ON QUASI-AFFINE TRANSFORMS OF READ’S OPERATOR

THOMAS SCHLUMPRECHT AND VLADIMIR G. TROITSKY

Abstract. We show that C. J. Read’s example [Read85, Read86] of an operator T on ℓ_{1} which does not have any non-trivial invariant subspaces is not the adjoint of an operator on a predual of ℓ_{1}. Furthermore, we present a bounded diagonal operator D such that even though D^{−1} is unbounded but D^{−1}T D is a bounded operator on ℓ_{1} with invariant subspaces, and is adjoint to an operator on c_{0}.

1. Introduction

In this note we deal with the Invariant Subspace Problem, the problem of the existence of a closed non-trivial invariant subspace for a given bounded operator on a Banach space. The problem was solved in the positive for certain classes of operators (see [RR73, AAB98] for details), however in the mid-seventies P. Enflo [Enf76, Enf87] constructed an example of a continuous operator on a Banach space with no invariant subspaces, thus answering the Invariant Subspace Problem for general Banach spaces in the negative. In [Read85] C. J. Read presented an example of a bounded operator T on ℓ_{1} with no invariant subspace. Recently V. Lomonosov suggested that every adjoint operator has an invariant subspace. In the first part of this note we show that the Read operator T is not an adjoint of any bounded operator defined on some predual of ℓ_{1}.

Suppose that A has a non-trivial invariant (or a hyperinvariant) subspace, and suppose that B is similar to A, that is, B = CAC^{−1} for some invertible operator C. Clearly, B also has a non-trivial invariant (respectively hyperinvariant) subspace. Moreover, it is known (see [RR73, Theorem 6.19]) that if A has a hyperinvariant subspace and B is quasi-similar to A (that is, CA = BC and AD = DB, where C and D are two bounded one-to-one operators with dense range), then B also has a hyperinvariant subspace. To our knowledge it is still unknown whether or not A has a non-trivial invariant subspace if and only if B has a non-trivial invariant subspace, assuming A and B are quasi-similar.

Recall (cf. [Sz-NF68]) that an operator A is said to be a a quasi-affine transform of B if CA = BC, for some injective operator C with dense range. In the second part of this paper we construct an injective diagonal operator D on ℓ_{1} such that even though D^{−1} is unbounded, the operator S = D^{−1}T D (T being Read’s operator) is bounded and has an invariant subspace. Thus, we show that a quasi-affine transform of an operator with no non-trivial invariant subspace might have a non-trivial invariant subspace. Furthermore, S is the adjoint of a bounded operator on c_{0}.

Proc. Amer. Math. Soc., 131(2003), 1405-1413.

1

2000 Mathematics Subject Classification. Primary 47A15; Secondary 47B37.

The first author was supported by the NSF. Most of the work on the paper was done during the Workshop on linear analysis and probability at Texas A&M University, College Station.

Although we prove our statement for a specific choice of D, it is true for a much more general choice, and it seems to be true for any diagonal operator D that S = D^{−1}T D has a non-trivial invariant subspace, whenever S is an adjoint of an operator on c_{0}. More generally, the following question is of interest in view of the above-mentioned conjecture by V. Lomonosov.

Question. Does every quasi-affine transform of Read’s operator, which is an adjoint of an operator on c_{0}, have a non-trivial invariant subspace?

We introduce the following notations. Following [Read86] we denote by F the vector space of all eventually vanishing scalar sequences, and by (f_{i}) the standard unit vector basis of F. For an x = ∑ a_{i}f_{i} ∈ F, we define the support of x to be the set {i ∈ ℕ : a_{i} ≠ 0} and denote it by supp(x). The linear span of some subset A of a vector space is denoted by lin A.

2. Read’s operator is not adjoint

We begin by reminding the reader of the construction of the operator T in [Read85, Read86]. It depends on a strictly increasing sequence d = (a_{1}, b_{1}, a_{2}, b_{2}, …) of positive integers which has to be chosen to be sufficiently rapidly increasing. Also let a_{0} = 1, v_{0} = 0, and v_{n} = n(a_{n} + b_{n}) for n ⩾ 1.

Read’s operator T is defined by prescribing the orbit (e_{i})_{i⩾0} of the first basis element f_{0}.

