#### Abstract

We show how a proof of J. Stampfli can be extended to prove that the operator − defined on the Hilbert-Schmidt class, when is an -hyponormal, -hyponormal, or log-hyponormal operator, has a closed range if and only if is finite.

#### 1. Introduction

Let be a complex, separable, infinite dimensional Hilbert space, and let denote the algebra of all linear bounded operators on The Hilbert-Schmidt class, denoted by is a Hilbert space with the -norm that arises from the inner product where is the scalar-valued trace. For define by and let denote the spectrum of Let the range of a linear operator be denoted by For a normal operator Anderson and Foiaş [1] proved that is norm closed if and only if is a finite set. In [2], Stampfli extended this result to the class of hyponormal operators.

Theorem 1 A ([2]). *Let be a hyponormal operator. Then is norm closed if and only if is finite. *

In fact, Stampfli provided a proof of the “only if" implication which extends to a larger class of operators than the class of hyponormal operators (see Proposition 2.2). For an operator let denote its * normal approximate point spectrum *, that is, the set of those for which there exists an orthonormal sequence in such that
Define the class as follows:
Some classes of hyponormal related operators, such as -hyponormal operators, that is,
-hyponormal operators, that is, for some or -hyponormal operators, that is, invertible operators such that have spectrum that is finite or they belong to Particularly, the hyponormal operators (i.e., -hyponormal) have this property.

In [3] Stampfli proved the following lemma which will be used in Section 2.

Lemma 1 B. *Let and let be a sequence of distinct points of Then for any sequence of positive numbers converging to zero, there exists an orthonormal sequence of vectors in such that
*

#### 2. The Closedness of the Range of

The operator defined on the Hilbert-Schmidt class will be denoted in the remainder of this note by that is, , Let denote the set of -hyponormal operators.

Proposition 2.1. *Let If is finite, then is closed. *

*Proof. *It is well known that an operator with finite spectrum is normal. Indeed, for such an operator, the restriction to an invariant subspace belongs to On the other hand, if with then (cf. [4]). Thus, we can write where 's are the spectral projections.

Let and be in such that Therefore in the norm, and according to Theorem A, there exists such that For an arbitrary let be the block-matrix representation of relative to the decomposition Thus

for all This implies that each is a Hilbert-Schmidt operator. Moreover can be chosen and thus

Proposition 2.2. *Let Then is not closed. *

*Proof. *We will use same notation and circle of ideas as in [2]. Let be sequence of distinct points of so that Let
and choose a nonincreasing sequence so that and According to Lemma B, there exists an orthonormal sequence that satisfies (1.4) and(1.5). Let and let such that
It results that
Define by and Let and let be the orthogonal projection onto and define A tedious calculation shows that
where Denoting by then for
Furthermore, from (2.3) it results that
and from (2.4)

We will show next that when thus there exists such that that is,

First, we will show that when Indeed,

The first sum of the right-hand side of the above can be majorized by
Since we have
According to (2.8),
and according to (2.4),
which implies
Therefore
where and are some constants. After a careful review of the proof, one can see that the sequence can be assumed to converge fast enough (otherwise choose a subsequence of it), more precisely when

We show next that when Indeed, we can write

Obviously, we can write with which implies
For an orthonormal basis of we will show that
For each write with Thus
Since we have and consequently, for
Since the sequence is orthonormal, we have
Therefore
For a fixed
Consequently, for

The operator is not in since, according to the proof of Theorem A in [2],

Theorem 2.3. *Let Then is closed if and only if is finite. *

*Proof. * If and are finite, then according to Proposition 2.1, is closed. Conversely, if has an infinite spectrum, then there are infinitely many distinct points that are either isolated points of the spectrum, in which case they are eigenvalues, or accumulation points of the spectrum, in which case they are in Since we have Thus and according to Proposition 2.2, is not closed.