Function classes related to Ruscheweyh derivatives

THEOREM 1. For any a, 0 < a < 1, Rn c K(a) for n>no = [32/(1 - a)]. PROOF. For f(z) = z + Y^,T=2akzk e ^4, a computation applied to (1) shows that (3) D»f(z) = z + k=2 If / e Rn, then Dnf e S"(0) and we must have (k+n n~l)\ak\ < k or, equiva- lently, (4) \ak\ < k (k + " ~ l \ for every k > 2. It is known [6] that / e K(a) if ££L2A:(fc - a)\ak\ < 1 - a. In view of (4) it thus suffices to show that £ £ 2 k2\ak\ < Y%Li k3(k+"n~lVl < 1 - " for n > n0. Since T,7=z(xlk2) < U we need only show that

THEOREM 2. The extreme points o/cl Rn(a), 0 < a < 1, are given by the functions / j ( x ) . x \x\ = 1, z € A, where (a)k = a(a + 1) • • • (a + k - 1). PROOF. In [3] it is shown that the extreme point of S* (a) are X Since D": f -* D"f is an isomorphism from Rn{a) to S*(a), and conse- quently preserves extreme points, we see from (3) that the extreme points of cl Rn(a) are given by n which simplifies to fx{z) denned in (6). REMARK. The special cases n = 0 and n = 1 in Theorem 2 reduce to the extr

Theorem 2 enables us to solve some extremal problems in Rn{a); for ex- ample, we have

COROLLARY 1. If'f(z) = z + J2T=2 akzk e Rn{a), then \ak\ ^ ~7T ~ ', k > 2, with equality for https://doi.org/10.1017/S1446788700033164 Published online by Cambridge University Press

COROLLARY 2. If f e Rn(a), then k=l v oo /» •I^+ETF k=2 ( ' " "'• with equality for fx(z) at z — xr. REMARK. It would be of interest to get fx(z) in (6) into closed form to obtain additional information and solutions to extremal problems. For exam- ple, we believe that the lower bounds for |/(z)| and |/'(z)| when / e Rn(a) occur for fx(z) at z = -xr. This is true for n — 0 and n = 1 (see [3]). The determination of the extreme points of cl Kn is an immediate conse-

THEOREM 3. The extreme points of cl Kn are {z/(l -xz); \x\ = 1, n e Wo}- PROOF. Note first from (1) that Z)"(z/(1 - xz)) = z/(l - xz)n+l so the family of functions {z/(l - xz)} is contained in Kn for every n. Thus we have the double inclusion {z/(l - xz)} cKncK0 = S*(l/2) (n = 0,1,2,...). Since the extreme points of cl S"(1/2) are {z/(l -xz): \x\ = 1} (see [3]) the result follows. 3. Convolution invariance For k + n- 1 we may express D"f as Z)"/ = hn* f. We also denote by h~l(z) the function nor

THEOREM 4. Iff,g e Rn(a), n>\, then (f*g)(z) € Rn(a). PROOF. With hn denned by (7) we must show that if / * hn e S*(a) and g*hn e S*(a), then (f*g)*hn e S*(a). Since / e Rn(a) c Ri(a) = K(a), we have {f*g)*hn = f*(g* hn) is the convolution of a convex function with a function in S*(a) and must therefore be in 5*(a). Hence (/* g){z) e i?n(a), and the proof is complete.

Theorem 4 may be put in an equivalent form.

THEOREM 4a. Ifz + ££°=2 {k+n-x)akzk and z + £ £ 2 {k+"-x)bkzk are both in S*{a), n > 1, then so is z + £j£ 2 (^-'Jafc^z*. Compare this with the following remarkable result of Suffridge. THEOREM A [9]. Define y(a,k), a < 1, by z °° V ' Jfc=2 ^/ z + Efe!=2 y(a,k)akzk and z + J2T=2 7(a,k)bkzk are both in S*(a), then so isz + J:k xL2y(a,k)akbkzk. Another equivalent form to Theorem 4a is

THEOREM 4b. Ifz + Y%L2 akzk and z + Y%L 2 bkzk are both in 5* (a), then so is z + £~2 ^ T T V f c Z * for n > 1. Setting bk — ak and n = 2 in Theorem 4b, we obtain the following COROLLARY. Ifz + ^=2aktk e S*(a), f/ie« z + 2 X 2 j^ryz* e S*(a). REMARK. Theorem 4 cannot be extended to include the case n = 0. The Koebe function fc(z) = z/(l - z)2 is in if0 = S*(0) but (it*A:)(z) = z + Z)m=2w2zm is n o t even univalent in A. We next show how to move to different classes of Rn(a) through convo- lutio

THEOREM 5. Let k=2^ n be a hypergeometric function. Then f e Rn{a) if and only iff* zH{z) belongs to the class Rn+m (a) for any m = 1,2, PROOF. For f{z) = J2T=i fl*z* G ^» ai = ^ a n d -^ defined in the lemma, we have that ( "V ) * ( ) \ k=0 '

THEOREM 6. The function f is in Rn(a) if and only if z + ,x(l+n)+n-l+2a\z2 PROOF. An application of (2) to (8) shows that / e Rn(a) if and only if https://doi.org/10.1017/S1446788700033164 Published online by Cambridge University Press