Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 5, Issue 2, Article 31, 2004 STARLIKENESS AND CONVEXITY CONDITIONS FOR CLASSES OF FUNCTIONS DEFINED BY SUBORDINATION R. AGHALARY, JAY M. JAHANGIRI, AND S.R. KULKARNI U NIVERSITY OF U RMIA , U RMIA , I RAN . raghalary@yahoo.com K ENT S TATE U NIVERSITY, O HIO , USA. jay@geauga.kent.edu F ERGUSSEN C OLLEGE , P UNE , I NDIA . srkulkarni40@hotmail.com Received 21 May, 2003; accepted 18 April, 2004 Communicated by H. Silverman A BSTRACT. We consider the family P(1, b), b > 0, consisting of functions p analytic in the open unit disc U with the normalization p(0) = 1 which have the disc formulation |p − 1| < b in U.P Applying the subordination properties to certain choices of p using the functions fn (z) = ∞ z + k=1+n ak z k , n = 1, 2, ..., we obtain inclusion relations, sufficient starlikeness and convexity conditions, and coefficient bounds for functions in these classes. In some cases our results improve the corresponding results appeared in print. Key words and phrases: Subordination, Hadamard product, Starlike, Convex. 2000 Mathematics Subject Classification. Primary 30C45. 1. I NTRODUCTION Let A denote the class of functions that are analytic in the open unit disc U = {z ∈ C : |z| < 1} and let An be the subclass of A consisting of functions fn of the form (1.1) fn (z) = z + ∞ X ak z k , n = 1, 2, 3, . . . . k=1+n The function p ∈ A and normalized by p(0) = 1 is said to be in P(1, b) if (1.2) |p(z) − 1| < b, b > 0, z ∈ U. The class P(1, b) which is defined using the disc formulation (1.2) was studied by Janowski [6] and has an alternative characterization in terms of subordination (see [5] or [14]), that is, for ISSN (electronic): 1443-5756 c 2004 Victoria University. All rights reserved. 069-03 2 R. AGHALARY , JAY M. JAHANGIRI , AND S.R. K ULKARNI z ∈ U , we have (1.3) p ∈ P(1, b) ⇐⇒ p(z)≺ 1 + bz. For the functions φ and ψ in A, we say that the φ is subordinate to ψ in U , denoted by φ ≺ ψ, if there exists a function w(z) in A with w(0) = 0 and |w(z)| < 1, such that φ(z) = ψ(w(z)) in U. For further references see Duren [3]. The family P(1, b) contains many interesting classes of functions which have close interrelations with different well-known classes of analytic univalent functions. For example, for fn ∈ An if 0 zfn ∈ P(1, 1 − α), 0 ≤ α ≤ 1, fn then fn is starlike of order α in U and if zfn00 1+ 0 ∈ P(1, 1 − α), 0 ≤ α ≤ 1, fn then fn is convex of order α in U. For 0 ≤ α ≤ 1 we let S ∗ (α) be the class of functions fn ∈ An which are starlike of order α in U , that is, 0 zfn ∗ S (α) ≡ fn ∈ An : < ≥ α, |z| < 1 , fn and let K(α) be the class of functions fn ∈ An which are convex of order α in U , that is, zfn00 ≥ α, |z| < 1 . K(α) ≡ fn ∈ An : < 1 + 0 fn Alexander [1] showed that fn is convex in U if and only if zfn0 is