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Prove that $f(x)=\sum_{n=1}^\infty 1/2^n(\cos3^nx)$ is continuous but nowhere differentiable on $\mathbb{R}$.

I have proved the continuity part, but unable to do the second one. Thanks for any help.

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I think this may be trickier than it looks at first sight. Weierstrass's example is $\sum_{n=1}^\infty a^n \cos(b^n \pi x)$ where $ab > 1 + 3 \pi/2$. The change of variables $t = \pi x$ gets rid of the first $\pi$, but you have only $ab = 3/2$, so Weierstrass's method is unlikely to work. – Robert Israel Dec 10 '12 at 21:36
His method won't work directly, but the concept behind it will still apply, mainly due to the self-similarity issue this curve will have concerning differentiation. More specifically the curve can at most be differentiated at a countable number of points (and in this case that will not happen due to the lack of smoothness). For an exploration into fractal differentiation see: – Vilid Dec 10 '12 at 22:46

I am not a proper mathematician, so I'll try to give a heuristic argument, why the function is not differentiable at all $x$.

We should start by applying the definition of the derivative to the function: $$f'(x) = \lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}$$

Since the limit and the sum are commutable, we can get the expression: $$f'(x) = \sum_{n=1}^{\infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$$

One can see that no proper limit can be found for this sum, as for large $n$ the value of some of the terms becomes indeterminate when the inequality $$3^n\delta \ll 1$$ is no longer satisfied. Since some of the terms in the sum don't have a proper limit, the whole sum doesn't have a proper limit.

This is by no means a proper, mathematically rigorous answer, but maybe it helps.


Now let's consider the 'last' terms in the series: $$\lim_{n \to \infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$$

or rather

$$\lim_{n \to \infty} 2^{-n} n{\left( \cos(3^nx) ( \cos(3^n/n) - 1 ) - \sin(3^nx)\sin(3^n/n) \right)}$$

which does not have a proper limit (I hope that the L'Hoptal's theorem can be applied here, but I suspect that is not the case)... Is this not sufficient for mathematicians? :)

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As you've pointed out this isn't mathematically rigorous, and in fact each term in the series does have a proper derivative with respect to x $\frac{-(3^n)\sin(3^nx)}{2^n}$ and the series will converge for some values of x. – Vilid Dec 10 '12 at 22:34
I was wondering if the non existence of the limit for the terms when n tends to infinity would be a valid proof? – gns-ank Dec 10 '12 at 22:56
No, as given any n in the natural numbers the derivative exists (provable easily by induction), if it were as set like $\mathbb{N}\cup\infty$ then there would be an issue. – Vilid Dec 10 '12 at 23:02
What I meant, is that the term $\cos(3^n\delta) - 1$ not necesserally will be zero due to the fact that n will tend to infinity and the value of the term becomes indeterminate. But as you say, for small $n$ this will have a derivative. Or am I missing something? – gns-ank Dec 10 '12 at 23:04
The issue is that you can pull the differentiation into the series and it becomes a special case of $\frac{d}{dx}a\cos(bx) = -ba\sin(bx)$ for every term in the sequence, and so if that series converges for any x the function is differentiated at some point, and that derivative exists, the function is differentiable. As a result you either have to prove that that series converges nowhere, or that there is some other property that creates an issue with linearity of derivatives in this case. If what you suggested in the edit worked, the initial series would be indeterminate as well. – Vilid Dec 10 '12 at 23:10

Think about the nature of symmetry of the curve here, mainly given an interval $(-a,a)$ (or a similar location about another point than 0) what does the function look like compared to the interval $(\frac{-a}{2}, \frac{a}{2})$, and what would that mean for intervals of the form $(\frac{-a}{2^n}, \frac{a}{2^n})$ as n gets larger?

Another point would be to direct you to look at the Weirstrass function, but that will likely lead to a give away of the answer, so depending on what approach you want to take you can either look there or take the hint I gave and run with it.

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