approximate identity element
$begingroup$
Let $Isubset prod_n B(H_{m_n})$ be a separable $C^*$ algebra,where $prod B(H_{m_n})$ denotes the $ell ^{infty}$direct sum of $B(H_n)$ and $dim(H_{m_n})<infty$. We suppose that there exits a strictly positive element $A=(A_1,cdots,A_n,cdots)$ in $I$ (each $A_i in B(H_{m_i})$) such that $lim tr_{m_n}(A_n)=0$ and $A^frac{1}{k},k=1,2,cdots$ is an approximate unit for $I$.
Let ${a_{1,n},cdots,a_{m_n,n}}$ denote the eigenvaluses of $A_n$,then $b_n={j:a_{j,n}leq2tr_n(A_n)}$ has cardiniality at least $frac{m_n}{2}$,
If $P=(P_1,cdots,P_n,cdots)$ is the orthogonal projection onto the span of the eigenvectors associated with ${a_{jn}:jin b_n}$then we can conculde that $0leq P_nA_nleq 2 tr_n(A_n)P_n$and hence$|P_nA_n|to 0$.
Can we use the fact that $A^frac{1}{k},k=1,2,cdots$ is an approximate identity element for $I$ to have the conclusion:
For any $B=(B_1,cdots,B_n,cdots)in I$,$|P_nB_n|to 0$
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
$endgroup$
add a comment |
$begingroup$
Let $Isubset prod_n B(H_{m_n})$ be a separable $C^*$ algebra,where $prod B(H_{m_n})$ denotes the $ell ^{infty}$direct sum of $B(H_n)$ and $dim(H_{m_n})<infty$. We suppose that there exits a strictly positive element $A=(A_1,cdots,A_n,cdots)$ in $I$ (each $A_i in B(H_{m_i})$) such that $lim tr_{m_n}(A_n)=0$ and $A^frac{1}{k},k=1,2,cdots$ is an approximate unit for $I$.
Let ${a_{1,n},cdots,a_{m_n,n}}$ denote the eigenvaluses of $A_n$,then $b_n={j:a_{j,n}leq2tr_n(A_n)}$ has cardiniality at least $frac{m_n}{2}$,
If $P=(P_1,cdots,P_n,cdots)$ is the orthogonal projection onto the span of the eigenvectors associated with ${a_{jn}:jin b_n}$then we can conculde that $0leq P_nA_nleq 2 tr_n(A_n)P_n$and hence$|P_nA_n|to 0$.
Can we use the fact that $A^frac{1}{k},k=1,2,cdots$ is an approximate identity element for $I$ to have the conclusion:
For any $B=(B_1,cdots,B_n,cdots)in I$,$|P_nB_n|to 0$
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
$endgroup$
add a comment |
$begingroup$
Let $Isubset prod_n B(H_{m_n})$ be a separable $C^*$ algebra,where $prod B(H_{m_n})$ denotes the $ell ^{infty}$direct sum of $B(H_n)$ and $dim(H_{m_n})<infty$. We suppose that there exits a strictly positive element $A=(A_1,cdots,A_n,cdots)$ in $I$ (each $A_i in B(H_{m_i})$) such that $lim tr_{m_n}(A_n)=0$ and $A^frac{1}{k},k=1,2,cdots$ is an approximate unit for $I$.
Let ${a_{1,n},cdots,a_{m_n,n}}$ denote the eigenvaluses of $A_n$,then $b_n={j:a_{j,n}leq2tr_n(A_n)}$ has cardiniality at least $frac{m_n}{2}$,
If $P=(P_1,cdots,P_n,cdots)$ is the orthogonal projection onto the span of the eigenvectors associated with ${a_{jn}:jin b_n}$then we can conculde that $0leq P_nA_nleq 2 tr_n(A_n)P_n$and hence$|P_nA_n|to 0$.
Can we use the fact that $A^frac{1}{k},k=1,2,cdots$ is an approximate identity element for $I$ to have the conclusion:
For any $B=(B_1,cdots,B_n,cdots)in I$,$|P_nB_n|to 0$
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
$endgroup$
Let $Isubset prod_n B(H_{m_n})$ be a separable $C^*$ algebra,where $prod B(H_{m_n})$ denotes the $ell ^{infty}$direct sum of $B(H_n)$ and $dim(H_{m_n})<infty$. We suppose that there exits a strictly positive element $A=(A_1,cdots,A_n,cdots)$ in $I$ (each $A_i in B(H_{m_i})$) such that $lim tr_{m_n}(A_n)=0$ and $A^frac{1}{k},k=1,2,cdots$ is an approximate unit for $I$.
Let ${a_{1,n},cdots,a_{m_n,n}}$ denote the eigenvaluses of $A_n$,then $b_n={j:a_{j,n}leq2tr_n(A_n)}$ has cardiniality at least $frac{m_n}{2}$,
If $P=(P_1,cdots,P_n,cdots)$ is the orthogonal projection onto the span of the eigenvectors associated with ${a_{jn}:jin b_n}$then we can conculde that $0leq P_nA_nleq 2 tr_n(A_n)P_n$and hence$|P_nA_n|to 0$.
Can we use the fact that $A^frac{1}{k},k=1,2,cdots$ is an approximate identity element for $I$ to have the conclusion:
For any $B=(B_1,cdots,B_n,cdots)in I$,$|P_nB_n|to 0$
operator-theory operator-algebras c-star-algebras von-neumann-algebras
operator-theory operator-algebras c-star-algebras von-neumann-algebras
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
asked Dec 8 '18 at 1:41
mathrookiemathrookie
832512
832512
add a comment |
add a comment |
1 Answer
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I think nothing prevents you from taking $B_n=P_n$, so $|B_nP_n|=1$.
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
$endgroup$
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
I think nothing prevents you from taking $B_n=P_n$, so $|B_nP_n|=1$.
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
$endgroup$
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
add a comment |
$begingroup$
I think nothing prevents you from taking $B_n=P_n$, so $|B_nP_n|=1$.
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
$endgroup$
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
add a comment |
$begingroup$
I think nothing prevents you from taking $B_n=P_n$, so $|B_nP_n|=1$.
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
$endgroup$
I think nothing prevents you from taking $B_n=P_n$, so $|B_nP_n|=1$.
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
answered Dec 8 '18 at 3:01
Martin ArgeramiMartin Argerami
125k1181179
125k1181179
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
add a comment |
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
If we also assume that $forall (x_n)in A,lim tr_{m_n}(x_n)=0$,can we have the above conclusion?
$endgroup$
– mathrookie
Dec 8 '18 at 9:50
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
$begingroup$
no,$tr(Pn)geqfrac{1}{2}$ was proved by J Anderson
$endgroup$
– mathrookie
Dec 8 '18 at 11:41
add a comment |
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