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THE GEOMETRIC MEAN IS A BERNSTEIN FUNCTION FENGQI,XIAO-JINGZHANG,ANDWEN-HUILI 3 1 0 Abstract. Inthepaper,theauthorsestablish,byusingCauchyintegralfor- 2 mula in the theory of complex functions, an integral representation for the n geometricmeanofnpositivenumbers. Fromthisintegralrepresentation, the a geometric mean is proved to be a Bernstein function and a new proof of the J wellknownAGinequalityisprovided. 9 2 ] 1. Introduction A C We recall some notions and definitions. h. Definition 1.1 ([15, 26]). A function f is said to be completely monotonic on an t interval I if f has derivatives of all orders on I and a m ( 1)nf(n)(t) 0 (1.1) − ≥ [ for x I and n 0. ∈ ≥ 1 The class of completely monotonic functions on (0, ) is characterized by the v ∞ famous Hausdorff-Bernstein-Widder Theorem below. 8 4 Proposition1.1([26,p.161,Theorem12b]). Anecessary andsufficientcondition 8 that f(x) should be completely monotonic for 0<x< is that 6 ∞ . ∞ 1 f(x)= e−xtdα(t), (1.2) 0 Z0 3 where α(t) is non-decreasing and the integral converges for 0<x< . 1 ∞ : Definition 1.2 ([18, 20]). A function f is said to be logarithmically completely v i monotonic on an interval I if its logarithm lnf satisfies X ( 1)k[lnf(t)](k) 0 (1.3) r − ≥ a for k N on I. ∈ It has been proved in [3, 8, 18, 20] that a logarithmically completely monotonic function on an interval I must be completely monotonic on I. Definition 1.3 ([24, 26]). A function f : I ( , ) [0, ) is called a ⊆ −∞ ∞ → ∞ Bernstein function on I if f(t) has derivatives of all orders and f′(t) is completely monotonic on I. The class of Bernstein functions can be characterized by 2010 Mathematics Subject Classification. Primary26E60, 30E20;Secondary26A48,44A20. Keywordsandphrases. Integralrepresentation;Geometricmean;Cauchyintegralformula;Bern- stein function; AG inequality; new proof; Completely monotonic function; Logarithmicallycom- pletelymonotonicfunction;Stieltjesfunction. Thispaperwastypeset usingAMS-LATEX. 1 2 F.QI,X.-J.ZHANG,ANDW.-H.LI Proposition 1.2 ([24, p. 15, Theorem 3.2]). A function f : (0, ) R is a ∞ → Bernstein function if and only if it admits the representation ∞ f(x)=a+bx+ 1 e−xt dµ(t), (1.4) − Z0 where a,b 0 and µ is a measure on (0, (cid:0)) satisfyin(cid:1)g ≥ ∞ ∞ min 1,t dµ(t)< . { } ∞ Z0 In [5, pp. 161–162, Theorem 3] and [24, p. 45, Proposition 5.17], it was dis- coveredthat the reciprocalof any Bernstein function is logarithmically completely monotonic. Definition 1.4 ([1]). Iff(k)(t)forsomenonnegativeintegerk iscompletelymono- tonic on an interval I, but f(k−1)(t) is not completely monotonic