Sharp Estimate of Fifth Coefficient for Ma Minda Starlike and Convex Functions

Lemma 1. [14, Lemma I] If the functions $1 + \sum_{n=1}^{\infty} b_n z^n$ and $1 + \sum_{n=1}^{\infty} c_n z^n$ are in $\mathcal{P}$ , then the same holds for the function $$1 + \frac{1}{2} \sum_{n=1}^{\infty} b_n c_n z^n.$$

Lemma 2. [14, Lemma II] Let $h(z) = 1 + u_1 z + u_2 z^2 + \cdots$ and $1 + G(z) = 1 + d_1 z + d_2 z^2 + \cdots$ be functions in $\mathcal{P}$ , and set $$\gamma_n = \frac{1}{2^n} \left[ 1 + \frac{1}{2} \sum_{k=1}^n \binom{n}{k} u_k \right], \quad \gamma_0 = 1.$$ If $A_n$ is defined by $$\sum_{n=1}^{\infty} (-1)^{n+1} \gamma_{n-1} G^n(z) = \sum_{n=1}^{\infty} A_n z^n,$$ then $|A_n| \leq 2$ . It is worth recalling the Möbius function $\Psi_{\zeta}$ which maps the unit disk onto the unit disk and given by $$\Psi_{\zeta}(z) = \frac{z - \zeta}{1 - \overline{\zeta}z}, \quad \zeta \in \mathbb{D}. \tag{5}$$

Lemma 3. [4, Lemma 2.4] If $p \in \mathcal{P}$ , then for some $\zeta_i \in \overline{\mathbb{D}}, i \in \{1, 2, 3\}$ , $$p_1 = 2\zeta_1, \tag{6}$$ $$p_2 = 2\zeta_1^2 + 2(1 - |\zeta_1|^2)\zeta_2,\tag{7}$$ $$p_3 = 2\zeta_1^3 + 4(1 - |\zeta_1|^2)\zeta_1\zeta_2 - 2(1 - |\zeta_1|^2)\overline{\zeta_1}\zeta_2^2 + 2(1 - |\zeta_1|^2)(1 - |\zeta_2|^2)\zeta_3.$$ (8) For $\zeta_1, \zeta_2 \in \mathbb{D}$ and $\zeta_3 \in \mathbb{T}$ , there is a unique function $p = L \circ \omega \in \mathcal{P}$ with $p_1, p_2$ and $p_3$ as in (6)-(8), where $$\omega(z) = z\Psi_{-\zeta_1}(z\Psi_{-\zeta_2}(\zeta_3 z)), \quad z \in \mathbb{D}, \tag{9}$$ that is $$p(z) = \frac{1 + (\overline{\zeta_2}\zeta_3 + \overline{\zeta_1}\zeta_2 + \zeta_1)z + (\overline{\zeta_1}\zeta_3 + \zeta_1\overline{\zeta_2}\zeta_3 + \zeta_2)z^2 + \zeta_3z^3}{1 + (\overline{\zeta_2}\zeta_3 + \overline{\zeta_1}\zeta_2 - \zeta_1)z + (\overline{\zeta_1}\zeta_3 - \zeta_1\overline{\zeta_2}\zeta_3 - \zeta_2)z^2 - \zeta_3z^3}, \quad z \in \mathbb{D}.$$ Conversely, if $\zeta_1, \zeta_2 \in \mathbb{D}$ and $\zeta_3 \in \overline{\mathbb{D}}$ are given, then we can construct a (unique) function $p \in \mathcal{P}$ of the form (3) so that $p_i, i \in \{1, 2, 3\}$ , satisfy the identities in (6)-(8). For this, we define $$\omega(z) = \omega_{\zeta_1, \zeta_2, \zeta_3}(z) = z\Psi_{-\zeta_1}(z\Psi_{-\zeta_2}(\zeta_3 z)), \quad z \in \mathbb{D}, \tag{10}$$ where $\Psi_{\zeta}$ is the function given as in (5). Then $\omega \in \mathcal{B}_0$ . Moreover, if we define $p(z) = (1+\omega(z))/(1-\omega(z))$ , $z \in \mathbb{D}$ , then p is represented by (3), where $p_1$ , $p_2$ and $p_3$ satisfy the identities in (6)-(8) (see the proof of [4, Lemma 2.4]).

