STRICTLY SEMI-TRANSITIVE OPERATOR ALGEBRAS

H. P. ROSENTHAL AND V. G. TROITSKY

Abstract. An algebra 𝔄 of operators on a Banach space X is called strictly semitransitive if for all non-zero x, y ∈ X there exists an operator A ∈ 𝔄 such that Ax = y or Ay = x. We show that if 𝔄 is norm-closed and strictly semi-transitive, then every 𝔄-invariant linear subspace is norm-closed. Moreover, Lat 𝔄 is totally and well ordered by reverse inclusion. If X is complex and 𝔄 is transitive and strictly semitransitive, then 𝔄 is WOT-dense in L(X). It is also shown that if 𝔄 is an operator algebra on a complex Banach space with no invariant operator ranges, then 𝔄 is WOT-dense in L(X). This generalizes a similar result for Hilbert spaces proved by Foiaş.

1. Introduction

Throughout the paper X is a real or complex Banach space, B_{X} stands for the closed unit ball of X. Let 𝔄 be a subalgebra of L(X), the algebra of all bounded linear operators on X. For x ∈ X the orbit of x under 𝔄 is defined by 𝔄x = {Ax | A ∈ 𝔄}. We say that x_{0} ∈ X is cyclic for 𝔄 if 𝔄x_{0} is dense in X; x_{0} is called strictly cyclic if 𝔄x_{0} = X. The algebra 𝔄 is called transitive (strictly transitive) if every non-zero element of X is cyclic (respectively, strictly cyclic). We write Lat 𝔄 for the lattice of all closed 𝔄-invariant subspaces of X. It is easy to see that 𝔄 is transitive if and only if Lat 𝔄 = {{0}, X}. Similarly, 𝔄 is strictly transitive if and only if it has no invariant linear subspaces other than {0} and X (by a linear subspace we mean a subspace which is not necessarily closed). A remarkable result due independently to Yood [Yood49] and Rickart [Ric50] yields that if 𝔄 is a strictly transitive algebra of operators on a complex Banach space, then it is WOT-dense in L(X). We refer the reader to [RR73] for a detailed introduction to transitive algebras.

An algebra 𝔄 is called semi-transitive if for all non-zero x, y ∈ X and ε > 0 there exists an operator A ∈ 𝔄 such that ∥Ax − y∥ < ε or ∥Ay − x∥ < ε. We say that 𝔄

Journal of Operator Theory, 53(2), 2005, 315–329.

1 2 H. P. ROSENTHAL, V. G. TROITSKY

Date: August 16, 2003.

2000 Mathematics Subject Classification. Primary: 47A15; Secondary: 47L10.

Key words and phrases. Strictly semi-transitive algebra, transitive algebra, invariant subspace, operator range.

The first author was partially supported by NSF Grant DMS 0070547.

is strictly semi-transitive if for all non-zero x, y ∈ X there is an operator A ∈ 𝔄 with Ax = y or Ay = x. One can show that a unital algebra 𝔄 is semi-transitive if and only if it is unicellular, that is, Lat 𝔄 is totally ordered by inclusion (see [RR73] for a treatment of unicellular algebras). Similarly, 𝔄 is strictly semi-transitive if and only if all the 𝔄-invariant linear subspaces are totally ordered by inclusion. In Section 3 we investigate the structure of strictly semi-transitive algebras. We show that if 𝔄 is norm closed and strictly semi-transitive, then every 𝔄-invariant linear subspace is norm-closed and Lat 𝔄 is well-ordered by reverse inclusion. We deduce that a transitive strictly semi-transitive operator algebra on a complex Banach space is WOT-dense in L(X). In the special case of CSL-algebras on a Hilbert space similar results were obtained in [Hop01, DHP01].