starlike in U. In this paper we investigate inclusion relations, starlikeness, convexity, and coefficient conditions on fn and its related classes for two choices of p(fn ) in P(1, b). In some cases, we improve the related known results appeared in the literature. Define F(1, b) be the subclass of P(1, b) consisting of functions p(f1 ) so that zf10 (z) zf100 (z) (1.4) p(f1 (z)) = 1+ 0 f1 (z) f1 (z) where f1 ∈ A1 is given by (1.1). For fixed v > −1, n ≥ 1, and for λ ≥ 0, define Mvλ (1, b) be the subclass of P(1, b) consisting of functions p(fn ) so that Dv fn (z) + λ(Dv fn (z))0 z where fn ∈ An and Dv f is the v-th order Ruscheweyh derivative [10]. The v-th order Ruscheweyh derivative Dv of a function fn in An is defined by ∞ X z v (1.6) D fn (z) = ∗ fn (z) = z + Bk (v)ak z k , 1+v (1 − z) k=1+n (1.5) p(fn (z)) = (1 − λ) where (1 + v)(2 + v) · · · (v + k − 1) (k − 1)! and the operator “ ∗ ” stands for the convolution or Hadamard product of two power series ∞ ∞ X X i f (z) = ai z and g(z) = bi z i Bk (v) = i=1 J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 i=1 http://jipam.vu.edu.au/ S TARLIKENESS AND C ONVEXITY C ONDITIONS defined by (f ∗ g)(z) = f (z) ∗ g(z) = ∞ X 3 ai b i z i . i=1 2. T HE FAMILY F(1, b) The class F(1, b) for certain values of b yields a sufficient starlikeness condition for the functions f1 ∈ A1 . Theorem 2.1. If 0 < b ≤ 9 4 and p(f1 ) ∈ F(1, b) then √ 3 − 9 − 4b zf10 . ∈ P 1, 2 f1 We need the following lemma, which is due to Jack [4]. Lemma 2.2. Let w(z) be analytic in U with w(0) = 0. If |w| attains its maximum value on the circle |z| = r at a points z0 , we can write z0 w0 (z0 ) = kw(z0 ) for some real k, k ≥ 1. √ zf 0 (z) Proof of Theorem 2.1. For b1 = 3− 29−4b write f11(z) = 1 + b1 w(z). Obviously, w is analytic in U and w(0) = 0. The proof is complete if we can show that |w| < 1 in U. On the contrary, suppose that there exists z0 ∈ U such that |w(z0 )| = 1. Then, by Lemma 2.2, we must have z0 w0 (z0 ) = kw(z0 ) for some real k, k ≥ 1 which yields 00 z0 f10 (z0 ) z f (z ) 0 0 1 = (1 + b1 w(z0 ))2 + b1 z0 w0 (z0 ) − 1 1 + − 1 0 f1 (z0 ) f1 (z0 ) = bk + 2b1 + b21 w(z0 ) ≥3b1 − b21 = b. This contradicts the hypothesis, and so the proof is complete. Corollary 2.3. For 0 < b ≤ 2 let p(f1 ) ∈ F(1, b). Then f1 ∈ S ∗ √ −1+ 9−4b 2 . Corollary 2.4. If p(f1 ) ∈ F(1, b) and 0 < b ≤ 2, then √ 0 zf (z) 3 − 9 − 4b 1 arg < arcsin . f1 (z) 2 It is not known if the above corollaries can be extended to the case when b > 2. √ f1 (z) > 12 then f1 ∈ S ∗ −1+2 5 . Corollary 2.5. If < zf 0 (z)+z 2 f 00 (z) 1 1 Remark 2.6. For 0 < b < 2, Theorem 2.1 is an improvement to Theorem 1 obtained by Obradović, Joshi, and Jovanović [8]. Corollary 2.7. If p(f1 ) ∈ F(1, b) then f1 is convex in U for 0 < b ≤ 0.935449. Proof. For p(f1 ) ∈ F(1, b) we can write | arg p(f1 )| < arcsin