Theorem 3.1
Let
\(\alpha \geq 1\), \(\beta \geq 1\). Then
\(E_{\alpha ,\beta } ( z ) \in \mathcal{R} [ \rho ,\mu ] \)
for
\(0\leq \mu <1\)
and
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq \textstyle\begin{cases} T_{1} ( \rho ,\mu ) , \quad 0\leq \rho \leq \rho _{1} ( \mu ) , \\ \max \{ T_{1} ( \rho ,\mu ) , T_{2} ( \rho , \mu ) \frac{2\varGamma ^{2} ( \alpha +\beta ) }{\varGamma ( \beta ) \varGamma ( 2\alpha +\beta ) }, T _{3} ( \rho ,\mu ) \frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+\beta ) }\frac{ \varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) } \} ,\\ \quad \rho _{1} ( \mu ) \leq \rho < 1, \end{cases} $$
where
$$\begin{aligned} T_{1} ( \rho ,\mu ) =&\frac{2 ( 2-\mu ) ( 1-\rho ) }{ ( 1-\mu ) },\qquad T_{2} ( \rho ,\mu ) =\frac{2^{\mu -1} ( 3-\mu ) ( 3-2\rho ) }{ ( 2-\mu ) }, \\ T_{3} ( \rho ,\mu ) =&\frac{2 ( 4-\mu ) ( 2-\rho ) }{ ( 2-\mu ) ( 3-\mu ) },\quad \rho _{1} ( \mu ) =1-\frac{2^{\mu } ( 3-\mu ) ( 1-\mu ) }{4 ( 2-\mu ) ^{2}-2.2^{\mu } ( 3-\mu ) ( 1-\mu ) }. \end{aligned}$$
Proof
Consider the function \(g ( z ) =z+\sum^{\infty }_{k=2} b_{k}z^{k}\), where \(b_{k}\) is given by
$$ b_{1}=1, \qquad b_{k+1}=\frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+\beta ) } \frac{ ( k+1-2\rho ) }{k}b_{k}, \quad \forall k\geq 1. $$
Now
$$\begin{aligned} & ( 1-\mu ) b_{1}- ( 2-\mu ) b_{2} \\ &\quad = ( 1-\mu ) -2 ( 2-\mu ) \frac{\varGamma ( \beta ) }{\varGamma ( \alpha +\beta ) } ( 1- \rho ) \\ &\quad =\frac{1}{\varGamma ( \alpha +\beta ) } \bigl[ ( 1- \mu ) \varGamma ( \alpha +\beta ) -2 ( 2-\mu ) ( 1-\rho ) \varGamma ( \beta ) \bigr] \geq 0 \end{aligned}$$
implies \(\varGamma ( \alpha +\beta ) \geq T_{1} ( \rho ,\mu ) \varGamma ( \beta ) \), where \(T_{1} ( \rho ,\mu ) =\frac{2 ( 2-\mu ) ( 1- \rho ) }{ ( 1-\mu ) }\). Again
$$\begin{aligned} & ( 2-\mu ) b_{2}-2^{\mu +1} ( 3-\mu ) b_{3} \\ &\quad = ( 2-\mu ) b_{2}-2^{\mu +1} ( 3-\mu ) \frac{ \varGamma ( \alpha +\beta ) }{\varGamma ( 2\alpha + \beta ) } \frac{ ( 3-2\rho ) }{2}b_{2} \\ &\quad =\frac{2 ( 2-\mu ) b_{2}}{\varGamma ( 2\alpha + \beta ) } \biggl[ \frac{\varGamma ( 2\alpha +\beta ) }{2}-T_{2} ( \rho , \mu ) \varGamma ( \alpha +\beta ) \biggr] \geq 0 \end{aligned}$$