Foiaş proved in [Foi72] that if 𝔄 is a WOT-closed algebra of operators on a Hilbert space H, and 𝔄 has no invariant operator ranges, then 𝔄 = L(H). In Section 4 we generalize the result of Foiaş to complex Banach spaces using a version of Arveson’s Lemma of [Arv67]. For Y ⊆ X, we say that Y is an injective operator range if Y = Range T⃗ for an injective bounded operator T⃗ ∈ L(M, X), where M is a closed subspace of X^{n} for some n ⩾ 1. We show that if 𝔄 has no invariant injective operator ranges in X, then 𝔄 is WOT-dense in L(X).

2. Preliminaries

We first formulate a version of the standard lemma for the Open Mapping Theorem, and then use it to deduce a results concerning strictly cyclic vectors.

Lemma 2.1 (Open Mapping Lemma). Suppose that Y is a normed space, and T: X → Y is a bounded operator. Assume that there exist K > 0 and 0 < ε < 1 such that

(2.1)  B_{Y} ⊆ KT (B_{X}) + εB_{Y} .

Then T is a surjective open map, and Y is complete.

Remark 2.2. In particular, (2.1) is satisfied if \overline{T (B_{X})} has non-empty interior.

Proof. Fix y ∈ B_{Y}, choose x_{1} ∈ KB_{X} such that ∥y − T x_{1}∥ ⩽ ε. Denote y_{1} = y − T x_{1}, then y_{1} ∈ εB_{Y}, so that there exists x_{2} ∈ εKB_{X} such that ∥y_{1} − T x_{2}∥ ⩽ ε^{2}, let y_{2} = y_{1} − T x_{2}. Continuing, we obtain sequences (x_{n}) in X and (y_{n}) in Y so that for all n we have

(2.2)  ∥x_{n}∥ ⩽ ε^{n−1}K, and

(2.3)  ∥y_{n}∥ ⩽ ε^{n}, and y_{n} = y − T ∑_{j=1}^{n} x_{j}.

Since X is complete, (2.2) yields that ∑_{j=1}^{∞} x_{j} converges to an element x of X. Since T is continuous, T x = y by (2.3). Thus T is surjective. Moreover, it follows from (2.2) that ∥x∥ ⩽ M where M = K ∑_{j=0}^{∞} ε^{j} = \frac{K}{1−ε}, so that B_{Y} ⊆ MT (B_{X}), hence T is open. Now let Z = X/ Null T and let π: X → Z be the canonical quotient map. Then Z is complete and there is a one-to-one operator T̃: Z → Y such that T̃π = T. Since T is open, the set T̃(U) = T (π^{−1}U) is open whenever U ⊆ Z is open, so that T̃^{−1} is continuous. Hence, T̃ is an isomorphism between Z and Y, so that Y is complete. □

Proposition 2.3. Let 𝔄 be norm closed and x_{0} ∈ X with x_{0} ≠ 0. The following are equivalent.

(i) x_{0} is strictly cyclic.

(ii) There is a constant C > 0 so that for all y ∈ B_{X} there is an operator A ∈ 𝔄

with ∥A∥ ⩽ C such that Ax_{0} = y. (iii) There are C > 0 and 0 < ε < 1 so that for all y ∈ B_{X} there is an operator

A ∈ 𝔄 with ∥A∥ ⩽ C and ∥y − Ax_{0}∥ < ε.

Proof. Define T: 𝔄 → X by T (A) = Ax_{0}, then T is a bounded linear operator from 𝔄 to X. Since 𝔄 is norm closed, it is a Banach space. Thus, (i)⇒(ii) follows immediately from the Open Mapping Theorem. The implication (ii)⇒(iii) is trivial. Finally, (iii)⇒(i) follows from the Open Mapping Lemma. □

Corollary 2.4. If 𝔄 is norm-closed, then the set of strictly cyclic vectors for 𝔄 is open.

