Proposition 17

Let E be a complex Hilbert space and let $u$ be a self-adjoint partial operator on E.

a) The spectrum of $u$ is contained in $\mathbf{R}$;

b) If $u$ is positive, then the spectrum of $u$ is contained in $\mathbf{R}_+$;

c) Suppose that E is nonzero. Let $\lambda \notin$ Sp($u$) and let $\delta >0$ be the distance from $\lambda$ to the spectrum of $u$. One has $\|R(u, \lambda )\|=\delta^{-1}$.

Let $(a, b)\in \mathbf{R}\times \mathbf{R}$ and $\lambda =a+ib$. Let $x\in$ dom($u$). Since $u$ is self-adjoint, one has $\langle x|u(x)\rangle \in \mathbf{R}$, whence

Suppose that $b\not = 0$. One then obtains

By Lemma 5, one therefore has $\lambda \notin$ Sp($u$), whence assertion a).

Suppose that $u$ is also positive. If $b= 0$ and $a <0$, one analogously obtains for $x\in$ dom($u$) the inequality

therefore $\lambda \notin$ Sp($u$) $($loc. cit.$)$, which proves b).

Finally, let us prove c). By Prop. 16, the resolvent $R(u, \lambda )$ is a normal endomorphism of E. Its norm is therefore equal to its spectral radius (Cor. 1 of I, p. 108), whence

$\|R(u, \lambda )\|=$ sup$_{\mu\in Sp(R(u,\lambda))}|\mu|$

(Th. 1 of I, p. 24). The spectrum of $R(u, \lambda )$ cannot be reduced to $\{0\}$, for in that case one would have $\|R(u, \lambda )\|= 0$, hence the image dom($u$) of $R(u, \lambda )$ would be zero, and E likewise. Prop. 15 therefore implies that

$\|R(u, \lambda )\|=$ sup$_{\mu\in Sp(u)}\frac{1}{|\lambda-\mu|}=\frac{1}{\delta}$.