About homeomorphisms on $[0,1]$

Question

I need some help with the following: Suppose that $T:[0,1] \rightarrow [0,1]$ is an homemorphism, which satisfies that $T(0)=0$. It is really intuitive that $$\int_0^1 |T(y) - y| dy = \int_0^1 |T^{-1}(y) - y| dy$$ since $T$ and $T^{-1}$ are symmetric with respect to the identity, and so the area between $T$ and identity will be the same as $T^{-1}$ and identity, which is the equality posted above.

For me, it's intuitive but I can't get a proof of that. Also, what happend if we change the absolute value by square (change the $L_1$ norm for $L_2$ norm).

Thanks a lot!

Answer 1

Here's a sketch of my solution:

Begin by proving that for any homeomorphism $f$ of $[0,1]$ we have $$\int_0^1 f(t)\, dt + \int_0^1 f^{-1}(t)\, dt = 1$$ That's easy. Then define $M(x) = \max\{x,T(x)\}$ and $m(x) = \min\{x,T^{-1}(x)\}$. Observe that $m(x)=M^{-1}(x)$

Then $$\int_0^1 |x-T(x)|\,dx = \int_0^1 2M(x)-x-T(x)\,dx = 2\int_0^1 M(x)\, dx -\frac{1}{2}-\int_0^1 T(x)\,dx$$ $$=2\left(1-\int_0^1 m(x)\, dx\right)-\frac{1}{2} - \left(1-\int_0^1 T^{-1}(x)\,dx\right) = \frac{1}{2}+\int_0^1 T^{-1}(x)\,dx - 2\int_0^1 m(x)\,dx$$ $$=\int_0^1 x + T^{-1}(x)-2 m(x)\,dx = \int_0^1 |x-T^{-1}(x)|\,dx$$

Answer 2

Here's a different proof:

The: Suppose $h:[a,b]\to [a,b]$ is a homeomorphism, with $h(a)=a, h(b)=b,$ and $h(x)> x$ on $(a,b).$ Then

$$\int_a^b(h(x)-x)\, dx = \int_a^b(y-h^{-1}(y))\, dy.$$

Proof: Over the interval $[a,b],$ let $R$ be the region bounded below by the graph of $y=x$ and above by the graph of $y=h(x).$ The area of $R$ is

$$\int_a^b(h(x)-x)\, dx.$$

But from the point of view of the $y$-axis, $R$ is the region bounded above by the graph of $x=y$ and below by the graph of $x=h^{-1}(y).$ Thus the area of $R$ is

$$\int_a^b(y-h^{-1}(y))\, dy.$$

This proves the theorem. Of course if $h(x)<x$ on $[a,b],$ then a similar result would hold. We would get

$$\int_a^b(x-h(x))\, dx = \int_a^b(h^{-1}(y)-y)\, dy.$$

Thus the following corollary holds: Suppose $h:[a,b]\to [a,b]$ is a homeomorphism, with $h(a)=a, h(b)=b,$ and $h(x)\ne x$ on $(a,b).$ Then

$$\int_a^b|h(x)-x|\, dx = \int_a^b|y-h^{-1}(y)|\, dy.$$

Now to our problem in complete generality: Consider the set $U=\{x\in [0,1]: h(x)\ne x\}.$ Then $U$ is an open subset of $(0,1),$ so can be written as the countable disjoint union of open intervals $(a_n,b_n).$ Thus

$$\int_a^b|h(x)-x|\, dx = \int_U |h(x)-x|\, dx = \sum_{n=1}^{\infty} \int_{a_n}^{b_n}|h(x)-x|\, dx.$$

Now for each $n$ we have $h(a_n)=a_n,h(b_n)=b_n.$ So we can use the corollary to see the last expression equals

$$\sum_{n=1}^{\infty} \int_{a_n}^{b_n}|y-h^{-1}(y)|\, dy = \int_U |y-h^{-1}(y)|\, dy = \int_a^b |y-h^{-1}(y)|\, dy.$$