Revisit Of Meromorphic Convex Functions

Theorem 2.7. Let $\alpha \in [0,1)$ and $q_{\alpha} \in \mathcal{P}((1+\alpha)/2)$ . Assume $f \in \mathcal{B}$ be a locally univalent function satisfying the condition $|S_f(z)| \leq 2q_{\alpha}(|z|)$ . Then $f \in \mathcal{BC}(\alpha)$ . Also, the constant $(1+\alpha)/2$ is the best possible. Remark 2.8. In particular for a fixed $\alpha \in [0, 1)$ , if we take $q_{\alpha}(z) = c_{\alpha}$ , a positive constant, then $q_{\alpha} \in \mathcal{P}(\alpha)$ if and only if the solution of the differential equation $$(2.2) y'' + c_{\alpha}y = 0, y(0) = 0, y'(0) = 1$$ satisfies y(x) > 0 in (0,1) with (2.3) $$\lim_{x \to 1^{-}} \frac{y'(x)}{y(x)} = \frac{1+\alpha}{2}.$$ Note that for the differential equation (2.2), the root is $$y(x) = \frac{\sin\sqrt{c_{\alpha}}x}{\sqrt{c_{\alpha}}}.$$ Next, we need to check whether the function y satisfies (2.3). It is simple to observe that $$\lim_{x \to 1} \frac{y'(x)}{y(x)} = \lim_{x \to 1^{-}} \sqrt{c_{\alpha}} \cot \sqrt{c_{\alpha}} x = \sqrt{c_{\alpha}} \cot \sqrt{c_{\alpha}}.$$ It leads to $$\lim_{x \to 1} \frac{y'(x)}{y(x)} = \frac{1+\alpha}{2} \Leftrightarrow \sqrt{c_{\alpha}} - \frac{1+\alpha}{2} \tan \sqrt{c_{\alpha}} = 0,$$ and so in this case, Theorem 2.7 coincides with [2, Theorem 2.1]. Next, we will illustrate Theorem 2.7 using the following example. Example 2.9. Consider the function f defined on $\mathbb{D}$ , given by $$f(z) = \sqrt{\frac{1-\alpha}{\pi}} \cot\left(\sqrt{\frac{1-\alpha}{\pi}}z\right).$$ As $\cot z$ is a meromorphic function with a simple pole at 0, f has the series expansion given by (1.1). It is easy to verify that $$S_f(z) = \frac{2 - 2\alpha}{\pi} \left( \csc^2 \left( \sqrt{\frac{1 - \alpha}{\pi}} z \right) - \cot^2 \left( \sqrt{\frac{1 - \alpha}{\pi}} z \right) \right) = \frac{2 - 2\alpha}{\pi} = 2 \left( \frac{1 - \alpha}{\pi} \right).$$ Next, we define the function $q_{\alpha}$ by $$q_{\alpha}(x) = \frac{2 - 2\alpha}{\pi(1 + x^2)}, \ x \in [0, 1).$$ Now note that for all $r \in (0,1)$ , we have $\pi(1+r^2) \leq 2\pi$ , and hence we deduce that $$\frac{2-2\alpha}{\pi(1+r^2)} \ge \frac{1-\alpha}{\pi}.$$ This shows that $$|S_f(z)| = \frac{4 - 4\alpha}{\pi} \le \frac{4 - 4\alpha}{\pi(1 + |z|^2)} = 2q_\alpha(|z|), \ z \in \mathbb{D}.$$ It is easy to verify that $$\int_0^1 q_{\alpha}(x)dx = \frac{2 - 2\alpha}{\pi} \tan^{-1}(1) = \frac{1 - \alpha}{2}.$$ Then by Lemma 3.4, we get $q_{\alpha} \in \mathcal{P}(1+\alpha)/2$ . Observe that f satisfies all the hypothesis in Theorem 2.7 and as a result, we conclude that $f \in \mathcal{BC}(\alpha)$ . Next, by the definition of the class $\mathcal{BC}(\alpha)$ , we are going to show that $f \in \mathcal{BC}(\alpha)$ . For that consider $$f(z) = b_{\alpha} \cot(b_{\alpha}z) = b_{\alpha} \frac{u(z)}{v(z)},$$ where $$b_{\alpha} = \sqrt{\frac{1-\alpha}{\pi}}, \ u(z) = \cos(b_{\alpha}z), \ v(z) = \sin(b_{\alpha}z).$$ It is easy to verify that u and v satisfies $$u(0) = 1$$ , $u'(0) = 0$ and $v(0) = 0$ , $v'(0) = 1$ , which implies that the Wronskian W(u,v)(z)=1 for all $z\in\mathbb{D}$ . Simple computations give $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) = 1 - 2\operatorname{Re}\left(\frac{zv'(z)}{v(z)}\right) = 1 - 2\operatorname{Re}\left(\frac{zb_{\alpha}\cos(b_{\alpha}z)}{\sin(b_{\alpha}z)}\right).$$ By Lemma 3.3, it is enough to prove (2.4) $$\operatorname{Re}\left(\frac{zb_{\alpha}\cos(b_{\alpha}z)}{\sin(b_{\alpha}z)}\right) \ge \frac{1+\alpha}{2} = \frac{2-\pi b_{\alpha}^{2}}{2}.$$ By taking $z = x + iy \in \mathbb{D}$ with $x \ge 0$ , $y \ge 0$ and performing a straightforward computation, one can see that $$\operatorname{Re}\left(\frac{zb_{\alpha}\cos(b_{\alpha}z)}{\sin(b_{\alpha}z)}\right) = \frac{b_{\alpha}x\cos(b_{\alpha}x)\sin(b_{\alpha}x) + b_{\alpha}y\sinh(b_{\alpha}y)\cosh(b_{\alpha}y)}{\sin^2(b_{\alpha}x)\cosh^2(b_{\alpha}y) + \sinh^2(b_{\alpha}y)\cos^2(b_{\alpha}x)}.$$ Then the inequality (2.4) will be equivalent to $$2b_{\alpha}x\cos(b_{\alpha}x)\sin(b_{\alpha}x) - (2 - \pi b_{\alpha}^{2})\cosh^{2}(b_{\alpha}y)\sin(b_{\alpha}x)\cos(b_{\alpha}x)\tan(b_{\alpha}x)$$ $$\geq (2 - \pi b_{\alpha}^{2})\cos^{2}(b_{\alpha}x)\sinh(b_{\alpha}y)\cosh(b_{\alpha}y)\tanh(b_{\alpha}y) - 2b_{\alpha}y\sinh(b_{\alpha}y)\cosh(b_{\alpha}y).$$ That is, $$\cos(b_{\alpha}x)\sin(b_{\alpha}x)[2b_{\alpha}x - (1+\alpha)\cosh^{2}(b_{\alpha}y)\tan(b_{\alpha}x)]$$ $$\geq \sinh(b_{\alpha}y)\cosh(b_{\alpha}y)[(1+\alpha)\cos^{2}(b_{\alpha}x)\tanh(b_{\alpha}y) - 2b_{\alpha}y].$$ From the fact that − cosh² (bαy) tan(bαx) ≤ − tan(bαx) and (1 + α) tan(bαx) ≥ tan(bαx), we get (2.5) $$\cos(b_{\alpha}x)\sin(b_{\alpha}x)[2b_{\alpha}x - \tan(b_{\alpha}x)]$$ $$\geq \sinh(b_{\alpha}y)\cosh(b_{\alpha}y)[(1+\alpha)\cos^{2}(b_{\alpha}x)\tanh(b_{\alpha}y) - 2b_{\alpha}y].