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If $f(x)$ is a real convex function, is it true that $\forall x<y$, $f(\sqrt{xy})f(\sqrt{xy})\leq f(x)f(y)$? $f(x)$ is strictly increasing.

I was writing a smoothing proof for another problem and came upon this step. Of course, it is well known that if $f(x)$ is a real convex function, then $\forall x<y, f(\frac{x+y}{2})+f(\frac{x+y}{2})\leq f(x)+f(y)$. However, what if we changed things to multiplication? This is clearly true by intuition but I want a rigorous proof.

My efforts to prove it:

Clearly, $x<\sqrt{xy}<y$. By the definitions of convexity, we have that $$f(\sqrt{xy})\leq \frac{(y-\sqrt{xy})f(x)+(\sqrt{xy}-x)f(y)}{y-x}$$

and the calculations got ugly and I got stuck.

How can we prove that $f(\sqrt{xy})f(\sqrt{xy})\leq f(x)f(y)$? Thanks in advance.

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    $\begingroup$ It’s false as can be seen using $f(x)=e^{-x}$. The inequality reduces to $1\leq e^{-(\sqrt{x}-\sqrt{y})^2}$ $\endgroup$
    – Clayton
    Jun 18 at 11:44
  • $\begingroup$ If $f$ is continuous, the condition given is called log-convexity. log-convex implies convex, but the converse is not true. $\endgroup$
    – robjohn ♦
    Jun 18 at 12:06
  • $\begingroup$ @robjohn the condition is not exactly equivalent to log-convex. May be you wanted to say that $f(e^u)$ is log convex. $\endgroup$
    – Kroki
    Jun 18 at 13:55
  • $\begingroup$ @Youem: yes, that is correct. $\endgroup$
    – robjohn ♦
    Jun 18 at 14:36

1 Answer 1

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The stattement is False for $f$ only convex (for example $e^{-x}$). However if you want conditions such that the statement is true you function $f$ can be non-decreasing log-convex i.e. $g(u) = \ln f\left(u\right)$ is a convex.

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    $\begingroup$ The answer is still viable. $\endgroup$
    – Kroki
    Jun 18 at 13:37

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