The motivation for this note is provided by the results obtained in [1-4]. Let T be a bounded linear operator on a complex Hilbert space H. The numerical range of T, denoted by W(T), is the subset of the complex plane and

The numerical radius of T is defined as,

The following lemma is known and is an easy consequence of the definitions involved.

Lemma 1.1., where T^* is the adjoint operator of T and is the complex conjugate of.

Berger and Stampfli in [2] have proved that if and, for some n, then. Also, they gave an example of an operator T and an element such that implies that and. In Theorem 2.1, we present a different proof of their result in [2] and show that is indeed the best constant.

Theorem 2.1 also generalizes the result in [4] and provides a partial converse to Theorem 1 in [1, p. 372].

Our next main result in Theorem 2.3 gives an alternative and shorter proof of Theorem 1 in [1].

Applying Lemma 2 and Proposition 2 of [1], a new result on the numerical range of nilpotent operators on H is obtained in Theorem 2.4. This gives a restricted version of Theorem 1 in [3].

Finally, two examples are discussed. Example 3.1 deals with the operator, where 1 is not the eigenvalue of if. Example 3.3 justifies why fails to increase until and unless.

Theorem 2.1. The following statements are true for a bounded linear operator T on a Hilbert space H with.

1) such that

,.

2) If for some integer n, then

and.

3) The set forms a nontrivial subspace of T so that its orthogonal complement is invariant.

Proof. 1) For each real number and a postive integer, n, let. Then the inner product relation implies that

That is,

Hence,

Since

it follows that

Dividing the above inequality by, we have

Let be the following block-diagonal matix of order n and

If γ_n denotes the determinant of such that then the value of γ_n is positive because all principal minors of are nonnegative. Suppose that

We consider the following cases:

Case 1. Iffor the leastthen

and converges to zero.

Case 2. Let for all. Then

and by induction

Further, the inequality

implies that converges to q as n goes to infinity for some q ≥ 0. Therefore from Equation (2.1), as. Thus. Obviously, q = 1 only if.

2) By the assumption, for some positive integer n. Now fom Equation (2.1), we obtain:

and so that. The equality,

now follows from (a) and thus. Also, which gives since.

3) To prove this case, we assume that if the vector is orthogonal to the spanning set then. Let

, for. Then

Hence, for and the spanning set is a non-trivial invariant subspace on T.

In [2, p. 1052], an example of an operator T on and an element x in H with, is given where

. Theorem 2.1 above establishes that is the best constant in this case.

Remark 2.2. An operator A on H is hyponormal if

. Let thenif A is a hyponormal operator. Hence, , and the set of vectors forms a reducing subspace of A.

A natural connection between Feijer’s inequality and the numerical radius of a nilpotent operator was estaplished by Haagerup and Harpe in [1]. They proved, using positive definite kernals, that for a bounded linear operator T on a Hilbert space H such that andthen. The external operator is shown to be a truncated shift with a suitable choice of the vector in H. The inequality is related to a result from Feijer about trigonometric polynomials of the form

with. Such a polynomial is positive if for all. Here, we present a simplified proof of Theorem 1 in [1].

Theorem 2.3. For an operator N on H with and, we have.

Proof. We will follow the notations of Theorem 1 in [1]. Let S be the operator on and, be the basis in. We define the operator S as follows:

and for

The matrix for S gives a dialation for T. Let A be the matrix for S and

If is a unitary operator on with diagonal then. By Lemma 1, we have:

This helps to define the characteristic function of a contraction.

For the operator N on H, let then

is a positive operator and depends on N. Let the range of be denoted by. Then the tensor product, , is a Hilbert space. We define the map so that F is an isometry.

For λ, let where I is the identity operator, and is an operator on.

Therefore and.

Now, we claim that, for we hope that By Lemma 1.1

That is,.

Since, we have:

and

where is the spectral radius of. By the definition of the spectral radius, we have the characteristic polynomial f such that by [5, p. 179, Example 9], the roots of are given by

, and and

.

Karaev in [3] has proved, using Theorem 1 in [1] and the Sz.-Nagy-Foias model in [6] that the numerical range of an arbitrary nilpotent operator N on a complex Hilbert space H is an open or closed disc centered at zero with radius less than or equal to,

Using Theorem 2 and the assumption that

, , we have as a closed or an open disc centered at zero with radius equal to. In fact, we have the following theorem.

Theorem 2.4. For a nilpotent operator N on H with, and, the numerical range is a disc centered at zero with radius.

Proof. For any we must claim that, for and is a vector in.

From [1, p. 374, Proposition 2], we have. Also, for some,

. Now by [1] [P.375, Lemma 2], we obtain:

and

Let. Then:

and the theorem follows from above since is arbitrarily chosen.

An operator A is a unilateral weghted shift if there is an orthonormal basis and a sequence of scalers such that for all. It is easy to see that where S is the unilateral shift and D is the diagonal operator with, for all n.

Thus, and for all n. So is the basis of eigenvectors for. Also, note that A is bounded if is bounded.

If A is a unilateral shift then and for. Consequently, for a hyponormal operator A, and for. A wighted shift is hyponormal if and only if its weight sequence is increasing.

Example 3.1. Let be an operator on such that and for and. Here, we show that is not an eigenvalue of if

. We prove our claim by contradiction Let be an eigenvalue of. Then, there exists with and, n = 2, 3, ···. It is not hard to see that:

For, we have and thus, which shows that, contrary to our assumption. Thus, is not an eigenvalue of if

.

Remark 3.2. Following [2], if then

Therefore, the numerical radius, is equal to 1.

The example below shows that there exists an operator such that for.

Example 3.3. Let be a unilateral shift. If is the orthogonal projection of onto the spanning set of vectors then and has the usual matrix representation. Let

Then the characteristic polynomial of is given by a Chebyshev polynomial of the first kind. Let where. Then:

(easily proven by trigonometric identities) and for is a linear combination of powers of x^k. Also, det. If then the roots are given by the Chebyshev polynomial of the first kind. The roots can be found by finding the eigenvalues of matrix B. By [2, p. 179, Example 9], the eigenvalues of B are given by

, for.

Suppose that

then. Hence, if.