SOME REVERSES OF THE

CAUCHY-BUNYAKOVSKY-SCHWARZ

INEQUALITY IN 2-INNER PRODUCT SPACES

Sever S. Dragomir, Yeol Je Cho and Seong Sik Kim

Abstract. In this paper, some reverses of the Cauchy-Bunyakovsky-Schwarz inequality in 2-inner product spaces are given. Using this framework, some 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 [3].

We recall here the basic definitions and the elementary properties of 2-inner product spaces that will be used in the sequel (see [4]).

Typeset by AMS-TEX Typeset by AMS-TEX

2000 AMS Subject Classification: Primary 46C05, 46C97. Secondary 26D15, 26D10.

Key Words and Phrases: 2-inner product spaces, Cauchy-Bunyakovsky-Schwarz inequality, Determinantal integral inequalities.

The corresponding author: yjcho@nongae.gsnu.ac.kr (Y. J. Cho) or sskim@dongeui.ac.kr (S. S. Kim).

Let X be a linear space of dimension greater than 1 over the number filed K, where K = R or K = C. Suppose that (·, ·|·) is a K-valued function on X × X × X satisfying the following conditions: (2I_{1}) (x, x|z) ≥ 0, (x, x|z) = 0 if and only if x and z are linearly

dependent, (2I_{2}) (x, x|z) = (z, z|x), (2I_{3}) (x, y|z) = \overline{(y, x|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 and (X, (·, ·|·)) a 2-inner product space (or 2-pre-Hilbert space).

Some basic properties of the 2-inner product spaces are as follows:

(1) If K = R, then (2I_{3}) reduces to (x, y|z) = (y, x|z).

(2) From (2I_{3}) and (2I_{4}), we have (0, y|z) = 0, (x, 0|y) = 0 and also

(1.1)  (x, αy|z) = \overline{α}(x, y|z).

(3) Using (2I_{3})∼(2I_{5}), we have

(z, z|x ± y) = (x ± y, x ± y|z) = (x, z|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 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)  \begin{aligned}(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)]. \end{aligned}

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 any α ∈ R, (1.6) reduces to (1.4). Also, it follows from (1.6) that

(x, y|0) = 0.

(4) For any given vectors x, y, z ∈ X, consider the vector u = (y, y|z)x − (x, x|z)y. By (2I_{1}), we know that (u, u|z) ≥ 0. 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.

If x = z, then (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.

Now, if y and z are linearly independent, then (y, y|z) > 0 and, from (1.2), it follows the Cauchy-Bunyakovsky-Schwarz inequality (shortly, the CBS-inequality) for 2-inner products:

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

Using (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 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 if and only if x and z are linearly dependent, (2N_{2}) ∥x|z∥ = ∥z|x∥, (2N_{3}) ∥αx|z∥ = |α|∥x|z∥ for all scalar α ∈ K, (2N_{4}) ∥x + x′|z∥ ≤ ∥x|z∥ + ∥x′|z∥.

Any function ∥·, ·∥ satisfying the conditions (2N_{1})∼(2N_{4}) is called a 2-norm on X and (X, ∥ · | · ∥) a linear 2-normed space.

For a systematic presentation of the recent results related to the theory of linear 2-normed spaces, see the book [8].

In terms of the 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 holds in (1.11) if and only if x, y and z are linearly dependent.

For recent inequalities in 2-inner product spaces, see the papers [1, 2], [4–7], [9–12] and the references therein.

2. Revereses of the CBS-Inequality

The following reverse of the CBS-inequality in 2-inner product spaces holds.

Theorem 2.1. Let A, a ∈ K(K = C, R) and x, y, z ∈ X, where, as above, (X, (·, ·|·)) is a 2-inner product space over 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 we have the inequality

(2.3)  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 (2.3) in the sense that it cannot be replaced by a smaller constant.

