Concavity of a function that is obtained from another concave function

Question

Let $f(x):[0,1]\rightarrow \mathbb{R}$ be a strictly concave function such that $f(0)=f(1)=0$. Let $x^*$ denote the maximizer of $f(x)$. For any value $x\in[0,x^*)$, there exists exactly one other point $y\in[x^*,1]$ such that $f(x)=f(y)$. Let $y$ be a function of $x$, $y(x)$, for $x\in[0,x^*)$, and consider the function $$G(x) = (y(x) - x)f(x).$$ Is $G(x)$ concave for $x\in[0,x^*)$? I cannot find a counterexample to this.

Answer 1

Assume that $f\in C^2$ then $f$ being strictly concave is equivalent to $f''(x)<0$. Thus for $G$ to not be concave we require $G''(x)>0$. You have an explicit equation for $G$ so work out it's second derivative.

Note that $g''(x)f(x) + 2g(x)f(x) + f(x)g''(x) = \left(\frac{d}{dx}\right)^2(g(x)f(x))$ and that $\frac{d}{dx}(x\frac{d}{dx}f(x) + f(x)) = 2f'(x) + xf''(x)$. Also you know that $y(0)=0$ and that $f(0)=f(1)=0$. Hence the inequality that is $G''(x)>0$ results in a differential inequality for some function of y (on the RHS) and some function independent of y (on the LHS) that have the same values at $x=0$.

This allows you to integrate the second order differential inequality to a first order inequality.

Funny thing is... the same technique can be applied to the resulting first order differential system. The result is an inequality where some cancellations can be made and you'll get an inequality where the RHS is a function of $y$ and the LHS is a function of $x$.

At that point you'll be able to determine under what condition $G''(x)>0$ can or cannot be achieved. You can use this condition to either inspire a counter-example or prove that no counter-example exists - which ever is appropriate.

I'm not providing all the details because:

1. This is a very good exercise,
2. I think Mathematics is best learn by doing,
3. Your question looks like piece of homework to me,
4. But it was a great distraction from my real work. Thank you.