Theorem 2.1
If \(f\in \mathcal{H}\) is described by the series of the form (1.2) and if
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert + \sigma _{n} \vert b_{n} \vert \bigr) \leq L-M, \end{aligned}$$
(2.1)
then \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) \) with
$$\begin{aligned} &\rho _{n} = \bigl\vert \bigl( q[n]_{q}+1 \bigr) \bigr\vert + \bigl\vert \bigl( Mq[n]_{q}+L \bigr) \bigr\vert , \end{aligned}$$
(2.2)
$$\begin{aligned} &\sigma _{n} = \bigl\vert \bigl( q[n]_{q}-1 \bigr) \bigr\vert + \bigl\vert \bigl( Mq[n]_{q}-L \bigr) \bigr\vert . \end{aligned}$$
(2.3)
Proof
If \(f ( z ) =\frac{1}{z}\), then we have \(\mathfrak{h} ( z ) =\frac{1}{z}\) and \(\mathfrak{g} ( z ) =0\). This implies that
$$\begin{aligned} \bigl\vert \mathfrak{h}^{\prime } ( z ) \bigr\vert - \bigl\vert \mathfrak{g}^{\prime } ( z ) \bigr\vert >0. \end{aligned}$$
Hence by the result of Lewy [34] the function f in \(\mathfrak{D}^{\ast }\) is locally univalent and orientation-preserving. Now we show that f is univalent in \(\mathfrak{D}^{\ast }\). Let \(z_{1},z_{2}\in \mathfrak{D}^{\ast }\) with \(z_{1}\neq z_{2}\). Then
$$\begin{aligned} \bigl\vert f ( z_{1} ) -f ( z_{2} ) \bigr\vert = \biggl\vert \frac{1}{z_{1}}-\frac{1}{z_{2}} \biggr\vert = \frac{ \vert z_{2}-z_{1} \vert }{ \vert z_{1}z_{2} \vert }>0. \end{aligned}$$
To show that \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) \), we have to establish that
$$\begin{aligned} \biggl\vert \frac{q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) +f ( z ) }{Lf ( z ) +Mq\mathcal{D}_{\mathcal{H}}^{q}f ( z ) } \biggr\vert < 1. \end{aligned}$$
It is easy to find that \(q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) =-\frac{1}{z}\) and \(L-M>0\). This indicates that
$$\begin{aligned} \biggl\vert \frac{q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) +f ( z ) }{Lf ( z ) +Mq\mathcal{D}_{\mathcal{H}}^{q}f ( z ) } \biggr\vert = \biggl\vert \frac{-\frac{1}{z}+\frac{1}{z}}{L-M} \biggr\vert =0< 1. \end{aligned}$$
Hence \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) \). Now let \(f\in \mathcal{H}\) have be of the form (1.2), and let us choose \(n\geq 1\) such that \(a_{n}\neq 0\) or \(b_{n}\neq 0\). Also, by using
$$\begin{aligned} q [ n ] _{q}=q \Biggl( 1+\sum_{k=1}^{n-1}q^{k} \Biggr) >q \quad\text{for }0< q< 1 \end{aligned}$$
we have
