The monotony of the q−Struve-Bessel functions

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08-12-2025
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Authors: Yücel Özkan; Semra Korkmaz; Erhan Deniz; Murat Çağlar
Published in: Journal of Inequalities and Applications | Issue 1/2026

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Abstract

This article delves into the fascinating world of q−Struve-Bessel functions, focusing on their monotony. It begins by introducing the basic definitions and properties of these functions, setting the stage for a deeper exploration. The article then examines the conditions under which these functions exhibit monotonic behavior, providing a comprehensive analysis of their derivatives and critical points. Additionally, it discusses the implications of these findings in various mathematical and physical contexts, highlighting the practical applications of q−Struve-Bessel functions. The article concludes with a summary of the key results and their significance, offering a clear understanding of the monotony of q−Struve-Bessel functions and their role in advanced mathematical research.

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1 Introduction and preliminaries

Let $\mathbb{D}=\left \{ z\in \mathbb{C} :\left \vert z\right \vert <1\right \} $ be the open unit disk. An analytic function f in $\mathbb{D}$ with $0\neq f^{\prime }(0)$ is convex, i.e., ${\ \Re }\left [ 1+zf^{\prime \prime }\left ( z\right ) /f^{\prime } \left ( z\right ) \right ] >0$ for all $z\in \mathbb{D}$, if and only if f is univalent in $\mathbb{D}$ with $f(\mathbb{D})$ being convex. The infinite q-shifted factorial (or q-Pochhammer symbol) $(a;q)_{\infty }$ is defined by

$$ \left ( a;q\right ) _{\infty }=\prod _{n=0}^{\infty }\left ( 1-aq^{n} \right ) \ \ \ (0< q< 1). $$

When n is a positive integer, we write

$$ \left ( a;q\right ) _{n}=\left ( 1-a\right ) \left ( 1-aq\right ) ... \left ( 1-aq^{n-1}\right ) . $$

Jackson [12] defined a q-analogue of the gamma function for $x>0$ by

$$ \Gamma _{q}\left ( x\right ) = \frac{\left ( q;q\right ) _{\infty }}{\left ( q^{x};q\right ) _{\infty }}\left ( 1-q\right ) ^{1-x}. $$

(1.1)

Throughout this section, we assume that $0< q<1$.

The q-gamma function satisfies the functional equation

$$ \Gamma _{q}(x+1)=[x]_{q}\Gamma _{q}(x), $$

(1.2)

where, for any $x\in \mathbb{R}$, the q-number $[x]_{q}$ is defined by

$$ \lbrack x]_{q}:=\frac{1-q^{x}}{1-q}. $$

The q-gamma function can be viewed as a natural q-analogue of the classical factorial function. In particular, $\Gamma _{q}(n+1)$ corresponds to the q-factorial for positive integers n.

The q-digamma function $\psi _{q}(z)$ is defined as the logarithmic derivative of the q-gamma function (see [10, 12])

$$ \psi _{q}(z)=\frac{d}{dz}\ln \Gamma _{q}(z)= \frac{\Gamma _{q}^{\prime }(z)}{\Gamma _{q}(z)}. $$

It also admits the following series representation

$$ \psi _{q}(z)=-\ln (1-q)+\ln q\sum _{k=0}^{\infty } \frac{q^{k+z}}{1-q^{k+z}}. $$

Thomae [18] and Jackson [12] showed that the q-beta function defined by

$$ B_{q}(x,y)=\frac{\Gamma _{q}(x)\Gamma _{q}(y)}{\Gamma _{q}(x+y)}, \ \ \ x,y>0 $$

has the q-integral representation

$$ B_{q}(x,y)=\int _{0}^{1}t^{x-1}(1-qt)_{q}^{y-1}\,d_{q}t, $$

which serves as a q-analogue of Euler’s classical beta integral formula.

They also introduced the q-integral by

$$ \int _{0}^{a}f(x)\,d_{q}x=(1-q)\sum _{n=0}^{\infty }aq^{n}f(aq^{n}). $$

Oraby and Mansour [15] introduced and studied the q-Struve-Bessel functions $H_{v}^{(k)}(z;q^{2}\!)$ for $k=2,3$ defined by

$$ H_{v}^{(2)}(z;q^{2})=\sum _{k=0}^{\infty } \frac{(-1)^{k}q^{2k^{2}+2kv+2k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) }\left ( \frac{z}{1+q}\right ) ^{2k+v+1} $$

and

$$ H_{v}^{(3)}(z;q^{2})=\sum _{k=0}^{\infty } \frac{(-1)^{k}q^{k^{2}+k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) }\left ( \frac{z}{1+q}\right ) ^{2k+v+1}, $$

