INEQUALITIES FOR THE NORM AND NUMERICAL RADIUS OF COMPOSITE OPERATORS IN HILBERT SPACES

S.S. DRAGOMIR

Abstract. Some new inequalities for the norm and the numerical radius of composite operators generated by a pair of operators are given.

1. Introduction

Let (H; ⟨·, ·⟩) be a complex Hilbert space. The numerical range of an operator T is the subset of the complex numbers ℂ given by [4, p. 1]:

(1.1)  W (T) = {⟨T x, x⟩ , x ∈ H, ∥x∥ = 1} .

It is well known that (see [4]):

(i) The numerical range of an operator is convex; (ii) The spectrum of an operator is contained in the closure of its numerical

range; (iii) T is self-adjoint if and only if W (T) is real. The numerical radius w (T) of an operator T on H is defined by [4, p. 8]

(1.2)  w (T) := sup {|λ| , λ ∈ W (T)} = sup {|⟨T x, x⟩| , ∥x∥ = 1} .

It is well known that w (·) is a norm on the Banach algebra B (H) of all bounded linear operators acting on H and the following inequality holds true:

(1.3)  w (T) ≤ ∥T ∥ ≤ 2w (T) .

We recall some classical results involving the numerical radius of two linear operators A, B.

The following general result for the product of two operators holds [4, p. 37]:

Theorem 1. If A, B are two bounded linear operators on the Hilbert space (H, ⟨·, ·⟩) , then

(1.4)  w (AB) ≤ 4w (A) w (B) .

In the case that AB = BA, then

(1.5)  w (AB) ≤ 2w (A) w (B) .

The following results are also well known [4, p. 38].

Date: 5 September, 2005.

2000 Mathematics Subject Classification. 47A12.

Key words and phrases. Numerical range, Numerical radius, Bounded linear operators, Hilbert spaces.

Theorem 2. If A is a unitary operator that commutes with another operator B, then

(1.6)  w (AB) ≤ w (B) .

If A is an isometry and AB = BA, then (1.6) also holds true.

We say that A and B double commute if AB = BA and AB^{∗} = B^{∗}A.

The following result holds [4, p. 38].

Theorem 3 (Double commute). If the operators A and B double commute, then

(1.7)  w (AB) ≤ w (B) ∥A∥ .

As a consequence of the above, we have [4, p. 39]:

Corollary 1. Let A be a normal operator commuting with B. Then

(1.8)  w (AB) ≤ w (A) w (B) .

For other results and historical comments on the above see [4, p. 39–41]. For more results on the numerical radius, see [5].

The main aim of this paper is to establish some new inequalities for composite operators generated by a pair of operators (A, B) under various assumptions. Namely, in one side, several inequalities involving the norm

∥\frac{A^{∗}A + B^{∗}B}{2} ∥

and the numerical radius w (B^{∗}A) are established. On the other side, upper bounds for the nonnegative quantities

∥A∥ ∥B∥ − w (B^{∗}A) and ∥A∥^{2} ∥B∥^{2} − w^{2} (B^{∗}A)

under special conditions for the operators involved are also given.

2. The Results

The following result may be stated:

Theorem 4. Let A, B : H → H be two bounded linear operators on the Hilbert space (H, ⟨·, ·⟩) . If r > 0 and

(2.1)  ∥A − B∥ ≤ r,

then

(2.2)  ∥\frac{A^{∗}A + B^{∗}B}{2} ∥ ≤ w (B^{∗}A) + \frac{1}{2}r^{2}.

Proof. For any x ∈ H, ∥x∥ = 1, we have from (2.1) that

(2.3)  ∥Ax∥^{2} + ∥Bx∥^{2} ≤ 2 Re ⟨Ax, Bx⟩ + r^{2}.