with \(\frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 2T_{2} ( \rho ,\mu ) \frac{\varGamma ^{2} ( \alpha +\beta ) }{\varGamma ( \beta ) \varGamma ( 2\alpha +\beta ) }\), where \(T_{2} ( \rho ,\mu ) =\frac{2^{\mu -1} ( 3-\mu ) ( 3-2 \alpha ) }{ ( 2-\mu ) }\). Also consider
$$\begin{aligned} & ( k-1-\mu ) ( k-\mu ) b_{k}-k ( k+1- \mu ) b_{k+1} \\ &\quad = ( k-1-\mu ) ( k-\mu ) b_{k}-k ( k+1- \mu ) \frac{ ( k+1-2\rho ) }{k} \frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+ \beta ) }b_{k} \\ &\quad =\frac{b_{k}}{\varGamma ( \alpha k+\beta ) } \\ &\qquad {}\times\bigl[ ( k-1-\mu ) ( k-\mu ) \varGamma ( \alpha k+ \beta ) - ( k+1-\mu ) ( k+1-2\rho ) \varGamma \bigl( \alpha ( k-1 ) +\beta \bigr) \bigr] \\ &\quad = A ( k ) M ( k ) , \end{aligned}$$
where \(A ( k ) =\frac{b_{k}}{\varGamma ( \alpha k+ \beta ) }\) and
$$\begin{aligned} M ( k ) =& ( k-1-\mu ) ( k-\mu ) \varGamma ( \alpha k+\beta ) - ( k+1-\mu ) ( k+1-2\rho ) \varGamma \bigl( \alpha ( k-1 ) + \beta \bigr) \\ =&A ( \rho ,\mu ) ( k-3 ) ^{2}+B ( \rho ,\mu ) ( k-3 ) +D ( \rho ,\mu ) . \end{aligned}$$
Here
$$\begin{aligned} &A ( \rho ,\mu ) =\varGamma ( \alpha k+\beta ) -\varGamma \bigl( \alpha ( k-1 ) + \beta \bigr) \geq 0, \\ &B ( \rho ,\mu ) = ( 5-2\mu ) \varGamma ( \alpha k+\beta ) - ( 8-2\rho -\mu ) \varGamma \bigl( \alpha ( k-1 ) +\beta \bigr) \geq 0, \\ &D ( \rho ,\mu ) = ( 2-\mu ) ( 3- \mu ) \varGamma ( \alpha k+\beta ) -2 ( 4- \mu ) ( 2-\rho ) \varGamma \bigl( \alpha ( k-1 ) +\beta \bigr) \geq 0. \end{aligned}$$
This implies that
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq T_{3} ( \rho ,\mu ) \frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+\beta ) } \frac{\varGamma ( \alpha +\beta ) }{ \varGamma ( \beta ) }, $$
where \(T_{3} ( \rho ,\mu ) =\frac{2 ( 4-\mu ) ( 2-\alpha ) }{ ( 2-\mu ) ( 3-\mu ) }\). It is clear that \(A ( \rho ,\mu )\), \(B ( \rho ,\mu )\), \(D ( \rho ,\mu )\) are non-negative. Since each coefficient of \(( k-3 ) \) and the constant term in \(M ( k ) \) are non-negative, therefore \(M ( k ) \) is an increasing function for \(k\geq 3\). Also, for \(M ( 3 ) >0\), we have \(( k-1-\mu ) ( k- \mu ) b_{k}\geq k ( k+1-\mu ) b_{k+1}\). Thus \(b_{k}\) satisfies the conditions of Lemma 2.1 and hence \(g\in \mathcal{S}^{\ast } ( \mu ) \). After simple computations, we observe that \(g ( z ) =E_{\alpha ,\beta } ( z ) \ast \frac{z}{ ( 1-z ) ^{2-2\rho }}\). Therefore, by the definition of \(\mathcal{R} [ \rho ,\mu ]\), we have \(E_{\alpha ,\beta } ( z ) \in \mathcal{R} [ \rho , \mu ] \). Now consider
