z ) ∣ ≥ 1 − j = 0 ∑ n − 1 a j ∣ z ∣ − j − 1 > f ( β ) = 0. This implies that the zeros of f f f β \beta β
Now let us assume that there exists φ ∈ ( 0 , π ] \varphi\in(0,\pi] φ ∈ ( 0 , π ] ρ = β e i φ \rho=\beta e^{i\varphi} ρ = β e i φ
0 = f ( β e i φ ) = 1 − ∑ j = 0 n − 1 a j β − j − 1 e − i ( j + 1 ) φ . 0=f(\beta e^{i\varphi})=1-\sum_{j=0}^{n-1}a_j\beta^{-j-1}e^{-i(j+1)\varphi}. 0 = f ( β e i φ ) = 1 − j = 0 ∑ n − 1 a j β − j − 1 e − i ( j + 1 ) φ . Taking the real part and using (2) we have
0 = 1 − ∑ j ≥ 0 a j β − j − 1 cos ( ( j + 1 ) φ ) = ∑ j ≥ 0 a j β − j − 1 ( 1 − cos ( ( j + 1 ) φ ) ) ≥ a 0 β − 1 ( 1 − cos ( φ ) ) \begin{aligned}
0=1-\sum_{j\geq0}a_j\beta^{-j-1}\cos((j+1)\varphi)=\sum_{j\geq0}a_j\beta^{-j-1}(1-\cos((j+1)\varphi)) \\
&\geq a_0\beta^{-1}(1-\cos(\varphi))
\end{aligned} 0 = 1 − j ≥ 0 ∑ a j β − j − 1 cos (( j + 1 ) φ ) = j ≥ 0 ∑ a j β − j − 1 ( 1 − cos (( j + 1 ) φ )) ≥ a 0 β − 1 ( 1 − cos ( φ )) which leads to φ = 0 \varphi=0 φ = 0 f f f β \beta β β \beta β
Now let us show that the geometric multiplicity of all eigenvalues equals one. We have
β ( P ψ j ) ( x ) = a j − 1 ψ 1 ( x ) + ψ j + 1 ( x ) , for j ∈ { 1 , … , n − 1 } , β ( P ψ n ) ( x ) = a n − 1 ψ 1 ( x ) . (52) \begin{aligned}
\beta(\mathcal{P}\psi_j)(x)&=a_{j-1}\psi_1(x)+\psi_{j+1}(x), &&\text{for }j\in\{1,\ldots,n-1\},\\
\beta(\mathcal{P}\psi_n)(x)&=a_{n-1}\psi_1(x). \tag*{(52)}
\end{aligned} β ( P ψ j ) ( x ) β ( P ψ n ) ( x ) = a j − 1 ψ 1 ( x ) + ψ j + 1 ( x ) , = a n − 1 ψ 1 ( x ) . for j ∈ { 1 , … , n − 1 } , (52) Any eigenfunction corresponding to an eigenvalue λ \lambda λ
u ( x ) = c 1 ψ 1 ( x ) + c 2 ψ 2 + ⋯ + c n ψ n ( x ) . u(x)=c_1\psi_1(x)+c_2\psi_2+\cdots+c_n\psi_n(x). u ( x ) = c 1 ψ 1 ( x ) + c 2 ψ 2 + ⋯ + c n ψ n ( x ) . Then β P u = λ u \beta\mathcal{P}u=\lambda u β P u = λ u 1 = ∑ k = 1 n a k − 1 λ k 1=\sum_{k=1}^n\frac{a_{k-1}}{\lambda^k} 1 = ∑ k = 1 n λ k a k − 1
u = c 1 ( ψ 1 ( x ) + 1 λ ψ 2 ( x ) + ⋯ + 1 λ n − 1 ψ n ( x ) ) = c 1 ∑ k = 0 n − 1 λ − k χ [ 0 , b k ] . (53) u=c_1\left(\psi_1(x)+\frac{1}{\lambda}\psi_2(x)+\cdots+\frac{1}{\lambda^{n-1}}\psi_n(x)\right)=c_1\sum_{k=0}^{n-1}\lambda^{-k}\chi_{[0,b_k]}. \tag*{(53)} u = c 1 ( ψ 1 ( x ) + λ 1 ψ 2 ( x ) + ⋯ + λ n − 1 1 ψ n ( x ) ) = c 1 k = 0 ∑ n − 1 λ − k χ [ 0 , b k ] . (53) which shows that this is a one-dimensional subspace. When λ = β > 1 \lambda=\beta>1 λ = β > 1 c 1 = C 0 c_1=C_0 c 1 = C 0
u 1 ( x ) = C 0 ( ψ 1 ( x ) + 1 β ψ 2 ( x ) + ⋯ + 1 β n − 1 ψ n ( x ) ) u_1(x)=C_0\left(\psi_1(x)+\frac{1}{\beta}\psi_2(x)+\cdots+\frac{1}{\beta^{n-1}}\psi_n(x)\right) u 1 ( x ) = C 0 ( ψ 1 ( x ) + β 1 ψ 2 ( x ) + ⋯ + β n − 1 1 ψ n ( x ) ) and
∫ 0 1 u 1 ( x ) d x = 1. (54) \int_0^1 u_1(x)\,\mathrm{d}x=1. \tag*{(54)} ∫ 0 1 u 1 ( x ) d x = 1. (54) Let us also compute a left eigenvector w 1 = [ d 1 , d 2 , … , d n ] w_1=[d_1,d_2,\ldots,d_n] w 1 = [ d 1 , d 2 , … , d n ] w 1 P n = β w 1 w_1P_n=\beta w_1 w 1 P n = β w 1 d 1 = 1 d_1=1 d 1 = 1
d 1 = 1 , d 2 = β − a 0 , d 3 = β d 2 − a 1 = β 2 − a 0 β − a 1 , … d_1=1,\qquad d_2=\beta-a_0,\qquad d_3=\beta d_2-a_1=\beta^2-a_0\beta-a_1,\ldots d 1 = 1 , d 2 = β − a 0 , d 3 = β d 2 − a 1 = β 2 − a 0 β − a 1 , … Comparing with (10) we see that w 1 = [ 1 , b 1 , b 2 , … , b n − 1 ] w_1=[1,b_1,b_2,\ldots,b_{n-1}] w 1 = [ 1 , b 1 , b 2 , … , b n − 1 ] C 0 C_0 C 0
C 0 ( 1 + b 1 β + ⋯ + b n − 1 β n − 1 ) = 1 , C_0\left(1+\frac{b_1}{\beta}+\cdots+\frac{b_{n-1}}{\beta^{n-1}}\right)=1, C 0 ( 1 + β b 1 + ⋯ + β n − 1 b n − 1 ) = 1 , which implies that the rank one Riesz projection Π β \Pi_\beta Π β β \beta β ψ k ( x ) \psi_k(x) ψ k ( x ) ( Π β ψ k ) ( x ) = b k − 1 u 1 ( x ) (\Pi_\beta\psi_k)(x)=b_{k-1}u_1(x) ( Π β ψ k ) ( x ) = b k − 1 u 1 ( x ) 1 ≤ k ≤ n 1\leq k\leq n 1 ≤ k ≤ n
( Π β ψ 1 ) ( x ) = u 1 ( x ) . (55) (\Pi_\beta\psi_1)(x)=u_1(x). \tag*{(55)} ( Π β ψ 1 ) ( x ) = u 1 ( x ) . (55) We also need to investigate how the restriction matrix looks. when we use the basis consisting of scaled Bernoulli polynomials H s n + j H_{sn+j} H s n + j s ∈ { 0 , … , ν − 1 } s\in\{0,\ldots,\nu-1\} s ∈ { 0 , … , ν − 1 } j ∈ { 1 , … , n } j\in\{1,\ldots,n\} j ∈ { 1 , … , n } k ∈ { 1 , … , ν n } k\in\{1,\ldots,\nu n\} k ∈ { 1 , … , ν n }
( P H s n + j ) ( x ) = ∑ k = 1 ν n p ~ k , s n + j H k ( x ) , (56) (\mathcal{P}H_{sn+j})(x)=\sum_{k=1}^{\nu n}\widetilde{p}_{k,sn+j}H_k(x), \tag*{(56)} ( P H s n + j ) ( x ) = k = 1 ∑ ν n p k , s n + j H k ( x ) , (56) where P ~ ν , n = { p ~ k , s n + j } \widetilde{P}_{\nu,n}=\{\widetilde{p}_{k,sn+j}\} P ν , n = { p k , s n + j } β − 1 P ν , n \beta^{-1}P_{\nu,n} β − 1 P ν , n H s n + j H_{sn+j} H s n + j Ψ s n + j = x s ψ j ( x ) \Psi_{sn+j}=x^s\psi_j(x) Ψ s n + j = x s ψ j ( x ) ν n × ν n \nu n\times\nu n ν n × ν n S S S Ψ \Psi Ψ H H H P ν , n \mathcal{P}_{\nu,n} P ν , n
S ν , n = ( S n , 1 ⋆ ⋆ ⋯ ⋆ 0 n S n , 2 ⋆ ⋯ ⋆ 0 n 0 n S n , 3 ⋯ ⋆ ⋮ ⋮ ⋱ ⋱ ⋆ 0 n 0 n 0 n ⋱ ⋆ 0 n 0 n 0 n ⋯ S n , ν ) , S ν , n − 1 = ( S n , 1 − 1 ⋆ ⋆ ⋯ ⋆ 0 n S n , 2 − 1 ⋆ ⋯ ⋆ 0 n 0 n S n , 3 − 1 ⋯ ⋆ ⋮ ⋮ ⋱ ⋱ ⋆ 0 n 0 n 0 n ⋱ ⋆ 0 n 0 n 0 n ⋯ S n , ν − 1 ) . S_{\nu,n} =
\begin{pmatrix}
S_{n,1} & \star& \star& \cdots& \star\\
0_n & S_{n,2} & \star& \cdots& \star\\
0_n & 0_n & S_{n,3} & \cdots& \star\\
\vdots& \vdots& \ddots& \ddots& \star\\
0_n & 0_n & 0_n & \ddots& \star\\
0_n & 0_n & 0_n & \cdots& S_{n,\nu}
\end{pmatrix},
\qquad
S_{\nu,n}^{-1} =
\begin{pmatrix}
S_{n,1}^{-1} & \star& \star& \cdots& \star\\
0_n & S_{n,2}^{-1} & \star& \cdots& \star\\
0_n & 0_n & S_{n,3}^{-1} & \cdots& \star\\
\vdots& \vdots& \ddots& \ddots& \star\\
0_n & 0_n & 0_n & \ddots& \star\\
0_n & 0_n & 0_n & \cdots& S_{n,\nu}^{-1}
\end{pmatrix}. S ν , n = S n , 1 0 n 0 n ⋮ 0 n 0 n ⋆ S n , 2 0 n ⋮ 0 n 0 n ⋆ ⋆ S n , 3 ⋱ 0 n 0 n ⋯ ⋯ ⋯ ⋱ ⋱ ⋯ ⋆ ⋆ ⋆ ⋆ ⋆ S n , ν , S ν , n − 1 = S n , 1 − 1 0 n 0 n ⋮ 0 n 0 n ⋆ S n , 2 − 1 0 n ⋮ 0 n 0 n ⋆ ⋆ S n , 3 − 1 ⋱ 0 n 0 n ⋯ ⋯ ⋯ ⋱ ⋱ ⋯ ⋆ ⋆ ⋆ ⋆ ⋆ S n , ν − 1 . Thus P ~ ν , n = S ν , n ( β − 1 P ν , n ) S ν , n − 1 \widetilde{\mathcal{P}}_{\nu,n}=S_{\nu,n}(\beta^{-1}\mathcal{P}_{\nu,n})S_{\nu,n}^{-1} P ν , n = S ν , n ( β − 1 P ν , n ) S ν , n − 1
P ~ ν , n = ( A n , 1 ⋆ ⋆ ⋯ ⋆ 0 n A n , 2 ⋆ ⋯ ⋆ 0 n 0 n A n , 3 ⋯ ⋆ ⋮ ⋮ ⋱ ⋱ ⋆ 0 n 0 n 0 n ⋱ ⋆ 0 n 0 n 0 n ⋯ A n , ν ) (57) \widetilde{\mathcal{P}}_{\nu,n} =
\begin{pmatrix}
\mathcal{A}_{n,1} & \star& \star& \cdots& \star\\
0_n & \mathcal{A}_{n,2} & \star& \cdots& \star\\
0_n & 0_n & \mathcal{A}_{n,3} & \cdots& \star\\
\vdots& \vdots& \ddots& \ddots& \star\\
0_n & 0_n & 0_n & \ddots& \star\\
0_n & 0_n & 0_n & \cdots& \mathcal{A}_{n,\nu}
\end{pmatrix} \tag*{(57)} P ν , n = A n , 1 0 n 0 n ⋮ 0 n 0 n ⋆ A n , 2 0 n ⋮ 0 n 0 n ⋆ ⋆ A n , 3 ⋱ 0 n 0 n ⋯ ⋯ ⋯ ⋱ ⋱ ⋯ ⋆ ⋆ ⋆ ⋆ ⋆ A n , ν (57) where A n , ℓ = S n , ℓ ( β − ℓ P n ) S n , ℓ − 1 \mathcal{A}_{n,\ell}=S_{n,\ell}(\beta^{-\ell}\mathcal{P}_n)S_{n,\ell}^{-1} A n , ℓ = S n , ℓ ( β − ℓ P n ) S n , ℓ − 1 1 ≤ ℓ ≤ ν 1\leq\ell\leq\nu 1 ≤ ℓ ≤ ν P ~ ν , n \widetilde{\mathcal{P}}_{\nu,n} P ν , n
σ ( P ~ ν , n ) = ⋃ ℓ = 1 ν β − ℓ σ ( P n ) . (58) \sigma(\widetilde{\mathcal{P}}_{\nu,n})=\bigcup_{\ell=1}^{\nu}\beta^{-\ell}\sigma(\mathcal{P}_n). \tag*{(58)} σ ( P ν , n ) = ℓ = 1 ⋃ ν β − ℓ σ ( P n ) . (58) Now assume that P ~ ν , n \widetilde{\mathcal{P}}_{\nu,n} P ν , n 2 ≤ E ≤ ν n 2\leq E\leq\nu n 2 ≤ E ≤ ν n π ( a ) \pi^{(a)} π ( a ) 1 ≤ a ≤ E 1\leq a\leq E 1 ≤ a ≤ E ζ a \zeta_a ζ a
∑ a = 1 E π ( a ) = I ν n × ν n and P ~ ν , n = π ( 1 ) + ∑ a = 2 E π ( a ) P ~ ν , n π ( a ) . \sum_{a=1}^{E}\pi^{(a)}=I_{\nu n\times\nu n}
\qquad\text{and}\qquad
\widetilde{\mathcal{P}}_{\nu,n}=\pi^{(1)}+\sum_{a=2}^{E}\pi^{(a)}\widetilde{\mathcal{P}}_{\nu,n}\pi^{(a)}. a = 1 ∑ E π ( a ) = I ν n × ν n and P ν , n = π ( 1 ) + a = 2 ∑ E π ( a ) P ν , n π ( a ) . where we recall that the projection π ( 1 ) \pi^{(1)} π ( 1 ) 1 1 1 1 ≤ m a 1\leq m_a 1 ≤ m a ζ a \zeta_a ζ a a ≥ 2 a\geq2 a ≥ 2 C > 0 C>0 C > 0 r ≥ 1 r\geq1 r ≥ 1
∥ ( π ( a ) P ~ ν , n π ( a ) ) r ∥ ≤ C r m a − 1 ∣ ζ a ∣ r . \left\|\left(\pi^{(a)}\widetilde{\mathcal{P}}_{\nu,n}\pi^{(a)}\right)^r\right\|\leq Cr^{m_a-1}|\zeta_a|^r. ( π ( a ) P ν , n π ( a ) ) r ≤ C r m a − 1 ∣ ζ a ∣ r . For the special case when the polynomial is Parry-Pisot, we recall from Definition 1.1 that all eigenvalues other than β \beta β s = 1 s=1 s = 1
