Differential Subordination for Starlike Functions Related to Domains Symmetric with Respect to the Real Axis

Abstract

We use a new approach to differential subordination to obtain some subordination implications for subclasses of the class of strongly starlike functions of order $\alpha$.

1. Introduction

Differential subordination is an important part of geometric function theory. This concept generalizes and extends the notion of differential inequalities to complex functions. Differential subordination was first proposed and developed by S. Miller and P. Mocanu. They wrote many research papers concerning this notion (see, e.g., [1,2,3,4]), and a survey of this theory can be found in their monograph [5].

Subordinate techniques are an effective tool for examining properties of analytic and univalent functions. Establishing that an unknown function is subordinate to a certain known function is used to derive results concerning geometric properties of functions (like starlikeness or convexity) and also, e.g., helps to determine the range of coefficients of functions or the range of their derivatives.

This is why there is still such great interest in this topic among many mathematicians. For some applications of differential subordination, we refer, e.g., to [6,7,8,9,10,11]. More recent results in the theory can be found, e.g., in [12,13,14,15,16,17].

Denote by U the unit disk in the complex plane $\mathbb{C}$. Let f and F be two functions holomorphic in the disk U. We say that a function f is subordinate to F in U and write $f\prec F$ in U, if there exists a holomorphic function $\omega$ such that $\left|\omega \right(z\left)\right|\le \left|z\right|$ and $f\left(z\right)=F\left(\omega \right(z\left)\right)$ for $z\in U$. Hence, $f\prec F$ in U implies $f\left(U\right)\subset F\left(U\right)$. Let $\mathcal{H}$ denote the class of all functions f normalized by $f\left(0\right)=f^{\prime }\left(0\right)-1=0$ that are analytic in U and let $\mathcal{S}$ be the subclass of $\mathcal{H}$ consisting of univalent functions.

By ${\mathcal{S}}_{\alpha }^{*}$ we denote the class of strongly starlike functions of order $\alpha$, introduced by Stankiewicz [18] and independently by Brannan and Kirwan [19]. Recall that $f\in \mathcal{S}$ is strongly starlike of order $\alpha$, $0<\alpha \le 1,$ if and only if

$$\left|\mathrm{Arg}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right|<{\displaystyle \frac{\alpha \pi }{2}},z\in U,$$

or, equivalently, in terms of subordination, if and only if

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }\mathrm{in}U.$$

For $\alpha =1$, we get the whole class ${\mathcal{S}}^{*}$ of starlike functions.

Geometrically, $f\in {\mathcal{S}}^{*}$ if and only if $f\left(U\right)$ is starlike with respect to the origin and $f\in {\mathcal{S}}_{\alpha }^{*}$ if and only if for every $w\in f\left(U\right)$, a certain lens-shaped region with end points 0 and w lies in $f\left(U\right)$.

Let $\mathcal{Q}$ denote the set of functions q that are analytic and univalent on $\overline{U}\setminus \mathcal{E}\left(q\right)$, where

$$\mathcal{E}\left(q\right)=\{\zeta \in \partial U:\underset{z\to \zeta }{lim}q\left(z\right)=\infty \}$$

such that $q^{\prime }\left(\zeta \right)\ne 0$ for $\zeta \in \partial U\setminus \mathcal{E}\left(q\right)$.

Definition 1 

([1]). Let Ω be a domain in $\mathbb{C}$ and $q\in \mathcal{Q}$. We define ${\Psi }_n(\Omega ,q)$ to be the class of functions $\psi :{\mathbb{C}}^3\to \mathbb{C}$ that satisfy the following:

(a) $\psi (r,s,t)$ is continuous in a domain $D\subset {\mathbb{C}}^3$,

(b) $\left(q\right(0),0,0)\in D$ and $\psi \left(q\right(0),0,0)\in \Omega$,

(c) $\psi (r_0,s_0,t_0)\notin \Omega$, when $(r_0,s_0,t_0)\in D,r_0=q\left(\zeta \right),s_0=m\zeta q^{\prime }\left(\zeta \right)$ and

$$\mathrm{Re}\left(1+{\displaystyle \frac{t_0}{s_0}}\right)\ge m\mathrm{Re}\left(1+{\displaystyle \frac{\zeta q^{″}\left(\zeta \right)}{q^{\prime }\left(\zeta \right)}}\right),$$

where $\left|\zeta \right|=1$, $q\left(\zeta \right)$ is finite and $m\ge n\ge 1$.

We write ${\Psi }_1(\Omega ,q)$ as $\Psi (\Omega ,q)$.