$$\begin{aligned} T_{3} ( \rho ,\mu ) -T_{1} ( \rho ,\mu ) =& \frac{2 ( 4-\mu ) ( 2-\rho ) }{ ( 2-\mu ) ( 3-\mu ) }-\frac{2 ( 2-\mu ) ( 1- \rho ) }{ ( 1-\mu ) } \\ =&\frac{2 ( 1-\mu ) ( 4-\mu ) ( 2- \rho ) -2 ( 2-\mu ) ^{2} ( 3-\mu ) ( 1-\rho ) }{ ( 1-\mu ) ( 2-\mu ) ( 3-\mu ) }. \end{aligned}$$
The numerator is negative for all μ and hence \(T_{3} ( \rho ,\mu ) \leq T_{1} ( \rho ,\mu ) \) for \(0\leq \rho \leq \rho _{0} ( \mu ) \). Similarly, if \(0\leq \rho \leq \rho _{1} ( \mu ) \), \(T_{1} ( \rho , \mu ) \geq T_{2} ( \rho ,\mu ) \) for all μ. Here,
$$\begin{aligned} \rho _{0} ( \mu ) =&1-\frac{ ( 4-\mu ) ( 1-\mu ) }{ ( 2-\mu ) ^{2} ( 3-\mu ) - ( 4-\mu ) ( 1-\mu ) }, \\ \rho _{1} ( \mu ) =&1-\frac{2^{\mu } ( 3-\mu ) ( 1-\mu ) }{4 ( 2-\mu ) ^{2}-2.2^{\mu } ( 3-\mu ) ( 1-\mu ) }. \end{aligned}$$
Clearly, we can investigate that, for \(0\leq \rho \leq \min \{ \rho _{0} ( \mu ) ,\rho _{1} ( \mu ) \} \),
$$ \max_{i=1,2,3} \bigl\{ T_{i} ( \rho ,\mu ) \bigr\} =T_{1} ( \rho ,\mu ) . $$
Now, we only need to check the \(\min \{ \rho _{0} ( \mu ) ,\rho _{1} ( \mu ) \} \). Consider
$$\begin{aligned} \rho _{1}-\rho _{0} =&-\frac{2^{\mu } ( 3-\mu ) ( 1- \mu ) }{4 ( 2-\mu ) ^{2}-2.2^{\mu } ( 3- \mu ) ( 1-\mu ) }+ \frac{ ( 4-\mu ) ( 1-\mu ) }{ ( 2-\mu ) ^{2} ( 3-\mu ) - ( 4-\mu ) ( 1-\mu ) } \\ =&\frac{N ( \mu ) }{ ( 4 ( 2-\mu ) ^{2}-2.2^{ \mu } ( 3-\mu ) ( 1-\mu ) ) ( ( 2-\mu ) ^{2} ( 3-\mu ) - ( 4-\mu ) ( 1-\mu ) ) }, \end{aligned}$$
where
$$\begin{aligned} N ( \mu ) =&2^{\mu } ( 3-\mu ) ( 1- \mu ) \bigl\{ ( 2-\mu ) ^{2} ( 3-\mu ) - ( 4-\mu ) ( 1-\mu ) \bigr\} \\ &{}+ ( 4-\mu ) ( 1-\mu ) \bigl\{ 4 ( 2- \mu ) ^{2}-2.2^{\mu } ( 3-\mu ) ( 1-\mu ) \bigr\} \\ < &0. \end{aligned}$$
This implies \(\rho _{0} ( \mu ) =\min \{ \rho _{0} ( \mu ) ,\rho _{1} ( \mu ) \} \), and the proof is complete. □
Theorem 3.2
Let
\(0\leq \mu <1\), \(\alpha \geq 1\), \(\beta \geq 1\). If
\(\frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) } \geq 2 ( 2-\mu ) \), then
\(E_{\alpha ,\beta } ( z ) \)
is pre-starlike of order
μ
in
\(\mathcal{U}\).
