A geometric investigation of a certain subclass of univalent functions

Lemma 2.1. [29, Theorem 4] Every $f \in \mathcal{M}(\lambda)$ has the representation $$\frac{z}{f(z)} = 1 - \frac{f''(0)}{2}z + \lambda \int_0^1 \frac{\omega(tz)}{t^2} \log(1/t)dt,$$ for some $\omega : \mathbb{D} \to \mathbb{D}$ with $\omega(0) = \omega'(0) = 0$ .

Lemma 2.2. [18, Area Theorem, Theorem 11 of Vol. 2] Let $\mu > 0$ and $f \in \mathcal{S}$ be in the form $(z/f(z))^{\mu} = 1 + \sum_{n=1}^{\infty} b_n z^n$ . Then, we have $\sum_{n=1}^{\infty} (n-\mu)|b_n|^2 \leq \mu$ .

Lemma 2.3. [29, Theorem 2, p. 171] Let $\phi(z) = 1 + \sum_{n=1}^{\infty} b_n z^n$ be a non-vanishing analytic function in $\mathbb{D}$ that satisfying the coefficient bound $\sum_{n=2}^{\infty} (n-1)^2 |b_n| \leq \lambda$ . Then the function f defined by $f(z) = z/\phi(z)$ is in $\mathcal{M}(\lambda)$ .

Lemma 2.4. [29, Corollary 1, p. 172] Let $f \in \mathcal{M}$ be of the form $z/f(z) = 1 + \sum_{n=1}^{\infty} b_n z^n$ . Then, we have $\sum_{n=2}^{\infty} (n-1)^4 |b_n|^2 \le 1$ . It is evident that the class $\mathcal{M}$ is not preserved under dilation. This motivates the determination of the largest disk in which the class $\mathcal{M}$ is preserved under dilation.

Lemma 2.5. If $f \in \mathcal{M}$ and g(z) = f(rz)/r, then $g \in \mathcal{M}$ for $0 < r \le r_0$ , where $r_0 = \sqrt{(\sqrt{5}-1)/2} \approx 0.786151$ is the unique root of the equation $r^4 + r^2 - 1 = 0$ in (0,1).

Theorem 3.1. Let $f \in \mathcal{S}$ and $\lambda \in (0,1]$ . Then the function f(rz)/r belongs to $\mathcal{M}(\lambda)$ for $0 < r \le r_0$ , where $r_0 \in (0,1)$ is the smallest root the smallest root of the equation $$r^{4}(r^{4} + 4r^{2} + 1) - \lambda^{2} (1 - r^{2})^{4} = 0.$$

Corollary 3.1. Suppose $f \in \mathcal{S}$ . Then, the function f(rz)/r belongs to $\mathcal{M}$ for $0 < r \le r_0$ , where $r_0 \approx 0.557384$ is the unique root of the equation $8r^6 - 5r^4 + 4r^2 - 1 = 0$ in (0,1), as shown in Figure 2.

Theorem 4.1. Suppose that $g, h \in \mathcal{M}$ . Then, the quotient function F defined by (4.1) belongs to $\mathcal{M}$ in the disk $|z| \leq r_0$ , where $r_0 (\approx 0.294876)$ is the smallest root of the equation $9r^4 + 16r^3 + 6r^2 - 1 = 0$ in (0,1). The result is sharp.

Corollary 4.1. Suppose that $g, h \in \mathcal{M}_2$ . Then, the quotient function F(z) = g(z)h(z)/z for $z \in \mathbb{D}$ belongs to $\mathcal{M}_2$ in the disk $|z| \leq r_0$ , where $r_0 = \sqrt{\sqrt{10} - 1/3}$ is the unique root of the equation $9r^4 + 2r^2 - 1 = 0$ in (0,1), as illustrated in Figure 4.

