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By Domingo A. Herrero

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But this is impo~ sible: if~ f Ak (for all k = 1,2, ••. -T) J J J case, this contradicts our assumptions on y. 17. LetT e L(H), and Let n 1 be a component ofpr(T)such that nul(A-T) 1, A e n1 • Fo~ any £ > 0 there exists a right resoLvent R of T on n1 except for an at most denumerabLe set s1 , which does not accumuZate in n1 , and satisfies s1 c ran 1 J£. = PROOF. -T)y f 0 for all A. E n 1 \s 1 , where s 1 is an at most denumerable subset which does not accumulate in n1 and such that sl c canlJ£.

Now a straightforward computation shows that L = VRV*, where V is the unitary operator defined by Vgt = { em+ 1-t' ft' m+l-r s t s 0 1 s t s s and em+1-t' 1 s t s m. • }), these two-dimensional subspaces are pa: wise orthogonal and (T-R)gt L (T-R)kt for 1 s t s m-1, we conclude that . n 2 (m+l) + (cos 2 (m+l) - 1) J , . n 2 (m+l) - 1) +bees 2 (m+l) max lal2+lbl2=1 ( 11 1) 2 ~ 2 . 'If ( ) . n (m+l) = s m • mw m1r Since IIVTV*-LII = IIT-V*LVII = IIT-RII, we are done. D The above proof admits a very simple geometric description: Th1 action of q r (e r + e r- 1 + e r- 2 + ••• + e 2 + e 1 + 0) can be described by an arrow of length r, and the action of T = qreqs by two arrows o1 lengths r and s, as follows m r s Similarly, R can be indicated as m n T maps em+l to em' the pair {em+l-j'fj} onto the pair {em-j'fj+: of "parallel" vectors (j = 1,2, ••• ,m-l) and the pair {e 1 ,fm} onto {0, fm+l}.

46 € M. \'le 0 \' n ~ n ~ l·k=l ker{~k-T) = ker ITk=l{~k-T) . PROOF. The inclusion 'c' is obvious and the converse inclusion is trivial for n = 1. We shall proceed by induction over n. Assume that we have n mk n ~ Lk=l ker{~k-T) ~ ker rrk=l{~k-T) and let . • ,~n} and mn+l be a natural number. : ker{Xk-Tfk = 1,2, ... n+l-T) n+l x = Lk~l yk. )m 1 {~k-T)~J n+ Yk' xk = { Li=l {~n+l-~k) k 1, 2, ••• , n, and follows. 14. -T)x• has an accumuZation point r;; e pr(T); then Pker(A-T)x = 0, foP alZ A in the component Qr, of pr(T) aontaining the point 1;.

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