Lemma 4. If −1 < σ < 1, then F(z) = (1 + 2σz + z 2 )/(1 − z 2 ) belongs to P. Proof. Let us consider $$\omega(z) = \frac{F(z) - 1}{F(z) + 1} = \frac{z(z + \sigma)}{1 + \sigma z}.$$ From [\(5\)](#page-2-2), we have $$\omega(z) = z\Psi_{-\sigma}(z), \quad z \in \mathbb{D}.$$ Since Ψ⁻σ(z) is a conformal automorphism of ^D, which gives ^|w(z)^| ^&lt; 1 and ^w(0) = 0. Therefore w is a Schwarz function and F ∈ P. To prove our next result, we require the following assumption: Assumption. Let ϕ(z) be as defined in [\(2\)](#page-0-0), whose coefficients satisfy the following conditions: $$\begin{split} \mathbf{C1} : & |B_1^2 + 2B_2| < 4B_1^2, \\ \mathbf{C2} : |B_1^3 - B_1^2B_2 + 18B_2^2 - 18B_1B_3| < 3|(B_1^2 + 2B_1 + 2B_2)(2B_1^2 - 3B_1 + 3B_2)|, \\ \mathbf{C3} : |30B_1^7 - 9B_1^8 - B_1^6(66B_2 - 5) - 648B_2^3 + 324B_2^4 + B_1^5(170B_2 - 126) \\ & - 648B_2B_3^2 + B_1^3(-180B_2 + 220B_2^2 + 108B_3 - 360B_2B_3) + B_1(1296B_2B_3 \\ & - 720B_2^2B_3) + 648B_2^2B_4 + B_1^4(108 + 10B_2 - 175B_2^2 + 90B_3 + 162B_4) \\ & + B_1^2(-144B_2^2 + 4B_2^3 + 180B_2B_3 - 324B_3^2 - 648B_4 + 648B_2B_4)| < 8|9B_1^6 \\ & + 9B_1^7 + B_1^4(-27 + 32B_2) + B_1^5(-52 + 63B_2) + 162B_2^2B_3 + B_1^3(81 - 189B_2 + 164B_2^2 + 9B_3) + B_1^2(18B_2^2 - 9B_2B_3) + B_1(-162B_2^2 + 198B_3^3 - 81B_3^2)|, \\ \mathbf{C4} : & 0 < (4B_1^2 + 6(B_2 - B_1))/((3B_1^2 + 6(B_2 - B_1)) < 1. \end{split}$$

Theorem 5. Let ϕ(z) be as defined in [\(2\)](#page-0-0), whose coefficients satisfy the above conditions C1 to C4. If ^f ∈ S^∗ (ϕ), then $$|a_5| \le \frac{B_1}{4}.$$ The inequality is sharp. Proof. Suppose ^f ∈ S^∗ (ϕ), then $$\frac{zf'(z)}{f(z)} = \varphi(\omega(z)),$$ where ω ∈ B0. If we choose ω(z) = (p(z) − 1)/(p(z) + 1), where p ∈ P given by [\(3\)](#page-1-0), then by comparing the coefficients obtained by series expansion of f(z) together with p(z) and ϕ(z) yields $$a_5 = \frac{B_1}{8}I, (11)$$ where $$I = p_4 + I_1 p_1^4 + I_2 p_1^2 p_2 + I_3 p_1 p_3 + I_4 p_2^2, (12)$$ with $$I_{1} = \frac{1}{48B_{1}} \left( B_{1}^{4} - 6B_{1}^{3} + 11B_{1}^{2} + 6B_{1}^{2}B_{2} - 6B_{1} + 3B_{2}^{2} - 22B_{1}B_{2} + 18B_{2} - 18B_{3} + 8B_{1}B_{3} + 6B_{4} \right), \tag{13}$$ $$I_2 = \frac{3B_1^3 - 11B_1^2 + 9B_1 - 18B_2 + 11B_1B_2 + 9B_3}{12B_1}, \quad I_3 = \frac{2B_1^2 - 3B_1 + 3B_2}{3B_1}$$ (14) and $$I_4 = \frac{B_1^2 - 2B_1 + 2B_2}{4B_1}. (15)$$ Let $q(z) = 1 + b_1 z + b_2 z^2 + \cdots$ be in $\mathcal{P}$ , then by Lemma 1, we have $$1 + \frac{1}{2}(p(z) - 1) * (q(z) - 1) = 1 + \frac{1}{2} \sum_{n=1}^{\infty} b_n p_n z^n \in \mathcal{P}.