Generalizations Of Kantorovich's Inequality Induced By Majorization Theory

Applied Mathematics E-Notes, 23(2023), 528-536 c Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/ ISSN 1607-2510 Generalizations Of Kantorovich’s Inequality Induced By Majorization Theory Huan-Nan Shiy, Dong-Sheng Wangz, Chun-Ru Fux Received 7 September 2022 Abstract In this paper, we apply Karamata’s theorem combined with majorization theory to establish a new inequality for the upper bound of the product of two …nite sums of convex functions. As applications, we derive some new generalizations of Kantorovich’s inequality. 1 Introduction and Preliminaries Let us start with some fundamental notations or de…nitions needed in this paper. The symbols R and N will denote the set of real numbers and the set of positive integers, respectively. For convenience, let Rn = R | and R R } {z n Rn++ = fx = (x1 ; : : : ; xn ) 2 Rn : xi > 0; i = 1; : : : ; ng: In particular, R1++ , simply denoted by R++ , is R++ := (0; 1). Rn is called convex if De…nition 1 ([14, 15, 16]) A set ( x1 + y1 ; : : : ; xn + yn ) 2 for any ; 2 [0; 1] with + = 1 and x = (x1 ; : : : ; xn ); y = (y1 ; : : : ; yn ) 2 . De…nition 2 ([14, 15, 16]) Let x = (x1 ; : : : ; xn ) and y = (y1 ; : : : ; yn ) 2 Rn . x is said to be majorized by y (in symbols x y) if k k X X x[i] y[i] for 1 k n 1, i=1 and i=1 n X i=1 where x[1] x[n] and y[1] xi = n X yi , i=1 y[n] are rearrangements of x and y in descending order. De…nition 3 ([2]) Let f : I ! R and g : I ! R be two real-valued functions on an interval I. f and g are said to be Mathematics Subject Classi…cations: 26D15, 26A51, 26A45. of Electronic Information, Teacher’s College, Beijing Union University, Beijing City, 100011, P.R.China z Basic courses department, Beijing Polytechnic, Beijing 100176, China x Institute of Mathematics and Physics, Beijing Union University, Beijing 100101, China, Corresponding author. y Department 528 529 Shi et al. (i) similarly ordered if (f (x) f (y))(g(x) g(y)) 0 for every x; y 2 I; (f (x) f (y))(g(x) g(y)) 0 for every x; y 2 I (f (x) f (y))(g(y) g(x)) 0 for every x; y 2 I: (ii) oppositely ordered if or, The estimation of the upper bound of the product of two …nite sum is a long lasting mathematical subject. Pólya-Szegö established a famous inequality as follows: Theorem 1 (Pólya-Szegö [1, 2]) Let 0 < m1 n X a2k k=1 n X 1 4 b2k k=1 r M1 and 0 < m2 ak M1 M2 + m1 m2 r m1 m 2 M1 M 2 !2 n X k=1 bk ak bk M2 (k = 1; : : : ; n). Then !2 (1) : In [3], an upper bound inequality equivalent to inequality (1) was introduced: Theorem 2 (Kantorovich inequality ) Let fxk g, k = 1; : : : ; n be any real number sequence. If 0 < m xk M , i = 1; : : : ; n, then ! ! n n (M + m)2 1X 1X 1 : xk n n xk 4M m k=1 k=1 Kantorovich inequality is a well-known inequality, this inequality is useful in numerical analysis and statistics, especially in the method of steepest descent. Therefore, it is valuable for its generalization and application. Over the years, various variations and extensions of this inequality have been investigated by many authors in several contexts. Reference [3]–[13] has many forms of generalizations and applications. In 2005, Xu [5] proved the following generalized Kantorovich inequality. Theorem 3 ([5]) Let > 0. If 0 < m xk M , i = 1; : : : ; n, then ! ! n n 1X 1X 1 (M +1 m (M m )(M m) xi n i=1 n i=1 xi 4(M m) +1 2 ) : The following lemmas are important and will be used for proving our main