ON AN INEQUALITY OF V. CSISZÁR AND T.F. MÓRI FOR CONCAVE FUNCTIONS OF TWO VARIABLES

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1 ON AN INEQUALITY OF V. CSISZÁR AND T.F. MÓRI FOR CONCAVE FUNCTIONS OF TWO VARIABLES BOŽIDAR IVANKOVIĆ Faculty of Transport and Traffic Engineering University of Zagreb, Vukelićeva Zagreb, Croatia ivankovb@fpz.hr JOSIP E. PEČARIĆ Faculty of Textile Technology University of Zagreb, Pierottijeva Zagreb, Croatia pecaric@mahazu.hazu.hr SAICHI IZUMINO Faculty of Education Toyama University Gofuku, Toyama , JAPAN s-izumino@h5.dion.ne.jp MASARU TOMINAGA Toyama National College of Technology 3, Hongo-machi, Toyama-shi , Japan mtommy@toyama-nct.ac.jp Received: 7 September, 006 Accepted: April, 007 Communicated by: I. Pinelis 000 AMS Sub. Class.: 6D5. Key words: Abstract: Diaz-Metcalf inequality, Hölder s inequality, Hadamard s inequality, Petrović s inequality, Giaccardi s inequality. V. Csiszár gave an extension of Diaz-Metcalf s inequality for concave functions. In this paper, we show its restatement. As its applications we first give a reverse inequality of Hölder s inequality. Next we consider two variable versions of Hadamard, Petrović and Giaccardi inequalities. Page of 8

2 Introduction 3 Reverse Hölder s Inequality 5 3 Hadamard s Inequality 9 4 Petrović s Inequality 5 Giaccardi s Inequality 4 Page of 8

3 . Introduction In this paper, let (X, Y be a random vector with P [(X, Y D] = where D := [a, A] [b, B] (0 a < A and 0 b < B. Let E[X] be the expectation of a random variable X with respect to P. For a function φ : D R, we put φ = φ(a, b, A, B := φ(a, b φ(a, B φ(a, b + φ(a, B. In [], V. Csiszár showed the following theorem as an extension of Diaz-Metcalf s inequality []. Theorem A. Let φ : D R be a concave function. We use the following notations: φ(a, b φ(a, b φ(a, B φ(a, b λ = λ 4 :=, µ = µ 3 :=, A a B b φ(a, B φ(a, B φ(a, B φ(a, b λ = λ 3 :=, µ = µ 4 :=, and A a B b AB ab b ν := φ(a, b φ(a, B a φ(a, b, (A a(b b B b A a ν := A φ(a, B + B AB ab φ(a, b φ(a, B, A a B b (A a(b b ν 3 := B φ(a, b a ab Ab φ(a, B + φ(a, B, B b A a (A a(b b ν 4 := A φ(a, b b ab Ab φ(a, B φ(a, b. A a B b (A a(b b a Suppose that φ 0. Page 3 of 8

4 a (i If (B be[x] + (A ae[y ] AB ab, then λ E[X] + µ E[Y ] + ν E[φ(X, Y ] ( φ(e[x], E[Y ]. a (ii If (B be[x] + (A ae[y ] AB ab, then λ E[X] + µ E[Y ] + ν E[φ(X, Y ] ( φ(e[x], E[Y ]. b Suppose that φ 0 b (iii If (B be[x] + (A ae[y ] ab Ab, then λ 3 E[X] + µ 3 E[Y ] + ν 3 E[φ(X, Y ] ( φ(e[x], E[Y ]. b (iv If (B be[x] + (A ae[y ] ab Ab, then λ 4 E[X] + µ 4 E[Y ] + ν 4 E[φ(X, Y ] ( φ(e[x], E[Y ]. Let us note that Theorem A can be given in the following form: Theorem.. Suppose that φ : D R is a concave function. a If φ 0, then (. max k=, {λ ke[x] + µ k E[Y ] + ν k } E[φ(X, Y ]( φ(e[x], E[Y ], where λ k, µ k and ν k (k =, are defined in Theorem A. b If φ 0,then (. max k=3,4 {λ ke[x] + µ k E[Y ] + ν k } E[φ(X, Y ]( φ(e[x], E[Y ], where λ k, µ k and ν k (k = 3, 4 are defined in Theorem A. Remark. The inequality E[φ(X, Y ] φ(e[x], E[Y ] is Jensen s inequality. So the inequalities in Theorem A represent reverse inequalities of it. In this note, we shall give some applications of these results. Page 4 of 8

