# Showing a sequence of convex functions admits a uniformly convergent subsequence.

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.

• At last it follows by the converse of Arzela-Ascoli that the original sequence must be equicontinuous ... – Hagen von Eitzen Sep 3 '16 at 15:39

## 1 Answer

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 $f(y)\le f(x)+(y-x)\cdot\frac{f(2)-f(x)}{2-x}\le f(x)+(y-x)$.

• Ah I see, and we can even give an explicit bound on the derivative on $[-1,1],$ namely that the magnitude of the derivative must not be more than 1. Hence the sequence is uniformly Lipschitz on [-1,1] and equicontinuous. – Merkh Sep 3 '16 at 16:01
• @Merkh Except that e cannot assume the existence of a derivative. But Lipschitz is of course good enough – Hagen von Eitzen Sep 4 '16 at 8:47