How to show this function involving min is continuous?

Question

Suppose I have a continuous function $F:\mathbb{R}^n \rightarrow \mathbb{R}$. I want to conider $G:\mathbb{R}^m \rightarrow \mathbb{R}$ defined by $$ G(x_1, ..., x_m) = \min_{y \in D} F(x_1, ..., x_m, y_1, ..., y_{n-m}) $$ where $D$ is a compact subset of $\mathbb{R}^{n-m}$. I am interested in showing $G$ is continous. Any comments would be appreciated. Thank you.

Answer 1

For simplicty in typing I will take the case $n=m=1$ but the proof is same in general. Let $N$ be a positive integer and consider $F$ restricted to $\{(x,y): |x| \leq N, y \in D\}$. This function is uniformly continuous because it is continuous and the domain is compact. Let $\epsilon >0$ and choose $\delta>0$ such that $|F(x,y)-F(x',y')| <\epsilon$ whenever $\|(x,y)-(x',y')\| <\delta$. Then $F(x,y) <F(x',y)+\epsilon$ whenever $|x| \leq N, |x'| \leq N$, $|x-x'|<\delta$ and $y \in D$. Conclude from this that $|G(x)-G(x')| <\epsilon$ whenever $|x| \leq N, |x'| \leq N$, and $|x-x'|<\delta$. This proves continuity of $G$ on $\{x: |x| \leq N\}$. Since $N$ is arbitrary we are done.