A REFINEMENT OF HÖLDER’S INEQUALITY AND APPLICATIONS XUEMEI GAO, MINGZHE GAO AND XIAOZHOU SHANG Department of Mathematics and Computer Science Normal College, Jishou University Jishou Hunan 416000, People’s Republic of China EMail: mingzhegao@163.com Received: 10 February, 2007 Accepted: 10 June, 2007 Communicated by: Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II S.S. Dragomir J I 2000 AMS Sub. Class.: 26D15, 46C99. Page 1 of 19 Key words: Inner product space, Gram matrix, Variable unit-vector, Minkowski’s inequality, Fan Ky’s inequality, Hardy’s inequality. Abstract: In this paper, it is shown that a refinement of Hölder’s inequality can be established using the positive definiteness of the Gram matrix. As applications, some improvements on Minkowski’s inequality, Fan Ky’s inequality and Hardy’s inequality are given. Acknowledgements: The authors would like to thank the anonymous referee for valuable comments that have been implemented in the final version of this paper. A Project Supported by scientific Research Fund of Hunan Provincial Education Department (06C657). Go Back Full Screen Close Contents 1 Introduction 3 2 Main Results 5 3 Applications 3.1 A Refinement of Minkowski’s Inequality . . . . . . . . . . . . . . . 3.2 A Strengthening of Fan Ky’s Inequality . . . . . . . . . . . . . . . 3.3 An Improvement of Hardy’s Inequality . . . . . . . . . . . . . . . . 9 9 12 15 Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 2 of 19 Go Back Full Screen Close 1. Introduction For convenience, we need to introduce the following notations which will be frequently used throughout the paper: (ar , bs ) = ∞ X arn bsn , kakr = n=1 r s Z (f , g ) = arn Z s f (x) g (x) dx, 0 and ! r1 , kak2 = kak , Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and n=1 ∞ r ∞ X ∞ kf kr = r f (x) dx r1 Xiaozhou Shang , kf k2 = kf k , vol. 8, iss. 2, art. 44, 2007 0 Sr (α, y) = αr/2 , y kαk−r/2 , r Title Page where a = (a1 , a2 , . . . ) are sequences of real numbers, f : [0, ∞) → [0, ∞) are measurable functions and α and y are elements of an inner product space E of real sequences. Let a = (a1 , a2 , . . . ) and b = (b1 , b2 , . . . ) be sequences of real numbers in Rn . Then Hölder’s inequality can be written in the form: Contents JJ II J I Page 3 of 19 (a, b) ≤ kakp kbkq . Go Back The equality in (1.1) holds if and only if api = kbqi , i = 1, 2, . . . , where k is a constant. This inequality is important in function theory, functional analysis, Fourier analysis and analytic number theory, etc. However, there are drawbacks in this inequality. For example, let Full Screen (1.1) a = (a1 , a2 , . . . , an , 0, . . . , 0) , b = (0, 0, . . . , bn+1 , bn+2 , . . . , b2n ) , a, b ∈ R2n . Close If we let ai = bj = 1, i = 1, 2, . . . , n; j = n + 1, n + 2, . . . , 2n, and substitute them into (1.1), then we have 0 ≤ n. In this case, Hölder’s inequality is meaningless. In the present paper we establish a new inequality that improves Hölder’s inequality and remedies the defect pointed out above. At the same time, some significant refinements for a number of the classical inequalities can be established. As space is limited, only several applications of the new inequality are given. Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 4 of 19 Go Back Full Screen Close 2. Main Results Let α and β be elements of an inner product space E. Then the inner p product of α and β is denoted by (α, β) and the norm of α is given by kαk = (α, α). In our previous papers ([1], [2]), the following result has been obtained by means of the positive definiteness of the Gram matrix. Lemma 2.1. Let α, β and γ be three arbitrary vectors of E. If kγk = 1, then (2.1) |(α, β)|2 ≤ kαk2 kβk2 − (kαk |x| − kβk |y|)2 , where x = (β, γ) , y = (α, γ). The equality in (2.1) holds if and only if α and β are linearly dependent, or γ is a linear combination of α and β, and xy = 0 but x and y are not simultaneously equal to zero. Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page For the sake of completeness, we give here a short proof of (2.1), which can also be found in [2]. Proof of Lemma 2.1. Consider the Gram determinant constructed by the vectors α, β and γ: (α, α) (α, β) (α, γ) G(α, β, γ) = (β, α) (β, β) (β, γ) . (γ, α) (γ, β) (γ, γ) According to the positive definiteness of Gram matrix we have G(α, β, γ) ≥ 0, and G(α, β, γ) = 0 if and only if the vectors α, β and γ are linearly dependent. Expanding this determinant and using the condition kγk = 1 we obtain G(α, β, γ) = kαk2 kβk2 − (α, β)2 − kαk2 x2 − 2(α, β)xy + kβk2 y 2 ≤ kαk2 kβk2 − (α, β)2 − kαk2 x2 − 2 |(α, β)xy| + kβk2 y 2 ≤ kαk2 kβk2 − (α, β)2 − {kαk |x| − kβk |y|}2 Contents JJ II J I Page 5 of 19 Go Back Full Screen Close where x = (β, γ) and y = (α, γ). It follows that the equality holds if and only if the vectors α and β are linearly dependent; or the vector γ is a linear combination of the vector α and β, and xy = 0 but x and y are not simultaneously equal to zero. Applying Lemma 2.1, we can now establish the following refinement of Hölder’s inequality. Theorem 2.2. Let an, bn ≥ 0, (n = 1, 2, . . . ), +∞ and 0 < kbkq < +∞, then 1 p 1 q + = 1 and p > 1. If 0 < kakp < Xiaozhou Shang m (a, b) ≤ kakp kbkq (1 − r) , (2.2) Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and vol. 8, iss. 2, art. 44, 2007 where 2 r = (Sp (a, c) − Sq (b, c)) , m = min 1 1 , p q Title Page , kck = 1 and ap/2 , c bq/2 , c ≥ 0. The equality in (2.2) holds if and only if ap/2 and bq/2 are linearly or if dependent; the vector c is a linear combination of ap/2 and bq/2 , and ap/2 , c bq/2 , c = 0, but the vector c is not simultaneously orthogonal to ap/2 and bq/2 . Proof. Firstly, we consider the case p 6= q. Without loss of generality, we suppose p that p > q > 1. Since p1 + 1q = 1, we have p > 2. Let R = p2 , Q = p−2 , then 1 1 + Q = 1. By Hölder’s inequality we obtain R (2.3) (a, b) = ∞ X k=1 ak b k = ∞ X k=1 q/p ak b k 1−q/p bk Contents JJ II J I Page 6 of 19 Go Back Full Screen Close ≤ ∞ X q/p ak b k R k=1 = a p/2 ! R1 ∞ X 1−q/p bk Q ! Q1 k=1 q/2 2/p ,b kbkq(1−2/p) q . The equality in (2.3) holds if and only if ap/2 and bq/2 are linearly dependent. In fact, the equality in (2.3) holds if and only if for any k, there exists c0 (c0 6= 0) such that R Q 1−q/p q/p ak b k = c 0 bk . p/2 q/2 It is easy to deduce that ak = c0 bk . If α, β and γ in (2.1) are replaced by ap/2 , bq/2 and c respectively, then we have 2 (2.4) ap/2 , bq/2 ≤ kakpp kbkqq (1 − r) , where r = (Sp (a, c) − Sq (b, c))2 . Substituting (2.4) into (2.3), we obtain after simplifications (2.5) 1 (a, b) ≤ kakp kbkq (1 − r) p . It is known from Lemma 2.1 that the equality in (2.5) holds if and only if ap/2 and bq/2 are linearly dependent; or if the vector c is a linear combination of ap/2 and bq/2 , and ap/2 , c bq/2 , c = 0, but the vector c is not simultaneously orthogonal to ap/2 and bq/2 . Note the symmetry of p and q. The