Lemma 1 

([1]). Let $q\left(0\right)=a$ and $\psi \in {\Psi }_n(\Omega ,q)$ with corresponding domain D. Let $p\left(z\right)=a+p_n{z}^n+p_{n+1}{z}^{n+1}+\dots$ be regular in U with $p\left(z\right)\not\equiv a$ and $n\ge 1$. If $(p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right))\in D$ when $z\in U$ and

$$\psi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right))\in \Omega whenz\in U,$$

then $p\left(z\right)\prec q\left(z\right)$.

In this paper, we use this principal lemma to obtain some subordination results connected with some angular regions that are symmetric with respect to the real axis. These regions are the images of the unit disk under functions that generate the class of strongly starlike functions of order $\alpha$ and some of its subclasses. In particular, we obtain some sufficient condition, in terms of differential subordination, for a function to be uniformly starlike in U. Applying the main results, we derive, in Section 3, some sufficient conditions under which analytic functions are subordinate to a given function closely connected with the class of Janowski starlike functions or with the class $k-\mathcal{ST}$ of starlike functions related to the class of k-uniformly convex functions. In general, it is hard to obtain the explicit form of a function that satisfies a given differential subordination. In Section 4, we find such functions and present some non-trivial examples illustrating some applications of our results.

2. Main Results

For fixed $0<\alpha \le 1$, let

$$q\left(z\right)={\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha },z\in U.$$

Then $q\left(U\right)=\Delta =\{w:|\mathrm{Arg}w|<\alpha \pi /2\}$. Moreover

$$q\left(e^{i\theta }\right)={\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }\mathrm{and}{e}^{i\theta }{q}^{\prime }\left(e^{i\theta }\right)={\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }{\displaystyle \frac{i\alpha }{sin\theta }}.$$

Hence, $\mathcal{E}\left(q\right)=\left\{1\right\}$ and $q\in \mathcal{Q}$.

Theorem 1. 

Let $0<\alpha \le 1$ and $0<\beta \le 1$ with $(\beta -\alpha )(\pi /2)\le arctan\alpha$. If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then $p\prec q$ in U.

Proof. 

Let $\psi (p\left(z\right),z{p}^{\prime }\left(z\right),z)=p\left(z\right)+z{p}^{\prime }\left(z\right)$ and $\Omega =\{w:|\mathrm{Arg}w|<\beta \pi /2\}$. Then $\psi$ satisfies the admissibility condition

$$\psi (q\left(\zeta \right),m\zeta q^{\prime }\left(\zeta \right),\zeta )\notin \Omega \mathrm{for}\zeta =e^{i\theta },\theta \in (0,2\pi ),m=1,$$

if and only if

$$\left|\mathrm{Arg}w\right|=\left|\mathrm{Arg}[q\left(e^{i\theta }\right)+e^{i\theta }{q}^{\prime }\left(e^{i\theta }\right)]\right|\ge {\displaystyle \frac{\beta \pi }{2}},\theta \in (0,2\pi ).$$

In view of Lemma 1, it suffices to show that $\left|\mathrm{Arg}w\right|\ge {\displaystyle \frac{\beta \pi }{2}}$ for $(\beta -\alpha )(\pi /2)\le arctan\alpha$. □

We have

$$\left|\mathrm{Arg}[q\left(e^{i\theta }\right)+e^{i\theta }{q}^{\prime }\left(e^{i\theta }\right)]\right|=\left|\mathrm{Arg}\left[{\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }+{\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }{\displaystyle \frac{i\alpha }{sin\theta }}\right]\right|$$

$$=\left|\mathrm{Arg}\left[{\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }\left(1+{\displaystyle \frac{i\alpha }{sin\theta }}\right)\right]\right|=\left|\alpha \mathrm{Arg}\left(icot{\displaystyle \frac{\theta }{2}}\right)+\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\theta }}\right)\right|.$$

It is sufficient to consider $\theta =\pi \pm \phi$, where $\phi \in (0,\pi )$. For $\theta =\pi -\phi$, we get

$$\mathrm{Arg}\left(icot{\displaystyle \frac{\theta }{2}}\right)={\displaystyle \frac{\pi }{2}}\mathrm{and}\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\theta }}\right)=\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\phi }}\right)\in (0,\pi /2).$$

For $\theta =\pi +\phi$, we get

$$\mathrm{Arg}\left(icot{\displaystyle \frac{\theta }{2}}\right)=-{\displaystyle \frac{\pi }{2}}\mathrm{and}\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\theta }}\right)=\mathrm{Arg}\left(1-{\displaystyle \frac{i\alpha }{sin\phi }}\right)=-\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\phi }}\right).$$

