If $T:X\to X$ is a compact linear operator, then for any bounded linear operator $S:X\to X$ we have that $S(I-T)=I$ if and only if $(I-T)S=I$. Where $X$ is a normed space, also $T$ is bounded.

With this result I should be able to show that $I-(I-T)^{-1}$ is a compact operator.

I'm not sure how to proceed with either. Any solutions or hints are greatly appreciated.

  • $\begingroup$ @TheoBendit: That is not true in general (although it holds for example in hilbert spaces). $\endgroup$ – PhoemueX Jul 11 '15 at 7:53

For the first part, see this question:

Assume $T$ is compact operator and $S(I- T) = I $.Is this true that $(I- T)S =I$?

For the second, let $A = I - (I-T)^{-1}$, then $A(I-T) = -T$ and so $A = -(I-T)^{-1}T$. Since $T$ is compact, so is this product.

  • $\begingroup$ Can you provide a link to the bounded inverse theorem cited in your linked post? Thanks. By the way, both answers endorsed! $\endgroup$ – Robert Lewis Jul 11 '15 at 21:16
  • $\begingroup$ The "bounded inverse theorem" is really the open mapping theorem for injective operators - see en.wikipedia.org/wiki/… $\endgroup$ – Prahlad Vaidyanathan Jul 11 '15 at 21:18
  • $\begingroup$ Got it. Thanks again! $\endgroup$ – Robert Lewis Jul 11 '15 at 21:19
  • $\begingroup$ Thank you for your post. One thing about the last sentence. How does $T$ being compact imply that the last product is compact? $\endgroup$ – 1233dfv Jul 11 '15 at 22:20
  • $\begingroup$ the compact operators form an ideal: if you compose a compact operator with a bounded operator on either side, the product is compact. $\endgroup$ – Prahlad Vaidyanathan Jul 11 '15 at 22:46

Note that:

$$I-(I-T)^{-1} = I-(I-T+T)(I-T)^{-1} = I-I-T(I-T)^{-1} = -T (I-T)^{-1}$$

A compact operator composed (on the right or on the left) with a bounded operator is still compact. But $-T$ is compact, and $(I-T)^{-1}$ is bounded...

  • $\begingroup$ @1233dfv: I've added a few details. $\endgroup$ – D. Thomine Jul 11 '15 at 21:14

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