Optimal lower and upper bounds for the geometric convex combination of the error function

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1 Li et al. Journal of Inequalities Applications (215) 215:382 DOI /s y R E S E A R C H Open Access Optimal lower upper bounds for the geometric convex combination of the error function Yong-Min Li 1, Wei-Feng Xia 2*, Yu-Ming Chu 3 Xiao-Hui Zhang 3 * Correspondence: xwf212@163.com 2 School of Automation, Nanjing University of Science Technology, Nanjing, 2194, China Full list of author information is available at the end of the article Abstract For x R, the error function erf(x)isdefinedas erf(x)= 2 x e t2 dt. π In this paper, we answer the question: what are the greatest value p the least value q, suchthatthedoubleinequality erf(m p (x, y; λ)) G(erf(x), erf(y); λ) erf(m q (x, y; λ)) holds for all x, y 1(or<x, y <1) λ (, 1)? Here, M r (x, y; λ)=(λx r +(1 λ)y r ) 1/r (r ),M (x, y; λ)=x λ y 1 λ G(x, y; λ)=x λ y 1 λ are the weighted power the weighted geometric mean, respectively. MSC: Primary 33B2; secondary 26D15 Keywords: error function; power mean; functional inequalities 1 Introduction For x R,theerrorfunctionerf(x)isdefinedas erf(x)= 2 x e t2 dt. π 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, 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]. In 1933, Aumann 19] introduced a generalized notion of convexity, the so-called MNconvexity, when M N aremeanvalues. Afunctionf :, ), ) ismn-convex if f (M(x, y)) N(f (x), f (y)) for x, y, ). The usual convexity is the special case when M 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 cosine function, to the special ones, such as the Ɣ 215 Li et al. This article is distributed under the terms of the Creative Commons Attribution 4. International License ( which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original author(s) the source, provide a link to the Creative Commons license, indicate if changes were made.

2 Li et al. Journal of Inequalities Applications (215) 215:382 Page 2 of 8 function, the Gaussian hypergeometric function, the Bessel function. For the details as regards MN-convexity its applications the reader is referred to 2 25]. Let λ (, 1), we define A(x, y; λ)=λx +(1 λ)y, G(x, y; λ)=x λ y 1 λ, H(x, y; λ)= xy λy+(1 λ)x M r (x, y; λ)=(λx r +(1 λ)y r ) 1/r (r ),M (x, y; λ)=x λ y 1 λ. These are commonly known as weighted arithmetic mean, weighted geometric mean, weighted harmonic mean, weighted power mean of two positive numbers x y, respectively. Then it is well known that the inequalities H(x, y; λ)=m 1 (x, y; λ)<g(x, y; λ)=m (x, y; λ)<a(x, y; λ)=m 1 (x, y; λ) hold for all λ (, 1) x, y >withx y. By elementary computations, one has r M r(x, y; λ)=min(x, y) (1.1) r + M r(x, y; λ)=max(x, y). In 26], Alzer proved that c 1 (λ)= such that the double inequality λ+(1 λ) erf(1) erf(1/(1 λ)) c 2 (λ) = 1 are the best possible factors c 1 (λ) erf ( H(x, y; λ) ) A ( erf(x), erf(y); λ ) c 2 (λ) erf ( H(x, y; λ) ) (1.2) holds for all x, y 1, + )λ (, 1/2). Inspired by (1.2), it is natural to ask: does the inequality erf(m(x, y)) N(erf(x), erf(y)) hold for other means M, N,suchasgeometric,harmonicorpowermeans? In 27, 28], the authors found the greatest values α 1, α 2 the least values β 1, β 2,such that the double inequalities erf ( M α1 (x, y; λ) ) A ( erf(x), erf(y); λ ) erf ( M β1 (x, y; λ) ) erf ( M α2 (x, y; λ) ) H ( erf(x), erf(y); λ ) erf ( M β2 (x, y; λ) ) hold for all x, y 1(or<x, y <1)λ (, 1). In the following we answer the question: what are the greatest value p the least value q, such that the double inequality erf ( M p (x, y; λ) ) G ( erf(x), erf(y); λ ) erf ( M q (x, y; λ) ) holds for all x, y 1(or<x, y <1)λ (, 1)? 2 Lemmas In this section we present two lemmas, which will be used in the proof of our main results.

