SOME RESULTS ON DOMINANT OPERATORS

We show that the Weyl spectrum of a dominant operator satisfies the spectral mapping theorem for analytic functions and then answer a question of Oberai.


INTRODUCTION
Throughout this paper H will denote an infinite dimensional Hilbert space and B(H) the space of all bounded linear operators on H.If T B(H), we write a(T) for the spectrum of T, r0(T) for the set of eigenvalues of T, and r00(T) for the isolated points of a(T) that are eigenvalues of finite multiplicity.If K is a subset of C, we write iso K for the set of isolated points of K.An operator T B(H) is said to FrvMAolm if its range ran T is closed and both the null space ker T and ker T" are finite dimensional.The index of a Fredholm operator T, denoted by i(T), is defined by i(T) dim ker T dim ker T'.
The essential spectrum of T, denoted by a(T), is defined by a(T) {A C: T-AI is not Fredholm}.,A Fredholm operator of index zero is called a Weyl operator.The Weyl spectrum of T, denoted by w(T), is defined by w(T) {A C T-AlisnotWeyl}.It was shown by Berberian [2] that w(T) is a nonempty compact subset of a(T).
An operator T B(H) is said to be dominant if for every z C there exists a real number Mz > 0 such that In this case, if sup._ec M M < o, T is said to be M-hyponormal, and if M 1, T is hyponormal.Evidently, We also note that an operator T need not be hyponormal even though T and T" are both M-hyponormal.To see this, consider the operator T= U 0 ."12 12 -*12 12, where U is the unilateral shift on 12 and K :12 l_ is given by g(xl, x2, xa,. (2x, 0, 0,.-. ).
Then a direct calculation shows that 1 for all z E C and for all x E 12 12, which says that T and T" are both dominant(even M-hyponormal).But since I 0a I +gK 0 I+K =T'TCTT'= 0 T is not normal(even hyponormal).
If T is redholm then by (1.1) T dominant == i(r) <_ 0. (1.2) It was known by Oberai [8] that the mappg T w(T) is upper semi-continuous, but not cohtinuous at T. However if T,, T with T,,T TT,, for all n N then lim w(T,.,)w(T). (1. 3) It was known that w(T) satisfies the one-way spectral mapping theorem for analytic functions: if f is analytic on a neighborhood of a(T) then w(f(T)) C f(w(T)). (1.4) The inclusion (1.4) may be proper(see Berberian [2, Example 3.3]).If T is normal then a(T) and w(T) coincide.Thus if T is normal since f(T) is also normal, it follows that co(T) satisfies the spectral mapping theorem for analytic functions.We say that Weyl's theorem holds for T if w(T) a(T) r00(T).
It was known (Berberian [1]) that Weyl's theorem holds for any hyponorm, al operator indeed, for any seminormal operator and for any Toeplitz operator.Oberai [9] has raised the following question: Does there exist a hyponormal operator T such that Weyl's theorem does not hold for T ?Note that T may not be hyponormal even if T is hyponormal(Halmos [5, Problem 209]).
In this paper we show that the Weyl spectrum of a dominant operator satisfies the spectral mapping theorem for analytic functions, and that Weyl's theorem holds for p(T) when T is hyponormal and p is any polynomial.The latter result answers the question of Oberai.2. SPECTRAL MAPPING THEOREM THEOREM 2.1.If S and T are dominant operators, then S,T Weyl ==> ST Weyl.
PROOF.If S, T are Weyl, then S, T are Fredholm and i(S) i(T) 0. By Conway [31, ST is Fredholm and by the index product theorem, i(ST) i(S) + i(T) 0. Hence ST is Weyl.
Conversely if ST is Weyl, then ST is Fredholm and i(ST) O. Since S and T are domi- nant, ker S C ker S" and ker T C ker T'.Since ker S" C ker(ST)', dim ker S _< dim ker S" _< dim ker(ST)* < oo.Thus ker S and ker S* are finite dimensional.By Schechter [10, Chap. 5   Theorem 3.5], S and T are Fredholm.Since 0 i(ST) i(S) + i(T) by the index product theorem, by (1.2) i(S) i(T) O. Hence S and T are Weyl.
If the "dominant" condition is dropped in the above theorem, then the backward implication may fail even though T1 and T2 commute: For example, if U is the unilateral shift on 12, consider the following operators on 12 12 T1 U $ I and T I U.
THEOREM 2.2.If T is dominant and f is analytic on a neighborhood of a(T), then w(f(T)) f(w(T)).
Let T be an M-hyponormal operator which satisfies the additional property that for all z in the complex plane, all integers n and all x in H, T is said to be an operator of M-power class (N) (Istrtescu [7]).The following M-hyponormal operator T which is not hyponormal is of M-power class (N) (Istrtescu [7]): Let {e,} be an orthonormal basis for H, and define if i=l Te, 2e3, if i=2 e,+, if i>_3 i.e., T is a weighted shift.From the definition of T we see that T is similar to the unilateral shift U (Halmos [5], Problem 90).Thus there exists an S such that T SUS-1.In our case IIS[[ 2, I[S-'[[ 1.Since U is the unilateral shift, U is a hyponormal operator, and thus for every n and z C the operator (U z)" is of class (N).It follows that for all x E H with ][x]l 1, and hence T is of M-power class with M 4. Thus our class is strictly larger than the class of hyponormal operators.Since w(T) w(U) D(the closed unit disc) and n0(T) 0, a(T) w(T) and so Weyl's theorem holds for T. THEOREM 2.4.If T B(H) is an operator of M-power class (N), then for any polyno- mial p on a neighborhood of a(T) Weyl's theorem holds for p(T).
Since every hyponormal operator is of 1-power class (N), we obtain the following result which answers the question of Oberai.
COROLLARY 2.5.If T B(H) is hyponormal, then for any polynomial p on a neighbor- hood of a(T) Weyl's theorem holds for p(T).

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