Arif et al. Journal of Inequalities and Alications 2015, 2015:5 R E S E A R C H Oen Access On roducts of multivalent close-to-star functions Muhammad Arif 1,JacekDiok 2*,MohsanRaa 3 and Janus Sokół 4 * Corresondence: jdiok@univ.resow.l 2 Faculty of Mathematics and Natural Sciences, University of Resów, Resów, 35-310, Poland Full list of author information is available at the end of the article Abstract In the resent aer we define a class of roducts of multivalent close-to-star functions and determine the set of airs a, r), a < r 1 a,suchthatevery function from the class mas the disk Da, r):={ : a < r} onto a domain starlike with resect to the origin. Some consequences of the obtained result are also considered. MSC: 30C45; 30C50; 30C55 Keywords: analytic functions; close-to-star functions; generalied starlikeness; radius of starlikeness 1 Introduction Let A denote the class of functions which are analytic in D = D0, 1), where Da, r)= { : a < r } and let A denote the class of functions f A of the form f )= + a n n D; N0 = {0,1,2,...} ). 1) n=+1 Afunctionf A is said to be starlike of order α in Dr):=D0, r)if f ) ) Re > α Dr); 0 α < ). f ) Afunctionf A 1 is said to be convex of order α in D if Re 1+ f ) ) > α D;0 α <1). f ) We denote by S c α) the class of all functions f A, which are convex of order α in D and by S α) wedenotetheclassofallfunctionsf A, which are starlike of order α in D. We also set S α)=s 1 0). Let H be a subclass of the class A.Wedefinethe radius of starlikeness of the class H by R { }) H)= inf su r 0, 1] : f is starlike of order 0 in Dr). f H 2015 Arif et al.; licensee Sringer. This is an Oen Access article distributed under the terms of the Creative Commons Attribution License htt://creativecommons.org/licenses/by/2.0), which ermits unrestricted use, distribution, and reroduction in any medium, rovided the original work is roerly cited.
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 2 of 14 We denote by Pβ), 0 < β 1, the class of functions h A such that h0) = 1 and { hd) β := w C \{0} : Arg w < β π }, 2 where Arg w denote the rincial argument of the comlex number w i.e. from the interval π, π]). The class P := P1) is the well-known class of Carathéodory functions. We say that a function f A belongs to the class CS α, β) ifthereexistsafunction g S α)suchthat f g Pβ). In articular, we denote CS α)=cs α,1), CS α)=cs 1 α), CS = CS 0). The class CS is the well-known class of close-to-star functions with argument 0. Silverman [1] introduced the class of functions F given by the formula F)= n ) fj ) aj n g j ) ) b j fj S α), g j S c β)j =1,2,...,n) ), where a j, b j j =1,2,...,n) are ositive real numbers satisfying the following conditions: a j = a, b j = b. Dimkov [2] studied the class of functions F given by the formula F)= n ) fj ) aj fj S α j ), j =1,2,...,n ), where a j j =1,2,...,n) are comlex numbers satisfying the condition 1 α j ) a j a. Let, n be ositive integer and let a, m, M, N be ositive real numbers, b [ m, m]. Moreover, let a =a 1, a 2,...,a n ), α =α 1, α 2,...,α n ), β =β 1, β 2,...,β n ) be fixed vectors, with a j R, 0 α j <, 0<β j 1 j =1,2,...,n).
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 3 of 14 We denote by H n a, α, β)theclassoffunctionsf given by the formula F)= n ) fj ) aj fj CS α j, β j ), j =1,...,n ). 2) By G nm, b, c) we denote union of all classes Hn a, α, β)forwhich α j ) a j = m, α j ) Re a j = b, β j a j = c. 3) Finally, let us denote G n M, N):= c [0,N] b [ m,m] m [0,M] G n m, b, c). 4) It is clear that the class G n M, N) contains functions F given by the formula 2)forwhich α j ) a j M, β j a j N. Aleksandrov [3] stated andsolvedthe followingroblem. Problem 1 Let H be the class of functions f A that are univalent in D and let D be a domain starlike with resect to an inner oint ω with smooth boundary given by the formula t)=ω + rt)e it 0 t 2π). Find conditions for the function rt) suchthatforeachf H the image domain f ) is starlike with resect to f ω). Świtoniak andstankiewic [4, 5], Dimkov and Diok [6]see also[7]) have investigated a similar roblem of generalied starlikeness. Problem 2 Let H A. Determine the set B H)ofallairsa, r) D R,suchthat a < r 1 a, 5) and every function f H mas the disk Da, r) onto a domain starlike with resect to the origin. The set B H) is called the set of generalied starlikeness of the class H. We note that R H)=su { r :0,r) B H) }. 6) In this aer we determine the sets B H n a, α, β)), B G n m, b, c)) and B G n M, N)). The sets of generalied starlikeness for some subclasses of the defined classes are also considered. Moreover, we obtain the radii of starlikeness of these classes of functions.
