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?

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 $x'=y$.

  • Maybe I’m missing something, how can the derivative be bounded (given that, e.g., $ \exp(i k x)$ are in the domain) ? – lcv Dec 7 at 18:12
  • $T$ is not a bounded operator. Unbounded operators can have closed graph. – Kavi Rama Murthy Dec 7 at 23:21
  • Absolutely. The OP misuses the closed graph theorem as the operator is not defined on the whole space. – lcv Dec 8 at 2:14

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