Monotonicity of Convex Conjugate Divided by Quadratic

Question

I came across the following (unproved) convex analysis fact in a paper.

Fact: Suppose $\varphi:[0,\infty)\to\mathbb{R}$ is a convex function. Define its convex conjugate $\varphi^*(y)=\sup_{x\ge 0}[xy-\varphi(x)]$. If $x\mapsto\varphi(x)/x^2$ is increasing on $(0,\infty)$, then $y\mapsto\varphi^*(y)/y^2$ is decreasing on $(0,\infty)$.

I am able to prove this fact, under two extra assumptions: (1) $\varphi$ is differentiable, and (2) the supremum in $\varphi^*$ is attained by a unique $x>0$. However, I do not know how to extend this proof to the general case.

My proof: Let $f(x)=\varphi(x)/x^2$, then $f'(x)=\left[x\varphi'(x) - 2\varphi(x)\right]/x^3\ge 0$ for all $x>0$. Let $g(y) = \varphi^*(y)/y^2$. Fix $y>0$, and assume that $\varphi^*(y)=zy-\varphi(z)$ for a unique $z > 0$. By Danskin's theorem, $(\varphi^*)'(y)=z$, hence $y=\varphi'(z)$. Therefore, $$ g'(y) = \frac{zy-2\varphi^*(y)}{y^3} = \frac{2\varphi(z) - zy}{y^3} = \frac{2\varphi(z) - z\varphi'(z)}{y^3} = -\frac{z^3}{y^3}f'(z)\le 0. $$

My questions: Could anyone give a general proof of this fact (without my assumptions) and the intuition behind? Moreover, does this result have a more general form, e.g., does it extend to $x\mapsto\varphi(x)/x^p$ for a general power $p$?

Answer 1

It is known that the convex conjugate of $x^p$ on $(0, \infty)$ is $y^q$, where $p, q$ are conjugate exponents, i.e. $1/p + 1/q = 1$. This suggests the following generalization:

Let $\varphi:[0,\infty)\to\Bbb R$ be a convex function and $p, q > 1$ be conjugate exponents. If $\phi(x)/x^p$ is increasing on $(0, \infty)$ then $\varphi^*(y)/y^q$ is decreasing on $(0, \infty)$.

This is indeed true and can be proven using only the definition of the complex conjugate.

For $0 < y_1 < y_2$ is $$ \begin{align} \frac{\varphi^*(y_1)}{y_1^q} &= \frac{1}{y_1^q}\sup \left\{ x_1y_1 - \varphi(x_1) : x_1 \ge 0\right\} \\ &= \sup \left\{ \frac{x_1}{y_1^{q-1}} - \frac{1}{y_1^q}\varphi(x_1) : x_1 \ge 0 \right\} \\ &= \sup \left\{ \frac{x_2}{y_2^{q-1}} - \frac{1}{y_1^{q}}\varphi\left(\frac{y_1^{q-1}}{y_2^{q-1}}x_2\right) : x_2 \ge 0 \right\} \end{align} $$ with the substitution $x_1 = x_2 y_1^{q-1}/y_2^{q-1}$.

$\phi(x)/x^p$ is increasing and $x_2 y_1^{q-1} /y_2^{q-1} \le x_2$, therefore is $$ \varphi\left(\frac{y_1^{q-1}}{y_2^{q-1}}x_2\right) \le \left(\frac{y_1^{q-1}}{y_2^{q-1}}\right)^p\varphi(x_2) = \frac{y_1^{q}}{y_2^{q}}\varphi(x_2)\, . $$ It follows that $$ \frac{\varphi^*(y_1)}{y_1^q} \ge \sup \left\{ \frac{x_2}{y_2^{q-1}} - \frac{1}{y_2^q}\varphi\left(x_2\right) : x_2 \ge 0 \right\} = \frac{\varphi^*(y_2)}{y_2^q} \, . $$