About closed graph of an unbounded operator












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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







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$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













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      edited Dec 1 '18 at 20:46









      mechanodroid

      27.1k62446




      27.1k62446










      asked Dec 1 '18 at 20:22









      ApocalypseApocalypse

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          2 Answers
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          0












          $begingroup$

          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 $X times Y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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





















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know where I got myself confused! Thank you!
            $endgroup$
            – Apocalypse
            Dec 1 '18 at 21:24











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          2 Answers
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          2 Answers
          2






          active

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          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 $X times Y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          0












          $begingroup$

          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 $X times Y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          0












          0








          0





          $begingroup$

          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 $X times Y$.






          share|cite|improve this answer









          $endgroup$



          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 $X times Y$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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













          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know where I got myself confused! Thank you!
            $endgroup$
            – Apocalypse
            Dec 1 '18 at 21:24
















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know where I got myself confused! Thank you!
            $endgroup$
            – Apocalypse
            Dec 1 '18 at 21:24














          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







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          answered Dec 1 '18 at 21:14









          DisintegratingByPartsDisintegratingByParts

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          • $begingroup$
            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




          $begingroup$
          I know where I got myself confused! Thank you!
          $endgroup$
          – Apocalypse
          Dec 1 '18 at 21:24


















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