Lemma A.5. Let ∣ λ ∗ ∣ < 1 |\lambda_*|<1 ∣ λ ∗ ∣ < 1 p ( z ) p(z) p ( z ) P ~ 2 , n \widetilde{\mathcal{P}}_{2,n} P 2 , n β − 1 \beta^{-1} β − 1 v 1 ( x ) ≠ 0 v_1(x)\ne0 v 1 ( x ) = 0 P \mathcal{P} P j ≥ 1 j\ge1 j ≥ 1
P j H n + 1 = β − j v 1 + O ( ( ∣ λ ∗ ∣ β ) j ) , ∫ 0 1 v 1 ( x ) d x = 0. (59) \mathcal{P}^j H_{n+1}=\beta^{-j}v_1+O\left(\left(\frac{|\lambda_*|}{\beta}\right)^j\right),\qquad\int_0^1 v_1(x)\,\mathrm{d}x=0. \tag*{(59)} P j H n + 1 = β − j v 1 + O ( ( β ∣ λ ∗ ∣ ) j ) , ∫ 0 1 v 1 ( x ) d x = 0. (59) Proof. The second largest in absolute value of the eigenvalues of the first n × n n\times n n × n A n , 1 \mathcal{A}_{n,1} A n , 1 ∣ λ ∗ ∣ / β < β − 1 |\lambda_*|/\beta<\beta^{-1} ∣ λ ∗ ∣/ β < β − 1 P j H n + 1 \mathcal{P}^jH_{n+1} P j H n + 1 A n , 2 \mathcal{A}_{n,2} A n , 2 β − 2 P n \beta^{-2}\mathcal{P}_n β − 2 P n β − 1 A n , 1 \beta^{-1}\mathcal{A}_{n,1} β − 1 A n , 1 1 / β 1/\beta 1/ β λ ∗ / β \lambda_*/\beta λ ∗ / β 1 / β 1/\beta 1/ β H n + 1 H_{n+1} H n + 1 v 1 v_1 v 1
Since ∫ 0 1 v 1 ( x ) d x = ∫ 0 1 ( P v 1 ) ( x ) d x = β − 1 ∫ 0 1 v 1 ( x ) d x \int_0^1v_1(x)\,\mathrm{d}x=\int_0^1(\mathcal{P}v_1)(x)\,\mathrm{d}x=\beta^{-1}\int_0^1v_1(x)\,\mathrm{d}x ∫ 0 1 v 1 ( x ) d x = ∫ 0 1 ( P v 1 ) ( x ) d x = β − 1 ∫ 0 1 v 1 ( x ) d x ∫ 0 1 v 1 ( x ) d x = 0 \int_0^1v_1(x)\,\mathrm{d}x=0 ∫ 0 1 v 1 ( x ) d x = 0
H n + 1 ( x ) = C 1 u 1 ( x ) + v 1 ( x ) + ψ ~ n + 1 ( x ) H_{n+1}(x)=C_1u_1(x)+v_1(x)+\widetilde{\psi}_{n+1}(x) H n + 1 ( x ) = C 1 u 1 ( x ) + v 1 ( x ) + ψ n + 1 ( x ) where P u 1 = u 1 \mathcal{P}u_1=u_1 P u 1 = u 1 P v 1 = β − 1 v 1 \mathcal{P}v_1=\beta^{-1}v_1 P v 1 = β − 1 v 1 ψ ~ n + 1 \widetilde{\psi}_{n+1} ψ n + 1 P ~ 2 , n \widetilde{\mathcal{P}}_{2,n} P 2 , n ∣ λ ∗ ∣ / β < 1 / β |\lambda_*|/\beta<1/\beta ∣ λ ∗ ∣/ β < 1/ β C 1 C_1 C 1 H n + 1 H_{n+1} H n + 1 v 1 v_1 v 1 P j \mathcal{P}^j P j
Splitting the interval [0,1]
We recall that the first splitting of the interval [ 0 , 1 ] [0,1] [ 0 , 1 ]
t k 1 j 1 = { j 1 β , 0 ≤ j 1 ≤ a 0 − 1 when k 1 = 1 , a 0 β + j 1 β 2 , 0 ≤ j 1 ≤ a 1 − 1 when k 1 = 2 , a 0 β + a 1 β 2 + j 1 β 3 , 0 ≤ j 1 ≤ a 2 − 1 when k 1 = 3 , ⋮ a 0 β + a 1 β 2 + ⋯ + a n − 2 β n − 1 + j 1 β n , 0 ≤ j 1 ≤ a n − 1 − 1 when k 1 = n . t_{k_1}^{j_1}=
\begin{cases}
\dfrac{j_1}{\beta}, & 0\le j_1\le a_0-1 \quad\text{when } k_1=1,\\
\dfrac{a_0}{\beta}+\dfrac{j_1}{\beta^2}, & 0\le j_1\le a_1-1 \quad\text{when } k_1=2,\\
\dfrac{a_0}{\beta}+\dfrac{a_1}{\beta^2}+\dfrac{j_1}{\beta^3}, & 0\le j_1\le a_2-1 \quad\text{when } k_1=3,\\
\vdots\\
\dfrac{a_0}{\beta}+\dfrac{a_1}{\beta^2}+\cdots+\dfrac{a_{n-2}}{\beta^{n-1}}+\dfrac{j_1}{\beta^n}, & 0\le j_1\le a_{n-1}-1 \quad\text{when } k_1=n.
\end{cases} t k 1 j 1 = ⎩ ⎨ ⎧ β j 1 , β a 0 + β 2 j 1 , β a 0 + β 2 a 1 + β 3 j 1 , ⋮ β a 0 + β 2 a 1 + ⋯ + β n − 1 a n − 2 + β n j 1 , 0 ≤ j 1 ≤ a 0 − 1 when k 1 = 1 , 0 ≤ j 1 ≤ a 1 − 1 when k 1 = 2 , 0 ≤ j 1 ≤ a 2 − 1 when k 1 = 3 , 0 ≤ j 1 ≤ a n − 1 − 1 when k 1 = n . where we use the convention that if a i − 1 = 0 a_{i-1}=0 a i − 1 = 0 i ∈ { 2 , … , n − 1 } i\in\{2,\ldots,n-1\} i ∈ { 2 , … , n − 1 } k 1 = i k_1=i k 1 = i β − i \beta^{-i} β − i
After m m m
t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = t k 1 k 2 ⋯ k m − 1 j 1 j 2 ⋯ j m − 1 + β − ( k 1 + k 2 + ⋯ + k m − 1 ) t k m j m . t_{k_1k_2\cdots k_m}^{j_1j_2\cdots j_m}
=
t_{k_1k_2\cdots k_{m-1}}^{j_1j_2\cdots j_{m-1}}
+
\beta^{-(k_1+k_2+\cdots+k_{m-1})}
t_{k_m}^{j_m}. t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = t k 1 k 2 ⋯ k m − 1 j 1 j 2 ⋯ j m − 1 + β − ( k 1 + k 2 + ⋯ + k m − 1 ) t k m j m . Furthermore, recall the shorthand notation
k m : = ( k 1 , k 2 , … , k m ) , j m : = ( j 1 , j 2 , … , j m ) , t k m j m : = t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m , ∣ k m ∣ : = k 1 + k 2 + ⋯ + k m \begin{aligned}
\mathbf{k}_m &:= (k_1,k_2,\ldots,k_m),\\
\mathbf{j}_m &:= (j_1,j_2,\ldots,j_m),\\
t^{\mathbf{j}_m}_{\mathbf{k}_m} &:= t^{j_1j_2\cdots j_m}_{k_1k_2\cdots k_m},\\
|\mathbf{k}_m| &:= k_1+k_2+\cdots+k_m
\end{aligned} k m j m t k m j m ∣ k m ∣ := ( k 1 , k 2 , … , k m ) , := ( j 1 , j 2 , … , j m ) , := t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m , := k 1 + k 2 + ⋯ + k m and that the points t k m j m t^{\mathbf{j}_m}_{\mathbf{k}_m} t k m j m t k m j m + β − ∣ k m ∣ t^{\mathbf{j}_m}_{\mathbf{k}_m}+\beta^{-|\mathbf{k}_m|} t k m j m + β − ∣ k m ∣
Lemma A.1. Let m ≥ 2 m\ge2 m ≥ 2 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n a i − 1 > 0 a_{i-1}>0 a i − 1 > 0 0 ≤ j 1 ≤ a i − 1 − 1 0\le j_1\le a_{i-1}-1 0 ≤ j 1 ≤ a i − 1 − 1
t 1 k 2 ⋯ k m j m = j 1 β + β − 1 t k 2 ⋯ k m j 2 ⋯ j m when i = 1 , (60) t^{\mathbf{j}_m}_{1k_2\cdots k_m}=\frac{j_1}{\beta}+\beta^{-1}t^{j_2\cdots j_m}_{k_2\cdots k_m}\qquad\text{when }i=1, \tag*{(60)} t 1 k 2 ⋯ k m j m = β j 1 + β − 1 t k 2 ⋯ k m j 2 ⋯ j m when i = 1 , (60) and if i ≥ 2 i\ge2 i ≥ 2
t i k 2 ⋯ k m j m = a 0 β + a 1 β 2 + ⋯ + a i − 2 β i − 1 + β − ( i − 1 ) t 1 k 2 ⋯ k m j m . (61) t^{\mathbf{j}_m}_{ik_2\cdots k_m}=\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\beta^{-(i-1)}t^{\mathbf{j}_m}_{1k_2\cdots k_m}. \tag*{(61)} t i k 2 ⋯ k m j m = β a 0 + β 2 a 1 + ⋯ + β i − 1 a i − 2 + β − ( i − 1 ) t 1 k 2 ⋯ k m j m . (61) Proof. We first prove (A.1) by induction in m ≥ 2 m\ge2 m ≥ 2 m = 2 m=2 m = 2 t k 1 k 2 j 1 j 2 t^{j_1j_2}_{k_1k_2} t k 1 k 2 j 1 j 2
t 1 k 2 ⋯ k m + 1 j m + 1 = t 1 k 2 ⋯ k m j m + β − ( 1 + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j 1 β − 1 + β − 1 t k 2 ⋯ k m j 2 ⋯ j m + β − ( 1 + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j 1 β − 1 + β − 1 t k 2 ⋯ k m + 1 j 2 ⋯ j m + 1 , \begin{aligned}
t^{\mathbf{j}_{m+1}}_{1k_2\cdots k_{m+1}}&=t^{\mathbf{j}_m}_{1k_2\cdots k_m}+\beta^{-(1+k_2+\cdots+k_m)}t^{j_{m+1}}_{k_{m+1}}\\
&=j_1\beta^{-1}+\beta^{-1}t^{j_2\cdots j_m}_{k_2\cdots k_m}+\beta^{-(1+k_2+\cdots+k_m)}t^{j_{m+1}}_{k_{m+1}}\\
&=j_1\beta^{-1}+\beta^{-1}t^{j_2\cdots j_{m+1}}_{k_2\cdots k_{m+1}},
\end{aligned} t 1 k 2 ⋯ k m + 1 j m + 1 = t 1 k 2 ⋯ k m j m + β − ( 1 + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j 1 β − 1 + β − 1 t k 2 ⋯ k m j 2 ⋯ j m + β − ( 1 + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j 1 β − 1 + β − 1 t k 2 ⋯ k m + 1 j 2 ⋯ j m + 1 , where in the second equality we used the induction step.
Now let us prove (A.2), again by induction in m ≥ 2 m\ge2 m ≥ 2 m = 2 m=2 m = 2
t i k 2 j 1 j 2 = t i j 1 + β − ( i − 1 ) t k 2 j 2 = ∑ j = 0 i − 2 a j β j + 1 + j 1 β i + β − i t k 2 j 2 = ∑ j = 0 i − 2 a j β j + 1 + β − ( i − 1 ) ( j 1 β + β − 1 t k 2 j 2 ) = ∑ j = 0 i − 2 a j β j + 1 + β − ( i − 1 ) t 1 k 2 j 1 j 2 . \begin{aligned}
t^{j_1j_2}_{ik_2}&=t^{j_1}_{i}+\beta^{-(i-1)}t^{j_2}_{k_2}
=\sum_{j=0}^{i-2}\frac{a_j}{\beta^{j+1}}+\frac{j_1}{\beta^i}+\beta^{-i}t^{j_2}_{k_2}\\
&=\sum_{j=0}^{i-2}\frac{a_j}{\beta^{j+1}}+\beta^{-(i-1)}\left(\frac{j_1}{\beta}+\beta^{-1}t^{j_2}_{k_2}\right)\\
&=\sum_{j=0}^{i-2}\frac{a_j}{\beta^{j+1}}+\beta^{-(i-1)}t^{j_1j_2}_{1k_2}.
\end{aligned} t i k 2 j 1 j 2 = t i j 1 + β − ( i − 1 ) t k 2 j 2 = j = 0 ∑ i − 2 β j + 1 a j + β i j 1 + β − i t k 2 j 2 = j = 0 ∑ i − 2 β j + 1 a j + β − ( i − 1 ) ( β j 1 + β − 1 t k 2 j 2 ) = j = 0 ∑ i − 2 β j + 1 a j + β − ( i − 1 ) t 1 k 2 j 1 j 2 . Now let m ≥ 2 m\ge2 m ≥ 2
t i k 2 ⋯ k m + 1 j 1 ⋯ j m + 1 = t i k 2 ⋯ k m j 1 ⋯ j m + β − ( i + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = ∑ j = 0 i − 2 a j β j + 1 + β − ( i − 1 ) t 1 k 2 ⋯ k m j 1 ⋯ j m + β − ( i + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = ∑ j = 0 i − 2 a j β j + 1 + β − ( i − 1 ) t 1 k 2 ⋯ k m + 1 j 1 ⋯ j m + 1 , \begin{aligned}
t^{j_1\cdots j_{m+1}}_{ik_2\cdots k_{m+1}}
&=t^{j_1\cdots j_m}_{ik_2\cdots k_m}+\beta^{-(i+k_2+\cdots+k_m)}t^{j_{m+1}}_{k_{m+1}}\\
&=\sum_{j=0}^{i-2}\frac{a_j}{\beta^{j+1}}+\beta^{-(i-1)}t^{j_1\cdots j_m}_{1k_2\cdots k_m}
+\beta^{-(i+k_2+\cdots+k_m)}t^{j_{m+1}}_{k_{m+1}}\\
&=\sum_{j=0}^{i-2}\frac{a_j}{\beta^{j+1}}+\beta^{-(i-1)}t^{j_1\cdots j_{m+1}}_{1k_2\cdots k_{m+1}},
\end{aligned} t i k 2 ⋯ k m + 1 j 1 ⋯ j m + 1 = t i k 2 ⋯ k m j 1 ⋯ j m + β − ( i + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j = 0 ∑ i − 2 β j + 1 a j + β − ( i − 1 ) t 1 k 2 ⋯ k m j 1 ⋯ j m + β − ( i + k 2 + ⋯ + k m ) t k m + 1 j m + 1 = j = 0 ∑ i − 2 β j + 1 a j + β − ( i − 1 ) t 1 k 2 ⋯ k m + 1 j 1 ⋯ j m + 1 , where in the second equality we used the induction step. □
We are now ready to prove Lemma 4.1.
