Chronic kidney disease report differently on change in sexual function dependent on treatment: a cohort study

1.   Introduction

Let $\mathcal{A}$ denote the class of functions $f$ which are analytic in the open unit disk $\Delta = \left\{ z\in \mathbb{C} :\left\vert z\right\vert < 1\right\}$, normalized by the conditions $f(0) = f'(0)-1 = 0$. So each $f\in\mathcal{A}$ has series representation of the form

For two analytic functions $f$ and $g$, $f$ is said to be subordinated to $g$ (written as $f\prec g)$ if there exists an analytic function $\omega$ with $\omega(0) = 0$ and $\left\vert{\omega(z)}\right\vert < 1$ for $z\in\Delta$ such that $f(z) = (g\circ\omega)(z)$.

A function $f\in\mathcal{A}$ is said to be in the class $\mathcal{S}$ if $f$ is univalent in $\Delta$. A function $f\in\mathcal{S}$ is in class $\mathcal{C}$ of normalized convex functions if $f\left(\Delta\right)$ is a convex domain. For $0\leq\alpha\leq1$, Mocanu ^[23] introduced the class $\mathcal{M} _{\alpha}$ of functions $f\in\mathcal{A}$ such that $\frac{f(z) f'(z)}{z}\neq0$ for all $z\in\Delta$ and

Geometrically, $f\in\mathcal{M}_{\alpha}$ maps the circle centred at origin onto $\alpha$-convex arcs which leads to the condition $\left(1.2\right)$. The class $\mathcal{M}_{\alpha}$ was studied extensively by several researchers, see ^{[1,10,11,12,24,25,26,27]} and the references cited therein.

A function $f\in\mathcal{S}$ is uniformly starlike if $f$ maps every circular arc $\Gamma$ contained in $\Delta$ with center at $\zeta$ $\in\Delta$ onto a starlike arc with respect to $f\left(\zeta\right)$. A function $f\in\mathcal{C}$ is uniformly convex if $f$ maps every circular arc $\Gamma$ contained in $\Delta$ with center $\zeta$ $\in\Delta$ onto a convex arc. We denote the classes of uniformly starlike and uniformly convex functions by $\mathcal{UST}$ and $\ \mathcal{UCV}$, respectively. For recent study on these function classes, one can refer to ^{[7,9,13,19,20,31]}.

In 1999, Kanas and Wisniowska ^[15]$\ $introduced the class $k$-$\mathcal{UCV}$ $\left(k\geq0\right)$ of $k$-uniformly convex functions. A function $f\in\mathcal{A}$ is said to be in the class $k$-$\mathcal{UCV}$ if it satisfies the condition

In recent years, many researchers investigated interesting properties of this class and its generalizations. For more details, see ^{[2,3,4,14,15,16,17,18,30,32,35]} and references cited therein.

In $2015$, Sokół and Nunokawa ^[33] introduced the class $\mathcal{MN}$, a function $f\in\mathcal{MN}$ if it satisfies the condition

In ^[28], it is proved that if ${\Re} (f') > 0$ in $\Delta$, then $f$ is univalent in $\Delta$. In 1972, MacGregor ^[21] studied the class $\mathcal{B}$ of functions with bounded turning, a function $f\in\mathcal{B}$ if it satisfies the condition ${\Re}(f') > 0$ for $z\in\Delta$. A natural generalization of the class $\mathcal{B}$ is $\mathcal{B}(\delta_1)\ (0\leq\delta_1 < 1)$, a function $f\in\mathcal{B}(\delta_1)$ if it satisfies the condition

for details associated with the class $\mathcal{B}(\delta_1)$ (see ^[5,6,34]).

Motivated essentially by the above work, we now introduce the following class $k$-$\mathcal{Q} (\alpha)$ of analytic functions.

