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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.

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    @MartinArgerami For the moment I thought that $f_i$ is only Borel function and we can't apply holomorphic functional calculus.2012-09-28

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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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    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 $A_i$).2015-01-30