Exchanging $\lim$ and $\inf$? Suppose we have a sequence of functions $f_n(x)$ that converge to a limiting function $f(x) = \lim_{n \to \infty} f_n(x)$ for $\forall x \in [a,b]$. I was wondering under what conditions the following holds?
\begin{equation*}
\lim_{n \to \infty} \inf_{x \in [a,b]} f_n(x) = \inf_{x \in [a,b]} f(x).
\end{equation*}

As already mentioned by sranthop, $\Gamma$-convergence is the precise notion to capture minima of function sequences.

Here is a sequence of functions, that does not have the desired property (here $[a,b]=[0,1]$):
$$
f_n(x) : = \begin{cases} 0 & x \in (0,\frac1n)\\ 1 & x\not\in (0,\frac1n) \end{cases}.
$$
Then $f_n(x) \to 1 =:f(x)$ for all $x$. Hence
$$
\lim_{n\to\infty} \inf_{x\in[0,1]} f_n(x)  = 0 \ne \inf_{x\in[0,1]} f(x) =1.
$$

However, you can deduce the inequality $$
\operatorname*{lim sup}_{n\to\infty}  \inf_{x\in[a,b]} f_n(x)  \le \inf_{x\in[a,b]} f(x)$$
for pointwise converging $f_n$ towards $f$:
Take $\epsilon>0$. Then there is $x_\epsilon$ such that $f(x_\epsilon)\le \inf_x f(x) +\epsilon$. Then we have
$$
\inf_x f_n(x) \le f_n(x_\epsilon),
$$
passing to the limit yields
$$
\operatorname*{lim sup}_{n\to\infty} f_n(x) \le f(x_\epsilon) \le \inf_x f(x) +\epsilon.
$$
Now, $\epsilon>0$ was arbitrary, the inequality is proven.
