Characterizations of inner product spaces by orthogonal ... that X is an inner product space if and

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  • JOURNAL OF c© 2006, Scientific Horizon FUNCTION SPACES AND APPLICATIONS http://www.jfsa.net

    Volume 4, Number 1 (2006), 1-6

    Characterizations of inner product spaces by

    orthogonal vectors

    Marco Baronti and Emanuele Casini

    (Communicated by Jürgen Appell)

    2000 Mathematics Subject Classification. 46C15, 46B99, 41A65.

    Keywords and phrases. Orthogonality, Characterizations of inner product

    spaces.

    Abstract. Let X be a real normed space with unit closed ball B . We prove

    that X is an inner product space if and only if it is true that whenever x, y are

    points in ∂B such that the line through x and y supports √

    2 2

    B then x ⊥ y in the sense of Birkhoff.

    1. Introduction

    Several known characterizations of inner product spaces are available in the literature [1]. This paper concerns a new characterization by means of orthogonal vectors. Let X be a real normed space (of dimension ≥ 2) and let SX denote its unit sphere. If x, y ∈ SX , we set:

    k(x, y) = inf{‖tx + (1 − t)y‖ : t ∈ [0, 1]}

    moreover if x, y ∈ SX then x ⊥ y denotes the orthogonality in the sense of Birkhoff [1] i.e. ‖x‖ ≤ ‖x + λy‖ ∀λ ∈ R . In this paper we consider

  • 2 Characterizations of inner product spaces

    the following properties

    (P1) : x, y ∈ SX x ⊥ y ⇒ k(x, y) = ‖x + y‖2 (P2) : x, y ∈ SX k(x, y) =

    √ 2

    2 ⇒ x ⊥ y

    We prove that these properties characterize inner product spaces. The definition of the constant k(x, y) is suggested by the following result by Benitez and Yanez [4]:{

    x, y ∈ SX k(x, y) = 12 ⇒ x + y ∈ SX } ⇔ X is an inner product space.

    It is easy to prove that every inner product spaces X satisfies the properties (P1) and (P2). Moreover from the particular structure of (P1) and (P2) we can suppose that X is to be a 2-dimensional real normed space.

    2. Some preliminary results

    We start with some preliminary observations on the constant k(x, y). If x, y ∈ SX and x ⊥ y we set:

    μ(x, y) = sup λ≥0

    1 + λ ‖x + λy‖

    and μ(X) = sup{μ(x, y); x, y ∈ SX x ⊥ y }.

    These constants were introduced in [6] and in [6] and [3] the following results were proved:

    μ(X) ≥ √

    2 for any space X

    and μ(X) =

    √ 2 ⇔ X is an inner product space

    Let x, y ∈ SX , x ⊥ y, we have

    μ(x, y) = sup λ≥0

    1 + λ ‖x + λy‖ = sup0≤t

  • M. Baronti and E. Casini 3

    and so

    (1) μ(X) = 1

    inf {k(x, y) : x, y ∈ SX x ⊥ y} From known results on the constant μ [3], [4] we obtain the following results:

    inf {k(x, y) :x, y∈SX x ⊥ y}≤ √

    2 2

    (2)

    inf {k(x, y) :x, y∈SX , x ⊥ y}= √

    2 2

    ⇔ X is inner product space.(3)

    Theorem 1. If X fulfills (P1) then X is an inner product space.

    Proof. Let x, y ∈ SX with x ⊥ y. From (P1) we have k(x, y) = ‖x+y‖2 . We define g : R → R by g(t) = ‖tx + (1 − t)y‖. g is a convex function and g(12 ) ≤ g(t) ∀ t ∈ [0, 1] . So we have g(12 ) ≤ g(t) ∀ t ∈ R. Hence

    ‖x + y‖ 2

    ≤ ∥∥∥∥y + t(x − y) + x + y2 − x + y2

    ∥∥∥∥ = ∥∥∥∥x + y2 + 2t − 12 (x − y)

    ∥∥∥∥ . So

    ||x + y|| 2

    ≤ ∥∥∥∥x + y2 + λ(x − y)

    ∥∥∥∥ ∀λ ∈ R that is equivalent to

    x + y ⊥ x − y. Therefore ∀ x, y ∈ SX x ⊥ y ⇒ x + y ⊥ x − y that is a characteristic property of inner product spaces [2]. �

    3. Main result

    To prove our main result we start with the following Lemma

    Lemma 2. Let X be a 2-dimensional real normed space and let x be in SX . Let γ be one of the two oriented arcs of SX joining x to −x. Then the function y ∈ γ �→ k(x, y) is nonincreasing.

    Proof. Let ȳ and y ∈ γ and we suppose that ȳ belongs to the part of the arc γ joining −x to y. We will prove that

    k(x, ȳ) ≤ k(x, y).

