ON BESSEL’S AND GRÜSS’ INEQUALITIES FOR ORTHORNORMAL FAMILIES IN 2-INNER PRODUCT SPACES

AND APPLICATIONS

S.S. DRAGOMIR, Y.J. CHO^{⋆}, S.S. KIM^{♦}, AND Y.-H. KIM

Abstract. A new counterpart of Bessel’s inequality for orthornormal families in real or complex 2-inner product spaces is obtained. Applications for some Grüss type results with applications for determinantal integral inequalities are also provided.

1. Introduction

The concepts of 2-inner products and 2-inner product spaces have been intensively studied by many authors in the last three decades. A systematic presentation of the recent results related to the theory of 2-inner product spaces as well as an extensive list of the related references can be found in [1]. Here we give the basic definitions and the elementary properties of 2-inner product spaces.

Let X be a linear space of dimension greater than 1 over the field 𝕂 = ℝ of real numbers or the field 𝕂 = ℂ of complex numbers. Suppose that (·, ·|·) is a 𝕂-valued function defined on X × X × X satisfying the following conditions:

(2I_{1}) (x, x|z) ≥ 0 and (x, x|z) = 0 if and only if x and z are linearly dependent, (2I_{2}) (x, x|z) = (z, z|x), (2I_{3}) (y, x|z) = \overline{(x, y|z)}, (2I_{4}) (αx, y|z) = α(x, y|z) for any scalar α ∈ 𝕂, (2I_{5}) (x + x′, y|z) = (x, y|z) + (x′, y|z). (·, ·|·) is called a 2-inner product on X and (X, (·, ·|·)) is called a 2-inner product

space (or 2-pre-Hilbert space). Some basic properties of 2-inner product spaces can be immediately obtained as follows [2]:

(1) If 𝕂 = ℝ, then (2I_{3}) reduces to

(y, x|z) = (x, y|z).

(2) From (2I_{3}) and (2I_{4}), we have

(0, y|z) = 0, (x, 0|z) = 0

and also

(1.1)  (x, αy|z) = ᾱ(x, y|z).

(3) Using (2I_{2})-(2I_{5}), we have

2000 Mathematics Subject Classification. 46C05, 46C99, 26D15, 26D10.

Key words and phrases. Bessel’s inequality, Grüss’ inequality, 2- Inner product spaces, Lebesgue integral, Synchronous functions.

2 S.S. DRAGOMIR, Y.J. CHO^{⋆}, S.S. KIM^{♦}, AND Y.-H. KIM

(z, z|x ± y) = (x ± y, x ± y|z) = (x, x|z) + (y, y|z) ± 2Re(x, y|z)

and

(1.2)  Re(x, y|z) = \frac{1}{4}[(z, z|x + y) − (z, z|x − y)].

In the real case 𝕂 = ℝ, (1.2) reduces to

(1.3)  (x, y|z) = \frac{1}{4}[(z, z|x + y) − (z, z|x − y)]

and, using this formula, it is easy to see that, for any α ∈ ℝ,

(1.4)  (x, y|αz) = α^{2}(x, y|z).

In the complex case, using (1.1) and (1.2), we have

Im(x, y|z) = Re[−i(x, y|z)] = \frac{1}{4}[(z, z|x + iy) − (z, z|x − iy)],

which, in combination with (1.2), yields

(1.5) (x, y|z) = \frac{1}{4}[(z, z|x + y) − (z, z|x − y)] + \frac{i}{4}[(z, z|x + iy) − (z, z|x − iy)].

Using the above formula and (1.1), we have, for any α ∈ ℂ,

(1.6)  (x, y|αz) = |α|^{2}(x, y|z).

However, for α ∈ ℝ, (1.6) reduces to (1.4).

Also, from (1.6) it follows that

(x, y|0) = 0.

(4) For any three given vectors x, y, z ∈ X, consider the vector u = (y, y|z)x − (x, y|z)y. By (2I_{1}), we know that (u, u|z) ≥ 0 with the equality if and only if u and z are linearly dependent. The inequality (u, u|z) ≥ 0 can be rewritten as,

(1.7)  (y, y|z)[(x, x|z)(y, y|z) − |(x, y|z)|^{2}] ≥ 0.

