Square roots of weighted shifts of multiplicity two

Abstract Given a weighted shift T of multiplicity two, we study the set 
$\sqrt {T}$
 of all square roots of T. We determine necessary and sufficient conditions on the weight sequence so that this set is non-empty. We show that when such conditions are satisfied, 
$\sqrt {T}$
 contains a certain special class of operators. We also obtain a complete description of all operators in 
$\sqrt {T}$
 .


Introduction
Let H be a complex Hilbert space. We use B(H) to denote the algebra of all bounded linear operators on H. For T ∈ B(H), we are interested in bounded operators Q for which Q 2 = T. If such an operator exists, we say that T has a square root and in that case we would like to describe √ T, the set of all possible square roots of T. It is known that while many operators have an abundance of square roots, others do not have any square root at all. Lebow (see [6,Solution 111]) showed that when H is infinitedimensional, the set of all square roots of zero is dense in B(H) in the strong operator topology. On the other hand, Halmos proved (see [5, p. 894]) that the unilateral shift S and more generally weighted shift operators do not have any square root. It was shown in [1] that the direct sum and the tensor product of S and its adjoint S * do not have square roots either. For properties of square and nth roots of normal and other classes of general operators, see the papers [3,[7][8][9][10][11][12]14].
Our work was motivated by a recent paper [13] in which the authors provide complete descriptions of the set of all square roots of certain well-known classical operators. More specifically, square roots of the square of the unilateral shift, the Volterra operator, certain compressed shifts, the unilateral shift plus its adjoint, the Hilbert matrix, and the Cesàro operator are discussed. Particularly interesting to us is the square of the unilateral shift. Let us discuss this case in more details.
Recall that the Hardy space H 2 consists of all holomorphic functions f (z) = ∑ ∞ n=0 a n z n on the unit disk for which The set {e n (z) = z n ∶ n = 0, 1, . . .} of monomials forms an orthonormal basis for H 2 . The unilateral shift on H 2 is defined as We see that S is the same as the operator M z of multiplication by the variable z: In [13, Section 2], a characterization of √ S 2 is given. In addition to the trivial square root, which is S itself, [13, Remark 2.19(iii)] provides another simple but interesting square root, which acts on the orthonormal basis as follows: for n ≥ 0, As it turns out later, weighted versions of S andS play important roles in our study.
The unilateral shift is a special case of (unilateral) weighted shift operators. Let {e n } ∞ n=0 be a fixed orthonormal basis for H. A weighted shift is a linear operator A on H such that for all n ≥ 0, where w n ∈ C. Weighted shift operators were investigated in great details in [15]. It was shown (see [15,Corollary 3]) that if A is an injective weighted shift, then A has no bounded kth root for any k ≥ 2.
In this paper, we study square roots of A 2 for a general injective weighted shift A. More generally, we shall be interested in square roots of weighted shift operators of multiplicity two.
for all n ≥ 0, where λ n ∈ C.
We alert the reader that there is a more general notion of weighted shift operators of multiplicity two, but we restrict our attention to only those defined above. Since we assume that T is bounded, the weight sequence {λ n } ∞ n=0 is bounded. We shall only consider the case T is injective, that is, λ n ≠ 0 for all n ≥ 0. Following the proof of [15,Corollary 1], it can be shown that any such T is unitarily equivalent to a weighted shift operator of multiplicity two with weight sequence {|λ n |} ∞ n=0 . Our goal is to find necessary and sufficient conditions on the weight sequence {λ n } ∞ n=0 for which T has a square root and to determine all possible such square roots. Examples illustrating various scenarios will be presented.

