Bounds on Hankel determinant for starlike and convex functions with respect to symmetric points

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1 PURE MATHEMATICS RESEARCH ARTICLE Bounds on Hankel determinant for starlike convex functions with respect to symmetric points Ambuj K. Mishra 1, Jugal K. Prajapat * Sudhana Maharana Received: 07 November 015 Accepted: 4 February 016 First Published: 09 March 016 *Corresponding author: Jugal K. Prajapat, Department of Mathematics, Central University of Rajasthan, NH-8, Barsindri, Kishangarh , Ajmer, Rajasthan, India jkprajapat@gmail.com Reviewing editor: Hari M. Srivastava, University of Victoria, Canada Additional information is available at the end of the article Abstract: In the present paper, we investigate upper bounds on the third Hankel determinants for the starlike convex functions with respect to symmetric points in the open unit disk. Subjects: Advanced Mathematics; Analysis - Mathematics; Complex Variables; Mathematics & Statistics; Science Keywords: analytic univalent functions; Hankel determinant; Toeplitz determinant; starlike convex functions with respect to symmetric points 1991 Mathematics subject classifications: 30C45; 30C50 1. Introduction 1.1. Hankel determinant Let denote the family of analytic functions in the open unit disk D ={z C:z < 1} of the form f (z) =z + a n z n, n= z D. (1) Jugal Kishore Prajapat ABOUT THE AUTHORS Ambuj K. Mishra is working as an assistant professor in the Department of Mathematics, GLA University, Mathura, Uttar Pradesh, India. He received his MSc degree from University Of Allahabad, Uttar Pradesh, in 00 currently pursuing his PhD in the field of Geometric function theory at GLA University, Mathura, Uttar Pradesh. His research interest includes Geometric function theory, Special functions. Jugal K. Prajapat is working as an associate professor in the Department of Mathematics, Central University of Rajasthan, Rajasthan, India. He received his PhD degree from University of Rajasthan. He has published 50 research articles in reputed international journals. His research interest includes Geometric function theory, Fractional calculus, Special functions. Sudhana Maharana is a research scholar in the Department of Mathematics, Central University of Rajasthan, Rajasthan, India. He received his MSc degree from Berhampur University, Odisha, in 01. His research interest includes the Geometric function theory Special functions. PUBLIC INTEREST STATEMENT The interplay between geometry analysis of function of complex variables is the most attracting part of complex analysis. From the beginning of the twentieth century, the work on the coefficients of the Taylor series expansion of analytic univalent function is of great importance in complex analysis. The bounds on Hankel determinants Fekete Szegö inequalities of coefficients of Taylor s series expansion of analytic univalent functions have been studied by many peoples. In this paper, we have studied bounds on third Hankel determinants for the functions which are starlike convex with respect to symmetric points. 016 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license. Page 1 of 9

