Determination of a symmetric convex region by parallel sections - MathOverflow

Determination of a symmetric convex region by parallel sections

2010-01-10T02:36:02Z

This question is partly inspired by a problem in Stewart's Calculus: "Find the area of the region enclosed by $y=x^2$ and $x=y^2$."

Suppose $f\colon [0,1]\to [0,1]$ is a convex increasing function that fixes 0 and 1. The graphs $y=f(x)$ and $x=f(y)$ enclose a convex region. To find its area, students will likely integrate the length of vertical sections: $f^{-1}(x)-f(x)$. They are unlikely to ask if two different functions can yield the same integrand. But I'll ask:

Does $f^{-1}-f$ determine $f$? (Among all convex increasing functions that fix 0 and 1).

Answer by Anton Petrunin for Determination of a symmetric convex region by parallel sections

2010-01-10T04:31:09Z

Yes.

Assume we have two distinct functions $f$ and $g$ such that $f^{-1}-f\equiv g^{-1}-g$. Take a sequence $x_n=f(x_{n-1})$. Clearly $f(x_n)-g(x_n)=0$ or $(-1)^n[f(x_n)-g(x_n)]$ has the same sign for all $n$.

Sinse $\int_0^1f=\int_0^1g$, there are two sequences $x_n$ and $y_n$ as above such that $f(x_n)=g(x_n)$, $f(y_n)=g(y_n)$ and say $(-1)^n[f(x)-g(x)]>0$ for any $x\in(x_n,y_n)$. Note that $x_n,y_n\to 0$ and $\int_{x_n}^{y_n}|f-g|=const>0$. It follows that $\limsup_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.