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Let $T:\mathscr{D}(T) \rightarrow H$ be a operator on a complex Hilbert space. If for some $\lambda$ we have that $S(I-\lambda S)^{-1}$ is a bounded inverse for $T-\lambda$ does it then follow that $S$ is an inverse for $T$?

simon
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1 Answers1

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I'm going to give it some further consideration, but I think this is substantially correct.

We have

$$( T - \lambda I ) S ( I - \lambda S )^{-1} = S ( I - \lambda S )^{-1} ( T - \lambda I ) = I$$

so

$$TS(I - \lambda S )^{-1} - \lambda S(I - \lambda S)^{-1} = S(I-\lambda S)^{-1} T - \lambda S(I - \lambda S)^{-1}$$

hence

$$TS (I - \lambda S)^{-1} = S(I - \lambda S)^{-1} T = I + \lambda S(I - \lambda S)^{-1}$$

set $R = S( I - \lambda S)^{-1}$. We have

$$ TR = RT = I + \lambda S(I - \lambda S)^{-1}$$

Applying $TR$ to $(I - \lambda S)$ we obtain $TS = I - \lambda S + \lambda S = I$. On the other hand,

$$ST = S ( I - \lambda S ) ( I - \lambda S )^{-1} ( T-\lambda I) + \lambda S = I - \lambda S + \lambda S = I, $$

so your claim is true.

Remark. I'm supposing all compositions can be done.

user91126
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    The first equation should be $I=( T - \lambda I ) S ( I - \lambda S )^{-1} \supseteq S ( I - \lambda S )^{-1} ( T - \lambda I ) $ etc. – Yurii Savchuk Jan 10 '14 at 12:52
  • @Federico very nice. As Yurri Savchuck noted there should some minor changes should I try to implement them. – simon Jan 10 '14 at 14:50
  • simon, you and @Yurii are right. This is what I was referring to above. As soon as possible, I'll improve my answer. I have little time at the moment, so in the meanwhile feel free to edit the answer. – user91126 Jan 11 '14 at 14:51