Definition 1. Let $k\geq0$ and $0\leq\alpha\leq1$. A function $f\in\mathcal{A}$ is said to be in the class $k$-$\mathcal{Q}\left(\alpha\right)$ if it satisfies the condition

It is worth mentioning that, for special values of parameters, one can obtain a number of well-known function classes, some of them are listed below:

1. $k$-$\mathcal{Q}\left(1\right) = k$-$\mathcal{UCV}$;

2. $0$-$\mathcal{Q}\left(\alpha\right) = \mathcal{C}$.

In what follows, we give an example for the class $k$-$\mathcal{Q}\left(\alpha\right)$.

Example 1. The function $f(z) = \frac{z}{1-Az}\, (A\neq0)$ is in the class $k$-$\mathcal{Q}\left(\alpha\right)$ with

The main purpose of this paper is to establish several interesting relationships between $k$-$\mathcal{Q}(\alpha)$ and the class $\mathcal{B}(\delta)$ of functions with bounded turning.

2.   Preliminaries

To prove our main results, we need the following lemmas.

Lemma 1. (^[8]) Let $h$ be analytic in $\Delta$ with $h(0) = 1$, $\beta > 0$ and $0\leq\gamma_1 < 1$. If

then

where

Lemma 2. Let $h$ be analytic in $\Delta$ and of the form

with $h(z)\neq0\ $in $\Delta$. If there exists a point $z_{0}\, (\left\vert z_{0}\right\vert < 1)$ such that $\left\vert \arg h(z)\right\vert < \frac{\pi\rho}{2}\, (\left\vert z\right\vert < \left\vert z_{0}\right\vert) \ $and $\left\vert \arg h\left(z_{0}\right) \right\vert = \frac{\pi\rho}{2}$ for some $\rho > 0$, then $\frac{z_{0}h'\left(z_{0}\right) }{h\left(z_{0}\right)} = i\ell\rho$, where

and $\left(h\left(z_{0}\right)\right)^{1/\rho} = \pm ic\, \left(c > 0\right)$.

This result is a generalization of the Nunokawa's lemma ^[29].

Lemma 3. (^[37]) Let $\varepsilon$ be a positive measure on $[0, 1]$. Let $\digamma$ be a complex-valued function defined on $\Delta\times\left[0, 1\right]$ such that $\digamma(., t)$ is analytic in $\Delta$ for each $t\in\left[0, 1\right]$ and $\digamma\left(z, .\right)$ is $\varepsilon$-integrable on $\left[0, 1\right]$ for all $z\in\Delta$. In addition, suppose that ${\Re}(\digamma\left(z, t\right)) > 0$, $\digamma(-r, t)$ is real and ${\Re}(1/\digamma(z, t))\geq1/\digamma(-r, t)$ for $\left\vert z\right\vert \leq r < 1\ $and $t\in[0, 1]$. If $\digamma(z) = \int_0^1\digamma\left(z, t\right)d\varepsilon(t)$, then ${\Re}\left(1/\digamma(z)\right) \geq1/\digamma(-r)$.

Lemma 4. (^[22]) If $-1\leq D < C\leq1, \ \lambda_1 > 0$ and ${\Re}\left(\gamma_2\right) \geq-\lambda_1\left(1-C\right)/\left(1-D\right)$, then the differential equation

has a univalent solution in $\Delta$ given by

If $r(z) = 1+c_{1}z+c_{2}z^{2}+\cdots$ satisfies the condition

then

and $s(z)$ is the best dominant.

Lemma 5. ([36,Chapter 14]) Let $w, \ x\ $and\ $y\neq0, -1, -2, \ldots$ be complex numbers. Then, for ${\Re}(y) > {\Re}(x) > 0$, one has

1. $_{2}G_{1}\left(w, x, y;z\right) = \frac {\Gamma\left(y\right)}{\Gamma\left(y-x\right) \Gamma\left(x\right) }\int_0^1 s^{x-1}\left(1-s\right)^{y-x-1}\left(1-sz\right) ^{-w}ds$;

2. $_{2}G_{1}\left(w, x, y;z\right) = \ _{2} G_{1}\left(x, w, y;z\right)$;

3. $_{2}G_{1}\left(w, x, y;z\right) = \left(1-z\right)^{-w}\, _{2}G_{1}\left(w, y-x, y;\frac{z}{z-1}\right)$.