However

∥Ax∥^{2} + ∥Bx∥^{2} = ⟨(A^{∗}A) x, x⟩ + ⟨(B^{∗}B) x, x⟩
= ⟨(A^{∗}A + B^{∗}B) x, x⟩

and by (2.3) we obtain

(2.4)  ⟨(A^{∗}A + B^{∗}B) x, x⟩ ≤ 2 |⟨(B^{∗}A) x, x⟩| + r^{2}

for any x ∈ H, ∥x∥ = 1.

Taking the supremum over x ∈ H, ∥x∥ = 1 in (2.4) we get

(2.5)  w (A^{∗}A + B^{∗}B) ≤ 2w (B^{∗}A) + r^{2}

and since the operator A^{∗}A + B^{∗}B is self-adjoint, hence

w (A^{∗}A + B^{∗}B) = ∥A^{∗}A + B^{∗}B∥

and by (2.5) we deduce the desired inequality (2.2).

Remark 1. We observe that, from the proof of the above theorem, we have the inequalities

(2.6)  0 ≤ ∥\frac{A^{∗}A + B^{∗}B}{2} ∥ − w (B^{∗}A) ≤ \frac{1}{2} ∥A − B∥^{2} ,

provided that A, B are bounded linear operators in H.

The second inequality in (2.6) is obvious while the first inequality follows by the fact that

⟨(A^{∗}A + B^{∗}B) x, x⟩ = ∥Ax∥^{2} + ∥Bx∥^{2}
≥ 2 ∥Ax∥ ∥Bx∥ ≥ 2 |⟨(B^{∗}A) x, x⟩|

for any x ∈ H.

The inequality (2.2) is obviously a reach source of particular inequalities of interest.

Indeed, if we assume, for λ ∈ ℂ and a bounded linear operator T, that we have

(2.7)  ∥T − λT^{∗}∥ ≤ r,

for a given positive number r, then by (2.6) we deduce the inequality

(2.8)  0 ≤ ∥\frac{T^{∗}T + |λ|^{2} T T^{∗}}{2} ∥ − |λ| w (T^{2}) ≤ \frac{1}{2}r^{2}.

Now, if we assume that for λ ∈ ℂ and a bounded linear operator V we have that

(2.9)  ∥V − λI∥ ≤ r,

where I is the identity operator on H, then by (2.2) we deduce the inequality

0 ≤ ∥\frac{V^{∗}V + |λ|^{2} I}{2} ∥ − |λ| w (V) ≤ \frac{1}{2}r^{2}.

As a dual approach, the following result may be noted as well:

Theorem 5. Let A, B : H → H be two bounded linear operators on the Hilbert space H. Then

(2.10)  ∥\frac{A + B}{2} ∥^{2} ≤ \frac{1}{2} [∥\frac{A^{∗}A + B^{∗}B}{2} ∥ + w (B^{∗}A)] .

Proof. We obviously have

∥Ax + Bx∥^{2} = ∥Ax∥^{2} + 2 Re ⟨Ax, Bx⟩ + ∥Bx∥^{2}
≤ ⟨(A^{∗}A + B^{∗}B) x, x⟩ + 2 |⟨(B^{∗}A) x, x⟩|

for any x ∈ H.

Taking the supremum over x ∈ H, ∥x∥ = 1, we get

∥A + B∥^{2} ≤ w (A^{∗}A + B^{∗}B) + 2w (B^{∗}A)
= ∥A^{∗}A + B^{∗}B∥ + 2w (B^{∗}A) ,

from where we get the desired inequality (2.10).

Remark 2. The inequality (2.10) can generate some interesting particular results such as the following inequality

(2.11)  ∥\frac{T + T^{∗}}{2} ∥^{2} ≤ \frac{1}{2} [∥\frac{T^{∗}T + T T^{∗}}{2} ∥ + w (T^{2})] ,

holding for each bounded linear operator T : H → H.

The following result may be stated as well.

Theorem 6. Let A, B : H → H be two bounded linear operators on the Hilbert space H and p ≥ 2. Then

(2.12)  ∥\frac{A^{∗}A + B^{∗}B}{2} ∥^{\frac{p}{2}} ≤ \frac{1}{4} [∥A − B∥^{p} + ∥A + B∥^{p}] .

