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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Determination of a symmetric convex region by parallel sections
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.


