1 Introduction

The conception of q-calculus model is a creative method for designs of the q-special functions. The procedure of q-calculus improves various kinds of orthogonal polynomials, operators, and special functions, which realize the form of their typical complements. The idea of q-calculus was principally realized by Carmichael [1], Jackson [2], Mason [3], and Trjitzinsky [4]. An analysis of this calculus for the early mechanism was offered by Ismail et al. [5]. Numerous integral and derivative features were formulated by using the convolution concept; for example, the Sàlàgean derivative [6], Al-Oboudi derivative (generalization of the Sàlàgean derivative) [7], and the symmetric Sàlàgean derivative [8]. It is significant to notify that the procedure of convolution finds its uses in different research, analysis, and study of the geometric properties of regular functions (see [9–11]). Here, we aim to study some geometric properties of a new quantum symmetric conformable differential operator (Q-SCDO). The classes of analytic functions are suggested by using the convolution product. The consequences are generalized classes in the open unit disk.

2 Methodology

This section provides the mathematical information that is used in this paper. Let ⋀ be the category of smooth functions given as follows:

$$ \curlyvee(\xi)= \xi+\sum^{\infty}_{n=2} \curlyvee_{n} \xi^{n}, \quad \xi\in\cup, $$
(2.1)

where \(\cup=\{\xi\in\mathbb{C}: |\xi| < 1\}\).

Definition 1

Two functions ⋎1 and ⋎2 in ⋀ are said to be subordinate, denoted by \(\curlyvee_{1} \prec\curlyvee_{2}\), if we can find a Schwarz function ⊺ with \(\intercal(0)=0\) and \(|\intercal(\xi)|<1\) such that \(\curlyvee_{1}(\xi) = \curlyvee _{2}(\intercal(\xi))\), \(\xi\in\cup\) (the details can be found in [12]). Obviously, \(\curlyvee_{1}(\xi) \prec\curlyvee_{2}(\xi) \) implies \(\curlyvee_{1}(0) = \curlyvee_{2}(0) \) and \(\curlyvee_{1} (\cup) \subset \curlyvee_{2}(\cup)\). In addition, the subordinate \(\curlyvee_{1}(\xi) \prec_{r} \curlyvee _{1}(\xi)\), \(\xi\in\cup(r)\) is written by

$$\curlyvee_{1}(r \xi) \prec\curlyvee_{2}(r \xi),\quad r< 1. $$

Definition 2

For two functions ⋎1 and ⋎2 in ⋀, the Hadamard or convolution product is defined as

$$ \begin{aligned}[b] \curlyvee_{1}(\xi)* \curlyvee_{2}(\xi)&= \Biggl( \xi+ \sum_{n=2}^{\infty}\curlyvee_{n} \xi^{n} \Biggr)* \Biggl( \xi+ \sum _{n=2}^{\infty}\flat_{n} \xi^{n} \Biggr) \\&= \Biggl( \xi+ \sum_{n=2}^{\infty}\curlyvee_{n} \flat_{n} \xi^{n} \Biggr), \quad \xi\in\cup. \end{aligned} $$
(2.2)

Definition 3

For each nonnegative integer n, the value of q-integer number, denoted by \([n]_{q} \), is defined by \([n]_{q} = \frac{1-q^{n}}{1-q}\), where \([0]_{q} =0\), \([1]_{q} =1\) and \(\lim_{q\rightarrow1^{-}} [n]_{q} =n\).

Example 2.1

\([1]_{0.5} =1\), \([2]_{0.5}=1.5\), \([3]_{0.5}=1.75\), \([2]_{0.75}=1.75\), \([3]_{0.5}=2.312\), \([2]_{0.99}= 1.99\), \([3]_{0.99}= 2.97\), \([3]_{1}= 3\).

Definition 4

The q-difference operator of ⋎ is written by the formula

$$ \Delta_{q} \curlyvee(\xi)= \frac {\curlyvee(q\xi)-\curlyvee(\xi)}{q \xi-\xi}, \quad\xi\in\cup. $$
(2.3)

Clearly, we have \(\Delta_{q} \xi^{n}= [n]_{q} \xi^{n-1}\). Consequently, for \(\curlyvee\in\bigwedge\), we have

$$ \Delta_{q} \curlyvee(\xi)= \sum _{n=1}^{\infty}\curlyvee_{n} [n]_{q} \xi ^{n-1}, \quad\xi\in\cup, \curlyvee_{1}=1. $$
(2.4)

For \(\curlyvee\in\bigwedge\), the Sàlàgean q-derivative factor [13] is formulated as follows:

$$ \begin{aligned} &S^{0}_{q} \curlyvee(\xi)=\curlyvee(\xi), \\& S^{1}_{q} \curlyvee(\xi)= \xi\Delta_{q} \curlyvee(\xi), \\& \dots \\& S^{k}_{q} \curlyvee(\xi)= \xi \Delta_{q} \bigl( S^{k-1}_{q} \curlyvee(\xi) \bigr), \end{aligned} $$
(2.5)

where k is a positive integer.

A computation based on the definition of \(\Delta_{q}\) implies that

$$ \begin{aligned}[b] S^{k}_{q} \curlyvee(\xi)&= \xi+\sum_{n=2}^{\infty}[n]_{q}^{k} \curlyvee _{n} \xi^{n} \\ &= \Biggl(\xi+\sum_{n=2}^{\infty}\curlyvee_{n} \xi^{n} \Biggr) * \Biggl( \xi+\sum _{n=2}^{\infty}[n]_{q}^{k} \xi^{n} \Biggr) \\ &:=\curlyvee(\xi)*\varPsi_{q}^{k}(\xi). \end{aligned} $$

Obviously,

$$ \lim_{q\rightarrow1^{-}}S^{k}_{q} \curlyvee(\xi)=\xi+\sum_{n=2}^{\infty}n^{k} \curlyvee_{n} \xi^{n}, $$
(2.6)

the Sàlàgean derivative factor [6].

