1 Introduction

For real numbers a, b, and c with \(c\ne 0,-1,-2\ldots \) , the Gaussian hypergeometric function is defined by

$$\begin{aligned} F(a,b;c;x)=_2F_1(a,b;c;x)\equiv \sum _{n=0}^\infty \frac{(a,n)(b,n)}{(c,n)}\frac{x^n}{n!},\quad |x|<1.\end{aligned}$$

Here \((a,0)=1\) for \(a\ne 0\), and (a, n) is the shifted factorial function

$$\begin{aligned} (a,n)\equiv a(a+1)(a+2)\cdots (a+n-1)\end{aligned}$$

for \(n\in {\mathbb {N}}\equiv \{k:k\ \text{ is } \text{ a } \text{ positive } \text{ integer }\}\). For some basic properties of F(a, b; c; x), the reader is referred to some standard handbooks, see e.g., [11]. For \(r\in (0,1)\), let \(r'=\sqrt{1-r^2}\). For \(a\in (0,1)\), the generalized elliptic integrals [3, 6, 9, 10] are defined as

$$\begin{aligned} \left\{ \begin{array}{l} {\mathcal {K}}_a={\mathcal {K}}_a(r)\equiv \frac{\pi }{2\sin (\pi a)}F(a,1-a;1;r^2),\\ {\mathcal {K}}_a'={\mathcal {K}}_a'(r)\equiv {\mathcal {K}}_a(r'),\\ {\mathcal {K}}_a(0)=\frac{\pi }{2\sin (\pi a)},\quad {\mathcal {K}}_a(1)=\infty , \end{array} \right. \end{aligned}$$
(1.1)

and

$$\begin{aligned} \left\{ \begin{array}{l} {\mathcal {E}}_a={\mathcal {E}}_a(r)\equiv \frac{\pi }{2\sin (\pi a)}F(a-1,1-a;1;r^2),\\ {\mathcal {E}}_a'={\mathcal {E}}_a'(r)\equiv {\mathcal {E}}_a(r'),\\ {\mathcal {E}}_a(0)=\frac{\pi }{2\sin (\pi a)},\quad {\mathcal {E}}_a(1)=\frac{1}{2(1-a)}. \end{array} \right. \end{aligned}$$

In the particular case \(a=1/2\), the functions \({\mathcal {K}}_a\) and \({\mathcal {E}}_a\) reduce to \({\mathcal {K}}\) and \({\mathcal {E}}\), respectively, which are the well-known complete elliptic integrals of the first and second kind, respectively. By symmetry of (1.1), it is obvious that \({\mathcal {K}}_a={\mathcal {K}}_{1-a}\) for \(a\in (0,1/2]\). Hence we may assume that \(a\in (0,1/2]\) in the sequel. In geometric function theory, the generalized elliptic integrals occur, e.g., in the study of Schwarz–Christoffel-type mapping problems, see [8, 10].

For \(a\in (0,1/2]\) and \(r\in (0,1),\) define the generalized modulus as

$$\begin{aligned} \mu _a(r)\equiv \frac{\pi }{2\sin (\pi a)}\frac{{\mathcal {K}}_a'(r)}{{\mathcal {K}}_a(r)},\qquad \mu (r)=\mu _{1/2}(r). \end{aligned}$$

The function \(\mu _a(r)\) plays a very important role in some fields of mathematics. For instance, it is indispensable in geometric function theory, quasiconformal theory, and the theory of Ramanujan’s modular equations (see [2–5, 10]). For the special case \(a=1/2\), the function \(\mu (r)=\mu _{1/2}(r)\) is the modulus of the Grötzsch ring domain in the plane, which has numerous applications in the conformal invariants and the theory of quasiconformal mappings [2]. Many noteworthy monotonicity and convexity properties of functions defined in terms of the modulus of the Grötzsch ring are presented in the excellent monograph [2]. Applications of these results lead to various sharp functional inequalities for the function \(\mu \). Some of these results have been extended to the generalized modulus, see e.g., [3, 7, 10, 13–15].

In this paper, we study the monotonicity and convexity properties of the generalized modulus and obtain sharp functional inequalities which extend and sharpen some results for the modulus of the Grötzsch ring in [1–3].