Thus

$$\left|\alpha \mathrm{Arg}\left(icot{\displaystyle \frac{\theta }{2}}\right)+\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\theta }}\right)\right|=\alpha {\displaystyle \frac{\pi }{2}}+\mathrm{Arg}\left(1+{\displaystyle \frac{i\alpha }{sin\phi }}\right),\phi \in (0,\pi )$$

and

$$\left|\mathrm{Arg}w\right|=\alpha {\displaystyle \frac{\pi }{2}}+arctan{\displaystyle \frac{\alpha }{sin\phi }}\ge \alpha {\displaystyle \frac{\pi }{2}}+arctan\alpha .$$

Hence, if $\alpha ,\beta$ are such that $\alpha {\displaystyle \frac{\pi }{2}}+arctan\alpha \ge \beta {\displaystyle \frac{\pi }{2}}$, then $\left|\mathrm{Arg}w\right|\ge \beta {\displaystyle \frac{\pi }{2}}$, and by Lemma 1, the result follows.

Remark 1. 

Our Theorem 1 coincides with the result obtained in[2], but the proofs are independent.

For $\alpha =\beta$, we get

Corollary 1. 

Let $0<\alpha \le 1$. If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $\beta =1$ in Theorem 1, we get

Corollary 2. 

Let $0<\alpha \le 1$ and $\alpha \ge cot(\alpha \pi /2)$. If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$\mathrm{Re}[p\left(z\right)+z{p}^{\prime }\left(z\right)]>0,z\in U,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f\left(z\right)/z$ in Theorem 1 we get

Corollary 3. 

Let $0<\alpha \le 1$ and $0<\beta \le 1$ with $(\beta -\alpha )(\pi /2)\le arctan\alpha$. If $f\in \mathcal{H}$ satisfies

$$f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then

$${\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f^{\prime }\left(z\right)$ in Theorem 1, we get

Corollary 4. 

Let $0<\alpha \le 1$ and $0<\beta \le 1$ with $(\beta -\alpha ){\displaystyle \frac{\pi }{2}}\le arctan\alpha$. If $f\in \mathcal{H}$ satisfies

$$f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then

$$f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

Let $\mathcal{UCV}$ and $\mathcal{UST}$ denote the classes of uniformly convex and uniformly starlike functions, respectively, introduced by Goodman ([20,21]). Recall that a function $f\in \mathcal{S}$ is in the class $\mathcal{UCV}$ ($\mathcal{UST}$) if for every circular arc $\gamma \subset U$ with center $\zeta \in U$, the arc $f\left(\gamma \right)$ is convex (starlike with respect to $f\left(\zeta \right)$). Goodman proved that

$$f\in \mathcal{UST}\iff \mathrm{Re}{\displaystyle \frac{f\left(z\right)-f\left(\zeta \right)}{(z-\zeta )f^{\prime }\left(z\right)}}\ge 0\mathrm{for}\mathrm{all}z\in U,\left|\zeta \right|\le 1.$$

In [22], the following sufficient condition for uniform starlikeness was proved.

Lemma 2. 

Let $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n$, $z\in U$. If

$$|\mathrm{Arg}{f}^{\prime }\left(z\right)|\le {\displaystyle \frac{\pi }{4}}forz\in U,$$

then $f\in \mathcal{UST}.$

Using Corollary 4 with $\alpha =1/2$ and applying Lemma 2, we immediately get

Corollary 5. 

Let $0<\beta <1$ with $tan{\displaystyle \frac{\beta \pi }{2}}\le 3$. If $f\in \mathcal{H}$ satisfies

$$f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then $f\in \mathcal{UST}$.

Theorem 2. 

Let $0<\alpha <1$ and $0<\beta <1$ be such that

$$tan{\displaystyle \frac{\beta \pi }{2}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Proof. 

Let $\psi (p\left(z\right),z{p}^{\prime }\left(z\right),z)=p\left(z\right)+z{p}^{\prime }\left(z\right)/p\left(z\right)$ and $\Omega =\{w:|\mathrm{Arg}w|<\beta \pi /2\}$. Then $\psi$ satisfies the admissibility condition

$$\psi (q\left(\zeta \right),m\zeta q^{\prime }\left(\zeta \right),\zeta )\notin \Omega \mathrm{for}\zeta =e^{i\theta },\theta \in (0,2\pi ),m=1,$$

if and only if

$$\left|\mathrm{Arg}w\left(\theta \right)\right|=\left|\mathrm{Arg}\left[q\left(e^{i\theta }\right)+{\displaystyle \frac{e^{i\theta }{q}^{\prime }\left(e^{i\theta }\right)}{q\left(e^{i\theta }\right)}}\right]\right|=\left|\mathrm{Arg}\left[{\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }+{\displaystyle \frac{i\alpha }{sin\theta }}\right]\right|\ge {\displaystyle \frac{\beta \pi }{2}}.$$

