Proving that the derivative operator on $L^p(0,1)$ has a closed graph
$begingroup$
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
edited Dec 7 '18 at 3:49
Andrews
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
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add a comment |
$begingroup$
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
edited Dec 7 '18 at 3:49
Andrews
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
$endgroup$
add a comment |
$begingroup$
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
edited Dec 7 '18 at 3:49
Andrews
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
$endgroup$
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
edited Dec 7 '18 at 3:49
Andrews
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
edited Dec 7 '18 at 3:49
Andrews
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
edited Dec 7 '18 at 3:49
Andrews
3831317
edited Dec 7 '18 at 3:49
Andrews
3831317
edited Dec 7 '18 at 3:49
Andrews
3831317
3831317
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
asked Dec 7 '18 at 3:31
Wyatt GregoryWyatt Gregory
706
706
add a comment |
add a comment |
1 Answer
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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 $x'=y$.
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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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 $x'=y$.
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x'=y$.
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x'=y$.
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
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 $x'=y$.
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Dec 7 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
54.4k32055
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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