SOME CAUCHY-BUNYAKOVSKY-SCHWARZ TYPE INEQUALITIES FOR SEQUENCES OF OPERATORS IN HILBERT SPACES

SEVER S. DRAGOMIR

Abstract. Some inequalities of Cauchy-Bunyakovsky-Schwarz type for sequences of bounded linear operators in Hilbert spaces and some applications are given.

1. Introduction

Let (H; (·, ·)) be a real or complex Hilbert space and B (H) the Banach algebra of all bounded linear operators that map H into H.

We recall that a self-adjoint operator A ∈ B (H) is positive in B (H) iff (Ax, x) ≥ 0 for any x ∈ H. The binary relation A ≥ B iff A − B is a positive self-adjoint operator, is an order relation on B (H) . We remark that for any A ∈ B (H) the operators U := AA^{∗} and V := A^{∗}A are positive self adjoint operators on H and ∥U∥ = ∥V ∥ = ∥A∥^{2} .

In [1], the author has proved the following inequality of Cauchy-Bunyakovsky-Schwarz type in the order of B (H) .

Theorem 1. Let A_{1}, …, A_{n} ∈ B (H) and z_{1}, …, z_{n} ∈ 𝕂 (ℝ, ℂ) . Then the following inequality holds:

(1.1)  ∑_{i=1}^{n} |z_{i}|^{2} ∑_{i=1}^{n} A_{i}A_{i}^{∗} ≥ (∑_{i=1}^{n} z_{i}A_{i}) (∑_{i=1}^{n} \overline{z_{i}}A_{i}^{∗}) ≥ 0.

Proof. For the sake of completeness, we give here a simple proof of this inequality. For any i, j ∈ {1, …, n} one has in the order of B (H) :

(\overline{z_{i}}A_{j} − \overline{z_{j}}A_{i}) (\overline{z_{i}}A_{j} − \overline{z_{j}}A_{i})^{∗} ≥ 0,

that is,

(\overline{z_{i}}A_{j} − \overline{z_{j}}A_{i}) (z_{i}A_{j}^{∗} − z_{j}A_{i}^{∗}) ≥ 0,

from where results

(1.2)  |z_{i}|^{2} A_{j}A_{j}^{∗} + |z_{j}|^{2} A_{i}A_{i}^{∗} ≥ \overline{z_{i}}z_{j}A_{j}A_{i}^{∗} + \overline{z_{j}}z_{i}A_{i}A_{j}^{∗}

for any i, j ∈ {1, …, n} .

If we sum (1.2) over i from 1 to n we deduce (1.3)

(∑_{i=1}^{n} |z_{i}|^{2}) A_{j}A_{j}^{∗} + |z_{j}|^{2} (∑_{i=1}^{n} A_{i}A_{i}^{∗}) ≥ z_{j}A_{j} (∑_{i=1}^{n} \overline{z_{i}}A_{i}^{∗}) + (∑_{i=1}^{n} z_{i}A_{i}) \overline{z_{j}}A_{j}^{∗},

1991 Mathematics Subject Classification. 26D15, 47A05, 46A05.

Key words and phrases. Cauchy-Bunyakovsky-Schwarz inequality, Hilbert spaces, Bounded linear operators.

for any j ∈ {1, …, n} .

If we sum (1.3) over j from 1 to n, we deduce

(1.4) ∑_{i=1}^{n} |z_{i}|^{2} ∑_{j=1}^{n} A_{j}A_{j}^{∗} + ∑_{j=1}^{n} |z_{j}|^{2} (∑_{i=1}^{n} A_{i}A_{i}^{∗})

≥ ∑_{j=1}^{n} z_{j}A_{j} ∑_{i=1}^{n} \overline{z_{i}}A_{i}^{∗} + (∑_{i=1}^{n} z_{i}A_{i}) (∑_{j=1}^{n} \overline{z_{j}}A_{j}^{∗}) ,

that is,

(1.5) ∑_{k=1}^{n} |z_{k}|^{2} ∑_{k=1}^{n} A_{k}A_{k}^{∗} ≥ ∑_{k=1}^{n} z_{k}A_{k} ∑_{k=1}^{n} \overline{z_{k}}A_{k}^{∗} = (∑_{k=1}^{n} z_{k}A_{k}) (∑_{k=1}^{n} z_{k}A_{k})^{∗} ≥ 0,

and the theorem is proved.