In view of Lemma 1, it suffices to show that

$$\left|\mathrm{Arg}w\left(\theta \right)\right|\ge {\displaystyle \frac{\beta \pi }{2}}\mathrm{for}tan{\displaystyle \frac{\beta \pi }{2}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

We have

$$w\left(\theta \right)={\left(icot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }+{\displaystyle \frac{i\alpha }{sin\theta }}={\left(cot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }{\left(e^{logi}\right)}^{\alpha }+{\displaystyle \frac{i\alpha }{sin\theta }}$$

$$={\left(cot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }{e}^{i{\displaystyle \frac{\alpha \pi }{2}}}+{\displaystyle \frac{i\alpha }{sin\theta }}={\left(cot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }cos{\displaystyle \frac{\alpha \pi }{2}}+i\left[{\left(cot{\displaystyle \frac{\theta }{2}}\right)}^{\alpha }sin{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha }{sin\theta }}\right].$$

Thus

$$\mathrm{Arg}w\left(\theta \right)=arctan{\displaystyle \frac{{\left(cot\frac{\theta }{2}\right)}^{\alpha }sin\frac{\alpha \pi }{2}+\frac{\alpha }{sin\theta }}{{\left(cot\frac{\theta }{2}\right)}^{\alpha }cos\frac{\alpha \pi }{2}}}=arctan\left[tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha }{sin\theta {\left(cot\frac{\theta }{2}\right)}^{\alpha }cos\frac{\alpha \pi }{2}}}\right].$$

It is sufficient to consider $\theta =\pi \pm \phi$, where $\phi \in (0,\pi ).$ Note that $\mathrm{Arg}w(\pi +\phi )=-\mathrm{Arg}w(\pi -\phi )$ and $\mathrm{Arg}w(\pi -\phi )\in (0,\pi /2).$ Hence

$$\left|\mathrm{Arg}w\left(\theta \right)\right|=\mathrm{Arg}w(\pi -\phi )=arctan\left[tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha }{sin\phi {\left(tan\frac{\phi }{2}\right)}^{\alpha }cos\frac{\alpha \pi }{2}}}\right],\phi \in (0,\pi ).$$

Let

$$G\left(\phi \right)=sin\phi {\left(tan{\displaystyle \frac{\phi }{2}}\right)}^{\alpha },\phi \in (0,\pi ).$$

We show that $G\left(\phi \right)\le G\left({\phi }_0\right)$ for $\phi \in (0,\pi )$, where $cos{\phi }_0=-\alpha .$ We have

$$G^{\prime }\left(\phi \right)={\left(tan{\displaystyle \frac{\phi }{2}}\right)}^{\alpha }\left[cos\phi +{\displaystyle \frac{\alpha sin\phi }{2\left(tan\frac{\phi }{2}\right){cos}^2\frac{\phi }{2}}}\right]={\left(tan{\displaystyle \frac{\phi }{2}}\right)}^{\alpha }(cos\phi +\alpha )$$

and

$$G^{″}\left(\phi \right)=\alpha {\left(tan{\displaystyle \frac{\phi }{2}}\right)}^{\alpha -1}{\displaystyle \frac{cos\phi +\alpha }{2{cos}^2\frac{\phi }{2}}}+{\left(tan{\displaystyle \frac{\phi }{2}}\right)}^{\alpha }(-sin\phi ).$$

We see that $G^{\prime }\left({\phi }_0\right)=0$ and $G^{″}\left({\phi }_0\right)<0$. Thus, for $\phi \in (0,\pi )$, we have $G\left(\phi \right)\le G\left({\phi }_0\right)$ with $cos{\phi }_0=-\alpha$, so

$$\left|\mathrm{Arg}w\left(\theta \right)\right|=arctan\left[tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha }{G\left(\phi \right)cos\frac{\alpha \pi }{2}}}\right]\ge arctan\left[tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha }{G\left({\phi }_0\right)cos\frac{\alpha \pi }{2}}}\right].$$

If $cos{\phi }_0=-\alpha$, then $sin{\phi }_0=\sqrt{1-{\alpha }^2}$ and $cos{\displaystyle \frac{{\phi }_0}{2}}=\sqrt{{\displaystyle \frac{1-\alpha }{2}}}$, $sin{\displaystyle \frac{{\phi }_0}{2}}=\sqrt{{\displaystyle \frac{1+\alpha }{2}}}$. Hence,

$$G\left({\phi }_0\right)=\sqrt{1-{\alpha }^2}{\left({\displaystyle \frac{\sqrt{1+\alpha }}{\sqrt{1-\alpha }}}\right)}^{\alpha }$$

and if $\alpha ,\beta$ are such that $tan{\displaystyle \frac{\beta \pi }{2}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}$, then $\left|\mathrm{Arg}w\right|\ge \beta {\displaystyle \frac{\pi }{2}}$. Therefore, by Lemma 1, the result follows. □

Remark 2. 

