REFINEMENTS OF SOME REVERSES OF SCHWARZ’S INEQUALITY IN 2−INNER PRODUCT SPACES AND APPLICATIONS FOR INTEGRALS

P. CERONE, Y.J. CHO^{⋆}, S.S. DRAGOMIR, AND S.S. KIM^{♦}

Abstract. Refinements of some recent reverse inequalities for the celebrated Cauchy-Bunyakovsky-Schwarz inequality in 2−inner product spaces are given. Using this framework, 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 the book [5]. We recall here the basic definitions and the elementary properties of 2−inner product spaces that will be used in the sequel (see also [3]).

Let X be a linear space of dimension greater than 1 over the number field 𝕂, when 𝕂 = ℝ or 𝕂 = ℂ. 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) , where x, x′, y, z ∈ X. The functional (·, ·|·) is called a 2−inner product on X and

(X, (·, ·|·)) is called a 2−inner product space (or 2-pre-Hilbert space) [5].

Some basic properties of the 2−inner product spaces can be immediately obtained as follows:

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

and also

(1.1)  (x, αy|z) = ᾱ (x, y|z) .

Date: August 28, 2003.

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

Key words and phrases. 2−Inner product spaces, Schwarz’s inequality, Determinantal integral inequalities ⋆, ♦ Corresponding authors.

2 P. CERONE, Y.J. CHO^{⋆}, S.S. DRAGOMIR, AND S.S. KIM^{♦}

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

(z, z|x ± y) = (x ± y, x ± y|z)
= (x, x|z) + (y, y|z) ± 2 Re (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, for any α ∈ ℝ, that

(1.4)  (x, y|αz) = α^{2} (x, y|z) .

In the complex case, 𝕂 = ℂ, 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 (1.5) 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. It is obvious that 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) also holds.

Now, if y and z are linearly independent, then (y, y|z) > 0, and from (1.7), it follows the Cauchy-Bunyakovsky-Schwarz inequality (CBS−inequality for short) for 2−inner products:

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

Utilizing (1.8), it is easy to see 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 [3].

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 [9]. In terms of 2-norms, the (CBS) −inequality (1.9) can be written as

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

The equality in (1.11) holds if and only if x, y and z are linearly dependent.

For recent inequalities in 2-inner products, see the recent works [1] - [13] and the references therein.

In [7], the authors pointed out the following reverses of the (CBS) −inequality in 2-inner product spaces.

Assume that x, y, z ∈ X and a, A ∈ 𝕂 are such that either

(1.12)  Re (Ay − x, x − ay|z) ≥ 0

or, equivalently

(1.13)  ∥x − \frac{a + A}{2}, y|z∥ ≤ \frac{1}{2} |A − a| ∥y|z∥

hold. Then one has the inequality [7]

(1.14)  0 ≤ ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2} ≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{4} .

The constant \frac{1}{4} is sharp in (1.14) in the sense that it cannot be replaced by a smaller constant.

With the same assumptions for x, y, z, a and A and, if moreover Re (āA) > 0, then [7]

(1.15)  ∥x|z∥ ∥y|z∥ ≤ \frac{1}{2} · \begin{matrix}Re & [(Ā + ā) (x, y|z)] \\ Re & [(āA)] & \frac{1}{2} \end{matrix}

≤ \frac{1}{2} · \frac{|A + a|}{Re [(āA)]^{\frac{1}{2}}} |(x, y|z)| .

Here the constant \frac{1}{2} is best possible in both inequalities.

As a consequence of (1.15) we may get the following additive reverse of the (CBS) −inequality as well [7]

(1.16)  0 ≤ ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2} ≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (āA) \end{matrix} |(x, y|z)|^{2} .

The constant \frac{1}{4} in (1.16) is best possible in the above sense.

2. Refinements of a Reverse (CBS) −Inequality

The following reverse of the (CBS) −inequality holds.