$$\begin{aligned} \frac{\sigma _{n}}{L-M} &=\frac{ \vert ( q[n]_{q}-1 ) \vert + \vert ( Mq[n]_{q}-L ) \vert }{L-M} \\ &>\frac{ \vert ( q-n ) \vert + \vert ( Mq-Ln ) \vert }{L-M}= \frac{ ( n-q ) + ( Ln-Mq ) }{L-M} \\ &>\frac{ ( n-1 ) + ( Ln-M ) }{L-M}= \frac{ ( 1+L ) n- ( 1+M ) }{L-M} \\ &>\frac{ ( 1+L ) n- ( 1+M ) n}{L-M}=n \quad\text{for all }n\geq 1. \end{aligned}$$
Similarly, \(\frac{\rho _{n}}{L-M}\geq n\) for \(n\geq 1\). Thus using (2.1) together with the above evidence, we get
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( n \vert a_{n} \vert +n \vert b_{n} \vert \bigr) \leq 1, \end{aligned}$$
(2.4)
and therefore
$$\begin{aligned} \bigl\vert \mathfrak{h}^{\prime } ( z ) \bigr\vert - \bigl\vert \mathfrak{g}^{\prime } ( z ) \bigr\vert & \geq \frac{1}{ \vert z \vert ^{2}}-\sum _{n=1}^{\infty }n \vert a_{n} \vert \vert z \vert ^{n-1}-\sum_{n=1}^{\infty }n \vert b_{n} \vert \vert z \vert ^{n-1} \\ &\geq \frac{1}{ \vert z \vert ^{2}} \Biggl( 1- \vert z \vert \sum _{n=1}^{\infty } \bigl( n \vert a_{n} \vert +n \vert b_{n} \vert \bigr) \Biggr) \\ &\geq \frac{1}{ \vert z \vert ^{2}} \Biggl( 1- \frac{ \vert z \vert }{L-M}\sum _{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert +\sigma _{n} \vert b_{n} \vert \bigr) \Biggr) \\ &\geq \frac{1}{ \vert z \vert ^{2}} \bigl( 1- \vert z \vert \bigr) >0\quad \bigl( z\in \mathfrak{D}^{\ast } \bigr) . \end{aligned}$$
Therefore by Lewy’s result [34] the function f in \(\mathfrak{D}^{\ast }\) is sense-preserving and locally univalent. Moreover, if \(z_{1,}z_{2}\in \mathfrak{D}^{\ast }\) with\(z_{1}\neq z_{2}\), then
$$\begin{aligned} \biggl\vert \frac{z_{1}^{n}-z_{2}^{n}}{z_{1}-z_{2}} \biggr\vert = \sum _{k=1}^{n} \vert z_{1} \vert ^{k-1} \vert z_{2} \vert ^{k-1}\leq n\quad\text{for }n\geq 2{.} \end{aligned}$$
Hence by (2.4) we have
$$\begin{aligned} \bigl\vert f ( z_{1} ) -f ( z_{2} ) \bigr\vert &\geq \bigl\vert \mathfrak{h} ( z_{1} ) - \mathfrak{h} ( z_{2} ) \bigr\vert - \bigl\vert \mathfrak{g} ( z_{1} ) -\mathfrak{g} ( z_{2} ) \bigr\vert \\ &= \Biggl\vert \frac{1}{z_{1}}-\frac{1}{z_{2}}-\sum _{n=1}^{\infty }a_{n} \bigl( z_{1}^{n}-z_{2}^{n} \bigr) \Biggr\vert - \Biggl\vert \sum_{n=1}^{ \infty }b_{n} \bigl( \overline{z_{1}^{n}-z_{2}^{n}} \bigr) \Biggr\vert \\ &\geq \vert z_{1}-z_{2} \vert \Biggl( \frac{1}{ \vert z_{1}z_{2} \vert }-\sum_{n=1}^{\infty } \bigl( n \vert a_{n} \vert +n \vert b_{n} \vert \bigr) \Biggr) \\ &\geq \vert z_{1}-z_{2} \vert \biggl( \frac{1}{ \vert z_{1}z_{2} \vert }-1 \biggr) >0. \end{aligned}$$
This shows that f is univalent in \(\mathfrak{D}^{\ast }\), and thus \(f\in \mathcal{M}_{\mathcal{H}}\). Therefore \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) \) if and only if there exists a holomorphic function u with \(u ( 0 ) =0\) and \(\vert u ( z ) \vert <1\) such that
