Successive Logarithmic Coefficients of Univalent Functions

Lemma 2.3. For all $r \in [-2, 2]$ and $\zeta \in \mathbb{T}$ , $\Phi_{r,\zeta} \in \mathcal{S}$ and $$\Phi_{r,\zeta}(\mathbb{D}) = \mathbb{C} \setminus \left( \left\{ \frac{\overline{\zeta}t}{2-r} : t \ge 1 \right\} \cup \left\{ -\frac{\overline{\zeta}t}{2+r} : t \ge 1 \right\} \right), \quad if |r| < 2, \quad (2.1)$$ as well as $$\Phi_{2,\zeta}(\mathbb{D}) = \mathbb{C} \setminus \left\{ -\overline{\zeta} \frac{t}{4} : t \ge 1 \right\} \quad and \quad \Phi_{-2,\zeta}(\mathbb{D}) = \mathbb{C} \setminus \left\{ \overline{\zeta} \frac{t}{4} : t \ge 1 \right\}, \quad (2.2)$$ where $$\mathbb{D}\ni z\mapsto \Phi_{r,\zeta}(z):=\frac{z}{1-r\zeta z+\zeta^2 z^2}.$$ (2.3) Proof Notice that for arbitrarily fixed $r \in [-2; 2]$ and $\zeta \in \mathbb{T}$ the function $\Phi_{r,\zeta}$ can be represented in the following manner $$\Phi_{r,\zeta}(z) = \overline{\zeta} \left( \zeta z + \frac{1}{\zeta z} - r \right)^{-1} = \overline{\zeta} \left( \Psi(\zeta z) - r \right)^{-1}, \quad z \in \mathbb{D} \setminus \{0\},$$ where $\Psi: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is the Joukowsky transform, i.e., $\Psi$ is a meromorphic function satisfying $\Psi(z) = z + 1/z$ for $z \in \mathbb{C} \setminus \{0\}$ . Thus the lemma follows from well known properties of $\Psi$ . 

Lemma 2.4. For every function $f: \mathbb{D} \to \mathbb{C}$ the following equivalence holds: $f \in \mathcal{S}$ and $|c_3(f) - c_2(f)^2| = 1$ if and only if $$f(z) = \frac{z}{1 - |c_2(f)|\zeta z + \zeta^2 z^2}, \quad z \in \mathbb{D},$$ (2.4) for a certain $\zeta \in \mathbb{T}$ . Proof Given $f \in \mathcal{S}$ suppose that $|c_3(f) - c_2(f)^2| = 1$ . Setting $\mathbb{D}^ := \{z \in \mathbb{C} : |z| > 1\} \cup \{\infty\}$ we see that $g \in \Sigma$ , where $g : \mathbb{D}^ \to \hat{\mathbb{C}}$ is the function satisfying $g(\infty) = \infty$ and $$g(z) = \frac{1}{f(1/z)} = z - c_2(f) + (c_2(f)^2 - c_3(f))\frac{1}{z} + \cdots, \quad z \in \mathbb{D}^*.$$ (2.5) Since $|c_3(f) - c_2(f)^2| = 1$ , we deduce from [23, Cor. 1.2 on p. 12] that there exist $\vartheta \in \mathbb{R}$ and $b_0 \in \mathbb{C}$ such that $$g(z) = z + b_0 + \frac{e^{2i\vartheta}}{z}, \quad z \in \mathbb{D}^*.$$ Comparing this with (2.5) we obtain $b_0 = -c_2(f)$ , which leads to the equality $$f(z) = \frac{z}{1 - c_2(f)z + \xi^2 z^2} = \frac{z}{P(z)}, \quad z \in \mathbb{D},$$ (2.6) where $\xi := e^{i\vartheta}$ and $$\mathbb{C}\ni z\mapsto P(z):=1-\eta\xi z+\xi^2 z^2 \tag{2.7}$$ with $\eta := c_2(f)\overline{\xi}$ . Hence there exist $z_1, z_2 \in \mathbb{C}$ such that $$P(z) = \xi^{2}(z - z_{1})(z - z_{2}), \quad z \in \mathbb{C}.