Definition 2.1. There is a unique sequence (e_{i})_{i=0}^{∞} ⊂ F with the following properties: (0) f_{0} = e_{0}; (A) if integers r, n, and i satisfy 0 < r ⩽ n, i ∈ [0, v_{n−r}] + ra_{n}, we have

f_{i} = a_{n−r}(e_{i} − e_{i−ra_{n}});

(B) if integers r, n, and i satisfy 1 ⩽ r < n, i ∈ (ra_{n}+v_{n−r}, (r+1)a_{n}), (respectively, 1 ⩽ n, i ∈ (v_{n−1}, a_{n})), then

f_{i} = 2^{(h−i)/\sqrt{a_{n}}}e_{i}, where h = (r + \frac{1}{2})a_{n} (respectively, h = \frac{1}{2}a_{n});

(C) if integers r, n, and i satisfy 1 ⩽ r ⩽ n, i ∈ [r(a_{n} + b_{n}), na_{n} + rb_{n}], then

f_{i} = e_{i} − b_{n}e_{i−b_{n}};

(D) if integers r, n, and i satisfy 0 ⩽ r < n, i ∈ (na_{n} + rb_{n}, (r + 1)(a_{n} + b_{n})), then

f_{i} = 2^{(h−i)/\sqrt{b_{n}}}e_{i}, where h = (r + \frac{1}{2})b_{n}.

Indeed, since f_{i} = ∑_{j=0}^{i} λ_{ij}e_{j} for each i ⩾ 0 and λ_{ii} is always nonzero, this linear relation is invertible. Further,

lin{e_{i} | i = 1, …, n} = lin{f_{i} | i = 1, …, n} for every n ⩾ 0.

In particular, all e_{i} are linearly independent and also span F. Then Read defines T: F → F to be the unique linear map such that T e_{i} = e_{i+1}. Read proves that T can be extended to a bounded operator on ℓ_{1} with no invariant subspaces provided d increases sufficiently rapidly.

Proposition 2.2. T is not the adjoint of an operator S : X → X where X is a Banach space whose dual is isometric to ℓ_{1}.

Proof. Assume that our claim is not true. Then there is a local convex topology τ on ℓ_{1} so that

(a) τ is weaker than the norm topology of ℓ_{1};

(b) B(ℓ_{1}) is sequentially compact with respect to τ;

(c) if (x_{n}) ⊂ ℓ_{1} converges with respect to τ to x, then lim inf_{n→∞}∥x_{n}∥ ⩾ ∥x∥;

(d) T is continuous with respect to τ. Note that with respect to any predual X of ℓ_{1} the weak^{∗} topology has properties (a)–(d). Let s ∈ ℕ be fixed, and n > s. Then f_{(n−s)a_{n}} = a_{s}(e_{(n−s)a_{n}} − e_{0}) by (A) above. It follows that T^{v_{s}+1}f_{(n−s)a_{n}} = a_{s}(e_{(n−s)a_{n}+v_{s}+1} − e_{v_{s}+1}). Further, it follows from (B) that e_{(n−s)a_{n}+v_{s}+1} equals 2^{(1+v_{s}− \frac{1}{2} a_{n})/\sqrt{a_{n}}}f_{(n−s)a_{n}+v_{s}+1} and converges to zero in norm (and, hence, in τ) as n → ∞. Therefore

(1)  τ-lim_{n→∞} T^{v_{s}+1}f_{(n−s)a_{n}} = −a_{s}e_{v_{s}+1} = T^{v_{s}+1}(−a_{s}e_{0}).

Notice that T^{v_{s}+1} is τ-continuous and one-to-one because its null space is T -invariant. By sequential compactness of B(ℓ_{1}), the sequence f_{(n−s)a_{n}} must have a τ-convergent subsequence. Then, by (1), the limit point has to be −a_{s}e_{0}. Since that argument applies to any subsequence, we deduce that

(2)  τ-lim_{n→∞} f_{(n−s)a_{n}} = −a_{s}e_{0}.

Since ∥f_{(n−s)a_{n}}∥ = 1 for each n and s while ∥a_{s}e_{0}∥ = a_{s} > 1, this contradicts (2). □

Remark. The statement of the theorem remains valid if we consider an equivalent norm on ℓ_{1}. Indeed, suppose \frac{1}{K} |||·||| ⩽ ∥·∥ ⩽ K|||·|||. Then |||f_{(n−s)a_{n}}||| ⩽ K for each n and s, but since lim_{n→∞} a_{n} = ∞, we can choose a_{s} in (2) so that |||a_{s}e_{0}||| > K.