Proof. Consider the vectors x, u, U, z ∈ X. We observe that

\begin{aligned}\frac{1}{4} & ∥U − u|z∥^{2} − ∥x − \frac{U + u}{2}|z∥^{2} \\ & = Re(U − x, x − u|z) \end{aligned}

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

Therefore, Re(U − x, z − u|z) ≥ 0 if and only if

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

If we apply this to the vectors U = Ay and u = ay, then we deduce that the inequalities (2.1) and (2.2) are equivalent, as stated.

Now, let us consider the real numbers

I_{1} = Re[(A∥y|z∥^{2} − (x, y|z))(\overline{(x, y|z)} − \overline{a}∥y|z∥^{2})]

and

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

A simple calculation shows that

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

and

\begin{aligned}I_{2} & = ∥y|z∥^{2}Re[A\overline{(x, y|z)} + \overline{a}(x, y|z)] \\ & − ∥x|z∥^{2}∥y, z∥^{2} − ∥y, z∥^{4}Re(A\overline{a}), \end{aligned}

which give

(2.4)  I_{1} − I_{2} = ∥x|z∥^{2}∥y|z∥^{2} − |(x, y|z)|^{2}

for any x, y, z ∈ X and a, A ∈ K. If (2.2) holds, then I_{2} ≥ 0 and thus

(2.5)  \begin{matrix}∥x|z∥^{2}∥y|z∥^{2} − |(x, y|z)|^{2} \\ ≤ Re[(A∥y|z∥^{2} − (x, y|z))(\overline{(x, y|z)} − \overline{a}∥y|z∥^{2}). \end{matrix}

If we use the elementary inequality Re(α\overline{β}) ≤ \frac{1}{4}|α + β|^{2} for any α, β ∈ K(K = R, C), then we have that

(2.6)  \begin{matrix}Re[(A∥y|z∥^{2} − (x, y|z))(\overline{(x, y|z)} − \overline{a}∥y, z∥^{2})] \\ ≤ \frac{1}{4}|A − a|^{2}∥y|z∥^{4}. \end{matrix}

Making use of the inequalities (2.5) and (2.6), we deduce the desired inequality (2.3).

To prove the sharpness of the constant \frac{1}{4}, assume that (2.4) holds with a constant C > 0, i.e.,

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

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

Consider y ∈ X with ∥y|z∥ = 1, a ≠ A, m ∈ X with ∥m|z∥ = 1 and (y, m|z) = 0 and define

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

Then we have

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

and then the condition (2.1) is fulfilled. From (2.7), we deduce

(2.8)  \begin{matrix}∥\frac{A + a}{2}y + \frac{A − a}{2}m|z∥^{2} − |(\frac{A + a}{2}y + \frac{A − a}{2}m, y|z)|^{2} \\ ≤ C|A − a|^{2} \end{matrix}

and, since

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

and

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

by (2.8), we have

|\frac{A − a}{2}|^{2} ≤ C|A − a|^{2}

for any A, a ∈ K with a ≠ A, which implies C ≥ \frac{1}{4}. This completes the proof.

Another reverse of the CBS-inequality in 2-inner product spaces is incorporated in the following theorem:

Theorem 2.2. Assume that x, y, z, a and A are the same as in Theorem 2.1. If Re(\overline{a}, A) > 0, then we have the inequalities

(2.9)  \begin{aligned}∥x|z∥∥y|z∥ & ≤ \frac{1}{2} · \frac{Re[(\overline{A} + \overline{a})(x, y|z)]}{[Re(\overline{a}A)]^{\frac{1}{2}}} \\ & ≤ \frac{1}{2} · \frac{|A + a|}{[Re(\overline{a}A)]^{\frac{1}{2}}} |(x, y|z)|. \end{aligned}

The constant \frac{1}{2} is best possible in both inequalities in the sense that it cannot be replaced by a smaller constant.

Proof. Define

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

We know that, for a complex number α ∈ C, Re(α) = Re(\overline{α}) and thus

Re[A \overline{(x, y|z)}] = Re[\overline{A}(x, y|z)],

which implies

(2.10)  I = Re[(\overline{A} + \overline{a})(x, y|z)] − ∥x|z∥^{2} − Re(\overline{a}A)∥y|z∥^{2}.