Definition 5

Let \(\curlyvee(\xi)\in\bigwedge\), and let \(\nu\in[0,1]\) be a constant. Then Q-SCDO has the following operations:

$$ \begin{aligned} &\bigl[ \mathcal{S}_{\nu}^{0} \bigr]_{q} \curlyvee(\xi)=\curlyvee(\xi), \\ &\bigl[ \mathcal{S}_{\nu}^{1}\bigr]_{q} \curlyvee(\xi)= \biggl( \frac{\kappa _{1}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa_{0}({\nu},\xi)} \biggr) \xi\Delta_{q} \curlyvee(\xi)- \biggl(\frac{\kappa_{0}({\nu},\xi )}{\kappa_{1}({\nu},\xi)+\kappa_{0}({\nu},\xi)} \biggr) \xi\Delta _{q} \curlyvee(-\xi)\hspace{-24pt} \\ &\phantom{\bigl[ \mathcal{S}_{\nu}^{1}\bigr]_{q} \curlyvee(\xi)} = \biggl( \frac{\kappa_{1}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr) \Biggl( \xi+\sum _{n=2}^{\infty}[n]_{q} \curlyvee _{n} \xi^{n} \Biggr)\\&\phantom{\bigl[ \mathcal{S}_{\nu}^{1}\bigr]_{q} \curlyvee(\xi)={}}- \biggl(\frac{\kappa_{0}({\nu},\xi)}{\kappa _{1}({\nu},\xi)+\kappa_{0}({\nu},\xi)} \biggr) \Biggl( - \xi+\sum_{n=2}^{\infty}[n]_{q} (-1)^{n} \curlyvee_{n} \xi^{n} \Biggr) \\ &\phantom{\bigl[ \mathcal{S}_{\nu}^{1}\bigr]_{q} \curlyvee(\xi)} = \xi+ \sum_{n=2}^{\infty}[n]_{q} \biggl(\frac{\kappa_{1}({\nu},\xi )+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr) \curlyvee_{n} \xi^{n}, \\ & \mathcal{S}_{\nu}^{2} \curlyvee(\xi)= \mathcal{S}_{\nu}^{1} \bigl[ \mathcal {S}_{\nu}^{1} \curlyvee(\xi)\bigr] \\ &\phantom{\mathcal{S}_{\nu}^{2} \curlyvee(\xi)}= \xi+ \sum_{n=2}^{\infty}[n]_{q}^{2} \biggl(\frac{\kappa_{1}({\nu },\xi)+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr)^{2} \curlyvee_{n} \xi^{n}, \\ & \vdots \\ & \bigl[ \mathcal{S}_{\nu}^{k} \bigr]_{q} \curlyvee(\xi)= \mathcal{S}_{\nu}^{1} \bigl[ \mathcal{S}_{\nu}^{k-1} \curlyvee(\xi)\bigr] \\ &\phantom{\bigl[ \mathcal{S}_{\nu}^{k} \bigr]_{q} \curlyvee(\xi)}= \xi+ \sum_{n=2}^{\infty}[n ]_{q}^{k} \biggl(\frac{\kappa_{1}({\nu },\xi)+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr)^{k} \curlyvee_{n} \xi^{n}, \end{aligned} $$
(2.7)

so that \(\kappa_{1}(\nu,\xi)\neq - \kappa_{0}(\nu,\xi)\),

$$\lim_{\nu\rightarrow0}\kappa_{1}(\nu,\xi)=1, \qquad \lim _{\nu \rightarrow1}\kappa_{1}(\nu,\xi)=0, \quad \kappa_{1}(\nu,\xi)\neq 0, \forall\xi\in\cup, {\nu} {\in} (0,1), $$

and

$$\lim_{\nu\rightarrow0}\kappa_{0}(\nu,\xi)=0, \qquad \lim _{\nu \rightarrow1}\kappa_{0}(\nu,\xi)=1, \quad \kappa_{0}(\nu,\xi)\neq 0, \forall\xi\in\cup{\nu} {\in} (0,1). $$

The value \(\nu=0\) indicates the Sàlàgean derivative

$$\lim_{q\rightarrow1^{-} }\mathcal{S}^{k} \curlyvee(\xi)= \xi+ \sum^{\infty}_{n=2}n^{k} \curlyvee_{n} \xi^{n}. $$

Moreover, the following operator can be located in [14], where

$$\lim_{q\rightarrow1^{-} } \bigl[ \mathcal{S}_{\nu}^{k} \bigr]_{q}\curlyvee(\xi)= \mathcal{S}_{\nu}^{k} \curlyvee(\xi). $$

3 Convolution classes

Based on the definition (2.7), we introduce the following classes. Denote the following functions:

$$\begin{aligned}& \varPsi^{k}_{q}(\xi):= \xi+ \sum _{n=2}^{\infty}[n]_{q}^{k} \xi^{n}; \end{aligned}$$
(3.1)
$$\begin{aligned}& \varPhi^{k}_{\nu}(\xi):= \xi+ \sum _{n=2}^{\infty}\biggl(\frac{\kappa _{1}({\nu},\xi)+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi )+\kappa_{0}({\nu},\xi)} \biggr)^{k} \xi^{n}. \end{aligned}$$
(3.2)

Thus, in terms of the convolution product, the factor (2.7) is formulated as follows:

$$ \bigl[ \mathcal{S}_{\nu}^{k} \bigr]_{q}\curlyvee(\xi )=\varPsi^{k}_{q}( \xi)*\varPhi_{\nu}^{k}(\xi)*\curlyvee(\xi), \quad \forall \curlyvee\in\bigwedge. $$
(3.3)

