SOME COMPANIONS OF GRÜSS INEQUALITY IN 2-INNER PRODUCT SPACES AND APPLICATIONS FOR DETERMINANTAL INTEGRAL INEQUALITIES

R.W. FREESE, S.S. DRAGOMIR, Y.J. CHO^{⋆}, AND S.S. KIM^{♦}

Abstract. Some companions of Grüss inequality in 2-inner product spaces and applications for determinantal integral inequalities are given.

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 the book [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 (·, ·|·) 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

1991 Mathematics Subject Classification. Primary 46C05, 46C99; Secondary 26D15, 26D10.

Key words and phrases. Grüss inequality, 2-Inner products, Integral inequality, Determinantal Inequalities ⋆, ♦ Corresponding authors.

2 R.W. FREESE, S.S. DRAGOMIR, Y.J. CHO^{⋆}, AND S.S. 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, (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, for any α ∈ ℝ, that

(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, that

(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 [6]. 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).

In the recent paper [5], the authors have established the following Grüss’ type inequality holding in 2-inner product spaces.

Theorem 1. Let (X, (., .|.)) be a 2-inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e, z ∈ X, with ∥e|z∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in X such that the conditions

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

hold, or equivalently, the following assumptions

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

are valid, then one has the inequality

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

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

The following result improving (1.13) holds (cf. [5, Theorem 1]).

Theorem 2. Let (X, (., .|.)) be an inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e, z ∈ X, with ∥e|z∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in X such that the conditions (1.11), or equivalently, (1.12) hold, then we have the inequality

|(x, y|z) − (x, e|z) (e, y|z)|

≤ \frac{1}{4} |Φ − φ| · |Γ − γ| − [Re (Φe − x, x − φe|z)]^{\frac{1}{2}} [Re (Γe − y, y − γe|z)]^{\frac{1}{2}}

(≤ \frac{1}{4} |Φ − φ| · |Γ − γ|) .

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

The following companion of Grüss inequality in inner product spaces also holds (see [5, Theorem 4]).

Theorem 3. Let (X, (., .|.)) be a 2-inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e, z ∈ X, ∥e|z∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ X are so that either

(1.14)  Re (Γe − \frac{x + y}{2} , \frac{x + y}{2} − γe|z) ≥ 0,

or equivalently,

(1.15)  ∥\frac{x + y}{2} − \frac{γ + Γ}{2} · e|z∥ ≤ \frac{1}{2} |Γ − γ| ,

holds, then we have the inequality

(1.16)  Re [(x, y|z) − (x, e|z) (e, y|z)] ≤ \frac{1}{4} |Γ − γ|^{2} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

The following corollary can be of interest if one wanted to evaluate the absolute value of

Re [(x, y|z) − (x, e|z) (e, y|z)]

(see for details [5]).

Corollary 1. Let (X, (., .|.)) be a 2-inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e, z ∈ X, ∥e, z∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ X are so that

(1.17)  Re (Γe − \frac{x ± y}{2} , \frac{x ± y}{2} − γe|z) ≥ 0

or equivalently,

(1.18)  ∥\frac{x ± y}{2} − \frac{γ + Γ}{2} · e|z∥ ≤ \frac{1}{2} |Γ − γ| ,

then we have the inequality

(1.19)  |Re [(x, y|z) − (x, e|z) (e, y|z)]| ≤ \frac{1}{4} |Γ − γ|^{2} .

If the inner product space X is real, then, for m, M ∈ ℝ, with M > m,

(1.20)  (Me − \frac{x ± y}{2} , \frac{x ± y}{2} − me|z) ≥ 0

or equivalently,

(1.21)  ∥\frac{x ± y}{2} − \frac{m + M}{2} · e|z∥ ≤ \frac{1}{2} (M − m) ,

implies

(1.22)  |(x, y|z) − (x, e|z) (e, y|z)| ≤ \frac{1}{4} (M − m)^{2} .

In both inequalities (1.19) and (1.22) , the constant \frac{1}{4} is best possible.

It is the main aim of this paper to provide other inequalities of Grüss type in the general setting of 2-inner product spaces over the real or complex number field 𝕂. Applications for determinantal integral inequalities are pointed out as well.

2. A Grüss Type Inequality for 2-Inner Products

The following Grüss type inequality in inner product spaces holds.

