Consider the subspace $\mathcal L_0$ of $L^2=L^2([0,1])$ consisting of functions of integral zero. Consider the map $T:L^2\to L^2$ given by $T(f)(x)=\int_0^x{\rm d}t \, f(t)$. Prove that it is one-to-one. Let $\mathcal D_0=T(\mathcal L_0)$. Prove that the operator $A=iT^{-1}$ with domain $\mathcal D_0$ is symmetric, that its graph is closed but that it is not self-adjoint.
Exercise 2.5.16 in What is a Quantum Field Theory by Michel Talagrand. I'm asking about the "graph is closed" part.
Attempt
For each $h\in\mathcal D_0$, the map $F_h:L^2 \times L^2 \to \mathbb C$ given by $F_h(f,g) = \langle g, h \rangle - \langle f, A(h)\rangle$ is continuous, so the set $X$ of all pairs $(f,g)\in L^2\times L^2$ such that $F_h(f,g)=0$ for all $h\in\mathcal D_0$, as an intersection of continuous preimages of the closed set $\{0\}$, is closed in $L^2\times L^2$.
For all $f$, $g\in L^2$, the pair $(f,g)$ is in the graph of $A$ if and only if $f\in\mathcal D_0$ and $g=A(f)$. But $g=A(f)$ if and only if $\langle g, h \rangle = \langle f, A(h)\rangle$ for all $h\in\mathcal D_0$ (since $A$ is symmetric). So the graph of $A$ is $(\mathcal D_0\times L^2)\cap X$, and it suffices to show that $\mathcal D_0$ is closed in $L^2$ to reach the desired conclusion.
Note $\mathcal L_0$ is closed in $L^2$, since the map $f\mapsto\int_0^1{\rm d}t \, f(t)$ is continuous and its kernel is $\mathcal L_0$.
Question
Can it be shown that $\mathcal D_0=T(\mathcal L_0)$ is also closed? (This would follow if $\mathcal L_0$ is compact, since then $T^{-1}$ would be continuous. But $\mathcal L_0$ is not compact.) If not, can you hint at another argument?
Thanks for your attention! But now I think I can answer this myself. Any comments would be very welcome.
For all $f$, $g\in L^2$, the pair $(f,g)$ is in the graph of $A$ if and only if: \begin{align} & f\in\mathcal D_0 \text{ and } g=A(f) \\ \Leftrightarrow{} & \text{there exists $k\in\mathcal L_0$ such that }f=T(k) \text{, and } g=A(f) \\ \Leftrightarrow{} & \text{there exists $k\in\mathcal L_0$ such that }f=T(k) \text{, and } g=ik \\ \Leftrightarrow{} & g \in \mathcal L_0 \text{ and } f=T(-ig)\\ \Leftrightarrow{} & g \in \mathcal L_0 \text{ and } g=A(f). \end{align} But $g=A(f)$ if and only if $\langle g, h \rangle = \langle f, A(h)\rangle$ for all $h\in\mathcal D_0$ (since $A$ is symmetric). So the graph of $A$ is $(L^2\times \mathcal L_0)\cap X$, and since $\mathcal L_0$ is closed in $L^2$, the desired conclusion follows.