# Is the composition of nn convex functions itself a convex function?

Is a set of convex functions closed under composition? I don’t necessarily need a proof, but a reference would be greatly appreciated.

There is no need for the first function in the composition to be nondecreasing. And here is a proof for the nondifferentiable case as well. The only assumptions are that the composition is well defined at the points involved in the proof for every $\alpha \in [0, 1]$ and that $f_n, f_{n - 1}, \dots, f_1$ are convex nondecreasing functions of one variable and that $f_0 : \mathbb R^n \to \mathbb R$ is a convex function.

First let $g : \mathbb R^m \to \mathbb R$ a convex function and $f : \mathbb R \to \mathbb R$ a convex nondecreasing function, then, by convexity of $g$:

So, using the fact that f is nondecreasing:

Therefore, again by convexity:

This reasoning can be used inductively in order to prove the result that

is convex under the stated hypothesis. And the composition will be nondecreasing if $f_0$ is nondecreasing.