Let ⋎ be a function from ⋀ and \(\sigma(\xi) \) be a convex univalent function in ∪ such that \(\sigma(0)=1\). The class \(\varXi^{k}_{q_{1},q_{2}}(\sigma) \) is defined by

$$ \varXi^{\nu,k}_{q_{1},q_{2}}(\sigma) = \biggl\{ \curlyvee\in\bigwedge: \frac{ [ \mathcal{S}_{\nu}^{k} ]_{q_{1}}\curlyvee(\xi)}{ [ \mathcal {S}_{\nu}^{k} ]_{q_{2}}\curlyvee(\xi)}=\frac{\varPsi^{k}_{q_{1}}(\xi )*\varPhi_{\nu}^{k}(\xi)*\curlyvee(\xi)}{\varPsi^{k}_{q_{2}}(\xi )*\varPhi_{\nu}^{k}(\xi)*\curlyvee(\xi)} \prec \sigma(\xi), \sigma(0)=1 \biggr\} . $$
(3.4)

Also, we define a special class involving the above functions when \(\nu \rightarrow0\), as follows:

$$ \varXi^{0,k}_{q_{1},q_{2}}(\sigma) = \biggl\{ \curlyvee\in\bigwedge: \frac{[ \mathcal{S}_{\nu}^{k} ]_{q_{1}}\curlyvee(\xi) }{ [ \mathcal {S}_{\nu}^{k} ]_{q_{2}}\curlyvee(\xi)}=\frac{\varPsi^{k}_{q_{1}}(\xi )*\curlyvee(\xi)}{\varPsi^{k}_{q_{2}}(\xi)*\curlyvee(\xi)} \prec \sigma(\xi), \sigma(0)=1 \biggr\} . $$
(3.5)

When \(k=0\), we have Dziok subclass [15].

We denote by \(\mathcal{S}^{*}(\sigma)\) the class of all functions given by

$$ \mathcal{S}^{*}(\sigma)= \biggl\{ \curlyvee \in\bigwedge \frac{\xi(\frac{\xi}{1-\xi})'*\curlyvee(\xi )}{(\frac{\xi}{1-\xi})*\curlyvee(\xi)} \prec\sigma(\xi), \sigma (0)=1 \biggr\} , $$
(3.6)

and by \(\mathcal{C}^{*}(\sigma)\) the class of all functions

$$ \mathcal{C}(\sigma)= \biggl\{ \curlyvee\in \bigwedge: \frac{\xi(\frac{\xi}{(1-\xi)^{2}})'*\curlyvee(\xi )}{(\frac{\xi}{(1-\xi)^{2}})*\curlyvee(\xi)} \prec\sigma(\xi), \sigma(0)=1 \biggr\} . $$
(3.7)

The following preliminary result can be found in [16, 17].

Lemma 3.1

IfKis smooth (analytic) in ∪, \(\curlyvee\in\mathcal{C}(\frac{1+\xi}{1-\xi})\)is convex and\(g \in S^{*}(\frac{1+\xi}{1-\xi})\)is starlike then

$$ \frac{\curlyvee *(Kg)}{\curlyvee *g}(\cup ) \subseteq\overline{\operatorname{co}}\bigl(K(\cup) \bigr), $$
(3.8)

where\(\overline{\operatorname{co}}(K(\cup))\)is the closed convex hull of\(K(\cup)\).

Lemma 3.2

For analytic functions\(h, \hbar \in\cup\), the subordination\(h\prec\hbar\)implies that

$$ \int_{0}^{2\pi} \bigl\vert h(\xi) \bigr\vert ^{p} \,d\theta\leq \int_{0}^{2\pi} \bigl\vert \hbar(\xi) \bigr\vert ^{p} \,d\theta, $$
(3.9)

where\(\xi=re^{i\theta}\), \(0< r<1\), andpis a positive number.

Some of the few studies in q-calculus are realized by comparison between two different values of calculus. Class \(\varXi^{\nu ,k}_{q_{1},q_{2}}(\sigma)\) shows the relation between the \(q_{1}\)- and \(q_{2}\)-calculus depending on the operator (2.7).

4 Inclusions

This section deals with the geometric representations of the class \(\varXi^{\nu,k}_{q_{1},q_{2}}(\sigma)\), \(q_{1}\neq q_{2}\) and their consequences.

Theorem 4.1

Let\(\curlyvee\in\bigwedge\)and let the function\(g:=\varPsi ^{k}_{q_{2}}*\curlyvee\in S^{*}(\frac{1+\xi}{1-\xi})\), \(\xi\in\cup\). If\(\curlyvee\in\varXi^{0,k}_{q_{1},q_{2}}(\sigma)\), \(q_{1}\neq q_{2}\)and the function\(\varPhi^{k}(\xi) \in\mathcal{C}(\frac{1+\xi}{1-\xi})\)then\(\curlyvee\in\varXi^{\nu,k}_{q_{1},q_{2}}(\sigma)\), \(\sigma(0)=1\).

Proof

Suppose that \(\curlyvee\in\varXi^{0,k}_{q_{1},q_{2}}(\sigma)\). This implies that there is a Schwarz function υ with \(\upsilon(0)=0\) and \(|\upsilon (\xi)|<1\) satisfying the following relation:

$$ \frac{\varPsi^{k}_{q_{1}}(\xi)*\curlyvee (\xi)}{\varPsi^{k}_{q_{2}}(\xi)*\curlyvee(\xi)}=\sigma\bigl(\upsilon(\xi )\bigr) \quad( \xi\in\cup). $$
(4.1)

This leads to

$$ \varPsi^{k}_{q_{1}}(\xi)*\curlyvee(\xi)= \bigl(\varPsi^{k}_{q_{2}}(\xi )*\curlyvee(\xi) \bigr)\sigma \bigl(\upsilon(\xi)\bigr) = g(\xi)\sigma\bigl(\upsilon(\xi)\bigr). $$
(4.2)

By employing the convolution’s properties, we arrive at

$$ \begin{aligned}[b] \frac{ (\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu}*\curlyvee )(\xi )}{ (\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu}*\curlyvee )(\xi )}&=\frac{\varPhi^{k}_{\nu}(\xi)* (\varPsi^{k}_{q_{1}}*\curlyvee )(\xi)}{\varPhi^{k}_{\nu}(\xi)* (\varPsi^{k}_{q_{2}}*\curlyvee )(\xi)} \\&= \frac{\varPhi^{k}_{\nu}(\xi)*[ g(\xi)\sigma(\upsilon(\xi ))]}{\varPhi^{k}_{\nu}(\xi)*g(\xi)}.\end{aligned} $$
(4.3)

