Uniformly convex functions II

ANNALES POLONICI MATHEMATICI LVIII. (199 Uniformly convex functions II by Wancang Ma and David Minda (Cincinnati, Ohio Abstract. Recently, A. W. Goodman introduced the class UCV of normalized uniformly convex functions. We present some sharp coefficient bounds for functions f(z = z + a z + a z +... UCV and their inverses f 1 (w = w + d w + d w +.... The series expansion for f 1 (w converges when w < ϱ f, where 0 < ϱ f depends on f. The sharp bounds on a n and all extremal functions were known for n = and ; the extremal functions consist of a certain function k UCV and its rotations. We obtain the sharp bounds on a n and all extremal functions for n = 4, 5, and 6. The same function k and its rotations remain the only extremals. It is known that k and its rotations cannot provide the sharp bound on a n for n sufficiently large. We also find the sharp estimate on the functional µa a for < µ <. We give sharp bounds on d n for n =, and 4. For n =, k 1 and its rotations are the only extremals. There are different extremal functions for both n = and n = 4. Finally, we show that k and its rotations provide the sharp upper bound on f (z over the class UCV. 1. Introduction. This is a continuation of our investigation of uniformly convex functions [MM]. Goodman [G] introduced the geometrically defined class UCV of uniformly convex functions in the unit disk D = {z : z < 1}. A function f is said to be uniformly convex in D if f is a normalized (f(0 = f (0 1 = 0 convex function and has the additional property that for every circular arc γ contained in D, with center also in D, the image arc f(γ is convex. The class CV of normalized (f(0 = f (0 1 = 0 convex univalent functions f is closely related to the class P of normalized holomorphic functions with positive real part. Recall that a holomorphic function p defined on D belongs to P provided that p(0 = 1 and Re{p(z} > 0, z D. Precisely, f CV if and only if p(z = 1 + zf (z/f (z P. In the earlier paper [MM], we introduced a subfamily PAR of P that plays an analogous role for the class UCV. Let PAR = {p(z P : p(d Ω}, 1991 Mathematics Subject Classification: Primary 0C45. Key words and phrases: convex functions, coefficient bounds. Research partially supported by a National Science Foundation Grant.

76 W. Ma and D. Minda where Ω = {w = u + iv : v < u 1} = {w : Re w > w 1 }. The characterization that f UCV if and only if p(z = 1 + zf (z/f (z PAR was proved independently by Ma and Minda [MM], and Rønning [Rø 1 ]. This characterization enabled us to derive some subordination results for the class UCV, from which we derived sharp distortion, growth, rotation and covering theorems, and sharp bounds on the second and third coefficients as well as the sharp order of growth for the coefficients [MM]. In all of these cases, the function k given below is the sole extremal function up to rotations. The sharp bound on the second coefficient and an estimate for all coefficients were also discovered by Rønning [Rø 1 ]. In [Rø ], Rønning presented a convolution theorem related to UCV. In [MM], we also defined holomorphic functions k n (z in D by k n (0 = k n(0 1 = 0 and 1 + zk n(z/k n(z = q(z n 1, where q(z is a normalized Riemann mapping function from D to Ω. Explicitly, q(z = 1 + ( π log 1 + z 1 = 1 + z = 1 + 8 π n=1 ( n 1 1 n k=0 1 k + 1 B n z n n=1 z n. Clearly k n (z UCV. We write k (z as k(z and set k(z = z + A z + A z +.... The function k cannot be extremal for the problem of maximizing the modulus of the nth coefficient of functions in UCV for n sufficiently large [MM]. In this paper, we first prove that k and its rotations are still extremal for the problem of maximizing the modulus of the nth coefficient of functions in UCV when n = 4, 5 and 6. For the inverse function f 1 (w = w +d w + d w +..., it is interesting to observe that extremal functions of d n are k n and its rotations when n =, and 4. Then we find the sharp estimate of the coefficient functional µa a, < µ <. Finally, in Section 4, we obtain the sharp upper bound on f (z.. Preliminaries. If f(z = z + a z + a z +... UCV, then there exists a function p(z = 1 + b 1 z + b z + b z +... PAR such that p(z = 1 + zf (z/f (z. Although the bounds b n B 1 = 8/π (n = 1,,... [MM] are sharp for p(z PAR, they do not yield sharp bounds for a n when n.