where $q\in (0,1)$, $\Re v > -\frac{1}{2}$ and $z\in \mathbb{C}$. An analytic function f is said to be normalized if it satisfies $f(0)=0$ and $f^{\prime }(0)=1$. In this sense, these functions are not normalized. Therefore, we define the normalized forms of the q-Struve-Bessel functions as follows

$$\begin{aligned}& \eta _{v}(z;q^{2})=(1+q)^{v}z^{\frac{1-v}{2}}H_{v}^{(2)}(\sqrt{z};q^{2})=\sum _{k=0}^{\infty } \frac{(-1)^{k}q^{2k^{2}+2kv+2k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) (1+q)^{2k+1}}z^{k+1}, \end{aligned}$$

(1.3)

$$\begin{aligned}& \xi _{v}(z;q^{2})=(1+q)^{v}z^{\frac{1-v}{2}}H_{v}^{(3)}(\sqrt{z};q^{2})=\sum _{k=0}^{\infty } \frac{(-1)^{k}q^{k^{2}+k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) (1+q)^{2k+1}}z^{k+1}, \end{aligned}$$

(1.4)

$$\begin{aligned}& \lambda _{v}(z;q^{2})=(1+q)^{v}z^{-v}H_{v}^{(3)}(z;q^{2})=\sum _{k=0}^{ \infty } \frac{(-1)^{k}q^{k^{2}+k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) (1+q)^{2k+1}}z^{2k+1}, \\& \end{aligned}$$

(1.5)

$$\begin{aligned}& \gamma _{v}(z;q^{2})=(1+q)^{v}z^{-v}H_{v}^{(2)}(z;q^{2})=\sum _{k=0}^{ \infty }\frac{(-1)^{k}q^{2k^{2}+2kv+2k}}{\Gamma _{q^{2}}\left ( k+\frac{3}{2}\right ) \Gamma _{q^{2}}\left ( k+v+\frac{3}{2}\right ) (1+q)^{2k+1}}z^{2k+1}. \\& \end{aligned}$$

(1.6)

In [16], Oraby and Mansour studied the starlikeness and convexity properties of the q-Struve-Bessel functions. More recently, András and Baricz [5], Cotîrlă and Szász [6, 7], Deniz and Szász [8], Mehrez [13], Alenazi and Mehrez [4] and Özkan et al. [17] investigated the monotonicity properties of certain special functions. Nowadays, the geometric behaviour of special functions has been widely studied by many authors, notably for its applications to univalent functions and the theory of special functions. In this direction, the interested reader can refer to some recent papers [1‐3] and the references therein.

Motivated by these studies, we examine the monotonicity properties of the normalized q-Struve-Bessel functions $\eta _{v}$, $\xi _{v}$, $\lambda _{v} $ and $\gamma _{v}$.

To establish our results, we first present the necessary definitions and lemmas in the next section. The aim of this paper is to prove the inclusion relations:

$$\begin{aligned}& \eta _{\mu }(\mathbb{D},q^{2})\subset \eta _{v}(\mathbb{D},q^{2}),\\& \xi _{\mu }(\mathbb{D},q^{2})\subset \xi _{v}(\mathbb{D},q^{2}),\\& \lambda _{\mu }(\mathbb{D},q^{2})\subset \lambda _{v}(\mathbb{D},q^{2}) \end{aligned}$$

and

$$ \gamma _{\mu }(\mathbb{D},q^{2})\subset \gamma _{v}(\mathbb{D},q^{2}) $$

regarding the normalized q-Struve-Bessel functions given by (1.3), (1.4), (1.5) and (1.6), respectively.

2 Preliminaries

In this section, we present several definitions and lemmas that will be used in the proofs of our main results.

Definition 2.1

[9] Let f, g, and h be analytic functions in $\mathbb{D}$. If the function h satisfies the conditions $h(0)=0$, $|h(z)|<1 $ for all $z\in \mathbb{D}$, and

$$ f(z)=g(h(z)) $$

then the function f is said to be 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

$$ f\prec g\quad \Longleftrightarrow \quad f(\mathbb{D})\subset g( \mathbb{D}). $$

Definition 2.2

[19] An infinite sequence $\left ( b_{n}\right ) _{n\geq 1}$ of complex numbers is called a subordination factor sequence if, for every convex function f defined by

$$ f(z)=\sum _{n=1}^{\infty }a_{n}z^{n}, $$

the function

$$ g(z)=\sum _{n=1}^{\infty }a_{n}b_{n}z^{n},\quad z\in \mathbb{D} $$

satisfies $g\prec f$.

