TITLE: Continuity of min function
QUESTION [5 upvotes]: I would like to ask a question about the continuity of this min function. 
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be continuous. Define
$$F(x):=\min\{f(t):t \in [-x,x]\}.$$ 
I believe that the function $F$ is continuous on $\mathbb{R}.$
However, I do not know how to use the $\delta$-$\epsilon$ definition to show its continuity. Could somebody help me?
Thank you.
Masih

REPLY [3 votes]: Given $\epsilon > 0$ and $x$, we want to show that there exists a $\delta>0$ such that $|F(x)-F(y)|<\epsilon$ whenever $|x-y|<\delta$.
Since $[-x,x]$ is compact, the minimum is attained at some point(s) $t_0\in[-x,x]$. 
Suppose first that $\pm x$ are not points of minimum, i.e. the minimum is attained strictly inside. Let $\delta$ be such that $|x-y|<\delta$ implies $|f(x)-f(y)| < \frac12|f(x) - f(t_0)|$, by continuity of $f$. This $\delta$ will work for $F$ for any $\epsilon>0$ since $|F(x) - F(y)| = 0$ in this case.
Now suppose the minimum is attained at $\pm x$. Then again by continuity of $f$ at $\pm x$, find $\delta$ such that $|f(x) - f(y)| < \epsilon$ if $|x-y|<\delta$. This $\delta$ will work as well.