Proof. We use the following inequality for vectors in inner product spaces obtained by Dragomir and Sándor in [2]:

(2.13)  2 (∥a∥^{p} + ∥b∥^{p}) ≤ ∥a + b∥^{p} + ∥a − b∥^{p}

for any a, b ∈ H and p ≥ 2.

Utilising (2.13) we may write

(2.14)  2 (∥Ax∥^{p} + ∥Bx∥^{p}) ≤ ∥Ax + Bx∥^{p} + ∥Ax − Bx∥^{p}

for any x ∈ H.

Now, observe that

∥Ax∥^{p} + ∥Bx∥^{p} = (∥Ax∥^{2})^{\frac{p}{2}} + (∥Bx∥^{2})^{\frac{p}{2}}

and by the elementary inequality:

\frac{α^{q} + β^{q}}{2} ≥ (\frac{α + β}{2} )^{q} , α, β ≥ 0 and q ≥ 1

we have

(2.15)  (∥Ax∥^{2})^{\frac{p}{2}} + (∥Bx∥^{2})^{\frac{p}{2}} ≥ 2^{1− \frac{p}{2}} (∥Ax∥^{2} + ∥Bx∥^{2})^{\frac{p}{2}}
= 2^{1− \frac{p}{2}} [⟨(A^{∗}A + B^{∗}B) x, x⟩]^{\frac{p}{2}} .

Combining (2.14) with (2.15) we get

(2.16)  \frac{1}{4} [∥Ax − Bx∥^{p} + ∥Ax + Bx∥^{p}] ≥ |⟨(\frac{A^{∗}A + B^{∗}B}{2} ) x, x⟩|^{\frac{p}{2}}

for any x ∈ H, ∥x∥ = 1. Taking the supremum over x ∈ H, ∥x∥ = 1, and taking into account that

w (\frac{A^{∗}A + B^{∗}B}{2} ) = ∥\frac{A^{∗}A + B^{∗}B}{2} ∥ ,

we deduce the desired result (2.12).

Remark 3. If p = 2, then we have the inequality:

(2.17)  ∥\frac{A^{∗}A + B^{∗}B}{2} ∥ ≤ ∥\frac{A − B}{2} ∥^{2} + ∥\frac{A + B}{2} ∥^{2} ,

for any A, B bounded linear operators. This result can also be obtained directly on utilising the parallelogram identity.

We also should observe that for A = T and B = T^{∗}, T a normal operator, the inequality (2.12) becomes

∥T ∥^{p} ≤ \frac{1}{4} [∥T − T^{∗}∥^{p} + ∥T + T^{∗}∥^{p}] ,

where p ≥ 2.

The following result may be stated as well.

Theorem 7. Let A, B : H → H be two bounded linear operators on the Hilbert space H and r ≥ 1. If A^{∗}A ≥ B^{∗}B in the operator order or, equivalently, ∥Ax∥ ≥ ∥Bx∥ for any x ∈ H, then:

(2.18) ∥\frac{A^{∗}A + B^{∗}B}{2} ∥^{r} ≤ ∥A∥^{r−1} ∥B∥^{r−1} w (B^{∗}A) + \frac{1}{2}r^{2} ∥A∥^{2r−2} ∥A − B∥^{2} .

Proof. We use the following inequality for vectors in inner product spaces due to Goldstein, Ryff and Clarke [3]:

(2.19)  ∥a∥^{2r} + ∥b∥^{2r} ≤ 2 ∥a∥^{r−1} ∥b∥^{r−1} Re ⟨a, b⟩ + r^{2} ∥a∥^{2r−2} ∥a − b∥^{2} ,

where r ≥ 1, a, b ∈ H and ∥a∥ ≥ ∥b∥ .

Utilising (2.19) we can state that:

(2.20) ∥Ax∥^{2r} + ∥Bx∥^{2r}

≤ 2 ∥Ax∥^{r−1} ∥Bx∥^{r−1} |⟨Ax, Bx⟩| + r^{2} ∥Ax∥^{2r−2} ∥Ax − Bx∥^{2} ,

for any x ∈ H.

