Radii of convexity associated with various subclasses of analytic functions for functions related to Kaplan classes

Lemma 2.1. [20] For $0 \le \alpha \le \beta$ , we have $f' \in \mathcal{K}(\alpha, \beta)$ if and only if there exists $g \in \mathcal{ST}((2-\beta+\alpha)/2)$ and $\phi \in \mathbb{R}$ such that $$\left|\arg e^{i\phi} \frac{zf'(z)}{g(z)}\right| \leqslant \frac{\alpha\pi}{2}, \qquad z \in \mathbb{D}.$$ (2.2) Define the function p by $p(z) = e^{i\phi}zf'(z)/g(z)$ where $f' \in \mathcal{K}(\alpha, \beta)$ and $g \in \mathcal{ST}((2 - \beta + \alpha)/2)$ . We note that $$\frac{zp'(z)}{p(z)} = 1 + \frac{zf''(z)}{f'(z)} - \frac{zg'(z)}{g(z)}.$$ (2.3) Since $g \in ST((2-\beta+\alpha)/2)$ , we have $zg'/g \in \mathcal{P}((2-\beta+\alpha)/2)$ , and by (2.1) $$\left| \frac{zg'(z)}{g(z)} - \frac{1 + (\beta - \alpha - 1)r^2}{1 - r^2} \right| \le \frac{(\beta - \alpha)r}{1 - r^2}, \quad |z| = r < 1.$$ (2.4) Note that for any $q \in \mathcal{P}$ , we have (see [15]) $$\left| \frac{zq'(z)}{q(z)} \right| \leqslant \frac{2r}{1 - r^2}, \quad |z| \leqslant r < 1. \tag{2.5}$$ Using (2.2), we have $p^{1/\alpha} \in \mathcal{P}$ . Hence, by (2.5) we have $$\left| \frac{zp'(z)}{p(z)} \right| \leqslant \frac{2\alpha r}{1 - r^2}, \quad |z| \leqslant r < 1. \tag{2.6}$$ From (2.3), (2.4) and (2.6), we get $$\left| 1 + \frac{zf''(z)}{f'(z)} - \frac{1 + (\beta - \alpha - 1)r^2}{1 - r^2} \right| \le \frac{(\alpha + \beta)r}{1 - r^2}, \quad |z| \le r < 1.$$ (2.7) From (2.7), it is evident that w = 1 + (zf''(z)/f'(z)) resides within the disc $|w - a(r)| \le r_a$ , where a(r) and $r_a(r)$ denote the center and the radius respectively: $$a(r) := \frac{1 + (\beta - \alpha - 1)r^2}{1 - r^2}$$ and $r_a(r) := \frac{(\alpha + \beta)r}{1 - r^2}$ . (2.8) We note that the center a(r) is an increasing function of r and $a(r) \ge 1$ as $\beta \ge \alpha \ge 0$ .

Theorem 3.1. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_{Ne}$ radius for functions whose derivative belongs to $\mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_{Ne}} := \frac{4}{3(\alpha+\beta) + \sqrt{9(\alpha+\beta)^2 + 8(3(\beta-\alpha)+2)}}.$$ Proof. We first note that $R = \mathcal{R}_{\mathcal{CV}_{Ne}}$ is the smallest positive root of the equation $(3(\beta - \alpha) + 2)r^2 + 3(\alpha + \beta)r - 2 = 0$ in the interval (0, 1). It then follows that $(3(\beta - \alpha) + 2)r^2 + 3(\alpha + \beta)r - 2 \leq 0$ for $0 \leq r \leq R$ and by rewriting it, we have $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{5}{3} - \frac{1+(\beta-\alpha-1)r^2}{1-r^2}, \quad r \leqslant R.$$ Consequently, for $0 \leqslant r \leqslant R = \Re_{\mathcal{CV}_{N_e}}$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le \frac{5}{3} - a(r)$$ (3.1) where a(r) is given by (2.8). By [27, Lemma 2.1], we have $\{w : |w-a| < 5/3 - a\} \subseteq \varphi_{Ne}(\mathbb{D}) = \Omega_{Ne}$ , for 1 < a < 5/3. The condition a < 5/3 ensures that the center lies inside the nephroid. Since the center a(r) given by (2.8) is always greater than or equal to 1, the disc in (3.1) lies inside the region $\Omega_{Ne}$ for $r \leqslant R$ proving that the $\mathcal{CV}_{Ne}$ radius for functions whose derivative belongs to $\mathcal{K}(\alpha, \beta)$ is at least R. To show that sharpness, consider the function $f_0$ defined by $$f_0'(z) = \frac{(1+z)^{\alpha}}{(1-z)^{\beta}}. (3.2)$$ This function $f_0$ has derivative $f_0' \in \mathcal{K}(\alpha, \beta)$ . Indeed, with $g_0(z) = z/(1-z)^{\beta-\alpha}$ , we have $\operatorname{Re}(zg_0'(z)/g_0(z)) = (1+(\beta-\alpha-1)z)/(1-z)$ and hence $g_0 \in \operatorname{ST}((2-\beta+\alpha)/2)$ . Also we have $|\operatorname{arg}(zf_0'(z)/g_0(z))| = |\operatorname{arg}(((1+z)/(1-z))^{\alpha})| < (\alpha\pi)/2$ . Since $f_0$ satisfies (2.2), it follows that $f_0' \in \mathcal{K}(\alpha, \beta)$ . From (3.2), we find $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)z^2 + (\alpha + \beta)z}{1 - z^2}.$$ (3.3) For z = R, the equation (3.3) gives $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R^2 + (\alpha + \beta)R}{1 - R^2} = \frac{5}{3} = \varphi_{Ne}(1).$$ Thus, the radius R is sharp for the function $f_0$ defined by (3.2) (See Fig.1) (This figure as well as other later figures show the image of the extremal function $f_0$ , and the disc given by (2.7) as well as the boundary of $\varphi(\mathbb{D})$ ).  Figure 1. Sharpness of $\Re_{\mathcal{CV}_{Ne}}$ radius when $\alpha = 0.1$ and $\beta = 1$

Corollary 3.2. Radius of convexity associated with the class $\mathcal{CV}_{Ne}$ for some special cases: (1) The $\mathcal{CV}_{Ne}$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_{Ne}} = \frac{4}{3(2-\sigma) + \sqrt{9\sigma^2 - 38\sigma + 54}}.$$ (2) The $\mathcal{CV}_{Ne}$ radius for the class $V_k$ is $$\mathcal{R}_{\mathcal{CV}_{Ne}} = \frac{4}{3k + \sqrt{9k^2 + 64}}.$$ (3) The $\mathcal{CV}_{Ne}$ radius for the class $\mathcal{SCC}(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_{Ne}} = \frac{2}{3(\alpha+1) + \sqrt{9(\alpha+1)^2 + 16}}.$$ Kumar and Arora [2] introduced the class which consists of convex functions associated with the petal shaped domain, given by $\mathcal{CV}_h = \mathcal{CV}(\varphi_h)$ , where $\varphi_h(z) = 1 + \sinh^{-1}(z)$ .

