> 
> **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?