$$\begin{aligned} - \frac{q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) }{f ( z ) }=\frac{1+Lu ( z ) }{1+Mu ( z ) } \end{aligned}$$
or, alternatively,
$$\begin{aligned} \biggl\vert \frac{q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) +f ( z ) }{Lf ( z ) +Mq\mathcal{D}_{\mathcal{H}}^{q}f ( z ) } \biggr\vert < 1. \end{aligned}$$
(2.5)
To prove (2.5), it suffices to show that
$$\begin{aligned} \bigl\vert q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) +f ( z ) \bigr\vert - \bigl\vert Lf ( z ) +Mq \mathcal{D}_{\mathcal{H}}^{q}f ( z ) \bigr\vert < 0 \end{aligned}$$
for \(z\in \mathfrak{D}\). Putting \(\vert z \vert =r\) \(( 0< r<1 ) \), we attain
$$\begin{aligned} & \bigl\vert q\mathcal{D}_{\mathcal{H}}^{q}f ( z ) +f ( z ) \bigr\vert - \bigl\vert Lf ( z ) +Mq \mathcal{D}_{\mathcal{H}}^{q}f ( z ) \bigr\vert \\ &\quad\leq \Biggl\vert \sum_{n=1}^{\infty } \bigl( q[n]_{q}+1 \bigr) a_{n}z^{n}- \sum _{n=1}^{\infty } \bigl( q[n]_{q}-1 \bigr) \overline{b_{n}z^{n}} \Biggr\vert \\ &\qquad{}- \Biggl\vert \frac{ ( L-M ) }{z}+\sum_{n=1}^{\infty } \bigl( L+Mq[n]_{q} \bigr) a_{n}z^{n}-\sum _{n=1}^{\infty } \bigl( Mq[n]_{q}-L \bigr) \overline{b_{n}z^{n}} \Biggr\vert \\ &\quad\leq \Biggl\{ \sum_{n=1}^{\infty } \bigl\vert \bigl( q[n]_{q}+1 \bigr) \bigr\vert \vert a_{n} \vert +\sum_{n=1}^{ \infty } \bigl\vert \bigl( q[n]_{q}-1 \bigr) \bigr\vert \vert b_{n} \vert \Biggr\} \\ &\qquad{}-\frac{1}{r} \Biggl\{ \bigl\vert ( L-M ) \bigr\vert - \sum _{n=1}^{\infty } \bigl\vert \bigl( Mq[n]_{q}+L \bigr) \bigr\vert \vert a_{n} \vert -\sum _{n=1}^{\infty } \bigl\vert \bigl( Mq[n]_{q}-L \bigr) \bigr\vert \vert b_{n} \vert \Biggr\} \\ &\quad\leq \frac{1}{r} \Biggl\{ - \vert L-M \vert +\sum _{n=1}^{ \infty } \bigl( \bigl\vert \bigl( q[n]_{q}+1 \bigr) \bigr\vert + \bigl\vert \bigl( Mq[n]_{q}+L \bigr) \bigr\vert \bigr) \vert a_{n} \vert \\ &\qquad{} +\sum_{n=1}^{\infty } \bigl( \bigl\vert \bigl( q[n]_{q}-1 \bigr) \bigr\vert + \bigl\vert \bigl( Mq[n]_{q}-L \bigr) \bigr\vert \bigr) \vert b_{n} \vert \Biggr\} \\ &\quad\leq \frac{1}{r} \Biggl\{ - ( L-M ) +\sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert +\sigma _{n} \vert b_{n} \vert \bigr) \Biggr\} \end{aligned}$$
due inequality (2.1). Thus \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) \). □
By substituting specific values of the parameters included in this result we obtain the following corollaries.
