Hadamard product of analytic functions and some special regions and curves

Lemma 2.1. The function $F_s$ given by (2.1) is a starlike univalent function for $s \in [0,1]$ . Proof We have $$\mathfrak{Re}\left\{\frac{zF_s'(z)}{F_s(z)}\right\} = \mathfrak{Re}\left\{\frac{1+sz^2}{1-sz^2}\right\} > 0 \quad \textit{for all } z \in \Delta.$$ Thus, it is obvious that $F_s$ is a starlike univalent function for $s \in [0,1]$ .

Lemma 2.2. Let the function $F_s$ be of the form (2.1), and let $\mathcal{K}(-1/2)$ be given by (1.4). If $|s| \le 6 - \sqrt{33} \approx 0.255437$ , then $F_s \in \mathcal{K}(-1/2)$ . Proof Assume that $sz^2 = r(\cos t + i\sin t)$ , $0 \le r < 1$ , $0 \le t \le 2\pi$ and set $$H(t,r) := \frac{zF_s''(z)}{F_s'(z)} + 2 = \frac{6(1+sz^2) - 4}{1 - (sz^2)^2} = \frac{2 + 6r\cos t + 6ir\sin t}{1 - r^2\cos 2t - ir^2\sin 2t}.$$ (2.2) By (1.4) if $\Re\{H(t,r)\} > 1/2$ for all $r \in [0,r_0)$ , $t \in [0,2\pi]$ , then $F_s \in \mathcal{K}(-1/2)$ for all $|s| \le r_0$ . After some calculations, we obtain $$\Re\left\{H(t,r)\right\} = \frac{-4r^2x^2 + 6r(1-r^2)x + 2(1+r^2)}{(1+r^2)^2 - 4r^2x^2} := \frac{n(x,r)}{d(x,r)},\tag{2.3}$$ where $x = \cos t$ . Thus we shall find $r_0$ in [0,1] such that $$\frac{n(x,r)}{d(x,r)} > 1/2$$ for all $r \in [0, r_0)$ and for all $x \in [-1, 1]$ . (2.4) It is easy to verify that d(x,r) > 0 for all $r \in [0,1)$ and for all $x \in [-1,1]$ , and that $n(-1,r) = 2(1-r^2)(1-3r) \le 0$ for all $r \in [1/3,1)$ . Hence, to find $r_0$ in [0,1] satisfying (2.4), we may restrict our considerations to the interval [0,1/3). Then, analyzing n(x,r) as a function of one variable x, we get $$\min\{n(x,r): -1 \le x \le 1\} = n(-1,r) = 2(1-r^2)(1-3r) > 0$$ for all $r \in [0, 1/3)$ . Moreover, we see that $$\min\{d(x,r): -1 \le x \le 1\} = d(-1,r) = (1-r^2)^2 > 0.$$ Therefore, if $0 \le r < 1/3$ , then $$\min\left\{\frac{n(x,r)}{d(x,r)}: x \in [-1,1]\right\} = \frac{2(1-3r)}{1-r^2}.$$ To find $r_0$ in [0, 1/3) satisfying (2.4), it is sufficient to solve the inequality $$\frac{2(1-3r)}{1-r^2} > \frac{1}{2},$$ which is true for $$0 \le r \le r_0 = 6 - \sqrt{33}$$ . Moreover, analogous consideration as in the above proof leads us to the fact that $F_s \notin \mathcal{K}(-1/2)$ for $s > 6 - \sqrt{33} \approx 0.255437$ . In Lemma 3.2 we show that $F_s \in \mathcal{K}$ for $0 \le s \le 3 - 2\sqrt{2} \approx 0.1715$ .

Theorem 2.3. If $0 \le s \le 6 - \sqrt{33} \approx 0.255437$ and $f \in \mathcal{H}$ , then $$f \prec F_s \implies L[f] \prec L[F_s],$$ (2.5) where $$L[f](z) = \frac{2}{z} \int_0^z f(t) \, \mathrm{d}t$$ is the Libera operator. Proof By Lemma 2.2 we obtain $$\Re \left\{1 + \frac{zF_s''(z)}{F_s'(z)}\right\} > -\frac{1}{2} \quad \text{for } z \in \Delta.$$ Therefore, using Theorem C, we get (2.5).