b. Therefore, √ 00 zf (z) 3 − 9 − 4b 1 arg 1 + < arcsin b + arcsin . f 0 (z) 2 1 Now the proof is complete upon noting that the right hand side of the above inequality is less than π2 for b = 0.935449 . Remark 2.8. It is not known if the above Corollary 2.7 is sharp but it is an improvement to Corollary 2 obtained by Obradovic, Joshi, and Jovanovic [8]. J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 http://jipam.vu.edu.au/ 4 R. AGHALARY , JAY M. JAHANGIRI , AND S.R. K ULKARNI Corollary 2.9. If p(f1 ) ∈ F(1, b) then f1 is convex in the disc |z| < b ≤ 1. 0.935449 b for 0.935449 ≤ Proof. We write p(f1 ) = 1+bw(z) where w is a Schwarz function. Let |z| ≤ ρ. Then |w(z)| ≤ ρ and so |p(f1 ) − 1| < bρ for |z| ≤ ρ. Upon choosing bρ = 0.935449 it follows from the above Corollary 2.7 that | arg(1 + zf100 /f10 )| < π/2 for |z| ≤ ρ = 0.935449/b . Therefore the proof is complete. In the following example we show that there exist functions f which are not necessarily starlike or univalent in U for p(f1 ) ∈ F(1, b) if b is sufficiently large. Example 2.1. For the spirallike function g(z) = z/(1 − z)1+i we have 0 1 1 − |z|2 − π4 i zg (z) < e =√ > 0, z ∈ U. g(z) 2 |1 − z|2 Since zg 0 (z) g(z) = 1+iz , 1−z we obtain < zg 0 (z) g(z) = 1 − r( cos θ + sin θ) 1 − 2r cos θ + r2 for z = reiθ . Thus g(z) is not starlike for |z| < t, √12 < t < 1. This means that f (z) = not starlike in U. Now set Z z g(ζ) h(z) = dζ = i((1 − z)−i − 1) ζ 0 g(rz) r is 2π −1 and let z0 = ee2π +1 ≈0.996 . Therefore, h(z0 ) = h(−z0 ) and so h is not univalent in U. Conseh(z0 z) quently, f (z) = z0 is not univalent in U for sufficiently large values of b. On the other hand, p(g) ∈ F(1, b) for sufficiemtly large b, since, 1 + 3iz z + − 1 < b |p(g(z)) − 1| = 2 (1 − z) 1−z for sufficiemtly large b. The following theorem is the converse of Theorem 2.1 for a special case. √ zf 0 Theorem 2.10. If f11 ∈ P 1, 3−2 5 then p(f1 ) ∈ F(1, 1) for |z| < r0 = 0.7851 . To prove our theorem, we need the following lemma due to Dieudonné [2]. Corollary 2.11. Let z0 and w0 be given points in U, with z0 6= 0. Then for all functions f analytic and satisfying |f (z)| < 1 in U, with f (0) = 0 and f (z0 ) = w0 , the region of values of f 0 (z0 ) is the closed disc 2 2 w − w0 ≤ |z0 | − |w0 | . z0 |z0 |(1 − |z0 |2 ) Proof of Theorem 2.10. Write zf 0 (z) q(z) = 1 =1+ f1 (z) J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 √ ! 3− 5 w(z), 2 http://jipam.vu.edu.au/ S TARLIKENESS AND C ONVEXITY C ONDITIONS 5 where w is a Schwarz function. We need to find the largest disc |z| < ρ for which " #2 √ ! √ ! 5 3 − 5 3 − 0 1+ w(z) + zw (z) − 1 2 2 √ !2 √ ! √ 3− 5 3− 5 2 0 = w (z) + (3 − 5)w(z) + zw (z) < 1. 2 2 For fixed r = |z| and R = |w(z)| we have R ≤ r. Therefore, by Lemma 2.11, we obtain |w0 (z)|≤ R r 2 − R2 + r r(1 − r2 ) and so 0 00 zf1 (z) zf (z) 1 |p(f1 ) − 1| = 1+ 0 − 1 f (z) f1 (z) 1 ! √ 2 √ 3− 5 = w2 (z) + (3 − 5)w(z) + 2 √ ! 