The following version of the Cauchy-Bunyakovsky-Schwarz inequality for norms also holds [1].

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

(1.6)  ∑_{k=1}^{n} |z_{k}|^{2} ∥∑_{k=1}^{n} A_{k}A_{k}^{∗}∥ ≥ ∥∑_{k=1}^{n} z_{k}A_{k}∥^{2} .

Proof. The operators:

A := ∑_{k=1}^{n} |z_{k}|^{2} ∑_{k=1}^{n} A_{k}A_{k}^{∗}, B := (∑_{k=1}^{n} z_{k}A_{k}) (∑_{k=1}^{n} \overline{z_{k}}A_{k})^{∗}

are obviously self-adjoint, positive and by (1.1), A ≥ B ≥ 0. Thus ∥A∥ ≥ ∥B∥ and since,

∥A∥ = ∑_{k=1}^{n} |z_{k}|^{2} ∥∑_{k=1}^{n} A_{k}A_{k}^{∗}∥

and

∥B∥ = ∥∑_{k=1}^{n} z_{k}A_{k}∥^{2}

the corollary is proved.

For other related results, see [2].

The main aim of this paper is to point out other inequalities similar to (1.6).

2. Norm Inequalities

The following result holds.

Theorem 2. Let α_{1}, …, α_{n} ∈ 𝕂 and A_{1}, …, A_{n} ∈ B (H) . Then one has the inequalities:

(2.1) ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ≤ \{ \begin{matrix}max_{i=\overline{1,n}} & |α_{i}| & 2 & ∑_{i=1}^{n} & ∥A_{i}∥ & 2 \\ (∑_{i=1}^{n} |α_{i}|^{2p})^{\frac{1}{p}} (∑_{i=1}^{n} ∥A_{i}∥^{2q})^{\frac{1}{q}} if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} & |α_{i}| & 2 & max_{i=\overline{1,n}} & ∥A_{i}∥ & 2 \end{matrix}

+ \{ \begin{matrix}max_{1≤i≠j≤n} & \{|α_{i}| |α_{j}|\} & ∑_{1≤i≠j≤n} & ∥A_{i}A_{j}^{∗}∥ \\ [(∑_{i=1}^{n} |α_{i}|^{r})^{2} − (∑_{i=1}^{n} |α_{i}|^{2r})^{\frac{1}{2}} ] & ( ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥^{s}) & \frac{1}{s} \\ if r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ [(∑_{i=1}^{n} |α_{i}|)^{2} − (∑_{i=1}^{n} |α_{i}|^{2})] max_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥ , \end{matrix}

where (2.1) should be seen as all the 9 possible configurations.

Proof. We have

(2.2)  0 ≤ (∑_{i=1}^{n} α_{i}A_{i}) (∑_{i=1}^{n} α_{i}A_{i})^{∗} = (∑_{i=1}^{n} α_{i}A_{i}) (∑_{j=1}^{n} \overline{α_{j}}A_{j}^{∗})

= ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}}A_{i}A_{j}^{∗} = ∑_{i=1}^{n} |α_{i}|^{2} A_{i}A_{i}^{∗} + ∑_{1≤i≠j≤n} α_{i}\overline{α_{j}}A_{i}A_{j}^{∗}.

