construct a sequence of operators [closed]
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Let $(H_n)$ be a sequence of different finite dimensional complex Hilbert spaces, $A_n in B(H_n),tr(A_n) to 0(n to infty)$,but the norm of $A_n$ does not converge to 0,where $tr()$ is the standard tracial state.
Can we construct a sequence of operators $(P_n)$ such that each $P_n in B(H_n)$ , $|P_n| to 0 $and $tr(P_nA_n)$ does not converge to 0?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 15 at 17:41
mathrookie
695512
closed as off-topic by Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos Nov 16 at 8:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Let $(H_n)$ be a sequence of different finite dimensional complex Hilbert spaces, $A_n in B(H_n),tr(A_n) to 0(n to infty)$,but the norm of $A_n$ does not converge to 0,where $tr()$ is the standard tracial state.
Can we construct a sequence of operators $(P_n)$ such that each $P_n in B(H_n)$ , $|P_n| to 0 $and $tr(P_nA_n)$ does not converge to 0?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 15 at 17:41
mathrookie
695512
closed as off-topic by Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos Nov 16 at 8:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $(H_n)$ be a sequence of different finite dimensional complex Hilbert spaces, $A_n in B(H_n),tr(A_n) to 0(n to infty)$,but the norm of $A_n$ does not converge to 0,where $tr()$ is the standard tracial state.
Can we construct a sequence of operators $(P_n)$ such that each $P_n in B(H_n)$ , $|P_n| to 0 $and $tr(P_nA_n)$ does not converge to 0?
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 15 at 17:41
mathrookie
695512
Let $(H_n)$ be a sequence of different finite dimensional complex Hilbert spaces, $A_n in B(H_n),tr(A_n) to 0(n to infty)$,but the norm of $A_n$ does not converge to 0,where $tr()$ is the standard tracial state.
Can we construct a sequence of operators $(P_n)$ such that each $P_n in B(H_n)$ , $|P_n| to 0 $and $tr(P_nA_n)$ does not converge to 0?
functional-analysis operator-theory operator-algebras c-star-algebras
functional-analysis operator-theory operator-algebras c-star-algebras
asked Nov 15 at 17:41
mathrookie
695512
asked Nov 15 at 17:41
mathrookie
695512
edited Nov 15 at 20:08
asked Nov 15 at 17:41
mathrookie
695512
asked Nov 15 at 17:41
mathrookie
695512
asked Nov 15 at 17:41
mathrookie
695512
695512
closed as off-topic by Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos Nov 16 at 8:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos Nov 16 at 8:43
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aweygan, Leucippus, jgon, Chinnapparaj R, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48
add a comment |
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes.
Take $H_n = mathbb{R}^2$ for all $n$, and
$$ A_n = begin{bmatrix}a_n& 0 \ 0 &-a_nend{bmatrix} quad P_n = begin{bmatrix}b_n& 0 \ 0 &-b_nend{bmatrix} $$
where $(a_n)_n$ and $(b_n)_n$ are two sequences we will choose in a second.
Then clearly $operatorname{tr}(A_n) = 0$ for each $n$, and $||P_n|| = |b_n|$, so if we choose $(b_n)_n$ such that $lim_{n} |b_n| = 0$ then we satisfy your assumptions.
Now $operatorname{tr}(P_nA_n) = 2a_n b_n$, so it is enough to choose $a_n = (b_n)^{-1}$ to have $operatorname{tr}(P_nA_n) = 2$ which clearly does not converge to zero.
answered Nov 15 at 18:49
Angelo Lucia
578213
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes.
Take $H_n = mathbb{R}^2$ for all $n$, and
$$ A_n = begin{bmatrix}a_n& 0 \ 0 &-a_nend{bmatrix} quad P_n = begin{bmatrix}b_n& 0 \ 0 &-b_nend{bmatrix} $$
where $(a_n)_n$ and $(b_n)_n$ are two sequences we will choose in a second.
Then clearly $operatorname{tr}(A_n) = 0$ for each $n$, and $||P_n|| = |b_n|$, so if we choose $(b_n)_n$ such that $lim_{n} |b_n| = 0$ then we satisfy your assumptions.