2 Weighted Hardy spaces and multipliers
One of the crucial ingredients used in [13,Section 2] is the fact that S 2 , the square of the unilateral shift, is unitarily equivalent to the direct sum S ⊕ S. It turns out that any weighted shift operator of multiplicity two is also unitarily equivalent to the direct sum of two weighted shifts. In order to establish this result, we need the notion of weighted Hardy spaces (see, for example, [2,Chapter 2] and [15,Section 4]).
Let β = {β n } ∞ n=0 be a sequence of positive real numbers. The weighted Hardy space H 2 β consists of all formal power series f = ∑ ∞ n=0f (n)z n for which The inner product of any two elements f , g in H 2 β is given by It is clear that H 2 β has {β −1 n z n ∶ n ≥ 0} as an orthonormal basis and hence the set of all polynomials, C[z], is dense in H 2 β . If β n = 1 for all n, then we obtain the Hardy space H 2 . In the case β n = 1 √ n+1 for all n, we have the standard Bergman space We shall use M z to denote the operator of multiplication on H 2 β by the function φ(z) = z. It is immediate that M z is a weighted shift with weight sequence Let T be a weighted shift operator of multiplicity two with weight sequence {λ n } ∞ n=0 such that λ n > 0 for all n ≥ 0. Define β 0 = ω 0 = 1 and for all k ≥ 1. We recall the direct sum on which the inner product is given as So W is an isometry. On the other hand, the range of W is dense in H 2 β ⊕ H 2 ω because it contains all pairs of monomials. As a result, W is a unitary operator. The inverse

Proposition 2.1 Let T be a weighted shift of multiplicity two on
for β and ω defined as in (2.1). In fact, we have the following commutative diagram: where W is given by (2.2).
Proof We first note that since T is assumed to be bounded, the weight sequence {λ n } ∞ n=0 is bounded and hence M z is bounded on both H 2 β and H 2 ω . For any since β n+1 /β n = λ 2n and ω n+1 /ω n = λ 2n+1 for all n ≥ 0. Therefore, we have Proposition 2.1 shows that in order to study the square roots of T, we need to investigate the square roots of

Remark 2.3
In the case β = ω, this result is well known (see [15,Theorem 3]). The proof for the general setting is quite similar, but for completeness, we provide here the details.
Proof For all integers n ≥ 0, we have M z R(z n ) = RM z (z n ), which gives z ⋅ R(z n ) = R(z n+1 ).

C. Kottegoda, T. Le, and T. M. Rodriguez
Define φ = R (1). It then follows that By linearity, for any polynomial p in z, From this identity and the boundedness of R, there exists B > 0 such that

Proof Consider the multiplication operator
It follows that for all integers n, m ≥ 0, we have provided thatφ(n) ≠ 0.

Square roots of weighted shifts of multiplicity two
797 Now, suppose f (z) = ∑ ∞ m=0f (m)z m ∈ H 2 β . We then have It follows that β , H 2 ω ) and by Property (M1) of multipliers, the operator M z k is bounded from H 2 β into H 2 ω . Thus, there exists C > 0 such that for all n ≥ 0,

C. Kottegoda, T. Le, and T. M. Rodriguez
Equivalently, for all n ≥ 0, we have Consequently, Now, suppose that the previous inequality holds. Then, for φ ∈ H 2 β , . By Property (M1), there exists C > 0 such that for all n ≥ 0, Conversely, if the above supremum is finite, then as we have proved in (a), the constant function 1 is a multiplier from H 2 β into H 2 ω . Recall that we assume M z is bounded on H 2 β , which implies that z k belongs to Mult(H 2 β ) for any k ≥ 0. Using Property (M2) of multipliers, we conclude that z k = 1 ⋅ z k is an element of Mult(H 2 β , H 2 ω ). By linearity, it follows that C[z] ⊆ Mult(H 2 β , H 2 ω ). ∎
As we shall see in Section 3, the characterization of √ M z ⊕ M z involves multipliers a, b, and c satisfying the equation a 2 + bc = z. We conclude this section with two results concerning such multipliers.
(n)z n , and c = ∑ ∞ n=0ĉ (n)z n be formal power series in z such that Then exactly one of the following statements is true.
In particular, both b and c are nonzero power series.