TITLE: Does $|f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|$ imply $f$ is a logarithm?
QUESTION [5 upvotes]: Let  $f:\mathbb R^+ \to \mathbb R$ be a continuous function, and suppose that 
$$ |f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|, \tag{1}$$
holds for every $x,y \in \mathbb R^+$.
Suppose also that $f(1)=0$, and that 
$$ x> y > 1 \Rightarrow |f(x)| \ge |f(y)| , \,\,\,0<x< y < 1 \Rightarrow |f(x)| \ge |f(y)| \tag{2}$$
(so $f$ penalizes when you get further away from $1$).

Does $f(x)=c\ln(x)$ for some constant $c$? If not, are there "many" such functions? 

Note that without condition $(2)$, every positive increasing convex function would do (would satisfy $(1)$):
$$ \sqrt{xy} \le\frac{x+y}{2} \Rightarrow f(\sqrt{xy}) \le f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2},$$
where the first inequality is due to monotonicity and the second one is due to convexity.
Finally, note that the "linear penalization" $f(x)=|x-1|$ does not satisfy condition $(1)$. (Take $x \to 0, y \to 1$).

REPLY [8 votes]: There are many such functions. 
Choose an arbitrary convex function $g: \Bbb R \to \Bbb R$ which is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ with $g(0) = 0$. Then $f(x) = g(\log x)$ satisfies condition (1):
$$
 f(\sqrt{xy}) = g\left( \frac{\log x + \log 
y}{2}\right) \le \frac{g(\log x) + g(\log y)}{2} = \frac{f(x) + f(y)}{2}
$$
and also condition (2).