1 Introduction
For \(x\in R\), the error function \(\operatorname{erf}(x)\) is defined as
$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x}e^{-t^{2}}\,dt. $$
The most important properties of this function are collected, for example, in [1, 2]. In the recent past, the error function has been a topic of recurring interest, and a great number of results on this subject have been reported in the literature [3‐16]. It might be surprising that the error function has application in the field of heat conduction besides probability [17, 18].
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In 1933, Aumann [19] introduced a generalized notion of convexity, the so-called MN-convexity, when M and N are mean values. A function \(f:[0,\infty)\to[0,\infty)\) is MN-convex if \(f(M(x,y))\leq N(f(x),f(y))\) for \(x,y\in[0,\infty)\). The usual convexity is the special case when M and N both are arithmetic means. Furthermore, the applications of MN-convexity reveal a new world of beautiful inequalities which involve a broad range of functions from the elementary ones, such as sine and cosine function, to the special ones, such as the Γ function, the Gaussian hypergeometric function, and the Bessel function. For the details as regards MN-convexity and its applications the reader is referred to [20‐25].
Let \(\lambda\in(0,1)\), we define \(A(x,y;\lambda)=\lambda x+(1-\lambda)y\), \(G(x,y;\lambda)=x^{\lambda}y^{1-\lambda}\), \(H(x,y;\lambda)=\frac{xy}{\lambda y+(1-\lambda)x}\) and \(M_{r}(x,y;\lambda)=(\lambda x^{r}+(1-\lambda)y^{r})^{1/r}\) (\(r\neq0\)), \(M_{0}(x,y;\lambda)=x^{\lambda}y^{1-\lambda}\). These are commonly known as weighted arithmetic mean, weighted geometric mean, weighted harmonic mean, and weighted power mean of two positive numbers x and y, respectively. Then it is well known that the inequalities hold for all \(\lambda\in(0,1)\) and \(x,y>0\) with \(x\neq y\).
$$H(x,y;\lambda)=M_{-1}(x,y;\lambda)< G(x,y;\lambda)=M_{0}(x,y; \lambda )< A(x,y;\lambda)=M_{1}(x,y;\lambda) $$
By elementary computations, one has and
$$ \lim_{r\to-\infty}M_{r}(x,y;\lambda)=\min(x,y) $$
(1.1)
$$\lim_{r\to+\infty}M_{r}(x,y;\lambda)=\max(x,y). $$
In [26], Alzer proved that \(c_{1}(\lambda)=\frac{\lambda+(1-\lambda )\operatorname{erf}(1)}{\operatorname{erf}(1/(1-\lambda))}\) and \(c_{2}(\lambda)=1\) are the best possible factors such that the double inequality holds for all \(x, y \in[1,+\infty)\) and \(\lambda\in(0,1/2)\).
$$ c_{1}(\lambda)\operatorname{erf}\bigl(H(x,y;\lambda) \bigr)\leq A\bigl(\operatorname{erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq c_{2}(\lambda)\operatorname{erf}\bigl(H(x,y;\lambda)\bigr) $$
(1.2)
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Inspired by (1.2), it is natural to ask: does the inequality \(\operatorname{erf}(M(x,y))\leq N(\operatorname{erf}(x),\operatorname {erf}(y))\) hold for other means M, N, such as geometric, harmonic or power means?
In [27, 28], the authors found the greatest values \(\alpha_{1}\), \(\alpha_{2}\) and the least values \(\beta_{1}\), \(\beta_{2}\), such that the double inequalities and hold for all \(x,y\geq1\) (or \(0< x,y<1\)) and \(\lambda\in(0,1)\).
$$\operatorname{erf}\bigl(M_{\alpha_{1}}(x,y;\lambda)\bigr)\leq A\bigl( \operatorname {erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq\operatorname {erf} \bigl(M_{\beta_{1}}(x,y;\lambda)\bigr) $$
$$\operatorname{erf}\bigl(M_{\alpha_{2}}(x,y;\lambda)\bigr)\leq H\bigl( \operatorname {erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq\operatorname {erf} \bigl(M_{\beta_{2}}(x,y;\lambda)\bigr) $$
In the following we answer the question: what are the greatest value p and the least value q, such that the double inequality holds for all \(x,y\geq1\) (or \(0< x,y<1\)) and \(\lambda\in(0,1)\)?
