Let $T = \mathbb{R}/\mathbb{Z}$ be the one dimensional-torus, and $$A:H^2(T)\to L^2(T)\;\;\\A u = iu_{xx}$$ I want to know if this operator is closed, and i'm having issues and not knowing how to start since i don't really know how to start working on Torus. For the record, if $u \in D'(T)$ we define $||.||_{H^2(t)}$ by $$||u||^2_{H^2(t)} = \displaystyle\sum_{\xi \in \mathbb{Z}} \langle\xi\rangle^4 |\widehat{u}(\xi)|^2$$ where $\widehat{u}$ is the laplace transform and $\langle\xi\rangle = (1+||\xi||^2)^{1/2}$.
Thank you in advance
Assume that $u_n\to u$ and $(A u_n)\to v$ in $L^2(T)$. By Parseval's identity, $\hat u_n\to \hat u$ and $\widehat{A u_n}(\xi)=-i\lvert\xi\rvert^2\hat u_n(\xi)\to \hat v$ in $\ell^2(\mathbb{Z})$. In particular, both sequences are bounded.
It follows from the Fatou's lemma that $$ \sum_{\xi\in\mathbb{Z}}\langle\xi\rangle^4\lvert \hat u(\xi)\rvert^2\leq\liminf_{n\to\infty}\sum_{\xi\in \mathbb{Z}}\langle \xi\rangle^4\lvert\hat u_n(\xi)\rvert^2\leq 2\sup_{n\in\mathbb{N}} \sum_{\xi\in\mathbb{Z}}(1+\lvert \xi\rvert^4)\lvert \hat u_n(\xi)\rvert^2<\infty. $$ Thus $u\in H^2(T)$. A very simlar application of Fatou's lemma also shows $\widehat{A u_n}\to \widehat{A u}$ in $\ell^2(\mathbb{Z})$, that is, $Au_n\to A u$ in $L^2(T)$.
A slightly more elegant way to see that $A$ is closed would be to show that multiplication by $-i\lvert \cdot\rvert^2$ with the natural domain is skew self-adjoint on $\ell^2(\mathbb{Z})$ - this follows more or less from the definition - and then to use that adjoints of operators are always closed.