Corollary 4.2. Suppose that $g \in \mathcal{M}_2(\lambda)$ and $h \in \mathcal{M}_2(\lambda')$ , where $\lambda, \lambda' \in (0,1]$ . Then, the quotient function F defined by (4.1) belongs to $\mathcal{M}_2(\mu)$ in the disk $|z| \leq r_1$ , where $$r_1 = \sqrt{\frac{-(\lambda + \lambda') + \sqrt{(\lambda + \lambda')^2 + 12\mu(\lambda^2 + \lambda\lambda' + \lambda'^2)}}{6(\lambda^2 + \lambda\lambda' + \lambda'^2)}}.$$

Theorem 4.2. Let $g, h \in \mathcal{S}$ . Then, the quotient function F defined by (4.1) belongs to the class $\mathcal{M}$ in the disk $|z| \leq r_0$ , where $r_0 (\approx 0.260985)$ is the smallest positive root of the equation $$6r^{2} + 4(\sqrt{2} + 4)r^{3} + \frac{2r^{4}\sqrt{-8r^{10} + 31r^{8} - 44r^{6} + 27r^{4}}}{(1 - r^{2})^{2}} + \frac{r^{4}(4r^{2} - 11r + 9)}{(1 - r)^{3}} + 4\left(\frac{r^{8}(15 - 20r^{2} + 15r^{4} - 4r^{6})}{(1 - r^{2})^{4}} - r^{4}\left(\log(1 - r^{2}) + r^{2}\right)\right)^{1/2} - 1 = 0$$ in (0,1).

Corollary 4.3. Let $g, h \in \mathcal{S}$ with g''(0) = 0. Then, the function F defined by (4.1) belongs to the class $\mathcal{M}$ in the disk $|z| \leq r_0$ , where $r_0 \approx 0.313967$ is the unique root of the equation $$B(r): = 2r^{2} + 4(\sqrt{2} + 2)r^{3} + \frac{2r^{2}\sqrt{-8r^{10} + 31r^{8} - 44r^{6} + 27r^{4}}}{(1 - r^{2})^{2}} + \frac{r^{4}(4r^{2} - 11r + 9)}{(1 - r)^{3}} + 2\left(\frac{r^{8}(15 - 20r^{2} + 15r^{4} - 4r^{6})}{(1 - r^{2})^{4}} - r^{4}\left(\log(1 - r^{2}) + r^{2}\right)\right)^{1/2} - 1 = 0$$ in (0,1), as illustrated in Figure 6.

Corollary 4.4. Let $g, h \in \mathcal{S}$ with g''(0) = 0 and h''(0) = 0. Then, the function F defined by (4.1) belongs to the class $\mathcal{M}$ in the disk $|z| \leq r_0$ , where $r_0 \approx 0.352049$ is the unique root of the equation $$C(r) := 2r^2 + 4\sqrt{2}r^3 + \frac{2r^2\sqrt{-8r^{10} + 31r^8 - 44r^6 + 27r^4}}{(1-r^2)^2} + \frac{r^4(4r^2 - 11r + 9)}{(1-r)^3} - 1 = 0$$ in (0,1), as illustrated in Figure 7.

