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

If $$f(x)$$ is a real convex function, is it true that $$\forall x, $$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. 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, $$x<\sqrt{xy}. By the definitions of convexity, we have that $$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.

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

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.