We check what will happen if we allow α to be equal to 1 in the thesis statement of Theorem 2. From the proof of Theorem 2, we see that for $\alpha =1$, we have

$$w\left(\theta \right)=i\left(cot{\displaystyle \frac{\theta }{2}}+{\displaystyle \frac{1}{sin\theta }}\right)$$

and

$$\left|\mathrm{Arg}w\left(\theta \right)\right|={\displaystyle \frac{\pi }{2}}.$$

Thus, the admissibility condition

$$\left|\mathrm{Arg}w\left(\theta \right)\right|\ge {\displaystyle \frac{\beta \pi }{2}}$$

is fulfilled only for $\beta =1$.

Corollary 6. 

Let $0<\alpha <1$ and $0<\beta \le \alpha$. If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Miller and Mocanu in [3] obtained the following general result.

Theorem 3. 

Let δ and γ be complex numbers with $\delta \ne 0$, and let p and h be analytic in U with $h\left(0\right)=p\left(0\right)$. If $Q\left(z\right)\equiv \delta h\left(z\right)+\gamma$ satisfies

(a)

$$\mathrm{Re}Q\left(z\right)>0,z\in U$$

and either

(b)

$$Qisconvex,or$$

(b’)

$$logQisconvex$$

then

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{\delta p\left(z\right)+\gamma }}\prec h\left(z\right)$$

implies that $p\left(z\right)\prec h\left(z\right)$.

Remark 3. 

We know that the function q is convex and $\mathrm{Re}q\left(z\right)>0$ for $z\in U$. In view of Theorem 3 (for $\delta =1$, $\gamma =0$), we have an implication as in Corollary 6 for $\alpha =\beta$

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{\delta p\left(z\right)+\gamma }}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }\Rightarrow p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }$$

so Corollary 6 does not represent anything new. However, if β is such that

$$tan{\displaystyle \frac{\beta \pi }{2}}=tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}$$

then $\beta >\alpha$. Hence, our result from Theorem 2 improves the result of Miller and Mocanu in Theorem 3.

Taking $p\left(z\right)=f\left(z\right)/z$ in Theorem 2, we get

Corollary 7. 

Let $0<\alpha <1$ and $0<\beta <1$ be such that

$$tan{\displaystyle \frac{\beta \pi }{2}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If $f\in \mathcal{H}$ satisfies

$${\displaystyle \frac{f\left(z\right)}{z}}+{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then

$${\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f^{\prime }\left(z\right)$ in Theorem 2, we get

Corollary 8. 

Let $0<\alpha <1$ and $0<\beta <1$ be such that

$$tan{\displaystyle \frac{\beta \pi }{2}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If $f\in \mathcal{H}$ satisfies

$$f^{\prime }\left(z\right)+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then

$$f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Using Corollary 8 with $\alpha =1/2$ and applying Lemma 2, we immediately get

Corollary 9. 

Let $0<\beta <1$ and $tan{\displaystyle \frac{\beta \pi }{2}}\le 1+\sqrt[4]{{\displaystyle \frac{4}{27}}}$. If $f\in \mathcal{H}$ satisfies

$$f^{\prime }\left(z\right)+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

then $f\in \mathcal{UST}$.

3. Some Applications

Let $k-\mathcal{UCV}$ denote the class of all k-uniformly convex functions introduced in [23]. This class was obtained by taking $\zeta$ such that $\left|\zeta \right|\le k,k\ge 0$, in Goodman’s definition of uniform convexity. Clearly, $1-\mathcal{UCV}=\mathcal{UCV}$, and for $k=0$, we obtain a whole class of convex functions. Let $k-\mathcal{ST}$, $k\ge 0$ denote the subclass of starlike functions introduced and investigated in [24], defined by

$$f\in k-\mathcal{UCV}\iff z{f}^{\prime }\left(z\right)\in k-\mathcal{ST}.$$

Then (see [24]) $f\in \mathcal{S}$ belongs to the class $k-\mathcal{ST}$ if and only if

$$\mathrm{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}>k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|,z\in U.$$

For $k=1$, it is the class ${\mathcal{S}}_p$ introduced and investigated by Rønning in [25].

