2

Find a measurable set $A \subset\mathbb R$ with positive Lebesgue measure $\mu(A)>0$ such that the Weierstrass function is Lipschitz continuous on $A$, or show that no such set exists.

Weierstrass function is the function $f:\mathbb R \to \mathbb R$ defined by the Fourier series $$ f(x) = \sum_{n=1}^\infty a^n cos(b^n \pi x), $$ where $a \in (0,1)$, $b$ is a positive odd integer, and $ab>1+\frac 3 2 \pi$. It is known to be continuous but not differentiable at any point.

Let me demonstrate the emphasize that $A$ can be any measurable set, not necessarily an interval. For example, the indicator function of the rational numbers is not Lipschitz continuous on any interval, but it is Lipschitz continuous on the set of irrational numbers (as it is constantly 0 on that domain).

A closely related question: A continuous function $f:[0,1]\to \mathbb R$ that is not Lipschitz continuous on any subset of a positive measure.

Alp Uzman
  • 10,742
Pavel Kocourek
  • 1,565
  • Take a look at Rademacher's theorem https://en.wikipedia.org/wiki/Rademacher%27s_theorem – Marco Cantarini Jan 11 '23 at 12:37
  • @MarcoCantarini Thanks for the reference. Rademacher's theorem is relevant for a function that is Lipschitz continuous on an open set. In contact, here the set $A$ can be any measurable set with positive measure, so it might have no nonempty open subset. – Pavel Kocourek Jan 11 '23 at 16:06
  • I know, but it could be a good start point. And maybe there are relevant generalizations, I would do a research on it. – Marco Cantarini Jan 12 '23 at 07:26
  • Also this work could be relevant https://arxiv.org/pdf/1508.02033.pdf – Marco Cantarini Jan 12 '23 at 08:06
  • This seems like a pretty hard problem. Perhaps studying the proof that the Weierstrass function isn't differentiable in detail will give insight to do with bounds – FShrike Jan 19 '23 at 22:58

0 Answers0