Definition 2.3

[9] Let f and g be analytic functions in $\mathbb{D}$, defined by the power series

$$ f(z) = \sum _{n=1}^{\infty} a_{n} z^{n} \quad \text{and} \quad g(z) = \sum _{n=1}^{\infty} b_{n} z^{n}. $$

The convolution (or Hadamard product) of the functions f and g, denoted by $f * g$, is defined by

$$ (f * g)(z) = \sum _{n=1}^{\infty} a_{n} b_{n} z^{n}. $$

Remark 2.4

Definitions 2.1 and 2.2 can be reformulated using convolution as follows. Let $\left ( b_{n}\right ) _{n\geq 1}$ be a sequence of complex numbers, and define the function ϕ by

$$ \phi (z)=\sum _{n=1}^{\infty }b_{n}z^{n}. $$

Then, the sequence $\left ( b_{n}\right ) _{n\geq 1}$ is called a subordination factor sequence if, for every convex function f, the following subordination relation holds

$$ f\ast \phi \prec f. $$

Lemma 2.5

[19] The following two properties are equivalent:

(i)

The infinite sequence $\left ( b_{n} \right )_{n \geq 1}$ of complex numbers is a subordination factor sequence.

(ii)

The inequality

$$ \frac{1}{2} + \Re \left ( \sum _{n=1}^{\infty} b_{n} z^{n} \right ) > 0 $$

holds for every $z \in \mathbb{D}$.

Lemma 2.6

[11] Let $\left ( f_{n}\right ) _{n\geq 0}$ be a sequence of real numbers such that

$$ f_{0}=1,\quad f_{n}-2f_{n+1}+f_{n+2}\geq 0,\quad \textit{and}\quad f_{n}-f_{n+1} \geq 0\quad \textit{for all }n\in \mathbb{N}_{0}. $$

Then, the inequality

$$ \Re \left ( 1+\sum _{n=1}^{\infty }f_{n}z^{n}\right ) >\frac{1}{2} $$

holds for every $z\in \mathbb{D}$.

Moreover, in Theorem 2 of [14], the authors show that for $n\geq 1$ and $z\in \mathbb{D}$, if a function of the form

$$ f(z)=z+a_{n+1}z^{n+1}+\cdots $$

satisfies $\left \vert f^{\prime \prime }(z)\right \vert \leq \frac{n}{n+1}$, then f is convex. The following lemma is a consequence of this result.

Lemma 2.7

[14] Let f be of the form

$$ f(z)=\sum _{n=1}^{\infty }a_{n}z^{n} $$

with $a_{1}\neq 0$. If

$$ \left \vert \frac{f^{\prime \prime }(z)}{f^{\prime }(0)}\right \vert < \frac{1}{2},\quad z\in \mathbb{D}, $$

(2.1)

then f is univalent, and its image $f(\mathbb{D})$ is a convex subset of $\mathbb{C}$.

Lemmas 2.8–2.10 presented below correspond to special cases of the respective Lemmas 2.8-2.10 in [17], under the transformations q → $q^{2}$, v $\rightarrow v+\frac{1}{2}$ and μ → $\mu +\frac{1}{2}$.

The following lemmas serve as key tools in the proofs of our main results.

Lemma 2.8

The following assessments are valid:

(i)

If $\mu > v > v_{0}$, then the following inequality holds:

$$ \frac{1 - q^{2v + 5}}{1 - q^{2\mu + 5}} > \frac{\Gamma _{q^{2}}\left (v + \frac{5}{2}\right )}{\Gamma _{q^{2}}\left (\mu + \frac{5}{2}\right )}, $$

(2.2)

where $v_{0}$ is the largest real root of the equation

$$ \frac{2q^{2v + 5}}{1 - q^{2v + 5}} \ln q + \psi _{q^{2}}\left (v + \tfrac{5}{2}\right ) = 0. $$

(ii)

If $\mu >v>v_{1}$, then the following inequality holds:

$$ \frac{q^{2v}}{q^{2\mu }}> \frac{\Gamma _{q^{2}}\left ( v+\frac{5}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +\frac{5}{2}\right ) }, $$

(2.3)

where $v_{1}$ is the largest real root of the equation

$$ 2\ln q-\psi _{q^{2}}\left ( v+\tfrac{5}{2}\right ) =0. $$

Lemma 2.9

For $\mu >v>\max \{v_{0},v_{1}\}$, the following inequality holds:

$$ \Re \left ( 1+\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}\left ( v+k+\tfrac{3}{2}\right ) q^{2k\mu }}{\Gamma _{q^{2}}\left ( \mu +k+\tfrac{3}{2}\right ) q^{2kv}}z^{k}\right ) >\frac{1}{2} $$

for all $z\in \mathbb{D}$.

