Is the composition of a strictly concave function with a concave function strictly concave?

Question

Let $h:\mathbb{R}\mapsto \mathbb{R}$ be a strictly concave and nondecreasing function and $g:\mathbb{R^n}\mapsto \mathbb{R}$ be a concave nondecreasing function. Is $f=h(g(x))$ strictly concave?

I thought that the answer would be "yes" since, for all $0\leq a\leq1$, \begin{align} h\left(g(ax+(1-a)y)\right)&\geq h\left(ag(x)+(1-a)g(y)\right)\\ & > ah(g(x))+(1-a)h(g(y)) \end{align} where the first inequality follows since $h$ is nondecreasing and $g$ is concave, and the second inequality follows because $h$ is strictly concave.

But then if I set $h(y)=\sqrt{y}$ and $g(x)=x_1+x_2$ I get $h(x)=\sqrt{x_1+x_2}$ which is not strictly concave. What is wrong with my argument?

Answer 1

A simpler counterexample, which helps shed some light: $g$ can be a constant function.

The issue in your argument is that you implicitly assume $g(x)\neq g(y)$ to obtain the strict inequality.

In you take $g(x) = x_1+x_2$, then you can have $g(x)=g(y)$ even if $x\neq y$: take for instance $x=(x_1,x_2)$ and $y=(x_2,x_1)$. You need injectivity somewhere for your argument to go through.