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The Equivalent conditions listed in wikipedia link: https://en.wikipedia.org/wiki/Logarithmically_convex_function.

I am struggling to find out why this is true. Quoting below from the link above:

If $f$ is a differentiable function defined on an interval $I ⊆ R$, then $f$ is logarithmically convex if and only if the following condition holds for all $x$ and $y$ in ${I}$: $$ \log{f(x)}\geq \log f(y)+{\frac {f'(y)}{f(y)}}(x-y)$$

Hope someone could help me a little to find out why this iff condition is true.

Thanks.

Aiden Chow
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Elder Yang
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1 Answers1

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You need to know the fact that if $f$ is differentiable, then $f$ is convex if and only if for each fixed $x$, we have $f(y)\geq f(x)+f'(x)(y-x)$ for any $y$ (This is the so-called subgradient inequality, you can search on the internet or any convex optimization textbook for a proof). Once you know this, then it is easy to see the equivalence.

William
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  • I do know about this subgradient inequality. The thing is log of a convex function does not guaranty its log convexity. The approach I tried to prove the equivalence is that assume $f$ is 1) convex, 2) concave, or 3) mixed of convex and concave. Then use the subgradient inequality and the equivalent condition above to prove that $f(x)^{1-t}f(y)^{t} \geq f((1-t)x+ty)$. I failed to prove this inequality. – Elder Yang Feb 14 '22 at 13:35
  • I am not saying the log of a convex function is convex, but by definition, $f$ is log-convex if and only if $log(f)$ is convex. Apply sub-gradient inequality to $log(f)$ you obtain exactly the same inequality in your post. – William Feb 14 '22 at 14:15