Matrix transformation of univalent power series

THEOREM 1. / / / is analytic in A and maps A one-to-one onto a convex domain, then the Cesaro transform g offis also analytic, univalent and convex in A. PROOF. A function / analytic in A and satisfying/'(0) ^ 0 is univalent and convex if and only if (12) R Such functions are known to be starlike of order \ as proved in [7] and [11]; that is, (13) Now, let / be analytic, univalent and convex in A and let g be the Cesaro transform of/. Then / and g are related through equation (10) from which we

THEOREM 2. Iff is analytic in A,/(0) = 0, and if f maps A one-to-one onto a starlike domain, then the Cesaro transformation offis also univalent in A. PROOF. Let g be the Cesaro transformation of/. Then, equation (10) holds and since w = 1 /(I — z) maps (one-to-one) onto Re w > \ this shows that (17) R< In particular, this implies that (18) R Equation (18) is the definition that g be close-to-convex in A, since/is univalent and starlike and /(0) = 0. Each close-to-convex function is univalent in

Theorem 2 provides us with numerous examples of univalent functions https://doi.org/10.1017/S1446788700029098 Published online by Cambridge University Press

LEMMA 1. Suppose that kk are real numbers and ek are complex numbers for k = 1,2, ••-,« so that n / k ^ 0 , Z lk<>2 and \ek\^ 1. k = l Then n c 1 - ^ is univalent and starlike in A. We also consider the examples (19) /(z)= (1 - z)2 which shall later be of importance in obtaining a counter-example with regard to

THEOREM 3. There is a function f that is analytic and univalent in A so that/(0) = 0 and the Cesdro transformation of f is not univalent in A. PROOF. Let / be given by (26) where a = 2e~'°'cos a, a| < n/2, a ¥= 0 and let g be the Cesaro transformation of/. Then, by (11) we may write (27) 9iz) = [ We shall show that g is not univalent by studying its local behavior at z = 1 and z = — 1 for points in A. The derivative of g is analytic for | z | :g 1 except at the points z = 1 and i = — 1 and at z

THEOREM 4. IfheU and h(z) = z + Y£=2cnzn, then (33) (34) (35) (36) Furthermore, if h is associated with f and f(z) = z + I anz\ then \h(z) h'(z) zh'(z) Hz) C2-C3 1 < z

LEMMA 2. If feSt and p is given (0 < p < 1), then f can be uniformly approximated in z\^p by functions of the form F(z) = , where \ek\ = 1, kk S: 0 n mi-skzf* and

THEOREM 5. IffeSt and g is the Euler-Knopp transformation off, then h = g Ire St. PROOF. IffeSt then/can be approximated by functions F given by Lemma 2. The Euler-Knopp transformation g off is thereby approximated by the functions G that are the Euler-Knopp transformations of the functions F. If we show that each such G is starlike, then Re zG'(z)/G(z) > 0 in A and, as a consequence, Re zg'{z) /g(z) S: 0 or, equivalently, Re zh'iz) \h{z) 2; 0. This implies that Re zh'(z)jh(z) > 0 in A as the eq

LEMMA 3. The set {w:w = zf'(z)jf(z), | z | < l , feS} consists of all complex numbers except w = 0.

LEMMA 4. The set {w: w = xf'(x)/f(x), 0 ^ x < 1, / e S } consists of all complex numbers except w = 0. These results are quite expected and probably are easy consequences of the more delicate information regarding the exact region of variability of zf'(z)/f(z) with z fixed and / varying over S. Our argument is quite direct and affords an explicit construction of a function/in S so that h $S. To prove Lemma 3 we consider the functions (19) where (20) holds so that / e S. From (19) we find that f($

Lemma 4 is a consequence of Lemma 3 and the fact that i f / e S and g(z) = (1 /e)/(ez) then g e S for each complex number s, | e | = 1. The relation between / and g is equivalent to/(z) = (I Id) g(5z) where 3 = 1/e and so zf'(z)/f(z) = 3zg'(8z)jg(3z). Thus, any number z/'(z)//(z) where / e S and zeA is also obtained in the form xg'(x)/g(x) where g eS and 0 ^ x < 1 simply by choosing 3 so that <5z = | z | = x.

THEOREM 6. T/rere is a function f in S so that the Euler-Knopp trans- formation g offis not univalent in A for some values ofr. In other words, [/ + S. PROOF. Since the derivative of a univalent function does not vanish, we obtain our conclusion by showing that there is a function / in S and a number r (0 < r < 1) so that g'(z) = 0 at some z (| z | < 1). Because of equation (32) the condition g'(z) = 0 is equivalent to l - ( l - r ) z If we let w = rz/[l — (1 — r)z], then (43) may be written in

THEOREM 7. ///(z) = T,^=oakzk is analytic and univalent in A then g(z) = X^°=o&nz"> the Taylor transformation of f where the bn's are given by (9), is univalent in A. PROOF. g(z) = £ bnz" = 2 ( I (1 - rf+1(k\rk-'ak )z" n = 0 n = 0 \k=n \ " / .' = (1 - r) • 2 a4 I 2 (1 - r)V-"z'1 = (1 - r) 2 ak{r + (1 - r)z]* The interchange in summation of the series involved is permissible since for