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 4 of 14 2 Main results We start from listing some lemmas which will be useful later on. Lemma 1 [5] A function f A mas the disk Da, r), a < r 1 a, onto a domain starlike with resect to the origin if and only if Re eiθ f a + re iθ ) f a + re iθ ) 0 0 θ 2π). 7) For a function f S α)itiseasytoverifythat f ) f ) α α) 1+ 2 1 2 2 α) 1 2 D). Thus, after some calculations we get the following lemma. Lemma 2 Let f S α), a, θ R, D 0 := D \{0}. Then [ f Re ae iθ ) f ) )] 2 α) 1 Re ae iθ a ). 2 Lemma 3 [8] If h Pβ), then h ) h) 2β D). 1 2 Theorem 1 Let m, b, c be defined by3) and set 0 r r 1, a < r, B = a, r) C R : r 1 < r < r 2, a ϕr),, 8) r 2 r < q, a q r B = { a, r) C R : a < r q a }, 9) where r 1 = r 2 = 4m + c), 10) m + c) m + c + m + c) 2 2b + 2 ) 2, 11) q = m + c + m + c) 2 2b +, 12) 2 ϕr)= r 2 1 2 rm + c)) 2. 13) 2b 1 Moreover, set { B for b > /2, B = for b /2. B 14)
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 5 of 14 If a, r) B, then a function F H n a, α, β) mas the disk Da, r) onto a domain starlike with resect to the origin. The result is shar for b /2 and for b > /2 the set B cannot be larger than B. It means that B B H n a, α, β)) B b > /2), 15) B H n a, α, β)) = B b /2). 16) Proof Let F belong to the class H n a, α, β)andlet = a+reiθ D satisfy 5). The functions g j,s )=e is f j e is ) D; j =1,2,...,n, s R) belong to the class CS α j, β j ) together with the functions f j ). Thus, by 2) the functions G s )=e is F e is ) D; s R) belong to the class H n a, α, β) together with the function F). In consequence, we have a, r) B H n a, α, β)) a, r ) B H n a, α, β)) a D, r 0). 17) Therefore, without loss of generality we may assume that a is nonnegative real number. Since f j CS α j, β j ), there exist functions g j S α j)andh j Pβ j )suchthat f j ) g j ) = h j) D) or equivalently f j )=g j )h j ) D). 18) After logarithmic differentiation of the equality 2)we obtain F ) F) = + fj ) a j f j ) ) D). Thus, using 18)wehave Re eiθ F ) F) = Re eiθ Re eiθ + + g Re a j e iθ j ) g j ) )) + g Re a j e iθ j ) g j ) )) ) Re a j e iθ h j) h j ) a j h j) h j ). By Lemma 2 and Lemma 3 we obtain Re eiθ F ) F) Re eiθ + 2 1 2 2 1 2 α j )a j Re e iθ) 2 α j ) a j 1 2 a j β j.