Lemma 2.10

For $\mu >v>v_{0}$, the following inequality holds:

$$ \Re \left ( 1+\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}\left ( k+v+\tfrac{3}{2}\right ) }{\Gamma _{q^{2}}\left ( k+\mu +\tfrac{3}{2}\right ) }z^{k} \right ) >\frac{1}{2} $$

for all $z\in \mathbb{D}$.

Lemma 2.11

If $\mu >v>-\tfrac{3}{2}$, then the following inequality holds:

$$ \Re \left ( 1+\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}\left ( v+1+\tfrac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k}{2}\right ) }z^{k} \right ) >\frac{1}{2} $$

for every $z\in \mathbb{D}$.

Proof

To prove this lemma, we apply Lemma 2.6. Let

$$ f_{0}=1,\quad f_{k}= \frac{\Gamma _{q^{2}}\left ( v+1+\tfrac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k}{2}\right ) }\quad \text{for }k \geq 1. $$

Clearly, since $\mu >v$, we have

$$ f_{0}=1> \frac{\Gamma _{q^{2}}\left ( v+1+\tfrac{1}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\tfrac{1}{2}\right ) }=f_{1}. $$

We now show that the sequence $(f_{k})$ is strictly decreasing, i.e.,

$$ f_{k}>f_{k+1}\quad \text{for all }k\geq 1. $$

This inequality is equivalent to proving that

$$ \frac{\Gamma _{q^{2}}\left ( v+1+\tfrac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( v+1+\tfrac{k+1}{2}\right ) }> \frac{\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k+1}{2}\right ) }. $$

(2.4)

To this end, define the function $u:(1,\infty )\rightarrow (0,\infty )$ by

$$ u(x)=\frac{\Gamma _{q^{2}}(x)}{\Gamma _{q^{2}}(x+\tfrac{1}{2})}. $$

Using the identity involving the q-Euler Beta function $B_{q^{2}}(x,y)$, we have

$$\begin{aligned} u(x)& = \frac{B_{q^{2}}(x,\tfrac{1}{2})}{\Gamma _{q^{2}}(\tfrac{1}{2})}= \frac{1}{\Gamma _{q^{2}}(\tfrac{1}{2})}\int _{0}^{1}t^{x-1}(1-q^{2}t)_{q^{2}}^{-1/2}\,d_{q^{2}}t \\ & =\frac{1-q^{2}}{\Gamma _{q^{2}}(\tfrac{1}{2})}\sum _{n=0}^{\infty } \frac{q^{2nx}}{\sqrt{(1-q^{2(n+1)})_{q^{2}}}}. \end{aligned}$$

(2.5)

Since, $\ln q<0$ for $0< q<1$ we have

$$ u^{\prime }(x)=\frac{1-q^{2}}{\Gamma _{q^{2}}(\tfrac{1}{2})}\sum _{n=0}^{\infty } \frac{2nq^{2nx}\ln q}{\sqrt{(1-q^{2(n+1)})_{q^{2}}}}< 0. $$

Thus, this representation implies that $u(x)$ is strictly decreasing on $(1,\infty )$, and hence

$$ u\left ( v+1+\tfrac{k}{2}\right ) >u\left ( \mu +1+\tfrac{k}{2} \right ) , $$

which proves (2.4). It remains to prove that the sequence $(f_{k})$ is log-concave, i.e.,

$$ f_{0}+f_{2}\geq 2f_{1}\quad \text{and}\quad f_{n}+f_{n+2}\geq 2f_{n+1} \quad \text{for all }n\geq 1. $$

(2.6)

To this end, define the function $\upsilon :(1,\infty )\rightarrow (0,\infty )$ by

$$ \upsilon (x)=\frac{\Gamma _{q^{2}}(x)}{\Gamma _{q^{2}}(x+\mu -v)}= \frac{B_{q^{2}}(x,\mu -v)}{\Gamma _{q^{2}}(\mu -v)}. $$

Then, inequality (2.6) is equivalent to the convexity condition

$$ \frac{\upsilon (x)+\upsilon (x+1)}{2}\geq \upsilon \left ( x+ \tfrac{1}{2}\right ) ,\quad \text{for all }x>1, $$

(2.7)

with $x=v+1+\tfrac{n}{2}$.

We verify the convexity of υ by differentiating twice:

$$ \upsilon ^{\prime \prime }(x)=\frac{1}{\Gamma _{q^{2}}(\mu -v)}\int _{0}^{1}t^{x-1}\ln ^{2}(t)(1-q^{2}t)_{q^{2}}^{\mu -v-1}\,d_{q^{2}}t \geq 0. $$

This shows that υ is convex on $(1,\infty )$. Thus the inequality (2.7) holds. Consequently the proof is completed. □

Lemma 2.12

Provided that $\mu >v>-\tfrac{3}{2}$, the following inequality holds:

$$ \Re \left ( 1+\sum _{k=1}^{\infty } \frac{q^{(k-1)\mu }\,\Gamma _{q^{2}}\left ( v+1+\tfrac{k}{2}\right ) }{q^{(k-1)v}\,\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k}{2}\right ) }z^{k}\right ) >\frac{1}{2} $$

for every $z\in \mathbb{D}$.

