Let $f:(0,\infty) \to [0,\infty)$ be a $C^1$ function satisfying $f(1)=0$. Suppose that $f(x)$ is strictly increasing on $[1,\infty)$ and strictly decreasing on $(0,1]$, and that $\lim_{x\to 0} f(x)$ is finite.

Define $F:(0,1) \to [0,\infty)$ by $$ F(s)=\min_{xy=s,x,y\in(0,\infty)} f(x)+ f(y). $$

The properties of $f$ imply that the minimum for $0<s<1$ is obtained at a point where $x,y \le 1$.

Question:Are there "natural" conditions on $f$ which imply that $F$ is convex?

The finiteness assumption of $\lim_{x\to 0} f(x)$ is mainly here to exclude the possibility that $(\sqrt s,\sqrt s)$ is a minimizer for every $s$, or $F(s)=2f(\sqrt s)$. (In this case $F$ is always convex).

Here are some examples and non-examples:

**Quadratic penalization:** $f(x)=(x-1)^2$.
$$
F(s) =
\begin{cases}
1-2s, & \text{ if }\, 0 \le s \le \frac{1}{4} \\
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4},
\end{cases}
$$

is convex as its derivative is non-decreasing. In this case, the minimum point $(x(s),y(s))$ for $s \le \frac{1}{4}$ satisfies $x(s)+y(s)=1$.

**Qubic penalization:** $f(x)=(1-x)^3$.
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2} &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2} &\text{ if } 1/9\le s<1,
\end{cases}
$$
is not convex, as $F''<0$ on $(0,1/9)$.

Similarly, for $f(x)=(x-1)^4$, $F$ is also non-convex.

So, interestingly, when we change from quadratic to either cubic or quartic the convexity properties of $F$ change. Is there any "conceptual" reason for this? Is the convexity of $F$ for $f(x)=(x-1)^2$ just a "mere miracle", without some hidden cause behind it?

**Edit:**

Since the question seems hard, I think that it's reasonable to start with an easier problem:

Characterize the situations where $F(s)$ has an

affinepart.

In the example where $f(x)=(x-1)^2$, $F$ is affine on a subinterval. This seems quite remarkable and unexpected. **Can we prove that this only happens when $f$ is quadratic?**

*Here is my attempt, which at the moment I don't know how to finish:*

Assume for simplicity that there exist a $C^1$ map $s \to (x(s),y(s))$ giving a minimizer to the problem, i.e. for any $s \in (0,1]$ $$ F(s)=f(x(s))+f(y(s)), \, \, \,x(s)y(s)=1. \tag{1} $$ The methods of Lagrange's multipliers gives $$ f'(x(s))=\lambda(s) y(s),f'(y(s))=\lambda(s) x(s). \tag{2} $$ We have $F'(s)=\lambda (s)$, so $F''(s)=0$ if and only if $\lambda(s) < 0$ is constant.

*I am not sure how to proceed from here.*

$\lambda $ must non-positive since $f' \le 0$ on $(0,1)$ by our assumption.

Note that our known solution for $f(x)=(x-1)^2$ satisfies equation $(2)$ for $\lambda(s)=-2$:

Indeed, the minimum is obtained at points where $x(s)+y(s)=1$, so $$ f'(x(s))=2(x(s)-1)=-2(1-x(s))=-2y(s). $$

Since multiplying $f$ by $\alpha>0$ results in a multiplication of $\lambda(s)$ by $\alpha$, it suffices to assume that $c=-2$ and prove that $f(x)=(x-1)^2$.