Corollary 2.4. If $0 \le s \le 6 - \sqrt{33}$ and $$f(z)=\sum_{n=1}^{\infty}a_nz^n,$$ then $$\sum_{n=1}^{\infty} a_n z^n < \sum_{n=1}^{\infty} s^{n-1} z^{2n-1} \quad \Longrightarrow \quad \sum_{n=1}^{\infty} \frac{a_n}{n+1} z^n < \sum_{n=1}^{\infty} \frac{s^{n-1}}{n+1} z^{2n-1}.$$ It is known [14] that if $$\sum_{n=0}^{\infty} \alpha_n z^n \prec \sum_{n=0}^{\infty} \beta_n z^n,$$ then $$\sum_{n=0}^{k} |\alpha_n|^2 \le \sum_{n=0}^{k} |\beta_n|^2.$$ Hence, for $0 \le s \le 6 - \sqrt{33}$ , Corollary 2.4 gives also that $$\sum_{n=1}^{\infty} a_n z^n \prec F_s(z) \quad \Rightarrow \quad \sum_{n=1}^{\infty} \frac{|a_n|^2}{(n+1)^2} \le \sum_{n=1}^{\infty} \frac{s^{2(n-1)}}{(n+1)^2} = \frac{-1}{s^2} + \frac{1}{s^2} \int_0^{\infty} \frac{t}{e^t - s^2} dt$$ $$= \frac{\Phi(s^2, 2, 1) - 1}{s^2},$$ because for $\Re \{a\} > 0$ , $-1 \le x < 1$ , $\Re \{c\} > 0$ , $$\Phi(x,c,a) = \sum_{k=0}^{\infty} \frac{x^k}{(k+a)^c} = \frac{1}{\Gamma(c)} \int_0^{\infty} \frac{t^{c-1}e^{-at}}{1-xe^{-t}} dt.$$

Theorem 2.5. If $0 \le s_i \le 6 - \sqrt{33}$ , i = 1, 2, and the functions $g_1$ and $g_2$ are in the class $\mathcal{H}$ , then $$g_1 \prec F_{s_1}$$ and $g_2 \prec F_{s_2} \implies L[g_1] * L[g_2] \prec L[L[F_{s_1 s_2}]].$ (2.6) Proof By Theorem 2.3 we have $L[g_1] \prec L[F_{s_1}]$ and $L[g_2] \prec L[F_{s_2}]$ . By Lemma 2.1 $F_s \in \mathcal{S}$ , $s \in [0,1]$ , so the functions $L[F_{s_1}]$ , $L[F_{s_2}]$ are in $\mathcal{K}$ because $L[\mathcal{S}^] = \mathcal{K}$ . Using Theorem D, we obtain $$L[g_1] L[g_2] \prec L[F_{s_1}] L[F_{s_2}]$$ but $L[F_{s_1}] * L[F_{s_2}] = L[L[F_{s_1s_2}]]$ because $$F_{s_1}(z) F_{s_2}(z) = \frac{z}{1 - s_1 z^2} \frac{z}{1 - s_2 z^2} = \frac{z}{1 - s_1 s_2 z^2} = F_{s_1 s_2}(z).$$

