About homeomorphisms on $[0,1]$












3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Randall
    Nov 29 '18 at 2:57


















3












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Randall
    Nov 29 '18 at 2:57
















3












3








3


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 0:26









Bastian Galasso-DiazBastian Galasso-Diaz

1856




1856












  • $begingroup$
    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.
    $endgroup$
    – Randall
    Nov 29 '18 at 2:57




















  • $begingroup$
    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.
    $endgroup$
    – Randall
    Nov 29 '18 at 2:57


















$begingroup$
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.
$endgroup$
– Randall
Nov 29 '18 at 2:57






$begingroup$
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.
$endgroup$
– Randall
Nov 29 '18 at 2:57












2 Answers
2






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is a very nice proof. +1
    $endgroup$
    – Matematleta
    Nov 29 '18 at 23:34










  • $begingroup$
    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!!
    $endgroup$
    – Bastian Galasso-Diaz
    Nov 30 '18 at 15:09





















1












$begingroup$

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



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
    $endgroup$
    – Bastian Galasso-Diaz
    Dec 2 '18 at 23:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is a very nice proof. +1
    $endgroup$
    – Matematleta
    Nov 29 '18 at 23:34










  • $begingroup$
    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!!
    $endgroup$
    – Bastian Galasso-Diaz
    Nov 30 '18 at 15:09


















2












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this is a very nice proof. +1
    $endgroup$
    – Matematleta
    Nov 29 '18 at 23:34










  • $begingroup$
    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!!
    $endgroup$
    – Bastian Galasso-Diaz
    Nov 30 '18 at 15:09
















2












2








2





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 4:57









jjagmathjjagmath

2057




2057












  • $begingroup$
    this is a very nice proof. +1
    $endgroup$
    – Matematleta
    Nov 29 '18 at 23:34










  • $begingroup$
    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!!
    $endgroup$
    – Bastian Galasso-Diaz
    Nov 30 '18 at 15:09




















  • $begingroup$
    this is a very nice proof. +1
    $endgroup$
    – Matematleta
    Nov 29 '18 at 23:34










  • $begingroup$
    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!!
    $endgroup$
    – Bastian Galasso-Diaz
    Nov 30 '18 at 15:09


















$begingroup$
this is a very nice proof. +1
$endgroup$
– Matematleta
Nov 29 '18 at 23:34




$begingroup$
this is a very nice proof. +1
$endgroup$
– Matematleta
Nov 29 '18 at 23:34












$begingroup$
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!!
$endgroup$
– Bastian Galasso-Diaz
Nov 30 '18 at 15:09






$begingroup$
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!!
$endgroup$
– Bastian Galasso-Diaz
Nov 30 '18 at 15:09













1












$begingroup$

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



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
    $endgroup$
    – Bastian Galasso-Diaz
    Dec 2 '18 at 23:54
















1












$begingroup$

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



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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
    $endgroup$
    – Bastian Galasso-Diaz
    Dec 2 '18 at 23:54














1












1








1





$begingroup$

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



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






share|cite|improve this answer









$endgroup$



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



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







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answered Dec 1 '18 at 19:41









zhw.zhw.

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  • $begingroup$
    Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
    $endgroup$
    – Bastian Galasso-Diaz
    Dec 2 '18 at 23:54


















  • $begingroup$
    Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
    $endgroup$
    – Bastian Galasso-Diaz
    Dec 2 '18 at 23:54
















$begingroup$
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
$endgroup$
– Bastian Galasso-Diaz
Dec 2 '18 at 23:54




$begingroup$
Thank you! I actually do something similar. Do you have some intuition about the same question but in $L_2$ norm instead $L_1$???
$endgroup$
– Bastian Galasso-Diaz
Dec 2 '18 at 23:54


















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