$$ If we assume $h(z) = 1 + \sum_{n=1}^{\infty} u_n z^n \in \mathcal{P}$ and take $1 + G(z) := 1 + \frac{1}{2} \sum_{n=1}^{\infty} b_n p_n z^n$ , then Lemma 2 gives $$|A_4| \le 2$$ , where $$A_4 = \frac{1}{2}\gamma_0 b_4 p_4 - \frac{1}{4}\gamma_1 b_2^2 p_2^2 - \frac{1}{2}\gamma_1 b_1 b_3 p_1 p_3 + \frac{3}{8}\gamma_2 b_1^2 b_2 p_1^2 p_2 - \frac{1}{16}\gamma_3 b_1^4 p_1^4$$ (16) and for $i \in \{0, 1, 2, 3\}$ , $\gamma_i$ 's are given by $\gamma_0 = 1$ , $$\gamma_1 = \frac{1}{2} \left( 1 + \frac{1}{2} u_1 \right), \quad \gamma_2 = \frac{1}{4} \left( 1 + u_1 + \frac{1}{2} u_2 \right), \quad \gamma_3 = \frac{1}{8} \left( 1 + \frac{3}{2} u_1 + \frac{3}{2} u_2 + \frac{1}{2} u_3 \right). \tag{17}$$ So from (12) and (16), we can observe that if there exist $q, h \in \mathcal{P}$ such that $$b_4 = 2, \quad I_1 = -\frac{1}{128} \left( 1 + \frac{3}{2}u_1 + \frac{3}{2}u_2 + \frac{1}{2}u_3 \right) b_1^4, \quad I_2 = \frac{3}{32} \left( 1 + u_1 + \frac{u_2}{2} \right) b_1^2 b_2,$$ $$I_3 = -\frac{1}{4} (1 + \frac{u_1}{2}) b_1 b_3 \quad \text{and} \quad I_4 = -\frac{1}{8} \left( 1 + \frac{u_1}{2} \right) b_2^2,$$ then we have $$I = A_4. (18)$$ The bound for $|A_4|$ can be obtained from Lemma 2, consequently, we can estimate the bound for |I| and thus we arrive at the desired bound by using (11). To prove the theorem, we construct the functions q and h in such a way that we obtain (18). From Lemma 3, suppose that the functions q and h are constructed by taking $\zeta_1, \zeta_2 \in \mathbb{D}, \zeta_3 \in \overline{\mathbb{D}}$ and $\xi_1, \xi_2 \in \mathbb{D}, \xi_3 \in \overline{\mathbb{D}}$ respectively, as follows: $$q = L \circ \omega_1 \quad \text{and} \quad h = L \circ \omega_2,$$ (19) where $$\omega_1(z) = z\Psi_{-\zeta_1}(z\Psi_{-\zeta_2}(\zeta_3 z)), \quad \omega_2(z) = z\Psi_{-\xi_1}(z\Psi_{-\xi_2}(\xi_3 z))$$ (20) and L(z) is given by (4). So again from Lemma 3, the $b_i$ 's and $u_i$ 's, $i \in \{1,2,3\}$ are given by $$b_1 = 2\zeta_1, \quad b_2 = 2\zeta_1^2 + 2(1 - |\zeta_1|^2)\zeta_2,$$ $$b_3 = 2\zeta_1^3 + 4(1 - |\zeta_1|^2)\zeta_1\zeta_2 - 2(1 - |\zeta_1|^2)\overline{\zeta_1}\zeta_2^2 + 2(1 - |\zeta_1|^2)(1 - |\zeta_2|^2)\zeta_3$$ and $$u_1 = 2\xi_1, \quad u_2 = 2\xi_1^2 + 2(1 - |\xi_1|^2)\xi_2,$$ $$u_3 = 2\xi_1^3 + 4(1 - |\xi_1|^2)\xi_1\xi_2 - 2(1 - |\xi_1|^2)\overline{\xi_1}\xi_2^2 + 2(1 - |\xi_1|^2)(1 - |\xi_2|^2)\xi_3.