Antinormality and m-isometry of shift on weighted trees

The shift operator and its various generalizations are amongst the most widely studied operators on a Hilbert space. In this paper, we characterize antinormal and m-isometric shift operator S on the Hilbert space L2(T,λ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^2({\mathcal {T}},\lambda ) $$\end{document} associated with a locally finite directed weighted tree T\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {T}}$$\end{document}. We also discuss interesting connections between antinormality and m-isometry of S.

weighted shift. We extend their result on the shift on a weighted directed tree. We also relate antinormality and m-isometry of the shift on a weighted directed tree. We now state various definitions and results relevant to our study. Definition 1.1 [8] An operator T ∈ B(H ) is said to be antinormal if d(T, N Observe that an operator T ∈ B(H ) is antinormal if and only if 0 ∈ B(H ) is the best normal approximation. We note that antinormality of an operator T ∈ B(H ) is necessary as well as sufficient for antinormality of T * . Definition 1.2 [7] An operator T ∈ B(H ) is said to be Fredholm operator if range(T ) is closed and dimension of both ker(T ) and ker(T * ) are finite. Definition 1.3 [2] The essential spectrum of an operator T ∈ B(H ) is defined as σ e (T ) = {α ∈ C : T − α I is not Fredholm}. Remark 1.6 [9] If i(T ) = 0 then T is not antinormal.
(2) m e (T ) = T . An element of V is called a vertex and an element of E is called an edge.
In this paper we restrict to the case when V is a countable set. In a directed graph (V, E), card({u ∈ V : (u, v) ∈ E}) and card({u ∈ V : (v, u) ∈ E}) are respectively called indegree of v and outdegree of v. A directed graph (V, E) is said to be locally finite if for every vertex in V both the indegree and the outdegree are finite. A directed graph (V, E) is said to be connected if its underlying graph is connected [17]. A finite sequence is called a directed tree if T does not have any directed circuits, T is connected and the indegree of every vertex in T is at most one.

Definition 1.11
In a directed tree T a vertex is called a root if its indegree is zero. If the outdegree of a vertex is zero then it is called a leaf.
We have adopted various notations introduced in [14]. Let T be a directed tree. For an edge (u, v) ∈ E, we define u = par(v) and call v as a child of u. For a vertex v ∈ V , par 2 (v) denotes par(par(v)) and for n ≥ 3 par n (v) = par(par n−1 (v)) whenever par n−1 (v) is not a root. For u ∈ V define chi(u) = {v ∈ V : u = par(v)} and for n ≥ 2, chi n (u) = {u ∈ V : par n (v) = u}. Also, for u ∈ V , γ (u) denotes the cardinality of chi(u).

Remark 1.12
Let u, v, w ∈ V . If chi(u) = {v}, then v is also denoted by chi(u). Similarly, if chi n (u) = {w} for some n ≥ 2, then w is also denoted by chi n (u). Definition 1.13 Two directed trees T = (V, E) andT = (Ṽ ,Ẽ) are said to be isomorphic if there exists a one-to-one correspondence ψ from V ontoṼ such that (u, v) ∈ E if and only if (ψ(u), ψ(v)) ∈Ẽ. Definition 1.14 Let G = (V, E) be a directed graph and W ⊆ V . Then, G \ W denotes the directed graph by removing all vertices of W from G and all edges whose at least one end vertex is in W .

Definition 1.15
Let T be a directed tree and λ = {λ v ∈ R : λ v > 0, v ∈ V } be a set. We call λ weight on the vertex set V . We denote by L 2 (T , λ) the space of complex valued functions f : This is a Hilbert space endow with the inner product Throughout this paper by "tree", we mean a locally finite directed tree. Also Z + T and Z − T , respectively, denote the trees (Z + , . For a tree T = (V, E) we define V leaves as the set of all leaves, V ≥2 as the set {u ∈ V : γ (u) ≥ 2} and chi(V ≥2 ) as the set v∈V ≥2 chi(v). Also, we assign a integer value δ T to the tree T as follows: , for every u ∈ V . And a sequence in is Definition 1. 16 Let T be a tree and λ be weight on V . Then the shift S : L 2 (T , λ) → L 2 (T , λ) is the operator defined as where f ∈ L 2 (T , λ).
Theorem 1.17 [14] Let T be a tree and λ be weight on V . The shift S : In either case, Hilbert adjoint of S [14]: The Hilbert adjoint S * of the shift operator S on L 2 (T , λ) is given by for each g ∈ L 2 (T , λ) and u ∈ V .

