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**Additional info for Completing the Riesz-Dunford Functional Calculus**

**Sample text**

Since C n meets A , this contradicts the condition t h a t U n f(E) = {X} . Therefore K is closed. Let N be a n o r m a l operator with a(N) = K e . 9, there exists a n operator S t h a t is similar to T1 0 N and also satisfies I I S - T x I I < 2e . 9) S = S x 0 • • • 8 S m 0 F , where F = T x | 9<(TY ; a(T x ) \ Kg) a n d for each j , a(Sj) is a component of K e . 13 ii), F G fAjf(E)) . By construction, for 1 < j < m, a(Sj) meets A . Therefore by Lemma 2 . 1 1 , each Sj e fAflflE)) . Aflf(E)) .

Since dim [ker T] > 1 and ker T p c ker T p + 1 , it m u s t be shown t h a t ker T p ^ ker T p + 1 for all integers p . If this is not the case, t h e n for some p , ker T p = ker T p + k for all k > 1 ; in particular, ker T p = ker T 2 p . So by replacing T with T p , it suffices to show t h a t if T is Fredholm with ind T = 0 a n d 0 is not a n isolated point of a(T) , t h e n ker T * ker T 2 . Suppose the contrary; so ker T = ker T 2 . Let ^ = r a n T ; since T is Fredholm, ^ is a Since ker T = ker T 2 , it follows t h a t ker T n ^ = (0).

14 ; also let m^ = ind (A - S) for A in Q k , r(k) =1, and (k) p x = p . Vmli ) : i < i < r(k) , m f)> o } k= 1 Q e0 r 0{A(Gi1kV"mh,:i*
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