Proof. Let x_{0} be a strictly cyclic vector for 𝔄, and choose C as in Proposition 2.3(ii). Let 0 < δ < 1/C, and suppose that x ∈ X with ∥x − x_{0}∥ < δ. Now if y ∈ B_{X}, we choose A ∈ 𝔄 with ∥A∥ ⩽ C and Ax_{0} = y. But then ∥Ax−y∥ = ∥Ax−Ax_{0}∥ ⩽ Cδ < 1. Hence x is strictly transitive by Proposition 2.3(iii). □

This yields an alternate proof of a result due to A. Lambert.

Corollary 2.5 ([Lam71]). If X is a complex Banach space and 𝔄 is a transitive operator algebra on X with a strictly cyclic vector, then 𝔄 is WOT-dense in L(X).

Proof. We may assume that 𝔄 is norm-closed by replacing 𝔄 by \overline{𝔄}. Let x_{0} be a strictly cyclic vector for 𝔄. By Corollary 2.4 there exists δ > 0 such that ∥x − x_{0}∥ < δ implies x is strictly cyclic. Let y ∈ X with y ≠ 0. Since 𝔄 is transitive we may choose A ∈ 𝔄 with ∥Ay − x_{0}∥ < δ. But then Ay is strictly cyclic, and so, of course, y is also strictly cyclic. Thus, 𝔄 is strictly transitive and so is WOT-dense in L(X). □

The next result shows that transitive algebras always have operators which are almost zero on prescribed vectors.

Proposition 2.6. Let 𝔄 be a transitive operator algebra on a complex Banach space X, then for all x ∈ X and ε > 0 there is an A ∈ 𝔄 with ∥A∥ = 1 and ∥Ax∥ < ε.

Proof. Without loss of generality, 𝔄 is WOT-closed. Indeed, suppose that the conclusion holds for \overline{𝔄}^{WOT}, show that it also holds for 𝔄. Fix x ∈ X and 0 < ε < 1, and let δ = ε/4. There exists A ∈ \overline{𝔄}^{WOT} with ∥A∥ = 1 and ∥Ax∥ < δ. Then ∥Ay∥ ⩾ 1 − δ for some y ∈ X with ∥y∥ = 1. Since \overline{𝔄}^{WOT} = \overline{𝔄}^{SOT}, there exists B ∈ 𝔄 such that ∥(A − B)x∥ < δ and ∥(A − B)y∥ < δ. Then ∥Bx∥ < 2δ and ∥By∥ ⩾ 1 − 2δ, so that ∥B∥ ⩾ 1 − 2δ. Let C = \frac{B}{∥B∥}, then ∥Cx∥ < \frac{2δ}{1−2δ} < ε.

If 𝔄 = L(X) then the conclusion is trivially satisfied. Suppose that 𝔄 is a WOTclosed proper subalgebra of L(X). Were the conclusion false, we could choose x of norm one and ε > 0 so that ∥Ax∥ > ε whenever A ∈ 𝔄 and ∥A∥ = 1. But then the operator T: 𝔄 → X defined by T (A) = Ax for A ∈ 𝔄 is an isomorphism. In particular, 𝔄x is norm-closed being isomorphic to 𝔄. Then 𝔄x = X since 𝔄 is transitive. This contradicts Corollary 2.5. □

The following generalization of this result is due to Jiaosheng Jiang, and we are grateful to him for giving us permission to present his proof here.

Theorem 2.7 (J. Jiang). Let 𝔄 be a commutative transitive operator algebra on a complex Banach space. Then for all x_{0}, x_{1}, … x_{n} ∈ X and ε > 0 there exists an operator A ∈ 𝔄 with ∥A∥ = 1 and ∥Ax_{i}∥ < ε for all 0 ⩽ i ⩽ n.