Theorem 3.3. Let $0 \le \alpha \le \beta$ . The $\mathcal{CV}_h$ radius for functions whose derivative belongs to $\mathcal{K}(\alpha, \beta)$ is given by $$\Re_{\mathcal{CV}_h} = \frac{2 \sinh^{-1}(1)}{(\alpha + \beta) + \sqrt{(\alpha + \beta)^2 + 4(\beta - \alpha + \sinh^{-1}(1))(\sinh^{-1}(1))}}.$$ The radius obtained is sharp. Proof. We first note that $R = \mathcal{R}_{\mathcal{CV}_h}$ is the smallest positive root of the equation $(\beta - \alpha + \sinh^{-1}(1))r^2 + (\alpha + \beta)r - \sinh^{-1}(1) = 0$ in the interval (0,1). Then it follows that $(\beta - \alpha + \sinh^{-1}(1))r^2 + (\alpha + \beta)r - \sinh^{-1}(1) \leq 0$ for $0 \leq r \leq R$ and hence $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant 1 + \sinh^{-1}(1) - \frac{1 + (\beta-\alpha-1)r^2}{1-r^2}, \quad r \leqslant R.$$ Consequently, for $0 \le r \le R = \Re_{\mathcal{CV}_h}$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le 1 + \sinh^{-1}(1) - a(r)$$ (3.4) where a(r) is given by (2.8). For $1 < a \leqslant 1 + \sinh^{-1}(1)$ , by [2, Lemma 2.1] we have $\{w : |w - a| < 1 + \sinh^{-1}(1) - a\} \subseteq \varphi_h(\mathbb{D}) = \Omega_h$ . The condition $a < 1 + \sinh^{-1}(1)$ tells us that the center lies inside the region $\Omega_h$ . Since the center a(r) is always greater than 1, the disc in (3.4) lies inside the region $\Omega_h$ for $r \leqslant R$ proving that $\mathcal{CV}_h$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha,\beta)$ is at least R. To prove the sharpness of the radius $R = \mathcal{R}_{\mathcal{CV}_h}$ , we consider the function $f_0$ defined by (3.2). For z = R in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R^2 + (\alpha + \beta)R}{1 - R^2} = 1 + \sinh^{-1}(1) = \varphi_h(1).$$ Therefore, the radius R is sharp for the function $f_0$ defined by (3.2) (See Fig.2).  Figure 2. Sharpness of $\mathcal{R}_{\mathcal{CV}_h}$ radius when $\alpha = 0.3$ and $\beta = 1.5$

Corollary 3.4. Radius of convexity associated with the class $\mathbb{CV}_h$ for some special cases: (1) The $\mathcal{CV}_h$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_h} = \frac{\sinh^{-1}(1)}{(2-\sigma) + \sqrt{(2-\sigma)^2 + (2(1-\sigma) + \sinh^{-1}(1))(\sinh^{-1}(1))}}.$$ (2) The $\mathcal{CV}_h$ radius for the class $V_k$ is $$\Re_{ev_h} = \frac{2\sinh^{-1}(1)}{k + \sqrt{k^2 + 4(\sinh^{-1}(1))(2 + \sinh^{-1}(1))}}.$$ (3) The $\mathcal{CV}_h$ radius for the class $\mathcal{SCC}(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_h} = \frac{\sinh^{-1}(1)}{(\alpha+1) + \sqrt{(\alpha+1)^2 + (2+\sinh^{-1}(1))(\sinh^{-1}(1))}}.$$ Goel and Kumar [6] introduced the class which consists of convex functions associated with the modified sigmoid, given by $\mathcal{CV}_{SG} = \mathcal{CV}(\varphi_{SG})$ , where $\varphi_{SG}(z) = 2/(1 + e^{-z})$ .

Theorem 3.5. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_{SG}$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\Re_{\mathcal{CV}_{SG}} = \frac{2(e-1)}{(\alpha+\beta)(e+1) + \sqrt{(e+1)^2(\alpha+\beta)^2 + 4(e-1)((e+1)(\beta-\alpha) + e-1)}}.$$ Proof. We first note that $R = \Re_{\mathcal{CV}_{SG}}$ is the smallest positive root of the equation $((e+1)(\beta-\alpha)+e-1)r^2+(e+1)(\alpha+\beta)r-e+1=0$ in (0,1). Then it follows that $((e+1)(\beta-\alpha)+e-1)r^2+(e+1)(\alpha+\beta)r-e+1\leqslant 0$ for $0\leqslant r\leqslant R$ and so $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{e-1}{e+1} + 1 - \frac{1 + (\beta-\alpha-1)r^2}{1-r^2}, \quad r \leqslant R.$$ Consequently, for $0 \le r \le R = \Re_{\mathcal{CV}_{SG}}$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant \frac{e - 1}{e + 1} + 1 - a(r)$$ (3.5) where a(r) is given by (2.8). By [6, Lemma 2.2], we have $\{w : |w-a| < ((e-1)/(e+1)) + 1-a\} \subseteq \varphi_{SG}(\mathbb{D}) = \Omega_{SG}$ , for 2/(1+e) < a < 2e/(1+e). The condition a < 2e/(1+e) assures that the center lies inside the region $\Omega_{SG}$ . As the center a(r) is always greater than 1, the disc in (3.5) lies inside the region $\Omega_{SG}$ for $r \leq R$ proving that $\mathcal{CV}_{SG}$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least R. To prove the sharpness of the radius $R = \mathcal{R}_{\mathcal{CV}_{SG}}$ , consider the function $f_0$ defined by (3.2). For z = R in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R^2 + (\alpha + \beta)R}{1 - R^2} = \frac{2e}{e + 1} = \varphi_{SG}(1),$$ which proves that the radius R is sharp for the function $f_0$ defined by (3.2) (See Fig.3). $\square$  Figure 3. Sharpness of $\Re_{\mathcal{CV}_{SG}}$ radius when $\alpha = 0.2$ and $\beta = 0.9$

Corollary 3.6. Radius of convexity associated with the class $\mathcal{CV}_{SG}$ for some special cases: (1) The $\mathcal{CV}_{SG}$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_{SG}} = \frac{(e-1)}{(2-\sigma)(e+1) + \sqrt{(e+1)^2(2-\sigma)^2 + ((3e+1) - 2\sigma(e+1))(e-1)}}.$$ (2) The $\mathcal{CV}_{SG}$ radius for the class $V_k$ $$\mathcal{R}_{\mathcal{CV}_{SG}} = \frac{2(e-1)}{k(e+1) + \sqrt{(e+1)^2k^2 + 4(3e+1)(e-1)}}.$$ (3) The $\mathcal{CV}_{SG}$ radius for the class $\mathcal{SCC}(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_{SG}} = \frac{(e-1)}{(\alpha+1)(e+1) + \sqrt{(e+1)^2(\alpha+1)^2 + (3e+1)(e-1)}}.$$ Cho et al. [3] introduced the class that consists of convex functions associated with the sine function denoted by $\mathcal{CV}_{\sin} = \mathcal{CV}(\varphi_{\sin})$ , where $\varphi_{\sin}(z) = 1 + \sin z$ .

