Banach Algebras and the peripheral spectrum - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:59:34Z http://mathoverflow.net/feeds/question/105627 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105627/banach-algebras-and-the-peripheral-spectrum Banach Algebras and the peripheral spectrum ChantelD 2012-08-27T13:45:44Z 2012-08-27T18:44:00Z <p>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.</p> <p>Denote by psp($a$) the peripheral spectrum of $a\in A$. </p> <p>$\underline{\textrm{Theorem}}$<br> Let $A$ be a Banach algebra with $a\in A$, $r(A)>0$ and psp($a$)={$\lambda_1,\lambda_2,...,\lambda_k$}.<br> Then $a=a_0+∑_1^k λ_i p_i$<br> where $a_0=a(1-p)$ and $p$ is the spectral projection relative to $a$ and {$λ_1,λ_2,⋯,λ_k$}.</p> <p>The following is as far as I got, and any suggestions would be much appreciated.</p> <p>$\underline{\textrm{Proof}}$:</p> <p>It is clear that $a=a(1-p)+ap=a_0+ap$. </p> <p>By Cauchy's Theorem for Multiply Connected Domains, we have that $p=p(λ_1,a)+p(λ_2,a)+...+p(λ_k,a)=p_1+p_2+...+p_k$<br> (if we denote $p(λ_i,a)$ by $p_i$).</p> <p>Thus, $a=a_0+\sum_1^k ap_i$.</p> <p>What's left to prove is that $ap_i=aλ_i$ for $i=1,...,k$.</p> <p>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.</p>