Theorem 4.3. Let $G(z)=z^2/f(z)$ for $z\in\mathbb{D}$ , where $f\in\mathcal{A}$ . Then, we have the following: - (i) If $f \in \mathcal{S}$ , then $G \in \mathcal{M}$ in the disk $|z| \leq 2 \sqrt{3}$ . The result is sharp. - (ii) If $f \in \mathcal{R}(1/2)$ , then $G \in \mathcal{M}$ in the disk $|z| \leq r_0$ , where $r_0 \approx 0.396608$ ) is the unique positive root of the equation $1 3r + 2r^2 2r^3 \geq 0$ in (0,1). The result is sharp. - (iii) If $f \in \mathcal{C}(-1/2)$ , then $G \in \mathcal{M}$ in the disk $|z| \leq r_1$ , where $r_1 (\approx 0.304725)$ is the unique positive root of the equation $2r^4 6r^3 + 4r^2 4r + 1 = 0$ in (0,1). The number $r_1$ is the best possible. - (iv) If $f \in \mathcal{G}$ , then $G \in \mathcal{M}$ in the disk $|z| \leq r_2$ , where $r_2 \approx 0.75085$ is the positive root of the equation $3r 2r^2 + (r^2 5r + 4)\log(1 r) = 0$ in (0, 1). Proof. (i) Let $f \in \mathcal{S}$ be such that $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ for $z \in \mathbb{D}$ . Then, we have $|a_n| \le n$ for $n \ge 2$ (see [8,44]). Since $f(z)/z \ne 0$ in $\mathbb{D}$ , therefore $G(z)/z \ne 0$ . Now $$z^{2} \left(\frac{z}{G(z)}\right)'' + G'(z) \left(\frac{z}{G(z)}\right)^{2} - 1 = z^{3} \left(\frac{1}{G(z)} - \frac{1}{z}\right)'' + z^{2} \left(\frac{1}{G(z)} - \frac{1}{z}\right)'$$ $$= z^{3} \left(\frac{f(z)}{z^{2}} - \frac{1}{z}\right)'' + z^{2} \left(\frac{f(z)}{z^{2}} - \frac{1}{z}\right)'.$$ It is evident that $$z^{2} \left( \frac{f(z)}{z^{2}} - \frac{1}{z} \right)' = z^{2} \left( \frac{1}{z} + \sum_{n=2}^{\infty} a_{n} z^{n-2} - \frac{1}{z} \right)' = \sum_{n=2}^{\infty} (n-2) a_{n} z^{n-1} \quad \text{and}$$ $$z^{3} \left( \frac{f(z)}{z} - \frac{1}{z} \right)'' = z^{3} \left( \sum_{n=2}^{\infty} (n-2) a_{n} z^{n-3} \right)' = \sum_{n=2}^{\infty} (n-2) (n-3) a_{n} z^{n-1}.$$ Therefore, we have <span id="page-11-0"></span> $$\left| z^2 \left( \frac{z}{G(z)} \right)'' + G'(z) \left( \frac{z}{G(z)} \right)^2 - 1 \right| \le \sum_{n=2}^{\infty} (n-2)^2 |a_n| r^{n-1} \quad \text{for} \quad |z| = r.$$ (4.9) As $|a_n| \le n$ for $n \ge 2$ , therefore, we have $G \in \mathcal{M}$ if $r^2 \left(3 + 4r - r^2\right)/(1 - r)^4 \le 1$ , i.e., if $r \le 2 - \sqrt{3}$ . To prove the sharpness of the radius $2 - \sqrt{3}$ , we consider the function $f(z) = z/(1-z)^2$ . Then the quotient function $G(z) = z(1-z)^2$ and $$\left| z^2 \left( \frac{z}{G(z)} \right)'' + G'(z) \left( \frac{z}{G(z)} \right)^2 - 1 \right|_{z=r} = \frac{3r^2 + 4r^3 - r^4}{(1-r)^4} > 1 \quad \text{for} \quad r > 2 - \sqrt{3},$$ which shows that the number $2 - \sqrt{3}$ is the best possible. (ii) Using similar argument as in (i), we have (4.9). Since $f \in \mathcal{R}(1/2)$ , so $|a_n| \leq 1$ for $n \geq 2$ . Therefore, $G \in \mathcal{M}$ if $r^2(1+r)/(1-r)^3 \leq 1$ , i.e., if $r \leq r_0$ , where $r_0 \approx 0.396608$ ) is the unique positive root of the equation $1 - 3r + 2r^2 - 2r^3 \geq 0$ in (0,1). To prove the sharpness of the radius $r_0$ , we consider the function f(z) = z/(1-z). Therefore, G(z) = z(1-z) and $$\left| z^2 \left( \frac{z}{G(z)} \right)'' + G'(z) \left( \frac{z}{G(z)} \right)^2 - 1 \right|_{z=r} = \frac{r^2 (1+r)}{(1-r)^3} > 1 \quad \text{for} \quad r > r_0,$$ which shows that the number $r_0$ is the best possible. (iii) Since $f \in C(-1/2)$ , so $|a_n| \le (n+1)/2$ for $n \ge 2$ . In view of (4.9), we have $G \in \mathcal{M}$ if $r^2(2+2r-r^2)/(1-r)^4 \le 1$ , i.e., $r \le r_1$ , where $r_1 (\approx 0.304725)$ is the unique positive root of the equation $2r^4 - 6r^3 + 4r^2 - 4r + 1 = 0$ . To prove the sharpness of the radius $r_0$ , we consider the function $f(z) = z(1 - z/2)/(1-z)^2$ . Then, we have $G(z) = z(1-z)^2/(1-z/2)$ and $$\left| z^2 \left( \frac{z}{G(z)} \right)'' + G'(z) \left( \frac{z}{G(z)} \right)^2 - 1 \right|_{z=r} = \frac{2r^2 + 2r^3 - r^4}{(1-r)^4} > 1 \quad \text{for} \quad r > r_1,$$ which shows that the number $r_1$ is the best possible. (iv) Since $f \in \mathcal{G}$ , so $|a_n| \le 1/(n(n-1))$ for $n \ge 2$ (see [36]). In view of (4.9), we have $G \in \mathcal{M}$ if $(4-3r)/(1-r) + (4-5r+r^2)\log(1-r)/(r(1-r)) \le 1$ , i.e., $r \le r_2$ , where $r_2 (\approx 0.75085)$ is the positive root of the equation $3r - 2r^2 + (r^2 - 5r + 4)\log(1-r) = 0$ in (0,1), as illustrated in Figure 8.  <span id="page-12-1"></span>

