Is a non-euclidean-norm preserving map necessarily linear?

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Let $V$ and $W$ be two normed vector spaces and let $f:V rightarrow W$ be a norm preserving map. I know that if both norms correspond to some inner product then $f$ is necessarily linear, but I can't find the answer for the more general case of normed vector spaces.

I suspect the answer is no, so I tried to come up with a counter-example involving "pseudo" rotations along non-euclideanly-spherical paths centered at the origin of $mathbb R^2$, unsuccessfully.

I'd most importantly like an answer that does not assume $f$ to be surjective. However, any additional information about that particular case would be appreciated as well.

asked Jul 30 at 12:01

user493048

686

1

The surjective case is also affine, that is the content of Mazur-Ulam.
â€“Â spiralstotheleft
Jul 30 at 12:05

1

Examples of non-affine can be found here for the non-surjective case.
â€“Â spiralstotheleft
Jul 30 at 12:07

add a commentÂ |Â

up vote
1
down vote

favorite

I suspect the answer is no, so I tried to come up with a counter-example involving "pseudo" rotations along non-euclideanly-spherical paths centered at the origin of $mathbb R^2$, unsuccessfully.

I'd most importantly like an answer that does not assume $f$ to be surjective. However, any additional information about that particular case would be appreciated as well.

asked Jul 30 at 12:01

user493048

686

1

The surjective case is also affine, that is the content of Mazur-Ulam.
â€“Â spiralstotheleft
Jul 30 at 12:05

1

Examples of non-affine can be found here for the non-surjective case.
â€“Â spiralstotheleft
Jul 30 at 12:07

add a commentÂ |Â

up vote
1
down vote

favorite

I suspect the answer is no, so I tried to come up with a counter-example involving "pseudo" rotations along non-euclideanly-spherical paths centered at the origin of $mathbb R^2$, unsuccessfully.

I'd most importantly like an answer that does not assume $f$ to be surjective. However, any additional information about that particular case would be appreciated as well.

asked Jul 30 at 12:01

user493048

686

I suspect the answer is no, so I tried to come up with a counter-example involving "pseudo" rotations along non-euclideanly-spherical paths centered at the origin of $mathbb R^2$, unsuccessfully.

I'd most importantly like an answer that does not assume $f$ to be surjective. However, any additional information about that particular case would be appreciated as well.

asked Jul 30 at 12:01

user493048

686

asked Jul 30 at 12:01

user493048

686

asked Jul 30 at 12:01

user493048

686

asked Jul 30 at 12:01

user493048

686

1

The surjective case is also affine, that is the content of Mazur-Ulam.
â€“Â spiralstotheleft
Jul 30 at 12:05

1

Examples of non-affine can be found here for the non-surjective case.
â€“Â spiralstotheleft
Jul 30 at 12:07

add a commentÂ |Â

1

The surjective case is also affine, that is the content of Mazur-Ulam.
â€“Â spiralstotheleft
Jul 30 at 12:05

1

Examples of non-affine can be found here for the non-surjective case.
â€“Â spiralstotheleft
Jul 30 at 12:07

The surjective case is also affine, that is the content of Mazur-Ulam.
â€“Â spiralstotheleft
Jul 30 at 12:05

Examples of non-affine can be found here for the non-surjective case.
â€“Â spiralstotheleft
Jul 30 at 12:07

add a commentÂ |Â

2 Answers
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1
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This is the example from J. A. Baker's paper for easy access.

Define $f:mathbbR^2tomathbbR$ by $$f(x,y)=begincasesy,&text if 0leq yleq xtext or xleq yleq 0\x,&text if 0leq xleq ytext or yleq xleq 0\0,&text otherwiseendcases$$

Then $f$ satisfies

$f(tx,ty)=tf(x,y)$

$|f(x,y)-f(u,v)|leqsqrt(x-u)^2+(y-v)^2$

$f$ is not linear

Put in $mathbbR^2$ the usual norm, and $mathbbR^3$ with the norm $$|(x,y,z)|=maxleft(sqrtx^2+y^2,|z|right)$$

Define $F:mathbbR^2tomathbbR^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

answered Jul 30 at 12:17

spiralstotheleft

30516

add a commentÂ |Â

up vote
0
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Take the map $mathbbR to mathbbR, r mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V setminus0$ to $W setminus0$ is norm preserving.

Or, you could take $V$ to have the trivial norm, and $W$ to be $mathbbC$ with the Euclidean norm. Then any map $V to W$ which maps $0$ to $0$ and maps $V setminus 0$ onto the unit circle is norm preserving. This certainly isn't surjective.

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

Â |Â
show 3 more comments

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2 Answers
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2 Answers
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active

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up vote
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This is the example from J. A. Baker's paper for easy access.

