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.


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .