Journal of Inequalities in Pure and Applied Mathematics REFINEMENTS OF THE SCHWARZ AND HEISENBERG INEQUALITIES IN HILBERT SPACES volume 5, issue 3, article 60, 2004. S.S. DRAGOMIR School of Computer Science and Mathematics Victoria University of Technology PO Box 14428 Melbourne VIC 8001 Australia. EMail: sever.dragomir@vu.edu.au URL: http://rgmia.vu.edu.au/SSDragomirWeb.html Received 24 August, 2004; accepted 17 September, 2004. Communicated by: S. Saitoh Abstract Contents JJ J II I Home Page Go Back Close c 2000 Victoria University ISSN (electronic): 1443-5756 156-04 Quit Abstract Some new refinements of the Schwarz inequality in inner product spaces are given. Applications for discrete and integral inequalities including the Heisenberg inequality for vector-valued functions in Hilbert spaces are provided. 2000 Mathematics Subject Classification: Primary 46C05; Secondary 26D15. Key words: Schwarz inequality, Triangle inequality, Heisenberg Inequality. Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Some New Refinements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 3 Discrete Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 4 Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5 Refinements of Heisenberg Inequality . . . . . . . . . . . . . . . . . . . 26 References Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 2 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au 1. Introduction Let (H; h·, ·i) be an inner product space over the real or complex number field K. One of the most important inequalities in inner product spaces with numerous applications, is the Schwarz inequality (1.1) |hx, yi|2 ≤ kxk2 kyk2 , x, y ∈ H with equality iff x and y are linearly dependent. In 1966, S. Kurepa [1] established the following refinement of the Schwarz inequality in inner product spaces that generalises de Bruijn’s result for sequences of real and complex numbers [2]. Theorem 1.1. Let H be a real Hilbert space and HC the complexification of H. Then for any pair of vectors a ∈ H, z ∈ HC (1.2) |hz, ai|2 ≤ 1 kak2 kzk2 + |hz, z̄i| ≤ kak2 kzk2 . 2 In 1985, S.S. Dragomir [3, Theorem 2] obtained a different refinement of (1.1), namely: Theorem 1.2. Let (H; h·, ·i) be a real or complex inner product space and x, y, e ∈ H with kek = 1. Then we have the inequality (1.3) kxk kyk ≥ |hx, yi − hx, ei he, yi| + |hx, ei he, yi| ≥ |hx, yi| . Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 3 of 30 In the same paper [3, Theorem 3], a further generalisation for orthonormal families has been given (see also [4, Theorem 3]). J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Theorem 1.3. Let {ei }i∈H be an orthonormal family in the Hilbert space H. Then for any x, y ∈ H X X kxk kyk ≥ hx, yi − (1.4) hx, ei i hei , yi + |hx, ei i hei , yi| i∈I i∈I X X ≥ hx, yi − hx, ei i hei , yi + hx, ei i hei , yi i∈I i∈I Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces ≥ |hx, yi| . The inequality (1.3) has also been obtained in [4] as a particular case of the following result. Theorem 1.4. Let x, y, a, b ∈ H be such that kak2 ≤ 2 Re hx, ai , kbk2 ≤ 2 Re hy, bi . Then we have: (1.5) 1 1 kxk kyk ≥ 2 Re hx, ai − kak2 2 2 Re hy, bi − kbk2 2 + |hx, yi − hx, bi − ha, yi + ha, bi| . Another refinement of the Schwarz inequality for orthornormal vectors in inner product spaces has been obtained by S.S. Dragomir and J. Sándor in [5, Theorem 5]. S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 4 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Theorem 1.5. Let {ei }i∈{1,...,n} be orthornormal vectors in the inner product space (H; h·, ·i). Then (1.6) kxk kyk − |hx, yi| ≥ n X n X |hx, ei i|2 |hy, ei i|2 i=1 i=1 ! 