As in the proof of Theorem 6, we also have

(2.21)  2^{1−r} [⟨(A^{∗}A + B^{∗}B) x, x⟩]^{r} ≤ ∥Ax∥^{2r} + ∥Bx∥^{2r} ,

for any x ∈ H.

Therefore, by (2.20) and (2.21) we deduce

(2.22) [⟨(\frac{A^{∗}A + B^{∗}B}{2} ) x, x⟩]^{r}

≤ ∥Ax∥^{r−1} ∥Bx∥^{r−1} |⟨Ax, Bx⟩| + \frac{1}{2}r^{2} ∥A∥^{2r−2} ∥Ax − Bx∥^{2}

for any x ∈ H.

Taking the supremum in (2.22) we obtain the desired result (2.18).

Remark 4. Following [4, p. 156], we recall that the bounded linear operator V is hyponormal, if

∥V^{∗}x∥ ≤ ∥V x∥ for all x ∈ H.

Now, if we choose in (2.18) A = V and B = V^{∗}, then, on taking into account that for hyponormal operators w (V^{2}) = ∥V ∥^{2} , we get the inequality

(2.23)  ∥\frac{V^{∗}V + V V^{∗}}{2} ∥^{r} ≤ ∥V ∥^{2r−2} [∥V ∥^{2} + \frac{1}{2}r^{2} ∥V − V^{∗}∥^{2}] ,

holding for any hyponormal operator V and any r ≥ 1.

3. Further Inequalities for an Invertible Operator

In this section we assume that B : H → H is an invertible bounded linear operator and let B^{−1} : H → H be its inverse. Then, obviously,

(3.1)  ∥Bx∥ ≥ \frac{1}{∥B^{−1}∥} ∥x∥ for any x ∈ H,

where ∥B^{−1}∥ denotes the norm of the inverse B^{−1}.

The following result holds true:

Theorem 8. Let A, B : H → H be two bounded linear operators on H and B is invertible such that, for a given r > 0,

(3.2)  ∥A − B∥ ≤ r.

Then:

(3.3)  ∥A∥ ≤ ∥B^{−1}∥ [w (B^{∗}A) + \frac{1}{2}r^{2}] .

Proof. The condition (3.2) is obviously equivalent to:

(3.4)  ∥Ax∥^{2} + ∥Bx∥^{2} ≤ 2 Re ⟨(B^{∗}A) x, x⟩ + r^{2}

for any x ∈ H, ∥x∥ = 1.

Since, by (3.1),

∥Bx∥^{2} ≥ \frac{1}{∥B^{−1}∥^{2}} ∥x∥^{2} , x ∈ H

and Re ⟨(B^{∗}A) x, x⟩ ≤ |⟨(B^{∗}A) x, x⟩| , hence by (3.4) we get

(3.5)  ∥Ax∥^{2} + \begin{matrix}∥x∥ & 2 \\ ∥B^{−1}∥ & 2 \end{matrix} ≤ 2 |⟨(B^{∗}A) x, x⟩| + r^{2}

for any x ∈ H, ∥x∥ = 1.

Taking the supremum over x ∈ H, ∥x∥ = 1 in (3.5), we have

(3.6)  ∥A∥^{2} + \frac{1}{∥B^{−1}∥^{2}} ≤ 2w (B^{∗}A) + r^{2}.

By the elementary inequality

(3.7)  \frac{2 ∥A∥}{∥B^{−1}∥} ≤ ∥A∥^{2} + \frac{1}{∥B^{−1}∥^{2}}

and by (3.6) we then deduce the desired result (3.3).