on I, then f(t) is called a completely monotonic function of k-th order on an interval I. It is obvious that a completely monotonic function of first order is a Bernstein function if and only if it is nonnegative on I. Definition 1.5 ([24, p. 19, Definition 2.1]). If f :(0, ) [0, ) can be written ∞ → ∞ in the form a ∞ 1 f(x)= +b+ dµ(s), (1.5) x s+x Z0 then it is called a Stieltjes function, where a,b 0 are nonnegative constants and ≥ µ is a nonnegative measure on (0, ) such that ∞ ∞ 1 dµ(s)< . 1+s ∞ Z0 The set of logarithmically completely monotonic functions on (0, ) contains ∞ all Stieltjes functions, see [3] or [22, Remark 4.8]. In other words, all the Stieltjes functions are logarithmically completely monotonic on (0, ). ∞ In the newly-published paper [7], a new notion “completely monotonic degree” of nonnegative functions was naturally introduced and initially studied. We also recall that the extended mean value E(r,s;x,y) may be defined as r(ys xs) 1/(s−r) E(r,s;x,y)= − , rs(r s)(x y)=0; (1.6) s(yr xr) − − 6 (cid:20) − (cid:21) yr xr 1/r E(r,0;x,y)= − , r(x y)=0; (1.7) r(lny lnx) − 6 (cid:20) − (cid:21) 1 xxr 1/(xr−yr) E(r,r;x,y)= , r(x y)=0; (1.8) e1/r yyr − 6 (cid:18) (cid:19) E(0,0;x,y)=√xy, x=y; (1.9) 6 E(r,s;x,x)=x, x=y; where x and y are positive numbers and r,s R. Because this mean was first ∈ definedin[25],soitisalsocalledStolarsky’smean. Manyspecialmeanvalueswith two variables are special cases of E, for example, E(r,2r;x,y)=M (x,y), (power mean or Ho¨lder mean) r E(1,p;x,y)=L (x,y), (generalized or extended logarithmic mean) p THE GEOMETRIC MEAN IS A BERNSTEIN FUNCTION 3 E(1,1;x,y)=I(x,y), (identric or exponential mean) E(1,2;x,y)=A(x,y), (arithmtic mean) E(0,0;x,y)=G(x,y), (geometric mean) E( 2, 1;x,y)=H(x,y), (harmonic mean) − − E(0,1;x,y)=L(x,y). (logarithmic mean) FormoreinformationonE,pleaserefertothemonograph[4],thepapers[9,10,11], and closely-related references therein. It is easy to see that the arithmetic mean A (t)=A(x+t,y+t)=A(x,y)+t x,y is a trivial Bernstein function of t ( min x,y , ) for x,y >0. ∈ − { } ∞ It is not difficult to see that the harmonic mean 2 H (t)=H(x+t,y+t)= (1.10) x,y 1 + 1 x+t y+t for t ( min x,y , ) and x,y >0 with x=y meets ∈ − { } ∞ 6 2 x2+y2+2(x+y)t+2t2 (x y)2 H′ (t)= =1+ − >1. (1.11) x,y (x+y+2t)2 (x+y+2t)2 (cid:2) (cid:3) It is obvious that the derivative H′ (t) is completely monotonic with respect x,y to t. As a result, the harmonic mean H (t) is a Bernstein function of t on x,y ( min x,y , ) for x,y >0 with x=y. − { } ∞ 6 In [21, Remark 6], it was pointed out that the reciprocal of the identric mean 1 (x+t)x+t 1/(x−y) I (t)=I(x+t,y+t)= (1.12) x,y e (y+t)y+t (cid:20) (cid:21) for x,y > 0 with x = y is a logarithmically completely monotonic function of 6 t ( min x,y , ) and that the identric mean I (t) for t > min x,y with x,y ∈ − { } ∞ − { } x = y is also a completely monotonic function of first order (that is, a Bernstein 6 function). In [17, p. 616], it was concluded that the logarithmic mean L (t)=L(x+t,y+t) (1.13) x,y isincreasingandconcaveint> min x,y forx,y >0withx=y. Morestrongly, − { } 6 itwasprovedin[19,Theorem1]thatthelogarithmicmeanL (t)forx,y >0with x,y x=yisacompletelymonotonicfunctionoffirstorderint ( min x,y , ),that 6 ∈ − { } ∞ is, the logarithmic mean L (t) is a Bernstein function of t ( min x,y , ). x,y ∈ − { } ∞ Recently, the geometric mean G (t)=G(x+t,y+t)= (x+t)(y+t) (1.14) x,y was proved in [23] to be a Bernstein function opf t on ( min x,y , ) for x,y >0 − { } ∞ with x=y, and its integral representation 6 x y ∞ ρ((x y)s) G (t)=G(x,y)+t+ − − e−ys 1 e−st ds (1.15) x,y 2π s − Z0 (cid:0) (cid:1) 4 F.QI,X.-J.ZHANG,ANDW.-H.LI for x>y >0 and t> y was discovered, where − 1/2 ρ(s)= q(u) 1 e−(1−2u)s e−usdu − Z0 1/2 1(cid:2) (cid:3) (1.16) = q u eus e−us e−s/2du 2 − − Z0 (cid:18) (cid:19) 0 (cid:0) (cid:1) ≥ on (0, ) and ∞ 1 1 q(u)= 1 (1.17) u − − 1/u 1 r − on (0,1). p Let a = (a ,a ,...,a ) for n N, the set of all positive integers, be a given 1 2 n ∈ sequence of positive numbers. Then the arithmetic and geometric means A (a) n and G (a) of the numbers a ,a ,...,a are defined respectively as n 1 2 n n 1 A (a)= a (1.18) n k n k=1 X and n 1/n G (a)= a . (1.19) n k ! k=1 Y It is general knowledge that G (a) A (a), (1.20) n n ≤ with equality if and only if a =a = =a . 1 2 n ··· Therehasbeenalargenumber,presumablyoveronehundred,ofproofsoftheAG inequality (1.20) in the mathematical literature. The most complete information, so far, can be found in the monographs [2, 4, 12, 13, 14, 16] and a lot of references therein. In this paper, we establish, by using Cauchy integral formula in the theory of complex functions, an integral representation of the geometric mean n 1/n G (a+z)= (a +z) , (1.21) n k " # k=1 Y where a=(a ,a ,...,a ) satisfies a >0 for 1 k n and 1 2 n k ≤ ≤ z C ( , min a ,1 k n ]. k ∈ \ −∞ − { ≤ ≤ } From this integral representation, it is immediately derived that the geometric mean G (a+t) for t ( min a ,1 k n , ) is a Bernstein function, where n k ∈ − { ≤ ≤ } ∞ a+t = (a +t,a +t,...,a +t), and a new proof of the AG inequality (1.20) is 1 2 n provided. 