3. An adjoint operator with invariant subspaces of the form D^{−1}T D

Define a sequence of positive reals (d_{i}) as follows:

(3)  d_{i} = \begin{cases}\frac{1}{r} & if ra_{m} ⩽ i ⩽ ra_{m} + v_{m−r} for some 0 < r ⩽ m, \\ 1 & otherwise. \end{cases}

Let D be the diagonal operator with diagonal (d_{i}), that is, Df_{i} = d_{i}f_{i} for every i. Define S = D^{−1}T D. Clearly, S is defined on F. Once we write S in matrix form it will be clear that it is bounded on F and, therefore, can be extended to ℓ_{1}. Let ê_{i} = D^{−1}e_{i}, in particular ê_{0} = e_{0}. Then Sê_{i} = D^{−1}T e_{i} = ê_{i+1}, so that the sequence (ê_{i}) is the orbit of e_{0} under S.

Next, we examine Definition 2.1 to represent the f_{i}’s in terms of ê_{i}’s.

(ˆ0) f_{0} = e_{0} = ê_{0};

(Â) if i satisfies i ∈ [0, v_{n−r}] + ra_{n} for some 0 < r ⩽ n, then

f_{i} = d_{i}D^{−1}f_{i} = d_{i}D^{−1}(a_{n−r}(e_{i} − e_{i−ra_{n}})) = \frac{a_{n−r}}{r} (ê_{i} − ê_{i−ra_{n}});

(B̂) if integers r, n, and i satisfy 1 ⩽ r < n, i ∈ (ra_{n}+v_{n−r}, (r+1)a_{n}), (respectively, 1 ⩽ n, i ∈ (v_{n−1}, a_{n})), then

f_{i} = d_{i}D^{−1}f_{i} = 2^{(h−i)/\sqrt{a_{n}}} ê_{i}, where h = (r + \frac{1}{2})a_{n} (respectively, h = \frac{1}{2}a_{n}); (Ĉ) if integers r, n, and i satisfy 1 ⩽ r ⩽ n, i ∈ [r(a_{n} + b_{n}), na_{n} + rb_{n}], then

f_{i} = d_{i}D^{−1}f_{i} = ê_{i} − b_{n}ê_{i−b_{n}};

(D̂) if integers r, n, and i satisfy 0 ⩽ r < n, i ∈ (na_{n} + rb_{n}, (r + 1)(a_{n} + b_{n})), then

f_{i} = d_{i}D^{−1}f_{i} = 2^{(h−i)/\sqrt{b_{n}}} ê_{i}, where h = (r + \frac{1}{2})b_{n}.

We see that it differs from Definition 2.1 only in case (Â). Now we can actually write the matrix of S:

Sf_{i} = \begin{cases}2^{(1− \frac{1}{2} a_{1})/\sqrt{a_{1}}}f_{1} & if i = 0 \\ f_{i+1} & if i ∈ [0, v_{n−r}) + ra_{n}, \\ & with r = 1, 2, …, n \\ f_{i+1} & if i ∈ [r(a_{n} + b_{n}), na_{n} + rb_{n}), \\ & with r = 1, 2, …, n \\ 2^{1/\sqrt{a_{n}}}f_{i+1} & if i ∈ (ra_{n} + v_{n−r}, (r + 1)a_{n} − 1), \\ & with r = 1, 2, …, n − 1 \\ & or i ∈ (v_{n−1}, a_{n} − 1) \\ 2^{1/\sqrt{b_{n}}}f_{i+1} & if i∈(na_{n}+rb_{n},(r+1)(a_{n}+b_{n})−1) \\ & with r = 0, 1, …, n − 1 \\ \frac{a_{n−r}}{r} (ε_{1}f_{i+1}−ε_{2}f_{v_{n−r}+1}) & if i = ra_{n} + v_{n−r}, \\ where & with r = 1, 2, …, n \\ ε_{2} =2^{(1+v_{n−r}− \frac{1}{2} a_{n−r+1})/\sqrt{a_{n−r+1}}} \\ ε_{1} =2^{(1+v_{n−r}− \frac{1}{2} a_{n})/\sqrt{a_{n}}} & if r < n and \\ ε_{1} =2^{(1+na_{n}− \frac{1}{2} b_{n})/\sqrt{b_{n}}} & if r = n, \\ 2^{(1− \frac{1}{2} a_{n})/\sqrt{a_{n}}}[f_{0} + \frac{(r+1)f_{i+1}}{a_{n−r−1}} ] & if i = (r + 1)a_{n} − 1 \\ & with r = 0, 1, …, n − 1 \\ ε_{1}f_{i+1} − b_{n}ε_{2}f_{i+1−b_{n}} & if i = na_{n} + rb_{n} \\ where & with r = 1, 2, …, n \\ ε_{2} =2^{(1+na_{n}− \frac{1}{2} b_{n})/\sqrt{b_{n}}} \\ ε_{1} = 2^{(1+na_{n}− \frac{1}{2} b_{n})/\sqrt{b_{n}}} & if r < n, and \\ ε_{1} = 2^{(v_{n}+1− \frac{1}{2} a_{n+1})/\sqrt{a_{n+1}}} & if r = n \\ 2^{−((r+1)a_{n}+ \frac{1}{2} b_{n}−1)/\sqrt{b_{n}}} & if i = (r + 1)(a_{n} + b_{n}) − 1 \\ ·[∑_{j=0}^{r} b_{n}^{j}f_{i−jb_{n}+1} \\ +b_{n}^{r+1}(f_{0} + \frac{(r+1)f_{(r+1)an}}{a_{n−r−1}} )] & with r = 0, 1, …, n − 1 \end{cases}