$$ Define the function g(y) := (1 + α) tanh(bαy) − 2bαy, y ≥ 0. We have sech² (y) ≤ 1, for all y ≥ 0, which implies that $$g'(y) = (1 + \alpha)b_{\alpha} \operatorname{sech}^{2}(b_{\alpha}y) - 2b_{\alpha} = b_{\alpha}[(1 + \alpha) \operatorname{sech}^{2}(b_{\alpha}y) - 2] < b_{\alpha}(-1 + \alpha) < 0.$$ This means that g is a decreasing function in [0,∞) and as a result we have g(y) ≤ g(0) = 0. Since cos² (bαx) ≤ 1, we have $$(1+\alpha)\cos^2(b_{\alpha}x)\tanh(b_{\alpha}y) - 2b_{\alpha}y \le (1+\alpha)\tanh(b_{\alpha}y) - 2b_{\alpha}y \le 0.$$ By the definition itself, for all x ≥ 0, we have sinh(bαy) cosh(bαy) ≥ 0. So the right-hand side of (2.5) is either negative or zero. It is obvious that 2bαx − tan(bαx) > 0 and cos(bαx) sin(bαx) ≥ 0. Hence, we obtain that the left-hand side of (2.5) is always non-negative. So we conclude that on the first quadrant of the unit disk, we established the inequality (2.4). Due to the symmetry of the unit disk about the real and imaginary axes, we obtained the desired inequality (2.4). 2.2. Meromorphically inverse convex functions of order α. We start this section by recalling the following. Let f ∈ S^∗ (α) and take g = 1/f. Then clearly g ∈ M and $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) = \operatorname{Re}\left(\frac{-zg'(z)}{g(z)}\right) \ge \alpha.$$ which implies g ∈ BS^∗ (α). That is, there is a one-to-one correspondence between the classes S ∗ (α) and BS^∗ (α). In other words, $$(2.6) f \in \mathcal{S}^(\alpha) \Leftrightarrow 1/f \in \mathcal{BS}^(\alpha).$$ However, it is important to note that the same argument is invalid for the case of convexity. The following example precisely discusses this. Example 2.10. Consider the function l(z) = z/(1−z), z ∈ D, clearly l ∈ C but l /∈ C(α), α ∈ (0, 1). Assume g(z) = 1/l(z) = (1 − z)/z, we get $$\operatorname{Re}\left(1 + \frac{zg''(z)}{g'(z)}\right) = -1.$$ This implies that g belongs to the class BC(α) for all α ≤ 1. In particular, g ∈ BC(1/2) but l is not in the class C(1/2). A reasonable question is whether the reciprocal of functions in the class C(α) exhibits any notable analytic properties. In [30], the case α = 0 is examined, leading to the introduction of a new class, namely the class of meromorphically inverse convex functions; which contains functions f ∈ B such that g(z)f(z) = 1, z ∈ D \ {0}, for some g ∈ C. Definition 2.11. A function f ∈ B is said to be meromorphically inverse convex of order α in D \ {0} if there exists a function g ∈ C(α) with g(z)f(z) = 1, for z ∈ D \ {0}. We denote such class of functions by BC^I (α). In this subsection, we discuss some of its analytic properties and also explore its association with existing classes. We also identify the largest disk on which a functions of the class M is in BC^I (α). Remark 2.12. Note that if f ∈ BC^I (α), then f = 1/g, where g ∈ C(α). As a result, every function in the class BC^I (α) is univalent. Example 2.13. Define g : D \ {0} → C by $$g(z) = -\frac{1}{\log(1-z)},$$ and f(z) = 1/g(z). Then $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) = \operatorname{Re}\left(\frac{1}{1-z}\right) > \frac{1}{2},$$ which implies f ∈ C(1/2) and hence g ∈ BC^I (1/2). Example 2.14. Fix α ̸= 1/2. Consider the function $$g(z) = \frac{\eta}{1 - (1 - z)^{\eta}},$$ where η = 2α − 1, and f(z) = 1/g(z). Note that $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) = \operatorname{Re}\left(1 + \frac{z(1-\eta)}{1-z}\right) = \operatorname{Re}\left(1 + \frac{2z(1-\alpha)}{1-z}\right) > \alpha,$$ which implies f ∈ C(α) and so g ∈ BC^I (α). Now we state the following result, which studies some inclusion properties and a characterization of the class BC^I (α). Theorem 2.15. The class BC^I (α) possesses the following properties. - (i) For α > β, BC^I (α) ⊂ BC^I (β). - (ii) BC^I (α) is invariant under any non zero linear transformation T(z) = λz. - (iii) For any α ∈ [0, 1), the class BC^I (α) is properly contained in BS^∗ (0). - (iv) g ∈ BC^I (α) if and only if g is univalent and (2.7) $$\operatorname{Re}\left(1 + \frac{zg''(z)}{g'(z)} - \frac{2zg'(z)}{g(z)}\right) \ge \alpha.$$ Next, we state the following result, which gives a radius for which every function in M is to be in BC^I (α). Theorem 2.16. Let g ∈ M. Then g is meromorphically inverse convex of order α on the disk of radius rα, where r^α is the unique zero in (0, 1) of the polynomial (2.8) $$P_{\alpha}(x) := (-1 - \alpha)x^2 + 4x + \alpha - 1.$$ This is false for every r > rα. Finally, we obtain the following relation between the classes $\mathcal{BC}_I(\alpha)$ and $\mathcal{BS}^*(\alpha)$ .