5 . Reverse Hölder s Inequality Let p, q > be real numbers with + =. Then φ(x, y := x p p q y q is a concave function on (0, (0,. For 0 < a < A and 0 < b < B, φ is represented as follows: φ = a p b q a p B q A p b q + A p B q = ( A p a p (B q b q (> 0. Moreover, putting A = B =, and replacing X, Y, a and b by X p, Y q, α p and β q, respectively, in Theorem A, we have the following result: Theorem.. Let p, q > be real numbers with + =. Let 0 < α X p q and 0 < β Y. (i If ( β q E[X p ] + ( α p E[Y q ] α p β q, then (. β( α α( β α p E[Xp ] + E[Y q ] β q + αβ( αp β q + α p β q + α p β q α p β q ( α p ( β q (ii If ( β q E[X p ] + ( α p E[Y q ] α p β q, then (. α α p E[Xp ]+ β β E[Y q ] α β + αβq + α p β α p β q q ( α p ( β q E[XY ]. E[XY ]. By Theorem. we have the following inequality related to Hölder s inequality: Page 5 of 8

6 Theorem.. Let p, q > be real numbers with + =. If 0 < α X and p q 0 < β Y, then (.3 p p q q (β αβ q p (α α p β q E[X p ] p E[Y q ] q Proof. We have by Young s inequality (β αβ q E[X p ] + (α α p βe[y q ] ( α p β q E[XY ]. (β αβ q E[X p ] + (α α p βe[y q ] = p p(β αβq E[X p ] + q q(α αp βe[y q ] {p(β αβ q E[X p ]} p {q(α α p βe[y q ]} q = p p q q (β αβ q p (α α p β q E[X p ] p E[Y q ] q. Hence the first inequality holds. Next, we see that (.4 and (.5 γ := αβ( αp β q + α p β q + α p β q α p β q ( α p ( β q γ := α β + αβq + α p β α p β q ( α p ( β q 0. 0 Indeed, we have ( α p ( β q > 0 and moreover by Young s inequality α p β q + α p β q + α p β q α p β q = α p β q α p ( β q β q ( α p ( α p β q p + q αp ( β q ( q + p βq ( α p = 0 Page 6 of 8

7 and and α β + αβ q + α p β α p β q = α p β q α( β q β( α p ( α p β q q + p αp ( β q ( p + q βq ( α p = 0. Multiplying both sides of (. by γ and those of (. by γ, respectively, and taking the sum of the two inequalities, we have β( α γ α( β α p γ β E[Xp ] + q E[Y q ] (γ γ E[XY ]. α α p γ β β q γ Here we note that from (.4 and (.5, β( α γ α p α γ α p = β( α( β( αβq, ( α p ( β q α( β γ β q = α( α( β( αp β ( α p ( β q Hence we have β β q γ β( α( β( αβ q ( α p ( β q γ γ = ( α( β( αp β q. ( α p ( β q E[X p ] + α( α( β( αp β E[Y q ] ( α p ( β q Page 7 of 8

8 ( α( β( αp β q E[XY ] ( α p ( β q and so the second inequality of (.3 holds. The second inequality is given in [5, p.4]. In (.3, the first and the third terms yield the following Gheorghiu inequality [4, p.84], [5, p.4]: Theorem B. Let p, q > be real numbers with + =. If 0 < α X and p q 0 < β Y, then (.6 E[X p ] p E[Y q ] q α p β q p p q q (β αβq p (α αp β q E[XY ]. We see that (.3 is a kind of a refinement of (.6. Theorem B gives us the next estimation. Corollary.3. Let X = {a i } and Y = {b j } be independent discrete random variables with distributions P (X = a i = w i and P (Y = b j = z j. Suppose 0 < α X and 0 < β Y. E[X p ], E[Y q ] and E[XY ] are given by n i= w ia p i, n i= z jb q j and n n i= j= w iz j a i b j, respectively. Then we have inequalities Page 8 of 8 i= ( w i z j a i b j w i a p i j= i= ( p z j b q j j= α p β q q p p q q (β αβq p (α αp β q w i z j a i b j. i= j=

