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.

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,fn1,,f1 are convex nondecreasing functions of one variable and that f0:RnR is a convex function.

First let g:RmR a convex function and f:RR a convex nondecreasing function, then, by convexity of g:
g(αx+(1α)y)αg(x)+(1α)g(y).
So, using the fact that f is nondecreasing:
f(g(αx+(1α)y))f(αg(x)+(1α)g(y)).
Therefore, again by convexity:
f(g(αx+(1α)y))αf(g(x))+(1α)f(g(y)).

This reasoning can be used inductively in order to prove the result that
fnfn1f0
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

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