A Subnormal Completion Problem for Weighted Shifts on Directed Trees, II

The subnormal completion problem on a directed tree is to determine, given a collection of weights on a subtree, whether the weights may be completed to the weights of a subnormal weighted shift on the directed tree. We study this problem on a directed tree with a single branching point, η\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\eta $$\end{document} branches and the trunk of length 1 and its subtree which is the “truncation” of the full tree to vertices of generation not exceeding 2. We provide necessary and sufficient conditions written in terms of two parameter sequences for the existence of a subnormal completion in which the resulting measures are 2-atomic. As a consequence, we obtain a solution of the subnormal completion problem for this pair of directed trees when η<∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\eta < \infty $$\end{document}. If η=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\eta =2$$\end{document}, we present a solution written explicitly in terms of initial data.


Introduction
The class of unilateral weighted shifts on Hilbert space has been a standard and important source of examples with which to study the properties of bounded linear operators on Hilbert space, including especially the investigation of subnormality (see [18] and [5]). A recently introduced class of weighted shifts on directed trees provides a more extensive collection of objects for study (see e.g., [1][2][3]14,15]). In [10], we initiated the study of a subnormal completion problem for weighted shift operators on directed trees. For a classical weighted shift, the subnormal completion problem is to be given an initial finite sequence of positive weights and to determine whether or not they may be extended to the weights of an injective, bounded, subnormal unilateral weighted shift; such a shift is called a subnormal completion of the initial weight sequence. It should be emphasized that the subnormal completion problem was initiated and solved explicitly for four initial weights by Stampfli in [19]. Almost three decades later the problem was abstractly solved by Curto and Fialkow in [7,8] (see also [6,9]). Relatively recently an explicit solution for five initial weights was given in [16].
In the present paper we continue the study of the subnormal completion problem for weighted shifts on directed trees. As in [10], we restrict our attention to the directed tree T η,κ with a single branching point, η branches and the trunk of length κ, and we consider the subnormal completion problem with respect to the subtree T η,κ,p , which is the "truncation" of T η,κ to vertices of generation not exceeding p. If η is finite and the p-generation subnormal completion problem on T η,κ has a solution for given initial data λ, then we can always find a subnormal completion Sλ such that the measures μλ i,1 , which are canonically associated with Sλ at vertices of the first generation, are at most κ+p+2 2 -atomic (see Theorem 2.2). So far as we know, there is no solution of the p-generation subnormal completion problem on T η,κ written in terms of initial data for p 2. The only explicit solution is that for the 1-generation subnormal completion problem on T η,1 (see [10,Theorem 5.1]). In view of the above discussion, if the 2-generation subnormal completion problem on T η,1 has a solution for given initial data λ, then we can always find a subnormal completion Sλ of λ on T η,1 with the property that each measure μλ i,1 is 1-or 2-atomic. In this paper we provide necessary and sufficient conditions written in terms of two parameter sequences for λ to admit a subnormal completion Sλ on T η,1 with 2-atomic measures μλ i, 1 . As a consequence, we solve the 2-generation subnormal completion problem on T η,1 for η < ∞. The results, taken as a whole, suggest that a complete answer to the subnormal completion problem even for this class of directed trees is at present out of reach.
The paper is organized as follows. In Sect. 2, we provide notation, terminology and results that are needed in this paper. In Sect. 3 we carry out an in-depth analysis of the first two "negative" moments of the Berger measure associated with the Stampfli completion of three increasing weights to the weight sequence for a subnormal unilateral weighted shift (see Lemma 3.6). In Sect. 4 we state and prove the solutions of the 2-generation subnormal completion problem on T η,1 with 2-atomic measures (the case η = ∞ is included, see Theorem 4.1) and without any restrictions on supports of the associated measures (only for η < ∞; see Theorem 4.2). In view of Proposition 4.4(iii), to solve the problem with 2-atomic measures we have to compute the infimum of a quadratic form in η variables subject to some constraints. In Sect. 5 we give a solution of the 2-generation subnormal completion problem on T 2,1 with 2-atomic measures written entirely in terms of the initial data (see Theorem 5.2). This confirms the scale of the complexity of this problem.

