Approximate identity in $ell _p$
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Show that for $1leq p<infty$, $ell _p$ with multiplication defined by $(a_n)_n(b_n)_n=(a_nb_n)_n$has an unbounded approximate identity but no bounded approximate identity, I don't know how to start the proof
Any help will be greatly appreciated
functional-analysis banach-algebras
asked Nov 17 at 10:40
user62498
1,889613
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up vote
0
down vote
favorite
Show that for $1leq p<infty$, $ell _p$ with multiplication defined by $(a_n)_n(b_n)_n=(a_nb_n)_n$has an unbounded approximate identity but no bounded approximate identity, I don't know how to start the proof
Any help will be greatly appreciated
functional-analysis banach-algebras
asked Nov 17 at 10:40
user62498
1,889613
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that for $1leq p<infty$, $ell _p$ with multiplication defined by $(a_n)_n(b_n)_n=(a_nb_n)_n$has an unbounded approximate identity but no bounded approximate identity, I don't know how to start the proof
Any help will be greatly appreciated
functional-analysis banach-algebras
asked Nov 17 at 10:40
user62498
1,889613
Show that for $1leq p<infty$, $ell _p$ with multiplication defined by $(a_n)_n(b_n)_n=(a_nb_n)_n$has an unbounded approximate identity but no bounded approximate identity, I don't know how to start the proof
Any help will be greatly appreciated
functional-analysis banach-algebras
functional-analysis banach-algebras
asked Nov 17 at 10:40
user62498
1,889613
asked Nov 17 at 10:40
user62498
1,889613
asked Nov 17 at 10:40
user62498
1,889613
asked Nov 17 at 10:40
user62498
1,889613
asked Nov 17 at 10:40
user62498
1,889613
1,889613
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
1
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The identity for $mathbb{C}$ is $1$, thus the identity for $ell_p$ with such multiplication should be $(1)_n$ (constant sequence) but this sequence is in $ell_p$ if and only if $p=infty$, which is not the case.
On the other hand, such sequence can be approximated by $(x^k)_ksubseteqell_p$, where $x^k = (x^k_n)_n$ is given by
$$x^k_n=begin{cases} 1, & text{ if } nleq k \ 0, & text{ if } n>k,
end{cases}
$$
in the sense that $limlimits_{ktoinfty}|x^ka-a|_p=0$ for $ainell_p$.
answered Nov 17 at 11:17
rldias
2,9301522
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The identity for $mathbb{C}$ is $1$, thus the identity for $ell_p$ with such multiplication should be $(1)_n$ (constant sequence) but this sequence is in $ell_p$ if and only if $p=infty$, which is not the case.
On the other hand, such sequence can be approximated by $(x^k)_ksubseteqell_p$, where $x^k = (x^k_n)_n$ is given by
$$x^k_n=begin{cases} 1, & text{ if } nleq k \ 0, & text{ if } n>k,
end{cases}
$$
in the sense that $limlimits_{ktoinfty}|x^ka-a|_p=0$ for $ainell_p$.
answered Nov 17 at 11:17
rldias
2,9301522
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
add a comment |
up vote
1
down vote
accepted
The identity for $mathbb{C}$ is $1$, thus the identity for $ell_p$ with such multiplication should be $(1)_n$ (constant sequence) but this sequence is in $ell_p$ if and only if $p=infty$, which is not the case.
On the other hand, such sequence can be approximated by $(x^k)_ksubseteqell_p$, where $x^k = (x^k_n)_n$ is given by
$$x^k_n=begin{cases} 1, & text{ if } nleq k \ 0, & text{ if } n>k,
end{cases}
$$
in the sense that $limlimits_{ktoinfty}|x^ka-a|_p=0$ for $ainell_p$.
answered Nov 17 at 11:17
rldias
2,9301522
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The identity for $mathbb{C}$ is $1$, thus the identity for $ell_p$ with such multiplication should be $(1)_n$ (constant sequence) but this sequence is in $ell_p$ if and only if $p=infty$, which is not the case.
On the other hand, such sequence can be approximated by $(x^k)_ksubseteqell_p$, where $x^k = (x^k_n)_n$ is given by
$$x^k_n=begin{cases} 1, & text{ if } nleq k \ 0, & text{ if } n>k,
end{cases}
$$
in the sense that $limlimits_{ktoinfty}|x^ka-a|_p=0$ for $ainell_p$.
answered Nov 17 at 11:17
rldias
2,9301522
The identity for $mathbb{C}$ is $1$, thus the identity for $ell_p$ with such multiplication should be $(1)_n$ (constant sequence) but this sequence is in $ell_p$ if and only if $p=infty$, which is not the case.
On the other hand, such sequence can be approximated by $(x^k)_ksubseteqell_p$, where $x^k = (x^k_n)_n$ is given by
$$x^k_n=begin{cases} 1, & text{ if } nleq k \ 0, & text{ if } n>k,
end{cases}
$$
in the sense that $limlimits_{ktoinfty}|x^ka-a|_p=0$ for $ainell_p$.
answered Nov 17 at 11:17
rldias
2,9301522
edited Nov 18 at 22:29
answered Nov 17 at 11:17
rldias
2,9301522
answered Nov 17 at 11:17
rldias
2,9301522
answered Nov 17 at 11:17
rldias
2,9301522
2,9301522
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
add a comment |
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
@Dear rldias, Thank you very much. This idea is very helpful to me. I am reading your answer. Thanks again
– user62498
Nov 17 at 12:13
add a comment |
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