3.   Main results

Firstly, we derive the following result.

Theorem 1. Let $0\leq\alpha < 1$ and $k\geq\frac{1}{1-\alpha}$. If $f\in k$-$\mathcal{Q}(\alpha)$, then $f\in{\mathcal{B}}\left(\delta\right)$, where

Proof. Let $f' = \hslash$, where $\hslash$ is analytic in $\Delta$ with $\hslash(0) = 1$. From inequality (1.5) which takes the form

we find that

which can be rewritten as

The above relationship can be written as the following Briot-Bouquet differential subordination

Thus, by Lemma 1, we obtain

where $\delta$ is given by (3.1). The relationship (3.2) implies that $f\in\mathit{\mathcal{B}}\left(\delta\right)$. We thus complete the proof of Theorem 3.1.

Theorem 2. Let $0 < \alpha\leq1$, $0 < \beta < 1$, $c > 0$, $k\geq1$, $n\geq m+1\, (m\in \mbox{ $\mathbb{N}$ })$, $\left\vert{\ell}\right\vert\geq\frac{n}{2}\left(c+\frac{1}{c}\right)$ and

If

and $f\in k$-$\mathcal{Q}(\alpha)$, then $f\in{\mathcal{B}}(\beta_{0})$, where

such that (3.3) holds.

Proof. By the assumption, we have

In view of (1.5) and (3.4), we get

If there exists a point $z_{0}\in\Delta$ such that

and

then from Lemma 2, we know that

where

and

For the case

we get

Moreover, we find from (3.3) that

By virtue of (3.5) and (3.6), we have

which is a contradiction to the definition of $k$-$\mathcal{Q}(\alpha)$. Since $\beta_{0} = {\min}\{\beta: \beta\in(0, 1)\}$ such that (3.3) holds, we can deduce that $f\in\mathcal{B}(\beta_0)$.

By using the similar method as given above, we can prove the case

is true. The proof of Theorem 2 is thus completed.

Theorem 3. If $0 < \beta < 1$ and $0\leq\nu < 1$. If $f\in k$-$\mathcal{Q}(\alpha)$, then

or equivalently, $k$-$\mathcal{Q}\left(\alpha\right)\subset{\mathcal{B}}\left(\nu_{0}\right)$, where

Proof. For

we define

To prove $k$-$\mathcal{Q}(\alpha)\subset\mathcal{B}\left(\nu _{0}\right)$, it suffices to prove that

which need to show that

By Lemma 3 and (3.7), it follows that

where

and

which is a positive measure on $\left[0, 1\right]$.

It is clear that ${\Re}(\digamma(z, t)) > 0$ and $\digamma(-r, t)$ is real for $\left\vert z\right\vert \leq r < 1$ and $t\in\left[0, 1\right]$. Also

for $\left\vert z\right\vert \leq r < 1$. Therefore, by Lemma 3, we get

If we let $r\rightarrow1^{-}$, it follows that

Thus, we deduce that $k$-$\mathcal{Q}\left(\alpha\right)\subset\mathcal{B}(\nu_{0})$.

Theorem 4. Let $0\leq\alpha < 1$ and $k\geq\frac{1}{1-\alpha}$. If $f\in k$-$\mathcal{Q}\left(\alpha\right)$, then

where

Proof. Suppose that $f' = \hslash$. From the proof of Theorem 1, we see that

If we set $\lambda_1 = \frac{1}{\lambda}$, $\gamma_2 = 0,$ $C = 1$ and $D = -1$ in Lemma 4, then

By putting $t = uz$, and using Lemma 5, we obtain

which is the desired result of Theorem 4.

Acknowledgments

The present investigation was supported by the Key Project of Education Department of Hunan Province under Grant no. 19A097 of the P. R. China. The authors would like to thank the referees for their valuable comments and suggestions, which was essential to improve the quality of this paper.

Conflict of interest

The authors declare no conflict of interest.