calculus - Every point takes local maximum value

If $f:\mathbb{R}\to\mathbb{R}$ and every point takes a local maximum value, it's a fact that the local maximum values of a real function can only have countable, so if we assume $f$ is continuous we have $f$ must be constant. My question is, if $f$ isn't continuous, can we prove there must be some interval that $f$ is constant on it?

Answer

I think your conclusion is right. I've written a proof, please help me check if it's right.

Since "local maximum values can only be countable", we assume they are $\{a_n\}_n$. And let $F_n=\{f=a_n\}$. Then $\mathbb{R}=\bigcup_{n\geq1}F_n$.

Due to Baire's theorem, there is a $n_0$ such that $F_{n_0}$ is dense in an open interval (expressed as $U$).

Because $\{f=a_{n_0}\}$ is dense in $U$, it's easy to prove that $f(x)\geq a_{n_0}$ in $U$.

Assume that $x_0\in\{f=a_{n_0}\}$ is not an interior point of $\{f=a_{n_0}\}$ in $U$. In other words, $\exists\{x_n\}_n\bigcap\{f=a_{n_0}\}=\emptyset$ such that $x_n\to x_0$. However, it can't be correct because $x_0$ is a local maximum.

Then we know $\{f=a_{n_0}\}$ has an interior point $x_0$ and we arrive at your conclusion. What's more, since $x_0$ is arbitrary, we know that $F_{n_0}\cap U$ is open too.