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Example of nonexpansive mapping.
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I am trying to construct some examples of the nonexpansive mapping $T$ from $R^2$ to $R^2$ such that $T$ should have fixed points more than one. But I could not construct. Can somebody help me? Please.
A mapping $T:X to Y$ is called nonexpansive if $||Tx-Ty|| leq ||x-y||$ for all $x,y$ in $X$.
functional-analysis nonlinear-analysis
asked Jul 17 at 6:53
Infinity
513213
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up vote
1
down vote
favorite
I am trying to construct some examples of the nonexpansive mapping $T$ from $R^2$ to $R^2$ such that $T$ should have fixed points more than one. But I could not construct. Can somebody help me? Please.
A mapping $T:X to Y$ is called nonexpansive if $||Tx-Ty|| leq ||x-y||$ for all $x,y$ in $X$.
functional-analysis nonlinear-analysis
asked Jul 17 at 6:53
Infinity
513213
The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to construct some examples of the nonexpansive mapping $T$ from $R^2$ to $R^2$ such that $T$ should have fixed points more than one. But I could not construct. Can somebody help me? Please.
A mapping $T:X to Y$ is called nonexpansive if $||Tx-Ty|| leq ||x-y||$ for all $x,y$ in $X$.
functional-analysis nonlinear-analysis
asked Jul 17 at 6:53
Infinity
513213
I am trying to construct some examples of the nonexpansive mapping $T$ from $R^2$ to $R^2$ such that $T$ should have fixed points more than one. But I could not construct. Can somebody help me? Please.
A mapping $T:X to Y$ is called nonexpansive if $||Tx-Ty|| leq ||x-y||$ for all $x,y$ in $X$.
functional-analysis nonlinear-analysis
asked Jul 17 at 6:53
Infinity
513213
edited Jul 17 at 7:10
asked Jul 17 at 6:53
Infinity
513213
asked Jul 17 at 6:53
Infinity
513213
asked Jul 17 at 6:53
Infinity
513213
513213
The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44
add a comment |Â
The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44
The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44
add a comment |Â
4 Answers
4
active
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up vote
1
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For instance, $f(x)=(arctan x_1,arctan x_2)$.
answered Jul 17 at 6:57
Saucy O'Path
2,716220
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Any linear map
$T:Bbb R^2 to Bbb R^2 tag 1$
will be nonexpansive by this definition provided
$Vert T Vert le 1, tag 2$
since then
$Vert Tx - Ty Vert = Vert T(x - y) Vert le Vert T Vert Vert x - y Vert = Vert x - y Vert. tag 2$
Examples include the identity map $I$ as well as
$J = beginbmatrix 0 & -1 \ 1 & 0 endbmatrix, tag 2$
$O(theta) = beginbmatrix cos theta & -sin theta \ sin theta & cos theta endbmatrix; tag 3$
there are many more.
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
add a comment |Â
up vote
0
down vote
Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A times A$.
answered Jul 18 at 11:05
Infinity
513213
add a comment |Â
up vote
0
down vote
If you only require $||Tx-Ty|| le ||x-y||quad forall x,y in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||quad forall x,y in X$.
answered Jul 18 at 12:08
gandalf61
5,689522
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For instance, $f(x)=(arctan x_1,arctan x_2)$.
answered Jul 17 at 6:57
Saucy O'Path
2,716220
add a comment |Â
up vote
1
down vote
For instance, $f(x)=(arctan x_1,arctan x_2)$.
answered Jul 17 at 6:57
Saucy O'Path
2,716220
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For instance, $f(x)=(arctan x_1,arctan x_2)$.
answered Jul 17 at 6:57
Saucy O'Path
2,716220
For instance, $f(x)=(arctan x_1,arctan x_2)$.
answered Jul 17 at 6:57
Saucy O'Path
2,716220
answered Jul 17 at 6:57
Saucy O'Path
2,716220
answered Jul 17 at 6:57
Saucy O'Path
2,716220
answered Jul 17 at 6:57
Saucy O'Path
2,716220
2,716220
add a comment |Â
add a comment |Â
up vote
0
down vote
Any linear map
$T:Bbb R^2 to Bbb R^2 tag 1$
will be nonexpansive by this definition provided
$Vert T Vert le 1, tag 2$
since then
$Vert Tx - Ty Vert = Vert T(x - y) Vert le Vert T Vert Vert x - y Vert = Vert x - y Vert. tag 2$
Examples include the identity map $I$ as well as
$J = beginbmatrix 0 & -1 \ 1 & 0 endbmatrix, tag 2$
$O(theta) = beginbmatrix cos theta & -sin theta \ sin theta & cos theta endbmatrix; tag 3$
there are many more.