results. Lemma 1 ([16, 17]) Let m xi M; i = 1; : : : ; n; n and unique integer k 2 f0; 1; : : : ; ng such that n X xi = (n 2, and m 6= M . Then there is a unique l 2 [m; M ) 1)m + l + kM; k i=1 where l, k is determined by (x1 ; : : : ; xn ) (M; : : : ; M ; l; m; : : : ; m): | {z } | {z } k Remark 1 (i) Because l = So we can determine k. Pn i=1 xi (n k 1)m Pn nm i=1 xi 1 M m n k 1 kM 2 [m; M ), we see that Pn nm i=1 xi k : M m 530 Generalizations of Kantorovich’s Inequality (ii) According to the proof of Lemma 1 in reference [17], we know that l is a variable that depends on x1 ; : : : ; xn and m l < M . Lemma 2 ([16, 17]) Let x = (x1 ; : : : ; xn ) 2 Rn , y = (y1 ; : : : ; yn ) 2 Rn and g is convex function on I If x y, then n n X X g(yi ): g(xi ) R. i=1 i=1 Kontenovich inequality is a famous inverse inequality of the famous Cauchy-Schwarz inequality, from Theorem 2 and Theorem 3 we observe that the functions corresponding to the two sequences are respectively: x and x1 , x and x1 ; ( > 0). They have the common oppositely ordered pproperty, and x1 , x1 is convex function. So this tells us, for two general functions, when they are oppositely ordered and convex, there may be results similar to Theorem 2 and Theorem 3. 2 New Inequalities for Di¤erentiable Convex Functions In this section, we establish the following new inequalities which will be applied to established new generalizations of Kantorovich’s inequality and other new results. Theorem 4 Let 0 < m xi M for i = 1; : : : ; n. Let f and g be two nonnegative convex functions and have second derivatives on [m, M ]. Suppose that (H1) (f (M ) f (m))(g(m) g(M )) 0, and (H2) [kf (M ) + (n k 1)f (m) + f (x)]g 00 (x) + [kg(M ) + (n m x M and 1 k n 1. k 1)g(m) + g(x)]f 00 (x) + 2f 0 (x)g 0 (x) 0 for Then n X i=1 ! f (xi ) n X i=1 ! g(xi ) (f (M ) f (m))(g(m) g(M )) n [f (M )g(m) f (m)g(M )] 2 2 : (2) Proof. In order to prove our conclusion, we consider the following two possible cases: Case 1. If m = M , then (2) is obvious. Case 2. Suppose m < M . If (f (M ) f (m))(g(m) g(M )) = 0, then the conclusion also holds immediately. Hence we may assume that (f (M ) f (m))(g(m) g(M )) > 0. By Lemma 1, there exists k 2 N with 1 k n 1, such that (x1 ; : : : ; xn ) (M; : : : ; M ; l; m; : : : ; m): | {z } | {z } n k 1 k According to the Remark 1, we know that l is a variable that depends on x1 ; : : : ; xn and m Because f and g are nonnegative convex functions, by Lemma 2, we have n X f (xi ) kf (M ) + (n k 1)f (m) + f (l); kg(M ) + (n k 1)g(m) + g(l): i=1 and n X g(xi ) i=1 Let ha (b) := [af (M ) + (n a 1)f (m) + f (b)] [ag(M ) + (n a 1)g(m) + g(b)] l M. 531 Shi et al. for (a; b) 2 N [m; M ]. So, we get n X f (xi ) n X g(xi ) hk (l): i=1 i=1 Clearly, for any …xed a, ha is a function of the variable b. Taking the derivative of ha with respect to the variable b, we have h0a (b) = f 0 (b)[ag(M ) + (n a 1)g(m) + g(b)] + g 0 (b)[af (M ) + (n 1)f (m) + f (b)]; a and h00a (b) = f 00 (b)[ag(M ) + (n a 1)g(m) + g(b)] +g 00 (b)[af (M ) + (n a 1)f (m) + f (b)] + 2f 0 (b)g 0 (b) 0: Thus for any a, ha is a convex function on [m; M ]. Since l 2 [m; M ], we see that hk (l) hk (l) hk (M ), where hk (m) = [kf (M ) + (n = [kf (M ) + (n k 1)f (m) + f (m)][kg(M ) + (n k)f (m)] [kg(M ) + (n k)g(m)] = f (M )g(M )k 2 + k(n = '(k); k)f (M )g(m) + k(n k hk (m) or 1)g(m) + g(m)] k)2 f (m)g(m) k)f (m)g(M ) + (n and '(k) = (f (M ) g(m))k 2 + n(f (M )g(m) + f (m)g(M ) f (m))(g(M ) 2f (m)g(m))k + n2 f (m)g(m): Next, we will …nd the maximum value of '. (f (M ) f (m))(g(m) g(M )) 0, ' has a maximum value. Since 0 ' (k) = 2(f (M ) f (m)) (g(M ) g(m))k + n(f (M )g(m) + f (m)g(M ) 2f (m)g(m)), 0 if ' (w) = 0, then we obtain w= 2nf (m)g(m) n(f (M )g(m) + f (m)g(M )) A = . 