1 Introduction

Let \(\mathbb {D}=\{z\in \mathbb {C}:|z|<1\}\) be the open unit disk. An analytic function f in \(\mathbb {D}\) with \(0 \ne f^\prime \left( 0\right) \) is convex, i.e. \({\text {Re}}\left[ 1+zf^{\prime \prime }\left( z\right) /f^\prime \left( z\right) \right] >0\) for all \(z\in \mathbb {D}, \) if and only if f is univalent in \(\mathbb {D}\) with \(f\left( \mathbb {D}\right) \) being convex.

The Struve functions, denoted by \(H_\nu \), are solutions of the inhomogeneous Bessel differential equation

$$\begin{aligned} z^2w''(z)+zw'(z)+(z^2-\nu ^2)w(z)=\frac{4\left( \frac{z}{2}\right) ^{\nu +1}}{\sqrt{\pi }\Gamma \left( \nu +\frac{1}{2}\right) }, \end{aligned}$$

and they have the power series form

$$\begin{aligned} H_\nu (z)=\sum _{n=0}^\infty \frac{\left( -1\right) ^n}{\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }\left( \frac{z}{2}\right) ^{2n+\nu +1}. \end{aligned}$$

We define the normalized form of Struve functions by the equality

$$\begin{aligned} h_\nu (z)=z^\frac{1-\nu }{2}2^\nu {H}_\nu (\sqrt{z})=\sum _{n=0}^\infty \frac{\left( -1\right) ^nz^{n+1}}{2^{2n+1}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }. \end{aligned}$$
(1.1)

An other normalized form of \(H_\nu \), which we are going to study is

$$\begin{aligned} g_\nu (z)=\left( \frac{2}{z}\right) ^\nu {H_\nu (z)}=\sum _{n=0}^\infty \frac{\left( -1\right) ^nz^{2n+1}}{2^{2n+1}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2} \right) }. \end{aligned}$$
(1.2)

After Louis de Branges proved the Bieberbach Conjecture by using the generalized hypergeometric function in 1984, special functions became popular in studies of geometric function theory. Recently, there has been great interest dealing with various properties of special functions such as Bessel, Struve and Lommel functions of the first kind [1,2,3,4,5,6,7,8,9,10, 12, 14, 15].

The aim of the paper is to prove the implications

$$\begin{aligned} \mu >\nu \ \Rightarrow h_\mu (\mathbb {D})\subset h_\nu (\mathbb {D})~~ \text {and} ~~g_\mu (\mathbb {D})\subset g_\nu (\mathbb {D}) \end{aligned}$$

regarding normalized Struve functions. In order to prove our result we need the definitions and lemmas presented in the next section.

2 Preliminaries

In this section, we are devoted to giving and proving our main results.

Definition 2.1

Let f, g and h be analytic functions in \(\mathbb {D}\). If the function h satisfies the conditions \(h(0)=0\), \(|h(z)|<1, \ z\in \mathbb {D}\) and \(f(z)=g(h(z)), \ z\in \mathbb {D}\), then we say that the function f is subordinate to the function g.

This subordination is denoted by \(f\prec {g}\).

If g is univalent and \(f(0)=g(0)\), then the following equivalence holds:

$$\begin{aligned} f\prec {g} \ \Leftrightarrow \ f(\mathbb {D})\subset {g}(\mathbb {D}). \end{aligned}$$

Definition 2.2

An infinite sequence \((b_n)_{n\ge 1}\) of complex numbers will be called a subordination factor sequence if for every convex function f defined by \(f(z)=\sum _{n=1}^\infty {a_n}z^n\) we have \(g\prec {f}\), where g is defined by \(g(z)=\sum _{n=1}^\infty {a_nb_n}z^n, z\in \mathbb {D}\).

