SOME BOMBIERI, SELBERG AND HEILBRONN TYPE INEQUALITIES IN 2-INNER PRODUCT SPACES

N.S. BARNETT, Y.J. CHO^{⋆}, S.S. DRAGOMIR, S.M. KANG, AND S.S. KIM^{♦}

Abstract. Some results related to the Bombieri type generalisation of Bessel’s inequality in 2-inner product spaces are given. The corresponding versions for Selberg and Heilbronn inequalities for 2-inner products and applications for determinantal integral inequalities are also pointed out.

1. Introduction

Let (X; (·, ·)) be an inner product space over the real or complex number field 𝕂. If (f_{i})_{1≤i≤n} are orthonormal vectors in the inner product space X, i.e., (f_{i}, f_{j}) = δ_{ij} for all i, j ∈ {1, …, n} where δ_{ij} is the Kronecker delta, then the following inequality is well known in the literature as Bessel’s inequality (see for example [10, p. 391]):

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

for any x ∈ X.

For other results related to Bessel’s inequality, see [6] – [8] and Chapter XV in the book [10].

In 1971, E. Bombieri [3] (see also [10, p. 394]) gave the following generalisation of Bessel’s inequality.

Theorem 1. If x, y_{1}, …, y_{n} are vectors in the inner product space (X; (·, ·)) , then the following inequality:

(1.2)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|} ,

holds.

It is obvious that if (y_{i})_{1≤i≤n} are supposed to be orthonormal, then from (1.2) one would deduce Bessel’s inequality (1.1).

Another generalisation of Bessel’s inequality was obtained by A. Selberg (see for example [10, p. 394]):

Theorem 2. Let x, y_{1}, …, y_{n} be vectors in X with y_{i} ≠ 0 (i = 1, …, n) , then

(1.3)  ∑_{i=1}^{n} \begin{matrix}|(x, y_{i})| & 2 \\ ∑ & {n \atop j=1} & |(y_{i}, y_{j})| \end{matrix} ≤ ∥x∥^{2} .

Date: 04 August, 2003. ⋆, ♦ Corresponding authors.

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

Key words and phrases. 2-Inner products, 2-Normed spaces, Bessel’s inequality in 2-inner product spaces, Bombieri, Selberg, Heilbronn type inequalities in 2-inner product spaces.

2 N.S. BARNETT, Y.J. CHO^{⋆}, S.S. DRAGOMIR, S.M. KANG, AND S.S. KIM^{♦}

In this case, also, if (y_{i})_{1≤i≤n} are orthonormal, then one may deduce Bessel’s inequality.

Another type of inequality related to Bessel’s result, was discovered in 1958 by H. Heilbronn [9] (see also [10, p. 395]).

Theorem 3. With the assumptions of Theorem 1,

(1.4)  ∑_{i=1}^{n} |(x, y_{i})| ≤ ∥x∥ ( ∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} .

If in (1.4) one chooses y_{i} = f_{i} (i = 1, …, n) , where (f_{i})_{1≤i≤n} are orthonormal vectors in X, then

(1.5)  ∑_{i=1}^{n} |(x, f_{i})| ≤ \sqrt{n} ∥x∥ , for any x ∈ X.

In 1992 J.E. Pečarić [12] (see also [10, p. 394]) proved the following general inequality in inner product spaces.

Theorem 4. Let x, y_{1}, …, y_{n} ∈ X and c_{1}, …, c_{n} ∈ 𝕂. Then

(1.6)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} (∑_{j=1}^{n} |(y_{i}, y_{j})|)

≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|} .

He showed that the Bombieri inequality (1.2) may be obtained from (1.6) for the choice c_{i} = \overline{(x, y_{i})} (using the second inequality), the Selberg inequality (1.3) may be obtained from the first part of (1.6) for the choice

c_{i} = \frac{\overline{(x, y_{i})}}{∑_{j=1}^{n} |(y_{i}, y_{j})|}, i ∈ \{1, …, n\};

while the Heilbronn inequality (1.4) may be obtained from the first part of (1.6) if one chooses c_{i} = \frac{\overline{(x,y_{i})}}{|(x,y_{i})|}, for any i ∈ \{1, …, n\} .

For other results connected with the above, see [7] and [8].

