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Here is a little theorem that I'm trying to prove. I haven't seen it in literature before, but I think the applications will be quite useful, particularly in the context of Banach algebras.

Denote by psp($a$) the peripheral spectrum of $a\in A$.

$\underline{\textrm{Theorem}}$
Let $A$ be a Banach algebra with $a\in A$, $r(A)>0$ and psp($a$)={$\lambda_1,\lambda_2,...,\lambda_k$}.
Then $a=a_0+∑_1^k λ_i p_i $
where $a_0=a(1-p)$ and $p$ is the spectral projection relative to $a$ and {$λ_1,λ_2,⋯,λ_k$}.

The following is as far as I got, and any suggestions would be much appreciated.

$\underline{\textrm{Proof}}$:

It is clear that $a=a(1-p)+ap=a_0+ap$.

By Cauchy's Theorem for Multiply Connected Domains, we have that $p=p(λ_1,a)+p(λ_2,a)+...+p(λ_k,a)=p_1+p_2+...+p_k$
(if we denote $p(λ_i,a)$ by $p_i$).

Thus, $a=a_0+\sum_1^k ap_i$.

What's left to prove is that $ap_i=aλ_i$ for $i=1,...,k$.

I have thought about using the notion of simple poles, which works very nicely if we consider $a$ to be quasi inessential (analogous to quasi compact). But we don't have this in our case.

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    $\begingroup$ In case $sp(a)=[-1,1]$, $psp(a)={-1,1}$. How do you compute $p(1,a)$, for instance ? $\endgroup$
    – Duchamp Gérard H. E.
    Commented Aug 27, 2012 at 16:05
  • $\begingroup$ What happens when your Banach algebra has no non-trivial idempotents? How are you constructing these supposed spectral idempotents? $\endgroup$
    – Yemon Choi
    Commented Aug 27, 2012 at 18:43
  • $\begingroup$ One way to repair your theorem would be to consider that your $\lambda_i$ are isolated. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Aug 27, 2012 at 23:40
  • $\begingroup$ Thanks for the comments. I realise I was not very clear in the theorem. I am actually assuming that psp(a) contains isolated points only and that $p(λ_i,a)= \frac{1}{2\pi i}\int_{\gamma_i}(a-\alpha)^{-1} d\alpha$ with $\gamma_i$ a circle around $\lambda_i$ which does not contain any other points of $\sigma(a)$. $\endgroup$
    – ChantelD
    Commented Sep 7, 2012 at 8:46
  • $\begingroup$ My ultimate aim is to "hopefully" show that under certain assumptions, the psp of an element in a Banach algebra consists entirely of Riezs points. This is just one step closer to that goal. I have seen it done by Schaefer for bounded linear operators, but I'm hoping to transport that result into Banach algebras. $\endgroup$
    – ChantelD
    Commented Sep 7, 2012 at 9:01

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