Remark 5. If we choose above B = λI, λ ≠ 0, then we get the inequality

(3.8)  (0 ≤) ∥A∥ − w (A) ≤ \frac{1}{2 |λ|}r^{2},

provided ∥A − λI∥ ≤ r. This result has been obtained in the earlier paper [1]. Also, if we assume that B = λA^{∗}, A is invertible, then we obtain

(3.9)  ∥A∥ ≤ ∥A^{−1}∥ [w (A^{2}) + \frac{1}{2 |λ|}r^{2}] ,

provided ∥A − λA^{∗}∥ ≤ r, λ ≠ 0.

The following result may be stated as well:

Theorem 9. Let A, B : H → H be two bounded linear operators on H. If B is invertible and for r > 0,

(3.10)  ∥A − B∥ ≤ r,

then

(3.11)  (0 ≤) ∥A∥ ∥B∥ − w (B^{∗}A) ≤ \frac{1}{2}r^{2} + \frac{∥B∥^{2} ∥B^{−1}∥^{2} − 1}{∥B^{−1}∥^{2}}.

Proof. The condition (3.10) is obviously equivalent to

∥Ax∥^{2} + ∥Bx∥^{2} ≤ 2 Re ⟨Ax, Bx⟩ + r^{2}

for any x ∈ H, which is clearly equivalent to

(3.12)  ∥Ax∥^{2} + ∥B∥^{2} ≤ 2 Re ⟨B^{∗}Ax, x⟩ + r^{2} + ∥B∥^{2} − ∥Bx∥^{2} .

Since

Re ⟨B^{∗}Ax, x⟩ ≤ |⟨B^{∗}Ax, x⟩| , ∥Bx∥^{2} ≥ \frac{1}{∥B^{−1}∥^{2}} ∥x∥^{2}

and

∥Ax∥^{2} + ∥B∥^{2} ≥ 2 ∥B∥ ∥Ax∥

for any x ∈ H, hence by (3.12) we get

(3.13)  2 ∥B∥ ∥Ax∥ ≤ 2 |⟨B^{∗}Ax, x⟩| + r^{2} + \frac{∥B∥^{2} ∥B^{−1}∥^{2} − 1}{∥B^{−1}∥^{2}}

for any x ∈ H, ∥x∥ = 1.

Taking the supremum over x ∈ H, ∥x∥ = 1 we deduce the desired result (3.11).

Remark 6. If we choose in Theorem 9, B = λA^{∗}, λ ≠ 0, A is invertible, then we get the inequality:

(3.14)  (0 ≤) ∥A∥^{2} − w (A^{2}) ≤ \frac{1}{2 |λ|}r^{2} + |λ| · \frac{∥A∥^{2} ∥A^{−1}∥^{2} − 1}{∥A^{−1}∥^{2}}

provided ∥A − λA^{∗}∥ ≤ r.

The following result may be stated as well.

Theorem 10. Let A, B : H → H be two bounded linear operators on H. If B is invertible and for r > 0 we have

(3.15)  ∥A − B∥ ≤ r < ∥B∥ ,

then

(3.16)  ∥A∥ ≤ \frac{1}{\sqrt{∥B∥^{2} − r^{2}}} (w (B^{∗}A) + \frac{∥B∥^{2} ∥B^{−1}∥^{2} − 1}{2 ∥B^{−1}∥^{2}}) .

Proof. The first part of condition (3.15) is obviously equivalent to

∥Ax∥^{2} + ∥Bx∥^{2} ≤ 2 Re ⟨Ax, Bx⟩ + r^{2}

for any x ∈ H, which is clearly equivalent to

(3.17)  ∥Ax∥^{2} + ∥B∥^{2} − r^{2} ≤ 2 Re ⟨B^{∗}Ax, x⟩ + ∥B∥^{2} − ∥Bx∥^{2} .