It is the main aim of the present paper to point out the corresponding versions of Bombieri, Selberg and Heilbronn inequalities in 2-inner product spaces. Some natural generalizations and related results are also given. Applications for determinantal integral inequalities are provided.

For a comprehensive list of fundamental results on 2-inner product spaces and linear 2-normed spaces, see the recent books [4] and [11] where further references are given.

2. Bessel’s Inequality in 2-Inner Product Spaces

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 [4]. 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 [5]:

(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

(2.1)  (x, αy|z) = ᾱ(x, y|z).

(3) Using (2I_{2})–(2I_{5}), we have

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

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

In the real case 𝕂 = ℝ, (2.2) reduces to

(2.3)  (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,

(2.4)  (x, y|αz) = α^{2}(x, y|z).

In the complex case, using (2.1) and (2.2), 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 (2.2), yields

(2.5) (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 (2.1), we have, for any α ∈ ℂ,

(2.6)  (x, y|αz) = |α|^{2}(x, y|z).

However, for α ∈ ℝ, (2.6) reduces to (2.4). Also, from (2.6) 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,

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

For x = z, (2.7) becomes

−(y, y|z)|(z, y|z)|^{2} ≥ 0,

which implies that

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

provided y and z are linearly independent. Obviously, when y and z are linearly dependent, (2.8) holds too. Thus (2.8) is true for any two vectors y, z ∈ X. Now, if y and z are linearly independent, then (y, y|z) > 0 and, from (2.7), it follows that

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

Using (2.8), it is easy to check that (2.9) is trivially fulfilled when y and z are linearly dependent. Therefore, the inequality (2.9) 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 (2.9) 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

(2.10)  ∥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 [11]. 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 (2.10).

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 [5]) for the z−orthonormal family (f_{i})_{1≤i≤n} in the 2-inner product space (X; (·, ·|·)):

(2.11)  ∑_{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 [5] and the references therein.

3. Some Inequalities for 2-Norms

We start with the following lemma which is also interesting in itself.

Lemma 1. Let (X, (·, ·|·)) be a 2-inner product space over 𝕂 and z_{1}, …, z_{n}, z ∈ X, a_{1}, …, a_{n} ∈ 𝕂, then

(3.1)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2}

≤ \{ \begin{matrix}max_{1≤k≤n} |a_{k}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j}|z)|; \\ max_{1≤k≤n} |a_{k}| (∑_{i=1}^{n} |a_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)^{s})^{\frac{1}{s}} , r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |a_{k}| ∑_{k=1}^{n} |a_{k}| max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|); \\ ( ∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} max_{1≤i≤n} |a_{i}| (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)^{q})^{\frac{1}{q}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ( ∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |a_{i}|^{t})^{\frac{1}{t}} [∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ ( ∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |a_{i}| max_{1≤i≤n} \{( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} \} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} |a_{k}| max_{1≤i≤n} |a_{i}| ∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j}|z)|]; \\ ∑_{k=1}^{n} |a_{k}| (∑_{i=1}^{n} |a_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j}|z)|]^{l})^{\frac{1}{l}} , m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ ( ∑_{k=1}^{n} |a_{k}|)^{2} max_{i,1≤j≤n} |(z_{i}, z_{j}|z)| . \end{matrix}

6 N.S. BARNETT, Y.J. CHO^{⋆}, S.S. DRAGOMIR, S.M. KANG, AND S.S. KIM^{♦}

Proof. We observe that

(3.2) ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} = (∑_{i=1}^{n} a_{i}z_{i}, ∑_{j=1}^{n} a_{j}z_{j}|z)

= ∑_{i=1}^{n} ∑_{j=1}^{n} a_{i}\overline{a_{j}} (z_{i}, z_{j}|z) = |∑_{i=1}^{n} ∑_{j=1}^{n} a_{i}\overline{a_{j}} (z_{i}, z_{j}|z)|

≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |a_{i}| |a_{j}| |(z_{i}, z_{j}|z)| = ∑_{i=1}^{n} |a_{i}| (∑_{j=1}^{n} |a_{j}| |(z_{i}, z_{j}|z)|)

:= M.

