Let F he a picccv ... isc continuous funclion (Sec. 42) of

Chapter 0, Problem 12.2

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Let F he a picccv ... isc continuous funclion (Sec. 42) of 8 on chc inlcrval 0 ~ A ::::: 2JT. The Poisson integral tra11sform of F is defined in tcm1s of the Poisson kernel P(ro. r. - 8 ). introduced in Sec. 134, hy means of the equation ( I ) I 12:-r U (r. H) = ~ P(ro.r. - H) F< - H ) I F ( ) - F ( 8) I d . 2n o For convenience. we let F be extended periodically. with period 2;rr. so that the integrand here is perilxlic in wilh that same period. Also. we agree that 0 < r < r0 because of the nature of the limit to be established. Next. we observe that since F is continuous al H. there is a small positive number ex such that (4) f; IFll - F(8)1 < ., whenever l-8l~ex. Assume now that I - HI ~ a and write (5) where I 1 111u /1 (r) = - P(r0 . r. - H) IF() - F({Jl) d. 2;rr II U h(r) = ~ { uil:r P(r0 r. - H) [F() - F(8)) d. _;rr Jn.1 u The fact that P is a positive function (Sec. 134). together with the first of inequalities (4) just above and properly (_{). Sec. 134. of that function. enables us to write I {111a l/1 (r)I ~ 'l;rr Jo u P(ro. r. - H) IF() - F(8)1 d f; f.l:r F. < - P(r0 r. - 8) d = -. 4;rr . 0 2 As for the integral /2 (r). one can sec from Fig. 192 in Sec. 134 that the denominator Is - :-f in expression (8) for P(ro. r. - H) in that section has a (positive) minimum value 111 as the argument of s v:.u-ics over the closed interval 8 + (X ~ ~ e - ex + 2;rr . So. if M denotes an upper bound of the piecewise continuous function I Fl) - F(Hll on the in le rval 0 ~ ~ 2;rr. it follows th al (r ~ - r 2 )M 2Mr11 21Wro F l/:~(r)I ~ 2:r < --(r0 - r) < -- 15 = - 2 :r Ill 111 111 ., 422 ll\TEGRAL fooRMt:L..\S or: TllE PoL~sol'\ TYPE whenever t(i - r < ,5 where ( 6) 111 F: D=--. 4M ro Finally. the results in the t\vo preceding paragraphs tell us that /-~ IU(r.8) - FI:::; l/i(r)I + lh (II= 0. J. 2 ... . ). iT () I ;2:r b,1 = - F(