Since x, y, z, α, A are assumed to satisfy the condition (2.1), by (2.10), we deduce the inequality

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

which gives

(2.11)  \begin{matrix}\frac{1}{[Re(\overline{a}A)]^{\frac{1}{2}}} ∥x|z∥^{2} + [Re(\overline{a}A)]^{\frac{1}{2}} ∥y|z∥^{2} \\ ≤ \frac{Re[(\overline{A} + \overline{a})(x, y|z)]}{[Re(\overline{a}A)]^{\frac{1}{2}}} \end{matrix}

since Re(\overline{a}A) > 0.

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq

for p, q ≥ 0 and α > 0, we have

(2.12)  2∥x|z∥∥y|z∥ ≤ \frac{1}{[Re(\overline{a}A)]^{\frac{1}{2}}} ∥x|z∥^{2} + [Re(\overline{a}A)]^{\frac{1}{2}} ∥y|z∥^{2}.

Using (2.11) and (2.12), we deduce the first inequality in (2.9). The last part is obvious by the fact that, for z ∈ C, |Re(z)| ≤ |z|.

To prove the sharpness of the constant \frac{1}{2} in the first inequality in (2.9), we assume that (2.9) holds with a constant C > 0, i.e.,

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

provided x, y, z, a and A satisfy (2.1). If we choose a = A = 1, y = x ≠ 0, then obviously (2.1) holds and, from (2.13), we obtain

(2.14)  ∥x|z∥^{2} ≤ 2C∥x|z∥^{2}

for any linearly independent vectors x, z ∈ X, which implies C ≥ \frac{1}{2}. This completes the proof.

When the constants involved are assumed to be positive, then we may state the following result:

Corollary 2.3. Let M ≥ m > 0 and assume that, for x, y, z ∈ X, we have

Re(My − x, x − my|z) ≥ 0

or, equivalently,

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

Then we have the following reverse of the CBS-inequality

(2.15) ∥x|z∥∥y|z∥ ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}}Re(x, y|z) ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}}|(x, y|z)|.

The constant \frac{1}{2} is sharp in (2.15).

Some additive versions of the above are obtained in the following:

Corollary 2.4. With the assumptions of Theorem 2.2, we have the inequalities

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

The constant \frac{1}{4} is best possible in (2.16).

Corollary 2.5. With the assumptions of Corollary 2.3, we have

(2.17) \begin{aligned}0 & ≤ ∥x|z∥∥y|z∥ − |(x, y|z)| ≤ ∥x|z∥∥y|z∥ − Re(x, y|z) \\ & ≤ \frac{1}{2} · \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} Re(x, y|z) ≤ \frac{1}{2} · \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} |(x, y|z)| \end{aligned}

and

(2.18) \begin{aligned}0 & ≤ ∥x|z∥^{2}∥y|z∥^{2} − |(x, y|z)|^{2} ≤ ∥x|z∥^{2}∥y|z∥^{2} − [Re(x, y|z)]^{2} \\ & ≤ \frac{1}{4} · \frac{(M − m)^{2}}{mM} [Re(x, y|z)]^{2} ≤ \frac{1}{4} · \frac{(M − m)^{2}}{mM} |(x, y|z)|^{2}. \end{aligned}

The constant \frac{1}{2} in (2.17) and the constant \frac{1}{4} in (2.18) are best possible.

The third inequality in (2.17) may be used to point out a reverse of the triangle inequality in 2-inner product spaces.

Corollary 2.6. Assume that x, y, z, m and M are the same as in Corollary 2.3. Then we have the following reverse of the triangle inequality

(2.19) 0 ≤ ∥x|z∥ + ∥y|z∥ − ∥x + y|z∥ ≤ \frac{\sqrt{M} − \sqrt{m}}{(mM)^{\frac{1}{4}}} \sqrt{Re(x, y|z)}.