Uniformly convex functions II 77 Here our idea is to obtain bounds on a n by first expressing a n in terms of coefficients of a function in the class P and then using coefficient bounds for functions in P. In this section, we derive the expressions needed and prove some coefficient inequalities for functions in P. The relationship between f(z UCV and p(z PAR implies that n 1 (1 n(n 1a n = ka k b n k. Furthermore, we have ( k=1 a = b 1, 6a = b + b 1, 1a 4 = b + b b 1 + 1 b 1, 0a 5 = b 4 + 4 b b 1 + 1 b + b b 1 + 1 6 b4 1, 0a 6 = b 5 + 5 4 b 4b 1 + 5 6 b b + 5 6 b b 1 + 5 8 b b 1 + 5 1 b b 1 + 1 4 b5 1. Since q(z is univalent in D and p q, the function p 1 (z = 1 + q 1 (p(z 1 q 1 (p(z = 1 + c 1z + c z + c z +... is holomorphic and has positive real part in D, that is, p 1 P. Equivalently, ( p1 (z 1 p(z = q. p 1 (z + 1 From the power series expansion of q, q(z = 1 + 8 π z + 16 π z + 184 45π z + 5 105π z4 + 4504 1575π z5 +..., we can express b n in terms of c n by direct calculation. Precisely, b 1 = 4 π c 1, b = 1 π (4c c 1, ( b = 1 π (4c 4 c 1c + 8 45 c 1, b 4 = 1 π (4c 4 4 c 1c c + 8 15 c 1c 5 c4 1, b 5 = 1 π (4c 5 4 c 1c 4 4 c c + 8 15 c 1c + 8 15 c 1c 8 5 c 1c + 1575 c5 1. The equalities in ( and ( then yield that (4.1 a = 4 π c 1,

78 W. Ma and D. Minda (4. 6a = 4 π c + ( 4 π π 1 c 1, (4. 1a 4 = 4 π c + 4 ( 18 π π 1 c c 1 + 8 ( 45π 1 45 π + 180 π 4 c 1, (4.4 0a 5 = 4 π c 4 + 4 ( 16 π π 1 c c 1 + ( 4 π π 1 c + 8 ( 1 π 15 11 9π + 8 c π 4 c 1 + π ( 1 5 + 79 15π 16 π 4 + 64 π 6 (4.5 0a 6 = 4 π c 5 + 4 ( 5 π π 1 c 4 c 1 + 4 ( 10 π π 1 c c + 4 ( π 5 0 π + 40 π 4 c c 1 + ( 4 π + 4 ( π 5 + 1 π 70 9π 4 + 80 π 6 c c 1 + ( 16 π 55 109 189π + 47 9π 4 80 π 6 + 64 π 8 15 5 9π + 0 π 4 c 5 1. c 4 1, c c 1 Hence coefficient estimates for the class UCV become non-linear coefficient problems for the class P. Note that if p 1 (z = (1 + z/(1 z, then p = q and f = k. Hence, if c n = for all n, then b n = B n and a n = A n. We list the explicit expressions for A n when n =,, 4, 5 and 6: A = 4 π, A = 8 9π + π 4, A 4 = 46 15π + 16 π 4 + 64 π 6, A 5 = 88 55π + 195 675π 4 + 56 15π 6 + 51 15π 8, A 6 = 5 65π + 14656 8505π 4 + 4864 405π 6 + 104 7π 8 + 048 45π 10. Now we recall some coefficient bounds for p 1 (z = 1 + c 1 z + c z + c z +... P. It is well known that c n (n = 1,,.... Livingston [L] proved that c n c m c n+m for n, m = 1,,.... We can also obtain the following lemma. (5 (6 and Lemma 1. If p 1 (z = 1 + c 1 z + c z + c z +... P, then c n 1 c n 1 c n, µc nc n c 4 n 8(µ (µ 4 (7 µc n c n c n 4(µ (µ 6.