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 6 of 14 Setting = a + re iθ and using 3) the above inequality yields Re eiθ F a + re iθ ) Fa + re iθ ) Re e iθ a + re iθ +2Rer + ae iθ )b m c 1 a + re iθ 2. By Lemma 1 it is sufficient to show that the right-hand side of the last inequality is nonnegative, that is, Re If we ut r + ae iθ +2Rer + ae iθ )b m c 1 r + ae iθ 2 0. 19) r + ae iθ = x + yi, then we obtain x x 2 + y +2bx m c 2 1 x 2 y 0. 2 Thus, using the equality x r) 2 + y 2 = a 2, 20) we obtain wx)=2r2b )x 2 2b ) r 2 a 2) +4rm + c) ) x +2m + c) r 2 a 2) 0. 21) The discriminant of wx)isgivenby = 2b ) r 2 a 2) +4rm + c) ) 2 16r2b )m + c) r 2 a 2) = A 1 A 2, 22) where A 1 = 2b) r 2 a 2) + +4rm + c)+4 rm + c), 23) A 2 = 2b) r 2 a 2) + +4rm + c) 4 rm + c). 24) Let D = { a, r) R 2 :0 a < r 1 a }. 25) First, we discuss the case b > /2. Thus, the inequality 21) issatisfiedforeveryx [r a, r + a] if one of the following conditions is fulfilled: 1 0, 2 >0, wr a) 0 and x 0 r a, 3 >0, wr + a) 0 and x 0 r + a,
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 7 of 14 where x 0 = 2b )r2 a 2 )+4rm + c). 26) 42b )r Ad 1.SinceA 1 >0,by22), the condition 0 is equivalent to the inequality A 2 0. Then B 1 := { a, r) D : 0 } = { a, r) D : A 2 0 } = { a, r) D : a ϕr) }, where ϕ is defined by 13). Let γ = { a, r) D : a = ϕr) }. Then γ is the curve which is tangent to the straight lines a = r and a = q r at the oints S 1 =r 1, r 1 ) and S 2 =q r 2, r 2 ), 27) where r 1, r 2, q are defined by 10), 11), 12), resectively. Moreover, γ cuts the straight line a =0attheoints r 3 = m + c + 2b )+ m + c ) 2, r 4 = m + c 2b )+ m + c ) 2. Since 0<r 3 < r 1 < r 2 < r 4 < q, we have γ = { a, r) R 2 : r 3 r r 4, a = ϕr) }, and consequently B 1 = { a, r) R 2 : r 3 r r 4,0 a ϕr) }, 28) where ϕ is defined by 13)seeFigure1). Ad 2.Let B 2 := { a, r) D : >0 wr a) 0 x 0 r a }. It is easy to verify that wr a)=r a) 2b 1)r a) 2 2m + c)r a)+1 ) =2b 1)r a) r a q ) r a q),
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 8 of 14 Figure 1 The set B 1. where q is defined by 12)and q = m + c m + c) 2 2b + 2) 1. 29) Since 0<q <1<q /2 < b m,a, r) D ), 30) we see that r a) r a q ) <0 a, r) D ). Thus, the inequality wr a) 0istrueifa r q. Theinequalityx 0 r a may be written in the form 2b )a 2 +32b )r 2 4m + c)r 42b )ar + 0. 31) The hyerbola h 1, which is the boundary of the set of all airs a, r) R 2 satisfying 31), cuts the boundary of the set D at the oint S 1 defined by 27)andattheointr 5,0),where r 5 = 2m + c)+ 4m + c) 2 32b ) ) 1. 32) It is easy to verify that r 3 < r 5 < r 4 < q. Thus we determine the set B 2 = { a, r) R 2 : { }} 0 r r 3,0 a < r,, 33) r 3 < r < r 1, ϕr)<a < r where ϕ is defined by 13)seeFigure2).