For x = z, (1.7) becomes

−(y, y|z)|(z, y|z)|^{2} ≥ 0,

which implies that

(1.8)  (z, y|z) = (y, z|z) = 0

provided y and z are linearly independent. Obviously, when y and z are linearly dependent, (1.8) holds too. Thus (1.8) is true for any two vectors y, z ∈ X. Now, if y and z are linearly independent, then (y, y|z) > 0 and, from (1.7), it follows that

(1.9)  |(x, y|z)|^{2} ≤ (x, x|z)(y, y|z).

Using (1.8), it is easy to check that (1.9) is trivially fulfilled when y and z are linearly dependent. Therefore, the inequality (1.9) holds for any three vectors x, y, z ∈ X and is strict unless the vectors u = (y, y|z)x−(x, y|z)y and z are linearly dependent. In fact, we have the equality in (1.9) if and only if the three vectors x, y and z are linearly dependent.

In any given 2-inner product space (X, (·, · | ·)), we can define a function ∥ · | · ∥ on X × X by

(1.10)  ∥x|z∥ = \sqrt{(x, x|z)}

for all x, z ∈ X.

It is easy to see that this function satisfies the following conditions:

(2N_{1}) ∥x|z∥ ≥ 0 and ∥x|z∥ = 0 if and only if x and z are linearly dependent, (2N_{2}) ∥z|x∥ = ∥x|z∥, (2N_{3}) ∥αx|z∥ = |α|∥x|z∥ for any scalar α ∈ 𝕂, (2N_{4}) ∥x + x′|z∥ ≤ ∥x|z∥ + ∥x′|z∥. Any function ∥ · | · ∥ defined on X × X and satisfying the conditions (2N_{1})-(2N_{4})

is called a 2-norm on X and (X, ∥ · | · ∥) is called a linear 2-normed space [5]. Whenever a 2-inner product space (X, (·, ·|·)) is given, we consider it as a linear 2-normed space (X, ∥ · | · ∥) with the 2-norm defined by (1.10).

Let (X; (·, ·|·)) be a 2-inner product space over the real or complex number field 𝕂. If (f_{i})_{1≤i≤n} are linearly independent vectors in the 2-inner product space X, and, for a given z ∈ X, (f_{i}, f_{j}|z) = δ_{ij} for all i, j ∈ {1, …, n} where δ_{ij} is the Kronecker delta (we say that the family (f_{i})_{1≤i≤n} is z−orthonormal), then the following inequality is the corresponding Bessel’s inequality (see for example [2]) for the z−orthonormal family (f_{i})_{1≤i≤n} in the 2-inner product space (X; (·, ·|·)):

(1.11)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ ∥x|z∥^{2}

for any x ∈ X. For more details on this inequality, see the recent paper [2] and the references therein.

The following reverse of Bessel’s inequality in 2-inner product spaces has been obtained in [3]:

Theorem 1. Let {e_{i}}_{i∈I} , F, ϕ_{i}, Φ_{i}, i ∈ F and x, z ∈ X so that either (i) Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0

or, equivalently,

(ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}}

holds. Then we have the inequality:

0 ≤ ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z)

(≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}) .

The constant \frac{1}{4} is best possible.

The main aim of the present paper is to establish a different reverse inequality for (1.11). Some companion results and applications for determinantal integral inequalities are also given.

2. Another Reverse of Bessel’s Inequality

The following lemma holds.

Lemma 1. Let {e_{i}}_{i∈I} be a family of z-orthornormal vectors in X, F a finite part of I, λ_{i} ∈ 𝕂, i ∈ F, r > 0 and x ∈ X. If

(2.1)  ∥x − ∑_{i∈F} λ_{i}e_{i}|z∥ ≤ r,

then we have the inequality

(2.2)  0 ≤ ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2} ≤ r^{2} − ∑_{i∈F} |λ_{i} − (x, e_{i}|z)|^{2} .

