Prove weighted ball in $l^2$ space is compact
$begingroup$
Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.
My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.
real-analysis general-topology compactness lp-spaces
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
$endgroup$
add a comment |
$begingroup$
Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.
My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.
real-analysis general-topology compactness lp-spaces
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
$endgroup$
add a comment |
$begingroup$
Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.
My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.
real-analysis general-topology compactness lp-spaces
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
$endgroup$
Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.
My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.
real-analysis general-topology compactness lp-spaces
real-analysis general-topology compactness lp-spaces
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
edited Jan 17 at 20:51
Davide Giraudo
126k16150261
126k16150261
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 18:26
Fluffy SkyeFluffy Skye
1148
1148
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
$$
limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
tag{1}$$
then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.
The theorem provides a characterization of totally bounded subsets of $ell_p$.
Let’s apply it to $B$.
For $xin B$, we have
$$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x in B$
$$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
Therefore
$$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
$$
limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
tag{1}$$
then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.
The theorem provides a characterization of totally bounded subsets of $ell_p$.
Let’s apply it to $B$.
For $xin B$, we have
$$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x in B$
$$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
Therefore
$$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
add a comment |
$begingroup$
Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
$$
limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
tag{1}$$
then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.
The theorem provides a characterization of totally bounded subsets of $ell_p$.
Let’s apply it to $B$.
For $xin B$, we have
$$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x in B$
$$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
Therefore
$$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
add a comment |
$begingroup$
Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
$$
limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
tag{1}$$
then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.
The theorem provides a characterization of totally bounded subsets of $ell_p$.
Let’s apply it to $B$.
For $xin B$, we have
$$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x in B$
$$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
Therefore
$$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
$endgroup$
Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
$$
limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
tag{1}$$
then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.
The theorem provides a characterization of totally bounded subsets of $ell_p$.
Let’s apply it to $B$.
For $xin B$, we have
$$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x in B$
$$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
Therefore
$$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
edited Jan 17 at 20:35
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 17 at 20:17
mathcounterexamples.netmathcounterexamples.net
26.7k22157
26.7k22157
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
add a comment |
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
Thank you so much!
$endgroup$
– Fluffy Skye
Jan 17 at 20:25
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:27
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@mathcounterexamples.net I think you made a typo in your second last inequality.
$endgroup$
– BigbearZzz
Jan 17 at 20:34
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
$begingroup$
@BigbearZzz Thanks for noticing. I corrected.
$endgroup$
– mathcounterexamples.net
Jan 17 at 20:35
add a comment |
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