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

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 '18 at 18:12
• $T$ is not a bounded operator. Unbounded operators can have closed graph. – Kavi Rama Murthy Dec 7 '18 at 23:21