Proof

Let $f_{0}=1$ and

$$ f_{k}= \frac{q^{(k-1)\mu }\Gamma _{q^{2}}\left ( v+1+\tfrac{k}{2}\right ) }{q^{(k-1)v}\Gamma _{q^{2}}\left ( \mu +1+\tfrac{k}{2}\right ) }. $$

The proof follows from the same arguments as in Lemma 2.11, based on the monotonicity and convexity properties of the q-gamma function. Therefore, we omit the details. □

3 The main results

Our first important result is the following theorem, which provides the monotonicity property of the function $\eta _{v}$ with respect to the parameters v.

Theorem 3.1

If $\mu >v\geq \max \{v_{0},v_{1},v_{2}\}$, then the following inclusion holds:

$$ \eta _{\mu }(\mathbb{D},q^{2})\subset \eta _{v}(\mathbb{D},q^{2}), $$

where $v_{0}$ and $v_{1}$ are as defined in Lemma 2.8, and $\eta _{v}$ is defined by (1.3). Moreover, $v_{2}$ is the largest real root of the equation $r_{q}(v)=0$, where the function $r_{q}$ is given by (3.2).

Proof

From straightforward calculations, we obtain

$$ \frac{\eta _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\eta _{v}^{\prime }\left ( 0;q^{2}\right ) }= \sum _{k=1}^{\infty } \frac{(-1)^{k}k(k+1)q^{2k^{2}+2kv+2k}(1+q)\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k+1}}z^{k-1}. $$

Then we obtain the following inequality

$$\begin{aligned} \left \vert \frac{\eta _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\eta _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert & \leq \sum _{k=1}^{\infty }\frac{k(k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & = \frac{2q^{4+2v}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}} \\ & +\sum _{k=2}^{\infty } \frac{k(k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}}. \end{aligned}$$

Also we have

$$ \frac{2q^{4+2v}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}}= \frac{2q^{4+2v}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})} $$

(3.1)

and

$$\begin{aligned} & \sum _{k=2}^{\infty } \frac{k(k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & \leq \frac{\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{q^{8}\Gamma _{q^{2}}(7/2)\Gamma _{q^{2}}(v+7/2)} \sum _{k=2}^{\infty }k(k+1)\left ( \frac{q^{10+2v}}{(1+q)^{2}}\right ) ^{k} \\ & = \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}} \end{aligned}$$

by using the inequality $k^{2}\geq 4k-4$ for $k\geq 2$.

Here we used the monotonity of the function

$$ f(k)=\frac{1}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)} $$

since

$$ \frac{f(k+1)}{f(k)}= \frac{\left ( 1-q^{2}\right ) ^{2}}{\left ( 1-q^{2k+3}\right ) \left ( 1-q^{2k+2v+3}\right ) }< 1. $$

Thus for $0< q<1$ and $v\geq v_{2}$, we conclude that

$$\begin{aligned} \left \vert \frac{\eta _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\eta _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert -\frac{1}{2}& \leq \frac{2q^{4+2v}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}-\frac{1}{2} \\ & + \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}} \\ & \leq 0. \end{aligned}$$

By Lemma 2.7, this implies that $\eta _{v}$ is a convex function for $0< q<1$ and $v\geq v_{2}$. Here, $v_{2}$ is the largest real root of the equation

$$\begin{aligned} r_{q}(v):=\ & \frac{2q^{4+2v}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})} \\ & + \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})\left ( (1+q)^{2}-q^{10+2v}\right ) ^{3}}-\frac{1}{2}=0. \end{aligned}$$

(3.2)

According to Lemma 2.9, the sequence

$$ \left ( \frac{\Gamma _{q^{2}}(v+k+3/2)q^{2k\mu }}{\Gamma _{q^{2}}(\mu +k+3/2)q^{2kv}} \right ) _{k\geq 1} $$

is a subordination factor sequence. Therefore, the subordination relation follows

$$ \eta _{\mu }=\chi \ast \eta _{v}\prec \eta _{v}, $$

where

$$ \chi (z)=\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}(v+k+3/2)q^{2k\mu }}{\Gamma _{q^{2}}(\mu +k+3/2)q^{2kv}}z^{k}. $$

Since $\eta _{\mu }(0)=\eta _{v}(0)$, the subordination $\eta _{\mu }\prec \eta _{v}$ is equivalent to the inclusion

$$ \eta _{\mu }(\mathbb{D},q^{2})\subset \eta _{v}(\mathbb{D},q^{2}). $$

Thus, the proof is completed. □

In Figure 1, it can be observed that $r_{q}(v)\leq 0$ for $v\geq v_{2}$ according to some special values of q.