Proof. Consider

I_{1} := ∥x − ∑_{i∈F} λ_{i}e_{i}|z∥^{2} = (x − ∑_{i∈F} λ_{i}e_{i}, x − ∑_{j∈F} λ_{j}e_{j}|z)

= ∥x|z∥^{2} − ∑_{i∈F} λ_{i}\overline{(x, e_{i}|z)} − ∑_{i∈F} \overline{λ_{i}} (x, e_{i}|z) + ∑_{i∈F} ∑_{j∈F} λ_{i}\overline{λ_{j}} (e_{i}, e_{j}|z)

= ∥x|z∥^{2} − ∑_{i∈F} λ_{i}\overline{(x, e_{i}|z)} − ∑_{i∈F} \overline{λ_{i}} (x, e_{i}|z) + ∑_{i∈F} |λ_{i}|^{2}

and

I_{2} := ∑_{i∈F} |λ_{i} − (x, e_{i}|z)|^{2} = ∑_{i∈F} (λ_{i} − (x, e_{i}|z)) (\overline{λ_{i}} − \overline{(x, e_{i}|z)})

= ∑_{i∈F} [|λ_{i}|^{2} + |(x, e_{i}|z)|^{2} − \overline{λ_{i}} (x, e_{i}|z) − λ_{i}\overline{(x, e_{i}|z)}]

= ∑_{i∈F} |λ_{i}|^{2} + ∑_{i∈F} |(x, e_{i}|z)|^{2} − ∑_{i∈F} \overline{λ_{i}} (x, e_{i}|z) − ∑_{i∈F} λ_{i}\overline{(x, e_{i}|z)}.

If we subtract I_{2} from I_{1}, we deduce an identity that is interesting in its own right:

(2.3)  ∥x − ∑_{i∈F} λ_{i}e_{i}|z∥^{2} − ∑_{i∈F} |λ_{i} − (x, e_{i}|z)|^{2} = ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2} ,

from which we easily deduce (2.2).

The following reverse of Bessel’s inequality holds.

Theorem 2. Let {e_{i}}_{i∈I} be a family of z-orthornormal vectors in X, F a finite part of I, ϕ_{i}, Φ_{i}, i ∈ I real or complex numbers. For x ∈ X, if either

(i) Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0,

or, equivalently,

(ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}}

holds, then the following reverse of Bessel’s inequality

(2.4)  0 ≤ ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{ϕ_{i} + Φ_{i}}{2} − (x, e_{i}|z)|^{2}

(≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}) ,

is valid. The constant \frac{1}{4} is best possible.

Proof. Firstly, we observe that, for y, a, A ∈ X, the following are equivalent

(2.5)  Re (A − y, y − a|z) ≥ 0

and

(2.6)  ∥y − \frac{a + A}{2}|z∥ ≤ \frac{1}{2} ∥A − a|z∥ .

Now, for a = ∑_{i∈F} ϕ_{i}e_{i}, A = ∑_{i∈F} Φ_{i}e_{i}, we have

∥A − a|z∥ = ∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}|z∥ = (∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}|z∥^{2})^{\frac{1}{2}}

= (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ∥e_{i}|z∥^{2})^{\frac{1}{2}} = (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

which gives, for y = x, the desired equivalence.

Now, if we apply Lemma 1 for λ_{i} = \frac{ϕ_{i}+Φ_{i}}{2} and

r := \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

then we deduce the first inequality in (2.4).

Let us prove that \frac{1}{4} is best possible in the second inequality in (2.4).

Assume that there is a c > 0 such that

(2.7) 0 ≤ ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2} ≤ c ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{ϕ_{i} + Φ_{i}}{2} − (x, e_{i}|z)|^{2} ,

provided that ϕ_{i}, Φ_{i}, x and F satisfy (i) or (ii).

Now, let F = {1} , e_{1} = e, ∥e|z∥ = 1 and m ∈ X so that ∥m|z∥ = 1 and (m, e|z) = 0. For Φ_{1} = Φ, ϕ_{1} = ϕ, Φ ≠ ϕ, define the vector

x := \frac{Φ + ϕ}{2}e + \frac{Φ − ϕ}{2}m.

A simple calculation shows that

(Φe − x, x − ϕe|z) = |\frac{Φ − ϕ}{2}|^{2} (e − x, x − e|z) = 0

and thus the condition (i) of the theorem holds true for F = {1} .