9 3. Hadamard s Inequality The following well-known inequality is due to Hadamard [5, p.]: For a concave function f : [a, b] R, f(a + f(b (3. b ( a + b f(tdt f. b a a Moreover, the following is an extension of the weighted version of Hadamard s inequality by Fejér ([3], [6, p.38]: Let g be a positive integrable function on [a, b] with g(a + t = g(b t for 0 t (a b. Then f(a + f(b b b ( a + b b (3. g(tdt f(tg(tdt f g(tdt. a Here we give an analogous result for a function of two variables. Theorem 3.. Let X and Y be independent random variables such that a (3.3 E[X] = a + A and E[Y ] = b + B for 0 < a X A and 0 < b Y B. If φ : D R is a concave function, then { } φ(a, b + φ(a, B φ(a, b + φ(a, B (3.4 min, E[φ(X, Y ] ( a + A φ, b + B. Proof. We only have to prove the case φ 0. Then with same notations as in Theorem A we have φ(a, b + φ(a, B λ E[X] + µ E[Y ] + ν = λ E[X] + µ E[Y ] + ν = a Page 9 of 8

10 by (3.3. Since φ 0, it is the same as the first expression in (3.4. Similarly calculation for φ 0 proves that the desired inequality (3.4 also holds. We can obtain the following result as an extension of Hadamard s inequality (3. from Theorem 3. by letting X and Y be independent, uniformly distrbuted radom variables on the intervals [a, A] and [b, B], respectively: Corollary 3.. Let 0 < a < A and 0 < b < B. If φ is a concave function, then { } φ(a, b + φ(a, B φ(a, b + φ(a, B A B min, φ(t, s dsdt (A a(b b a b ( a + A φ, b + B. By Theorem 3., we have the following analogue of (3. for a function of two variables: Corollary 3.3. Let w : D R be a nonnegative integrable function such that w(s, t = u(sv(t where u : [a, A] R is an integrable function with u(s = u(a + A s, A u(sds = and v : [b, B] R is an integrable function such that a v(tdt =, v(t = v(b + B t. If φ is a concave function, then B b { φ(a, b + φ(a, B min, } φ(a, b + φ(a, B A B a b ( a + A φ w(s, tφ(s, t dsdt., b + B Page 0 of 8

11 4. Petrović s Inequality The following is called Petrović s inequality for a concave function f : [0, c] R: ( f p i x i p i f(x i + ( P n f(0, i= i= where x = (x,..., x n and p = (p,..., p n are n-tuples of nonnegative real numbers such that n i= p ix i x k for k =,..., n, n i= p ix i [0, c] and P n := n i= p i (see [5, p.] and [6]. We give an analogous result for a function of two variables. Theorem 4.. Let p = (p,..., p n and q = (q,..., q n be n-tuples of nonnegative real numbers and put P n := n i= p i (> 0 and Q n := n j= q j (> 0. Suppose that x = (x,..., x n and y = (y,..., y n are n-tuples of nonnegative real numbers with 0 x k n i= p ix i c and 0 y k n j= q jy j d for k =,,..., n. Let φ : [0, c] [0, d] R be a concave function. a Suppose ( ( ( φ(0, 0 + φ p i x i, q j y j φ p i x i, 0 + φ 0, q j y j. i= j= a (i If P n + Q n, then ( ( (4. φ p i x i, 0 + φ 0, P n Q n i= i= q j y j + j= j= ( Pn Qn φ(0, 0 P n Q n i= p i q j φ(x i, y j. j= Page of 8

12 a (ii If P n + Q n, then ( + ( φ p i x i, P n Q n i= ( + ( Qn φ b Suppose ( φ(0, 0 + φ p i x i, i= i= q j y j j= p i x i, 0 + ( ( Pn φ 0, P n Q n i= q j y j j= p i q j φ(x i, y j. j= ( ( q j y j φ p i x i, 0 + φ 0, j= b (iii If P n Q n, then ( φ p i x i, q j y j P n i= j= ( + ( φ 0, Q n P n i= q j y j + j= q j y j. j= ( Qn φ(0, 0 P n Q n i= p i q j φ(x i, y j. j= Page of 8

13 b (iv If Q n P n, then ( φ p i x i, Q n i= j= ( q j y j Q n P n ( φ p i x i, 0 i= + ( Pn φ(0, 0 P n Q n i= p i q j φ(x i, y j. Proof. We put a = b = 0, A = n i= p ix i and B = n j= q jy j in Theorem A. Let X = {a i } and Y = {b j } be independent discrete random variables with distributions P (X = x i = p i P n and P (Y = y j = q i Q n, i n, respectively. So we have the desired inequalities. Specially, if p i = q j = (i, j =,..., n in Theorem 4., then we have the following: Corollary 4.. Suppose that x = (x,..., x n and y = (y,..., y n are n-tuples of nonnegative real numbers for n with n i= x i [0, c] and n i= y i [0, d]. If φ : [0, c] [0, d] R is a concave function, then ( ( (4. φ x i, 0 + φ 0, y j + (n φ(0, 0 φ(x i, y j. n i= j= j= i= j= Page 3 of 8