2. Lemmas In order to prove our main results, we need the following lemmas. THE GEOMETRIC MEAN IS A BERNSTEIN FUNCTION 5 Lemma2.1(Cauchyintegralformula[6,p.113]). LetD beaboundeddomain with piecewise smooth boundary. If f(z) is analytic on D, and f(z) extends smoothly to the boundary of D, then 1 f(w) f(z)= dw, z D. (2.1) 2πi w z ∈ I∂D − Lemma 2.2. For z C ( , min a ,1 k n ] with a = (a ,a ,...,a ) k 1 2 n ∈ \ −∞ − { ≤ ≤ } and a >0, the principal branch of the complex function k f (z)=G (a+z) z, (2.2) a,n n − where a+z =(a +z,a +z,...,a +z), meets 1 2 n lim f (z)=A (a). (2.3) a,n n z→∞ Proof. By L’Hoˆspital’s rule in the theory of complex functions, we have a lim f (z)= lim z G 1+ 1 a,n n z→∞ z→∞ z − (cid:26) (cid:20) (cid:18) (cid:19) (cid:21)(cid:27) n 1/n G (1+az) 1 d n = lim − = lim (1+a z) =A (a), k n z→0 z z→0dz" # k=1 Y where1+a = 1+a1,1+a2,...,1+an and1+az=(1+a z,1+a z,...,1+a z). z z z z 1 2 n Lemma 2.2 is thus proved. (cid:3) (cid:0) (cid:1) Lemma 2.3. Let a = (a ,a ,...,a ) with a > 0 for 1 k n and let [a] 1 2 n k ≤ ≤ be the rearrangement of the positive sequence a in an ascending order, that is, [a]= a ,a ,...,a and a a a . For z C ( ,0], let [1] [2] [n] [1] [2] [n] ≤ ≤···≤ ∈ \ −∞ (cid:0) (cid:1) h (z)=G [a] a +z z, (2.4) n n [1] − − where [a] a +z = z,a a +z,(cid:0)...,a a +(cid:1) z . Then the principal branch [1] [2] [1] [n] [1] − − − of h (z) satisfies n (cid:0) (cid:1) lim h ( t+iε)= n ε→0+ℑ − n 1/n ℓπ a a t sin , t a a ,a a " [k]− [1]− # n ∈ [ℓ]− [1] [ℓ+1]− [1] (2.5) 0,kY=1(cid:12)(cid:12) (cid:12)(cid:12) t (cid:0)a[n] a[1] (cid:3) ≥ − for 1 ℓ n 1. ≤ ≤ − Proof. For t (0, ) a a ,1 ℓ n 1 and ε>0, we have [ℓ+1] [1] ∈ ∞ \ − ≤ ≤ − (cid:8) (cid:9) h ( t+iε)=G [a] a t+iε +t iε n n [1] − − − − n (cid:0) 1 (cid:1) =exp ln a a t+iε +t iε "n [k]− [1]− # − k=1 X (cid:0) (cid:1) n 1 =exp ln a a t+iε +iarg a a t+iε +t iε (n k− [1]− [k]− [1]− ) − k=1 X(cid:2) (cid:12) (cid:12) (cid:0) (cid:1)(cid:3) (cid:12) (cid:12) 6 F.QI,X.-J.ZHANG,ANDW.-H.LI n 1 ℓπ exp ln a a t + i +t,t a a ,a a  n [k]− [1]− n ! ∈ [ℓ]− [1] [ℓ+1]− [1] k=1 →exp 1 Xn ln(cid:12)(cid:12)a a t(cid:12)(cid:12)+πi +t,t>a(cid:0) a (cid:1) n [k]− [1]− ! [n]− [1] k=1  n X (cid:12)(cid:12) 1/n (cid:12)(cid:12) ℓπ a a t exp i +t,t a a ,a a = kY=n1(cid:12)(cid:12)a[k]−a[1]−t(cid:12)(cid:12)!1/nexp((cid:18)πin)+(cid:19)t,t>a∈(cid:0) [ℓa]− [1] [ℓ+1]− [1](cid:1) [k] [1] [n] [1] − − ! − as ε→0+.kYA=s1(cid:12)(cid:12)a result, we h(cid:12)(cid:12)ave lim h ( t+iε)= n ε→0+ℑ − n 1/n ℓπ a a t sin , t a a ,a a ;  [k]− [1]− ! n ∈ [ℓ]− [1] [ℓ+1]− [1] 0,kY=1(cid:12)(cid:12) (cid:12)(cid:12) t>(cid:0)a[n] a[1]. (cid:1) − For t=a[ℓ+1]−a[1] for 1≤ℓ≤n−1, we have n 1 1 h ( t+iε)=exp ln a a t+iε + ln(iε) +t iε n − "n [k]− [1]− n # − k6=ℓ+1 X (cid:0) (cid:1) n 1 1 π =exp ln a a t+iε exp ln ε + i +t iε "nk6=ℓ+1 [k]− [1]− # (cid:20)n(cid:18) | | 2 (cid:19)(cid:21) − X (cid:0) (cid:1) n 1 1 π exp ln a a t lim exp ln ε + i +t → "nk6=ℓ+1 [k]− [1]− #ε→0+ (cid:20)n(cid:18) | | 2 (cid:19)(cid:21) X (cid:0) (cid:1) =t as ε 0+. Hence, when t=a a for 1 ℓ n 1, we have [ℓ+1] [1] → − ≤ ≤ − lim h ( t+iε)=0. n ε→0+ℑ − The proof of Lemma 2.3 is completed. (cid:3) 3. The geometric mean is a Bernstein function We now turn our attention to establishing an integralrepresentationof the geometric mean G (a+z) and to showing that the geometric mean is a Bernstein n function. Theorem 3.1. Let a = (a ,a ,...,a ) with a > 0 for 1 k n and let [a] 1 2 n k ≤ ≤ denote the rearrangement of the sequence a in an ascending order, that is, [a] = a ,a ,...,a and a a a . For z C ( , min a ,1 [1] [2] [n] [1] [2] [n] k ≤ ≤ ··· ≤ ∈ \ −∞ − { ≤ k n ], the principal branch of the geometric mean G (a +z) has the integral n (cid:0) ≤ } (cid:1) representation 1 n−1 ℓπ a[ℓ+1] n 1/n dt G (a+z)=A (a)+z sin (a t) , (3.1) n n k − π Xℓ=1 n Za[ℓ] (cid:12)(cid:12)kY=1 − (cid:12)(cid:12) t+z (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) THE GEOMETRIC MEAN IS A BERNSTEIN FUNCTION 7 wherea+z =(a +z,a +z,...,a +z). Consequently,thegeometricmeanG (a+t) 1 2 n n is a Bernstein function on ( min a ,1 k n , ). k − { ≤ ≤ } ∞ Proof. By standard arguments, it is not difficult to see that lim [zh (z)]=0 and h (z)=h (z), (3.2) n n n z→0+ where h (z) is defined by (2.4). n For any fixed point z C ( ,0], choose 0 < ε < 1 and r > 0 such that 0<ε< z <r, and consid∈er th\e−po∞sitively oriented contour C(ε,r) in C ( ,0] | | \ −∞ consisting of the half circle z = εeiθ for θ π,π and the half lines z = x iε ∈ −2 2 ± for x 0 until they cut the circle z = r, which close the contour at the points ≤ | | (cid:2) (cid:3) r(ε) iε, where 0<r(ε) r as ε 0. See Figure 1. − ± → → y |z|=r −r(ε) +εi z=x+εi ε z=εeiθ x O −r(ε)−εi z=x−εi −ε −r Figure 1. The contour C(ε,r) By Cauchy integral formula, that is, Lemma 2.1, we have 1 h (w) n h (z)= dw n 2πi w z IC(ε,r) − 1 −π/2 iεeiθh εeiθ arg[−r(ε)+iε] ireiθh reiθ = dθ+ dθ (3.3) 2πi εeiθ z reiθ z (cid:20)Zπ/2 (cid:0)− (cid:1) Zarg[−r(ε)−iε] (cid:0)− (cid:1) 0 h (x+iε) −r(ε) h (x iε) n n + dx+ − dx . x+iε z x iε z Z−r(ε) − Z0 − − (cid:21) By the limit in (3.2), it follows that −π/2 iεeiθh εeiθ n lim dθ =0. (3.4) ε→0+Zπ/2 εeiθ−(cid:0) z (cid:1) 8 F.QI,X.-J.ZHANG,ANDW.