Fredholm Weighted Composition Operators on Dirichlet Space

Let H be a Hilbert space of analytic functions on the unit disk D. For an analytic function ψ on D, we can define the multiplication operator Mψ : f → ψf, f ∈ H. For an analytic selfmapping φ of D, the composition operator Cφ defined on H as Cφf f ◦ φ, f ∈ H. These operators are two classes of important operators in the study of operator theory in function spaces 1–3 . Furthermore, for ψ and φ, we define the weighted composition operator Cψ,φ on H as


Introduction
Let H be a Hilbert space of analytic functions on the unit disk D. For an analytic function ψ on D, we can define the multiplication operator M ψ : f → ψf, f ∈ H.For an analytic selfmapping ϕ of D, the composition operator C ϕ defined on H as C ϕ f f • ϕ, f ∈ H.These operators are two classes of important operators in the study of operator theory in function spaces 1-3 .Furthermore, for ψ and ϕ, we define the weighted composition operator C ψ,ϕ on H as 1.1

International Journal of Mathematics and Mathematical Sciences
Recall the Dirichlet space D that consists of analytic function f on D with finite Dirichlet integral: where dA is the normalized Lebesgue area measure on D. It is well known that D is the only m öbius invariant Hilbert space up to an isomorphism 10 .Endow D with norm Furthermore D is a reproducing function space with reproducing kernel Denote M {ψ : ψ is analytic on D, ψf ∈ D for f ∈ D}.M is called the multiplier space of D. By the closed graph theorem, the multiplication operator M ψ defined by ψ ∈ M is bounded on D. For the characterization of the element in M, see 11 .
For analytic function ψ on D and analytic self-mapping ϕ of D, the weighted composition operator C ψ,ϕ on D is not necessarily bounded.Even the composition operator C ϕ is not necessarily bounded on D, which is different from the cases in Hardy space and Bergman space.See 12 for more information about the properties of composition operators acting on the Dirichlet space.
The main result of the paper reads as the following.If ψ z 1, then the result above gives the characterization of bounded Fredholm composition operator C ϕ on D, which was obtained in 12 .
As corollaries, in the end of this paper one gives the characterization of bounded invertible and unitary weighted composition operator on D, respectively.Some idea of this paper is derived from 4, 13 , which characterize normal and bounded invertible weighted composition operator on the Hardy space, respectively.

Proof of the Main Result
In the following, ψ and ϕ denote analytic functions on D with ϕ D ⊂ D. It is easy to verify that ψ ∈ D if C ψ,ϕ is defined on D. where I is the identity operator.Since where k w K w / K w is the normalization of reproducing kernel function K w .Since S is compact and k w weakly converges to 0 as |w| → 1, Sk w → 0 as |w| → 1.It follows that there exists constant r, 0 < r < 1, such that Sk w < 1/2 for all w with r < |w| < 1. Inequality 2.3 shows that which implies that ψ has no zeroes in {w ∈ D, r < |w| < 1}, and, hence, ψ has at most finite zeroes in {w ∈ D, |w| ≤ r}.
Since k w weakly converges to 0 as |w| → 1, ψ, k w → 0 as |w| → 1, that is, For the proof of the following lemma, we cite Carleson's formula for the Dirichlet integral 14 .

International Journal of Mathematics and Mathematical Sciences
Let f ∈ D, f BSF be the canonical factorization of f as a function in the Hardy space, where B ∞ j 1 a j /|a j | a j − z / 1 − a j z , is a Blaschke product, S is the singular part of f and F is the outer part of f.Then  The following lemmas is well-known.It is easy to verify by the fact M * ψ K w ψ w K w also.Proof.Since a bounded invertible operator is a bounded Fredholm operator, the proof is similar to the proof of Theorem 1.1.Proof.If C ψ,ϕ is a unitary operator, then it must be an invertible operator.By Corollary 2.7, ϕ is an automorphism of D and ψ is invertible in M.
Let n be nonnegative integer, e n z z n , z ∈ D. A unitary is also an isometry, so we have ψ C ψ,ϕ e 0 e 0 1, 2.9 ψϕ n C ψ,ϕ e n e n √ n, n ≥ 1.