Figure 1

Graph of $r_{q}(v)$

Full size image

Theorem 3.2

If $\mu >v\geq \max \{v_{0},v_{3}\}$, then the following inclusion holds:

$$ \xi _{\mu }(\mathbb{D},q^{2})\subset \xi _{v}(\mathbb{D},q^{2}), $$

where $v_{0}$ is as defined in Lemma 2.8, and $\xi _{v}$ is given by (1.4). Moreover, $v_{3}$ is the largest real root of the $\rho _{q}(v)=0 $, where the function $\rho _{q}$ is defined by (3.4).

Proof

From straightforward calculations, we obtain

$$ \frac{\xi _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\xi _{v}^{\prime }\left ( 0;q^{2}\right ) }= \sum _{k=1}^{\infty } \frac{(-1)^{k}k(k+1)q^{k^{2}+k}(1+q)\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k+1}}z^{k-1}. $$

(3.3)

Thus, we derive the inequality

$$\begin{aligned} \left \vert \frac{\xi _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\xi _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert & \leq \sum _{k=1}^{\infty }\frac{k(k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & = \frac{2q^{2}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}} \\ & +\sum _{k=2}^{\infty } \frac{k(k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}}. \end{aligned}$$

By using the identity

$$ \frac{2q^{2}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}}= \frac{2q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})} $$

and the inequality $k^{2}\geq 4k-4$ for $k\geq 2$, we get

$$\begin{aligned} & \sum _{k=2}^{\infty } \frac{k(k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & \leq \frac{\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{q^{4}\Gamma _{q^{2}}(7/2)\Gamma _{q^{2}}(v+7/2)} \sum _{k=2}^{\infty }k(k+1)\left ( \frac{q^{5}}{(1+q)^{2}}\right ) ^{k} \\ & = \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}} \end{aligned}$$

hence, we have

$$\begin{aligned} \left \vert \frac{\xi _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\xi _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert -\frac{1}{2}& \leq \frac{2q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}-\frac{1}{2} \\ & + \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}} \\ & \leq 0 \end{aligned}$$

for $0< q<1$ and $v\geq v_{3}$. By Lemma 2.7, this implies that $\xi _{v}$ is a convex function for $0< q<1$ and $v\geq v_{3}$. Here, $v_{3}$ is the largest root of the equation:

$$\begin{aligned} \rho _{q}(v):=\ & \frac{2q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}- \frac{1}{2} \\ & + \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{6}-((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}}=0. \end{aligned}$$

(3.4)

According to Lemma 2.10, the sequence

$$ \left ( \frac{\Gamma _{q^{2}}(k+v+3/2)}{\Gamma _{q^{2}}(k+\mu +3/2)} \right ) _{k\geq 1} $$

is a subordination factor sequence. Hence, the subordination relation

$$ \xi _{\mu }=\varkappa \ast \xi _{v}\prec \xi _{v} $$

holds, where

$$ \varkappa (z)=\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}(k+v+3/2)}{\Gamma _{q^{2}}(k+\mu +3/2)}z^{k}. $$

Since $\xi _{\mu }(0)=\xi _{v}(0)$, the subordination $\xi _{\mu }\prec \xi _{v}$ is equivalent to the inclusion

$$ \xi _{\mu }(\mathbb{D},q^{2})\subset \xi _{v}(\mathbb{D},q^{2}). $$

Thus, the proof is completed. □

In Figure 2, it can be observed that $\rho _{q}(v)\leq 0$ for $v\geq v_{3}$ according to some special values of q.

Figure 2

Graph of $\rho _{q}(v)$

Full size image

Theorem 3.3

If $\mu >v\geq \max \left \{ -\tfrac{3}{2},v_{0},v_{4}\right \} $, then the following inclusion holds:

$$ \lambda _{\mu }(\mathbb{D},q^{2})\subset \lambda _{v}(\mathbb{D},q^{2}) $$

where $v_{0}$ is as defined in Lemma 2.8, and $\lambda _{v}$ is given by (1.5). Moreover, $v_{4}$ is the largest real root of the $\theta _{q}(v)=0$, where the function $\theta _{q}$ is defined by (3.6).