Taking the norm in (2.2) and observing that ∥UU^{∗}∥ = ∥U∥^{2} for any U ∈ B (H) , one has the inequality

(2.3)  ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} = ∥∑_{i=1}^{n} |α_{i}|^{2} A_{i}A_{i}^{∗} + ∑_{1≤i≠j≤n} α_{i}\overline{α_{j}}A_{i}A_{j}^{∗}∥

≤ ∑_{i=1}^{n} |α_{i}|^{2} ∥A_{i}A_{i}^{∗}∥ + ∑_{1≤i≠j≤n} |α_{i}| |α_{j}| ∥A_{i}A_{j}^{∗}∥

= ∑_{i=1}^{n} |α_{i}|^{2} ∥A_{i}∥^{2} + ∑_{1≤i≠j≤n} |α_{i}| |α_{j}| ∥A_{i}A_{j}^{∗}∥ .

Using Hölder’s inequality, we may write that:

(2.4) ∑_{i=1}^{n} |α_{i}|^{2} ∥A_{i}∥^{2} ≤ \{ \begin{matrix}max_{i=\overline{1,n}} & |α_{i}| & 2 & ∑_{1≤i≠j≤n} & ∥A_{i}∥ & 2 \\ (∑_{i=1}^{n} |α_{i}|^{2p})^{\frac{1}{p}} (∑_{i=1}^{n} ∥A_{i}∥^{2q})^{\frac{1}{q}} if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{i=\overline{1,n}} ∥A_{i}∥^{2} . \end{matrix}

Also, Hölder’s inequality for double sums produces

∑_{1≤i≠j≤n} |α_{i}| |α_{j}| ∥A_{i}A_{j}^{∗}∥ ≤ \{ \begin{matrix}max_{1≤i≠j≤n} & \{|α_{i}| |α_{j}|\} & ∑_{1≤i≠j≤n} & ∥A_{i}A_{j}^{∗}∥ \\ ( ∑_{1≤i≠j≤n} |α_{i}|^{r} |α_{j}|^{r}) & \frac{1}{r} & ( ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥^{s}) & \frac{1}{s} \\ if r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ ∑_{1≤i≠j≤n} |α_{i}| |α_{j}| max_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥ , \end{matrix} (2.5)

= \{ \begin{matrix}max_{1≤i≠j≤n} & \{|α_{i}| |α_{j}|\} & ∑_{1≤i≠j≤n} & ∥A_{i}A_{j}^{∗}∥ \\ [(∑_{i=1}^{n} |α_{i}|^{r})^{2} − (∑_{i=1}^{n} |α_{i}|^{2r})^{\frac{1}{2}} ] & (∑_{i=1}^{n} ∥A_{i}A_{j}^{∗}∥^{s}) & \frac{1}{s} \\ if r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ [(∑_{i=1}^{n} |α_{i}|)^{2} − (∑_{i=1}^{n} |α_{i}|^{2})] max_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥ , \end{matrix}

Using (2.3) and (2.4), (2.5) one deduces the desired inequality (2.1).

The following corollaries are natural consequences. Corollary 2. With the assumptions of Theorem 2, one has the inequality

(2.6)  ∥∑_{i=1}^{n} α_{i}A_{i}∥ ≤ max_{i=\overline{1,n}} |α_{i}| ( ∑_{i,j=1}^{n} ∥A_{i}A_{j}^{∗}∥)^{\frac{1}{2}} .

Proof. Follows by the first line in (2.1) on taking into account that

max_{1≤i≠j≤n} \{|α_{i}| |α_{j}|\} ≤ max_{i=\overline{1,n}} |α_{i}|^{2} ,

and

∑_{i,j=1}^{n} ∥A_{i}A_{j}^{∗}∥ = ∑_{i=1}^{n} ∥A_{i}∥^{2} + ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥ .

Corollary 3. With the assumptions in Theorem 2, one has the inequality:

(2.7) ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2}

≤ (∑_{i=1}^{n} |α_{i}|^{2p})^{\frac{1}{p}} [(∑_{i=1}^{n} ∥A_{i}∥^{2q})^{\frac{1}{q}} + (n − 1) ( ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥^{q})^{\frac{1}{q}} ] ,

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

Proof. Using the Cauchy-Bunyakovsky-Schwarz inequality for positive numbers

(∑_{i=1}^{n} a_{i})^{2} ≤ n ∑_{i=1}^{n} a_{i}^{2}

we may write that

(∑_{i=1}^{n} |α_{i}|^{p})^{2} − ∑_{i=1}^{n} |α_{i}|^{2p} ≤ n ∑_{i=1}^{n} |α_{i}|^{2p} − ∑_{i=1}^{n} |α_{i}|^{2p}

= (n − 1) ∑_{i=1}^{n} |α_{i}|^{2p} .

Now, using the second line in (2.1) for r = p, s = q, we deduce the desired result (2.7).

Corollary 4. With the assumptions in Theorem 2, one has the inequality

(2.8)  ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} [max_{i=\overline{1,n}} ∥A_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥] .

Proof. Follows by the third line of (2.1) on taking into account that

[(∑_{i=1}^{n} |α_{i}|)^{2} − ∑_{i=1}^{n} |α_{i}|^{2}]^{\frac{1}{2}} ≤ (n − 1) ∑_{i=1}^{n} |α_{i}|^{2} .

Another interesting particular case is embodied in the following corollary as well.

Corollary 5. With the assumptions in Theorem 2, one has the inequality

(2.9)  ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} [max_{i=\overline{1,n}} ∥A_{i}∥^{2} + ( ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥^{2})^{\frac{1}{2}} ] .

Proof. It is obvious that

[(∑_{i=1}^{n} |α_{i}|^{2})^{2} − ∑_{i=1}^{n} |α_{i}|^{4}]^{\frac{1}{2}} ≤ ∑_{i=1}^{n} |α_{i}|^{2} .

Thus, combining the third line in the first bracket in (2.1) with the second line for r = s = 2 in the second bracket, the inequality (2.9) is obtained.

Remark 1. If one is interested in obtaining bounds in terms of ∑_{i=1}^{n} |α_{i}|^{2} , there are other possibilities as shown below. Obviously, since

max_{1≤i≠j≤n} \{|α_{i}| |α_{j}|\} ≤ max_{i=\overline{1,n}} |α_{i}|^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} .

then, by (2.1), in choosing the third line in the first bracket with the first line in the second bracket, one would obtain

(2.10)  ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} [max_{i=\overline{1,n}} ∥A_{i}∥^{2} + ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥] .

Also, it is evident that

[(∑_{i=1}^{n} |α_{i}|^{r})^{2} − (∑_{i=1}^{n} |α_{i}|^{2r})^{\frac{1}{2}} ]^{\frac{1}{r}} ≤ (∑_{i=1}^{n} |α_{i}|^{r})^{\frac{2}{r}} .

By the monotonicity of the power mean (\frac{1}{n} ∑_{i=1}^{n} a_{i}^{m})^{\frac{1}{m}} as a function of m ∈ ℝ, we have

(\frac{∑_{i=1}^{n} |α_{i}|^{r}}{n} )^{\frac{1}{r}} ≤ (\frac{∑_{i=1}^{n} |α_{i}|^{2}}{n} )^{\frac{1}{2}} , 1 < r ≤ 2,

giving

(∑_{i=1}^{n} |α_{i}|^{r})^{\frac{2}{r}} ≤ n^{\frac{2}{r} −1} ∑_{i=1}^{n} |α_{i}|^{2} .

Thus, using the third line in the first bracket of (2.1) combined with the second line in the second bracket for 1 ≤ r ≤ 2 , \frac{1}{s} + \frac{1}{r} = 1, we deduce

(2.11) ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} [max_{i=\overline{1,n}} ∥A_{i}∥^{2} + n^{\frac{2}{r} −1} ( ∑_{1≤i≠j≤n} ∥A_{i}A_{j}^{∗}∥^{s})^{\frac{1}{s}} ] .

