Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(0)=0$, 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?

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?

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

In other words, $|f|$ penalizes the dsitance from $1$.

For any $s \in (0,1)$, define $$ F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y) $$

Question: Is it true that $s \to F(s)$ is convex or that there exist a convex function $g(s)$ such that $g^2=F$?

(Actually if $F$ can be represented as $F=g^p$ for some $p \ge 1$ and convex $g$ this would also be useful for me).

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?

Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(0)=0$, 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?

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

In other words, $|f|$ penalizes the dsitance from $1$.

For any $s \in (0,1)$, define $$ F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y) $$

Question:

Is it true that $s \to F(s)$ is convex or that there exist a convex function $g(s)$ such that $g^2=F$?

I am asking whether at least one of these possibilities always occurs.

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?

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

In other words, $|f|$ penalizes the dsitance from $1$.

For any $s \in (0,1)$, define $$ F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y) $$

Question: Is it true that $s \to F(s)$ is convex or that there exist a convex function $g(s)$ such that $g^2=F$?

(Actually if $F$ can be represented as $F=g^p$ for some $p \ge 1$ and convex $g$ this would also be useful for me).

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?