Theorem 3.7. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_{\sin}$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_{\sin}} = \frac{2(\sin 1)}{(\alpha + \beta) + \sqrt{(\alpha + \beta)^2 + 4(\beta - \alpha + \sin 1)(\sin 1)}}.$$ Proof. We first note that $R = \mathcal{R}_{\mathcal{CV}_{\sin}}$ is the smallest positive root of the equation $(\beta - \alpha + (\sin 1))r^2 + (\alpha + \beta)r - (\sin 1) = 0$ in the interval (0,1). Then it follows that $(\beta - \alpha + (\sin 1))r^2 + (\alpha + \beta)r - (\sin 1) \leq 0$ for $0 \leq r \leq R$ and so $$\frac{(\alpha + \beta)r}{1 - r^2} \leqslant (\sin 1) + 1 - \frac{1 + (\beta - \alpha - 1)r^2}{1 - r^2}, \quad r \leqslant R.$$ Consequently, for $0 \leqslant r \leqslant R = \Re_{\mathcal{CV}_{\sin}}$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant (\sin 1) + 1 - a(r) \tag{3.6}$$ where a(r) is given by (2.8). By [3, Lemma 3.3], we have $\{w : |w-a| < (\sin 1) + 1 - a\} \subseteq \varphi_{\sin}(\mathbb{D}) = \Omega_{\sin}$ , for $1 - \sin 1 < a < 1 + \sin 1$ . The condition $a < 1 + \sin 1$ assures that the center lies inside the region $\Omega_{\sin}$ . As the center a(r) is always greater than 1, the disc in (3.6) lies inside the region $\Omega_{\sin}$ for $r \leqslant R$ proving that $\mathcal{CV}_{\sin}$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least R. In order to prove the sharpness of the radius $R = \Re_{\mathcal{CV}_{\sin}}$ , we consider the function $f_0$ defined by (3.2). For z = R in (3.3),we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R^2 + (\alpha + \beta)R}{1 - R^2} = 1 + (\sin 1) = \varphi_{\sin}(1),$$ which proves that the radius R is sharp for the function $f_0$ defined by (3.2) (See Fig.4). $\square$  Figure 4. Sharpness of $\Re_{\mathcal{CV}_{\sin}}$ radius when $\alpha = 0.1$ and $\beta = 0.9$

Corollary 3.8. Radius of convexity associated with the class $even{equation} vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex v$ (1) The $\mathcal{CV}_{\sin}$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_{\sin}} = \frac{\sin 1}{(2-\sigma) + \sqrt{(2-\sigma)^2 + (2-2\sigma + \sin 1)(\sin 1)}}.$$ (2) The $\mathcal{CV}_{sin}$ radius for the class $V_k$ is $$\mathcal{R}_{\mathcal{CV}_{\sin}} = \frac{2\sin 1}{k + \sqrt{k^2 + 4(2 + \sin 1)(\sin 1)}}.$$ (3) The $\mathcal{CV}_{\sin}$ radius for the class $SCC(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_{\sin}} = \frac{\sin 1}{(\alpha+1) + \sqrt{(\alpha+1)^2 + (2+\sin 1)(\sin 1)}}.$$ Sokól and Stankiewicz [26] introduced the class of convex functions associated with a lemniscate domain, given by $\mathcal{CV}_L = \mathcal{CV}(\varphi_L)$ , where $\varphi_L(z) = \sqrt{1+z}$ consisting of functions $f \in \mathcal{A}$ such that for each $z \in \mathbb{D}$ , w = 1 + (zf''(z)/f'(z)) lies in the region bounded by right half of the lemniscate of Bernoulli given by $|w^2 - 1| < 1$ .

Theorem 3.9. The radius of lemniscate convexity for the functions whose derivative belongs to $K(\alpha, \beta)$ , $0 \le \alpha \le \beta$ is given by $$\mathcal{R}_{\mathcal{CV}_L} = \frac{2(\sqrt{2}-1)}{(\alpha+\beta) + \sqrt{(\alpha+\beta)^2 + 4(\sqrt{2}-1)(\sqrt{2}+(\beta-\alpha-1))}}.$$ Proof. Firstly, we note that $R = \mathcal{R}_{\text{CV}_L}$ is the smallest positive root of the equation $(\beta - \alpha + \sqrt{2} - 1)r^2 + (\alpha + \beta)r - (\sqrt{2} - 1) = 0$ in the interval (0, 1). Then it follows that $(\beta - \alpha + \sqrt{2} - 1)r^2 + (\alpha + \beta)r - (\sqrt{2} - 1) \leqslant 0$ for $0 \leqslant r \leqslant R$ and so $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \sqrt{2} - \frac{1+(\beta-\alpha-1)r^2}{1-r^2}, \quad r \leqslant R.$$ Consequently, for $0 \le r \le R = \Re_{\mathcal{CV}_L}$ , the disc in (2.7) becomes $$\left|1 + \frac{zf''(z)}{f'(z)} - a(r)\right| \leqslant \sqrt{2} - a(r) \tag{3.7}$$ where a(r) is given by (2.8). By [1, Lemma 2.2], we have $\{w: |w-a| < \sqrt{2} - a\} \subseteq \varphi_L(\mathbb{D}) = \Omega_L$ for $(2\sqrt{2})/3 < a < \sqrt{2}$ . The condition $a < \sqrt{2}$ ensures that the center lies inside the region $\Omega_L$ . As the center a(r) is always greater than 1, the disc in (3.7) lies inside the region $\Omega_L$ for $r \leq R$ proving that $\mathcal{CV}_L$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_3$ . To show the sharpness of the radius $R = \Re_{\mathcal{CV}_L}$ , we consider the function $f_0$ defined by (3.2). For z = R in (3.3),we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R^2 + (\alpha + \beta)R}{1 - R^2} = \sqrt{2} = \varphi_L(1),$$ which proves that the radius R is sharp for the function $f_0$ defined by (3.2) (See Fig.5). $\square$  Figure 5. Sharpness of $\Re_{\mathcal{CV}_L}$ radius when $\alpha = 2$ and $\beta = 3$

