Weighted P-norm
$begingroup$
It is known that $ l^p subseteq l^r $ if $ 1 le p le r < infty $. Also, $ l_w^p $ are spaces where we consider a collection of positive weights, $ w_n > 0 $ for each $ n in mathbb{N} $, that given sequence $ x = (x_n) $, the weighted $p$-norm is
$$ ||x||_{w,p} = big (sum_{n=1}^infty w_n |x_n|^p big )^frac{1}{p} text{.} $$
Prove that if $ sum_{n=1}^infty w_n < infty $, then $ l^1_w not subseteq l^2_w $.One way to prove this is to find a subsequence $ (w_{n_k}), $ so that $ w_{n_k} le frac{1}{2^k} $ and consider elements which are nonzero only at the $ n_k $ coordinates.
the subsequence is probably very easy to think up, but I can just not figure it out... :(
functional-analysis norm lp-spaces
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
$endgroup$
add a comment |
$begingroup$
It is known that $ l^p subseteq l^r $ if $ 1 le p le r < infty $. Also, $ l_w^p $ are spaces where we consider a collection of positive weights, $ w_n > 0 $ for each $ n in mathbb{N} $, that given sequence $ x = (x_n) $, the weighted $p$-norm is
$$ ||x||_{w,p} = big (sum_{n=1}^infty w_n |x_n|^p big )^frac{1}{p} text{.} $$
Prove that if $ sum_{n=1}^infty w_n < infty $, then $ l^1_w not subseteq l^2_w $.One way to prove this is to find a subsequence $ (w_{n_k}), $ so that $ w_{n_k} le frac{1}{2^k} $ and consider elements which are nonzero only at the $ n_k $ coordinates.
the subsequence is probably very easy to think up, but I can just not figure it out... :(
functional-analysis norm lp-spaces
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
$endgroup$
add a comment |
$begingroup$
It is known that $ l^p subseteq l^r $ if $ 1 le p le r < infty $. Also, $ l_w^p $ are spaces where we consider a collection of positive weights, $ w_n > 0 $ for each $ n in mathbb{N} $, that given sequence $ x = (x_n) $, the weighted $p$-norm is
$$ ||x||_{w,p} = big (sum_{n=1}^infty w_n |x_n|^p big )^frac{1}{p} text{.} $$
Prove that if $ sum_{n=1}^infty w_n < infty $, then $ l^1_w not subseteq l^2_w $.One way to prove this is to find a subsequence $ (w_{n_k}), $ so that $ w_{n_k} le frac{1}{2^k} $ and consider elements which are nonzero only at the $ n_k $ coordinates.
the subsequence is probably very easy to think up, but I can just not figure it out... :(
functional-analysis norm lp-spaces
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
$endgroup$
It is known that $ l^p subseteq l^r $ if $ 1 le p le r < infty $. Also, $ l_w^p $ are spaces where we consider a collection of positive weights, $ w_n > 0 $ for each $ n in mathbb{N} $, that given sequence $ x = (x_n) $, the weighted $p$-norm is
$$ ||x||_{w,p} = big (sum_{n=1}^infty w_n |x_n|^p big )^frac{1}{p} text{.} $$
Prove that if $ sum_{n=1}^infty w_n < infty $, then $ l^1_w not subseteq l^2_w $.One way to prove this is to find a subsequence $ (w_{n_k}), $ so that $ w_{n_k} le frac{1}{2^k} $ and consider elements which are nonzero only at the $ n_k $ coordinates.
the subsequence is probably very easy to think up, but I can just not figure it out... :(
functional-analysis norm lp-spaces
functional-analysis norm lp-spaces
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
edited Dec 22 '18 at 10:01
Davide Giraudo
127k17154268
127k17154268
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
asked Dec 17 '18 at 3:34
Chase SariaslaniChase Sariaslani
805
805
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $(n_k)$ be the increasing sequence of integers mentioned in the opening post and define
$x_{n_k}=left(w_{n_k}k^2right)^{-1}$ and $x_l=0$ if $l$ is not equal to $n_k$ for some $k$. The series $sum_{k}w_{k}leftlvert x_{k}rightrvert$ is convergent and $sum_{k}w_{n_k}x_{n_k}^2$ is divergent since $w_{n_k}x_{n_k}^2geqslant 2^k/k^4$.
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Let $(n_k)$ be the increasing sequence of integers mentioned in the opening post and define
$x_{n_k}=left(w_{n_k}k^2right)^{-1}$ and $x_l=0$ if $l$ is not equal to $n_k$ for some $k$. The series $sum_{k}w_{k}leftlvert x_{k}rightrvert$ is convergent and $sum_{k}w_{n_k}x_{n_k}^2$ is divergent since $w_{n_k}x_{n_k}^2geqslant 2^k/k^4$.
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Let $(n_k)$ be the increasing sequence of integers mentioned in the opening post and define
$x_{n_k}=left(w_{n_k}k^2right)^{-1}$ and $x_l=0$ if $l$ is not equal to $n_k$ for some $k$. The series $sum_{k}w_{k}leftlvert x_{k}rightrvert$ is convergent and $sum_{k}w_{n_k}x_{n_k}^2$ is divergent since $w_{n_k}x_{n_k}^2geqslant 2^k/k^4$.
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
add a comment |
$begingroup$
Let $(n_k)$ be the increasing sequence of integers mentioned in the opening post and define
$x_{n_k}=left(w_{n_k}k^2right)^{-1}$ and $x_l=0$ if $l$ is not equal to $n_k$ for some $k$. The series $sum_{k}w_{k}leftlvert x_{k}rightrvert$ is convergent and $sum_{k}w_{n_k}x_{n_k}^2$ is divergent since $w_{n_k}x_{n_k}^2geqslant 2^k/k^4$.
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
$endgroup$
Let $(n_k)$ be the increasing sequence of integers mentioned in the opening post and define
$x_{n_k}=left(w_{n_k}k^2right)^{-1}$ and $x_l=0$ if $l$ is not equal to $n_k$ for some $k$. The series $sum_{k}w_{k}leftlvert x_{k}rightrvert$ is convergent and $sum_{k}w_{n_k}x_{n_k}^2$ is divergent since $w_{n_k}x_{n_k}^2geqslant 2^k/k^4$.
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
answered Dec 18 '18 at 17:46
Davide GiraudoDavide Giraudo
127k17154268
127k17154268
add a comment |
add a comment |
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