The class $k-\mathcal{ST}$ is related to conic domains (see [23,24]):

$$f\in k-\mathcal{ST}\iff {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\in {\Omega }_k,z\in U,$$

where $1\in {\Omega }_k$ and the boundary of ${\Omega }_k$ is the conic curve

$$\{w=u+iv:\mathrm{Re}w=k|w-1\left|\right\}=\{u+iv:u^2=k^2{(u-1)}^2+k^2{v}^2\},k\ge 0.$$

Note that $\partial {\Omega }_k$ is the right branch of the hyperbola for $k\in (0,1)$, the parabola $v^2=2u-1$ for $k=1$, and the ellipse for $k>1$.

The function $p_k$, analytic and univalent in U, such that $p_k\left(0\right)=1$ and $p_k\left(U\right)={\Omega }_k$, designates the function that is extremal for many problems in the class $k-\mathcal{ST}$. It is known that [25,26]

$$p_0\left(z\right)={\displaystyle \frac{1+z}{1-z}},p_1\left(z\right)=1+{\displaystyle \frac{2}{{\pi }^2}}{\left(log{\displaystyle \frac{1+\sqrt{z}}{1-\sqrt{z}}}\right)}^2,z\in U.$$

For the explicit form of functions $p_k$ for $k>0$ and $k\ne 1$, we refer to [23,24].

Observe that ${\Omega }_k\subset \{w:|\mathrm{Arg}w|<arctan{\displaystyle \frac{1}{k}}\}$ for $k>0$. Using this relation and Theorem 1 with $\beta ={\displaystyle \frac{2}{\pi }}arctan{\displaystyle \frac{1}{k}}$, we immediately get

Corollary 10. 

Let $0<\alpha \le 1$ and $k>0$ be such that $arctan(1/k)\le \pi \alpha /2+arctan\alpha$. If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$\mathrm{Re}[p\left(z\right)+z{p}^{\prime }\left(z\right)]>k|p\left(z\right)+z{p}^{\prime }\left(z\right)-1|,z\in U,$$

or equivalently

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\in {\Omega }_{k}forz\in U,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f\left(z\right)/z$ or $p\left(z\right)=f^{\prime }\left(z\right)$ in Corollary 9, we get

Corollary 11. 

Let $0<\alpha \le 1$ and $k>0$ be such that $arctan(1/k)\le \pi \alpha /2+arctan\alpha$. If $f\in \mathcal{H}$, then

$$[f^{\prime }\left(z\right)\in {\Omega }_{k}forz\in U]\Rightarrow {\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU,$$

$$[f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\in {\Omega }_{k}forz\in U]\Rightarrow f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Using the relation ${\Omega }_k\subset \{w:|\mathrm{Arg}w|<arctan{\displaystyle \frac{1}{k}}\}$ for $k>0$ and Theorem 2 with $\beta ={\displaystyle \frac{2}{\pi }}arctan{\displaystyle \frac{1}{k}}$, we immediately get

Corollary 12. 

Let $0<\alpha <1$ and $k>0$ be such that

$${\displaystyle \frac{1}{k}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}}\in {\Omega }_{k}forz\in U,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f\left(z\right)/z$ or $p\left(z\right)=f^{\prime }\left(z\right)$ in Corollary 12, we get

Corollary 13. 

Let $0<\alpha <1$ and $k>0$ be such that

$${\displaystyle \frac{1}{k}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If $f\in \mathcal{H}$, then

$$\left[\left({\displaystyle \frac{f\left(z\right)}{z}}+{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right)\in {\Omega }_{k}forz\in U\right]\Rightarrow {\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU,$$

$$\left[\left(f^{\prime }\left(z\right)+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\in {\Omega }_{k}forz\in U\right]\Rightarrow f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Denote by ${\mathcal{S}}^{*}(A,B)$ the class of Janowski starlike functions [27]. A function $f\in \mathcal{S}$ belongs to the class ${\mathcal{S}}^{*}(A,B)$ for $-1<B\le 0<A\le 1$ if and only if

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec {\displaystyle \frac{1+Az}{1+Bz}}\mathrm{in}U,$$

or, equivalently, if and only if

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\in {\Omega }_{A,B}=\left\{w:\left|w-{\displaystyle \frac{1-AB}{1-B^2}}\right|<{\displaystyle \frac{A-B}{1-B^2}}\right\}\mathrm{for}z\in U.$$

Observe that ${\Omega }_{A,B}$ is the disk with the center at ${\displaystyle \frac{1-AB}{1-B^2}}$ and the radius ${\displaystyle \frac{A-B}{1-B^2}}$. We are looking for the smallest $\beta$ such that the disk ${\Omega }_{A,B}$ is contained in the angle $\{w:|\mathrm{Arg}w|<{\displaystyle \frac{\beta \pi }{2}}\}$. Simple geometry shows that $sin{\displaystyle \frac{\beta \pi }{2}}={\displaystyle \frac{A-B}{1-AB}}.$

Using Theorem 1 with $\beta ={\displaystyle \frac{2}{\pi }}arcsin{\displaystyle \frac{A-B}{1-AB}}$, we get

Corollary 14. 

