Question about Normed vector space. The 2019 Stack Overflow Developer Survey Results Are Indifference between normed linear space and inner product spaceProof that multiplying by the scalar 1 does not change the vector in a normed vector space.Show that a normed Vector space is complete, need smart help.Must a normed vector space be over $mathbbR$ or $mathbbC$?Is there such thing as an unnormed vector space?Why in normed vector spaces we can define infinite series but in metric space we can not?Definition of a bounded subset in a normed vector spaceWhat definition of continuity does one use in proving continuity of vector space operations in a normed space?Relation between metric spaces, normed vector spaces, and inner product space.Problem about strictly normed spaces.
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Question about Normed vector space.
The 2019 Stack Overflow Developer Survey Results Are Indifference between normed linear space and inner product spaceProof that multiplying by the scalar 1 does not change the vector in a normed vector space.Show that a normed Vector space is complete, need smart help.Must a normed vector space be over $mathbbR$ or $mathbbC$?Is there such thing as an unnormed vector space?Why in normed vector spaces we can define infinite series but in metric space we can not?Definition of a bounded subset in a normed vector spaceWhat definition of continuity does one use in proving continuity of vector space operations in a normed space?Relation between metric spaces, normed vector spaces, and inner product space.Problem about strictly normed spaces.
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
real-analysis vector-spaces metric-spaces normed-spaces
edited Mar 30 at 20:51
Glorfindel
3,41381930
3,41381930
asked Mar 24 '14 at 0:08
user119615user119615
3,96531850
3,96531850
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 |
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
1
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
$begingroup$
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 |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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 |
$begingroup$
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.
$endgroup$
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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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited Mar 24 '14 at 4:15
answered Mar 24 '14 at 0:21
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
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
$begingroup$
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
1
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered Mar 24 '14 at 0:14
ncmathsadistncmathsadist
43.1k261103
43.1k261103
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
$begingroup$
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
1
1
$begingroup$
This would be appropriate as a comment.
$endgroup$
– user122283
Mar 24 '14 at 0:14
$begingroup$
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
$begingroup$
What is norm 1 and what is norm 2 in your example?
$endgroup$
– user119615
Mar 24 '14 at 0:17
add a comment |
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$begingroup$
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