Observe also that

∥x|z∥^{2} = ∥\frac{Φ + ϕ}{2}e + \frac{Φ − ϕ}{2}m|z∥^{2}

= |\frac{Φ + ϕ}{2}|^{2} + |\frac{Φ − ϕ}{2}|^{2}

and

(x, e|z) = (\frac{Φ + ϕ}{2}e + \frac{Φ − ϕ}{2}m, e|z) = \frac{Φ + ϕ}{2}.

Consequently, by (2.7), we deduce

|\frac{Φ − ϕ}{2}|^{2} ≤ c |Φ − ϕ|^{2} ,

which gives c ≥ \frac{1}{4}, and the proof is completed.

Remark 1. If F = {1} , e_{1} = e, ∥e|z∥ = 1 and, for ϕ, Φ ∈ 𝕂 and x ∈ X, one has either

(2.8)  Re (Φe − x, x − ϕe|z) ≥ 0

or, equivalently,

(2.9)  ∥x − \frac{ϕ + Φ}{2}e|z∥ ≤ \frac{1}{2} |Φ − ϕ| ,

then

(2.10)  0 ≤ ∥x|z∥^{2} − |(x, e|z)|^{2}

≤ \frac{1}{4} |Φ − ϕ|^{2} − |\frac{ϕ + Φ}{2} − (x, e|z)|^{2}

(≤ \frac{1}{4} |Φ − ϕ|^{2}) .

The constant \frac{1}{4} is best possible.

3. A Refinement of the Grüss Inequality

The following result holds.

Theorem 3. Let {e_{i}}_{i∈I} be a family of z-orthornormal vectors in X, F a finite part of I, ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂, i ∈ F and x, y ∈ X. If either

(3.1)  Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0,

Re (∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}|z) ≥ 0

or, equivalently,

(3.2)  ∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,
∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

hold, then we have the inequalities

(3.3)  0 ≤ |(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (x, e_{i}|z)| |\frac{Γ_{i} + γ_{i}}{2} − (y, e_{i}|z)|

(≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ) .

The constant \frac{1}{4} is best possible. Proof. Using Schwarz’s inequality in the 2-inner product space (X, (·, ·|·)) , one has

|(x − ∑_{i∈F} (x, e_{i}) e_{i}, y − ∑_{i∈F} (y, e_{i}) e_{i}|z)|^{2} (3.4)

≤ ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}|z∥^{2} ∥y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}|z∥^{2}

and, since a simple calculation shows that

(x − ∑_{i∈F} (x, e_{i}) e_{i}, y − ∑_{i∈F} (y, e_{i}) e_{i}|z) = (x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)

and

∥x − ∑_{i∈F} (x, e_{i}) e_{i}|z∥^{2} = ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2}

for any x, y ∈ X, then, by (3.4) and the reverse of Bessel’s inequality in Theorem 2, we have

|(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)|^{2} (3.5)

≤ (∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2}) (∥y|z∥^{2} − ∑_{i∈F} |(y, e_{i}|z)|^{2})

≤ [\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (x, e_{i}|z)|^{2}]

× [\frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} − ∑_{i∈F} |\frac{Γ_{i} + γ_{i}}{2} − (y, e_{i}|z)|^{2}]

:= K.

Using Aczél’s inequality for real numbers, i.e., we recall that

(3.6)  (a^{2} − ∑_{i∈F} a_{i}^{2}) (b^{2} − ∑_{i∈F} b_{i}^{2}) ≤ (ab − ∑_{i∈F} a_{i}b_{i})^{2} ,

provided that a, b, a_{i}, b_{i} > 0, i ∈ F, (originally, Aczél proved it under more restrictive assumptions for a, b, a_{i}, b_{i}, i.e., either a^{2} − ∑_{i∈F} a_{i}^{2} or b^{2} − ∑_{i∈F} b_{i}^{2} are nonnegative, but those conditions are not necessary), we may state that

(3.7)  K ≤ [\frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (x, e_{i}|z)| |\frac{Γ_{i} + γ_{i}}{2} − (y, e_{i}|z)|]^{2} .

Using (3.5) and (3.7) we conclude that

|(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)|^{2} (3.8)

≤ [\frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (x, e_{i}|z)| |\frac{Γ_{i} + γ_{i}}{2} − (y, e_{i}|z)|]^{2} .

