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I am working on the following problem:

Let $f:\mathbb{R}\to\mathbb{R}$ satisfies that for all $x_0\in\mathbb{R}$, there exists some $\delta>0$ such that $f(x)\le f(x_0)$ for all $x\in (x_0-\delta, x_0+\delta)$. Does there exists some subinterval $I$ of $\mathbb{R}$ such that $f$ is a constant on $I$?

The problem can be done when $f$ is assumed to be continuous, in which case $f$ is a constant. I have known that $f$ need not to be a constant in my case, for instance, consider $f(x)=\chi_{(0,1)}+2\chi_{(-\infty,0]\cup[0,+\infty)}$, but I'm not sure whether such $I$ always exists.

EDIT: Seems that this answer offers a proof involving Baire's Theorem, however I didn't see how Baire's Theorem works in this case. As I know Baire's theorem works only when $F_n$ are closed. If we take the closure more elements are added to $F_n$ and the proof seems to fail to me. Can someone kindly elaborate this to me? Thanks in advance.

MathLearner
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  • Related: https://math.stackexchange.com/q/1402071/42969 – Martin R Sep 04 '24 at 11:23
  • I was wondering if there are some of such functions that it is not a constant on any subinterval of $\mathbb{R}$, an indicator function must be identical on some intervals. – MathLearner Sep 04 '24 at 11:33
  • Did you see the comments to that answer? It is not asssumed that the $F_n$ are closed sets. – Martin R Sep 04 '24 at 12:29
  • I do, but I don't know a version of Baire Theorem that applies when $F_n$ are not closed. – MathLearner Sep 04 '24 at 12:30
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    As mentioned in those comments, you can work with the closures of $F_n$. One $\overline F_n$ contains an open interval $U$, which means that $F_n \cap U $ is dense in $U$. – Martin R Sep 04 '24 at 12:34
  • I think I got it, thanks. I'm voting to close this question now. – MathLearner Sep 04 '24 at 12:37

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