  • 4 Characterizations of inner product spaces

    We can suppose ȳ = ±x so there exist a, b ∈ R such that y = ax + bȳ. From the assumption about ȳ we have a ≥ 0 and b ≥ 0 and so

    ‖y‖ = 1 = ‖ax + bȳ‖ ≤ a + b .

    If a + b = 1, then y = ax + (1 − a)ȳ hence x, y and ȳ are collinear and this implies:

    1 = k(x, ȳ) = k(x, y) .

    Now we suppose a + b > 1 and k(x, ȳ) > k(x, y). Then there exists t1 ∈ (0, 1) such that ||t1x + (1 − t1)y|| < k(x, ȳ) ≤ 1. Let

    λ = 1

    t1(1 − a − b) + a + b .

    It is easy to verify that λ ∈ (0, 1). Let qλ = λ(t1x + (1 − t1)y)

    = t1

    t1(1 − a − b) + a + bx + (1 − t1)(ax + bȳ)

    t1(1 − a − b) + a + b

    = (t1 + a − at1)

    t1(1 − a − b) + a + bx + (1 − t1)b

    t1(1 − a − b) + a + b ȳ .

    So qλ belongs to the segment joining x to ȳ and

    ‖qλ‖ = λ‖t1x + (1 − t1)y‖ ≥ k(x, ȳ) > ‖t1x + (1 − t1)y‖ .

    So λ > 1. This contradiction implies the thesis. �

    Theorem 3. If X fulfills (P2) then X is an inner product space.

    Proof. We suppose that X is not an inner product space. So by using results (2) and (3) there exist x, y ∈ SX , x ⊥ y such that k(x, y) <

    √ 2

    2 . Let γ be one of the two oriented arcs of SX joining x to −x. Let z1 and z2 be in γ such that x ⊥ z1 , x ⊥ z2 and the part of γ joining z1 to z2 is the arc containing all points z such that x ⊥ z and moreover we suppose that z1 belongs to the part of γ joining x to z2 ; we will write z1 ∈ x̂z2. Note that y ∈ ẑ1z2. Then using a result by Precupanu [7] we can suppose that there exists a sequence (xn) such that:

    1. xn ∈ x̂z1 2. xn = x 3. xn → x 4. xn are smooth.

  • M. Baronti and E. Casini 5

    Let yn be in γ such that xn ⊥ yn. From the monotony of the Birkhoff orthogonality it follows that yn ∈ −̂xz2 Let � be a positive number such that

    k(x, y) + � < √

    2 2

    .

    From Lemma 2 and the continuity of the function k(· , y) there exists n̄ ∈ N such that for n > n̄

    k(xn, yn) ≤ k(xn, y) ≤ k(x, y) + � < √

    2 2

    .

    Now if we fix n0 > n̄, we have

    k(xn0 , yn0) < √

    2 2

    ; xn0 ⊥ yn0 ; xn0 is smooth .

    The continuous function k(xn0 , ·) assumes values between k(xn0 , yn0) < √

    2 2

    and k(xn0 , xn0) = 1. So there exists y∗ = yn0 , y∗ ∈ x̂n0yn0 , k(xn0 , y∗) =√ 2

    2 . From (P2) we have xn0 ⊥ y∗ . But xn0 is a smooth point and so y∗ = yn0 . This contradiction completes the proof. �

    References

    [1] D. Amir, Characterizations of Inner Product Spaces, Birkhauser Verlag, Basel, 1986.

    [2] M. Baronti, Su alcuni parametri degli spazi normati, B.U.M.I.(5), 18-B (1981), 1065–1085.

    [3] C. Benitez and M. Del Rio, The rectangular constant for two- dimensional spaces, J. Approx. Theory, 19 (1977), 15–21.

    [4] C. Benitez and D. Yanez, Two characterizations of inner product spaces, to appear.

    [5] R. C. James, Orthogonality in normed linear spaces, Duke Math. J., 12 (1945), 291–302.

    [6] J. L. Joly, Caracterisations d’espaces hilbertiens au moyen de la constante rectangle, J. Approx. Theory, 2 (1969), 301–311.

    [7] T. Precupanu, Characterizations of hilbertian norms, B.U.M.I.(5), 15- B (1978), 161–169.

  • 6 Characterizations of inner product spaces

    Dipartimento di Ingegneria della Produzione e Metodi e Modelli Matematici Università degli studi di Genova Piazzale Kennedy, Pad. D 16129 Genova Italy (E-mail : baronti@dimet.unige.it)

    Dipartimento di Fisica e Matematica Università dell’Insubria Via Valleggio 11 22100 Como Italy (E-mail : emanuele.casini@uninsubria.it)

    (Received : July 2004 )

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