Theorem 4. Let x, y, z, e ∈ X with ∥e, z∥ = 1 and the scalars a, A, b, B ∈ 𝕂 (𝕂 = ℂ, ℝ) such that Re (āA) > 0 and Re (b̄B) > 0. If

(2.1)  Re (Ae − x, x − ae|z) ≥ 0 and Re (Be − y, y − be|z) ≥ 0,

or equivalently,

(2.2)  ∥x − \frac{a + A}{2}e|z∥ ≤ \frac{1}{2} |A − a| and ∥y − \frac{b + B}{2} e|z∥ ≤ \frac{1}{2} |B − b| ,

then we have the inequality

(2.3)  |(x, y|z) − (x, e|z) (e, y|z)| ≤ \frac{1}{4} \begin{matrix}|A − a| & |B − a| \\ √ & \overline{Re (āA) Re (b̄B)} \end{matrix} |(x, e|z) (e, y|z)| .

The constant \frac{1}{4} in (2.3) is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Let u, x, z, U be vectors in the 2-inner product space (X, (., .|)) over 𝕂 (𝕂 = ℝ,ℂ) with u ≠ U. We claim (see also [5]), that

Re (U − x, x − u|z) ≥ 0

if and only if

∥x − \frac{u + U}{2} |z∥ ≤ \frac{1}{2} ∥U − u|z∥ .

Define

I_{1} := Re (U − x, x − u|z) , I_{2} := \frac{1}{4} ∥U − u|z∥^{2} − ∥x − \frac{u + U}{2} |z∥^{2} .

A simple calculation shows that

I_{1} = I_{2} = Re [(x, u|z) + (U, x|z)] − Re (U, u|z) − ∥x|z∥^{2}

and thus, obviously, I_{1} ≥ 0 if and only if I_{2} ≥ 0 showing, for U := Ae, u := ae (U := Be, u := be) , that the required equivalence holds true.

Apply Schwarz’s inequality in (X; (·, ·|·)) for the vectors x − (x, e|z) e and y − (y, e|z) e, to get (see also [3]) that

(2.4) |(x, y|z) − (x, e|z) (e, y|z)|^{2} ≤ (∥x|z∥^{2} − |(x, e|z)|^{2}) (∥y|z∥^{2} − |(y, e|z)|^{2}) .

Now, assume that u, v ∈ X, and c, C ∈ 𝕂 with Re (c̄C) > 0 and Re (Cv − u, u − cv|z) ≥ 0. This last inequality is equivalent to

(2.5)  ∥u|z∥^{2} + Re (c̄C) ∥v|z∥^{2} ≤ Re [C\overline{(u, v|z)} + c̄(u, v|z)]
(2.6)  = Re [(C̄ + c̄) (u, v|z)] ,

since it is obvious that

Re [C\overline{(u, v|z)}] = Re [C̄ (u, v|z)] .

Dividing this inequality by [Re (Cc̄)]^{\frac{1}{2}} > 0, we deduce

(2.7)  \frac{1}{[Re (c̄C)]^{\frac{1}{2}}} ∥u|z∥^{2} + [Re (c̄C)]^{\frac{1}{2}} ∥v|z∥^{2} ≤ \begin{matrix}Re & [(C̄ + c̄) (u, v|z)] \\ [Re (c̄C)] & \frac{1}{2} \end{matrix}.

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq, α > 0, p, q ≥ 0,

we deduce

(2.8)  2 ∥u|z∥ ∥v|z∥ ≤ \frac{1}{[Re (c̄C)]^{\frac{1}{2}}} ∥u|z∥^{2} + [Re (c̄C)]^{\frac{1}{2}} ∥v|z∥^{2} .

Making use of (2.7) and (2.8) and the fact that, for any z ∈ ℂ, Re (z) ≤ |z| , we get

∥u|z∥ ∥v|z∥ ≤ \frac{Re [(C̄ + c̄) (u, v|z)]}{2 [Re (c̄C)]^{\frac{1}{2}}} ≤ \frac{|c + C|}{2 [Re (c̄C)]^{\frac{1}{2}}} |(u, v|z)| .

Consequently, we have

(2.9)  ∥u|z∥^{2} ∥v|z∥^{2} − |(u, v|z)|^{2} ≤ [\frac{|c + C|^{2}}{4 [Re (c̄C)]} − 1] |(u, v|z)|^{2}

= \frac{1}{4} \begin{matrix}|C − c| & 2 \\ Re & (c̄C) \end{matrix} |(u, v|z)|^{2} .

Now, if we write (2.9) for the choices u = x, v = e and u = y, v = e respectively and use (2.4), we deduce the desired result (2.2). The sharpness of the constant will be proved in the case where X is a real 2-inner product space.

The following corollary which provides a simpler Grüss type inequality for real constants (and, in particular, for real 2-inner product spaces) holds.