Define $f:mathbbR^2tomathbbR$ by $$f(x,y)=begincasesy,&text if 0leq yleq xtext or xleq yleq 0\x,&text if 0leq xleq ytext or yleq xleq 0\0,&text otherwiseendcases$$

Then $f$ satisfies

$f(tx,ty)=tf(x,y)$

$|f(x,y)-f(u,v)|leqsqrt(x-u)^2+(y-v)^2$

$f$ is not linear

Put in $mathbbR^2$ the usual norm, and $mathbbR^3$ with the norm $$|(x,y,z)|=maxleft(sqrtx^2+y^2,|z|right)$$

Define $F:mathbbR^2tomathbbR^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

answered Jul 30 at 12:17

spiralstotheleft

30516

add a commentÂ |Â

up vote
1
down vote

This is the example from J. A. Baker's paper for easy access.

Define $f:mathbbR^2tomathbbR$ by $$f(x,y)=begincasesy,&text if 0leq yleq xtext or xleq yleq 0\x,&text if 0leq xleq ytext or yleq xleq 0\0,&text otherwiseendcases$$

Then $f$ satisfies

$f(tx,ty)=tf(x,y)$

$|f(x,y)-f(u,v)|leqsqrt(x-u)^2+(y-v)^2$

$f$ is not linear

Put in $mathbbR^2$ the usual norm, and $mathbbR^3$ with the norm $$|(x,y,z)|=maxleft(sqrtx^2+y^2,|z|right)$$

Define $F:mathbbR^2tomathbbR^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

answered Jul 30 at 12:17

spiralstotheleft

30516

add a commentÂ |Â

up vote
1
down vote

This is the example from J. A. Baker's paper for easy access.

Define $f:mathbbR^2tomathbbR$ by $$f(x,y)=begincasesy,&text if 0leq yleq xtext or xleq yleq 0\x,&text if 0leq xleq ytext or yleq xleq 0\0,&text otherwiseendcases$$

Then $f$ satisfies

$f(tx,ty)=tf(x,y)$

$|f(x,y)-f(u,v)|leqsqrt(x-u)^2+(y-v)^2$

$f$ is not linear

Put in $mathbbR^2$ the usual norm, and $mathbbR^3$ with the norm $$|(x,y,z)|=maxleft(sqrtx^2+y^2,|z|right)$$

Define $F:mathbbR^2tomathbbR^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

answered Jul 30 at 12:17

spiralstotheleft

30516

This is the example from J. A. Baker's paper for easy access.

Define $f:mathbbR^2tomathbbR$ by $$f(x,y)=begincasesy,&text if 0leq yleq xtext or xleq yleq 0\x,&text if 0leq xleq ytext or yleq xleq 0\0,&text otherwiseendcases$$

Then $f$ satisfies

$f(tx,ty)=tf(x,y)$

$|f(x,y)-f(u,v)|leqsqrt(x-u)^2+(y-v)^2$

$f$ is not linear

Put in $mathbbR^2$ the usual norm, and $mathbbR^3$ with the norm $$|(x,y,z)|=maxleft(sqrtx^2+y^2,|z|right)$$

Define $F:mathbbR^2tomathbbR^3$ by $F(x,y)=(x,y,f(x,y))$.

Then $F$ is a non-affine isometry.

answered Jul 30 at 12:17

spiralstotheleft

30516

answered Jul 30 at 12:17

spiralstotheleft

30516

answered Jul 30 at 12:17

spiralstotheleft

30516

answered Jul 30 at 12:17

spiralstotheleft

30516

add a commentÂ |Â

up vote
0
down vote

Take the map $mathbbR to mathbbR, r mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V setminus0$ to $W setminus0$ is norm preserving.

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

Â |Â
show 3 more comments

up vote
0
down vote

Take the map $mathbbR to mathbbR, r mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V setminus0$ to $W setminus0$ is norm preserving.

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

Â |Â
show 3 more comments

up vote
0
down vote

Take the map $mathbbR to mathbbR, r mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V setminus0$ to $W setminus0$ is norm preserving.

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

Take the map $mathbbR to mathbbR, r mapsto |r|$. This certainly preserves the Euclidean norm, but is not even additive, so not linear.

Another kind of dumb example is the trivial norm taking $0$ to $0$ and anything else to $1$: then any map taking $0_V$ to $0_W$ and $V setminus0$ to $W setminus0$ is norm preserving.

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

edited Jul 30 at 12:19

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

answered Jul 30 at 12:12

Ashwin Iyengar

1,082515

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

Â |Â
show 3 more comments

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

The distance between $-1$ and $1$ is $2$, but their images by $|cdot|$ have distance $0$.
â€“Â spiralstotheleft
Jul 30 at 12:16

The question says norm-preserving, not distance preserving.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I incorrectly wrote isometry before.
â€“Â Ashwin Iyengar
Jul 30 at 12:20

I know. But I am pretty sure the question is really about isometries, given that it is tagged as such, and the conclusion of Mazur-Ulam is assumed true, which requires the assumption of isometry, not norm preserving.
â€“Â spiralstotheleft
Jul 30 at 12:23

Fair enough: I'll leave this here anyway.
â€“Â Ashwin Iyengar
Jul 30 at 12:24

Â |Â
show 3 more comments

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