Proving that the derivative operator on $L^p(0,1)$ has a closed graph












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Let $Tx(t) = x'(t)$ on $L_p(0,1)$, and let the domain of $T$ be only the absolutely continuous functions on $[0,1]$ whose derivatives are in $L^p(0,1)$. I need to show that this operator has a closed graph. Naturally, by the closed graph theorem, it would be sufficient to show that $T$ is a continuous operator. This seems like a standard exercise, but I can't seem to be able to do it. I don't make much progress with the definition of continuity. Is it best to try and show that $T$ is bounded?






real-analysis functional-analysis







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    Let $Tx(t) = x'(t)$ on $L_p(0,1)$, and let the domain of $T$ be only the absolutely continuous functions on $[0,1]$ whose derivatives are in $L^p(0,1)$. I need to show that this operator has a closed graph. Naturally, by the closed graph theorem, it would be sufficient to show that $T$ is a continuous operator. This seems like a standard exercise, but I can't seem to be able to do it. I don't make much progress with the definition of continuity. Is it best to try and show that $T$ is bounded?






    real-analysis functional-analysis







    share|cite|improve this question











    $endgroup$















      0












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      0





      $begingroup$


      Let $Tx(t) = x'(t)$ on $L_p(0,1)$, and let the domain of $T$ be only the absolutely continuous functions on $[0,1]$ whose derivatives are in $L^p(0,1)$. I need to show that this operator has a closed graph. Naturally, by the closed graph theorem, it would be sufficient to show that $T$ is a continuous operator. This seems like a standard exercise, but I can't seem to be able to do it. I don't make much progress with the definition of continuity. Is it best to try and show that $T$ is bounded?






      real-analysis functional-analysis







      share|cite|improve this question











      $endgroup$




      Let $Tx(t) = x'(t)$ on $L_p(0,1)$, and let the domain of $T$ be only the absolutely continuous functions on $[0,1]$ whose derivatives are in $L^p(0,1)$. I need to show that this operator has a closed graph. Naturally, by the closed graph theorem, it would be sufficient to show that $T$ is a continuous operator. This seems like a standard exercise, but I can't seem to be able to do it. I don't make much progress with the definition of continuity. Is it best to try and show that $T$ is bounded?






      real-analysis functional-analysis




      real-analysis functional-analysis






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      edited Dec 7 '18 at 3:49









      Andrews

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      asked Dec 7 '18 at 3:31









      Wyatt GregoryWyatt Gregory

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          $begingroup$

          I suppose you are taking $p>1$. Suppose $x_n to x$ in $L^{p}$ and $x_n' to y$ in $L^{p}$. We have to show that $x$ is absolutely continuous, $x' in L^{p}$ and $y=x'$. By absolute continuity we have $x_n(t)=x_n(0)+int_0^{t} x_n'(s) , ds$ $,,, (1)$. Let $n_k$ be a subseqeunce of the integers such that $x_{n_k}$ converges almost everywhere to $x$. Since convergence in $L^{p}$ implies convergence in $L^{1}$ we get $x(t)=a+int_0^{t} y(s), ds$ for some constant $a$. [In (1) LHS converges and the second term on RHS converges. Hence the first term must converge]. Since $y$ is integrable this last equation implies that $x$ is absolutely continuous [more precisely it can be modified on a null set to make it absolutely continuous] and $x'=y$.






          share|cite|improve this answer









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          • $begingroup$
            Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
            $endgroup$
            – lcv
            Dec 7 '18 at 18:12










          • $begingroup$
            $T$ is not a bounded operator. Unbounded operators can have closed graph.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 23:21










          • $begingroup$
            Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
            $endgroup$
            – lcv
            Dec 8 '18 at 2:14











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          1 Answer
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          active

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          0












          $begingroup$

          I suppose you are taking $p>1$. Suppose $x_n to x$ in $L^{p}$ and $x_n' to y$ in $L^{p}$. We have to show that $x$ is absolutely continuous, $x' in L^{p}$ and $y=x'$. By absolute continuity we have $x_n(t)=x_n(0)+int_0^{t} x_n'(s) , ds$ $,,, (1)$. Let $n_k$ be a subseqeunce of the integers such that $x_{n_k}$ converges almost everywhere to $x$. Since convergence in $L^{p}$ implies convergence in $L^{1}$ we get $x(t)=a+int_0^{t} y(s), ds$ for some constant $a$. [In (1) LHS converges and the second term on RHS converges. Hence the first term must converge]. Since $y$ is integrable this last equation implies that $x$ is absolutely continuous [more precisely it can be modified on a null set to make it absolutely continuous] and $x'=y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
            $endgroup$
            – lcv
            Dec 7 '18 at 18:12










          • $begingroup$
            $T$ is not a bounded operator. Unbounded operators can have closed graph.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 23:21