Preliminaries
In this section we sketch briefly the notation and results necessary for the present discussion, but the reader is encouraged to consult [10] for a considerably more complete presentation. Given a complex Hilbert space H, we denote by B(H) the C * -algebra of all bounded linear operators on H. Recall that an operator T ∈ B(H) is said to be subnormal if there exists a complex Hilbert space K containing H and a normal operator N ∈ B(K) such that T h = Nh for all h ∈ H. We refer the reader to [4,5,12] for the foundations of the theory of subnormal operators.
Denote by Z + , N, R, R + and C the sets of nonnegative integers, positive integers, real numbers, nonnegative real numbers and complex numbers, respectively. Set using the convention that J 0 = ∅. We write card(X) for the cardinality of a set X. In what follows, δ x stands for the Dirac Borel measure on R + at the point x ∈ R + . The closed support of a Borel probability measure μ on R + is denoted by supp μ.
is a directed tree if T is a directed graph, which is connected, has no circuits and has the property that for each vertex u ∈ V there exists at most one vertex v ∈ V , called the parent of u and denoted here by par(u), such that (v, u) ∈ E. The reader is referred to [15,17] for more information on directed trees needed in this paper.
We will consider a certain class of directed trees with a single branching point obtained as follows: given η ∈N 2 and κ ∈ N, we define the directed tree T η,κ = (V η,κ , E η,κ ) by (see Fig. 1): Define as well a subtree of T η,κ on which we may be given "some of" the weights of a proposed shift as initial data: for η ∈N 2 , κ ∈ N and p ∈ N, let the directed tree T η,κ,p = (V η,κ,p , E η,κ,p ) be defined by (see Fig. 2) Given a directed tree T = (V, E), let 2 (V ) be the Hilbert space of all square summable complex functions on V equipped with the usual inner product. The family {e u } u∈V defined by is clearly an orthonormal basis of 2 (V ).
We call S λ the weighted shift on the directed tree T with weights {λ v } v∈V • . Throughout this paper it is assumed that the resulting operator S λ is bounded (see [10] or more generally [15] for an approach suitable even for unbounded shifts, and for discussions of when S λ is indeed bounded). If S λ ∈ B( 2 (V )), in view of [15, (3.1.4)], one may more easily express S λ by (We adopt the convention that v∈∅ x v = 0.) The weighted shifts desired are those which are subnormal operators. According to [15, Theorem 6.1.3 and Notation 6.1.9], the following assertion holds. The characterizations of subnormality of S λ on the directed tree T η,κ can be found in [15,Corollary 6.2.2]. Matters are in hand for the statement of the fundamental problem considered. Let T = (V, E) be a subtree of a directed treeT = (V ,Ê) and let λ = {λ v } v∈V • be a system of positive real numbers. We say that a weighted shift and Sλ is subnormal. If such a completion exists, we may sometimes say λ admits a subnormal completion onT . The subnormal completion problem for (T ,T ) consists of seeking necessary and sufficient conditions for a system λ = {λ v } v∈V • ⊆ (0, ∞) to have a subnormal completion onT .
Following [10], the subnormal completion problem for (T η,κ,p , T η,κ ), where p ∈ N, is called the p-generation subnormal completion problem on T η,κ (sometimes abbreviated to p-generation SCP on T η,κ ). In this particular case, our initial data takes the form The following result, which gives the measure-theoretic way of solving the p-generation subnormal completion problem on T η,κ , is a consequence of [ In [10, Theorem 5.1] we gave an explicit solution of the 1-generation subnormal completion problem on T η,1 written in terms of initial data. As shown in [10, Theorem 6.2], the 1-generation subnormal completion problem on T η,κ , where κ ∈ N, reduces to seeking necessary and sufficient conditions to have a 2-generation flat subnormal completion on T η,κ . It was also proved in [10,Theorem 8.3] that the problem of finding a p-generation flat subnormal completion on T η,κ can be solved by using the well-known solutions of the subnormal completion problem for unilateral weighted shifts given in [6][7][8][9]16,19]. However, in most cases these solutions are not written explicitly in terms of initial data. This means that we have no explicit (i.e., written in terms of initial data) solution of the p-generation subnormal completion problem on T η,κ for p 2, even when κ = 1. It is the right moment to make the following important observation which is implicitly contained in the proof of [10, Theorem 7.1].
