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Let $f: [0, 1] \to \mathbb R$ be a function on the unit interval. We say $y \in \mathbb R$ is a local maximum value of $f$ if $y = f(x)$ for some strict local maximum $x$ of $f$.

Question: Does there exist a $C^1$ function $f$ on the unit interval such that the set of local maximum values of $f$ is dense in some (nontrivial) open interval $(a, b)$?

Remark: Note that a $C^0$ example is provided by a Brownian motion sample path, since a Brownian motion almost surely achieves a strict maximum in every interval.

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Such a function cannot exist. Let $M \subset [0,1]$ be the set of strict local maxima of $f$, and $C \subset [0,1]$ be the critical points, that is the set of points $x \in [0,1]$ so that $f'(x) = 0$. Then $f(C)$ is a closed subset of $\mathbf{R}$. As a consequence, if $f(M) \cap (a,b)$ were dense in some interval $(a,b)$ then necessarily $(a,b) \subset f(C)$. This is impossible because $f(C)$ has zero measure by Sard's theorem.

Edit. To see why $f(C)$ is closed, we can argue as follows. Let $(y_k \mid k \in \mathbf{N})$ be any convergent sequence in $f(C)$, with $y_k \to y$ as $k \to \infty$. We may pick a sequence $(x_k \mid k \in \mathbf{N})$ of points in $C$ so that $f(x_k) = y_k$. After extracting a subsequence we may additionally assume that $x_{k'} \to x \in [0,1]$ say, which is a critical point. Therefore $f(x_{k'}) = y_{k'} \to f(x) = y$, which in conclusion belongs to $f(C)$.

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  • $\begingroup$ Unless I'm misunderstanding the question, the OP doesn't want the maxima $x_k$ to be dense, but the maximum values $f(x_k)$. $\endgroup$
    – gmvh
    May 10, 2021 at 11:15
  • $\begingroup$ @gmvh You're right, I had misread the question. I've changed the argument - hopefully it works now. $\endgroup$
    – Leo Moos
    May 10, 2021 at 12:13
  • $\begingroup$ Sorry, how do we know that $f(C)$ is closed? And what is the difference between $M$ and $C$? $\endgroup$
    – Nate River
    May 10, 2021 at 12:17
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    $\begingroup$ About the Edit: this is just the fact that the image of a compact set is compact $\endgroup$
    – Nicolast
    May 10, 2021 at 12:42
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    $\begingroup$ Ye, I seem to have forgotten that critical point means $f' = 0$, and so by $C^1$-ness of f, the set of critical points is closed... $\endgroup$
    – Nate River
    May 10, 2021 at 12:43

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