Given a sequence of continuous, real-valued functions $(f_n)_{n\in\mathbb{N}}$ and a compact set $C$, $f_n$ converge monotonically (increasing) and pointwise to an extended-real valued lower-semicontinuous function $f$ which may take values in $\mathbb{R}\cup\{+\infty\}$ ($f$ is $+\infty$ at those points for which $f_n(x)\nearrow +\infty$; we can assume there is a nonempty compact subset of $C$ where $f$ is finite). I am curious if $$\inf_{x\in C}f_n(x)\to \inf_{x\in C} f(x)?$$
If $f$ were real and continuous on $C$, we could just use Dini's theorem to get uniform convergence. However, we do not need (or get) uniform convergence here; we only need convergence of the infima. But I cannot find a proof (or counterexample) of this statement.
Edit: This link discusses a related problem on the real number line. Additionally assuming convexity of all functions here would be okay as well, I suppose.
Let me show that $\inf_C f_n \to \inf_C f$ indeed holds. We will adapt the proof of Dini's theorem for this.
Let $A = \inf_{C} f$. Fix $a < A$ and set $U_n = \{x \in C : f_n(x) > a\}$. Then for each $x \in C$,
$$ \lim_{n\to\infty} f_n(x) = f(x) > a, $$
hence $f_n(x) > a$ (i.e., $x \in U_n$) for all sufficiently large $n$. Since this is true for any $x \in C$, $ \bigcup_n U_n = C $ and hence $(U_n)$ is an open cover of $C$. Then by the compactness of $C$ and monotonicity of $(U_n)$, there exists $N$ such that $U_N = C$. This then implies that
$$ n \geq N \qquad\implies\qquad \inf_{C} f_n \geq a, $$
from which we get $ \lim_{n\to\infty} \inf_{C} f_n \geq a $. Since the left-hand side does not depend on $a$, letting $a \uparrow A$ yields
$$ \lim_{n\to\infty} \inf_{C} f_n \geq A. $$
On the other hand, the fact $f_n \leq f$ immediately gives $\inf_C f_n \leq \inf_C f$, hence
$$ \lim_{n\to\infty} \inf_{C} f_n \leq A. $$
Combining these two inequalities, we conclude the desired equality.