Corollary 3.10. Radius of convexity associated with the class $\mathfrak{CV}_L$ for some special cases: (1) The $\mathcal{CV}_L$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\Re_{\mathcal{CV}_L} = \frac{(\sqrt{2} - 1)}{(2 - \sigma) + \sqrt{(2 - \sigma)^2 - 2(\sqrt{2} - 1)\sigma + 1}}.$$ (2) The $\mathcal{CV}_L$ radius for the class $V_k$ is $$\mathcal{R}_{\mathcal{CV}_L} = \frac{2(\sqrt{2} - 1)}{k + \sqrt{k^2 + 4}}.$$ (3) The $\mathcal{CV}_L$ radius for the class $\mathcal{SCC}(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_L} = \frac{(\sqrt{2} - 1)}{(\alpha + 1) + \sqrt{(\alpha + 1)^2 + 1}}.$$ The class $\mathcal{CV}_R = \mathcal{CV}(\varphi_R)$ consisting of convex functions associated with the rational function $\varphi_R(z) = 1 + ((z^2 + kz)/k^2 - kz)$ for $k = \sqrt{2} + 1$ , was introduced by Kumar and Ravichandran [10].

Theorem 3.11. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_R$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_R} = \begin{cases} R_2, & R_2 \leqslant R_1 \\ R_3, & R_2 \geqslant R_1 \end{cases}$$ where $$R_{1} = \left(\frac{\sqrt{2} - 1}{\beta - \alpha + \sqrt{2} - 1}\right)^{1/2},$$ $$R_{2} = \frac{2(3 - 2\sqrt{2})}{(\alpha + \beta) + \sqrt{(\alpha + \beta)^{2} + 4(\beta - \alpha + 2\sqrt{2} - 3)(2\sqrt{2} - 3)}}$$ and $$R_3 = \frac{2}{(\alpha+\beta) + \sqrt{(\alpha+\beta)^2 + 4(\beta-\alpha+1)}}.$$ Proof. We first consider the case where $R_2 \leq R_1$ . Note that $R_1$ is the positive root of the equation $(\beta - \alpha + \sqrt{2} - 1)r^2 + 1 - \sqrt{2} = 0$ or equivalently $a(r) = \sqrt{2}$ where a(r) given by (2.8). For $r \leq R_2 \leq R_1$ , we see that the increasing function a(r) given by (2.8) satisfies $$2(\sqrt{2}-1) < 1 \leqslant a(r) \leqslant a(R_2) = \frac{1+(\beta-\alpha-1)R_2^2}{1-R_2^2} \leqslant a(R_1) = \sqrt{2}.$$ Since $R_2$ is the smallest positive root of the equation $(\beta - \alpha + 2\sqrt{2} - 3)r^2 - (\alpha + \beta)r - 2\sqrt{2} + 3 = 0$ in the interval (0,1), it follows that $(\beta - \alpha + 2\sqrt{2} - 3)r^2 - (\alpha + \beta)r - 2\sqrt{2} + 3 \le 0$ for $0 \le r \le R_2$ , and hence, upon rewriting, we have $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{1+(\beta-\alpha-1)r^2}{1-r^2} - 2(\sqrt{2}-1), \quad r \leqslant R_2.$$ Therefore, for $r \leq R_2$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le a(r) - 2(\sqrt{2} - 1) \tag{3.8}$$ where a(r) is given by (2.8). For $2(\sqrt{2}-1) < a < \sqrt{2}$ , [10, Lemma 2.2] establishes that $$\{w \in \mathbb{C} : |w - a| < a - 2(\sqrt{2} - 1)\} \subseteq \varphi_R(\mathbb{D}) = \Omega_R. \tag{3.9}$$ Using the inclusion in (3.9), it follows that the disc in (3.8) is contained in $\Omega_R$ , proving that the $\mathcal{CV}_R$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_2$ . To prove the sharpness of $R_2$ , we consider the function $f_0$ defined by (3.2). For $z = -R_2$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_2^2 - (\alpha + \beta)R_2}{1 - R_2^2} = 2\sqrt{2} - 2 = \varphi_R(-1).$$ which proves that the radius $R_2$ is sharp for the function $f_0$ defined by (3.2) (See Fig.6(a)). We now consider the case where $R_1 \leq R_2$ . Clearly, $R_3$ is the smallest positive root of the equation $(\beta - \alpha + 1)r^2 - (\alpha + \beta)r - 1 = 0$ in the interval (0, 1) and hence $(\beta - \alpha + 1)r^2 - (\alpha + \beta)r - 1 \leq 0$ for $0 \leq r \leq R_3$ . Rewriting this, we get $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant 2 - \frac{1 + (\beta - \alpha - 1)r^2}{1-r^2}, \quad r \leqslant R_3.$$ Therefore, the disc in (2.7) becomes $$\left|1 + \frac{zf''(z)}{f'(z)} - a(r)\right| \le 2 - a(r)$$ (3.10) where a(r) is given by (2.8). For $\sqrt{2} < a < 2$ , [10, Lemma 2.2] establishes the containment: $$\{w \in \mathbb{C} : |w - a| < 2 - a\} \subseteq \varphi_R(\mathbb{D}) = \Omega_R. \tag{3.11}$$ It is easy to see that $R_2 \leqslant R_3$ . Since $R_1 \leqslant R_2$ , it follows that $R_1 \leqslant R_3$ . For a(r) given by (2.8), we have $$\sqrt{2} = a(R_1) \leqslant a(r) \leqslant a(R_3) = \frac{1 + (\beta - \alpha - 1)R_3^2}{1 - R_3^2} < 2.$$ Using the inclusion in (3.11), it follows that the disc in (3.10) lies inside $\Omega_R$ proving that the $\mathcal{CV}_R$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_3$ . Consider the function $f_0$ defined by (3.2). For $z = R_3$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_3^2 + (\alpha + \beta)R_3}{1 - R_3^2} = 2 = \varphi_R(1),$$ which proves the sharpness of the radius $R_3$ for the function $f_0$ defined by (3.2) (See Fig.6(b)).  (a). Sharpness of $\Re_{\mathfrak{CV}_R} = R_2$ radius when $\alpha = 0.015$ and $\beta = 0.35$  (b). Sharpness of $\mathcal{R}_{\mathfrak{CV}_R}=R_3$ radius when $\alpha=0.002$ and $\beta=0.05$ Figure 6. Sharpness $\mathcal{R}_{\mathcal{CV}_R}$ radius

