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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.

edited Oct 5 2011 at 4:10

François G. Dorais♦
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asked Oct 4 2011 at 12:49

Bati
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I would say the thing to do is consult Hardy's paper ... Hardy, G. H. (1916) "Weierstrass's nondifferentiable function," Transactions of the American Mathematical Society, vol. 17, pages 301–325. – Gerald Edgar Oct 4 2011 at 14:59

GH · Accepted Answer · 2011-10-05 04:06:59Z UTC

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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.

answered Oct 5 2011 at 4:06

GH
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Note that $W$ can have a local maximum (at $0$, for example) or local minimum. And thus one-sided points of increase or decrease. – Gerald Edgar Oct 5 2011 at 11:44

Unfortunately I can't get a preview of the book but this seems like a nice way to prove the statement. Thanks. – Bati Oct 5 2011 at 12:11

@Gerald: I agree, and I believe it is difficult to find the local minima of $W(x)$. At any rate, Weierstrass' proof gives the following: (1) if $\lfloor 1/2-b^n x_0 \rfloor$ is even infinitely often, then the function is not locally decreasing at $x_0$ from the left and not locally increasing at $x_0$ from the right, in particular $x_0$ is not a local minimum; (2) if $\lfloor 1/2-b^n x_0 \rfloor$ is odd infinitely often, then the function is not locally increasing at $x_0$ from the left and not locally decreasing at $x_0$ from the right, in particular $x_0$ is not a local maximum. – GH Oct 5 2011 at 14:26

Gerry Myerson · Answer 2 · 2011-10-05 05:14:07Z UTC

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A similar function is proved to be nowhere monotonic in Gelbaum and Olmsted, Counterexamples in Analysis, Chapter 2, Example 21.

answered Oct 5 2011 at 5:14

Gerry Myerson
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	Actually nowhere monotonic is a weaker property (to me) than having no point of increase or decrease. – GH Oct 5 2011 at 5:49
	Nowhere monotonic (ie, no interval on which it is monotonic) is weaker than nowhere differentiable. – George Lowther Oct 5 2011 at 8:37
	Point taken. But Bati still might get something out of Example 21, and, anyway, I never pass up a chance to promote the Gelbaum and Olmsted book. – Gerry Myerson Oct 5 2011 at 22:52

Does the Weierstrass function have a point of increase?

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