Bounds of the Neuman-Sándor Mean Using Power and Identric Means

Abstract

In this paper we find the best possible lower power mean bounds for the Neuman-Sándor mean and present the sharp bounds for the ratio of the Neuman-Sándor and identric means.

1. Introduction

For 𝑝∈ℝ the 𝑝th power mean 𝑀_𝑝(𝑎,𝑏), Neuman-Sándor Mean 𝑀(𝑎,𝑏) [1], and identric mean 𝐼(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 are defined by 𝑀_𝑝()=𝑎𝑎,𝑏^𝑝+𝑏^𝑝2^1/𝑝√,𝑝≠0,𝑀()=𝑎𝑏,𝑝=0,𝑎,𝑏𝑎−𝑏2sinh⁻¹)/(𝐼()=1((𝑎−𝑏𝑎+𝑏)),𝑎≠𝑏,𝑎,𝑎=𝑏,𝑎,𝑏𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)},𝑎≠𝑏,𝑎,𝑎=𝑏, (1)(2)(3) respectively, where sinh⁻¹(𝑥)=log(𝑥+√1+𝑥²) is the inverse hyperbolic sine function.

The main properties for 𝑀_𝑝(𝑎,𝑏) and 𝐼(𝑎,𝑏) are given in [2]. It is well known that 𝑀_𝑝(𝑎,𝑏) is continuously and strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎,𝑏>0 with 𝑎≠𝑏. Recently, the power, Neuman-Sándor, and identric means have been a subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [3–26].

Let 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)=(𝑏−𝑎)/(log𝑏−log𝑎), 𝑃(𝑎,𝑏)=(𝑎−𝑏)/[4arctan(𝑎/𝑏)−𝜋], 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝑇(𝑎,𝑏)=(𝑎−𝑏)/[2arctan((𝑎−𝑏)/(𝑎+𝑏))], 𝑄(𝑎,𝑏)=(𝑎²+𝑏²)/2, and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of two positive numbers 𝑎 and 𝑏 with 𝑎≠𝑏, respectively. Then, it is well known that the inequalities 𝐻(𝑎,𝑏)=𝑀₋₁(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝑀₀()()()()()𝑎,𝑏<𝐿𝑎,𝑏<𝑃𝑎,𝑏<𝐼𝑎,𝑏<𝐴𝑎,𝑏=𝑀₁()()()()𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏<𝑄𝑎,𝑏=𝑀₂(𝑎,𝑏)<𝐶(𝑎,𝑏), (4) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

The following sharp bounds for 𝐿, 𝐼, (𝐼𝐿)^1/2, and (𝐼+𝐿)/2 in terms of power means are presented in [27–32]: 𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀_1/3(),𝑀𝑎,𝑏_2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀_log2(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐼^1/2(𝑎,𝑏)𝐿^1/2(𝑎,𝑏)<𝑀_1/2(),1𝑎,𝑏2[𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏)]<𝑀_1/2(𝑎,𝑏), (5) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Pittenger [31] found the greatest value 𝑟₁ and the least value 𝑟₂ such that the double inequality 𝑀_𝑟1(𝑎,𝑏)≤𝐿_𝑝(𝑎,𝑏)≤𝑀_𝑟2(𝑎,𝑏), (6) holds for all 𝑎,𝑏>0, where 𝐿_𝑟(𝑎,𝑏) is the 𝑟th generalized logarithmic means which is defined by 𝐿_𝑟()=𝑏𝑎,𝑏^𝑟+1−𝑎^𝑟+1(𝑟+1)(𝑏−𝑎)^1/𝑟1,𝑎≠𝑏,𝑟≠−1,𝑟≠0,𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)},𝑎≠𝑏,𝑟=0,𝑏−𝑎log𝑏−log𝑎,𝑎≠𝑏,𝑟=−1,𝑎,𝑎=𝑏. (7)

The following sharp power mean bounds for the first Seiffert mean 𝑃(𝑎,𝑏) are given in [10, 33]: 𝑀_log2/log𝜋(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝑀_2/3(𝑎,𝑏), (8) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [17], the authors answered the question: for 𝛼∈(0,1), what are the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝑀_𝑝(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_𝑞(𝑎,𝑏) (9) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏?