Proof of Lemma 4.1
The functions appearing below were introduced in Definition 4.1. We will use induction and for the initial step we put m = 1 m=1 m = 1
First, let k 1 = 1 k_1=1 k 1 = 1 0 ≤ j 1 ≤ a 0 − 1 0\le j_1\le a_0-1 0 ≤ j 1 ≤ a 0 − 1
G s , 1 j 1 ( x ) = β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ( x − t 1 j 1 ) ) χ [ t 1 j 1 , t 1 j 1 + 1 ] ( x ) = β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ( x − j 1 β ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( x ) . G^{j_1}_{s,1}(x)=\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta\left(x-t^{j_1}_1\right)\right)\chi_{[t^{j_1}_1,t^{j_1+1}_1]}(x)
=\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta\left(x-\frac{j_1}{\beta}\right)\right)\chi_{[j_1\beta^{-1},(j_1+1)\beta^{-1}]}(x). G s , 1 j 1 ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( β ( x − t 1 j 1 ) ) χ [ t 1 j 1 , t 1 j 1 + 1 ] ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( β ( x − β j 1 ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( x ) . Thus by applying the operator P \mathcal{P} P x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ]
β ( P G s , 1 j 1 ) ( x ) = ∑ j = 0 a 0 − 1 G s , 1 j 1 ( x + j β ) χ [ 0 , 1 ] ( x ) + G s , 1 j 1 ( x + a 0 β ) χ [ 0 , 1 ] ( x ) = ∑ j = 0 a 0 − 1 β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ( x + j β − j 1 β ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( x + j β ) χ [ 0 , 1 ] ( x ) + β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ( x + a 0 β − j 1 β ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( x + a 0 β ) χ [ 0 , 1 ] ( x ) = ∑ j = 0 a 0 − 1 β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( x + j − j 1 ) χ [ j 1 − j , j 1 − j + 1 ] ( x ) χ [ 0 , 1 ] ( x ) + β ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( x + a 0 − j 1 ) χ [ j 1 − a 0 , j 1 − a 0 + 1 ] ( x ) χ [ 0 , 1 ] ( x ) . \begin{aligned}
\beta(\mathcal{P}G^{j_1}_{s,1})(x)
&=\sum_{j=0}^{a_0-1}G^{j_1}_{s,1}\left(\frac{x+j}{\beta}\right)\chi_{[0,1]}(x)+G^{j_1}_{s,1}\left(\frac{x+a_0}{\beta}\right)\chi_{[0,1]}(x)\\
&=\sum_{j=0}^{a_0-1}\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta\left(\frac{x+j}{\beta}-\frac{j_1}{\beta}\right)\right)\chi_{[j_1\beta^{-1},(j_1+1)\beta^{-1}]}\left(\frac{x+j}{\beta}\right)\chi_{[0,1]}(x)\\
&\quad+\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta\left(\frac{x+a_0}{\beta}-\frac{j_1}{\beta}\right)\right)\chi_{[j_1\beta^{-1},(j_1+1)\beta^{-1}]}\left(\frac{x+a_0}{\beta}\right)\chi_{[0,1]}(x)\\
&=\sum_{j=0}^{a_0-1}\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s(x+j-j_1)\chi_{[j_1-j,j_1-j+1]}(x)\chi_{[0,1]}(x)\\
&\quad+\frac{\beta}{\lVert B_s\rVert_{L^1([0,1])}}B_s(x+a_0-j_1)\chi_{[j_1-a_0,j_1-a_0+1]}(x)\chi_{[0,1]}(x).
\end{aligned} β ( P G s , 1 j 1 ) ( x ) = j = 0 ∑ a 0 − 1 G s , 1 j 1 ( β x + j ) χ [ 0 , 1 ] ( x ) + G s , 1 j 1 ( β x + a 0 ) χ [ 0 , 1 ] ( x ) = j = 0 ∑ a 0 − 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( β ( β x + j − β j 1 ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( β x + j ) χ [ 0 , 1 ] ( x ) + ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( β ( β x + a 0 − β j 1 ) ) χ [ j 1 β − 1 , ( j 1 + 1 ) β − 1 ] ( β x + a 0 ) χ [ 0 , 1 ] ( x ) = j = 0 ∑ a 0 − 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( x + j − j 1 ) χ [ j 1 − j , j 1 − j + 1 ] ( x ) χ [ 0 , 1 ] ( x ) + ∥ B s ∥ L 1 ([ 0 , 1 ]) β B s ( x + a 0 − j 1 ) χ [ j 1 − a 0 , j 1 − a 0 + 1 ] ( x ) χ [ 0 , 1 ] ( x ) . Since 0 ≤ j 1 ≤ a 0 − 1 0\le j_1\le a_0-1 0 ≤ j 1 ≤ a 0 − 1 j = j 1 j=j_1 j = j 1
( P G s , 1 j 1 ) ( x ) = B s ( x ) ∥ B s ∥ L 1 ( [ 0 , 1 ] ) χ [ 0 , 1 ] ( x ) = H s n + 1 ( x ) . (62) (\mathcal{P}G^{j_1}_{s,1})(x)=\frac{B_s(x)}{\lVert B_s\rVert_{L^1([0,1])}}\chi_{[0,1]}(x)=H_{sn+1}(x). \tag*{(62)} ( P G s , 1 j 1 ) ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) B s ( x ) χ [ 0 , 1 ] ( x ) = H s n + 1 ( x ) . (62) Next, let k 1 = i k_1=i k 1 = i i ∈ { 2 , … , n } i\in\{2,\ldots,n\} i ∈ { 2 , … , n } a i − 1 ≠ 0 a_{i-1}\ne0 a i − 1 = 0 k 1 = i k_1=i k 1 = i 0 ≤ j 1 ≤ a i − 1 − 1 0\le j_1\le a_{i-1}-1 0 ≤ j 1 ≤ a i − 1 − 1
G s , i j 1 ( x ) = β i ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i ( x − t i j 1 ) ) χ [ t i j 1 , t i j 1 + 1 ] ( x ) = β i ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i ( x − ∑ k = 0 i − 2 a k β k + 1 − j 1 β i ) ) χ [ ∑ k = 0 i − 2 a k β k + 1 + j 1 β i , ∑ k = 0 i − 2 a k β k + 1 + j 1 + 1 β i ] ( x ) G^{j_1}_{s,i}(x)=\frac{\beta^i}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^i\left(x-t^{j_1}_i\right)\right)\chi_{[t^{j_1}_i,t^{j_1+1}_i]}(x)
=\frac{\beta^i}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^i\left(x-\sum_{k=0}^{i-2}\frac{a_k}{\beta^{k+1}}-\frac{j_1}{\beta^i}\right)\right)\chi_{\left[\sum_{k=0}^{i-2}\frac{a_k}{\beta^{k+1}}+\frac{j_1}{\beta^i},\sum_{k=0}^{i-2}\frac{a_k}{\beta^{k+1}}+\frac{j_1+1}{\beta^i}\right]}(x) G s , i j 1 ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β i B s ( β i ( x − t i j 1 ) ) χ [ t i j 1 , t i j 1 + 1 ] ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β i B s ( β i ( x − k = 0 ∑ i − 2 β k + 1 a k − β i j 1 ) ) χ [ ∑ k = 0 i − 2 β k + 1 a k + β i j 1 , ∑ k = 0 i − 2 β k + 1 a k + β i j 1 + 1 ] ( x ) and
β ( P G s , i j ) ( x ) = ∑ j = 0 a 0 − 1 G s , i j ( x + j β ) χ [ 0 , 1 ] ( x ) + G s , i j ( x + a 0 β ) χ [ 0 , 1 ] ( x ) = ∑ j = 0 a 0 − 1 β i ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i ( x + j β − a 0 β − a 1 β 2 − ⋯ − a i − 2 β i − 1 − j 1 β i ) ) \begin{aligned}
\beta\left(\mathcal{P}G^{j}_{s,i}\right)(x)=\sum_{j=0}^{a_0-1}G^{j}_{s,i}\left(\frac{x+j}{\beta}\right)\chi_{[0,1]}(x)+G^{j}_{s,i}\left(\frac{x+a_0}{\beta}\right)\chi_{[0,1]}(x) \\
&=\sum_{j=0}^{a_0-1}\frac{\beta^i}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^i\left(\frac{x+j}{\beta}-\frac{a_0}{\beta}-\frac{a_1}{\beta^2}-\cdots-\frac{a_{i-2}}{\beta^{i-1}}-\frac{j_1}{\beta^i}\right)\right)
\end{aligned} β ( P G s , i j ) ( x ) = j = 0 ∑ a 0 − 1 G s , i j ( β x + j ) χ [ 0 , 1 ] ( x ) + G s , i j ( β x + a 0 ) χ [ 0 , 1 ] ( x ) = j = 0 ∑ a 0 − 1 ∥ B s ∥ L 1 ([ 0 , 1 ]) β i B s ( β i ( β x + j − β a 0 − β 2 a 1 − ⋯ − β i − 1 a i − 2 − β i j 1 ) ) ⋅ χ [ a 0 β + a 1 β 2 + ⋯ + a i − 2 β i − 1 + j 1 β i , a 0 β + ⋯ + a i − 2 β i − 1 + j 1 + 1 β i ] ( x ) χ [ 0 , 1 ] ( x ) + β i ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i − 1 ( x − a 1 β − ⋯ − a i − 2 β i − 2 − j 1 β i − 1 ) ) \begin{aligned}
\qquad\cdot\chi_{\left[\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\frac{j_1}{\beta^i},\frac{a_0}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\frac{j_1+1}{\beta^i}\right]}(x)\chi_{[0,1]}(x) \\
&+\frac{\beta^i}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{i-1}\left(x-\frac{a_1}{\beta}-\cdots-\frac{a_{i-2}}{\beta^{i-2}}-\frac{j_1}{\beta^{i-1}}\right)\right)
\end{aligned} ⋅ χ [ β a 0 + β 2 a 1 + ⋯ + β i − 1 a i − 2 + β i j 1 , β a 0 + ⋯ + β i − 1 a i − 2 + β i j 1 + 1 ] ( x ) χ [ 0 , 1 ] ( x ) + ∥ B s ∥ L 1 ([ 0 , 1 ]) β i B s ( β i − 1 ( x − β a 1 − ⋯ − β i − 2 a i − 2 − β i − 1 j 1 ) ) ⋅ χ [ a 1 β + ⋯ + a i − 2 β i − 2 + j 1 β i − 1 , a 1 β + ⋯ + a i − 2 β i − 2 + j 1 + 1 β i − 1 ] ( x ) χ [ 0 , 1 ] ( x ) . \qquad\cdot\chi_{\left[\frac{a_1}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-2}}+\frac{j_1}{\beta^{i-1}},\frac{a_1}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-2}}+\frac{j_1+1}{\beta^{i-1}}\right]}(x)\chi_{[0,1]}(x). ⋅ χ [ β a 1 + ⋯ + β i − 2 a i − 2 + β i − 1 j 1 , β a 1 + ⋯ + β i − 2 a i − 2 + β i − 1 j 1 + 1 ] ( x ) χ [ 0 , 1 ] ( x ) . Since 0 ≤ j ≤ a 0 − 1 0\leq j\leq a_0-1 0 ≤ j ≤ a 0 − 1 j j j
( P G s , i j 1 ) ( x ) = β i − 1 ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i − 1 ( x − a 1 β − ⋯ − a i − 2 β i − 2 − j 1 β i − 1 ) ) \left(\mathcal{P}G^{j_1}_{s,i}\right)(x)=\frac{\beta^{i-1}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{i-1}\left(x-\frac{a_1}{\beta}-\cdots-\frac{a_{i-2}}{\beta^{i-2}}-\frac{j_1}{\beta^{i-1}}\right)\right) ( P G s , i j 1 ) ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β i − 1 B s ( β i − 1 ( x − β a 1 − ⋯ − β i − 2 a i − 2 − β i − 1 j 1 ) ) × χ [ a 1 β + ⋯ + a i − 2 β i − 2 + j 1 β i − 1 , a 1 β + ⋯ + a i − 2 β i − 2 + j 1 + 1 β i − 1 ] ( x ) . \qquad\times\chi_{\left[\frac{a_1}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-2}}+\frac{j_1}{\beta^{i-1}},\frac{a_1}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-2}}+\frac{j_1+1}{\beta^{i-1}}\right]}(x). × χ [ β a 1 + ⋯ + β i − 2 a i − 2 + β i − 1 j 1 , β a 1 + ⋯ + β i − 2 a i − 2 + β i − 1 j 1 + 1 ] ( x ) . Another ( i − 1 ) (i-1) ( i − 1 ) P \mathcal{P} P a ν < a 0 a_\nu<a_0 a ν < a 0 a ν = a 0 a_\nu=a_0 a ν = a 0 ν ∈ { 1 , … , n − 1 } \nu\in\{1,\ldots,n-1\} ν ∈ { 1 , … , n − 1 } a i − 1 ≤ a 0 a_{i-1}\leq a_0 a i − 1 ≤ a 0
( P i G s , i j 1 ) ( x ) = H s n + 1 ( x ) , \left(\mathcal{P}^{i}G^{j_1}_{s,i}\right)(x)=H_{sn+1}(x), ( P i G s , i j 1 ) ( x ) = H s n + 1 ( x ) , which finishes the basis step with m = 1 m=1 m = 1
Now we assume that the equality is true for m = ℓ m=\ell m = ℓ m = ℓ + 1 m=\ell+1 m = ℓ + 1 k 1 = 1 k_1=1 k 1 = 1
G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x ) = β 1 + k 2 + ⋯ + k ℓ + 1 ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β 1 + k 2 + ⋯ + k ℓ + 1 ( x − t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) χ [ t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 , t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 + 1 β 1 + k 2 + ⋯ + k ℓ + 1 ] ( x ) = β 1 + k 2 + ⋯ + k ℓ + 1 ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β 1 + k 2 + ⋯ + k ℓ + 1 ( x − j 1 β − β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 ) ) \begin{aligned}
G^{j_{\ell+1}}_{s,1k_2\cdots k_{\ell+1}}(x)=\frac{\beta^{1+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{1+k_2+\cdots+k_{\ell+1}}\left(x-t^{j_{\ell+1}}_{1k_2\cdots k_{\ell+1}}\right)\right)\chi_{\left[t^{j_{\ell+1}}_{1k_2\cdots k_{\ell+1}},t^{j_{\ell+1}}_{1k_2\cdots k_{\ell+1}}+\frac{1}{\beta^{1+k_2+\cdots+k_{\ell+1}}}\right]}(x) \\
&=\frac{\beta^{1+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{1+k_2+\cdots+k_{\ell+1}}\left(x-\frac{j_1}{\beta}-\beta^{-1}t^{j_2\cdots j_{\ell+1}}_{k_2\cdots k_{\ell+1}}\right)\right)
\end{aligned} G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β 1 + k 2 + ⋯ + k ℓ + 1 B s ( β 1 + k 2 + ⋯ + k ℓ + 1 ( x − t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) χ [ t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 , t 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 + β 1 + k 2 + ⋯ + k ℓ + 1 1 ] ( x ) = ∥ B s ∥ L 1 ([ 0 , 1 ]) β 1 + k 2 + ⋯ + k ℓ + 1 B s ( β 1 + k 2 + ⋯ + k ℓ + 1 ( x − β j 1 − β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 ) ) ⋅ χ [ j 1 β + β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 , j 1 β + β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 + 1 β 1 + k 2 + ⋯ + k ℓ + 1 ] ( x ) , \qquad\cdot\chi_{\left[\frac{j_1}{\beta}+\beta^{-1}t^{j_2\cdots j_{\ell+1}}_{k_2\cdots k_{\ell+1}},\frac{j_1}{\beta}+\beta^{-1}t^{j_2\cdots j_{\ell+1}}_{k_2\cdots k_{\ell+1}}+\frac{1}{\beta^{1+k_2+\cdots+k_{\ell+1}}}\right]}(x), ⋅ χ [ β j 1 + β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 , β j 1 + β − 1 t k 2 ⋯ k ℓ + 1 j 2 ⋯ j ℓ + 1 + β 1 + k 2 + ⋯ + k ℓ + 1 1 ] ( x ) , for 0 ≤ j 1 ≤ a 0 − 1 0\leq j_1\leq a_0-1 0 ≤ j 1 ≤ a 0 − 1 x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ]
β ( P 1 + k 2 + ⋯ + k ℓ + 1 G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ( x ) = ( P k 2 + ⋯ + k ℓ + 1 ( β P G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) ( x ) = P k 2 + ⋯ + k ℓ + 1 ( ∑ j = 0 a 0 − 1 G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x + j β ) χ [ 0 , 1 ] ( x ) + G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x + a 0 β ) χ [ 0 , 1 ] ( x ) ) = ( P k 2 + ⋯ + k ℓ + 1 G s , k 2 ⋯ k m + 1 j 2 ⋯ j m + 1 ) ( x ) = H s n + 1 ( x ) . \begin{aligned}
\beta\left(\mathcal{P}^{1+k_2+\cdots+k_{\ell+1}}G^{j_{\ell+1}}_{s,1k_2\cdots k_{\ell+1}}\right)(x)=\left(\mathcal{P}^{k_2+\cdots+k_{\ell+1}}\left(\beta\mathcal{P}G^{j_{\ell+1}}_{s,1k_2\cdots k_{\ell+1}}\right)\right)(x) \\
&=\mathcal{P}^{k_2+\cdots+k_{\ell+1}}\left(\sum_{j=0}^{a_0-1}G^{j_{\ell+1}}_{s,1k_2\cdots k_{\ell+1}}\left(\frac{x+j}{\beta}\right)\chi_{[0,1]}(x)+G^{j_{\ell+1}}_{s,1k_2\cdots k_{\ell+1}}\left(\frac{x+a_0}{\beta}\right)\chi_{[0,1]}(x)\right) \\
&=\left(\mathcal{P}^{k_2+\cdots+k_{\ell+1}}G^{j_2\cdots j_{m+1}}_{s,k_2\cdots k_{m+1}}\right)(x) \\
&=H_{sn+1}(x).