Proof. Applying [6, Theorems 5.1(iii) and 5.3(iii)] and arguing as in the proof of the implication (iii)⇒(iv) of [10, Theorem 7.1], we may assume without loss of generality that for every i ∈ J η , the measure ρ i appearing in the proof of the implication (iv)⇒(v) of [10, Theorem 7.1] satisfies the following condition: Next, by arguing as in the proof of the implication (iv)⇒(v) of [10, Theorem 7.1], we get a subnormal completion with the desired property.
It follows from Theorem 2.2 that if η ∈ N 2 and the 2-generation subnormal completion problem on T η,1 has a solution for a given data λ = , then one can always find a subnormal completion Sλ of λ on T η,1 with the property that each measure μλ i,1 is 1-or 2-atomic.

Preparatory Lemmas
One of the goals of this paper is to explore when initial data λ = {λ v } v∈V • η,1,2 on T η,1,2 admits a subnormal completion Sλ on T η,1 such that each of the measures μλ i,1 (see (2.1)) is 2-atomic (under present circumstances, the measures μλ i,1 may be taken to be 1-or 2-atomic due to Theorem 2.2). We first require some background on the Stampfli completion of three increasing weights to the weight sequence for a subnormal unilateral weighted shift (cf. [19]).
In [19] the author gives an explicit construction of the completion of an initial finite sequence of three weights x, y, z satisfying 0 < x < y < z to the  Figure 3. An illustration of a 2-generation SCP on T η,1 with λ 0 , {λ i,1 } i∈Jη and {λ i,2 } i∈Jη as given data sequence of positive weights {α n } ∞ n=0 for a (bounded, injective) subnormal unilateral weighted shift W α . This includes a construction of the associated Berger measure of W α (i.e., a unique Borel probability measure ξ on R + such that for any n ∈ N, γ n := α 2 0 . . . α 2 n−1 = R+ t n dξ(t)), which turns out to be 2-atomic. The completion sequence {α n } ∞ n=0 , which is called the Stampfli completion of (x, y, z), is customarily denoted by (x, y, z) ∧ ; it is known that the moment sequence {γ n } ∞ n=0 (with γ 0 = 1) for W α satisfies a recursion. Definition 3.1. The Berger measure of W α will be called the Berger measure associated with the Stampfli completion (x, y, z) ∧ of (x, y, z) and denoted by ξ x,y,z .
One may see [6][7][8] for an alternative approach to these same results. We note also that one may consider the cases 0 < x < y = z (yielding a 2-atomic measure with an atom at zero) and 0 < x = y = z, and this last case yields a completion whose Berger measure is 1-atomic.
Our technique will be to try to choose y i and z i , which will become respectivelyλ i,3 andλ i,4 of the completion Sλ , in such a way that the Berger measures associated to (x i , y i , z i ) ∧ will become the μλ i,1 and have good properties (and of course they will automatically be 2-atomic). Figure 3 summarizes the task and our notation.
We begin by calculating the first two "negative" moments of the Berger measure associated with the Stampfli completion (x, y, z) ∧ of (x, y, z). and Straightforward computations now yield (3.1) and (3.2).
Next we investigate the function f which comes from the expression appearing on the right-hand side of the second equality in (3.1).
Proof. It is a routine matter to verify that for any u ∈ (1, ∞), the function ϕ u is a well-defined bijection from (1, ∞) to (u, ∞). Clearly, the function f is well defined. Since (u, ϕ u (r)) ∈ Ω and f (u, ϕ u (r)) = r for all r ∈ (1, ∞) and It is a computation to show that (3.3) holds.
Corollary 3.4 below provides more information on the behavior of the first "negative" moment of the Berger measure appearing in Lemma 3.2.
Conversely, for any r ∈ (1, ∞), there exists a unique z ∈ (y, ∞) such that (3.5) holds; the number z is determined by the formula v = ϕ u (r) with u = y 2 Proof. Making the substitutions u = y 2 The expression on the right-hand side of the second equality in (3.2) leads to the function g which appears in a lemma below. The proof of this lemma follows from straightforward computations via Lemma 3.3. The details are left to the reader.

Lemma 3.5.
Let Ω be as in Lemma 3.3 and g be a real function on Ω given by For r ∈ (1, ∞), let h r be a real function on (1, ∞) defined by 1 where ϕ u is as in (3.4). Then the following statements hold for each r ∈ (1, ∞): Putting together the last three lemmas, we obtain the following crucial lemma.
(3.8) 1 In view of Lemma 3.3, the definition of hr is correct.