Neuman and Sándor [1] established that 𝐴()()<𝐴()𝑎,𝑏<𝑀𝑎,𝑏𝑎,𝑏log1+√2,𝜋4𝑇()()(),𝑀()<()()𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏𝑎,𝑏2𝐴𝑎,𝑏+𝑄𝑎,𝑏3, (10) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Let 0<𝑎,𝑏≤1/2 with 𝑎≠𝑏, 𝑎^′=1−𝑎 and 𝑏^′=1−𝑏. Then, the Ky Fan inequalities 𝐺(𝑎,𝑏)𝐺𝑎^′,𝑏^′<𝐿(𝑎,𝑏)𝐿𝑎^′,𝑏^′<𝑃(𝑎,𝑏)𝑃𝑎^′,𝑏^′<𝐴(𝑎,𝑏)𝐴𝑎^′,𝑏^′<𝑀(𝑎,𝑏)𝑀𝑎^′,𝑏^′<𝑇(𝑎,𝑏)𝑇𝑎^′,𝑏^′ (11) were presented in [1].

In [24], Li et al. found the best possible bounds for the Neuman-Sándor mean 𝑀(𝑎,𝑏) in terms of the generalized logarithmic mean 𝐿_𝑟(𝑎,𝑏). Neuman [25] and Zhao et al. [26] proved that the inequalities ()+()𝐴()()()+𝐴(),()+()𝐴()()()+𝐴(),𝛼𝛼𝑄𝑎,𝑏1−𝛼𝑎,𝑏<𝑀𝑎,𝑏<𝛽𝑄𝑎,𝑏1−𝛽𝑎,𝑏𝜆𝐶𝑎,𝑏1−𝜆𝑎,𝑏<𝑀𝑎,𝑏<𝜇𝐶𝑎,𝑏1−𝜇𝑎,𝑏₁𝐻(𝑎,𝑏)+1−𝛼₁𝑄(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽₁𝐻(𝑎,𝑏)+1−𝛽₁𝑄(),𝛼𝑎,𝑏₂𝐺(𝑎,𝑏)+1−𝛼₂𝑄(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽₂𝐺(𝑎,𝑏)+1−𝛽₂𝑄(𝑎,𝑏) (12) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤[1−log(1+√2)]/[(√2−1)log(1+√2)], 𝛽≥1/3, 𝜆≤[1−log(1+√2)]/log(1+√2), 𝜇≥1/6, 𝛼₁≥2/9, 𝛽₁≤1−1/[√2log(1+√2)], 𝛼₂≥1/3, and 𝛽₂≤1−1/[√2log(1+√2)].

In [7], Sándor and Trif proved that the inequalities 𝑒^{((𝑎−𝑏)2/6(𝑎+𝑏)2)}<𝐴(𝑎,𝑏)𝐼(𝑎,𝑏)<𝑒^{((𝑎−𝑏)2/24𝑎𝑏)},𝑒^{((𝑎−𝑏)2/3(𝑎+𝑏)2)}<𝐼(𝑎,𝑏)𝐺(𝑎,𝑏)<𝑒^{((𝑎−𝑏)2/12𝑎𝑏)},𝑒^{((𝑎−𝑏)4/30(𝑎+𝑏)4)}<𝐼(𝑎,𝑏)𝐴^2/3(𝑎,𝑏)𝐺^1/3(𝑎,𝑏)<𝑒^{((𝑎−𝑏)4/120𝑎𝑏(𝑎+𝑏)4)} (13) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Neuman and Sándor [15] and Gao [20] proved that 𝛼₁=1, 𝛽₁=𝑒/2, 𝛼₂=1, 𝛽₂=2√2/𝑒, 𝛼₃=1, 𝛽₃=3/𝑒, 𝛼₄=𝑒/𝜋, 𝛽₄=1, 𝛼₅=1, and 𝛽₅=2𝑒/𝜋 are the best possible constants such that the double inequalities 𝛼₁<𝐴(𝑎,𝑏)/𝐼(𝑎,𝑏)<𝛽₁,  𝛼₂<𝐼(𝑎,𝑏)/𝑀_2/3(𝑎,𝑏)<𝛽₂, 𝛼₃<𝐼(𝑎,𝑏)/𝐻𝑒(𝑎,𝑏)<𝛽₃, 𝛼₄<𝑃(𝑎,𝑏)/𝐼(𝑎,𝑏)<𝛽₄, and 𝛼₅<𝑇(𝑎,𝑏)/𝐼(𝑎,𝑏)<𝛽₅ hold for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐻𝑒(𝑎,𝑏)=(𝑎+√𝑎𝑏+𝑏)/3=(2𝐴(𝑎,𝑏)+𝐺(𝑎,𝑏))/3 is the Heronian mean of 𝑎 and 𝑏.

