Why is $d(x,0)$ not a norm?

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If $|x|$ is a norm, then we can define $d(x,y):=|x-y|$ and it will be a metric. Now, if $d$ is a metric, why is $|x|:= d(x,0)$ not a norm? I think it fail for the sub-linearity, but I don't see how.

functional-analysis

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edited Jul 15 at 17:56

Shaun

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asked Jul 15 at 15:52

Peter

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up vote
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down vote

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If $|x|$ is a norm, then we can define $d(x,y):=|x-y|$ and it will be a metric. Now, if $d$ is a metric, why is $|x|:= d(x,0)$ not a norm? I think it fail for the sub-linearity, but I don't see how.

functional-analysis

share|cite|improve this question

edited Jul 15 at 17:56

Shaun

7,41492972

asked Jul 15 at 15:52

Peter

358112

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

If $|x|$ is a norm, then we can define $d(x,y):=|x-y|$ and it will be a metric. Now, if $d$ is a metric, why is $|x|:= d(x,0)$ not a norm? I think it fail for the sub-linearity, but I don't see how.

functional-analysis

share|cite|improve this question

edited Jul 15 at 17:56

Shaun

7,41492972

asked Jul 15 at 15:52

Peter

358112

If $|x|$ is a norm, then we can define $d(x,y):=|x-y|$ and it will be a metric. Now, if $d$ is a metric, why is $|x|:= d(x,0)$ not a norm? I think it fail for the sub-linearity, but I don't see how.

functional-analysis

share|cite|improve this question

edited Jul 15 at 17:56

Shaun

7,41492972

asked Jul 15 at 15:52

Peter

358112

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 17:56

Shaun

7,41492972

edited Jul 15 at 17:56

Shaun

7,41492972

edited Jul 15 at 17:56

Shaun

7,41492972

7,41492972

asked Jul 15 at 15:52

Peter

358112

asked Jul 15 at 15:52

Peter

358112

asked Jul 15 at 15:52

Peter

358112

358112

add a comment | 

add a comment | 

3 Answers
3

active

oldest

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up vote
5
down vote



accepted

Consider $$d(x,y)=lvert x-yrvert^3.$$ You can prove that this is a metric. However, if you define $lVert xrVert:=d(x,0),$ then in general, for $xneq 0,$ we will have $$lVertalpha xrVertneqlvertalpharvertlVert xrVert,$$ with a few exceptions.

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answered Jul 15 at 15:57

Cameron Buie

83.5k771153

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up vote
1
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Homogeneity is not verified for the discrete metric for example. You can check the answer at here

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answered Jul 15 at 15:55

ippiki-ookami

350217

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up vote
1
down vote

Take for example $d$ as the discrete distance, then for $xnot=0$ and $|lambda|not=0,1$, we
$$1=d(lambda x,0)=|lambda x|not=|lambda||x|=|lambda|d(x,0)=|lambda|$$
and the absolute homogeneity property does not hold.

share|cite|improve this answer

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Consider $$d(x,y)=lvert x-yrvert^3.$$ You can prove that this is a metric. However, if you define $lVert xrVert:=d(x,0),$ then in general, for $xneq 0,$ we will have $$lVertalpha xrVertneqlvertalpharvertlVert xrVert,$$ with a few exceptions.

share|cite|improve this answer

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

add a comment | 

up vote
5
down vote



accepted

Consider $$d(x,y)=lvert x-yrvert^3.$$ You can prove that this is a metric. However, if you define $lVert xrVert:=d(x,0),$ then in general, for $xneq 0,$ we will have $$lVertalpha xrVertneqlvertalpharvertlVert xrVert,$$ with a few exceptions.

share|cite|improve this answer

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Consider $$d(x,y)=lvert x-yrvert^3.$$ You can prove that this is a metric. However, if you define $lVert xrVert:=d(x,0),$ then in general, for $xneq 0,$ we will have $$lVertalpha xrVertneqlvertalpharvertlVert xrVert,$$ with a few exceptions.

share|cite|improve this answer

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

Consider $$d(x,y)=lvert x-yrvert^3.$$ You can prove that this is a metric. However, if you define $lVert xrVert:=d(x,0),$ then in general, for $xneq 0,$ we will have $$lVertalpha xrVertneqlvertalpharvertlVert xrVert,$$ with a few exceptions.

share|cite|improve this answer

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

answered Jul 15 at 15:57

Cameron Buie

83.5k771153

83.5k771153

add a comment | 

add a comment | 

up vote
1
down vote

Homogeneity is not verified for the discrete metric for example. You can check the answer at here

share|cite|improve this answer

answered Jul 15 at 15:55

ippiki-ookami

350217

add a comment | 

up vote
1
down vote

Homogeneity is not verified for the discrete metric for example. You can check the answer at here

share|cite|improve this answer

answered Jul 15 at 15:55

ippiki-ookami

350217

add a comment | 

up vote
1
down vote

up vote
1
down vote

Homogeneity is not verified for the discrete metric for example. You can check the answer at here

share|cite|improve this answer

answered Jul 15 at 15:55

ippiki-ookami

350217

Homogeneity is not verified for the discrete metric for example. You can check the answer at here

share|cite|improve this answer

answered Jul 15 at 15:55

ippiki-ookami

350217

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 15:55

ippiki-ookami

350217

answered Jul 15 at 15:55

ippiki-ookami

350217

answered Jul 15 at 15:55

ippiki-ookami

350217

350217

add a comment | 

add a comment | 

up vote
1
down vote

Take for example $d$ as the discrete distance, then for $xnot=0$ and $|lambda|not=0,1$, we
$$1=d(lambda x,0)=|lambda x|not=|lambda||x|=|lambda|d(x,0)=|lambda|$$
and the absolute homogeneity property does not hold.

share|cite|improve this answer

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

add a comment | 

up vote
1
down vote

Take for example $d$ as the discrete distance, then for $xnot=0$ and $|lambda|not=0,1$, we
$$1=d(lambda x,0)=|lambda x|not=|lambda||x|=|lambda|d(x,0)=|lambda|$$
and the absolute homogeneity property does not hold.

share|cite|improve this answer

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

add a comment | 

up vote
1
down vote

up vote
1
down vote

Take for example $d$ as the discrete distance, then for $xnot=0$ and $|lambda|not=0,1$, we
$$1=d(lambda x,0)=|lambda x|not=|lambda||x|=|lambda|d(x,0)=|lambda|$$
and the absolute homogeneity property does not hold.

share|cite|improve this answer

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

Take for example $d$ as the discrete distance, then for $xnot=0$ and $|lambda|not=0,1$, we
$$1=d(lambda x,0)=|lambda x|not=|lambda||x|=|lambda|d(x,0)=|lambda|$$
and the absolute homogeneity property does not hold.

share|cite|improve this answer

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 16:06

edited Jul 15 at 16:06

edited Jul 15 at 16:06

answered Jul 15 at 15:58

Robert Z

84.2k955123

answered Jul 15 at 15:58

Robert Z

84.2k955123

answered Jul 15 at 15:58

Robert Z

84.2k955123

84.2k955123

add a comment | 

add a comment | 

 

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