Theorem 1. Let (X, (·, ·|·)) be a 2-inner product space on 𝕂, x, y, z ∈ X and a, A ∈ 𝕂. If

(2.1)  Re (Ay − x, x − ay|z) ≥ 0,

or, equivalently,

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

holds, then one has the inequality

(2.3)  0 ≤ ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2}
≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{4} − |\frac{a + A}{2} ∥y|z∥^{2} − (x, y|z)|^{2}

(≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{4}) .

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

Proof. Observe , for x, u, U ∈ X, that we have

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

Therefore

Re (U − u, x − u|z) ≥ 0,

if and only if

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

If we choose above U = Ay and u = ay, we deduce that the conditions (2.1) and (2.3) are equivalent.

Now, if we consider x, y, z ∈ X and λ ∈ 𝕂, then we may state that

(2.4)  ∥x − λy|z∥^{2} = ∥x|z∥^{2} − 2 Re [λ (x, y|z)] + |λ|^{2} ∥y|z∥^{2}

and

(2.5) |λ ∥y|z∥^{2} − (x, y|z)|^{2} = |λ|^{2} ∥y|z∥^{2} − 2 ∥y|z∥^{2} Re [λ (x, y|z)] + |(x, y|z)| .

If we multiply (2.4) by ∥x|z∥^{2} ≥ 0 and then subtract equation (2.5), we deduce the following equality, that is of interest in itself,

(2.6) ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2} = ∥x − λy|z∥^{2} ∥y|z∥^{2} − |λ ∥y|z∥^{2} − (x, y|z)|^{2} .

If we now use (2.6) for λ = \frac{a+A}{2} and take into account (2.2), then we deduce the desired inequality (2.3).

To prove the sharpness of the constant \frac{1}{4} in the second inequality in (2.3), assume that, this inequality holds with a constant C > 0. That is,

(2.7) ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2} ≤ C |A − a|^{2} ∥y|z∥^{4} − |\frac{a + A}{2} ∥y|z∥^{2} − (x, y|z)|^{2} ,

where x, y, z, a and A satisfy the hypothesis of the theorem.

Consider y, z ∈ X with ∥y|z∥ = 1, a ≠ A, a, A ∈ 𝕂 and m ∈ X with ∥m|z∥ = 1 and (y, m|z) = 0. Define the vector

x := \frac{a + A}{2}y + \frac{A − a}{2}m.

Then a simple calculation shows that

(Ay − x, x − ay|z) = \frac{|A − a|^{2}}{4} (y − m, y + m|z) = 0,

and thus the condition (2.1) is fulfilled.

Observe also that

∥x|z∥^{2} = ∥ \frac{a + A}{2}y + \frac{A − a}{2}m| z∥ = |\frac{A + a}{2}|^{2} + |\frac{A − a}{2}|^{2} ,

and

(x, y|z) = (\frac{a + A}{2}y + \frac{A − a}{2}m, y|z) = \frac{a + A}{2}.

Consequently, by (2.7), we deduce

\frac{(A − a)^{2}}{4} ≤ C |A − a|^{2} ,

giving C ≥ \frac{1}{4}, and the theorem is proved.

Another reverse for the (CBS)-inequality is incorporated in the following theorem.

Theorem 2. With the assumptions of Theorem 1, one has the inequality

(2.8)  0 ≤ ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2}
≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{4} − Re (Ay − x, x − ay|z) ∥y|z∥^{2}

(≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{4}) .

The constant \frac{1}{4} is sharp in (2.8).

Proof. We use the following identity that has been obtained in [7] and can be proved by direct computation

(2.9) ∥x|z∥^{2} ∥y|z∥^{2} − |(x, y|z)|^{2}

= Re [(A ∥y|z∥^{2} − (x, y|z)) (\overline{(x, y|z)} − ā ∥y|z∥^{2})]
− ∥y|z∥^{2} Re (Ay − x, x − ay|z) .

By the elementary inequality

Re (αβ̄) ≤ \frac{1}{4} |α + β|^{2} , α, β ∈ 𝕂

applied for

α := A ∥y|z∥^{2} − (x, y|z) and β = (x, y|z) − a ∥y|z∥^{2} ,

we deduce the required inequality (2.8).

The sharpness of the constant may be proved as above in Theorem 1 and we omit the details.

3. Another Reverse for the (CBS) −Inequality

The following result also holds.

Theorem 3. Let (X; (·, ·|·)) be a 2-inner product space over 𝕂 (𝕂 = ℂ, ℝ) and x, y, z ∈ X, a, A ∈ 𝕂. If A ≠ −a and either

(3.1)  Re (Ay − x, x − ay|z) ≥ 0

or, equivalently,

(3.2)  ∥x − \frac{a + A}{2}y| z∥ ≤ \frac{1}{2} |A − a| ∥y|z∥ ,

holds, then we have the inequality

(3.3)  0 ≤ ∥x|z∥ ∥y|z∥ − Re [sgn (\frac{a + A}{2}) (x, y|z)]

≤ ∥x|z∥ ∥y|z∥ − |(x, y|z)|

≤ \frac{1}{4} \frac{|A − a|^{2}}{|A + a|} ∥y|z∥^{2} ,

where sgn (α) := \frac{α}{|α|}, α ∈ ℂ\ \{0\} .

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

Proof. We observe that the condition (3.2) is equivalent with

∥x|z∥^{2} − 2 Re [(\frac{a + A}{2}) (x, y|z)] + |\frac{a + A}{2}|^{2} ∥y|z∥^{2} ≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{2}

giving

(3.4) ∥x|z∥^{2} + |\frac{a + A}{2}|^{2} ∥y|z∥^{2} ≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{2} + 2 Re [(\frac{a + A}{2}) (x, y|z)]

≤ \frac{1}{4} |A − a|^{2} ∥y|z∥^{2} + 2 |\frac{a + A}{2}| |(x, y|z)| .

By the elementary inequality

α^{2} + β^{2} ≥ 2αβ, α, β ≥ 0,

we have

(3.5)  2 |\frac{a + A}{2}| ∥x|z∥ ∥y|z∥ ≤ ∥x|z∥^{2} + |\frac{a + A}{2}|^{2} ∥y|z∥^{2} .

By making use of (3.4) and (3.5), we deduce

0 ≤ |\frac{a + A}{2}| ∥x|z∥ ∥y|z∥ − Re [(\frac{a + A}{2}) (x, y|z)]

≤ |\frac{a + A}{2}| [∥x|z∥ ∥y|z∥ − |(x, y|z)|]

≤ \frac{1}{8} |A − a|^{2} ∥y|z∥^{2} ,

which is clearly equivalent to the desired inequality (3.3).

To prove the sharpness of the constant \frac{1}{4} in (3.3), let us assume that there is a constant D > 0 such that

(3.6)  ∥x|z∥ ∥y|z∥ − |(x, y|z)| ≤ D · \frac{|A − a|^{2}}{|A + a|} ∥y|z∥^{2} ,

provided x, y, z and a, A satisfy the hypotheses of the theorem.

Assume now, x, y, z, e ∈ X are such that ∥y, z∥ = 1, ∥e, z∥ = 1 and (e, y|z) = 0. For a, A ∈ 𝕂 with a ≠ −A, define

x = \frac{a + A}{2}y + \frac{A − a}{2}e.

Then

∥x − \frac{a + A}{2}, y|z∥ = \frac{1}{2} |A − a| ,

and thus the condition (3.2) is satisfied with equality.

Observe that, with the above choices for x, y, z and e we have

∥x|z∥ = \sqrt{\frac{|A + a|^{2}}{4} + \frac{|A − a|^{2}}{4}} = \sqrt{\frac{|A|^{2} + |a|^{2}}{2}} ,
|(x, y|z)| = |\frac{a + A}{2}| ,

and thus, from (3.6), we deduce the inequality

(3.7)  \sqrt{\frac{|A|^{2} + |a|^{2}}{2}} − |\frac{a + A}{2}| ≤ D · \frac{|A − a|^{2}}{|A + a|}

for a, A ∈ ℂ, a ≠ −A.