Lemma 3.1. Let $s \in [0,1]$ , then $D(s) = F_s(\Delta)$ , where $$F_s(z) = \frac{z}{1 - sz^2}$$ $(z \in \Delta).$ Proof The function $F_s$ is analytic in the unit disc, so for the proof we need to find an image of the circle |z| = 1 under the function $F_s$ . For $s \in [0,1)$ and for $\varphi \in [0,2\pi)$ , we have $$F_{s}(e^{i\varphi}) = \frac{e^{i\varphi}}{1 - se^{2i\varphi}} \cdot \frac{1 - se^{-2i\varphi}}{1 - se^{-2i\varphi}} = \frac{e^{i\varphi} - se^{-i\varphi}}{1 + s^{2} - s(e^{2i\varphi} + e^{-2i\varphi})}$$ $$= \frac{(1 - s)\cos\varphi + i(1 + s)\sin\varphi}{1 + s^{2} - 2s\cos2\varphi}.$$ (3.3) Let us define $$x = \Re\left\{F_s\left(e^{i\varphi}\right)\right\} = \frac{(1-s)\cos\varphi}{1+s^2-2s\cos2\varphi} \tag{3.4}$$ and $$y = \mathfrak{Im}\left\{F_s(e^{i\varphi})\right\} = \frac{(1+s)\sin\varphi}{1+s^2-2s\cos2\varphi}.$$ (3.5) Then, after some calculations, we can obtain from (3.4) and (3.5) that $$x^{2} + y^{2} = \frac{1}{1 + s^{2} - 2s\cos 2\varphi} \quad \text{and} \quad (x, y) \neq (0, 0).$$ (3.6) Therefore, using the relations (3.3)-(3.5), we can find that $F_s(e^{i\varphi})$ is an algebraic curve of order four whose equation in orthogonal Cartesian coordinates is $$\left(x^2 + y^2\right)^2 - \frac{x^2}{(1-s)^2} - \frac{y^2}{(1+s)^2} = 0 \quad \text{and} \quad (x,y) \neq (0,0).$$ (3.7) Thus, because $F_s(0) = 0$ for $s \in [0,1)$ , the set $F_s(\Delta)$ is bounded by the curve (3.7). Moreover, for s = 1, the function $F_s$ becomes $$F_1(z) = \frac{z}{1 - z^2}$$ and it is easy to see that $F_1(\Delta) = D(1)$ , see Figure 1. Then the proof is completed.  Notice that a curve described by $$(x^2 + y^2)^2 - (n^4 + 2m^2)x^2 - (n^4 - 2m^2)y^2 = 0 \quad \text{and} \quad (x, y) \neq (0, 0)$$ (3.8) is called the Booth lemniscate, named after Booth [15, 16]. The Booth lemniscate is called elliptic if $n^4 > 2m^2$ , while for $n^4 < 2m^2$ it is termed hyperbolic. Thus it is clear that the curve (3.7) is the Booth lemniscate of elliptic type. The Booth lemniscate is a special case of a Persian curve. The Booth lemniscate of elliptic type can be described geometrically in the following equivalent ways. 1. Suppose that $s \in [0,1)$ . Let $\mathfrak{C}(S,R)$ be a circle with the center S and the length of the radius R such that $$S = \left(\frac{\sqrt{s}}{1 - s^2}, 0\right)$$ and $R = \frac{1}{2(1 - s)}$ . A ray is drawn from the point x on the circle $\mathfrak{C}(S,R)$ through the origin o that cuts also the circle $\mathfrak{C}(S,R)$ at the point z. A point v is also on the ray and satisfies $$|ov| = |xz|. \tag{3.9}$$ If the point x goes along the circle $\mathfrak{C}(S, R)$ , then the point $\nu$ describes the curve (3.7), see Figure 2. 2. Let $s \in [0,1)$ . Then the curve (3.7) consists of points M such that $$|F_1M|^2|F_2M|^2=\frac{1+s^2}{(1-s^2)^2}|OM|^2+\frac{s^2}{(1-s^2)^4},$$ where O = (0,0) and the points $F_1$ , $F_2$ are the focuses $$F_1\left(\frac{-\sqrt{s}}{1-s^2},0\right), \qquad F_2\left(\frac{\sqrt{s}}{1-s^2},0\right),$$ see Figure 3. We say that a closed curve $\gamma$ is convex when it is boundary of a convex bounded domain. Otherwise, we say that the curve $\gamma$ is concave.  