Proof
Consider \(T_{i} ( \rho ,\mu ) ,i=1,2,3\), as in Theorem 3.1. Replacing ρ by μ, we get
$$\begin{aligned} T_{1} ( \mu ) =&2 ( 2-\mu ) , \\ T_{2} ( \mu ) =&\frac{2^{\mu -1} ( 3-\mu ) ( 3-2\mu ) }{ ( 2-\mu ) }, \\ T_{3} ( \mu ) =&\frac{2 ( 4-\mu ) }{ ( 3- \mu ) }. \end{aligned}$$
It is noticed that, for \(0\leq \mu <1\),
$$\begin{aligned} T_{1} ( \mu ) -T_{2} ( \mu ) =&2 ( 2- \mu ) - \frac{2^{\mu -1} ( 3-\mu ) ( 3-2 \mu ) }{ ( 2-\mu ) } \\ =&\frac{2 ( 2-\mu ) ^{2}-2^{\mu -1} ( 3-\mu ) ( 3-2\mu ) }{ ( 2-\mu ) }>0. \end{aligned}$$
Similarly,
$$\begin{aligned} T_{1} ( \mu ) -T_{3} ( \mu ) =&2 ( 2- \mu ) - \frac{2 ( 4-\mu ) }{ ( 3-\mu ) } \\ =&\frac{2 ( 2-\mu ) ( 3-\mu ) -2 ( 4- \mu ) }{ ( 3-\mu ) }>0. \end{aligned}$$
Therefore, \(T_{1} ( \mu ) \) is maximum. Hence
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq \max \left\{ \textstyle\begin{array}{c} T_{1} ( \mu ) , T_{2} ( \mu ) \frac{2\varGamma ^{2} ( \alpha +\beta ) }{\varGamma ( \beta ) \varGamma ( 2\alpha +\beta ) }, \\ T_{3} ( \mu ) \frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+\beta ) }\frac{ \varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) } \end{array}\displaystyle \right \} =T_{1} ( \mu ) . $$
This is equivalent to
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 2 ( 2-\mu ) . $$
(3.1)
□
Corollary 3.3
Let
\(\alpha \geq 1\), \(\beta \geq 1\). Then
\(E_{\alpha ,\beta } ( z ) \in \mathcal{C}\)
if
\(\frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 4\).
Proof
It is noticed that, for \(\mu =0\), we have \(\frac{z}{ ( 1-z ) ^{2}}\ast f ( z ) \in \mathcal{S}^{\ast }\). By using the definition of convolution, it is easy to see that \(zf^{\prime } ( z ) \in \mathcal{S}^{\ast }\). Therefore, by Alexander relation it follows that \(f\in \mathcal{C}\). We also see from Theorem 3.2 that
$$ T_{1} ( 0 ) =4,\qquad T_{2} ( 0 ) =\frac{9}{4}, \qquad T_{3} ( 0 ) =\frac{8}{3}. $$
Now
$$ T_{1} ( 0 ) -T_{2} ( 0 ) =4-\frac{9}{4}= \frac{7}{4}>0, $$
similarly,
$$ T_{1} ( 0 ) -T_{3} ( 0 ) =4-\frac{8}{3}= \frac{4}{3}>0. $$
Therefore, \(T_{1} ( 0 ) \) is maximum. Hence
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 4. $$
This shows that \(E_{\alpha ,\beta } ( z ) \in \mathcal{C}\). □
Example 3.4
For \(\alpha =3\), \(\beta =1\), we have \(\frac{\varGamma ( \alpha + \beta ) }{\varGamma ( \beta ) }\geq 4\), therefore the function
$$ E_{3,1} ( z ) =\frac{z}{2} \biggl[ e^{z^{1/3}}+2e^{- \frac{1}{2}z^{1/3}} \cos \biggl( \frac{\sqrt{3}}{2}z^{1/3} \biggr) \biggr] $$
is in \(\mathcal{C}\).
Example 3.5
For \(\alpha =1\), \(\beta =4\), we have \(\frac{\varGamma ( \alpha + \beta ) }{\varGamma ( \beta ) }=4\), therefore the function
$$ E_{1,4} ( z ) =\frac{6 ( e^{z}-z-1 ) -3z^{2}}{z ^{2}} $$
is in \(\mathcal{C}\).
The mappings of these functions are given in Fig. 1.
Corollary 3.6
Let
\(\alpha \geq 1\), \(\beta \geq 1\). Then
\(E_{\alpha ,\beta } ( z ) \in \mathcal{S}^{\ast } ( \frac{1}{2} ) \)
if
\(\frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 3\).