For ε ∈ (0, 1) , consider A = 1 + \sqrt{ε}, a = 1 − \sqrt{ε}. Then a ≠ −A and by (3.9) we deduce

\sqrt{1 + ε} − 1 ≤ 2Dε,

giving by multiplication by \sqrt{1 + ε} + 1 > 0 that

ε ≤ 2ε (\sqrt{1 + ε} + 1) D.

Since ε ∈ (0, 1) , we may divide by ε and thus we get

(3.8)  D ≥ \frac{1}{2 (\sqrt{1 + ε} + 1)}, ε ∈ (0, 1) .

Letting ε → 0+ in (3.8), we obtain D ≥ \frac{1}{4}, and the sharpness of the constant is proved.

When the constants A, a are real, we can point out the following reverse of the triangle inequality.

Corollary 1. Let (X; (·, ·|·)) be a 2-inner product space over 𝕂, x, y, z ∈ X, and m, M ∈ (0, ∞) with M > m. If either

(3.9)  Re (My − x, x − my|z) ≥ 0

or, equivalently,

(3.10)  ∥x − \frac{m + M}{2} y| z∥ ≤ \frac{1}{2} (M − m) ∥y|z∥

holds, then we have the inequality

(3.11)  0 ≤ ∥x|z∥ + ∥y|z∥ − ∥x + y|z∥ ≤ \frac{1}{2} · \frac{(M − m)}{\sqrt{M + m}} ∥y|z∥ .

Proof. A simple computation shows that

(∥x|z∥ + ∥y|z∥)^{2} − ∥x + y|z∥^{2} = 2 (∥x|z∥ ∥y|z∥ − Re (x, y|z)) .

Using the inequality (3.3), we may state that

(3.12)  (∥x|z∥ + ∥y|z∥)^{2} ≤ ∥x + y|z∥^{2} + \frac{1}{4} \frac{(M − m)^{2}}{(M + m)} ∥y|z∥^{2} .

Taking the square root of (3.12), we get

∥x|z∥ + ∥y|z∥ ≤ \sqrt{∥x + y|z∥^{2} + \frac{1}{4} \frac{(M − m)^{2}}{(M + m)} ∥y|z∥^{2}}

≤ ∥x + y|z∥ + \frac{1}{2} · \frac{(M − m)}{\sqrt{M + m}} ∥y|z∥

and the inequality (3.11) is proved.

Remark 1. Firstly, let us observe that from the inequality (1.15) in the Introduction, we may state the following additive reverse of the (CBS) −inequality

(3.13)  0 ≤ ∥x|z∥ ∥y|z∥ − |(x, y|z)| ≤ \frac{1}{2} · \frac{|A + a| − 2 [Re (āA)]^{\frac{1}{2}}}{[Re (āA)]^{\frac{1}{2}}} |(x, y|z)| ,

provided x, y, z ∈ X, a, A ∈ 𝕂 with Re (Aā) > 0 and either the condition (2.1) or, equivalently (2.2), is valid.

If M > m > 0 and either (3.9) or, equivalently, (3.10) holds, then from (3.13) we may state the following simpler form

(3.14)  0 ≤ ∥x|z∥ ∥y|z∥ − |(x, y|z)| ≤ \frac{1}{2} · \begin{matrix}(\sqrt{M} − \sqrt{m}) & 2 \\ √ & \overline{Mm} \end{matrix} |(x, y|z)| .

If, for the same M, m we write the inequality (3.3), then we have another bound, namely:

(3.15)  0 ≤ ∥x|z∥ ∥y|z∥ − |(x, y|z)| ≤ \frac{1}{4} · \frac{(M − m)^{2}}{(M + m)} ∥y|z∥^{2} ,

provided (3.9), or equivalently, (3.10) holds.

4. Integral Inequalities

Let (Ω, Σ, μ) be a measure space consisting of a set Ω, a σ−algebra Σ of parts of Ω and 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. That is,

∫_{Ω} ρ (t) |f (s)|^{2} dμ (s) < ∞,

where ρ : Ω → (0, ∞) is a measurable function on Ω.