2 A function f is said to be univalent in a domain D, if it is one-to-one in D. Let denote the subclass of consisting of functions which are univalent in D. The Hankel determinant H q,n (f ) of Taylor s coefficients of function f of the form (1), is defined by H q,n (f )= a n a n+1 a n+q 1 a n+1 a n+ a n+q a n+q 1 a n+q a n+(q 1) (a 1 = 1; n, q N ={1,, }). () The Hankel determinent is useful in showing that a function of bounded characteristic in D, i.e. a function which is a ratio of two bounded analytic functions with its Laurent series around the origin having integral coefficients, is rational (see Cantor, 1963). Pommerenke (1967) proved that the Hankel determinant of univalent functions satisfy H q,n (f ) < Kn ( 1 +β)q+ 3, where β > K depends only on q. Later, Hayman (1968) proved that H,n (f ) < An 1 (A is an absolute constant) for a really mean univalent functions. The study of H q,n (f ) for various subfamilies of are of interest for many researchers (see Ehrenborg, 000; Noonan & Thomas, 1976; Noor, 199; Pommerenke, 1966). Note that, the H,1 (f )= a is the classical Fekete Szegö functional. Fekete Szegö (1933) found the maximum value of H,1 (f ) over the function f. The problem of calculating max f H,1 (f ) for various compact subfamilies of, was considered by many authors (see Bhowmik, Ponnusamy, & Wirths, 011; Keogh & Merkes, 1969; Koepf, 1987; Mishra & Gochhayat, 008a, 010, 011; Srivastava & Mishra, 000; Srivastava, Mishra, & Das, 001). Further, for the second Hankel determinant H, (f ), the max f H, (f ) has been studied by many researchers (see Janteng, Halim, & Darus, 006; Lee, Ravichran, & Supramaniam, 013; Mishra & Gochhayat, 008b; Mishra & Kund, 013; Patel & Sahoo, 014) upper bound on the third Hankel determinant H 3,1 (f ) studied recently by Babalola (010), Bansal, Maharana, Prajapat (015), Prajapat, Bansal, Singh, Mishra (015), Raza Malik (013), Vamshee Krishna, Venkateswarlu, RamReddy (015). 1.. Toeplitz determinant Let denote the class of analytic functions p in D with R(p(z)) > 0 p(0) =1. If p is of the form p(z) =1 + c n z n, n=1 z D, (3) then c n, n N: ={1,, }. This inequality is sharp the equality holds for the function φ(z) =(1 + z) (1 z) (see Duren, 1983). The power series (3) converges in D to a function in, if only if the Toeplitz determinants T n (p) = c n c 1 c n 1 c c 1 c n c n c n+1 c n+, n N are positive, where c n = c n. The only exception is when f(z) has the form Page of 9

3 f (z) = m ν=1 where ρ ν > 0, ε ν = 1, ε k ε l if k l; k, l = 1,,, m; we have then T n (p) > 0 for n (m 1) T n (p) =0 for n m (see Grener & Szegö, 1984). Recently, in an article (Janteng et al., 006) the Toeplitz determinant found to be useful to estimate upper bound on the coefficients functional for various subfamilies of analytic functions. Note that for n = T (p) = is equivalent to ρ ν 1 + ε ν z 1 ε ν z, m 1, = 8 + R{ c } 4 0, = 1 + x(4 c 1 ) (4) for some x with x 1. Similarly, if T 3 (p) = c 3 c 3, then T 3 (p) 0 is equivalent to (4c c 3 1 )(4 c 1 )+ ( c 1 ) (4 1 ) ( 1 ). (5) Solving (5) with the help of (4), we get 4c 3 = c x(4 c 1 ) x (4 1 )+(4 c 1 )(1 x )z, (6) for some x z with x 1 z 1. Conditions (4) (6) are due to Libera Zlotkiewicz (198, 1983) Starlike convex functions with respect to symmetric points A function f is called starlike, if f is univalent in D f (D) is a starlike domain with respect to the origin. Analytically, f is called starlike, denoted by f, if R(zf (z) f (z)) > 0, z D. A function f is called convex, denoted by f, if only if zf (z). A function f is said to be starlike with respect to symmetric points (see Sakaguchi, 1959) if for every r less than sufficiently close to one every η on the circle z = r, the angular velocity of f(z) about the point f ( η) is positive at z = η as z traverses the circle z = r in the positive direction, i.e. ( zf ) (z) R > 0 for z = η, η = r. (7) f (z) f( η) We denote by s, the class of all functions in which are starlike with respect to symmetric points. A function f in the class s is characterized by ( zf ) (z) R > 0, z D. (8) f (z) f( z) Page 3 of 9