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 9 of 14 Figure 2 The sets B 2 and B 3. Ad 3.Let B 3 := { a, r) D : >0 wr + a) 0 x 0 r + a } and let q and q be defined by 12)and29), resectively. Then wa + r)=r + a) [ 2b )r + a) 2 2m + c)r + a)+ ] =2b )r + a) r + a q ) r + a q). Moreover, by 30) wehave r + a) r + a q ) <0 a, r) D ). Thus, we conclude that the inequality wr + a) 0istrueifa q r. Theinequality x 0 r + a may be written in the form 2b )a 2 +32b )r 2 4m + c)r +42b )ar + 0. 34) The hyerbola h 2, which is the boundary of the set of all airs a, r) R 2 satisfying 34), cuts the boundary of the set D at the oint S 2 defined by 27)andattheointr 5,0),where r 5 is defined by 32). Thus, we describe the set { { }} B 3 = a, r) R 2 r 2 < r < r 4, ϕr)<a q r, :, 35) r 4 < r < q,0 a q r where ϕ is defined by 13)seeFigure2). The union of the sets B 1, B 2, B 3 defined by 28), 33), and 35)givestheset 0 r r 1,0 a < r, B = a, r) R2 : r 1 < r < r 2,0 a ϕr),. r 2 r < q,0 a q r
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 10 of 14 Thus, by 17)wehave B B H n a, α, β)) b > /2), 36) where B is defined by 8). Now, let b < /2. Then the inequality 21)issatisfiedforeveryx [r a, r + a]if wr a) 0 and wr + a) 0. 37) We see that wa + r)=2b )r + a) r + a q ) r + a q), wr a)=2b )r a) r a q ) r a q), where q and q are defined by 12)and29), resectively. Since q <0<q <1 b < /2), the condition 37)issatisfiedifa, r) D and a q r. 38) Let b = 1/2. Then by 21)weobtain 4rm + c) ) x +2m + c) r 2 a 2) 0. The above inequality holds for every x [r a, r + a]ifa, r) D and r a 2m + c) or equivalently 38). Thus, by 17)wehave B B H n a, α, β)) b /2), 39) where B is defined by 9). Because the function F)= n 1 1 1 + sgna j )) 2 α) ) 1 βj sgna j )) aj D) 40) 1+ belongs to the class H n a, α, β), and for = a + r, θ =0,a + r > q we have Re eiθ F ) F) Lemma 1 yields = 2 2m + c)a + r)+2b )a + r) <0. a + r)1 a + r) 2 ) B H n a, α, β)) B. 41)
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 11 of 14 From 36) and41) wehave15), while 39) and41) give16), which comletes the roof. Since the set B defined by 14) is deendent only of m, b, c, the following result is an immediate consequence of Theorem 1. Theorem 2 Let B be defined by 14). If a, r) B, then a function F G n m, b, c) mas the disk Da, r) onto a domain starlike with resect to the origin. The obtained result is shar for b /2 and for b > /2 the set B cannot be larger than B, where B is defined by 8). It means that B B G n m, b, c)) B B G n m, b, c)) = B b /2). b > /2), The functions described by 40),with 3) are the extremal functions. Theorem 3 B G n M, N) ) = { a, r) C R : a < r q 1 a }, 42) where q 1 = M + N + M + N) 2 +2M + 2. The equality in 42) is realied by the function F of the form 1 )2M+N F)= D). 43) 1 + ) N Proof Let M, N be ositive real numbers and let B = B m, b, c), B = B m, b, c), q = qm, b, c) andϕr) =ϕr; m, b, c) bedefinedby8), 9), 12), and 13), resectively. It is easy to verify that ϕr; m, b, c) qm, /2, c) r 1/ 2qm, /2, c) ) r qm, /2, c), /2 < b m ). Moreover, the function q = qm, b, c) is decreasing with resect to m and c, and increasing with resect to b.thus,fromtheorems1 and 2 we have see Figure 3) B G n m, /2, c) ) = B m, /2, c) B m, b, c) B G n m, b, c) ) m [0, M], c [0, N], b /2, m]) and B G n M, M, N) ) B G n m, b, c) ) B G n m, /2, c) ) m [0, M], c [0, N], b [ m, /2] ). Therefore, by 4) weobtain B G n M, N) ) = B G n M, M, N) ) 44)
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 12 of 14 Figure 3 The sets B and B. and by Theorem 2 we get 42). Putting m = M, b = M in 3)weseethata 1, a 2,...,a n are negative real numbers. Thus, the extremal function 40) hasthe form F)= or equivalently n 1 1 ) 21 α j) ) 1+ βj ) aj D) 1 ) 1+ n β j a j F)= 1 ) 2 n 1 α j ) a j 1 D). Consequently, using 3)weobtain F)= ) 1+ N D), 1 ) 2M 1 that is, we have the function 43) and the roof is comleted. Since H a1), α), β)) = CS α, β), by Theorem 1 we obtain the following theorem. Theorem 4 Let 0 α <,0<β 1, and 0 r r 1, a < r, B = a, r) C R : r 1 < r < r 2, a ϕr),, r 2 r < q, a q r B = { a, r) C R : a < r q a }, where r 1 = r 2 = 1 4β α + ), β + α) β + α + β α) 2 +2β) 2,