In [34], Sándor established that 𝐻𝑒(𝑎,𝑏)<𝑀_2/3(𝑎,𝑏), (14) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

It is not difficult to verify that the inequality 2𝐴(𝑎,𝑏)+𝑄(𝑎,𝑏)3<𝐻𝑒𝑎²,𝑏^21/2 (15) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

From inequalities (10), (14), and (15), one has 𝑀(𝑎,𝑏)<𝑀_2/3𝑎²,𝑏^21/2=𝑀_4/3(𝑎,𝑏), (16) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

It is the aim of this paper to find the best possible lower power mean bound for the Neuman-Sándor mean 𝑀(𝑎,𝑏) and to present the sharp constants 𝛼 and 𝛽 such that the double inequality 𝑀()𝛼<𝑎,𝑏𝐼(𝑎,𝑏)<𝛽 (17) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

2. Main Results

Theorem 1. 𝑝₀=(log2)/log[2log(1+√2)]=1.224… is the greatest value such that the inequality 𝑀(𝑎,𝑏)>𝑀_𝑝0(𝑎,𝑏) (18) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Proof. From (1) and (2), we clearly see that both 𝑀(𝑎,𝑏) and 𝑀_𝑝(𝑎,𝑏) are symmetric and homogenous of degree one. Without loss of generality, we assume that 𝑏=1 and 𝑎=𝑥>1.
Let 𝑝₀=(log2)/log[2log(1+√2)], then from (1) and (2) one has log𝑀(𝑥,1)−log𝑀_𝑝0(𝑥,1)=log𝑥−12sinh⁻¹)/(−1((𝑥−1𝑥+1))𝑝₀𝑥log^𝑝0+12. (19) Let 𝑓(𝑥)=log𝑥−12sinh⁻¹)/(−1((𝑥−1𝑥+1))𝑝₀𝑥log^𝑝0+12. (20) Then, simple computations lead to lim_𝑥→1+𝑓(𝑥)=0,lim_𝑥→+∞𝑓(𝑥)=1𝑝₀log2−log2sinh⁻¹(1𝑓)=0,_′(𝑥)=1+𝑥^𝑝0−1𝑓₁(𝑥)(𝑥−1)(𝑥^𝑝0+1)sinh⁻¹)/(,((𝑥−1𝑥+1)) (21)(22)(23) where 𝑓₁(𝑥)√=−2(𝑥−1)𝑥^𝑝0+1(𝑥+1)𝑥^𝑝0−1+1√1+𝑥²+sinh⁻¹𝑥−1,𝑓𝑥+1₁(1)=0,lim_𝑥→+∞𝑓₁(𝑥)=−√2+sinh⁻¹(1)𝑓=−0.5328⋯<0,_′1(𝑥)=√2(𝑥−1)𝑓₂(𝑥)(𝑥+1)²𝑥^𝑝0−1+1²1+𝑥^23/2, (24)(25)(26) where 𝑓₂(𝑥)=1+𝑥+2𝑥²+𝑝₀−1𝑥^𝑝0−2−𝑥^𝑝0−1+𝑥^𝑝0+1−𝑝₀−1𝑥^𝑝0+2−2𝑥^2𝑝0−2−𝑥^2𝑝0−1−𝑥^2𝑝0,𝑓₂(1)=0,lim_𝑥→+∞𝑓₂(𝑥)𝑓=−∞,_′2(𝑥)=1+4𝑥+𝑝₀−1𝑝₀−2𝑥^𝑝0−3−𝑝₀−1𝑥^𝑝0−2+𝑝₀+1𝑥^𝑝0−𝑝₀−1𝑝₀+2𝑥^𝑝0+1−4𝑝₀−1𝑥^2𝑝0−3−2𝑝₀−1𝑥^2𝑝0−2−2𝑝₀𝑥^2𝑝0−1,𝑓^′₂(1)=44−3𝑝₀>0,lim_𝑥→+∞𝑓^′₂(𝑥)𝑓=−∞,_2′′(𝑥)=4+𝑝₀−1𝑝₀−2𝑝₀−3𝑥^𝑝0−4−𝑝₀−1𝑝₀−2𝑥^𝑝0−3+𝑝₀𝑝₀+1𝑥^𝑝0−1−𝑝₀−1𝑝₀+2𝑝₀+1𝑥^𝑝0−4𝑝₀−12𝑝₀−3𝑥^2𝑝0−4−22𝑝₀−1𝑝₀−1𝑥^2𝑝0−3−2𝑝₀2𝑝₀−1𝑥^2𝑝0−2,𝑓₂^′′(1)=42𝑝₀−14−3𝑝₀>0,lim_𝑥→+∞𝑓₂^′′(𝑥)𝑓=−∞,_2′′′(𝑥)=𝑝₀−1𝑥^𝑝0−5𝑓₃(𝑥), (27)(28)(29)(30)(31)(32)(33) where 𝑓₃(𝑥)=−2−𝑝₀3−𝑝₀4−𝑝₀−2−𝑝₀3−𝑝₀𝑥+𝑝₀𝑝₀+1𝑥³−𝑝₀𝑝₀+1𝑝₀+2𝑥⁴−83−2𝑝₀2−𝑝₀𝑥^𝑝0+22𝑝₀−13−2𝑝₀𝑥^𝑝0+1−4𝑝₀2𝑝₀−1𝑥^𝑝0+2<−2−𝑝₀3−𝑝₀4−𝑝₀−2−𝑝₀3−𝑝₀𝑥+𝑝₀𝑝₀+1𝑥⁴−𝑝₀𝑝₀+1𝑝₀+2𝑥⁴−83−2𝑝₀2−𝑝₀𝑥^𝑝0+22𝑝₀−13−2𝑝₀𝑥^𝑝0+2−4𝑝₀2𝑝₀−1𝑥^𝑝0+2=−2−𝑝₀3−𝑝₀4−𝑝₀−2−𝑝₀3−𝑝₀𝑥−𝑝₀𝑝₀+1²𝑥⁴−83−2𝑝₀2−𝑝₀𝑥^𝑝0−22𝑝₀−14𝑝₀−3𝑥^𝑝0+2<0, (34) for 𝑥>1.
Equation (33) and inequality (34) imply that 𝑓₂^′′(𝑥) is strictly decreasing on [1,+∞). Then, the inequality (31) and (32) lead to the conclusion that there exists 𝑥₁>1, such that 𝑓^′₂(𝑥) is strictly increasing on [1,𝑥₁] and strictly decreasing on [𝑥₁,+∞).
From (29) and (30) together with the piecewise monotonicity of 𝑓^′₂(𝑥), we clearly see that there exists 𝑥₂>𝑥₁>1, such that 𝑓₂(𝑥) is strictly increasing on [1,𝑥₂] and strictly decreasing on [𝑥₂,+∞).
It follows from (26)–(28) and the piecewise monotonicity of 𝑓₂(𝑥) that there exists 𝑥₃>𝑥₂>1, such that 𝑓₁(𝑥), is strictly increasing on [1,𝑥₃] and strictly decreasing on [𝑥₃,+∞).
From (23)–(25) and the piecewise monotonicity of 𝑓₁(𝑥) we see that there exists 𝑥₄>𝑥₃>1, such that 𝑓(𝑥) is strictly increasing on (1,𝑥₄] and strictly decreasing on [𝑥₄,+∞).
Therefore, 𝑀(𝑥,1)>𝑀_𝑝0(𝑥,1) for 𝑥>1 follows easily from (19)–(22) and the piecewise monotonicity of 𝑓(𝑥).
Next, we prove that 𝑝₀=(log2)/log[2log(1+√2)]=1.224… is the greatest value such that 𝑀(𝑥,1)>𝑀_𝑝0(𝑥,1) for all 𝑥>1.
For any 𝜀>0 and 𝑥>1, from (1) and (2), one has lim_𝑥→+∞𝑀_𝑝0+𝜀(𝑥,1)𝑀(𝑥,1)=lim_𝑥→+∞1+𝑥^𝑝0+𝜀2^{1/(𝑝0+𝜀)}2sinh⁻¹((𝑥−1)/(𝑥+1))𝑥−1=2^{−1/(𝑝0+𝜀)}×2sinh⁻¹(1)=2^{𝜀/𝑝0(𝑝0+𝜀)}>1. (35)
Inequality (35) implies that for any 𝜀>0, there exists 𝑋=𝑋(𝜀)>1, such that 𝑀(𝑥,1)<𝑀_𝑝0+𝜀(𝑥,1) for 𝑥∈(𝑋,+∞).

