Let $X$ be an infinite dimensional Banach space and $A:X\to X$ be a compact operator with the operator norm $\|A\|<1$. Then $I-A$ is invertible and the Neumann series $$ S_N = \sum_{k=0}^N A^k $$ converges in the operator norm to $(I-A)^{-1}$: $$ \|S_N-(I-A)^{-1}\| \to 0, \ \text{ as } \ N\to \infty $$ Now I think all $S_N$ are compact operators hence the limit $(I-A)^{-1}$ is also compact. However this cannot be the case because then $$ I=(I-A)(I-A)^{-1} $$ would be compact, which is not possible for infinite dimension $X$. What is wrong in my argument?

  • 6
    $\begingroup$ $A^0 = I$ is not compact (except in trivial cases). Since $A^n$ is compact for $n > 0$, none of the $S_N$ is compact (they are all compact perturbations of the identity). $\endgroup$ – Daniel Fischer Apr 22 '14 at 13:22

As pointed out by Daniel Fischer, it is true that for each $n\geqslant 1$, $A^n$ is compact hence so is $\sum_{n=1}^NA^n$. Since a norm-limit of compact operators is compact and $\sum_{n=1}^\infty\lVert A^n\rVert$ is convergent, we obtain that $\color{red}{A}(I-A)^{-1}$ is compact.

But $(I-A)^{-1}$ is not compact, otherwise so would be $A(I-A)^{-1}-(I-A)^{-1}=-I$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.