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SO, I have been reading on convex functions and came across the property that the function h(x) = g(f(x)) is convex. given g(x) and f(x) are convex, could someone give me the proof? given g is non-decresing

Riya
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    This is false if $g$ is not a non-decreasing function: e.g. if $f(x) = g(x) = e^{-x}$ then $f$ and $g$ are convex but $g \circ f$ is not, see here: https://math.stackexchange.com/questions/108393/is-the-composition-of-n-convex-functions-itself-a-convex-function – Keeley Hoek Nov 28 '21 at 06:18
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    If $g : \Bbb{R}^n \to \Bbb{R}$ and $h : \Bbb{R} \to \Bbb{R}$ and $f(x) = h(g(x))$ then $f$ is convex if $g$ is convex, $h$ is convex and $\tilde{h}$ is non-decreasing – p_square Nov 28 '21 at 06:23

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This is not true. $g(x)=e^{-x}, f(x)=x^{2}$ is a counter-example. If add the condition that $g$ is also increasing then the result is true.

[ $g\circ f$ is concave on $(0,\frac 1 {\sqrt 2})$ and convex on $(\frac 1 {\sqrt 2}, \infty)$].

Kavi Rama Murthy
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