Gelfand transformation of $l^p$












2












$begingroup$


I would like to describe Gelfand transofrmation of commutative Banach algebra $l^p(mathbb{N}),p in [1,infty)$ with multiplication define by $(a_n)_n(b_n)_n=(a_n b_n)_n$, but I have no idea, how to do it. Any hints ? Thanks






lp-spaces banach-algebras gelfand-representation







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$endgroup$












  • $begingroup$
    It doesn't seem like $l^p$ is closed under your multiplication operation.
    $endgroup$
    – Daniel Schepler
    Dec 18 '18 at 0:05










  • $begingroup$
    @DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
    $endgroup$
    – lojdmoj
    Dec 30 '18 at 2:20


















2












$begingroup$


I would like to describe Gelfand transofrmation of commutative Banach algebra $l^p(mathbb{N}),p in [1,infty)$ with multiplication define by $(a_n)_n(b_n)_n=(a_n b_n)_n$, but I have no idea, how to do it. Any hints ? Thanks






lp-spaces banach-algebras gelfand-representation







share|cite|improve this question











$endgroup$












  • $begingroup$
    It doesn't seem like $l^p$ is closed under your multiplication operation.
    $endgroup$
    – Daniel Schepler
    Dec 18 '18 at 0:05










  • $begingroup$
    @DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
    $endgroup$
    – lojdmoj
    Dec 30 '18 at 2:20
















2












2








2





$begingroup$


I would like to describe Gelfand transofrmation of commutative Banach algebra $l^p(mathbb{N}),p in [1,infty)$ with multiplication define by $(a_n)_n(b_n)_n=(a_n b_n)_n$, but I have no idea, how to do it. Any hints ? Thanks






lp-spaces banach-algebras gelfand-representation







share|cite|improve this question











$endgroup$




I would like to describe Gelfand transofrmation of commutative Banach algebra $l^p(mathbb{N}),p in [1,infty)$ with multiplication define by $(a_n)_n(b_n)_n=(a_n b_n)_n$, but I have no idea, how to do it. Any hints ? Thanks






lp-spaces banach-algebras gelfand-representation




lp-spaces banach-algebras gelfand-representation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 0:36







lojdmoj

















asked Dec 17 '18 at 15:56









lojdmojlojdmoj

877




877












  • $begingroup$
    It doesn't seem like $l^p$ is closed under your multiplication operation.
    $endgroup$
    – Daniel Schepler
    Dec 18 '18 at 0:05










  • $begingroup$
    @DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
    $endgroup$
    – lojdmoj
    Dec 30 '18 at 2:20




















  • $begingroup$
    It doesn't seem like $l^p$ is closed under your multiplication operation.
    $endgroup$
    – Daniel Schepler
    Dec 18 '18 at 0:05










  • $begingroup$
    @DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
    $endgroup$
    – lojdmoj
    Dec 30 '18 at 2:20


















$begingroup$
It doesn't seem like $l^p$ is closed under your multiplication operation.
$endgroup$
– Daniel Schepler
Dec 18 '18 at 0:05




$begingroup$
It doesn't seem like $l^p$ is closed under your multiplication operation.
$endgroup$
– Daniel Schepler
Dec 18 '18 at 0:05












$begingroup$
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
$endgroup$
– lojdmoj
Dec 30 '18 at 2:20






$begingroup$
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
$endgroup$
– lojdmoj
Dec 30 '18 at 2:20












1 Answer
1






active

oldest

votes


















0












$begingroup$

The Gelfand transform in this case is nothing but the inclusion map $ell_pto c_0$ because the point-evaluations are the only non-zero characters on $ell_p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:09












  • $begingroup$
    the question is if there exist any other element of $Delta(l^p)$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:13










  • $begingroup$
    @lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
    $endgroup$
    – Tomek Kania
    Jan 4 at 20:29












  • $begingroup$
    Many thanks for your help :)
    $endgroup$
    – lojdmoj
    Jan 4 at 20:39






  • 1




    $begingroup$
    @lojdmoj, but $f$ is linear...
    $endgroup$
    – Tomek Kania
    Jan 4 at 22:35











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The Gelfand transform in this case is nothing but the inclusion map $ell_pto c_0$ because the point-evaluations are the only non-zero characters on $ell_p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:09












