Timeline for Banach Algebras and the peripheral spectrum
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2012 at 9:01 | comment | added | ChantelD | 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. | |
| Sep 7, 2012 at 8:46 | comment | added | ChantelD | 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 $\sigma(a)$. | |
| Aug 27, 2012 at 23:40 | comment | added | Duchamp Gérard H. E. | One way to repair your theorem would be to consider that your $\lambda_i$ are isolated. | |
| Aug 27, 2012 at 18:44 | history | edited | Yemon Choi |
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| Aug 27, 2012 at 18:43 | comment | added | Yemon Choi | What happens when your Banach algebra has no non-trivial idempotents? How are you constructing these supposed spectral idempotents? | |
| Aug 27, 2012 at 16:05 | comment | added | Duchamp Gérard H. E. | In case $sp(a)=[-1,1]$, $psp(a)={-1,1}$. How do you compute $p(1,a)$, for instance ? | |
| Aug 27, 2012 at 13:45 | history | asked | ChantelD | CC BY-SA 3.0 |