Note that for r = s = 2, we recapture (2.9).

The following particular result also holds.

Proposition 1. Let α_{1}, …, α_{n} ∈ 𝕂 and A_{1}, …, A_{n} ∈ B (H) with the property that A_{i}A_{j}^{∗} = 0 for any i ≠ j, i, j ∈ {1, …, n} . Then one has the inequality;

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

If by M (α, A) we denote any of the bounds provided by (2.1), (2.6), (2.7), (2.8), (2.9), (2.10) or (2.11), then we may state the following proposition as well.

CAUCHY-BUNYAKOVSKY-SCHWARZ TYPE INEQUALITIES 7

Proposition 2. Under the assumptions of Theorem 2, we have:

(i) For any x ∈ H

(2.13) ∥∑_{i=1}^{n} α_{i}A_{i}x∥^{2} ≤ ∥x∥^{2} M (α, A) .
(ii) For any x, y ∈ H,
(2.14) |∑_{i=1}^{n} α_{i} ⟨A_{i}x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} M (α, A) .

Proof. (i) Obviously,

∥∑_{i=1}^{n} α_{i}A_{i}x∥^{2} = ∥(∑_{i=1}^{n} α_{i}A_{i}) (x)∥^{2} ≤ ∥∑_{i=1}^{n} α_{i}A_{i}∥^{2} ∥x∥^{2}
≤ M (α, A) ∥x∥^{2} .

(ii) We have

|∑_{i=1}^{n} α_{i} ⟨A_{i}x, y⟩|^{2} = |⟨∑_{i=1}^{n} α_{i}A_{i}x, y⟩|^{2} = ∥∑_{i=1}^{n} α_{i}A_{i}x∥^{2} ∥y∥^{2} ,

which, by (i), gives the desired result (2.14).

3. Inequalities for Vectors in Hilbert Spaces

We consider the non zero vectors y_{1}, …, y_{n} ∈ H. Define the operators

A_{i} : H → H, A_{i}x = \frac{(x, y_{i})}{∥y_{i}∥} · y_{i}, i ∈ \{1, …, n\} .

Since

(3.1)  ∥A_{i}∥ = sup_{∥x∥=1} ∥A_{i}x∥ = sup_{∥x∥=1} |(x, y_{i})| = ∥y_{i}∥ , i ∈ {1, …, n}

then A_{i} are bounded linear operators in H. Also, since

(3.2)  (A_{i}x, x) = (\frac{(x, y_{i}) y_{i}}{∥y_{i}∥} , x) = \frac{|(x, y_{i})|^{2}}{∥y_{i}∥} ≥ 0, x ∈ H, i ∈ \{1, …, n\}

and

(A_{i}x, z) = (\frac{(x, y_{i}) y_{i}}{∥y_{i}∥} , z) = \frac{(x, y_{i}) (y_{i}, z)}{∥y_{i}∥},
(x, A_{i}z) = (x, \frac{(z, y_{i}) y_{i}}{∥y_{i}∥} ) = \frac{(x, y_{i}) \overline{(z, y_{i})}}{∥y_{i}∥} = \frac{(x, y_{i}) (y_{i}, z)}{∥y_{i}∥},

giving

(3.3)  (A_{i}x, z) = (x, A_{i}z) , x, z ∈ H, i ∈ {1, …, n} ,

we may conclude that A_{i} (i = 1, …, n) are positive self-adjoint operators on H.