12 n X − hx, ei i hei , yi ≥ 0 i=1 and (1.7) Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces kxk kyk − Re hx, yi ≥ n X n X 2 |hx, ei i| |hy, ei i|2 i=1 i=1 ! 12 − n X S.S. Dragomir Re [hx, ei i hei , yi] ≥ 0. Title Page i=1 For some properties of superadditivity, monotonicity, strong superadditivity and strong monotonicity of Schwarz’s inequality, see [6]. Here we note only the following refinements of the Schwarz inequality in its different variants for linear operators [6]: a) Let H be a Hilbert space and A, B : H → H two selfadjoint linear operators with A ≥ B ≥ 0, then we have the inequalities (1.8) 1 Contents JJ J II I Go Back Close Quit 1 hAx, xi 2 hAy, yi 2 − |hAx, yi| 1 1 Page 5 of 30 ≥ hBx, xi 2 hBy, yi 2 − |hBx, yi| ≥ 0 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au and (1.9) hAx, xi hAy, yi − |hAx, yi|2 ≥ hBx, xi hBy, yi − |hBx, yi|2 ≥ 0 for any x, y ∈ H. b) Let A : H → H be a bounded linear operator on H and let kAk = sup {kAxk , kxk = 1} the norm of A. Then one has the inequalities (1.10) kAk2 (kxk kyk − |hx, yi|) ≥ kAxk kAyk − |hAx, Ayi| ≥ 0 and (1.11) kAk4 kxk2 kyk2 − |hx, yi|2 ≥ kAxk2 kAyk2 −|hAx, Ayi|2 ≥ 0. c) Let B : H → H be a linear operator with the property that there exists a constant m > 0 such that kBxk ≥ m kxk for any x ∈ H. Then we have the inequalities (1.12) 2 kBxk kByk − |hBx, Byi| ≥ m (kxk kyk − |hx, yi|) ≥ 0 and 2 2 2 2 2 2 (1.13) kBxk kByk − |hBx, Byi| ≥ m4 kxk kyk − |hx, yi| ≥ 0. For other results related to Schwarz’s inequality in inner product spaces, see Chapter XX of [8] and the references therein. Motivated by the results outlined above, it is the aim of this paper to explore other avenues in obtaining new refinements of the celebrated Schwarz inequality. Applications for vector-valued sequences and integrals in Hilbert spaces are mentioned. Refinements of the Heisenberg inequality for vector-valued functions in Hilbert spaces are also given. Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 6 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au 2. Some New Refinements The following result holds. Theorem 2.1. Let (H; h·, ·i) be an inner product space over the real or complex number field K and r1 , r2 > 0. If x, y ∈ H satisfy the property kx − yk ≥ r2 ≥ r1 ≥ |kxk − kyk| , (2.1) then we have the following refinement of Schwarz’s inequality kxk kyk − Re hx, yi ≥ (2.2) 1 2 r2 − r12 (≥ 0) . 2 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir 1 2 The constant is best possible in the sense that it cannot be replaced by any larger quantity. Title Page Proof. From the first inequality in (2.1) we have Contents (2.3) kxk2 + kyk2 ≥ r22 + 2 Re hx, yi . Subtracting in (2.3) the quantity 2 kxk kyk , we get (2.4) (kxk − kyk)2 ≥ r22 − 2 (kxk kyk − Re hx, yi) . Since, by the second inequality in (2.1) we have (2.5) r12 ≥ (kxk − kyk)2 , hence from (2.4) and (2.5) we deduce the desired inequality (2.2). JJ J II I Go Back Close Quit Page 7 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au To prove the sharpness of the constant a constant C > 0 such that (2.6) 1 2 in (2.2), let us assume that there is kxk kyk − Re hx, yi ≥ C r22 − r12 , provided that x and y satisfy (2.1). Let e ∈ H with kek = 1 and for r2 > r1 > 0, define (2.7) x= r 2 + r1 r1 − r2 · e and y = · e. 2 2 Then kx − yk = r2 and |kxk − kyk| = r1 , showing that the condition (2.1) is fulfilled with equality. If we replace x and y as defined in (2.7) into the inequality (2.6), then we get r22 − r12 ≥ C r22 − r12 , 2 which implies that C ≤ 12 , and the theorem is completely proved. The following corollary holds. Corollary 2.2. With the assumptions of Theorem 2.1, we have the inequality: √ √ q 2 2 (2.8) kxk + kyk − kx + yk ≥ r22 − r12 . 2 2 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 8 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Proof. We have, by (2.2), that (kxk + kyk)2 − kx + yk2 = 2 (kxk kyk − Re hx, yi) ≥ r22 − r12 ≥ 0 which gives (2.9) 2 2 (kxk + kyk) ≥ kx + yk + 2 q r22 − r12 . Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces By making use of the elementary inequality 2 α2 + β 2 ≥ (α + β)2 , α, β ≥ 0; S.S. Dragomir we get (2.10) 2 kx + yk + 2 q r22 − r12 1 ≥ 2 kx + yk + q Title Page 2 r22 − r12 . Utilising (2.9) and (2.10), we deduce the desired inequality (2.8). If (H; h·, ·i) is a Hilbert space and {ei }i∈I is an orthornormal family in H, i.e., we recall that hei , ej i = δij for any i, j ∈ I, where δij is Kronecker’s delta, then we have the following inequality which is well known in the literature as Bessel’s inequality X (2.11) |hx, ei i|2 ≤ kxk2 for each x ∈ H. Contents JJ J II I Go Back Close Quit Page 9 of 30 i∈I J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Here, the meaning of the sum is ( ) X X 2 2 |hx, ei i| = sup |hx, ei i| , F is a finite part of I . F ⊂I i∈I i∈F The following result providing a refinement of the Bessel inequality (2.11) holds. Theorem 2.3. Let (H; h·, ·i) be a Hilbert space and {ei }i∈I an orthornormal family in H. If x ∈ H, x 6= 0, and r2 , r1 > 0 are such that: ! 12 X X (2.12) x − (≥ 0) , hx, ei i ei ≥ r2 ≥ r1 ≥ kxk − |hx, ei i|2 i∈I Title Page Contents ! 12 kxk − X |hx, ei i|2 ≥ i∈I The constant 1 · 2 P r22 i∈I − r12 1 2 2 (≥ 0) . |hx, ei i| 1 2 is best possible. P Proof. Consider y := i∈I hx, ei i ei . Obviously, since H is a Hilbert space, y ∈ H. We also note that v 2 s X u X X u t kyk = hx, ei i ei = hx, ei i ei = |hx, ei i|2 , i∈I S.S. Dragomir i∈I then we have the inequality (2.13) Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces i∈I i∈I JJ J II I Go Back Close Quit Page 10 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au and thus (2.12) is in fact (2.1) of Theorem 2.1. Since ! 21 * + X X 2 kxk kyk − Re hx, yi = kxk |hx, ei i| − Re x, hx, ei i ei i∈I i∈I ! 21 = X |hx, ei i|2 kxk − i∈I ! 12 X |hx, ei i|2 , i∈I hence, by (2.2), we deduce the desired result (2.13). We will prove the sharpness of the constant for the case of one element, i.e., I = {1} , e1 = e ∈ H, kek = 1. For this, assume that there exists a constant D > 0 such that r2 − r12 kxk − |hx, ei| ≥ D · 2 |hx, ei| (2.14) provided x ∈ H\ {0} satisfies the condition (2.15) kx − hx, ei ek ≥ r2 ≥ r1 ≥ kxk − |hx, ei| . Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Assume that x = λe + µf with e, f ∈ H, kek = kf k = 1 and e ⊥ f. We wish to see if there exists positive numbers λ, µ such that Close Quit (2.16) kx − hx, ei ek = r2 > r1 = kxk − |hx, ei| . Page 11 of 30 Since (for λ, µ > 0) kx − hx, ei ek = µ J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au and kxk − |hx, ei| = p λ2 + µ2 − λ hence, by (2.16), we get µ = r2 and q λ2 + r22 − λ = r1 giving λ2 + r22 = λ2 + 2λr1 + r12 from where we get r22 − r12 > 0. 