A Study of Fourth-Order Hankel Determinants for Starlike Functions Connected with the Sine Function

<jats:p>In this paper, upper bounds for the fourth-order Hankel determinant <jats:inline-formula>
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                        <mfenced open="(" close=")">
                           <mrow>
                              <mn>1</mn>
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                  </jats:inline-formula> for the function class <jats:inline-formula>
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                        <msubsup>
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                              <mi mathvariant="script">S</mi>
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                           </mrow>
                           <mrow>
                              <mo>∗</mo>
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                  </jats:inline-formula> associated with the sine function are given.</jats:p>


Introduction
Let A denote the class of functions f which are analytic in the open unit disk D = fz : |z|<1g of the form f z ð Þ = z + a 2 z 2 + a 3 z 3 +⋯ z ∈ D ð Þ, ð1Þ and let S denote the subclass of A consisting of univalent functions. Suppose that P is the class of analytic functions p normalized by and satisfying the condition Assume that f and g are two analytic functions in D. Then, we say that the function g is subordinate to the function f , and we write if there exists a Schwarz function ωðzÞ with ωð0Þ = 0 and |ω ðzÞ | <1, such that (see [1]) In 2018, Cho et al. [2] introduced the following function class S * s : which implies that the quantity ðzf ′ ðzÞÞ/ð f ðzÞÞ lies in an eight-shaped region in the right-half plane.
In recent years, many papers have been devoted to finding upper bounds for the second-order Hankel determinant H 2 ð2Þ and the third-order Hankel determinant H 3 ð1Þ, whose elements are various classes of analytic functions; it is worth mentioning that [7][8][9][10][11][12][13][14][15][16][17][18][19][20]. For instance, Murugusundaramoorthy and Bulboacă [21] defined a new subclass of analytic functions ML a c ðλ, ϕÞ and got upper bounds for the Fekete-Szegö functional and the Hankel determinant of order two for f ∈ ML a c ðλ, ϕÞ: Islam et al. [22] examined the q-analog of starlike functions connected with a trigonometric sine function and discussed some interesting geometric properties, such as the well-known problems of Fekete-Szegö, the necessary and sufficient condition, the growth and distortion bound, closure theorem, and convolution results with partial sums for this class. Zaprawa et al. [23] obtained the bound of the third Hankel determinant for the univalent starlike functions. Very recently, Arif et al. [24]

Main Results
By proving our desired results, we need the following lemmas.

Theorem 4.
If the function f ðzÞ ∈ S * s and of the form ((1)), then Proof. Since f ðzÞ ∈ S * s , according to subordination relationship, thus there exists a Schwarz function ωðzÞ with ωð0Þ = 0 and |ωðzÞ | <1, satisfying 2 Journal of Function Spaces Here, It is easy to see that pðzÞ ∈ P and On the other hand, Comparing the coefficients of z, z 2 , z 3 , z 4 , z 5 , z 6 between equations (15) and (18), we obtain 144 , Applying Lemma 2, we easily get Let c 1 = c, c ∈ ½0, 2; by using Lemma 3, we show also, let obviously, we find Setting F′ðcÞ = 0, we have c = 2 ffiffi ffi 3 p /3, and so, FðcÞ has a maximum value attained at c = 2 ffiffi ffi 3 p /3, also which is Let c 1 = c, c ∈ ½0, 2, according to Lemma 3, we obtain Putting 3 Journal of Function Spaces we get Therefore, the function FðcÞ has a maximum value attained at c = 0, also which is we obtain Thus, c = 0 is the root of the function F ′ ðcÞ = 0 and F ′ ′ð 0Þ < 0; we are easy to see that the function FðcÞ has a maximum value attained at c = 0, also which is so we get Thus, the function FðcÞ has a maximum value attained at c = 2, also which is Hence, the proof is complete.

Theorem 5.
If the function f ðzÞ ∈ S * s and of the form ((1)), then we have Proof. Applying equation (21), we have Then, by applying Lemma 1, we get Suppose that jxj = t, t ∈ ½0, 1, c 1 = c, c ∈ ½0, 2: Then, using the triangle inequality, we obtain Suppose then for any t ∈ ð0, 1Þ and c ∈ ð0, 2Þ, we get 4 Journal of Function Spaces which means that Fðc, tÞ is an increasing function on the closed interval [0,1] about t. Therefore, the function Fðc, tÞ can get the maximum value at t = 1, that is, So, obviously, Hence, the proof is complete.

Theorem 6.
If the function f ðzÞ ∈ S * s and of the form ( (1)), then we have Proof. From (21), we have Now, in view of Lemma 1, we get Let jxj = t, t ∈ ½0, 1, c 1 = c, c ∈ ½0, 2: Then, using the triangle inequality, we deduce that Assume that Therefore, for any t ∈ ð0, 1Þ and c ∈ ð0, 2Þ, we have that is, Fðc, tÞ is an decreasing function on the closed interval [0,1] about t. This implies that the maximum value of Fðc, tÞ occurs at t = 0, which is Define we clearly see that the function GðcÞ has a maximum value attained at c = 0, also which is Hence, the proof is complete.

Theorem 7.
If the function f ðzÞ ∈ S * s and of the form ( (1)), then we have Proof. Let f ðzÞ ∈ S * s , then by equation (21), we get Now, in terms of Lemma 1, we obtain Let jxj = t, t ∈ ½0, 1, c 1 = c, c ∈ ½0, 2: Then, using the triangle inequality, we get Setting then, for any t ∈ ð0, 1Þ and c ∈ ð0, 2Þ, we have Putting then we have If G ′ ðcÞ = 0, then the root is c = 0: Also, since G ′ ′ð0Þ = −1/12 < 0, so the function GðcÞ can take the maximum value at c = 0, which is Hence, the proof is complete.

Theorem 8.
If the function f ðzÞ ∈ S * s and of the form ( (1)), then we have Proof. Let f ðzÞ ∈ S * s , then by using (21), we have Let c 1 = c, c ∈ ½0, 2, according to Lemma 3, we obtain Taking Then, ∀c ∈ ð0, 2Þ, we have which implies that FðcÞ increases on the closed interval [0,2] about c. Namely, the maximum value of FðcÞ attains at c = 2, also which is The proof of Theorem 8 is completed.

Theorem 9.
If the function f ðzÞ ∈ S * s and of the form ( (1)), then we have |a 5 − a 2 a 4 | ≤ 13 32 : Proof. Assume that f ðzÞ ∈ S * s , then from (21), we obtain Let F ′ ðcÞ = 0, we get c = 0 or c = ffiffi ffi 2 p and F ′ ð ffiffi ffi 2 p Þ < 0, which implies that the maximum value of FðcÞ attains at c = ffiffi ffi 2 p , also which is Hence, the proof is complete.

Theorem 10.
If the function f ðzÞ ∈ S * s and of the form ((1)), then we have