$$\operatorname{erf}\bigl(M_{p}(x,y;\lambda)\bigr)\leq G\bigl( \operatorname {erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq\operatorname {erf} \bigl(M_{q}(x,y;\lambda)\bigr) $$
2 Lemmas
In this section we present two lemmas, which will be used in the proof of our main results.
Lemma 2.1
Let
\(r\neq0\), \(r_{0}=-1-\frac{2}{e\sqrt{\pi}\operatorname{erf}(1)}=-1.4926\ldots \) , and
\(u(x)=\log\operatorname{erf}(x^{1/r})\). Then the following statements are true:
(1)
if
\(r< r_{0}\), then
\(u(x)\)
is strictly convex on
\([1,+\infty)\);
(2)
if
\(r_{0}\leq r<0\), then
\(u(x)\)
is strictly concave on
\((0,1]\);
(3)
if
\(r>0\), then
\(u(x)\)
is strictly concave on
\((0,+\infty)\).
Proof
Simple computations lead to and where
$$ u'(x)=\frac{2e^{-x^{2/r}}x^{1/r-1}}{r\sqrt{\pi}\operatorname{erf}(x^{1/r})} $$
(2.1)
$$ u''(x)=\frac{2e^{-x^{2/r}}x^{1/r-2}}{r^{2}\sqrt{\pi}\operatorname {erf}^{2}(x^{1/r})}g(x), $$
(2.2)
$$ g(x)=\bigl(-2x^{2/r}+1-r\bigr)\operatorname{erf} \bigl(x^{1/r}\bigr)-\frac{2}{\sqrt{\pi }}e^{-x^{2/r}}x^{1/r}. $$
(2.3)
Then
and
$$\begin{aligned}& g'(x)=4x^{2/r-1}g_{1}(x), \end{aligned}$$
(2.4)
$$\begin{aligned}& g_{1}(x)=-\frac{1}{r}\operatorname{erf}\bigl(x^{1/r} \bigr)-\frac{1}{2\sqrt {\pi}}e^{-x^{2/r}}x^{-1/r}, \end{aligned}$$
(2.5)
$$ g_{1}'(x)=\frac{1}{2r^{2}\sqrt{\pi}}e^{-x^{2/r}}x^{-1/r-1} \bigl[(2r-4)x^{2/r}+r\bigr]. $$
(2.6)
We divide the proof into two cases.
Case 1. If \(r<0\), then (2.6), (2.5), and (2.3) lead to
and
$$\begin{aligned}& g_{1}'(x)< 0, \end{aligned}$$
(2.7)
$$\begin{aligned}& \lim_{x\to0^{+}}g_{1}(x)>0, \qquad \lim _{x\to+\infty}g_{1}(x)=-\infty, \end{aligned}$$
(2.8)
$$\begin{aligned}& \lim_{x\to0^{+}}g(x)=-\infty,\qquad \lim_{x\to+\infty}g(x)=0, \end{aligned}$$
(2.9)
$$ g(1)=(-1-r)\operatorname{erf}(1)-\frac{2}{e\sqrt{\pi}}. $$
(2.10)
It follows from the monotonicity of \(g_{1}(x)\) and (2.8) that there exists \(x_{1}\in(0,+\infty)\), such that \(g(x)\) is strictly increasing on \([0,x_{1}]\) and strictly decreasing on \([x_{1},+\infty)\).
From the piecewise monotonicity of \(g(x)\) and (2.9) we clearly see that there exists \(x_{2}\in(0,+\infty)\), such that \(g(x)<0\) for \(x\in(0,x_{2})\) and \(g(x)>0\) for \(x\in(x_{2},+\infty)\).