Using Hölder’s inequality, we may write, (3.3)

∑_{j=1}^{n} |a_{j}| |(z_{i}, z_{j}|z)| ≤ \{ \begin{matrix}max_{1≤k≤n} & |a_{k}| & ∑_{j=1}^{n} & |(z_{i}, z_{j}|z)| \\ ( ∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} & |a_{k}| & max_{1≤j≤n} & |(z_{i}, z_{j}|z)| \end{matrix}

for any i ∈ {1, …, n} , giving

(3.4)  M ≤ \{ \begin{matrix}max_{1≤k≤n} |a_{k}| ∑_{i=1}^{n} |a_{i}| ∑_{j=1}^{n} |(z_{i}, z_{j}|z)| =: M_{1}; \\ ( ∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |a_{i}| ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} := M_{p}, \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} |a_{k}| ∑_{i=1}^{n} |a_{i}| max_{1≤j≤n} |(z_{i}, z_{j}|z)| =: M_{∞}. \end{matrix}

By Hölder’s inequality we also have:

(3.5) ∑_{i=1}^{n} |a_{i}| (∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)

≤ \{ \begin{matrix}max_{1≤i≤n} |a_{i}| ∑_{i,j=1}^{n} |(z_{i}, z_{j}|z)|; \\ (∑_{i=1}^{n} |a_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)^{s})^{\frac{1}{s}} , r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ ∑_{i=1}^{n} |a_{i}| max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|); \end{matrix}

and thus

M_{1} ≤ \{ \begin{matrix}max_{1≤k≤n} |a_{k}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j}|z)|; \\ max_{1≤k≤n} |a_{k}| (∑_{i=1}^{n} |a_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)^{s})^{\frac{1}{s}} , r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |a_{k}| ∑_{i=1}^{n} |a_{i}| max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|); \end{matrix}

and the first 3 inequalities in (3.1) are obtained. By Hölder’s inequality we also have:

M_{p} ≤ (∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}}

× \{ \begin{matrix}max_{1≤i≤n} |a_{i}| ∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |a_{i}|^{t})^{\frac{1}{t}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{u}{q}} )^{\frac{1}{u}} , t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ ∑_{i=1}^{n} |a_{i}| max_{1≤i≤n} \{( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} \}; \end{matrix}

and the next 3 inequalities in (3.1) are proved. Finally, by the same Hölder inequality we may state that:

M_{∞} ≤ ∑_{k=1}^{n} |a_{k}|×\{ \begin{matrix}max_{1≤i≤n} |a_{i}| ∑_{i=1}^{n} ( max_{1≤j≤n} |(z_{i}, z_{j}|z)|); \\ (∑_{i=1}^{n} |a_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} ( max_{1≤j≤n} |(z_{i}, z_{j}|z)|)^{l})^{\frac{1}{l}} , m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ ∑_{i=1}^{n} |a_{i}| max_{1≤i,j≤n} |(z_{i}, z_{j}|z)|; \end{matrix}

and the last 3 inequalities in (3.1) are proved.

To obtain some bounds for ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} in terms of ∑_{i=1}^{n} |a_{i}|^{2} , then the following corollaries may be used.

8 N.S. BARNETT, Y.J. CHO^{⋆}, S.S. DRAGOMIR, S.M. KANG, AND S.S. KIM^{♦}

Corollary 1. Let z_{1}, …, z_{n}, z and a_{1}, …, a_{n} be as in Lemma 1. If 1 < p ≤ 2, 1 < t ≤ 2, then one has the inequality

(3.6)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{k=1}^{n} |a_{k}|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}}

where \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1. Proof. By the monotonicity of power means, we may write,

(\frac{∑_{k=1}^{n} |a_{k}|^{p}}{n} )^{\frac{1}{p}} ≤ (\frac{∑_{k=1}^{n} |a_{k}|^{2}}{n} )^{\frac{1}{2}} ; 1 < p ≤ 2,
(\frac{∑_{k=1}^{n} |a_{k}|^{t}}{n} )^{\frac{1}{t}} ≤ (\frac{∑_{k=1}^{n} |a_{k}|^{2}}{n} )^{\frac{1}{2}} ; 1 < t ≤ 2,

from which we get

(∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} ≤ n^{\frac{1}{p} − \frac{1}{2}} (∑_{k=1}^{n} |a_{k}|^{2})^{\frac{1}{2}} ,
(∑_{k=1}^{n} |a_{k}|^{t})^{\frac{1}{t}} ≤ n^{\frac{1}{t} − \frac{1}{2}} (∑_{k=1}^{n} |a_{k}|^{2})^{\frac{1}{2}} .