Corollary 2. With the assumptions of Theorem 4 and if m, M, n, N ∈ ℝ are such that M > m > 0, N > n > 0 and either

(Me − x, x − me|z) ≥ 0 and (Ne − y, y − ne|z) ≥ 0

or equivalently,

(2.10) ∥x − \frac{m + M}{2} e|z∥ ≤ \frac{1}{2} (M − m) and ∥y − \frac{n + N}{2} e|z∥ ≤ \frac{1}{2} (N − n) ,

holds, then we have the inequality

(2.11)  |(x, y|z) − (x, e|z) (e, y|z)| ≤ \frac{1}{4} · \begin{matrix}(M − m) & (N − n) \\ √ & \overline{mnMN} \end{matrix} |(x, e|z) (e, y|z)| .

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

Proof. We will prove the best constant in (2.11) for x = y. To do that, let assume there is a constant k > 0 so that

(2.12)  ∥x|z∥^{2} − (x, e|z)^{2} ≤ k · \frac{(M − m)^{2}}{mM} (x, e|z)^{2}

provided

(2.13)  (Me − x, x − me|z) ≥ 0,

where M > m > 0, e, z ∈ X with ∥e|z∥ = 1 and x ∈ X.

For ε > 0, consider M := 2 + ε, m := ε > 0, y ∈ X with ∥y|z∥ = 1 and (e, y|z) = 0. Define x := (1 + ε) e + y. With these choices, we have

(Me − x, x − me|z) = ((2 + ε) e − (1 + ε) e − y, (1 + ε) e + y − εe|z)
= (e − y, e + y|z) = ∥e|z∥^{2} − ∥y|z∥^{2} = 0

and thus the condition (2.13) is obviously satisfied.

On the other hand,

∥x|z∥^{2} = ∥(1 + ε) e + y|z∥^{2} = (1 + ε)^{2} + 1

and

(x, e|z)^{2} = ((1 + ε) e + y, e|z)^{2} = (1 + ε)^{2} .

Replacing the above values in the inequality (2.12), we get

(2.14)  1 ≤ \frac{4k (ε + 1)^{2}}{ε (ε + 2)}

for any ε > 0. Letting ε → +∞, we deduce k ≥ \frac{1}{4}, proving the fact that \frac{1}{4} is the best constant in (2.11).

Remark 1. If (x, e|z) , (y, e|z) are assumed not to be zero, then the inequality (2.3) is equivalent to

(2.15)  |\frac{(x, y|z)}{(x, e|z) (e, y|z)} − 1| ≤ \frac{1}{4} \begin{matrix}|A − a| & |B − a| \\ √ & \overline{Re (āA) Re (b̄B)} \end{matrix},

while the inequality (2.11) is equivalent to

(2.16)  |\frac{(x, y|z)}{(x, e|z) (e, y|z)} − 1| ≤ \frac{1}{4} · \begin{matrix}(M − m) & (N − n) \\ √ & \overline{mnMN} \end{matrix}.

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

3. Some Related Results

The following result which provides a generalisation of Theorem 3 holds.

Theorem 5. Let (X; (·, ·|·)) be a 2-inner product space over 𝕂 (𝕂 = ℂ, ℝ) . If γ, Γ ∈ 𝕂, e, x, y, z ∈ X with ∥e, z∥ = 1 and λ ∈ (0, 1) are such that

(3.1)  Re (Γe − (λx + (1 − λ) y) , (λx + (1 − λ) y) − γe|z) ≥ 0,

or equivalently,

(3.2)  ∥λx + (1 − λ) y − \frac{γ + Γ}{2}e|z∥ ≤ \frac{1}{2} |Γ − γ| ,

then we have the inequality

(3.3)  Re [(x, y|z) − (x, e|z) (e, y|z)] ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

The constant \frac{1}{16} is the best possible constant in (3.3) in the sense that it cannot be replaced by a smaller one.

Proof. We know that for any z, u ∈ X one has

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

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

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

Since

(x, y|z) − (x, e|z) (e, y|z) = (x − (x, e|z) e, y − (y, e|z) e|z) (as ∥e|z∥ = 1), using (3.4), we have

Re [(x, y|z) − (x, e|z) (e, y|z)] (3.5)

= Re [(x − (x, e|z) e, y − (y, e|z) e|z)]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − (x, e|z) e) + (1 − λ) (y − (y, e|z) e) |z∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − (λx + (1 − λ) y, e|z) e|z∥^{2} .

