Let $f_n : [-2,2] \to [0,1]$ be a sequence of convex functions. Show that there is a sub-sequence ${f_{n_k}}$ that converges uniformly on $[-1,1]$.
When I see this problem, I immediately think Arzela-Ascoli because it asks to show that a sequence of functions has a uniformly convergent sub-sequence. It gives us that they are uniformly bounded, but for some reason I can't convince myself (or prove) that a family of bounded convex functions on a compact interval must be equicontinuous. Perhaps this is the wrong approach and there is something more immediate. Any suggestions? Thanks in advance.
Note that we are given convexity and boundedness on $[-2,2]$ and want to show uniform convergence only on $[-1,1]$.
The only obstacle against equicontinuity is the possible existence of very steep parts, i.e., $x,y\in[-1,1]$, $n\in\Bbb N$ with $\frac{|f_n(x)-f_n(y)|}{|x-y|}\gg 0$. But by convexity, such steep parts are only possible near the boundary of the domain $[-2,2]$, hence not in $[-1,1]$ (in other words, such a steep part inside $[-1,1]$ would require $f_n$ to go above $1$ at $-2$ or $+2$).
Formally, if $-1\le x<y\le 1$ then $f(y)\le f(x)+(y-x)\cdot\frac{f(2)-f(x)}{2-x}\le f(x)+(y-x)$.