Case 1.1. If \(r< r_{0}\), then from (2.10) we know that \(g(1)>0\). This leads to \(g(x)>0\) for \(x\in[1,+\infty)\). Therefore (2.2) leads to the conclusion that \(u(x)\) is strictly convex on \([1,+\infty)\).
Case 1.2. If \(r_{0}\leq r<0\), then (2.10) implies that \(g(1)\leq0\). This leads to \(g(x)\leq0\) for \(x\in(0,1]\). Therefore (2.2) leads to the conclusion that \(u(x)\) is strictly concave on \((0,1]\).
Lemma 2.2
The function
\(h(x)=2x^{2}+\frac{xe^{-x^{2}}}{\int_{0}^{x}e^{-t^{2}}\,dt}\)
is strictly increasing on
\((0,+\infty)\).
Proof
Simple computations lead to where and for \(x\in(0,+\infty)\).
$$ h'(x)=\frac{h_{1}(x)}{(\int_{0}^{x}e^{-t^{2}}\,dt)^{2}}, $$
(2.13)
$$\begin{aligned}& h_{1}(x)=4x \biggl( \int_{0}^{x}e^{-t^{2}}\,dt \biggr)^{2}+\bigl(1-2x^{2}\bigr)e^{-x^{2}} \int_{0}^{x}e^{-t^{2}}\,dt-xe^{-2x^{2}}, \\& \lim_{x\to0^{+}}h_{1}(x)=0, \end{aligned}$$
(2.14)
$$ h_{1}'(x)=4 \biggl( \int_{0}^{x}e^{-t^{2}}\,dt \biggr)^{2}+\bigl(4x^{3}+2x\bigr)e^{-x^{2}} \int_{0}^{x}e^{-t^{2}}\,dt+2x^{2}e^{-2x^{2}}>0 $$
(2.15)
3 Main results
Theorem 3.1
Let
\(\lambda\in(0,1)\)
and
\(r_{0}=-1-\frac {2}{e\sqrt{\pi}\operatorname{erf}(1)}=-1.4926\ldots\) . Then the double inequality
holds for all
\(x,y\geq1\)
if and only if
\(p=-\infty\)
and
\(q\geq r_{0}\).
$$ \operatorname{erf}\bigl(M_{p}(x,y;\lambda)\bigr)\leq G \bigl(\operatorname{erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq \operatorname{erf}\bigl(M_{q}(x,y;\lambda)\bigr) $$
(3.1)
Proof
First of all, we prove that inequality (3.1) holds if \(p=-\infty\) and \(q\geq r_{0}\). It follows from (1.1) that the first inequality in (3.1) is true if \(p=-\infty\). Since the weighted power mean \(M_{t}(x,y;\lambda)\) is strictly increasing with respect to t on R, thus we only need to prove that the second inequality in (3.1) is true if \(r_{0}\leq q<0\).
If \(r_{0}\leq q<0\), \(u(z)=\log\operatorname{erf}(z^{1/q})\), then Lemma 2.1(2) leads to for \(\lambda\in(0,1)\) and \(s,t\in(0,1]\).
$$ \lambda u(s)+(1-\lambda)u(t)\leq u\bigl(\lambda s+(1-\lambda)t \bigr) $$
(3.2)
Second, we prove that the second inequality in (3.1) implies \(q\geq r_{0}\).
Let \(x\geq1\) and \(y\geq1\). Then the second inequality in (3.1) leads to
$$ D(x,y)=:\operatorname{erf}\bigl(M_{q}(x,y;\lambda) \bigr)-G\bigl(\operatorname {erf}(x),\operatorname{erf}(y);\lambda\bigr)\geq0. $$
(3.3)
It follows from (3.3) that and
$$D(y,y)=\frac{\partial}{\partial x}D(x,y)|_{x=y}=0 $$
$$ \frac{\partial^{2}}{\partial x^{2}}D(x,y)|_{x=y}=\frac{\lambda(1-\lambda)y}{\operatorname {erf}'(y)} \biggl[q-1+ \biggl(2y^{2}+\frac{ye^{-y^{2}}}{\int ^{y}_{0}e^{-t^{2}}\,dt} \biggr) \biggr]. $$
(3.4)
Therefore, follows from (3.3) and (3.4) together with Lemma 2.2.
$$q\geq\lim_{y\to 1^{+}}\biggl(1-2y^{2}-\frac{ye^{-y^{2}}}{\int^{y}_{0}e^{-t^{2}}\,dt} \biggr)=r_{0} $$
Finally, we prove that the first inequality in (3.1) implies \(p=-\infty\). We distinguish two cases.