$$ (2.8) Since $f \in S$ , $|z_1| \ge 1$ and $|z_2| \ge 1$ . On the other hand $|z_1||z_2| = |z_1z_2| = 1$ . Therefore $|z_1| = |z_2| = 1$ , and so $\overline{z_1}z_1 = 1$ . Hence by (2.7), $$0 = P(z_1) = 1 - \eta \xi z_1 + \xi^2 z_1^2 = (\xi z_1)^2 \left( (\overline{\xi} z_1)^2 - \eta (\overline{\xi} z_1) + 1 \right)$$ $$= (\xi z_1)^2 \left( \xi^2 (\overline{\xi^2 z_1})^2 - \eta \xi (\overline{\xi^2 z_1}) + 1 \right) = (\xi z_1)^2 P(\overline{\xi^2 z_1}),$$ which gives $z_1 = \overline{\xi^2 z_1}$ or $z_2 = \overline{\xi^2 z_1}$ . In the first case $\xi z_1 = \overline{\xi z_1}$ . Hence $\xi z_1 \in \mathbb{R}$ , and in view of (2.7) and (2.8), $\eta \in \mathbb{R}$ . In the second case $\xi z_2 = \overline{\xi z_1}$ . Hence by (2.7) and (2.8), $$\eta = \xi z_1 + \xi z_2 = \xi z_1 + \overline{\xi z_1} = 2\operatorname{Re}(\xi z_1) \in \mathbb{R}.$$  Thus in both the cases $\eta \in \mathbb{R}$ . Since $|\eta| = |c_2(f)|$ , we see that $\eta = |c_2(f)|$ or $\eta = -|c_2(f)|$ . Setting $\zeta := \xi$ in the first case and $\zeta := -\xi$ in the second case we derive from (2.6) and (2.7) the equality (2.4). This proves the lemma in the direction ( $\Rightarrow$ ). Conversely, suppose that the equality (2.4) holds for a certain $\zeta \in \mathbb{T}$ . Since $\infty \notin f(\mathbb{D})$ , we see that $r := |c_2(f)| \le 2$ . Then by the formula (2.3), $f = \Phi_{r,\zeta}$ , which in view of Lemma 2.3 leads to $f \in \mathcal{S}$ . The equality $|c_3(f) - c_2(f)^2| = 1$ follows directly from (2.4), which completes the

Theorem 3.1. For every $f \in \mathcal{S}$ , $$-\frac{\sqrt{2}}{2} \le |\gamma_2(f)| - |\gamma_1(f)| \le \frac{1}{2}.$$ (3.5) Moreover, both inequalities are sharp and for every function $f: \mathbb{D} \to \mathbb{C}$ the following two equivalences hold: (i) $f \in \mathcal{S}$ and the equality in the second inequality in (3.5) holds if and only if $$f(z) = \frac{z}{1 + \zeta^2 z^2}, \quad z \in \mathbb{D},\tag{3.6}$$ for a certain $\zeta \in \mathbb{T}$ ; (ii) $f \in S$ and the equality in the first inequality in (3.5) holds if and only if $$f(z) = \frac{z}{1 - \sqrt{2}\zeta_z + \zeta^2 z^2}, \quad z \in \mathbb{D},\tag{3.7}$$ for a certain $\zeta \in \mathbb{T}$ . Proof We will show first that the inequalities (3.5) hold in the class $S_0$ . Let $f \in S_0$ be arbitrarily fixed. Then f can be represented in the form (3.4), where F is given by (3.2). Now expanding each function $\mathbb{D} \ni z \mapsto f_t(z) := F(z, t)$ , $t \in [0; +\infty)$ , into a power series we obtain $$F(z,t) = e^{-t} \sum_{n=0}^{+\infty} c_n(f_t) z^n, \quad z \in \mathbb{D}, \ t \in [0; +\infty),$$ with $c_0(f_t) = 0$ and $c_1(f_t) = 1$ for $t \in [0; +\infty)$ . Using this representation we can rewrite Loewner's differential equation (3.3) as an infinite system of differential equations with unknown coefficients functions $[0; +\infty) \ni t \mapsto c_n(f_t)$ for $n \in \mathbb{Z}_2$ . This system can be recursively solved by integration with respect to t. In addition, from (3.4) and obvious equalities $c_n(f_0) = 0$ for $n \in \mathbb{Z}_2$ , it follows that $$c_n(f) = \lim_{t \to +\infty} [c_n(f_t) - c_n(f_0)] = \int_0^{+\infty} \frac{\mathrm{d}}{\mathrm{d}t} c_n(f_t) \mathrm{d}t, \quad n \in \mathbb{Z}_2,$$ which leads to $$c_{2}(f) = -2 \int_{0}^{+\infty} e^{-t} \kappa(t) dt,$$ $$c_{3}(f) = -2 \int_{0}^{+\infty} e^{-2t} (\kappa(t))^{2} dt + 4 \left( \int_{0}^{+\infty} e^{-t} \kappa(t) dt \right)^{2}$$  $$= -2 \int_0^{+\infty} e^{-2t} (\kappa(t))^2 dt + c_2(f)^2$$ (3.8) where $\kappa:[0;+\infty)\to\mathbb{T}$ is a certain continuous function; cf. [8, pp. 93–94]. Since $|c_2(f)|\leq 2$ , we conclude from (3.1) and (3.8) that $$2(|\gamma_{2}(f)| - |\gamma_{1}(f)|) = \left| c_{3}(f) - \frac{1}{2} c_{2}(f)^{2} \right| - |c_{2}(f)|$$ $$= \left| \left( -2 \int_{0}^{+\infty} e^{-2t} (\kappa(t))^{2} dt + c_{2}(f)^{2} \right) - \frac{1}{2} c_{2}(f)^{2} \right| - |c_{2}(f)|$$ $$\leq 2 \int_{0}^{+\infty} e^{-2t} |\kappa(t)|^{2} dt + \frac{1}{2} |c_{2}(f)|^{2} - |c_{2}(f)|$$ $$= 1 - \frac{1}{2} |c_{2}(f)|(2 - |c_{2}(f)|) \leq 1, \tag{3.9}$$ which proves the second inequality in (3.5) for every $f \in S_0$ . The proof of the first inequality in (3.5) is more sophisticated. There exists $\theta \in \mathbb{R}$ such that $$c_3(f) - \frac{1}{2}c_2(f)^2 = \left|c_3(f) - \frac{1}{2}c_2(f)^2\right|e^{2i\theta}.$$ Thus the function $$\mathbb{D} \ni z \mapsto \tilde{f}(z) := e^{i\theta} f(e^{-i\theta}z)$$ satisfies $$c_3(\tilde{f}) - \frac{1}{2}c_2(\tilde{f})^2 = \left(c_3(f) - \frac{1}{2}c_2(f)^2\right)e^{-2i\theta} = \left|c_3(f) - \frac{1}{2}c_2(f)^2\right|.$$ Setting $a_2 := c_2(\tilde{f})$ and $a_3 := c_3(\tilde{f})$ we obtain $$\operatorname{Re}\left(a_3 - \frac{1}{2}a_2^2\right) = \left|c_3(f) - \frac{1}{2}c_2(f)^2\right| \ge 0 \text{ and } \operatorname{Im}\left(a_3 - \frac{1}{2}a_2^2\right) = 0.$$ (3.10) By definition, $\tilde{f} \in \mathcal{S}_0$ . Therefore, the equalities (3.8) hold with $f := \tilde{f}$ for a certain continuous function $\kappa : [0; +\infty) \to \mathbb{T}$ . Then there exists a continuous function $\varphi : [0, +\infty) \to \mathbb{R}$ satisfying $\kappa(t) = \mathrm{e}^{\mathrm{i}\varphi(t)}$ for $t \in [0; +\infty)$ . Setting  $$\alpha := \int_0^{+\infty} e^{-2t} \cos(\varphi(t))^2 dt \tag{3.11}$$ we obtain $0 \le \alpha \le 1/2$ , $$\operatorname{Re}\left(a_{3} - \frac{1}{2}a_{2}^{2}\right) = \operatorname{Re}\left\{\left(-2\int_{0}^{+\infty} e^{-2t} e^{2i\varphi(t)} dt + a_{2}^{2}\right) - \frac{1}{2}a_{2}^{2}\right\}$$ $$= \frac{1}{2}\operatorname{Re}a_{2}^{2} - 2\operatorname{Re}\int_{0}^{+\infty} e^{-2t} e^{2i\varphi(t)} dt$$ $$= \frac{1}{2}\operatorname{Re}a_{2}^{2} - 2\int_{0}^{+\infty} e^{-2t} \cos(2\varphi(t)) dt$$ $$= \frac{1}{2}\operatorname{Re}a_{2}^{2} - 2\int_{0}^{+\infty} e^{-2t} \left(2(\cos(\varphi(t)))^{2} - 1\right) dt$$ $$= \frac{1}{2}\operatorname{Re}a_{2}^{2} + 1 - 4\alpha \tag{3.12}$$ as well as $$\operatorname{Im}\left(a_{3} - \frac{1}{2}a_{2}^{2}\right) = \operatorname{Im}\left\{\left(-2\int_{0}^{+\infty} e^{-2t} e^{2i\varphi(t)} dt + a_{2}^{2}\right) - \frac{1}{2}a_{2}^{2}\right\}$$ $$= \frac{1}{2}\operatorname{Im}a_{2}^{2} - 2\operatorname{Im}\int_{0}^{+\infty} e^{-2t} e^{2i\varphi(t)} dt$$ $$= \frac{1}{2}\operatorname{Im}a_{2}^{2} - 2\int_{0}^{+\infty} e^{-2t} \sin(2\varphi(t)) dt$$ $$= \frac{1}{2}\operatorname{Im}a_{2}^{2} - 4\int_{0}^{+\infty} e^{-2t} \sin(\varphi(t)) \cos(\varphi(t)) dt. \tag{3.13}$$ From (3.10) and (3.12) it follows that $$\frac{1}{2} \operatorname{Re} a_2^2 \ge 4\alpha - 1. \tag{3.14}$$ Combining (3.10) with (3.13) and using the Schwarz integral inequality we obtain $$\begin{split} \left| \frac{1}{2} \operatorname{Im} a_2^2 \right| &= 4 \left| \int_0^{+\infty} e^{-2t} \sin(\varphi(t)) \cos(\varphi(t)) dt \right| \\ &\leq 4 \int_0^{+\infty} \left| e^{-t} \sin(\varphi(t)) \right| \left| e^{-t} \cos(\varphi(t)) \right| dt \\ &\leq 4 \left( \int_0^{+\infty} e^{-2t} \sin(\varphi(t))^2 dt \right)^{1/2} \left( \int_0^{+\infty} e^{-2t} \cos(\varphi(t))^2 dt \right)^{1/2} \\ &= 4 \left( \int_0^{+\infty} e^{-2t} (1 - \cos(\varphi(t))^2) dt \right)^{1/2} \alpha^{1/2} \end{split}$$  $$= 4 \left( \int_0^{+\infty} e^{-2t} dt - \int_0^{+\infty} e^{-2t} \cos(\varphi(t))^2 dt \right)^{1/2} \alpha^{1/2}$$ $$= 4 \left( \frac{1}{2} - \alpha \right)^{1/2} \alpha^{1/2}.$$ (3.15) Setting $$u := \frac{1}{2} \operatorname{Re} a_2^2$$ and $v := \frac{1}{2} \operatorname{Im} a_2^2$ we conclude from (3.14) and (3.15) that $$u \ge 4\alpha - 1$$ and $|v| \le 4\left(\frac{1}{2} - \alpha\right)^{1/2} \alpha^{1/2}$ (3.16) as well as, by (3.12), that $$|a_2| - \text{Re}\left(a_3 - \frac{1}{2}a_2^2\right) = \sqrt{2}(u^2 + v^2)^{1/4} - u - 1 + 4\alpha.$$ (3.17) From the first inequality in (3.16) it follows that $$\tau := u - 4\alpha + 1 > 0. \tag{3.18}$$ Since $0 \le \alpha \le 1/2$ , we deduce from (3.17) and by the second inequality in (3.16) that $$|a_{2}| - \operatorname{Re}\left(a_{3} - \frac{1}{2}a_{2}^{2}\right) \leq \sqrt{2}\left((\tau + 4\alpha - 1)^{2} + 16\left(\frac{1}{2} - \alpha\right)\alpha\right)^{1/4} - \tau$$ $$= \sqrt{2}\left((\tau^{2} + 2(4\alpha - 1)\tau + (4\alpha - 1)^{2} + 8\alpha - 16\alpha^{2}\right)^{1/4} - \tau$$ $$= \sqrt{2}\left((\tau^{2} + 2(4\alpha - 1)\tau + 1\right)^{1/4} - \tau$$ $$\leq \sqrt{2}\left(\tau^{2} + 2\tau + 1\right)^{1/4} - \tau$$ $$= \sqrt{2}\left(\tau + 1\right)^{1/2} - \tau$$ $$\leq \sqrt{2} - \left(1 - \frac{1}{\sqrt{2}}\right)\tau \leq \sqrt{2}.$$ Combining this with (3.18) we deduce from (3.10) and (3.12) that for every $f \in S_0$ there exists $\tau \ge 0$ such that $$\left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| - \left| c_2(f) \right| \ge -\sqrt{2} + \left( 1 - \frac{1}{\sqrt{2}} \right) \tau \ge -\sqrt{2}$$ (3.19)  as well as $$\left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| = \tau.$$ (3.20) By (3.1) and (3.19) we obtain the first inequality in (3.5) for every $f \in S_0$ . Thus both inequalities in (3.5) hold for every $f \in S_0$ . As shown in [8, Thm. 3.2], the class $S_0$ is dense in the class $S_0$ in the space H. Then by Remarks 2.1 and 2.2, the inequalities in (3.5) hold for every $f \in S$ . It remains to prove that both inequalities in (3.5) are sharp and determine extremal functions. Let $f: \mathbb{D} \to \mathbb{C}$ be arbitrarily fixed. Suppose that $f \in \mathcal{S}$ satisfies $$\left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| - \left| c_2(f) \right| = 1.