Let $R_z=(A-zI)^{-1}$ be compact and let $w$ be another resolvent point. The resolvent identity gives
$$
R_w-R_z=(w-z)R_wR_z,
$$
or
$$
\boxed{R_w=[I+(w-z)R_w]R_z}.
$$
The bracket is bounded and the product of a bounded operator with a <compact operator> is compact. Thus compactness at one resolvent point implies compactness at every resolvent point.

Fix such a $z$. Spectral mapping for the bounded compact operator $R_z$ gives
$$
\lambda\in\sigma(A)
\quad\Longleftrightarrow\quad
(\lambda-z)^{-1}\in\sigma(R_z)\setminus\{0\}.
$$
Every nonzero spectral point of a compact operator is an isolated eigenvalue of finite multiplicity, and zero is its only possible accumulation point. Hence a <compact resolvent> operator has only isolated eigenvalues of finite multiplicity, with no finite accumulation point. The spectrum is allowed to be empty.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 / 1 / a / Solution

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A closed densely defined operator has compact resolvent when $(A-zI)^{-1}$ is compact at one, equivalently every, resolvent point. Its spectrum consists only of isolated eigenvalues of finite multiplicity, with possible accumulation only at infinity.


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