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
add a comment |Â
up vote
0
down vote
Any linear map
$T:Bbb R^2 to Bbb R^2 tag 1$
will be nonexpansive by this definition provided
$Vert T Vert le 1, tag 2$
since then
$Vert Tx - Ty Vert = Vert T(x - y) Vert le Vert T Vert Vert x - y Vert = Vert x - y Vert. tag 2$
Examples include the identity map $I$ as well as
$J = beginbmatrix 0 & -1 \ 1 & 0 endbmatrix, tag 2$
$O(theta) = beginbmatrix cos theta & -sin theta \ sin theta & cos theta endbmatrix; tag 3$
there are many more.
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Any linear map
$T:Bbb R^2 to Bbb R^2 tag 1$
will be nonexpansive by this definition provided
$Vert T Vert le 1, tag 2$
since then
$Vert Tx - Ty Vert = Vert T(x - y) Vert le Vert T Vert Vert x - y Vert = Vert x - y Vert. tag 2$
Examples include the identity map $I$ as well as
$J = beginbmatrix 0 & -1 \ 1 & 0 endbmatrix, tag 2$
$O(theta) = beginbmatrix cos theta & -sin theta \ sin theta & cos theta endbmatrix; tag 3$
there are many more.
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
Any linear map
$T:Bbb R^2 to Bbb R^2 tag 1$
will be nonexpansive by this definition provided
$Vert T Vert le 1, tag 2$
since then
$Vert Tx - Ty Vert = Vert T(x - y) Vert le Vert T Vert Vert x - y Vert = Vert x - y Vert. tag 2$
Examples include the identity map $I$ as well as
$J = beginbmatrix 0 & -1 \ 1 & 0 endbmatrix, tag 2$
$O(theta) = beginbmatrix cos theta & -sin theta \ sin theta & cos theta endbmatrix; tag 3$
there are many more.
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
answered Jul 17 at 7:06
Robert Lewis
37.1k22256
37.1k22256
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A times A$.
answered Jul 18 at 11:05
Infinity
513213
add a comment |Â
up vote
0
down vote
Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A times A$.
answered Jul 18 at 11:05
Infinity
513213
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A times A$.
answered Jul 18 at 11:05
Infinity
513213
Let $P(x)$ be a nonexpansive mapping from $R$ to $R$ with the fixed points set $A$. Then $T(x,y)=(P(x),P(y))$ is a nonexpansive mapping from $R^2$ to $R^2$ with the fixed points set $A times A$.
answered Jul 18 at 11:05
Infinity
513213
answered Jul 18 at 11:05
Infinity
513213
answered Jul 18 at 11:05
Infinity
513213
answered Jul 18 at 11:05
Infinity
513213
513213
add a comment |Â
add a comment |Â
up vote
0
down vote
If you only require $||Tx-Ty|| le ||x-y||quad forall x,y in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||quad forall x,y in X$.
answered Jul 18 at 12:08
gandalf61
5,689522
add a comment |Â
up vote
0
down vote
If you only require $||Tx-Ty|| le ||x-y||quad forall x,y in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||quad forall x,y in X$.
answered Jul 18 at 12:08
gandalf61
5,689522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you only require $||Tx-Ty|| le ||x-y||quad forall x,y in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||quad forall x,y in X$.
answered Jul 18 at 12:08
gandalf61
5,689522
If you only require $||Tx-Ty|| le ||x-y||quad forall x,y in X$, you can take T to be the identity map. Then every point in $X$ is a fixed pioint and $||Tx-Ty|| = ||x-y||quad forall x,y in X$.
answered Jul 18 at 12:08
gandalf61
5,689522
answered Jul 18 at 12:08
gandalf61
5,689522
answered Jul 18 at 12:08
gandalf61
5,689522
answered Jul 18 at 12:08
gandalf61
5,689522
5,689522
add a comment |Â
add a comment |Â
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The identity? A scaling of it by something smaller than 1?
– AlgebraicsAnonymous
Jul 17 at 6:57
@ AlgebraicsAnonymous Yes, I have that one, but I need some nonlinear mapping so that I can have fixed points more than one.
– Infinity
Jul 17 at 7:06
You don't need a nonlinear mapping to have multiple fixed points. Consider $(x,y)mapsto(x,0)$.
– Rahul
Jul 17 at 10:44