$$ There may be many solutions for the above set of equations. For our purpose, we impose some restrictions on the parameters. We take all $\xi_i \in \mathbb{R}$ , then $$u_1 = 2\xi_1, \quad u_2 = 2\xi_1^2 + 2(1 - \xi_1^2)\xi_2,$$ $$u_3 = 2\xi_1^3 + 4(1 - \xi_1^2)\xi_1\xi_2 - 2(1 - \xi_1^2)\xi_1\xi_2^2 + 2(1 - \xi_1^2)(1 - \xi_2^2)\xi_3.$$ (21) Further, if we define $$\xi_{1} = -\frac{B_{1}^{2} + 2B_{2}}{2B_{1}}, \quad \xi_{2} = \frac{B_{1}^{3} - B_{1}^{2}B_{2} + 18B_{2}^{2} - 18B_{1}B_{3}}{3(B_{1}^{2} + 2B_{1} + 2B_{2})(2B_{1}^{2} - 3B_{1} + 3B_{2})},$$ $$\xi_{3} = \left(-9B_{1}^{8} + 30B_{1}^{7} - B_{1}^{6}(66B_{2} - 5) + 2B_{1}^{5}(85B_{2} - 63) + 4B_{1}^{3}(5B_{2}(11B_{2} - 18B_{3} - 9) + 27B_{3}) + 4B_{1}^{2}(B_{2}^{3} - 36B_{2}^{2} - 81B_{3}^{2} + 45B_{2}B_{3} + 162(B_{2} - 1)B_{4}) + 144B_{1}(5B_{2} - 9)B_{2}B_{3} + 324B_{2}(-2B_{3}^{2} + B_{2}((B_{2} - 2)B_{2} + 2B_{4})) + 18B_{1}^{4}(9B_{4} + 5B_{3} + 6) - 5B_{1}^{4}B_{2}(35B_{2} - 2)\right) / \left(8(3B_{1}^{4} + 2B_{1}^{3} + 18B_{2}^{2} + B_{1}^{2}(10B_{2} - 9) - 9B_{1}B_{3})(B_{1}(3B_{1}^{2} + B_{1} + 11B_{2} - 9) + 9B_{3})\right),$$ then the conditions C1, C2 and C3 on the coefficients $B_1, B_2, B_3$ and $B_4$ yield $$|\xi_1| < 1$$ , $|\xi_2| < 1$ , $|\xi_3| < 1$ respectively. Using these $\xi_i$ 's in (21), we can obtain $u_i$ 's, which in turn by using (17) gives $$\gamma_{1} = \frac{1}{4} \left( 2 - B_{1} - \frac{2B_{2}}{B_{1}} \right), \gamma_{2} = \frac{\left( B_{1}^{2} + 2B_{2} - 2B_{1} \right) \left( 3B_{1}^{3} - 11B_{1}^{2} + B_{1} \left( 11B_{2} + 9 \right) + 9 \left( B_{3} - 2B_{2} \right) \right)}{24B_{1} \left( 2B_{1}^{2} + 3B_{2} - 3B_{1} \right)}, \gamma_{3} = -\frac{1}{64B_{1} \left( 2B_{1}^{2} + 3B_{2} - 3B_{1} \right)^{2}} \left( 3 \left( B_{1}^{2} + 2B_{2} - 2B_{1} \right)^{2} \left( B_{1}^{4} - 6B_{1}^{3} + B_{1}^{2} \left( 6B_{2} + 11 \right) + B_{1} \left( 8B_{3} - 22B_{2} - 6 \right) + 3 \left( B_{2}^{2} + 6B_{2} - 6B_{3} + 2B_{4} \right) \right).$$ (22) Let us consider $$q(z) = \frac{1 + 2\sigma z + z^2}{1 - z^2},$$ with $\sigma = \sqrt{(4B_1^2 + 6(B_2 - B_1)/((3B_1^2 + 6(B_2 - B_1)))}$ , then $$b_1 = b_3 = 2\sigma$$ and $b_2 = b_4 = 2$ . (23) If we choose $B_1$ and $B_2$ such that $0 < \sigma < 1$ , which is equivalent to condition C4. Then by Lemma 4, we have $q \in \mathcal{P}$ . On putting the values of $b_i's$ and $\gamma_i's$ obtained from (22) and (23) respectively in (16), we get (18), which together with (11) gives the desired bound for $|a_5|$ . Let the function $H: \mathbb{D} \longrightarrow \mathbb{C}$ be given by $$H(z) = z \exp \int_0^z \frac{\varphi(t^4) - 1}{t} dt = z + \frac{B_1}{4} z^5 + \frac{1}{32} (B_1^2 + 4B_2) z^9 + \cdots,$$ where coefficients of $\varphi(z)$ satisfy the conditions C1 to C4. Then H(0) = 0, H'(0) = 1 and $zH'(z)/H(z) = \varphi(z^4)$ and hence the function $H \in \mathcal{S}^*(\varphi)$ , proving the result to be sharp for the function H. Choose $\varphi(z) = 1 + \sin z$ , whose coefficients satisfy C1 to C4, to obtain the following result:

