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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.

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Let me know what you have tried. –  Prasad G Jun 5 '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? –  Thomas E. Jun 5 '12 at 13:06
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What is $ X $? A compact metric space? –  Jonas Meyer Jun 5 '12 at 13:07
    
I'm sorry, Thomas E. and Jonas Meyer. X is a compact metric space. Thanks by your help. –  rodrigo Jun 5 '12 at 13:15
    
A tiny nitpick: $X$ should also be non-empty. –  Egbert Jun 5 '12 at 13:33

1 Answer 1

up vote 1 down vote accepted

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 $f(x,t)=u(t)$.

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Why is $f(t,x)=\max_{y\in X}f(t,y)$? –  Thomas E. Jun 5 '12 at 13:38
    
Yeah, that does the trick. +1 for nice answer. –  Thomas E. Jun 5 '12 at 13:54

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