How can it be proved that the function $$f(x)=\sum_{n=1}^{\infty}2^{-n\alpha}\cos(2^nx)$$ for $\alpha \in ]0,1[$ is $\alpha$-Hölder continuous but not differentiable at any point of $[0,1]$?

I tried to write down $f(x+h)-f(x)$ and use addition formulas for the cosine, but I don't obtain anything..


1 Answer 1


This is one of the well know examples of Weierstrass function. Hardy studied Hölder-continuity of such functions in Weierstrass's nondifferentiable function, G.H. Hardy, Trans. Amer. Math. Soc., 17 (1916), 301–325.. Here is a proof of Hölder-continuity for your case.

Theorem. Let $0<a<1$, $b>1$ and $ab>1$ then the function $$ f(x)=\sum\limits_{n=1}^\infty a^n\cos(b^n x) $$ is $(-\log_b a)$-Hölder continuous.

Proof. Consider $x\in\mathbb{R}$ and $h\in(-1,1)$, then $$ f(x+h)-f(x)= \sum\limits_{n=1}^\infty a^{n}(\cos(b^n(x+h))-\cos(b^nx))= $$ $$ -2\sum\limits_{n=1}^\infty a^{n}\sin(2^{-1}b^n(2x+h))\sin(2^{-1}b^{n}h)= $$ Since $b>1$ and $h\in(-1,1)$ there exist $p\in\mathbb{N}$ such that $2^{-1}b^{p}|h|\leq 1<b^p|h|$, so $$ |f(x+h)-f(x)|\leq 2\sum\limits_{n=1}^\infty a^{n}|\sin(2^{-1}b^{n}(2x+h))||\sin(2^{-1}b^{n}h)|\leq 2\sum\limits_{n=1}^\infty a^{n}|\sin(2^{-1}b^{n}h)|= $$ $$ 2\sum\limits_{n=1}^p a^{n}|\sin(2^{-1}b^{n}h)|+ 2\sum\limits_{n=p+1}^\infty a^{n}|\sin(2^{-1}b^{n}h)|\leq 2\sum\limits_{n=1}^p a^{n}|2^{-1}b^{n}h|+ 2\sum\limits_{n=p+1}^\infty a^{n}= $$ $$ \frac{ab|h|}{ab-1}(a^p b^p-1)+\frac{2a}{1-a}a^p\leq \frac{ab|h|}{ab-1}a^p b^p+\frac{2a}{1-a}a^p $$ Since $2^{-1}b^{p}|h|\leq 1<b^p|h|$ and $0<a<1$, then we have $b^p|h|<2$ and $a^p\leq |h|^{-\log_b a}$. Hence $$ |f(x+h)-f(x)|\leq \frac{ab|h|}{ab-1}a^p b^p+\frac{2a}{1-a}a^p\leq \frac{2ab}{ab-1}a^p +\frac{2a}{1-a}a^p\leq $$ $$ \left(\frac{2ab}{ab-1} +\frac{2a}{1-a}\right)|h|^{-\log_b a}= \frac{b-1}{(ab-1)(1-a)}|h|^{-\log_b a} $$ This means that $f$ is $(-\log_b a)$-Hölder continuous.

Corollary. For $\alpha\in(0,1)$ the function $$ f(x)=\sum\limits_{n=1}^\infty 2^{-n\alpha}\cos(2^n x) $$ is $\alpha$-Hölder continuous.

Proof Apply previous theorem with $a=2^{-\alpha}$ and $b=2$.

As for the proof of nowhere differentiability, I don't know a short proof. The problem is that the standard Weierstrass argument is not applicable here - parameters $a$, $b$ must satisfy inequality $ab>1+\frac{3\pi}{2}$. So it seems to me that one should repeat all the steps of Hardy's proof.

  • 1
    $\begingroup$ In general it is a good idea, I think, to give the title of the paper you're referring to (in this case Hardy's paper). For example, the link you gave does not work (anymore?). $\endgroup$
    – user2093
    Sep 29, 2014 at 21:11
  • $\begingroup$ @William fixed. $\endgroup$
    – Norbert
    Sep 29, 2014 at 21:24

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