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
27.1k62446
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
27.1k62446
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
27.1k62446
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
27.1k62446
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
edited Dec 1 '18 at 20:46
mechanodroid
27.1k62446
asked Dec 1 '18 at 20:22
ApocalypseApocalypse
1378
edited Dec 1 '18 at 20:46
mechanodroid
27.1k62446
edited Dec 1 '18 at 20:46
mechanodroid
27.1k62446
edited Dec 1 '18 at 20:46
mechanodroid
27.1k62446
27.1k62446
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
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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
27.1k62446
$endgroup$
$begingroup$
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
58.9k42580
$endgroup$
$begingroup$
I know where I got myself confused! Thank you!
$endgroup$
– Apocalypse
Dec 1 '18 at 21:24
add a comment |
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2 Answers
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active
oldest
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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
27.1k62446
$endgroup$
$begingroup$
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
27.1k62446
$endgroup$
$begingroup$
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
27.1k62446
$endgroup$
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
27.1k62446
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.1k62446
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.1k62446
answered Dec 1 '18 at 20:45
mechanodroidmechanodroid
27.1k62446
27.1k62446
$begingroup$
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
58.9k42580
$endgroup$
$begingroup$
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
58.9k42580
$endgroup$
$begingroup$
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
58.9k42580
$endgroup$
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
58.9k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
58.9k42580
answered Dec 1 '18 at 21:14
DisintegratingByPartsDisintegratingByParts
58.9k42580
58.9k42580
$begingroup$
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 |
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