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Is there a continuous function $f:[0,1]\to \mathbb R$ that is not Lipschitz continuous on any measurable subset of a positive Lebesgue measure.

There are functions that are continuous but not Lipschitz continuous on any interval: Continuous or Differentiable but Nowhere Lipschitz Continuous Function I can't tell if the Weierstrass function mentioned there is not Lipschitz continuous on any subset of $[0,1]$ with a positive measure.

Let me demonstrate the difference between the Lipschitz continuity on an interval and a measurable set of positive Lebesgue measure, consider the following example of a function which is however not continuous:

The indicator function of the rational numbers $f(x)=\mathbb 1_{\mathbb Q}(x)$ is not continuous on any interval, but it is Lipschitz continuous on the set $A = [0,1]\setminus \mathbb Q$ which has Lebesgue measure $1$.

Pavel Kocourek
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  • Hello :) I don't get, why $f=1_{\mathbb Q}(x)$ would be continuous on $A$? – Jochen Jan 06 '23 at 08:18
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    According to my (quick) understanding of this question, the Weierstrass function is an example. – nicomezi Jan 06 '23 at 08:18
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    I think most continuous functions $f:[0,1]\to \mathbb R$ in the sense of Baire category (sup norm) have this property, which was probably first shown in Approximation of non-parametric surfaces of finite area by Goffman (1963; see Lemma 5 on p. 743), and also given in First Course in Functional Analysis by Goffman/Pedrick (1965; see Section 3.18 on pp. 155-159). Note that their result is for all functions having a specified modulus of continuity, which is much stronger than (continued) – Dave L. Renfro Jan 06 '23 at 08:57
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    Lipschitz continuity (linear modulus of continuity). Goffman's result was greatly strengthened in Strong porosity features of typical continuous functions by Bruckner/Hausserman (1985; summary here), which in turn was strengthened in Typical intersections of continuous functions with monotone functions by Hejný (1993; see end of paper). (continued) – Dave L. Renfro Jan 06 '23 at 08:58
  • The reason I say "I think" is that I believe for these to apply to your question you need to know that every Lipschitz function defined on a subset of $[0,1]$ of positive measure has a unique extension to a Lipschitz function defined on $[0,1].$ Note: The strengthenings by Bruckner/Hausserman and Hejný involve replacing "positive measure" (i.e. "not measure zero") with "not $\mathcal P$", where $\mathcal P$ represents various proper subcollections of measure zero subsets of $[0,1]$ (the sets in $\mathcal P$ are also nowhere dense subsets of $[0,1],$ and in fact much smaller than this). – Dave L. Renfro Jan 06 '23 at 09:10
  • Regarding my last comment about "you need to know", possibly Extension of range of functions by McShane (1934) will take care of things, except for the "unique" part, but maybe this isn't needed? I don't have time now to think further about this $\dots$ – Dave L. Renfro Jan 06 '23 at 09:16
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    OK, I'm back (from an early AM gym workout). I was thinking about the above and I realized that we don't need "unique". Let $f$ be such a (Baire) typical continuous function and assume $E \subseteq [0,1]$ has positive measure and assume the restriction of $f$ to $E$ is Lipschitz on $E.$ Let $g$ be any Lipschitz extension of $f|_E$ to $[0,1].$ Then ${x \in [a,b]:; f(x) = g(x)}$ has positive measure (because this set contains $E),$ which contradicts Goffman's result for $f.$ By the way, I realize that "not measure zero" is not the same as "positive measure", but all sets here are measurable. – Dave L. Renfro Jan 06 '23 at 10:50
  • @Jochen The indicator function of rational numbers is Lipschitz continuous on the set of irrational numbers as it is constantly equal to 0 on that domain. – Pavel Kocourek Jan 11 '23 at 06:11
  • @nicomezi There is no reference to the Weierstrass function in the question you refer to, can you be more specific on how you concluded that the Weierstrass function is an example? – Pavel Kocourek Jan 11 '23 at 06:13
  • @DaveL.Renfro Thank you kindly for sharing your thoughts and references. I think that your comments would make for a perfect answer to the question, and I believe that they would be even more useful to future visitors of this post. – Pavel Kocourek Jan 11 '23 at 06:26
  • The question I quoted is about proving that a Lipschitz continuous function has a derivative almost everywhere. Since the Weierstrass function is continuous but nowhere differentiable ... – nicomezi Jan 11 '23 at 06:37
  • @nicomezi I wanted to know if Weierstrass function is not Lipschitz continuous on any set with a positive measure, I just edited the question to clarify it. – Pavel Kocourek Jan 11 '23 at 07:03
  • Well, according to my previous comment, if there is, then the Weierstrass function would be differentiable on that set of positive measure, which is not true. – nicomezi Jan 11 '23 at 08:23
  • @nicomezi The post you referred to discuses $f$ that is Lipschitz continuous on $\mathbb R$, whilst this question is about Lipschitz continuity on some set of positive measure. However, based on Dave's comments, a Lipschitz continuous function $f:A\to \mathbb R$ can be extended into a Lipschtiz continuous function $g:\mathbb R \to \mathbb R$, and so we $g$ would have to be differentiable almost everywhere, thus also even on set $A$. Yet it does not imply that $f$ is differentiable at some point in $A$, or does it? – Pavel Kocourek Jan 11 '23 at 08:38
  • I see what you mnea. I unfortunately do not have an answer to that. – nicomezi Jan 11 '23 at 09:14

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