Since, for m, e, z ∈ X with ∥e|z∥ = 1, one has the equality

(3.6)  ∥m − (m, e|z) e|z∥^{2} = ∥m|z∥^{2} − |(m, e|z)|^{2} ,

then, by (3.5), we deduce the inequality

Re [(x, y|z) − (x, e|z) (e, y|z)] (3.7)

≤ \frac{1}{4λ (1 − λ)} [∥λx + (1 − λ) y|z∥^{2} − |(λx + (1 − λ) y, e|z)|^{2}] .

Now, if we apply Grüss’ inequality

0 ≤ ∥a|z∥^{2} − |(a, e|z)|^{2} ≤ \frac{1}{4} |Γ − γ|^{2}

provided

Re (Γe − a, a − γe|z) ≥ 0,

for a = λx + (1 − λ) y, then we have

(3.8)  ∥λx + (1 − λ) y|z∥^{2} − |(λx + (1 − λ) y, e|z)|^{2} ≤ \frac{1}{4} |Γ − γ|^{2} .

Utilising (3.7) and (3.8) we deduce the desired inequality (3.3).

To prove the sharpness of the constant \frac{1}{16}, assume that (3.3) holds with a constant C > 0, provided (3.1) is valid, i.e.,

(3.9)  Re [(x, y|z) − (x, e|z) (e, y|z)] ≤ C · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

If in (3.9) we choose x = y, provided (3.1) holds with x = y and λ ∈ (0, 1) , then

(3.10)  ∥x|z∥^{2} − |(x, e|z)|^{2} ≤ C · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} ,

provided

Re (Γe − x, x − γe|z) ≥ 0.

Since we know, in Grüss’ inequality, that the constant \frac{1}{4} is best possible, then by (3.10), one has

\frac{1}{4} ≤ \frac{C}{λ (1 − λ)}

for λ ∈ (0, 1) , which gives, for λ = \frac{1}{2}, that C ≥ \frac{1}{16}.

The theorem is completely proved.

The following corollary is a natural consequence of the above result:

Corollary 3. Assume that γ, Γ, e, x, y and λ are as in Theorem 5. If

(3.11)  Re (Γe − (λx ± (1 − λ) y) , (λx ± (1 − λ) y) − γe|z) ≥ 0,

or equivalently,

(3.12)  ∥λx ± (1 − λ) y − \frac{γ + Γ}{2}e|z∥ ≤ \frac{1}{2} |Γ − γ|^{2} ,

then we have the inequality

(3.13)  |Re [(x, y|z) − (x, e|z) (e, y|z)]| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

The constant \frac{1}{16} is best possible in (3.13).

Proof. Using Theorem 5 for (−y) instead of y, we have that

Re (Γe − (λx − (1 − λ) y) , (λx − (1 − λ) y) − γe|z) ≥ 0,

which implies that

Re [− (x, y|z) + (x, e|z) (e, y|z)] ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

This last inequality shows that

(3.14)  Re [(x, y|z) − (x, e|z) (e, y|z)] ≥ −\frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

Consequently, by (3.3) and (3.14) we deduce the desired inequality (3.13).

Remark 2. If M, m ∈ ℝ with M > m and, for λ ∈ (0, 1) ,

(3.15)  ∥λx + (1 − λ) y − \frac{M + m}{2}e|z∥ ≤ \frac{1}{2} (M − m)

then

(x, y|z) − (x, e|z) (e, y|z) ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} (M − m)^{2} .

If (3.15) holds with “±” instead of “+” , then

(3.16)  |(x, y|z) − (x, e|z) (e, y|z)| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} (M − m)^{2} .

Remark 3. If λ = \frac{1}{2} in (3.1) or (3.2), then we obtain the result from [5] mentioned in Theorem 3.

For λ = \frac{1}{2}, Corollary 3 and Remark 2 will produce the corresponding results obtained in [5]. We omit the details.

The following similar result with the one incorporated in Theorem 5 may be stated as well.

Theorem 6. Assume that γ, Γ, e, x, y and λ are as in Theorem 5. If Re (Γ\overline{γ}) > 0, then we have the inequality:

Re [(x, y|z) − (x, e|z) (e, y|z)] (3.17)

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} · \begin{matrix}|Γ − γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |(λx + (1 − λ) y, e|z)|^{2} .

The constant \frac{1}{16} is best possible in (3.17).

Proof. If we apply the companion of Grüss inequality incorporated in Theorem 4, then we may state that

∥λx + (1 − λ) y|z∥^{2} − |(λx + (1 − λ) y, e|z)|^{2} (3.18)

≤ \frac{|Γ − γ|^{2}}{4 Re (Γ\overline{γ})} |(λx + (1 − λ) y, e|z)|^{2} .

Utilising (3.7) and (3.18) we deduce the desired inequality (3.17). The sharpness of the constant may be shown in a similar way to the one used in Theorem 5 and we omit the details.

