Why is this transformation convex?

Question

Let $f:\mathbb{R}^n \to \mathbb{R}\cup\{ \infty \}$ be convex. It's claimed that this implies $g:\mathbb{R}^n \times (0,\infty) \to \mathbb{R}\cup\{ \infty \},(x,y)\mapsto yf(\frac{x}{y})$ is convex. This claim was made in some notes on convex fns I was reading, and I'd like to convince myself of this fact.
Some preceding facts about convex fns I know: the defn, f is convex iff its epigraph is convex, composition w/ affine fn is convex, p/w max of convex fns is convex, partial min of a jointly convex fn is convex. I'm trying to use any of these to prove the result.

Answer 1

Take $(x_0,y_0), (x_1,y_1) \in \mathbb R^n \times (0,\infty)$ and $\lambda \in [0,1]$. You have

\begin{align} g[(1-\lambda)(x_0,y_0)+\lambda (x_1,y_1)] &=[(1-\lambda)y_0+\lambda y_1]f\left(\frac{(1-\lambda)x_0+\lambda x_1}{(1-\lambda)y_0+\lambda y_1}\right)\\ &=[(1-\lambda)y_0+\lambda y_1]f\left(\frac{(1-\lambda)y_0(x_0/y_0)+\lambda y_1 (x_1/y_1)}{(1-\lambda)y_0+\lambda y_1}\right)\\ &\le [(1-\lambda)y_0+\lambda y_1][\frac{(1-\lambda) y_0}{(1-\lambda)y_0+\lambda y_1}f\left(\frac{x_0}{y_0}\right)+\frac{\lambda y_1}{(1-\lambda)y_0+\lambda y_1}f\left(\frac{x_1}{y_1}\right)]\\ &=(1-\lambda) y_0 f\left(\frac{x_0}{y_0}\right)+\lambda y_1 f\left(\frac{x_1}{y_1}\right)\\ &=(1-\lambda) g(x_0,y_0)+\lambda g(x_1,y_1) \end{align}

Proving that $g$ is convex.