$$ (3.21) Since $S_0$ is dense in S in H there exists a sequence $\mathbb{N} \ni n \mapsto f_n \in S_0$ which is convergent to f in H. By (3.9) we have $$\left| c_3(f_n) - \frac{1}{2} c_2(f_n)^2 \right| - |c_2(f_n)| \le 1 - \frac{1}{2} |c_2(f_n)| (2 - |c_2(f_n)|) \le 1. \quad (3.22)$$ Applying now Remark 2.2, we deduce from (3.21) and (3.22) that $|c_2(f)|(2 - |c_2(f)|) = 0$ , which yields $|c_2(f)| = 0$ or $|c_2(f)| = 2$ . In the second case f is a rotation of the Koebe function by Bieberbach's Theorem (cf. [3], [8, Thm. 2.2]), and so the left side of the equality (3.21) is equal to -1, which is impossible. Therefore $|c_2(f)| = 0$ , and by (3.21), $|c_3(f) - c_2(f)^2| = 1$ . Then Lemma 2.4 implies that f satisfies the equality (3.6) for a certain $\xi \in \mathbb{T}$ . Conversely, if f satisfies the equality (3.6) for a certain $\xi \in \mathbb{T}$ , then by Lemma 2.3, $f = \Phi_{0,\xi} \in \mathcal{S}$ and the equality (3.21) holds, which together with (3.1) proves the condition (i) of the theorem. Suppose that $f \in \mathcal{S}$ satisfies the equality $$\left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| - \left| c_2(f) \right| = -\sqrt{2}.$$ (3.23) Since the class $S_0$ is dense in the class S in the space H, there exists a sequence $\mathbb{N} \ni n \mapsto f_n \in S_0$ which is convergent to f in H. From (3.19) and (3.20) there exists a sequence $\mathbb{N} \ni n \mapsto \tau_n \ge 0$ such that $$\left| c_3(f_n) - \frac{1}{2} c_2(f_n)^2 \right| - |c_2(f_n)| \ge -\sqrt{2} + \left( 1 - \frac{1}{\sqrt{2}} \right) \tau_n \ge -\sqrt{2}, \quad n \in \mathbb{N},$$ (3.24) as well as $$\left| c_3(f_n) - \frac{1}{2} c_2(f_n)^2 \right| = \tau_n, \quad n \in \mathbb{N}.$$ (3.25)  Applying now Remark 2.2, we deduce from (3.23) and (3.24) that $\tau_n \to 0$ as $n \to +\infty$ , which together with (3.25) leads to $$\left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| = 0.$$ Combining this with (3.23) we have $|c_2(f)| = \sqrt{2}$ , which yields $$\left| c_3(f) - c_2(f)^2 \right| \le \left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| + \frac{1}{2} |c_2(f)|^2 = 1$$ as well as $$\left| c_3(f) - c_2(f)^2 \right| \ge \frac{1}{2} |c_2(f)|^2 - \left| c_3(f) - \frac{1}{2} c_2(f)^2 \right| = 1.$$ Therefore $|c_3(f)-c_2(f)^2|=1$ , and consequently Lemma 2.4 implies that f satisfies the equality (3.7) for a certain $\zeta \in \mathbb{T}$ . Conversely, from Lemma 2.3 it follows that if f satisfies the equality (3.7) for a certain $\zeta \in \mathbb{T}$ , then $f = \Phi_{\sqrt{2},\zeta} \in \mathcal{S}$ and the equality (3.23) holds, which together with (3.1) proves the condition (ii) of the theorem. Recall that a function $f \in \mathcal{H}$ is called starlike provided $f \in \mathcal{S}$ and the image $f(\mathbb{D})$ is a starlike set with respect to the origin. The class of starlike functions, usually denoted by $\mathcal{S}^*$ , was introduced by Alexander in [1]. It is well known that starlike functions are characterized by the condition $$\operatorname{Re} \frac{zf'(z)}{f(z)} > 0, \quad z \in \mathbb{D} \setminus \{0\}; \tag{3.26}$$ cf. e.g. [1], [8, p. 41], [18]. From Lemma 2.3 and Theorem 3.1 we easily derive the following result.

