Mixing
Proof completenes of $ {x in mathbb{C}^mathbb{N} | sum_{n=1}^infty s_n |x_n|^p < infty }$
$begingroup$
Let $(s_n)_{ninmathbb{N}} subseteq mathbb{R}$ such that for all $n$: $0 < s_n leq frac{1}{n} $. Let $p>1$.
How to show that the space of sequences
$ l^p_s := {x in mathbb{C}^mathbb{N} | sum_{n=1}^infty s_n |x_n|^p < infty } $ is complete in respect to the norm $ ||x||:= (sum_{n=1}^infty s_n |x_n|^p)^{frac{1}{p} }$.
I tried to fix a $kinmathbb{N}$ and show that $(x_n(k))_{ninmathbb{N}}$ is a Cauchy Sequence, but it doesn't work.
sequences-and-series functional-analysis convergence cauchy-sequences complete-spaces
asked Jan 18 at 11:51
SlyderSlyder
316
$endgroup$
add a comment |
$begingroup$
Let $(s_n)_{ninmathbb{N}} subseteq mathbb{R}$ such that for all $n$: $0 < s_n leq frac{1}{n} $. Let $p>1$.
How to show that the space of sequences
$ l^p_s := {x in mathbb{C}^mathbb{N} | sum_{n=1}^infty s_n |x_n|^p < infty } $ is complete in respect to the norm $ ||x||:= (sum_{n=1}^infty s_n |x_n|^p)^{frac{1}{p} }$.
I tried to fix a $kinmathbb{N}$ and show that $(x_n(k))_{ninmathbb{N}}$ is a Cauchy Sequence, but it doesn't work.
sequences-and-series functional-analysis convergence cauchy-sequences complete-spaces
asked Jan 18 at 11:51
SlyderSlyder
316
$endgroup$
add a comment |
$begingroup$
Let $(s_n)_{ninmathbb{N}} subseteq mathbb{R}$ such that for all $n$: $0 < s_n leq frac{1}{n} $. Let $p>1$.
How to show that the space of sequences
$ l^p_s := {x in mathbb{C}^mathbb{N} | sum_{n=1}^infty s_n |x_n|^p < infty } $ is complete in respect to the norm $ ||x||:= (sum_{n=1}^infty s_n |x_n|^p)^{frac{1}{p} }$.
I tried to fix a $kinmathbb{N}$ and show that $(x_n(k))_{ninmathbb{N}}$ is a Cauchy Sequence, but it doesn't work.
sequences-and-series functional-analysis convergence cauchy-sequences complete-spaces
asked Jan 18 at 11:51
SlyderSlyder
316
$endgroup$
Let $(s_n)_{ninmathbb{N}} subseteq mathbb{R}$ such that for all $n$: $0 < s_n leq frac{1}{n} $. Let $p>1$.
How to show that the space of sequences
$ l^p_s := {x in mathbb{C}^mathbb{N} | sum_{n=1}^infty s_n |x_n|^p < infty } $ is complete in respect to the norm $ ||x||:= (sum_{n=1}^infty s_n |x_n|^p)^{frac{1}{p} }$.
I tried to fix a $kinmathbb{N}$ and show that $(x_n(k))_{ninmathbb{N}}$ is a Cauchy Sequence, but it doesn't work.
sequences-and-series functional-analysis convergence cauchy-sequences complete-spaces
sequences-and-series functional-analysis convergence cauchy-sequences complete-spaces
asked Jan 18 at 11:51
SlyderSlyder
316
asked Jan 18 at 11:51
SlyderSlyder
316
asked Jan 18 at 11:51
SlyderSlyder
316
asked Jan 18 at 11:51
SlyderSlyder
316
asked Jan 18 at 11:51
SlyderSlyder
316
316
add a comment |
add a comment |
1 Answer
1
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Use the fact that $ell^{p}(mathbb C ^{mathbb N})$ is complete and the map $(x_n) to (s_n^{1/p}x_n)$ is an isometric isomorphism. The condition $s_n leq frac 1 n$ plays no role in this. Alternatively, you can imitate the proof of completeness of $ell^{p}(mathbb C ^{mathbb N})$.
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Use the fact that $ell^{p}(mathbb C ^{mathbb N})$ is complete and the map $(x_n) to (s_n^{1/p}x_n)$ is an isometric isomorphism. The condition $s_n leq frac 1 n$ plays no role in this. Alternatively, you can imitate the proof of completeness of $ell^{p}(mathbb C ^{mathbb N})$.
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
add a comment |
$begingroup$
Use the fact that $ell^{p}(mathbb C ^{mathbb N})$ is complete and the map $(x_n) to (s_n^{1/p}x_n)$ is an isometric isomorphism. The condition $s_n leq frac 1 n$ plays no role in this. Alternatively, you can imitate the proof of completeness of $ell^{p}(mathbb C ^{mathbb N})$.
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
add a comment |
$begingroup$
Use the fact that $ell^{p}(mathbb C ^{mathbb N})$ is complete and the map $(x_n) to (s_n^{1/p}x_n)$ is an isometric isomorphism. The condition $s_n leq frac 1 n$ plays no role in this. Alternatively, you can imitate the proof of completeness of $ell^{p}(mathbb C ^{mathbb N})$.
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
Use the fact that $ell^{p}(mathbb C ^{mathbb N})$ is complete and the map $(x_n) to (s_n^{1/p}x_n)$ is an isometric isomorphism. The condition $s_n leq frac 1 n$ plays no role in this. Alternatively, you can imitate the proof of completeness of $ell^{p}(mathbb C ^{mathbb N})$.
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:08
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
63.2k42362
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
add a comment |
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
The imitation of the completeness of $l^p$ yields the result. But eventhough the idea of using an isometry is interesting, this ismorphism actually is not a bijection. Hence this approach does not work. Thank you :)
$endgroup$
– Slyder
Jan 23 at 12:37
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
$begingroup$
@Slyder It is a bijection. If $(y_n) in ell_s^{p}$ then $(x_n) =(s_n^{-1/p}y_n) in ell^{p}$ and $(y_n)$ is the image of $(x_n)$. Of course, the map is also one-to-one.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 12:41
add a comment |
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