# Distributional derivative of Weierstrass function

How can we compute the distributional derivative of the Weierstrass function $$W(x) =\sum_{k=1}^\infty \lambda^{(s-2)k}\sin(\lambda^k x)$$ where $$s \in (0,2)$$ and $$\lambda$$ are fixed parameters?

We know that the Weierstrass function is nowhere differentiable. This implies that it does not have a weak derivative. However, since $$W \in L^1_{loc}$$, we can consider the associated distribution $$T_W$$ and compute its distributional derivative.

I'm having troubles doing that computation because of the series representation of $$W$$.

• Your question doesn't make sense as written. What are you actually trying to do, what tools do you have available, and what is the motivation? – user296602 Apr 6 '19 at 1:10
• @T.Bongers Sorry, there was a typo (that is, I had mistakenly deleted the main verb of the sentence). – Riku Apr 6 '19 at 1:16
• I'm glad you edited it to fix that, but it doesn't address 2 of the 3 issues in my comment; since this is not a do-my-homework site, please edit the question to actually improve it. – user296602 Apr 6 '19 at 1:16
• @T.Bongers Done. – Riku Apr 6 '19 at 1:25
• You have two parameters: $s$ and $\lambda$. How is $W$ a function just of $x$? – Ted Shifrin Apr 6 '19 at 1:26

For some $$r \in (1/2,1)$$ let

$$W(x) = \sum_{m=0}^\infty r^m e^{2i\pi 2^m x}$$

The series converges uniformly to a continuous function and for any $$\phi \in C^\infty(\Bbb{R/Z})$$ smooth $$1$$-periodic $$\langle W,\phi \rangle = \int_0^1 W(x)\phi(x)dx = \sum_{m=0}^\infty r^m \hat{\phi}(-2^m), \qquad \hat{\phi}(n) = \langle e^{-2i \pi n x},\phi \rangle$$ Which defines a distribution $$W \in D'(\Bbb{R/Z})$$.

Then

$$W^{(k)}(x) = \lim_{M\to \infty}(2i\pi)^k\sum_{m=0}^M (2^k r)^m e^{2i\pi 2^m x}$$ (with the limit taken in the sense of distribution)

defines a distribution

$$\langle W^{(k)},\phi \rangle = (2i\pi)^k\sum_{m=0}^\infty (2^k r)^m \hat{\phi}(-2^m)$$ the series converges absolutely since $$\hat{\phi}(n) = O(n^{-l})$$ for every $$l$$.

• Thank you. 1. Why can we rewrite $W(x)$ in that way for some $r \in (1/2, 1)$? 2. Why can we take $\phi \in C^\infty(\mathbb{R}/\mathbb{Z})$ instead of just $\phi \in C^\infty(\mathbb{R})$? 3. What did you deduce the decay of $\hat \phi$ from? 4. Can we rewrite the series in the last formula in a more explicit way? – Riku Apr 6 '19 at 1:31
• If you want to consider $W$ as a distribution on $\Bbb{R}$ then $\langle W,\varphi \rangle_\Bbb{R} = \langle W, P \varphi \rangle_\Bbb{R/Z}$ where $P \varphi(x) = \sum_n \varphi(x+n)$, so it doesn't change anything (as $P$ is a continuous linear map $C^\infty_c(\Bbb{R}) \to C^\infty(\Bbb{R/Z})$), equivalently replace the Fourier series coefficients of $\phi$ by the Fourier transform of $\varphi$. – reuns Apr 6 '19 at 1:39
• The faster-than-polynomial decay of Fourier coefficients of smooth functions is well-known : $|\hat{\phi}(n) (2i\pi n)^l| \le \|\phi^{(l)}\|_{L^1(0,1)}$ – reuns Apr 6 '19 at 1:41
• anything unclear ? – reuns Apr 7 '19 at 0:07
• Thank you for your comments. Why is it true that there exists $r \in (1/2,1)$ such that $W(x) = \sum_{m=0}^\infty r^m e^{2i\pi 2^m x}$? Could you add some details on that? – Riku Apr 7 '19 at 13:07