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**Sample text**

Step IV. 2). 7) yields that Ige(z)1 = 1 Ife(z)1 for z E aD almost everywhere (cf. [Du, Chapter 2]). 1). If z E e, then (cf. II). Hence h, h- I and fe are in A. Moreover 0 _1__ Ih(z)1 Hence remembering that B 11 - fiz)1 ~ 1 v'w 2 (z) + v2 (z) _1_ _ - I "'" lw(z)1 - B > 1, we get = 11 - exp h-l(z)1 ~ explh-l(z)1 - 1 ~ Ih-l(z)lexplh-l(z)1 ~ B-Iexp(I) = 0(0 1 / 2 ). 2) is slightly harder. To treat simultaneously the case where e is a closed set and e is not closed we put e* = e" if e is closed, and e* = e otherwise.

Du, Chapter 2]). 1). If z E e, then (cf. II). Hence h, h- I and fe are in A. Moreover 0 _1__ Ih(z)1 Hence remembering that B 11 - fiz)1 ~ 1 v'w 2 (z) + v2 (z) _1_ _ - I "'" lw(z)1 - B > 1, we get = 11 - exp h-l(z)1 ~ explh-l(z)1 - 1 ~ Ih-l(z)lexplh-l(z)1 ~ B-Iexp(I) = 0(0 1 / 2 ). 2) is slightly harder. To treat simultaneously the case where e is a closed set and e is not closed we put e* = e" if e is closed, and e* = e otherwise. Let X = {z E aD: Iv(t)1 ~ 77} for some 77 > O. lwldm ~Bo + b. lfel dm ~ m(e*) + CI(B77-IO + b77- I ) + b + 17 2 b- 1 ~ 0(0 1 / 6 ), provided we put 77 = This completes Step IV.

T(x) E Xl. 8), 1. The construc- tion of Q is more complicated. Given f E C(aD),! denotes the continuous function on the unit disc D which extends f and is harmonic for Iz I < I. Now fix n = I, 2, . . Un(s))v(ds) Js for fE C(aD). fn is well defined and if f E A [regarded as a subspace of C(aD)), then! is the analytic extension of f onto the interior of the unit disc; hence it admits uniform approximation by polynomials in z. This implies that! 0 fn E X; hence f aDf da = 0 (because v E Xl). Thus by the M.