4 This can be easily seen that the function (f (z) f ( z)) is a starlike function in D, therefore functions satisfying (8) are close-to-convex ( hence univalent) in D. Further, the class of all functions in, which are convex with respect to symmetric points is denoted by s. The necessary sufficient condition for the function f to be univalent convex with respect symmetric points in D is characterized by (see Das & Singh, 1977, Theorem 1) ( (zf (z)) ) R > 0, z D. (f (z) f( z)) (9) In the present paper, we aim to investigate the upper bounds on the third Hankel determinant H 3,1 (f ) for the functions belonging to the classes s s defined above in (8) (10). For this purpose, we shall use Equations (4 6) the following known results. Lemma 1.1 (Sakaguchi, 1959) If f of the form (1), then a s n 1, n. Equality holds for the function f (z) =z(1 + εz) 1, ε = 1. Lemma 1. (Das & Singh, 1977) If f s of the form (1), then a n 1 n, n. Equality holds for the function f (z) =(1 ε) log(1 + εz), ε = 1. Lemma 1.3 (Shanmugam, Ramachran, & Ravichran, 006, Example.3) If f s of the form (1), then a 1. This inequality is sharp the equality is attained for the function f (z) =z(1 + εz ) 1, ε = 1. Lemma 1.4 (Shanmugam et al., 006, Example.5) If f s of the form (1), then a 1 3. This inequality is sharp.. Main results Theorem.1 Let the function f given by (1) be in the class s. Then a 1 a a 3 1. (10) The inequalities in (10) are sharp. Proof Let f s, then by (8), we have zf (z) f (z) f( z) = p(z), where p is of the form (3). Substituting the series expansion of f(z) p(z) equating the coefficients, we get a = 1, = 1 = 1 8 (c 3 + ). (11) Hence a = 1 8 c 3 a a = c c cc 1 4c. (1) Using (4) (6) in (1) for some x z such that x 1 z 1, we get Page 4 of 9

5 a = 1 16 c x(4 1 c)+ 1 x (4 ) (4 1 c)(1 1 x )z a a = cx(4 1 c 1 ) x (4 1 ) 1 x (4 )+c (4 1 1 c)(1 1 x )z. As, therefore, letting = c, we may assume without restriction that c [0,. Thus applying the triangle inequality with μ = x, we obtain a 1 [ cμ(4 )+cμ (4 )+(4 )(1 μ ) 16 : = F(c, μ) a a 1 [ 3 μ(4 )+8μ (4 ) μ (4 )+c(4 )(1 μ ) 3 : = G(c, μ). Now we need to find the maximum value of F G over the region Ω = { (c, μ):0 c,0 μ 1 }. For this, first differentiating F with respect to μ c, we get F μ = 1 16 F c = 1 16 [ (4 )(c + cμ 4μ) [ 4μ + 4μ 3 μ 3 μ 4c(1 μ ). A critical point of F(c, μ) must satisfy F F F = 0 = 0. The condition = 0 gives c = or μ c μ μ = c. Points (c, μ) satisfying such conditions are not interior points of Ω. So the function F(c, μ) c 4 cannot have a maximum in the interior of Ω. Since Ω is closed bounded F is continuous on Ω, the maximum shall be attained on the boundary of Ω. It is easy to see that on the boundary line c = 0, 0 μ 1, we have F(0, μ) =(1 μ ) its maximum on this line is equal to 1/. On the boundary line c =, 0 μ 1, we have F(, μ) =0. Similarly, on the boundary line μ = 0, 0 c, we have F(c,0)=(4 ) 8 the maximum on this line is 1/. Lastly, on the boundary line μ = 1, 0 c, we have F(c,1)=c(4 ) 8 the maximum on this line is 7. Comparing the four maxima we get that the maximum value of F(c, μ) on Ω is 1/. To show the sharpness of first inequality in (10), by setting = x = 0, z = 1 in (4) (6), we get = 0 c 3 =. Using these values in (1), we find that the first inequality in (10) is sharp. (13) Further, to find the maximum value of G over Ω, differentiating G with respect to μ, we get G μ = 1 [ (4 ){ + μ(8 c)} > 0 if 0 < c < 0 <μ<1. 3 Note that, G is a non-decreasing function of μ on [0, 1, hence max 0 μ 1 G(c, μ) =G(c,1)= 1 4 (4 c )= (c). Further, it is clear that (c) is a decreasing function on [0,, hence it attain maximum value at c = 0. Therefore the maximum of G(c, μ) is at the point (0, 1). Further, Ω is closed bounded G is continuous on Ω, the maximum shall be attained on the boundary of Ω. Hence, we look on the boundary of Ω as we have done with the function F, it is easy to see that on the boundary line c = 0, 0 μ 1, we have G(0, μ) =μ its maximum on this line is equal to 1. On the boundary line c =, 0 μ 1, we have G(, μ) =0. Similarly, on the boundary line μ = 0, 0 c, we have G(c,0)=c(4 ) 16 the maximum on this line is 1 7. Lastly, on the boundary line μ = 1, 0 c, we have G(c,1)=(4 ) 4 the maximum on this line is 1. Comparing the four maxima we get that the maximum value of Page 5 of 9

