A linear operator with a closed graph that is not bounded. [closed]
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Let $D$ be a linear operator on the subspace $ S={(x_n):sum_n n^2 vert x_n vert^2 < infty } subset l_2 $ such that $ D(x_n) = (n x_n) $. How to prove that the graph of $D$ is closed but is not bounded.
functional-analysis operator-theory closed-graph
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
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closed as off-topic by Aweygan, Saad, jgon, Leucippus, Cesareo Nov 26 '18 at 3:00
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$begingroup$
Let $D$ be a linear operator on the subspace $ S={(x_n):sum_n n^2 vert x_n vert^2 < infty } subset l_2 $ such that $ D(x_n) = (n x_n) $. How to prove that the graph of $D$ is closed but is not bounded.
functional-analysis operator-theory closed-graph
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
$endgroup$
closed as off-topic by Aweygan, Saad, jgon, Leucippus, Cesareo Nov 26 '18 at 3:00
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, Saad, jgon, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $D$ be a linear operator on the subspace $ S={(x_n):sum_n n^2 vert x_n vert^2 < infty } subset l_2 $ such that $ D(x_n) = (n x_n) $. How to prove that the graph of $D$ is closed but is not bounded.
functional-analysis operator-theory closed-graph
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
$endgroup$
Let $D$ be a linear operator on the subspace $ S={(x_n):sum_n n^2 vert x_n vert^2 < infty } subset l_2 $ such that $ D(x_n) = (n x_n) $. How to prove that the graph of $D$ is closed but is not bounded.
functional-analysis operator-theory closed-graph
functional-analysis operator-theory closed-graph
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
asked Nov 25 '18 at 17:54
togemoxevatogemoxeva
41
41
closed as off-topic by Aweygan, Saad, jgon, Leucippus, Cesareo Nov 26 '18 at 3:00
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, Saad, jgon, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Aweygan, Saad, jgon, Leucippus, Cesareo Nov 26 '18 at 3:00
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, Saad, jgon, Leucippus, 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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$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
add a comment |
$begingroup$
$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
$endgroup$
$S+I$ is closed iff $S$ is closed. $S+I$ has a bounded inverse $(S+I)^{-1}$, which is closed because it is bounded. So $S+I$ is closed because its graph is the transpose of the graph of $(S+I)^{-1}$. Therefore $S$ is closed.
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Nov 25 '18 at 18:48
DisintegratingByPartsDisintegratingByParts
58.9k42580
58.9k42580
add a comment |
add a comment |