3 Li et al. Journal of Inequalities Applications (215) 215:382 Page 3 of 8 2 Lemma 2.1 Let r,r = 1 e π erf(1) = , u(x) =log erf(x1/r ). Then the following statements are true: (1) if r < r, then u(x) is strictly convex on 1, + ); (2) if r r <, then u(x) is strictly concave on (, 1]; (3) if r >, then u(x) is strictly concave on (, + ). Proof Simple computations lead to u (x)= 2e x2/r x 1/r 1 r π erf(x 1/r ) (2.1) where 2e x2/r x 1/r 2 u (x)= r 2 π erf 2 g(x), (2.2) (x 1/r ) g(x)= ( 2x 2/r +1 r ) erf ( x 1/r) 2 π e x2/r x 1/r. (2.3) Then g (x)=4x 2/r 1 g 1 (x), (2.4) g 1 (x)= 1 r erf( x 1/r) 1 2 e x2/r x 1/r, π (2.5) g 1 (x)= 1 2r 2 π e x2/r x 1/r 1 (2r 4)x 2/r + r ]. (2.6) We divide the proof into two cases. Case 1. If r <,then(2.6), (2.5), (2.3)leadto g 1 (x)<, (2.7) g 1(x)>, x + g(x)=, x + g 1(x)=, (2.8) x + g(x)=, (2.9) x + g(1) = ( 1 r) erf(1) 2 e π. (2.1) Inequality (2.7)impliesthatg 1 (x) is strictly decreasing on, + ). It follows from the monotonicity of g 1 (x) (2.8) that there exists x 1 (, + ), such that g(x) is strictly increasing on, x 1 ] strictly decreasing on x 1,+ ). From the piecewise monotonicity of g(x) (2.9) we clearly see that there exists x 2 (, + ), such that g(x)<forx (, x 2 )g(x)>forx (x 2,+ ).

4 Li et al. Journal of Inequalities Applications (215) 215:382 Page 4 of 8 Case 1.1. If r < r,thenfrom(2.1) weknowthatg(1) >. This leads to g(x) >for x 1, + ). Therefore (2.2) leads to the conclusion that u(x) is strictly convex on 1, + ). Case 1.2. If r r <,then(2.1)impliesthatg(1). This leads to g(x) forx (, 1]. Therefore (2.2) leads to the conclusion that u(x) is strictly concave on (, 1]. Case 2. If r >,then(2.5)(2.3)implythat g 1 (x)< (2.11) g(x)= (2.12) x + for x (, + ). It follows from (2.11), (2.4), (2.12)thatg(x) <. Therefore (2.2) leads to the conclusion that u(x) is strictly concave on (, + ). Lemma 2.2 The function h(x)=2x 2 + xe x2 x is strictly increasing on (, + ). e t2 dt Proof Simple computations lead to where h (x)= h 1 (x) ( x e t2 dt) 2, (2.13) ( x ) 2 h 1 (x)=4x e t2 dt + ( 1 2x 2) x e x2 e t2 dt xe 2x2, h 1(x)=, (2.14) x + h 1 (x)=4 ( x ) 2 e t2 dt + ( 4x 3 +2x ) x e x2 e t2 dt +2x 2 e 2x2 > (2.15) for x (, + ). Hence, h(x) isstrictlyincreasingon(,+ ), as follows from (2.15), (2.14), (2.13). 3 Main results Theorem 3.1 Let λ (, 1) r = 1 2 e π erf(1) = Then the double inequality erf ( M p (x, y; λ) ) G ( erf(x), erf(y); λ ) erf ( M q (x, y; λ) ) (3.1) holds for all x, y 1 if only if p = q r. Proof First of all, we prove that inequality (3.1)holdsifp = q r. It follows from (1.1) that the first inequality in (3.1) istrueifp =. Since the weighted power mean

5 Li et al. Journal of Inequalities Applications (215) 215:382 Page 5 of 8 M t (x, y; λ) is strictly increasing with respect to t on R, thus we only need to prove that the second inequality in (3.1)istrueifr q <. If r q <,u(z)=log erf(z 1/q ), then Lemma 2.1(2) leads to λu(s)+(1 λ)u(t) u ( λs +(1 λ)t ) (3.2) for λ (, 1) s, t (, 1]. Let s = x q, t = y q,x, y 1. Then (3.2) leads to the second inequality in (3.1). Second, we prove that the second inequality in (3.1)impliesq r. Let x 1y 1. Then the second inequality in (3.1)leadsto D(x, y)=:erf ( M q (x, y; λ) ) G ( erf(x), erf(y); λ ). (3.3) It follows from (3.3)that D(y, y)= x D(x, y) x=y = 2 x 2 D(x, y) x=y = Therefore, )] λ(1 λ)y erf q 1+ (2y 2 + ye y2 (y) y. (3.4) e t2 dt ) q (1 2y 2 ye y2 y 1 + y = r e t2 dt follows from (3.3)(3.4) together with Lemma 2.2. Finally, we prove that the first inequality in (3.1) impliesp =. We distinguish two cases. Case I. p. Then for any fixed y 1, + )wehave x + erf( M p (x, y; λ) ) =1 x + G( erf(x), erf(y); λ ) = erf 1 λ (y)<1, which contradicts the first inequality in (3.1). Case II. < p <.Letx 1, α = λ 1/p y +. Then the first inequality in (3.1) leads to E(x)=:erf λ (x) erf(αx). (3.5) It follows from (3.5)that E(x)= (3.6) x +