Let $0<\alpha \le 1$ and $A,B$ be such that $-1<B\le 0<A\le 1$ and

$$arcsin{\displaystyle \frac{A-B}{1-AB}}\le {\displaystyle \frac{\alpha \pi }{2}}+arctan\alpha .$$

If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\displaystyle \frac{1+Az}{1+Bz}}inU$$

or, equivalently,

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\in {\Omega }_{A,B}forz\in U,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f\left(z\right)/z$ or $p\left(z\right)=f^{\prime }\left(z\right)$ in Corollary 14, we get

Corollary 15. 

Let $0<\alpha \le 1$ and $A,B$ be such that $-1<B\le 0<A\le 1$ and

$$arcsin{\displaystyle \frac{A-B}{1-AB}}\le {\displaystyle \frac{\alpha \pi }{2}}+arctan\alpha .$$

If $f\in \mathcal{H}$, then

$$\left[f^{\prime }\left(z\right)\prec {\displaystyle \frac{1+Az}{1+Bz}}inU\right]\Rightarrow {\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU,$$

$$\left[f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\prec {\displaystyle \frac{1+Az}{1+Bz}}inU\right]\Rightarrow f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Applying the relation ${\Omega }_{A,B}\subset \{w:|\mathrm{Arg}w|<arcsin{\displaystyle \frac{A-B}{1-AB}}\}$ and Theorem 2 with $\beta ={\displaystyle \frac{2}{\pi }}arcsin{\displaystyle \frac{A-B}{1-AB}}$, we immediately get

Corollary 16. 

Let $0<\alpha <1$ and $A,B$ be such that $-1<B\le 0<A\le 1$ and

$${\displaystyle \frac{A-B}{\sqrt{(1-A^2)(1-B^2)}}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If p is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}}\prec {\displaystyle \frac{1+Az}{1+Bz}}inU,$$

then

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

Taking $p\left(z\right)=f\left(z\right)/z$ or $p\left(z\right)=f^{\prime }\left(z\right)$ in Corollary 16, we get

Corollary 17. 

Let $0<\alpha <1$ and $A,B$ be such that $-1<B\le 0<A\le 1$ and

$${\displaystyle \frac{A-B}{\sqrt{(1-A^2)(1-B^2)}}}\le tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}.$$

If $f\in \mathcal{H}$, then

$$\left[{\displaystyle \frac{f\left(z\right)}{z}}+{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\prec {\displaystyle \frac{1+Az}{1+Bz}}inU\right]\Rightarrow {\displaystyle \frac{f\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU,$$

$$\left[f^{\prime }\left(z\right)+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\prec {\displaystyle \frac{1+Az}{1+Bz}}inU\right]\Rightarrow f^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU.$$

4. Concluding Remarks

We gave some differential subordination results for classes of strongly starlike functions of order $\alpha$ and their various subclasses. Using Miller–Mocanu’s lemma, we derived sufficient conditions under which analytic functions are subordinate to a given function closely related to the class of strongly starlike functions of order $\alpha$.

Obtained results can be used to generate non-trivial examples of functions illustrating the application of the main theorems.

Example 1. 

Let

$$f_1\left(z\right)=2arctan\sqrt{{\displaystyle \frac{1+z}{1-z}}}-\sqrt{1-z^2}+1-{\displaystyle \frac{\pi }{2}},z\in U.$$

Then

$$f_1^{\prime }\left(z\right)=\sqrt{{\displaystyle \frac{1+z}{1-z}}}={\left({\displaystyle \frac{1+z}{1-z}}\right)}^{{\displaystyle \frac{1}{2}}},z\in U.$$

Thus, in view of Corollary 3

$${\displaystyle \frac{f_1\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha \le 1$ that $\alpha \pi /2+arctan\alpha \le \pi /4$.

Example 2. 