Definition 2.3

Let f and g be two analytic functions on \(\mathbb {D}\) defined by the power series \(f(z)=\sum ^\infty _{n=1}a_nz^n\) and \(g(z)=\sum ^\infty _{n=1}b_nz^n\). The convolution of the functions f and g denoted by \(f*g\) is defined by the equality

$$\begin{aligned} (f*g)(z)=\sum ^\infty _{n=1}a_nb_nz^n. \end{aligned}$$

Remark 2.1

Definition 2.1 and 2.2 can be reformulated using convolution as follows: Let \((b_n)_{n\ge 1}\) be a sequence of complex numbers and the function \(\phi \) be defined by the equality \(\phi (z)=\sum ^\infty _{n=1}b_nz^n\). An infinite sequence \((b_n)_{n\ge 1}\) of complex numbers will be called a subordination factor sequence if for every convex function f we have

$$\begin{aligned} f*\phi \prec {f}. \end{aligned}$$

Lemma 2.1

[16] The following two properties of sequence of complex numbers are equivalent:

  1. (I)

    The infinite sequence \((b_n)_{n\ge 1}\) of complex numbers is a subordination factor sequence.

  2. (II)

    The inequality \(\frac{1}{2}+{\text {Re}}\sum _{n=1}^\infty {b_n}z^n>0\), holds for every \(z\in \mathbb {D}\).

Lemma 2.2

[11] Let \((f_n)_{n\ge 0}\) be a sequence of real numbers such that \(f_0=1\), \(f_n-2f_{n+1}+f_{n+2}\ge 0\) and \(f_{n}-f_{n+1} \ge 0\) for all \(n\in \{0,1,2,3,\ldots \}\), then the inequality holds

$$\begin{aligned} {\text {Re}}\left( 1+\sum ^\infty _{n=1} f_nz^{n}\right) >\frac{1}{2}, \ \text {for every} \ z\in \mathbb {D}. \end{aligned}$$

In Theorem 2 of the study [13], the authors show that for \( n\ge 1\) and \(z\in \mathbb {D}\), if \(f(z)=z+a_{n+1}z^{n+1}+\dots \) and \(|f^{\prime \prime }(z)|\le \frac{n}{n+1}\), then f is convex. Then, the following lemma is a consequence of this result.

Lemma 2.3

[13] If f is of the form \(f(z)=\sum ^\infty _{n=1}a_nz^n\) with \(a_1\ne 0\) and

$$\begin{aligned} \left| \frac{f''(z)}{f'(0)}\right| <\frac{1}{2}, \ z\in \mathbb {D},\end{aligned}$$
(2.1)

then f is univalent with \(f(\mathbb {D})\) convex set in \(\mathbb {C}\).

The following lemmas are the key tool in the proof of our main results.

Lemma 2.4

If \(y>x>\frac{5}{2}\) then the inequality holds

$$\begin{aligned} \frac{x}{y}\ge \frac{\Gamma (x)}{\Gamma (y)}. \end{aligned}$$

Proof

The inequality we have to prove is equivalent to

$$\begin{aligned} \frac{x}{\Gamma (x)}\ge \frac{y}{\Gamma (y)}. \end{aligned}$$

In order to prove this inequality we will prove that the function \(u:(\frac{3}{2},\infty )\rightarrow \mathbb {R}, \ u(t)=\frac{1+t}{\Gamma (1+t)}\) is decreasing.

We have

$$\begin{aligned} \frac{u'(t)}{u(t)}=\frac{1}{1+t} -\frac{\Gamma '(1+t)}{\Gamma (1+t)}=\frac{1}{1+t} +\gamma -\sum _{n=1}^\infty \frac{t}{n(n+t)}. \end{aligned}$$
(2.2)

The sum \(\sum _{n=1}^\infty \frac{t}{n(n+t)}\) is increasing with respect to t and \(t>1\). Thus it follows

$$\begin{aligned} \frac{u'(t)}{u(t)}\le \frac{1}{1+\frac{3}{2}}+\gamma -\sum _{n=1}^\infty \frac{1}{n(n+1)}=\frac{2}{5}+\gamma -1<0, \ (\forall ) t\in \left( \frac{3}{2},\infty \right) . \end{aligned}$$

Consequently the mapping \(u:(\frac{3}{2},\infty )\rightarrow \mathbb {R}, \ u(t)=\frac{1+t}{\Gamma (1+t)}\) is decreasing. On the other hand the condition \(y>x>\frac{5}{2}\) is equivalent to \(y-1>x-1>\frac{3}{2}\) and the monotony of u implies \(u(y-1)\le {u}(x-1)\) and this inequality is equivalent to \(\frac{x}{\Gamma (x)}\ge \frac{y}{\Gamma (y)}\). The proof is completed \(\square \)

Lemma 2.5

Provided that \(\mu>\nu >0\) the inequality

$$\begin{aligned} {\text {Re}}\left( 1+\sum ^\infty _{n=1}\frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) }z^{n}\right) >\frac{1}{2} \end{aligned}$$

holds for every \(z\in \mathbb {D}\).