The 2-Generation SCP on T η ,1
We begin by solving the 2-generation subnormal completion problem on T η,1 with 2-atomic measures (the case η = ∞ is included).
given. Then the following statements are equivalent: Moreover, if (i) holds, then μλ i,1 ({0}) = 0 for all i ∈ J η and Proof. We concentrate on proving the case when η = ∞. If η < ∞, the proof simplifies. In particular, the statement (4.1) can be dropped.
(ii)⇒(i) Suppose now that (ii) holds. Applying the "moreover" part of Lemma 3.6 to x = x i := λ i,2 , r = r i and ϑ = ϑ i , we deduce that for every i ∈ N, there exists (y i , z i ) ∈ R 2 with x i < y i < z i such that (4.7) and (4.8) hold with μ i = ξ xi,yi,zi . According to the proof of Lemma 3.6, the sequences are constructed via the procedure (4.9). It follows from (4.1) and (4.10) that sup i∈N sup supp μ i < ∞. (4.16) Since in general R+ sdξ x,y,z (s) = x 2 whenever 0 < x < y < z, we get Putting together the conditions (4.16), (4.17), (4.18) and (4.19), and applying Lemma 2.1, we conclude that λ has a subnormal completion Sλ on T ∞,1 such that μλ i,1 = μ i = ξ xi,yi,zi for every i ∈ N. As a consequence, each μλ i,1 is 2atomic, which gives (i). Now we prove the "moreover" part. Suppose (i) is satisfied. The fact that μλ i,1 ({0}) = 0 for every i ∈ N is a direct consequence of [10, Theorem 3.5(i)]. Notice that the second inequality in (4.4) follows from (4.3) while the third and the fourth can be deduced from (4.2). Thus, it remains to prove the first inequality in (4.4). It follows from [10,Theorem 4.9 Then, by the "moreover" part of [10, Propositon 7.6], we see that 1 , which contradicts our assumption that each μλ i,1 is 2-atomic. Now we are ready to solve the 2-generation subnormal completion problem on T η,1 for η ∈ N 2 without any restrictions on the supports of the resulting measures μλ i,1 .
Proof. In view of Theorem 2.2, the statement (i) is equivalent to the fact that λ admits a subnormal completion Sλ on T η,1 such that each measure μλ i,1 is 1-or 2-atomic. Hence, in the proof of the implication (i)⇒(ii), we can define the partition (A, B) of J η as follows Since ϑ i > 1 for any i ∈ B, we see that If i ∈ A , then we set μ i = δ λ 2 i,2 . If i ∈ B , then we define μ i as in the proof of the implication (ii)⇒(i) of Theorem 4.1 with ϑ i = τ . As in that proof, we verify that   ∞) is arbitrary and the remaining weights are given by (see Fig. 4) We are looking for λ 0 for which λ admits a subnormal completion Sλ on T 2,1 such that μλ 1,1 is 1-atomic while μλ 2,1 is 2-atomic, and all these atoms are different. In view of [10,Theorem 4.9], any subnormal completion Sλ of the above λ on T 2,1 comes from compactly supported Borel probability measures μ 1 and μ 2 on R + which satisfy the following conditions (4.28) Let μ 1 and μ 2 be Borel probability measures on R + given by It is a matter of routine to verify that μ 1 and μ 2 with x = 2, y = 3 − √ 3 and z = 3 + √ 3 satisfy (4.26) and (4.27). Hence (4.28) holds if and only if λ 2 0 12 7 , meaning that for such λ 0 's the corresponding λ admits a subnormal completion on T 2,1 with the desired properties.
If η ∈ N 2 , Theorem 4.1 takes a much simpler form which is a direct consequence of Theorem 4.2.
given. Then the following statements are equivalent: Below we discuss the 2-generation SCP on T η,1 with 2-atomic measures under some constraints. Let us suppose temporarily that η ∈ N 2 . According to the "moreover" part of Theorem 4.1, if Sλ is a subnormal completion of λ on T η,1 such that each measure μλ i,1 is 2-atomic, then there exists i ∈ J η such that η j=1 λ 2 j,1 < λ 2 i,2 (use the fourth inequality in (4.4)). If the last inequality holds for all i ∈ J η , then the solution of the 2-generation subnormal completion problem on T η,1 with 2-atomic measures takes a simple form. Namely, the first inequality in (4.4), which is a necessary condition for solving the 2-generation subnormal completion problem on T η,1 , now becomes sufficient (see Corollary 4.5(ii) below). It is worth pointing out that in general Then the following statements are equivalent: Proof. (i)⇒(ii) This is a direct consequence of the first inequality in (4.4).