Proof. It is easy to see that

(2.20) 0 ≤ (∥x|z∥ + ∥y|z∥)^{2} − ∥x + y|z∥^{2} = 2[∥x|z∥∥y∥ − Re(x, y|z)]

for any x, y, z ∈ X. If the assumptions of Corollary 2.3 hold, then (2.17) is valid and, by (2.20), we deduce

0 ≤ (∥x|z∥ + ∥y|z∥)^{2} − ∥x + y|z∥^{2} ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} Re(x, y|z),

which gives

(2.21) (∥x|z∥ + ∥y|z∥)^{2} ≤ ∥x + y|z∥^{2} + \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} Re(x, y|z).

Taking the square root in (2.21), we have

∥x|z∥ + ∥y|z∥ ≤ \sqrt{∥x + y|z∥^{2} + \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} Re(x, y|z)}

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

from where we deduce the desired inequality (2.21). This completes the proof.

3. Integral Inequalities

Let (Ω, Σ, μ) be a measure space consisting of a set Ω, a σ-algebra Σ of subsets of Ω and a countably additive and positive measure μ on Σ with valued in R ∪ {∞}. Denote by L_{ϕ}^{2}(Ω) the Hilbert space of all real-valued functions f defined on Ω that are 2-ϕ-integrable on Ω, i.e., ∫_{Ω} ϕ(s)|ϕ(s)|^{2}dμ(s) < ∞, where ϕ : Ω → [0, ∞) is a measurable function on Ω.

We can introduce the following 2-inner product on L_{ϕ}^{2}(Ω)

(3.1) \begin{matrix}(f, g|h)_{ϕ} \\ = \frac{1}{2} ∫_{Ω} ∫_{Ω} ϕ(x)ϕ(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), \end{matrix}

where by | \begin{matrix}f(x) & f(y) \\ h(x) & h(y) \end{matrix} | we understand the determinant of the matrix

[ \begin{matrix}f(x) & f(y) \\ h(x) & h(y) \end{matrix} ], generating the 2-norm

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

and

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

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

(q(x) − q(y))(p(x) − p(y)) ≥ 0

for almost every x, y ∈ Ω. If Ω = [a, b] is an interval of real numbers, then a sufficient condition of synchronicity for (p, q) is that they are monotonic in the same sense, i.e., both of them are increasing or decreasing on [a, b].

Now, suppose that h ∈ L_{ϕ}^{2}(Ω) is such that h(x) ≠ 0 for almost every x ∈ Ω. Then, by (3.1), we have

(3.3)  \begin{matrix}(f, g|h)_{ϕ} \\ = \frac{1}{2} ∫_{Ω} ∫_{Ω} ϕ(x)ϕ(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) \end{matrix}

and thus a sufficient condition for the inequality

(3.4)  (f, g|h)_{ϕ} ≥ 0

to hold is that the pair of function (\frac{f}{h}, \frac{g}{h}) to be synchronous.

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

Proposition 3.1. Let M > m > 0 and f, g, h ∈ L_{ϕ}^{2}(Ω) so that the functions

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

are synchronous. Then we have the inequalities

(3.6)

0 ≤ | \begin{matrix}∫_{Ω} ϕf^{2} & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕfh & ∫_{Ω} ϕh^{2} \end{matrix} | · | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |

− | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{2}

≤ \frac{1}{4}(M − m)^{2} | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{2} .

The constant \frac{1}{4} is best possible in (3.6).

The proof is obvious by Theorem 2.1 and we omit the details.

Proposition 3.2. With the assumptions of Proposition 3.1, we have the inequality

(3.7)

0 ≤ | \begin{matrix}∫_{Ω} ϕf^{2} & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕfh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}} | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}}

≤ \frac{1}{2} · \frac{(M − m)}{\sqrt{mM}} | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} | .

The constant \frac{1}{2} is best possible in (3.7).

