Point-wise converging convex functions on $[0,1]$ Suppose we have a sequence of continuous convex functions $\{f_n\}$ defined on $[0,1]$ which converge point-wise to a limit $f$ on $[0,1]$, i.e. for all $x \in [0,1]$
$$\lim_n  f_n(x) = f(x).$$
Let $G \subset [0,1]$ be an open set.
Is it true that 
$$\inf_{x \in G} f(x) = \lim_n \left[\inf_{x \in G} f_n(x)\right] ~~? $$
 A: Yes, this is true. The direction
$$\limsup_{n\to\infty} \inf_G f_n \le \inf_G f$$
is easy and does not require convexity. For the reverse direction, begin by observing that 
$$M:=\max_{x\in \{0,1/2,1\}}\sup_n |f_n(x)|$$
is finite due to pointwise convergence. Convexity implies that $f_n\le M$ and $f_n\ge -3M$ on $[0,1]$, the latter coming from the fact that $f(x)<-3M$ together with $f(0),f(1)\le M$ would make $f(-1/2)<-M$. 
Fix $\epsilon>0$. 
For each $n$, pick $x_n\in G$ such that $f_n(x_n)\le \epsilon + \inf_G f_n$. 
Every bounded sequence contains a monotone subsequence. 
Without loss of generality, assume there is an increasing subsequence, also denoted $x_n$. So, $x_n\uparrow x_*$. If $x_*=0$ then all $x_n=0$ and the claim holds. Assume  $x_*>0$. Truncate the sequence so that $x_n\ge x_*/2$ for all $n$.   
Using the convexity and boundedness of $f_n$ on $[0,x_n]$ we find that for all $n$, 
$$f_n(x)\le f_n(x_n)+\epsilon\quad \text{ when } \ x_n-C\epsilon \le x\le x_n $$
where $C$ depends on $M$ and $x_0$ but not on $n$.  
It follows that $f\le 2\epsilon+\liminf_n \inf_G f_n $ on the interval $(x_*,x_*-C\epsilon)$. Since this interval intersects $G$, and $\epsilon$ was arbitrary, the claim follows.
