A Reverse HÄolder Type Inequality for the Logarithmic Mean and Generalizations John Maloney ¤ Jack Heidel y Josip Pe¸cari¶c z December 13, 1994 Abstract An inequality involving the logarithmic mean is established. Speci¯- cally, we show that ln(c=x) ln(x=a) L(c; x) ln(c=a) L(x; a) ln(c=a) < L(c; a) (1) y¡x where 0 < a < x < c and L(x; y) = ln y¡ln x ; 0 < x < y: Then several generalizations are given. Key words Logarithmic mean, inequality ¤Department of Mathematics University of Nebraska at Omaha 68182 yDepartment of Mathematics University of Nebraska at Omaha 68182 zFaculty of Textile Technology, Zagreb Croatia 1 AMS(MOS) subjeAct classi¯cation 15, 65, 68 1 Introduction The logarithmic mean y x L(y; x) = ¡ 0 < x < y; ln y ln x ¡ has many applications in statistics and economics [8]. It is well known, and easily established [1,3,6,9]that G(y; x) L(y; x) A(y; x) · · where G(y; x) = pxy is the geometric mean and A(y; x) = (x + y)=2 is the arithmetic mean. In fact, writing A(y; x) = M1(y; x) where yp + xp 1=p M (y; x) = p 2 µ ¶ it is known [6] that Mp1(y; x) Mp2(y; x) for p1 p2 It is also known, [4,5,8,11,13], that · · L(y; x) M (y; x) · 1=3 On the other hand, Hlder's inequality states that M (y y ; x x ) M (y ; x )M (y ; x ) 1 1 2 1 2 · p 1 1 q 2 2 if 1=p + 1=q = 1 with p; q > 0: . It is thus curious that the logarithmic mean L(y; x) satis¯es the inequality ln(c=x) ln(x=a) L(c; x) ln(c=a) L(x; a) ln(c=a) < L(c; a) (2) where 0 < a < x < c and it is noted that ln(c=x) ln(x=a) + = 1 ln(c=a) ln(c=a) It is the reverse HÄolder type inequality (1) which is the subject of this note and will be established below. (1) arises in a parameter identi¯cation problem for a fractal Michaelis-Mention equation [7]. In the following, use will be made of Jensen's inequality [10] which we now state for the reader's convenience: 2 1.1. Jensen's Inequality 1). if w > 0 i = 1; 2; : : : ; n i 8 2). ®0; ®1; ®2; : : : ; ®n R 3). © : [0; ) R is 2a strictly convex function then 1 ! n n w ® n w © i=1 i i w ©(® ) i n w · i i Ãi=1 ! i=1 i i=1 X µP ¶ X and the inequality is strict unless ®P0 = ®1 = ®2 = = ®n: ¢ ¢ ¢ 2 Main Result nu Lemma 2.1. Let g(u) = u 1 where g(1) = 1: Then u > 0 i). g is a strictly decreasing¡functin of u 8 ii). iii). lim g(u) = ; lim g(u) = 0; lim g(u) = 1 u 0+ 1 u u 1 iv). g!(1=u) = ug(u). !1 ! Proof: 1 1 Set z(u) = 1 1=u ln u then z 0(u) = u u 1 which is positive for 0 < u < 1 and negativ¡ e for¡ u > 1: Thus z(u) increase¡ from to 0 at u = 1 ¡ ¢ ¡1 and then decreases to as u tends to : Thus g 0(u) is negative except at u = 1. This establishes (i). The¡1 limits in (ii) can1 be computed in the usual fashion using L'hopital's rule. For (iii) we have ln(1=u) g(1=u) = = ug(u): 1=u 1 ¡ ¤ Lemma 2.2. Let f(x) = x ln x, then i). f is decreasing on (0,1) and¡ increasing on (1, ) ii). lim f(x) = ; f(1) = 1; and lim f(x) =1 x 0+ 1 x 1 iii). !if ® > 0; x > 0 then f(®x) =!1f(x) for x = g(®) so that f(®g(®)) = f(g(®)). Proof: (i) and (ii) can be established in the usual way. For (iii) we have f(® x) = f(x) ® x ln(® x) = x ln x (® 1)x = ln ® x = g(®): ) ¡ ¡ ) ¡ ) ¤ Let y(x) denote the left hand side of (1) and set ® = ln c ln a: Note that y(x) > 0 a < x < c. Then ¡ 8 ® ln y = [ln c ln x] [ln (c x) ln(ln c ln