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 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:
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 f0 is nondecreasing.

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

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