Corollary 3.12. Radius of convexity associated with the class $\mathfrak{CV}_R$ for some special cases: (1) The $\mathcal{CV}_R$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\Re_{\mathcal{CV}_R} = \frac{3 - 2\sqrt{2}}{2(2 - \sigma) + \sqrt{4((2 - \sigma)^2 + (6 - 4\sqrt{2})\sigma - 8\sqrt{2} + 11)}}.$$ (2) The $\mathcal{CV}_R$ radius for the class $V_k$ is $$\Re_{\mathcal{CV}_R} = \frac{3 - 2\sqrt{2}}{k + \sqrt{k^2 + 44 - 32\sqrt{2}}}.$$ (3) The $\mathcal{CV}_R$ radius for the class $\mathcal{SCC}(\alpha)$ is $$\Re_{\mathcal{CV}_R} = \frac{3 - 2\sqrt{2}}{2(\alpha + 1) + \sqrt{4((\alpha + 1)^2 + 44 - 32\sqrt{2})}}.$$ Raina and Sokól [19] studied the class $\mathcal{CV}_{\mathbb{Q}} = \mathcal{CV}(\varphi_{\mathbb{Q}})$ , where $\varphi_{\mathbb{Q}}(z) = z + \sqrt{1 + z^2}$ and proved that $f \in \mathcal{CV}_{\mathbb{Q}}$ if and only if $1 + (zf''(z)/f'(z)) \in \Omega_{\mathbb{Q}}$ , where $\Omega_{\mathbb{Q}}$ is the interior of a lune given by $\Omega_{\mathbb{Q}} := \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$ .

Theorem 3.13. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_{\mathbb{C}}$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_{\mathcal{C}}} = \begin{cases} R_2, & R_2 \leqslant R_1 \\ R_3, & R_2 \geqslant R_1 \end{cases}$$ where $$R_1 = \left(\frac{\sqrt{2} - 1}{\beta - \alpha + \sqrt{2} - 1}\right)^{1/2}, \quad R_2 = \frac{4 - 2\sqrt{2}}{(\alpha + \beta) + \sqrt{(\alpha + \beta)^2 + 4(\sqrt{2} - 2)(\beta - \alpha + \sqrt{2} - 2)}}$$ and $$R_3 = \frac{2\sqrt{2}}{(\alpha+\beta) + \sqrt{(\alpha+\beta)^2 + 4\sqrt{2}(\beta-\alpha+\sqrt{2})}}.$$ Proof. Firstly, we consider the case where $R_2 \leq R_1$ . Note that $R_1$ is the positive root of the equation $(\beta - \alpha + \sqrt{2} - 1)r^2 + 1 - \sqrt{2} = 0$ or equivalently $a(r) = \sqrt{2}$ where a(r) is given by (2.8). For $r \leq R_2 \leq R_1$ , we note that the increasing function a(r) given by (2.8) satisfies $$(\sqrt{2}-1) < 1 \le a(r) \le a(R_2) = \frac{1+(\beta-\alpha-1)R_2^2}{1-R_2^2} \le a(R_1) = \sqrt{2}.$$ Since $R_2$ is the smallest positive root of the equation $(\beta - \alpha + \sqrt{2} - 2)r^2 - (\alpha + \beta)r - \sqrt{2} + 2 = 0$ in the interval (0,1), it follows that $(\beta - \alpha + \sqrt{2} - 2)r^2 - (\alpha + \beta)r - \sqrt{2} + 2 \le 0$ for $0 \le r \le R_2$ and hence upon rewriting, we have $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{1+(\beta-\alpha-1)r^2}{1-r^2} - (\sqrt{2}-1), \quad r \leqslant R_2.$$ Therefore, for $r \leq R_2$ the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant a(r) - (\sqrt{2} - 1). \tag{3.12}$$ where a(r) is given by (2.8). For $2(\sqrt{2}-1) < a < \sqrt{2}$ , [4, Lemma 2.1] provides the containment $$\{w \in \mathbb{C} : |w - a| < a - (\sqrt{2} - 1)\} \subseteq \varphi_{\mathcal{C}}(\mathbb{D}) = \Omega_{\mathcal{C}}. \tag{3.13}$$ Using the inclusion in (3.13), it follows that the disc in (3.12) lies within $\Omega_{\mathbb{C}}$ proving that the $\mathcal{CV}_{\mathbb{C}}$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_2$ . To examine the sharpness, we consider the function $f_0$ defined by (3.2). For $z = -R_2$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_2^2 - (\alpha + \beta)R_2}{1 - R_2^2} = \sqrt{2} - 1 = \varphi_{\mathcal{C}}(-1),$$ this proves the sharpness of the radius $R_2$ for the function $f_0$ defined by (3.2) (See Fig.7(a)). We now consider the case where $R_1 \leq R_2$ . Clearly, $R_3$ is the smallest positive root of the equation $(\beta - \alpha + 1)r^2 - (\alpha + \beta)r - 1 = 0$ in the interval (0,1) and hence $(\beta - \alpha + 1)r^2 - (\alpha + \beta)r - 1 \leq 0$ for $0 \leq r \leq R_3$ . Rewriting this, we get $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \sqrt{2} + 1 - \frac{1 + (\beta - \alpha - 1)r^2}{1-r^2}, \quad r \leqslant R_3.$$ Therefore, for $r \leq R_3$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant \sqrt{2} + 1 - a(r)$$ (3.14) where a(r) is given by (2.8). For $\sqrt{2} < a < \sqrt{2} + 1$ , [4, Lemma 2.1] establishes the inclusion: $$\{w \in \mathbb{C} : |w - a| < \sqrt{2} + 1 - a\} \subseteq \varphi_{\mathbb{C}}(\mathbb{D}) = \Omega_{\mathbb{C}}. \tag{3.15}$$ It is easy to that $R_2 \leqslant R_3$ . Since $R_1 \leqslant R_2$ , it follows that $R_1 \leqslant R_3$ . For a(r) given by (2.8), we have $$\sqrt{2} = a(R_1) \leqslant a(r) \leqslant a(R_3) = \frac{1 + (\beta - \alpha - 1)R_3^2}{1 - R_3^2} < \sqrt{2} + 1.$$ Using the inclusion (3.15), it follows that the disc in (3.14) lies inside $\Omega_{\mathbb{C}}$ proving that the $\mathcal{CV}_{\mathbb{C}}$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_3$ . Consider the function $f_0$ defined by (3.2) for proving the sharpness of the radius $R_3$ . For $z = R_3$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_3^2 + (\alpha + \beta)R_3}{1 - R_3^2} = \sqrt{2} + 1 = \varphi_{\mathbb{Q}}(1),$$ this establishes the sharpness of the radius $R_3$ for the function $f_0$ defined by (3.2) (See Fig.7(b)).  (a). Sharpness of $\Re_{\mathbb{CV}_{\mathbb{Q}}} = R_2$ radius when $\alpha = 0.1$ and $\beta = 2$  (b). Sharpness of $\Re ev_{\mathbb{Q}} = R_3$ radius when $\alpha = 0.015$ and $\beta = 0.4$ Figure 7. Sharpness of $\mathcal{R}_{\mathcal{CV}_{\mathcal{I}}}$ radius

