the uniform norm of a positive linear map on von Neumann algebras

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

2

Let $H$ be a Hilbert space. Let $phi:B(H)to B(H)$ be a positive linear map.

Q. Do we have $|phi|=sup : 0leq xleq1$?

functional-analysis operator-algebras c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Aug 3 at 16:22

Martin Argerami

115k1071164

asked Aug 3 at 6:33

Ali Bagheri

765412

add a comment | 

up vote
1
down vote

favorite

2

Let $H$ be a Hilbert space. Let $phi:B(H)to B(H)$ be a positive linear map.

Q. Do we have $|phi|=sup : 0leq xleq1$?

functional-analysis operator-algebras c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Aug 3 at 16:22

Martin Argerami

115k1071164

asked Aug 3 at 6:33

Ali Bagheri

765412

add a comment | 

up vote
1
down vote

favorite

2

up vote
1
down vote

favorite

2

2

Let $H$ be a Hilbert space. Let $phi:B(H)to B(H)$ be a positive linear map.

Q. Do we have $|phi|=sup : 0leq xleq1$?

functional-analysis operator-algebras c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Aug 3 at 16:22

Martin Argerami

115k1071164

asked Aug 3 at 6:33

Ali Bagheri

765412

Let $H$ be a Hilbert space. Let $phi:B(H)to B(H)$ be a positive linear map.

Q. Do we have $|phi|=sup : 0leq xleq1$?

functional-analysis operator-algebras c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Aug 3 at 16:22

Martin Argerami

115k1071164

asked Aug 3 at 6:33

Ali Bagheri

765412

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 16:22

Martin Argerami

115k1071164

edited Aug 3 at 16:22

Martin Argerami

115k1071164

edited Aug 3 at 16:22

Martin Argerami

115k1071164

115k1071164

asked Aug 3 at 6:33

Ali Bagheri

765412

asked Aug 3 at 6:33

Ali Bagheri

765412

asked Aug 3 at 6:33

Ali Bagheri

765412

765412

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

Yes. More, in fact.

If $A,B$ are unital C$^*$-algebras, and $phi:Ato B$ is positive, then $$ |phi|=|phi(I_A)|.$$

Proof. Fix $ain A$, with $|a|leq1$. Assume $Asubset B(H)$. Consider the operator system $Ssubset C(mathbb T)$ given by $$ S=p(e^it)+overlineq(e^it): p,qin mathbb C[z].$$ Define $psi:Sto B(H)$ by $psi(p+bar q)=p(a)+q(a)^*$. Then $psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $psi$ is bounded and $S$ is dense, we can extend $psi$ to all of $C(mathbb T)$, and it will still be unital and positive.

Now, as $a=psi(e^it)$
$$
|phi(a)|=|phicircpsi(e^it)|.
$$
The map $phicircpsi:C(mathbb T)to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $|phicircpsi|=|phicircpsi(I)|=|phi(I)|$, and
$$
|phi(a)|=|phicircpsi(e^it)|leq |phi(I_A)|,|e^it|=|phi(I_A)|.
$$
Thus $|phi(a)|leq|phi(I_A)|$, and it follows that $|phi|=|phi(I_A)|$.

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870806%2fthe-uniform-norm-of-a-positive-linear-map-on-von-neumann-algebras%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Yes. More, in fact.

If $A,B$ are unital C$^*$-algebras, and $phi:Ato B$ is positive, then $$ |phi|=|phi(I_A)|.$$

Proof. Fix $ain A$, with $|a|leq1$. Assume $Asubset B(H)$. Consider the operator system $Ssubset C(mathbb T)$ given by $$ S=p(e^it)+overlineq(e^it): p,qin mathbb C[z].$$ Define $psi:Sto B(H)$ by $psi(p+bar q)=p(a)+q(a)^*$. Then $psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $psi$ is bounded and $S$ is dense, we can extend $psi$ to all of $C(mathbb T)$, and it will still be unital and positive.

Now, as $a=psi(e^it)$
$$
|phi(a)|=|phicircpsi(e^it)|.
$$
The map $phicircpsi:C(mathbb T)to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $|phicircpsi|=|phicircpsi(I)|=|phi(I)|$, and
$$
|phi(a)|=|phicircpsi(e^it)|leq |phi(I_A)|,|e^it|=|phi(I_A)|.
$$
Thus $|phi(a)|leq|phi(I_A)|$, and it follows that $|phi|=|phi(I_A)|$.