6 Li et al. Journal of Inequalities Applications (215) 215:382 Page 6 of 8 E (x)= 2λ π e x2 erf λ 1 (x) α λ e(1 α2 )x 2 ]. (3.7) Note that α >1,then erf λ 1 (x) αλ ] x + e(1 α2 )x 2 =1. (3.8) It follows from (3.7)(3.8) that there exists a sufficiently large η 1 1, + ), such that E (x)>forx (η 1,+ ). Hence E(x) is strictly increasing on η 1,+ ). From the monotonicity of E(x)onη 1,+ )(3.6) we conclude that there exists η 2 1, + ), such that E(x)<forx (η 2,+ ), this contradicts (3.5). Theorem 3.2 Let λ (, 1), then the double inequality erf ( M μ (x, y; λ) ) G ( erf(x), erf(y); λ ) erf ( M ν (x, y; λ) ) (3.9) holds for all <x, y <1if only if μ r ν. Proof First of all, we prove that (3.9)holdsifμ r ν. If μ r, u(z)=log erf(z 1/μ ), then Lemma 2.1(1) leads to u ( λs +(1 λ)t ) λu(s)+(1 λ)u(t) (3.1) for λ (, 1), s, t >1. Let s = x μ, t = y μ,<x, y <1.Then(3.1) leads to the first inequality in (3.9). If ν, u(z)=log erf(z 1/ν ), then Lemma 2.1(3) leads to λu(s)+(1 λ)u(t) u ( λs +(1 λ)t ) (3.11) for λ (, 1), < s, t <1. Therefore, the second inequality in (3.9) follows from s = x ν, t = y ν,<x, y <1together with (3.11). Second, we prove that the second inequality in (3.9)impliesν. Let < x, y < 1. Then the second inequality in (3.9)leadsto J(x, y)=:erf ( M ν (x, y; λ) ) G ( erf(x), erf(y); λ ). (3.12) It follows from (3.12)that J(y, y)= x J(x, y) x=y = 2 x 2 J(x, y) x=y = )] λ(1 λ)y erf ν 1+ (2y 2 + ye y2 (y) y. (3.13) e t2 dt

7 Li et al. Journal of Inequalities Applications (215) 215:382 Page 7 of 8 Hence, from (3.12)(3.13)together with Lemma 2.2we know that )] ν 1 (2y 2 + ye y2 y + y =. e t2 dt Finally, we prove that the first inequality in (3.9)impliesμ r. Let y 1. Then the first inequality in (3.9)leadsto L(x)=:G ( erf(x), erf(1); λ ) erf ( M μ (x,1;λ) ) (3.14) for < x <1. It follows from (3.14)that L(1) = (3.15) L (x)= 2λe x2 π erf 1 λ (1) erf λ 1 (x) x μ 1( λx μ +1 λ ) 1/μ 1 e x 2 (λx μ +1 λ) 2/μ ]. (3.16) Let L 1 (x)=log erf 1 λ (1) erf λ 1 (x) ] log x μ 1( λx μ +1 λ ) 1/μ 1 e x 2 (λx μ +1 λ) 2/μ ]. (3.17) Then x 1 L 1(x)=, (3.18) L 1 (x)=(λ (x) (μ 1)(1 λ) 1)erf erf(x) x(λx μ +1 λ) 2x +2λxμ 1( λx μ +1 λ ) 2/μ 1, ] x 1 L 1 (x)=(1 λ) 2 μ 1 e. (3.19) π erf(1) If μ > r,thenfrom(3.19) we clearly see that there exists a small δ 1 >,suchthatl 1 (x)< for x (1 δ 1, 1). Therefore, L 1 (x) is strictly decreasing on 1 δ 1,1]. The monotonicity of L 1 (x)on1 δ 1,1](3.18) imply that there exists δ 2 >,suchthat L 1 (x)>forx (1 δ 2,1). Hence, (3.16) (3.17) leadtol(x) being strictly increasing on 1 δ 2, 1]. It follows from the monotonicity of L(x) (3.15) that there exists δ 3 >,suchthatl(x) <for x (1 δ 3, 1), this contradicts (3.14). 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 approved the final manuscript.

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