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

In the present paper, we investigate upper bounds on the third Hankel determinants for the starlike and convex functions with respect to symmetric points in the open unit disk. Subjects: Advanced Mathematics; Analysis Mathematics; Complex Variables; Mathematics & Statistics; Science


Hankel determinant
Let  denote the family of analytic functions in the open unit disk = {z ∈ ℂ:|z| < 1} of the form (1) f (z) = z + ∞ ∑ n=2 a n z n , z ∈ .

PUBLIC INTEREST STATEMENT
The interplay between geometry and 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 and 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 and convex with respect to symmetric points.
Recently, in an article (Janteng et al., 2006) 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 = 2 is equivalent to for some x with |x| ≤ 1. Similarly, if then T 3 (p) ≥ 0 is equivalent to Solving (5) with the help of (4), we get for some x and z with |x| ≤ 1 and |z| ≤ 1. Conditions (4) and (6) are due to Zlotkiewicz (1982, 1983).

Starlike and convex functions with respect to symmetric points
A function f ∈  is said to be starlike with respect to symmetric points (see Sakaguchi, 1959) if for every r less than and sufficiently close to one and every on the circle |z| = r, the angular velocity of f(z) about the point f (− ) is positive at z = as z traverses the circle |z| = r in the positive direction, i.e.
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 This can be easily seen that the function (f (z) − f (−z))∕2 is a starlike function in , therefore functions satisfying (8) are close-to-convex (and hence univalent) in .
Further, the class of all functions in , which are convex with respect to symmetric points is denoted by  s . The necessary and sufficient condition for the function f ∈  to be univalent and convex with respect symmetric points in is characterized by (see Das & Singh, 1977, Theorem 1) 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 and  s defined above in (8) and (10). For this purpose, we shall use Equations (4 and 6) and the following known results.

. This inequality is sharp and the equality is attained for the function
. This inequality is sharp.

Main results
Theorem 2.1 Let the function f given by (1) be in the class  * s . Then The inequalities in (10) are sharp.
Proof Let f ∈  * s , then by (8), we have where p ∈  is of the form (3). Substituting the series expansion of f(z) and p(z) and equating the coefficients, we get Hence Using (4) and (6) in (12) for some x and z such that |x| ≤ 1 and |z| ≤ 1, we get As |c 1 | ≤ 2, therefore, letting c 1 = c, we may assume without restriction that c ∈ [0, 2]. Thus applying the triangle inequality with = |x|, we obtain and Now we need to find the maximum value of F and G over the region Ω = (c, ):0 ≤ c ≤ 2,0 ≤ ≤ 1 . we have F(c, 0) = (4 − c 2 )∕8 and the maximum on this line is 1/2. Lastly, on the boundary line = 1, 0 ≤ c ≤ 2, we have F(c, 1) = c(4 − c 2 )∕8 and the maximum on this line is 2∕ √ 27. Comparing the four maxima we get that the maximum value of F(c, ) on Ω is 1/2.
To show the sharpness of first inequality in (10), by setting c 1 = x = 0, z = 1 in (4) and (6), we get c 2 = 0 and c 3 = 2. Using these values in (12), we find that the first inequality in (10) is sharp.
Further, to find the maximum value of G over Ω, differentiating G with respect to , we get Note that, G is a non-decreasing function of on [0, 1], hence Further, it is clear that (c) is a decreasing function on [0, 2], hence it attain maximum value at c = 0. Therefore the maximum of G(c, ) is at the point (0, 1). Further, Ω is closed and bounded and 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 ≤ ≤ 1, we have G(0, ) = 2 and its maximum on this line is equal to 1. On the boundary line c = 2, 0 ≤ ≤ 1, we have G(2, ) = 0. Similarly, on the boundary line = 0, 0 ≤ c ≤ 2, we have G(c, 0) = c(4 − c 2 )∕16 and the maximum on this line is 1∕ √ 27. Lastly, on the boundary line = 1, 0 ≤ c ≤ 2, we have G(c, 1) = (4 − c 2 )∕4 and the maximum on this line is 1. Comparing the four maxima we get that the maximum value of G(c, ) on Ω is 1.
To show the sharpness in the second inequality of (10), by setting c 1 = 0, x = 1 in (4) and (6), we get c 2 = 2 and c 3 = 0. Using these values in (12), we find that the second inequality in (10)

.3 Let the function f given by (1) be in the class  s . Then
The second inequality in (14) is sharp.
Proof Let f ∈  s , then by (9), we have where p ∈  is of the form (3). From the definitions of the class  * s and  s , it follows that the function f (z) ∈  s if and only if zf � (z) ∈  * s . Thus replacing a n by na n in (11), we get Hence Using (4) and (6) in (16) for some x and z such that |x| ≤ 1 and |z| ≤ 1, we get As |c 1 | ≤ 2, therefore, letting c 1 = c, we may assume without restriction that c ∈ [0, 2]. Thus applying the triangle inequality with = |x|, we obtain and |H 3,1 (f )| ≤ 5 2 .
(16) Further, to find the maximum value of Y over Ω, differentiating Y with respect to , we get Note that, Y is a non-decreasing function of on [0, 1], hence It is clear that (c) is a decreasing function on [0, 2] and it attained maximum value at c = 0. Therefore, the maximum of Y(c, ) is at the point (0, 1). Further, Ω is closed and bounded and 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 ≤ ≤ 1, we have Y(0, ) = 2 ∕9 and its maximum on this line is equal to 1/9. On the boundary line c = 2, 0 ≤ ≤ 1, we have Y(2, ) = 1∕72. Similarly, on the boundary line = 0, 0 ≤ c ≤ 2, we have Y(c, 0) = c(c 3 − 9c 2 + 36)∕1152 and the maximum on this line is less than 1/9. Lastly, on the boundary line = 1, 0 ≤ c ≤ 2, we have Y(c, 1) = (2c 4 − 36c 2 + 128)∕1152 and the maximum on this line is 1/9. Comparing the four maxima we get that the maximum value of Y(c, ) on Ω is 1/9.