Lemma 3.2. Suppose that $F_s$ is given by (2.1). If $0 \le s \le 3 - 2\sqrt{2} \approx 0.1715$ , then the curve $F_s(e^{i\varphi})$ , $\varphi \in [0, 2\pi)$ is convex and $$\max\left\{\mathfrak{Im}\left\{F_s\left(e^{i\varphi}\right)\right\}:\varphi\in[0,2\pi)\right\} \tag{3.10}$$ is attained at one point only. If $s \in (3-2\sqrt{2},1)$ , then the curve $F_s(e^{i\varphi})$ , $\varphi \in [0,2\pi)$ is concave and (3.10) is attained twice. Moreover, in both cases this curve is symmetric with respect to both axes. Proof If s=0, then the curve $F_s(e^{i\varphi})$ , $\varphi\in[0,2\pi)$ becomes a circle and it is clear that (3.10) is attained one time only. Suppose, in the sequel, that $s\in(0,1)$ . From (2.2) and (2.3) we have $$\mathfrak{Re}\left\{\frac{e^{i\varphi}F_s''(e^{i\varphi})}{F_s'(e^{i\varphi})}+1\right\} = \frac{1-s^4+6s(1-s^2)\cos\varphi}{(1+s^2)^2-4s^2\cos^2\varphi}.$$ The denominator is positive, hence for the convexity of $F_s(e^{i\varphi})$ we need the nominator to be positive. Since $s^2 - 6s + 1 > 0$ for $0 \le s \le 3 - 2\sqrt{2}$ , then the nominator is positive because $$1 - s^4 + 6s(1 - s^2)\cos\varphi \ge 1 - s^4 - 6s(1 - s^2)$$ $$= (1 - s^2)(s^2 - 6s + 1).$$ From (3.5) we have $$\mathfrak{Im}\left\{F_s\left(e^{i\varphi}\right)\right\} = \frac{(1+s)\sin\varphi}{1+s^2-2s\cos2\varphi} := \frac{(1+s)x}{1+s^2-2s(1-2x^2)} := f(x).$$ Then f'(x) = 0 if and only if $x^2 = (1 - s)^2/(4s)$ . That is possible when $(1 - s)^2/(4s) \le 1$ because $x = \sin \varphi$ . Therefore, using the elementary considerations, we can find that if $s \in (3 - 2\sqrt{2}, 1)$ , then the curve $F_s(e^{i\varphi})$ , $\varphi \in [0, 2\pi)$ is concave and (3.10) is attained twice, when $\sin \varphi = (1 - s)/(2\sqrt{s})$ . In this case, $$\max\left\{\mathfrak{Im}\left\{F_s\left(e^{i\varphi}\right)\right\}:\varphi\in[0,2\pi)\right\}=\frac{1+s}{4\sqrt{s}(1-s)},$$ see Figure 3. Moreover, if $s \in [0, 3 - 2\sqrt{2}]$ , then (3.10) is attained at one point only such that $$\max\left\{\mathfrak{Im}\left\{F_s\left(e^{i\varphi}\right)\right\}:\varphi\in[0,2\pi)\right\}=\mathfrak{Im}\left\{F_s\left(e^{i\pi/2}\right)\right\}=\frac{1}{1+s},$$ see Figures 2 and 3.

Corollary 3.3. If $0 \le s \le 3 - 2\sqrt{2} \approx 0.1715$ , then $F_s \in \mathcal{K}$ . By Lemma 2.1 the functions $F_s$ , $s \in [0,1]$ , are starlike univalent, hence by the geometric interpretation (1.2) of a subordination under univalent functions, we obtain from Theorem 2.3 the following corollary.

Corollary 3.4. If $0 \le s \le 6 - \sqrt{33}$ and $f \in \mathcal{H}$ , then $$f(\Delta) \subset D(s) \implies L[f](\Delta) \subset L[F_s](\Delta).$$ (3.11)

Theorem 3.5. Assume that $0 < s_i \le 3 - 2\sqrt{2} \approx 0.1715$ , i = 1, 2, and that the functions $g_1$ and $g_2$ are in the class $\mathcal{H}$ with $g_1(0) = g_2(0) = 0$ . If $$g_1(\Delta) \subset D(s_1)$$ and $g_2(\Delta) \subset D(s_2)$ , (3.12) then $$(g_1 * g_2)(\Delta) \subset D(s_1 s_2). \tag{3.13}$$ Proof Using Lemma 3.1, we may rewrite (3.12) as $$g_1(\Delta) \subset F_{s_1}(\Delta)$$ and $g_2(\Delta) \subset F_{s_2}(\Delta)$ . (3.14) By Corollary 3.3, $F_s$ is convex univalent in $\Delta$ for $s \in [0, 3 - 2\sqrt{2}]$ , so $F_{s_1}$ , $F_{s_2}$ are convex univalent in $\Delta$ . From (1.2) and from (3.14), we obtain $$g_1 \prec F_{s_1}$$ and $g_2 \prec F_{s_2}$ . (3.15) By Theorem D, we get from (3.15) $$g_1 g_2 \prec F_{s_1} F_{s_2}.$$ (3.16) But $$F_{s_1}(z) F_{s_2}(z) = \frac{z}{1 - s_1 z^2} \frac{z}{1 - s_2 z^2} = \frac{z}{1 - s_1 s_2 z^2} = F_{s_1 s_2}(z)$$ and hence by Lemma 3.1 the subordination (3.16) is equivalent to (3.13). $\Box$