Using the fifth inequality in (3.1), we deduce (3.6). Remark 1. An interesting particular case is the one for p = q = t = u = 2, giving

(3.7)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ ∑_{k=1}^{n} |a_{k}|^{2} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j}|z)|^{2})^{\frac{1}{2}} .

Corollary 2. With the assumptions of Lemma 1 and if 1 < p ≤ 2, then

(3.8)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ n^{\frac{1}{p}} ∑_{k=1}^{n} |a_{k}|^{2} max_{1≤i≤n} [(∑_{j=1}^{n} |(z_{i}, z_{j}|z)|^{q})^{\frac{1}{q}} ] ,

where \frac{1}{p} + \frac{1}{q} = 1. Proof. Since

(∑_{k=1}^{n} |a_{k}|^{p})^{\frac{1}{p}} ≤ n^{\frac{1}{p} − \frac{1}{2}} (∑_{k=1}^{n} |a_{k}|^{2})^{\frac{1}{2}} ,

and

∑_{k=1}^{n} |a_{k}| ≤ n^{\frac{1}{2}} (∑_{k=1}^{n} |a_{k}|^{2})^{\frac{1}{2}} ,

then by the sixth inequality in (3.1) we deduce (3.8).

In a similar fashion, one may prove the following two corollaries.

Corollary 3. With the assumptions of Lemma 1 and if 1 < m ≤ 2, then

(3.9)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ n^{\frac{1}{m}} ∑_{k=1}^{n} |a_{k}|^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j}|z)|]^{l})^{\frac{1}{l}} ,

where \frac{1}{m} + \frac{1}{l} = 1. Corollary 4. With the assumptions of Lemma 1, we have:

(3.10)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ n ∑_{k=1}^{n} |a_{k}|^{2} max_{1≤i,j≤n} |(z_{i}, z_{j}|z)| .

The following lemma may be of interest as well. Lemma 2. With the assumptions of Lemma 1, one has the inequalities,

(3.11)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ ∑_{i=1}^{n} |a_{i}|^{2} ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|

≤ \{ \begin{matrix}∑_{i=1}^{n} |a_{i}|^{2} max_{1≤i≤n} [ ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|]; \\ (∑_{i=1}^{n} |a_{i}|^{2p})^{\frac{1}{p}} (( ∑_{j=1}^{n} |(z_{i}, z_{j}|z)|)^{q})^{\frac{1}{q}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ max_{1≤i≤n} |a_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j}|z)| . \end{matrix}

Proof. As in Lemma 1, we know that,

(3.12)  ∥∑_{i=1}^{n} a_{i}z_{i}|z∥^{2} ≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |a_{i}| |a_{j}| |(z_{i}, z_{j}|z)| .

Using the simple observation that (see also [5, p. 394])

|a_{i}| |a_{j}| ≤ \frac{1}{2} (|a_{i}|^{2} + |a_{j}|^{2}) , i, j ∈ \{1, …, n\}

we have,

∑_{i=1}^{n} ∑_{j=1}^{n} |a_{i}| |a_{j}| |(z_{i}, z_{j}|z)| ≤ \frac{1}{2} ∑_{i,j=1}^{n} (|a_{i}|^{2} + |a_{j}|^{2}) |(z_{i}, z_{j}|z)|

= \frac{1}{2} [ ∑_{i,j=1}^{n} |a_{i}|^{2} |(z_{i}, z_{j}|z)| + ∑_{i,j=1}^{n} |a_{j}|^{2} |(z_{i}, z_{j}|z)|]

= ∑_{i,j=1}^{n} |a_{i}|^{2} |(z_{i}, z_{j}|z)| ,

which proves the first inequality in (3.11).

The second part follows by Hölder’s inequality and we omit the details.

4. Some Inequalities for Fourier Coefficients in 2-Inner Product Spaces

We are now able to point out the following result.

