# 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.

• – user987907
Commented Nov 12, 2021 at 13:56
• @LouisPan It doesn't seem to be a Weierstrass function since $ab=1$ . Does this mean it's differentiable everywhere and if yes is there any "easy" way to show it is? Commented Nov 12, 2021 at 14:03
• The identity $f(x) = \sin(x) + f(2x)/2$ shows (a) $f$ is not differentiable at 0 (if it were we would have $f'(0) = 1 + f'(0)$), and hence (b) $f$ is not differentiable at any dyadic rational. Commented Nov 12, 2021 at 14:40
• @bringradical "G. H. Hardy showed that the function of the above construction is nowhere differentiable with the assumptions 0 < a < 1, ab ≥ 1." Commented Nov 15, 2021 at 12:30
• This may help: on pp. 114-118 and p. 174 (Problem 8) in Stein&Shakarchi "Fourier Analysis" $f(x) = \sum_{n=0}^\infty 2^{n\alpha}e^{i2^n x}$ ($0 < \alpha <1$) is shown to be nowhere differentiable. Commented Dec 10, 2021 at 3:46

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.

• Well, for what it's worth, I'm not sure I deserve the green tick, just almost all of it. Commented Nov 16, 2021 at 22:27

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 $$f$$.

• Problem aside, how could you just write all those things on the phone?! wow Commented Nov 25, 2021 at 9:11
• I am hiking for 3 month now, without computer, I was obliged to get used to it aha
– jvc
Commented Nov 25, 2021 at 9:15
• I use "text replacement" on my phone's keyboard, so that when I type "bmat", my phone types \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, when I type "iint", my phone types \int_{ }^{ }, when I type "rrn", my phone types \mathbb{R}^{n}, etc.
– Joe
Commented Nov 26, 2021 at 12:41