Proof

We can easily see that

$$ \frac{\lambda _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\lambda _{v}^{\prime }\left ( 0;q^{2}\right ) }= \sum _{k=1}^{\infty } \frac{(-1)^{k}2k(2k+1)q^{k^{2}+k}(1+q)\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k+1}}z^{2k-1}. $$

(3.5)

Hence, we derive the inequality

$$\begin{aligned} \left \vert \frac{\lambda _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\lambda _{v}^{\prime }\left ( 0;q^{2}\right ) }\right \vert & \leq \sum _{k=1}^{\infty } \frac{2k(2k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & = \frac{6q^{2}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}} \\ & +\sum _{k=2}^{\infty } \frac{2k(2k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}}. \end{aligned}$$

We also have

$$ \frac{6q^{2}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}}= \frac{6q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{2v+3})}. $$

Using the inequality $k^{2}\geq 4k-4$ for $k\geq 2$, we obtain

$$\begin{aligned} & \sum _{k=2}^{\infty } \frac{2k(2k+1)q^{k^{2}+k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & \leq \frac{\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{q^{4}\Gamma _{q^{2}}(7/2)\Gamma _{q^{2}}(v+7/2)} \sum _{k=2}^{\infty }2k(2k+1)\left ( \frac{q^{5}}{(1+q)^{2}}\right ) ^{k} \\ & = \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{5}]-3((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \left \vert \frac{\lambda _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\lambda _{v}^{\prime }\left ( 0;q^{2}\right ) }\right \vert - \frac{1}{2}& \leq \frac{6q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{2v+3})}- \frac{1}{2} \\ & + \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{5}]-3((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}} \\ & \leq 0 \end{aligned}$$

for $0< q<1$ and $v\geq v_{4}$. By Lemma 2.7, this implies that $\lambda _{v}$ is a convex function for $0< q<1$ and $v\geq v_{4}$. Here, $v_{4}$ is the largest real root of the equation

$$\begin{aligned} \theta _{q}(v):=\ & \frac{6q^{2}(1-q)^{2}}{(1-q^{3})(1-q^{2v+3})}- \frac{1}{2} \\ & + \frac{2q(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{5}]-3((1+q)^{2}-q^{5})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{5})^{3}}=0. \end{aligned}$$

(3.6)

According to Lemma 2.11, the sequence

$$ \left ( \frac{\Gamma _{q^{2}}\left ( v+1+\frac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\frac{k}{2}\right ) } \right ) _{k\geq 1} $$

is a subordination factor sequence. Consequently, the subordination

$$ \lambda _{\mu }=\kappa \ast \lambda _{v}\prec \lambda _{v} $$

holds, where

$$ \kappa (z)=\sum _{k=1}^{\infty } \frac{\Gamma _{q^{2}}\left ( v+1+\frac{k}{2}\right ) }{\Gamma _{q^{2}}\left ( \mu +1+\frac{k}{2}\right ) }z^{k}. $$

Since $\lambda _{\mu }(0)=\lambda _{v}(0)$, the subordination $\lambda _{\mu }\prec \lambda _{v}$ is equivalent to

$$ \lambda _{\mu }(\mathbb{D},q^{2})\subset \lambda _{v}(\mathbb{D},q^{2}), $$

which completes the proof. □

In Figure 3, it can be observed that $\theta _{q}(v)\leq 0$ for $v\geq v_{4}$ according to some special values of q.

Figure 3

Graph of $\theta _{q}(v)$

Full size image

Theorem 3.4

If $\mu >v\geq \max \left \{ -\tfrac{3}{2},v_{0},v_{5}\right \} $, then the following inclusion holds:

$$ \gamma _{\mu }(\mathbb{D},q^{2})\subset \gamma _{v}(\mathbb{D},q^{2}), $$

where $v_{0}$ is as defined in Lemma 2.8, and $\gamma _{v}$ is given by (1.6). Moreover, $v_{5}$ is the largest real root of the equation $\phi _{q}(v)=0$, where the function $\phi _{q}$ is defined by (3.8).

Proof

We have

$$ \frac{\gamma _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\gamma _{v}^{\prime }\left ( 0;q^{2}\right ) }= \sum _{k=1}^{\infty } \frac{(-1)^{k}2k(2k+1)q^{2k^{2}+2kv+2k}(1+q)\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k+1}}z^{2k-1}. $$

(3.7)

This yields

$$\begin{aligned} \left \vert \frac{\gamma _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\gamma _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert & \leq \sum _{k=1}^{\infty }\frac{2k(2k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & = \frac{6q^{4+2v}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}} \\ & +\sum _{k=2}^{\infty } \frac{2k(2k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}}. \end{aligned}$$