Corollary 3.14. Radius of convexity associated with the class $ev_{\ell}$ for some special cases: (1) The $\mathfrak{CV}_{\mathfrak{C}}$ radius for the class $\mathfrak{CC}(\sigma,0)$ is $$\Re_{\operatorname{CV}_{\operatorname{\mathbb{C}}}} = \frac{(2 - \sqrt{2})}{(2 - \sigma) + \sqrt{(2 - \sigma)^2 + (\sqrt{2} - 2\sigma)(\sqrt{2} - 2)}}.$$ (2) The $ev_{\mathcal{C}}$ radius for the class $V_k$ is $$\Re_{\text{CV}_{\text{C}}} = \frac{2(2 - \sqrt{2})}{k + \sqrt{k^2 + 8(1 - \sqrt{2})}}.$$ (3) The $\mathfrak{CV}_{\mathfrak{C}}$ radius for the class $\mathfrak{SCC}(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_{\mathbb{Q}}} = \frac{(2 - \sqrt{2})}{(\alpha + 1) + \sqrt{(\alpha + 1)^2 + 2(2 - \sqrt{2})}}.$$ Sharma et al. [24] studied the class $\mathcal{CV}_C = \mathcal{CV}(\varphi_C)$ which consists of convex functions associated with a cardiod, where $\varphi_C(z) = 1 + (4/3)z + (2/3)z^2$ .

Theorem 3.15. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_C$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_C} = \begin{cases} R_2, & R_2 \leqslant R_1 \\ R_3, & R_2 \geqslant R_1 \end{cases}$$ where $$R_{1} = \left(\frac{2}{3(\beta - \alpha) + 2}\right)^{1/2},$$ $$R_{2} = \frac{4}{3(\alpha + \beta) + \sqrt{9(\alpha + \beta)^{2} - 8(3(\beta - \alpha) - 2)}}$$ and $$R_3 = \frac{4}{(\alpha+\beta) - \sqrt{(\alpha+\beta)^2 + 8(\beta-\alpha+2)}}.$$ Proof. We first consider the case where $R_2 \leq R_1$ . Note that $R_1$ is the positive root of the equation $(3(\beta - \alpha) + 2)r^2 - 2 = 0$ which is equivalent to setting a(r) = 5/3 where a(r) is given by (2.8). For $r \leq R_2 \leq R_1$ , we see that the increasing function a(r) given by (2.8) satisfies $$\frac{1}{3} < 1 \leqslant a(r) \leqslant a(R_2) \leqslant \frac{1 + (\beta - \alpha - 1)R_2^2}{1 - R_2^2} \leqslant a(R_1) = \frac{5}{3}.$$ Since $R_2$ is the smallest positive root of the equation $(3(\beta - \alpha) - 2)r^2 - 3(\alpha + \beta)r + 2 = 0$ in the interval (0, 1), it follows that $(3(\beta - \alpha) - 2)r^2 - 3(\alpha + \beta)r + 2 \leq 0$ for $0 \leq r \leq R_2$ and hence upon rewriting, we have $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{1+(\beta-\alpha-1)r^2}{1-r^2} - \frac{1}{3}, \quad r \leqslant R_2.$$ Therefore, for $r \leq R_2$ , the disc in (2.7) becomes $$\left|1 + \frac{zf''(z)}{f'(z)} - a(r)\right| \le a(r) - \frac{1}{3}$$ (3.16) where a(r) is given by (2.8). For 1/3 < a < 5/3, [24, Lemma 2.5] provides the inclusion: $$\{w \in \mathbb{C} : |w - a| < a - 1/3\} \subseteq \varphi_C(\mathbb{D}) = \Omega_C. \tag{3.17}$$ Using the inclusion (3.17), it follows that the disc in (3.16) resides inside $\Omega_C$ proving that the $\mathcal{CV}_C$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_2$ . The function $f_0$ defined by (3.2) proves the sharpness of the radius $R_2$ . For $z = -R_2$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_2^2 - (\alpha + \beta)R_2}{1 - R_2^2} = \frac{1}{3} = \varphi_C(-1),$$ which proves that the radius $R_2$ is sharp for the function $f_0$ defined by (3.2) (See Fig.8(a)). We now consider the case where $R_1 \leq R_2$ . Clearly, $R_3$ is the smallest positive root of the equation $(\beta - \alpha + 2)r^2 + (\alpha + \beta)r - 2 = 0$ in the interval (0, 1) and hence $(\beta - \alpha + 2)r^2 + (\alpha + \beta)r - 2 \leq 0$ for $0 \leq r \leq R_3$ . Upon rewriting, we get $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant 3 - \frac{1 + (\beta-\alpha-1)r^2}{1-r^2}, \quad r \leqslant R_3.$$ Therefore, the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le 3 - a(r)$$ (3.18) where a(r) is given by (2.8). For 5/3 < a < 3, by [24, Lemma 2.5] the following inclusion relation holds: $$\{w \in \mathbb{C} : |w - a| < 3 - a\} \subseteq \varphi_C(\mathbb{D}) = \Omega_C. \tag{3.19}$$ It is easy to see that $R_2 \leq R_3$ . Since $R_1 \leq R_2$ , it follows that $R_1 \leq R_3$ . For a(r) given by (2.8), we have $$\frac{5}{3} = a(R_1) \leqslant a(r) \leqslant a(R_3) = \frac{1 + (\beta - \alpha - 1)R_3^2}{1 - R_3^2} < 3.$$ Using the inclusion in (3.19), it follows that the disc in (3.18) lies inside $\Omega_C$ proving that the $\mathcal{CV}_C$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ is at least $R_3$ . To prove the sharpness, we consider the function $f_0$ defined by (3.2). For $z = R_3$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_3^2 + (\alpha + \beta)R_3}{1 - R_3^2} = 3 = \varphi_C(1).$$ which proves the sharpness of the radius $R_3$ for the function $f_0$ defined by (3.2) (See Fig.8(b)).  (a). Sharpness of $\Re_{\mathcal{CV}_C} = R_2$ radius when $\alpha = 0.3$ and $\beta = 2$  (b). Sharpness of $\Re_{\mathcal{CV}_C} = R_3$ radius when $\alpha = 0.02$ and $\beta = 0.5$ Figure 8. Sharpness of $\mathcal{R}_{\mathcal{CV}_C}$ radius

Corollary 3.16. Radius of convexity associated with the class $\text{CV}_C$ for some special cases: (1) The $\mathcal{CV}_C$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_C} = \frac{2}{3(2-\sigma) + \sqrt{9\sigma^2 - 24\sigma + 28}}.$$ (2) The $\mathcal{CV}_C$ radius for the class $V_k$ is $$\mathcal{R}_{\mathcal{CV}_C} = \frac{4}{3k + \sqrt{9k^2 - 32}}.$$ (3) The $\mathcal{CV}_C$ radius for the class $SCC(\alpha)$ is $$\mathcal{R}_{\mathcal{CV}_C} = \frac{2}{3(\alpha+1) + \sqrt{9(\alpha+1)^2 - 8}}.$$ Mendiratta et al. [16], introduced the class $\mathcal{CV}_e = \mathcal{CV}(\varphi_e)$ where $\varphi_e(z) = e^z$ , which consists of all functions $f \in \mathcal{A}$ such that $1 + (zf''(z)/f'(z)) \prec e^z$ or equivalently $|\log(1 + (zf''(z)/f'(z)))| < 1$ .