6 G(c, μ) on Ω is 1. To show the sharpness in the second inequality of (10), by setting = 0, x = 1 in (4) (6), we get = c 3 = 0. Using these values in (1), we find that the second inequality in (10) is sharp. Theorem. Let the function f given by (1) be in the class s. Then H 3,1 (f ) 5. Proof Using Lemma 1.1, Lemma 1.3, Theorem.1 applying the triangle inequality, we get H 3,1 (f ) a a + a a a a + a a a = 5. Theorem.3 Let the function f given by (1) be in the class s. Then a 4 7 a a (14) The second inequality in (14) is sharp. Proof Let f s, then by (9), we have (zf (z)) = p(z), (f (z) f( z)) where p is of the form (3). From the definitions of the class s s, it follows that the function f (z) s if only if zf (z). Thus replacing a by na s n n in (11), we get a = 1 4, = 1 6 = 1 3 (c 3 + ). (15) Hence a = c 3 a a = c c cc 1 3c. (16) Using (4) (6) in (16) for some x z such that x 1 z 1, we get a = 1 19 c3 +(4 1 c){ 5c x x 6(1 x )z} a a = c 4 1 +(4 c 1 ) { 5 1 x 16x (4 1 ) 9c 1 x + 18 (1 x )z}. As, therefore, letting = c, we may assume without restriction that c [0,. Thus applying the triangle inequality with μ = x, we obtain a 1 [ c 3 +(4 ){5cμ + 3cμ + 6(1 μ )} 19 : = X(c, μ) a a 1 [ 3 c 4 +(4 ){5 μ + 64μ 7 μ + 18c(1 μ )} 304 : = Y(c, μ). Page 6 of 9

7 Now to find the maximum value of X Y over the region Ω. First differentiating X with respect to μ c, we get X μ = 1 [ (4 )(5c + 6cμ 1μ) 19 X c = 1 [ 6 +(4 )(5μ + 3μ ) c(5cμ + 3cμ + 6 6μ ). 19 A critical point of X(c, μ) must satisfy X X X = 0 = 0. The condition = 0 gives c = or μ c μ μ = 5c. Points (c, μ) satisfying such conditions are not interior points of Ω. So the function X(c, μ) 6c 1 cannot have a maximum in the interior of Ω. Since Ω is closed bounded X is continuous the maximum shall be attained on the boundary of Ω. It is easy to see that on the boundary line c = 0, 0 μ 1, we have X(0, μ) =(1 μ ) 8 its maximum on this line is equal to 1/8. On the boundary line c =, 0 μ 1, we have X(, μ) =1 1. Similarly, on the boundary line μ = 0, 0 c, we have X(c,0)=(c ) 96 the maximum on this line is 1/8. Lastly, on the boundary line μ = 1, 0 c, we have X(c,1)=c(16 3 ) 96 the maximum on this line is 4/7. Comparing the four maxima we get that the maximum value of X(c, μ) on Ω is 4/7. Further, to find the maximum value of Y over Ω, differentiating Y with respect to μ, we get Y μ = 1 [ (4 ){5 + μ( c)} > 0 if 0 < c < 0 <μ< Note that, Y is a non-decreasing function of μ on [0, 1, hence 1 max Y(c, μ) =Y(c,1)= 0 μ (13c ) = (c). It is clear that (c) is a decreasing function on [0, it attained maximum value at c = 0. Therefore, the maximum of Y(c, μ) is at the point (0, 1). Further, Ω is closed bounded Y is continuous, the maximum shall be attained on the boundary of Ω. Hence, we look on the boundary of Ω, it is easy to see that on the line c = 0, 0 μ 1, we have Y(0, μ) =μ 9 its maximum on this line is equal to 1/9. On the boundary line c =, 0 μ 1, we have Y(, μ) =1 7. Similarly, on the boundary line μ = 0, 0 c, we have Y(c,0)=c(c ) 115 the maximum on this line is less than 1/9. Lastly, on the boundary line μ = 1, 0 c, we have Y(c,1)=(c ) 115 the maximum on this line is 1/9. Comparing the four maxima we get that the maximum value of Y(c, μ) on Ω is 1/9. To show the sharpness in the second inequality of (14), by setting = 0, x = 1 in (4) (6), we get = c 3 = 0. Using these values in (16), we find that the second inequality in (14) is sharp. This completes the proof of the theorem. Theorem.4 Let the function f given by (1) be in the class s. Then H 3,1 (f ) Proof Using Lemma 1., Lemma 1.4, Theorem.3 applying the triangle inequality, we get H 3,1 (f ) a a + a a a a + a a a = Page 7 of 9

