image of some ideal under the quotient map
1
$begingroup$
!
I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.
Is my understanding correct?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
$endgroup$
add a comment |
1
$begingroup$
!
I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.
Is my understanding correct?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
$endgroup$
add a comment |
1
1
1
$begingroup$
!
I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.
Is my understanding correct?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
$endgroup$
!
I am still confused about the range of $sigma(I_{omega})$.Since $|x_n|_2to 0,$we have $|x_n|to 0$,$sigma(I_{omega})=0,$then $J=0$,it is trivial.
Is my understanding correct?
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
asked Dec 7 '18 at 16:16
mathrookiemathrookie
899512
899512
add a comment |
add a comment |
1 Answer
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No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.
For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
|
show 1 more comment
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1 Answer
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1 Answer
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active
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1
$begingroup$
No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.
For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
|
show 1 more comment
1
$begingroup$
No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.
For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
|
show 1 more comment
1
1
1
$begingroup$
No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.
For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
No, it is not true that if $|x_n|_2to0$, then $|x_n|to0$.
For instance, suppose that $k(n)=n$, $x_n=E_{11}$ for all $n$. Then $|x_n|_2=1/sqrt n$, while $|x_n|=1$.
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 7 '18 at 17:26
Martin ArgeramiMartin Argerami
127k1182182
127k1182182
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
|
show 1 more comment
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
Your example is right.But we have the inequlity $|x|leq |x|_2$.I am quite puzzeld.
$endgroup$
– mathrookie
Dec 7 '18 at 17:36
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
No, you don't. You are mixing the normalized and non-normalized cases.
$endgroup$
– Martin Argerami
Dec 7 '18 at 17:42
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Suppose $Tin K(H)$, is the 2–norm of $T$ defined as follows:$|T|_2=(sum_{ein E}|Te|^2)^frac{1}{2}$,where $E$ is an orthonormal basis of H?
$endgroup$
– mathrookie
Dec 7 '18 at 18:46
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
Yes, and $|T|_2=infty$ if $T$ is not Hilbert-Schmidt.
$endgroup$
– Martin Argerami
Dec 7 '18 at 19:00
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
$begingroup$
So the inequality $|x|leq|x|_2$ holds when in the non-normalized case.
$endgroup$
– mathrookie
Dec 8 '18 at 9:44
|
show 1 more comment
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