Theorem 5. Let x, y_{1}, …, y_{n}, z be vectors of a 2-inner product space (X; (·, ·|·)) and c_{1}, …, c_{n} ∈ 𝕂. Then one has the inequalities:

(4.1)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2}

≤ ∥x|z∥^{2}×\{ \begin{matrix}max_{1≤k≤n} |c_{k}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|; \\ max_{1≤k≤n} |c_{k}| (∑_{i=1}^{n} |c_{i}|^{r})^{\frac{1}{r}} [∑_{i=1}^{n} ( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)^{s}]^{\frac{1}{s}} , r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |c_{k}| ∑_{k=1}^{n} |c_{k}| max_{1≤i≤n} ( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|); \\ ( ∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} max_{1≤i≤n} |c_{i}| (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)^{q})^{\frac{1}{q}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ( ∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |c_{i}|^{t})^{\frac{1}{t}} [∑_{i=1}^{n} ( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ ( ∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |c_{i}| max_{1≤i≤n} \{( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{1}{q}} \} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} |c_{k}| max_{1≤i≤n} |c_{i}| ∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|]; \\ ∑_{k=1}^{n} |c_{k}| (∑_{i=1}^{n} |c_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|]^{l})^{\frac{1}{l}} , m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ ( ∑_{k=1}^{n} |c_{k}|)^{2} max_{i,1≤j≤n} |(y_{i}, y_{j}|z)| . \end{matrix}

Proof. We note that

∑_{i=1}^{n} c_{i} (x, y_{i}|z) = (x, ∑_{i=1}^{n} \overline{c_{i}}y_{i}|z) .

BOMBIERI, SELBERG AND HEILBRONN TYPE INEQUALITIES 11

Using Schwarz’s inequality in 2-inner product spaces, we have

(4.2)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} ∥∑_{i=1}^{n} \overline{c_{i}}y_{i}|z∥^{2} .

Finally, using Lemma 1 with a_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , we deduce the desired inequality (4.1). We omit the details.

The following corollaries may be useful if one needs bounds in terms of ∑_{i=1}^{n} |c_{i}|^{2} . Corollary 5. With the assumptions in Theorem 5 and if 1 < p ≤ 2, 1 < t ≤ 2, \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1, one has the inequality:

(4.3) |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{i=1}^{n} |c_{i}|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

and, in particular, for p = q = t = u = 2,

(4.4)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} ( ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|^{2})^{\frac{1}{2}} .

The proof is similar to that of Corollary 1. Corollary 6. With the assumptions in Theorem 5 and if 1 < p ≤ 2, then,

(4.5)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} n^{\frac{1}{p}} ∑_{k=1}^{n} |c_{k}|^{2} max_{1≤i≤n} [∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q}]^{\frac{1}{q}} ,

where \frac{1}{p} + \frac{1}{q} = 1.

The proof is similar to that of Corollary 2.

The following two inequalities also hold. Corollary 7. With the above assumptions for X, y_{i}, c_{i} and if 1 < m ≤ 2, then,

(4.6)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} n^{\frac{1}{m}} ∑_{k=1}^{n} |c_{k}|^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|]^{l})^{\frac{1}{l}} ,

where \frac{1}{m} + \frac{1}{l} = 1. Corollary 8. With the above assumptions for X, y_{i}, c_{i}, one has

(4.7)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} n ∑_{k=1}^{n} |c_{k}|^{2} max_{i,1≤j≤n} |(y_{i}, y_{j}|z)| .

Using Lemma 2, we may state the following result as well. Remark 2. With the assumptions of Theorem 5,

(4.8)  |∑_{i=1}^{n} c_{i} (x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|

≤ ∥x|z∥^{2} × \{ \begin{matrix}∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} [ ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|]; \\ (∑_{i=1}^{n} |c_{i}|^{2p})^{\frac{1}{p}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)^{q})^{\frac{1}{q}} , p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ max_{1≤i≤n} |c_{i}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)| . \end{matrix}

5. Bombieri, Selberg and Heilbronn Inequalities in 2-Inner Product Spaces

We first note the following Bombieri type inequality for 2-inner products as an important consequence of the second part of (4.8),

(5.1)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j}|z)|} .

This result can be easily derived from the first branch of that inequality for the choice c_{i} = \overline{(x, y_{i}|z)} (i = 1, …, n) .

It is obvious that if (y_{i})_{1≤i≤n} is a z−orthonormal family in the 2-inner product space (X; (·, ·|·)), then (5.1) will produce Bessel’s inequality (2.11).

