How to show that $(lambda_n x_n)$ is compact iff $(lambda_n)in c_0$? [closed]
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Let $(lambda_n) in l_infty$, and $A:l_2 rightarrow l_2$ a linear operator defined as $$A(x_n)=(lambda_n x_n), ,, (x_n) in l_2$$ How to show that $A$ is compact if and only if ($lambda_n)in c_0$?
functional-analysis compact-operators
asked Nov 27 at 19:17
nehaci
31
closed as off-topic by Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo Nov 28 at 1:19
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Let $(lambda_n) in l_infty$, and $A:l_2 rightarrow l_2$ a linear operator defined as $$A(x_n)=(lambda_n x_n), ,, (x_n) in l_2$$ How to show that $A$ is compact if and only if ($lambda_n)in c_0$?
functional-analysis compact-operators
asked Nov 27 at 19:17
nehaci
31
closed as off-topic by Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo Nov 28 at 1:19
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." – Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(lambda_n) in l_infty$, and $A:l_2 rightarrow l_2$ a linear operator defined as $$A(x_n)=(lambda_n x_n), ,, (x_n) in l_2$$ How to show that $A$ is compact if and only if ($lambda_n)in c_0$?
functional-analysis compact-operators
asked Nov 27 at 19:17
nehaci
31
Let $(lambda_n) in l_infty$, and $A:l_2 rightarrow l_2$ a linear operator defined as $$A(x_n)=(lambda_n x_n), ,, (x_n) in l_2$$ How to show that $A$ is compact if and only if ($lambda_n)in c_0$?
functional-analysis compact-operators
functional-analysis compact-operators
asked Nov 27 at 19:17
nehaci
31
asked Nov 27 at 19:17
nehaci
31
asked Nov 27 at 19:17
nehaci
31
asked Nov 27 at 19:17
nehaci
31
asked Nov 27 at 19:17
nehaci
31
31
closed as off-topic by Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo Nov 28 at 1:19
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." – Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo Nov 28 at 1:19
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." – Adrian Keister, Trevor Gunn, Davide Giraudo, KReiser, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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One way: if $lambda_n to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $lambda_n$ for $n > N$ by $0$).
Other way: if $lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.
answered Nov 27 at 19:25
Robert Israel
316k23206457
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
One way: if $lambda_n to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $lambda_n$ for $n > N$ by $0$).
Other way: if $lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.
answered Nov 27 at 19:25
Robert Israel
316k23206457
add a comment |
up vote
0
down vote
accepted
One way: if $lambda_n to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $lambda_n$ for $n > N$ by $0$).
Other way: if $lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.
answered Nov 27 at 19:25
Robert Israel
316k23206457
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
One way: if $lambda_n to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $lambda_n$ for $n > N$ by $0$).
Other way: if $lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.
answered Nov 27 at 19:25
Robert Israel
316k23206457
One way: if $lambda_n to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $lambda_n$ for $n > N$ by $0$).
Other way: if $lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.
answered Nov 27 at 19:25
Robert Israel
316k23206457
answered Nov 27 at 19:25
Robert Israel
316k23206457
answered Nov 27 at 19:25
Robert Israel
316k23206457
answered Nov 27 at 19:25
Robert Israel
316k23206457
316k23206457
add a comment |
add a comment |
