analysisreal-analysis
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..
The plot looks like this. It's hard to tell by a casual look that this is nowhere differentiable.
For a better view of what's going on, here's an animation that zooms into the graph. You can see that the picture looks similar at different length scales. A differentiable curve, on the other hand, would look more and more like a straight line as you zoomed in.
EDIT: I plotted this in Maple. I don't have the code any more, but it might have been something like this.
h:= x -> abs(x - 2*round(x/2)):
g:= unapply(add(1/2^n*h(2^n*x), n=0..1000),x):
x0:= 0.23456: y0:= evalf(g(x0)): r:= 1.13:
for i from 0 to 99 do
frame[i]:= plot(r^i*(g(x/r^i + x0) - y0), x=-1..1, axes=none,
title = sprintf("Zoom factor %7.1f",r^i))
od:
plots[display]([seq(frame[i],i=0..99)],insequence=true);
Your example of a function continuous at irrationals but discontinuous at rationals can be rephrased as follows: writing $\mathbb Q\cap[0,1] = \{q_0,q_1,q_2,\dots\}$ as before, define $$ f_n(x) = \begin{cases} 0, &\text{if } x\le q_n, \\ 2^{-n}, &\text{if } x>q_n.\end{cases} $$ Then your example $f(x)$ is simply $f(x) = \sum_{n=0}^\infty f_n(x)$. In this formulation, it's easy to see where the discontinuity at some $q_k$ comes from: all the functions being added are continuous there except for $f_k$.
So to construct a function differentiable at irrationals but nondifferentiable at rationals, I suggest trying $F(x) = \sum_{n=0}^\infty F_n(x)$, where $$ F_n(x) = \begin{cases} 0, &\text{if } x\le q_n, \\ 2^{-n}(x-q_n), &\text{if } x>q_n.\end{cases} $$ (And yes, this $F$ really is $\int f$, although it's not necessary to know that: one can deduce the differentiability properties of $F$ directly from the above definition.)
Best Answer
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.