Let $0<A<1$ and

$$f_2\left(z\right)=-z-{\displaystyle \frac{2}{A}}log(1-Az)=z+A{z}^2+{\displaystyle \frac{2}{3}}{A}^2{z}^3+{\displaystyle \frac{1}{2}}{A}^3{z}^4+\dots ,z\in U.$$

Then

$$f_2^{\prime }\left(z\right)={\displaystyle \frac{1+Az}{1-Az}}$$

and it maps U onto the disk

$$\left\{w\in \mathbb{C}:\left|w-{\displaystyle \frac{1+A^2}{1-A^2}}\right|<{\displaystyle \frac{2A}{1-A^2}}\right\}$$

contained in the region $\{w\in \mathbb{C}:|\mathrm{Arg}w|<arcsin{\displaystyle \frac{2A}{1+A^2}}\}.$ Thus

$$f_2^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }inU,$$

where $\beta ={\displaystyle \frac{2}{\pi }}arcsin{\displaystyle \frac{2A}{1+A^2}}$. Hence, by Corollary 3

$${\displaystyle \frac{f_2\left(z\right)}{z}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha \le 1$ that $\alpha \pi /2+arctan\alpha \le arcsin{\displaystyle \frac{2A}{1+A^2}}.$

Example 3. 

Let

$$f_3\left(z\right)=2-2\sqrt{1-z}=z+{\displaystyle \frac{1·1}{1·4}}{z}^2+{\displaystyle \frac{1·1·3}{1·4·6}}{z}^3+{\displaystyle \frac{1·1·3·5}{1·4·6·8}}{z}^4+\dots ,z\in U.$$

Then

$$f_3^{\prime }\left(z\right)={\displaystyle \frac{1}{\sqrt{1-z}}},z\in U.$$

This function maps the unit disk onto the domain bounded by the hyperbola

$$\{w=u+iv:u^2-v^2=1/2,u>0\}$$

and $f_3^{\prime }\left(U\right)\subset \{w\in \mathbb{C}:|\mathrm{Arg}w|<\pi /4\}.$ Thus

$$f_3^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{{\displaystyle \frac{1}{2}}}inU$$

and by Corollary 3

$${\displaystyle \frac{f_3\left(z\right)}{z}}={\displaystyle \frac{2}{1+\sqrt{1-z}}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha \le 1$ that $\alpha \pi /2+arctan\alpha \le \pi /4$.

Example 4. 

Let

$$f_4\left(z\right)={\displaystyle \frac{2}{3}}\left[{(1+z)}^{3/2}-1\right]=z+{\displaystyle \frac{1}{4}}{z}^2-{\displaystyle \frac{1·1}{4·6}}{z}^3+{\displaystyle \frac{1·1·3}{4·6·8}}{z}^4-\dots ,z\in U.$$

Then

$$f_4^{\prime }\left(z\right)=\sqrt{1+z},z\in U.$$

The image of the unit disk U under $f_4^{\prime }$ is the region bounded by the lemniscate

$$\{w=u+iv:{(u^2+v^2)}^2-2(u^2-v^2)=0,u>0\}$$

and $f_4^{\prime }\left(U\right)\subset \{w\in \mathbb{C}:|\mathrm{Arg}w|<\pi /4\}.$ Thus

$$f_4^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{{\displaystyle \frac{1}{2}}}inU$$

and by Corollary 3

$${\displaystyle \frac{f_4\left(z\right)}{z}}={\displaystyle \frac{2(z^2+3z+3)}{3\left[{(1+z)}^{3/2}+1\right]}}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha \le 1$ that $\alpha \pi /2+arctan\alpha \le \pi /4$.

Example 5. 

Let $f_5$ be such that $f_5\in \mathcal{H}$ and it satisfies

$$f_5^{\prime }\left(z\right)+{\displaystyle \frac{z{f}_5^{″}\left(z\right)}{f_5^{\prime }\left(z\right)}}=\sqrt{1+z},z\in U.$$

We know (see the previous example) that

$$\sqrt{1+z}\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{{\displaystyle \frac{1}{2}}}inU.$$

Thus, by Corollary 8

$$f_5^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha <1$ that

$$tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}\ge 1.$$

Example 6. 

Let $f_6$ be such that $f_6\in \mathcal{H}$ and it satisfies

$$f_6^{\prime }\left(z\right)+{\displaystyle \frac{z{f}_6^{″}\left(z\right)}{f_6^{\prime }\left(z\right)}}={\left({\displaystyle \frac{1+z}{1-z}}\right)}^{{\displaystyle \frac{1}{2}}},z\in U.$$

Then by Corollary 8

$$f_6^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }inU$$

for such $0<\alpha <1$ that

$$tan{\displaystyle \frac{\alpha \pi }{2}}+{\displaystyle \frac{\alpha {\left(\sqrt{1-\alpha }\right)}^{\alpha -1}}{cos\frac{\alpha \pi }{2}{\left(\sqrt{1+\alpha }\right)}^{\alpha +1}}}\ge 1.$$