In what way is convergence in operator norm 'better' than other forms of convergence?
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If we show a sequence of operators $T_n$ converges uniformly/in operator norm to some operator $T$ this implies the sequence converges strongly/pointwise, which in term means it converges weakly.
I know in some cases we can only show weak convergence so we have to be content with that. So what have we 'lost' when we can only show weak convergence? In contrast, what do we gain by having convergence in operator norm as opposed to the other forms of convergence? Actually what do we gain as we move up the hierarchy of convergences:
- Weakly
- Strongly
- Uniformly
I am interested in what we gain in a mathematical sense, and also how how it relates to applications in physics or numerical analysis.
real-analysis functional-analysis operator-theory uniform-convergence weak-convergence
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
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add a comment |
$begingroup$
If we show a sequence of operators $T_n$ converges uniformly/in operator norm to some operator $T$ this implies the sequence converges strongly/pointwise, which in term means it converges weakly.
I know in some cases we can only show weak convergence so we have to be content with that. So what have we 'lost' when we can only show weak convergence? In contrast, what do we gain by having convergence in operator norm as opposed to the other forms of convergence? Actually what do we gain as we move up the hierarchy of convergences:
- Weakly
- Strongly
- Uniformly
I am interested in what we gain in a mathematical sense, and also how how it relates to applications in physics or numerical analysis.
real-analysis functional-analysis operator-theory uniform-convergence weak-convergence
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
$endgroup$
1
$begingroup$
Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
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– Giuseppe Negro
Sep 29 '18 at 10:40
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That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
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– eurocoder
Sep 30 '18 at 10:07
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13
add a comment |
$begingroup$
If we show a sequence of operators $T_n$ converges uniformly/in operator norm to some operator $T$ this implies the sequence converges strongly/pointwise, which in term means it converges weakly.
I know in some cases we can only show weak convergence so we have to be content with that. So what have we 'lost' when we can only show weak convergence? In contrast, what do we gain by having convergence in operator norm as opposed to the other forms of convergence? Actually what do we gain as we move up the hierarchy of convergences:
- Weakly
- Strongly
- Uniformly
I am interested in what we gain in a mathematical sense, and also how how it relates to applications in physics or numerical analysis.
real-analysis functional-analysis operator-theory uniform-convergence weak-convergence
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
$endgroup$
If we show a sequence of operators $T_n$ converges uniformly/in operator norm to some operator $T$ this implies the sequence converges strongly/pointwise, which in term means it converges weakly.
I know in some cases we can only show weak convergence so we have to be content with that. So what have we 'lost' when we can only show weak convergence? In contrast, what do we gain by having convergence in operator norm as opposed to the other forms of convergence? Actually what do we gain as we move up the hierarchy of convergences:
- Weakly
- Strongly
- Uniformly
I am interested in what we gain in a mathematical sense, and also how how it relates to applications in physics or numerical analysis.
real-analysis functional-analysis operator-theory uniform-convergence weak-convergence
real-analysis functional-analysis operator-theory uniform-convergence weak-convergence
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
asked Sep 29 '18 at 10:38
eurocodereurocoder
1,126415
1,126415
1
$begingroup$
Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
$endgroup$
– Giuseppe Negro
Sep 29 '18 at 10:40
$begingroup$
That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
$endgroup$
– eurocoder
Sep 30 '18 at 10:07
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13
add a comment |
1
$begingroup$
Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
$endgroup$
– Giuseppe Negro
Sep 29 '18 at 10:40
$begingroup$
That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
$endgroup$
– eurocoder
Sep 30 '18 at 10:07
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13
1
1
$begingroup$
Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
$endgroup$
– Giuseppe Negro
Sep 29 '18 at 10:40
$begingroup$
Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
$endgroup$
– Giuseppe Negro
Sep 29 '18 at 10:40
$begingroup$
That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
$endgroup$
– eurocoder
Sep 30 '18 at 10:07
$begingroup$
That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
$endgroup$
– eurocoder
Sep 30 '18 at 10:07
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13
add a comment |
1 Answer
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Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.
On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
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add a comment |
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$begingroup$
Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.
On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.
On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.
On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.
On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
edited Jan 11 at 22:16
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
answered Sep 30 '18 at 16:16
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
add a comment |
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Operator norm convergence preserves compact operators. (That is, a sequence of compact operators converging in norm converges to a compact operator). This is often used to prove compactness. The other modes of convergence do not have that.
$endgroup$
– Giuseppe Negro
Sep 29 '18 at 10:40
$begingroup$
That's definitely a useful property! Are there any others? What does strong convergence gives us that weak convergence doesn't?
$endgroup$
– eurocoder
Sep 30 '18 at 10:07
$begingroup$
You know, "often" weak convergence implies strong convergence. It is the case of semigroup theory, which is more or less the application of these things to PDEs. So I do not have meaningful examples.
$endgroup$
– Giuseppe Negro
Sep 30 '18 at 10:13