Some applications of differential subordinations in the geometric function theory

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1 Sokół Journal of Inequalities and Applications 2013, 2013:389 R E S E A R C H Open Access Some applications of differential subordinations in the geometric function theory Janus Sokół * Dedicated to Professor Hari M Srivastava * Correspondence: jsokol@pr.edu.pl Department of Mathematics, Resów University of Technology, al. Powstańców Warsawy 12, Resów, , Poland Abstract Let A denote the class of functions f, which are analytic in the open unit disc and normalied by f(0) = f (0) 1 = 0. We investigate the connection of the quantity (f () 1)/f()withf()/. Obtained results are the extensions of those presented by Tuneski and Obradović (Comput. Math. Appl. 62: , 2011). Moreover, we solve the open problems posed in the paper above. MSC: Primary 30C45; secondary 30C80 Keywords: differential subordination; starlike function; convex function; order of starlikeness; order of convexity; best dominant 1 Introduction Let A(n), n N, denote the class of functions of the form = + a n+1 n+1 + a n+2 n+2 +, which are analytic in the open unit disc U = { : < 1} on the complex plane C. For two analytic functions f, g, wesaythatf is subordinate to g, writtenasf g, ifandonlyif there exists an analytic function ω with property ω() in U such that =g(ω()). In particular, if g isunivalent in U, wehavethefollowingequivalence g() f (0) = g(0) and f (U) g(u). (1.1) The idea of subordination was used for defining many of classes of functions studied in the geometric function theory. For obtaining the main result, we shall use the method of differential subordinations. The main result in the theory of differential subordinations was introduced by Miller and Mocanu in [1, 2]. A function p, analytic in U,issaidtosatisfy a first order differential subordination if φ ( p(), p () ) h(), (1.2) where (p(), p ()) D C 2, φ : C 2 C is analytic in D, h is analytic and univalent in U. The function q is said to be a dominant of the differential subordination (1.2)ifp q for 2013 Sokół; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 2 of 11 all p satisfying (1.2). If q is a dominant of (1.2)and q q for all dominants q of (1.2), we say that q is the best dominant of the differential subordination (1.2). By using a Miller-Mocanu lemma on the first order differential subordination, the authors of [3] proved the following theorem. Theorem 1.1 [3] Let f A(1), / 0for all U and 0<μ 1. If (f () 1) 2μ 1+μ h 1() ( U), (1.3) 1 μ (1.4) and μ is the best dominant of (1.4). Furthermore, 1 < μ ( U), (1.5) andthisconclusionissharp, i.e., in inequality (1.5), μ cannot be replaced by a smaller number such that the implications holds. 2 Main results It is easy to verify that under the assumptions of Theorem 1.1, the function p()=/is analytic in U, and that the subordination (1.3) becomes [ ( ) ] p()+ p () +1 h 1 ()= 2μ ( U). p() 1+μ Thus, (1.3) can be rewritten as the Briot-Bouquet-type differential subordination s()+ s () βs()+γ h() ( U) (2.1) with s()=p(), β =1,γ =0,h()=1 h 1 (). One of the basic results in the theory of Briot-Bouquet differential subordinations is a theorem from [4], which says (in its particular case) that if h is convex univalent with positive real part in U and the functions s, h satisfy (2.1), s h. It can sometimes be improved if we know more about the function h. A better subordination will be derived by applying the results from the book [2]. After some adaptation, Theorem 3.2j [2, p.97] becomes the following lemma. Lemma 2.1 [2] Suppose that β >0and n is a positive integer. Let us denote R β,n ()=β n 1 2 ( U), (2.2) and let h be a convex univalent function in U with h(0) = 1 such that βh() R β,n () ( U). (2.3)