Accordingly, by virtue of Lemma 3.1, we obtain

$$ \frac{ (\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu}*\curlyvee )(\xi )}{ (\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu}*\curlyvee )(\xi)} \in\overline{\operatorname{co}}\bigl(\sigma\bigl( \upsilon(\cup)\bigr)\bigr) \subset\overline {\operatorname{co}}\bigl(\sigma(\cup)\bigr). $$
(4.4)

Since \(\sigma(\xi) \) is a convex univalent function in ∪ with \(\sigma(0)=1\), by the concept of subordination, we conclude that

$$ \frac{\varPsi^{k}_{q_{1}}(z)*\varPhi^{k}_{\nu}(\xi)*\curlyvee(\xi )}{\varPsi^{k}_{q_{2}}(\xi)*\varPhi^{k}_{\nu}(\xi)*\curlyvee(\xi)} \prec \sigma(\xi), $$
(4.5)

which means that \(\curlyvee\in\varXi^{\nu,k}_{q_{1},q_{2}}(\sigma)\). This completes the proof. □

In this place, we note that the conclusion of Theorem 4.1 yields the following consequence:

Corollary 4.2

Let ⋎ be a function from ⋀ and\(\sigma(\xi) \)be a convex univalent function in ∪ such that\(\sigma(0)=1\). Then

$$\varXi^{0,k}_{q_{1},q_{2}}(\sigma) \subset\varXi^{\nu ,k}_{q_{1},q_{2}}( \sigma). $$

In general, we have the following result:

Theorem 4.3

Let\(\curlyvee\in\bigwedge\)and let the function\(G:=\varPsi ^{k}_{q_{2}}*\varPhi^{k}_{\nu}*\curlyvee\in S^{*}(\frac{1+\xi}{1-\xi})\), \(\xi \in\cup\). If\(\rho_{1}:=\varPsi_{q_{1}}*\varPhi_{\nu}\prec_{r} \rho _{2}:=\varPsi_{q_{2}}*\varPhi_{\nu}\)for some\(r<1\)and the function\(\rho _{2} \in\mathcal{C}(\frac{1+\xi}{1-\xi})\)then

$$ \varXi^{\nu,k}_{q_{1},q_{2}}(\sigma )\subset \varXi^{\nu,k+1}_{q_{1},q_{2}}(\sigma). $$
(4.6)

Proof

Suppose that \(\curlyvee\in\varXi^{\nu,k}_{q_{1},q_{2}}(\sigma)\). Then there is a Schwarz transform ω with \(\omega(0)=0\) and \(|\omega (\xi)|<1\) such that

$$ \frac{(\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu}*\curlyvee)(\xi)}{(\varPsi ^{k}_{q_{2}}*\varPhi^{k}_{\nu}*\curlyvee)(\xi)}=\sigma\bigl(\omega(\xi)\bigr), \quad\xi \in\cup. $$
(4.7)

This yields the following equality:

$$ \bigl(\varPsi^{k}_{q_{1}}* \varPhi^{k}_{\nu}*\curlyvee\bigr) (\xi)= \bigl(\varPsi ^{k}_{q_{2}}*\varPhi^{k}_{\nu}* \curlyvee \bigr) (\xi)\sigma\bigl(\omega(\xi)\bigr) = G(\xi)\sigma\bigl(\omega( \xi)\bigr). $$
(4.8)

By considering the convolution’s properties, we obtain

$$ \frac{(\varPsi^{k+1}_{q_{1}}*\varPhi^{k+1}_{\nu}*\curlyvee)(\xi) }{(\varPsi^{k+1}_{q_{2}}*\varPhi^{k+1}_{\kappa}*\curlyvee)(\xi) }= \frac{\rho_{1}(\xi)*(G(\xi)\sigma(\xi))}{\rho_{2}(\xi)*G(\xi)}. $$
(4.9)

Since \(\rho_{1}\prec_{r} \rho_{2}\), by letting \(r\rightarrow1\), we obtain \(\rho_{1}(\xi)= \rho_{2}(\xi)\). As a result, by Lemma 3.1, we deduce that

$$ \frac{(\varPsi^{k+1}_{q_{1}}*\varPhi ^{k+1}_{\nu}*\curlyvee)(\xi) }{(\varPsi^{k+1}_{q_{2}}*\varPhi ^{k+1}_{\nu}*\curlyvee)(\xi) }= \frac{\rho_{1}(\xi)*(G(\xi)\sigma (\xi))}{\rho_{2}(\xi)*G(\xi)} \in\overline{\operatorname{co}} \bigl(\sigma\bigl(\omega(\cup )\bigr)\bigr) \subset\overline{\operatorname{co}}\bigl(\sigma(\cup) \bigr). $$
(4.10)

Since \(\sigma(\xi) \) is a convex univalent function in ∪ with \(\sigma(0)=1\), then by the definition of subordination, we obtain

$$ \frac{(\varPsi^{k+1}_{q_{1}}*\varPhi^{k+1}_{\nu}*\curlyvee)(\xi) }{(\varPsi^{k+1}_{q_{2}}*\varPhi^{k+1}_{\nu}*\curlyvee)(\xi) } \prec \sigma(\xi) \quad\Rightarrow\quad\curlyvee\in \varXi^{\nu,k+1}_{q_{1},q_{2}}(\sigma), $$
(4.11)

which completes the proof. □

We note that if we replace the condition of Theorem 4.3 by \(\rho_{2} \prec_{r} \rho_{1}\) such that \(\rho_{1} \in\mathcal{C}(\frac {1+\xi}{1-\xi})\) then we obtain the same conclusion.

