# Is Weierstrass function nowhere pointwise Lipschitz?

Consider the classical Weierstrass function $$W(x)=\sum_{n=1}^\infty \frac{e^{i2^nx}}{2^n}.$$ It is a well-known result that this function is nowhere differentiable (Hardy, TAMS 1916, Thm 1.31). In particular, by Rademacher theorem, it does not belong to the class $Lip(T)$. The question is whether there may exist some point $x_0$ such that $W\in Lip(x_0)$, that is, for some $\delta>0$ it holds $$|W(x_0+h)-W(x_0)|\leq C |h|$$ when $|h|<\delta$.

The argument in Zygmund's paper is valid for any continuous function $f$ in the class $\Lambda^*$, that is with $$\|f\|_*=\sup_{x\in{\mathbf T},\; h\not=0}\Big|\frac{f(x+h)+f(x-h)-2f(x)}h\Big|<\infty.$$ Assume first that $f$ has a local minimum at some point $x_0\in\mathbb T$. Then for some $\delta>0$ and all $h\in(0,\delta)$ one has $$\frac{f(x_0+h)-f(x_0)}h + \frac{f(x_0-h)-f(x_0)}h =\frac{\Delta^2_hf(x_0)}h\leq \|f\|_*.$$ Since the first two summands are non-negative, one deduces that $$0\leq \frac{f(x_0\pm h)-f(x_0)}h \leq \|f\|_*,$$ and so $f\in Lip(x_0)$. The same applies if $f$ has a local maximum (replacing $f$ by $-f$).
In the case of $W$, this already implies the existence of at least 2 points with the local Lipschitz condition.
To obtain an everywhere dense set, for a general $f\in\Lambda^*$, pick any interval $(a,b)\subset\mathbb T$ and a line function $L(x)$ with $L(a)=f(a)$ and $L(b)=f(b)$. Then $F=f-L$ belongs to $\Lambda^*$, is continuous in $[a,b]$ and vanishes at the end-points. So it must have either a maximum or minimum at some $x_0\in(a,b)$. Then $F$, and hence $f$, must belong to $Lip(x_0)$.
Sorry, I apologize for posing this question. I just realized that the answer is yes, by an elementary argument in Zygmund's paper (Duke Math, 1945, proof of Theorem 1). The function $W$ is pointwise Lipschitz at an everywhere dense set of points.