7

Let $A$ be a Banach Space and $T$ be a bounded operator on $A$. Given that, spectrum of $T$, $\sigma[T]$, is $ F_1 \cup F_2;$ where $ F_1, F_2$ are disjoint closed set in complex plane. 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 $i=1,2.$

Until now what I have done is, taking 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 the 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.

ViktorStein
  • 4,838
Timon
  • 2,401

1 Answers1

3

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 $0\not\in\sigma(T|_{A_i})$.

Martin Argerami
  • 205,756
  • I have just one doubt that how i will show F_1 is a subset of spectrum of T|A_1. – Timon Sep 28 '12 at 07:09
  • 2
    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. – Martin Argerami Sep 28 '12 at 18:21
  • 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. – JZS Jan 28 '15 at 21:27
  • 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 $A_i$). – Martin Argerami Jan 30 '15 at 03:44