Proof

We use Lemma 2.2 and Lemma 2.4 in this proof. We have \(f_0=1\) and \(f_n=\frac{\Gamma (n+\nu +\frac{3}{2})}{\Gamma (n+\mu +\frac{3}{2})}\) in case \(n\ge 1\). Thus \(f_0-f_1=1-\frac{\Gamma (\nu +\frac{5}{2})}{\Gamma (\mu +\frac{5}{2})}\ge 0\) because \(\mu>\nu >0\) and if \(n\ge 1\), then

$$\begin{aligned} f_n-f_{n+1}= & {} \frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) } -\frac{\Gamma \left( n+\nu +\frac{5}{2}\right) }{\Gamma \left( n+\mu +\frac{5}{2}\right) }\\= & {} \frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) }\left( 1-\frac{n+\nu +\frac{3}{2}}{n+\mu +\frac{3}{2}}\right) \ge 0. \end{aligned}$$

In order to prove the inequality \(f_n-2f_{n+1}+f_{n+2}\ge 0\), we distinguish two cases exactly as before. In case \(n=0\), we have to show \(f_0-2f_{1}+f_{2}\ge 0\).

We have

$$\begin{aligned} f_0-2f_{1}+f_{2}= & {} 1-2\frac{\Gamma \left( \nu +\frac{5}{2}\right) }{\Gamma \left( \mu +\frac{5}{2}\right) }+\frac{\Gamma \left( 1+\nu +\frac{5}{2}\right) }{\Gamma \left( 1+\mu +\frac{5}{2}\right) }\\= & {} 1+\frac{\left( \nu +\frac{5}{2}\right) \Gamma \left( \nu +\frac{5}{2}\right) }{\left( \mu +\frac{5}{2}\right) \Gamma \left( \mu +\frac{5}{2}\right) }-2\frac{\Gamma \left( \nu +\frac{5}{2}\right) }{\Gamma \left( \mu +\frac{5}{2}\right) }\\{} & {} \quad \ge 2\sqrt{\frac{\left( \nu +\frac{5}{2}\right) \Gamma \left( \nu +\frac{5}{2}\right) }{\left( \mu +\frac{5}{2}\right) \Gamma \left( \mu +\frac{5}{2}\right) }}-2\frac{\Gamma \left( \nu +\frac{5}{2}\right) }{\Gamma \left( \mu +\frac{5}{2}\right) }\\= & {} 2\sqrt{\frac{\Gamma \left( \nu +\frac{5}{2}\right) }{\Gamma \left( \mu +\frac{5}{2}\right) }}\left( \sqrt{\frac{\nu +\frac{5}{2}}{\mu +\frac{5}{2}}}-\sqrt{\frac{\Gamma \left( \nu +\frac{5}{2}\right) }{\Gamma \left( \mu +\frac{5}{2}\right) }}\right) . \end{aligned}$$

On the other hand Lemma 2.4 implies \(\sqrt{\frac{\nu +\frac{5}{2}}{\mu +\frac{5}{2}}}-\sqrt{\frac{\Gamma (\nu +\frac{5}{2})}{\Gamma (\mu +\frac{5}{2})}}\ge 0\). Thus we get \(f_0-2f_{1}+f_{2}\ge 0\).

The proof of the inequality \(f_n-2f_{n+1}+f_{n+2}\ge 0\) in case \(n\ge 1\) is much easier than the particular case for \(n=0\).