\end{aligned} β ( P 1 + k 2 + ⋯ + k ℓ + 1 G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ( x ) = ( P k 2 + ⋯ + k ℓ + 1 ( β P G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) ( x ) = P k 2 + ⋯ + k ℓ + 1 ( j = 0 ∑ a 0 − 1 G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( β x + j ) χ [ 0 , 1 ] ( x ) + G s , 1 k 2 ⋯ k ℓ + 1 j ℓ + 1 ( β x + a 0 ) χ [ 0 , 1 ] ( x ) ) = ( P k 2 + ⋯ + k ℓ + 1 G s , k 2 ⋯ k m + 1 j 2 ⋯ j m + 1 ) ( x ) = H s n + 1 ( x ) . where in the third equality we have used that the second term is zero a.e., and that in the sum, only the term j = j 1 j=j_1 j = j 1
Next let i ∈ { 2 , … , n } i\in\{2,\ldots,n\} i ∈ { 2 , … , n } k 1 = i k_1=i k 1 = i a i − 1 ≠ 0 a_{i-1}\ne0 a i − 1 = 0
\begin{aligned}
G_{s,ik_2\cdots k_{\ell+1}}^{j_{\ell+1}}(x)=\frac{\beta^{i+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{i+k_2+\cdots+k_{\ell+1}}\left(x-\frac{j_{\ell+1}_{ik_2\cdots k_{\ell+1}}}{\beta^{i+k_2+\cdots+k_{\ell+1}}}\right)\right)\chi_{\left[\frac{j_{\ell+1}_{ik_2\cdots k_{\ell+1}}}{\beta^{i+k_2+\cdots+k_{\ell+1}}},\frac{j_{\ell+1}_{ik_2\cdots k_{\ell+1}}+1}{\beta^{i+k_2+\cdots+k_{\ell+1}}}\right]}(x) \\
&=\frac{\beta^{i+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{i+k_2+\cdots+k_{\ell+1}}\left(x-\sum_{m=0}^{i-2}\frac{a_m}{\beta^{m+1}}-\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}\right)\right)
\cdot\chi_{\left[\sum_{m=0}^{i-2}\frac{a_m}{\beta^{m+1}}+\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1},\sum_{m=0}^{i-2}\frac{a_m}{\beta^{m+1}}+\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}+1/\beta^{i+k_2+\cdots+k_{\ell+1}}\right]}(x),
\end{aligned}
for 0 ≤ j 1 ≤ a i − 1 − 1 0\le j_1\le a_{i-1}-1 0 ≤ j 1 ≤ a i − 1 − 1 x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ]
β ( P i + k 2 + ⋯ + k ℓ + 1 G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ( x ) = ( P ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( β P G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) ( x ) = P ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( ∑ j = 0 a 0 − 1 G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x + j β ) χ [ 0 , 1 ] ( x ) + G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ( x + a 0 β ) χ [ 0 , 1 ] ( x ) ) = P ( i − 1 ) + k 2 + ⋯ + k m + 1 ( β i + k 2 + ⋯ + k ℓ + 1 ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β i + k 2 + ⋯ + k ℓ + 1 ( x β − ∑ m = 1 i − 2 a m β m + 1 − β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 ) ) \begin{aligned}
\beta\left(\mathcal{P}^{i+k_2+\cdots+k_{\ell+1}}G_{s,ik_2\cdots k_{\ell+1}}^{j_{\ell+1}}\right)(x)=\left(\mathcal{P}^{(i-1)+k_2+\cdots+k_{\ell+1}}\left(\beta\mathcal{P}G_{s,ik_2\cdots k_{\ell+1}}^{j_{\ell+1}}\right)\right)(x) \\
&=\mathcal{P}^{(i-1)+k_2+\cdots+k_{\ell+1}}\left(\sum_{j=0}^{a_0-1}G_{s,ik_2\cdots k_{\ell+1}}^{j_{\ell+1}}\left(\frac{x+j}{\beta}\right)\chi_{[0,1]}(x)+G_{s,ik_2\cdots k_{\ell+1}}^{j_{\ell+1}}\left(\frac{x+a_0}{\beta}\right)\chi_{[0,1]}(x)\right) \\
&=\mathcal{P}^{(i-1)+k_2+\cdots+k_{m+1}}\left(\frac{\beta^{i+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{i+k_2+\cdots+k_{\ell+1}}\left(\frac{x}{\beta}-\sum_{m=1}^{i-2}\frac{a_m}{\beta^{m+1}}-\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}\right)\right)\right.
\end{aligned} β ( P i + k 2 + ⋯ + k ℓ + 1 G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ( x ) = ( P ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( β P G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ) ) ( x ) = P ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( j = 0 ∑ a 0 − 1 G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ( β x + j ) χ [ 0 , 1 ] ( x ) + G s , i k 2 ⋯ k ℓ + 1 j ℓ + 1 ( β x + a 0 ) χ [ 0 , 1 ] ( x ) ) = P ( i − 1 ) + k 2 + ⋯ + k m + 1 ( ∥ B s ∥ L 1 ([ 0 , 1 ]) β i + k 2 + ⋯ + k ℓ + 1 B s ( β i + k 2 + ⋯ + k ℓ + 1 ( β x − m = 1 ∑ i − 2 β m + 1 a m − β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 ) ) ⋅ χ [ a 0 β + ⋯ + a i − 2 β i − 1 + β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 , a 0 β + ⋯ + a i − 2 β i − 1 + β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 + 1 / β i + k 2 + ⋯ + k ℓ + 1 ] ( x + a 0 β ) ) = P ( i − 1 ) + k 2 + ⋯ + k m + 1 ( β i + k 2 + ⋯ + k ℓ + 1 ∥ B s ∥ L 1 ( [ 0 , 1 ] ) B s ( β ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( x − ∑ m = 1 i − 2 a m β m − β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 ) ) \begin{aligned}
\left.\cdot\chi_{\left[\frac{a_0}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1},\frac{a_0}{\beta}+\cdots+\frac{a_{i-2}}{\beta^{i-1}}+\beta^{-(i-1)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}+1/\beta^{i+k_2+\cdots+k_{\ell+1}}\right]}\left(\frac{x+a_0}{\beta}\right)\right) \\
&=\mathcal{P}^{(i-1)+k_2+\cdots+k_{m+1}}\left(\frac{\beta^{i+k_2+\cdots+k_{\ell+1}}}{\lVert B_s\rVert_{L^1([0,1])}}B_s\left(\beta^{(i-1)+k_2+\cdots+k_{\ell+1}}\left(x-\sum_{m=1}^{i-2}\frac{a_m}{\beta^m}-\beta^{-(i-2)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}\right)\right)\right.