Theorem 3.17. Let $0 \le \alpha \le \beta$ . The sharp $\mathfrak{CV}_e$ radius for functions f such that $f' \in \mathfrak{K}(\alpha,\beta)$ is given by $$\mathcal{R}_{\mathcal{CV}_e} = \begin{cases} R_2, & R_2 \leqslant R_1 \ R_3, & R_2 \geqslant R_1 \end{cases}$$ where $$R_1 = \left(\frac{e + e^{-1} - 2}{2(\beta - \alpha) + e + e^{-1} - 2}\right)^{1/2},$$ $$R_2 = \frac{2(e - 1)}{e(\alpha + \beta) + \sqrt{e^2(\alpha + \beta)^2 - 4(e(\beta - \alpha) - (e - 1))(e - 1)}}$$ and $$R_3 = \frac{2(e-1)}{(\alpha+\beta) + \sqrt{(\alpha+\beta)^2 + 4(\beta-\alpha+e-1)(e-1)}}.$$ Proof. We first consider the case where $R_2 \leq R_1$ . Note that $R_1$ is the positive root of the equation $(2(\beta - \alpha) + e + e^{-1} - 2)r^2 - (e + e^{-1} - 2) = 0$ which is equivalent to setting $a(r) = (e + e^{-1})/2$ where a(r) is given by (2.8). For $r \leq R_2 \leq R_1$ , we see that the increasing function a(r) satisfies $$\frac{1}{e} < 1 \leqslant a(r) \leqslant a(R_2) = \frac{1 + (\beta - \alpha - 1)R_2^2}{1 - R_2^2} \leqslant a(R_1) = \frac{e + e^{-1}}{2}.$$ Since $R_2$ is the smallest positive root of the equation $(e(\beta-\alpha)-e+1)r^2-e(\alpha+\beta)r+e-1=0$ in the interval (0,1), it follows that $(e(\beta-\alpha)-e+1)r^2-e(\alpha+\beta)r+e-1 \le 0$ for $0 \le r \le R_2$ , and hence upon rewriting, we have $$\frac{(\alpha+\beta)r}{1-r^2} \le \frac{1+(\beta-\alpha-1)r^2}{1-r^2} - \frac{1}{e}, \quad r \le R_2.$$ Therefore, $r \leq R_2$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le a(r) - \frac{1}{e}$$ (3.20) where a(r) is given by (2.8). For $1/e < a \le (e + e^{-1})/2$ , [16, Lemma 2.5] provides the inclusion: $$\{w \in \mathbb{C} : |w - a| < a - 1/e\} \subseteq \varphi_e(\mathbb{D}) = \Omega_e. \tag{3.21}$$ Using the inclusion (3.21), it follows that the disc in (3.20) resides inside $\Omega_e$ proving that the $\mathcal{CV}_e$ radius for functions whose derivatives belongs to $\mathcal{K}(\alpha,\beta)$ is at least $R_2$ . In order to prove the sharpness consider the function $f_0$ defined by (3.2). For $z = -R_2$ in (3.3), we have $$\left|\log\left(1+\frac{zf_0''(z)}{f_0'(z)}\right)\right| = \left|\log\left(\frac{1+(\beta-\alpha-1)R_2^2-(\alpha+\beta)R_2}{1-R_2^2}\right)\right| = \left|\log\left(\frac{1}{e}\right)\right| = 1,$$ which proves the sharpness of the radius $R_3$ for the function $f_0$ defined by (3.2) (See Fig.9(a)).    (b). Sharpness of $\Re_{\mathrm{CV}_e} = R_3$ radius when $\alpha = 0.03$ and $\beta = 0.5$ Figure 9. Sharpness of $\mathcal{R}_{\mathcal{CV}_e}$ radius We next consider the case where $R_1 \leq R_2$ . Clearly, $R_3$ is the smallest positive root of the equation $(\beta - \alpha + e - 1)r^2 + (\alpha + \beta)r - (e - 1) = 0$ in the interval (0, 1) and hence $(\beta - \alpha + e - 1)r^2 + (\alpha + \beta)r - (e - 1) \leq 0$ for $0 \leq r \leq R_3$ . Rewriting this, we get $$\frac{(\alpha + \beta)r}{1 - r^2} \le e - \frac{1 + (\beta - \alpha - 1)r^2}{1 - r^2}, \quad r \le R_3.$$ Therefore, the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \le e - a(r).$$ (3.22) where a(r) is given by (2.8). By [16, Lemma 2.5], for $(e + e^{-1})/2 < a < e$ , the following inclusion relation holds: $$\{w \in \mathbb{C} : |w - a| < e - a\} \subseteq \varphi_e(\mathbb{D}) = \Omega_e. \tag{3.23}$$ It is easy to see that $R_2 \leq R_3$ . Since $R_1 \leq R_2$ , it follows that $R_1 \leq R_3$ . For a(r) given by (2.8), we have $$\frac{e+e^{-1}}{2} = a(R_1) \leqslant a(r) \leqslant a(R_3) = \frac{1+(\beta-\alpha-1)R_3^2}{1-R_2^2} < e.$$ Using the inclusion in (3.23), it follows that the disc in (3.22) lies inside $\Omega_e$ proving that the $\mathcal{CV}_e$ radius is for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ at least $R_3$ . To prove the sharpness, we consider the function $f_0$ defined by (3.2). For $z = R_3$ in (3.3), we have $$\left| \log \left( 1 + \frac{z f_0''(z)}{f_0'(z)} \right) \right| = \left| \log \left( \frac{1 + (\beta - \alpha - 1)R_3^2 + (\alpha + \beta)R_3}{1 - R_3^2} \right) \right| = \left| \log e \right| = 1.$$ which proves the sharpness of the radius $R_3$ for the function $f_0$ defined by (3.2) (See Fig.9(b)).

