SOME BOMBIERI TYPE INEQUALITIES IN INNER PRODUCT SPACES

S.S. DRAGOMIR

Abstract. Companion results to the Bombieri generalisation of Bessel’s inequality in inner product spaces are given.

1. Introduction

In 1971, E. Bombieri [1], has given the following generalisation of Bessel’s inequality:

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

where x, y_{1}, …, y_{n} are vectors in the inner product space (H; (·, ·)) .

It is obvious that if (y_{i})_{1≤i≤n} = (e_{i})_{1≤i≤n} , where (e_{i})_{1≤i≤n} are orthornormal vectors in H, i.e., (e_{i}, e_{j}) = δ_{ij} (i, j = 1, …, n) , where δ_{ij} is the Kronecker delta, then (1.1) provides Bessel’s inequality

(1.2)  ∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥^{2} , x ∈ H.

In this paper we point out other Bombieri type inequalities and show that, some times, the new ones may provide better bounds for ∑_{i=1}^{n} |(x, y_{i})|^{2} .

2. The Results

The following lemma which is of interest in itself holds.

Date: 11 June, 2003.

2000 Mathematics Subject Classification. 26D15, 46C05.

Key words and phrases. Bessel’s inequality, Bombieri inequality.

2 S.S. DRAGOMIR

Lemma 1. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequalities:

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} (2.1)

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

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} (∑_{i=1}^{n} ∥z_{i}∥)^{2} ; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{2}{p}} (∑_{i=1}^{n} ∥z_{i}∥^{q})^{\frac{2}{q}} , where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |α_{i}|)^{2} max_{1≤i≤n} ∥z_{i}∥^{2} . \end{matrix}

Proof. We observe that

(2.2) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} = (∑_{i=1}^{n} α_{i}z_{i}, ∑_{j=1}^{n} α_{j}z_{j}) = ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j})

= |∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} |(z_{i}, z_{j})|| ≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| =: M.

Firstly, we have

M ≤ max_{1≤i,j≤n} {|α_{i}| |α_{j}|} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|

= max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})| .

Secondly, by the Hölder inequality for double sums, we have

M ≤ [ ∑_{i,j=1}^{n} (|α_{i}| |α_{j}|)^{p}]^{\frac{1}{p}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}}

= (∑_{i=1}^{n} |α_{i}|^{p} ∑_{j=1}^{n} |α_{j}|^{p})^{\frac{1}{p}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}}

= (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{2}{p}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} ,

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

Finally, we have

M ≤ max_{1≤i,j≤n} |(z_{i}, z_{j})| ∑_{i,j=1}^{n} |α_{i}| |α_{j}| = (∑_{i=1}^{n} |α_{i}|)^{2} max_{1≤i,j≤n} |(z_{i}, z_{j})|

and the first part of the lemma is proved.

The second part is obvious on taking into account, by Schwarz’s inequality in H, that we have

|(z_{i}, z_{j})| ≤ ∥z_{i}∥ ∥z_{j}∥ ,

for any i, j ∈ {1, ..., n} . We omit the details.

Corollary 1. With the assumptions in Lemma 1, one has

(2.3)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{2})^{\frac{1}{2}}

≤ ∑_{i=1}^{n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} .

The proof follows by Lemma 1 on choosing p = q = 2.

Note also that (2.3) provides a refinement of the well known Cauchy-Bunyakovsky-Schwarz inequality for sequences of vectors in inner product spaces, namely

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} .

The following lemma also holds.

Lemma 2. Let x, y_{1}, …, y_{n} ∈ H and c_{1}, …, c_{n} ∈ 𝕂. Then one has the inequalities:

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} (2.4)

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

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

Proof. We have, by Schwarz’s inequality in the inner product (H; (·, ·)) , that

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

Now, applying Lemma 1 for α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , the inequality (2.4) is proved. Corollary 2. With the assumptions in Lemma 2, one has

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

The proof follows by Lemma 2, on choosing p = q = 2. Remark 1. The inequality (2.5) was firstly obtained in [2] (see inequality (7)).

The following theorem incorporating three Bombieri type inequalities holds. Theorem 1. Let x, y_{1}, …, y_{n} ∈ H. Then one has the inequalities:

(2.6) ∑_{i=1}^{n} |(x, y_{i})|^{2}

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

Proof. Choosing c_{i} = \overline{(x, y_{i})} (i = 1, …, n) in (2.4) we deduce

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

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

which, by taking the square root, is clearly equivalent to (2.6).

Remark 2. If (y_{i})_{1≤i≤n} = (e_{i})_{1≤i≤n} where (e_{i})_{1≤i≤n} are orthornormal vectors in H, then by (2.6) we deduce

(2.8) ∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥×\{ \begin{matrix}\sqrt{n} max_{1≤i≤n} |(x, e_{i})|; \\ n^{\frac{1}{2q}} (∑_{i=1}^{n} |(x, e_{i})|^{p})^{\frac{1}{p}} , where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} |(x, e_{i})| . \end{matrix}

If in (2.7) we take p = q = 2, then we obtain the following inequality that was formulated in [2, p. 81].

Corollary 3. With the assumptions in Theorem 1, we have:

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

Remark 3. Observe, that by the monotonicity of power means, we may write

(2.10)  (\frac{∑_{i=1}^{n} |(x, y_{i})|^{p}}{n} )^{\frac{1}{p}} ≤ (\frac{∑_{i=1}^{n} |(x, y_{i})|^{2}}{n} )^{\frac{1}{2}} , 1 < p ≤ 2.

Taking the square in both sides, one has

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

giving

(2.11)  (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{2}{p}} ≤ n^{\frac{2}{p} −1} ∑_{i=1}^{n} |(x, y_{i})|^{2} .

Using (2.11) and the second inequality in (2.7) we may deduce the following result

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

for 1 < p ≤ 2, \frac{1}{p} + \frac{1}{q} = 1.

Note that for p = 2 (q = 2) we recapture (2.9).

Remark 4. Let us compare Bombieri’s result

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

with our general result (2.12).

To do that, denote

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

and

M_{2} := n^{\frac{2}{p} −1} ( ∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} , 1 < p ≤ 2, \frac{1}{p} + \frac{1}{q} = 1.

Consider the inner product space H = ℝ, (x, y) = x · y, n = 2 and y_{1} = a > 0, y_{2} = b > 0. Then

M_{1} = max {a^{2} + ab, ab + b^{2}} = (a + b) max {a, b} ,
M_{2} = 2^{\frac{2}{p} −1} (a^{q} + b^{q})^{\frac{2}{q}} = 2^{\frac{2}{p} −1} (a^{\frac{p}{p−1}} + b^{\frac{p}{p−1}} )^{\frac{2(p−1)}{p}} , 1 < p ≤ 2.

Assume a = 1, b ∈ [0, 1] , p ∈ (1, 2]. Utilizing Maple 6, one may easily see by plotting the function

f (b, p) := M_{2} − M_{1} = 2^{\frac{2}{p} −1} (1 + b^{\frac{p}{p−1}} )^{\frac{2(p−1)}{p}} − 1 − b

that it has positive and negative values in the box [0, 1] × [1, 2], showing that the inequalities (2.12) and (2.13) cannot be compared. This means that one is not always better than the other.

References

[1] E. BOMBIERI, A note on the large sieve, Acta Arith., 18 (1971), 401-404.

[2] 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.

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