SUBCLASSES OF UNIFORMLY STARLIKE FUNCTIONS

We study subclasses of the class of uniformly starlike functions which were recently introduced by A.W. Goodman. One new subclass is defined and it is shown that it shares many properties of the parent class.

who proved in particular the following analytic characterization.THEOREM A. Let f be analytic in D, f(0) f'(0)-1 0. Then f is in UST if and only if Q(z,) f() f'() (z-)f'(z) (z # () and Q(z,z)= 1 (1.1) has positive real part for all z and ( in D. Properties of the class UST are difficult to establish.One reason is that the usual transformations of univalent function theory generally do not preserve the UST class.The only known exceptions are rotations, e-iaf(eiaz) for some real or, and the transformation t-lf(tz), 0<t<l.
In order to obtain a coefficient bound an <_ 2In (n 2,3,...) for f(z) z + an zn in UST, n=2 Goodman proves that class UST is embedded in a larger subclass of starlike functions UST*.A function f E ST is in UST* if there is a real c, such that Re{eiaf'(z)} > 0 for z E D. (Goodman credits the result to Charles Horowitz.)This suggests that information about the UST class might be generated through the study of subclasses of UST as well.In this paper we study one such class that shares and extends some known properties of UST. 2. SOME SUBCLASSES OF UST.
These identities lead to an equivalent form for Theorem A. THEOREM 1.Let f be a normalized analytic function in D. Then f E UST if and only if for all complex numbers ct, , c -< 1, I/l _< 1, and for all z e D we have ( -'/( / > 0 (..

J'*(1 -z) 2
In this form we can appeal to the extensive work on Hadamard products initiated by the proof of the Polya-Schoenberg conjecture by Ruscheweyh and Sheil-Small [4].The fundamental result in this proof was the following theorem [4].whenever F is an analytic function with positive real part in D.
These results, along with the following elementary observation about certain linear fractional transformations enable us to generate functions in UST from functions in the convex subclass K of ST.
The function z/(1pz) 2 is starlike of order (1-p)/(1 + p).Now a normMized Mytic function f is sMd to be in the cls Re of prestfike functions if f.z/(1-z) 2-2a is in the cls STa of stlike functions of order a when a < 1 or Re{f(z)/z} > 1/2 when a 1. Ruscheweyh [5,  p. 54] proves a generMization of Threm B to the ce where E Ra d g E STa.By gument that is simil to the prf of Threm 2, we obtMn the following result.THEOREM 3. If f is in the cls Re of prestlike functions of order a, then F(z) f(pz)/p is in UST whenever p (1 )/(1 +) d (-1)/(+ 1) Except for the shpness, Threm 2 is a speciM ce of Theorem 3 since R 0 K C Ra for 0 < a 1.The link betwn the convex ce d the fundentM Theorem B is our justification of first proving the less generM result.
It is interesting to note that for a > 1/2, the cls Ra contMns functions that e not univMent in D [6].The function F of Theorem 3 is, of course, univalent and starlike in D.
To obtain another subset of UST, notice that 1 I.f,((z,+(l_)C)at f'(z)f'(z) (2.4) THEOREM 4. Let f be a normalized analytic function in D. Then f E UST if for all w,z in If f E UST, then for all w,z in D Re f'(w)] I/2 l,i'(:)J _>o and the I/2 is best possible.PROOF.The first part of the theorem follows from Theorem A and the real part of (2.4).
For the second statement, we note that for f The function f(z) z/(1 pz) with p 1/ proves the exponent 1/2 is best possible.
The sharp bounds for the coefficients of f 6-UST is an elusive open problem.Some information for this problem is contained in the next theorem.

expression becomes
The above l+A(l+c,+c,2+...+an-1)z l+A(z+Az where ( a + c, 2 + + c, n-1.The image of zl 1 by this linear fractional transformation is a circle with center c and radius R given by 1 nA2(C + 1), The image is in the right half plane if I(n-1)-CI A< 1-nA2(Re '-I-1) Since CI a + a 2 / / '-s n 1, the inequality holds when A212(n 1) 2(n )x] < ( nA2) 2 2nA2(1 nA2)x + n2A2x 2 where x Re , that is, The minimum of the function of z on the ]eft of this inequality occurs when x (n2A 2 1)/n2A 2 If we substitute this for x and simplify we obtain (n 1)(n + 1 2n3A2) > 0 or n + 1 > A2.QED.This improves the bound AI < 1/(ln) of Goodman [1].It does not appear to be the best possible except when n 2. Godman [1] states that z + Az 2 6_ UST if and only if AI < 1-/4.We prove this result for a subclass of UST in the next section.A rather natural way to construct a subclass of UST is to replace the derivative in (1.1) by a difference quotient.This generates a family of functions that shares most of the known sharp properties of the class UST.To be precise, we define UST. as the class of normalized analytic functions f in D such that ff*z/(1 > 0 (3.1) -az)(1 --z)Jfor all z { D, and c,/J,-in }.Sincec reduces (3.1) to (2.2), we have UST, C UST.
There is other wy to chacteri UST, (d UST) that c usef.
for all l=l 1 and % D,c in D. This result follows directly from (3.1).The expression in braces in (3.1) cannot equal (z+ 1)/(-1), [xl I, when its real part is positive.This yields (3.2) upon algebraic simplification.COROLLARY 2. z + Az 2 .UST,i f and only if AI _< /4.PROOF.If a 2 is the coeffcient of z 2 in the power series expansion of the second function in the Hadarnard product (3.2), then with f(z) z + Az 2 this product is not zero in D if Aa2l <_ 1.Now, with a 1, the second coefficient is 1% (1 )-/ (1 / )].Now la21 < 11/1/2{//D)I and we seek a maximum of the fight hand side when I1 _< 1, I1 _< 1.A computation shows a maximum occurs when 3' 1, eit, t 2 arcsin 1/,.This gives us [2[ -< 4/ and proves that [AI _< /4.Since there is a choice of z, [z 1, such that l2[ 1 + (-+/)[ + 1/2(- 1,   we conclude that the result is sharp. 4. ARC LENGTH FOR THE CLASS UST. In the final section we state a result on the length of images of circles under UST mappings which extends a well-known result of Keogh for bounded starlike functions [3].Let C((',r) be a THEOREM B. If o is a normalized convex univalent function in D and g .ST,t hen for all z.D o.