By Philip E. Gill

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**Example text**

Proof. 2 (ii), observing that = xg(x) where Using the Cauchy integral representation of that if x £ A f(O) = 0 => f(x) g e Hol ((S (x)) . f £ Hol(0(x)), and one immediately verifies f(x) f £ Hol(0(x + K)) then since 0(x+K)CQ(x) and f(x+K) = f(x) +K. (ii) Since f (x) £ R, 0(f(x) + K) NOW (iii) 6(x + K) C Q(x), Q(f(x + K)) 0 = = {O}. U(x + K) = {O}, so, by hypothesis, 0(x + K), f(0(x + K)) f((J (x + K)) = x £ mK(A) _> 0 j o(x + K). not vanish on thus = Now 0(x + K) G 0(x), hence x E R. 4 THEOREM.

Is equivalent to invertibility is an open semigroup of A which is stable under perturbations by elements of I(A). 5 we have Inv(A)' If Inv(A'); = then P £ II (A) A/P _ (D(A)' $(A'); = I(A'). is a primitive unital Banach algebra and this fact enables us to develop Fredholm theory in the structure space of I(A)' A. 3, h(I(A)) = h(psoc(A)). 3 s LEMMA. P, If further (P £ h(soc(A))). there exists a unique s' £ Min(A') P E 11(A) Proof. 5. s 0 P, s+ P # 0, so if a c A (s + P) (a + P) (s + P) But S' £ Min(A') such that s + P E Min(A/P).

F £ Hol(0(x)), and one immediately verifies f(x) f £ Hol(0(x + K)) then since 0(x+K)CQ(x) and f(x+K) = f(x) +K. (ii) Since f (x) £ R, 0(f(x) + K) NOW (iii) 6(x + K) C Q(x), Q(f(x + K)) 0 = = {O}. U(x + K) = {O}, so, by hypothesis, 0(x + K), f(0(x + K)) f((J (x + K)) = x £ mK(A) _> 0 j o(x + K). not vanish on thus = Now 0(x + K) G 0(x), hence x E R. 4 THEOREM. K. Q(A) is due to Zemanek (104). 3). Let A be a unitaZ Banach algebra, then rad(A) _ {x £ A : x + Inv(A) C Inv(A)} = {x £ A : x + Q(A) C Q(A)}.