If one chooses in the first inequality of (4.8),

c_{i} = \frac{\overline{(x, y_{i}|z)}}{∑_{j=1}^{n} |(y_{i}, y_{j}|z)|}, i = 1, …, n

then one can state the following inequality,

(5.2)  ∑_{i=1}^{n} \begin{matrix}|(x, y_{i}|z)| & 2 \\ ∑ & {n \atop j=1} & |(y_{i}, y_{j}|z)| \end{matrix} ≤ ∥x|z∥^{2} , z ∈ X,

provided that ∑_{j=1}^{n} |(y_{i}, y_{j}|z)| ≠ 0.

When (y_{i})_{1≤i≤n} is a z−orthonormal family in the 2-inner product space (X; (·, ·|·)), then (5.1) will produce Bessel’s inequality (2.11) as well.

The inequality (5.2) is the corresponding version for 2-inner product spaces of the Selberg inequality.

Finally, if one considers

c_{i} = \frac{\overline{(x, y_{i}|z)}}{|(y_{i}, y_{j}|z)|}, i = 1, …, n

in the first inequality of (4.8), then after simple computation we deduce the following result,

(5.3)  ∑_{i=1}^{n} |(x, y_{i}|z)| ≤ ∥x|z∥ ( ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|)^{\frac{1}{2}} ,

which is the corresponding version for 2-inner products of Heilbronn’s result.

6. More Inequalities of the Bombieri Type in 2-Inner Product Spaces

Further, we point out other inequalities of Bombieri type that may be obtained from (4.1) on choosing c_{i} = \overline{(x, y_{i}|z)} (i = 1, …, n) .

If the above choice is made in the first inequality of (4.1), then one obtains:

(∑_{i=1}^{n} |(x, y_{i}|z)|^{2})^{2} ≤ ∥x|z∥^{2} max_{1≤i≤n} |(x, y_{i}|z)|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|

giving,

(6.1)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥ max_{1≤i≤n} |(x, y_{i}|z)| ( ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|)^{\frac{1}{2}} , x ∈ X.

If the same choice for c_{i} is made in the second inequality of (4.1), then

(∑_{i=1}^{n} |(x, y_{i}|z)|^{2})^{2}

≤ ∥x|z∥^{2} max_{1≤i≤n} |(x, y_{i}|z)| (∑_{i=1}^{n} |(x, y_{i}|z)|^{r})^{\frac{1}{r}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)^{s}]^{\frac{1}{s}} ,

implying that,

(6.2) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2}

≤ ∥x|z∥ max_{1≤i≤n} |(x, y_{i}|z)|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i}|z)|^{r})^{\frac{1}{2r}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)^{s}]^{\frac{1}{2s}} ,

where \frac{1}{r} + \frac{1}{s} = 1, s > 1.

The other inequalities in (4.1) will produce the following results, respectively

∑_{i=1}^{n} |(x, y_{i}|z)|^{2}

≤ ∥x|z∥ max_{1≤i≤n} |(x, y_{i}|z)|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i}|z)|)^{\frac{1}{2}} [ max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|)];

(6.3) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2}

≤ ∥x|z∥ max_{1≤i≤n} |(x, y_{i}|z)|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i}|z)|^{p})^{\frac{1}{2p}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{1}{q}} ]^{\frac{1}{2}} ,

14 N.S. BARNETT, Y.J. CHO^{⋆}, S.S. DRAGOMIR, S.M. KANG, AND S.S. KIM^{♦}

where p > 1, \frac{1}{p} + \frac{1}{q} = 1;

(6.4) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2}

≤ ∥x|z∥ (∑_{i=1}^{n} |(x, y_{i}|z)|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, y_{i}|z)|^{t})^{\frac{1}{2t}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{2u}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1, t > 1, \frac{1}{t} + \frac{1}{u} = 1;

(6.5) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2}

≤ ∥x|z∥ (∑_{i=1}^{n} |(x, y_{i}|z)|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, y_{i}|z)|)^{\frac{1}{2}} max_{1≤i≤n} \{(∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{1}{2q}} \} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1;

∑_{i=1}^{n} |(x, y_{i}|z)|^{2}
≤ ∥x|z∥ [∑_{i=1}^{n} |(x, y_{i}|z)|]^{\frac{1}{2}} max_{1≤i≤n} |(x, y_{i}|z)|^{\frac{1}{2}} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|])^{\frac{1}{2}} ;
(6.6) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥ [∑_{i=1}^{n} |(x, y_{i}|z)|^{m}]^{\frac{1}{2m}} [∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|^{l}]]^{\frac{1}{2l}} ,

where m > 1, \frac{1}{m} + \frac{1}{l} = 1; and

(6.7) ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥ ∑_{i=1}^{n} |(x, y_{i}|z)| max_{i,1≤j≤n} |(y_{i}, y_{j}|z)|^{\frac{1}{2}} .