Inspecting the matrix line by line we observe that, assuming (a_{n}) and (b_{n}) are increasing sufficiently rapidly, it follows that ∥S∥ ⩽ 2. Again by inspecting each line of the matrix, we deduce that if f_{j}^{∗} is the j-th coordinate functional on ℓ_{1}, j ⩾ 0, it follows that lim_{i→∞} f_{j}^{∗}(S(f_{i})) = 0. In other words, the rows of the matrix converge to zero. Therefore S is the adjoint of a linear bounded operator on c_{0}.

Theorem 3.1. S has a non-trivial closed invariant subspace.

We shall show that S has an invariant subspace by producing a vector x_{∞} such that the linear span of the orbit of x_{∞} stays away from e_{0}, hence its closure is a non-trivial S-invariant subspace.

We will introduce the following notations.

First we choose two sequences of positive integers (m_{i}) and (r_{i}) as follows. Let m_{0} ⩾ 2 be arbitrary, put r_{0} = 1. Once m_{i} and r_{i} are defined, choose r_{i+1} ∈ ℕ so that

(4)  r_{i+1} ∈ [a_{m_{i}−1} · max_{ℓ⩽v_{m_{i}−1}}∥ê_{ℓ}∥, 1 + a_{m_{i}−1} · max_{ℓ⩽v_{m_{i}−1}}∥ê_{ℓ}∥]

and let

(5)  m_{i+1} = m_{i} + r_{i+1}.

Define an increasing sequence (j_{i}) of positive integers inductively: pick any

(6)  j_{0} ∈ [r_{0}a_{m_{0}}, r_{0}a_{m_{0}} + v_{m_{0}−r_{0}}],

and once j_{i} is defined, put

(7)  j_{i+1} = j_{i} + r_{i}b_{m_{i}} + r_{i+1}a_{m_{i+1}}.

Finally, for each i ⩾ 0 define

(8)  p_{i} = ∏_{k=0}^{i} b_{m_{k}}^{−r_{k}},

(9)  z_{i} = f_{j_{i}+r_{i}b_{m_{i}}} + b_{m_{i}}f_{j_{i}+(r_{i}−1)b_{m_{i}}} + ⋯ + b_{m_{i}}^{r_{i}−1}f_{j_{i}+b_{m_{i}}} + \frac{r_{i+1}f_{j_{i+1}}}{a_{m_{i}}} ,
(10)  x_{i} = p_{i−1}ê_{j_{i}}.

We note the following easy-to-prove properties for our choices.

Proposition 3.2. For each i ⩾ 0 the following statements hold:

(a) j_{i} ∈ [r_{i}a_{m_{i}}, r_{i}a_{m_{i}} + v_{m_{i}−r_{i}}];

(b) x_{i+1} = x_{i} + p_{i}z_{i}, and thus x_{i} = ê_{j_{0}} + ∑_{k=0}^{i−1} p_{k}z_{k};

(c) if i and i+ℓ both belong to [ra_{n}, ra_{n} +v_{n−r}] or both belong to [r(a_{n} +b_{n}), na_{n} + rb_{n}], then S^{ℓ}f_{i} = f_{i+ℓ};

(d) if ℓ < m_{i}a_{m_{i}} − j_{i}, then min supp S^{ℓ}z_{k} ⩾ j_{i} + b_{m_{i}} whenever k ⩾ i.