Theorem 4.4. Let $H(z) = z^2 / \int_0^z (t/f(t)) dt$ for $z \in \mathbb{D}$ , where $f \in \mathcal{A}$ . Then, we have the following: (i) If $f \in \mathcal{S}$ , then $H \in \mathcal{M}$ in the disk $|z| \leq r_3$ , where $r_3 (\approx 0.7829)$ is the unique positive root of the equation $$\frac{-4+4r^2+r^4}{(1-r^2)^2} - \frac{12}{r^2}\log(1-r^2) - \frac{8}{r^2}Li_2(r^2) - 1 = 0 \quad in \quad (0,1).$$ (ii) If $f \in \mathcal{M}$ , then $H \in \mathcal{M}$ in the unit disk $\mathbb{D}$ . Proof. (i) Let $f \in \mathcal{S}$ . Then, f have the power series representation of the form $z/f(z) = 1 + \sum_{n=1}^{\infty} b_n z^n$ . In view of Lemma 2.2, we have $\sum_{n=2}^{\infty} (n-1)|b_n|^2 \le 1$ . Let $g(z) = \int_0^z (t/f(t))dt = z + \sum_{n=1}^{\infty} (b_n/(n+1))z^{n+1} = z + \sum_{n=2}^{\infty} (b_{n-1}/n)z^n$ . Thus, $$\frac{1}{H(z)} - \frac{1}{z} = \sum_{n=2}^{\infty} \frac{b_{n-1}}{n} z^{n-2} = \sum_{n=1}^{\infty} \frac{b_n}{n+1} z^{n-1}.$$ Therefore, <span id="page-13-0"></span> $$z^{2} \left(\frac{z}{H(z)}\right)'' + H'(z) \left(\frac{z}{H(z)}\right)^{2} - 1 = z^{3} \left(\frac{1}{H(z)} - \frac{1}{z}\right)'' + z^{2} \left(\frac{1}{H(z)} - \frac{1}{z}\right)'$$ $$= \sum_{n=2}^{\infty} \frac{(n-1)^{2} b_{n}}{n+1} z^{n}. \tag{4.10}$$ In view of the Cauchy-Schwarz inequality and for |z| = r, we have $$\left| z^{2} \left( \frac{z}{H(z)} \right)'' + H'(z) \left( \frac{z}{H(z)} \right)^{2} - 1 \right| \leq \sum_{n=2}^{\infty} \frac{(n-1)^{2} |b_{n}|}{n+1} r^{n} \\ \leq \left( \sum_{n=2}^{\infty} (n-1) |b_{n}|^{2} \right)^{1/2} \left( \sum_{n=2}^{\infty} \frac{(n-1)^{3}}{(n+1)^{2}} r^{2n} \right)^{1/2} \\ \leq \left( \sum_{n=2}^{\infty} \frac{(n-1)^{3}}{(n+1)^{2}} r^{2n} \right)^{1/2} .$$ It is evident that $$\sum_{n=2}^{\infty} \frac{(n-1)^3}{(n+1)^2} r^{2n} = \sum_{n=2}^{\infty} \left( -5 + n + \frac{12}{(1+n)} - \frac{8}{(1+n)^2} \right) r^{2n}$$ $$= -\frac{5r^4}{1-r^2} + \frac{r^4(2-r^2)}{(1-r^2)^2} - 6(r^2+2) - \frac{12}{r^2} \log(1-r^2)$$ $$+2(4+r^2) - \frac{8}{r^2} \text{Li}_2(r^2) = \frac{1}{r^2} A_1(r),$$ where $\text{Li}_2(r^2)$ is the dilogarithm function such that $$\operatorname{Li}_{2}(r^{2}) = \sum_{n=1}^{\infty} \frac{r^{2n}}{n^{2}} = -\int_{0}^{r^{2}} \frac{\log(1-u)}{u} du \text{ and}$$ $$A_{1}(r) = \frac{-4r^{2} + 4r^{4} + r^{6}}{(1-r^{2})^{2}} - 12\log(1-r^{2}) - 8\operatorname{Li}_{2}(r^{2}).