Theorem 4.4

Let\(\curlyvee\in\bigwedge\)and let the function\(H:=\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu_{1}}*\curlyvee\in S^{*}(\frac{1+\xi }{1-\xi})\), \(\xi\in\cup\). If\(\varPhi^{k}_{\nu_{1}}\prec_{r} \varPhi^{k}_{\nu_{2}}\)for some\(r<1\)then

$$ \varXi^{\nu_{1},k}_{q_{1},q_{2}}(\sigma)\subset \varXi^{\nu _{2},k}_{q_{1},q_{2}}(\sigma). $$
(4.12)

Proof

Suppose that \(\curlyvee\in\varXi^{\nu_{1},k}_{q_{1},q_{2}}(\sigma)\). Consequently, a Schwarz function ϑ exists with \(\vartheta (0)=0\) and \(|\vartheta(z)|<1\) such that

$$ \frac{(\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu_{1}}*\psi)(z)}{(\varPsi ^{k}_{q_{2}}*\varPhi^{k}_{\nu_{1}}*\curlyvee)(\xi)}=\sigma\bigl(\vartheta(\xi )\bigr), \quad \xi\in\cup. $$
(4.13)

This yields

$$ \bigl(\varPsi^{k}_{q_{1}}* \varPhi^{k}_{\nu _{1}}*\curlyvee\bigr) (\xi)= \bigl( \varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu _{1}}* \curlyvee \bigr) (\xi)\sigma\bigl(\omega(\xi)\bigr) = H(\xi)\sigma \bigl( \vartheta(\xi)\bigr). $$
(4.14)

But the condition \(\varPhi^{k}_{\nu_{1}}\prec_{r} \varPhi^{k}_{\nu_{2}}\) implies that \(\varPhi^{k}_{\nu_{1}}(r \xi)= \varPhi^{k}_{\nu_{2}}(r \xi)\) (for some r). It is clear that \(\eta(\xi)=\xi\in\mathcal{C}(\frac{1+\xi}{1-\xi })\); therefore, by the convolution’s properties, we attain

$$ \frac{(\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu_{2}}*\curlyvee)(\xi) }{(\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu_{2}}*\curlyvee)(\xi) }= \frac {\eta(\xi)*(H(\xi)\vartheta(\xi))}{\eta(\xi)*H(\xi)}, \quad\xi \in{ \cup}. $$
(4.15)

Thus, in view of Lemma 3.1, we get

$$ \frac{(\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu_{2}}*\curlyvee)(\xi) }{(\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu_{2}}*\curlyvee)(\xi) } \in \overline{\operatorname{co}}\bigl(\sigma\bigl( \vartheta({\cup})\bigr)\bigr) \subset\overline {\operatorname{co}}\bigl(\sigma({\cup})\bigr). $$
(4.16)

Since \(\sigma(\xi) \) is a convex univalent function in ∪ with \(\sigma(0)=1\), then by the definition of subordination, we obtain

$$ \frac{(\varPsi^{k}_{q_{1}}*\varPhi^{k}_{\nu_{2}}*\curlyvee)(\xi) }{(\varPsi^{k}_{q_{2}}*\varPhi^{k}_{\nu_{2}}* \curlyvee)(\xi) } \prec \sigma(\xi) \quad\Rightarrow\quad\curlyvee\in \varXi^{\nu_{2},k}_{q_{1},q_{2}}(\sigma), $$
(4.17)

which completes the proof. □

We record that if we change the condition of Theorem 4.4 by \(\varPhi^{k}_{\nu_{2}}\prec_{r} \varPhi^{k}_{\nu_{1}}\), we have

$$ \varXi^{\nu_{2},k}_{q_{1},q_{2}}(\sigma)\subset \varXi^{\nu _{1},k}_{q_{1},q_{2}}(\sigma). $$
(4.18)

5 Integral inequalities

The following section deals with some inequalities containing the operator (2.7). For two functions \(h(\xi)=\sum a_{n}\xi^{n}\) and \(\hbar(\xi)=\sum b_{n}\xi^{n}\), we have \(h \ll\hbar\) if and only if \(|a_{n}| \leq|b_{n}|\), ∀n. This inequality is known as the majorization of two analytic functions.

Theorem 5.1

Consider the operator\([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)\), \(\curlyvee\in\bigwedge\). If the coefficients of ⋎ satisfy the inequality\(|\curlyvee_{n}| \leq(\frac{1}{n \nu})^{k}\), \(\nu\in (0,1)\)then

$$ \int_{0}^{2\pi} \biggl\vert \frac{ [ \mathcal {S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}{\xi} \biggr\vert ^{p} \,d\theta\leq \int _{0}^{2\pi} \biggl\vert \biggl( \frac{1+\xi}{1-\xi} \biggr)^{\delta}\biggr\vert ^{p} \,d \theta, \quad p>0. $$
(5.1)

Proof

Let

$$ \sigma(\xi,\delta)= \biggl( \frac{1+\xi }{1-\xi} \biggr)^{\delta}, \quad\xi\in{\cup}, \delta\geq1. $$
(5.2)

Then, a straightforward computation implies that

$$ \begin{aligned} & \sigma(\xi,1)= 1+\sum _{n=1} (2n)\xi^{n}, \\ & \sigma(\xi,2)= 1+\sum_{n=1} (4n) \xi^{n}= 1+4\xi+8 \xi^{2}+12\xi ^{3}+16 \xi^{4}+20\xi^{5}+\cdots, \\ & \sigma(\xi,3)=1+\sum_{n=1} \bigl(2+4n^{2} \bigr)\xi^{n}=1+6\xi+18\xi^{2}+38\xi^{3}+ \cdots, \\ & \sigma(\xi,4)=1+\sum_{n=1}\frac{1}{3} \bigl(8n\bigl(2+n^{2}\bigr)\bigr)\xi^{n}=1+8\xi +16 \xi^{2}+24\xi^{3}+\cdots, \\ & \vdots \end{aligned} $$
(5.3)