8 Acknowledgements The authors express their sincere thanks to the editor referees for their valuable suggestions to improve the manuscript. Funding The authors received no direct funding for this research. Author details Ambuj K. Mishra 1 ambuj_math@rediffmail.com ORCID ID: Jugal K. Prajapat jkprajapat@gmail.com Sudhana Maharana snmmath@gmail.com 1 Department of Mathematics, Institute of Applied Sciences Humanities, GLA University, Mathura, Uttar Pradesh, India. Department of Mathematics, Central University of Rajasthan, NH-8, Barsindri, Kishangarh, Ajmer, Rajasthan, India. Citation information Cite this article as: Bounds on Hankel determinant for starlike convex functions with respect to symmetric points, Ambuj K. Mishra, Jugal K. Prajapat & Sudhana Maharana, Cogent Mathematics (016), 3: References Babalola, K. O. (010). On third order Hankel determinant for some classes of univalent functions. Inequality Theory Applications, 6, 1 7. Bansal, D., Maharana, S., & Prajapat, J. K. (015). Third order Hankel determinant for certain univalent functions. Journal of the Korean Mathematical Society, 5, Bhowmik, B., Ponnusamy, S., & Wirths, K.-J. (011). On the Fekete Szegö problem for the concave univalent functions. Journal of Mathematical Analysis Applications, 373, Cantor, D. G. (1963). Power series with the integral coefficients. Bulletin of the American Mathematical Society, 69, Das, R. N., & Singh, P. (1977). On subclasses of schlicht mapping. Indian Journal of Pure Applied Mathematics, 8, Duren, P. L. (1983). Univalent functions. Grundlehren der Mathematischen Wissenschaften ( B 59). New York, NY: Springer-Verlag. Ehrenborg, R. (000). The Hankel determinant of exponential polynomials. American Mathematical Monthly, 107, Fekete, M., & Szegö, G. (1933). Eine Benberkung uber ungerada Schlichte funktionen. Journal of the London Mathematical Society, 8, Grener, U., & Szegö, G. (1984). Toeplitz forms their applications (nd ed.). New York, NY: Chelsea. Hayman, W. K. (1968). On second Hankel determinant of mean univalent functions. Proceedings of the London Mathematical Society, 18, Janteng, A., Halim, S., & Darus, M. (006). Coefficient inequality for a function whose derivative has a positive real part. Journal of Inequalities in Pure Applied Mathematics, 7(), 1 5. Keogh, F. R., & Merkes, E. P. (1969). A coefficient Inequality for certain classes of analytic functions. Proceedings of the American Mathematical Society, 0, 8 1. Koepf, W. (1987). On the Fekete Szegö problem for closeto-convex functions. Proceedings of the American Mathematical Society, 101, Lee, S. K., Ravichran, V., & Supramaniam, S. (013). Bounds for the second Hankel determinant of certain univalent functions. Journal of Inequalities Applications, 013, Article 81. Libera, R. J., & Zlotkiewicz, E. J. (198). Early coefficients of the inverse of a regular convex function. Proceedings of the American Mathematical Society, 85, Libera, R. J., & Zlotkiewicz, E. J. (1983). Coefficient bounds for the inverse of a function with derivatives in ρ. Proceedings of the American Mathematical Society, 87, Mishra, A. K., & Gochhayat, P. (008a). The Fekete Szegö problem for k-uniformly convex functions for a class defined by the Owa Srivastava operator. Journal of Mathematical Analysis Applications, 347, Mishra, A. K., & Gochhayat, P. (008b). Second Hankel determinant for a class of analytic functions defined by fractinal