Continuity of $x\mapsto \underset{K}{\mathrm{min}}f(x,\cdotp)$ Let $f:U\times K\to\mathbb{R}$ be a continuous function, where $U^{\mathrm{open}},K^{\mathrm{compact}}\subset\mathbb{R}^n$. We define for each $x\in U$: $$f_x:K\to\mathbb{R}$$ $$f_x(y)=f(x,y)$$
How to prove that $$h:U\to\mathbb{R}$$ $$h(x)=\underset{K}{\mathrm{min}}f_x$$
is a continuous function. 
Any hint would be appreciated.

It suffices to show that, for all $x\in U$, we have $h(x) \ge \lim\limits_{t\rightarrow x}{h(t)}$ and $h(x)\le \lim\limits_{t\rightarrow x}{h(t)}$.
Use the continuity of $f$ (in $x$) to show that $h(x)\ge\lim\limits_{t\rightarrow x}{h(t)}$.
Conversely, let $x_k$ be a sequence in $U$ converging to $x$, and let $y_k$ be the corresponding sequence in $U$ such that $f(x_k,y_k) = h(x_k)$. Use the compactness of $K$ to show that $y_k$ converges to some $y$ along a subsequence, and use the continuity of $f$ to show that $\lim\limits_{k\rightarrow \infty}{h(x_k)} = f(x,y) \ge h(x)$.
Let me know if you need more details...
