Raza and Malik Journal of Inequalities and Applications 2013, 2013:412

Lemma 1.1. [16] Let $p \in P$ and of the form (1.2). Then $$\left| p_2 - \nu p_1^2 \right| \le \left\{ \begin{array}{ll} -4\nu + 2, & \nu < 0, \\ 2, & 0 \le \nu \le 1, \\ 4\nu - 2, & \nu > 1. \end{array} \right.$$ When v < 0 or v > 1, the equality holds if and only if p(z) is $\frac{1+z}{1-z}$ or one of its rotations. If 0 < v < 1, then the equality holds if and only if $p(z) = \frac{1+z^2}{1-z^2}$ or one of its rotations. If v = 0, the equality holds if and only if $p(z) = (\frac{1}{2} + \frac{\eta}{2})\frac{1+z}{1-z} + (\frac{1}{2} - \frac{\eta}{2})\frac{1-z}{1+z}$ ( $0 \le \eta \le 1$ ) or one of its rotations. If v = 1, the equality holds if and only if p is the reciprocal of one of the functions such that the equality holds in the case of v = 0. Although the above upper bound is sharp, when 0 < v < 1, it can improved as follows: $$|p_2 - \nu p_1^2| + \nu |p_1|^2 \le 2 \quad (0 < \nu \le 1/2)$$ and $$|p_2 - \nu p_1^2| + (1 - \nu)|p_1|^2 \le 2 \quad (1/2 < \nu \le 1).$$

Lemma 1.2. [16] If $p(z) = 1 + p_1 z + p_2 z^2 + \cdots$ is a function with positive real part in E, then for v a complex number <span id="page-2-2"></span> $$|p_2 - \nu p_1^2| \le 2 \max(1, |2\nu - 1|).$$ This result is sharp for the functions $$p(z) = \frac{1+z^2}{1-z^2}, \qquad p(z) = \frac{1+z}{1-z}.$$

Lemma 1.3. [17] Let $p \in P$ and of the form (1.2). Then $$2p_2 = p_1^2 + x(4 - p_1^2)$$ for some x, $|x| \le 1$ , and $$4p_3 = p_1^3 + 2(4 - p_1^2)p_1x - (4 - p_1^2)p_1x^2 + 2(4 - p_1^2)(1 - |x|^2)z$$ for some z, $|z| \leq 1$ .

Theorem 2.1. Let $f \in SL^*$ and of the form (1.1). Then $$\left|a_3 - \mu a_2^2\right| \le \begin{cases} \frac{1}{16}(1 - 4\mu), & \mu < -\frac{3}{4}, \\ \frac{1}{4}, & -\frac{3}{4} \le \mu \le \frac{5}{4}, \\ \frac{1}{16}(4\mu - 1), & \mu > \frac{5}{4}. \end{cases}$$ Furthermore, for $-\frac{3}{4} < \mu \leq \frac{1}{4}$ , $$|a_3 - \mu a_2^2| + \frac{1}{4} (4\mu + 3)|a_2|^2 \le \frac{1}{4},$$ and for $\frac{1}{4} < \mu \leq \frac{5}{4}$ , <span id="page-2-0"></span> $$|a_3 - \mu a_2^2| + \frac{1}{4}(5 - 4\mu)|a_2|^2 \le \frac{1}{4}.$$ These results are sharp. Proof If $f \in SL^*$ , then it follows from (1.3) that $$\frac{zf'(z)}{f(z)} \prec \phi(z),\tag{2.1}$$ <span id="page-3-2"></span><span id="page-3-1"></span><span id="page-3-0"></span> where $\phi(z) = \sqrt{1+z}$ . Define a function $$p(z) = \frac{1 + w(z)}{1 - w(z)} = 1 + p_1 z + p_2 z^2 + \cdots$$ It is clear that $p \in P$ . This implies that $$w(z) = \frac{p(z)-1}{p(z)+1}.$$ From (2.1), we have $$\frac{zf'(z)}{f(z)} = \phi(w(z)),$$ with $$\phi(w(z)) = \left(\frac{2p(z)}{p(z)+1}\right)^{\frac{1}{2}}.$$ Now $$\left(\frac{2p(z)}{p(z)+1}\right)^{\frac{1}{2}} = 1 + \frac{1}{4}p_1z + \left[\frac{1}{4}p_2 - \frac{5}{32}p_1^2\right]z^2 + \left[\frac{1}{4}p_3 - \frac{5}{16}p_1p_2 + \frac{13}{128}p_1^3\right]z^3 + \cdots$$ Similarly, $$\frac{zf'(z)}{f(z)} = 1 + a_2 z + \left[2a_3 - a_2^2\right]z^2 + \left[3a_4 - 3a_2a_3 + a_2^3\right]z^3 + \cdots$$ Therefore $$a_2 = \frac{1}{4}p_1,\tag{2.2}$$ $$a_3 = \frac{1}{8}p_2 - \frac{3}{64}p_1^2,\tag{2.3}$$ $$a_4 = \frac{1}{12}p_3 - \frac{7}{96}p_1p_2 + \frac{13}{768}p_1^2. \tag{2.4}$$ This implies that $$|a_3 - \mu a_2^2| = \frac{1}{8} |p_2 - \frac{1}{8} (4\mu + 3)p_1^2|.$$ Now, using Lemma 1.1, we have the required result. The results are sharp for the functions $K_i(z)$ , i = 1, 2, 3, 4, such that $$\frac{zK_1'(z)}{K_1(z)} = \sqrt{1+z} \quad \text{if } \mu < -\frac{3}{4} \text{ or } \mu > \frac{5}{4},$$ $$\frac{zK_2'(z)}{K_2(z)} = \sqrt{1+z^2} \quad \text{if } -\frac{3}{4} < \mu < \frac{5}{4},$$ $$\frac{zK_3'(z)}{K_3(z)} = \sqrt{1 + \Phi(z)}$$ if $\mu = -\frac{3}{4}$ , $$\frac{zK_4'(z)}{K_4(z)} = \sqrt{1 - \Phi(z)}$$ if $\mu = \frac{5}{4}$ , where $\Phi(z) = \frac{z(z+\eta)}{1+\eta z}$ with $0 \le \eta \le 1$ .