\end{aligned} ⋅ χ [ β a 0 + ⋯ + β i − 1 a i − 2 + β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 , β a 0 + ⋯ + β i − 1 a i − 2 + β − ( i − 1 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 + 1/ β i + k 2 + ⋯ + k ℓ + 1 ] ( β x + a 0 ) ) = P ( i − 1 ) + k 2 + ⋯ + k m + 1 ( ∥ B s ∥ L 1 ([ 0 , 1 ]) β i + k 2 + ⋯ + k ℓ + 1 B s ( β ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ( x − m = 1 ∑ i − 2 β m a m − β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 ) ) ⋅ χ [ ∑ m = 1 i − 2 a m β m + β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 , ∑ m = 1 i − 2 a m β m + β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 + 1 / β ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ] ( x ) ) , \left.\cdot\chi_{\left[\sum_{m=1}^{i-2}\frac{a_m}{\beta^m}+\beta^{-(i-2)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1},\sum_{m=1}^{i-2}\frac{a_m}{\beta^m}+\beta^{-(i-2)}j_{\!1k_2\cdots k_{\ell+1}}^{\ell+1}+1/\beta^{(i-1)+k_2+\cdots+k_{\ell+1}}\right]}(x)\right), ⋅ χ [ ∑ m = 1 i − 2 β m a m + β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 , ∑ m = 1 i − 2 β m a m + β − ( i − 2 ) j 1 k 2 ⋯ k ℓ + 1 ℓ + 1 + 1/ β ( i − 1 ) + k 2 + ⋯ + k ℓ + 1 ] ( x ) ) , where in the third equality, we have used that the sum over 0 ≤ j ≤ a 0 − 1 0\le j\le a_0-1 0 ≤ j ≤ a 0 − 1
Applying P \mathcal{P} P ( i − 2 ) (i-2) ( i − 2 ) a ν < a 0 a_\nu<a_0 a ν < a 0 a ν = a 0 a_\nu=a_0 a ν = a 0 ν ∈ { 1 , … , n − 1 } \nu\in\{1,\ldots,n-1\} ν ∈ { 1 , … , n − 1 } k 1 = 1 k_1=1 k 1 = 1 □ \square □
Proof of Lemma 4.6
By definition, there exists j 1 < a k 1 − 1 j_1<a_{k_1-1} j 1 < a k 1 − 1
t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = a 0 β + a 1 β 2 + ⋯ + a k 1 − 2 β k 1 − 1 + j 1 β k 1 + β − k 1 t k 2 ⋯ k m j 2 ⋯ j m . t_{k_1k_2\cdots k_m}^{j_1j_2\cdots j_m}
=\frac{a_0}{\beta}+\frac{a_1}{\beta^2}+\cdots+\frac{a_{k_1-2}}{\beta^{k_1-1}}+\frac{j_1}{\beta^{k_1}}+\beta^{-k_1}t_{k_2\cdots k_m}^{j_2\cdots j_m}. t k 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = β a 0 + β 2 a 1 + ⋯ + β k 1 − 1 a k 1 − 2 + β k 1 j 1 + β − k 1 t k 2 ⋯ k m j 2 ⋯ j m . If k 1 = 2 k_1=2 k 1 = 2
t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m = a 0 β + j 1 β 2 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m t_{2k_2\cdots k_m}^{j_1j_2\cdots j_m}
=\frac{a_0}{\beta}+\frac{j_1}{\beta^2}+\beta^{-2}t_{k_2\cdots k_m}^{j_2\cdots j_m} t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m = β a 0 + β 2 j 1 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m hence t ~ = t 1 k 2 ⋯ k m j 1 j 2 ⋯ j m \widetilde{t}=t_{1k_2\cdots k_m}^{j_1j_2\cdots j_m} t = t 1 k 2 ⋯ k m j 1 j 2 ⋯ j m k 1 ≥ 3 k_1\ge3 k 1 ≥ 3
If a 1 , … , a k 1 − 2 < a 0 a_1,\ldots,a_{k_1-2}<a_0 a 1 , … , a k 1 − 2 < a 0
t ~ = a 1 β + ⋯ + a k 1 − 2 β k 1 − 2 + j 1 β k 1 − 1 + β − ( k 1 − 1 ) t k 2 ⋯ k m j 2 ⋯ j m = t 1 × ( k 1 − 1 ) k 2 ⋯ k m a 1 ⋯ a k 1 − 2 j 1 ⋯ j m \widetilde{t}=\frac{a_1}{\beta}+\cdots+\frac{a_{k_1-2}}{\beta^{k_1-2}}+\frac{j_1}{\beta^{k_1-1}}+\beta^{-(k_1-1)}t_{k_2\cdots k_m}^{j_2\cdots j_m}
=t_{1\times(k_1-1)k_2\cdots k_m}^{a_1\cdots a_{k_1-2}j_1\cdots j_m} t = β a 1 + ⋯ + β k 1 − 2 a k 1 − 2 + β k 1 − 1 j 1 + β − ( k 1 − 1 ) t k 2 ⋯ k m j 2 ⋯ j m = t 1 × ( k 1 − 1 ) k 2 ⋯ k m a 1 ⋯ a k 1 − 2 j 1 ⋯ j m where again the notation 1 × ( k 1 − 1 ) 1 \times(k_1-1) 1 × ( k 1 − 1 ) k 1 − 1 k_1-1 k 1 − 1 1 ≤ r ≤ k 1 − 2 1 \le r \le k_1-2 1 ≤ r ≤ k 1 − 2 a r = a 0 a_r=a_0 a r = a 0 r 1 r_1 r 1 a 1 , a 2 , … , a r 1 − 1 < a 0 a_1,a_2,\ldots,a_{r_1-1}<a_0 a 1 , a 2 , … , a r 1 − 1 < a 0
a 1 β + ⋯ + a r 1 − 1 β r 1 − 1 = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 . \frac{a_1}{\beta}+\cdots+\frac{a_{r_1-1}}{\beta^{r_1-1}}=t_{1\times(r_1-1)}^{a_1\cdots a_{r_1-1}}. β a 1 + ⋯ + β r 1 − 1 a r 1 − 1 = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 . Then with a r 1 = a 0 a_{r_1}=a_0 a r 1 = a 0
t ~ = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 + β − ( r 1 − 1 ) ( a 0 β + a r 1 + 1 β 2 + ⋯ + a k 1 − 2 β k 1 − r 1 − 1 + j 1 β k 1 − r 1 ) + β − ( k 1 − 1 ) t k 2 ⋯ k m j 2 ⋯ j m . \widetilde{t}=t_{1\times(r_1-1)}^{a_1\cdots a_{r_1-1}}+\beta^{-(r_1-1)}\left(\frac{a_0}{\beta}+\frac{a_{r_1+1}}{\beta^2}+\cdots+\frac{a_{k_1-2}}{\beta^{k_1-r_1-1}}+\frac{j_1}{\beta^{k_1-r_1}}\right)+\beta^{-(k_1-1)}t_{k_2\cdots k_m}^{j_2\cdots j_m}. t = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 + β − ( r 1 − 1 ) ( β a 0 + β 2 a r 1 + 1 + ⋯ + β k 1 − r 1 − 1 a k 1 − 2 + β k 1 − r 1 j 1 ) + β − ( k 1 − 1 ) t k 2 ⋯ k m j 2 ⋯ j m . We need to control the parenthesis. Parry’s lexicographic condition implies:
[ a r 1 , a r 1 + 1 , … , a k 1 − 2 , j 1 , 0 , … ] ≺ [ a 0 , a 1 , … , a n − 1 , 0 , … ] . [a_{r_1},a_{r_1+1},\ldots,a_{k_1-2},j_1,0,\ldots]\prec[a_0,a_1,\ldots,a_{n-1},0,\ldots]. [ a r 1 , a r 1 + 1 , … , a k 1 − 2 , j 1 , 0 , … ] ≺ [ a 0 , a 1 , … , a n − 1 , 0 , … ] . In the “shortest case scenario", we have a “best saturation" of the above inequality where a r 1 + 1 = a 1 a_{r_1+1}=a_1 a r 1 + 1 = a 1 a r 1 + 2 = a 2 a_{r_1+2}=a_2 a r 1 + 2 = a 2 a k 1 − 2 = a k 1 − r 1 − 2 a_{k_1-2}=a_{k_1-r_1-2} a k 1 − 2 = a k 1 − r 1 − 2 j 1 < a k 1 − 1 ≤ a k 1 − r 1 − 1 j_1<a_{k_1-1}\le a_{k_1-r_1-1} j 1 < a k 1 − 1 ≤ a k 1 − r 1 − 1
a 0 β + a r 1 + 1 β 2 + ⋯ + a s − 2 β k 1 − r 1 − 1 + j 1 β k 1 − r 1 = t k 1 − r 1 a 0 a r 1 + 1 ⋯ a s − 2 j 1 \frac{a_0}{\beta}+\frac{a_{r_1+1}}{\beta^2}+\cdots+\frac{a_{s-2}}{\beta^{k_1-r_1-1}}+\frac{j_1}{\beta^{k_1-r_1}}=t_{k_1-r_1}^{a_0a_{r_1+1}\cdots a_{s-2}j_1} β a 0 + β 2 a r 1 + 1 + ⋯ + β k 1 − r 1 − 1 a s − 2 + β k 1 − r 1 j 1 = t k 1 − r 1 a 0 a r 1 + 1 ⋯ a s − 2 j 1 and
t ~ = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 + β − ( r 1 − 1 ) ( t k 1 − r 1 j 1 + β − ( k 1 − r 1 ) t k 2 ⋯ k m j 2 ⋯ j m ) = t 1 × ( r 1 − 1 ) , k 1 − r 1 , k 2 ⋯ k m a 1 ⋯ a r 1 − 1 j 1 j 2 ⋯ j m . \widetilde{t}=t_{1\times(r_1-1)}^{a_1\cdots a_{r_1-1}}+\beta^{-(r_1-1)}\left(t_{k_1-r_1}^{j_1}+\beta^{-(k_1-r_1)}t_{k_2\cdots k_m}^{j_2\cdots j_m}\right)=t_{1\times(r_1-1),k_1-r_1,k_2\cdots k_m}^{a_1\cdots a_{r_1-1}j_1j_2\cdots j_m}. t = t 1 × ( r 1 − 1 ) a 1 ⋯ a r 1 − 1 + β − ( r 1 − 1 ) ( t k 1 − r 1 j 1 + β − ( k 1 − r 1 ) t k 2 ⋯ k m j 2 ⋯ j m ) = t 1 × ( r 1 − 1 ) , k 1 − r 1 , k 2 ⋯ k m a 1 ⋯ a r 1 − 1 j 1 j 2 ⋯ j m . If the best saturation does not happen, for example when a r 1 + 1 < a 1 a_{r_1+1}<a_1 a r 1 + 1 < a 1
a 0 β + a r 1 + 1 β 2 + ⋯ + a k 1 − 2 β k 1 − r 1 − 1 + j 1 β k 1 − r 1 = t 2 a r 1 + 1 + β − 2 ( a r 1 + 2 β + ⋯ + j 1 β k 1 − r 1 − 2 ) \frac{a_0}{\beta}+\frac{a_{r_1+1}}{\beta^2}+\cdots+\frac{a_{k_1-2}}{\beta^{k_1-r_1-1}}+\frac{j_1}{\beta^{k_1-r_1}}=t_{2}^{a_{r_1+1}}+\beta^{-2}\left(\frac{a_{r_1+2}}{\beta}+\cdots+\frac{j_1}{\beta^{k_1-r_1-2}}\right) β a 0 + β 2 a r 1 + 1 + ⋯ + β k 1 − r 1 − 1 a k 1 − 2 + β k 1 − r 1 j 1 = t 2 a r 1 + 1 + β − 2 ( β a r 1 + 2 + ⋯ + β k 1 − r 1 − 2 j 1 ) and we continue by induction using Parry’s condition until the parenthesis is exhausted. Now let us prove the other way around, that is if a 0 / β + β − 1 t k m j m < 1 a_0/\beta+\beta^{-1}t_{k_m}^{j_m}<1 a 0 / β + β − 1 t k m j m < 1 t k ~ p j ~ p t_{\widetilde{k}_p}^{\widetilde{j}_p} t k p j p k 1 = 1 k_1=1 k 1 = 1
a 0 β + β − 1 t 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = a 0 β + j 1 β 2 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m . \frac{a_0}{\beta}+\beta^{-1}t_{1k_2\cdots k_m}^{j_1j_2\cdots j_m}=\frac{a_0}{\beta}+\frac{j_1}{\beta^2}+\beta^{-2}t_{k_2\cdots k_m}^{j_2\cdots j_m}. β a 0 + β − 1 t 1 k 2 ⋯ k m j 1 j 2 ⋯ j m = β a 0 + β 2 j 1 + β − 2 t k 2 ⋯ k m j 2 ⋯ j m . If j 1 > a 1 j_1>a_1 j 1 > a 1 j 1 < a 1 j_1<a_1 j 1 < a 1 t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m t_{2k_2\cdots k_m}^{j_1j_2\cdots j_m} t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m j 1 = a 1 j_1=a_1 j 1 = a 1 t 3 k 2 ⋯ k m 0 j 2 ⋯ j m t_{3k_2\cdots k_m}^{0j_2\cdots j_m} t 3 k 2 ⋯ k m 0 j 2 ⋯ j m
If k 1 = 2 k_1=2 k 1 = 2
a 0 β + β − 1 t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m = a 0 β + a 0 β 2 + j 1 β 3 + β − 3 t k 2 ⋯ k m j 2 ⋯ j m , j 1 < a 1 . \frac{a_0}{\beta}+\beta^{-1}t_{2k_2\cdots k_m}^{j_1j_2\cdots j_m}=\frac{a_0}{\beta}+\frac{a_0}{\beta^2}+\frac{j_1}{\beta^3}+\beta^{-3}t_{k_2\cdots k_m}^{j_2\cdots j_m},\qquad j_1<a_1. β a 0 + β − 1 t 2 k 2 ⋯ k m j 1 j 2 ⋯ j m = β a 0 + β 2 a 0 + β 3 j 1 + β − 3 t k 2 ⋯ k m j 2 ⋯ j m , j 1 < a 1 . First of all, if a 1 < a 0 a_1<a_0 a 1 < a 0 a 1 = a 0 a_1=a_0 a 1 = a 0 j 1 < a 2 j_1<a_2 j 1 < a 2 t 3 k 2 ⋯ k m j 1 j 2 ⋯ j m t_{3k_2\cdots k_m}^{j_1j_2\cdots j_m} t 3 k 2 ⋯ k m j 1 j 2 ⋯ j m j 1 > a 2 j_1>a_2 j 1 > a 2 j 1 = a 2 j_1=a_2 j 1 = a 2 t 4 k 2 ⋯ k m 0 j 2 ⋯ j m t_{4k_2\cdots k_m}^{0j_2\cdots j_m} t 4 k 2 ⋯ k m 0 j 2 ⋯ j m k 1 ≥ 3 k_1\ge3 k 1 ≥ 3
Euler-Bernoulli approximation formula
We refer the reader to [3 ] for proofs of the results in this section. Define B 0 ( x ) = 1 B_0(x)=1 B 0 ( x ) = 1 B 1 ( x ) = x − 1 / 2 B_1(x)=x-1/2 B 1 ( x ) = x − 1/2 x ∈ [ 0 , 1 ) x\in[0,1) x ∈ [ 0 , 1 ) Z \mathbb{Z} Z B 0 B_0 B 0 B 1 B_1 B 1 R \mathbb{R} R B ~ 0 \widetilde{B}_0 B 0 B ~ 1 \widetilde{B}_1 B 1 r ≥ 2 r \ge2 r ≥ 2