3− 5 0 zw (z) 2 r 2 − R2 ≤ t R + 3tR + t 1 − r2 t ψ(R), = 1 − r2 2 2 where √ 3− 5 ψ(R) = R (t − tr − 1) + 3R(1 − r ) + r and t = . 2 2 3(1−r ) We note that ψ(R) attains its maximum at R0 = 2(1+tr 2 −t) . So the theorem follows for r0 ≈0.7851 t which is the root of the equation 1−r2 ψ(R0 ) = 1. 2 2 2 2 Letting z0 and w0 in Lemma 2.11 be so that |z0 | = r0 and |w0 | = the bound given by Theorem 2.10 is sharp. 3(1−r02 ) 2(1+tr02 −t) we conclude that 3. T HE FAMILY Mvλ (1, b) We begin with stating and proving some properties of the family Mvλ (1, b). Theorem 3.1. If p(fn ) ∈ Mvλ (1, b) then Dv fn (z) b ∈ P(1, ). z 1 + λn We need the following lemma, which is due to Miller and Mocanu [7]. Lemma 3.2. Let q(z) = 1+qn z n +· · · (n ≥ 1) be analytic in U and let h(z) be convex univalent in U with h(0) = 1. If q(z) + 1c zq 0 (z) ≺ h(z) for c > 0, then Z c c −c/n z q(z) ≺ z h(t)t n −1 dt. n 0 Proof of Theorem 3.1. For p(fn ) ∈ Mvλ (1, b) set q(z) = D λzq 0 (z) ≺ 1 + bz. Now, applying Lemma 3.2, we obtain b q(z) ≺ +1 + z. 1 + λn J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 vf n (z) z . Then we can write q(z) + http://jipam.vu.edu.au/ 6 R. AGHALARY , JAY M. JAHANGIRI , AND S.R. K ULKARNI Substituting back for q(z) and choosing w(z) to be analytic in U with |w(z)| ≤ |z|n , by the definition of subordination we have Dv fn (z) b =1+ w(z). z (1 + λn) Now the theorem follows using the necessary and sufficient condition (1.3). The estimates in Theorem 3.1 are sharp for p(fn ) where fn is given by Dv fn (z) b =1+ zn. z (1 + λn) Corollary 3.3. If p(fn ) ∈ Mvλ (1, b) then v D fn (z) b ≤1+ |z|n . z 1 + λn Corollary 3.4. If |fn0 (z) + λzfn00 (z) − 1| < b then fn0 (z) ≺ 1 + b z. 1 + λn Corollary 3.5. If (1 − λ) fnz(z) + λfn0 (z) − 1 < b then fn (z) b ≺1+ z. z 1 + λn In the next two theorems we investigate the inclusion relations for classes of Mvλ . Theorem 3.6. For 0 ≤ λ1 < λ and v ≥ 0, let b1 = 1+λ1 n b. 1+nλ Then Mvλ (1, b) ⊂ Mvλ1 (1, b1 ). Proof. The case for λ1 = 0 is trivial. For λ1 6= 0 suppose that p(fn ) ∈ Mvλ (1, b). Therefore, we can write (1 − λ1 ) Dv fn (z) + λ1 (Dv fn (z))0 z v λ1 Dv fn (z) λ1 D fn (z) v 0 (1 − λ) + λ(D fn (z)) + 1 − . = λ z λ z Now, by definition, p(fn ) ∈ Mvλ1 (1, b1 ) and so the proof is complete. b(1+v) . n+1+v Theorem 3.7. Let v ≥ 0 and b1 = Then v Mv+1 λ (1, b) ⊂ Mλ (1, b1 ). Proof. For fn ∈ An suppose that p1 (fn ) ∈ Mv+1 λ (1, b) where p1 (fn (z)) = (1 − λ) D1+v fn (z) + λ(Dv+1 fn (z))0 . z Set p2 (fn (z)) = (1 − λ) J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 Dv fn (z) + λ(Dv fn (z))0 . z http://jipam.vu.edu.au/ S TARLIKENESS AND C ONVEXITY C ONDITIONS 7 An elementary differentiation yields D1+v fn (z) + λ(Dv+1 fn (z))0 z 1 = p2 (fn (z)) + zp0 (fn (z)). 