Lemma 5

Let E be a complex Hilbert space and let $u$ be a closed partial operator with dense domain on E. Let $\lambda \in \mathbf{C}$. Suppose that there exists a real number $c >0$ such that

(10) $\|u(x)-\lambda x\|\geqslant c\|x\|$ for every $x\in$ dom($u$),

(11) $\|u^*(x)-\lambda x\|\geqslant c\|x\|$ for every $x\in$ dom($u^*$).

Then $\lambda$ belongs to the resolvent set of $u$ and $\|R(u, \lambda )\|\leqslant c^{-1}$.

The assumption implies that $u-\lambda 1_E$ and $u^*-\overline{\lambda}1_E$ are injective. Let F be the image of $u-\lambda 1_E$. The space F is dense in E, since its orthogonal is equal to Ker($u^*-\overline{\lambda}1_E$) (prop. 7, c) of IV, p. 236), which is zero.

Let us prove that the space F is closed. Let $(x_n)_{n\in\mathbf{N}}$ be a sequence in dom($u$) such that the sequence $(u(x_n)-\lambda x_n)_{n\in\mathbf{N}}$ converges to $y\in F$. Inequality (10) implies that the sequence $(x_n)_{n\in\mathbf{N}}$ is a Cauchy sequence in E. Let $x\in E$ be its limit. The sequence $(x_n, u(x_n))$ converges to $(x, y+\lambda x)$ in $E\times E$; since the graph of $u$ is closed, we therefore have $x\in$ dom($u$) and $u(x) =y+\lambda x$, which proves that $y\in F$.

We conclude that F = E. Thus, the partial operator $u-\lambda 1_E$ is bijective, whence $\lambda \notin$ Sp($u$) (remark 3). Inequality (10) then implies that $\|R(u, \lambda )\|\leqslant c^{-1}$.