derivatives. International Journal of Mathematics Mathematical Sciences, 008, 10, Article ID Mishra, A. K., & Gochhayat, P. (010). Fekete Szegö problem for a class defined by an integral operator. Kodai Mathematical Journal, 33, Mishra, A. K., & Gochhayat, P. (011). A coefficient inequality for a subclass of Cathéodory functions defined using a conical domain. Computers & Mathematics with Applications, 61, Mishra, A. K., & Kund, S. N. (013). Second Hankel determinant for a class of analytic functions defined by Carlson Shaffer operator. Tamkang Journal of Mathematics, 44, Noonan, J. W., & Thomas, D. K. (1976). On the second Hankel determinant of areally mean p-valent functions. Transactions of the American Mathematical Society, 3, Noor, K. I. (199). Higer order close-to-convex functions. Mathematica Japonica, 37(1), 1 8. Patel, J., & Sahoo, A. K. (014). On certain subclasses of analytic functions associated with the Carlson--Shaffer operator. Annales Universitatis Mariae Curie-Sklodowska Lublin-Polonia, LXVIII, Pommerenke, C. (1966). On the coefficients Hankel determinant of univalent functions. Journal of the London Mathematical Society, 41, Pommerenke, C. (1967). On the Hankel determinant of univalent functions. Mathematika, 14, Prajapat, J. K., Bansal, D., Singh, A., & Mishra, A. K. (015). Bounds on third Hankel determinant for close-to-convex functions. Acta Universitatis Sapientiae, Mathematica, 7, Raza, M., & Malik, S. N. (013). Upper bound of the third Hankel determinant for a class of analytic functions related with Lemniscate of Bernoulli. Journal of Inequalities Applications, 013, Article 41. Sakaguchi, K. (1959). On a certain univalent mapping. Journal of the Mathematical Society of Japan, 11, Shanmugam, T. N., Ramachran, C., & Ravichran, V. (006). Fekete Szegö problem for subclasses of starlike functions with respect to symmetric points. Bulletin of the Korean Mathematical Society, 43, Srivastava, H. M., & Mishra, A. K. (000). Applications of fractional calculus to parabolic starlike uniformly convex functions. Computers & Mathematics with Applications, 39, Srivastava, H. M., Mishra, A. K., & Das, M. K. (001). Fekete Szegö problem for a subclass of close-to-convex functions. Complex Variables, Theory Application, 44, Vamshee Krishna, D., Venkateswarlu, B., & RamReddy, T. (015). Third Hankel determinant for certain subclass of p-valent functions. Complex Variables Elliptic Equations, 60, Page 8 of 9

9 016 The Author(s). This open access article is distributed under a Creative Commons Attribution (CC-BY) 4.0 license. You are free to: Share copy redistribute the material in any medium or format Adapt remix, transform, build upon the material for any purpose, even commercially. The licensor cannot revoke these freedoms as long as you follow the license terms. Under the following terms: Attribution You must give appropriate credit, provide a link to the license, indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. No additional restrictions You may not apply legal terms or technological measures that legally restrict others from doing anything the license permits. Cogent Mathematics (ISSN: ) is published by Cogent OA, part of Taylor & Francis Group. Publishing with Cogent OA ensures: Immediate, universal access to your article on publication High visibility discoverability via the Cogent OA website as well as Taylor & Francis Online Download citation statistics for your article Rapid online publication Input from, dialog with, expert editors editorial boards Retention of full copyright of your article Guaranteed legacy preservation of your article Discounts waivers for authors in developing regions Submit your manuscript to a Cogent OA journal at Page 9 of 9