Is $\sum_{n=1}^{\infty} \frac{\sin(2^nx)}{2^n}$ continuous? or differentiable anywhere?

Question

Does it make sense to call $\sum_{n=1}^{\infty} \frac{\sin(2^nx)}{2^n}$ continuous? For any finite n it is easy to declare that the function is continuous. Now when we let $n \to \infty$ we get a function that would seem pretty continuous, with each value very very close to its neighbors. However when we attempt to take the derivative anywhere on the function (except for a few key points) we find due the infinitesimal and ever present oscillations inherent in the function it doesn't really make sense to take the derivative.

My hunch is to say that $\sum_{n=1}^{\infty} \frac{\sin(2^nx)}{2^n}$ is continuous but not differentiable. But how can one justify that?

PS:
if you want to see an interactive, zoomable version of the function

PPS: Could the function possibly be differentiable at $f(2\pi k)$ for $k =$ some integer. At $2\pi k $ the derivative of $ \frac{\sin(2^nx)}{2^n}$ is $\frac{1}{2^n}$ for all n. Does that imply diferentiablitity at $2 \pi k$?

Answer 1

Since $$\left|\frac{\sin(2^nx)}{2^n}\right|\le\frac1{2^n}$$ and $\sum\frac1{2^n}$ converges, Weierstrass $M$-test implies that the function $$f(x)=\sum_{n=1}^\infty \frac{\sin(2^nx)}{2^n}$$ is continuous.

However, if we define $$f_n(x)=\frac{\sin(2^nx)}{2^n}$$ then $$f_n'(x)=\cos(2^nx)$$ and we can't apply the Weierstrass $M$-test anymore. This does not mean that $f$ is not differentiable, of course. (My guts say that $f$ is not differentiable at any point, but proving it seems difficult).

Answer 2

Claim: $f'(0)$ doesn't exist.

Proof: Because $\lim_{x\to 0} (\sin x)/x = 1,$ we can regard $(\sin x)/x$ as positive and continuous on $[0,1].$ Hence there exists $c>0$ such that $(\sin x)/x \ge c$ for $x\in [0,1],$ which gives $\sin x \ge cx$ there.

Let's see what happens to the difference quoitents of $f$ along the sequence $2^{-N}:$

$$\frac{f(2^{-N})-f(0)}{2^{-N}-0} = 2^Nf(2^{-N}) =2^N\left (\sum_{n=1}^{N} \frac{\sin (2^{n-N})}{2^n} + \sum_{n=N+1}^{\infty} \frac{\sin (2^{n-N})}{2^n}\right)$$ $$ \ge 2^N\left (cN2^{-N} - \sum_{n=N+1}^{\infty} \frac{1}{2^n}\right ).$$

That last sum adds to $1/2^{N}.$ Thus the last expression equals $cN - 1,$ which $\to \infty$ as $N\to \infty.$ This shows $f'(0)$ doesn't exist.