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Is $g(x^n)$ a convex function of $x$, if $g$ is a convex function of $x$ and $n>2$; given $x$ is nonnegative?

like $g(x^2)$ or $g(x^3)$

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Jingjings
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    First, try to prove that is $f$ is convex nondecreasing and $h$ is convex then $f\circ h$ is convex. Then try to see why this is related to your problem. – user37238 Dec 17 '13 at 09:53

2 Answers2

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Try $g(x) = -x$.

You might do better if $g$ is also nondecreasing.

Robert Israel
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  • I got it.But reading some unedited literature out there misled me. – DeepSea Dec 17 '13 at 10:10
  • Thanks a lot. so, the answer is probably that for nonnegative x, x^n is convex, if g is non-decreasing convex, then g(x^n) is convex, right? – Jingjings Dec 17 '13 at 10:41
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A standard fact about convexity of composition is:

If $f$ is a convex function and $g$ is a convex non-decreasing function of one variable, then $g\circ f$ is convex.

For a proof, see The composition of two convex functions is convex.

This applies to your situation, if $g$ is also nondecreasing.

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