To prove an inequality regarding to the resolvent of a self-adjoint operator.
$begingroup$
I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).
Show that for a self-adjoint operator $A$ we have $Vert AR_A(z)Vert le frac{|z|}{|Im(z)|}, $ where $R_A(z) = (A-z)^{-1}$ is the resolvent.
It seems radical to me. I do not know where to start. I tried $AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$, but it did not proceed further. I guess the resolvent formula $R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$ will play the significant role in solving the problem.
Can someone give me a help? Any idea or hint would be aprreciated a lot.
functional-analysis hilbert-spaces spectral-theory self-adjoint-operators
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
$endgroup$
add a comment |
$begingroup$
I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).
Show that for a self-adjoint operator $A$ we have $Vert AR_A(z)Vert le frac{|z|}{|Im(z)|}, $ where $R_A(z) = (A-z)^{-1}$ is the resolvent.
It seems radical to me. I do not know where to start. I tried $AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$, but it did not proceed further. I guess the resolvent formula $R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$ will play the significant role in solving the problem.
Can someone give me a help? Any idea or hint would be aprreciated a lot.
functional-analysis hilbert-spaces spectral-theory self-adjoint-operators
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
$endgroup$
add a comment |
$begingroup$
I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).
Show that for a self-adjoint operator $A$ we have $Vert AR_A(z)Vert le frac{|z|}{|Im(z)|}, $ where $R_A(z) = (A-z)^{-1}$ is the resolvent.
It seems radical to me. I do not know where to start. I tried $AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$, but it did not proceed further. I guess the resolvent formula $R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$ will play the significant role in solving the problem.
Can someone give me a help? Any idea or hint would be aprreciated a lot.
functional-analysis hilbert-spaces spectral-theory self-adjoint-operators
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
$endgroup$
I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).
Show that for a self-adjoint operator $A$ we have $Vert AR_A(z)Vert le frac{|z|}{|Im(z)|}, $ where $R_A(z) = (A-z)^{-1}$ is the resolvent.
It seems radical to me. I do not know where to start. I tried $AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$, but it did not proceed further. I guess the resolvent formula $R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$ will play the significant role in solving the problem.
Can someone give me a help? Any idea or hint would be aprreciated a lot.
functional-analysis hilbert-spaces spectral-theory self-adjoint-operators
functional-analysis hilbert-spaces spectral-theory self-adjoint-operators
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
asked Dec 8 '18 at 8:22
EuduardoEuduardo
1638
1638
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So
$$
|AR_A(z)|=max{|lambda|: lambdain sigma(AR_A(z))}=maxleft{frac{|lambda|}{|lambda-z|}: lambdainsigma(A)right}.
$$
Since $A$ is selfadjoint, we know that $sigma(A)subsetmathbb R$. Write $z=a+ib$. For your formula to work we need $bne0$. Thus we want to maximize
$$
frac{|t|}{|t-a+ib|}=frac{|t|}{sqrt{(t-a)^2+b^2}}.
$$
Consider the square, we want to maximize the function
$$
tlongmapsto frac{t^2}{(t-a)^2+b^2}.
$$
Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So
$$
|AR_A(z)|leqsqrt{frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=sqrt{frac{a^2+b^2}{b^2}}=frac{|z|}{|operatorname{Im} z|}
$$
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
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%2f3030825%2fto-prove-an-inequality-regarding-to-the-resolvent-of-a-self-adjoint-operator%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
$begingroup$
I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So
$$
|AR_A(z)|=max{|lambda|: lambdain sigma(AR_A(z))}=maxleft{frac{|lambda|}{|lambda-z|}: lambdainsigma(A)right}.
$$
Since $A$ is selfadjoint, we know that $sigma(A)subsetmathbb R$. Write $z=a+ib$. For your formula to work we need $bne0$. Thus we want to maximize
$$
frac{|t|}{|t-a+ib|}=frac{|t|}{sqrt{(t-a)^2+b^2}}.
$$
Consider the square, we want to maximize the function
$$
tlongmapsto frac{t^2}{(t-a)^2+b^2}.
$$
Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So
$$
|AR_A(z)|leqsqrt{frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=sqrt{frac{a^2+b^2}{b^2}}=frac{|z|}{|operatorname{Im} z|}
$$
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
add a comment |
$begingroup$
I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So
$$
|AR_A(z)|=max{|lambda|: lambdain sigma(AR_A(z))}=maxleft{frac{|lambda|}{|lambda-z|}: lambdainsigma(A)right}.
$$
Since $A$ is selfadjoint, we know that $sigma(A)subsetmathbb R$. Write $z=a+ib$. For your formula to work we need $bne0$. Thus we want to maximize
$$
frac{|t|}{|t-a+ib|}=frac{|t|}{sqrt{(t-a)^2+b^2}}.
$$
Consider the square, we want to maximize the function
$$
tlongmapsto frac{t^2}{(t-a)^2+b^2}.
$$
Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So
$$
|AR_A(z)|leqsqrt{frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=sqrt{frac{a^2+b^2}{b^2}}=frac{|z|}{|operatorname{Im} z|}
$$
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
add a comment |
$begingroup$
I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So
$$
|AR_A(z)|=max{|lambda|: lambdain sigma(AR_A(z))}=maxleft{frac{|lambda|}{|lambda-z|}: lambdainsigma(A)right}.
$$
Since $A$ is selfadjoint, we know that $sigma(A)subsetmathbb R$. Write $z=a+ib$. For your formula to work we need $bne0$. Thus we want to maximize
$$
frac{|t|}{|t-a+ib|}=frac{|t|}{sqrt{(t-a)^2+b^2}}.
$$
Consider the square, we want to maximize the function
$$
tlongmapsto frac{t^2}{(t-a)^2+b^2}.
$$
Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So
$$
|AR_A(z)|leqsqrt{frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=sqrt{frac{a^2+b^2}{b^2}}=frac{|z|}{|operatorname{Im} z|}
$$
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
$endgroup$
I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $AR_A(z)$ is normal. So
$$
|AR_A(z)|=max{|lambda|: lambdain sigma(AR_A(z))}=maxleft{frac{|lambda|}{|lambda-z|}: lambdainsigma(A)right}.
$$
Since $A$ is selfadjoint, we know that $sigma(A)subsetmathbb R$. Write $z=a+ib$. For your formula to work we need $bne0$. Thus we want to maximize
$$
frac{|t|}{|t-a+ib|}=frac{|t|}{sqrt{(t-a)^2+b^2}}.
$$
Consider the square, we want to maximize the function
$$
tlongmapsto frac{t^2}{(t-a)^2+b^2}.
$$
Differentiating and equating to zero gives you that the above function has critical points at $t=0$ and $t=(a^2+b^2)/a$. So
$$
|AR_A(z)|leqsqrt{frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=sqrt{frac{a^2+b^2}{b^2}}=frac{|z|}{|operatorname{Im} z|}
$$
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
answered Dec 8 '18 at 19:23
Martin ArgeramiMartin Argerami
127k1182181
127k1182181
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
add a comment |
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
$begingroup$
This is insane. Surprised by your insight and computation. Thanks Argerami.
$endgroup$
– Euduardo
Dec 9 '18 at 5:43
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%2f3030825%2fto-prove-an-inequality-regarding-to-the-resolvent-of-a-self-adjoint-operator%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