2(f (M ) f (m))(g(M ) g(m)) 2B Hence ' has maximum value '(w) = ' A 2B =B A 2B 2 A A 4BC A2 +C = ; 2B 4B where A = 2nf (m)g(m) B = (f (M ) n(f (M )g(m) + f (m)g(M )); f (m))(g(M ) g(m)); 2 C = n f (m)g(m): On the other hand, since hk (M ) = [kf (M ) + (n k = [(k + 1)f (M ) + (n = hk+1 (m); 1)f (m) + f (M )] [kg(M ) + (n k 1)g(m) + g(M )] k 1)f (m)] [(k + 1)g(M ) + (n k the maximum value of hk (M ) is the same as that of hk (m). So, we have n X i=1 f (xi ) n X i=1 g(xi ) hk (l) hk (m) 4BC A2 : 4B 1)g(m)] 532 Generalizations of Kantorovich’s Inequality Since 4BC = 4(f (M )g(M ) f (M )g(m) = 4n2 f (m)g(m)f (M )g(M ) f (m)g(M ) + f (m)g(m))n2 f (m)g(m) 4n2 f (m)f (M )g 2 (m) 4n2 f 2 (m)g(m)g(M ) + 4n2 f 2 (m)g 2 (m); and A2 = 4n2 f 2 (m)g 2 (m) 4n2 f (m)f (M )g 2 (m) 4n2 f 2 (m)g(m)g(M ) + n2 f 2 (M )g 2 (m) + n2 f 2 (m)g 2 (M ) + 2n2 f (m)f (M )g(m)g(M ); we get 4BC and hence A2 = n2 (f (M )g(m) f (m)g(M ))2 n2 (f (M )g(m) f (m)g(M ))2 4BC A2 = 4B 4(f (M ) f (m))(g(M ) g(m)) 0: Therefore we show that n X f (xi ) n X g(xi ) i=1 i=1 n2 (f (M )g(m) f (m)g(M ))2 ; 4(f (M ) f (m))(g(M ) g(m)) or, equivalence, n X i=1 ! f (xi ) n X ! g(xi ) (f (M ) i=1 f (m))(g(m) n (f (M )g(m) f (m)g(M )) 2 g(M )) 2 : The proof is completed. As a consequence of Theorem 4, we can obtain the following generalized Kantorovich’s inequality. Corollary 1 Let 0 < m xi M for i = 1; : : : ; n, and let f and g be two nonnegative convex functions and have second derivatives on [m, M ]. If f; g are oppositely ordered on [m, M ] and (f g)00 0 on [m, M ], then ! ! n n 2 X X n (f (M )g(m) f (m)g(M )) : g(xi ) (f (M ) f (m))(g(m) g(M )) f (xi ) 2 i=1 i=1 Proof. Since f; g are oppositely ordered on [m, M ], we have (f (M ) f (m))(g(m) g(M )) 0 and condition (H1) as in Theorem 4 is proved. We conclude that condition (H2) as in Theorem 4 holds. In fact note that (f g)00 0 on [m; M ] if and only if f (x)g 00 (x) + f 00 (x)g(x) + 2f 0 (x)g 0 (x) For m x M and 1 k n 0 for x 2 [m; M ]. 1, we obtain [kf (M ) + (n k 1)f (m) + f (x)]g 00 (x) +[kg(M ) + (n k 1)g(m) + g(x)]f 00 (x) + 2f 0 (x)g 0 (x) = [kf (M ) + (n k 1)f (m)]g 00 (x) + [kg(M ) + (n k 1)g(m)]f 00 (x) +f (x)g 00 (x) + g(x)f 00 (x) + 2f 0 (x)g 0 (x) 0: Therefore all the conditions of Theorem 4 are satis…ed and the conclusion follows from Theorem 4 immediately. 533 Shi et al. 3 New generalizations of Kantorovich’s inequality Applying Theorem 4 or Corollary 1, we can obtain new exponential generalizations of Kantorovich’s inequality. Theorem 5 Let 0 < m (i) = 1 and (ii) > 1 and xi M for i = 1; : : : ; n. If one of the following conditions is satis…ed: < 0; < 0 with + 1 or + 0, then n 1X x n i=1 i ! n 1X x n i=1 i ! (M m )(m M m M ) m M 2 : 2 (3) Proof. (i) Suppose that = 1 and < 0. Let f (x) = x and g(x) = x for x > 0. Then f (x) is an increasing convex function and g(x) is a decreasing convex function. Clearly, (f (M ) f (m))(g(m) g(M )) 0. Let m x M and 1 k n 1. Set u = kM + (n k 1)m. Thus u M . Since [kf (M ) + (n k 1)f (m) + f (x)]g 00 (x) +[kg(M ) + (n k 1)g(m) + g(x)]f 00 (x) + 2f 0 (x)g 0 (x) = [kM + (n = = (u + x) ( 1)x 2 x [(u + x)(1 = and m x M 2 x [u( k 1)m + x] ( 2 1)l +2 x ) 2x] + 1) + ( 2 +2 x 1 1 1)x] u, we get [kf (M ) + (n k 1)f (m) + f (x)]g 00 (x) + [kg(M ) + (n k 1)g(m) + g(x)]f 00 (x) + 2f 0 (x)g 0 (x) 0. Hence all the conditions of Theorem 4 are satis…ed. By Theorem 4, we have ! ! n n M m m M 1X 1X m )(m M ) x x (M n i=1 i n i=1 i 2 2 : 00 (ii) Let '(x) = x for x > 0. Then ' (x) = ( 1)x 2 for x > 0. It is easy to see that when 1 or < 0, h(x) is a convex function on R++ . Suppose that > 1and < 0 with + 1 or + 0. Let f (x) = x and g(x) = x for x > 0. Then f (x) is an increasing convex function and g(x) is a decreasing convex function. Obviously, (f (M ) f (m))(g(m) g(M )) 0. Direct calculation gives (f (x)g(x))00 = f (x)g 00 (x) + g(x)f 00 (x) + 2f 0 (x)g 0 (x) = ( 1)x + = ( + )( + 2 + ( 1)x 1)x + 2 + 2 +2 x + 2 0: Therefore all the conditions of Corollary 1 are satis…ed and the conclusion follows immediately from Corollary 1. The proof is completed. Remark 2 (a) Taking = 1 and < 0 in Theorem 5, we can get Theorem 3. 534 Generalizations of Kantorovich’s Inequality (b) Let 0 < m M for i = 1; : : : ; n. For any xi An (x)(M > 0, by Theorem 5, we obtain m ) m)(M (M +1 m 4M m +1 2 ) Hn (x ); where An (x) is the arithmetic mean of x1 ; ; xn and Hn (x ) is the harmonic mean of x1 ; In particular, if we take = 1 in last inequality, then (M + m)2 Hn (x): 4M m An (x) Theorem 6 Let 0 < m M for i = 1; : : : ; n. If one of the following conditions is satis…ed: xi 2M p 1; (i) p > 1 and (ii) p > 1 and (iii) p ; xn . 0; m< 1 and 0, then n 1X p x n i=1 i ! n 1X 1 n i=1 ( + xi )p ! (M p mp ) [( + M )p ( + m)p ] 2 [M p ( + M )p mp ( + m)p ] : 4( + m)p ( + M )p Proof. We only verify (i) and (ii), and a similar argument could be made for proving (iii). Let f (x) = xp 2M m < 0, then direct and g(x) = ( + x) p for 0 < x M . If p > 1 and p 1 , or p > 1 and calculation give f 0 (x) = pxp 1 > 0, g 0 (x) = p( + x) p 1 < 0, f 00 (x) = p(p 1)xp 2 > 0, g 00 (x) = p(p + 1)( + x) p 2 > 0, and (f (x)g(x))00 = f 00 (x)g(x) + g 00 (x)f (x) + 2f 0 (x)g 0 (x) = p(p 1) xp 2 xp + p(p + 1) ( + x)p ( + x)p+2 xp 2 (p 1)( + x)2 + (p + 1)x2 ( + x)p+2 xp 2 =p [(p 1) 2x]: ( + x)p+2 =p If p > 1 and m< So, we get (f (x)g(x))00 = p 2M p 1, xp 1 ( + x)p+1 2px( + x) 0, we have ( + x)p+2 > 0, p If p > 1 and 2p2 since x (f (x)g(x))00 = p 0 and (p 1) 2x < 0: xp 2 [(p ( + x)p+2 1) 2x] 0: M , we obtain xp 2 [(p ( + x)p+2 1) 2x] p xp 2 (2M ( + x)p+2 2x) 0: 535 Shi et al. On the other hand, since f (M ) = M p , f (m) = mp , g(M ) = it is not hard to show that (f (M ) immediately from Corollary 1. The proof is completed. Remark 3 In fact, by taking p = f (m))(g(m) 1 1 and g(m) = ; p ( + M) ( + m)p g(M )) 0. Therefore the desired conclusion follows 1 and = 0 in Theorem 6, we can obtain ! ! n n 1X 1 (M + m)2 1X xk ; n n xk 4M m k=1 k=1 which is the original Kantorovich inequality. Hence, Theorem 6 is a real generalization of Kantorovich inequality. 4 Conclusions In this paper, we use Karamata’s theorem combined with majorization theory to establish an inequality for the upper bound of the product of two …nite sums about convex function. As applications, we derive some new generalizations of Kantorovich’s inequality. Acknowledgements. The authors are indebted to professor Wei-Shih Du and the referees for many helpful and valuable comments and suggestions. References should be typed as follows: References [1] D. S. Mitrinović, Analytic Inequalities, Springer-Verlag Berlin. Heidelberg. New York, 1970. [2] J. C. 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