2r1 With these values for λ and µ, we have λ= kxk − |hx, ei| = r1 , r2 − r12 |hx, ei| = 2 2r1 giving D ≤ 1 . 2 This proves the theorem. The following corollary is obvious. r22 − r12 r22 −r12 2r1 S.S. Dragomir Title Page Contents JJ J and thus, from (2.14), we deduce r1 ≥ D · Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces , II I Go Back Close Quit Page 12 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Corollary 2.4. Let x, y ∈ H with hx, yi = 6 0 and r2 ≥ r1 > 0 such that hx, yi (2.17) kyk x − kyk · y ≥ r2 kyk ≥ r1 kyk ≥ kxk kyk − |hx, yi| (≥ 0) . Then we have the following refinement of the Schwarz’s inequality: kyk2 1 2 2 kxk kyk − |hx, yi| ≥ r − r1 (≥ 0) . 2 2 |hx, yi| (2.18) The constant 1 2 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces is best possible. S.S. Dragomir The following lemma holds. Lemma 2.5. Let (H; h·, ·i) be an inner product space and R ≥ 1. For x, y ∈ H, the subsequent statements are equivalent: (i) The following refinement of the triangle inequality holds: (2.19) kxk + kyk ≥ R kx + yk ; (ii) The following refinement of the Schwarz inequality holds: (2.20) kxk kyk − Re hx, yi ≥ 1 2 R − 1 kx + yk2 . 2 Proof. Taking the square in (2.19), we have (2.21) 2 kxk kyk ≥ R2 − 1 kxk2 + 2R2 Re hx, yi + R2 − 1 kyk2 . Title Page Contents JJ J II I Go Back Close Quit Page 13 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Subtracting from both sides of (2.21) the quantity 2 Re hx, yi , we obtain 2 (kxk kyk − Re hx, yi) ≥ R2 − 1 kxk2 + 2 Re hx, yi + kyk2 = R2 − 1 kx + yk2 , which is clearly equivalent to (2.20). By the use of the above lemma, we may now state the following theorem concerning another refinement of the Schwarz inequality. Theorem 2.6. Let (H; h·, ·i) be an inner product space over the real or complex number field and R ≥ 1, r ≥ 0. If x, y ∈ H are such that (2.22) 1 (kxk + kyk) ≥ kx + yk ≥ r, R then we have the following refinement of the Schwarz inequality (2.23) kxk kyk − Re hx, yi ≥ 1 2 R − 1 r2 . 2 The constant 21 is best possible in the sense that it cannot be replaced by a larger quantity. Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Proof. The inequality (2.23) follows easily from Lemma 2.5. We need only prove that 21 is the best possible constant in (2.23). Assume that there exists a C > 0 such that (2.24) kxk kyk − Re hx, yi ≥ C R2 − 1 r2 Quit Page 14 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au provided x, y, R and r satisfy (2.22). Consider r = 1, R > 1 and choose x = 1−R e, y = 1+R e with e ∈ H, 2 2 kek = 1. Then kxk + kyk x + y = e, =1 R and thus (2.22) holds with equality on both sides. From (2.24), for the above choices, we have 12 (R2 − 1) ≥ C (R2 − 1) , which shows that C ≤ 21 . Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces Finally, the following result also holds. Theorem 2.7. Let (H; h·, ·i) be an inner product space over the real or complex number field K and r ∈ (0, 1]. For x, y ∈ H, the following statements are equivalent: S.S. Dragomir Title Page (i) We have the inequality Contents |kxk − kyk| ≤ r kx − yk ; (2.25) JJ J (ii) We have the following refinement of the Schwarz inequality (2.26) The constant kxk kyk − Re hx, yi ≥ 1 2 Go Back 1 1 − r2 kx − yk2 . 