Proposition 20

Let $u$ be a self-adjoint partial operator on E. Then $u$ has compact resolvent if and only if the canonical injection $j$ of the Hilbert space $E_u$ into E is compact.

Suppose that $u$ has compact resolvent. There exists $\lambda \notin$ Sp($u$) such that $R(u, \lambda )$ is compact (prop. 19, (iii)). Let B be the unit ball of the Hilbert space $E_u$. Since $u$ is a continuous linear mapping of $E_u$ into E, the subset $B'= (\lambda 1_E-u)(B)$ of E is bounded. Since $R(u, \lambda )$ is compact, the subset $B = R(u, \lambda )(B')$ is relatively compact in E (remark 1 of III, p. 2). This proves that $j$ is compact.

Conversely, suppose that the linear mapping $j$ is compact. The complex number $i$ belong to the resolvent set of $u$ (Proposition 17 of IV, p. 248). We have $u\circ R(u, i) =-1_E+iR(u, i)$. Let B be the unit ball of E. For every $x\in B$, we have

and consequently

The image C of B under $R(u, i)$ is therefore bounded in $E_u$; since $j$ is compact by hypothesis, the set $C =j(C)$ is relatively compact in E (III, loc. cit.). Consequently, the resolvent $R(u, i)$ is compact and $u$ has compact resolvent (Proposition 19, (iii)).