About closed graph of an unbounded operator
$begingroup$
I am working on problems related to the closed graph of an unbounded operator. There is a proposition:
Let $X,Y$ be Banach spaces and let $A:mathrm{dom}(A)to Y$ be linear and defined on a linear subspace $mathrm{dom}(A)subset X$. Prove that the graph of $A$ is a closed subspace of $Xtimes Y$ if and only if $mathrm{dom}(A)$ is Banach with respect to the graph norm.
I finished one direction. Suppose $mathrm{graph}(A)$ is closed. We take any Cauchy sequence $x_n$ in $mathrm{dom}(A)$, and since the norm is graph norm, we know $x_n$ and $Ax_n$ will both be Cauchy. Then we have a Cauchy sequence $(x_n,Ax_n)$ in the graph, so the pair converges to a certain $(x_0,y_0)$ since the graph is closed. Therefore $x_0inmathrm{dom}(A)$, which means that $mathrm{dom}(A)$ is Banach.
However I encountered some trouble on the other direction. Suppose $operatorname{dom}(A)$ is Banach with respect to the graph norm. If we take a Cauchy sequence $(x_n,Ax_n)$ in $mathrm{graph}(A)$, since $X,Y$ are both Banach, it converges to a pair $(x_0,y_0)in Xtimes Y$. Then we know $x_n$ converges to $x_0$ in the graph norm and so $x_0inmathrm{dom}(A)$, but this only tells $(x_0,Ax_0)in Xtimes Y$. We still don't know whether $Ax_0=y_0$.
functional-analysis
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
$endgroup$
add a comment |
$begingroup$
I am working on problems related to the closed graph of an unbounded operator. There is a proposition:
Let $X,Y$ be Banach spaces and let $A:mathrm{dom}(A)to Y$ be linear and defined on a linear subspace $mathrm{dom}(A)subset X$. Prove that the graph of $A$ is a closed subspace of $Xtimes Y$ if and only if $mathrm{dom}(A)$ is Banach with respect to the graph norm.
I finished one direction. Suppose $mathrm{graph}(A)$ is closed. We take any Cauchy sequence $x_n$ in $mathrm{dom}(A)$, and since the norm is graph norm, we know $x_n$ and $Ax_n$ will both be Cauchy. Then we have a Cauchy sequence $(x_n,Ax_n)$ in the graph, so the pair converges to a certain $(x_0,y_0)$ since the graph is closed. Therefore $x_0inmathrm{dom}(A)$, which means that $mathrm{dom}(A)$ is Banach.
However I encountered some trouble on the other direction. Suppose $operatorname{dom}(A)$ is Banach with respect to the graph norm. If we take a Cauchy sequence $(x_n,Ax_n)$ in $mathrm{graph}(A)$, since $X,Y$ are both Banach, it converges to a pair $(x_0,y_0)in Xtimes Y$. Then we know $x_n$ converges to $x_0$ in the graph norm and so $x_0inmathrm{dom}(A)$, but this only tells $(x_0,Ax_0)in Xtimes Y$. We still don't know whether $Ax_0=y_0$.
functional-analysis
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
$endgroup$
add a comment |
$begingroup$
I am working on problems related to the closed graph of an unbounded operator. There is a proposition:
Let $X,Y$ be Banach spaces and let $A:mathrm{dom}(A)to Y$ be linear and defined on a linear subspace $mathrm{dom}(A)subset X$. Prove that the graph of $A$ is a closed subspace of $Xtimes Y$ if and only if $mathrm{dom}(A)$ is Banach with respect to the graph norm.
I finished one direction. Suppose $mathrm{graph}(A)$ is closed. We take any Cauchy sequence $x_n$ in $mathrm{dom}(A)$, and since the norm is graph norm, we know $x_n$ and $Ax_n$ will both be Cauchy. Then we have a Cauchy sequence $(x_n,Ax_n)$ in the graph, so the pair converges to a certain $(x_0,y_0)$ since the graph is closed. Therefore $x_0inmathrm{dom}(A)$, which means that $mathrm{dom}(A)$ is Banach.
However I encountered some trouble on the other direction. Suppose $operatorname{dom}(A)$ is Banach with respect to the graph norm. If we take a Cauchy sequence $(x_n,Ax_n)$ in $mathrm{graph}(A)$, since $X,Y$ are both Banach, it converges to a pair $(x_0,y_0)in Xtimes Y$. Then we know $x_n$ converges to $x_0$ in the graph norm and so $x_0inmathrm{dom}(A)$, but this only tells $(x_0,Ax_0)in Xtimes Y$. We still don't know whether $Ax_0=y_0$.
functional-analysis
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
$endgroup$
I am working on problems related to the closed graph of an unbounded operator. There is a proposition:
Let $X,Y$ be Banach spaces and let $A:mathrm{dom}(A)to Y$ be linear and defined on a linear subspace $mathrm{dom}(A)subset X$. Prove that the graph of $A$ is a closed subspace of $Xtimes Y$ if and only if $mathrm{dom}(A)$ is Banach with respect to the graph norm.
