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I'm trying to solve this problem, but I don't have any idea. Can you help me?

Let X a compact metric space and $f:X\times\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function. Consider $m(t_0)=\max_x (f(x,t_0))$. Show that $m$ is continuous.

Thanks.

rodrigo
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    Let me know what you have tried. – Prasad G Jun 05 '12 at 13:06
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    Are you sure that $m$ is well defined? Should there be $\sup_{x}$ instad of $\max_{x}$? Or are you assuming that $X$ is compact? – T. Eskin Jun 05 '12 at 13:06
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    What is $ X $? A compact metric space? – Jonas Meyer Jun 05 '12 at 13:07
  • I'm sorry, Thomas E. and Jonas Meyer. X is a compact metric space. Thanks by your help. – rodrigo Jun 05 '12 at 13:15
  • A tiny nitpick: $X$ should also be non-empty. – Egbert Jun 05 '12 at 13:33
  • If you are familiar with the topological definition of continuity, then another presentation of the result can be found here. Note in particular that the supremum of any family of continuous functions is automatically lower semicontinuous. The compactness of $X$ is only used to guarantee that the family (which we think of as indexed by $X$) is equicontinuous which allows us to get also upper semicontinuoity. – Willie Wong Jun 05 '12 at 14:07

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Fix $t\in\Bbb R$. We just have to show sequential continuity, since we are working in a metric space. Let $\{t_n\}\subset\Bbb R$ a sequence which converges to $t$. Since $X$ is compact, we can find $x_n$ such that $m(t_n)=f(x_n,t_n)$. We show that for each subsequence of $\{t_n\}$ we can find a further subsequence $\{t_{n_k}\}$ such that $m(t_{n_k})\to m(t)$. It will show that $m(t_n)\to m(t)$.

Let $\{t_{n'}\}$ a subsequence of $\{t_n\}$. The sequence $\{x_{n'}\}$ admits a converging subsequence $\{x_{n_k}\}$, say to $x$. Then $(t_{n_k},x_{n_k})\to (t,x)$ and we conclude using the continuity of $f$ (and the fact that $f(x_n,t_n)\geq f(y,t_n)$ for all $y\in Y$, to show that $f(x,t)=u(t)$.

Davide Giraudo
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