Proof. (a) The proof is by induction. For i = 0 the required inclusion follows from the choice of j_{0}, and if this condition holds for j_{i}, then

j_{i+1} = j_{i} + r_{i}b_{m_{i}} + r_{i+1}a_{m_{i+1}}
∈ [r_{i}a_{m_{i}} + r_{i}b_{m_{i}} + r_{i+1}a_{m_{i+1}}, r_{i}a_{m_{i}} + v_{m_{i}−r_{i}} + r_{i}b_{m_{i}} + r_{i+1}a_{m_{i+1}}]
⊆ [r_{i+1}a_{m_{i+1}}, r_{i+1}a_{m_{i+1}} + m_{i}(a_{m_{i}} + b_{m_{i}})] = [r_{i+1}a_{m_{i+1}}, r_{i+1}a_{m_{i+1}} + v_{m_{i}}].

(b) First note that by using (D̂) we obtain for a i ∈ [r(a_{n} + b_{n}), na_{n} + rb_{n}], with 1 ⩽ r ⩽ n in ℕ, that

(11)  ê_{i} = b_{n}ê_{i−b_{n}} + f_{i}
= b_{n}^{2}ê_{i−2b_{n}} + b_{n}f_{i−b_{n}} + f_{i}
⋮
= b_{n}^{r}ê_{i−rb_{n}} + b_{n}^{r−1}f_{i−(r−1)b_{n}} + … + b_{n}f_{i−b_{n}} + f_{i}.

Note that j_{i} + r_{i}b_{m_{i}} ∈ [r_{i}(a_{m_{i}} + b_{m_{i}}), m_{i}a_{m_{i}} + r_{i}b_{m_{i}}]. By using first (Â) and then (11) we obtain

ê_{j_{i+1}} = ê_{j_{i}+r_{i}b_{m_{i}}+r_{i+1}a_{m_{i}}}

= ê_{j_{i}+r_{i}b_{m_{i}}} + \frac{r_{i+1}}{a_{m_{i}}} f_{j_{i}+r_{i}b_{m_{i}}+r_{i+1}a_{m_{i}}}

= b_{m_{i}}^{r_{i}} ê_{j_{i}} + b_{m_{i}}^{r_{i}−1}f_{j_{i}+b_{m_{i}}} + … + b_{m_{i}}f_{j_{i}+(r_{i}−1)b_{m_{i}}} + \frac{r_{i+1}}{a_{m_{i}}} f_{j_{i}+r_{i}b_{m_{i}}+r_{i+1}a_{m_{i}}}

= b_{m_{i}}^{r_{i}} ê_{j_{i}} + z_{i}.

Thus, x_{i+1} = p_{i}ê_{j_{i+1}} = p_{i−1}ê_{j_{i}} + p_{i}z_{i} = x_{i} + p_{i}z_{i}.

(c) If i and i + ℓ are both in [ra_{n}, ra_{n} + v_{n−r}], it follows from (Â) that

S^{ℓ}(f_{i}) = \frac{a_{n−r}}{r} S^{ℓ}(ê_{i} − ê_{i−ra_{n}}) = \frac{a_{n−r}}{r} (ê_{i+ℓ} − ê_{i−ra_{n}+ℓ}) = f_{i+ℓ}.

The second part of (c) can be deduced in a similar way using (Ĉ).

(d) First note that for k ⩾ i it follows that (recall that m_{k} ⩾ m_{0} ⩾ 2)

m_{k}a_{m_{k}} − j_{k} > (m_{k} − r_{k} − 1)a_{m_{k}} = (m_{k−1} − 1)a_{m_{k}} ⩾ m_{k−1}a_{m_{k−1}} − j_{k−1}.