Finally, we may state the following

Corollary 4. Assume that γ, Γ, e, x, y and λ are as in Corollary 3. If Re (Γ\overline{γ}) > 0, then we have the inequality:

|Re [(x, y|z) − (x, e|z) (e, y|z)]| (3.19)

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} · \begin{matrix}|Γ − γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |(λx + (1 − λ) y, e|z)|^{2} .

The constant \frac{1}{16} is best possible in (3.19).

Remark 4. The particular case λ = \frac{1}{2} may produce some particular inequalities of interest, but we omit the details.

4. 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 (4.1)

(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 by

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

we denote the determinant of the matrix

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

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

(4.2) ∥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

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

and

(4.4)  ∥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 a.e. x ∈ Ω. Then, by the definition of 2-inner product (f, g|h)_{ρ} , we have

(f, g|h)_{ρ} (4.5)

= \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

(4.6)  (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 (4.3), (4.4) and the inequalities for 2-inner products and 2-norms established in the previous sections, one may state some interesting determinantal integral inequalities as follows:

Proposition 1. Let f, g, h, u ∈ L_{ρ}^{2} (Ω) with h ≠ 0 a.e. and

(4.7)  ∫_{Ω} ρu^{2}dμ ∫_{Ω} ρh^{2}dμ − (∫_{Ω} ρuhdμ)^{2} = 1.

If M > m and N > n are real numbers with the property that the functions

(4.8)  (M · \frac{u}{h} − \frac{f}{h} , \frac{f}{h} − m · \frac{u}{h}) and (N · \frac{u}{h} − \frac{g}{h}, \frac{g}{h} − n · \frac{u}{h})

are synchronous on Ω, then we have the following determinantal integral Grüss type inequality

|det [ \begin{matrix}∫_{Ω} ρfgdμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρghdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]

− det [ \begin{matrix}∫_{Ω} ρfudμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] · det [ \begin{matrix}∫_{Ω} ρgudμ & ∫_{Ω} ρghdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]|

≤ \frac{1}{4} \begin{matrix}(M − m) & (N − n) \\ √ & \overline{mMnM} \end{matrix}

× |det [ \begin{matrix}∫_{Ω} ρfudμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] · det [ \begin{matrix}∫_{Ω} ρgudμ & ∫_{Ω} ρghdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]|

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

The proof follows by Theorem 4 applied for the 2-inner product (., .|.)_{ρ} defined in (4.1).

Proposition 2. Let f, g, h, u ∈ L_{ρ}^{2} (Ω) with h ≠ 0 a.e. and

∫_{Ω} ρu^{2}dμ ∫_{Ω} ρh^{2}dμ − (∫_{Ω} ρuhdμ)^{2} = 1.

If M > m and N > n and λ ∈ (0, 1) are real numbers with the property that the functions

(4.9)  (M · \frac{u}{h} − (λ\frac{f}{h} + (1 − λ) \frac{g}{h}) , λ\frac{f}{h} + (1 − λ) \frac{g}{h} − m · \frac{u}{h})

are synchronous on Ω, then we have the following determinantal integral Grüss type inequality

J := det [ \begin{matrix}∫_{Ω} ρfgdμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρghdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]

− det [ \begin{matrix}∫_{Ω} ρfudμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] · det [ \begin{matrix}∫_{Ω} ρgudμ & ∫_{Ω} ρghdμ \\ ∫_{Ω} ρuhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ]

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} · (M − m)^{2} .

If (4.9) holds with ” ± ” instead of ” + ”, then

|J| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} · (M − m)^{2} .

The proof is obvious by the inequality (3.15) and we omit the details.

Remark 5. It is obvious that if one chooses the discrete measure above, then all the inequalities in this section may be written for sequences of real or complex numbers. We omit the details.

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, P.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, A generalization of Grüss’ inequality in inner product spaces and applications, J. Math. Anal. Appl., 237(1999), 74-82.

[4] S.S. DRAGOMIR and I. GOMM, Some integral and discrete versions of the Grüss inequality for real and complex functions and sequences, RGMIA Res. Rep. Coll., 5(2003), No. 3, Article

   9 [ON LINE http://rgmia.vu.edu.au/v5n3.html]

[5] S.S. DRAGOMIR, Y.J. CHO, S.M. KANG, S.S. KIM and J.S. JUNG, Some Grüss’ type inequalities in 2-inner product spaces and applications for determinantal integral inequalities, Preprint

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

Department of Mathematics and Computer Science Saint Louis University Saint Louis, MO 63103, USA

   E-mail address: freeserw@slu.edu

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