Theorem 2.2. Let $f \in SL^*$ and of the form (1.1). Then for a complex number $\mu$ , $$\left|a_3 - \mu a_2^2\right| \le \frac{1}{4} \max\left\{1; \left|\mu - \frac{1}{4}\right|\right\}.$$ Proof Since $$|a_3 - \mu a_2^2| = \frac{1}{8} |p_2 - \frac{1}{8} (4\mu + 3)p_1^2|,$$ therefore, using Lemma 1.2, we get the result. This result is sharp for the functions $$\frac{zf'(z)}{f(z)} = \sqrt{1+z}$$ <span id="page-4-0"></span>or $$\frac{zf'(z)}{f(z)} = \sqrt{1+z^2}.$$ <span id="page-4-1"></span>For $\mu = 1$ , we have $H_2(1)$ .

Corollary 2.3. Let $f \in SL$ and of the form (1.1). Then* $$\left|a_3-a_2^2\right|\leq \frac{1}{4}.$$

Theorem 2.4. Let $f \in SL^*$ and of the form (1.1). Then $$\left|a_2a_4-a_3^2\right|\leq \frac{1}{16}.$$ Proof From (2.2), (2.3) and (2.4), we obtain $$a_2 a_4 - a_3^2 = \frac{1}{48} \left( p_1 p_3 - \frac{7}{8} p_1^2 p_2 + \frac{13}{64} p_1^4 \right) - \left( \frac{1}{8} p_2 - \frac{3}{64} p_1^2 \right)^2$$ $$= \frac{1}{48} p_1 p_3 - \frac{1}{64} p_2^2 - \frac{5}{768} p_1^2 p_2 + \frac{25}{12,288} p_1^4$$ $$= \frac{1}{12,288} \left( 256 p_1 p_3 - 192 p_2^2 - 80 p_1^2 p_2 + 25 p_1^4 \right).$$ Putting the values of $p_2$ and $p_3$ from Lemma 1.3, we assume that p > 0, and taking $p_1 = p \in [0, 2]$ , we get $$|a_2a_4 - a_3^2| = \frac{1}{12,288} |64p_1\{p_1^3 + 2(4 - p_1^2)p_1x - (4 - p_1^2)p_1x^2 + 2(4 - p_1^2)(1 - |x|^2)z\}$$ $$-48\{p_1^2 + x(4 - p_1^2)\}^2 - 40p_1^2\{p_1^2 + x(4 - p_1^2)\} + 25p_1^4|.$$ After simple calculations, we get $$|a_2a_4 - a_3^2| = \frac{1}{12,288} |41p^4 - 8(4-p^2)p^2x - 128(4-p^2)(1-|x|^2)z$$ $$+ x^2(4-p^2)(64p^2 + 48)(4-p^2)|.$$ Now, applying the triangle inequality and replacing |x| by $\rho$ , we obtain $$\begin{aligned} \left| a_2 a_4 - a_3^2 \right| &\leq \frac{1}{12,288} \left[ 41 p^4 + 128 \left( 4 - p^2 \right) + 8 \left( 4 - p^2 \right) p^2 \rho + \rho^2 \left( 4 - p^2 \right) \left( 16 p^2 + 64 \right) \right] \\ &= F(p,\rho) \quad \text{(say)}. \end{aligned}$$ Differentiating with respect to $\rho$ , we have $$\frac{\partial F(p,\rho)}{\partial \rho} = \frac{1}{12,288} \Big[ 8 \big( 4 - p^2 \big) p^2 + 2 \rho \big( 4 - p^2 \big) \big( 16 p^2 + 64 \big) \Big].$$ It is clear that $\frac{\partial F(p,\rho)}{\partial \rho} > 0$ , which shows that $F(p,\rho)$ is an increasing function on the closed interval [0,1]. This implies that maximum occurs at $\rho=1$ . Therefore max $F(p,\rho)=F(p,1)=G(p)$ (say). Now $$G(p) = \frac{1}{12.288} \left[ 17p^4 - 96p^2 + 768 \right].$$ Therefore $$G'(p) = \frac{1}{12.288} [68p^3 - 192p]$$ and $$G''(p) = \frac{1}{12,288} [204p^2 - 192] < 0$$ for p = 0. This shows that maximum of G(p) occurs at p = 0. Hence, we obtain $$\left| a_2 a_4 - a_3^2 \right| \le \frac{768}{12,288}$$ $$= \frac{1}{16}.$$ <span id="page-5-0"></span>This result is sharp for the functions $$\frac{zf'(z)}{f(z)} = \sqrt{1+z}$$ or $$\frac{zf'(z)}{f(z)} = \sqrt{1+z^2}.$$