We can therefore assume that k = i. Furthermore, note that for any 1 ⩽ r ⩽ r_{i} it follows that

r(a_{m_{i}} + b_{m_{i}}) ⩽ j_{i} + rb_{m_{i}} ⩽ j_{i} + rb_{m_{i}} + ℓ ⩽ m_{i}a_{m_{i}} + rb_{m_{i}}

and

r_{i+1}a_{m_{i+1}} ⩽ j_{i+1} ⩽ j_{i+1} + ℓ ⩽ j_{i+1} + m_{i}a_{m_{i}} − j_{i}
= r_{i+1}a_{m_{i+1}} + r_{i}b_{m_{i}} + m_{i}a_{m_{i}}
⩽ r_{i+1}a_{m_{i+1}} + v_{m_{i}}
= r_{i+1}a_{m_{i+1}} + v_{m_{i+1}−r_{i+1}}.

Therefore the claim follows from the definition of z_{i}, (9) and part (c). □

Notice that

∥z_{i}∥ = 1 + b_{m_{i}} + b_{m_{i}}^{2} + ⋯ + b_{m_{i}}^{r_{i}−1} + \frac{r_{i+1}}{a_{m_{i}}} ⩽ m_{i}b_{m_{i}}^{r_{i}−1} + \frac{r_{i+1}}{a_{m_{i}}} .

Further, since p_{i} ⩽ \frac{1}{b_{m_{i}}^{r_{i}}}, we have

∥p_{i}z_{i}∥ ⩽ \frac{m_{i}}{b_{m_{i}}} + \frac{r_{i+1}}{a_{m_{i}}b_{m_{i}}^{r_{i}}} .

The series ∑_{i=0}^{∞} \frac{m_{i}}{b_{m_{i}}} converges because (b_{i}) increases sufficiently rapidly. Secondly, it follows from the definition of (r_{i}) that

a_{m_{i}}^{−1}r_{i+1} ⩽ a_{m_{i}}^{−1}[1 + a_{m_{i}−1} · max_{ℓ⩽v_{m_{i}−1}}∥ê_{ℓ}∥].

Thus, again since (b_{i}) is increasing fast enough, it follows that the series

∑_{i=0}^{∞} \frac{r_{i+1}}{a_{m_{i}}b_{m_{i}}^{r_{i}}}

converges. Therefore the ∑_{i=0}^{∞} p_{i}z_{i} converges, and the following definition is justified.

Definition 3.3. Define x_{∞} = lim_{i} x_{i} = lim_{i} p_{i−1}ê_{j_{i}} = ê_{j_{0}} + ∑_{i=0}^{∞} p_{i}z_{i}.

Now we can state and prove the key result for proving Theorem 3.1.

Lemma 3.4. There exists a constant C > 0 such that dist(y, e_{0}) ⩾ C for every i and every vector of the form y = ∑_{j=j_{i}}^{m_{i}a_{m_{i}}} γ_{j}ê_{j}.

Proof. Let C = inf{dist(y, e_{0}) | y = ∑_{j=j_{0}}^{m_{0}a_{m_{0}}} γ_{j}ê_{j}}. Since the infimum is taken over a finite-dimensional set, it must be attained at some y_{0}. However since all ê_{j} are linear independent, it follows that C = dist(y_{0}, e_{0}) > 0.

We shall prove the statement of the lemma by induction on i. The way we defined C guarantees that the base of the induction holds. Suppose y = ∑_{j=j_{i}}^{m_{i}a_{m_{i}}} γ_{j}ê_{j}. Write y = y_{1} + y_{2} + y_{3}, where

y_{1} = ∑_{j=j_{i}}^{r_{i}a_{m_{i}}+v_{m_{i−1}}} γ_{j}ê_{j}, y_{2} = ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}} γ_{j}ê_{j}, and y_{3} = ∑_{r=r_{i}}^{m_{i}−1} ∑_{j=ra_{m_{i}}+v_{m_{i}−r}+1}^{(r+1)a_{m_{i}}−1} γ_{j}ê_{j}.

Notice that by (B̂)

y_{3} = ∑_{r=r_{i}}^{m_{i}−1} ∑_{j=ra_{m_{i}}+v_{m_{i}−r}+1}^{(r+1)a_{m_{i}}−1} γ_{j}2^{−(r+ \frac{1}{2} −j)/\sqrt{a_{m_{i}}}}f_{j},

so that supp y_{3} ⊆ ⋃_{r=r_{i}}^{m_{i}−1}(ra_{m_{i}} + v_{m_{i}−r}, (r + 1)a_{m_{i}}). Furthermore, using (Â), we write y_{2} = y′_{2} + y′′_{2} where

y′_{2} = ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}} γ_{j}ê_{j−ra_{m_{i}}} = ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=0}^{v_{m_{i}−r}} γ_{j+ra_{m_{i}}} ê_{j}

and y′′_{2} = ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}} \frac{γ_{j}r}{a_{m_{i}−r}} f_{j}.

