By Yves Diers

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**Extra info for Categories of Boolean Sheaves of Simple Algebras**

**Example text**

D For certain applications, it is convenient to weaken the formal definition of a spectral capacity. Given an operator T E L(X) on a Banach space X, a mapping E from the collection of all closed subsets of C into the collection of all T-invariant closed linear subspaces of X is called a 2 -spectral capacity for T if E preserves countable intersections and satisfies the conditions E(0) = {O} , E(C) = X, a(T I E(F)) <;;; F for every closed set F <;;; C , and X = E(U) + E(V) for every open cover { U, V} of C.

By the Riesz functional calculus, it follows that f (T/ E(U) ) Q I Xr (F) = Q f(T I Xr (F)) for every complex-valued analytic function f on some open neighbourhood of F. Since V n F = 0, there is an analytic function f for which f = 1 on an open neighbourhood of V and f = 0 on an open neighbourhood of F. For this function, it follows that f(T/E(U)) is the identity operator on X/E(U) , while f(T I Xr (F)) is the zero operator on Xr (F) . We conclude that Q I Xr (F) = 0, and hence that Xr (F) <;;; E(U) .

Thus <:;; and therefore x E Xr (F) . (f) First suppose that T has SVEP. Then, for every E there is an analytic function f : C -t for which (T - >.. ) j(>.. ) = for all >.. E C . Since for >.. E p(T) , and l l (T - >.. ) - 1 I I -t 0 as -t oo , it follows that J(>.. 3, f is constant. Because (T -t 0 as j >.. i -t oo , we conclude that f = 0 on C, and hence = 0. This proves that = whenever T has SVEP. Conversely, suppose that Xr (0) = and consider an analytic function f : U -t on an open set U <:;; C such that (T - >..