Gelfand transformation of $l^p$
2
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
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
add a comment |
2
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
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20
add a comment |
2
2
2
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
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
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
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
edited Dec 20 '18 at 0:36
lojdmoj
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
asked Dec 17 '18 at 15:56
lojdmojlojdmoj
877
877
It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20
add a comment |
It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20
It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20
add a comment |
1 Answer
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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 $ell_p$.
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
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 $ell_p$.
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
|
show 2 more comments
0
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 $ell_p$.
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
|
show 2 more comments
0
0
0
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 $ell_p$.
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
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 $ell_p$.
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
answered Jan 4 at 19:37
Tomek KaniaTomek Kania
12.1k11943
12.1k11943
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
|
show 2 more comments
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
we know, that $Delta (l^p)={e_n:nin mathbb{N}}$ so then GT of $e_n$ is projection to n-th coordinate of $xin l^p$ ...$widehat{x}(n)=x_n$
– lojdmoj
Jan 4 at 20:09
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
the question is if there exist any other element of $Delta(l^p)$
– lojdmoj
Jan 4 at 20:13
@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.
– Tomek Kania
Jan 4 at 20:29
@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.
– Tomek Kania
Jan 4 at 20:29
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
Many thanks for your help :)
– lojdmoj
Jan 4 at 20:39
1
1
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
@lojdmoj, but $f$ is linear...
– Tomek Kania
Jan 4 at 22:35
|
show 2 more comments
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It doesn't seem like $l^p$ is closed under your multiplication operation.
– Daniel Schepler
Dec 18 '18 at 0:05
@DanielSchepler It is closed - we consider space $l^p$, not $L^p$.
– lojdmoj
Dec 30 '18 at 2:20