Determination of a symmetric convex region by parallel sections

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).

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That is a good problem for math competition. –  Anton Petrunin Jan 10 '10 at 16:24
It would be better to write "strictly increasing" in place of "increasing", to guarantee that $f^{-1}$ exists. –  Bjorn Poonen Jan 10 '10 at 19:45

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.
Great. It took me some thought to figure out why the integral is constant towards the end. In case others have the same difficulty: Write the integral as $\pm\int_{x_n}^{y_n}(f^{-1}(y)-g^{-1}(y))\,dy$ and deal with the integral of each addend separately. Substitute $y=f(x)$ in the first integral and perform a partial integration to deal with the resulting factor $x$. Ditto in the second integral, with $y=g(x)$. Subtract. –  Harald Hanche-Olsen Jan 10 '10 at 14:08