2 Close Quit in (2.26) is best possible. Page 15 of 30 Proof. Taking the square in (2.25), we have kxk2 − 2 kxk kyk + kyk2 ≤ r2 kxk2 − 2 Re hx, yi + kyk2 II I J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au which is clearly equivalent to 1 − r2 kxk2 − 2 Re hx, yi + kyk2 ≤ 2 (kxk kyk − Re hx, yi) or with (2.26). Now, assume that (2.26) holds with a constant E > 0, i.e., (2.27) kxk kyk − Re hx, yi ≥ E 1 − r2 kx − yk2 , provided (2.25) holds. Define x = r+1 e, y = 2 r−1 e 2 with e ∈ H, kek = 1. Then |kxk − kyk| = r, kx − yk = 1 showing that (2.25) holds with equality. If we replace x and y in (2.27), then we get E (1 − r2 ) ≤ 21 (1 − r2 ) , implying that E ≤ 12 . Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 16 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au 3. Discrete Inequalities Assume that (K; (·, ·)) is a Hilbert space P∞over the real or complex number field. Assume also that pi ≥ 0, i ∈ H with i=1 pi = 1 and define ( ) ∞ X `2p (K) := x := (xi ) xi ∈ K, i ∈ N and pi kxi k2 < ∞ . i∈N i=1 It is well known that `2p (K) endowed with the inner product h·, ·ip defined by hx, yip := ∞ X pi (xi , yi ) Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir i=1 and generating the norm Title Page kxkp := ∞ X ! 12 pi kxi k 2 i=1 is a Hilbert space over K. We may state the following discrete inequality improving the CauchyBunyakovsky-Schwarz classical result. Proposition 3.1. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with P∞ 2 i=1 pi = 1. Assume that x, y ∈ `p (K) and r1 , r2 > 0 satisfy the condition (3.1) kxi − yi k ≥ r2 ≥ r1 ≥ |kxi k − kyi k| Contents JJ J II I Go Back Close Quit Page 17 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au for each i ∈ N. Then we have the following refinement of the Cauchy-BunyakovskySchwarz inequality (3.2) ∞ X ∞ X pi kxi k2 pi kyi k2 i=1 i=1 The constant 1 2 ! 21 − ∞ X pi Re (xi , yi ) ≥ i=1 1 2 r2 − r12 ≥ 0. 2 is best possible. Proof. From the condition (3.1) we simply deduce (3.3) ∞ X 2 pi kxi − yi k ≥ r22 ≥ r12 i=1 ≥ ∞ X Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces pi (kxi k − kyi k)2 S.S. Dragomir i=1 ≥ ∞ X ! 12 pi kxi k 2 i=1 − ∞ X ! 1 2 2 2 pi kyi k . i=1 Contents In terms of the norm k·kp , the inequality (3.3) may be written as (3.4) kx − ykp ≥ r2 ≥ r1 ≥ kxkp − kykp . Title Page Utilising Theorem 2.1 for the Hilbert space `2p (K) , h·, ·ip , we deduce the desired inequality (3.2). For n = 1 (p1 = 1) , the inequality (3.2) reduces to (2.2) for which we have shown that 21 is the best possible constant. By the use of Corollary 2.2, we may state the following result as well. JJ J II I Go Back Close Quit Page 18 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Corollary 3.2. With the assumptions of Proposition 3.1, we have the inequality (3.5) ∞ X ! 21 2 pi kxi k + i=1 ∞ X ! 12 pi kyi k 2 i=1 √ 2 − 2 ∞ X ! 12 pi kxi + yi k i=1 2 √ q 2 ≥ r22 − r12 . 2 The following proposition also holds. Proposition 3.3. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with P∞ p = 1. Assume that x, y ∈ `2p (K) and R ≥ 1, r ≥ 0 satisfy the condition i=1 i Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir 1 (3.6) (kxi k + kyi k) ≥ kxi + yi k ≥ r R for each i ∈ N. Then we have the following refinement of the Schwarz inequality (3.7) ∞ X pi kxi k i=1 ∞ X 2 ! 21 pi kyi k 2 − ∞ X i=1 i=1 1 2 pi Re (xi , yi ) ≥ R − 1 r2 . 2 1 2 The constant is best possible in the sense that it cannot be replaced by a larger quantity. Proof. By (3.6) we deduce "∞ # 21 X 1 pi (kxi k + kyi k)2 ≥ (3.8) R i=1 Title Page Contents JJ J II I Go Back Close Quit ∞ X i=1 Page 19 of 30 ! 