2 Lemmas

For \(a\in (0,1/2]\) and \(0<r<1\), the derivative formula can be found in [3, Theorem 4.1] or [10, Theorem 4.15]):

$$\begin{aligned} \frac{{\text {d}}\mu _a}{{\text {d}}r}=-\frac{\pi ^2}{4\sin ^2(\pi a)}\frac{1}{rr'^2{\mathcal {K}}_a(r)^2}. \end{aligned}$$
(2.1)

For \(r\in (0,1)\), let

$$\begin{aligned} m_a(r)=\frac{2\sin (\pi a)}{\pi }r'^2{\mathcal {K}}_a(r){\mathcal {K}}_a'(r). \end{aligned}$$

This function plays an important role in the study of the distortion functions in the theory of quasiconformal mappings.

We also need the function

$$\begin{aligned} R(a)=-2\gamma -\psi (a)-\psi (1-a),\quad R(1/2)=\log 16, \end{aligned}$$

where \(\gamma =0.577215\ldots \) is the Euler-Mascheroni constant and \(\psi \) is the psi function.

The following Lemma 2.1 is from [3, Theorem 5.5] and [12, Theorem 2.5].

Lemma 2.1

Let \(a\in (0,1/2]\). Then

  1. 1.

    the function \(f_1(r)=\mu _a(r)/\log (1/r)\) is strictly increasing from (0, 1) onto \((1,\infty )\);

  2. 2.

    the function \(f_2(r)=\mu _a(r)\mathrm{{arth}}r\) is strictly increasing from (0, 1) onto \((0,\pi ^2/(4\sin ^2(\pi a)))\);

  3. 3.

    the function \(f_3(r)=\mu _a(r)+\log r\) is strictly decreasing and concave from (0, 1) onto (0, R(a) / 2);

  4. 4.

    the function \(f_5(r)=m_a(r)/\log (1/r)\) is strictly increasing from (0, 1) onto \((1,\infty )\).

It follows from Lemma 2.1 (3) that for all \(r\in (0,1)\),

$$\begin{aligned} \log \frac{1}{r}<\mu _a(r)<\log \frac{1}{r}+\frac{R(a)}{2}. \end{aligned}$$

This inequality can be improved as the following lemma shows.

Lemma 2.2

Let \(a\in (0,1/2]\). The function

$$\begin{aligned} f(r)=\mu _a(r)-\log \frac{1+r'}{r}\end{aligned}$$

is strictly decreasing from (0, 1) onto \((0,R(a)/2-\log 2)\). In particular, for all \(r\in (0,1)\),

$$\begin{aligned} \log \frac{1+r'}{r}<\mu _a(r)<\log \frac{1+r'}{2r}+\frac{R(a)}{2}. \end{aligned}$$

Proof

By differentiation, we have

$$\begin{aligned} f'(r)=\frac{1}{rr'}\left( 1-\frac{\pi ^2}{4\sin ^2{\pi a}}\frac{1}{r'{\mathcal {K}}_a(r)^2}\right) \end{aligned}$$

which is negative by [3, Lemma 5.4(1)]. Hence the monotonicity of f follows. By Lemma 2.1 (3), the limiting values of f are clear. \(\square \)

3 Monotonicity, convexity, and inequalities

In this section, we study the monotonicity and convexity of some functions defined in terms of the generalized modulus and obtain sharp functional inequalities for the generalized modulus.

It follows from [2, Lemma 5.11(1)] that the function \(x\mapsto \mu (1/\cosh x)\) is an increasing and concave automorphism of \((0,\infty )\). In particular, for \(s,t\in (0,1)\),

$$\begin{aligned} \mu (s)+\mu (t)\le 2\mu \left( \sqrt{\frac{2st}{1+st+s't'}}\right) \end{aligned}$$

with equality if and only if \(s=t\). This result has been extended to the generalized modulus in [3, Theorem 5.6]. The next two theorems give reverse inequalities for the generalized modulus.

Theorem 3.1

Let \(a\in (0,1/2]\). The function \(f(x)=\mu _a(1-1/\cosh x)\) is strictly decreasing and convex from \((0,\infty )\) onto \((0,\infty )\). In particular, for \(s,t\in (0,1)\),

$$\begin{aligned} 2\mu _a\left( 1-\sqrt{\frac{2st}{1+st+s't'}}\right) \le \mu _a(1-s)+\mu _a(1-t), \end{aligned}$$
(3.1)

with equality if and only if \(s=t\).

