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Let $f:\mathbb R \to \mathbb R$ has Intermediate value property. If f has local extremum at every point of $\mathbb R$, can we say f is constant? We know $$f(x)=\begin{cases}1 & x \in \Bbb{Q} \\ 0 & x \not \in \Bbb{Q}\end{cases}$$ has I.V.P and every point of $\mathbb R$ is local extremum but f is not constant. We know If f is continuous and have local extremum at every point then we can conclude f is constant if all extremum are maximum or minimum then proof is easy and is as follow:

If f is continuous and has a local maximum everywhere, then, let $a \in \mathbb R$. By continuity, $\{x:f(x)≤f(a)\}$ is closed, and by the hypothesis, $\{x:f(x)≤f(a)\}$ is open. The set $\{x:f(x)≤f(a)\}$ is nonempty, open and closed set of $\mathbb R$ so it is all of $\mathbb R$ by connectedness of $\mathbb R$. Therefore, for all $x$ and $y$ in $\mathbb R$, $f(b)≤f(a)$ and similarly $f(a)≤f(b)$.

Semiclassical
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Fin8ish
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1 Answers1

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Lemma: For any function $f$ on $\mathbb R$, there are at most countably many values $y$ such that for some local extremum $x$ of $f$, $f(x) = y$.

Proof: For each local maximum $x$, take an interval $I(x)$ with rational endpoints that contains $x$, such that $f(x) = \sup \{f(t): t \in I(x)\}$. There are only countably many intervals with rational endpoints, and if $I(x_1) = I(x_2)$ then $f(x_1) = f(x_2)$. Similarly for local minimum.

Corollary: A function $f$ that has a local extremum at every point of $\mathbb R$ takes only countably many values. In particular, it can't have the IVP unless it is constant.

Robert Israel
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