Taking the square root in (3.8) and taking into account that the quantity in the last square brackets is nonnegative (see for example (2.4)), we deduce the second inequality in (3.3).

The fact that \frac{1}{4} is the best possible constant follows by Theorem 2 and we omit the details.

The following corollary may be stated.

BESSEL AND GRÜSS INEQUALITIES FOR ORTHORNORMAL FAMILIES 9

Corollary 1. Let e ∈ X, ∥e|z∥ = 1, ϕ, Φ, γ, Γ ∈ 𝕂 and x, y ∈ X such that either

(3.9)  Re (Φe − x, x − ϕe|z) ≥ 0, Re (Γe − y, y − γe|z) ≥ 0

or, equivalently,

(3.10) ∥x − \frac{ϕ + Φ}{2}e|z∥ ≤ \frac{1}{2} |Φ − ϕ| , ∥y − \frac{γ + Γ}{2}e|z∥ ≤ \frac{1}{2} |Γ − γ| .

Then we have the following refinement of Grüss’ inequality

(3.11)  0 ≤ |(x, y|z) − (x, e|z) (e, y|z)|

≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| − |\frac{ϕ + Φ}{2} − (x, e|z)| |\frac{γ + Γ}{2} − (y, e|z)|

(≤ \frac{1}{4} |Φ − ϕ| |Γ − γ|) .

The constant \frac{1}{4} is best possible.

4. Some Companion Inequalities

The following companion of the Grüss inequality also holds. Theorem 4. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in X, F a finite part of I and ϕ_{i}, Φ_{i} ∈ 𝕂, i ∈ F, x, y ∈ X and λ ∈ (0, 1) such that either

(4.1)  Re (∑_{i∈F} Φ_{i}e_{i} − (λx + (1 − λ) y) , λx + (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0

or, equivalently,

(4.2)  ∥λx + (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}}

holds. Then we have the inequality

Re [(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)] (4.3)

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}

− \frac{1}{4} \frac{1}{λ (1 − λ)} ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (λx + (1 − λ) y, e_{i}|z)|^{2}

(≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}) .

The constant \frac{1}{16} is the best possible constant in (4.3) in the sense that it cannot be replaced by a smaller constant. Proof. We know, for any z, u, v ∈ X, that one has

Re (z, u|v) ≤ \frac{1}{4} ∥z + u|v∥^{2} .

Then, for any a, b, z ∈ X and λ ∈ (0, 1) , one has

(4.4)  Re (a, b|z) ≤ \frac{1}{4λ (1 − λ)} ∥λa + (1 − λ) b|z∥^{2} .

Since

(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z) = (x − ∑_{i∈F} (x, e_{i}) e_{i}, y − ∑_{i∈F} (y, e_{i}) e_{i}|z) ,

for any x, y ∈ X, then, by (4.4), we get

Re [(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)] (4.5)

= Re [(x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}|z)]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − ∑_{i∈F} (x, e_{i}) e_{i}) + (1 − λ) (y − ∑_{i∈F} (y, e_{i}) e_{i}) |z∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − ∑_{i∈F} (λx + (1 − λ) y, e_{i}) e_{i}|z∥^{2}

= \frac{1}{4λ (1 − λ)} [∥λx + (1 − λ) y|z∥^{2} − ∑_{i∈F} |(λx + (1 − λ) y, e_{i}|z)|^{2}] .

If we apply the reverse of Bessel’s inequality in Theorem 2 for λx + (1 − λ) y, we may state that

∥λx + (1 − λ) y|z∥^{2} − ∑_{i∈F} |(λx + (1 − λ) y, e_{i}|z)|^{2} (4.6)

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (λx + (1 − λ) y, e_{i}|z)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Now, by making use of (4.5) and (4.6), we deduce (4.3).

The fact that \frac{1}{16} is the best possible constant in (4.3) follows by the fact that if in (4.1) we choose x = y, then it becomes (i) of Theorem 2, which implies for λ = \frac{1}{2} the inequality (2.4), and so we have shown that \frac{1}{4} is the best constant.