Proof

Let \(r=1-1/\cosh x\). Then

$$\begin{aligned} f'(x)=-\frac{\pi ^2}{4\sin ^2(\pi a)}\sqrt{\frac{2}{r}-1}\frac{1}{(1+r){\mathcal {K}}_a(r)^2},\end{aligned}$$

which is negative and strictly increasing in r, and hence increasing in x. Therefore, f is decreasing and convex on \((0,\infty )\). In particular, we have \(f((x+y)/2)\le (f(x)+f(y))/2\), with equality if and only if \(x=y\). Set \(s=1/\cosh x\) and \(t=1/\cosh y\). Now

$$\begin{aligned} \cosh \left( \frac{x+y}{2}\right) =\sqrt{\frac{1+st+s't'}{2st}}. \end{aligned}$$

Hence

$$\begin{aligned} f\left( \frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2} \end{aligned}$$

gives

$$\begin{aligned} 2\mu _a\left( 1-\sqrt{\frac{2st}{1+st+s't'}}\right) \le \mu _a(1-s)+\mu _a(1-t) \end{aligned}$$

with equality if and only if \(s=t\). \(\square \)

Theorem 3.2

Let \(a\in (0,1/2]\). The function \(f(x)=\mu _a(\tanh x)\) is strictly decreasing and convex from \((0,\infty )\) onto \((0,\infty )\). In particular, for \(s,t\in (0,1)\),

$$\begin{aligned} 2\mu _a\left( \frac{s(1+t')+t(1+s')}{st+(1+s')(1+t')}\right) \le \mu _a(s)+\mu _a(t), \end{aligned}$$
(3.2)

with equality if and only if \(s=t\).

Proof

Let \(s=\tanh x\) and \(t=\tanh y\). Then

$$\begin{aligned} f'(x)=-\frac{\pi ^2}{4\sin ^2(\pi a)}\frac{1}{s{\mathcal {K}}_a(s)^2},\end{aligned}$$

which is negative and strictly increasing in s, and hence increasing in x. Therefore, f is decreasing and convex on \((0,\infty )\). In particular, we have \(f((x+y)/2)\le (f(x)+f(y))/2\), with equality if and only if \(x=y\). Now

$$\begin{aligned} \tanh \left( \frac{x+y}{2}\right) =\frac{s(1+t')+t(1+s')}{st+(1+s')(1+t')}. \end{aligned}$$

Hence

$$\begin{aligned} f\left( \frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2} \end{aligned}$$

gives

$$\begin{aligned} 2\mu _a\left( \frac{s(1+t')+t(1+s')}{st+(1+s')(1+t')}\right) \le \mu _a(s)+\mu _a(t), \end{aligned}$$

with equality if and only if \(s=t\). \(\square \)

Remark 3.1

1. In [2, Exercise 5.62(4)] (see also [9, Theorem 4.3] for the generalized modulus with two parameters), it is proved that

$$\begin{aligned} 2\mu (1-\sqrt{(1-s)(1-t)})\le \mu (s)+\mu (t). \end{aligned}$$
(3.3)

Since

$$\begin{aligned} (st+s't')^2&=s^2 t^2+ 2st's't + s'^2 t'^2 \\&\le s^2 t^2 + s^2 t'^2 + s'^2 t^2 + s'^2 t'^2 \\&= (s^2+s'^2)(t^2 + t'^2)=1, \end{aligned}$$

we have that

$$\begin{aligned} \sqrt{\frac{2st}{1+st+s't'}}\ge \sqrt{st}. \end{aligned}$$

Thus, by the monotonicity of \(\mu _a\),

$$\begin{aligned} \mu _a\left( 1-\sqrt{\frac{2st}{1+st+s't'}}\right) \ge \mu _a(1-\sqrt{st}). \end{aligned}$$

It follows that the inequality (3.1) with parameter \(a=1/2\) is a refinement of the inequality (3.3). One can check that the inequality (3.2) with parameter \(a=1/2\) is also a refinement of the inequality (3.3).

2. The inequalities (3.1) and (3.2) are not comparable on the whole domain \((0,1)\times (0,1)\).

The next theorem, when \(a=1/2\), reduces to Lemma 5.11(2) in [2].

