Mixing
$underset{_{arightarrow 0}}{lim }L_{a}^{ast }left( L_{a}L_{a}^{ast }right) ^{-1}L_{a}=?$
$begingroup$
Let $varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{lambda
}=L-lambda I$ is surjective of every $lambda $ such that $varepsilon
>leftvert lambda rightvert >0$.
I'm trying, by applying the Cauchy criterion, to show that the limit
$underset{_{lambda rightarrow 0}}{lim }L_{lambda }^{ast }left(
L_{lambda }L_{lambda }^{ast }right) ^{-1}L_{lambda }$
exists, where $L_{lambda }^{ast }=L^{ast }-overline{lambda }I.$ If this
is not true, can you give me some counter-example ?
Thank you !
functional-analysis limits operator-theory hilbert-spaces adjoint-operators
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
$endgroup$
add a comment |
$begingroup$
Let $varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{lambda
}=L-lambda I$ is surjective of every $lambda $ such that $varepsilon
>leftvert lambda rightvert >0$.
I'm trying, by applying the Cauchy criterion, to show that the limit
$underset{_{lambda rightarrow 0}}{lim }L_{lambda }^{ast }left(
L_{lambda }L_{lambda }^{ast }right) ^{-1}L_{lambda }$
exists, where $L_{lambda }^{ast }=L^{ast }-overline{lambda }I.$ If this
is not true, can you give me some counter-example ?
Thank you !
functional-analysis limits operator-theory hilbert-spaces adjoint-operators
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
$endgroup$
add a comment |
$begingroup$
Let $varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{lambda
}=L-lambda I$ is surjective of every $lambda $ such that $varepsilon
>leftvert lambda rightvert >0$.
I'm trying, by applying the Cauchy criterion, to show that the limit
$underset{_{lambda rightarrow 0}}{lim }L_{lambda }^{ast }left(
L_{lambda }L_{lambda }^{ast }right) ^{-1}L_{lambda }$
exists, where $L_{lambda }^{ast }=L^{ast }-overline{lambda }I.$ If this
is not true, can you give me some counter-example ?
Thank you !
functional-analysis limits operator-theory hilbert-spaces adjoint-operators
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
$endgroup$
Let $varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{lambda
}=L-lambda I$ is surjective of every $lambda $ such that $varepsilon
>leftvert lambda rightvert >0$.
I'm trying, by applying the Cauchy criterion, to show that the limit
$underset{_{lambda rightarrow 0}}{lim }L_{lambda }^{ast }left(
L_{lambda }L_{lambda }^{ast }right) ^{-1}L_{lambda }$
exists, where $L_{lambda }^{ast }=L^{ast }-overline{lambda }I.$ If this
is not true, can you give me some counter-example ?
Thank you !
functional-analysis limits operator-theory hilbert-spaces adjoint-operators
functional-analysis limits operator-theory hilbert-spaces adjoint-operators
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
asked Jan 28 at 12:17
Djalal OunadjelaDjalal Ounadjela
29818
29818
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Setting $L=0$ shows $L_lambda = -lambda I$, and the operator $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda =I$ holds for finite-dimensional $H$ or normal $L$.
Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^perpto R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(mu)$.
Then
$$
L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda
= VT_{bar f-barlambda}U^* (UT_{f-lambda}V^* VT_{bar f-barlambda}U^*)^{-1}UT_{f-lambda}V^*
=VT_{bar f-barlambda}U^* (UT_{|f|^2+|lambda|^2}U^*)^{-1}UT_{f-lambda}V^*.
$$
If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.
I do not know how to show this for the general case. Maybe someone else has an idea?
answered Jan 31 at 7:58
dawdaw
24.8k1745
$endgroup$
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
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%2f3090783%2funderset-a-rightarrow-0-lim-l-a-ast-left-l-al-a-ast-right%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
$begingroup$
Setting $L=0$ shows $L_lambda = -lambda I$, and the operator $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda =I$ holds for finite-dimensional $H$ or normal $L$.
Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^perpto R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(mu)$.
Then
$$
L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda
= VT_{bar f-barlambda}U^* (UT_{f-lambda}V^* VT_{bar f-barlambda}U^*)^{-1}UT_{f-lambda}V^*
=VT_{bar f-barlambda}U^* (UT_{|f|^2+|lambda|^2}U^*)^{-1}UT_{f-lambda}V^*.
$$
If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.
I do not know how to show this for the general case. Maybe someone else has an idea?
answered Jan 31 at 7:58
dawdaw
24.8k1745
$endgroup$
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
add a comment |
$begingroup$
Setting $L=0$ shows $L_lambda = -lambda I$, and the operator $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda =I$ holds for finite-dimensional $H$ or normal $L$.
Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^perpto R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(mu)$.
Then
$$
L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda
= VT_{bar f-barlambda}U^* (UT_{f-lambda}V^* VT_{bar f-barlambda}U^*)^{-1}UT_{f-lambda}V^*
=VT_{bar f-barlambda}U^* (UT_{|f|^2+|lambda|^2}U^*)^{-1}UT_{f-lambda}V^*.
$$
If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.
I do not know how to show this for the general case. Maybe someone else has an idea?
answered Jan 31 at 7:58
dawdaw
24.8k1745
$endgroup$
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
add a comment |
$begingroup$
Setting $L=0$ shows $L_lambda = -lambda I$, and the operator $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda =I$ holds for finite-dimensional $H$ or normal $L$.
Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^perpto R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(mu)$.
Then
$$
L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda
= VT_{bar f-barlambda}U^* (UT_{f-lambda}V^* VT_{bar f-barlambda}U^*)^{-1}UT_{f-lambda}V^*
=VT_{bar f-barlambda}U^* (UT_{|f|^2+|lambda|^2}U^*)^{-1}UT_{f-lambda}V^*.
$$
If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.
I do not know how to show this for the general case. Maybe someone else has an idea?
answered Jan 31 at 7:58
dawdaw
24.8k1745
$endgroup$
Setting $L=0$ shows $L_lambda = -lambda I$, and the operator $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.
It is quite easy to show that $L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda =I$ holds for finite-dimensional $H$ or normal $L$.
Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^perpto R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(mu)$.
Then
$$
L_lambda^*(L_lambda L^*_lambda)^{-1}L_lambda
= VT_{bar f-barlambda}U^* (UT_{f-lambda}V^* VT_{bar f-barlambda}U^*)^{-1}UT_{f-lambda}V^*
=VT_{bar f-barlambda}U^* (UT_{|f|^2+|lambda|^2}U^*)^{-1}UT_{f-lambda}V^*.
$$
If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.
I do not know how to show this for the general case. Maybe someone else has an idea?
answered Jan 31 at 7:58
dawdaw
24.8k1745
answered Jan 31 at 7:58
dawdaw
24.8k1745
answered Jan 31 at 7:58
dawdaw
24.8k1745
answered Jan 31 at 7:58
dawdaw
24.8k1745
24.8k1745
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
add a comment |
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
$begingroup$
Thank you so much @daw
$endgroup$
– Djalal Ounadjela
Jan 31 at 23:59
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%2f3090783%2funderset-a-rightarrow-0-lim-l-a-ast-left-l-al-a-ast-right%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
