Timeline for Is the sum of spectral projections a projection?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2015 at 9:55 | comment | added | M.González | If $T$ is a closed operator with non-empty resolvent set $\rho(T)$ and you take $\alpha\in \rho(T)$, then $A:=(T-\alpha I)^{-1}$ is in the Banach algebra of bounded operator on $X$, and given an analytic function $f$ on a neighborhood of $\sigma(T)$ you can find an analytic function $g$ on a neighborhood of $\sigma(A)$ so that $f(T)=g(A)$. | |
| Oct 10, 2015 at 16:33 | comment | added | Frank | Hang on... But unbounded operators don't form a Banach Algebra, do they? | |
| Oct 9, 2015 at 20:37 | comment | added | Alain Valette | @Frank: It is rather the holomorphic functional calculus that is well-behaved... This holds in any Banach algebra! | |
| Oct 9, 2015 at 15:49 | comment | added | Frank | Thanks for these answers. The book by Taylor and Lay was a very helpful reference! It really surprised me that non-normal operaors are so well-behaved... | |
| Oct 9, 2015 at 14:36 | comment | added | M.González | For finite sums and a closed operator $T$ acting on a complex Banach space $X$, the formula is valid in general. This is a consequence of the properties of the analytic functional calculus. See Theorem 9.1 in Chapter V of the book A.E. Taylor and D.C. Lay "Introduction to functional analysis, 2nd ed". Wiley 1980. | |
| Oct 9, 2015 at 14:15 | history | answered | Chris Ramsey | CC BY-SA 3.0 |