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

Proposition 3.3. With the assumptions of Proposition 3.1, we have the inequalities

(3.8)

0 ≤ | \begin{matrix}∫_{Ω} ϕf^{2} & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕfh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}} | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}}

− | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |

≤ \frac{1}{2} \frac{(\sqrt{M} − \sqrt{m})^{2}}{\sqrt{mM}} | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |

and

(3.9)

0 ≤ | \begin{matrix}∫_{Ω} ϕf^{2} & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕfh & ∫_{Ω} ϕh^{2} \end{matrix} | | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |

− | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{2}

≤ \frac{1}{4} \frac{(M − m)^{2}}{\sqrt{mM}} | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{2} .

The constants \frac{1}{2} in (3.8) and \frac{1}{4} in (3.9) are best possible.

Finally, we may state the following reverse of the triangle inequality for determinants:

Proposition 3.4. With the assumptions of Proposition 3.1, we have the inequalities

(3.10)

0 ≤ | \begin{matrix}∫_{Ω} ϕf^{2} & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕfh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}} + | \begin{matrix}∫_{Ω} ϕg^{2} & ∫_{Ω} ϕgh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}}

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

≤ \frac{\sqrt{M} − \sqrt{m})}{(mM)^{\frac{1}{4}}} | \begin{matrix}∫_{Ω} ϕfg & ∫_{Ω} ϕfh \\ ∫_{Ω} ϕgh & ∫_{Ω} ϕh^{2} \end{matrix} |^{\frac{1}{2}} .

Acknowledgement. S. S. Dragomir and Y. J. Cho greatly acknowledge the financial support from 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] C. Budimir, Y. J. Cho, M. Matić and J. Pečarić, Cebysev’s inequality in n-inner product space, “Inequality Theory and Applications, Vol. 2” (Editors: Y. J. Cho, J. K. Kim and S. S. Dragomir), Nova Science Publishers, Inc., New York, 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, Paul C. S Lin, S. S. Kim and A. Misiak, Theory of 2-Inner Product Spaces, Nova Science Publishers, Inc., New York, 2001.

[4] Y. J. Cho, M. Matić and J. Pečarić, On Gram’s determinant in 2-inner product spaces, J. Korean Math. Soc., 38(6)(2001), 1125– 1156.

[5] Y. J. Cho, M. Matić and J. Pečarić, Inequalities of Hlawka’s type in G-inner product spaces, “Inequality Theory and Applications, Vol. 1” (Editors: Y. J. Cho, J. K. Kim and S. S. Dragomir), Nova Science Publishers, Inc., New York, 2001, pp. 95–102.

[6] Y. J. Cho, M. Matić and J. Pečarić, On Gram’s determinant in n-inner product spaces, preprint.

[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. 29(1)(1998), 13–32.

[8] R. W. Freese and Y. J. Cho, Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Inc., New York, 2001.

[9] H. Gunawan, On n-inner product, n-norms and the Cauchy-Schwarz inequality, Sci. Math. Japon. 55(1)(2002), 53–60.

[10] C. S. Lin, On inequalities in inner product and 2-inner product spaces, Internat. J. Pure and Appl. Math. 3(3)(2002), 287–298.

[11] S. S. Kim, S. S. Dragomir, A. White and Y. J. Cho, On the Grüss type inequality in 2-inner product spaces and applications, PanAmer. Math. J. 71(3)(2001), 89–97.

[12] S. S. Kim and S. S. Dragomir, Inequalities involuding Gram’s determinant in 2-inner product spaces, “Inequality Theory and Applications, Vol. 1” (Editors: Y. J. Cho, J. K. Kim and S. S. Dragomir), Nova Science Publishers, Inc., New York, 2001, pp.183–192.

Sever S. Dragomir School of Computer Science and Mathematics Victoria University P. O. Box 14428, Melbourne City MC Victoria 8001, Australia E-mail: sever.drgomir@vu.edu.au

Yeol Je Cho Department of Mathematics Education The Research Institute of Natural Sciences College of Education Gyeongsang National University Chinju 660-701, Korea E-mail: yjcho@nongae.gsnu.ac.kr

Seong Sik Kim Department of Mathematics Dongeui University Pusan 614-714, Korea sskim@dongeui.ac.kr