12 2 pi kxi + yi k ≥ r. J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au By the classical Minkowsky inequality for nonnegative numbers, we have (3.9) ∞ X i=1 ! 12 pi kxi k2 + ∞ X i=1 ! 21 pi kyi k2 " ≥ ∞ X # 12 pi (kxi k + kyi k)2 , i=1 and thus, by utilising (3.8) and (3.9), we may state in terms of k·kp the following inequality (3.10) 1 kxkp + kykp ≥ kx + ykp ≥ r. R Employing Theorem 2.6 for the Hilbert space `2p (K) and the inequality (3.10), we deduce the desired result (3.7). Since, for p = 1, n = 1, (3.7) is reduced to (2.23) for which we have shown that 21 is the best constant, we conclude that 12 is the best constant in (3.7) as well. Finally, we may state and prove the following result incorporated in Proposition 3.4. Let (K; (·, ·)) be a Hilbert space and pi ≥ 0 (i ∈ N) with P∞ p = 1. Assume that x, y ∈ `2p (K) and r ∈ (0, 1] such that i=1 i (3.11) |kxi k − kyi k| ≤ r kxi − yi k for each i ∈ N, Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 20 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au holds true. Then we have the following refinement of the Schwarz inequality (3.12) ∞ X pi kxi k i=1 2 ∞ X ! 12 pi kyi k 2 − i=1 ∞ X pi Re (xi , yi ) i=1 ∞ ≥ The constant 1 2 X 1 1 − r2 pi kxi − yi k2 . 2 i=1 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces is best possible in (3.12). Proof. From (3.11) we have S.S. Dragomir " ∞ X i=1 # 12 pi (kxi k − kyi k)2 " ≤r ∞ X # 12 pi kxi − yi k2 . Title Page i=1 Utilising the following elementary result ! 21 ! 12 ∞ ∞ X X 2 2 ≤ p kx k − p ky k i i i i i=1 i=1 Contents ∞ X ! 12 pi (kxi k − kyi k)2 , i=1 JJ J II I Go Back Close we may state that kxkp − kykp ≤ r kx − ykp . Now, by making use of Theorem 2.7, we deduce the desired inequality (3.12) and the fact that 21 is the best possible constant. We omit the details. Quit Page 21 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au 4. Integral Inequalities Assume that (K; (·, ·)) is a Hilbert space over the real or complex number field K. If ρ : [a, b] ⊂ R → [0, ∞) is a Lebesgue integrable function with Rb ρ (t) dt = 1, then we may consider the space L2ρ ([a, b] ; K) of all functions a Rb f : [a, b] → K, that are Bochner measurable and a ρ (t) kf (t)k2 dt < ∞. It is known that L2ρ ([a, b] ; K) endowed with the inner product h·, ·iρ defined by b Z hf, giρ := ρ (t) (f (t) , g (t)) dt a and generating the norm Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Z kf kρ := b 21 ρ (t) kf (t)k dt 2 a is a Hilbert space over K. Now we may state and prove the first refinement of the Cauchy-BunyakovskySchwarz integral inequality. Title Page Contents JJ J II I Proposition 4.1. Assume that f, g ∈ L2ρ ([a, b] ; K) and r2 , r1 > 0 satisfy the condition Go Back kf (t) − g (t)k ≥ r2 ≥ r1 ≥ |kf (t)k − kg (t)k| Quit (4.1) Close Page 22 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au for a.e. t ∈ [a, b] . Then we have the inequality b Z 2 Z b 21 ρ (t) kg (t)k dt 2 ρ (t) kf (t)k dt a Z b 1 2 − ρ (t) Re (f (t) , g (t)) dt ≥ r2 − r12 (≥ 0) . 2 a (4.2) a The constant 1 2 is best possible in (4.2). Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces Proof. Integrating (4.1), we get b Z (4.3) 12 ρ (t) (kf (t) − g (t)k)2 dt S.S. Dragomir a Z ≥ r2 ≥ r 1 ≥ b 12 ρ (t) (kf (t)k − kg (t)k) dt . 