1. Introduction
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
.
2. Main Results
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 γn denotes the determinant of 
such that 
then the value of γn is positive because all principal minors of 
are nonnegative. Suppose that 
(2.1)
We consider the following cases:
Case 1. If
for the least
then
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 
then
if 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 
and
then
. 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.
3. An Application
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 xk. 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
.