Let $H=\mathcal{L}^2(\mathbb R^2,dxdy)$ and let $A$ the operator defined by: $$ A[f](x,y)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+i(y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial y}) $$ let $D(A)=C_0^\infty (\mathbb R^2)$ be the domain of $A$, show that the operator $A$ is closable.
I tried to use the criterion of closable operators, and so I took a sequence $f_n$ such that $f_n\to 0$ as $n\to +\infty$ and $\lim_{n\to +\infty}A[f_n]=g$ and I tried to prove that $g=0$, for that I took an other function $\varphi$ of $H$ then tried compute the scalar product $<A[f_n],\varphi>$. My problem is that when I express this scalar product I'm confused with the double integral and how to compute it and then making an integration by parts if ever it's going to lead to something ! is my way correct? if it's so how to calculate the scalar product ? if not what are the other ways to solve this problem ? thank you for your help and time.
Hint: Use integration by parts to show that $\langle A f, \varphi \rangle = \langle f , A \varphi \rangle$ for all $f, \varphi \in C^\infty_0(\mathbb{R}^2)$.
Use this to show that $\langle g, \varphi \rangle = 0$ for all $\varphi \in C^\infty_0(\mathbb{R}^2)$. By the density of $C^\infty_0$, conclude that $g=0$.
This argument proves a more general result: every symmetric operator is closable.