Theorem 5 (Kato-Rellich)

Let $u$ be a self-adjoint partial operator on E and $v$ a symmetric partial operator bounded relatively to $u$. If the relative norm $\|v\|_u$ is $<1$, then the partial operator $u+v$ with domain dom($u$) is self-adjoint.

Since $\|v\|_u<1$, there exist by definition positive real numbers $a$ and $b$ such that $a <1$ and $\|v(x)\|\leqslant a\|u(x)\|+b\|x\|$ for every $x\in$ dom($u$).

Let $t\in \mathbf{R}^*$. Put $w_t=v\circ R(u, it)$. We have dom($w_t$) $= E$ since the image of $R(u, it)$ is contained in the domain of $u$, which is contained in the domain of $v$ by hypothesis. Let $x\in$ dom($u$). Since $u$ is self-adjoint, we have

whence the inequalities $\|u(x)\|\leqslant \|(it-u)x\|$ and $\|x\|\leqslant |t|^{-1}\|(it-u)x\|$. But then we get

In particular, put $x= R(u, it)y$ for $y\in E$; we obtain

It follows that $w_t\in \mathscr{L}(E)$ and that $\|w_t\|\leqslant a+|t|^{-1}b$.

Since $a <$ 1, there exists $t\in \mathbf{R}^*_+$ such that $a+b|t|^{-1}<1$. Then $\|w_t\|<1$ and $\|w_{-t}\|<1$; the endomorphisms $1_E-w_t$ and $1_E-w_{-t}$ of E are therefore invertible (Prop. 2 of I, p. 22). Since we have the formula $(1_E-w_t)\circ (it-u) =it-(u+v)$, this implies that $u+v-it$ is surjective; analogously, the partial operator $u+v+it$ is surjective. It follows that $u+v$ is self-adjoint (Prop. 11 of IV, p. 240).