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I posted the question about continuity, Let me assume that the function $f(x,y,z,w)$ is continuous. Is the $\max_{w} f(x,y,z,w)$ continuous?

and got the answer that the function is continuous when the domain of the function is a compact set.

Now I want to know how to prove it, "If $f(x,y,z,w)$ is continuous and domains of $x,y,z,w$ are all compact set, then $\max_{w} f(x,y,z,w)$ is continuous."

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You can't prove it, because it's not true. A two-dimensional counterexample: Let $f(x,w)=w$ on the cross-shaped set $\{(x,w): -2\le x,w\le 2\text{ and }\min(|x|,|w|)\le 1\}$. For this clearly continuous function, $$\max_w f(x,w)=\begin{cases}1&-2\le x< 1\\ 2&-1\le x\le 1\\ 1& 1<x\le 2\end{cases}$$ What additional conditions would we need to make this work? Convexity of the domain should do it.

Addendum: As seen in supinf's answer here, convexity isn't enough. I was hoping to avoid the overkill of making the domain a compact Cartesian product, but it looks like we can't do that.

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  • $\begingroup$ Could you give me the proof for the case? $\endgroup$ – Heasung Kim Mar 20 at 8:55
  • $\begingroup$ @jmerry: Convexity of the domain also doesnt work, see my answer here. $\endgroup$ – supinf Mar 20 at 12:41
  • $\begingroup$ It should work if the domain is a cartesian product, $\{(x,w) : x \in A, w \in B\}$, $B$ is compact, and $A$ locally compact. Continuity in $x$ is a local property and global properties of $f$ wrt. $x$ should not be needed. $\endgroup$ – Hans Engler Mar 20 at 13:21

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