Theorem 2.17. A function $$g \in \mathcal{BC}_I(\alpha)$$ if and only if $\frac{1}{z(1/g)'} \in \mathcal{BS}^*(\alpha)$ .

Lemma 3.1. Let p be an analytic function and f be a function with the Schwarzian derivative $S_f = 2p$ . Then f is of the form $$(3.1) f(z) = \frac{u(z)}{v(z)},$$ where u and v are arbitrary linearly independent solutions of the ordinary linear differential equation y'' + py = 0. By adopting a similar method of proof as outlined in [11, Lemma II], we arrive at the following result.

Lemma 3.2. Let $\alpha \in [0,1)$ and g be a continuously differentiable function on [0,1) satisfying the conditions g(0) = 0, $g'(0) \neq 0$ . Then for given $\epsilon > 0$ , there exists a $\delta > 0$ such that for any $r \in (1 - \delta, 1)$ , $$r \int_0^r (g'(x))^2 dx \ge r \int_0^r q_\alpha(x) g^2(x) dx + \left(\frac{1+\alpha}{2} - \epsilon\right) g^2(r),$$ where $q_{\alpha} \in \mathcal{P}((1+\alpha)/2)$ .

Definition 2.5. For $\alpha \geq 0$ , the class $\mathcal{P}(\alpha)$ is defined as the class of continuous functions $q_{\alpha}$ on [0,1) with the following properties - (a) For $r \in [0, 1), q_{\alpha}(r) \geq 0$ . - (b) For the differential equation $$(2.1) y'' + q_{\alpha} y = 0$$ satisfying y(0) = 0 and y'(0) = 1, the solution is positive on (0,1) with $$\lim_{x \to 1^{-}} \frac{y'(x)}{y(x)} \ge \alpha.$$ Remark 2.6. From the definition itself, it is obvious that the class $\mathcal{P}(\alpha)$ satisfies the inclusion $$\mathcal{P}(\alpha) \subset \mathcal{P}(\beta)$$ , for $\alpha > \beta$ . Next, we state our first main result, which gives a sufficient condition for a meromorphic function to be in the class $\mathcal{BC}(\alpha)$ .