By Stanislaw Saks
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Extra resources for Analytic Functions
10). Next choose p_ < p and replaced by p p? and 2. 5) holds with p replaced by p i = 1 p or p~. By the Riesz Thorin theorem  vol. II, p. 5) also holds for p. 10) and a > X and 0 < r < 1 $ > X assuming the hypotheses of instead of a > -1 and b > -1. 10) on 3 > X, be written a and b |a-b| < pX, u < a < v and b < p-1. can, in the case and The a _> X and u < b < v, where u = u(p) = max(-l-p(X-l/s), -l-p(X-l/2), -Xp) and v = v(p) = min(-l+p(X+l-l/s), -l+p(X+l/2), Xp). 5) follows from the dual. Therefore, assume that v < p-1, u < -1 and v > p-1.
1). 3) holds f and r. 2) l e t theorem ( 7 . 3 ) . | f (x) | P w(x)dx. J h 3TT/4 2TT/3 L ( r , x , y ) f (y)dy| P w(x)dx <_ C Because by M = [A]. Since A is M < A < M-Hl. a . 2). (r,x)((); ' (y)]f(y)dy|Pw(x)dx. 4) bounded by C The condition r3ir/4 [ 0 M+1 > A f(y)|y2M+Y+1/2dy]pxa-p(2M^+3/2)dx. 2). 5); the inequality holds for any value of a. 6). 10) is greater than - 1 . 2). x/2 = u x is greater than - 1 . 7). 4), and replace M-l Y,o; (Y>6), I A. (r,x)()): (y). 3) is bounded by r 2 T r/3 J 0 |f(y)|y2M+Y+1/2dy]P(^-x)b+p(3+1/2)dx.
8) to replace | <(>. 6). 3) holds, then there is a j , such that |f(x)(j>fY ' 6) (x)|dx£C(j+l) C. 8). - 0 3 r ][ rnd (p n=0 (x)dx = r d. 8). x | d. | <_ C(j+l) C and (x) 45 with converges for every for every nonnegative integer in (0,TT) x j. 8); the dominated convergence theorem then shows that g (x)c()JY'6)(x)dx = lim 3 0 W**> ri\ N [ I rn d cj)(Y'6)(x)]cj)^'6)(x)dx= rjd.. 2) is finite. a < -l+p(y+3/2) and T f(x). 3). 3) holds because of the hypothesized existence of 0 f (x)c))^Y'