3
$\begingroup$

Let $\{I_n\}_{n\in\mathbb{N}}$ be a sequence of intervals in the form

$ I_n = \Big [ \frac{q_n}{b_n}, \frac{q_n + 1}{b_n} \Big),$

where $q_{n}$ is some integer, for all $n\in\mathbb{N}$. Define the sequences of real numbers $\{ y_n \}_{n\in\mathbb{N}}$ and $\{ z_n \}_{n\in\mathbb{N}}$ by

$y_n = \frac{q_n + 1}{b_n}\;\;\;\;\text{ and }\;\;\;\; z_n = \frac{q_n + 3/2}{b_n}.$

I want to to find a upper bound for

$\Delta_n = \Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big|$

Where $\{a_n\}_{n\in\mathbb{N}}$ an $\{b_n\}_{n\in\mathbb{N}}$ are real number sequences such that

$ \sum a_n < \infty\;\;\;\;\text{and }\;\;\;\; b_n = \prod_{i=0}^{n}p_i,$

$\{p_n\}_{n\in\mathbb{N}}$ is a prime sequence such that

$ \lim_{n\to\infty}\frac{2^n}{a_n p_n} = 0.$

Johan Thim proved [2] that $\Delta_n \leq \frac{b\pi}{p_n}2^{n-2},$

by the following: Define the constants

$ a = \prod_{i=0}^{\infty}(1 - a_i)\;\;\;\;\text{ and }\;\;\;\; b=\prod_{i=0}^{\infty}(1+a_i).$

and the partial products $ w_n(x) = \prod_{i=0}^{n}[1+a_i\sin(b_i\pi x)].$

Given a natural $k>n$, the quotient $b_k/b_n$ is a even integer. Thus, for some $r_k\in\mathbb{Z}$,

$ \sin(b_k\pi y_n) = \sin(2r_k\pi(q_n+1)) = 0 \;\text{ and }\; \sin(b_k\pi z_n) = \sin(2r_k\pi(q_n + 3/2)) = 0.$

And $k=n$, we have $ \sin(b_k \pi y_n) = 0\;\;\text{ and }\;\; \sin(b_k\pi z_n) = -(-1)^{q_n}.$

Therefore, $ \Delta_n = W_{n-1}(z_n)[(1+a_n\sin(b_n\pi z_n))] - W_{n-1}(y_n)[(1+a_n\sin(b_n\pi y_n))]=$ $ = W_{n-1}(z_n) - W_{n-1}(y_n) - (-1)^{q_n}(z_n)[1+a_n\sin(b_n \pi z_n)]$

I understood everything until here. The trouble comes with the following steps:

Thus, there is a $\sigma_k\in\mathbb{R}$ such that

$a_k\sin(b_k\pi z_n) = a_k\sin(b_k\pi y_n) + \sigma_k.$

and for $|W_{n-1}(z_n) - W_{n-1}(y_n)|$, it's true that

$|W_{n-1}(z_n) - W_{n-1}(y_n)| = \Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big| = $ $ = \Big | \sum_{i=0}^{2^{n-1}-1}\sigma_ {l_i}\big( \prod_{j\in I_i}\sigma_j\big)\big( \prod_{j\in J_i}[1+a_j\sin(b_j\pi y_n)]\big)\Big | \leq \frac{\pi}{2p_n} \sum_{i=0}^{2^{n-1}-1} \Big (\prod_{j\in J_i}|1+a_j|\Big)\leq \frac{b\pi}{2p_n}(2^{n-1}-1)\leq\frac{b\pi}{p_n}2^{n-2} $

For $I_i, J_i\subseteq \mathbb{N}$ and some natural $l_i$. Therefore $ |\Delta_n|\leq a_n\Big(a - \frac{2^{n-2}}{a_n p_n}b\pi\Big) $

My question is concerning the step in which Thim states the equality

$\Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big| = $ $ = \Big | \sum_{i=0}^{2^{n-1}-1}\sigma_ {l_i}\big( \prod_{j\in I_i}\sigma_j\big)\big( \prod_{j\in J_i}[1+a_j\sin(b_j\pi y_n)]\big)\Big |.$

Why is that true?


[1] - Wen, Liu : A Nowhere Differentiable Continuous Function Constructed by Infinite Products - The American Mathematical Monthly, Vol. 109, No. 4 (Apr., 2002), pp. 378-380.

[2] - Thim, Johan: Continuous Nowhere Differentiable Functions - December 2003, Departament of Mathematics, Luleå University of Technology.

  • 0
    Sorry! Try [this one](http://www.math.cmu.edu/~bobpego/21132/nowhdiff.pdf).2012-03-24

0 Answers 0