Proof
Consider \(T_{i} ( \rho ,\mu ) \), \(i=1,2,3\), as in Theorem 3.1. Replacing ρ by μ, we get
$$\begin{aligned} T_{1} ( \mu ) =&2 ( 2-\mu ) , \\ T_{2} ( \mu ) =&\frac{2^{\mu -1} ( 3-\mu ) ( 3-2\mu ) }{ ( 2-\mu ) }, \\ T_{3} ( \mu ) =&\frac{2 ( 4-\mu ) }{ ( 3- \mu ) }. \end{aligned}$$
For \(\mu =\frac{1}{2}\),
$$ T_{1} \biggl( \frac{1}{2} \biggr) =3,\qquad T_{2} \biggl( \frac{1}{2} \biggr) =\frac{5\sqrt{2}}{3},\qquad T_{3} \biggl( \frac{1}{2} \biggr) =\frac{14}{5}. $$
Now
$$ T_{1} \biggl( \frac{1}{2} \biggr) -T_{2} \biggl( \frac{1}{2} \biggr) =3-\frac{5 \sqrt{2}}{3}=\frac{9-5\sqrt{2}}{3}>0. $$
Similarly,
$$ T_{1} \biggl( \frac{1}{2} \biggr) -T_{3} \biggl( \frac{1}{2} \biggr) =3- \frac{14}{5}=\frac{1}{5}>0. $$
Therefore, \(T_{1} ( \frac{1}{2} ) \) is maximum. Hence
$$ \frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) }\geq 3. $$
Therefore, for \(\alpha \geq 1\), \(\beta \geq 1\), and \(\frac{\varGamma ( \alpha +\beta ) }{\varGamma ( \beta ) } \geq 3\), \(E_{\alpha ,\beta } ( z ) \in \mathcal{S}^{\ast } ( \frac{1}{2} ) \). □
Example 3.7
For \(\alpha =1\), \(\beta =3\), we have \(\frac{\varGamma ( \alpha + \beta ) }{\varGamma ( \beta ) }=3\), therefore the function
$$ E_{1,3} ( z ) =\frac{2 ( e^{z}-z-1 ) }{z} $$
is in \(\mathcal{S}^{\ast } ( \frac{1}{2} ) \).
Example 3.8
For \(\alpha =2\), \(\beta =2\), we have \(\frac{\varGamma ( \alpha + \beta ) }{\varGamma ( \beta ) }=6\), therefore the function
$$ E_{2,2} ( z ) =\sqrt{z}\sinh \sqrt{z} $$
is in \(\mathcal{S}^{\ast } ( \frac{1}{2} ) \).
The mappings of these functions are given in Fig. 2.
Theorem 3.9
Let
\(\mu \geq 1\), \(\eta \geq \frac{-\mu +\sqrt{\mu ^{2}+4\mu -4}}{2}\)
with
\(\alpha \geq 1\), \(\beta \geq 2\)
If
\(M_{1}=(1+\eta )(1-\mu - \eta )>0\)
and
\(M_{2}=+2\mu +\eta -1>0\), then
\(E_{\alpha ,\beta }(z)\)
is starlike of order
\(2\mu +\eta -1\).
Proof
It is observed that \(( E_{\alpha ,\beta } ) _{n}(z)= \sum_{k=1}^{\infty }a_{k}z^{k}\) gives \(a_{1}=1\) and \(a_{k}=\frac{ \varGamma ( \beta ) }{\varGamma ( \alpha ( k-1 ) +\beta ) }\) for \(k\geq 2\). The relation between \(a_{k}\) and \(a_{k+1}\) is
$$ a_{k+1}=\frac{\varGamma ( \alpha ( k-1 ) +\beta ) }{\varGamma ( \alpha k+\beta ) }a_{k}\quad \text{for }k \geq 1. $$
To prove this theorem, it is enough to show that \(\{ a_{k} \} \) satisfies conditions (2.1) and (2.3) of Lemma 2.4. Using the above relation and simple computations yields
$$\begin{aligned} & \bigl[ k(1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a_{k}- \bigl[ (k+1) (1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a_{k+1} \\ &\quad =\frac{\varGamma ( \alpha (k-1)+\beta ) a_{k}}{\varGamma ( \alpha k+\beta ) }h(k), \end{aligned}$$
where \(h(k)\) is defined as follows:
