$Prove Holder-continuity for $\mu_\lambda (x) = \sum\limits_{n=1}^\infty \frac{ \cos(2^n x)}{2^{n \lambda} }$$

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Consider $\lambda \in (0,1]$ and the function $\mu_\lambda$ defined by $\mu_\lambda (x) = \sum\limits_{n=1}^\infty \frac{ \cos(2^n x)}{2^{n \lambda} }$ for $x \in [0,\pi]$. Show that $\mu_\lambda$ is $\alpha$-Holder-continuous on $[0, \pi]$ for all $0 < \alpha < \lambda$.

Functional Analysis Real Analysis

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Savionf

This is a tricky question. The offered bounty is low.
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I would double the bounty.

Savionf

This is a tricky question. The offered bounty is low.

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Assume

$$2^{-k-1}\le |x-y|\le 2^{-k}, \text{for some} k \in \mathbb{Z}.$$

Then by mean value theorem, there exists $c$ between $x$ and $y$ such that
$$|\cos(2^nx)-\cos(2^ny)|=|\sin (c)(2^nx-2^ny)| \le 2^{n}|x-y|\le 2^{n-k}.$$
For $0<\alpha<\lambda =1$, we have
$$|f(x)-f(y)|\le \sum_{n=1}^k\frac{|\cos(2^nx)-\cos(2^ny)|}{2^{n\lambda}}+\sum_{n=k+1}^\infty\frac{ |\cos(2^nx)-\cos(2^ny)|}{2^{n\lambda}}$$$$\le \sum_{n=1}^k2^{n-k-\lambda n}+2\sum_{n=k+1}^\infty 2^{-\lambda n}$$ $$ \le \sum_{n=1}^k2^{n-k-\alpha n}+2\sum_{n=k+1}^\infty 2^{-\alpha n}$$ $$ \le {2^{-k}2^{(1-\alpha)(k+1)}\over 2^{1-\alpha}-1}+2\cdot {2^{-\alpha(k+1)}\over 1-2^{-\alpha}}$$ $$ ={2\over 2^{1-\alpha}-1}2^{-\alpha(k+1)}+{2\over 1-2^{-\lambda}}2^{-\alpha(k+1)}$$ $$ =({2\over 2^{1-\alpha}-1}+{2\over 1-2^{-\alpha}})2^{-\alpha(k+1)}$$ $$ \le c_\alpha|x-y|^\alpha.$$

Thus $f$ is Holder continuous for any $0<\alpha <\lambda \leq 1.$

Nirenberg

Nirenberg

This took me over an hour to figure out the solution and and write it down. Please offer higher bounties in future, otherwise you may not get a response.

$Prove Holder-continuity for $\mu_\lambda (x) = \sum\limits_{n=1}^\infty \frac{ \cos(2^n x)}{2^{n \lambda} }$$

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