Is this function nowhere differentiable? I was looking at the following function,
$$\displaystyle f(x) := \sum_{n=0}^\infty {\sin(2^nx) \over 2^n}.$$
It is pretty obvious that $f$ is continous everywhere in $\mathbb{R}$. But I can't figure out where it is differentiable. Differentiating term by term would lead me to believe it is differentiable nowhere but I'm not sure if I can do that.
 A: I suspect $f$ is differentiable nowhere, because its derivative is "trying" to be $\sum_{n=0}^\infty \cos(2^n x)$, which converges nowhere. Below I show that $f$ is differentiable almost nowhere.
Claim 1: If $f$ is differentiable at $x$ then the sequence of partial sums $\sum_{n=0}^N \cos(2^n x)$ is bounded.
Proof: There is a uniform estimate of the form
$$\frac{\sin(x) - \sin(y)}{x - y} = \cos(x) + O(|x-y|).$$
Plugging this into the definition of $f$ and truncating the sum gives $$\frac{f(x) - f(y)}{x-y} = \sum_{n=0}^N \cos(2^n x) + O(2^N |x-y|+ 2^{-N} |x-y|^{-1}).$$
Now pick $y$ so that $|x-y| = 2^{-N}$. It follows that
$$\sum_{n=0}^N \cos(2^n x) = \frac{f(x) - f(y)}{x - y} + O(1) = f'(x) + O(1)$$
if $f$ is differentiable at $x$.
Claim 2: For almost all $x$ (in the sense of either measure or category), the sequence of partial sums $\sum_{n=0}^N \cos(2^n x)$ is not bounded.
Proof: Consider the binary expansion of $x / (2 \pi)$. Almost surely there will be a long stretch of zeros somewhere. That means there is some $n$ such that $2^n x \approx 0$ mod $2 \pi$, so the partial sums cannot be bounded.
There are some $x$ for which the partial sums $\sum_{n=0}^N \cos(2^n x)$ are bounded, e.g., $x = 2\pi / q$ for any proper prime power $q$ such that $2$ is a primitive root (e.g., $q = 9$). I still doubt that $f$ is differentiable at these points.
A: I apologize because I am force to write it on phone. I hope it will be not to hard to read.
In fact, there is a powerful theorem that give conditions for which a "lacunary trigonometric serie" is nowhere differenriable. The proof can be found in the french book, Analyse pour l'agrégation, Zuily-Quéffelec.
Let $a_n$, be a complex sequence with $\sum |a_n| < \infty$. Let $b_n$ be a real sequence. Define $d_n = dist(b_n, (b_k)_{k \neq n}) $. We suppose that $d_n>0$ and $d_n \to \infty$.
Then, if $f(x) = \sum a_n e^{i b_n t} $ is differentiable at one point (at least), then $a_n d_n \to 0$.
By contradiction, it show that the function of this question is nowhere differentiable ($a_n = 1/2^n$ or $- i/2^n$ and $b_n = 2^n$ or $-2^n$, depending of the parity of n. Thus $a_n d_n = 1/2$).
The proof of the theorem goes as follow. Let $\phi$ be a Schwarz function with $\hat{\phi}(0)=1$ and with support of $\hat{\phi} $ included in $[-1,1]$ (we can construct it by taking the Fourier transfort of the a test function with support included in $[-1,1]$).
By Fubini theorem :
$a_n = \int f(x/d_n) \phi(x) e^{-i b_n/d_n x} dx$
(just expand f, intervert sum and integral and use the fact that the support of $\hat{\phi}$ is included in $[-1,1] $.
The next step is to handle the case where f is differentiable at 0 with $f(0)=f'(0)=0$.
Then by considering "small x" (main argument : differentiability at 0) and "big x" (main argument : $\sum |a_n| < \infty$) we can prove that for all x :
$|f(x)| \leq C| x|$
with $C > 0$
The can now easily conclude, in that particular case, using the expression of $a_n$, the previous estimate of $f$ and the fact that $\phi$ is Schwarz, that $a_n d_n \to 0$ (Lebesgue theorem can be applied).
For the general case, if f is differentiable at $x_0$, we consider $g(x) = f(x + x_0) + a e^{i b_1 x} + b e^{ i b_2 x}$ with $a, b$ chosen such that $g$ satisfies the particular case above. Thus $g$ is nowhere differentiable, and thus also $f$.