Therefore,

supp(y_{1} + y_{2}) ⊆ [0, r_{i}a_{m_{i}} + v_{m_{i−1}}] ∪ ⋃_{r=r_{i}+1}^{m_{i}}[ra_{m_{i}}, ra_{m_{i}} + v_{m_{i}−r_{i}}].

One observes that the vectors y_{1} + y_{2} and y_{3} have disjoint supports; it follows that dist(y, e_{0}) ⩾ dist(y_{1} + y_{2}, e_{0}).

Furthermore,

∥y′_{2}∥ = ∥ ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}} γ_{j}ê_{j−ra_{m_{i}}} ∥ ⩽ ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}}|γ_{j}| · max_{k⩽v_{m_{i−1}−1}}∥ê_{k}∥.

By choice of (r_{i}) (4), we have max_{k⩽v_{m_{i−1}−1}}∥ê_{k}∥ ⩽ \frac{r_{i}}{a_{m_{i}−r_{i}−1}} ⩽ \frac{r}{a_{m_{i}−r}} when r_{i} < r ⩽ m_{i}.

This yields

∥y′_{2}∥ ⩽ ∥ ∑_{r=r_{i}+1}^{m_{i}} ∑_{j=ra_{m_{i}}}^{ra_{m_{i}}+v_{m_{i}−r}} \frac{γ_{j}r}{a_{m_{i}−r}} f_{j}∥ = ∥y′′_{2}∥.

Since the support of y′′_{2} is disjoint from that of y_{1} + y′_{2} and doesn’t contain 0, we have

dist(y_{1}, e_{0}) ⩽ dist(y_{1} + y′_{2}, e_{0}) + ∥y′_{2}∥
= dist(y_{1} + y′_{2} + y′′_{2}, e_{0}) − ∥y′′_{2}∥ + ∥y′_{2}∥
⩽ dist(y_{1} + y_{2}, e_{0}) ⩽ dist(y, e_{0}).

It is left to show that dist(y_{1}, e_{0}) ⩾ C. Since j_{i} ⩾ r_{i}a_{m_{i}}, it follows from (Â) that y_{1} = y′_{1} + y′′_{1} where

y′_{1} = ∑_{j=j_{i}}^{r_{i}a_{m_{i}}+v_{m_{i−1}}} γ_{j}ê_{j−r_{i}a_{m_{i}}} and y′′_{1} = ∑_{j=j_{i}}^{r_{i}a_{m_{i}}+v_{m_{i−1}}} \frac{γ_{j}r}{a_{m_{i}−r_{i}}} f_{j}.

Since j_{i} = j_{i−1} + r_{i−1}b_{m_{i−1}} + r_{i}a_{m_{i}}, we have y′_{1} = ∑_{j=j_{i−1}+r_{i−1}b_{m_{i−1}}}^{v_{m_{i−1}}} β_{j}ê_{j}, where β_{j} = γ_{j+r_{i}a_{m_{i}}}. In particular this means, that supp y′_{1} ⊆ [0, v_{m_{i−1}}], while min supp y′′_{1} ⩾ j_{i} ⩾ r_{i}a_{m_{i}}. Thus, the supports are disjoint, which yields dist(y_{1}, e_{0}) ⩾ dist(y′_{1}, e_{0}).

Split the index set of y′_{1} into two disjoint subsets: let

A=[j_{i−1}+r_{i−1}b_{m_{i−1}}, v_{m_{i−1}}] ∩ ⋃_{r=r_{i−1}}^{m_{i−1}}(m_{i−1}a_{m_{i−1}} +rb_{m_{i−1}}, (r+1)(a_{m_{i−1}} +b_{m_{i−1}})),

B =[j_{i−1}+r_{i−1}b_{m_{i−1}}, v_{m_{i−1}}] ∩ ⋃_{r=r_{i−1}}^{m_{i−1}}[r(a_{m_{i−1}} +b_{m_{i−1}}), m_{i−1}a_{m_{i−1}} +rb_{m_{i−1}}].

