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Let $X$ be a compact metric space. Does there always exist a continuous $f : X \to \mathbb{R}$ such that $\{ x \in X : f(x) = \| f \|_{\infty} \}$ is at most countable?

Certainly this is true for $[0,1]^n$, take $f(x_1,\ldots,x_n) = x_1 + \cdots + x_n$. It seems it should hold in general but I'm not sure how to attempt to prove that.

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    $\begingroup$ Since $X$ is compact $m:=\max_{x,y \in X} d(x,y)$ exists. Fix any $x_0 \in X$ and set $f(x)=m-d(x,x_0)$. Then $0 \le f \le m$ on $X$ and $f(x)=m$ iff $x=x_0$. Hence $\{x \in X: f(x)=\|f\|_\infty\}=\{x_0\}$. $\endgroup$
    – Gerd
    Oct 23, 2023 at 20:27
  • $\begingroup$ Actually the difficulty is in the "metric space" condition, i.e. defining a distance: that is probably not possible to get separation on sets $X$ that have a greater cardinal than $\mathbb R$. $\endgroup$ Oct 23, 2023 at 21:10
  • $\begingroup$ It is always possible in a perfectly normal space, regardless of its cardinality. $\endgroup$
    – freakish
    Oct 23, 2023 at 21:38
  • $\begingroup$ @freakish You are right, I was dumb. There is the trivial distance $d(x,y)=1$ if $x \ne y$, $d(x,y)=0$ if $x = y$. $\endgroup$ Oct 24, 2023 at 9:20

1 Answer 1

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Take any point $x$ in $X$ and let $f(y) = \frac 1 {1+d(x,y)}$, where $d$ is the distance of metric space $X$.

Then $f$ is continuous, because $d$ is continuous in $y$ (a distance is always continuous).

$f$ reaches its maximum $1$ only on $x$, because of the separation property of a distance.

The compactness condition is not needed.

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  • $\begingroup$ The set $\{y \in X: f(y)=\|f\|_\infty\}$ is unclear to me in this example. If $X$ is not compact even $\|f\|_\infty = \infty$ is possible. $\endgroup$
    – Gerd
    Oct 23, 2023 at 20:43
  • $\begingroup$ One could take $(1-d(x,y))_+$ instead. $\endgroup$
    – LL 3.14
    Oct 23, 2023 at 20:46
  • $\begingroup$ @Gerd You are right. I forgot how the infinite norm was defined. So I have to change $f$ definition. $\endgroup$ Oct 23, 2023 at 20:51
  • $\begingroup$ @Gerd Now it should be correct. $\endgroup$ Oct 23, 2023 at 20:58
  • $\begingroup$ @LL3.14 Yes that would work too. $\endgroup$ Oct 23, 2023 at 20:59

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