  • $begingroup$
    the question is if there exist any other element of $Delta(l^p)$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:13










  • $begingroup$
    @lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
    $endgroup$
    – Tomek Kania
    Jan 4 at 20:29












  • $begingroup$
    Many thanks for your help :)
    $endgroup$
    – lojdmoj
    Jan 4 at 20:39






  • 1




    $begingroup$
    @lojdmoj, but $f$ is linear...
    $endgroup$
    – Tomek Kania
    Jan 4 at 22:35
















0












$begingroup$

The Gelfand transform in this case is nothing but the inclusion map $ell_pto c_0$ because the point-evaluations are the only non-zero characters on $ell_p$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:09












  • $begingroup$
    the question is if there exist any other element of $Delta(l^p)$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:13










  • $begingroup$
    @lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
    $endgroup$
    – Tomek Kania
    Jan 4 at 20:29












  • $begingroup$
    Many thanks for your help :)
    $endgroup$
    – lojdmoj
    Jan 4 at 20:39






  • 1




    $begingroup$
    @lojdmoj, but $f$ is linear...
    $endgroup$
    – Tomek Kania
    Jan 4 at 22:35














0












0








0





$begingroup$

The Gelfand transform in this case is nothing but the inclusion map $ell_pto c_0$ because the point-evaluations are the only non-zero characters on $ell_p$.






share|cite|improve this answer









$endgroup$



The Gelfand transform in this case is nothing but the inclusion map $ell_pto c_0$ because the point-evaluations are the only non-zero characters on $ell_p$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 19:37









Tomek KaniaTomek Kania

12.2k11945




12.2k11945












  • $begingroup$
    we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:09












  • $begingroup$
    the question is if there exist any other element of $Delta(l^p)$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:13










  • $begingroup$
    @lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
    $endgroup$
    – Tomek Kania
    Jan 4 at 20:29












  • $begingroup$
    Many thanks for your help :)
    $endgroup$
    – lojdmoj
    Jan 4 at 20:39






  • 1




    $begingroup$
    @lojdmoj, but $f$ is linear...
    $endgroup$
    – Tomek Kania
    Jan 4 at 22:35


















  • $begingroup$
    we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:09












  • $begingroup$
    the question is if there exist any other element of $Delta(l^p)$
    $endgroup$
    – lojdmoj
    Jan 4 at 20:13










  • $begingroup$
    @lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
    $endgroup$
    – Tomek Kania
    Jan 4 at 20:29












  • $begingroup$
    Many thanks for your help :)
    $endgroup$
    – lojdmoj
    Jan 4 at 20:39






  • 1




    $begingroup$
    @lojdmoj, but $f$ is linear...
    $endgroup$
    – Tomek Kania
    Jan 4 at 22:35
















$begingroup$
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
$endgroup$
– lojdmoj
Jan 4 at 20:09






$begingroup$
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
$endgroup$
– lojdmoj
Jan 4 at 20:09














$begingroup$
the question is if there exist any other element of $Delta(l^p)$
$endgroup$
– lojdmoj
Jan 4 at 20:13




$begingroup$
the question is if there exist any other element of $Delta(l^p)$
$endgroup$
– lojdmoj
Jan 4 at 20:13












$begingroup$
@lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
$endgroup$
– Tomek Kania
Jan 4 at 20:29






$begingroup$
@lojdmoj, certainly not. Suppose that $f$ is a non-zero character. Then $f(e_n) = f(e_n)^2 in {0,1}$. If $f(e_n) = 1 = f(e_m)$ and $nneq m$, then $2 = f(e_n + e_m) = f(e_n + e_m)^2$ because $e_n+e_m$ is an idempotent; a contradiction. Thus $f$ is a point-evaluation.
$endgroup$
– Tomek Kania
Jan 4 at 20:29














$begingroup$
Many thanks for your help :)
$endgroup$
– lojdmoj
Jan 4 at 20:39




$begingroup$
Many thanks for your help :)
$endgroup$
– lojdmoj
Jan 4 at 20:39




1




1




$begingroup$
@lojdmoj, but $f$ is linear...
$endgroup$
– Tomek Kania
Jan 4 at 22:35




$begingroup$
@lojdmoj, but $f$ is linear...
$endgroup$
– Tomek Kania
Jan 4 at 22:35


















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