Real and complex analysis

90 REAL AND COMPLEX ANALYSISIf we replace s by - s (using Theorem 2.20( e» and then by s - t, the periodicity ofJ and Qk shows that the value of the integral is not affected. HencePk(t) = 2n: 1 f" _/(S)Qk(t - s) ds (k = 1, 2, 3, ...). (2)Since each Qk is a trigonometric polynomial, Qk is of the formNtQ k (t) = ~ L. a e illt ft, k , (3),,= -Ntand if we replace t by t - s in (3) and substitute the result in (2), we see that eachP k is a trigonometric polynomial.Let E > 0 be given. Since J is uniformly continuous on T, there exists a fJ > 0such that I J(t) - J(s) I < E whenever It - s I < fJ. By (b), we havePk(t) - J(t) = 2n: 1 f" -"{J(t - s) - J(t)} Qk(S) ds,and (a) implies, for all t, that1 f"I Pk(t) - J(t) I ~ 2n: _"I J(t - s) - J(t) I Qk(S) ds = A1 + A 2,where A1 is the integral over [ -fJ, fJ] and A2 is the integral over [-n:, -fJ] u[fJ, n:]. In Ab the integrand is less than EQk(S), so A1 < E, by (b). In A2, we haveQk(S) ~ '1k(fJ), hencefor sufficiently large k, by (c). Since these estimates are independent of t, we haveproved thatlim IIJ - Pkll co = o. (5)k"'coIt remains to construct the Qk. This can be done in many ways. Here is asimple one. Put{I + cos t}kQk(t) = Ck 2 ' (6)where Ck is chosen so that (b) holds. Since (a) is clear, we only need to show (c).Since Qk is even, (b) shows that1 Ck I" {I + cos t}kd Ck I" {I + cos t}k. d 2ck=- t>- smt t= .n: 0 2 n: 0 2 n:(k + 1)Since Qk is decreasing on [0, n:], it follows thatQk(t) ~ Qk(fJ) ~ n:(k; 1) C + ~os fJy (7)(4)