We have

$$\begin{aligned}{} & {} f_n-2f_{n+1}+f_{n+2}\\{} & {} \quad = \frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) }-2\frac{\Gamma \left( n+1+\nu +\frac{3}{2}\right) }{\Gamma \left( n+1+\mu +\frac{3}{2}\right) }+\frac{\Gamma \left( n+2+\nu +\frac{3}{2}\right) }{\Gamma \left( n+2+\mu +\frac{3}{2}\right) } \\{} & {} \quad = \frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) }\left( 1+\frac{\left( n+\nu +\frac{3}{2}\right) \left( n+\nu +\frac{5}{2}\right) }{\left( n+\mu +\frac{3}{2}\right) \left( n+\mu +\frac{5}{2}\right) }-2\frac{n+\nu +\frac{3}{2}}{n+\mu +\frac{3}{2}}\right) \\{} & {} \ge 2\frac{\Gamma \left( n+\nu +\frac{3}{2}\right) }{\Gamma \left( n+\mu +\frac{3}{2}\right) }\left( \sqrt{\frac{\left( n+\nu +\frac{3}{2}\right) \left( n+\nu +\frac{5}{2}\right) }{\left( n+\mu +\frac{3}{2}\right) \left( n+\mu +\frac{5}{2}\right) }}-\frac{n+\nu +\frac{3}{2}}{n+\mu +\frac{3}{2}}\right) \ge 0. \end{aligned}$$

Thus, the proof is completed. \(\square \)

Lemma 2.6

Provided that \(\mu>\nu >0\) the inequality holds

$$\begin{aligned} {\text {Re}}\left( 1+\sum ^\infty _{n=1}\frac{\Gamma \left( \nu +1+\frac{n}{2}\right) }{\Gamma \left( \mu +1+\frac{n}{2}\right) }z^{n}\right) >\frac{1}{2}, \ \text {for every} \ z\in \mathbb {D}. \end{aligned}$$
(2.3)

Proof

We use for the second time 2.2 in order to prove this lemma. This time \(f_0=1\) and \(f_n=\frac{\Gamma (\nu +1+\frac{n}{2})}{\Gamma (\mu +1+\frac{n}{2})}\).

Since \(\mu>\nu >0\) it follows that \(f_0=1>\frac{\Gamma (\nu +1+\frac{1}{2})}{\Gamma (\mu +1+\frac{1}{2})}=f_1\). We have to show \(f_n=\frac{\Gamma (\nu +1+\frac{n}{2})}{\Gamma (\mu +1+\frac{n}{2})}>\frac{\Gamma (\nu +1+\frac{n+1}{2})}{\Gamma (\mu +1+\frac{n+1}{2})}=f_{n+1}\) in case \(n\ge 1\). This inequality can be rewritten as follows

$$\begin{aligned} \frac{\Gamma \left( \nu +1+\frac{n}{2}\right) }{\Gamma \left( \nu +1+\frac{n+1}{2}\right) }>\frac{\Gamma \left( \mu +1+\frac{n}{2}\right) }{\Gamma \left( \mu +1+\frac{n+1}{2}\right) }. \end{aligned}$$
(2.4)

We define the function \(u:(1,\infty )\rightarrow (0,\infty )\) by \(u(x)=\frac{\Gamma (x)}{\Gamma (x+\frac{1}{2})}\). We have

$$\begin{aligned} u(x)=\frac{\Gamma (x)}{\Gamma \left( x+\frac{1}{2}\right) }=\frac{B\left( x,\frac{1}{2}\right) }{\Gamma \left( \frac{1}{2}\right) }. \end{aligned}$$
(2.5)

B(x, y) denotes the Euler’s Beta function defined by \(B(x,y)=\int _0^1t^{x-1}(1-t)^{y-1}dt\). Thus equality (2.5) implies that the mapping u is strictly decreasing and consequently \(u(\nu +1+\frac{n}{2})>u(\mu +1+\frac{n}{2})\), which is equivalent to (2.4). In order to finish the proof we have to show that

$$\begin{aligned} 1+f_2\ge 2f_1 \ \text {and} \ \ f_n+f_{n+2}\ge 2f_{n+1}, \ n\ge 1.\end{aligned}$$
(2.6)

