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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.2012-06-05
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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?2012-06-05
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    What is $ X $? A compact metric space?2012-06-05
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    I'm sorry, Thomas E. and Jonas Meyer. X is a compact metric space. Thanks by your help.2012-06-05
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    A tiny nitpick: $X$ should also be non-empty.2012-06-05
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    If you are familiar with the topological definition of continuity, then another presentation of the result can be found [here](http://williewong.wordpress.com/2011/11/01/continuity-of-the-infimum/). 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.2012-06-05

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

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    Why is $f(t,x)=\max_{y\in X}f(t,y)$?2012-06-05
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    Yeah, that does the trick. +1 for nice answer.2012-06-05