Proof that for a strongly continuous contraction resolvent, there is exactly one linear operator that...
$begingroup$
I have questions about the proof of the following Proposition from the book Introduction to the Theory of Non-Symmetric Dirichlet forms.
First, how do we get the independence of $G_alpha (B)$ of $alpha$ by 1.4(iii) and that each $G_alpha$ is one-to-one by 1.4(i) and 1.4(iii)?
Second, once we conclude that $alpha - G_alpha^{-1} = beta - G_beta^{-1}$, how do we get that $D(L)=G_alpha (B)$ is dense in $B$ by 1.4 (i)?
I would greatly appreciate any help.
functional-analysis analysis semigroup-of-operators
$endgroup$
add a comment |
$begingroup$
I have questions about the proof of the following Proposition from the book Introduction to the Theory of Non-Symmetric Dirichlet forms.
First, how do we get the independence of $G_alpha (B)$ of $alpha$ by 1.4(iii) and that each $G_alpha$ is one-to-one by 1.4(i) and 1.4(iii)?
Second, once we conclude that $alpha - G_alpha^{-1} = beta - G_beta^{-1}$, how do we get that $D(L)=G_alpha (B)$ is dense in $B$ by 1.4 (i)?
I would greatly appreciate any help.
functional-analysis analysis semigroup-of-operators
$endgroup$
add a comment |
$begingroup$
I have questions about the proof of the following Proposition from the book Introduction to the Theory of Non-Symmetric Dirichlet forms.
First, how do we get the independence of $G_alpha (B)$ of $alpha$ by 1.4(iii) and that each $G_alpha$ is one-to-one by 1.4(i) and 1.4(iii)?
Second, once we conclude that $alpha - G_alpha^{-1} = beta - G_beta^{-1}$, how do we get that $D(L)=G_alpha (B)$ is dense in $B$ by 1.4 (i)?
I would greatly appreciate any help.
functional-analysis analysis semigroup-of-operators
$endgroup$
I have questions about the proof of the following Proposition from the book Introduction to the Theory of Non-Symmetric Dirichlet forms.
First, how do we get the independence of $G_alpha (B)$ of $alpha$ by 1.4(iii) and that each $G_alpha$ is one-to-one by 1.4(i) and 1.4(iii)?
Second, once we conclude that $alpha - G_alpha^{-1} = beta - G_beta^{-1}$, how do we get that $D(L)=G_alpha (B)$ is dense in $B$ by 1.4 (i)?
I would greatly appreciate any help.
functional-analysis analysis semigroup-of-operators
functional-analysis analysis semigroup-of-operators
edited Feb 1 at 12:10
Pedro
10.9k23475
10.9k23475
asked Nov 15 '18 at 14:49
takecaretakecare
2,38321541
2,38321541
add a comment |
add a comment |
0
active
oldest
votes
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%2f2999794%2fproof-that-for-a-strongly-continuous-contraction-resolvent-there-is-exactly-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2999794%2fproof-that-for-a-strongly-continuous-contraction-resolvent-there-is-exactly-one%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