3 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 3 of 11 Then the Briot-Bouquet differential equation q()+ nq () βq() = h() ( U) has a univalent solution q n analytic in U. Furthermore, if the functions h and s() =1+ a n n + a n+1 n+1 + satisfy the Briot-Bouquet differential subordination s()+ s () βs() h() ( U), s() q n () h() ( U), (2.4) and the function q n isthebestdominantofthesubordination(2.4) in the sense that if there exists a function p such that s() p(), q n () p(). Notice that the function R β,n is called the open door function, and it is univalent in U, R β,n (0) = β. Thus,by(1.1), in order to verify (2.3), it is sufficient to show that h(0) = 1 and βh(u) R β,n (U). The set R β,n (U) is the complex plane with slits along the half-lines Rew =0and Im w γ = n 1+2β/n. ForR 4,1, these slits are placed in Figure 1 with respect to the circle centered at w = 1 and the radius 10. If h is a special bilinear transformation h()= 1+A, A C, B [ 1, 0], A B, 1+B and if we replace condition (2.3) by a simpler condition Re{h()} >0,itwillbesatis- fied (see [2,p.108])ifandonlyif Re [ β(1 + AB) ] βa + βb (2.5) Figure 1 R β,n (U).

4 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 4 of 11 when B ( 1, 0], or Reβ(1 + A)>0 and Reβ(1 A) 0, (2.6) when B = 1. Then the univalent solution q n of the Briot-Bouquet differential equation q()+ nq () βq() = 1+A 1+B ( U) (2.7) has the form where q n ()= g n ()= 1 βg n (), (2.8) 1 1 n 0 [ 1+Bt 1 n 1+B ] β n ( A B 1) t β n 1 dt, ifb 0, 0 e βa(t 1) n t β n 1 dt, ifb =0. Theorem 2.2 Let f A(n), / 0for all U. If / =1+a n n + a n+1 n+1 + satisfies the Briot-Bouquet differential subordination (f () 1) h() ( U), (2.9) with a convex univalent function h, such that h(0)=0 and 1 h() R 1,n () ( U), where R 1,n is the open door function given in (2.2), where and q n ()= 1 n q n () ( U), (2.10) 0 H()= exp [ H(t) H() 0 ] 1/n t 1 dt ( U) (2.11) h(t)/t dt, and q n is the best dominant of (2.10). Proof The function p()=/=1 a n n + is analytic in U and (f () 1) ( ) = p()+ p () +1 ( U), p()

5 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 5 of 11 hence subordination (2.9)becomes p()+ p () p() 1 h() ( U). By Lemma 2.1,theequation q()+ nq () q() =1 h() ( U) has the univalent solution of the form {q n ()} 1,whereq n is in (2.11) (for more details see [2, p.86]). Therefore, again by Lemma 2.1, the function {q n ()} 1 is the best dominant of the subordination p()= { q n () } 1 ( U), (2.12) which is equivalent to (2.10) with the best dominant q n. Theorem 2.3 Let f A(n), / 0for all U. Assume that A C, B [ 1, 0], A B and that A and B satisfy either (2.5) or (2.6) with β =1.If / =1+a n n + a n+1 n+1 + satisfies the Briot-Bouquet differential subordination (f () 1) (B A) 1+B ( U), (2.13) g n ()= { 1 n 0 [ 1+Bt 1 n 1+B ] 1 n ( A B 1) t 1 n 1 dt, if B 0, 0 e A(t 1) n t 1 n 1 dt, if B =0, (2.14) and g n () is the best dominant of (2.14). Proof Subordination (2.13)withp()=/=1 a n n + becomes p()+ p () p() = 1+A 1+B ( U). If we return to Lemma 2.1 with β = 1 and to the remarks below Figure 1,wecansee that under assumptions (2.5)or(2.6)thereexistsaunivalentsolutionq n of equation (2.7). Moreover, this function q n has the form (2.8) and is best dominant of the subordination p q n.italsoimplies1/p g n,whichisequivalentto(2.14). Theorem 2.3 with n =1,A = μand B = μ becomes the earlier Theorem 1.1, cited in the first section. Notice that for B ( 1, 0] the function on the right hand side of (2.13)maps the open unit disc U onto the disc D(C, R)= { w C : w C < R },

