Corollary

Let $u$ be a closed symmetric operator on E. Then $u$ is self-adjoint if and only if the deficiency index of $u$ is equal to $(0,0)$.

In fact, the symmetric operator $u$ is self-adjoint if and only if dom($u^*$) $=$ dom($u$); now the proposition proves that dom($u^*$) is the Hilbert sum of dom($u$) and of the deficiency subspaces of $u$.

Let $u$ be a closed symmetric partial operator. The spectrum of $u$ is contained in $\mathbf{R}$ if and only if $u$ is self-adjoint. In fact, it is known that Sp($u$)$\subset \mathbf{R}$ if $u$ is self-adjoint (Prop. 17 of IV, p. 248); if $u$ is not self-adjoint, one of the subspaces Ker($u^*+i$) or Ker($u^*-i$) is nonzero, hence the image of $u+i$ or of $u-i$ is a proper subspace of E (Prop. 25 of IV, p. 256), so that $i$ or $-i$ belongs to Sp($u$).