I finished one direction. Suppose $mathrm{graph}(A)$ is closed. We take any Cauchy sequence $x_n$ in $mathrm{dom}(A)$, and since the norm is graph norm, we know $x_n$ and $Ax_n$ will both be Cauchy. Then we have a Cauchy sequence $(x_n,Ax_n)$ in the graph, so the pair converges to a certain $(x_0,y_0)$ since the graph is closed. Therefore $x_0inmathrm{dom}(A)$, which means that $mathrm{dom}(A)$ is Banach.
However I encountered some trouble on the other direction. Suppose $operatorname{dom}(A)$ is Banach with respect to the graph norm. If we take a Cauchy sequence $(x_n,Ax_n)$ in $mathrm{graph}(A)$, since $X,Y$ are both Banach, it converges to a pair $(x_0,y_0)in Xtimes Y$. Then we know $x_n$ converges to $x_0$ in the graph norm and so $x_0inmathrm{dom}(A)$, but this only tells $(x_0,Ax_0)in Xtimes Y$. We still don't know whether $Ax_0=y_0$.
functional-analysis
functional-analysis
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
edited Dec 1 '18 at 20:46
mechanodroid
27.6k62447
27.6k62447
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
1378
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $(x_n, Ax_n)$ be a Cauchy sequence in $operatorname{graph}(A)$. Then, by definition of the graph norm, $(x_n)_n$ is a Cauchy sequence in $operatorname{dom}(A)$.
Since $operatorname{dom}(A)$ is a Banach space w.r.t. the graph norm, $(x_n)_n$ converges to some $x in operatorname{dom}(A)$ w.r.t. the graph norm. This precisely means $(x_n, Ax_n) to (x, Ax)$ in $X times Y$. Hence, the sequence $(x_n, Ax_n)$ converges in $operatorname{graph}(A)$ so $operatorname{graph}(A)$ is a Banach space. In particular, it is a closed subspace of $X times Y$.
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
$endgroup$
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
add a comment |
$begingroup$
Suppose $mathcal{D}(A)$ is a Banach space in the graph norm of $A$. To show that $A$ is closed, suppose that $x_n rightarrow x$ and $Ax_n rightarrow y$, where ${ x_n}subsetmathcal{D}(A)$. We want to show that $xinmathcal{D}(A)$ and $Ax=y$. Under these assumptions, ${ x_n }$ is a Cauchy sequence in the graph norm of $A$, which means that ${ x_n }$ converges in the graph norm to some $x'inmathcal{D}(A)$. So $x_nrightarrow x'$ in $X$ and $Ax_nrightarrow Ax'$ in $Y$. It follows that $x=x'$ and $y=Ax'$, which proves that $A$ is closed.
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
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%2f3021781%2fabout-closed-graph-of-an-unbounded-operator%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $(x_n, Ax_n)$ be a Cauchy sequence in $operatorname{graph}(A)$. Then, by definition of the graph norm, $(x_n)_n$ is a Cauchy sequence in $operatorname{dom}(A)$.
Since $operatorname{dom}(A)$ is a Banach space w.r.t. the graph norm, $(x_n)_n$ converges to some $x in operatorname{dom}(A)$ w.r.t. the graph norm. This precisely means $(x_n, Ax_n) to (x, Ax)$ in $X times Y$. Hence, the sequence $(x_n, Ax_n)$ converges in $operatorname{graph}(A)$ so $operatorname{graph}(A)$ is a Banach space. In particular, it is a closed subspace of $X times Y$.
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
$endgroup$
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
add a comment |
$begingroup$
Let $(x_n, Ax_n)$ be a Cauchy sequence in $operatorname{graph}(A)$. Then, by definition of the graph norm, $(x_n)_n$ is a Cauchy sequence in $operatorname{dom}(A)$.
Since $operatorname{dom}(A)$ is a Banach space w.r.t. the graph norm, $(x_n)_n$ converges to some $x in operatorname{dom}(A)$ w.r.t. the graph norm. This precisely means $(x_n, Ax_n) to (x, Ax)$ in $X times Y$. Hence, the sequence $(x_n, Ax_n)$ converges in $operatorname{graph}(A)$ so $operatorname{graph}(A)$ is a Banach space. In particular, it is a closed subspace of $X times Y$.
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
$endgroup$
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
add a comment |
$begingroup$
Let $(x_n, Ax_n)$ be a Cauchy sequence in $operatorname{graph}(A)$. Then, by definition of the graph norm, $(x_n)_n$ is a Cauchy sequence in $operatorname{dom}(A)$.
Since $operatorname{dom}(A)$ is a Banach space w.r.t. the graph norm, $(x_n)_n$ converges to some $x in operatorname{dom}(A)$ w.r.t. the graph norm. This precisely means $(x_n, Ax_n) to (x, Ax)$ in $X times Y$. Hence, the sequence $(x_n, Ax_n)$ converges in $operatorname{graph}(A)$ so $operatorname{graph}(A)$ is a Banach space. In particular, it is a closed subspace of $X times Y$.
