Is continuous $f$ constant if every point of $\mathbb{R}$ is local minimum of $f$?

Question

Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous. Is $f$ constant if every point of $\mathbb{R}$ is local minimum of $f$?

What metric spaces we can use instead of $\mathbb{R}$? I guess we have same result for $f:\mathbb{R}^n \rightarrow \mathbb{R}$.

In general, we can set $U$ to be any open subset of a space $X$ and the function: $$f = \begin{cases} 1 & \text{ if } x \in U \ 0 & \text{ if } x \in X - U \end{cases}$$ has the property that every point is a local minimum. This is of course only continuous if $U$ is clopen. — jxnh, Dec 29 '14 at 20:23
@JHance Conversely, the result is clearly true for a path-connected space. Is it true for a connected space? — hunter, Dec 29 '14 at 20:28
possible duplicate of Can non-constant functions have the IVP and have local extremum everywhere? — PhoemueX, Dec 29 '14 at 20:30
@PhoemueX The duplicate does not address the second part of the question, though. On the other hand, it deals with extrema of either kind, so not every answer here would work there. — , Dec 29 '14 at 21:34

score 6 · Answer 1 · edited Apr 13 '17 at 12:21

This holds for $f:X\to\mathbb R$ if $X$ is a connected space. For each $x\in X$, $f^{-1}([f(x),\infty))$ is closed by continuity, and open by the condition on local minima. This set is nonempty because it contains $x$, hence it equals $X$ by connectedness. Thus for all $y\in X$, $f(y)\geq f(x)$. Because $x$ and $y$ were arbitrary, this implies that $f$ is constant.

See also Continuous function with local maxima everywhere but no global maxima.

Is continuous $f$ constant if every point of $\mathbb{R}$ is local minimum of $f$?

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