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I am stuck in a problem regarding functional calculus. Can anyone help me? Here is the problem.

Let $A$ be a Banach Space and $T$ be a bounded operator on $A$. Given that, $\sigma[T]-$ spectrum of $T$ is $ F_1 \cup F_2;$ where $ F_1, F_2$ are disjoint closed set in complex plane. Then show that there exist topologically complemented subspace $A_1,A_2$ of $A$ such that $A_1,A_2$ are invariant subspace for $T$ and $\sigma(T|A_i)=F_i$ for $i=1,2.$

Till now what I have done is, I have taken disjoint open set $G_i$ containing $F_i$ and have chosen $f_i= 1_{G_i}-$ the characteristic function on $G_i$, (which are actually analytic on $G_1\cup G_2$). Then I have taken $A_i$ as range of the projection $f_i(T)$, using the functional calculus for $T$.

using spectral mapping theorem one can tell that, $\sigma(Tf_i(T))= F_i\cup$ {$0$}. If my guess is correct then I have to show $\sigma(Tf_i(T)|A_i)= F_i$. At this stage I need help.

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  • $\begingroup$ I think you assume that $A$ is Hilbert and $T$ is normal. Right? $\endgroup$ – Norbert Sep 27 '12 at 21:02
  • $\begingroup$ No, A is Banach space and T is any bounded operator on A. Use functional calculus to define f(T) for any holomorphic function f on the spectrum of T. $\endgroup$ – Timon Sep 27 '12 at 21:05
  • $\begingroup$ Where have learned about Borel functional calculus for arbitrary Banach spaces and arbitrary bounded operators? $\endgroup$ – Norbert Sep 27 '12 at 21:07
  • $\begingroup$ learned from Conway's A course in a Functional Analysis. $\endgroup$ – Timon Sep 27 '12 at 21:09
  • $\begingroup$ In this case see proposition 4.11 Chepter VII. Here you can find a mathod how to "extract" $\{0\}$ $\endgroup$ – Norbert Sep 27 '12 at 21:41
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If $0\in F_i$, you are done. Otherwise, as $F_i$ is compact, there exists $\delta>0$ such that $|z|>\delta$ for all $z\in F_i$. Then the function $g:z\mapsto zf_i(z)$ satisfies $|g(z)|\geq\delta$ for all $z\in F_i$ and so $h=1/g$ is well-defined and analytic on $F_i$. This implies that $Tf_i(T)|_{A_i}$ is invertible (with inverse $h(T)$), and so $0\not\in\sigma(T|_{A_i})$.

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  • $\begingroup$ I have just one doubt that how i will show F_1 is a subset of spectrum of T|A_1. $\endgroup$ – Timon Sep 28 '12 at 7:09
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    $\begingroup$ If $\lambda\in F_1$, then you can repeat the proof in the answer to show that $T-\lambda I$ is invertible when restricted to $A_2$. This forces $T-\lambda I$ to be non-invertible on $A_1$, as otherwise it would be invertible everywhere contradicting that $\lambda$ is in the spectrum. $\endgroup$ – Martin Argerami Sep 28 '12 at 18:21
  • $\begingroup$ I am now working on this problem several years after your answer has been posted. I appreciate your insight. But I am wondering, could you clarify why it is enough for us to know merely that $h$ is well-defined and analytic on $F_i$? To use $h$ with functional calculus, I thought that we need to know that $h$ is analytic on a neighborhood which contains the whole spectrum of some particular operator. So I'm wondering which operator has just $F_i$ as its spectrum? Any clarification you can give me is greatly appreciated. $\endgroup$ – JZS Jan 28 '15 at 21:27
  • $\begingroup$ Properly, $h$ is defined as $1/g$ on $F_i$ and as for example as the identity $h(z)=z$ on the other one. As $F_1\cap F_2=\emptyset$, $h$ is analytic. Then you have $$h(T)\,T|_{A_i}=h(T)\,Tf_i(T)|_{A_i}=f_i(T), $$ which is the identity on $A_i$. So $T|_{A_i}$ is invertible (as an operator on $A_i$). $\endgroup$ – Martin Argerami Jan 30 '15 at 3:44

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