# Notation: Show that the function $f^2/g$ is convex.

I am afraid I am confused about notation in the following question.

Suppose that $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ is nonnegative and convex, and $g:\mathbb{R}^n \longrightarrow \mathbb{R}$ is positive and concave.

Show that the function $f^2/g$ is convex.

Does $f^2/g$ mean $f \circ f \circ g^{-1}$ or does it mean $f(x)^2/g(x)$ where $x\in\mathbb{R}^n$?

The confusion is rather silly, but I am really not sure how to interpret this notation. I do not know what to prove if I cannot interpret correctly.

Thank you for the help.

• $f \circ f$ doesn't even make sense here, because $f : \mathbb{R}^n \to \mathbb{R}$. So it's multiplication and division.
– user296602
Commented Sep 18, 2018 at 20:25
• Oh my. I didn't notice that. Thank you! Commented Sep 18, 2018 at 20:27
• Another giveaway: $g$ is assumed positive (i.e., constant strict sign) for a reason -- no division by zero. Commented Sep 18, 2018 at 21:15
• Commented Sep 19, 2018 at 23:13

The solution is pretty straightforward, though calculations are tedious. Basically you need to show that $$\frac{\left(f(\alpha x + (1-\alpha)y)\right)^2}{g(\alpha x + (1-\alpha)y)}\leq \alpha\frac{\left(f(x)\right)^2}{g(x)} + (1-\alpha) \frac{\left(f(y)\right)^2}{g(y)}$$ for all $x,y\in\mathbb{R}^n$ and for all $\alpha\in(0,1)$.
From the convexity and non-negativity of $f$ and concavity and positivity of $g$ we conclude that: $$\frac{\left(f(\alpha x + (1-\alpha)y)\right)^2}{g(\alpha x + (1-\alpha)y)} \leq \frac{\left(\alpha f(x) + (1-\alpha)f(y)\right)^2}{\alpha g(x) + (1-\alpha)g(y)}.$$ Then, it remains to prove the following inequality: $$\frac{\left(f(\alpha x + (1-\alpha)y)\right)^2}{g(\alpha x + (1-\alpha)y)} \leq \alpha\frac{\left(f(x)\right)^2}{g(x)} + (1-\alpha) \frac{\left(f(y)\right)^2}{g(y)}.$$ However, after simple transformations most of the terms cancel out and you should obtain: $$0 \leq\left(f(x)g(y) -f(y)g(x)\right)^2,$$ which is obviously true.