Comparing Eq. (5.3) and the coefficients of \([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)\), which are satisfying

$$ \lim_{q\rightarrow1^{-}} \biggl\vert [n ]_{q}^{k} \biggl(\frac{\kappa_{1}({\nu},\xi)+(-1)^{n+1}\kappa_{0}({\nu},\xi )}{\kappa_{1}({\nu},\xi)+\kappa_{0}({\nu},\xi)} \biggr)^{k} \curlyvee _{n} \biggr\vert \leq1, $$
(5.4)

we conclude that \([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)\) is majorized by the function \(\sigma(\xi,\delta)\) for all \(\delta\geq1\). By the properties of majorization [18], we have

$$ \frac{ [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}{\xi} \prec\sigma (\xi,\delta), \quad\xi\in{ \cup}. $$
(5.5)

Thus, according to Lemma 3.2, we conclude that

$$ \int_{0}^{2\pi} \biggl\vert \frac{ [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi )}{\xi} \biggr\vert ^{p} \,d\theta\leq \int_{0}^{2\pi} \biggl\vert \biggl( \frac{1+\xi }{1-\xi} \biggr)^{\delta}\biggr\vert ^{p} \,d \theta, \quad p>0. $$
(5.6)

 □

In the same manner as in the proof of Theorem 5.1, one can get the next result:

Theorem 5.2

Consider the operator\([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)\), \(\curlyvee\in\bigwedge\). If the coefficients of ⋎ satisfy the inequality\(|\curlyvee_{n}| \leq(\frac{1}{n \nu})^{k}\), \(\nu\in (0,1)\)then

$$\int_{0}^{2\pi} \bigl\vert \bigl( \bigl[ \mathcal{S}_{\nu}^{k} \bigr]_{q}\curlyvee(\xi) \bigr)' \bigr\vert ^{p} \,d\theta\leq \int_{0}^{2\pi} \biggl\vert \biggl( \frac {1+\xi}{1-\xi} \biggr)^{\delta}\biggr\vert ^{p} \,d \theta, \quad p>0. $$

Moreover, the inequality in Theorem 5.1 can be studied in the following result:

Theorem 5.3

Consider the operator\(\mathcal{S}_{q}^{\kappa,k} \psi(z)\), \(\psi\in \varLambda\). If the coefficients ofψsatisfy the inequality\(|\vartheta_{n}| \leq(\frac{1}{n\kappa})^{k}\), \(\kappa\in(0,\infty)\)then there is a probability measureμon\((\partial U)^{2}\), for all\(\delta>1\).

Proof

Let \(\epsilon,\varepsilon\in\partial{\cup.}\) Then we have

$$ \begin{aligned}[b] \biggl( \frac{1+\epsilon\xi}{1+\varepsilon\xi} \biggr)^{\delta}& = \frac{(1+\epsilon\xi)^{\delta}}{1+\varepsilon\xi}. \frac{1}{(1+ \varepsilon\xi)^{\delta-1}} \\ & \ll \frac{(1+ \xi)^{\delta}}{1-\xi}. \frac{1}{(1- \xi)^{\delta -1}} \\ &= \biggl( \frac{1+\xi}{1-\xi} \biggr)^{\delta}, \quad\delta>1. \end{aligned} $$
(5.7)

By virtue of Theorem 1.11 in [19], the functional \(( \frac{1+\epsilon\xi}{1+\varepsilon\xi} )^{\delta}\) defines a probability measure μ in \((\partial{\cup})^{2}\) fulfilling

$$ \chi(\xi)= \int_{(\partial{\cup})^{2}} \biggl( \frac{1+\epsilon\xi}{1+\varepsilon\xi} \biggr)^{\delta}\, d \mu (\epsilon,\varepsilon), \quad\xi\in{\cup}. $$
(5.8)

Then there is a constant c (diffusion constant) such that

$$ \int_{(\partial{\cup})^{2}} \biggl( \frac {1+\epsilon\xi}{1+\varepsilon\xi} \biggr)^{\delta}\,d \mu(\epsilon ,\varepsilon) =c \int_{(\partial{\cup})^{2}} \biggl( \frac{ [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi\epsilon)}{\varepsilon\xi} \biggr)^{\delta}\,d \mu(\epsilon,\varepsilon), \quad\xi\in{\cup}. $$
(5.9)

This completes the proof. □

6 A class of differential equations

This section deals with an application of the operator (2.7) in a class of differential equations (for recent work see [20]). The class of quantum III-Painlevé differential equations has been studied recently in [21–23]. This class takes the formula

$$ { \xi\curlyvee(\xi){\frac{d^{2}\curlyvee(\xi)}{d\xi^{2}}}=\xi \biggl({ \frac{d\curlyvee(\xi)}{d\xi}} \biggr)^{2}-\curlyvee(\xi ){ \frac{d\curlyvee(\xi)}{d \xi}}},\quad \xi\in\cup, \curlyvee\in \bigwedge. $$
(6.1)

Rearranging Eq. (6.1), we have

$$ \biggl(1+\frac{\xi\curlyvee''(\xi)}{\curlyvee'(\xi)} \biggr)-\frac{\xi\curlyvee'(\xi) }{\curlyvee(\xi)}=0, \quad\xi\in \cup, $$
(6.2)

subjected to the boundary conditions

$$ \curlyvee(\xi)= \xi+\curlyvee_{2} \xi^{2}+O\bigl(\xi^{3}\bigr), \qquad \vert \curlyvee _{n} \vert \leq\frac{1}{[Q_{n} ]_{q}^{k}},\quad n \geq2, \xi\in\cup= \bigl\{ \xi \in\mathbb{C}: \vert \xi \vert < 1 \bigr\} , $$
(6.3)

where

$$[Q_{n} ]_{q}^{k}:= [n ]_{q}^{k} \biggl(\frac{\kappa_{1}({\nu},\xi )+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr)^{k}. $$

Now by employing the operator (2.7), Eq. (6.2) becomes (called q-Painlevé differential equation of type III)

$$ \biggl(1+\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )''}{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'} \biggr)-\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )}=0, \quad\xi\in\cup, $$
(6.4)

subjected to (6.3). Our aim is to study the geometric solution of (6.4) satisfying the boundary condition (6.3). For this purpose, we define the following analytic class:

Definition 6

For a function \(\curlyvee\in\bigwedge\) and a convex function \(\psi \in\cup\) with \(\psi(0)=0\), the function ⋎ is said to be in the class \(\mathbf{V}_{q} (\psi)\) if and only if

$$ \mathsf{P}(\xi):= \biggl(1+\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )''}{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'} \biggr)- \frac{\xi ( [ \mathcal {S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )} \prec\psi(\xi), \quad\xi\in\cup, $$
(6.5)

where \(\psi(\xi)\in\bigwedge\).