If we denote by

| \begin{matrix}a & b \\ c & d \end{matrix} | , a, b, c, d ∈ ℝ

the determinant associated with the matrix

( \begin{matrix}a & b \\ c & d \end{matrix} ) , a, b, c, d ∈ ℝ;

then we can introduce on L_{ρ}^{2} (Ω) the following 2-inner product

(4.1) (f, g|h)_{ρ} := \frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (x) ρ (y) | \begin{matrix}f (x) & f (y) \\ h (x) & h (y) \end{matrix} |

× | \begin{matrix}g (x) & g (y) \\ h (x) & h (y) \end{matrix} | dμ (x) dμ (y) ,

generating the 2-norm

(4.2)  ∥f|h∥_{ρ} = (\frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (x) ρ (y) | \begin{matrix}f (x) & f (y) \\ h (x) & h (y) \end{matrix} |^{2} dμ (x) dμ (y))^{\frac{1}{2}} .

A simple computation with integrals shows that

(f, g|h)_{ρ} = | \begin{matrix}∫ _{Ω} ρ (x) f (x) g (x) dμ (x) & ∫ _{Ω} ρ (x) f (x) h (x) dμ (x) \\ ∫ _{Ω} ρ (x) g (x) h (x) dμ (x) & ∫ _{Ω} ρ (x) h^{2} (x) dμ (x) \end{matrix} |

and

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

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

(q (x) − q (y)) (p (x) − p (y)) ≥ 0

for a.e. x, y ∈ Ω.

Now, suppose that h ∈ L_{ρ}^{2} (Ω) is such that h (x) ≠ 0 for a.e. x ∈ Ω. Then by (4.1) we have the obvious identit,

(4.3) (f, g|h)_{ρ} = \frac{1}{2} ∫_{Ω} ∫_{Ω} ρ (x) ρ (y) h^{2} (x) h^{2} (y)

× (\frac{f (x)}{h (x)} − \frac{f (y)}{h (y)}) (\frac{g (x)}{h (x)} − \frac{g (y)}{h (y)}) dμ (x) dμ (y)

and thus, a sufficient condition for the inequality

(4.4)  (f, g|h)_{ρ} ≥ 0

to hold, is that the pair of functions (\frac{f}{h}, \frac{g}{h}) be synchronous. This condition is not necessary.

If Ω = [a, b] ⊂ ℝ (a < b) and μ is the Lebesgue measure, then a sufficient condition for the functions (\frac{f(x)}{h(x)}, \frac{g(x)}{h(x)}) , x ∈ [a, b] to be synchronous is that they

are monotonic in the same sense, i.e. \frac{f}{h} and \frac{f}{g} are both increasing or decreasing on [a, b] . Obviously, this condition is not necessary.

We are able now to state some integral inequalities that can be derived using the general framework presented above.

Proposition 1. Let M > m > 0 and f, g, h ∈ L_{ρ}^{2} (Ω) , h ≠ 0, such that the functions

(4.5)  M · \frac{g}{h} − \frac{f}{h} , \frac{f}{h} − m · \frac{g}{h}

are synchronous on Ω. Then we have the inequalities

(4.6) 0 ≤ | \begin{matrix}∫ _{Ω} ρf^{2} & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρfh & ∫ _{Ω} ρh^{2} \end{matrix} | · | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |

− | \begin{matrix}∫ _{Ω} ρfg & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2}

≤ \frac{1}{4} (M − m)^{2} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2}

− |\frac{m + M}{2} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} | − | \begin{matrix}∫ _{Ω} ρfg & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} ||

(≤ \frac{1}{4} (M − m)^{2} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2}) .

The proof is obvious by Theorem 1 and we omit the details.

The following counterpart of the (CBS) −inequality for determinants also holds.

Proposition 2. With the assumptions of Proposition 1, we have the inequality

(4.7)  0 ≤ | \begin{matrix}∫ _{Ω} ρf^{2} & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρfh & ∫ _{Ω} ρh^{2} \end{matrix} | · | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |

− | \begin{matrix}∫ _{Ω} ρfg & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2}

≤ (\frac{1}{4} (M − m)^{2} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |

− | \begin{matrix}∫ _{Ω} (Mg − f) (f − mg) & ∫ _{Ω} ρ (Mg − f) h \\ ∫ _{Ω} ρ (f − mg) h & ∫ _{Ω} ρh^{2} \end{matrix} |)

× | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |

(≤ \frac{1}{4} (M − m)^{2} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2}) .