inequality (2.2) follows from (2.5). Secondly, consider the case p = 2. By Lemma 2.1, we obtain: (2.6) 1 (a, b) ≤ kak kbk (1 − r) 2 , 2 (b,c) where r = (a,c) − , kck = 1 and (a, c) (b, c) ≥ 0. The equality in (2.6) holds kak kbk if and only if a and b are linearly dependent, or the vector c is a linear combination Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 7 of 19 Go Back Full Screen Close of a and b, and (a, c)(b, c) = 0, but (a, c) and (b, c) are not simultaneously equal to zero. The proof of the theorem is thus completed. Consider the example given in the Introduction. Let c = (c1 , c2 , . . . , c2n ), c ∈ R , where ci = √1n , i = 1, 2, . . . , n and cj = 0, j = n + 1, n + 2, . . . , 2n. It is easy to deduce that kck = 1 and r = 1. Substituting them into (2.2), it follows that the equality is valid. The following theorem provides a similar result to Theorem 2.2. 2n Theorem 2.3. Let f (x) , g (x) ≥ 0 (x ∈ (0, +∞)), 0 < kf kp < +∞ and 0 < kgkq < +∞, then 1 p + 1 q = 1 and p > 1. If Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page (f, g) ≤ kf kp kgkq (1 − r)m , (2.7) Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Contents where 2 r = (Sp (f, h) − Sq (g, h)) , m = min Z khk = 1, i.e. khk = ∞ h2 (x)dx 1 1 , p q , 12 =1 0 and f p/2 , h g q/2 , h ≥ 0. The equality in (2.3) holds if and only if f p/2 and g q/2 are linearly dependent; or the q/2 p/2 q/2 p/2 vector h is a linear combination of f and g , and f , h g , h = 0, but the vector h is not simultaneously orthogonal to f p/2 and g q/2 . Its proof is similar to that of Theorem 2.2. Hence it is omitted. JJ II J I Page 8 of 19 Go Back Full Screen Close 3. Applications 3.1. A Refinement of Minkowski’s Inequality We firstly give a refinement of Minkowski’s inequality for the discrete form. Theorem 3.1. Let ak , bk ≥ 0, p > 1. If 0 < kakp < +∞ and 0 < kbkp < +∞, then (3.1) Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and ka + bkp < kakp + kbkp (1 − r)m , Xiaozhou Shang where vol. 8, iss. 2, art. 44, 2007 ka + bkp = ∞ X ! p1 (ak + bk )p , Title Page k=1 1 1 ,1 − p p r = min {r (a) , r (b)} , m = min , ( )2 xp/2 , c (a + b)p/2 , c r (x) = − , x = a, b; kxkp/2 ka + bkp/2 p p ∞ X p/2 (a + b) , c = (ak + bk )p/2 ck , JJ II J I Page 9 of 19 Go Back Full Screen k=1 and c is a variable unit-vector. n o Proof. Let m = min p1 , 1 − p1 , ka + bkp = Contents Close ∞ X k=1 ! p1 p (ak + bk ) . By Theorem 2.2, we have (3.2) ∞ X !1− p1 ∞ X ak (ak + bk )p−1 ≤ kakp k=1 k=1 ∞ X ∞ X (1 − r(a))m (ak + bk )p and (3.3) bk (ak + bk )p−1 ≤ kbkp !1− p1 (ak + bk )p (1 − r(b))m , k=1 k=1 Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 where ( r (x) = xp/2 , c kxkp/2 p − (a + b)p/2 , c (a + b) ,c = ∞ X Title Page , ka + bkp/2 p ka + bkp/2 = p p/2 )2 x = a, b, ! 12 (ak + bk )p k=1 ∞ X , Contents JJ II J I Page 10 of 19 p/2 (ak + bk ) ck , Go Back k=1 and c is a variable unit-vector. Adding (3.5) and (3.3) we obtain, after simplifying: (3.4) ka + bkp ≤ kakp (1 − r(a))m + kbkp (1 − r(b))m . Let r = min {r (a) , r (b)}, then the inequality (3.1) follows. This completes the proof of Theorem 3.1. Full Screen Close If we choose a unit-vector c such that its ith component is 1 and the rest is zero, i.e. c = (0, 0, . . . , 0, 1 , 0, . . . ), then (i) ( r (x) = p/2 xi kxkp/2 p − (ai + bi )p/2 )2 x = a, b. ka + bkp/2 p Similarly, we can establish a refinement of Minkowski’s integral inequality. Theorem 3.2. Let f (x) , g (x) ≥ 0, p > 1. If 0 < kf kp < +∞ and 0 < kgkp < +∞, then (3.5) kf + gkp < kf kp + kgkp (1 − r)m , Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page where ∞ Z p kf + gkp = (f (x) + g (x)) dx , 1 1 r = min {r (f ) , r (g)} , m = min ,1 − , p p ( )2 (f + g)p/2 , h tp/2 , h r (t) = − , t = f, g, p/2 ktkp/2 kf + gk p p Z ∞ (f + g)p/2 , h = (f (x) + g (x))p/2 h (x) dx, 0 0 and h is a variable unit-vector, i.e. Z khk = 0 ∞ 2 h (x) dx 12 = 1. Contents p1 JJ II J I Page 11 of 19 Go Back Full Screen Close Its proof is similar to that of Theorem 3.1. Hence it is omitted. Remark 1. The variable unit-vector h can be chosen in accordance with our requirements. For example, we may choose h such that s 2 h (x) = . π (1 + x2 ) 3.2. A Strengthening of Fan Ky’s Inequality Xiaozhou Shang Theorem 3.3. Let A, B and C be three positive definite matrices of order n, 0 ≤ λ ≤ 1. Then (3.6) vol. 8, iss. 2, art. 44, 2007 Title Page |A|λ |B|1−λ 1 4 1 4 m = min {λ, 1 − λ}. Proof. When λ = 0, 1, the inequality (3.3) is obviously valid. Hence we need only consider the case 0 < λ < 1. If D is a positive definite matrix of order n, then it is known from [4] that Z +∞ Z +∞ π n/2 (3.7) Jn = ··· e−(x,Dx) dx = 1 , −∞ −∞ |D| 2 where x = (x1 , x2 , . . . , xn ), and dx = dx1 dx2 · · · dxn . Contents 2 m |BC| |AC| ≤ |λA + (1 − λ) B| 1 − 1 − 1 , 1 (B + C) 2 1 (A + C) 2 2 2 where |C| = π n , Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and JJ II J I Page 12 of 19 Go Back Full Screen Close Let F (x) = e−λ(x,Ax) and G (x) = e−(1−λ)(x,Bx) . If p = ing to (3.4) and (2.7) we have 1 λ and q = 1 , 1−λ accord- π n/2 (3.8) 1 |λA + (1 − λ) B| 2 Z +∞ Z +∞ = ··· F (x) G (x) dx −∞ Z +∞ ≤ Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and −∞ Z ··· −∞ n/2 p1 Z +∞ p +∞ Z +∞ q ··· F (x) dx −∞ m −∞ 1q G (x) dx (1 − r)m −∞ π (1 − r) = 1 , λ 1−λ 2 |A| |B| Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents where 2 r = S 1 (F, H) − S 1 (G, H) λ 1−λ 1 1 1 − 2(1−λ) 1 − 2λ 2(1−λ) = F 2λ , H kF k 1 − G , H kGk 1 , λ 1−λ 1 where H = e− 2 (x,Cx) , C is a positive definite matrix of order n, and Z +∞ kHk = Z +∞ ··· −∞ 2 H (x) dx 21 JJ II J I Page 13 of 19 Go Back Full Screen Close = 1. −∞ By the definition of the variable unit-vector H, it is easy to deduce that |C| = π n . Hence we have 1 2λ F Z +∞ Z +∞ ··· ,H = −∞ 1 F 2λ (x) H (x) dx −∞ ( π n/2 = 1 = 1 (A + C) 2 2 |C| 1 (A + C) 2 ) 12 Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and and Xiaozhou Shang 1/2λ kF k1/λ Z +∞ Z ··· = −∞ 21 +∞ F 1/λ (x) dx ( = π n/2 ) 12 |A|1/2 −∞ = |C| |A| 14 vol. 8, iss. 2, art. 44, 2007 , Title Page whence Contents 1 |AC| 4 S1/λ (F, H) = 1 . 1 (A + C) 2 2 Similarly, 1 4 |BC| S1/(1−λ) (G, H) = 1 , 1 (B + C) 2 2 therefore we obtain II J I Page 14 of 19 Go Back Full Screen (3.9) JJ |AC| 1 4 |BC| 1 4 2 r = 1 − 1 1 . 1 (A + C) 2 (B + C) 2 2 2 It follows from (3.8) and (3.9) that the inequality (3.3) is valid. Close 3.3. An Improvement of Hardy’s Inequality We give firstly a refinement of Hardy’s inequality for the discrete form. P Theorem 3.4. Let an ≥ 0, βn = n1 nk=1 ak , p1 + 1q = 1 and p > 1. If 0 < kakp < +∞, then p kakp (1 − r)m , (3.10) kβkp ≤ p−1 Xiaozhou Shang where r= (ap/2 , c) kakp/2 p − (β p/2 , c) kβkp/2 p n o c is a variable unit-vector and m = min p1 , 1q . !2 vol. 8, iss. 2, art. 44, 2007 , Proof. Firstly, we estimate the difference of the following two terms: (3.11) βnp − p p βnp−1 an = βnp − (nβn − (n − 1)βn−1 ) βnp−1 p−1 p−1 p1 (n − 1)p np p p = βn 1 − + (βnp )p−1 βn−1 . p−1 p−1 Applying the arithmetic-geometric mean inequality to the second term on the righthand side of (3.11) we get (3.12) Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and p (βnp )p−1 βn−1 p1 ≤ 1 p (p − 1)βnp + βn−1 . p Title Page Contents JJ II J I Page 15 of 19 Go Back Full Screen Close It follows from (3.11) and (3.12) that p np (n − 1) p p p−1 p βn − βn an ≤ βn 1 − + (p − 1)βnp + βn−1 p−1 p−1 p−1 1 p = (n − 1)βn−1 − nβnp . p−1 Summing the above inequality with respect to n, we have N X n=1 N βnp 1 p X p−1 p βn an ≤ − − (N βN ) ≤ 0. p − 1 n=1 p−1 Hence N X N βnp n=1 p X p−1 ≤ β an . p − 1 n=1 n Letting N → ∞, we get (3.13) ∞ X n=1 ∞ βnp p X p−1 ≤ β an . p − 1 n=1 n Applying the inequality (2.2) to the right-hand side of (3.13) we obtain ! p1 ∞ ! 1q ∞ ∞ X X p X p an βnp−1 ≤ ap (3.14) βn(p−1)q (1 − r)m p − 1 n=1 p − 1 n=1 n n=1 1q p = kakp kβkpp (1 − r)m , p−1 n o 2 1 1 p−1 where r = (Sp (a, c) − Sq (β , c)) , c is a variable unit-vector and m = min p , q . Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 16 of 19 Go Back Full Screen Close We obtain from (3.13) and (3.14) after simplification p kakp (1 − r)m . (3.15) kβkp ≤ p−1 It is easy to deduce that ap/2 , c Sp (a, c) = kakp/2 p Sq (β p−1 , c) = and (β (p−1)q/2 , c) kβ p−1 kq/2 q = (β p/2 , c) kβkp/2 p . Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang Hence r= a p/2 ,c kakp−p/2 − β p/2 ,c kβk−p/2 p 2 vol. 8, iss. 2, art. 44, 2007 , where c is a variable unit-vector. The proof of the theorem is completed. Title Page A variable unit-vector c can be chosen in accordance with our requirements. For example, we may choose c ∈ R∞ such that c = (1, 0, 0, . . . ). Obviously, kck = 1 and 2 −p/2 r = ap1 kak−p/2 − kβk . p p Contents Similarly, we can establish a refinement of Hardy’s integral inequality. Rx Theorem 3.5. Let f (x) ≥ 0, g(x) = x1 0 f (t)dt, p1 + 1q = 1 and p > 1. If R∞ 0 < 0 f (t)dt < +∞, then (3.16) kgkp < p kf kp (1 − r)m , p−1 where r= (f p/2 , h) kf kp/2 p − (g p/2 , h) kgkp/2 p JJ II J I Page 17 of 19 Go Back Full Screen Close !2 , h is a variable unit-vector, i.e. Z ∞ 12 1 1 2 khk = h (t)dt = 1 and m = min , . p q 0 Proof. Using integration by parts and then applying (2.2) we obtain that Z ∞ p p (3.17) f, g p−1 kgkp = g p (t)dt = p−1 0 p−1 p ≤ kf kp g q (1 − r)m p−1 p = kf kp kgkp−1 (1 − r)m , p p−1 o n 2 where r = (Sp (f, h) − Sq (g p−1 , h)) , m = min p1 , 1q and h is a variable unitvector. It is easy to deduce that g p/2 , h f p/2 , h p−1 Sp (f, h) = and Sq g , h = . p/2 kf kp/2 kgk p p It follows that the inequality (3.16) is valid. The theorem is thus proved. A variable unit-vector h can be chosen in accordance with our requirements. For example, we may choose h such that h (x) = e−x/2 . Obviously, we then have Z khk = 0 ∞ 12 h2 (t)dt = 1. Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 18 of 19 Go Back Full Screen Close References [1] MINGZHE GAO, On Heisenberg’s inequality, J. Math. Anal. Appl., 234(2) (1999), 727–734. [2] TIAN-XIAO HE, J.S. SHIUE AND ZHONGKAI LI, Analysis, Combinatorics and Computing, Nova Science Publishers, Inc. New York, 2002, 197-204. [3] JICHANG KUANG, Applied Inequalities, Hunan Education Press, 2nd ed. 1993. MR 95j: 26001. [4] E.F. BECKENBACK 1965. AND R. BELLMAN, Inequalities, 2nd ed., Springer, Hölder’s Inequality and Applications Xuemei Gao, Mingzhe Gao and Xiaozhou Shang vol. 8, iss. 2, art. 44, 2007 Title Page Contents JJ II J I Page 19 of 19 Go Back Full Screen Close