1+v 2 From this and Lemma 3.2, we conclude that p1 (fn ) ∈ Mvλ (1, b1 ). p1 (fn (z)) = (1 − λ) Corollary 3.8. fn0 (z) + λzfn00 (z) fn (z) b 0 ∈ P(1, b) =⇒ (1 − λ) + λfn (z) ∈ P 1, . z 1+n Theorem 3.9. For v ≥ 0 and λ > 0 let b < 1 + λn. If p(fn ) ∈ Mvλ (1, b) then z(Dv fn (z))0 b(2 + λn) Dv fn (z) − 1 < λ[(1 + λn) − b] . Proof. First note that, we can write v (1 − λ) D fn (z) + λ(Dv fn (z))0 − 1 < b ; z For b1 = b(2+λn) λ[(1+λn)−b] v D fn (z) b < − 1 . z 1 + λn we define w(z) by [z(Dv fn (z))0 ] . [Dv fn (z)] One can easily verify that w(z) is analytic in U and w(0) = 0. To conclude the proof, it suffices to show that |w(z)| < 1 in U. If this is not the case, then by Lemma 2.2, there exists a point z0 ∈ U such that |w(z0 )| = 1 and z0 w0 (z0 ) = kw(z0 ). Therefore v D f (z ) 0 v 0 |p(fn (z0 )) − 1| = (1 − λ) + λ(D f (z0 )) − 1 z0 v v 0 D fn (z0 ) z (D f (z )) 0 n 0 (1 − λ) + λ − 1 = z0 Dv fn (z0 ) z0 (Dv fn (z0 ))0 Dv fn (z0 ) Dv fn (z0 ) = λ − 1 + − 1 Dv fn (z0 ) z0 z0 b b ≥ λb1 1 − − = b. 1 + nλ 1 + nλ 1 + b1 w(z) = This is a contradiction to the hypothesis and so |w(z)| < 1 in U. Corollary 3.10. 1+n i) If fn0 (z) ∈ P(1, 3+n ) then ) then ii) If fn0 (z) + zfn00 (z) ∈ P(1, 1+n 3+n zfn00 (z) fn0 (z) zfn0 (z) fn (z) ∈ P(1, 1). ∈ P(1, 1). Theorem 3.11. Let p(fn ) ∈ Mvλ (1, b) for some λ > 0. If √ λ(1 + λn) (n − 3) + n2 + 2n + 9 0<λ≤ 2 + λ(n − 1) ; 2n b= √ r 2λ − 1 (n − 3) + n2 + 2n + 9 ; ≤λ≤1 (1 + λn) λ2 n2 + 2λ(1 + n) 2n then v+1 D fn (z) v < > . v D fn (z) 1+v J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 http://jipam.vu.edu.au/ 8 R. AGHALARY , JAY M. JAHANGIRI , AND S.R. K ULKARNI We need the following lemma, which is due to Ponnusamy and Singh [9]. Lemma 3.12. Let 0 < λ1 < λ < 1 and let Q be analytic in U satisfying Q(z) ≺ 1 + λ1 z, and Q(0) = 1. If q(z) is analytic in U, q(0) = 1 and satisfies Q(z)[c + (1 − c)q(z)] ≺ 1 + λz, where c= 1−λ , 1 + λ1 0 < λ + λ1 ≤ 1 1 − (λ2 + λ21 ) , λ2 + λ21 ≤ 1 ≤ λ + λ1 2(1 − λ21 ) then Re{q(z)} > 0, z ∈ U . Proof of Theorem 3.11. From Theorem 3.1 and the fact 0 < b < 1 < 1 + λn we conclude that Dv fn (z) b ≺ 1 + b1 z, 0 < b1 = < b < 1. z 1 + nλ On the other hand, we may write Dv fn (z) z(Dv fn (z))0 (1 − λ) + λ ≺ 1 + bz. z Dv fn (z) v v 0 fn (z)) Letting Q(z) = D fzn (z) , q(z) = z(D , and c = 1 − λ, we see that all conditions in Lemma Dv fn (z) 3.12 are satisfied. This implies that Re q(z) > 0 and so the proof is complete. Corollary 3.13. Let p(fn ) ∈ Mvλ (1, b) for some λ > 0. Then Dv fn is starlike in the disc λ(1 + nλ) if 0 < λ < λ1 and b1 ≤ b ≤ 1 (2 + λ(n − 1))b r |z| ≤ 2λ − 1 (1 + λn) if λ1 ≤ λ ≤ 1 and b2 ≤ b ≤ 1, 2 2 b λ n + 2λ(1 + n) where λ1 = (n − 3) + √ n2 + 2n + 9 λ(1 + nλ) , b1 = , and 2n [2 + λ(n − 1)] s b2 = (1 + λn ) fn0 ∈ P 1, √ λ2 n 2 2λ − 1 . + 2λ(1 + n) (1+n) then fn is starlike in U . (1+n) 0 00 ii) If fn + zfn ∈ P 1, √ then fn is convex in U . 