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

add a comment | 

up vote
1
down vote

Yes. More, in fact.

If $A,B$ are unital C$^*$-algebras, and $phi:Ato B$ is positive, then $$ |phi|=|phi(I_A)|.$$

Proof. Fix $ain A$, with $|a|leq1$. Assume $Asubset B(H)$. Consider the operator system $Ssubset C(mathbb T)$ given by $$ S=p(e^it)+overlineq(e^it): p,qin mathbb C[z].$$ Define $psi:Sto B(H)$ by $psi(p+bar q)=p(a)+q(a)^*$. Then $psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $psi$ is bounded and $S$ is dense, we can extend $psi$ to all of $C(mathbb T)$, and it will still be unital and positive.

Now, as $a=psi(e^it)$
$$
|phi(a)|=|phicircpsi(e^it)|.
$$
The map $phicircpsi:C(mathbb T)to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $|phicircpsi|=|phicircpsi(I)|=|phi(I)|$, and
$$
|phi(a)|=|phicircpsi(e^it)|leq |phi(I_A)|,|e^it|=|phi(I_A)|.
$$
Thus $|phi(a)|leq|phi(I_A)|$, and it follows that $|phi|=|phi(I_A)|$.

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

add a comment | 

up vote
1
down vote

up vote
1
down vote

Yes. More, in fact.

If $A,B$ are unital C$^*$-algebras, and $phi:Ato B$ is positive, then $$ |phi|=|phi(I_A)|.$$

Proof. Fix $ain A$, with $|a|leq1$. Assume $Asubset B(H)$. Consider the operator system $Ssubset C(mathbb T)$ given by $$ S=p(e^it)+overlineq(e^it): p,qin mathbb C[z].$$ Define $psi:Sto B(H)$ by $psi(p+bar q)=p(a)+q(a)^*$. Then $psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $psi$ is bounded and $S$ is dense, we can extend $psi$ to all of $C(mathbb T)$, and it will still be unital and positive.

Now, as $a=psi(e^it)$
$$
|phi(a)|=|phicircpsi(e^it)|.
$$
The map $phicircpsi:C(mathbb T)to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $|phicircpsi|=|phicircpsi(I)|=|phi(I)|$, and
$$
|phi(a)|=|phicircpsi(e^it)|leq |phi(I_A)|,|e^it|=|phi(I_A)|.
$$
Thus $|phi(a)|leq|phi(I_A)|$, and it follows that $|phi|=|phi(I_A)|$.

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

Yes. More, in fact.

If $A,B$ are unital C$^*$-algebras, and $phi:Ato B$ is positive, then $$ |phi|=|phi(I_A)|.$$

Proof. Fix $ain A$, with $|a|leq1$. Assume $Asubset B(H)$. Consider the operator system $Ssubset C(mathbb T)$ given by $$ S=p(e^it)+overlineq(e^it): p,qin mathbb C[z].$$ Define $psi:Sto B(H)$ by $psi(p+bar q)=p(a)+q(a)^*$. Then $psi$ is unital positive (this requires a few computations, it is not immediately obvious). Because $psi$ is bounded and $S$ is dense, we can extend $psi$ to all of $C(mathbb T)$, and it will still be unital and positive.

Now, as $a=psi(e^it)$
$$
|phi(a)|=|phicircpsi(e^it)|.
$$
The map $phicircpsi:C(mathbb T)to B(H)$ is unital and positive; but as the domain is a commutative C$^*$-algebra, it is completely positive. Thus $|phicircpsi|=|phicircpsi(I)|=|phi(I)|$, and
$$
|phi(a)|=|phicircpsi(e^it)|leq |phi(I_A)|,|e^it|=|phi(I_A)|.
$$
Thus $|phi(a)|leq|phi(I_A)|$, and it follows that $|phi|=|phi(I_A)|$.

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 17:11

Martin Argerami

115k1071164

answered Aug 3 at 17:11

Martin Argerami

115k1071164

answered Aug 3 at 17:11

Martin Argerami

115k1071164

115k1071164

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870806%2fthe-uniform-norm-of-a-positive-linear-map-on-von-neumann-algebras%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email