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

**Answer**

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 α∈[0,1] and that fn,fn−1,…,f1 are convex nondecreasing functions of one variable and that f0:Rn→R is a convex function.

First let g:Rm→R a convex function and f:R→R a convex nondecreasing function, then, by convexity of g:

g(αx+(1−α)y)≤αg(x)+(1−α)g(y).

So, using the fact that f is nondecreasing:

f(g(αx+(1−α)y))≤f(αg(x)+(1−α)g(y)).

Therefore, again by convexity:

f(g(αx+(1−α)y))≤αf(g(x))+(1−α)f(g(y)).

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

fn∘fn−1∘⋯∘f0

is convex under the stated hypothesis. And the composition will be nondecreasing if f0 is nondecreasing.

**Attribution***Source : Link , Question Author : Phillip Cloud , Answer Author : Elias*