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

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?

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.
• Of course, thank you! So it looks like the proof is correct if $g$ is injective as well? Commented Feb 26, 2021 at 2:22