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 $W$.
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 $l$.