Proving difficulty in MASA [closed]
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Prove that $ L^infty{(mathbb{R},mu)}$ is a masa in $B(L^2(mathbb{R},mu))$, $mu$ is sigma finite measure in particular Lebesgue, and what is the cyclic vector of it?
operator-algebras von-neumann-algebras
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
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closed as off-topic by Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin Dec 25 '18 at 11:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that $ L^infty{(mathbb{R},mu)}$ is a masa in $B(L^2(mathbb{R},mu))$, $mu$ is sigma finite measure in particular Lebesgue, and what is the cyclic vector of it?
operator-algebras von-neumann-algebras
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
$endgroup$
closed as off-topic by Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin Dec 25 '18 at 11:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that $ L^infty{(mathbb{R},mu)}$ is a masa in $B(L^2(mathbb{R},mu))$, $mu$ is sigma finite measure in particular Lebesgue, and what is the cyclic vector of it?
operator-algebras von-neumann-algebras
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
$endgroup$
Prove that $ L^infty{(mathbb{R},mu)}$ is a masa in $B(L^2(mathbb{R},mu))$, $mu$ is sigma finite measure in particular Lebesgue, and what is the cyclic vector of it?
operator-algebras von-neumann-algebras
operator-algebras von-neumann-algebras
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
asked Dec 25 '18 at 5:01
mathlovermathlover
150110
150110
closed as off-topic by Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin Dec 25 '18 at 11:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin Dec 25 '18 at 11:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, Lee David Chung Lin, Math_QED, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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Any function that is essentially nonzero is both cyclic and separating. In particular this proves that $L^infty(mathbb R,mu)$ is a masa.
For instance, let
$$
f(t)=sum_{ninmathbb Z} tfrac1{|n|},1_{[n,n+1]}.
$$
Then, for any $gin L^2$ bounded with compact support contained in $[-m,m]$, let
$$
h=sum_{n=-m}^m|n|,g,1_{[n,n+1]}in L^infty;
$$
then $hf=g$. As bounded functions of compact support are dense in $L^2$, $f$ is cyclic. And if $hin L^infty$ and $hf=0$, then $h=0$ a.e., so $f$ is separating.
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any function that is essentially nonzero is both cyclic and separating. In particular this proves that $L^infty(mathbb R,mu)$ is a masa.
For instance, let
$$
f(t)=sum_{ninmathbb Z} tfrac1{|n|},1_{[n,n+1]}.
$$
Then, for any $gin L^2$ bounded with compact support contained in $[-m,m]$, let
$$
h=sum_{n=-m}^m|n|,g,1_{[n,n+1]}in L^infty;
$$
then $hf=g$. As bounded functions of compact support are dense in $L^2$, $f$ is cyclic. And if $hin L^infty$ and $hf=0$, then $h=0$ a.e., so $f$ is separating.
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
add a comment |
$begingroup$
Any function that is essentially nonzero is both cyclic and separating. In particular this proves that $L^infty(mathbb R,mu)$ is a masa.
For instance, let
$$
f(t)=sum_{ninmathbb Z} tfrac1{|n|},1_{[n,n+1]}.
$$
Then, for any $gin L^2$ bounded with compact support contained in $[-m,m]$, let
$$
h=sum_{n=-m}^m|n|,g,1_{[n,n+1]}in L^infty;
$$
then $hf=g$. As bounded functions of compact support are dense in $L^2$, $f$ is cyclic. And if $hin L^infty$ and $hf=0$, then $h=0$ a.e., so $f$ is separating.
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
add a comment |
$begingroup$
Any function that is essentially nonzero is both cyclic and separating. In particular this proves that $L^infty(mathbb R,mu)$ is a masa.
For instance, let
$$
f(t)=sum_{ninmathbb Z} tfrac1{|n|},1_{[n,n+1]}.
$$
Then, for any $gin L^2$ bounded with compact support contained in $[-m,m]$, let
$$
h=sum_{n=-m}^m|n|,g,1_{[n,n+1]}in L^infty;
$$
then $hf=g$. As bounded functions of compact support are dense in $L^2$, $f$ is cyclic. And if $hin L^infty$ and $hf=0$, then $h=0$ a.e., so $f$ is separating.
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
Any function that is essentially nonzero is both cyclic and separating. In particular this proves that $L^infty(mathbb R,mu)$ is a masa.
For instance, let
$$
f(t)=sum_{ninmathbb Z} tfrac1{|n|},1_{[n,n+1]}.
$$
Then, for any $gin L^2$ bounded with compact support contained in $[-m,m]$, let
$$
h=sum_{n=-m}^m|n|,g,1_{[n,n+1]}in L^infty;
$$
then $hf=g$. As bounded functions of compact support are dense in $L^2$, $f$ is cyclic. And if $hin L^infty$ and $hf=0$, then $h=0$ a.e., so $f$ is separating.
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
answered Dec 25 '18 at 5:23
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
add a comment |
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
$begingroup$
i know what nasa is. what's masa
$endgroup$
– mathworker21
Dec 25 '18 at 5:28
1
1
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Maximal abelian sub-algebra. It's a completely standard term in von Neumann algebras, and likely also in C$^*$-algebras.
$endgroup$
– Martin Argerami
Dec 25 '18 at 5:30
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
$begingroup$
Sir, if I decompose the $mathbb{R}$ into disjoint unions of $E_{n}$ such that $mu_{E_n}$ is finite naming $mu_{n}=text{ restriction of }mu text{ on }E_{n}$, can I tell that the algebra $L^infty{(mathbb{R},mu)}$ is direct sum of masas hense masa?
$endgroup$
– mathlover
Dec 25 '18 at 6:29
1
1
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
$begingroup$
Yes, that should work if you prove that a direct sum of masas is a masa.
$endgroup$
– Martin Argerami
Dec 25 '18 at 6:33
add a comment |