Since

Re ⟨B^{∗}Ax, x⟩ ≤ |⟨B^{∗}Ax, x⟩| ,

∥Bx∥^{2} ≥ \frac{1}{∥B^{−1}∥^{2}} ∥x∥^{2}

and, by the second part of (3.15),

∥Ax∥^{2} + ∥B∥^{2} − r^{2} ≥ 2\sqrt{∥B∥^{2} − r^{2}} ∥Ax∥ ,

for any x ∈ H, hence by (3.17) we get

(3.18)  2 ∥Ax∥ \sqrt{∥B∥^{2} − r^{2}} ≤ 2 |⟨B^{∗}Ax, x⟩| + \frac{∥B∥^{2} ∥B^{−1}∥^{2} − 1}{∥B^{−1}∥^{2}}

for any x ∈ H, ∥x∥ = 1.

Taking the supremum over x ∈ H, ∥x∥ = 1 in (3.18), we deduce the desired inequality (3.16).

Remark 7. The above Theorem 10 has some particular cases of interest. For instance, if we choose B = λI, with |λ| > r, then (3.15) is obviously fulfilled and by (3.16) we get

(3.19)  ∥A∥ ≤ \frac{w (A)}{\sqrt{1 − (\frac{r}{|λ|})^{2}}} ,

provided ∥A − λI∥ ≤ r. This result has been obtained in the earlier paper [1].

On the other hand, if in the above we choose B = λA^{∗} with ∥A∥ ≥ \frac{r}{|λ|} (λ ≠ 0) , then by (3.16) we get

(3.20)  ∥A∥ ≤ \frac{1}{\sqrt{∥A∥^{2} − (\frac{r}{|λ|})^{2}}} [w (A^{2}) + |λ| · \frac{∥A∥^{2} ∥A^{−1}∥^{2} − 1}{2 ∥A^{−1}∥^{2}}] ,

provided ∥A − λA^{∗}∥ ≤ r.

The following result may be stated as well.

Theorem 11. Let A, B and r be as in Theorem 8. Moreover, if

(3.21)  ∥B^{−1}∥ < \frac{1}{r},

then

(3.22)  ∥A∥ ≤ \frac{∥B^{−1}∥}{\sqrt{1 − r^{2} ∥B^{−1}∥^{2}}} w (B^{∗}A) .

Proof. Observe that, by (3.6) we have

(3.23)  ∥A∥^{2} + \frac{1 − r^{2} ∥B^{−1}∥^{2}}{∥B^{−1}∥^{2}} ≤ 2w (B^{∗}A) .

Utilising the elementary inequality

(3.24)  2\frac{∥A∥}{∥B^{−1}∥}\sqrt{1 − r^{2} ∥B^{−1}∥^{2}} ≤ ∥A∥^{2} + \frac{1 − r^{2} ∥B^{−1}∥^{2}}{∥B^{−1}∥^{2}} ,

which can be stated since (3.21) is assumed to be true, hence by (3.23) and (3.24) we deduce the desired result (3.22).

Remark 8. If we assume that B = λA^{∗} with λ ≠ 0 and A an invertible operator, then, by applying Theorem 11, we get the inequality:

(3.25)  ∥A∥ ≤ \frac{∥A^{−1}∥ w (A^{2})}{\sqrt{|λ|^{2} − r^{2} ∥A^{−1}∥^{2}}} ,

provided ∥A − λA^{∗}∥ ≤ r and ∥A^{−1}∥ ≤ \frac{|λ|}{r}.

The following result may be stated as well.

Theorem 12. Let A, B : H → H be two bounded linear operators. If r > 0 and B is invertible with the property that ∥A − B∥ ≤ r and

(3.26)  \frac{1}{\sqrt{r^{2} + 1}} ≤ ∥B^{−1}∥ < \frac{1}{r},

then

(3.27)  ∥A∥^{2} ≤ w^{2} (B^{∗}A) + 2w (B^{∗}A) · \frac{∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}}}{∥B^{−1}∥} .

Proof. Let x ∈ H, ∥x∥ = 1. Then by (3.5) we have

(3.28)  ∥Ax∥^{2} + \frac{1}{∥B^{−1}∥^{2}} ≤ 2 |⟨B^{∗}Ax, x⟩| + r^{2},

and since

\frac{1}{∥B^{−1}∥^{2}} − r^{2} > 0,

we can conclude that |⟨B^{∗}Ax, x⟩| > 0 for any x ∈ H, ∥x∥ = 1.

