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Question about Normed vector space.

6

1

$begingroup$

Here is the definition of a normed vector space my book uses:

And here is a remark I do not understand:

I do not understand that a sequence can converge to a vector in one norm, and not the other. For instance: Lets say $s_n$ converges to $u$ with the $||_1$-norm. From definition 4.5.2 (i) we must have that $s_n$ becomes closer and closer to $u$. Why is it that it could fail in the other norm, when it can become as close as we want in the first norm?Are there any simple examples of this phenomenon?

PS:I know that they say we will see examples of this later in the book, but what comes later is too hard for me to udnerstand now.

real-analysis vector-spaces metric-spaces normed-spaces

share|cite|improve this question

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It should be mentioned that this never happens in finite dimensions. In finite dimensions, every norm is "equivalent" meaning that if a sequence converges in one norm, in converges in the other (which implies a series converges in one norm if it converges in the other). This should give you an idea of how bizarre things get when you have an infinite dimensional vector space.
  $endgroup$
  – Cameron Williams
  Mar 24 '14 at 0:24
  

  
  
  
  

add a comment | 

6

1

$begingroup$

Here is the definition of a normed vector space my book uses:

And here is a remark I do not understand:

I do not understand that a sequence can converge to a vector in one norm, and not the other. For instance: Lets say $s_n$ converges to $u$ with the $||_1$-norm. From definition 4.5.2 (i) we must have that $s_n$ becomes closer and closer to $u$. Why is it that it could fail in the other norm, when it can become as close as we want in the first norm?Are there any simple examples of this phenomenon?

PS:I know that they say we will see examples of this later in the book, but what comes later is too hard for me to udnerstand now.

real-analysis vector-spaces metric-spaces normed-spaces

share|cite|improve this question

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It should be mentioned that this never happens in finite dimensions. In finite dimensions, every norm is "equivalent" meaning that if a sequence converges in one norm, in converges in the other (which implies a series converges in one norm if it converges in the other). This should give you an idea of how bizarre things get when you have an infinite dimensional vector space.
  $endgroup$
  – Cameron Williams
  Mar 24 '14 at 0:24
  

  
  
  
  

add a comment | 

$begingroup$

Here is the definition of a normed vector space my book uses:

And here is a remark I do not understand:

I do not understand that a sequence can converge to a vector in one norm, and not the other. For instance: Lets say $s_n$ converges to $u$ with the $||_1$-norm. From definition 4.5.2 (i) we must have that $s_n$ becomes closer and closer to $u$. Why is it that it could fail in the other norm, when it can become as close as we want in the first norm?Are there any simple examples of this phenomenon?

PS:I know that they say we will see examples of this later in the book, but what comes later is too hard for me to udnerstand now.

real-analysis vector-spaces metric-spaces normed-spaces

share|cite|improve this question

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

$endgroup$

share|cite|improve this question

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

share|cite|improve this question

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

edited Mar 30 at 20:51

Glorfindel

3,41381930

edited Mar 30 at 20:51

Glorfindel

3,41381930

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

asked Mar 24 '14 at 0:08

user119615user119615

3,96531850

2 Answers
2

active

oldest

votes

4

$begingroup$

Consider the following sequence of elements in the space $V$ of finite sequences:
$$
u_1=(1,0,0,ldots), u_2=(0,frac12,0,ldots), u_3=(0,0,frac13,0,ldots)
$$
Then
$$
sum_k=1^nu_k=(1,frac12,frac13,ldots,frac1n,0,ldots)
$$
Now consider these two norms on $u=(a_1,a_2,ldots)$:
$$
|u|_1=sum_k=1^infty|a_k|, |u|_2=left(sum_k=1^infty|a_k|^2right)^1/2
$$
Then, for the $u_n$ defined above,
$$
left|sum_k=M^nu_kright|_1=sum_k=M^Nfrac1k, left|sum_k=M^Nu_kright|_2=sum_k=M^Nfrac1k^2
$$
So, in $|cdot|_1$, the tails of the series $sum_k=1^infty u_k$ are unbounded, which means that the series diverges; while in $|cdot|_2$, the tails go to zero, so the series converges.

share|cite|improve this answer

edited Mar 24 '14 at 4:15

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, so I guess in the first case we have that the distance between |(1,1/2,1/3...,1/n,0,0,...)-(1/1,1/2,1/3,1/4,1/5,1/6.......)|=|(0,0,0,0,0,0,0,1/(n+1),1/(n+1),1/(n+3),.....| will never be as small as we want, but it will be that in the second case? Thank you for your help!
  $endgroup$
  – user119615
  Mar 24 '14 at 1:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes indeed. The point, as your book states, is that the notion of series depends on the choice of a topology, as the series is a limit of the sequence of partial sums.
  $endgroup$
  – Martin Argerami
  Mar 24 '14 at 4:16
  

  
  
  
  

add a comment | 

1

$begingroup$

You can get in trouble if the convergence is not absolute, i.e. $$sum_n |x_n| = +infty$$ but
$$lim_Ntoinfty sum_nle N x_n $$
exists.

share|cite|improve this answer

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This would be appropriate as a comment.
  $endgroup$
  – user122283
  Mar 24 '14 at 0:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is norm 1 and what is norm 2 in your example?
  $endgroup$
  – user119615
  Mar 24 '14 at 0:17
  

  
  