6 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 6 of 11 where C = B(A B) A B, R = 1 B 2 1 B. 2 This functionis also univalent in U whence, by (1.1), Theorem 2.3can be written in the following form. Corollary 2.4 Let f A(n), / 0for all U. If / =1+a n n + a n+1 n+1 + satisfies the inequality (f () 1) B(A B) 1 B 2 < A B 1 B 2 ( U), where A C, B ( 1, 0], A Bsatisfy(2.5) with β =1, g n ()= { 1 n 0 [ 1+Bt 1 n 1+B ] 1 n ( A B 1) t 1 n 1 dt, if B 0, 0 e A(t 1) n t 1 n 1 dt, if B =0 ( U), (2.15) and it is the best dominant of (2.15). Moreover, if A satisfies (2.6) with β =1,and if / = 1+a n n + a n+1 n+1 + satisfies condition (2.13) with B = 1,i.e., { (f } () 1) Re < 1, ( U), (1 + A) 2 g n ()= 1 n and g n () is the best dominant of (2.16). 0 [ ] 1+A 1 n 1 t n 1 dt, ( U), (2.16) 1 t If B =0,A = λ (0, 1], n =1,(2.5) is satisfied, and Corollary 2.4 becomes the following one, which is a generaliation of Corollary 1 in [3]. Corollary 2.5 Let λ (0, 1], f A(1). Suppose that / 0for all U. If f satisfies the inequality (f () 1) < λ ( U), (2.17) 1 eλ 1 λe λ 1= ( 1) k (λ) k (k +1)! ( U), (2.18) and it is the best dominant of (2.18). Note that instead of (2.18), in [3]istheinequality 1 < μ ( U), (2.19)

7 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 7 of 11 where μ = λ, under the same assumption as in Corollary 2.5. In[3], the authors posed 2 λ also the problem of finding the smallest number μ such that (2.19) holdsundertheassumptions of Corollary 2.5. Inviewof(2.18), solving this problem is equivalent to finding { max λe λ 1 }. : U (2.20) To find (2.20), we use the Taylor series (2.18)with = e it. λe λ 1 ( 1) k+1 λ k e kit (k +1)! = 1 λ λ k (k +1)! λ k+1 (k +1)! = 1 λ( λ ). Moreover, hence λe λ 1 { max λe = 1 e = 1 λ( λ 1 λ ), λ 1 } : U = λ( 1 λ ). (2.21) For all λ (0, 1], the number (2.21) is better than the bound in (2.19), because it is smaller than μ = λ, given in [3]. This is because 2 λ λ 2 λ = 1 λ 2 /2 λ 1 λ/2 = 1 λ 2 k λ k+1 1 λ k+1 λ (k +1)! { } = 1 λ k 1 λ + λ k! = 1 λ( λ ). k=0 Therefore, the following corollary contains the solution of the first open problem posed in [3].

8 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 8 of 11 Corollary 2.6 Let λ (0, 1], f A(1). Suppose that / 0for all U. If f satisfies the inequality (f () 1) < λ ( U), 1 < eλ 1 λ λ ( U), and this bound is the best possible. The second open problem posed in [3] istofind thesharpversion ofthe followingcorollary. Corollary 2.7 [3] Let λ (0, 1], f A(1). If f satisfies the inequality f () 1 < λ ( U), (2.22) f () 1 < 2λ 2 λ ( U) (2.23) and { f } () Re >1 3λ 2 ( U). (2.24) Using the earlier results, we can improve the corollary above. Corollary 2.8 Under the assumptions of Corollary 2.7, we have f () 1 < ( U) (2.25) and Re f () > λ(2 eλ ) ( U). (2.26) Proof Inequality (2.22) follows that / 0forall U. Hence(2.22) isthesameas (2.17), thus,from (2.18), we obtain eλ 1 λe λ ( 1) k+1 (λ) k =1+ (k +1)! ( U).