Corollary 3.2. For every class $S' \subset S$ , if $S^* \subset S'$ , then Theorem 3.1 holds with S replaced by the class S'. We recall that a function $f \in \mathcal{H}$ is close-to-convex if f(0) = 0 = f'(0) - 1 and there exist a function $g \in \mathcal{S}^*$ and $\delta \in (-\pi/2; \pi/2)$ such that $$\operatorname{Re} \frac{e^{i\delta}zf'(z)}{g(z)} > 0, \quad z \in \mathbb{D} \setminus \{0\}. \tag{3.27}$$ The class $\mathcal{C}$ of all close-to-convex functions was introduced by Kaplan [15]. By the condition (3.26), each $f \in \mathcal{S}$ satisfies the condition (3.27) with g := f and $\delta := 0$ , and so $\mathcal{S}^ \subset \mathcal{C}$ . As was shown by Ozaki [22] and Kaplan [15] the condition (3.27) implies the injectivity of f. Lewandowski [19, 20] showed that the class of close-to-convex functions is identical with the class of linearly accessible functions introduced by Biernacki [4]. A function $f \in \mathcal{S}$ is said to be linearly accessible if the complement of $f(\mathbb{D})$ is the union of non-intersecting half-lines; cf. [8, p. 51].  Remark 3.3 Since $S^* \subset C \subset S$ , we may apply Corollary 3.2 with S' := C. As a result we see that Theorem 3.1 holds with S replaced by the class C. For any $\zeta \in \mathbb{T}$ a non-empty set $\Omega \subset \mathbb{C}$ is said to be convex in the direction $\zeta$ provided the intersection $\{\zeta t + a : t \in \mathbb{R}\} \cap \Omega$ is a connected set for each $a \in \mathbb{C}$ . We denote by $\mathcal{CV}(\zeta)$ the class of all $f \in \mathcal{S}$ such that the image $f(\mathbb{D})$ is convex in the direction $\zeta \in \mathbb{T}$ ; cf. e.g. [11, Vol. I., pp. 193–194] and [16]. Notice that $\mathcal{CV}(\zeta) = \mathcal{CV}(-\zeta)$ for $\zeta \in \mathbb{T}$ . In particular, the class $\mathcal{CV}(i)$ coincides with the class of all convex functions in the direction of the imaginary axis, introduced by Robertson [24]; cf. [11, Vol. I., pp. 193–206]. The following result is an immediate consequence of Lemma 2.3 and Theorem 3.1.

Corollary 3.4. For all $\zeta \in \mathbb{T}$ and $f \in \mathcal{CV}(\zeta)$ the inequalities (3.5) hold. Moreover, both the inequalities are sharp and for every function $f : \mathbb{D} \to \mathbb{C}$ : - (i) $f \in \mathcal{CV}(\zeta)$ and the equality in the second inequality in (3.5) holds if and only if $f = \Phi_0 \overline{\zeta}$ or $f = \Phi_0 \overline{\zeta}$ ; - (ii) $f \in \mathcal{CV}(\zeta)$ and the equality in the first inequality in (3.5) holds if and only if $f = \Phi_{\sqrt{2},\overline{\zeta}}$ or $f = \Phi_{\sqrt{2},-\overline{\zeta}}$ . Funding Not applicable. Data Availability Statement Not applicable. Code Availability Not applicable.