$$\begin{aligned} h(k) =&\frac{\varGamma ( \alpha k+\beta ) }{\varGamma ( \alpha (k-1)+\beta ) } \bigl[ k(1+\eta ) (1-\mu -\eta )-1+2 \mu +\eta \bigr] \\ &{}- \bigl[ (k+1) (1+\eta ) (1-\mu -\eta )-1+2\mu +\alpha \bigr] \\ =&\frac{\varGamma ( \alpha k+\beta ) }{\varGamma ( \alpha (k-1)+\beta ) } ( kM_{1}+M_{2} ) - \bigl[ (k+1)M _{1}+M_{2} \bigr] \\ =& \biggl[ \frac{k\varGamma ( \alpha k+\beta ) }{\varGamma ( \alpha (k-1)+\beta ) }-(k+1) \biggr] M_{1}+ \biggl[ \frac{ \varGamma ( \alpha k+\beta ) }{\varGamma ( \alpha (k-1)+ \beta ) }-1 \biggr] M_{2} >0. \end{aligned}$$
(3.2)
It is observed that under the conditions \(\mu \geq 1\), \(\eta \geq \frac{- \mu +\sqrt{\mu ^{2}+4\mu -4}}{2}\), \(\alpha \geq 1\), and \(\beta \geq 3\), expression (3.2) is positive for \(k\geq 1\). It remains to verify (2.3). That is,
$$\begin{aligned} &(k+\mu +\eta -1) \bigl[ 2k(1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a_{2k} \\ &\quad \geq k \bigl[ (2k+1) (1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a _{2k+1}. \end{aligned}$$
Clearly,
$$\begin{aligned} &(k+\mu +\eta -1) \bigl[ 2k(1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a_{2k} \\ &\quad {}-k \bigl[ (2k+1) (1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] a_{2k+1}= \frac{ \varGamma ( \alpha (2k-1)+\beta ) a_{2k}}{2\varGamma ( \alpha (2k)+\beta ) }g(k), \end{aligned}$$
where \(g(k)\) is defined as follows:
$$\begin{aligned} g(k) =&\frac{\varGamma ( \alpha (2k)+\beta ) }{\varGamma ( \alpha (2k-1)+\beta ) }\bigl(k+\mu +\eta ^{\prime}-1\bigr) \bigl[ 2k(1+ \eta ) (1-\mu -\eta )-1+2\mu +\eta ) \bigr] \\ &{}- \bigl[ (2k+1) (1+\eta ) (1-\mu -\eta )-1+2\mu +\eta \bigr] \\ =&\frac{\varGamma ( \alpha (2k)+\beta ) }{\varGamma ( \alpha (2k-1)+\beta ) }(k+\mu +\eta -1) [ 2kM_{1}+M_{2} ] -k \bigl[ (2k+1)M_{1}+M_{2} \bigr] \\ =& \biggl[ \frac{\varGamma ( \alpha (2k)+\beta ) }{\varGamma ( \alpha (2k-1)+\beta ) }2k(k+\mu +\eta -1)-k(2k+1) \biggr] M_{1} \\ &{}+ \biggl[ \frac{\varGamma ( \alpha (2k)+\beta ) }{\varGamma ( \alpha (2k-1)+\beta ) }(k+\mu +\eta -1)-k \biggr] M _{2}>0. \end{aligned}$$
(3.3)
It is observed that under the conditions \(\mu \geq 1\), \(\eta \geq \frac{- \mu +\sqrt{\mu ^{2}+4\mu -4}}{2}\), \(2\mu +\eta >1\), \(\alpha \geq 1\), and \(\beta \geq 2\), expression (3.3) is positive for \(k\geq 1\), which completes the proof. □
Theorem 3.10
Let
\(\mu \geq 1\), \(\eta \geq \frac{-\mu +\sqrt{\mu ^{2}+4\mu -4}}{2}\), \(2\mu +\eta >1\), \(\alpha \geq 1\), \(\beta \geq 2\), \(a_{1}=1\), \(a_{k}\geq 0\)
satisfy
$$\begin{aligned} \begin{aligned} & ka_{k}-(k+1)a_{k+1} \geq 0,\quad k=1,2,3,\ldots,n-1, \\ & (n-k+1) (k+\mu +\eta -1) (2k-1)a_{2k-1} \geq 2k^{2}(n-k+1+ \eta )a_{2k},\\ &\quad k=4,5,\ldots, \biggl[ \frac{n+3}{2} \biggr] , \end{aligned} \end{aligned}$$
(3.4)
for
\(k\geq 4\). Then
\(( E_{\alpha ,\beta } ) _{n}(z)= \sum_{k=4}^{n}a_{k}z^{k}\)
is convex in the direction of imaginary axis.