If in the above inequalities we assume that (y_{i})_{1≤i≤n} = (f_{i})_{1≤i≤n} , where (f_{i})_{1≤i≤n} are z−orthonormal vectors in the 2-inner product space (X, (·, ·|·)) , then from (6.1)

– (6.7) we may deduce the following inequalities similar, in a sense to Bessel’s inequality:

(6.8)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ \sqrt{n} ∥x|z∥ max_{1≤i≤n} \{|(x, f_{i}|z)|\};
(6.9)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ n^{\frac{1}{2s}} ∥x|z∥ max_{1≤i≤n} \{|(x, f_{i}|z)|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, f_{i}|z)|^{r})^{\frac{1}{2r}} ,

where r > 1, \frac{1}{r} + \frac{1}{s} = 1;

(6.10)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ ∥x|z∥ max_{1≤i≤n} \{|(x, f_{i}|z)|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, f_{i}|z)|)^{\frac{1}{2}} ;

(6.11)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ \sqrt{n} ∥x|z∥ max_{1≤i≤n} \{|(x, f_{i}|z)|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, f_{i}|z)|^{p})^{\frac{1}{2p}} ,

where p > 1;

(6.12)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ n^{\frac{1}{2u}} ∥x|z∥ (∑_{i=1}^{n} |(x, f_{i}|z)|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, f_{i}|z)|^{t})^{\frac{1}{2t}} ,

where p > 1, t > 1, \frac{1}{t} + \frac{1}{u} = 1;

(6.13) ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ ∥x|z∥ (∑_{i=1}^{n} |(x, f_{i}|z)|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, f_{i}|z)|)^{\frac{1}{2}} , p > 1;

(6.14)  ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ \sqrt{n} ∥x|z∥ (∑_{i=1}^{n} |(x, f_{i}|z)|)^{\frac{1}{2}} max_{1≤i≤n} \{|(x, f_{i}|z)|^{\frac{1}{2}} \};

(6.15) ∑_{i=1}^{n} |(x, f_{i}|z)|^{2} ≤ n^{\frac{1}{2l}} ∥x|z∥ [∑_{i=1}^{n} |(x, f_{i}|z)|^{m}]^{\frac{1}{m}} , m > 1, \frac{1}{m} + \frac{1}{l} = 1;

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

Corollaries 5 – 8 will produce the following results which do not contain the Fourier coefficients in the right side of the inequality.

Indeed, if one chooses c_{i} = \overline{(x, y_{i}|z)} in (4.3), then,

(∑_{i=1}^{n} |(x, y_{i}|z)|^{2})^{2} ≤ ∥x|z∥^{2} n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

giving the following Bombieri type inequality:

(6.17)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ n^{\frac{1}{p} + \frac{1}{t} −1} ∥x|z∥^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j}|z)|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

where 1 < p ≤ 2, 1 < t ≤ 2, \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1.

If in this inequality we consider p = q = t = u = 2, then

(6.18)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} ( ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|^{2})^{\frac{1}{2}} .

In a similar way, if c_{i} = \overline{(x, y_{i}|z)} in (4.6), then,

(6.19)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ n^{\frac{1}{m}} ∥x|z∥^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j}|z)|]^{l})^{\frac{1}{l}} ,

where m > 1, \frac{1}{m} + \frac{1}{l} = 1.

Finally, if c_{i} = \overline{(x, y_{i}|z)} (i = 1, …, n) , is taken in (4.7), then

(6.20)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ n ∥x|z∥^{2} max_{1≤i,j≤n} |(y_{i}, y_{j}|z)| .

Remark 3. We now compare,

(6.21)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j}|z)|}

with,

(6.22)  ∑_{i=1}^{n} |(x, y_{i}|z)|^{2} ≤ ∥x|z∥^{2} \{ ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|^{2}\}^{\frac{1}{2}} .