2.10
Let α ∈ D such that ϕ α 0. Since ϕ is an automorphism of D, ϕ n is a finite Blaschke product with zero α of order n.By Carleson's formula for Dirichlet integral, we have

International Journal of Mathematics and Mathematical Sciences
That is, Let n → ∞, then 1 T P α ξ |ψ ξ | 2 |dξ|/2π .By 2.12 , we have D ψ 0 and |ψ 0 ϕ 0 | 0. By 2.9 , we obtain ψ is a constant with |ψ| 1, which implies that ϕ 0 0, that is, ϕ is a rotation of D. The sufficiency is easy to verify.Proof.If ψ a 0 for a ∈ D, then C * ψ,ϕ K a ψ a K ϕ a 0, which implies that K a is in the kernel of C * ψ,ϕ .Thus if ψ had infinitely many zeros, the kernel of C * ψ,ϕ would be infinite dimensional and hence this operator would not be Fredholm. If

Proposition 2 . 1 .
Let C ψ,ϕ be a bounded Fredholm operator on D. Then ψ has at most finite zeroes in D and ϕ is an inner function.Proof.If C ψ,ϕ is a bounded Fredholm operator, then there exist a bounded operator T and a compact operator S on D such that T C ψ,ϕ *

Lemma 2 . 5 .Lemma 2 . 6 .Corollary 2 . 7 .
Let ψ ∈ M. Then M ψ is an invertible operator on D if and only if ψ is invertible in M. Let ψ ∈ M. Then M ψ is a Fredholm operator on D if and only if ψ is bounded away from the unit circle.Now we give the proof of Theorem 1.1.Proof of Theorem 1.1.If C ψ,ϕ is a bounded Fredholm operator on D, by Corollary 2.4, ψ ∈ M and ϕ is an automorphism of D. Since C ϕ is invertible, M ψ is a Fredholm operator.So ψ is bounded away form the unit circle follows from Lemma 2.6.On the other hand, if ψ ∈ M and bounded away from the unit circle, then M ψ is a bounded Fredholm operator on D. If ϕ is an automorphism of D, then C ϕ is invertible.Hence C ψ,ϕ M ψ C ϕ is a bounded Fredholm operator on D. As corollaries, in the following, we characterize bounded invertible and unitary weighted composition operators on D. Let ψ and ϕ be analytic functions on D with ϕ D ⊂ D. Then C ψ,ϕ is a bounded invertible operator on D if and only if ψ ∈ M, invertible in M, and ϕ is an automorphism of D.

Corollary 2 . 8 .
Let ψ and ϕ be analytic functions on D with ϕ D ⊂ D. C ψ,ϕ is a bounded operator on D. Then C ψ,ϕ is a unitary operator if and only if ψ is a constant with |ψ| 1 and ϕ is a rotation of D.

Remark 2 . 9 .
The key step in the proof of the main result is to analyze zeros of the symbol ψ and univalency of ϕ.The following result pointed out by the referee gives a simple characterization of the symbols ψ and ϕ for the bounded Fredholm operator C ψ,ϕ on D. Proposition 2.10.Let ψ and ϕ be analytic functions on D with ϕ D ⊂ D. C ψ,ϕ is a bounded Fredholm operator on D. Then ψ has only finitely many zeros in D and ϕ is univalent.
ϕ a ϕ b for a, b ∈ D with a / b, by a similar reasoning as 1, Lemma 3.26 , there exist infinite sets {a n } and {b n } in D which is disjoint such that ϕ a n ϕ b n .Since ψ has only finitely many zeros in D, we can choose infinitely many a n and b n such that ψ a n / 0, ψ b n / 0. Hence,

Theorem 1.1. Let ψ and ϕ be analytic functions on D with
|, P α ξ is the Poisson kernel, and μ is the singular measure corresponding to S. Let C ψ,ϕ be a bounded operator on D, ψ BF with B a finite Blaschke product.Then C F,ϕ is bounded.Proof.Let M B be the multiplication operator on D. Then C ψ,ϕ M B C F,ϕ .Since B is a finite Blaschke product, by the Carleson's formula, we have By Lemma 2.2, C F,ϕ is a bounded operator on D. Since C ψ,ϕ M B C F,ϕ and M B is a Fredholm operator, C F,ϕ is a Fredholm operator also.By Proposition 2.1 and Lemma 2.3, by the inequality above it is easy to verify that C F,ϕ is bounded on D if C ψ,ϕ is bounded.Lemma 2.3.Let F be an analytic function on D with zero-free.If C F,ϕ is a bounded Fredholm operator on D, then ϕ is univalent.Proof.If ϕ a ϕ b for a, b ∈ D with a / b, by a similar reasoning as 1, Lemma 3.26 , there exist infinite sets {a n } and {b n } in D which is disjoint such that ϕ a n ϕ b n .Hence, * is finite dimensional.Corollary 2.4.If C ψ,ϕ is a bounded Fredholm operator on D, then ϕ is an automorphism of D and ψ ∈ M. Proof.By Proposition 2.1, ψ has the factorization of BF with B a finite Blaschke product and F zero free in D.