Compact linear operators $S$ and $T$, show that $S(I-T)=I$ if and only if $(I-T)S=I$ and deduce that $I-(I-T)^{-1}$ is a compact operator

Question

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.

@TheoBendit: That is not true in general (although it holds for example in hilbert spaces). — PhoemueX, Jul 11 '15 at 07:53

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 11 '15 at 21:06

Prahlad Vaidyanathan

31,554

Can you provide a link to the bounded inverse theorem cited in your linked post? Thanks. By the way, both answers *endorsed!* – Robert Lewis Jul 11 '15 at 21:16
The "bounded inverse theorem" is really the open mapping theorem for injective operators - see https://en.wikipedia.org/wiki/Open_mapping_theorem_%28functional_analysis%29 – Prahlad Vaidyanathan Jul 11 '15 at 21:18
Got it. Thanks again! – Robert Lewis Jul 11 '15 at 21:19
Thank you for your post. One thing about the last sentence. How does $T$ being compact imply that the last product is compact? – 1233dfv Jul 11 '15 at 22:20
the compact operators form an ideal: if you compose a compact operator with a bounded operator on either side, the product is compact. – Prahlad Vaidyanathan Jul 11 '15 at 22:46

D. Thomine · Answer 2 · 2015-07-11T21:13:48.347

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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...

edited Jul 11 '15 at 21:13

answered Jul 11 '15 at 13:37

D. Thomine

10,870

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

Compact linear operators $S$ and $T$, show that $S(I-T)=I$ if and only if $(I-T)S=I$ and deduce that $I-(I-T)^{-1}$ is a compact operator

2 Answers2