Denote

M_{1} := max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j}|z)|}

and

M_{2} := [ ∑_{i,j=1}^{n} |(y_{i}, y_{j}|z)|^{2}]^{\frac{1}{2}} .

If we choose n = 2, then for a := |(y_{1}, y_{1}|z)| , b := |(y_{1}, y_{2}|z)| , c := |(y_{2}, y_{2}|z)| , a, b, c > 0, we have

M_{1} = max {a + b, b + c} ,
M_{2} = (a^{2} + 2b^{2} + c^{2})^{\frac{1}{2}} .

Assume that a = 2, b = 1 and c = 3. Then M_{1} = 4 > \sqrt{15} = M_{2}, showing that, in this case, the bound provided by (6.22) is better than the bound provided by (6.21). If (y_{i})_{1≤i≤n} are z−orthonormal vectors, then M_{1} = 1, M_{2} = \sqrt{n}, showing that in this case the Bombieri type inequality (which becomes Bessel’s inequality) provides a better bound than (6.22).

7. Applications for Determinantal 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 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

(7.1) (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

(7.2)  ∥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

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

and

(7.4)  ∥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.

Using the representations (7.3), (7.4) 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 f, g_{1}, ..., g_{n}, h ∈ L_{ρ}^{2} (Ω) , where ρ : Ω → [0, ∞) is a measurable function on Ω, then we have the inequality,

(7.5)  ∑_{i=1}^{n} | \begin{matrix}∫_{Ω} ρfg_{i}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |^{2}

≤ | \begin{matrix}∫_{Ω} ρf^{2}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} | max_{1≤i≤n} \{∑_{j=1}^{n} | det [ \begin{matrix}∫_{Ω} ρg_{j}g_{i}dμ & ∫_{Ω} ρg_{j}hdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] |\} .

The proof follows by the Bombieri type inequality for 2-inner products incorporated in 5.1.

Proposition 2. Let f, g_{1}, ..., g_{n}, h ∈ L_{ρ}^{2} (Ω) , where ρ : Ω → [0, ∞) is a measurable function on Ω, then,

(7.6)  ∑_{i=1}^{n} \frac{| \begin{matrix}∫_{Ω} ρfg_{i}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |^{2}}{∑_{j=1}^{n} | det [ \begin{matrix}∫_{Ω} ρg_{j}g_{i}dμ & ∫_{Ω} ρg_{j}hdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] |}

≤ | \begin{matrix}∫_{Ω} ρf^{2}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |

provided that

∑_{j=1}^{n} | det [ \begin{matrix}∫_{Ω} ρg_{j}g_{i}dμ & ∫_{Ω} ρg_{j}hdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] | ≠ 0

for each i ∈ {1, ..., n} .

This result follows by the Selberg type inequality (5.2).

Finally, by the use of the Heilbronn type inequality (5.3), we have:

Proposition 3. With the above assumptions for f, g_{1}, ..., g_{n}, h,

(7.7)  ∑_{i=1}^{n} | det [ \begin{matrix}∫_{Ω} ρfg_{i}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] |

≤ | \begin{matrix}∫_{Ω} ρf^{2}dμ & ∫_{Ω} ρfhdμ \\ ∫_{Ω} ρfhdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} |^{1/2} \{ ∑_{i,j=1}^{n} | det [ \begin{matrix}∫_{Ω} ρg_{j}g_{i}dμ & ∫_{Ω} ρg_{j}hdμ \\ ∫_{Ω} ρg_{i}hdμ & ∫_{Ω} ρh^{2}dμ \end{matrix} ] |\}^{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

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[7] S.S. DRAGOMIR and B. MOND, On the Boas-Bellman generalisation of Bessel’s inequality in inner product spaces, Italian J. of Pure & Appl. Math., 3 (1998), 29–35.

[8] S.S. DRAGOMIR, B. MOND and J.E. PEČARIĆ, Some remarks on Bessel’s inequality in inner product spaces, Studia Univ. Babeş-Bolyai, Mathematica, 37(4) (1992), 77–86.

[9] H. HEILBRONN, On the averages of some arithmetical functions of two variables, Mathematica, 5(1958), 1-7.

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

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

Department of Mathematics, College of Education, 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, Victoria 8001, Australia.

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

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

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

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

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