Corollary 5.1. If $f \in \mathcal{S}_{\sin}^*$ , where $\varphi(z) = 1 + \sin z$ , then $$|a_5| \le \frac{1}{4}.$$ The bound is sharp. For all below mentioned choices of $\varphi$ conditions C1 to C4 are valid. Therefore, the bounds of $|a_5|$ for some of the known classes are obtained from our result as a special case. Remark. 1. If $\varphi(z) = 2/(1 + e^{-z})$ , then $f \in \mathcal{S}_{SG}^*$ and $|a_5| \le 1/8$ [7, Theorem 4.1]. - 2. If $\varphi(z) = \sqrt{1+z}$ , then $f \in \mathcal{S}_L^*$ and $|a_5| \le 1/8$ [15, Theorem 3.1]. - 3. If $\varphi(z) = \sqrt{1+bz}$ , then $f \in \mathcal{S}_{a_1}^*$ , and $|a_5| \leq b/8$ , where $b \in (0,1]$ [5, Theorem 3.1]. - 4. If $f \in \mathcal{S}_{RL}^*$ , then $|a_5| \le (5 3\sqrt{2})/8$ [15, Theorem 3.1]. - 5. If $\varphi(z) = ((1+z)/(1-z))^{\delta}$ , then the conditions of Theorem 5 are satisfied only for $0 < \delta \le \delta_0 \approx 0.350162$ . Therefore, $|a_5| \le \delta/2$ for $f \in S^*(\varphi)$ where $0 < \delta \le \delta_0$ [3].

Theorem 6. Let $\varphi(z)$ be as defined in (2), whose coefficients satisfy the conditions C1 to C4. If $f \in \mathcal{C}(\varphi)$ , then $$|a_5| \le \frac{B_1}{20}.$$ The inequality is sharp. Proof. Since f ∈ C(ϕ), therefore $$1 + \frac{zf''(z)}{f'(z)} = \varphi((p(z) - 1)/(p(z) + 1)), \tag{24}$$ where p ∈ P is given by [\(3\)](#page-1-0). By comparison of the coefficients of z, z 2 , z 3 in [\(24\)](#page-7-0) with the series expansion of f, ϕ and p, we get $$a_5 = \frac{B_1}{40}I, (25)$$ where $$I = p_4 + I_1 p_1^4 + I_2 p_1^2 p_2 + I_3 p_1 p_3 + I_4 p_2^2,$$ with I1, I2, I³ and I⁴ given as in [\(13\)](#page-4-3), [\(14\)](#page-4-4) and [\(15\)](#page-4-5). Using the same method as in Theorem [5,](#page-3-3) we obtain $$|I| \leq 2,$$ when B1, B2, B³ and B⁴ satisfy all the conditions C1, C2, C3 and C4. Thus bound of |a⁵| follows from [\(25\)](#page-7-1). Let H(z) = z + a2z ² + a3z ³ ⁺ · · · ∈ S be given by $$1 + \frac{zH''(z)}{H'(z)} = \varphi(z^4),$$ where coefficients of ϕ(z) satisfy the conditions C1 to C4. Clearly, H ∈ C(ϕ) and for the function H, we have a² = a³ = a⁴ = 0 and a⁵ = B1/20. Thus bound is sharp for H.