Theorem 3.3

Let \(a\in (0,1/2]\) and \(r,s\in (0,1)\). The function \(f(r)=\mu _a(rs)-\mu _a(r)\) is strictly increasing from (0, 1) onto \((-\log s,\mu _a(s))\). In particular, for \(r,s\in (0,1)\),

$$\begin{aligned} \mu _a(r)-\log s<\mu _a(rs)<\mu _a(r)+\mu _a(s). \end{aligned}$$
(3.4)

Proof

Let \(t=rs\). Then differentiation gives

$$\begin{aligned} rf'(r)=-\frac{\pi ^2}{4\sin ^2(\pi a)}\left( \frac{1}{t'^2{\mathcal {K}}_a(t)^2}-\frac{1}{r'^2{\mathcal {K}}_a(r)^2}\right) \end{aligned}$$

which is positive, since the function \(r\mapsto r'{\mathcal {K}}_a(r)\) is strictly decreasing (see [3, Lemma 5.4(1)]). Hence the function f is strictly increasing in r. It is clear that \(f(1^-)=\mu _a(s)\). By Lemma 2.1 (3), we have that

$$\begin{aligned} f(0^+)=\lim _{r\rightarrow 0^+}\mu _a(rs)+\log (rs)-(\mu _a(r)+\log r)-\log s=-\log s. \end{aligned}$$

\(\square \)

In the next theorem, when \(a=1/2\), parts (1) and (2) reduce to Theorems 1 and 2, respectively, in [1].

Theorem 3.4

For each \(a\in (0,1/2]\) and \(r\in (0,1)\), the function

  1. 1.

    \(f_1(\alpha )=\mu _a(r^\alpha )/\alpha \) is strictly decreasing and strictly log-convex from \((0,\infty )\) onto \((-\log r,\infty )\).

  2. 2.

    \(f_2(\alpha )=1/f_1(\alpha )\) is strictly concave on \((0,\infty )\).

Proof

(1) Let \(t=r^\alpha \). By logarithmic differentiation, we obtain

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}\alpha }\log f_1(\alpha )=\frac{1}{m_a(t)}\left( \frac{m_a(t)}{\log (1/t)}-1\right) \log r \end{aligned}$$

which is negative and strictly decreasing in t, and hence strictly increasing in \(\alpha \) by Lemma 2.1 (4). Therefore, we get the monotonicity and log-convexity of \(f_1\). For the limiting values, by Lemma 2.1 (1),

$$\begin{aligned} f_1(\infty )=\lim _{\alpha \rightarrow \infty }\frac{\mu _a(t)}{\log (1/t)}\frac{\log (1/t)}{\alpha }=-\log r. \end{aligned}$$

By Lemma 2.1 (2),

$$\begin{aligned} f_1(0^+)=\lim _{\alpha \rightarrow 0^+}\frac{\mu _a(t)\mathrm{{arth}}t}{\alpha \,\mathrm{{arth}}t}=\infty , \end{aligned}$$

since it is easy to see that \(\lim \nolimits _{\alpha \rightarrow 0^+}\alpha \,\mathrm{{arth}}t=0\).

(2) Let \(t=r^\alpha \). Then, by differentiation,

$$\begin{aligned} f_2'(\alpha )=\frac{1}{\mu _a(t)}\left( 1-\frac{\log (1/t)}{m_a(t)}\right) \end{aligned}$$

which is positive and strictly increasing in t, and hence strictly decreasing in \(\alpha \) by Lemma 2.1 (4). This implies the concavity of \(f_2\). \(\square \)

The inequalities in the following corollary extend the corresponding inequalities in [1].

Corollary 3.1

Let \(a\in (0,1/2]\).

  1. 1.

    Let \(\alpha \) and \(\beta \) be real numbers with \(\alpha >\beta >0\). Then for all \(r\in (0,1)\),

    $$\begin{aligned} 1<\frac{\mu _a(r^\alpha )}{\mu _a(r^\beta )}<\frac{\alpha }{\beta }. \end{aligned}$$
    (3.5)

    Both bounds are sharp.

  2. 2.

