A Similarity Invariant and the Commutant of some Multiplication Operators on the Sobolev Disk Algebra

Let R D be the algebra generated in Sobolev space W22 D by the rational functions with poles outside the unit disk D. In this paper, we study the similarity invariant of the multiplication operators Mg in L R D , when g is univalent analytic on D or Mg is strongly irreducible. And the commutants of multiplication operators whose symbols are composite functions, univalent analytic functions, or entire functions are studied.


Introduction
Let Ω be an analytic Cauchy domain in the complex plane and let W 22 Ω denote the Sobolev space, the distributional partial derivatives of first and second order of f belong to L 2 Ω, dA , 1.1 dA denotes the planar Lebesgue measure.For f, g ∈ W 22 Ω , we define f, g |α|≤2 D α fD α gdA.

International Journal of Mathematics and Mathematical Sciences
Then W 22 Ω is a Hilbert space and a Banach algebra with identity under an equivalent norm.
W 22 Ω can be continuously embedded in the space C Ω of continuous functions on Ω by Sobolev embedding theorem.Let R Ω be the subalgebra generated by the rational functions with poles outside Ω.When Ω D, the unit disc, we call R D Sobolev disk algebra.For f ∈ R D , the multiplication operator M f on R D is defined by M f g fg, g ∈ R D .Then R D A M z e 0 , where e 0 is the identity in R D and A M z is the algebra generated by M z and identity.In fact, R D consists of all analytic functions in W 22 D .We have the following properties of the space R D and the multiplication operators on it., n 0, 1, 2, . . . .1.3 (ii) As a functional Hilbert space, R D has reproducing kernel which is Then for z 0 ∈ D, (iv) The operator M f admits the following matrix representation with respect to {e n } ∞ n 0 : Note that R D is a subset of the disk algebra A D , hence a subset of H ∞ .Because of the special definition of the inner product and the complex behavior of the boundary value, the structure of the space R D is much more complicated than H ∞ or H 2 .For more details about the Sobolev disk algebra, the reader refers to 1-3 .
Let H be a complex separable Hilbert space and L H denote the collection of bounded linear operators on H.One of the basic problems in operator theory is to determine when two operators A and B in L H are similar.A quantity quantities or a property properties P is similarity invariant invariants if A has P and A ∼ B implies that B has P 2 .From this point of view, one of the basic problems in operator theory mentioned above is to determine the similarity invariants.There have already been a lot of results on the similarity invariants of operators, especially that of nonadjoint operators, which can be found in, for example, 4-6 .In 7 , Wang et al. proved that in R D , M f is similar to M z n if and only if f is an n-Blaschke product.In this paper, we study the similarity invariant of the multiplication operators M g in L R D , when g is univalent analytic on D or M g is strongly irreducible.
It is well known that the commutant of a bounded linear operator or operators on a complex, separable Hilbert space plays an important role in determining the structure of this operator or these operators.The commutant of a multiplication operator on Sobolev disk algebra has been studied in the literature see 1-3 .In this paper, we describe the commutant of the multiplication operator M fg when g is an n-Blaschke product.And by this result, we generalize the result which is obtained by Liu and Wang in 3 .Moreover, we study the commutants of the multiplication operators whose symbols are composite functions, univalent analytic functions, or entire functions.

Similarity Invariant of Some Multiplication Operators
In this section, we will characterize the similarity invariant of some multiplication operators on Sobolev disk algebra.Here, we briefly recall some background information.
For T in L H , let σ T , σ p T , and σ e T be the spectrum, point spectrum, and essential spectrum of a bounded linearly operator T , respectively.An operator A in L H is said to be a Cowen-Douglas operator with index n if there exists Ω, a connected open subset of complex plane C, and n, a positive integer, such that where iv is equivalent to iv 8 ; iv there exists λ 0 in Ω, such that {ker B n Ω denotes the collection of Cowen-Douglas operators with index n.
For T ∈ L H , the set of operators which commute with it is A T .That is A T {A ∈ L H : AT TA}.Operator T is strongly irreducible if there is no nontrivial idempotent in A T 8, 9 .Denote SI the set of all strongly irreducible operators on H.
where T is the unit circle.Since f and g are univalent and analytic on D, then Because g is univalent analytic from D to Ω, g −1 : Ω → D is also univalent analytic.Then g −1 • f is injective and surjective analytic function on D. If g −1 • f z 0 0, there exists a M öbius transform ϕ with ϕ z 0 z − z 0 / 1 − z 0 z and a complex number c with |c| 1 such that g −1 • f z cϕ z 0 see 11 .Therefore f z g cϕ z 0 z .By Lemma 2.3, M f ∼ M g .Lemma 2.5 see 3 .Given f ∈ R D , the following are equivalent: Proof."⇒": Suppose that g is not univalent on D. There exists some λ ∈ g D such that the number of zeros of This contradicts to M f that is a strongly irreducible operator see Lemma 2.5 .So g is univalent analytic on D. By the proof of Theorem 2.4, there exists a function χ c z − z 0 / 1 − z 0 z such that f g • χ, where z 0 ∈ D and c ∈ C, |c| 1.
"⇐": By the conditions of the theorem, g is univalent analytic on For any operator T on Hilbert space H and any integer n, , where n 1 , n 2 are natural numbers.Then for each maximal ideal J of A T , J must be one of the following two forms: where The following statements are equivalent: where id R D denotes the identity operator of R D .
Suppose that M * f / ∼M * g .By Lemma 2.7, each maximal ideal J of A M * f ⊕ M * g must be one of the following two forms where J 11 and J 22 are the maximal ideals of A M * f and A M * g , respectively.We can assume that J admits the first form.Then This contradicts that J is a maximal ideal.So M * f ∼ M * g and M f ∼ M g .