For the functions in the class \(\mathbf{V}_{q}(\psi)\), the following result holds.

Theorem 6.1

If the function\(\curlyvee\in\mathbf{V}_{q}(\psi)\)is given by (2.1), then

$$ \vert \curlyvee_{2} \vert \leq \frac{1}{[Q_{2} ]_{q}^{k}}, \qquad \vert \curlyvee_{3} \vert \leq \frac{1}{[Q_{3} ]_{q}^{k}}. $$
(6.6)

Proof

Let \(\curlyvee\in\mathbf{V}_{q} (\psi)\) have the expansion

$$\curlyvee(\xi)= \xi+\curlyvee_{2} \xi^{2}+ \curlyvee_{3} \xi^{3}+ \cdots , \quad\xi\in\cup. $$

Moreover, we let

$$[Q_{n} ]_{q}^{k}:= [n ]_{q}^{k} \biggl(\frac{\kappa_{1}({\nu},\xi )+(-1)^{n+1}\kappa_{0}({\nu},\xi)}{\kappa_{1}({\nu},\xi)+\kappa _{0}({\nu},\xi)} \biggr)^{k}. $$

Then by the definition of subordination, there is a Schwarz function ⊤ with \(\top(0)=0\) and \(|\top(\xi)|<1\) satisfying \(\mathsf {P}(\xi) = \psi(\top(\xi))\), \(\xi\in\cup\). Furthermore, if we assume that \(|\top(\xi)|=|\xi|<1\), then, in view of Schwarz lemma, there is a complex number τ with \(|\tau|=1\) satisfying \(\top(\xi)=\tau\xi\). Consequently, we obtain

$$ \begin{aligned}[b] & \mathsf{P}(\xi) =\psi\bigl(\top(\xi)\bigr) \\ &\quad\Longrightarrow\quad \biggl(1+\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )''}{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'} \biggr)-\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )} = \psi\bigl(\tau( \xi)\bigr) \\ &\quad\Longrightarrow\quad 1+ \bigl( 2\curlyvee_{2}[Q_{2} ]_{q}^{k} \xi+ \bigl( 6\curlyvee_{3}[Q_{3} ]_{q}^{k}-4 \bigl(\curlyvee_{2}[Q_{2} ]_{q}^{k} \bigr)^{2} \bigr) \xi^{2}+\cdots \bigr) \\ &\phantom{\quad\Longrightarrow\quad\quad\quad} - \bigl( 1+ \curlyvee_{2}[Q_{2} ]_{q}^{k} \xi+ \bigl(2\curlyvee_{3}[Q_{3} ]_{q}^{k}- \bigl(\curlyvee_{2}[Q_{2} ]_{q}^{k} \bigr)^{2}\bigr) \xi^{2}+\cdots \bigr) \\ & \phantom{\quad\Longrightarrow\quad\quad}= \tau \bigl( \xi+\psi_{2}\xi^{2}+ \psi_{3}\xi^{3}+\cdots \bigr) \\ &\quad\Longrightarrow \quad\curlyvee_{2}[Q_{2} ]_{q}^{k} \xi+ \bigl(4\curlyvee_{3}[Q_{3} ]_{q}^{k}-3 \bigl(\curlyvee_{2}[Q_{2} ]_{q}^{k} \bigr)^{2} \bigr) \xi^{2}+\cdots\\ &\phantom{\quad\Longrightarrow\quad\quad}=\tau \xi+\tau \psi_{2}\xi^{2}+\tau\psi_{3}\xi ^{3}+\cdots. \end{aligned} $$

It follows that

$$\bigl\vert \curlyvee_{2}[Q_{2} ]_{q}^{k} \bigr\vert = \vert \tau \vert =1, \qquad \bigl\vert \curlyvee_{3}[Q_{3} ]_{q}^{k} \bigr\vert \leq\frac{1}{4} \bigl( \vert \psi_{2} \vert +3 \bigr). $$

Since ψ is convex univalent in ∪, \(|\psi_{n}|\leq1\), ∀n; this implies that

$$\vert \curlyvee_{2} \vert \leq\frac{1}{[Q_{2} ]_{q}^{k}}, \qquad \vert \curlyvee_{3} \vert \leq \frac{1}{[Q_{3} ]_{q}^{k}}. $$

Hence, the proof is complete. □

We need the following fact, which can be located in [12].

Lemma 6.2

Consider functions\(f_{1}, f_{2}, f_{3}: \cup\rightarrow\mathbb{C}\)such that\(\Re(f_{1}) \geq a\geq0\). If\(f\in\mathbb{H}[1,n]\) (the set of analytic functions having the expansion\(f(\xi)= 1+\varphi_{1}\xi+\cdots \)) and

$$\Re \bigl( a\xi^{2} f''( \xi)+f_{1}(\xi) \xi f'(\xi) + f_{2}(\xi) f(\xi)+f_{3}(\xi) \bigr)>0, \quad a\geq0, \xi\in\cup, $$

then\(\Re(f(\xi))>0\).

Lemma 6.3

Let ♭ be convex in ∪ and suppose\(f_{1}, f_{2}, f_{3}: \cup \rightarrow\mathbb{C}\)are analytic functions such that\(\Re(f_{1}) \geq a\geq0\). If\(g \in\mathbb{H}[0,m]\) (the set of analytic functions with the expansion\(g(\xi)= g_{1}\xi^{m}+\cdots\)), \(m\geq1\)and

$$a\xi^{2} g''(\xi)+f_{1}( \xi) \xi g'(\xi) + f_{2}(\xi) g(\xi)+f_{3}( \xi) \prec \flat(\xi),\quad a\geq0, \xi\in\cup, $$

then\(g(\xi) \prec \flat(\xi)\).

