About homeomorphisms on $[0,1]$
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!
calculus real-analysis analysis measure-theory lebesgue-integral
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
add a comment |
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!
calculus real-analysis analysis measure-theory lebesgue-integral
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57
add a comment |
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!
calculus real-analysis analysis measure-theory lebesgue-integral
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
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!
calculus real-analysis analysis measure-theory lebesgue-integral
calculus real-analysis analysis measure-theory lebesgue-integral
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
asked Nov 29 at 0:26
Bastian Galasso-Diaz
1856
1856
Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57
add a comment |
Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57
Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57
Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57
add a comment |
2 Answers
2
active
oldest
votes
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
$$int_0^1 |x-T(x)|,dx = int_0^1 2M(x)-x-T(x),dx = 2int_0^1 M(x), dx -frac{1}{2}-int_0^1 T(x),dx$$
$$=2left(1-int_0^1 m(x), dxright)-frac{1}{2} - left(1-int_0^1 T^{-1}(x),dxright) = frac{1}{2}+int_0^1 T^{-1}(x),dx - 2int_0^1 m(x),dx$$
$$=int_0^1 x + T^{-1}(x)-2 m(x),dx = int_0^1 |x-T^{-1}(x)|,dx$$
answered Nov 29 at 4:57
jjagmath
2307
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
add a comment |
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 $U={xin [0,1]: h(x)ne x}.$ 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
$$sum_{n=1}^{infty} int_{a_n}^{b_n}|y-h^{-1}(y)|, dy = int_U |y-h^{-1}(y)|, dy = int_a^b |y-h^{-1}(y)|, dy.$$
answered Dec 1 at 19:41
zhw.
71.4k43075
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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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
$$int_0^1 |x-T(x)|,dx = int_0^1 2M(x)-x-T(x),dx = 2int_0^1 M(x), dx -frac{1}{2}-int_0^1 T(x),dx$$
$$=2left(1-int_0^1 m(x), dxright)-frac{1}{2} - left(1-int_0^1 T^{-1}(x),dxright) = frac{1}{2}+int_0^1 T^{-1}(x),dx - 2int_0^1 m(x),dx$$
$$=int_0^1 x + T^{-1}(x)-2 m(x),dx = int_0^1 |x-T^{-1}(x)|,dx$$
answered Nov 29 at 4:57
jjagmath
2307
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
add a comment |
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
$$int_0^1 |x-T(x)|,dx = int_0^1 2M(x)-x-T(x),dx = 2int_0^1 M(x), dx -frac{1}{2}-int_0^1 T(x),dx$$
$$=2left(1-int_0^1 m(x), dxright)-frac{1}{2} - left(1-int_0^1 T^{-1}(x),dxright) = frac{1}{2}+int_0^1 T^{-1}(x),dx - 2int_0^1 m(x),dx$$
$$=int_0^1 x + T^{-1}(x)-2 m(x),dx = int_0^1 |x-T^{-1}(x)|,dx$$
answered Nov 29 at 4:57
jjagmath
2307
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
add a comment |
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
$$int_0^1 |x-T(x)|,dx = int_0^1 2M(x)-x-T(x),dx = 2int_0^1 M(x), dx -frac{1}{2}-int_0^1 T(x),dx$$
$$=2left(1-int_0^1 m(x), dxright)-frac{1}{2} - left(1-int_0^1 T^{-1}(x),dxright) = frac{1}{2}+int_0^1 T^{-1}(x),dx - 2int_0^1 m(x),dx$$
$$=int_0^1 x + T^{-1}(x)-2 m(x),dx = int_0^1 |x-T^{-1}(x)|,dx$$
answered Nov 29 at 4:57
jjagmath
2307
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
$$int_0^1 |x-T(x)|,dx = int_0^1 2M(x)-x-T(x),dx = 2int_0^1 M(x), dx -frac{1}{2}-int_0^1 T(x),dx$$
$$=2left(1-int_0^1 m(x), dxright)-frac{1}{2} - left(1-int_0^1 T^{-1}(x),dxright) = frac{1}{2}+int_0^1 T^{-1}(x),dx - 2int_0^1 m(x),dx$$
$$=int_0^1 x + T^{-1}(x)-2 m(x),dx = int_0^1 |x-T^{-1}(x)|,dx$$
answered Nov 29 at 4:57
jjagmath
2307
answered Nov 29 at 4:57
jjagmath
2307
answered Nov 29 at 4:57
jjagmath
2307
answered Nov 29 at 4:57
jjagmath
2307
2307
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
add a comment |
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
this is a very nice proof. +1
– Matematleta
Nov 29 at 23:34
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
Thanks a lot! I actually do a proof which was really a mess! Do you have some ideas to how extend this result for $L_2$? What I try to say is, $$int_0^1(T(x) - x)^2dx = int_0^1 (T^{-1}(x)-x)^2dx$$ ??? Of course, if it is true!!
– Bastian Galasso-Diaz
Nov 30 at 15:09
add a comment |
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 $U={xin [0,1]: h(x)ne x}.$ 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
$$sum_{n=1}^{infty} int_{a_n}^{b_n}|y-h^{-1}(y)|, dy = int_U |y-h^{-1}(y)|, dy = int_a^b |y-h^{-1}(y)|, dy.$$
answered Dec 1 at 19:41
zhw.
71.4k43075
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
add a comment |
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 $U={xin [0,1]: h(x)ne x}.$ 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
$$sum_{n=1}^{infty} int_{a_n}^{b_n}|y-h^{-1}(y)|, dy = int_U |y-h^{-1}(y)|, dy = int_a^b |y-h^{-1}(y)|, dy.$$
answered Dec 1 at 19:41
zhw.
71.4k43075
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
add a comment |
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 $U={xin [0,1]: h(x)ne x}.$ 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
$$sum_{n=1}^{infty} int_{a_n}^{b_n}|y-h^{-1}(y)|, dy = int_U |y-h^{-1}(y)|, dy = int_a^b |y-h^{-1}(y)|, dy.$$
answered Dec 1 at 19:41
zhw.
71.4k43075
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 $U={xin [0,1]: h(x)ne x}.$ 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
$$sum_{n=1}^{infty} int_{a_n}^{b_n}|y-h^{-1}(y)|, dy = int_U |y-h^{-1}(y)|, dy = int_a^b |y-h^{-1}(y)|, dy.$$
answered Dec 1 at 19:41
zhw.
71.4k43075
answered Dec 1 at 19:41
zhw.
71.4k43075
answered Dec 1 at 19:41
zhw.
71.4k43075
answered Dec 1 at 19:41
zhw.
71.4k43075
71.4k43075
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
add a comment |
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
– Bastian Galasso-Diaz
Dec 2 at 23:54
add a comment |
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Does it help to know that any self-homeo $T: [0,1] to [0,1]$ must (a) map endpoints to endpoints and (b) be strictly increasing or decreasing? In your case, this means $T(1)=1$ and $T$ is strictly increasing.
– Randall
Nov 29 at 2:57