It is easily seen that with this choice of we obtain (4.30). Applying Proposition 4.4 completes the proof. appearing in (4.30). Finally, we require the value so obtained to be strictly less than 1 It is worth mentioning that the procedure described in Remark 4.6 has been applied in the proof of Corollary 4.5 (see (4.33)). Below we give a few more examples illustrating this procedure. ∞) are given. Then λ has a subnormal completion Sλ on T η,1 such that each measure μλ i,1 is 2-atomic provided any of the following conditions holds: in the cases (i), (ii) and (iii), respectively.
The above discussion can be applied to solve the 1-generation subnormal completion problem on T η,1 with 2-atomic measures. ∞) are given. Then the following statements are equivalent: Proof. (i)⇒(ii) Let Sλ be a subnormal completion of λ on T η,1 such that each measure μλ i,1 is 2-atomic. Then clearly this Sλ is a subnormal completion of λ ∪ {λ i,2 } i∈Jη on T η,1 . Hence, by the condition (4.4), (ii) is valid.
In concluding this section, we consider the 2-generation subnormal completion problem on T η,1 from the point of view of the condition (4.32).
Remark 4.9. Suppose that η ∈ N 2 . Let us consider the situation in which we are given initial data Notice that under the above assumption, if λ admits a subnormal completion on T η,1 , then it admits a 2-generation flat subnormal completion on T η,1 (see [10,Theorem 8.3]), and any 2-generation flat subnormal completion Sλ of λ on T η,1 has the property that μλ i,1 = μλ 1,1 for all i ∈ J η (apply (2.1) to u = e i,1 ). Below, we discuss a few cases related to the condition (4.32).
to a subnormal unilateral weighted shift, because it is well known that a subnormal unilateral weighted shift is hyponormal and so its weights must be monotonically increasing. Therefore, by [10, Theorem 8.3(iii)] there is no subnormal completion of λ on T η,1 .  1 , then the "moreover" part of Theorem 4.1 shows that there is no subnormal completion Sλ of λ on T η,1 with 2atomic measures μλ i,1 since we lack the third inequality in (4.4). Alternatively, we may use the "moreover" part of [10,Proposition 7.6]. 1 we may apply Corollary 4.5 to deduce that there is a subnormal completion Sλ of λ on T η,1 with 2-atomic measures μλ i,1 . 1 we may apply the condition (4.4) of Theorem 4.1 to deduce that there is no subnormal completion Sλ of λ on T η,1 with the μλ i,1 2-atomic. Alternatively, again using the "moreover" part of [10, Proposition 7.6] we may deduce that the measures μλ i,1 for any subnormal completion Sλ of λ on T η,1 (provided it exists) must be 1-atomic.

An Explicit Solution of the 2-Generation SCP on T 2,1
In this section we give necessary and sufficient conditions for solving the 2generation subnormal completion problem on T 2,1 with 2-atomic measures explicitly in terms of the initial data. In view of This task is complicated. The main difficulty comes from the requirement that the numbers r i should belong to the open interval (1, ∞). Even in the case of η = 2, there are three different formulas for the infimum depending on weights in question (see Theorem 5.2 below). However under some additional restrictive assumptions, the infimum of the above quadratic form can be computed explicitly. Proof. Suppose {r i } η i=1 ⊆ (0, ∞) and η i=1 a i r i = 1. It follows from the Cauchy-Schwarz inequality that Substituting {r i } η i=1 as in (5.3) shows that the inequality in (5.4) becomes equality. This proves (5.1). The condition (5.2) is a consequence of (5.1).
As shown below, a solution of the 2-generation subnormal completion problem on T 2,1 with 2-atomic measures can be written entirely in terms of the initial data. This is done by computing the infimum β(2). , a j = λ 2 1,j and b j = λ 2 2,j for j = 1, 2.
Proof. In view of Proposition 4.4, it is enough to compute β(2). If we replace r 1 and r 2 by x and y, respectively, then (4.31) with η = 2 takes the form and φ(x, y) = x a 1 a 2 + y b 1 b 2 for x, y ∈ (1, ∞).