B ~ r ( x ) : = − r ! ∑ m ≠ 0 e 2 π i m x ( 2 π i m ) − r , d B ~ r d x ( x ) = r B ~ r − 1 ( x ) , 0 < x < 1. \widetilde{B}_r(x):=-r!\sum_{m\ne0}e^{2\pi imx}(2\pi im)^{-r},\qquad\frac{d\widetilde{B}_r}{dx}(x)=r\widetilde{B}_{r-1}(x),\qquad0<x<1. B r ( x ) := − r ! m = 0 ∑ e 2 π im x ( 2 π im ) − r , d x d B r ( x ) = r B r − 1 ( x ) , 0 < x < 1. Then B ~ r \widetilde{B}_r B r Z \mathbb{Z} Z B r B_r B r ( 0 , 1 ) (0,1) ( 0 , 1 ) r ∈ N r\in\mathbb{N} r ∈ N F ~ ∈ C N ( [ 0 , 1 ] ) \widetilde{F}\in C^N([0,1]) F ∈ C N ([ 0 , 1 ]) N ≥ 1 N\ge1 N ≥ 1
F ~ ( x ) = ∫ 0 1 F ~ ( t ) d t + ∑ s = 1 N ( F ~ ( s − 1 ) ( 1 ) − F ~ ( s − 1 ) ( 0 ) ) B ~ s ( x ) s ! \widetilde{F}(x)=\int_0^1\widetilde{F}(t)\,dt+\sum_{s=1}^N\left(\widetilde{F}^{(s-1)}(1)-\widetilde{F}^{(s-1)}(0)\right)\frac{\widetilde{B}_s(x)}{s!} F ( x ) = ∫ 0 1 F ( t ) d t + s = 1 ∑ N ( F ( s − 1 ) ( 1 ) − F ( s − 1 ) ( 0 ) ) s ! B s ( x ) − ∫ 0 1 F ~ ( N ) ( u ) B ~ N ( x − u ) N ! d u , 0 < x < 1. (63) -\int_0^1\widetilde{F}^{(N)}(u)\frac{\widetilde{B}_N(x-u)}{N!}\,du,\qquad0<x<1. \tag*{(63)} − ∫ 0 1 F ( N ) ( u ) N ! B N ( x − u ) d u , 0 < x < 1. (63) This can be rescaled to general intervals. If F : [ a , b ] → R F:[a,b]\to\mathbb{R} F : [ a , b ] → R C N ( [ a , b ] ) C^N([a,b]) C N ([ a , b ]) F ~ ( x ) = F ( a ( 1 − x ) + b x ) \widetilde{F}(x)=F(a(1-x)+bx) F ( x ) = F ( a ( 1 − x ) + b x ) 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 a < y < b a<y<b a < y < b
F ( y ) = ( b − a ) − 1 ∫ a b F ( u ) d u + ∑ s = 1 N ( b − a ) s − 1 ( F ( s − 1 ) ( b ) − F ( s − 1 ) ( a ) ) B ~ s ( y − a b − a ) s ! F(y)=(b-a)^{-1}\int_a^bF(u)\,du+\sum_{s=1}^N(b-a)^{s-1}\left(F^{(s-1)}(b)-F^{(s-1)}(a)\right)\frac{\widetilde{B}_s\left(\frac{y-a}{b-a}\right)}{s!} F ( y ) = ( b − a ) − 1 ∫ a b F ( u ) d u + s = 1 ∑ N ( b − a ) s − 1 ( F ( s − 1 ) ( b ) − F ( s − 1 ) ( a ) ) s ! B s ( b − a y − a ) − ( b − a ) N ∫ 0 1 F ( N ) ( a ( 1 − u ) + b u ) B ~ N ( y − a b − a − u ) N ! d u . (64) -(b-a)^N\int_0^1F^{(N)}(a(1-u)+bu)\frac{\widetilde{B}_N\left(\frac{y-a}{b-a}-u\right)}{N!}\,du. \tag*{(64)} − ( b − a ) N ∫ 0 1 F ( N ) ( a ( 1 − u ) + b u ) N ! B N ( b − a y − a − u ) d u . (64) By this rescaled Bernoulli approximation formula applied to each interval in the splitting, i.e. b = t k m j m + β − ∣ k m ∣ b=t_{k_m}^{j_m}+\beta^{-|k_m|} b = t k m j m + β − ∣ k m ∣ a = t k m j m a=t_{k_m}^{j_m} a = t k m j m
F ( y ) = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + ∑ s = 1 N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) B ~ s ( β ∣ k m ∣ ( y − t k m j m ) ) s ! \begin{aligned}
F(y)=\beta^{|k_m|}\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-|k_m|}}F(u)\,du \\
&+\sum_{s=1}^N\beta^{-(s-1)|k_m|}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-|k_m|}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{\widetilde{B}_s\left(\beta^{|k_m|}\left(y-t_{k_m}^{j_m}\right)\right)}{s!}
\end{aligned} F ( y ) = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + s = 1 ∑ N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! B s ( β ∣ k m ∣ ( y − t k m j m ) ) − β − N ∣ k m ∣ ∫ 0 1 F ( N ) ( t k m j m ( 1 − u ) + t k m j m + β − ∣ k m ∣ u ) B ~ N ( β ∣ k m ∣ ( y − t k m j m ) − u ) N ! d u = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + ∑ s = 1 N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) B ~ s ( β ∣ k m ∣ ( y − t k m j m ) ) s ! + O ( β − ∣ k m ∣ N ∥ F ( N ) ∥ C ( [ 0 , 1 ] ) ) , ∀ y ∈ [ t k m j m , t k m j m + β − ∣ k m ∣ ] . (65) \begin{aligned}
-\beta^{-N|k_m|}\int_0^1F^{(N)}\left(t_{k_m}^{j_m}(1-u)+t_{k_m}^{j_m}+\beta^{-|k_m|}u\right)\frac{\widetilde{B}_N\left(\beta^{|k_m|}\left(y-t_{k_m}^{j_m}\right)-u\right)}{N!}\,du \\
&=\beta^{|k_m|}\int_{t_{k_m}^{j_m}}^{t_{k_m}^{j_m}+\beta^{-|k_m|}}F(u)\,du \\
&+\sum_{s=1}^N\beta^{-(s-1)|k_m|}\left(F^{(s-1)}\left(t_{k_m}^{j_m}+\beta^{-|k_m|}\right)-F^{(s-1)}\left(t_{k_m}^{j_m}\right)\right)\frac{\widetilde{B}_s\left(\beta^{|k_m|}\left(y-t_{k_m}^{j_m}\right)\right)}{s!} \\
&+O\left(\beta^{-|k_m|N}\|F^{(N)}\|_{C([0,1])}\right),\qquad\forall y\in\left[t_{k_m}^{j_m},t_{k_m}^{j_m}+\beta^{-|k_m|}\right]. \tag*{(65)}
\end{aligned} − β − N ∣ k m ∣ ∫ 0 1 F ( N ) ( t k m j m ( 1 − u ) + t k m j m + β − ∣ k m ∣ u ) N ! B N ( β ∣ k m ∣ ( y − t k m j m ) − u ) d u = β ∣ k m ∣ ∫ t k m j m t k m j m + β − ∣ k m ∣ F ( u ) d u + s = 1 ∑ N β − ( s − 1 ) ∣ k m ∣ ( F ( s − 1 ) ( t k m j m + β − ∣ k m ∣ ) − F ( s − 1 ) ( t k m j m ) ) s ! B s ( β ∣ k m ∣ ( y − t k m j m ) ) + O ( β − ∣ k m ∣ N ∥ F ( N ) ∥ C ([ 0 , 1 ]) ) , ∀ y ∈ [ t k m j m , t k m j m + β − ∣ k m ∣ ] . (65) Proofs of lemmas regarding bounded variation
Proof of Lemma 4.2
Let us denote by X 1 = P f − g 1 X_1=\mathcal{P}f-g_1 X 1 = P f − g 1 0 = x 0 < x 1 < ⋯ < x m = 1 0=x_0<x_1<\cdots<x_m=1 0 = x 0 < x 1 < ⋯ < x m = 1 (17) . There exists some 1 ≤ j 0 ≤ m − 1 1\le j_0\le m-1 1 ≤ j 0 ≤ m − 1 x j 0 ≤ b 1 < x j 0 + 1 x_{j_0}\le b_1<x_{j_0+1} x j 0 ≤ b 1 < x j 0 + 1
∑ j = 0 m − 1 ∣ X 1 ( x j + 1 ) − X 1 ( x j ) ∣ = ∑ j = 0 j 0 − 1 ∣ ( P f ) ( x j + 1 ) − ( P f ) ( x j ) ∣ + ∣ X 1 ( x j 0 + 1 ) − X 1 ( x j 0 ) ∣ + ∑ j = j 0 + 1 m − 1 ∣ ( P f ) ( x j + 1 ) − ( P f ) ( x j ) ∣ . \sum_{j=0}^{m-1} \lvert X_1(x_{j+1})-X_1(x_j)\rvert
=
\sum_{j=0}^{j_0-1} \lvert(\mathcal{P}f)(x_{j+1})-(\mathcal{P}f)(x_j)\rvert
+\lvert X_1(x_{j_0+1})-X_1(x_{j_0})\rvert
+\sum_{j=j_0+1}^{m-1} \lvert(\mathcal{P}f)(x_{j+1})-(\mathcal{P}f)(x_j)\rvert. j = 0 ∑ m − 1 ∣ X 1 ( x j + 1 ) − X 1 ( x j )∣ = j = 0 ∑ j 0 − 1 ∣( P f ) ( x j + 1 ) − ( P f ) ( x j )∣ + ∣ X 1 ( x j 0 + 1 ) − X 1 ( x j 0 )∣ + j = j 0 + 1 ∑ m − 1 ∣( P f ) ( x j + 1 ) − ( P f ) ( x j )∣ . We have f ( ( a 0 + x j 0 + 1 ) / β ) = g 1 ( x j 0 + 1 ) = 0 f((a_0+x_{j_0+1})/\beta)=g_1(x_{j_0+1})=0 f (( a 0 + x j 0 + 1 ) / β ) = g 1 ( x j 0 + 1 ) = 0 g 1 ( x j 0 ) = f ( 1 ) β − 1 g_1(x_{j_0})=f(1)\beta^{-1} g 1 ( x j 0 ) = f ( 1 ) β − 1
X 1 ( x j 0 + 1 ) − X 1 ( x j 0 ) = ( P f ) ( x j 0 + 1 ) − ( P f ) ( x j 0 ) + f ( 1 ) β − 1 = β − 1 ∑ k = 0 a 0 − 1 ( f ( ( k + x j 0 + 1 ) / β ) − f ( ( k + x j 0 ) / β ) ) + β − 1 ( f ( 1 ) − f ( ( a 0 + x j 0 ) / β ) ) . \begin{aligned}
X_1(x_{j_0+1})-X_1(x_{j_0})
=
(\mathcal{P}f)(x_{j_0+1})-(\mathcal{P}f)(x_{j_0})+f(1)\beta^{-1} \\
&=
\beta^{-1}\sum_{k=0}^{a_0-1}\left(f((k+x_{j_0+1})/\beta)-f((k+x_{j_0})/\beta)\right)
+\beta^{-1}\left(f(1)-f((a_0+x_{j_0})/\beta)\right).
\end{aligned} X 1 ( x j 0 + 1 ) − X 1 ( x j 0 ) = ( P f ) ( x j 0 + 1 ) − ( P f ) ( x j 0 ) + f ( 1 ) β − 1 = β − 1 k = 0 ∑ a 0 − 1 ( f (( k + x j 0 + 1 ) / β ) − f (( k + x j 0 ) / β ) ) + β − 1 ( f ( 1 ) − f (( a 0 + x j 0 ) / β ) ) . In the sum over j > j 0 j>j_0 j > j 0 k = a 0 k=a_0 k = a 0
∑ j = 0 m − 1 ∣ X 1 ( x j + 1 ) − X 1 ( x j ) ∣ ≤ β − 1 ∑ j = 0 j 0 − 1 ∑ k = 0 a 0 ∣ f ( ( k + x j + 1 ) / β ) − f ( ( k + x j ) / β ) ∣ + β − 1 ∑ k = 0 a 0 − 1 ∣ f ( ( k + x j 0 + 1 ) / β ) − f ( ( k + x j 0 ) / β ) ∣ + β − 1 ∣ f ( 1 ) − f ( ( a 0 + x j 0 ) / β ) ∣ + β − 1 ∑ j = j 0 + 1 m − 1 ∑ k = 0 a 0 − 1 ∣ f ( ( k + x j + 1 ) / β ) − f ( ( k + x j ) / β ) ∣ . \begin{aligned}
\sum_{j=0}^{m-1}\lvert X_1(x_{j+1})-X_1(x_j)\rvert
&\leq\beta^{-1}\sum_{j=0}^{j_0-1}\sum_{k=0}^{a_0}\lvert f((k+x_{j+1})/\beta)-f((k+x_j)/\beta)\rvert\\
&\quad+\beta^{-1}\sum_{k=0}^{a_0-1}\lvert f((k+x_{j_0+1})/\beta)-f((k+x_{j_0})/\beta)\rvert\\
&\quad+\beta^{-1}\lvert f(1)-f((a_0+x_{j_0})/\beta)\rvert\\
&\quad+\beta^{-1}\sum_{j=j_0+1}^{m-1}\sum_{k=0}^{a_0-1}\lvert f((k+x_{j+1})/\beta)-f((k+x_j)/\beta)\rvert.
\end{aligned} j = 0 ∑ m − 1 ∣ X 1 ( x j + 1 ) − X 1 ( x j )∣ ≤ β − 1 j = 0 ∑ j 0 − 1 k = 0 ∑ a 0 ∣ f (( k + x j + 1 ) / β ) − f (( k + x j ) / β )∣ + β − 1 k = 0 ∑ a 0 − 1 ∣ f (( k + x j 0 + 1 ) / β ) − f (( k + x j 0 ) / β )∣ + β − 1 ∣ f ( 1 ) − f (( a 0 + x j 0 ) / β )∣ + β − 1 j = j 0 + 1 ∑ m − 1 k = 0 ∑ a 0 − 1 ∣ f (( k + x j + 1 ) / β ) − f (( k + x j ) / β )∣ . The points y k , j : = ( k + x j ) / β y_{k,j}:=(k+x_j)/\beta y k , j := ( k + x j ) / β [ k / β , ( k + 1 ) / β ] [k/\beta,(k+1)/\beta] [ k / β , ( k + 1 ) / β ] [ 0 , ( a 0 + 1 ) / β ] [0,(a_0+1)/\beta] [ 0 , ( a 0 + 1 ) / β ] ( a 0 + x j 0 ) / β (a_0+x_{j_0})/\beta ( a 0 + x j 0 ) / β k ≤ a 0 − 1 k\leq a_0-1 k ≤ a 0 − 1 k = a 0 k=a_0 k = a 0
β − 1 ∑ k = 0 a 0 − 1 ∑ j = 0 m − 1 ∣ f ( ( k + x j + 1 ) / β ) − f ( ( k + x j ) / β ) ∣ + β − 1 ∑ j = 0 j 0 − 1 ∣ f ( ( a 0 + x j + 1 ) / β ) − f ( ( a 0 + x j ) / β ) ∣ + β − 1 ∣ f ( 1 ) − f ( ( a 0 + x j 0 ) / β ) ∣ = β − 1 ( ∑ k = 0 a 0 − 1 ∑ j = 0 m − 1 ∣ f ( y k , j + 1 ) − f ( y k , j ) ∣ + ∑ j = 0 j 0 − 1 ∣ f ( y a 0 , j + 1 ) − f ( y a 0 , j ) ∣ + ∣ f ( 1 ) − f ( y a 0 , j 0 ) ∣ ) ≤ β − 1 Var 0 1 ( f ) . \begin{aligned}
\beta^{-1}\sum_{k=0}^{a_0-1}\sum_{j=0}^{m-1}\lvert f((k+x_{j+1})/\beta)-f((k+x_j)/\beta)\rvert\\
&+\beta^{-1}\sum_{j=0}^{j_0-1}\lvert f((a_0+x_{j+1})/\beta)-f((a_0+x_j)/\beta)\rvert
+\beta^{-1}\lvert f(1)-f((a_0+x_{j_0})/\beta)\rvert\\
&=
\beta^{-1}\left(\sum_{k=0}^{a_0-1}\sum_{j=0}^{m-1}\lvert f(y_{k,j+1})-f(y_{k,j})\rvert
+\sum_{j=0}^{j_0-1}\lvert f(y_{a_0,j+1})-f(y_{a_0,j})\rvert
+\lvert f(1)-f(y_{a_0,j_0})\rvert\right) \\
&\leq\beta^{-1}\operatorname{Var}_0^1(f).