-H.LI By virtue of the limit (2.3) in Lemma 2.2, we deduce that arg[−r(ε)+iε] ireiθh reiθ π ireiθh reiθ n n lim dθ = lim dθ εr→→0∞+Zarg[−r(ε)−iε] reiθ −(cid:0) z (cid:1) r→∞Z−π reiθ −(cid:0) z (cid:1) (3.5) =2A [a] a πi, n [1] − where[a] a[1] = 0,a[2] a[1],...,a[n] a[1] . Uti(cid:0)lizingthe(cid:1)secondformulain(3.2) − − − and the limit (2.5) in Lemma 2.3 results in (cid:0) (cid:1) 0 h (x+iε) −r(ε) h (x iε) n n dx+ − dx x+iε z x iε z Z−r(ε) − Z0 − − 0 h (x+iε) h (x iε) n n = − dx x+iε z − x iε z Z−r(ε)(cid:20) − − − (cid:21) 0 (x iε z)h (x+iε) (x+iε z)h (x iε) n n = − − − − − dx (x+iε z)(x iε z) Z−r(ε) − − − 0 (x z)[h (x+iε) h (x iε)] iε[h (x iε)+h (x+iε)] n n n n = − − − − − dx (x+iε z)(x iε z) Z−r(ε) − − − 0 (x z) h (x+iε) ε h (x+iε) n n =2i − ℑ − ℜ dx (x+iε z)(x iε z) Z−r(ε) − − − 0 lim h (x+iε) 2i ε→0+ℑ n dx → x z Z−r − r lim h ( t+iε) = 2i ε→0+ℑ n − dt − t+z Z0 ∞ lim h ( t+iε) 2i ε→0+ℑ n − dt →− t+z Z0 n−1 ℓπ a[ℓ+1]−a[1] n 1/n dt = 2i sin a a t (3.6) − ℓ=1 n Za[ℓ]−a[1] "k=1 [k]− [1]− # t+z X Y(cid:12) (cid:12) asε 0+ andr . Substituting e(cid:12)quations(3.4)(cid:12),(3.5), and(3.6)into (3.3)and → →∞ simplifying generate 1 n−1 ℓπ a[ℓ+1]−a[1] n 1/n dt h (z)=A [a] a sin a a t . (3.7) n n − [1] −π ℓ=1 n Za[ℓ]−a[1] "k=1 [k]− [1]− # t+z (cid:0) (cid:1) X Y(cid:12) (cid:12) From (2.2) and (2.4), it is easy to obtain that (cid:12) (cid:12) f (z)=h z+a +a . a,n n [1] [1] Combining this with (3.7) and changing(cid:0)the var(cid:1)iables of integrals, it is immediate to deduce that f (z)=A [a] a +a a,n n [1] [1] − 1(cid:0)n−1 ℓ(cid:1)π a[ℓ+1]−a[1] n 1/n dt sin a a t − π ℓ=1 n Za[ℓ]−a[1] "k=1 [k]− [1]− # t+z+a[1] X Y(cid:12) (cid:12) 1 n−1 ℓπ a[ℓ+1] (cid:12) n 1(cid:12)/n dt =A ([a]) sin a t , n − π ℓ=1 n Za[ℓ] "k=1| [k]− |# t+z X Y THE GEOMETRIC MEAN IS A BERNSTEIN FUNCTION 9 from which and the facts that n n A ([a])=A (a) and a t = a t, n n [k] k | − | | − | k=1 k=1 Y Y the integral representation (3.1) follows. Differentiating with respect to z on both sides of (3.1) yields 1 n−1 ℓπ a[ℓ+1] n 1/n dt G′ (a+z)=1+ sin a t , n π ℓ=1 n Za[ℓ] "k=1| k− |# (t+z)2 X Y which implies that G′ (a+t) is completely monotonic, and so the geometric mean n G (a+t) is a Bernstein function. Theorem 3.1 is proved. (cid:3) n 4. A new proof of the AG inequality As an application of the integral representation (3.1) in Theorem 3.1, we can easily deduce the AG inequality (1.20) as follows. Taking z =0 in the integral representation (3.1) yields 1 n−1 ℓπ a[ℓ+1] n 1/ndt G (a)=A (a) sin a t A (a), (4.1) n n k n − π ℓ=1 n Za[ℓ] "k=1| − |# t ≤ X Y from which the inequality (1.20) follows. From (4.1), it is also immediate that the equality in (1.20) is valid if and only if a = a = = a , that is, a = a = = a . The proof of the AG [1] [2] [n] 1 2 n ··· ··· inequality (1.20) is complete. References [1] R.D.Atanassov andU. 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