2 a Utilising the obvious fact Z (4.4) a b 1 2 ρ (t) (kf (t)k − kg (t)k) dt Z 12 12 Z b b ≥ ρ (t) kf (t)k2 dt − ρ (t) kg (t)k2 dt , a a 2 we can state the following inequality in terms of the k·kρ norm: (4.5) kf − gkρ ≥ r2 ≥ r1 ≥ kf kρ − kgkρ . Title Page Contents JJ J II I Go Back Close Quit Page 23 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Employing Theorem 2.1 for the Hilbert space L2ρ ([a, b] ; K) , we deduce the desired inequality (4.2). To prove the sharpness of 12 in (4.2), we choose a = 0, b = 1, f (t) = 1, t ∈ [0, 1] and f (t) = x, g (t) = y, t ∈ [a, b] , x, y ∈ K. Then (4.2) becomes kxk kyk − Re hx, yi ≥ 1 2 r2 − r12 2 provided kx − yk ≥ r2 ≥ r1 ≥ |kxk − kyk| , which, by Theorem 2.1 has the quantity 1 2 as the best possible constant. The following corollary holds. Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Corollary 4.2. With the assumptions of Proposition 4.1, we have the inequality Title Page Z (4.6) a b 12 Z b 12 2 2 ρ (t) kf (t)k dt + ρ (t) kg (t)k dt a √ Z b √ q 12 2 2 2 − ρ (t) kf (t) + g (t)k dt ≥ r22 − r12 . 2 2 a Contents JJ J II I Go Back The following two refinements of the Cauchy-Bunyakovsky-Schwarz (CBS) integral inequality also hold. Proposition 4.3. If f, g ∈ L2ρ ([a, b] ; K) and R ≥ 1, r ≥ 0 satisfy the condition Close Quit Page 24 of 30 (4.7) 1 (kf (t)k + kg (t)k) ≥ kf (t) + g (t)k ≥ r R J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au for a.e. t ∈ [a, b] , then we have the inequality Z b (4.8) a 21 ρ (t) kf (t)k2 dt ρ (t) kg (t)k2 dt a Z b 1 2 − ρ (t) Re (f (t) , g (t)) dt ≥ R − 1 r2 . 2 a Z 1 2 The constant b is best possible in (4.8). The proof follows by Theorem 2.6 and we omit the details. Proposition 4.4. If f, g ∈ L2ρ ([a, b] ; K) and ζ ∈ (0, 1] satisfy the condition Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir |kf (t)k − kg (t)k| ≤ ζ kf (t) − g (t)k (4.9) for a.e. t ∈ [a, b] , then we have the inequality Title Page Contents Z (4.10) a The constant b Z 12 b ρ (t) kg (t)k2 dt ρ (t) kf (t)k2 dt a Z b − ρ (t) Re (f (t) , g (t)) dt a Z b 1 2 1−ζ ρ (t) kf (t) − g (t)k2 dt. ≥ 2 a 1 2 is best possible in (4.10). The proof follows by Theorem 2.7 and we omit the details. JJ J II I Go Back Close Quit Page 25 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au 5. Refinements of Heisenberg Inequality It is well known that if (H; h·, ·i) is a real or complex Hilbert space and f : [a, b] ⊂ R →H is an absolutely continuous vector-valued function, then f is differentiable almost everywhere on [a, b] , the derivative f 0 : [a, b] → H is Bochner integrable on [a, b] and Z t (5.1) f (t) = f 0 (s) ds for any t ∈ [a, b] . a The following theorem provides a version of the Heisenberg inequalities in the general setting of Hilbert spaces. Theorem 5.1. Let ϕ : [a, b] → H be an absolutely continuous function with the property that b kϕ (b)k2 = a kϕ (a)k2 . Then we have the inequality: Z (5.2) a 2 Z b Z b b 2 2 2 2 kϕ (t)k dt ≤ 4 t kϕ (t)k dt · kϕ0 (t)k dt. a a The constant 4 is best possible in the sense that it cannot be replaced by any smaller constant. Proof. Integrating by parts, we have successively b Z b Z b d 2 2 (5.3) kϕ (t)k dt = t kϕ (t)k − t kϕ (t)k2 dt dt a a a Z b d 2 2 = b kϕ (b)k − a kϕ (a)k − t hϕ (t) , ϕ (t)i dt dt a Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 26 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Z b =− t [hϕ0 (t) , ϕ (t)i + hϕ (t) , ϕ0 (t)i] dt a Z b = −2 t Re hϕ0 (t) , ϕ (t)i dt a Z b =2 Re hϕ0 (t) , (−t) ϕ (t)i dt. a If we apply the Cauchy-Bunyakovsky-Schwarz integral inequality Z b Z Re hg (t) , h (t)i dt ≤ b kg (t)k2 dt a a Z 21 b 2 kh (t)k dt a for g (t) = ϕ0 (t) , h (t) = −tϕ (t) , t ∈ [a, b] , then we