    For all \(s,t\in (0,1)\),

    $$\begin{aligned} \frac{\mu _a(\sqrt{st})}{\log (1/\sqrt{st})}\le \sqrt{\frac{\mu _a(s)\mu _a(t)}{\log (1/s)\log (1/t)}} \end{aligned}$$
    (3.6)

    with equality if and only if \(s=t\).

  3. 3.

    For all \(s,t\in (0,1)\),

    $$\begin{aligned} \frac{\log (st)}{\mu _a(\sqrt{st})}\le \frac{\log s}{\mu _a(s)}+\frac{\log t}{\mu _a(t)} \end{aligned}$$
    (3.7)

    with equality if and only if \(s=t\).

Proof

(1) Since \(\mu _a\) is strictly decreasing on (0, 1), we derived from \(0<r^\alpha <r^\beta \) that the left-hand side of (3.5) is valid. The right-hand side of (3.5) follows from Theorem 3.4 (1).

Next we show the sharpness of the bounds 1 and \(\alpha /\beta \). Making use of Lemma 2.1 (1), we obtain

$$\begin{aligned} \lim _{r\rightarrow 0+}\frac{\mu _a(r^\alpha )}{\mu _a(r^\beta )}=\lim _{r\rightarrow 0+}\frac{\mu _a(r^\alpha )}{\log (1/r^\alpha )}\frac{\log (1/r^\beta )}{\mu _a(r^\beta )}\frac{\log (1/r^\alpha )}{\log (1/r^\beta )}=\frac{\alpha }{\beta }. \end{aligned}$$

By Lemma 2.1 (2),

$$\begin{aligned} \lim _{r\rightarrow 1-}\frac{\mu _a(r^\alpha )}{\mu _a(r^\beta )}=\lim _{r\rightarrow 1-}\frac{\mu _a(r^\alpha )\mathrm{{arth}}(r^\alpha )}{\mu _a(r^\beta )\mathrm{{arth}}(r^\beta )}\frac{\mathrm{{arth}}(r^\beta )}{\mathrm{{arth}}(r^\alpha )}=\lim _{r\rightarrow 1-}\frac{\mathrm{{arth}}(r^\beta )}{\mathrm{{arth}}(r^\alpha )}=1. \end{aligned}$$

(2) Let \(r\in (0,1)\). It follows from Theorem 3.4 (1) that the function \(\alpha \mapsto \mu _a(r^\alpha )/\alpha \) is strictly log-convex. Hence the log-convexity implies that

$$\begin{aligned} \frac{\mu _a\left( r^{(\alpha +\beta )/2}\right) }{(\alpha +\beta )/2}\le \sqrt{\frac{\mu _a(r^\alpha )\mu _a(r^\beta )}{\alpha \beta }} \end{aligned}$$
(3.8)

with the equality if and only if \(\alpha =\beta \). Set \(s=r^\alpha \) and \(t=r^\beta \). Then

$$\begin{aligned} r^{(\alpha +\beta )/2}=\sqrt{st},\end{aligned}$$
$$\begin{aligned} (\alpha +\beta )/2=\log (1/\sqrt{st})/\log (1/r), \end{aligned}$$

and

$$\begin{aligned} \sqrt{\alpha \beta }=\sqrt{\log (1/s)\log (1/t)}/\log (1/r). \end{aligned}$$

Then from (3.8) we conclude that

$$\begin{aligned} \frac{\mu _a(\sqrt{st})}{\log (1/\sqrt{st})}\le \sqrt{\frac{\mu _a(s)\mu _a(t)}{\log (1/s)\log (1/t)}} \end{aligned}$$
(3.9)

with equality if and only if \(s=t\).

(3) Let \(r\in (0,1)\). Using Theorem 3.4 (2) gives that for \(\alpha ,\beta >0\),

$$\begin{aligned} \frac{1}{2}\left( \frac{\alpha }{\mu _a(r^\alpha )}+\frac{\beta }{\mu _a(r^\beta )}\right) \le \frac{(\alpha +\beta )/2}{\mu _a(r^{(\alpha +\beta )/2})} \end{aligned}$$
(3.10)

with equality if and only if \(\alpha =\beta \). Set \(s=r^\alpha \) and \(t=r^\beta \). Then we conclude from (3.10) that the inequality (3.7) is valid with equality if and only if \(s=t\). \(\square \)