Write y′_{1} = z_{a} + z_{b} where z_{a} = ∑_{j∈A} β_{j}ê_{j} and z_{b} = ∑_{j∈B} β_{j}ê_{j}. For j ∈ A we have

ê_{j} = 2^{((r+1/2)b_{m_{i−1}}−j)/\sqrt{b_{m_{i−1}}}}f_{j}, so that supp z_{a} ⊆ A. In view of (11) we can write z_{b} = z′_{b} + z′′_{b}, where

z′_{b} = ∑_{j∈B} ∑_{k=0}^{r−1} β_{j}b_{m_{i−1}}^{k}f_{j−kb_{m_{i−1}}} and z′′_{b} = ∑_{j∈B} β_{j}b_{m_{i−1}}^{r}ê_{j−rb_{m_{i−1}}}.

We first note that supp z′_{b} ⊆ B and determine the support of z′′_{b} as follows. If j ∈ B, then j ⩾ j_{i−1} + r_{i−1}b_{m_{i−1}} and j ∈ [r(a_{m_{i−1}} + b_{m_{i−1}}), m_{i−1}a_{m_{i−1}} + rb_{m_{i−1}}] for some r ∈ [r_{i−1}, m_{i−1}]. If r = r_{i−1}, then j − rb_{m_{i−1}} ⩾ j_{i−1}. If r > r_{i−1}, then j − rb_{m_{i−1}} ⩾ ra_{m_{i−1}} > r_{i−1}a_{m_{i−1}} + v_{m_{i−2}} ⩾ j_{i−1} by (7). We see that z′′_{b} is a linear combination of ê_{j}’s with j_{i−1} ⩽ j ⩽ m_{i−1}a_{m_{i−1}}. Hence its support is contained in [0, m_{i−1}a_{m_{i−1}}] and, therefore, is disjoint from that of z_{a} and z′_{b}. It follows that dist(y, e_{0}) ⩾ dist(y′_{1}, e_{0}) ⩾ dist(z′′_{b}, e_{0}). Finally, dist(z′′_{b}, e_{0}) ⩾ C by the induction hypothesis. □

Proof of Theorem 3.1. We will prove that the linear span of the orbit of x_{∞} under S is at least distance C from e_{0}, hence its closure is a non-trivial invariant subspace for S. Consider a linear combination ∑_{ℓ=0}^{N} α_{ℓ}S^{ℓ}x_{∞}. It follows from (7) that the sequence (m_{i}a_{m_{i}} − j_{i}) is unbounded, so that N < m_{i}a_{m_{i}} − j_{i} for some i ⩾ 0. Recall that x_{∞} = x_{i} + ∑_{k=i}^{∞} p_{k}z_{k}; then

∑_{ℓ=0}^{N} α_{ℓ}S^{ℓ}x_{∞} = ∑_{s=0}^{N} α_{ℓ}S^{ℓ}x_{i} + ∑_{ℓ=0}^{N} ∑_{k=i}^{∞} α_{ℓ}S^{ℓ}(p_{k}z_{k}).

Notice that the two sums have disjoint supports, and the support of the second one does not contain 0. Indeed, since x_{i} = p_{i−1}ê_{j_{i}} then S^{ℓ}x_{i} = p_{i−1}ê_{j_{i}+ℓ} for ℓ = 1, …, N. Furthermore,

j_{i} ⩽ j_{i} + ℓ ⩽ j_{i} + N < j_{i} + (m_{i}a_{m_{i}} − j_{i}) = m_{i}a_{m_{i}}.

It follows that ∑_{ℓ=0}^{N} S^{ℓ}x_{i} is a linear combination of ê_{j}’s with j_{i} ⩽ j ⩽ m_{i}a_{m_{i}}. In particular, its support is contained in [0, m_{i}a_{m_{i}}]. On the other hand, Proposition 3.2 (d) implies that

min supp(∑_{ℓ=0}^{N} ∑_{k=i}^{∞} S^{ℓ}(p_{k}z_{k})) ⩾ j_{i} + b_{m_{i}}.

Therefore, by Lemma 3.4

dist(∑_{ℓ=0}^{N} S^{ℓ}x_{∞}, e_{0}) ⩾ dist(∑_{ℓ=0}^{N} S^{ℓ}x_{i}, e_{0}) ⩾ C.

□

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Department of Mathematics, Texas A&M University, College Station, TX 77843. USA.

E-mail address: schlump@math.tamu.edu

Department of Mathematics, University of Alberta, Edmonton, AB T6G 2G1. Canada.

E-mail address: vtroitsky@math.ualberta.ca