Lemma 3

Suppose that $K =\mathbf{R}$. Let $u$ be a partial operator with dense domain from E into F.

a) The adjoint of $u_{(\mathbf{C})}$ is $(u^*)_{(\mathbf{C})}$;

b) Suppose that E = F; the partial operator $u$ is symmetric (resp. self-adjoint) if and only if the partial operator $u_{(\mathbf{C})}$ is symmetric (resp. self-adjoint).

Let us prove a). Let $y\in F_{(\mathbf{C})}$ and write $y=y_1+iy_2$ with $y_1,y_2\in F$. For all $(x_1, x_2)\in E\times E$, we have

If $y\in$ dom($u^*$)$_{(\mathbf{C})}$, it follows that $y\in$ dom(($u_{(\mathbf{C})}$)$^*$) and that $u^*_{(\mathbf{C})}(y) = (u_{(\mathbf{C})})^*(y)$, hence $u^*_{(\mathbf{C})}\subset (u_{(\mathbf{C})})^*$.

Conversely, suppose that $y$ belongs to dom(($u_{(\mathbf{C})}$)$^*$). Taking $x_2= 0$ (resp. $x_1= 0$) in the above formula, we verify that $y_1\in$ dom($u^*$) (resp. that $y_2\in$ dom($u^*$)), whence $y\in$ dom($u^*$)$_{(\mathbf{C})}$.

Assertion a) implies that $u_{(\mathbf{C})}$ is symmetric (resp. self-adjoint) if $u$ is.

Conversely, suppose that $u_{(\mathbf{C})}$ is symmetric. The relation $\langle u(x)|y\rangle =\langle x|u(y)\rangle$ for all $(x, y)\in$ dom($u_{(\mathbf{C})}$)$\times$ dom($u_{(\mathbf{C})}$) implies that $u$ is symmetric on taking $x$ and $y$ in the subspace dom($u$) of dom($u_{(\mathbf{C})}$). If $u_{(\mathbf{C})}$ is self-adjoint, what precedes proves that $u$ is symmetric; since dom($u^*$) $=$ dom($u^*_{(\mathbf{C})}$)$\cap F$, assertion a) implies that dom($u^*$) $=$ dom($u_{(\mathbf{C})}$)$\cap F =$ dom($u$), therefore $u$ is self-adjoint.