If an operator in $X$ with compact resolvent is bounded, $X$ must be finite dimensional.

Question

In Kato's book, this is problem 6.30. I have been looking at different ways of proving this. Here is my attempt;

Theorem If an operator $A$ in $X$ with compact resolvent is bounded, $X$ must be finite dimensional.

Proof Let $X$ be an infinite dimensional, and let $\{e_i\}_{n=1}^{\infty}$ be a basis for $X$. As $A$ has compact resolvent, its spectrum is of isolated eigenvalues with finite multiplicities. Therefore, let $x \in X$ be an eigenfunction of $A$, with $Ax= \lambda x$, where $\lambda$ is a constant. Then we have that \begin{equation} ||Ax||_X = ||\lambda x ||_X = |\lambda| ||x||_X =|\lambda| ||\sum_{n=0}^\infty a_i e_i||_X \leq |\lambda| (\max |a_i|) ||\sum_{n=0}^\infty \ e_i||_X. \end{equation}

As $X$ is infinite dimensional $||\sum_{n=0}^\infty \ e_i||_X$ is infinite. This contradicts that $A$ us bounded.

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Now, I think my last sentence is dodgy. Can anyone offer any clarification or a more natural way to prove it?

Kindest regards, C

Answer 1

There exists $c$ such that $B=(A-cI)^{-1}$ is compact, (definition of resolvent) since $A$ is bounded, $A-cI$ is bounded and $(A-cI)(A-cI)^{-1}=I$ is compact. We deduce that the space is finite dimensional.

Composition of bounded operator and compact operators