Lemma 6.4

Let\(a,b,c \in\mathbb{R}\)be such that\(a\geq0\), \(b\geq-a\), \(c\geq-b\). If\(q \in\mathbb{H}[0,1]\), where\(q(\xi)= q_{1}\xi+\cdots\)and

$$a \xi^{2} q''(\xi)+b \xi q'(\xi) + c q(\xi) \prec \xi,\quad a\geq0, \xi\in\cup, $$

then\(q(\xi) \prec \frac{\xi}{b+c}\), which is the best dominant.

Theorem 6.5

Let\(\curlyvee\in\mathbf{V}_{q}(\xi)\)and\(F(\xi)= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )}\). If\(\Re(\xi F (\xi ))>-1\), \(\xi\in\cup\), then\([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee\in\mathfrak{S}^{*}\) (starlike with respect to the origin).

Proof

Let \(F(\xi)=\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )}\). Then a straightforward computation implies that

$$ \begin{aligned}[b] \xi F'(\xi)&= \xi \biggl( \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )} \biggr)' \\ &= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )''}{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}\frac {\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}+\frac{\xi ( [ \mathcal {S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}- \biggl( \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi )} \biggr)^{2} \\ &= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{\curlyvee(\xi)} \biggl(1+\frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )''}{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}- \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi )} \biggr) \\ &= F(\xi) P(\xi). \end{aligned} $$

Hence, we obtain

$$ \begin{aligned}[b] P(\xi)&= \frac{\xi F'(\xi)}{F(\xi)}\prec\xi, \qquad\psi(\xi ):=\xi. \end{aligned} $$

It is clear that \(F \in\mathbb{H}[1,1]\) and

$$\Re\biggl(\frac{\xi F'(\xi)}{F(\xi)}\biggr)= \Re(\xi) \quad\Rightarrow\quad 1+ \Re \bigl( \xi F'(\xi) \bigr)= 1+ \Re\bigl(\xi F(\xi)\bigr)>0. $$

Then in view of Lemma 6.2, with \(a=0\), \(f_{1}=1\), \(f_{2}=0\) and \(f_{3}=1\), we have

$$\Re \bigl(F(\xi) \bigr)= \Re \biggl( \frac{\xi ( [ \mathcal {S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )'}{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)} \biggr) >0; $$

that is, \([ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee\in\mathfrak{S}^{*}\) with respect to the origin. □

Theorem 6.6

Let\(\curlyvee\in\bigwedge\)and\(F(\xi)= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}\). If

$$\frac{\xi F'(\xi)}{F(\xi)} \biggl( 2+ \frac{\xi F''(\xi)}{F'(\xi )}-\frac{\xi F'(\xi)}{F(\xi)} \biggr)\prec\psi(\xi), $$

whereψis convex in ∪, then\(\curlyvee\in\mathbf {V}_{q}(\psi)\).

Proof

Let \(\curlyvee\in\bigwedge\) and \(F(\xi)= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}\). As in the proof of Theorem 6.5, we have \(P(\xi)= \frac{\xi F'(\xi)}{F(\xi)}\). Then a calculation gives

$$ \begin{aligned}[b] \xi P'(\xi)+P(\xi)&= \xi \biggl( \frac{\xi F'(\xi)}{F(\xi)} \biggr)'+ \biggl(\frac{\xi F'(\xi)}{F(\xi)} \biggr) \\ &= \frac{\xi F'(\xi)}{F(\xi)} \biggl( 2+ \frac{\xi F''(\xi)}{F'(\xi )}-\frac{\xi F'(\xi)}{F(\xi)} \biggr) \\ &\prec\psi(\xi). \end{aligned} $$

Obviously, \(P\in\mathbb{H}[0,m]\), \(m=1\), and, by letting \(a=0\), \(f_{1}(\xi )=1\), and \(f_{2}(\xi)=1\), in view of Lemma 6.3, we have

$$P(\xi) =1+\frac{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )'' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }- \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)} \prec\psi(\xi). $$

Consequently, we get \(\curlyvee\in\mathbf{V}_{q}(\psi)\). □

Theorem 6.7

Let\(\curlyvee\in\bigwedge\)and\(F(\xi)= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}\). If

$$\frac{\xi F'(\xi)}{F(\xi)} \biggl( 1+ \frac{\xi F''(\xi)}{F'(\xi )}-\frac{\xi F'(\xi)}{F(\xi)} \biggr) \prec\xi, $$

whereψis convex in ∪, then\(\curlyvee\in\mathbf {V}_{q}(\xi)\).

Proof

Let \(\curlyvee\in\bigwedge\) and \(F(\xi)= \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)}\). As in the proof of Theorem 6.5, we have \(P(\xi)= \frac{\xi F'(\xi)}{F(\xi)}\). Then a straightforward calculation gives

$$ \begin{aligned}[b] \xi P'(\xi)&= \xi \biggl( \frac{\xi F'(\xi)}{F(\xi)} \biggr)' \\ &= \frac{\xi F'(\xi)}{F(\xi)} \biggl( 1+ \frac{\xi F''(\xi)}{F'(\xi )}-\frac{\xi F'(\xi)}{F(\xi)} \biggr) \\ &\prec\xi. \end{aligned} $$

Obviously, \(P\in\mathbb{H}[0,1]\) and, by letting \(a=0\), \(b=1\), and \(c=0\), where \(c\geq-b\), in view of Lemma 6.4, we have

$$P(\xi) =1+\frac{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )'' }{ ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi) )' }- \frac{\xi ( [ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi ) )' }{[ \mathcal{S}_{\nu}^{k} ]_{q}\curlyvee(\xi)} \prec\xi. $$

Consequently, we obtain \(\curlyvee\in\mathbf{V}_{q}(\xi)\). □

7 Conclusion

In this paper, we presented different types of integral inequalities based on q-calculus and conformable differential operator. These inequalities described the relations between the quantum conformable differential operators for different orders.