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 13 of 14 q = β + α + β α) 2 +2β, ϕr)= r 2 1 2 rβ α + )) 2. 2 2α 1 Moreover, let us ut { B for α < /2, B = for α /2. B If a, r) B, then the function f CS α, β) mas the disk Da, r) onto a domain starlike with resect to the origin. The obtained result is shar for α /2 and for α < /2 the set B cannot be larger then B. It means that B B CS α)) B α < /2), B CS α)) = B α /2). The function F of the form F)= 1 + ) β D) 1 ) 2 2α+β is the extremal function. Using 6)andTheorems1-4, we obtain the radii of starlikeness for the classes H n a, α, β), G nm, b, c), Gn M, N), CS α, β). Corollary 1 TheradiusofstarlikenessoftheclassesG nm, b, c) and Hn a, α, β) is given by R G n m, b, c)) = R H n a, α, β)) = m + c + m + c) 2 2b + 2. Corollary 2 TheradiusofstarlikenessoftheclassG n M, N) is given by R G n M, N)) = M + N + M + N) 2 +2M + 2. Corollary 3 TheradiusofstarlikenessoftheclassCS α, β) is given by R CS α, β)) = β + α + β α) 2 +2β. Remark 1 Putting β =1inCorollary3 we get the radius of starlikeness of the class CS α)=cs α, 1) obtained by Diok [7]. Putting = β = 1 we get the radius of starlikeness of the class CS α)=cs 1 α, 1) obtained by Ratti [9]. Putting, moreover, α =0weget the radius of starlikeness of the class CS = CS 1 0, 1) obtained by MacGregor [10]. Cometing interests The authors declare that they have no cometing interests.
Arif et al. Journal of Inequalities and Alications 2015, 2015:5 Page 14 of 14 Authors contributions All authors jointly worked on the results and they read and aroved the final manuscrit. Author details 1 Deartment of Mathematics, Abdul Wali Khan University, Mardan, Pakistan. 2 Faculty of Mathematics and Natural Sciences, University of Resów, Resów, 35-310, Poland. 3 Deartment of Mathematics, GC University Faisalabad, Faisalabad, Pakistan. 4 Deartment of Mathematics, Resów University of Technology, Resów, 35-959, Poland. Acknowledgements This work is artially suorted by the Centre for Innovation and Transfer of Natural Sciences and Engineering Knowledge, University of Resów. Received: 11 October 2014 Acceted: 3 December 2014 Published: 06 Jan 2015 References 1. Silverman, H: Products of starlike and convex functions. Ann. Univ. Mariae Curie-Skłodowska, Sect. A 29, 109-116 1975) 2. Dimkov, G: On roducts of starlike functions. I. Ann. Pol. Math. 55, 75-79 1991) 3. Aleksandrov, IA: On the star-shaed character of the maings of a domain by functions regular and univalent in the circle. Iv.Vysš.Učebn. Zaved., Mat. 411),9-15 1959) 4. Stankiewic, J, Świtoniak, B: Generalied roblems of convexity and starlikeness. In: Comlex Analysis and Alications 85 Varna, 1985),. 670-675. Publ. House Bulgar. Acad. Sci., Sofia 1986) 5. Świtoniak, B: On a starlikeness roblem in the class of functions S α. Folia Sci. Univ. Tech. Resov. 14, 17-27 1984) 6. Dimkov, G, Diok, J: Generalied roblem of starlikeness for roducts of -valent starlike functions. Serdica Math. J. 24, 339-344 1998) 7. Diok, J: Generalied roblem of starlikeness for roducts of close-to-star functions. Ann. Pol. Math. 107, 109-121 2013) 8. Nunokawa, M, Causey, WM: On certain analytic functions bounded argument. Sci. Re. Fac. Educ., Gunma Univ. 34, 1-3 1985) 9. Ratti, JS: The radius of univalence of certain analytic functions. Math. Z. 107, 241-248 1968) 10. MacGregor, TH: The radius of univalence of certain analytic functions. Proc. Am. Math. Soc. 14, 514-520 1963) 10.1186/1029-242X-2015-5 Cite this article as: Arif et al.: On roducts of multivalent close-to-star functions. Journal of Inequalities and Alications 2015, 2015:5