Remark 2. In practical applications, we may use only the inequality between the first and the last term in (4.3).

BESSEL AND GRÜSS INEQUALITIES FOR ORTHORNORMAL FAMILIES 11

Remark 3. If, in Theorem 4, we choose λ = \frac{1}{2}, then we get

Re [(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)] (4.7)

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − (\frac{x + y}{2} , e_{i}|z)|^{2}

(≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}) ,

provided

Re (∑_{i∈F} Φ_{i}e_{i} − \frac{x + y}{2} , \frac{x + y}{2} − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0

or, equivalently,

∥\frac{x + y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

Corollary 2. With the assumptions of Theorem 4 and if

(4.8)  Re (∑_{i∈F} Φ_{i}e_{i} − (λx ± (1 − λ) y) , λx ± (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}|z) ≥ 0

or, equivalently,

(4.9) ∥λx ± (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}|z∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

then we have the inequality

(4.10) |Re [(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)]| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{16} is best possible in (4.10). Remark 4. If X is a real inner product space and m_{i}, M_{i} ∈ ℝ with the property

(4.11)  (∑_{i∈F} M_{i}e_{i} − (λx ± (1 − λ) y) , λx ± (1 − λ) y − ∑_{i∈F} m_{i}e_{i}|z) ≥ 0

or, equivalently,

(4.12) ∥λx ± (1 − λ) y − ∑_{i∈F} \frac{M_{i} + m_{i}}{2} · e_{i}|z∥ ≤ \frac{1}{2} [∑_{i∈F} (M_{i} − m_{i})^{2}]^{\frac{1}{2}} ,

then we have the Grüss type inequality

(4.13)  |(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} (M_{i} − m_{i})^{2} .

5. Applications for Determinantal Integral Inequalities

Let (Ω, Σ, μ) be a measure space consisting of a set Ω, Σ a σ−algebra of subsets of Ω and μ a countably additive and positive measure on Σ with values in ℝ ∪ {∞}.

Denote by L_{ρ}^{2} (Ω) the Hilbert space of all real-valued functions f defined on Ω that are 2−ρ−integrable on Ω, i.e., ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) < ∞, where ρ : Ω → [0, ∞) is a measurable function on Ω.

We can introduce the following 2-inner product on L_{ρ}^{2} (Ω) by formula (5.1)

(f, g|h)_{ρ} := \frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (s) ρ (t) | \begin{matrix}f (s) & f (t) \\ h (s) & h (t) \end{matrix} | | \begin{matrix}g (s) & g (t) \\ h (s) & h (t) \end{matrix} | dμ (s) dμ (t) ,

where

| \begin{matrix}f (s) & f (t) \\ h (s) & h (t) \end{matrix} |

denotes the determinant of the matrix

[ \begin{matrix}f (s) & f (t) \\ h (s) & h (t) \end{matrix} ] ,

which generates the 2-norm on L_{ρ}^{2} (Ω) expressed by

(5.2) ∥f|h∥_{ρ} := (\frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (s) ρ (t) | \begin{matrix}f (s) & f (t) \\ h (s) & h (t) \end{matrix} |^{2} dμ (s) dμ (t))^{1/2} .

A simple calculation with integrals reveals that

(5.3)  (f, g|h)_{ρ} = | \begin{matrix}∫_{Ω} ρfgdμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρghdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |

and

(5.4)  ∥f|h∥_{ρ} = | \begin{matrix}∫_{Ω} ρf^{2}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |^{1/2} ,

where, for simplicity, instead of ∫_{Ω} ρ (s) f (s) g (s) dμ (s) , we have written ∫_{Ω} ρfgdμ.

We recall that the pair of functions (q, p) ∈ L_{ρ}^{2} (Ω)×L_{ρ}^{2} (Ω) is called synchronous if

(q (x) − q (y)) (p (x) − p (y)) ≥ 0

for a.e. x, y ∈ Ω.

We note that, if Ω = [a, b] , then a sufficient condition for synchronicity is that the functions are both monotonic increasing or decreasing. This condition is not necessary.