We define the mapping \(v:(1,\infty )\rightarrow (0,\infty )\) by \(v(x)=\frac{\Gamma (x)}{\Gamma (x+\mu -\nu )}=\frac{B(x,\mu -\nu )}{\Gamma (\mu -\nu )}\). The equivalence holds

$$\begin{aligned} f_n+f_{n+2}\ge 2f_{n+1} \end{aligned}$$

if and only if

$$\begin{aligned} \frac{v\left( \nu +1+\frac{n}{2}\right) +v\left( \nu +1+\frac{n+2}{2}\right) }{2} \ge v \left( \frac{\left( \nu +1+\frac{n}{2}\right) +\left( \nu +1+\frac{n+2}{2}\right) }{2}\right) , \ \ n\ge 1. \end{aligned}$$

Since \(f_0=1\ge {v}(\nu +1)\), it follows that in order to prove (2.6) it is enough to prove

$$\begin{aligned}{} & {} \frac{v\left( \nu +1+\frac{n}{2}\right) +v\left( \nu +1+\frac{n+2}{2}\right) }{2}\nonumber \\{} & {} \quad \ge v \left( \frac{\left( \nu +1+\frac{n}{2}\right) +\left( \nu +1+\frac{n+2}{2}\right) }{2}\right) , \ \ n\ge 0. \end{aligned}$$
(2.7)

On the other hand we have \(v''(x)=\frac{1}{\Gamma (\mu -\nu )}\int _0^1t^{x-1}\ln ^2(t)(1-t)^{\mu -\nu -1}dt\ge 0\). Thus the function v is convex and consequently inequality (2.7) hods. \(\square \)

3 The Main Result

Our first main result is the following theorem, which gives the monotonicity result for the function \(h_\nu \).

Theorem 3.1

If \(\mu>\nu >\nu _0=\frac{\sqrt{8585}}{90}-\frac{5}{3}\approx -0.63716\), then the following inclusion holds

$$\begin{aligned} h_\mu (\mathbb {D})\subset {h_\nu }(\mathbb {D}), \end{aligned}$$
(3.1)

where \(\nu _0\) is the biggest real root of the equation \(1620\nu ^2+5400\nu +2783=0\) and \(h_\nu \) is defined by (1.1).

Proof

We have

$$\begin{aligned} \left| \frac{h_\nu ''(z)}{h_\nu '(0)}\right|{} & {} \le \sum _{n=1}^\infty \frac{n(n+1)\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) } =\frac{\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2\Gamma \left( 1+\frac{3}{2}\right) \Gamma \left( 1+\nu +\frac{3}{2}\right) }\nonumber \\{} & {} \quad + \sum _{n=2}^\infty \frac{n\left( n+1\right) \Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }. \end{aligned}$$
(3.2)

Since the following inequalities hold

$$\begin{aligned} \frac{\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2\Gamma \left( 1+\frac{3}{2}\right) \Gamma \left( 1+\nu +\frac{3}{2}\right) }<\frac{2}{3\left( 2\nu _0+3\right) }, \end{aligned}$$

and

$$\begin{aligned} \sum _{n=2}^\infty \frac{n\left( n+1\right) \Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }{} & {} \le \left( \frac{16}{15\left( 2\nu _0+5\right) \left( 2\nu _0+3\right) }\right) \sum _{n=2}^\infty \frac{n\left( n+1\right) }{2^{2n}}\\{} & {} =\frac{296}{405}\frac{1}{\left( 2\nu _0+5\right) \left( 2\nu _0+3\right) }, \end{aligned}$$

we get

$$\begin{aligned} \left| \frac{h_\nu ''(z)}{h_\nu '(0)}\right| \le \frac{2}{3\left( 2\nu _0+3\right) }+\frac{296}{405}\frac{1}{\left( 2\nu _0+5\right) \left( 2\nu _0+3\right) }=\frac{1}{2}. \end{aligned}$$

Thus Lemma 2.3 implies that \(h_\nu \) is a convex function. According to Lemma 2.5 the sequence \(\left( \frac{\Gamma (n+\nu +\frac{3}{2})}{\Gamma (n+\mu +\frac{3}{2})}\right) _{n\ge 1}\) is a subordination factor sequence and the function \(h_\nu \) is convex, consequently the subordination follows

$$\begin{aligned} h_\mu =\chi *h_\nu \prec {h_\nu }, \end{aligned}$$

where \(\chi (z)=\sum ^\infty _{n=1}\frac{\Gamma (n+\nu +\frac{3}{2})}{\Gamma (n+\mu +\frac{3}{2})}z^n\). Since \(h_\mu (0)=h_\nu (0)\), the subordination \(h_\mu \prec {h_\nu }\) is equivalent to \(h_\mu (\mathbb {D})\subset {h_\nu }(\mathbb {D})\). Thus, the proof is completed. \(\square \)

Our second result is stated as follows.

