Canonical Decomposition Theorem by Pilgrim K. M.

By Pilgrim K. M.

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Is a Banach algebra hence there exists and our next task is to Clearly then x1 6 p1Ap1 such that The hypothesis implies that zero ap Ap (x1), hence this set contains then g2p1 = plg2 with similar relations holding for p2 6 p1Ap1, we so the associated spectral p e J. 1,x1) e piAp1, so, at least one of then and x 6 J p ¢ soc(A). islthe only possible accumulation point of Set a(x), there exists contains at least two points. O. lO, an isolated point such that zero is the only (x c J). is semisimple.

En+l A. Proof. p = e1 + ... + en. Write fA ¢` pA f = pf + (1 - p)f, and (1 - p)fA idempotent Then and R = pA. p2 = p it follows that g, say. 3), = then g = gpgp = 0 en+l £ Min(A). (1 < k < n), eken+l = ekpen+l = 0 = en+lpek = en+lek is an orthogonal subset of Min(A). 3) contains a minimal en+l - en+12. which is false. en+1 Since (1 - p)fA C fA + pA, so the set Further e A + ... + en+lA C R + fA. 3) en+l = g(l - p), = = g A g(1 - p) A = en+l A. n+l Therefore R + fA CIE ekA S giving fA C pA + en+1A 1 Let LEMMA.

A - x + K E Inv(A/K), hence 3 Let A be a unitaZ Banach (Ruston characterisation) x E A. is a Riesz point of n {Ak}l, say. K. x, the The spectral n idempotent associated with this set, p = Z p(}k,x) E K, and r(x - px) < S. 1 57 r (x + K) < r (x - px) < S, Thus and since r(x + K) = 0 0 is arbitrarily small, 6 In terms of the Browder spectrum this result states that x c R <=> SW C {O}. and dote that if then T 6 P(X) is a finite dimensional linear space X is empty. 3. closed nor two-sided. 6 THEOREM.

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