$$ We claim that $A_1(r) \geq 0$ for $r \in [0,1]$ . Differentiating $A_1(r)$ with respect to r, we have $$A'_{1}(r) = \frac{2}{r(1-r^{2})^{3}} \left(8r^{2} - 20r^{4} + 15r^{6} - r^{8} + (8 - 24r^{2} + 24r^{4} - 8r^{6})\log(1-r^{2})\right)$$ $$= \frac{1}{4r(1-r^{2})^{6}} A_{2}(r),$$ where $$A_2(r) = (8r^2 - 20r^4 + 15r^6 - r^8)/(8 - 24r^2 + 24r^4 - 8r^6) + \log(1 - r^2)$$ . Therefore $$A_2'(r) = \frac{r^9 + 4r^7 + r^5}{4(1 - r^2)^4} \ge 0,$$ which shows that $A_2(r)$ is a monotonically increasing function of $r \in [0, 1]$ and it follows that $A_2(r) \geq A_2(0) = 0$ . Thus, $A'_1(r) \geq 0$ , i.e., $A_1(r)$ is a monotonically increasing function of $r \in [0, 1]$ , and hence, we have $A_1(r) \geq A_1(0) = 0$ for $r \in [0, 1]$ . Therefore, $$\left| z^2 \left( \frac{z}{H(z)} \right)'' + H'(z) \left( \frac{z}{H(z)} \right)^2 - 1 \right| \le 1 \quad \text{for} \quad r \le r_3,$$ where $r_3 \approx 0.7829$ is the unique positive root of the equation $$A_3(r) := \frac{-4 + 4r^2 + r^4}{(1 - r^2)^2} - \frac{12}{r^2}\log(1 - r^2) - \frac{8}{r^2}\operatorname{Li}_2(r^2) - 1 = 0$$ in (0,1) and it is shown in Figure 9. (ii) Let $f \in \mathcal{M}$ . Then, f can be written as $z/f(z) = 1 + \sum_{n=1}^{\infty} b_n z^n$ . In view of Lemma 2.4, we have $\sum_{n=2}^{\infty} (n-1)^4 |b_n|^2 \le 1$ . Using similar argument as in (i) with the Cauchy-Schwarz inequality and from (4.10), we have $$\left| z^{2} \left( \frac{z}{H(z)} \right)'' + H'(z) \left( \frac{z}{H(z)} \right)^{2} - 1 \right| \leq \sum_{n=2}^{\infty} \frac{(n-1)^{2} |b_{n}|}{n+1} r^{n}$$ $$\leq \left( \sum_{n=2}^{\infty} (n-1)^{4} |b_{n}|^{2} \right)^{1/2} \left( \sum_{n=2}^{\infty} \frac{r^{2n}}{(n+1)^{2}} \right)^{1/2}$$ $$\leq \left( \sum_{n=2}^{\infty} \frac{r^{2n}}{(n+1)^{2}} \right)^{1/2}.$$ It is evident that $$\sum_{n=2}^{\infty} \frac{r^{2n}}{(n+1)^2} \le \sum_{n=2}^{\infty} \frac{1}{(n+1)^2} = \frac{\pi^2}{6} - \frac{5}{4} < 1.$$ Therefore, $H \in \mathcal{M}$ in the unit disk $\mathbb{D}$ .