9 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 9 of 11 Therefore, for = e it,weobtain λe λ 1+ ( 1) k+1 λ k e kit (k +1)! λ k 1+ (k +1)! { } = 1 λ k+1 λ + λ (k +1)! = 1 λ( ). (2.27) Moreover, λe λ = 1 e = 1 λ( λ 1 ). So the bound (2.27) is the best possible. Applying (2.27)in(2.22), we directly obtain (2.25). Making use of (2.22)andof(2.27), we get the inequality Re f () = Re (f () 1) = λ(2 eλ ) + Re > λ + ( U), λ which is (2.26). For all λ (0, 1] the bound (2.25) is smaller than the bound 2λ, given in (2.24). To show 2 λ this, observe that 2λ 2 λ = λ 1 λ/2 = λ k 2 > k 1 λ k k! = 1+ k=0 λ k k! = eλ 1. (2.28) Similarly, for all λ (0, 1], the bound (2.26) is greater than the bound 1 3λ/2, given in (2.23). To show this, observe that the inequality λ(2 e λ ) >1 3λ 2, that has to be proved, after some calculations, becomes 2λ 2 λ > eλ 1, which was proved above in (2.28). The function =()/λ shows that bound (2.25) delivers the sharp version of (2.23), so this is the solution of the first part of the second

10 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 10 of 11 open problem posed in [3]. The bound (2.26) does not seem to be the best possible. We conjecture that the best possible bound is Re f () > λ ( U), (2.29) which is suggested by the function =()/λ. In [5], Robertson introduced the classes Sα, K α of starlike and convex functions of order α 1, which are defined by { Sα := f A : Re f () { K α := f A : Re ( } > α, D, 1+ f () f () ) } > α, D = { f A : f () Sα}. If α [0; 1), a function in each of these sets is univalent, if α <0,itmayfailtobe univalent. In particular, we have S0 = S, K 0 = K, the usual classes of starlike and of convex functions, respectively. In Corollary 2.8, weobtainedtheorder of starlikenessfor functions satisfying (2.22), so we have where f () 1 < λ f Sα, α = λ(2 eλ ), λ (0, 1]. (2.30) Therefore, if λ (0, log 2], condition (2.22)issufficientforf to be starlike, namely, f () 1 < λ f S. Notice that the number log is better than 2/3 given in this place in [3]. Recall that in (2.29), we conjectured that for all λ (0, 1] a function satisfying (2.22) isstarlike, even more, starlike of order λ/().usingf () instead of f in Corollary 2.8,weobtain the order of convexity for a function satisfying (2.22), so we have f ()+f () 1 < λ f () ( 1+ f ) () > α, f () where α is described in (2.30), which means that f K α. For some similar conditions for starlikeness of order α, we also refer to [6, 7]andto[8]. Competing interests The author declares that they have no competing interests. Received: 3 April 2013 Accepted: 6 August 2013 Published: 20 August 2013

11 Sokół Journal of Inequalities and Applications 2013, 2013:389 Page 11 of 11 References 1. Miller, SS, Mocanu, PT: Differential subordinations and univalent functions. Mich. Math. J. 28, (1981) 2. Miller, SS, Mocanu, PT: Differential Subordinations: Theory and Applications. Series of Monographs and Textbooks in Pure and Applied Mathematics, vol Dekker, New York (2000) 3. Tuneski, N, Obradović, M: Some properties of certain expressions of analytic functions. Comput. Math. Appl. 62, (2011) 4. Eenigenburg, P, Miller, SS, Mocanu, PT, Reade, MO: General Inequalities 3. International Series of Numerical Mathematics, vol. 64, pp Birkhäuser, Basel (1983) 5. Robertson, MS: On the theory of univalent functions. Ann. Math. 37, (1936) 6. Nunokawa, M, Kuroki, K, Sokół, J, Owa, S: New extensions concerned with results by Ponnusamy and Karunakaran. Adv. Differ. Equ. 2013, 134 (2013) 7. Sokół, J: On a condition for starlikeness. J. Math. Anal. Appl. 352, (2009) 8. Sokół, J, Nunokawa, M: On some sufficient conditions for univalence and starlikeness. J. Inequal. Appl. 2012, 282 (2012) doi: / x Cite this article as: Sokół: Some applications of differential subordinations in the geometric function theory. Journal of Inequalities and Applications :389.