Proof
To show that Mittag-Leffler function is convex in the direction of imaginary axis, we will prove that \(z ( E_{\alpha ,\beta } ) _{n}^{\prime }(z) \) is a typically real function. Also \(( E_{ \alpha ,\beta } ) _{n}(z)\) has real coefficients. Set
$$\begin{aligned} z ( E_{\alpha ,\beta } ) _{n}^{\prime }(z) =&z+\sum _{k=2} ^{n}\frac{\varGamma ( \beta ) }{\varGamma ( \alpha ( k-1 ) +\beta ) }z^{k} \\ =&z+\sum_{k=2}^{n}b_{k}z^{k}, \end{aligned}$$
where \(b_{k}=\frac{\varGamma ( \beta ) }{\varGamma ( \alpha ( k-1 ) +\beta ) }\). To get the result, it is required that \(\{ b_{k} \} \) must satisfy the conditions mentioned in Lemma 2.3. Consider
$$\begin{aligned} kb_{k}-(k+1)b_{k+1} =&\varGamma ( \beta ) \biggl[ \frac{k}{ \varGamma ( \alpha ( k-1 ) +\beta ) }-\frac{k+1}{ \varGamma ( \alpha k+\beta ) } \biggr] \\ =&\varGamma ( \beta ) \biggl[ \frac{k\varGamma ( \alpha k+\beta ) -(k+1)\varGamma ( \alpha ( k-1 ) + \beta ) }{\varGamma ( \alpha k+\beta ) \varGamma ( \alpha ( k-1 ) +\beta ) } \biggr] >0 \end{aligned}$$
for \(k=1,2,3,\ldots,n-1\). Take
$$ (n-k+1) (k+\mu +\eta -1)b_{2k-1}-k(n-k+1+\eta )b_{2k}= \frac{b_{2k} \varGamma ( \alpha ( 2k-1 ) +\beta ) }{\varGamma ( \alpha ( 2k-2 ) +\beta ) }q(k), $$
here \(q(k)\) is defined as
$$ q(k)=(k+\mu +\eta -1) (n-k+1)-k(n-k+1+\eta )\frac{\varGamma ( \alpha ( 2k-2 ) +\beta ) }{\varGamma ( \alpha ( 2k-1 ) +\beta ) }. $$
(3.5)
Since Γ is an increasing function in \([ \frac{3}{2}, \infty )\), therefore (3.5) becomes positive when \(\mu \geq 1\), \(\eta \geq \frac{-\mu +\sqrt{\mu ^{2}+4 \mu -4}}{2}\), \(2\mu +\eta >1 \), \(\alpha \geq 1\), and \(\beta \geq 2\). Thus \(\{ b_{k} \} \) satisfies the conditions of Lemma 2.3. Therefore, by using the minimum principle for harmonic functions under the conditions \(\mu +\eta \in ( 0,\frac{1+\eta }{2} ] \),
$$ \operatorname{Im} \bigl( zf_{n}^{\prime } ( z ) \bigr) =\sum _{k=1}^{n}b_{k}r^{k} \sin k\theta >0,\quad \text{where }\theta \in \,] 0,\pi [ \text{ and }r\in \,] 0,1 [ $$
and
$$ \operatorname{Im} \bigl( zf_{n}^{\prime } ( z ) \bigr) =0\quad \text{for }z \in ( 0,1 ) . $$
The Schwarz reflection principle yields that \(\operatorname{Im} ( zf_{n}^{ \prime } ( z ) ) <0\) for \(\theta \in ( \pi ,2 \pi ) \). So \(zf_{n}^{\prime } ( z ) \) is a typically real function, which is equivalent to saying that \(f_{n}(z)\) is convex in the direction of imaginary axis. □
Theorem 3.11
Let
\(\mu \in \mathbb{R} \)
and
\(\eta \geq 0\)
such that
\(2\mu +\eta >1\), and let
\(a_{1}=1\)
and
\(a_{k}\geq 0\)
satisfy
$$ 0\leq na_{n}\leq \cdots\leq (k+1)a_{k+1}\leq ka_{k}\leq \cdots\leq 3a_{3}\leq 2a _{2}\leq \frac{\mu +\eta }{\mu _{0}^{\ast }},\quad \mu +\eta \in \bigl( 0,\mu _{0}^{\ast } \bigr] , $$
(3.6)
and
$$ 2(n-k+1) (k+\mu +\eta -1)a_{2k}\geq (2k+1) (n-k+1+\eta )a_{2k+1},\quad 1\leq k\leq \biggl[ \frac{n}{2} \biggr] . $$
(3.7)
If
\(\alpha \geq 1\)
and
\(\beta \geq 2\), then
\(( E_{\alpha , \beta } ) _{n}(z)=z+\sum_{k=2}^{n}a_{k}z^{k}\)
satisfies
\(\operatorname{Re}(f_{n}^{\prime }(z))>1-\frac{\mu +\eta }{\mu _{0}^{\ast }}\).