Dividing in (3.28) with |⟨B^{∗}Ax, x⟩| > 0, we obtain

(3.29)  \begin{matrix}∥Ax∥ & 2 \\ | & ⟨B^{∗}Ax, x⟩ & | \end{matrix} ≤ 2 + \frac{r^{2}}{|⟨B^{∗}Ax, x⟩|} − \frac{1}{∥B^{−1}∥^{2} |⟨B^{∗}Ax, x⟩|}.

Subtracting |⟨B^{∗}Ax, x⟩| from both sides of (3.29), we get

\begin{matrix}∥Ax∥ & 2 \\ | & ⟨B^{∗}Ax, x⟩ & | \end{matrix} − |⟨B^{∗}Ax, x⟩| (3.30)

≤ 2 − |⟨B^{∗}Ax, x⟩| − \frac{1 − r^{2} ∥B^{−1}∥^{2}}{|⟨B^{∗}Ax, x⟩| ∥B^{−1}∥^{2}}

= 2 − \frac{2\sqrt{1 − r^{2} ∥B^{−1}∥^{2}}}{∥B^{−1}∥} − (\sqrt{|⟨B^{∗}Ax, x⟩|} − \begin{matrix}√ & \overline{1 − r^{2} ∥B^{−1}∥^{2}} \\ ∥B^{−1}∥ & \sqrt{|⟨B^{∗}Ax, x⟩|} \end{matrix})^{2}

≤ 2 (\frac{∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}}}{∥B^{−1}∥} ) ,

which gives:

(3.31)  ∥Ax∥^{2} ≤ |⟨B^{∗}Ax, x⟩|^{2} + 2 |⟨B^{∗}Ax, x⟩| \frac{∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}}}{∥B^{−1}∥} .

10 S.S. DRAGOMIR

We also remark that, by (3.26) the quantity

∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}} ≥ 0,

hence, on taking the supremum in (3.31) over x ∈ H, ∥x∥ = 1, we deduce the desired inequality. Remark 9. It is interesting to remark that if we assume λ ∈ ℂ with 0 < r ≤ |λ| ≤ \sqrt{r^{2} + 1} and ∥A − λI∥ ≤ r, then by (3.2) we can state the following inequality:

(3.32)  ∥A∥^{2} ≤ |λ|^{2} w (A^{2}) + 2 |λ| (1 − \sqrt{|λ|^{2} − r^{2}}) w (A) .

Also, if ∥A − A^{∗}∥ ≤ r, A is invertible and \frac{1}{\sqrt{r^{2}+1}} ≤ ∥A^{−1}∥ ≤ \frac{1}{r}, then, by (3.27) we also have

(3.33)  ∥A∥^{2} ≤ w^{2} (A^{2}) + 2w (A^{2}) · \frac{∥A^{−1}∥ − \sqrt{1 − r^{2} ∥A^{−1}∥^{2}}}{∥A^{−1}∥} .

One can also prove the following result. Theorem 13. Let A, B : H → H be two bounded linear operators. If r > 0 and B is invertible with the property that ∥A − B∥ ≤ r and ∥B^{−1}∥ ≤ \frac{1}{r}, then

(3.34)  (0 ≤) ∥A∥^{2} ∥B∥^{2} − w^{2} (B^{∗}A)

≤ 2w (B^{∗}A) · \frac{∥B∥}{∥B^{−1}∥} (∥B∥ ∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}}) .