2 i) If 1+(1+n)2 1+(1+n) If we let λ = 1 and v = 0, 1 in Corollary 3.13, then we obtain Corollary 3.14. Let √ (1+n) 1+(1+n)2 ≤ b ≤ 1 and fn ∈ An . i) If fn0 ∈ P(1, b) then f is starlike for |z| < √ (1+n) ii) If fn0 + zfn00 ∈ P(1, b) then f is convex for J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 . 1+(1+n)2 |z| < √ 1+n 2 . b 1+(1+n) b http://jipam.vu.edu.au/ S TARLIKENESS AND C ONVEXITY C ONDITIONS 9 4. C OEFFICIENT B OUNDS Sufficient coefficient conditions for F(1, b) and Mvλ (1, b) are given next. Theorem 4.1. Let p(f1 ) be given by (1.4) for f1 as in (1.1). If ∞ X (4.1) (k 2 + b − 1)|ak | < b, k=2 then p(f1 ) ∈ F(1, b). Proof. We need to show that if (4.1) then |p(f1 (z)) − 1| < b. For p(f1 ) we can write 0 zf1 zf100 1 + 0 − 1 |p(f1 (z)) − 1| = f1 f1 P∞ 2 k=2 (k − 1)ak z k P = k z+ ∞ k=2 ak z P∞ 2 − 1)|ak ||z|k−1 k=2 (k P∞ ≤ 1 − k=2 |ak ||z|k−1 P∞ 2 − 1)|ak | k=2 (k P∞ . < 1 − k=2 |ak | The above right hand inequality is less than b by (4.1) and so p(f1 ) ∈ F(1, b). Theorem 4.2. Let p(fn ) be given by (1.5) for fn as in (1.1). If ∞ X (4.2) (λk − λ + 1)Bk (v)|ak | < b, k=1+n then p(fn ) ∈ Mvλ (1, b). Proof. Apply the Ruscheweyh derivative (1.6) to the function fn (z) and substitute in (1.5) to obtain v D f (z) n |p(fn (z)) − 1| = (1 − λ) + λ(Dv fn (z))0 − 1 z ∞ X = (λk − λ + 1)Bk (v)ak z k−1 < k=1+n ∞ X (λk − λ + 1)Bk (v)|ak |. k=1+n Now this latter inequality is less than b by (4.2) and so p(fn ) ∈ Mvλ (1, b). Next, by judiciously varying the arguments of the coefficients of the functions fn given by (1.1), we shall show that the sufficient coefficient conditions (4.1) and (4.2) are also necessary for their respective classes with varying arguments. A function fn given by (1.1) is said to be in V(θk ) if arg(ak ) = θk for all k. If, further, there exists a real number β such that θk + (k − 1)β ≡ π(mod 2π) then fn is said to be in V(θk ; β). The union of V(θk ; β) taken over all possible {θk } and all possible real β is denoted by V. For more details see Silverman [13]. Some examples of functions in V are i) T ≡ V(π; 0) ⊂ V where T is the class of analytic univalent functions with negative coefficients studied by Schild [11] and Silverman [12]. J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 http://jipam.vu.edu.au/ 10 R. AGHALARY , JAY M. JAHANGIRI , AND S.R. K ULKARNI P iθk k ii) Univalent functions of the form z + ∞ k=2 |ak |e z are in V(θk ; 2π/k) ⊂ V for θk = π − 2(k − 1)π/k. Note that the family V is rotationally invariant since fn ∈ V(θk ; β) implies that e−iγ fn (zeiγ ) ∈ V(θk + (k − 1)γ; β − γ). Finally, we let VF(1, b) ≡ V ∩ F(1, b) and VMvλ (1, b) ≡ V ∩ Mvλ (1, b). Theorem 4.3. p(f1 ) ∈ VF(1, b) ⇐⇒ ∞ X (k 2 + b − 1)|ak | < b. k=2 Proof. In light of Theorem 4.1, we only need to prove the “only if ” part