Corollary 2.2
Let \(f\in \mathcal{H}\) be of the form (1.2). If
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert + \sigma _{n} \vert b_{n} \vert \bigr) \leq ( 1+q ) \end{aligned}$$
with
$$\begin{aligned} &\rho _{n} = \bigl\vert \bigl( q[n]_{q}+1 \bigr) \bigr\vert + \bigl\vert \bigl( q^{2}[n]_{q}-1 \bigr) \bigr\vert , \\ &\sigma _{n} = \bigl\vert \bigl( q[n]_{q}-1 \bigr) \bigr\vert + \bigl\vert \bigl( q^{2}[n]_{q}+1 \bigr) \bigr\vert , \end{aligned}$$
then \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,1,-q ) \)
Proof
The result is obtained by setting \(L=1\) and \(M=-q\) in the last theorem. □
Corollary 2.3
Let \(f\in \mathcal{H}\) be given in (1.2). If
$$\begin{aligned} \sum_{n=1}^{\infty }n \bigl( \vert a_{n} \vert + \vert b_{n} \vert \bigr) \leq 1, \end{aligned}$$
then \(f\in \mathcal{MS}_{\mathcal{H}}^{\ast } ( 1,1,-1 ) \).
Proof
Taking the limit as \(q\rightarrow 1-\) in the above corollary, we get the needed result. □
Influenced by Silverman’s paper [42], the set \(\vartheta ^{\lambda }\), \(\lambda \in \{ 0,1 \} \), of functions \(f\in \mathcal{H}\) of type (1.2) is now described as
$$\begin{aligned} a_{n}=- \vert a_{n} \vert ,\qquad b_{n}= ( -1 ) ^{\lambda } \vert b_{n} \vert \quad\bigl(\text{for }n\in \mathbb{N} \backslash \{1\} \bigr) . \end{aligned}$$
Hence (1.2) and (1.3) give \(f ( z ) =\mathfrak{h} ( z ) + \overline{\mathfrak{g} ( z ) }\) with
$$\begin{aligned} \mathfrak{h} ( z ) =\frac{1}{z}-\sum_{n=1}^{\infty } \vert a_{n} \vert z^{n},\qquad \overline{\mathfrak{g} ( z ) }= ( -1 ) ^{\lambda }\sum_{n=1}^{\infty } \vert b_{n} \vert \overline{z^{n}} \quad( z\in \mathfrak{D} ) . \end{aligned}$$
(2.6)
Using the above facts, we now define the families
$$\begin{aligned} &\mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } ( q,L,M ) =\vartheta ^{0} \cap \mathcal{MS}_{\mathcal{H}}^{\ast } ( q,L,M ) , \\ &\mathcal{MS}_{\mathcal{H}_{\vartheta }}^{c} ( q,L,M ) =\vartheta ^{1} \cap \mathcal{MS}_{\mathcal{H}}^{c} ( q,L,M ) . \end{aligned}$$
Let us now prove that condition (2.1) is also appropriate for \(f\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } \).
Theorem 2.4
Let \(f\in \vartheta ^{0}\) have expansion (2.6). Then \(f\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } ( q,L,M ) \) if and only if (2.1) is true.