Remark 2. 4/3 is the least value such that inequality (16) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏, namely, 𝑀_4/3(𝑎,𝑏) is the best possible upper power mean bound for the Neuman-Sándor mean 𝑀(𝑎,𝑏).
In fact, for any 𝜀∈(0,4/3) and 𝑥>0, one has 𝑀_4/3−𝜀()()=()1+𝑥,1−𝑀1+𝑥,11+𝑥^4/3−𝜀+12^{1/(4/3−𝜀)}−𝑥2sinh⁻¹((.𝑥/2+𝑥)) (36)
Letting 𝑥→0 and making use of Taylor expansion, we get (1+𝑥)^4/3−𝜀+12^{1/(4/3−𝜀)}−𝑥2sinh⁻¹((=𝑥/2+𝑥))1+4−3𝜀6()𝑥+4−3𝜀)(1−3𝜀𝑥36²+𝑜𝑥^{21/(4/3−𝜀)}−𝑥𝑥−(1/2)𝑥²+(5/24)𝑥³+𝑜𝑥³=11+2𝑥+1−3𝜀𝑥24²+𝑜𝑥²−11+21𝑥+𝑥24²+𝑜𝑥²𝜀=−8𝑥²+𝑜𝑥². (37)
Equations (36) and (37) imply that for any 𝜀∈(0,4/3) there exists 𝛿=𝛿(𝜀)>0, such that 𝑀(1+𝑥,1)>𝑀_(4/3)−𝜀(1+𝑥,1) for 𝑥∈(0,𝛿).