The Commutant Algebra of Multiplication Operator
In 3 , Liu and Wang give the following proposition.
. ., n be n-Blaschke product.Considering z n is a special n-Blaschke product, we study the commutant of M f where f z B n z h z .The following theorem is obtained, and by this result, the above proposition is generalized.
To prove the above theorem, we need the following lemmas.

So we only need to prove
International Journal of Mathematics and Mathematical Sciences 7 we have where

3.11
It follows from ker We are now in need to prove that A A ⊂ A ⊕ n 1 M * z .Suppose that Q ∈ A A .Since h does not vanish on D, ranM h R D and so ker T {0}.

3.12
It follows from QA AQ that ker A and ker A k are both the invariant subspaces of Q.Since

3.13
A admits the matrix representation 3.5 with the above decomposition.So From QA AQ, we have

3.15
Comparing the n, n 1 entries of both sides, we have Comparing the n, n 2 entries of QA and AQ, we have

3.16
It follows from Lemma 3.4 that

International Journal of Mathematics and Mathematical Sciences
Setting Comparing the n, n k 1 entries of QA and AQ, we have

3.19
Therefore, by Lemma 3.4, By the following lemma, we discuss the commutant of the multiplication operators whose symbols are composite functions in R D .Lemma 3.6.For T in L R D and f in R D the following are equivalent:

3.22
Since J is not a Blaschke sequence, this means Proof.By the Embedding Theory of Sobolev space, g

3.26
International Journal of Mathematics and Mathematical Sciences Therefore,

3.27
By 3.24 , 3.25 , and 3.27 , we have g ⊕ n 1 M f .Therefore, we will only prove that Let f be an injective function in R D and Ω f D .Then for each z 0 ∈ D, it is obvious that f z 0 is not in f T .Denote the component of ρ s−F M f containing f z 0 as Ω, then z 0 is the only zero point of f z − f z 0 in D. By Lemma 2.2, M * f is a Cowen-Douglas operator with index 1.By Lemma 2.5, we have So the following corollary is obtained.Corollary 3.11.Let f be a univalent analytic function in R D and B n be an n-Blaschke product.
Easy examples show that ker 33

3.34
Then there exists a linearly independent set Let n ≥ 2 and ω be the nth root of 1, that is, ω ∈ C and ω n 1.Let Δ n denote the Vandermonde determinant of order n:

3.36
For 1 ≤ i, j ≤ n, the i, j -cofactor will be denoted by Δ ij .
by Lemma 3.12, with a 1 , a 2 , . . ., a n ∈ C. For all g ∈ R D , we define an operator T : R D → R D as follows: Tg z a 1 g z a 2 g ωz • • • a n g ω n−1 z .

3.42
International Journal of Mathematics and Mathematical Sciences 15 Therefore, Easy examples show that, in general, the converse of Proposition 3.8 is false.But the following case is an exception.To prove it, we need the following lemma.Proof.By Proposition 3.14, for all z 0 ∈ D, we have

3.47
For each λ ∈ D, we can find z 0 ∈ D such that z n 0 λ.We define h on D by h λ f z 0 and h is well defined.Indeed, set z n

Proposition 1 . 1
see 1 .(i) Hilbert space R D has an orthogonal basis {e n } ∞ n 0 , where e n β n z n , β n n 1 3n 4 − n 2 2n 1 π 1/2 Then X and X −1 are what we want.ii ⇒ i : Since M f and M g are in SI , we have M * f and M * g that are strongly irreducible and in B 1 Ω by Lemma 2.5.Computations show

Lemma 3 .
15. Let f h z n be in R D and h, analytic onD.Then h is in R D .Proof.Since h is analytic on D, we have h z ∞ m 0 h m z m , hence, {β n } is monotonically decreasing, β m ≥ β nm for all m ≥ 0. So that h is in R D .Proposition 3.16.If f ∈ R D and A M z n ⊂ A M f , then there exists h being in R D such that f h z n .

→ λ 0 h λ − h λ 0 λ − λ 0 lim z → z 0 f z − f z 0 z n − z n 0 limz → z 0 f z − f z 0 z − z 0 z
Thenk z 1 ∈ z n − z n 0 R D ⊥ ⊂ f − f z 0 R D ⊥ .3.48Hence f z 1 f z 0 .International Journal of Mathematics and Mathematical SciencesFor 0 / λ 0 ∈ D, we have z 0 / 0. Therefore, lim λ λ 0, z 0 0. Sincez n R D ⊃ f − f 0 R D , 3.50there exists g ∈ R D such that f − f 0 z n g.Hence, lim h is analytic on D. By Lemma 3.15, we have h ∈ R D and f z h z n .For each f ∈ R D and a / ∈ f ∂D , η f ∂D , a denote the winding number of f ∂D at a. Define s k f inf η f ∂D , a : η f ∂D , a / 0 .3.52 Proposition 3.17.If f ∈ R D is a nonconstant entire function and s k f , then A M f A M z s .Proof.By Theorem 1 in 14 , there exists an entire function h such that f z h z s and k h 1.Since h is an entire function, h, h , and h are all bounded and analytic on D. So h ∈ R D .By k h 1, there is only one zero of h − λ in D for some λ.By Lemma 2.2, M * h ∈ B 1 Ω .By Proposition 3.10, A M f A M h z s A M z s .
iii let z 0 ∈ D and f z 0 / ∈ f ∂D .Denote the component of ρ s−F M f containing f z 0 as Ω, then M * f ∈ B n Ω , where n is the number of the zeros of f z − f z 0 in D.
International Journal of Mathematics and Mathematical SciencesLemma 3.13 see 1 .A ∈ A M z n if and only if for all g ∈ R D and z / 0,