14 5. Giaccardi s Inequality In 955, Giaccardi (cf. [5, p.] proved the following inequality for a convex function f : [a, A] R, ( p i f(x i C f p i x i + D (P n f(x 0, where i= C = i= n i= p i(x i x 0 n i= p ix i x 0 and D = n i= p ix i n i= p ix i x 0 for a nonnegative n-tuple p = (p,..., p n with P n := n i= p i and a real (n + - tuple x = (x 0, x,..., x n such that for k = 0,,..., n ( a x i A, (x k x 0 p i x i x 0 0, a < p i x i < A i= and i= p i x i x 0. In this section, we discuss a generalization of Giaccardi s inequality to a function of two variables under similar conditions. Let x = (x 0, x,..., x n and y = (y 0, y,..., y n be nonnegative (n + -tuples, and p = (p, p,..., p n and q = (q, q,..., q n be nonnegative n-tuples with (5. x 0 x k i= p i x i and y 0 y k i= q j y j for k =,..., n. j= Page 4 of 8

15 We use the following notations: P n := p i ( 0, Q n := i= q j ( 0, j= and := M(X, Y := N(X, Y K(X := n i= p ix i P n x 0 n i= p ix i x 0, K(Y := n j= q jy j Q n y 0 n j= q jy j y 0, L(X := (P n n i= p ix i n i= p, L(Y := (Q n n j= q jy j ix i x n 0 j= q, jy j y 0 { (P n Q n n p i x i i= (P n Q n P n Q n p i x i i= + Q n y 0 p i x i + P n x 0 i= q j y j j= } q j y j P n Q n x 0 y 0 j= ( ( n i= p n ix i x 0 j= q jy j y 0 q j y j (P n Q n p i x i y 0 + P n (Q n x 0 q j y j i= j= ( n (. p i x i x 0 q j y j y 0 i= j= j= Page 5 of 8

16 Then we have the following theorem: Theorem 5.. Let φ : [x 0, n i= p ix i ] [y 0, n j= q jy j ] R be a concave function. a If ( ( ( φ(x 0, y 0 + φ p i x i, q j y j φ x 0, q j y j + φ p i x i, y 0, i= j= j= then { ( ( max Q n K(Xφ p i x i, y 0 + P n K(Y φ x 0, b If i= i= q j y j + M(X, Y φ(x 0, y 0, j= ( ( P n L(Y φ p i x i, y 0 + Q n L(Xφ x 0, ( φ(x 0, y 0 + φ p i x i, i= ( M(X, Y φ p i x i, i= i= } q j y j j= ( q j y j φ x 0, j= q j y j j= i= p i q j φ(x i, y j. j= ( q j y j + φ p i x i, y 0, j= i= Page 6 of 8

17 then { ( max Q n K(Xφ p i x i, i= q j y j j= + P n L(Y φ (x 0, y 0 + N(X, Y φ ( P n K(Y φ p i x i, i= ( x 0, q j y j, j= q j y j + Q n L(Xφ (x 0, y 0 j= ( } N(X, Y φ p i x i, y 0 i= i= p i q j φ(x i, y j. Proof. Let X and Y be as they were in the proof of Theorem 4., and put a = x 0, A = n i= p ix i, b = y 0 and B = n j= q jy j, and use Theorem A. Then we have the desired inequalities of this theorem. j= Page 7 of 8

18 References [] V. CSISZÁR AND T.F. MÓRI, The convexity method of proving moment-type inequalities, Statist. Probab. Lett., 66 (004, [] J.B. DIAZ AND F.T. METCALF, Stronger forms of a class of inequalities of G. Pólya-G. Szegö, and L. V. Kantorovich, Bull. Amer. Math. Soc., 69 (963, [3] L. FEJÉR, Über die Fourierreihen, II., Math. Naturwiss, Ant. Ungar. Acad. Wiss, 4 (906, (in Hungarian. [4] S. IZUMINO AND M. TOMINAGA, Estimations in Hölder s type inequalities, Math. Inequal. Appl., 4 (00, [5] D. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer. Acad. Pub., Boston, London 993. [6] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Mathematics in Science and Engineering, Georgia Institute of Technology, 99. Page 8 of 8