          • $begingroup$
            Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
            $endgroup$
            – lcv
            Dec 8 '18 at 2:14
















          0












          $begingroup$

          I suppose you are taking $p>1$. Suppose $x_n to x$ in $L^{p}$ and $x_n' to y$ in $L^{p}$. We have to show that $x$ is absolutely continuous, $x' in L^{p}$ and $y=x'$. By absolute continuity we have $x_n(t)=x_n(0)+int_0^{t} x_n'(s) , ds$ $,,, (1)$. Let $n_k$ be a subseqeunce of the integers such that $x_{n_k}$ converges almost everywhere to $x$. Since convergence in $L^{p}$ implies convergence in $L^{1}$ we get $x(t)=a+int_0^{t} y(s), ds$ for some constant $a$. [In (1) LHS converges and the second term on RHS converges. Hence the first term must converge]. Since $y$ is integrable this last equation implies that $x$ is absolutely continuous [more precisely it can be modified on a null set to make it absolutely continuous] and $x'=y$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
            $endgroup$
            – lcv
            Dec 7 '18 at 18:12










          • $begingroup$
            $T$ is not a bounded operator. Unbounded operators can have closed graph.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 23:21










          • $begingroup$
            Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
            $endgroup$
            – lcv
            Dec 8 '18 at 2:14














          0












          0








          0





          $begingroup$

          I suppose you are taking $p>1$. Suppose $x_n to x$ in $L^{p}$ and $x_n' to y$ in $L^{p}$. We have to show that $x$ is absolutely continuous, $x' in L^{p}$ and $y=x'$. By absolute continuity we have $x_n(t)=x_n(0)+int_0^{t} x_n'(s) , ds$ $,,, (1)$. Let $n_k$ be a subseqeunce of the integers such that $x_{n_k}$ converges almost everywhere to $x$. Since convergence in $L^{p}$ implies convergence in $L^{1}$ we get $x(t)=a+int_0^{t} y(s), ds$ for some constant $a$. [In (1) LHS converges and the second term on RHS converges. Hence the first term must converge]. Since $y$ is integrable this last equation implies that $x$ is absolutely continuous [more precisely it can be modified on a null set to make it absolutely continuous] and $x'=y$.






          share|cite|improve this answer









          $endgroup$



          I suppose you are taking $p>1$. Suppose $x_n to x$ in $L^{p}$ and $x_n' to y$ in $L^{p}$. We have to show that $x$ is absolutely continuous, $x' in L^{p}$ and $y=x'$. By absolute continuity we have $x_n(t)=x_n(0)+int_0^{t} x_n'(s) , ds$ $,,, (1)$. Let $n_k$ be a subseqeunce of the integers such that $x_{n_k}$ converges almost everywhere to $x$. Since convergence in $L^{p}$ implies convergence in $L^{1}$ we get $x(t)=a+int_0^{t} y(s), ds$ for some constant $a$. [In (1) LHS converges and the second term on RHS converges. Hence the first term must converge]. Since $y$ is integrable this last equation implies that $x$ is absolutely continuous [more precisely it can be modified on a null set to make it absolutely continuous] and $x'=y$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 5:55









          Kavi Rama MurthyKavi Rama Murthy

          56.2k42158




          56.2k42158












          • $begingroup$
            Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
            $endgroup$
            – lcv
            Dec 7 '18 at 18:12










          • $begingroup$
            $T$ is not a bounded operator. Unbounded operators can have closed graph.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 23:21










          • $begingroup$
            Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
            $endgroup$
            – lcv
            Dec 8 '18 at 2:14


















          • $begingroup$
            Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
            $endgroup$
            – lcv
            Dec 7 '18 at 18:12










          • $begingroup$
            $T$ is not a bounded operator. Unbounded operators can have closed graph.
            $endgroup$
            – Kavi Rama Murthy
            Dec 7 '18 at 23:21










          • $begingroup$
            Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
            $endgroup$
            – lcv
            Dec 8 '18 at 2:14
















          $begingroup$
          Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
          $endgroup$
          – lcv
          Dec 7 '18 at 18:12




          $begingroup$
          Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ exp(i k x)$ are in the domain) ?
          $endgroup$
          – lcv
          Dec 7 '18 at 18:12












          $begingroup$
          $T$ is not a bounded operator. Unbounded operators can have closed graph.
          $endgroup$
          – Kavi Rama Murthy
          Dec 7 '18 at 23:21




          $begingroup$
          $T$ is not a bounded operator. Unbounded operators can have closed graph.
          $endgroup$
          – Kavi Rama Murthy
          Dec 7 '18 at 23:21












          $begingroup$
          Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
          $endgroup$
          – lcv
          Dec 8 '18 at 2:14




          $begingroup$
          Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space.
          $endgroup$
          – lcv
          Dec 8 '18 at 2:14


















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