Case I. \(p\geq0\). Then for any fixed \(y\in[1,+\infty)\) we have and which contradicts the first inequality in (3.1).
$$\lim_{x\to+\infty}\operatorname{erf}\bigl(M_{p}(x,y; \lambda)\bigr)=1 $$
$$\lim_{x\to+\infty}G\bigl(\operatorname{erf}(x),\operatorname {erf}(y); \lambda\bigr)=\operatorname{erf}^{1-\lambda}(y)< 1, $$
Case II. \(-\infty< p<0\). Let \(x\geq1\), \(\alpha=\lambda ^{1/p}\) and \(y\to +\infty\). Then the first inequality in (3.1) leads to
$$ E(x)=:\operatorname{erf}^{\lambda}(x)-\operatorname{erf}( \alpha x)\geq0. $$
(3.5)
It follows from (3.5) that and Note that \(\alpha>1\), then
$$ \lim_{x\to+\infty}E(x)=0 $$
(3.6)
$$ E'(x)=\frac{2\lambda}{\sqrt{\pi}}e^{-x^{2}} \biggl[ \operatorname {erf}^{\lambda-1}(x)-\frac{\alpha}{\lambda}e^{(1-\alpha ^{2})x^{2}} \biggr]. $$
(3.7)
$$ \lim_{x\to +\infty} \biggl[\operatorname{erf}^{\lambda-1}(x)- \frac{\alpha }{\lambda}e^{(1-\alpha^{2})x^{2}} \biggr]=1. $$
(3.8)
Theorem 3.2
Let
\(\lambda\in(0,1)\), then the double inequality
holds for all
\(0< x,y<1\)
if and only if
\(\mu\leq r_{0}\)
and
\(\nu\geq 0\).
$$ \operatorname{erf}\bigl(M_{\mu}(x,y;\lambda)\bigr)\leq G \bigl(\operatorname{erf}(x),\operatorname{erf}(y);\lambda\bigr)\leq \operatorname{erf}\bigl(M_{\nu}(x,y;\lambda)\bigr) $$
(3.9)
Proof
If \(\mu\leq r_{0}\), \(u(z)=\log\operatorname{erf}(z^{1/\mu})\), then Lemma 2.1(1) leads to for \(\lambda\in(0,1)\), \(s,t>1\).
$$ u\bigl(\lambda s+(1-\lambda)t\bigr)\leq\lambda u(s)+(1- \lambda)u(t) $$
(3.10)
Let \(s=x^{\mu}\), \(t=y^{\mu}\), and \(0< x,y<1\). Then (3.10) leads to the first inequality in (3.9).
If \(\nu\geq0\), \(u(z)=\log\operatorname{erf}(z^{1/\nu})\), then Lemma 2.1(3) leads to for \(\lambda\in(0,1)\), \(0< s,t<1\).
$$ \lambda u(s)+(1-\lambda)u(t)\leq u\bigl(\lambda s+(1-\lambda)t \bigr) $$
(3.11)
Therefore, the second inequality in (3.9) follows from \(s=x^{\nu}\), \(t=y^{\nu}\), and \(0< x,y<1\) together with (3.11).
Second, we prove that the second inequality in (3.9) implies \(\nu\geq0\).