Proof
Let \(\sigma -1=-\frac{\mu +\eta }{\mu _{0}^{\ast }}\) and \(( E _{\alpha ,\beta } ) _{n}(z)=z+\sum_{k=2}^{n}\frac{\varGamma ( \beta ) }{\varGamma ( \alpha ( k-1 ) + \beta ) }z^{k}\), where \(a_{k}=\frac{\varGamma ( \beta ) }{\varGamma ( \alpha ( k-1 ) +\beta ) }\). Then
$$ \frac{ ( E_{\alpha ,\beta } ) _{n}^{\prime }(z)-\sigma }{1- \sigma }=\sum_{k=0}^{n-1}c_{k}z^{k}, $$
where \(c_{k}=\frac{(k+1)b_{k+1}}{1-\sigma } \) and \(c_{0}=1\) for \(1 \leq k\leq n-1\). It is observed that under the conditions \(\alpha \geq 1\) and \(\beta \geq 2\) the coefficients \(a_{k}\) are positive. Therefore, \(c_{k}>0\) for \(k\geq 1\). To prove this theorem, we will show that the coefficients \(\{ c_{k} \} \) are decreasing and satisfy (2.5). Now, for this, consider
$$\begin{aligned} (k+1)a_{k+1}-(k+2)a_{k+2} =&\varGamma ( \beta ) \biggl[ \frac{k+1}{ \varGamma ( \alpha k+\beta ) }-\frac{k+2}{\varGamma ( \alpha ( k+1 ) +\beta ) } \biggr] \\ =&\varGamma ( \beta ) \biggl[ \frac{ ( k+1 ) \varGamma ( \alpha ( k+1 ) +\beta ) -(k+2) \varGamma ( \alpha k+\beta ) }{\varGamma ( \alpha k+ \beta ) \varGamma ( \alpha ( k+1 ) +\beta ) } \biggr] >0 \end{aligned}$$
for \(k=1,2,3,\ldots,n-2\). This shows that the coefficients of Mittag-Leffler function are decreasing and \(c_{1}< c_{0}\Rightarrow 2b _{2}<1-\sigma \). Now to have (2.5), consider
$$\begin{aligned} & ( n-k+1 ) ( \mu +\eta +k-1 ) a_{2k}-(n-k+1+ \eta ) (2k+1)a_{2k+1} \\ &\quad =\frac{ ( n-k+1 ) ( \mu +\eta +k-1 ) \varGamma ( \beta ) }{\varGamma ( \alpha ( 2k-1 ) + \beta ) }-\frac{(n-k+1+\eta )(2k+1)\varGamma ( \beta ) }{\varGamma ( \alpha ( 2k ) +\beta ) } \\ &\quad =\varGamma ( \beta ) \\ &\qquad {}\times \biggl[ \frac{ ( n-k+1 ) ( \mu +\eta +k-1 ) \varGamma ( \alpha ( 2k ) +\beta ) -(n-k+1+\eta )(2k+1)\varGamma ( \alpha ( 2k-1 ) +\beta ) }{\varGamma ( \alpha ( 2k ) +\beta ) \varGamma ( \alpha ( 2k-1 ) +\beta ) } \biggr] \\ &\quad >0,\quad \text{for }1\leq k\leq \biggl[ \frac{n}{2} \biggr] . \end{aligned}$$
It is clear that the above relation is positive for \(n\geq k\), \(\alpha \geq 1 \), and \(\beta \geq 2\). Also, Γ is an increasing function in \([ \frac{3}{2},\infty ) \). This yields (3.7). Using similar arguments and the minimum principle for harmonic function gives the required result. □