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
$endgroup$
Let $(x_n, Ax_n)$ be a Cauchy sequence in $operatorname{graph}(A)$. Then, by definition of the graph norm, $(x_n)_n$ is a Cauchy sequence in $operatorname{dom}(A)$.
Since $operatorname{dom}(A)$ is a Banach space w.r.t. the graph norm, $(x_n)_n$ converges to some $x in operatorname{dom}(A)$ w.r.t. the graph norm. This precisely means $(x_n, Ax_n) to (x, Ax)$ in $X times Y$. Hence, the sequence $(x_n, Ax_n)$ converges in $operatorname{graph}(A)$ so $operatorname{graph}(A)$ is a Banach space. In particular, it is a closed subspace of $X times Y$.
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.6k62447
27.6k62447
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
add a comment |
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded.
$endgroup$
– Apocalypse
Dec 1 '18 at 20:57
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $|x|_A = |x|_X + |Ax|_Y$ for $x in operatorname{dom}(A)$. If $x_n to x$ w.r.t. the graph norm, this means that $$|x-x_n|_X + |Ax - Ax_n|_Y = |x-x_n|_{A} to 0$$ so in particular $x_n to x$ in $X$ and $Ax_n to Ax$ in $Y$.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:06
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
@Apocalypse Indeed, if only $x_n to x$ in $X$ (meaning $|x_n - x|_X to 0$), we cannot conclude that $Ax_n to Ax$ in $Y$ even if $x in operatorname{dom}(A)$ precisely because $A$ is unbounded.
$endgroup$
– mechanodroid
Dec 1 '18 at 21:08
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
$begingroup$
Ohhhh! Indeed! I forgot that the second part $|Ax-Ax_n|$ in graph norm already requires $Ax_nto Ax$. Thank you so much!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:15
add a comment |
$begingroup$
Suppose $mathcal{D}(A)$ is a Banach space in the graph norm of $A$. To show that $A$ is closed, suppose that $x_n rightarrow x$ and $Ax_n rightarrow y$, where ${ x_n}subsetmathcal{D}(A)$. We want to show that $xinmathcal{D}(A)$ and $Ax=y$. Under these assumptions, ${ x_n }$ is a Cauchy sequence in the graph norm of $A$, which means that ${ x_n }$ converges in the graph norm to some $x'inmathcal{D}(A)$. So $x_nrightarrow x'$ in $X$ and $Ax_nrightarrow Ax'$ in $Y$. It follows that $x=x'$ and $y=Ax'$, which proves that $A$ is closed.
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
add a comment |
$begingroup$
Suppose $mathcal{D}(A)$ is a Banach space in the graph norm of $A$. To show that $A$ is closed, suppose that $x_n rightarrow x$ and $Ax_n rightarrow y$, where ${ x_n}subsetmathcal{D}(A)$. We want to show that $xinmathcal{D}(A)$ and $Ax=y$. Under these assumptions, ${ x_n }$ is a Cauchy sequence in the graph norm of $A$, which means that ${ x_n }$ converges in the graph norm to some $x'inmathcal{D}(A)$. So $x_nrightarrow x'$ in $X$ and $Ax_nrightarrow Ax'$ in $Y$. It follows that $x=x'$ and $y=Ax'$, which proves that $A$ is closed.
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
add a comment |
$begingroup$
Suppose $mathcal{D}(A)$ is a Banach space in the graph norm of $A$. To show that $A$ is closed, suppose that $x_n rightarrow x$ and $Ax_n rightarrow y$, where ${ x_n}subsetmathcal{D}(A)$. We want to show that $xinmathcal{D}(A)$ and $Ax=y$. Under these assumptions, ${ x_n }$ is a Cauchy sequence in the graph norm of $A$, which means that ${ x_n }$ converges in the graph norm to some $x'inmathcal{D}(A)$. So $x_nrightarrow x'$ in $X$ and $Ax_nrightarrow Ax'$ in $Y$. It follows that $x=x'$ and $y=Ax'$, which proves that $A$ is closed.
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
Suppose $mathcal{D}(A)$ is a Banach space in the graph norm of $A$. To show that $A$ is closed, suppose that $x_n rightarrow x$ and $Ax_n rightarrow y$, where ${ x_n}subsetmathcal{D}(A)$. We want to show that $xinmathcal{D}(A)$ and $Ax=y$. Under these assumptions, ${ x_n }$ is a Cauchy sequence in the graph norm of $A$, which means that ${ x_n }$ converges in the graph norm to some $x'inmathcal{D}(A)$. So $x_nrightarrow x'$ in $X$ and $Ax_nrightarrow Ax'$ in $Y$. It follows that $x=x'$ and $y=Ax'$, which proves that $A$ is closed.
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
add a comment |
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
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%2f3021781%2fabout-closed-graph-of-an-unbounded-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
