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If $f(x)$ is concave, $g(x)= e^{f(x)}$ is quasiconcave, but not necessarily concave. For instance, $f(x)= -x^2/2$ is concave but $g(x)= e^{-x^2/2}$ has the shape of the probability density function of the normal distribution (bell-shaped), hence, neither concave nor convex.

Is there an example that would show that $g(x)= e^{f(x)}$ is not necessarily convex if $f(x)$ is convex?

You can find the proof for $g(x)$ being convex here but I was wondering if there are counterexamples to it. Show that $e^{f(x)}$ is convex.

Rodrigo de Azevedo
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Valeria Ryabchina
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2 Answers2

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In general the composition of two convex functions is not convex but if we insist one of them must be non-decreasing it will be. Since $e^x$ is non-decreasing $e^{f(x)}$ will be convex if $f(x)$ is convex. You can find a proof here.

CyclotomicField
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$g'(x)=f'(x)e^{f(x)}$ $g''(x)=e^{f(x)}f"(x)+e^{f(x)}(f'(x))^2=e^{f(x)}(f"(x)+(f'(x))^2)$

Since $f$ is convex $f''(x)\geq 0$ Which implies $g''(x) \geq 0$ so $g$ is convex

Nikos127
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