Theorem 4.5. If $f, g \in \mathcal{M}(\lambda)$ for some $0 < \lambda \leq 1$ , then $$\frac{fg}{(1-t)f+tg} \in \mathcal{M}(\lambda) \quad for \quad 0 \le t \le 1.$$

Lemma 6.1. [40] If $f \in \Omega$ , then $$|z| - \frac{1}{2}|z|^2 \le |f(z)| \le |z| + \frac{1}{2}|z|^2$$ and $1 - |z| \le |f'(z)| \le 1 + |z|$ . For each $z \in \mathbb{D}$ , $z \neq 0$ , equality occurs in both estimates if and only if $f(z) = z + \frac{1}{2}\lambda z^2$ , where $|\lambda| = 1$ . The following result provides a sufficient condition for functions to be members of the class $\Omega$ .

Lemma 6.2. Let $f \in \mathcal{A}$ be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ and $\sum_{n=2}^{\infty} (n-1)|a_n| \le 1/2$ . Then, $f \in \Omega$ .

Theorem 7.1. Let $f \in \Omega_A$ be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ for $z \in \mathbb{D}$ . Then, we have $$|f(z)| + \sum_{n=2}^{\infty} |a_n| r^n \le d\left(f(0), \partial f(\mathbb{D})\right) \quad \text{for} \quad r \le (\sqrt{3} - 1)/2,$$ The radius $(\sqrt{3}-1)/2$ is the best possible.

Theorem 7.2. Let $f \in \Omega$ be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ for $z \in \mathbb{D}$ . Then, we have $$r + \sum_{n=2}^{\infty} |a_n| r^n \le d\left(f(0), \partial f(\mathbb{D})\right) \quad \text{for} \quad r \le \sqrt{2} - 1.$$ The radius $\sqrt{2} - 1$ is the best possible.

Theorem 7.3. Let $f \in \Omega$ be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ for $z \in \mathbb{D}$ . Then, we have $$|f(z)| + |f'(z)||z| + \sum_{n=2}^{\infty} |a_n|r^n \le d(f(0), \partial f(\mathbb{D})) \quad \text{for} \quad r \le (\sqrt{2} - 1)/2,$$ The number $(\sqrt{2}-1)/2$ is the best possible.

Definition 1.1. The polylogarithm $Li_k(z)$ is a special function of order k and argument z defined in the complex plane over the unit disk $\mathbb{D}$ as $$Li_k(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^k} = z + \frac{z^2}{2^k} + \frac{z^3}{3^k} + \cdots$$ and it is also known as Jonquière's function. The special cases k=2 and k=3 are called the dilogarithm and trilogarithm respectively. The remaining sections are organized as follows: In section 2, we present a number of lemmas and a proof of one of these lemmas, which are essential for the proof of our main theorems. In section 3, we investigate a radius property of the class $\mathcal{M}(\lambda)$ and a number of associated results pertaining to $\mathcal{M}$ . In section 4, we examine the largest disks for which the functions defined by the quotient of analytic functions belongs to $\mathcal{M}$ . Sections 5 and 6 comprise the introductory sections of a certain subclass of starlike functions and the Bohr phenomenon regarding the class of bounded analytic functions, respectively. In section 7, we obtain the sharp Bohr radius, Bohr-Rogosinski radius and improved Bohr radius for a certain subclass of starlike functions.