x)]+[ln x ln a] [ln(x a) ln(ln x ln a)] ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ 3 and so ® y 0 1 1 1=x = [ln(c x) ln(ln c ln x)] + [ln c ln x] ¡ ¡ + y ¡x ¡ ¡ ¡ ¡ c x ¡ ln c ln x · ¡ ¡ ¸ 1 1 1=x [ln(x a) ln(ln x ln a)] + [ln x ln a] x ¡ ¡ ¡ ¡ x a ¡ ln x ln a · ¡ ¡ ¸ 1 x a ln x ln a 1 1 ln c ln x 1 c x = ln ¡ + ¡ + ¡ ln ¡ x ln x ln a x a ¡ x x ¡ c x ¡ x ln c ln x · · ¡ ¸¸ · ¡ ¸ ¡ · ¡ ¸ 1 a=x 1 1 ln(a=x) 1 ln(c=x) 1 c=x 1 = ln x ¡ + ln x ¡ x ln(a=x) x a=x 1 ¡ x c=x 1 ¡ x ln(c=x) · ¸ · ¸ ¡ ¡ (3) or ® y 1 1 0 = [f(g(a=x)) f(g(c=x))] = h(x) (4) y x ¡ x Now f(g(a=x)) is an increasing function of x while f(g(c=x)) is a decreasing function of x so that h(x) is an increasing function of x. Clearly ®y 0=y is zero at exactly one point which implies that y 0 is zero at exactly one point. Lemma 2.3. y 0 is zero at the point x = pac. Proof: a Now f(g(c=x)) = f(g(a=x)) = f x g(a=x)) ; from lemma 2.3 (iii), so that g(c=x) = (a=x)g(a=x) = g(x=a) by lemma 2.2 (iii). Thus c=x = x=a giving ¡ ¢ x = pac. ¤ Theorem 2.1. For all values of 0 < a < x < c ln c ln x ln x ln a ln c ln a c x ¡ x a ¡ c a ¡ ¡ ¡ < ¡ (5) ln c ln x ln x ln a ln c ln a µ ¡ ¶ µ ¡ ¶ µ ¡ ¶ Proof: The results hold i® c x x a c a (ln c ln x) ¡ +(ln x ln a) ¡ < (ln c ln a) ¡ ¡ ln c ln x ¡ ln x ln a ¡ ln c ln a µ ¡ ¶ µ ¡ ¶ µ ¡ ¶ xi xi 1 Set x = a; x = x; x = c and let w = ln x ln x ; ® = ¡ ¡ 0 1 2 i i i 1 i ln xi ln xi 1 ¡ ¡ ¡ and let © (x) = ln x, the result follows from the Jensen's inequalit¡ y with rather than¡ <. But · y ® y 0 = [f(g(a=x)) f(g(c=x))] x ¡ so that y 0 is negative on [a; pac] and positive on [pac; c]: Strict inequal- ity in Theorem 2.4 now follows from the previous results since the derivative is strictly negative on [a; pac] and positive on the interval [pac; c]: Thus equality holds only at a and c. ¤ 4 3 Convexity Theorem 3.1. The function ln c ln x ln x ln a ln c¡ln a ln c¡ln a c x ¡ x a ¡ y(x) = ¡ ¡ (6) ln c ln x ln x ln a µ ¡ ¶ µ ¡ ¶ is log-convex, and hence convex, on the interval pac: Proof: Let w = ® ln y, then w 0 = ®y 0=y and hence from (5) xw 0 = f(g(a=x)) ¡ f(g(c=x)) is an increasing function so that w 0 + xw 00 0: Thus xw 00 ¸ ¸ w 0: Now on [a; pac], w 0 0, and so w 00 0 so that w is convex (and ¡ · ¸ hence log convex) on [a; pac]: ¤ Lemma 3.1. The curve ln c ln x ln x ln a ln c¡ln a ln c¡ln a c x ¡ x a ¡ y(x) = ¡ ¡ (6) ln c ln x ln x ln a µ ¡ ¶ µ ¡ ¶ is invariant under the transformation x ac=x: à Proof: ln c ln ac ln ac ln a ¡ ( x ) ( x )¡ ac ln c ln a ac ln c ln a c ¡ a ¡ z(x) = ¡ x x ¡ ln c ln ac ln ac ln a à ¡ x ! à x ¡ ! ln c ln(ac)+ln x ln(ac) ln x ln a c¡(x ¢a) ¡ln c¡ ln a¢ a(c x) ln¡c ln ¡a ¡ ¡ ¡ ¡ = x x ln c ln a ln c + ln x ln(ac) ln x ln a à ¡ ¡ ! à ¡ ¡ ! ln x ln a ln c ln x c(x a) ln c¡ln a a(c x) ln c¡ln a ¡ ¡ ¡ ¡ = x x ln x ln a ln c ln x à ¡ ! à ¡ ! ln x ln a ln c ln x c ln c¡ln a a ln c¡ln a = ¡ ¡ y(x): x x ³ ´ ³ ´ (7) Now ln x ln a ln c ln x ¡ + ¡ = 1 ln c ln a ln c ln a ¡ ¡ Thus from (7) ln x ln a ln c ln x ln x ln a ln x ln a c ln c¡ln a a ln c¡ln a c ln c¡ln a a 1 ln c¡ln a ¡ ¡ = ¡ ¡ ¡ x x x x ³ ´ ³ ´ ³ ´ln x ln a ³ ´ c ln c¡ln a x ¡ a = ln x ln a a ln c¡ln a x ¡ x ¢ ¡ a c ln(x=a)= ln(c=a) = ¡ ¢ x a ³a x´ ³ ´ = = 1 since bx = ex ln b: x a 5 Thus z(x) = y(x) and the lemma is proved.