  
  

add a comment | 

Your Answer

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4

$begingroup$

Consider the following sequence of elements in the space $V$ of finite sequences:
$$
u_1=(1,0,0,ldots), u_2=(0,frac12,0,ldots), u_3=(0,0,frac13,0,ldots)
$$
Then
$$
sum_k=1^nu_k=(1,frac12,frac13,ldots,frac1n,0,ldots)
$$
Now consider these two norms on $u=(a_1,a_2,ldots)$:
$$
|u|_1=sum_k=1^infty|a_k|, |u|_2=left(sum_k=1^infty|a_k|^2right)^1/2
$$
Then, for the $u_n$ defined above,
$$
left|sum_k=M^nu_kright|_1=sum_k=M^Nfrac1k, left|sum_k=M^Nu_kright|_2=sum_k=M^Nfrac1k^2
$$
So, in $|cdot|_1$, the tails of the series $sum_k=1^infty u_k$ are unbounded, which means that the series diverges; while in $|cdot|_2$, the tails go to zero, so the series converges.

share|cite|improve this answer

edited Mar 24 '14 at 4:15

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, so I guess in the first case we have that the distance between |(1,1/2,1/3...,1/n,0,0,...)-(1/1,1/2,1/3,1/4,1/5,1/6.......)|=|(0,0,0,0,0,0,0,1/(n+1),1/(n+1),1/(n+3),.....| will never be as small as we want, but it will be that in the second case? Thank you for your help!
  $endgroup$
  – user119615
  Mar 24 '14 at 1:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes indeed. The point, as your book states, is that the notion of series depends on the choice of a topology, as the series is a limit of the sequence of partial sums.
  $endgroup$
  – Martin Argerami
  Mar 24 '14 at 4:16
  

  
  
  
  

add a comment | 

4

$begingroup$

Consider the following sequence of elements in the space $V$ of finite sequences:
$$
u_1=(1,0,0,ldots), u_2=(0,frac12,0,ldots), u_3=(0,0,frac13,0,ldots)
$$
Then
$$
sum_k=1^nu_k=(1,frac12,frac13,ldots,frac1n,0,ldots)
$$
Now consider these two norms on $u=(a_1,a_2,ldots)$:
$$
|u|_1=sum_k=1^infty|a_k|, |u|_2=left(sum_k=1^infty|a_k|^2right)^1/2
$$
Then, for the $u_n$ defined above,
$$
left|sum_k=M^nu_kright|_1=sum_k=M^Nfrac1k, left|sum_k=M^Nu_kright|_2=sum_k=M^Nfrac1k^2
$$
So, in $|cdot|_1$, the tails of the series $sum_k=1^infty u_k$ are unbounded, which means that the series diverges; while in $|cdot|_2$, the tails go to zero, so the series converges.

share|cite|improve this answer

edited Mar 24 '14 at 4:15

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, so I guess in the first case we have that the distance between |(1,1/2,1/3...,1/n,0,0,...)-(1/1,1/2,1/3,1/4,1/5,1/6.......)|=|(0,0,0,0,0,0,0,1/(n+1),1/(n+1),1/(n+3),.....| will never be as small as we want, but it will be that in the second case? Thank you for your help!
  $endgroup$
  – user119615
  Mar 24 '14 at 1:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes indeed. The point, as your book states, is that the notion of series depends on the choice of a topology, as the series is a limit of the sequence of partial sums.
  $endgroup$
  – Martin Argerami
  Mar 24 '14 at 4:16
  

  
  
  
  

add a comment | 

$begingroup$

Consider the following sequence of elements in the space $V$ of finite sequences:
$$
u_1=(1,0,0,ldots), u_2=(0,frac12,0,ldots), u_3=(0,0,frac13,0,ldots)
$$
Then
$$
sum_k=1^nu_k=(1,frac12,frac13,ldots,frac1n,0,ldots)
$$
Now consider these two norms on $u=(a_1,a_2,ldots)$:
$$
|u|_1=sum_k=1^infty|a_k|, |u|_2=left(sum_k=1^infty|a_k|^2right)^1/2
$$
Then, for the $u_n$ defined above,
$$
left|sum_k=M^nu_kright|_1=sum_k=M^Nfrac1k, left|sum_k=M^Nu_kright|_2=sum_k=M^Nfrac1k^2
$$
So, in $|cdot|_1$, the tails of the series $sum_k=1^infty u_k$ are unbounded, which means that the series diverges; while in $|cdot|_2$, the tails go to zero, so the series converges.

share|cite|improve this answer

edited Mar 24 '14 at 4:15

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

share|cite|improve this answer

edited Mar 24 '14 at 4:15

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

answered Mar 24 '14 at 0:21

Martin ArgeramiMartin Argerami

129k1184185

1

$begingroup$

You can get in trouble if the convergence is not absolute, i.e. $$sum_n |x_n| = +infty$$ but
$$lim_Ntoinfty sum_nle N x_n $$
exists.

share|cite|improve this answer

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This would be appropriate as a comment.
  $endgroup$
  – user122283
  Mar 24 '14 at 0:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is norm 1 and what is norm 2 in your example?
  $endgroup$
  – user119615
  Mar 24 '14 at 0:17
  

  
  
  
  

add a comment | 

1

$begingroup$

You can get in trouble if the convergence is not absolute, i.e. $$sum_n |x_n| = +infty$$ but
$$lim_Ntoinfty sum_nle N x_n $$
exists.

share|cite|improve this answer

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This would be appropriate as a comment.
  $endgroup$
  – user122283
  Mar 24 '14 at 0:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is norm 1 and what is norm 2 in your example?
  $endgroup$
  – user119615
  Mar 24 '14 at 0:17
  

  
  
  
  

add a comment | 

$begingroup$

You can get in trouble if the convergence is not absolute, i.e. $$sum_n |x_n| = +infty$$ but
$$lim_Ntoinfty sum_nle N x_n $$
exists.

share|cite|improve this answer

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

$endgroup$

share|cite|improve this answer

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103

answered Mar 24 '14 at 0:14

ncmathsadistncmathsadist

43.1k261103