Uniformly convex functions II 79 P r o o f. For p 1 (z = 1 + c 1 z + c z + c z +... P, we define h(z = 1 n p 1 (e ikπ/n z. n k=1 Then h(z = 1 + c n z n + c n z n + c n z n +... P, so that the function h 1 (z = 1 + c n z + c n z + c n z +... P. Hence it is enough to show the desired inequalities for n = 1. Consider w(z = p 1(z 1 p 1 (z + 1 = 1 c 1z + 1 (c 1 c 1z +..., which is holomorphic in D with w(0 = 0 and w(z < 1. As w (0/ 1 w (0 [A, p. 16], we have If µ, then c 1 c 1 1 c 1. µc 1c c 4 1 = c 1 µ(c 1 c 1 + ( 1 µ 1c 1 c 1 {µ( 1 c 1 + ( 1 µ 1 c 1 } = c 1 (µ c 1. As 0 c 1 and the right-hand side of the above inequality is an increasing function of c 1 for µ 4, we have inequality (6 when n = 1. If µ, then µc 1 c c 1 = c 1 µ(c 1 c 1 + ( 1 µ 1c 1 c 1 {µ( 1 c 1 + ( 1 µ 1 c 1 } = c 1 (µ c 1. Again the right-hand side of this inequality is an increasing function of c 1 for 0 c 1 and µ 6. This completes the proof of Lemma 1.. Coefficient bounds. Now we are ready to prove coefficient bounds for functions in UCV. Theorem 1. Let f(z = z + a z + a z +... UCV. Then a n A n (n = 4, 5, 6. Equality holds if and only if f(z is k(z or one of its rotations. P r o o f. When n = 4, we see that the coefficients in the expression of a 4 in (4. are all positive. So we get a 4 A 4 from c k (k = 1,,. When n = 5, from (4.4 we have 0a 5 = 4 π c 4 + 4 π + 16 π ( 16 π 1 c c 1 + ( 4 π π 1 c c 1 ( 5 4 45π + 4 π 4 + π 6 + π ( 1 5 79 15π + 16 π 4 64 π 6 c (4c 1c c 4 1.

80 W. Ma and D. Minda In this expression, all coefficients are positive. By using (6 of Lemma 1 for n = 1, µ = 4 and c k (k = 1,,, 4, we see that the upper bound of a 5 is given if we replace all c k by. That is, we have a 5 A 5. When n = 6, we rearrange the expression for a 6 given in (4.5 as follows: 0a 6 = 4 π c 5 + 4 ( 5 π π 1 c 4 c 1 + 4 ( 10 π π 1 c c + ( 16 π 55 109 189π + 47 9π 4 80 π 6 + 64 π 8 (c c c 1 + c 1c 1 + ( 404 c c 1 π + 4 π 55 411 189π + 67 9π 4 + 80 π 6 64 π ( 8 74 55 458 189π + 16 9π 4 160 π 6 64 π 8 (6c c 1 c 1c + ( π 5 175 + 1097 189π 47 9π 4 + 640 π 6 + 56 π 8 c c 1. Now, all coefficients in this expression are positive. This time we use (7 of Lemma 1 for n = 1 and µ = 6, c c c 1 + c 1 [LZ] and c k (k = 1,..., 5 to derive that the upper bound of 0 a 6 is achieved when we replace all c k by. Thus we have a 6 A 6. In each case, we have used the inequality c 1 in our proof. Hence equality holds only if p 1 (z = (1 + z/(1 z or one of its rotations, which implies that f = k or one of its rotations. On the other hand, it is clear that inequalities become equalities for k and its rotations. This completes the proof of Theorem 1. Next we discuss the coefficient functional µa a. But first, we introduce the following functions in UCV. For 0 λ 1, define h λ and g λ by h λ (0 = h λ (0 1 = g λ(0 = g λ (0 1 = 0, 1 + zh λ (z ( z(z + λ h λ (z = q, 1 + zg λ (z ( 1 + λz g λ (z = q z(z + λ. 1 + λz Then it is clear that both h λ and g λ belong to UCV. Also notice that h 1 = k, h 0 = k, g 1 (z = k( z and g 0 (z = k ( z. Theorem. Let f(z = z + a z + a z +... UCV. Then ( 8 6 π π µ 4 π 1 if + 5π 6 µ, µa 4 a π if π 6 µ + 5π 6, ( 8 π 6 π µ + 4 π + 1 if µ π 6.