Antinormality of the shift operator
Prior to our investigations on antinormality of S, we state some results. From the definition of S and S * , we have ker(S) = { f ∈ L 2 (T , λ) : f (u) = 0, f or each u ∈ V \V leaves } and ker(S * ) = f ∈ L 2 (T , λ) : Hence, dim(ker(S)) is finite if and only if card(V leaves ) is finite. Further, if card(V leaves ) is finite then dim(ker(S)) = card(V leaves ).
We now consider the case when the tree has a root.
Preceding arguments also implies that if card(V ≥2 ) is finite and the tree has no root then ker( (2) In view of preceding definitions and the above lemma, the following statements follow easily.
Proof First consider the case when the tree T is finite. In this case, using induction on number of vertices we prove that the above inequality is an equality. Clearly equality holds for the tree with 1 and 2 vertices. Let the assertion be true for the tree with k(≥ 2) vertices. Now, consider a tree T with k + 1 vertices. Then T has a leaf v * . Consequently, the treeT = (Ṽ ,Ẽ) = T \{v * } has k vertices with two possibilities: Further, in this case, we have following two possibilities: Since equality holds for a tree with k vertices, therefore This implies Thus, Which gives Thus, equality also holds for a tree with k + 1 vertices. Hence, it holds for any finite tree. Now consider the case when the tree T is not finite. Let us split this case into two parts as follows: (1) If card(V ≥2 ) is not finite then card(chi(V ≥2 )) is also not finite. Thus in this case the result holds.  (b) T \W does not contain any linear tree which is isomorphic to Z + T (in this case T \W is isomorphic to Z − T )and card(W leaves ) = card(V leaves ). Thus, we have card(V leaves ) ≤ card(W leaves ). Therefore, Hence, the result holds for every tree T .

Remark 2.4
Above proposition can be visualized in the following trees. Figure 1 is a finite tree for which the inequality of Proposition 2.3 becomes an equality. In Figure 2 tree T is not finite but card(V ≥2 ) is finite. In this tree, if we take W as collection of all the vertices such that V ≥2 , chi(V ≥2 ) and V leaves are all contained in W , then T \W is collection of finite number of linear trees each of which is isomorphic to Z + T or Z − T .

Corollary 2.5 For the S :
Proof Consider the case when dim(ker(S)) and dim(ker(S * )) are both infinite then by definition, i(S) = 0. The Remark 2.2 precludes the possibility that dim(ker(S)) is infinite and dim(ker(S * )) is finite. If dim(ker(S)) is finite but dim(ker(S * )) is infinite then i(S) = −∞. Now if dim(ker(S)) and dim(ker(S * )) both are finite, then if the tree is without a root 0, otherwise.

Remark 2.6
If i(S) = 1, then tree is not finite and by Corollary 2.5 δ T = 0. This implies that tree has no root. Again by Corollary 2.5 card(V ≥2 ) is finite and card(V leaves ) = card(chi(V ≥2 ))−card(V ≥2 )+1. Now repeating the arguments of Proposition 2.3 above, it follows that there exists W ⊆ V such that T \W is isomorphic to Z − T . In all other cases, the i(S) ≤ 0.
We now have the following propositions.

Proposition 2.7 If i(S) < 0 then S is antinormal if and only if for every
Proof Let i(S) < 0. Then, V leaves must be finite. For if, V leaves is an infinite set then V ≥2 is also an infinite set. Hence both the dim(ker(S)) and dim(ker(S * )) are infinite. Consequently i(S) = 0, which is a contradiction. Now consider T = (S * S) Thus, T is a diagonal operator. Hence, essential minimum modulus of S is is not finite or range(T − α I ) is not closed} = inf{α ≥ 0 : α is a limit of some convergent sequence in }, is equivalent to has a convergent sequence and every convergent sequence in converge to S . As sup v∈V (v) = S , therefore S is antinormal if and only if for every > 0, there exists a finite subset U ⊂ V such that (v) − S < for each v ∈ V \U .

Proposition 2.8 If i(S) = 1 then S is antinormal if and only for every > 0, there exists a finite subset U
Proof Let i(S) = 1. Then by Remark 2.6, the tree T has no root and there exists a set of finite vertices W = {w 1 , w 2 , . . . , w l } such that the tree T \W is isomorphic to the Z − T . Hence any sequence in

. A simple computation shows that
Therefore, The above equation can also be expressed as Using the fact that W is finite set and T is diagonal operator on V \W , we get Combining above two propositions and Remark 1.6 we obtain the following result.

m-isometry of the shift operator
In this section, we investigate the m-isometry of the shift operator and relate it to its antinormality. From the definition of m-isometry we have the following result.

Remark 3.2 If
S is an m-isometry for any m ≥ 1, then the tree is without leaves.