Theorem 3. For all 𝑎,𝑏>0 with 𝑎≠𝑏, one has 𝑀()1<𝑎,𝑏𝐼()<𝑒𝑎,𝑏2log1+√2, (38) with the best possible constants 1 and 𝑒/[2log(1+√2)]=1.5419….

Proof. From (2) and (3), we clearly see that both 𝑀(𝑎,𝑏) and 𝐼(𝑎,𝑏) are symmetric and homogenous of degree one. Without loss of generality, we assume that 𝑏=1 and 𝑎=𝑥>1. Let 𝑓(𝑥)=𝑀(𝑥,1)𝐼()=𝑒()𝑥,1𝑥−12𝑥^{𝑥/(𝑥−1)}sinh⁻¹)/(.((𝑥−1𝑥+1)) (39) Then, simple computations lead to 𝑓^′(𝑥)𝑓(𝑥)=log𝑥(𝑥−1)²sinh⁻¹)/(𝑓((𝑥−1𝑥+1))¹(𝑥), (40) where 𝑓₁(𝑥)=sinh⁻¹𝑥−1−√𝑥+12(𝑥−1)²(𝑥+1)√1+𝑥²,log𝑥lim_𝑥→1+𝑓₁(𝑥)𝑓=0,_′1(𝑥)=√2𝑓₂(𝑥)𝑥(𝑥+1)²1+𝑥^23/2log²𝑥, (41)(42) where 𝑓₂(𝑥)=𝑥(𝑥+1)1+𝑥²log²𝑥−𝑥3𝑥³−𝑥²+()+𝑥−3log𝑥𝑥−1²(𝑥+1)1+𝑥²,𝑓₂(1)𝑓=0,_′2(𝑥)=4𝑥³+3𝑥²+2𝑥+1log²𝑥+5−2𝑥³+𝑥²+1log𝑥+5𝑥⁴−7𝑥³+𝑥²𝑓−𝑥+2,_′2(1)𝑓=0,_2′′(𝑥)=26𝑥²+3𝑥+1log²𝑥+2−11𝑥²+8𝑥+2+𝑥⁻¹log𝑥+20𝑥³−31𝑥²+7𝑥−1+5𝑥⁻¹,𝑓₂^′′(1)𝑓=0,_2′′′(𝑥)=6(4𝑥+1)log²𝑥+2−10𝑥+14+2𝑥⁻¹−𝑥⁻²log𝑥+60𝑥²−84𝑥+23+4𝑥⁻¹−3𝑥⁻²,𝑓₂^′′′(1)𝑓=0,₂₍₄₎(𝑥)=24log²𝑥+47+3𝑥⁻¹−𝑥⁻²+𝑥⁻³log𝑥+120𝑥−104+28𝑥⁻¹+4𝑥⁻³>0 (43)(44)(45)(46)(47) for 𝑥>1.
From (46) and (47), we clearly see that 𝑓₂^′′(𝑥) is strictly increasing on [1,+∞). Then, (45) leads to the conclusion that 𝑓^′₂(𝑥) is strictly increasing on [1,+∞).
Equations (43) and (44) together with the monotonicity of 𝑓^′₂(𝑥) impliy that 𝑓₂(𝑥)>0 for 𝑥>1. Then, (42) leads to the conclusion that 𝑓₁(𝑥) is strictly increasing on [1,+∞).
It follows from equations (40) and (41) together with the monotonicity of 𝑓₁(𝑥) that 𝑓(𝑥) is strictly increasing on (1,+∞).
Therefore, Theorem 3 follows from (39) and the monotonicity of 𝑓(𝑥) together with the facts that lim_𝑥→+∞𝑓(𝑥)=𝑒2log1+√2,lim_𝑥→1+𝑓(𝑥)=1. (48)

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants nos. 11071069 and 11171307, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant no. T200924.