Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(1)=0$, and suppose that $|f|$ grows with the distance from $1$: $|f(x)|$ is strictly increasing when $x \ge 1$, and strictly decreasing when $x \le 1$.

Suppose also that $\lim_{x \to \infty} |f(x)| = \infty$. For any $s \in (0,1)$, define $$ F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y) $$

(The minimum exists since $|f|$ diverges at infinity.)

Question:Does there exist a convex function $g(s)$ such that $F=g^p$ for some $p \ge 1$? I do not require $g$ to be positive.

Here are two examples where this happens:

**Linear penalization:** $f(x)=x-1$. In that case
$$F(s) =
\begin{cases}
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\
1-2s, & \text{ if }\, s \le \frac{1}{4}
\end{cases}
$$
is convex, since $F'(s)$ is non-decreasing.

**Logarithmic penalization:** $f(x)=\log x$. In that case
$$ F(s)=2f^2(\sqrt s)=\frac{1}{2}(\log s)^2$$ is **not** convex.

However, we have $F(s)=g^2(s)$ where $g(s)=-\frac{1}{\sqrt 2}\log s$ which is convex.

Is there a general phenomena lying behind these two examples?