Theorem 2.5. Let $f \in SL^*$ and of the form (1.1). Then $$|a_2a_3 - a_4| \le \frac{1}{6}.$$ Proof Since $$a_2 = \frac{1}{4}p_1,$$ $$a_3 = \frac{1}{8}p_2 - \frac{3}{64}p_1^2,$$ $$a_4 = \frac{1}{12}p_3 - \frac{7}{96}p_1p_2 + \frac{13}{768}p_1^2.$$ Therefore, by using Lemma 1.3, we can obtain <span id="page-6-0"></span> $$|a_2a_3 - a_4| \le \frac{1}{768} \left\{ 7p^3 + 8p\rho \left( 4 - p^2 \right) + 32\left( 4 - p^2 \right) + 16\rho^2 (p - 2)\left( 4 - p^2 \right) \right\}.$$ Let $$F_1(p,\rho) = \frac{1}{768} \left\{ 7p^3 + 8p\rho \left(4 - p^2\right) + 32\left(4 - p^2\right) + 16\rho^2 (p - 2)\left(4 - p^2\right) \right\}. \tag{2.5}$$ We assume that the upper bound occurs at the interior point of the rectangle $[0,2] \times [0,1]$ . Differentiating (2.5) with respect to $\rho$ , we get $$\frac{\partial F_1}{\partial \rho} = \frac{1}{768} \left\{ 8p(4-p^2) + 32\rho(p-2)(4-p^2) \right\}.$$ For $0 < \rho < 1$ and fixed $p \in (0,2)$ , it can easily be seen that $\frac{\partial F_1}{\partial \rho} < 0$ . This shows that $F_1(p,\rho)$ is a decreasing function of $\rho$ , which contradicts our assumption; therefore, $\max F_1(p,\rho) = F_1(p,0) = G_1(p)$ . This implies that $$G_1'(p) = \frac{1}{768} \{21p^2 - 64p\}$$ <span id="page-6-1"></span>and $$G_1''(p) = \frac{1}{768} \{42p - 64\} < 0$$ for p = 0. Therefore p = 0 is a point of maximum. Hence, we get the required result. $\Box$

Lemma 2.6. If the function $f(z) = \sum_{n=1}^{\infty} a_n z^n$ belongs to the class $SL^*$ , then $$|a_2| < 1/2$$ , $|a_3| < 1/4$ , $|a_4| < 1/6$ , $|a_5| < 1/8$ . These estimations are sharp. The first three bounds were obtained by Sokół [3] and the bound for $|a_5|$ can be obtained in a similar way.

Theorem 2.7. Let $f \in SL^*$ and of the form (1.1). Then $$|H_3(1)| \leq \frac{43}{576}$$ Proof Since $$H_3(1) = a_3(a_2a_4 - a_3^2) - a_4(a_4 - a_2a_3) + a_5(a_1a_3 - a_2^2).$$ Now, using the triangle inequality, we obtain $$|H_3(1)| \le |a_3| |a_2a_4 - a_3^2| + |a_4| |a_2a_3 - a_4| + |a_5| |a_1a_3 - a_2^2|.$$ Using the fact that $a_1 = 1$ with the results of Corollary 2.3, Theorem 2.4, Theorem 2.5 and Lemma 2.6, we obtain <span id="page-7-1"></span> $$\left|H_3(1)\right| \le \frac{43}{576}.$$