Consider $f(x)=2^x-1: \mathbb{R}\to\mathbb{R}$ which is a convex function. Then its perspective function is $g(x,y) = xf\left(\frac{y}{x}\right): \mathbb{R}^2\to\mathbb{R}$. According to S. Boyd's book "Convex optimization", we know that $g(x,y)$ is also a convex function.
Assume that $h(z): \mathbb{R}\to\mathbb{R}$ is convex. Is the function $g(x,h(z))=xf\left(\frac{h(z)}{x}\right)$ a convex function of $(x,z)$?
Appreciate any comments on this problem.
It appears that over all of $(x,z)\in\mathbb{R}^2$, the answer is no. If we take $h(z)=z^2$, then we get a function that looks like this:
which is clearly not convex (in the image above, $y$ should be replaced with $z$).
Perhaps you need to restrict the signs of $x$?