A REVERSE OF BESSEL’S INEQUALITY IN 2-INNER PRODUCT SPACES AND SOME GRÜSS TYPE RELATED

RESULTS WITH APPLICATIONS

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

Abstract. A reverse of Bessel’s inequality in 2-inner product spaces and companions of Grüss inequality with 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 [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

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

(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 [4]. 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).

2. A Reverse of Bessel’s Inequality

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; (·, ·|·)):

(2.1)  ∑_{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 lemma holds.

Lemma 1. Let {e_{i}}_{i∈I} be a family of z-orthornormal vectors in X, F a finite part of I and ϕ_{i}, Φ_{i} (i ∈ F) , real or complex numbers. The following statements are equivalent for x ∈ X :

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

Proof. It is easy to see that for y, a, A ∈ X, the following are equivalent

(a) Re (A − y, y − a|z) ≥ 0 and (aa) ∥y − \frac{a+A}{2}|z∥ ≤ \frac{1}{2} ∥A − a|z∥ .

4 S.S. DRAGOMIR, Y.J. CHO^{⋆}, S.S. KIM^{♦}, AND J. ROUMELIOTIS

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.

The following reverse of Bessel’s inequality holds. Theorem 1. Let {e_{i}}_{i∈I} , F, ϕ_{i}, Φ_{i}, i ∈ F and x, z ∈ X so that either (i) or (ii) of Lemma 1 holds. Then we have the inequality:

(2.2)  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 in both inequalities. Proof. Define

I_{1} := ∑_{i∈X} Re [(Φ_{i} − (x, e_{i}|z)) (\overline{(x, e_{i}|z)} − \overline{ϕ_{i}})]

and

I_{2} := Re [(∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}|z)] .

Observe that

I_{1} = ∑_{i∈H} Re [Φ_{i}\overline{(x, e_{i}|z)}] + ∑_{i∈H} Re [\overline{ϕ_{i}} (x, e_{i}|z)]

− ∑_{i∈H} Re [Φ_{i}\overline{ϕ_{i}}] − ∑_{i∈H} |(x, e_{i}|z)|^{2}

and

I_{2} = Re [∑_{i∈H} Φ_{i}\overline{(x, e_{i}|z)} + ∑_{i∈H} \overline{ϕ_{i}} (x, e_{i}|z) − ∥x|z∥^{2} − ∑_{i∈H} ∑_{j∈H} Φ_{i}\overline{ϕ_{i}} (e_{i}, e_{j}|z)]

= ∑_{i∈H} Re [Φ_{i}\overline{(x, e_{i}|z)}] + ∑_{i∈H} Re [\overline{ϕ_{i}} (x, e_{i}|z)] − ∥x|z∥^{2} − ∑_{i∈H} Re [Φ_{i}\overline{ϕ_{i}}] .

Consequently, subtracting I_{2} from I_{1}, we deduce the following equality that is useful in its turn

(2.3)  ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2} = ∑_{i∈H} Re [(Φ_{i} − (x, e_{i}|z)) (\overline{(x, e_{i}|z)} − \overline{ϕ_{i}})]

− Re [(∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}|z)] .

Using the following elementary inequality for complex numbers

Re [a\overline{b}] ≤ \frac{1}{4} |a + b|^{2} , a, b ∈ 𝕂,

for the choices a = Φ_{i} − (x, e_{i}|z) , b = (x, e_{i}|z) − ϕ_{i} (i ∈ F) , we deduce

(2.4)  ∑_{i∈H} Re [(Φ_{i} − (x, e_{i}|z)) (\overline{(x, e_{i}|z)} − \overline{ϕ_{i}})] ≤ \frac{1}{4} ∑_{i∈H} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (2.3), (2.4) and the assumption (i), we deduce (2.2).

To prove the sharpness of the constant \frac{1}{4}, assume that there is a c > 0 such that

(2.5)  0 ≤ ∥x|z∥^{2} − ∑_{i∈F} |(x, e_{i}|z)|^{2}

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

provided 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 end (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 Lemma 1 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 (2.5), we deduce

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

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

6 S.S. DRAGOMIR, Y.J. CHO^{⋆}, S.S. KIM^{♦}, AND J. ROUMELIOTIS

3. A Refinement of the Grüss Inequality

The following result holds. Theorem 2. Let {e_{i}}_{i∈I} be a family of z-orthornormal vectors in X, F a finite part of I and 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

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

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

− [Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z)]^{\frac{1}{2}} [Re (∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}|z)]^{\frac{1}{2}}

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

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

and since a simple calculation shows that

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

and

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

for any x, y ∈ X, then, by (3.4) and by the counterpart of Bessel’s inequality in Theorem 1, 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} − Re (∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}|z)]

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

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

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

where, for the last inequality, we have made use of the elementary inequality

(m^{2} − n^{2}) (p^{2} − q^{2}) ≤ (mp − nq)^{2}

valid for any m, n, p, q > 0.

Taking the square root in (3.5) and observing that the quantity in the last square bracket is nonnegative (see for example (2.2)), we deduce the desired result (3.3).

The best constant follows by Theorem 1 and we omit the details.

4. Some Companion Inequalities

The following companion of the Grüss inequality also holds.

Theorem 3. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in X, F a finite part of I, ϕ_{i}, Φ_{i} ∈ 𝕂, i ∈ F and x, y ∈ X such that

(4.1)  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,

(4.2)  ∥\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}} .

Then we have the inequality

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

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

Proof. Start with the well known inequality

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

Since

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

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

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

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

If we apply the reverse of Bessel’s inequality from Theorem 1 for \frac{x+y}{2} , we may state that

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

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

The fact that \frac{1}{4} is the best constant in (4.3) follows by the fact that if in (4.1) we choose x = y, then it becomes (i) of Lemma 1, implying (2.2), for which, we have shown that \frac{1}{4} is the best constant.

The following corollary may be of interest if we wish to evaluate the absolute value of

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

Corollary 1. With the assumptions of Theorem 3 and if

(4.7)  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,

(4.8)  ∥\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}} ,

then we have the inequality

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

Proof. We only remark that, if

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

holds, then by Theorem 3 for (−y) instead of y, we have

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

which shows that

(4.10)  Re [(x, y|z) − ∑_{i∈F} (x, e_{i}|z) (e_{i}, y|z)] ≥ −\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (4.3) and (4.10), we deduce the desired inequality (4.9).

Remark 1. If X is a real inner product space and m_{i}, M_{i} ∈ ℝ with the property that

(4.11)  (∑_{i∈F} M_{i}e_{i} − \frac{x ± y}{2} , \frac{x ± y}{2} − ∑_{i∈F} m_{i}e_{i}|z) ≥ 0

or equivalently,

(4.12)  ∥\frac{x ± y}{2} − ∑_{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}{4} ∑_{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} ] ,

generating 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}

− | \begin{matrix}∫_{Ω} ρ (∑_{i∈F} M_{i}f_{i} − f) (f − ∑_{i∈F} m_{i}f_{i}) dμ & ∫_{Ω} ρ (∑_{i∈F} M_{i}f_{i} − f) hdμ \\ ∫_{Ω} ρ (f − ∑_{i∈F} m_{i}f_{i}) hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |

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

The proof follows by Theorem 1 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] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993.

[4] 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

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

E-mail address: johnr@matilda.vu.edu.au