Now, suppose that h ∈ L_{ρ}^{2} (Ω) is such that h (x) ≠ 0 for μ − a.e. x ∈ Ω. Then, by the definition of 2-inner product (f, g|h)_{ρ} , we have

(f, g|h)_{ρ} (5.5)

= \frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (s) ρ (t) h^{2} (s) h^{2} (t) (\frac{f (s)}{h (s)} − \frac{f (t)}{h (t)}) (\frac{g (s)}{h (s)} − \frac{g (t)}{h (t)}) dμ (s) dμ (t)

and thus a sufficient condition for the inequality

(5.6)  (f, g|h)_{ρ} ≥ 0

to hold, is that, the functions (\frac{f}{h}, \frac{g}{h}) are synchronous. It is obvious that, this condition is not necessary.

Using the representations (5.3), (5.4) and the inequalities for 2-inner products and 2-norms established in the previous sections, we have some interesting determinantal integral inequalities.

Proposition 1. Let h ∈ L_{ρ}^{2} (Ω) be such that h (x) ≠ 0 for μ − a.e. x ∈ Ω and (f_{i})_{i∈I} a family of functions in L_{ρ}^{2} (Ω) with the property that

| \begin{matrix}∫_{Ω} ρf_{i}f_{j}dμ & ∫_{Ω} ρf_{i}hdμ \\ ∫_{Ω} ρf_{j}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} | = δ_{i,j}

for any i, j ∈ I, where δ_{i,j} is the Kronecker delta.

If we assume that there exists the real numbers M_{i}, m_{i}, i ∈ F, where F is a given finite part of I, such that the functions

∑_{i∈F} M_{i} · \frac{f_{i}}{h} − \frac{f}{h} , \frac{f}{h} − ∑_{i∈F} m_{i} · \frac{f_{i}}{h}

are synchronous on Ω, then we have the inequalities

0 ≤ | \begin{matrix}∫_{Ω} ρf^{2}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} | − ∑_{i∈F} | \begin{matrix}∫_{Ω} ρf_{i}fdμ & ∫_{Ω} ρf_{i}hdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |^{2}

≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2}

− ∑_{i∈F} |\frac{M_{i} + m_{i}}{2} − det [ \begin{matrix}∫_{Ω} ρf_{i}fdμ & ∫_{Ω} ρf_{i}hdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]|^{2}

(≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2}) .

The proof follows by Theorem 2 applied for the 2-inner product (·, ·|·)_{ρ} and we omit the details.

Similar determinantal integral inequalities may be stated if one uses the other results for 2-inner products obtained above, but we do not present them here.

Acknowledgement: S. S. Dragomir and Y. J. Cho greatly acknowledge the financial support from the Brain Pool Program (2002) of the Korean Federation of Science and Technology Societies. The research was performed under the ”Memorandum of Understanding” between Victoria University and Gyeongsang National University.

References

[1] Y.J. CHO, C.S. LIN, S.S. KIM and A. MISIAK, Theory of 2-Inner Product Spaces, Nova Science Publishers, Inc., New York, 2001.

[2] Y.J. CHO, M. MATIĆ and J.E. PEČARIĆ, On Gram’s determinant in 2-inner product spaces, J. Korean Math. Soc., 38(2001), No. 6, pp. 1125-1156.

[3] S.S. DRAGOMIR, Y.J. CHO, S.S. KIM and J. ROUMELIOTIS, A reverse of Bessel’s inequality in 2-inner product spaces and some Grüss type related results with applications, Preprint on line: http://www.mathpreprints.com/math/Preprint/Sever/20030923.1/2

[4] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993.

[5] R.W. FREESE and Y.J. CHO, Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Inc., New York, 2001.

School of Computer Science & Mathematics, Victoria University, Melbourne, Victoria, Australia

E-mail address: sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au.SSDragomirWeb.html

Department of Mathematics Education, The Research Institute of Natural Sciences, College of Education, Gyeongsang National University, Chinju 660-701, Korea

E-mail address: yjcho@nongae.gsnu.ac.kr

Department of Mathematics, Dongeui University, Pusan, 614-714, Korea

E-mail address: sskim@dongeui.ac.kr

Department of Applied Mathematics, Changwon National University, Changwon, Kyung-Nam 641-773, Korea.

E-mail address: yhkim@sarim.changwon.ac.kr