Let $K\subset \mathbb R^n$ be a compact subset and consider a continuous function $g:K\times K\longrightarrow \mathbb R$. Define $f:K\longrightarrow \mathbb R$ by, $$f(x)=\sup_{y\in K}g(x, y).$$ Is $f$ a continuous function?
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Yes. Since $K$ is compact, $K\times K$ is also compact. Since $g$ is continuous, $g$ is uniformly continuous.
Let $x_0\in K$. Given $\epsilon>0$ there is a $\delta>0$ such that $$|x-x_0|<\delta\implies |g(x_0,y)-g(x,y)|<\epsilon\quad\forall y\in K.$$ Then $$ g(x_0,y)\le g(x,y)+\epsilon\le f(x)+\epsilon\quad\forall y\in K. $$ Taking the supreme with respect to $y\in K$ we get $$ f(x_0)\le f(x)+\epsilon. $$ The same argument shows that $$ f(x)\le f(x_0)+\epsilon. $$ Putting it all together we get $$ |x-x_0|<\delta\implies|f(x_0)-f(x)|\le\epsilon. $$
Julián Aguirre
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