ON PEČARIĆ’S INEQUALITY IN INNER PRODUCT SPACES

S.S. DRAGOMIR

Abstract. Some related results to Pečarić’s inequality in inner product spaces that generalises Bombieri’s inequality, are given.

1. Introduction

In 1992, J.E. Pečarić [3] proved the following inequality for vectors in complex inner product spaces (H; (·, ·)).

Theorem 1. Suppose that x, y_{1}, …, y_{n} are vectors in H and c_{1}, …, c_{n} are complex numbers. Then the following inequalities

(1.1)  |∑_{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})|) ,

hold.

He also showed that for c_{i} = \overline{(x, y_{i})}, i ∈ \{1, ..., n\} , one gets

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

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

which improves Bombieri’s result [1] (see also [2, p. 394])

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

Note that (1.3) is in its turn a natural generalisation of Bessel’s inequality

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

which holds for the orthornormal vectors (e_{i})_{1≤i≤n} .

Date: 12 June, 2003.

2000 Mathematics Subject Classification. 26D15, 46C05.

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

2 S.S. DRAGOMIR

In this paper we point out some related results to Pečarić’s inequality (1.1). Some results of Bombieri type are also mentioned.

2. Preliminary Results

We start with the following lemma that is interesting in its own right. Lemma 1. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequalities:

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

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{γq})^{\frac{1}{γq}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} ; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{αp})^{\frac{1}{αp}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{βq}} , \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{αp} & (∑_{i=1}^{n} |α_{i}|^{γq}) & \frac{1}{γq} & (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{pβ} \\ × (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} (∑_{i=1}^{n} |α_{i}|^{αp})^{\frac{1}{αp}} max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{pβ}} , \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} ( ∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |α_{i}|^{γq})^{\frac{1}{γq}} max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} (∑_{i=1}^{n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} ( ∑_{j=1}^{n} |(z_{i}, z_{j})|) , \end{matrix}

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

PEČARIĆ INEQUALITY 3

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.

If one uses the Hölder inequality for double sums, i.e., we recall it

(2.3) ∑_{i,j=1}^{n} m_{ij}a_{ij}b_{ij} ≤ ( ∑_{i,j=1}^{n} m_{ij}a_{ij}^{p})^{\frac{1}{p}} ( ∑_{i,j=1}^{n} m_{ij}b_{ij}^{q})^{\frac{1}{q}} ,

where m_{ij}, a_{ij}, b_{ij} ≥ 0, \frac{1}{p} + \frac{1}{q} = 1, p > 1; then

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

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

and the first inequality in (2.1) is proved.

Observe that

∑_{i=1}^{n} |α_{i}|^{p} (∑_{j=1}^{n} |(z_{i}, z_{j})|) ≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{p} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{α} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{β} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{p} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|); \end{matrix}

giving

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

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}| (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} ; \\ (∑_{i=1}^{n} |α_{i}|^{αp})^{\frac{1}{αp}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{βp}} if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} . \end{matrix}

Similarly, we have

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

≤ \{ \begin{matrix}max & 1≤i≤n & |α_{i}| & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{q} \\ (∑_{i=1}^{n} |α_{i}|^{γq})^{\frac{1}{γq}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} . \end{matrix}

Using (2.1) and (2.5) – (2.6), we deduce the 9 inequalities in the second part of (2.2).

If we choose p = q = 2, then the following result holds.

Corollary 1. If z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂, then one has

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

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{2γ})^{\frac{1}{2γ}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{2})^{\frac{1}{2}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} ; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{2α})^{\frac{1}{2α}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{2β}} , \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{2α}) & \frac{1}{2α} & (∑_{i=1}^{n} |α_{i}|^{2γ}) & \frac{1}{2γ} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{2β} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} & (∑_{i=1}^{n} |α_{i}|^{2α}) & \frac{1}{2α} & max & 1≤i≤n & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{2β}} ,  if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} |α_{i}|^{2})^{\frac{1}{2}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} ; \\ (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} & (∑_{i=1}^{n} |α_{i}|^{2γ}) & \frac{1}{2γ} & max & 1≤i≤n & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} ,  if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|) . \end{matrix}

3. Some Pečarić Type Inequalities

We are now able to point out the following result which complements and generalises the inequality (1.1 ) due to J. Pečarić.

PEČARIĆ INEQUALITY 5

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

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

≤ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≤n} |c_{i}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|; \\ max & 1≤i≤n & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} ,  if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max & 1≤i≤n & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{q}) & \frac{1}{q} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{q}} ; \\ max & 1≤i≤n & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β}) & \frac{1}{pβ} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} if α > 1, \frac{1}{α} + \frac{1}{β} = 1 \\ and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{q}) & \frac{1}{q} & (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} ,  if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ max & 1≤i≤n & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{p}) & \frac{1}{p} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |c_{i}|^{p}) & \frac{1}{p} & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} ,  if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |c_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|); \end{matrix}

where p > 1, \frac{1}{p} + \frac{1}{q} = 1. Proof. We note that

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

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

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

6 S.S. DRAGOMIR

Finally, using Lemma 1 with α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , we deduce the desired inequality (3.1). Remark 1. If in (3.1) we choose p = q = 2, we obtain amongst others, the result (1.1) due to J. Pečarić.

4. Some Results of Bombieri Type

The following results of Bombieri type hold. Theorem 3. Let x, y_{1}, …, y_{n} ∈ H. Then one has the inequality:

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

≤ ∥x∥×\{ \begin{matrix}max_{1≤i≤n} |(x, y_{i})| (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} ; \\ max & 1≤i≤n & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}} ,  if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max & 1≤i≤n & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{2q} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2q}} ; \\ max & 1≤i≤n & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{αp}) & \frac{1}{2αβ} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} ,  if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{αp}) & \frac{1}{2αp} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β}) & \frac{1}{2pβ} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}}  if α > 1, \frac{1}{α} + \frac{1}{β} = 1 \\ and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{2q} & (∑_{i=1}^{n} |(x, y_{i})|^{αp}) & \frac{1}{2αp} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{2pβ}} ,  if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ max & 1≤i≤n & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{p}) & \frac{1}{2p} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2q}} ; \\ (∑_{i=1}^{n} |(x, y_{i})|^{p}) & \frac{1}{2p} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} & max & 1≤i≤n & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}} ,  if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, y_{i})|^{q})^{\frac{1}{2q}} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} , \end{matrix}

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

Proof. The proof follows by Theorem 2 on choosing c_{i} = \overline{(x, y_{i})}, i ∈ \{1, ..., n\} and taking the square root in both sides of the inequalities involved. We omit the details.

Remark 2. We observe, by the last inequality in (4.1), we get

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

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

If in this inequality we choose p = q = 2, then we recapture Bombieri’s result (1.3) .

References

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

[2] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993.

[3] J.E. PEČARIĆ, On some classical inequalities in unitary spaces, Mat. Bilten (Skopje), 16 (1992), 63-72.

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