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Suppose $f:[a,b] \to [0,\infty]$ is convex and nonincreasing. Are there sufficient conditions for $f$ such that $g(x) := f(|x|)$ is convex as well?

The fact that $f$ is nonincreasing seems to work against convexity. But since $[a,b]$ is compact, I think the statement is true if $f$ is "very convex and only decreasing a little"? Do you see how I could translate that into a proper condition? Maybe some inequality relating $f'$ to $f''$?

Elias Strehle
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    Codomain issues aside, if $f$ is convex considering $x$ and $-x$ gives $f(0)=f(\left|\frac{-x+x}{2}\right|)\leq f(|x|)$, so $f$ should be constant – Del Nov 24 '16 at 17:27
  • @Del I guess you meant "$f$ should be constant in $[0,b)$"... –  Nov 24 '16 at 20:06

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If $a\geq0$ then $g=f$ and your claim holds trivially.

Otherwise it must be $a<0$ and $b>0$. If there are some $x\in(0,b)$ such that $f(x)<f(0)$ then $g(-x)=g(x)<g(0)$, which by definition would be a contradiction if we assume that $g$ is convex. Thus if $g$ is convex then (having into account the assumption that $f$ is nonincreasing) it has to be $f(x)= f(0)$ for $x\in(0,b)$, but in this case $g$ must be constant.