The proof follows by Theorem 2 applied for the 2-inner product defined in (4.3). A different reverse of the (CBS) −inequality for determinants is incorporated in

the following proposition.

Proposition 3. With the assumptions of Proposition 1, we have the inequality

(4.8)  0 ≤ | \begin{matrix}∫ _{Ω} ρf^{2} & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρfh & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}} · | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}}

− |det ( \begin{matrix}∫ _{Ω} ρfg & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} )|

≤ \frac{1}{4} \frac{(M − m)^{2}}{M + m} | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{2} .

The constant \frac{1}{4} is best possible in (4.8).

The proof follows from Theorem 3 applied for the 2-inner product defined in (4.3).

Finally, by the use of Corollary 1, we may state the following reverse of the triangle inequality for determinants.

Proposition 4. With the assumptions of Proposition 1, we have the inequality:

(4.9)  0 ≤ | \begin{matrix}∫ _{Ω} ρf^{2} & ∫ _{Ω} ρfh \\ ∫ _{Ω} ρfh & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}} + | \begin{matrix}∫ _{Ω} ρgh & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}}

− | \begin{matrix}∫ _{Ω} ρ (f + g)^{2} & ∫ _{Ω} ρ (f + g) h \\ ∫ _{Ω} ρ (f + g) h & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}}

≤ \frac{1}{2} · \frac{M − m}{\sqrt{M + m}} · | \begin{matrix}∫ _{Ω} ρg^{2} & ∫ _{Ω} ρgh \\ ∫ _{Ω} ρgh & ∫ _{Ω} ρh^{2} \end{matrix} |^{\frac{1}{2}} .

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] I. BUDIMIR, Y.J. CHO, M. MATIĆ and J.E. PEČARIĆ, Čebyšev’s inequality in n−inner product spaces, in Inequality Theory and Applications, Vol. 1, Y.J. Cho, J.K. Kim and S.S. Dragomir (Eds.), Nova Science Publishers, N.Y., 2001, pp. 87-94.

[2] Y.J.CHO, S.S. DRAGOMIR, A. WHITE and S.S. KIM, Some inequalities in 2-inner product spaces, Demonstratio Math., 32(3) (1999), 485-493.

[3] Y.J. CHO, M. MATIĆ and J.E. PEČARIĆ, On Gram’s determinant in 2-inner product spaces, J. Korean Math. Soc., 38(6) (2001), 1125-1156.

[4] Y.J. CHO, M. MATIĆ and J.E. PEČARIĆ, Inequalities of Hlawka’s type in G−inner product spaces, in Inequality Theory and Applications, Vol. 1, Y.J. Cho, J.K. Kim and S.S. Dragomir (Eds.), Nova Science Publishers, N.Y., 2001, pp. 95-102.

[5] Y.J. CHO, P.C.S. LIN, S.S. KIM and A. MISIAK, Theorey of 2−inner product spaces, Nova Science Publishers, 2001.

[6] Y.J. CHO, M. MATIĆ and J.E. PEČARIĆ, On Gram’s determinant in n−inner product spaces, in preparation.

[7] S.S. DRAGOMIR, Y.J. CHO and S.S. KIM, Superadditivity and monotonicity of 2-norms generated by inner products and related results, Soochow J. Math., 24(1) (1998), 13-32.

[8] S.S. DRAGOMIR, Y.J. CHO and S.S. KIM, Reverse of the Cauchy-Bunyakovsky-Schwarz inequality in 2-inner product spaces, submitted.

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School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, MCMC 8001, Victoria, Australia.

E-mail address: pietro.cerone@vu.edu.au
URL: http://rgmia.vu.edu.au/cerone

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

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

School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, MCMC 8001, Victoria, Australia.

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

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

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