Distance and origin preserving map between two normed vector spaces

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Let $V$ and $W$ be two normed vector spaces and let $f: V rightarrow W$ be a map that preserves distance as well as the origin, that is :
$ forall x,y in V quad |f(x)-f(y)|_W=|x-y|_V quad $ and $quad f(0_V)=0_W $

Is $f$ necessarily linear ?

I know that if the norm satisfies the parallelogram identity then the answer is yes.

By the Mazur–Ulam theorem, I know that if $f$ is surjective then the answer is also yes. Here I'm asking about the consequences of requiring origin preservation instead of surjectivity.

norm normed-spaces isometry

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edited Jul 30 at 22:59

Michael Hardy

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asked Jul 30 at 20:10

James Well

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up vote
1
down vote

favorite

Let $V$ and $W$ be two normed vector spaces and let $f: V rightarrow W$ be a map that preserves distance as well as the origin, that is :
$ forall x,y in V quad |f(x)-f(y)|_W=|x-y|_V quad $ and $quad f(0_V)=0_W $

Is $f$ necessarily linear ?

I know that if the norm satisfies the parallelogram identity then the answer is yes.

By the Mazur–Ulam theorem, I know that if $f$ is surjective then the answer is also yes. Here I'm asking about the consequences of requiring origin preservation instead of surjectivity.

norm normed-spaces isometry

share|cite|improve this question

edited Jul 30 at 22:59

Michael Hardy

204k23185460

asked Jul 30 at 20:10

James Well

501310

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $V$ and $W$ be two normed vector spaces and let $f: V rightarrow W$ be a map that preserves distance as well as the origin, that is :
$ forall x,y in V quad |f(x)-f(y)|_W=|x-y|_V quad $ and $quad f(0_V)=0_W $

Is $f$ necessarily linear ?

I know that if the norm satisfies the parallelogram identity then the answer is yes.

By the Mazur–Ulam theorem, I know that if $f$ is surjective then the answer is also yes. Here I'm asking about the consequences of requiring origin preservation instead of surjectivity.

norm normed-spaces isometry

share|cite|improve this question

edited Jul 30 at 22:59

Michael Hardy

204k23185460

asked Jul 30 at 20:10

James Well

501310

Let $V$ and $W$ be two normed vector spaces and let $f: V rightarrow W$ be a map that preserves distance as well as the origin, that is :
$ forall x,y in V quad |f(x)-f(y)|_W=|x-y|_V quad $ and $quad f(0_V)=0_W $

Is $f$ necessarily linear ?

I know that if the norm satisfies the parallelogram identity then the answer is yes.

By the Mazur–Ulam theorem, I know that if $f$ is surjective then the answer is also yes. Here I'm asking about the consequences of requiring origin preservation instead of surjectivity.

norm normed-spaces isometry

share|cite|improve this question

edited Jul 30 at 22:59

Michael Hardy

204k23185460

asked Jul 30 at 20:10

James Well

501310

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 22:59

Michael Hardy

204k23185460

edited Jul 30 at 22:59

Michael Hardy

204k23185460

edited Jul 30 at 22:59

Michael Hardy

204k23185460

204k23185460

asked Jul 30 at 20:10

James Well

501310

asked Jul 30 at 20:10

James Well

501310

asked Jul 30 at 20:10

James Well

501310

501310

add a comment | 

add a comment | 

2 Answers
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up vote
1
down vote



accepted

No. Take $V = mathbbR$ with the standard norm and $W = mathbbR^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

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answered Jul 30 at 20:25

B. Mehta

11.6k21844

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No, it may well not be linear. Consider the map$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbbR^2\&x&mapsto&(x,|x|).endarray$$On $mathbbR^2$, consider the norm $bigl|(x,y)bigr|=maxbiglx$ and on $mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

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answered Jul 30 at 20:24

José Carlos Santos

112k1696172

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

No. Take $V = mathbbR$ with the standard norm and $W = mathbbR^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

share|cite|improve this answer

answered Jul 30 at 20:25

B. Mehta

11.6k21844

add a comment | 

up vote
1
down vote



accepted

No. Take $V = mathbbR$ with the standard norm and $W = mathbbR^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

share|cite|improve this answer

answered Jul 30 at 20:25

B. Mehta

11.6k21844

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

No. Take $V = mathbbR$ with the standard norm and $W = mathbbR^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

share|cite|improve this answer

answered Jul 30 at 20:25

B. Mehta

11.6k21844

No. Take $V = mathbbR$ with the standard norm and $W = mathbbR^2$ with the maximum norm. Then let $f(x) = (|x|,x)$, clearly distance and origin-preserving, but not linear.

share|cite|improve this answer

answered Jul 30 at 20:25

B. Mehta

11.6k21844

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 20:25

B. Mehta

11.6k21844

answered Jul 30 at 20:25

B. Mehta

11.6k21844

answered Jul 30 at 20:25

B. Mehta

11.6k21844

11.6k21844

add a comment | 

add a comment | 

up vote
1
down vote

No, it may well not be linear. Consider the map$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbbR^2\&x&mapsto&(x,|x|).endarray$$On $mathbbR^2$, consider the norm $bigl|(x,y)bigr|=maxbiglx$ and on $mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

share|cite|improve this answer

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

add a comment | 

up vote
1
down vote

No, it may well not be linear. Consider the map$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbbR^2\&x&mapsto&(x,|x|).endarray$$On $mathbbR^2$, consider the norm $bigl|(x,y)bigr|=maxbiglx$ and on $mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

share|cite|improve this answer

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

add a comment | 

up vote
1
down vote

up vote
1
down vote

No, it may well not be linear. Consider the map$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbbR^2\&x&mapsto&(x,|x|).endarray$$On $mathbbR^2$, consider the norm $bigl|(x,y)bigr|=maxbiglx$ and on $mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

share|cite|improve this answer

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

No, it may well not be linear. Consider the map$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbbR^2\&x&mapsto&(x,|x|).endarray$$On $mathbbR^2$, consider the norm $bigl|(x,y)bigr|=maxbiglx$ and on $mathbb R$ the usual norm. Then $f$ preserves distances and the origin, but it is not linear.

share|cite|improve this answer

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

answered Jul 30 at 20:24

José Carlos Santos

112k1696172

112k1696172

add a comment | 

add a comment | 

 

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