\end{aligned} β − 1 k = 0 ∑ a 0 − 1 j = 0 ∑ m − 1 ∣ f (( k + x j + 1 ) / β ) − f (( k + x j ) / β )∣ + β − 1 j = 0 ∑ j 0 − 1 ∣ f (( a 0 + x j + 1 ) / β ) − f (( a 0 + x j ) / β )∣ + β − 1 ∣ f ( 1 ) − f (( a 0 + x j 0 ) / β )∣ = β − 1 ( k = 0 ∑ a 0 − 1 j = 0 ∑ m − 1 ∣ f ( y k , j + 1 ) − f ( y k , j )∣ + j = 0 ∑ j 0 − 1 ∣ f ( y a 0 , j + 1 ) − f ( y a 0 , j )∣ + ∣ f ( 1 ) − f ( y a 0 , j 0 )∣ ) ≤ β − 1 Var 0 1 ( f ) . Proof of Lemma 4.3
Let g 1 ( x ) = f ( 1 ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ( x ) ) g_1(x)=f(1)\beta^{-1}(-b_1+\chi_{[0,b_1]}(x)) g 1 ( x ) = f ( 1 ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ( x )) g 1 g_1 g 1 Lemma 4.2 , while here ∫ 0 1 g 1 ( x ) d x = 0 \int_0^1 g_1(x)\,\mathrm{d}x=0 ∫ 0 1 g 1 ( x ) d x = 0 Var 0 1 ( P f − g 1 ) ≤ β − 1 Var 0 1 ( f ) \operatorname{Var}_0^1(\mathcal{P}f-g_1)\leq\beta^{-1}\operatorname{Var}_0^1(f) Var 0 1 ( P f − g 1 ) ≤ β − 1 Var 0 1 ( f ) r ≥ 1 r\geq1 r ≥ 1
g r + 1 ( x ) : = ( P g r ) ( x ) + ( ( P r f ) ( 1 ) − g r ( 1 ) ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ( x ) ) . (66) g_{r+1}(x):=(\mathcal{P}g_r)(x)+\left((\mathcal{P}^r f)(1)-g_r(1)\right)\beta^{-1}(-b_1+\chi_{[0,b_1]}(x)). \tag*{(66)} g r + 1 ( x ) := ( P g r ) ( x ) + ( ( P r f ) ( 1 ) − g r ( 1 ) ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ( x )) . (66) We have
P r + 1 f − g r + 1 = P ( P r f − g r ) − ( ( P r f ) ( 1 ) − g r ( 1 ) ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ) \mathcal{P}^{r+1}f-g_{r+1}=\mathcal{P}(\mathcal{P}^r f-g_r)-\left((\mathcal{P}^r f)(1)-g_r(1)\right)\beta^{-1}(-b_1+\chi_{[0,b_1]}) P r + 1 f − g r + 1 = P ( P r f − g r ) − ( ( P r f ) ( 1 ) − g r ( 1 ) ) β − 1 ( − b 1 + χ [ 0 , b 1 ] ) hence the induction step implies
Var 0 1 ( P r + 1 f − g r + 1 ) ≤ β − 1 Var 0 1 ( P r f − g r ) ≤ β − ( r + 1 ) Var 0 1 ( f ) . \operatorname{Var}_0^1(\mathcal{P}^{r+1}f-g_{r+1})\leq\beta^{-1}\operatorname{Var}_0^1(\mathcal{P}^r f-g_r)\leq\beta^{-(r+1)}\operatorname{Var}_0^1(f). Var 0 1 ( P r + 1 f − g r + 1 ) ≤ β − 1 Var 0 1 ( P r f − g r ) ≤ β − ( r + 1 ) Var 0 1 ( f ) . In conclusion, we have just proved that the function X r = P r f − g r X_r=\mathcal{P}^r f-g_r X r = P r f − g r r ≥ 1 r\geq1 r ≥ 1
Var 0 1 ( X r ) ≤ β − r Var 0 1 ( f ) and ∫ 0 1 X r ( x ) d x = 0 , (67) \operatorname{Var}_0^1(X_r)\leq\beta^{-r}\operatorname{Var}_0^1(f)\quad\text{and}\quad\int_0^1 X_r(x)\,\mathrm{d}x=0, \tag*{(67)} Var 0 1 ( X r ) ≤ β − r Var 0 1 ( f ) and ∫ 0 1 X r ( x ) d x = 0 , (67) where the last equality comes from the fact that P \mathcal{P} P ∫ 0 1 X r ( x ) d x = 0 \int_0^1X_r(x)\,\mathrm{d}x=0 ∫ 0 1 X r ( x ) d x = 0 X r X_r X r ∣ X r ( x ) ∣ ≤ Var 0 1 ( X r ) ≤ β − r Var 0 1 ( f ) \lvert X_r(x)\rvert\leq\operatorname{Var}_0^1(X_r)\leq\beta^{-r}\operatorname{Var}_0^1(f) ∣ X r ( x )∣ ≤ Var 0 1 ( X r ) ≤ β − r Var 0 1 ( f ) x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ] c r = ( P r f ) ( 1 ) − g r ( 1 ) c_r=(\mathcal{P}^r f)(1)-g_r(1) c r = ( P r f ) ( 1 ) − g r ( 1 )
∣ c r ∣ ≤ β − r Var 0 1 ( f ) , r ≥ 1. (68) \lvert c_r\rvert\leq\beta^{-r}\operatorname{Var}_0^1(f),\quad r\geq1. \tag*{(68)} ∣ c r ∣ ≤ β − r Var 0 1 ( f ) , r ≥ 1. (68) Let h : = β − 1 ( − b 1 + χ [ 0 , b 1 ] ) h:=\beta^{-1}(-b_1+\chi_{[0,b_1]}) h := β − 1 ( − b 1 + χ [ 0 , b 1 ] )
g r + 1 = P g r + c r h , r ≥ 1. g_{r+1}=\mathcal{P}g_r+c_rh,\quad r\geq1. g r + 1 = P g r + c r h , r ≥ 1. Then for all r ≥ 2 r\geq2 r ≥ 2
g r = P g r − 1 + c r − 1 h = P 2 g r − 2 + c r − 2 P h + c r − 1 h = P r − 1 g 1 + ∑ k = 1 r − 1 c r − k P k − 1 h . (69) \begin{aligned}
g_r&=\mathcal{P}g_{r-1}+c_{r-1}h=\mathcal{P}^2g_{r-2}+c_{r-2}\mathcal{P}h+c_{r-1}h\\
&=\mathcal{P}^{r-1}g_1+\sum_{k=1}^{r-1}c_{r-k}\mathcal{P}^{k-1}h. \tag*{(69)}
\end{aligned} g r = P g r − 1 + c r − 1 h = P 2 g r − 2 + c r − 2 P h + c r − 1 h = P r − 1 g 1 + k = 1 ∑ r − 1 c r − k P k − 1 h . (69) We observe that h h h g r g_r g r χ [ 0 , b j ] \chi_{[0,b_j]} χ [ 0 , b j ] 0 ≤ j ≤ n − 1 0\leq j\leq n-1 0 ≤ j ≤ n − 1 P \mathcal{P} P § 5 . The restriction of β P \beta\mathcal{P} β P n n n
Since ∫ 0 1 g 1 ( x ) d x = ∫ 0 1 h ( x ) d x = 0 \int_0^1g_1(x)\,\mathrm{d}x=\int_0^1h(x)\,\mathrm{d}x=0 ∫ 0 1 g 1 ( x ) d x = ∫ 0 1 h ( x ) d x = 0 g 1 g_1 g 1 h h h β \beta β P n P_n P n h ( x ) = c 1 u 1 ( x ) + h ~ ( x ) h(x)=c_1u_1(x)+\widetilde{h}(x) h ( x ) = c 1 u 1 ( x ) + h ( x ) h ~ ( x ) \widetilde{h}(x) h ( x ) P n P_n P n P m h − c 1 u 1 = P m h ~ → 0 \mathcal{P}^m h-c_1u_1=\mathcal{P}^m\widetilde{h}\to0 P m h − c 1 u 1 = P m h → 0 m → ∞ m\to\infty m → ∞ P \mathcal{P} P h h h c 1 = 0 c_1=0 c 1 = 0
Let δ = 1 \delta=1 δ = 1 λ ∗ \lambda_* λ ∗ n − 1 n-1 n − 1 δ = 0 \delta=0 δ = 0 λ ∗ \lambda_* λ ∗ C > 0 C>0 C > 0 j ≥ 1 j\geq1 j ≥ 1
max { ∥ P j g 1 ∥ B V , ∥ P j h ∥ B V } ≤ C j δ ( n − 1 ) ( ∣ λ ∗ ∣ / β ) j . (70) \max\left\{\lVert\mathcal{P}^j g_1\rVert_{BV},\lVert\mathcal{P}^j h\rVert_{BV}\right\}\leq Cj^{\delta(n-1)}\left(\lvert\lambda_*\rvert/\beta\right)^j. \tag*{(70)} max { ∥ P j g 1 ∥ B V , ∥ P j h ∥ B V } ≤ C j δ ( n − 1 ) ( ∣ λ ∗ ∣ / β ) j . (70) This together with (A.3) applied to (A.4) leads to (remember that ∣ f ( 1 ) ∣ ≤ ∥ f ∥ L ∞ ≤ Var 0 1 ( f ) |f(1)| \le\|f\|_{L^\infty} \le\operatorname{Var}_{0}^{1}(f) ∣ f ( 1 ) ∣ ≤ ∥ f ∥ L ∞ ≤ Var 0 1 ( f ) f f f
∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) ∑ k = 1 r − 1 β − ( r − k ) k δ ( n − 1 ) ( ∣ λ ∗ ∣ β ) k . (71) \|g_r-(\beta^{-1}\mathcal{P}_n)^{r-1}g_1\|_{BV}\le C\operatorname{Var}_{0}^{1}(f)\sum_{k=1}^{r-1}\beta^{-(r-k)}k^{\delta(n-1)}\left(\frac{|\lambda_*|}{\beta}\right)^k. \tag*{(71)} ∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) k = 1 ∑ r − 1 β − ( r − k ) k δ ( n − 1 ) ( β ∣ λ ∗ ∣ ) k . (71) We have the following alternatives:
If ∣ λ ∗ ∣ < 1 |\lambda_*|<1 ∣ λ ∗ ∣ < 1 p ( z ) p(z) p ( z ) δ = 0 \delta=0 δ = 0 ∑ k = 1 r − 1 β − ( r − k ) ( ∣ λ ∗ ∣ β ) k ≤ β − r ( 1 − ∣ λ ∗ ∣ ) − 1 \sum_{k=1}^{r-1}\beta^{-(r-k)}\left(\frac{|\lambda_*|}{\beta}\right)^k\le\beta^{-r}(1-|\lambda_*|)^{-1} ∑ k = 1 r − 1 β − ( r − k ) ( β ∣ λ ∗ ∣ ) k ≤ β − r ( 1 − ∣ λ ∗ ∣ ) − 1
∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) ( 1 − ∣ λ ∗ ∣ ) − 1 β − r ; \|g_r-(\beta^{-1}\mathcal{P}_n)^{r-1}g_1\|_{BV}\le C\operatorname{Var}_{0}^{1}(f)(1-|\lambda_*|)^{-1}\beta^{-r}; ∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) ( 1 − ∣ λ ∗ ∣ ) − 1 β − r ; ∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) r n − 1 ( ∣ λ ∗ ∣ / β ) r ∣ λ ∗ ∣ − 1 . \|g_r-(\beta^{-1}\mathcal{P}_n)^{r-1}g_1\|_{BV}\le C\operatorname{Var}_{0}^{1}(f)\frac{r^{n-1}(|\lambda_*|/\beta)^r}{|\lambda_*|-1}. ∥ g r − ( β − 1 P n ) r − 1 g 1 ∥ B V ≤ C Var 0 1 ( f ) ∣ λ ∗ ∣ − 1 r n − 1 ( ∣ λ ∗ ∣/ β ) r . Going back to (A.2) and using that X r = P r f − g r X_r=\mathcal{P}^r f-g_r X r = P r f − g r
∥ P r f − g r ∥ L ∞ ≤ Var 0 1 ( P r f − g r ) ≤ Var 0 1 ( f ) β − r \|\mathcal{P}^r f-g_r\|_{L^\infty}\le\operatorname{Var}_{0}^{1}(\mathcal{P}^r f-g_r)\le\operatorname{Var}_{0}^{1}(f)\beta^{-r} ∥ P r f − g r ∥ L ∞ ≤ Var 0 1 ( P r f − g r ) ≤ Var 0 1 ( f ) β − r which together with the above estimates on (A.6) and the triangle inequality we obtain (35) . □
Proof of Lemma 4.4
Let g = f − ∫ 0 1 f ( x ) d x g=f-\int_{0}^{1}f(x)\,\mathrm{d}x g = f − ∫ 0 1 f ( x ) d x ∫ 0 1 g ( x ) d x = 0 \int_{0}^{1}g(x)\,\mathrm{d}x=0 ∫ 0 1 g ( x ) d x = 0 g r g_r g r r ≥ 1 r\ge1 r ≥ 1 χ [ 0 , b j ] \chi_{[0,b_j]} χ [ 0 , b j ] Var 0 1 ( X r ) ≤ β − r Var 0 1 ( g ) \operatorname{Var}_{0}^{1}(X_r)\le\beta^{-r}\operatorname{Var}_{0}^{1}(g) Var 0 1 ( X r ) ≤ β − r Var 0 1 ( g ) h r = g r + χ [ 0 , 1 ] ∫ 0 1 f ( x ) d x h_r=g_r+\chi_{[0,1]}\int_{0}^{1}f(x)\,\mathrm{d}x h r = g r + χ [ 0 , 1 ] ∫ 0 1 f ( x ) d x
Various properties of β \beta β
Criteria for p ( z ) p(z) p ( z )
We start by giving here a short alternative proof for a result of Brauer [1 ] which holds for all n ≥ 2 n\ge2 n ≥ 2
Lemma A.1. Assume that n ≥ 2 n\ge2 n ≥ 2 a 0 ≥ ⋯ ≥ a n − 1 ≥ 1 a_0\ge\cdots\ge a_{n-1}\ge1 a 0 ≥ ⋯ ≥ a n − 1 ≥ 1
p ( z ) = z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) . p(z)=z^n-(a_0z^{n-1}+\cdots+a_{n-1}). p ( z ) = z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) . Then p ( z ) p(z) p ( z ) Definition 1.1 .
Proof. We already know from the analysis of (1) that f ( x ) = x − n p ( x ) f(x)=x^{-n}p(x) f ( x ) = x − n p ( x ) p p p β > 1 \beta>1 β > 1 (5) we infer that p ( z ) p(z) p ( z ) p ( z ) p(z) p ( z )
q ( z ) = ( 1 − z ) p ( z ) = ( 1 − z ) ( z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) ) = ( a 0 + 1 ) z n − ( z n + 1 + ( a 0 − a 1 ) z n − 1 + ⋯ + ( a n − 2 − a n − 1 ) z + a n − 1 ) = : H ( z ) − M ( z ) . (72) \begin{aligned}
q(z)=(1-z)p(z)=(1-z)\left(z^n-(a_0z^{n-1}+\cdots+a_{n-1})\right) \\
&=(a_0+1)z^n-\left(z^{n+1}+(a_0-a_1)z^{n-1}+\cdots+(a_{n-2}-a_{n-1})z+a_{n-1}\right) \\
&=:H(z)-M(z). \tag*{(72)}
\end{aligned} q ( z ) = ( 1 − z ) p ( z ) = ( 1 − z ) ( z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) ) = ( a 0 + 1 ) z n − ( z n + 1 + ( a 0 − a 1 ) z n − 1 + ⋯ + ( a n − 2 − a n − 1 ) z + a n − 1 ) =: H ( z ) − M ( z ) . (72) Let us show that if ∣ z ∣ = 1 + r |z|=1+r ∣ z ∣ = 1 + r r > 0 r>0 r > 0 ∣ H ( z ) ∣ > ∣ M ( z ) ∣ |H(z)|>|M(z)| ∣ H ( z ) ∣ > ∣ M ( z ) ∣ q q q r r r
∣ H ( z ) ∣ = ( a 0 + 1 ) ( 1 + n r ) + O ( r 2 ) |H(z)|=(a_0+1)(1+nr)+O(r^2) ∣ H ( z ) ∣ = ( a 0 + 1 ) ( 1 + n r ) + O ( r 2 ) and by the triangle inequality
∣ M ( z ) ∣ ≤ 1 + ( n + 1 ) r + ( a 0 − a 1 ) ( 1 + ( n − 1 ) r ) + ⋯ + ( a n − 2 − a n − 1 ) ( 1 + r ) + a n − 1 + O ( r 2 ) = 1 + a 0 + r ( n + 1 + ( a 0 − a 1 ) ( n − 1 ) + ( a 1 − a 2 ) ( n − 2 ) + ⋯ + ( a n − 2 − a n − 1 ) ) + O ( r 2 ) . \begin{aligned}
|M(z)|\leq1+(n+1)r+(a_0-a_1)(1+(n-1)r)+\cdots+(a_{n-2}-a_{n-1})(1+r)+a_{n-1}+O(r^2) \\
&=1+a_0+r\left(n+1+(a_0-a_1)(n-1)+(a_1-a_2)(n-2)+\cdots+(a_{n-2}-a_{n-1})\right)+O(r^2).
