Consider functions $f_n, f$ from $\mathbb{R}$ to $\mathbb{R}$. Suppose $f_n(y)$ converges pointwise to $f(y)$ for all $y$ as $n \rightarrow \infty$. I would like to know under what conditions is the following statement correct.
$\lim_{n \rightarrow \infty} \min_y f_n(y) = \min_y f(y)$.
Uniform convergence implies the result. Let's first substitute $\min$ with $\inf$.
First take a $y_0$ such that $$f(y_0)\leq \inf _y f(y) +\epsilon$$
Then since $$ f(y_0)= \lim _n f_n(y_0) \geq \lim _n \inf_y f_n(y)$$
we get
$$\inf _y f(y) \geq \lim _n \inf_y f_n(y)\fbox{1}.$$
Now take $N$ such that $ |f_n(y) -f(y)| < \epsilon$ $\forall n \geq N$.
Pick $y_n$ such that $$ f_n(y_n)\leq \inf _y f_n(y) +\delta$$
This gives that $-\epsilon<f_n(y_n)-f(y_n) < \epsilon $ therefore 4 $$-\epsilon < \inf _y f_n(y) +\delta - f(y_n)\Rightarrow \inf_y f(y)\leq f_n(y_n)-\epsilon < \inf _y f_n(y) +\delta $$
taking $n\rightarrow \infty$
We get $$\inf _y f(y) \leq \lim _n \inf_y f_n(y) \fbox{2}$$
So we conclude.
Remark Wecould have been more rigorous if in $\fbox {1}$ we had taken $\limsup$, and in $\fbox{2}$ $\liminf $ instead of $\lim$.