Corollary 3.18. Radius of convexity associated with the class $even{equation} vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex vertex$ (1) The $\mathcal{CV}_e$ radius for the class $\mathcal{CC}(\sigma,0)$ is $$\mathcal{R}_{\mathcal{CV}_e} = \frac{e-1}{e(2-\sigma)+\sqrt{e^2(2-\sigma)^2-(e(1-2\sigma)+1)(e-1)}}.$$ (2) The $\mathcal{CV}_e$ radius for the class $V_k$ is $$\Re_{\mathcal{CV}_e} = \frac{2(e-1)}{ek + \sqrt{e^2(k^2 - 4) + 4}}.$$ (3) The $\mathcal{CV}_e$ radius for the class $SCC(\alpha)$ is $$\Re_{\mathcal{CV}_e} = \frac{e - 1}{e(\alpha + 1) + \sqrt{e^2(\alpha + 1)^2 - (e^2 - 1)}}.$$ The class uniformly convex functions was introduced by Goodman [7]. A function $f \in \mathcal{A}$ is said to be uniformly convex if and only if $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > \left|\frac{zf''(z)}{f'(z)}\right|$$ for all $z \in \mathbb{D}$ and the class of all uniformly convex functions is denoted by $\mathcal{UCV}$ . The following theorem was proved earlier by Ravichandran et al. [20] for the case $R_2 \leq R_1$ . However, the proof given here is different. The result in the case $R_2 > R_1$ is not necessarily sharp.

Theorem 3.19. Let $0 < \gamma \le 1$ and $0 \le \alpha \le \beta$ . The radius of uniform convexity for functions whose derivative belongs to $\mathcal{K}(\alpha, \beta)$ is given by $\mathcal{R}_{UCV} = R_2$ if $R_2 \le R_1$ and $\mathcal{R}_{UCV} \ge R_3$ , if $R_2 \ge R_1$ where $$R_{1} = \left(\frac{1}{2(\beta - \alpha) + 1}\right)^{1/2},$$ $$R_{2} = \frac{1}{(\alpha + \beta) + \sqrt{(\alpha + \beta)^{2} - (2(\beta - \alpha) - 1)}}$$ and $$R_3 = \left(1 - \frac{(\alpha + \beta)^2}{2(\beta - \alpha)}\right)^{1/2}.$$ Proof. We first consider the case where $R_2 \leq R_1$ . Note that $R_1$ is the positive root of the equation $(2(\beta - \alpha) + 1)r^2 - 1 = 0$ which is equivalent to setting a(r) = 3/2 where a(r) is given by (2.8). For $r \leq R_2 \leq R_1$ , we see that the increasing function a(r) satisfies $$\frac{1}{2} < 1 \leqslant a(r) \leqslant a(R_2) = \frac{1 + (\beta - \alpha - 1)R_2^2}{1 - R_2^2} \leqslant a(R_1) = \frac{3}{2}.$$ Since $R_2$ is the smallest positive root of the equation $(2(\beta - \alpha) - 1)r^2 - 2(\alpha + \beta)r + 1 = 0$ in the interval (0, 1), it follows that $(2(\beta - \alpha) - 1)r^2 - 2(\alpha + \beta)r + 1 \leq 0$ for $0 \leq r \leq R_2$ , and hence upon rewriting, we have $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \frac{1+(\beta-\alpha-1)r^2}{1-r^2} - \frac{1}{2}, \quad r \leqslant R_2.$$ Therefore, $r \leq R_2$ , the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant a(r) - \frac{1}{2}$$ (3.24) where a(r) is given by (2.8). For $1/2 < a \le 3/2$ , [23, Lemma 1] provides the inclusion: $$\{w \in \mathbb{C} : |w - a| < a - 1/2\} \subseteq \varphi(\mathbb{D}) = \Omega \tag{3.25}$$ where $\Omega$ is a parabolic region symmetric with respect to the real axis and (1/2,0) is its vertex. Using the inclusion (3.25), it follows that the disc in (3.24) resides inside $\Omega$ proving that the UCV radius for functions whose derivatives belongs to $K(\alpha, \beta)$ is at least $R_2$ . In order to prove the sharpness consider the function $f_0$ defined by (3.2). For $z = -R_2$ in (3.3), we have $$1 + \frac{zf_0''(z)}{f_0'(z)} = \frac{1 + (\beta - \alpha - 1)R_2^2 - (\alpha + \beta)R_2}{1 - R_2^2} = \frac{1}{2},$$ which proves the sharpness of the radius $R_2$ for the function $f_0$ defined by (3.2). We next consider the case where $R_1 \leq R_2$ . Clearly, $R_3$ is the smallest positive root of the equation $2(\beta - \alpha)r^2 + (\alpha + \beta)^2 - 2(\beta - \alpha) = 0$ in the interval (0,1) and hence $2(\beta - \alpha)r^2 + (\alpha + \beta)^2 - 2(\beta - \alpha) \leq 0$ for $0 \leq r \leq R_3$ . Rewriting this, we get $$\frac{(\alpha+\beta)r}{1-r^2} \leqslant \left(\frac{2(\beta-\alpha)r^2}{1-r^2}\right)^{1/2}, \quad r \leqslant R_3.$$ Therefore, the disc in (2.7) becomes $$\left| 1 + \frac{zf''(z)}{f'(z)} - a(r) \right| \leqslant (2a(r) - 2)^{1/2}. \tag{3.26}$$ where a(r) is given by (2.8). By [23, Lemma 1], for $3/2 \le a$ , the following inclusion relation holds: $$\{w \in \mathbb{C} : |w - a| < \sqrt{2a - 2}\} \subseteq \varphi(\mathbb{D}) = \Omega. \tag{3.27}$$ It is easy to see that $R_2 \leqslant R_3$ . Since $R_1 \leqslant R_2$ , it follows that $R_1 \leqslant R_3$ . For a(r) given by (2.8), we have $$\frac{3}{2} = a(R_1) \leqslant a(r) \leqslant a(R_3) = \frac{1 + (\beta - \alpha - 1)R_3^2}{1 - R_2^2}.$$ Using the inclusion in (3.27), it follows that the disc in (3.26) lies inside $\Omega$ proving that the UCV radius is for functions whose derivatives belongs to $\mathcal{K}(\alpha, \beta)$ at least $R_3$ . A function $f \in \mathcal{A}$ is said to be strongly convex of order $\gamma$ , $0 < \gamma \leq 1$ , if $$\left| \arg \left( 1 + \frac{zf''(z)}{f'(z)} \right) \right| \leqslant \frac{\pi \gamma}{2}.$$ The set of all such functions is denoted by $SeV(\gamma)$ . We state the following theorem without proof. Finding the sharp radius constant is open.

Theorem 3.20. Let $0 < \gamma \le 1$ and $0 \le \alpha \le \beta$ . The radius of strong convexity of order $\gamma$ for functions whose derivative belongs to $\mathcal{K}(\alpha, \beta)$ is given by $$\Re_{\text{SCV}(\gamma)} \geqslant \frac{2\sin\frac{\pi\gamma}{2}}{(\alpha+\beta) + \sqrt{(\alpha+\beta)^2 - 4(\beta-\alpha-1)(\sin\frac{\pi\gamma}{2})^2}}.$$