of the theorem. Suppose p(f1 ) ∈ VF(1, b), then P∞ k=2 (k 2 − 1)ak z k−1 <b P |p(f1 ) − 1| = k−1 1+ ∞ a z k=2 k or ∞ ∞ X X 2 k−1 k−1 (4.3) ak z . (k − 1)ak z < b 1 + k=2 k=2 The condition (4.3) must hold for all values of z in U. Therefore, for f1 ∈ V(θk ; β) we set z = reiβ in (4.3) and let r −→ 1− . Upon clearing the inequality (4.3) we obtain the condition ! ∞ ∞ X X (k 2 − 1)|ak | < b 1 − |ak | k=2 k=2 as required. Corollary 4.4. If 0 < b ≤ 1 and p(f1 ) ∈ VF(1, b) then f1 is convex in U. Corollary 4.5. If 1 < b ≤ 3 and p(f1 ) ∈ VF(1, b) then f1 is starlike in U. The above two corollaries can be justifed using Theorem 4.3 and the following lemma due to Silverman [12]. Lemma 4.6. For f1 of the form (1.1) and univalent in U we have P 2 i) If P∞ k=2 k |ak | ≤ 1, then f1 is convex in U. ∞ ii) If k=2 k|ak | ≤ 1, then f1 is starlike in U. Next, we show that the above sufficient coefficient condition (4.2) is also necessary for functions in VMvλ (1, b). Theorem 4.7. p(fn ) ∈ VMvλ (1, b) ⇐⇒ ∞ X (λk − λ + 1)Bk (v)|ak | < b. k=1+n Proof. Suppose that p(fn ) ∈ VMvλ (1, b). Then, by (1.5), we have Dv fn (z) v 0 |p(fn (z)) − 1| = (1 − λ) + λ(D fn (z)) − 1 < b. z J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 http://jipam.vu.edu.au/ S TARLIKENESS AND C ONVEXITY C ONDITIONS 11 On the other hand, for fn ∈ V(θk ; β) we have fn (z) = z + ∞ X |ak |eiθk z k . k=1+n VMvλ (1, b) The condition required for p(fn ) ∈ must hold for all values of z in U. Setting z = reiβ yields ∞ X (λk − λ + 1)Bk (v)|ak |rk−1 < b. k=1+n The required coefficient condition follows upon letting z −→ 1− . From the above Theorem 4.7 and Lemma 4.6.ii, we obtain Corollary 4.8. If λ ≥ 2b − 1 and p(fn ) ∈ VMvλ (1, b) then f is starlike in U. R EFERENCES [1] J.W. ALEXANDER, Functions which map the interior of the unit circle upon simple regions, Ann. of Math. Soc., 17 (1915-1916), 12–22. [2] J. DIEUDONNÉ, Recherches sur quelques problèms relatifs aux polynômes et aux functions bornées d’une variable complexe, Ann. Sci. Ecole Norm. Sup., 48 (1931), 247–358. [3] P.L. DUREN, Univalent Functions, Grundlehren der Mathematischen Wissenschaften 259, A series of Comprehensive Studies in Mathematics, Springer-Verlag New York Inc, 1983. [4] I.S. JACK, Functions starlike and convex of order α, J. London Math. Soc., 3 (1971), 469–474. [5] J.M. JAHANGIRI, H. SILVERMAN AND E.M. SILVIA, Inclusion relations between classes of functions defined by subordination, J. Math. Anal. 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SILVERMAN, Univalent functions with varying arguments, Houston J. Math., 7 (1981), 283– 287. [14] H. SILVERMAN AND E. M. SILVIA, Subclasses of starlike functions subordinate to convex functions, Canad. J. Math., 37 (1985), 48–61. J. Inequal. Pure and Appl. Math., 5(2) Art. 31, 2004 http://jipam.vu.edu.au/