Uniformly convex functions II 81 Equality holds if and only if f is k or one of its rotations when µ < / π /6 or / + 5π /6 < µ. For / π /6 < µ < / + 5π /6, equality holds if and only if f is equal to k or one of its rotations. If µ = / π /6, then equality holds if and only if f is h λ or one of its rotations. Finally, equality holds if and only if f is g λ or one of its rotations when µ = / + 5π /6. R e m a r k. When / π /6 < µ < / + 5π /6, the above inequality can be improved as follows: µa a + 1 ( π 1 + 1µ 8 a 4 π if π 6 < µ + π 18 and µa a + 1 ( 5π 1 1µ + 8 a 4 π In particular, by setting µ = 1, we get ( 1 a a + + π a 4 6 π. if + π 18 µ < + 5π 6. This clearly improves a a + 1 a normalized convex functions (see [T]. 1, which was established for P r o o f o f T h e o r e m. From (4.1 and (4. we obtain µa a = 1 (( 1 π π µ 8 π + 1 c 1 c. If µ / + 5π /6, then 1µ/π 8/π 5/ 0 and µa a = 1 ( 1 π π µ 8 π 5 c 1 + (c 1 c 1 ( ( 1 π 4 π µ 8 π 5 + 4 = 8 π ( 6 π µ 4 π 1 Here we have used c 1 c and c 1. If µ / π /6, then 1µ/π +8/π 1/ 0 and c 1, c imply that µa a = 1 ( 1 c 1 + c π 8 π π µ + 8 π 1 ( 6 π µ + 4 π + 1..

8 W. Ma and D. Minda Now we assume / π /6 µ / + π /18, then 1µ/π + 8/π + / 0. By using (5 of Lemma 1 for n = 1, we have µa a + 1 ( π 1 + 1µ 8 a = 1 { ( π c 1 ( c 1 + 1 π µ + 8 π + c 1 ( 1 + π µ 8 π + 1 } c 1 1 { ( π 4 c 1 + 1 π µ + 8 π + ( 1 c 1 + π µ 8 π + 1 } c 1 = 4 π. This is the stronger result in the remark above. Finally, we consider the case when / + π /18 µ / + 5π /6. Note that in this case 1µ/π 8/π / 0. Once again we use (5 for n = 1 to get µa a + 1 ( 5π 1 1µ + 8 a = 1 { ( π c 1 ( 1 c 1 π µ 8 π c 1 ( + 1 π µ + 8 π + 5 } c 1 1 { ( 1 π 4 c 1 + π µ 8 π ( c 1 + 1 π µ + 8 π + 5 } c 1 = 4 π. It is the stronger result stated in the above remark. From our proof, we see that when µ < / π /6 or µ > / + 5π /6, equality holds if and only if c 1 =, equivalently, f is k or one of its rotations. For / π /6 < µ < / + 5π /6, equality holds if and only if c = and c 1 = 0, that is, f is equal to k or one of its rotations. If µ = / π /6, then equality holds if and only if c =, or up to rotation [P, p. 41], p 1 (z = 1 + λ So f is h λ or one of its rotations. 1 + z 1 z + 1 λ 1 z 1 + z, 0 λ 1.

