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Some Improvements of the Cauchy-Schwarz Inequality Using the Tapia Semi-Inner-Product

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Some Improvements of the Cauchy-Schwarz Inequality Using the Tapia Semi-Inner-Product mathematics Article Some Improvements of the Cauchy-Schwarz Inequality Using the Tapia Semi-Inner-Product Nicu¸sorMinculete 1,* and Hamid Reza Moradi 2 1 Department of Mathematics and Computer Science, Transilvania University of Bra¸sov, 500091 Bra¸sov, Romania 2 Department of Mathematics, Payame Noor University (PNU), Tehran P.O. Box 19395-4697, Iran; [email protected] * Correspondence: [email protected] Received: 24 October 2020; Accepted: 23 November 2020; Published: 26 November 2020 Abstract: The aim of this article is to establish several estimates of the triangle inequality in a normed space over the field of real numbers. We obtain some improvements of the Cauchy–Schwarz inequality, which is improved by using the Tapia semi-inner-product. Finally, we obtain some new inequalities for the numerical radius and norm inequalities for Hilbert space operators. Keywords: inner product space; triangle inequality; Cauchy–Schwarz inequality JEL Classification: Primary 46C05; secondary 26D15; 26D10 1. Introduction In an inner product space an important inequality is the inequality of Cauchy–Schwarz [1,2], namely: jhx, yij ≤ kxkkyk, (1) for all x, y 2 X, where X is a complex inner product space. Aldaz [3] and Dragomir [4] studied the Cauchy–Schwarz inequality in the complex case. Another inequality that plays a central role in a normed space is the triangle inequality, kx + yk ≤ kxk + kyk, (2) for all x, y 2 X, where X is a complex normed space. Peˇcari´cand Raji´cin [5] proved other results about the triangle inequality. In [6], Maligranda showed the following inequality: A · minfkxk, kykg ≤ kxk + kyk − kx + yk ≤ A · maxfkxk, kykg, (3) x y where A = 2 − + , and x and y are nonzero vectors in a normed space X = (X, k · k). kxk kyk Using this inequality we obtain an estimate for the norm-angular distance or Clarkson distance x y (see, e.g., [7]) between nonzero vectors x and y, a[x, y] = − , thus [8]: kxk kyk kx − yk − jkxk − kykj kx − yk + jkxk − kykj ≤ a[x, y] ≤ . (4) minfkxk, kykg maxfkxk, kykg Mathematics 2020, 8, 2112; doi:10.3390/math8122112 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 2112 2 of 13 In [6], Maligranda generalized the norm-angular distance to the p-angular distance in a normed space given by: p−1 p−1 ap[x, y] := kkxk x − kyk yk, where p 2 R. In [9], Dragomir studied new bounds for this distance. Other results for bounds for the angular distance, named Dunkl–Williams type theorems (see [10]), are given by Moslehian et al. [11] and Krni´cand Minculete [12,13]. Dehghan [14] presented a new refinement of the triangle inequality and defined the skew angular x y distance between nonzero vectors x and y by b[x, y] = − . In [15], we remarked on kyk kxk several estimates of the triangle inequality using integrals and in [16] a characterization is given for a generalized triangle inequality in normed spaces. The aim of this article is to establish several estimates of the triangle inequality in a normed space over the field of real numbers. This study is presented in Section2. We also obtain, in Section3, some improvements of the Cauchy–Schwarz inequality, which is improved by using the Tapia semi-inner-product. In Section4, we obtain some new inequalities for the numerical radius and norm inequalities for Hilbert space operators. 2. Inequalities Related to the Triangle Inequality We present some results regarding the several estimates of the triangle inequality in a normed space over the field of real numbers R. Theorem 1. If X = (X, k · k) is a normed vector space over the field R, then the following inequality holds a b minf , g(ckxk + dkyk − kcx + dyk) ≤ akxk + bkyk − kax + byk c d a b ≤ maxf , g(ckxk + dkyk − kcx + dyk), (5) c d for all vectors x and y in X and a, b, c, d 2 R+, c, d 6= 0. a b Proof. Without loss of generality, we may assume that 0 ≤ ≤ . Then, by calculations and using c d the triangle inequality, we have a b akxk + bkyk − minf , g(ckxk + dkyk − kcx + dyk) = c d ad ad akxk + bkyk − akxk − kyk + kax + yk) = c c ad ad ad ad (b − )kyk + kax + yk = k(b − )yk + kax + yk ≥ kax + byk. c c c c