TITLE: Differentiation operator is closed on $L^2[0,1]$?
QUESTION [7 upvotes]: Proposition. Let $\mathcal{H}=L^2[0,1]$ with a differentiation operator $T=i\frac{d}{dt}$ on it, whose domain $D(T)$ consists of all the absolutely continuous functions in $\mathcal{H}$, and the derivatives of those functions is still $L^2$. Then $T$ is a closed operator. (which means its graph is closed in $\mathcal{H}\times\mathcal{H}$)

Here is my try so far.
Let $\{u_n\}$ be a sequence of $L^2$ functions in $\mathcal{H}$ converging to $u\in\mathcal{H}$. Also, $Tu_n$ converge to another $v\in\mathcal{H}$.
It suffices to show that $Tu=v$, which is to say, $\| Tu-v\|=0$. Since
$$\|Tu-v\|\le\|Tu-Tu_n\|+\|Tu_n-v\|$$
and $\|Tu_n-v\|\rightarrow 0$, we only need to prove that $\|Tu-Tu_n\|\rightarrow 0$.
$$\lim_{n\rightarrow\infty}\|Tu-Tu_n\|^2=\lim_{n\rightarrow\infty}\int \left|i\frac{d}{dt}(u-u_n)\right|^2dx.$$
Ideally, we “may” swap the “limits” with the “integral” and “derivative” and the “proof” is finished. However, it seems to me not very clear that theorems from real analysis allow me to do so.
I’ve been stuck here for a long time. Any hints?

REPLY [3 votes]: There is a simple solution provided that you are vaguely familiar with the notion of Sobolev space and weak derivative.
Here $D(T)$ is nothing but the Sobolev space $W^{1,2}(0,1)$, which embeds into the space of absolutely continuous functions on $[0,1]$. So, your problem (after eliminating the irrelevant $i$ factor) is equivalent to proving, for a sequence $\left\{u_n\right\}\subset W^{1,2}(0,1)$,
$$ (\|u_n-u\|_{L^2}\to 0\;\land\;\|u_n'-v\|_{L^2}\to 0)\Rightarrow v=u'$$
Solution: The relations $\|u_n-u\|_{L^2}\to 0$ and $\|u_n'-v\|_{L^2}\to 0$ yield respectively
\begin{align*}\int_0^1 u_n\psi&\to \int_0^1u\psi,\qquad \forall \psi\in C_c^{\infty}([0,1])\\
\int_0^1u_n'\varphi&\to \int_0^1v\varphi,\qquad \forall \varphi \in C_c^{\infty}([0,1]) \end{align*}
On the other hand, integrating by parts and using the first relation we obtain 
$$\int_0^1u_n'\varphi=-\int_0^1u_n\varphi'\to -\int_0^1u\varphi',\qquad \forall \varphi \in C_c^{\infty}([0,1])$$
and hence 
$$\int_0^1u\varphi'=-\int_0^1v\varphi,\qquad \forall \varphi \in C_c^{\infty}([0,1]) $$
which tells us precisely that $v$ is the weak derivative of $u$, i.e. $u'=v$ 
(and since $u$ is a.e. differentiable, the weak derivative is equal to the classical derivative wherever the latter is defined).