Consider the classical Weierstrass function $$ W(x)=\sum_{n=1}^\infty \frac{e^{i2^nx}}{2^n}. $$ It is a wellknown 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 $h<\delta$.
2 Answers
The argument in Zygmund's paper is valid for any continuous function $f$ in the class $\Lambda^*$, that is with $$ \f\_*=\sup_{x\in{\mathbf T},\; h\not=0}\Big\frac{f(x+h)+f(xh)2f(x)}h\Big<\infty. $$ 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 $$ \frac{f(x_0+h)f(x_0)}h + \frac{f(x_0h)f(x_0)}h =\frac{\Delta^2_hf(x_0)}h\leq \f\_*. $$ Since the first two summands are nonnegative, 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 $F=fL$ belongs to $\Lambda^*$, is continuous in $[a,b]$ and vanishes at the endpoints. 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.

$\begingroup$ will you give a quick sketch of Zygmund's argument? $\endgroup$ May 10, 2018 at 16:22

1$\begingroup$ Which paper of Zygmund? He had several papers in Duke in 1945. $\endgroup$ May 10, 2018 at 21:59