Since, for any x ∈ H, one has

∥(A_{i}A_{j}) (x)∥ = ∥(A_{i}) (A_{j}x)∥ = ∥A_{i} (\frac{(x, y_{j}) y_{j}}{∥y_{j}∥} )∥

= \frac{|(x, y_{j})|}{∥y_{j}∥} ∥A_{i}y_{j}∥ = \frac{|(x, y_{j})|}{∥y_{j}∥} · \frac{|(y_{j}, y_{i})| ∥y_{j}∥}{∥y_{i}∥}

= \frac{|(x, y_{j})| |(y_{j}, y_{i})|}{∥y_{i}∥}, i, j ∈ \{1, …, n\} ,

we deduce that

(3.4)  ∥A_{i}A_{j}∥ = sup_{∥α∥=1} \frac{|(x, y_{j})| |(y_{j}, y_{i})|}{∥y_{i}∥} = |(y_{i}, y_{j})|; i, j ∈ \{1, …, n\} .

If (y_{i})_{i=\overline{1,n}} is an orthogonal family on H, then ∥A_{i}∥ = 1 and A_{i}A_{j} = 0 for i, j ∈ {1, …, n} , i ≠ j.

The following inequality for vectors holds.

Theorem 3. Let x, y_{1}, …, y_{n} ∈ H and α_{n}, …, α_{n} ∈ 𝕂. Then one has the inequalities:

(3.5) ∥∑_{i=1}^{n} α_{i} \frac{(x, y_{i})}{∥y_{i}∥}y_{i}∥^{2}

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

+ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≠j≤n} & \{|α_{i}| |α_{j}|\} & ∑_{1≤i≠j≤n} & |(y_{i}, y_{j})| \\ [(∑_{i=1}^{n} |α_{i}|^{r})^{2} − ∑_{i=1}^{n} |α_{i}|^{2r}] & r & ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{s}) & \frac{1}{s} \\ if r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ [(∑_{i=1}^{n} |α_{i}|)^{2} − ∑_{i=1}^{n} |α_{i}|^{2}] max_{1≤i≠j≤n} |(y_{i}, y_{j})| . \end{matrix}

Proof. Follows by Theorem 2 and Proposition 2, (i) on choosing A_{i} = \frac{(·,y_{i})}{∥y_{i}∥}y_{i} and taking into account that ∥A_{i}∥ = ∥y_{i}∥ ,

∥A_{i}A_{j}^{∗}∥ = |(y_{i}, y_{j})| , i, j ∈ {1, …, n} .

We omit the details.

Using Corollaries 2–5 and Remark 1, we may state the following particular inequalities:

(3.6)  ∥∑_{i=1}^{n} α_{i} \frac{(x, y_{i})}{∥y_{i}∥}y_{i}∥ ≤ ∥x∥ max_{i=\overline{1,n}} |α_{i}| ( ∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} ;

(3.7) ∥∑_{i=1}^{n} α_{i} \frac{(x, y_{i})}{∥y_{i}∥}y_{i}∥^{2}

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

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

(3.8) ∥∑_{i=1}^{n} α_{i} \frac{(x, y_{i})}{∥y_{i}∥}y_{i}∥^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |α_{i}|^{2} [max_{i=\overline{1,n}} ∥y_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} |(y_{i}, y_{j})|];

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

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

(3.11) ∥∑_{i=1}^{n} α_{i} \frac{(x, y_{i})}{∥y_{i}∥}y_{i}∥^{2}

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

where 1 < r ≤ 2, \frac{1}{s} + \frac{1}{r} = 1.

Remark 2. The choice α_{i} = ∥y_{i}∥ (i = 1, …, n) will produce some interesting bounds for

∥∑_{i=1}^{n} (x, y_{i}) y_{i}∥^{2} .

We omit the details.

References

[1] S.S. Dragomir, On some operatorial inequalities in Hilbert spaces, Bull. Inst. Pol. Cluj-Napoca (Romania), 36(1987), 23-28.

[2] S.S. Dragomir and B. Mond, On Cauchy-Buniakowski-Schwarz’s inequality for sequences of linear operators in Hilbert space, Huston J. Math., 22(1996), No. 2, 329-339.

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

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