Therefore, we obtain the first inequality of the statement. a b In the same way, according to inequality ≤ , we have ( bc − a)kxk = k bc x − axk and c d d d we deduce a b maxf , g(ckxk + dkyk − kcx + dyk) + kax + byk c d bc bc = kxk + bkyk − k x + byk + kax + byk d d bc bc = akxk + bkyk + − a kxk + kax + byk − k x + byk d d Mathematics 2020, 8, 2112 3 of 13 bc bc = akxk + bkyk + k x − axk + kax + byk − k x + byk ≥ akxk + bkyk. d d Because, using the triangle inequality, we have bc bc k x − axk + kax + byk ≥ k x + byk, d d for all vectors x and y in X and a, b, c, d 2 R+, c, d 6= 0. b a For 0 ≤ ≤ , we make the similar calculations. Therefore, the inequalities of the statement d c are true. Remark 1. (a) If c = d in inequality (5), then we have minfa, bg(kxk + kyk − kx + yk) ≤ akxk + bkyk − kax + byk ≤ maxfa, bg(kxk + kyk − kx + yk), (6) for all vectors x and y in X and a, b 2 R+. Another extension of this result can be found in [13]. ! x ! y (b) For x kxk and y kyk in relation (6), we obtain the following inequalities: x y a b minfa, bg 2 − + ≤ a + b − x + y kxk kyk kxk kyk x y ≤ maxfa, bg 2 − + , (7) kxk kyk for all nonzero vectors x and y in X and a, b 2 R+. For a = kxk and b = kyk in inequality (7), we find inequality (3). (c) For the nonzero vectors x and y in X, in Theorem 1, we make the following substitutions: a = d = 1 = = 1 kxk , b c kyk . We obtain kxk kyk kxk kyk x y x y minf , g + − + ≤ 2 − + ≤ kyk kxk kyk kxk kyk kxk kxk kyk kxk kyk kxk kyk x y maxf , g + − + . (8) kyk kxk kyk kxk kyk kxk In relation (8), if we replace y by −y, then we have an inequality which uses the angular distance and the skew angular distance, thus: 2 kxk kyk kxk − kyk minf , g − b[x, y] ≤ 2 − a[x, y] ≤ kyk kxk kxkkyk 2 kxk kyk kxk − kyk maxf , g − b[x, y] , (9) kyk kxk kxkkyk for all nonzero vectors x and y in X. Theorem 2. If X = (X, k · k) is a normed vector space over the field of real numbers R, with kxk = kyk = 1, then we have ax + by − ja − bj ax + by + ja − bj ≤ x + y ≤ (10) minfa, bg maxfa, bg Mathematics 2020, 8, 2112 4 of 13 ∗ for every vector x and y in X and a, b 2 R+. Proof. From relation (6), using the first inequality, we find, a + b − ja − bj − minfa, bg x + y ≤ a + b − ax + by , which implies ax + by − ja − bj ≤ minfa, bg x + y . Using the second inequality of relation (6), we find the inequality a + b − ax + by ≤ a + b + ja − bj − maxfa, bg x + y , which becomes maxfa, bg x + y ≤ ax + by + ja − bj, ∗ for all vectors x and y in X and a, b 2 R+. Therefore, we deduce the relation of the statement. Remark 2. If we replace y by −y in Theorem 2, then we obtain an inequality for Clarkson distance, in the case kxk = kyk = 1, given by: ax − by − ja − bj ax − by + ja − bj ≤ x − y ≤ . (11) minfa, bg maxfa, bg x Inequality (11) represents a generalization of Maligranda’s inequality, because, replacing x by kxk , y by y = k k = k k kyk and taking a x , b y , we deduce the inequalities from relation (3). 3. Some Inequalities Related to the Tapia Semi-Inner-Product Next, we study estimates of the Cauchy–Schwarz inequality using the Tapia semi-inner-product. The Tapia semi-inner-product on the normed space X (see [17]) is the function (·, ·)T : X × X ! R, defined by j(x + ty) − j(x) (x, y)T := lim , t!0 t t>0 1 2 where j(x) = 2 kxk , x 2 X. The above limit exists for any pair of elements x, y 2 X. The semi-inner-product have been used by many authors in several contexts (see, e.g., [18]). The Tapia semi-inner-product is positive homogeneous in each argument and satisfies inequality j(x, y)Tj ≤ kxkkyk, (12) for all x, y 2 X. In the case when the norm k · k is generated by an inner product h·, ·i, then (x, y)T = hx, yi, for all x, y 2 X, thus we find the Cauchy–Schwarz inequality. Theorem 3. If X = (X, k · k) is a normed vector space over the field of real numbers R, then we have kxkkyk + (x, y)T ≤ xkyk + ykxk . (13) Mathematics 2020, 8, 2112 5 of 13 for every vector x and y in X. Proof. If x = 0 or y = 0, then the inequality of the statement is true. We consider x 6= 0 and y 6= 0. If in inequality (7) we replace a and b by kxk and tkyk with t > 0, then we obtain x y 2 − + minfkxk, tkykg ≤ kxk + tkyk − kx + tyk kxk kyk x y ≤ 2 − + maxfkxk, tkykg, kxk kyk which is equivalent to x y 1 kx + tyk − kxk 2 − + minfkxk, tkykg ≤ kyk − kxk kyk t t x y 1 ≤ 2 − + maxfkxk, tkykg.
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