Theorem 3.6. Assume that $0 < s \le 3 - 2\sqrt{2} \approx 0.1715$ and that the functions $g_1$ and $g_2$ are in the class $\mathcal{H}$ with $g_1(0) = g_2(0) = 0$ . Let $\mathfrak{K}(c,R)$ be the disc with center $c \in \mathbb{R}$ and radius R with |c| < R. If $$g_1(\Delta) \subset D(s)$$ and $g_2(\Delta) \subset \mathfrak{K}(c,R)$ , (3.17) then $$(g_1 * g_2)(\Delta) \subset \frac{R^2 - c^2}{R} D\left(\frac{cs^2}{R^2}\right). \tag{3.18}$$ Proof Using Lemma 3.1, we may rewrite (3.17) as $$g_1(\Delta) \subset F_{s_1}(\Delta)$$ and $g_2(\Delta) \subset H_{c,R}(\Delta)$ , (3.19) where $$H_{c,R}(z) = \frac{(R^2 - c^2)z}{R - cz} \quad (z \in \Delta)$$ maps $\Delta$ onto $\mathfrak{K}(c,R)$ . By Corollary 3.3, $F_s$ is convex univalent in $\Delta$ , also $H_{c,R}$ is convex univalent in $\Delta$ . Therefore, by Theorem D we obtain $$(g_1 g_2)(z) \prec \frac{z}{1 - sz^2} \frac{(R^2 - c^2)z}{R - cz}$$ $$= \frac{R^2 - c^2}{R} \frac{z}{1 - sc^2 z^2 / R^2}$$ $$= \frac{R^2 - c^2}{R} F_{sc^2 / R^2}(z).$$ The function $F_{sc^2/R^2}$ is univalent as the convolution of a convex univalent function, so by (1.2) we obtain (3.18).

Theorem 3.7. Assume that $0 < s \le 3 - 2\sqrt{2} \approx 0.1715$ and that the functions $g_1$ and $g_2$ are in the class $\mathcal{H}$ with $g_1(0) = g_2(0) = 0$ . Let $\mathfrak{L}(\varphi, d) = \{z \in \mathbb{C} : \mathfrak{Re}\{ze^{-i\varphi}\} > -d\}$ , $\varphi \in \mathbb{R}$ , d > 0, be the half-plane. If $$g_1(\Delta) \subset D(s)$$ and $g_2(\Delta) \subset \mathfrak{L}(\varphi, d)$ , (3.20) then $$(g_1 * g_2)(\Delta) \subset 2de^{-i\varphi}D(s). \tag{3.21}$$ Proof Using Lemma 3.1, we may rewrite (3.20) as $$g_1(\Delta) \subset F_{s_1}(\Delta)$$ and $g_2(\Delta) \subset G_{\varphi,d}(\Delta)$ , (3.22) where $$G_{\varphi,d}(z) = \frac{2de^{-i\varphi}z}{1-z} \quad (z \in \Delta)$$ maps $\Delta$ onto $\mathfrak{L}(\varphi,d)$ . By Corollary 3.3, $F_s$ is convex univalent in $\Delta$ , also $G_{\varphi,d}$ is convex univalent in $\Delta$ . Therefore, by Theorem D we obtain $$(g_1 g_2)(z) \prec \frac{z}{1 - sz^2} \frac{2de^{-i\varphi}z}{1 - z}$$ = $\frac{2de^{-i\varphi}z}{1 - sz^2} = 2de^{-i\varphi}F_s(z)$ . Hence by (1.2) we obtain (3.21).