\end{aligned} ∣ M ( z ) ∣ ≤ 1 + ( n + 1 ) r + ( a 0 − a 1 ) ( 1 + ( n − 1 ) r ) + ⋯ + ( a n − 2 − a n − 1 ) ( 1 + r ) + a n − 1 + O ( r 2 ) = 1 + a 0 + r ( n + 1 + ( a 0 − a 1 ) ( n − 1 ) + ( a 1 − a 2 ) ( n − 2 ) + ⋯ + ( a n − 2 − a n − 1 ) ) + O ( r 2 ) . Replacing n − k n-k n − k n − 1 n-1 n − 1
( a 0 − a 1 ) ( n − 1 ) + ( a 1 − a 2 ) ( n − 2 ) + ⋯ + ( a n − 2 − a n − 1 ) ≤ ( n − 1 ) ( a 0 − a n − 1 ) (a_0-a_1)(n-1)+(a_1-a_2)(n-2)+\cdots+(a_{n-2}-a_{n-1})\leq(n-1)(a_0-a_{n-1}) ( a 0 − a 1 ) ( n − 1 ) + ( a 1 − a 2 ) ( n − 2 ) + ⋯ + ( a n − 2 − a n − 1 ) ≤ ( n − 1 ) ( a 0 − a n − 1 ) and thus
∣ M ( z ) ∣ ≤ 1 + a 0 + r ( n + 1 + ( a 0 − a n − 1 ) ( n − 1 ) ) + O ( r 2 ) . |M(z)|\leq1+a_0+r\left(n+1+(a_0-a_{n-1})(n-1)\right)+O(r^2). ∣ M ( z ) ∣ ≤ 1 + a 0 + r ( n + 1 + ( a 0 − a n − 1 ) ( n − 1 ) ) + O ( r 2 ) . Hence for small r > 0 r>0 r > 0
r − 1 ( ∣ H ( z ) ∣ − ∣ M ( z ) ∣ ) ≥ ( 1 + a 0 ) n − ( n + 1 ) − ( a 0 − a n − 1 ) ( n − 1 ) + O ( r ) = a 0 − 1 + a n − 1 ( n − 1 ) + O ( r ) ≥ 1 + O ( r ) > 0. \begin{aligned}
r^{-1}\left(|H(z)|-|M(z)|\right)\geq(1+a_0)n-(n+1)-(a_0-a_{n-1})(n-1)+O(r) \\
&=a_0-1+a_{n-1}(n-1)+O(r)\geq1+O(r)>0.
\end{aligned} r − 1 ( ∣ H ( z ) ∣ − ∣ M ( z ) ∣ ) ≥ ( 1 + a 0 ) n − ( n + 1 ) − ( a 0 − a n − 1 ) ( n − 1 ) + O ( r ) = a 0 − 1 + a n − 1 ( n − 1 ) + O ( r ) ≥ 1 + O ( r ) > 0. Now fix such an r r r
q t ( z ) = H ( z ) − t M ( z ) , ∣ z ∣ ≤ 1 + r , 0 ≤ t ≤ 1 , q ( z ) = q 1 ( z ) , H ( z ) = q 0 ( z ) . q_t(z)=H(z)-tM(z),\qquad|z|\leq1+r,\qquad0\leq t\leq1,\qquad q(z)=q_1(z),\qquad H(z)=q_0(z). q t ( z ) = H ( z ) − tM ( z ) , ∣ z ∣ ≤ 1 + r , 0 ≤ t ≤ 1 , q ( z ) = q 1 ( z ) , H ( z ) = q 0 ( z ) . As in Rouché’s theorem, the number of zeros (counting algebraic multiplicities) of q t q_t q t ∣ z ∣ ≤ 1 + r |z|\leq1+r ∣ z ∣ ≤ 1 + r
N t = 1 2 π i ∮ ∣ z ∣ = 1 + r q t ′ ( z ) q t ( z ) d z N_t=\frac{1}{2\pi i}\oint_{|z|=1+r}\frac{q_t'(z)}{q_t(z)}\,dz N t = 2 π i 1 ∮ ∣ z ∣ = 1 + r q t ( z ) q t ′ ( z ) d z which is integer valued and (Lipschitz) continuous on 0 ≤ t ≤ 1 0\leq t\leq1 0 ≤ t ≤ 1 N 0 = n N_0=n N 0 = n r r r q ( z ) = q 1 ( z ) q(z)=q_1(z) q ( z ) = q 1 ( z ) n n n ∣ z ∣ ≤ 1 |z|\leq1 ∣ z ∣ ≤ 1
By construction, z = 1 z=1 z = 1 q q q p p p q q q ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 z = 1 z=1 z = 1 z = e i θ z=e^{i\theta} z = e i θ − π < θ ≤ π -\pi<\theta\leq\pi − π < θ ≤ π z − n q ( z ) = 0 z^{-n}q(z)=0 z − n q ( z ) = 0
a 0 + 1 = cos ( θ ) + ( a 0 − a 1 ) cos ( θ ) + ⋯ + ( a n − 2 − a n − 1 ) cos ( ( n − 1 ) θ ) + a n − 1 cos ( n θ ) ≤ cos ( θ ) + a 0 \begin{aligned}
a_0+1=\cos(\theta)+(a_0-a_1)\cos(\theta)+\cdots+(a_{n-2}-a_{n-1})\cos((n-1)\theta)+a_{n-1}\cos(n\theta) \\
&\leq\cos(\theta)+a_0
\end{aligned} a 0 + 1 = cos ( θ ) + ( a 0 − a 1 ) cos ( θ ) + ⋯ + ( a n − 2 − a n − 1 ) cos (( n − 1 ) θ ) + a n − 1 cos ( n θ ) ≤ cos ( θ ) + a 0 which demands θ = 0 \theta=0 θ = 0 q q q p p p n − 1 n-1 n − 1
Lemma A.2. Assume that { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1 n ≥ 2 n \ge2 n ≥ 2
a n − 1 ≥ 1 and a 0 ≥ 2 + a 1 + ⋯ + a n − 1 . a_{n-1} \ge1 \qquad\text{and}\qquad a_0 \ge2+a_1+\cdots+a_{n-1}. a n − 1 ≥ 1 and a 0 ≥ 2 + a 1 + ⋯ + a n − 1 . Then they form a finite Parry sequence and p ( z ) = z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) p(z)=z^n-(a_0z^{n-1}+\cdots+a_{n-1}) p ( z ) = z n − ( a 0 z n − 1 + ⋯ + a n − 1 ) Definition 1.1 .
Proof. The sequence { a j } j = 0 n − 1 \{a_j\}_{j=0}^{n-1} { a j } j = 0 n − 1 (12) with j = 0 j=0 j = 0 p ( z ) p(z) p ( z ) p ( z ) = H ( z ) + M ( z ) p(z)=H(z)+M(z) p ( z ) = H ( z ) + M ( z )
H ( z ) = − a 0 z n − 1 and M ( z ) = z n − a 1 z n − 2 − ⋯ − a n − 1 . H(z)=-a_0z^{n-1} \qquad\text{and}\qquad M(z)=z^n-a_1z^{n-2}-\cdots-a_{n-1}. H ( z ) = − a 0 z n − 1 and M ( z ) = z n − a 1 z n − 2 − ⋯ − a n − 1 . If ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 ∣ H ( z ) ∣ − ∣ M ( z ) ∣ ≥ a 0 − ( 1 + a 1 + ⋯ + a n − 1 ) ≥ 1 |H(z)|-|M(z)|\ge a_0-(1+a_1+\cdots+a_{n-1})\ge1 ∣ H ( z ) ∣ − ∣ M ( z ) ∣ ≥ a 0 − ( 1 + a 1 + ⋯ + a n − 1 ) ≥ 1 p ( z ) p(z) p ( z ) H ( z ) H(z) H ( z ) n − 1 n-1 n − 1 β > 1 \beta>1 β > 1 p ( z ) p(z) p ( z ) β \beta β □ \square □
It turns out that if n ≤ 3 n\le3 n ≤ 3 p ( z ) p(z) p ( z )
Lemma A.3. For any finite Parry sequence with n ≤ 3 n\le3 n ≤ 3 p ( z ) p(z) p ( z )
Proof. If n = 2 n=2 n = 2 a 0 ≥ a 1 a_0\ge a_1 a 0 ≥ a 1 z 2 = − a 1 / β z_2=-a_1/\beta z 2 = − a 1 / β ∣ z 2 ∣ < 1 |z_2|<1 ∣ z 2 ∣ < 1 n = 3 n=3 n = 3 a 0 ≥ max { a 1 , a 2 } a_0\ge\max\{a_1,a_2\} a 0 ≥ max { a 1 , a 2 } a 1 < a 2 a_1<a_2 a 1 < a 2 Lemma 1.3 but do not satisfy Brauer’s criterion. Nevertheless, let us show that
p ( z ) = z 3 − a 0 z 2 − a 1 z − a 2 p(z)=z^3-a_0z^2-a_1z-a_2 p ( z ) = z 3 − a 0 z 2 − a 1 z − a 2 is Parry-Pisot. Denote its roots with β > 1 \beta>1 β > 1 z 2 z_2 z 2 z 3 z_3 z 3 z 2 z_2 z 2 z 3 = z 2 ‾ z_3=\overline{z_2} z 3 = z 2 β z 2 z 3 = a 2 \beta z_2z_3=a_2 β z 2 z 3 = a 2 ∣ z 2 ∣ = ∣ z 3 ∣ = a 2 / β ≤ a 0 / β < 1 |z_2|=|z_3|=\sqrt{a_2/\beta}\le\sqrt{a_0/\beta}<1 ∣ z 2 ∣ = ∣ z 3 ∣ = a 2 / β ≤ a 0 / β < 1 z 2 z_2 z 2 z 3 z_3 z 3 β \beta β
β + z 2 + z 3 = a 0 \beta+z_2+z_3=a_0 β + z 2 + z 3 = a 0 hence 0 < − z 2 − z 3 = β − a 0 < 1 0<-z_2-z_3=\beta-a_0<1 0 < − z 2 − z 3 = β − a 0 < 1 z 2 , 3 ∈ ( − 1 , 0 ) z_{2,3}\in(-1,0) z 2 , 3 ∈ ( − 1 , 0 ) a 1 < a 2 a_1<a_2 a 1 < a 2 z ∈ [ − 1 , 0 ] z\in[-1,0] z ∈ [ − 1 , 0 ]
p ( z ) < − a 1 z − a 2 ≤ a 1 − a 2 < 0. p(z)<-a_1z-a_2\le a_1-a_2<0. p ( z ) < − a 1 z − a 2 ≤ a 1 − a 2 < 0. □ \square □
Lemma A.4. If p ( z ) p(z) p ( z )
Proof. Let us first prove that p ( z ) p(z) p ( z ) β \beta β 0 < ∣ λ ∣ < 1 0<|\lambda|<1 0 < ∣ λ ∣ < 1
From Gauss’ Lemma we have that p ( z ) p(z) p ( z ) d ( z ) d(z) d ( z ) p ( z ) p(z) p ( z ) p ′ ( z ) p'(z) p ′ ( z ) d ( z ) d(z) d ( z ) p ( z ) p(z) p ( z ) p ′ ( z ) p'(z) p ′ ( z ) p ( z ) p(z) p ( z ) d ( z ) d(z) d ( z ) d ( z ) d(z) d ( z ) p ( z ) p(z) p ( z ) □ \square □
All roots of p ( z ) p(z) p ( z )
The interest of this result is more that there exist Parry polynomials whose roots can get arbitrarily close to 1 in absolute value, that is given ε \varepsilon ε z z z ∣ 1 − ∣ z ∣ ∣ < ε |1-|z||<\varepsilon ∣1 − ∣ z ∣∣ < ε p ( z ) = z n − z n − 1 − 1 p(z)=z^n-z^{n-1}-1 p ( z ) = z n − z n − 1 − 1 n ≥ 3 n\ge3 n ≥ 3 a 0 = a n − 1 = 1 a_0=a_{n-1}=1 a 0 = a n − 1 = 1 a 1 = ⋯ = a n − 2 = 0 a_1=\cdots=a_{n-2}=0 a 1 = ⋯ = a n − 2 = 0
Let f ( z ) = − p ( z ) / z n − 1 = 1 + 1 / z n − 1 − z f(z)=-p(z)/z^{n-1}=1+1/z^{n-1}-z f ( z ) = − p ( z ) / z n − 1 = 1 + 1/ z n − 1 − z f f f p p p β \beta β f ( 1 ) > 0 f(1)>0 f ( 1 ) > 0
f ( 1 + 1 n − 1 ) = e − ( n − 1 ) ln ( 1 + 1 / n − 1 ) − 1 n − 1 = − 1 n − 1 + O ( e − n − 1 ) < 0 f\left(1+\frac{1}{\sqrt{n-1}}\right)=e^{-(n-1)\ln(1+1/\sqrt{n-1})}-\frac{1}{\sqrt{n-1}}=-\frac{1}{\sqrt{n-1}}+O\left(e^{-\sqrt{n-1}}\right)<0 f ( 1 + n − 1 1 ) = e − ( n − 1 ) l n ( 1 + 1/ n − 1 ) − n − 1 1 = − n − 1 1 + O ( e − n − 1 ) < 0 if n n n β ∈ ( 1 , 1 + 1 / n − 1 ) \beta\in(1,1+1/\sqrt{n-1}) β ∈ ( 1 , 1 + 1/ n − 1 ) n n n
Since a 0 = 1 a_0=1 a 0 = 1 β < a 0 + 1 < 2 \beta<a_0+1<2 β < a 0 + 1 < 2 z z z p ( z ) p(z) p ( z ) ∣ z ∣ ≤ β ≤ 2 |z|\le\beta\le2 ∣ z ∣ ≤ β ≤ 2
∣ 1 − z ∣ = 1 / ∣ z ∣ n − 1 ≤ 3 , ∣ z ∣ ≥ 1 / 3 1 / ( n − 1 ) . |1-z|=1/|z|^{n-1}\le3,\qquad|z|\ge1/3^{1/(n-1)}. ∣1 − z ∣ = 1/∣ z ∣ n − 1 ≤ 3 , ∣ z ∣ ≥ 1/ 3 1/ ( n − 1 ) . This means that for large enough n n n p p p
1 3 1 / ( n − 1 ) ≤ ∣ z ∣ ≤ 1 + 1 n − 1 , \frac{1}{3^{1/(n-1)}}\le|z|\le1+\frac{1}{\sqrt{n-1}}, 3 1/ ( n − 1 ) 1 ≤ ∣ z ∣ ≤ 1 + n − 1 1 , thus ∣ z ∣ |z| ∣ z ∣ n n n
Acknowledgements. We are grateful for the technical assistance of A. Author.
References 