Suppose that function $h : [0,1]^n \to \mathbb R$ is concave and non-decreasing and that function $g:[0,1] \to [0,1]$ is concave. Define:
$$ f(x_1,\dots, x_n)=h\left(g(x_1),\dots,g(x_n)\right) $$
Is it true that $f$ is concave?
According to page 86 of Boyd and Vandenberghe Convex Optimization, vector decomposition exhibits:
However, here, function $g$ need not be twice differentiable. Any hint on how to prove this claim?
Best Answer
Use the definition of concave: $$f((1-\alpha )x+\alpha y)\geq (1-\alpha ) f(x)+\alpha f(y)$$ So in this case you would show $$h(g((1-\alpha)x_1+\alpha y_1),...,g((1-\alpha)x_n+\alpha y_n))$$ $$\ge h((1-\alpha)g(x_1)+\alpha g(y_1),...,(1-\alpha)g(x_n)+\alpha g(y_n))$$ $$\ge (1-\alpha)h(g(x_1),...,g(x_n))+\alpha h(g(y_1),...,g(y_n)))$$ using the facts that h is non-decreasing and concave, respectively.