Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$
$begingroup$
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
$endgroup$
add a comment |
$begingroup$
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
$endgroup$
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
1
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
$begingroup$
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
$endgroup$
Let $H$ be a Hilbert space and let $T:Hto H$ be a bounded self-adjoint linear operator.
Show that there exists $x in H$ with $|x|=1$ and $|langle Tx,xrangle |=|T|$.
I know that $|T|=sup{|langle Tx,xrangle| : |x|=1}$. I think the completeness can produce such $x$, but I don't know how to prove this.
linear-algebra general-topology hilbert-spaces
linear-algebra general-topology hilbert-spaces
edited Nov 2 '17 at 14:10
amWhy
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
edited Nov 2 '17 at 14:10
amWhy
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
edited Nov 2 '17 at 14:10
amWhy
1
edited Nov 2 '17 at 14:10
amWhy
1
edited Nov 2 '17 at 14:10
amWhy
1
1
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
asked Nov 2 '17 at 8:24
NYRAHHHNYRAHHH
35629
35629
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
1
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
1
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
1
1
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
$endgroup$
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2501010%2fshow-that-there-exists-x-in-h-with-x-1-and-langle-tx-x-rangle-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
$endgroup$
add a comment |
$begingroup$
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
$endgroup$
add a comment |
$begingroup$
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
$endgroup$
This is not true. Consider for instance $Tin B(ell^2(mathbb N))$ given by
$$
Tx=(x_1/2, 2x_2/3, 3x_3/4,ldots).
$$
As $Te_n=tfrac{n}{n+1},e_n$, we have that $|T|=1$. But for any nonzero $x$ with $|x|=1$ we have
$$
|langle Tx,xrangle|=sum_ntfrac{n}{n+1},|x_n|^2<sum_n|x_n|^2=1.
$$
So no such maximum exists.
When $T$ is compact, though, the answer is affirmative. For a selfadjoint compact $T$, we have via the Spectral Theorem that
$$
T=sum_nlambda_nP_n,
$$
where $lambda_ninmathbb R$ for all $n$, and $lambda_nsearrow0$. In this case $|T|=max_n|lambda_n|$. If we take $|lambda_j|=|T|$, then put $x$ with $P_jx=x$ and $|x|=1$, and
$$
|langle Tx,xrangle|=|lambda_j|=|T|.
$$
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
answered Dec 4 '18 at 1:57
Martin ArgeramiMartin Argerami
125k1179178
125k1179178
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2501010%2fshow-that-there-exists-x-in-h-with-x-1-and-langle-tx-x-rangle-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Can't you just consider a sequence of $x$ values so that $langle Tx, xrangle$ tends to $Vert TVert$, and then by continuity of the inner product you can just move the limit inside the inner product, and now the limit of $x$ is in $H$?
$endgroup$
– Harambe
Nov 2 '17 at 8:31
$begingroup$
I suppose you would need that the set ${|langle Tx,xrangle| : |x|=1}$ is compact, but this isn't so easy to prove for general self-adjoint operators. Are you sure your exercise isn't talking about a compact self-adjoint operator? That would make it a lot easier.
$endgroup$
– Demophilus
Nov 2 '17 at 13:22
$begingroup$
Isn't it just false? Consider, for example, diagonal operator in l2, with eigenvalues: (1-1/n). It is self—adjoint, norm is 1, and clearly no vector realises unity.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 16:52
$begingroup$
For compact self—adjoint operator it is true, if you look how its spectrum looks like. Ofcourse, in my previous example operator is not compact.
$endgroup$
– Fedor Goncharov
Nov 2 '17 at 17:03
1
$begingroup$
This question should not be closed as a duplicate, the linked question does not contain an answer whether there exists $x in H$ such that $|x| = 1$ and $|langle Tx,xrangle| = |T|$. It only shows this: $$|T|=sup_{|x|=1}|langle x,Txrangle|$$
$endgroup$
– mechanodroid
Nov 4 '17 at 10:53