Uniformly convex functions II 8 Finally, when µ = /+5π /6, equality holds if and only if c 1 c =. Thus up to rotation, p 1 is given by p 1 (z 1 = 1 + λ 1 + z 1 z + 1 λ 1 z 1 + z, 0 λ 1. This implies that f is g λ or one of its rotations. This completes the proof of Theorem. The known sharp bounds on a and a can easily be obtained from Theorem. To discuss coefficient bounds for the inverses of functions in UCV, we first observe that for the inverse function K n (w of k n (z, K n (w = w 8 (n 1nπ wn +.... Hence for f UCV with F (w = f 1 (w = w + d w + d w +..., max{ d n : f UCV} 8 (n 1nπ. For n =,, 4, we can prove that equality holds. Note that the series expansion for f 1 (w converges when w < ϱ f, where 0 < k( 1 ϱ f [MM] depends on f. Theorem. Let f UCV and F (w = f 1 (w = w+d w +d w +.... Then 8 d n (n 1nπ (n =,, 4. Equality holds if and only if f is equal to k n or one of its rotations. P r o o f. As F (f(z = z, we have d = a, d = a a, d 4 = a 4 + 5a a 5a. By using (4.1 (4., we can express d n in terms of c n as follows: d = π c 1, d = 1 (( 16 π π + 1 c 1 c and d 4 = 1 ( ( 1 π c + 14 ( π c c 1 + 45 + 7 π + 48 π 4 c 1. Now it is clear that d 4/π. Equality holds if and only if c 1 =, that is, f is k or one of its rotations. Also, d = 1 ( 16 π π + 1 ( 5 (c c 1 + 16 π c 4 π.

84 W. Ma and D. Minda Here we have used c c 1 and c. Equality holds if and only if c = and c 1 = 0, or equivalently, f is k or one of its rotations. By using c c c 1 + c 1 [LZ], c c c 1 and c, we get d 4 = 1 ( (c c c 1 + c 1 π + 45 + 7 π + 48 π ( 4 11 45 + 8 π 96 π 4 (c c c 1 + ( 45 5 π + 48 π 4 c π. In this case, the inequality becomes equality if and only if c = and c = c 1 = 0, which is the same as saying f is k 4 or one of its rotations. The proof of Theorem is now complete. 4. Upper bound on f (z. Finally in this section, we derive the sharp upper bound on f (z for functions in UCV. Theorem 4. Let f UCV and z = r < 1. Then f (z k (r. Equality holds for any z D if and only if f is k or one of its rotations. P r o o f. Let p(z=1+zf (z/f (z. Then p q implies that p 1 q 1. As all coefficients of q 1 are positive, the subordination principle yields that for z = r, p(z 1 q(r 1. This is the same as f (z/f (z k (r/k (r. From f (z k (r [MM], we see that f (z k (r f (z/f (z k (r. We also know that equality holds in f (z k (r for some z 0 if and only if f is a rotation of k [MM]. Moreover, equality in Theorem 4 at z = 0 is equivalent to a = A. Hence equality holds if and only if f is k or one of its rotations. This completes our proof. References [A] L. V. Ahlfors, Complex Analysis, rd ed., McGraw-Hill, New York, 1979. [G] A. W. Goodman, On uniformly convex functions, Ann. Polon. Math. 56 (1991, 87 9. [L] A. E. L i v i n g s t o n, The coefficients of multivalent close-to-convex functions, Proc. Amer. Math. Soc. 1 (1969, 545 55. [LZ] R. J. Libera and E. J. Z lotkiewicz, Early coefficients of the inverse of a regular convex function, Proc. Amer. Math. Soc. 85 (198, 5 0.

Uniformly convex functions II 85 [MM] W. M a and D. M i n d a, Uniformly convex functions, Ann. Polon. Math. 57 (199, 165 175. [P] Ch. Pommerenke, Univalent Functions, Vandenhoeck & Ruprecht, Göttingen, 1975. [Rø 1 ] F. Rønning, Uniformly convex functions and a corresponding class of starlike functions, Proc. Amer. Math. Soc. 118 (199, 189 196. [Rø ], On starlike functions associated with parabolic regions, Ann. Univ. Mariae Curie-Sk lodowska Sect. A 45 (1991, 117 1. [T] S. Y. T r i m b l e, A coefficient inequality for convex univalent functions, Proc. Amer. Math. Soc. 48 (1975, 66 67. DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF CINCINNATI CINCINNATI, OHIO 451-005 U.S.A. Reçu par la Rédaction le 8.8.199 Révisé le 1..199