Proposition 3.3 S is an m-isometry if and only if for every u ∈ V , there exists a polynomial P u of degree at
More generally for every n ≥ 0, we have Now using theory of difference equations, the auxiliary equation of the recursive Eq. (3.2) is Above equation has 1 as the only zero with multiplicity m. Therefore, the general solution of Eq. (3.2) is b u n = A 0 (1) n + A 1 n(1) n + A 2 n 2 (1) n + · · · + A m−1 n m−1 (1) n . Thus, where A 0 , A 1 , . . . , A m−1 are real constants which can be determined by m initial conditions Conversely, let u ∈ V . Then, there exists a polynomial P u of degree at most m − 1 such that b u n = v∈chi n (u) λ v = P u (n) for every n ∈ N ∪ {0}.Let P u (n) = C 0 + C 1 n + C 2 n 2 + · · · + C k n k , C k = 0 and 0 ≤ k ≤ m − 1.
This implies b u n = C 0 + C 1 n + C 2 n 2 + · · · + C k n k , C k = 0 and 0 ≤ k ≤ m − 1. Then, the sequence {b u n } ∞ n=0 can be recursively obtained by In particular, Thus, S is an m-isometry.

Corollary 3.4 Suppose S is an m-isometry. Then, S is strictly m-isometry if and only if max{deg(P
Now onwards, whenever we talk about a polynomial P u of a vertex u then it is presumed that S is an m-isometry for some m ≥ 1.  (1) Tree is rootless and has no leaves, (2) for every u ∈ V, γ (u) = 1 and (3) for every u ∈ V, λ u λ par(u) = β for some positive real number β.
Remark 3.7 Suppose S is an m-isometry and for a vertex u, deg(P u ) = k. Then, the following statements hold.
(1) For every n ∈ N and for every v ∈ chi n (u), deg(P v ) ≤ k. (2) For every n ∈ N there exists a v ∈ chi n (u), deg(P v ) = k.
Using the above remarks, we have the following result.

Proposition 3.8
If S is an m-isometry and there exists u 0 ∈ V such that P u 0 is constant polynomial, then there are infinitely many u in V such that P u is constant polynomial (in other words we have a sequence { (μ(n))} ∞ n=1 in with (μ(n)) = 1 for each n ≥ 1). We now investigate a connection between antinormality and m-isometry of S. Recall that, if S is antinormal then i(S) = 0. Hence, either i(S) = 1 or i(S) < 0. If i(S) = 1, then the tree must have a leaf. Consequently S can not be m-isometry for any m ≥ 1.

Theorem 3.9 If S is antinormal and i(S) = −k for some k ∈ N then S is an m-isometry for some m ≥ 2 if and only if it is an isometry.
Proof One way the implication is trivial. On the other hand, suppose that S is antinormal and an m-isometry for some m ≥ 2. As S is antinormal, there is a sequence { (μ(n))} ∞ n=1 in which converges to S . Also, since i(S) = −k, therefore card(V ≥2 ) is finite, and the tree has no leaves. This implies that we can choose a finite set W ⊆ V such that T \W is finite collection of trees each of which is isomorphic to Z + , therefore for each n ∈ N there exists a m n ∈ N such that P u 0 (m n +1) P u 0 (m n ) = (μ(n)) 2 . Further, choose a sequence (n l ) ∞ l=1 in N such that (m n l ) ∞ l=1 is an increasing sequence in N. Since (μ(n))(n l ) 2 converges to S 2 , so P u 0 (m n l +1) P u 0 (m n l ) converge to S 2 . But P u 0 is a non-zero polynomial, hence P u 0 (m n l +1) P u 0 (m n l ) converges to 1. Therefore, S = 1. Now consider the case thatT is isomorphic to Z − T with leaf w 0 . Since for each vertex w inT , P w (n) > 0 therefore using Remark 3.5 we get P par(w) (n) = P w (n − 1). Consequently P w 0 (−n) = λ par n (w 0 ) for every n ∈ N. Hence, S = 1. Thus S is an isometry.

Remark 3.10
It is worth noting that, if i(S) = −k, then tree must be locally finite. Therefore in the above theorem, we can relax the condition that tree is locally finite.
Finally, we give some examples and counterexamples. In the following Figs. 3, 4, 5 and 6, a "•" represents a vertex labeled by their corresponding weights and a line joining two vertices represent an edge with an arrow pointing from vertex par(u) to vertex u. Example 3.11 The shift operator on the following weighted tree is strictly 2-isometry but is not antinormal. This follows from Corollary 3.4 and Theorem 2.9.
Example 3.12 Consider the shift operator S on the following weighted tree. It can be readily seen from Theorem 2.9 that S is antinormal. However, since weights of vertices grow exponentially, so S is not an m-isometry for any m ∈ N.   13 The shift operator on the following weighted tree is antinormal as well as an isometry. This is because v∈chi(u) λ v = λ u , for every vertex u in the following weighted tree.
The following example shows that when i(S) = −∞ and S is antinormal as well as an m-isometry then it need not be an isometry.
Example 3.14 In the following figure a vertex u ∈ V is labeled by P u (n), where P u is the polynomial corresponding to u and q j denotes the jth prime number.
Finally, we obtain the following proposition followed by its corollary.

Proposition 3.15
If S is antinormal and strictly m-isometry for m ≥ 2 then for every u ∈ V deg(P u ) ≥ 1.