Theorem 3.2

If \(\mu>\nu >\nu _1=\frac{\sqrt{28385}}{90}-1\approx 0.87198\), then the following inclusion holds

$$\begin{aligned} g_\mu (\mathbb {D})\subset {g_\nu }(\mathbb {D}), \end{aligned}$$
(3.3)

where \(\nu _1\) is the biggest real root of the equation \(1620\nu ^2+3200\nu -4057=0\) and \(g_\nu \) is defined by (1.2).

Proof

We have

$$\begin{aligned} \left| \frac{g_\nu ''(z)}{g_\nu '(0)}\right|{} & {} \le \sum _{n=1}^\infty \frac{2n(2n+1)\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }=\frac{3\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2\Gamma \left( 1+\frac{3}{2}\right) \Gamma \left( 1+\nu +\frac{3}{2}\right) }\nonumber \\{} & {} \quad + \sum _{n=2}^\infty \frac{2n(2n+1)\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }. \end{aligned}$$
(3.4)

After simple calculations, the inequalities

$$\begin{aligned} \frac{3\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2\Gamma \left( 1+\frac{3}{2}\right) \Gamma \left( 1+\nu +\frac{3}{2}\right) }<\frac{2}{2\nu _1+3}, \end{aligned}$$

and

$$\begin{aligned} \frac{\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }\le \frac{16}{15}\frac{1}{\left( 2\nu _1+3\right) \left( 2\nu _1+5\right) }, \end{aligned}$$

true for \(\ n\ge 2, \ \nu >\nu _1\). Finally

$$\begin{aligned} \sum _{n=2}^\infty \frac{2n(2n+1)\Gamma \left( \frac{3}{2}\right) \Gamma \left( \nu +\frac{3}{2}\right) }{2^{2n}\Gamma \left( n+\frac{3}{2}\right) \Gamma \left( n+\nu +\frac{3}{2}\right) }{} & {} \le \left( \frac{16}{15}\frac{1}{\left( 2\nu _1+3\right) \left( 2\nu _1+5\right) }\right) \sum _{n=2}^\infty \frac{2n(2n+1)}{2^{2n}}\nonumber \\{} & {} =\frac{16}{15}\frac{1}{\left( 2\nu _1+3\right) \left( 2\nu _1+5\right) }\frac{127}{54}.\end{aligned}$$
(3.5)

Thus, we get

$$\begin{aligned} \left| \frac{g^{\prime \prime }_\nu (z)}{g^\prime _\nu (0)} \right| \le \frac{2}{2\nu _1+3}+\frac{1016}{405} \frac{1}{\left( 2\nu _1+3\right) \left( 2\nu _1+5\right) } =\frac{1}{2}. \end{aligned}$$

Thus Lemma 2.3 implies that \(g_\nu \) is a convex function. According to Lemma 2.6 the sequence \(\left( \frac{\Gamma (\nu +1+\frac{n}{2})}{\Gamma (\mu +1+\frac{n}{2})}\right) _{n\ge 1}\) is a subordination factor sequence and the function \(g_\nu \) is convex, consequently the subordination follows

$$\begin{aligned} g_\mu =\psi *g_\nu \prec {g_\nu }, \end{aligned}$$

where \(\psi (z)=\sum ^\infty _{n=1}\frac{\Gamma (\nu +1+\frac{n}{2})}{\Gamma (\mu +1+\frac{n}{2})}z^n\). Since \(g_\mu (0)=g_\nu (0)\), the subordination \(g_\mu \prec {g_\nu }\) is equivalent to \(g_\mu (\mathbb {D})\subset {g_\nu }(\mathbb {D})\). Thus, the proof is completed. \(\square \)