deduce the desired inequality (4.5). The fact that 4 is the best constant in (4.5) follows from the fact that in the (CBS) inequality, the case of equality holds iff g (t) = λh (t) for a.e. t ∈ [a, b] and λ a given scalar in K. We omit the details. For details on the classical Heisenberg inequality, see, for instance, [7]. Utilising Proposition 4.1, we can state the following refinement of the Heisenberg inequality obtained above in (5.2): Proposition 5.2. Assume that ϕ : [a, b] → H is as in the hypothesis of Theorem 5.1. In addition, if there exist r2 , r1 > 0 so that 0 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 27 of 30 0 kϕ (t) + tϕ (t)k ≥ r2 ≥ r1 ≥ |kϕ (t)k − |t| kϕ (t)k| J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au for a.e. t ∈ [a, b] , then we have the inequality Z b 2 2 Z b t kϕ (t)k dt · a 21 Z 1 b kϕ (t)k dt − kϕ (t)k2 dt 2 a 1 ≥ (b − a) r22 − r12 (≥ 0) . 2 0 a 2 The proof follows by Proposition 4.1 on choosing f (t) = ϕ0 (t) , g (t) = 1 −tϕ (t) and ρ (t) = b−a , t ∈ [a, b] . On utilising Proposition 4.3 for the same choices of f, g and ρ, we may state the following results as well: Proposition 5.3. Assume that ϕ : [a, b] → H is as in the hypothesis of Theorem 5.1. In addition, if there exist R ≥ 1 and r > 0 so that 1 (kϕ0 (t)k + |t| kϕ (t)k) ≥ kϕ0 (t) − tϕ (t)k ≥ r R for a.e. t ∈ [a, b] , then we have the inequality Z b t2 kϕ (t)k2 dt · Z a a b 12 Z 1 b 0 kϕ (t)k dt − kϕ (t)k2 dt 2 a 1 ≥ (b − a) R2 − 1 r2 (≥ 0) . 2 2 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I Go Back Close Quit Page 28 of 30 Finally, we can state J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au Proposition 5.4. Let ϕ : [a, b] → H be as in the hypothesis of Theorem 5.1. In addition, if there exists ζ ∈ (0, 1] so that |kϕ0 (t)k − |t| kϕ (t)k| ≤ ζ kϕ0 (t) + tϕ (t)k for a.e. t ∈ [a, b] , then we have the inequality Z b 2 2 Z t kϕ (t)k dt · a a b 21 Z 1 b kϕ (t)k2 dt kϕ (t)k dt − 2 a Z b 0 1 2 2 kϕ (t) + tϕ (t)k dt (≥ 0) . 1−ζ ≥ 2 a 0 2 Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir This follows by Proposition 4.4 and we omit the details. Title Page Contents JJ J II I Go Back Close Quit Page 29 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au References [1] S. KUREPA, On the Buniakowsky-Cauchy-Schwarz inequality, Glasnik Mat. Ser III, 1(21) (1966), 147–158. [2] N.G. DE BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12–13. [3] S.S. DRAGOMIR, Some refinements of Schwarz inequality, Suppozionul de Matematică şi Aplicaţii, Polytechnical Institute Timişoara, Romania, 1-2 November 1985, 13–16. [4] S.S. DRAGOMIR AND J. SÁNDOR, Some inequalities in prehilbertian spaces, Studia Univ., Babeş-Bolyai, Mathematica, 32(1) (1987), 71–78 MR 89h: 46034. [5] S.S. DRAGOMIR AND J. SÁNDOR, On Bessels’ and Gram’s inequalities in prehilbertian spaces, Periodica Math. Hungarica, 29(3) (1994), 197– 205. [6] S.S. DRAGOMIR AND B. MOND, On the superadditivity and monotonicity of Schwarz’s inequality in inner product spaces, Contributions, Macedonian Acad. of Sci. and Arts, 15(2) (1994), 5–22. Refinements of the Schwarz and Heisenberg Inequalities in Hilbert Spaces S.S. Dragomir Title Page Contents JJ J II I [7] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, Cambridge, United Kingdom, 1952. Go Back [8] D.S. MITRINOVĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993. Quit Close Page 30 of 30 J. Ineq. Pure and Appl. Math. 5(3) Art. 60, 2004 http://jipam.vu.edu.au