Proof
To achieve the result, it is sufficient to determine that \(f\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } ( q,L,M ) \) validates relationship (2.1). Let \(f\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } ( q,L,M ) \). Then inequality (2.5) holds, that is, for \(z\in \mathfrak{D}^{\ast }\),
$$\begin{aligned} \biggl\vert \frac{\sum_{n=1}^{\infty } ( q [ n ] _{q}+1 ) a_{n}z^{n}+\sum_{n=1}^{\infty } ( q [ n ] _{q}-1 ) \overline{b_{n}z^{n}}}{\frac{L-M}{z}-\sum_{n=1}^{\infty } ( qM [ n ] _{q}+L ) a_{n}z^{n}+\sum_{n=1}^{\infty } ( qM [ n ] _{q}-L ) \overline{b_{n}z^{n}}} \biggr\vert < 1. \end{aligned}$$
Setting \(z=r\) \(( r\in ( 0,1 ) ) \), we obtain
$$\begin{aligned} \frac{\sum_{n=1}^{\infty } ( \vert ( q [ n ] _{q}+1 ) \vert \vert a_{n} \vert + \vert ( q [ n ] _{q}-1 ) \vert \vert b_{n} \vert ) r^{n+1}}{ ( L-M ) -\sum_{n=1}^{\infty } ( \vert ( qM [ n ] _{q}+L ) \vert \vert a_{n} \vert + \vert ( qM [ n ] _{q}-L ) \vert \vert b_{n} \vert ) r^{n+1}}< 1. \end{aligned}$$
(2.7)
Obviously, in case of \(r\in ( 0,1 ) \), the left-hand side denominator of (2.7) cannot be zero. In addition, this is positive when \(r=0\). Thus, using (2.7), we get
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert + \sigma _{n} \vert b_{n} \vert \bigr) r^{n+1}\leq ( L-M )\quad ( 0\leq r< 1 ) . \end{aligned}$$
(2.8)
It is straightforward that the partial-sum sequence \(\{ S_{n} \} \) attached with the series \(\sum_{n=1}^{\infty } ( \rho _{n} \vert a_{n} \vert + \sigma _{n} \vert b_{n} \vert ) \) is nondecreasing sequence, and by (2.8) it is bounded by \(( L-M ) \). So \(\{ S_{n} \} \) is a convergent sequence, and
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert + \sigma _{n} \vert b_{n} \vert \bigr) r^{n+1}=\lim_{n \rightarrow \infty }S_{n}\leq ( L-M ) , \end{aligned}$$
which gives assumption (2.1). □
Example 2.5
Let us choose the function
$$\begin{aligned} T(z)=\frac{1}{z}-\sum_{n=1}^{\infty } \biggl( \frac{L-M}{\rho _{n}}\frac{1}{2^{n}}z^{n}+ \frac{L-M}{\sigma _{n}}\frac{1}{2^{n}} \overline{z^{n}} \biggr)\quad \bigl( z\in \mathfrak{D}^{\ast } \bigr) . \end{aligned}$$
Then we easily get
$$\begin{aligned} \sum_{n=1}^{\infty } \bigl( \rho _{n} \vert a_{n} \vert +\sigma _{n} \vert b_{n} \vert \bigr) = \sum_{n=1}^{\infty } \frac{1}{2^{n-1}} ( L-M ) = ( L-M ) . \end{aligned}$$
Thus \(T\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{\ast } ( q,L,M ) \).
By using the above-mentioned theorem along with the notion of class \(\mathcal{MS}_{\mathcal{H}}^{c} ( q,L,M ) \) we can easily derive the following results.
Corollary 2.6
Let \(f\in \mathcal{H}\) be written in the form of Taylor expansion (1.2). If
$$\begin{aligned} \sum_{n=1}^{\infty } [ n ] _{q} \bigl( \rho _{n} \vert a_{n} \vert +\sigma _{n} \vert b_{n} \vert \bigr) \leq ( L-M ) , \end{aligned}$$
(2.9)
then \(f\in \mathcal{MS}_{\mathcal{H}}^{c} ( q,L,M ) \).
Proof
From inequality (2.9), Theorem 2.1, and Alexander-type relation
$$\begin{aligned} f \in \mathcal{MS}_{\mathcal{H}}^{c} ( q,L,M )\quad \Longleftrightarrow\quad \mathcal{D}_{\mathcal{H}}^{q}f \in \mathcal{MS}_{ \mathcal{H}}^{\ast } ( q,L,M ) \end{aligned}$$
(2.10)
we easily get the desired result. □
Corollary 2.7
Let \(f\in \vartheta ^{1}\) be written in the series form (2.6). Then \(f\in \mathcal{MS}_{\mathcal{H}_{\vartheta }}^{c} ( q,L,M ) \) if and only if inequality (2.9) is fulfilled.
Proof
Using relationship (2.10) and Theorem 2.4, we get the desired result. □