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Recall that the Weierstrass function is given by

$f(x) = \sum_{n=0}^{\infty} a^n \cos(b^n\cdot\pi\cdot x)$ where $0<a<1$, $b$ is a positive odd integer, and $ab>1+\frac{3}{2}\cdot\pi$.

It is well known that this function is continuous everywhere but differentiable nowhere.

My question: is $f$ is uniformly continuous on $\mathbb{R}$? Further, Are there any general techniques to tackle such a problem?

Thanks!

Edit: as noted above, as asked, the question is silly, simply because since $b$ is an integer, the function is periodic, and hence uniformly continuous. But what if we remove the restriction that $b$ is an integer? as far as I know, the function is still continuous everywhere but differentiable nowhere. Is it still uniformly continuous?

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  • $\begingroup$ The series converges uniformly, and the terms are uniformly continuous. Is that enough to conclude $f(x)$ is uniformly continuous? $\endgroup$ – GEdgar Feb 29 '12 at 15:05
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If $b$ is an odd integer, then $f$ is periodic of period $2$, since $$ \cos(b^n\cdot\pi\cdot (x+2))=\cos(b^n\cdot\pi\cdot x+b^n\cdot2\cdot\pi)=\cos(b^n\cdot\pi\cdot x). $$ Since $[0,2]$ is compact, $f$ is uniformly continuous on $[0,2]$, and by periodicity, $f$ is uniformly continuous on $\mathbb{R}$.

Added to answer the new question

In this paper, Hardy proved that $f$ is nowhere differentiable if $0<a<1$ and $a\,b>1$. Moreover, he proved that $f$ is Hölder of order $\alpha=\log(1/a)/\log b<1$, that is, there exists a constant $C>0$ such that $$ |f(x)-f(y)|\le C\,|x-y|^\alpha\qquad\forall x,y\in\mathbb{R}. $$ Uniform continuity follows.

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  • $\begingroup$ yeah, you are right of course. This is silly. But what if I remove the demend that $b$ is an integer? from what I understand, the function is still continuous everywhere but differentiable nowhere. Is it still uniformly continuous? $\endgroup$ – the L Feb 29 '12 at 9:34
  • $\begingroup$ Yes. See the edited answer. $\endgroup$ – Julián Aguirre Feb 29 '12 at 14:10

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