Now $operatorname{tr}(P_nA_n) = 2a_n b_n$, so it is enough to choose $a_n = (b_n)^{-1}$ to have $operatorname{tr}(P_nA_n) = 2$ which clearly does not converge to zero.
answered Nov 15 at 18:49
Angelo Lucia
578213
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
add a comment |
up vote
2
down vote
accepted
Yes.
Take $H_n = mathbb{R}^2$ for all $n$, and
$$ A_n = begin{bmatrix}a_n& 0 \ 0 &-a_nend{bmatrix} quad P_n = begin{bmatrix}b_n& 0 \ 0 &-b_nend{bmatrix} $$
where $(a_n)_n$ and $(b_n)_n$ are two sequences we will choose in a second.
Then clearly $operatorname{tr}(A_n) = 0$ for each $n$, and $||P_n|| = |b_n|$, so if we choose $(b_n)_n$ such that $lim_{n} |b_n| = 0$ then we satisfy your assumptions.
Now $operatorname{tr}(P_nA_n) = 2a_n b_n$, so it is enough to choose $a_n = (b_n)^{-1}$ to have $operatorname{tr}(P_nA_n) = 2$ which clearly does not converge to zero.
answered Nov 15 at 18:49
Angelo Lucia
578213
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes.
Take $H_n = mathbb{R}^2$ for all $n$, and
$$ A_n = begin{bmatrix}a_n& 0 \ 0 &-a_nend{bmatrix} quad P_n = begin{bmatrix}b_n& 0 \ 0 &-b_nend{bmatrix} $$
where $(a_n)_n$ and $(b_n)_n$ are two sequences we will choose in a second.
Then clearly $operatorname{tr}(A_n) = 0$ for each $n$, and $||P_n|| = |b_n|$, so if we choose $(b_n)_n$ such that $lim_{n} |b_n| = 0$ then we satisfy your assumptions.
Now $operatorname{tr}(P_nA_n) = 2a_n b_n$, so it is enough to choose $a_n = (b_n)^{-1}$ to have $operatorname{tr}(P_nA_n) = 2$ which clearly does not converge to zero.
answered Nov 15 at 18:49
Angelo Lucia
578213
Yes.
Take $H_n = mathbb{R}^2$ for all $n$, and
$$ A_n = begin{bmatrix}a_n& 0 \ 0 &-a_nend{bmatrix} quad P_n = begin{bmatrix}b_n& 0 \ 0 &-b_nend{bmatrix} $$
where $(a_n)_n$ and $(b_n)_n$ are two sequences we will choose in a second.
Then clearly $operatorname{tr}(A_n) = 0$ for each $n$, and $||P_n|| = |b_n|$, so if we choose $(b_n)_n$ such that $lim_{n} |b_n| = 0$ then we satisfy your assumptions.
Now $operatorname{tr}(P_nA_n) = 2a_n b_n$, so it is enough to choose $a_n = (b_n)^{-1}$ to have $operatorname{tr}(P_nA_n) = 2$ which clearly does not converge to zero.
answered Nov 15 at 18:49
Angelo Lucia
578213
answered Nov 15 at 18:49
Angelo Lucia
578213
answered Nov 15 at 18:49
Angelo Lucia
578213
answered Nov 15 at 18:49
Angelo Lucia
578213
578213
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
add a comment |
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
the standard tracial state needs to multiply $1/2$
– mathrookie
Nov 15 at 18:57
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
If $H_n$ are different,can we have a similar construction?
– mathrookie
Nov 15 at 18:59
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
I think you need to specify something more concrete than "the $H_n$ are different". You can definitely construct a lot of similar "easy" examples.
– Angelo Lucia
Nov 15 at 19:03
add a comment |
What is the connection between your two sentences? You introduce these $A_n$, then never mention them. Additionally, the norm of a projection is either $0$ or $1$, so this sequence would be exceptionally boring.
– Aweygan
Nov 15 at 17:44
Aweygan,Imade a mistake.I have reedited it
– mathrookie
Nov 15 at 17:48
Is the question about whether this can be done for some sequence $(A_n)_n$, or for any sequence $(A_n)_n$ on which the trace tends to zero?
– Angelo Lucia
Nov 15 at 18:45
@Angelo Lucia，for some sequence!
– mathrookie
Nov 15 at 18:48