Problem
The Weierstrass function $W(x)$ is given by
$W(x)=\sum_{n\geq 0} a^n \cos(b^n \pi x)$
where $0< a <1$ and $b$ is an odd integer such that $ab > 1+3\pi/2$.
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is said to have a point of increase if there exists a $t \in \mathbb{R}$ and $\delta>0$ such that
$f(t-s)\leq f(t) \leq f(t+s) \quad \forall s \in [0,\delta]$.
So my question is does the Weierstrass function have a point of increase?
Motivation
In Burdzy's paper there is a proof that a Brownian motion does not have a point of increase. There are examples of nowhere differentiable functions which have a point of increase that one could construct but I have been having difficulty seeing if the Weierstrass function does.
I would be grateful for any references or heuristics regarding this problem, or any comments as to the difficulty.
Best Answer
The original proof of Weierstrass (see pages 4 to 7 in Elgar (ed.): Classics on Fractals, Westview Press, 2004) constructs, for any $x_0\in\mathbb{R}$, two sequences $(x'_n)$ and $(x''_n)$ such that $$x'_n < x_0 < x''_n,\qquad x'_n\to x_0,\qquad x''_n\to x_0,$$ but $$\frac{W(x'_n)-W(x)}{x'_n-x}\qquad\text{and}\qquad \frac{W(x''_n)-W(x)}{x''_n-x}$$ are of opposite signs and their absolute values tend to infinity. This shows that $W(x)$ has no point of increase and no point of decrease.