Proof. We subtract the quantity \begin{matrix}| & ⟨B^{∗}Ax,x⟩ & | \\ ∥B∥ & ^{2} \end{matrix} from both sides of (3.29) to obtain

(3.35) 0 ≤ \begin{matrix}∥Ax∥ & 2 \\ | & ⟨B^{∗}Ax, x⟩ & | \end{matrix} − \begin{matrix}| & ⟨B^{∗}Ax, x⟩ & | \\ ∥B∥ & 2 \end{matrix}

≤ 2 − \begin{matrix}| & ⟨B^{∗}Ax, x⟩ & | \\ ∥B∥ & 2 \end{matrix} − \frac{1 − r^{2} ∥B^{−1}∥^{2}}{|⟨B^{∗}Ax, x⟩| ∥B^{−1}∥^{2}}

= 2 − 2 · \begin{matrix}√ & \overline{1 − r^{2} ∥B^{−1}∥^{2}} \\ ∥B∥ & ∥B^{−1}∥ \end{matrix} − (\frac{\sqrt{|⟨B^{∗}Ax, x⟩|}}{∥B∥} − \begin{matrix}√ & \overline{1 − r^{2} ∥B^{−1}∥^{2}} \\ \sqrt{|⟨B^{∗}Ax, x⟩|} & ∥B^{−1}∥ \end{matrix})^{2}

≤ 2 · \frac{(∥B∥ ∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}})}{∥B∥ ∥B^{−1}∥},

which is equivalent with

(3.36)  (0 ≤) ∥Ax∥^{2} ∥B∥^{2} − |⟨B^{∗}Ax, x⟩|^{2}

≤ 2\frac{∥B∥}{∥B^{−1}∥} |⟨B^{∗}Ax, x⟩| (∥B∥ ∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}})

for any x ∈ H, ∥x∥ = 1.

The inequality (3.36) also shows that ∥B∥ ∥B^{−1}∥ ≥ \sqrt{1 − r^{2} ∥B^{−1}∥^{2}} and then, by (3.36), we get

(3.37) ∥Ax∥^{2} ∥B∥^{2} ≤ |⟨B^{∗}Ax, x⟩|^{2}

+ 2\frac{∥B∥}{∥B^{−1}∥} |⟨B^{∗}Ax, x⟩| (∥B∥ ∥B^{−1}∥ − \sqrt{1 − r^{2} ∥B^{−1}∥^{2}})

for any x ∈ X, ∥x∥ = 1.

Taking the supremum in (3.37) we deduce the desired inequality (3.34).

Remark 10. The above Theorem 13 has some particular instances of interest as follows. If, for instance, we choose B = λI with |λ| ≥ r > 0 and ∥A − λI∥ ≤ r, then by (3.34) we obtain the inequality

(3.38)  (0 ≤) ∥A∥^{2} − w^{2} (A)

≤ 2 |λ| w (A) (1 − \sqrt{1 − \frac{r^{2}}{|λ|^{2}}} ) .

Also, if A is invertible, ∥A − λA^{∗}∥ ≤ r and ∥A^{−1}∥ ≤ \frac{|λ|}{r}, then by (3.34) we can state:

(3.39)  (0 ≤) ∥A∥^{4} − w^{2} (A^{2})

≤ 2 |λ| w (A^{2}) · \frac{∥A∥}{∥A^{−1}∥} (∥A∥ ∥A^{−1}∥ − \sqrt{1 − \frac{r^{2}}{|λ|^{2}} ∥A^{−1}∥^{2}}) .

References

[1] S.S. DRAGOMIR, Reverse inequalities for the numerical radius of linear operators in Hilbert spaces, RGMIA Res. Rep. Coll., 8(2005), Supplement, Article 9 [ONLINE: http://rgmia.vu.edu.au/v8(E).html].

[2] S.S. DRAGOMIR and J. SÁNDOR, Some inequalities in prehilbertian spaces, Studia Univ. “Babeş-Bolyai”- Mathematica, 32(1) (1987), 71-78.

[3] A. GOLDSTEIN, J.V. RYFF and L.E. CLARKE, Problem 5473, Amer. Math. Monthly, 75(3) (1968), 309.

[4] K.E. GUSTAFSON and D.K.M. RAO, Numerical Range, Springer-Verlag, New York, Inc., 1997.

[5] P.R. HALMOS, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, Chelsea Pub. Comp, New York, N.Y., 1972.

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

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