Let \(0< x,y<1\). Then the second inequality in (3.9) leads to
$$ J(x,y)=:\operatorname{erf}\bigl(M_{\nu}(x,y;\lambda) \bigr)-G\bigl(\operatorname {erf}(x),\operatorname{erf}(y);\lambda\bigr)\geq 0. $$
(3.12)
It follows from (3.12) that and
$$J(y,y)=\frac{\partial}{\partial x}J(x,y)|_{x=y}=0 $$
$$ \frac{\partial^{2}}{\partial x^{2}}J(x,y)|_{x=y}=\frac{\lambda(1-\lambda)y}{\operatorname {erf}'(y)} \biggl[\nu-1+ \biggl(2y^{2}+\frac{ye^{-y^{2}}}{\int _{0}^{y}e^{-t^{2}}\,dt} \biggr) \biggr]. $$
(3.13)
Hence, from (3.12) and (3.13) together with Lemma 2.2 we know that
$$\nu\geq\lim_{y\to0^{+}} \biggl[1- \biggl(2y^{2}+ \frac{ye^{-y^{2}}}{\int _{0}^{y}e^{-t^{2}}\,dt} \biggr) \biggr]=0. $$
Finally, we prove that the first inequality in (3.9) implies \(\mu\leq r_{0}\).
Let \(y\to1\). Then the first inequality in (3.9) leads to for \(0< x<1\).
$$ L(x)=:G\bigl(\operatorname{erf}(x),\operatorname{erf}(1);\lambda \bigr)-\operatorname{erf}\bigl(M_{\mu}(x,1;\lambda)\bigr)\geq 0 $$
(3.14)
It follows from (3.14) that and Let Then and
$$ L(1)=0 $$
(3.15)
$$ L'(x)=\frac{2\lambda e^{-x^{2}}}{\sqrt{\pi}} \bigl[\operatorname{erf}^{1-\lambda }(1) \operatorname{erf}^{\lambda-1}(x)-x^{\mu-1}\bigl(\lambda x^{\mu}+1-\lambda\bigr)^{1/\mu-1}e^{x^{2}-(\lambda x^{\mu}+1-\lambda)^{2/\mu}} \bigr]. $$
(3.16)
$$ L_{1}(x)=\log \bigl[\operatorname{erf}^{1-\lambda}(1) \operatorname {erf}^{\lambda-1}(x) \bigr]-\log \bigl[x^{\mu-1}\bigl( \lambda x^{\mu}+1-\lambda\bigr)^{1/\mu-1}e^{x^{2}-(\lambda x^{\mu}+1-\lambda)^{2/\mu}} \bigr]. $$
(3.17)
$$\begin{aligned}& \lim_{x\to1^{-}}L_{1}(x)=0, \\& L'_{1}(x)=(\lambda-1)\frac{\operatorname{erf}'(x)}{\operatorname {erf}(x)}- \frac{(\mu-1)(1-\lambda)}{x(\lambda x^{\mu}+1-\lambda)}-2x+2\lambda x^{\mu-1}\bigl(\lambda x^{\mu}+1- \lambda\bigr)^{2/\mu-1}, \end{aligned}$$
(3.18)
$$ \lim_{x\to1^{-}}L_{1}'(x)=(1- \lambda) \biggl[-\mu-1-\frac{2}{e \sqrt{\pi}\operatorname{erf}(1)} \biggr]. $$
(3.19)
If \(\mu>r_{0}\), then from (3.19) we clearly see that there exists a small \(\delta_{1}>0\), such that \(L_{1}'(x)<0\) for \(x\in(1-\delta_{1},1)\). Therefore, \(L_{1}(x)\) is strictly decreasing on \([1-\delta_{1},1]\).
The monotonicity of \(L_{1}(x)\) on \([1-\delta_{1},1]\) and (3.18) imply that there exists \(\delta_{2}>0\), such that \(L_{1}(x)>0\) for \(x\in(1-\delta_{2},1)\).
Acknowledgements
This research was supported by the Natural Science Foundation of China under Grants 61174076, 61374086, 11371125, and 11401191, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004. The authors wish to thank the anonymous referees for their careful reading of the manuscript and their fruitful comments and suggestions.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.