From simple calculation we obtain

$$ \frac{6q^{4+2v}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(5/2)\Gamma _{q^{2}}(v+5/2)(1+q)^{2}}= \frac{6q^{2(2+v)}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}. $$

Using $k^{2}\geq 4k-4$ for $k\geq 2$, we get

$$\begin{aligned} & \sum _{k=2}^{\infty } \frac{2k(2k+1)q^{2k^{2}+2kv+2k}\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{\Gamma _{q^{2}}(k+3/2)\Gamma _{q^{2}}(k+v+3/2)(1+q)^{2k}} \\ & \leq \frac{\Gamma _{q^{2}}(3/2)\Gamma _{q^{2}}(v+3/2)}{q^{8}\Gamma _{q^{2}}(7/2)\Gamma _{q^{2}}(v+7/2)} \sum _{k=2}^{\infty }2k(2k+1)\left ( \frac{q^{10+2v}}{(1+q)^{2}}\right ) ^{k} \\ & = \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{10+2v}]-3((1+q)^{2}-q^{10+2v})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{10+2v})^{3}}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} &\left \vert \frac{\gamma _{v}^{\prime \prime }\left ( z;q^{2}\right ) }{\gamma _{v}^{\prime }\left ( 0;q^{2}\right ) } \right \vert -\frac{1}{2}\\ &\quad \leq \frac{6q^{2(2+v)}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}-\frac{1}{2} \\ &\qquad + \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{10+2v}]-3((1+q)^{2}-q^{10+2v})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{10+2v})^{3}} \\ &\quad \leq 0 \end{aligned}$$

for $0< q<1$ and $v\geq v_{5}$. By Lemma 2.7, this implies that $\gamma _{v}$ is a convex function for $0< q<1$ and $v\geq v_{5}$. Here, $v_{5}$ is the largest real root of the equation:

$$\begin{aligned} \phi _{q}(v):=\ & \frac{6q^{2(2+v)}(1-q)^{2}}{(1-q^{3})(1-q^{3+2v})}- \frac{1}{2} \\ & + \frac{2q^{2(1+v)}(1-q)^{4}(1+q)^{2}\left ( (1+q)^{4}[3(1+q)^{2}+q^{10+2v}]-3((1+q)^{2}-q^{10+2v})^{3}\right ) }{(1-q^{3})(1-q^{5})(1-q^{2v+3})(1-q^{2v+5})((1+q)^{2}-q^{10+2v})^{3}} \\ &=0. \end{aligned}$$

(3.8)

According to Lemma 2.12, the sequence

$$ \left ( \frac{q^{(k-1)\mu }\Gamma _{q^{2}}\left ( v+1+\frac{k}{2}\right ) }{q^{(k-1)v}\Gamma _{q^{2}}\left ( \mu +1+\frac{k}{2}\right ) }\right ) _{k \geq 1} $$

is a subordination factor sequence. Consequently, the subordination

$$ \gamma _{\mu }=\tau \ast \gamma _{v}\prec \gamma _{v} $$

holds, where

$$ \tau (z)=\sum _{k=1}^{\infty } \frac{q^{(k-1)\mu }\Gamma _{q^{2}}\left ( v+1+\frac{k}{2}\right ) }{q^{(k-1)v}\Gamma _{q^{2}}\left ( \mu +1+\frac{k}{2}\right ) }z^{k}. $$

Since $\gamma _{\mu }(0)=\gamma _{v}(0)$, the subordination $\gamma _{\mu }\prec \gamma _{v}$ is equivalent to

$$ \gamma _{\mu }(\mathbb{D},q^{2})\subset \gamma _{v}(\mathbb{D},q^{2}), $$

which completes the proof. □

In Figure 4, it can be observed that $\phi _{q}(v)\leq 0$ for $v\geq v_{5}$ according to some special values of q.

Figure 4

Graph of $\phi _{q}(v)$

Full size image

Remark 3.5

In all of the above theorems, taking the limit $q\rightarrow 1$ recovers the corresponding results of Deniz and Szász in [8].

Acknowledgements

Not applicable.

Declarations

Competing interests

The authors declare no competing interests.

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Title: The monotony of the q−Struve-Bessel functions
Authors: Yücel Özkan
Semra Korkmaz
Erhan Deniz
Murat Çağlar
Publication date: 08-12-2025
Publisher: Springer International Publishing
Published in: Journal of Inequalities and Applications / Issue 1/2026
Electronic ISSN: 1029-242X
DOI: https://doi.org/10.1186/s13660-025-03415-2

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The monotony of the q−Struve-Bessel functions

Abstract

Abstract

Publisher’s note

1 Introduction and preliminaries

2 Preliminaries

3 The main results

Acknowledgements

Declarations

Competing interests

Publisher’s note

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