Proof. We may assume without loss of generality that x_{0} ≠ 0. By the transitivity of 𝔄, for each 1 ⩽ i ⩽ n we may choose B_{i} ∈ 𝔄 so that ∥B_{i}x_{0} − x_{i}∥ ⩽ ε/2. Let δ = min\{ε, \frac{ε}{2·max_{i}∥B_{i}∥}\}. By Proposition 2.6, we may choose A ∈ 𝔄 with ∥A∥ = 1 and ∥Ax_{0}∥ < δ. Then ∥Ax_{0}∥ < ε and for all 1 ⩽ i ⩽ n,

∥Ax_{i}∥ ⩽ ∥A(x_{i} − B_{i}x_{0})∥ + ∥AB_{i}x_{0}∥ ⩽ ∥x_{i} − B_{i}x_{0}∥ + ∥B_{i}Ax_{0}∥ ⩽ ε/2 + ∥B_{i}∥ · δ < ε.

□

STRICTLY SEMI-TRANSITIVE OPERATOR ALGEBRAS 5

We do not know if the conclusion of this proposition holds for general transitive algebras. The following result gives a partial answer. Recall that 𝔄 is said to be ntransitive for some n ⩾ 1 if every linearly independent n-tuple in X^{n} is cyclic for the algebra 𝔄^{(n)} = {A ⊕ ⋯ ⊕ A | A ∈ 𝔄}.

Proposition 2.8. Let 𝔄 be an operator algebra on a complex Banach space. If 𝔄 is n-transitive for some n ⩾ 1, then for all x_{1}, …, x_{n} ∈ X and ε > 0 there exists an operator A ∈ 𝔄 with ∥A∥ = 1 and ∥Ax_{i}∥ < ε for all i.

Proof. As in the proof of Proposition 2.6, we may assume that 𝔄 is a WOT-closed proper subalgebra of L(X). Were the conclusion false, we could find x_{1}, …, x_{n} ∈ X and ε > 0 such that max_{1⩽i⩽n}∥Ax_{i}∥ ⩾ ε for all A ∈ 𝔄 with ∥A∥ = 1. Without loss of generality, we may assume that x_{1}, …, x_{k} are linearly independent for some k ⩽ n, and x_{j} ∈ span{x_{1}, …, x_{k}} whenever k < j ⩽ n. It follows that there is a C > 1 such that

max_{1⩽i⩽n}∥Ax_{i}∥ ⩽ C · max_{1⩽i⩽k}∥Ax_{i}∥

whenever A ∈ 𝔄 with ∥A∥ = 1, so that max_{1⩽i⩽k}∥Ax_{i}∥ ⩾ ε/C. Then the map T: 𝔄 → X^{k} defined by T (A) = (Ax_{1}, …, Ax_{k}) for A ∈ 𝔄 is an isomorphism, and hence T (𝔄) is closed. Since 𝔄 is n-transitive, it is also k-transitive. Then T (𝔄) = X^{k} because x_{1}, …, x_{k} are linearly independent. This implies, in particular, that x_{1} is strictly cyclic for 𝔄, and, since 𝔄 is transitive, 𝔄 is WOT-dense in L(X) by Corollary 2.5, a contradiction. □

Recall that T ∈ L(X, Y) is called a semi-embedding if it is one-to-one and T (B_{X}) is closed. The following result is certainly known, and we include a proof for the sake of completeness.

Proposition 2.9. Let Y be a normed space, and T: Y → X a semi-embedding. Then Y is complete.

Proof. Let S: Range(T) → Y be the inverse of T, and let Ỹ be the completion of Y. Then T may be extended to a continuous operator T̃: Ỹ → X. Since T (B_{Y}) is closed in X we have T̃(BỸ) = T (B_{Y}), and hence Range T̃ = Range T. Consider the operator ST̃: Ỹ → Y. If y ∈ Y then ST̃y = ST y = y. Moreover, ST̃(BỸ) = S(T (B_{Y})) = B_{Y} ⊆ BỸ. It follows that ST̃ is a bounded projection of Ỹ onto Y, and thus Y is complemented in Ỹ, so that it is closed in Ỹ, hence Ỹ = Y. □

6 H. P. ROSENTHAL, V. G. TROITSKY

We are now prepared for the final result of this section. For W ⊆ X and ε > 0, let W_{ε} = {x ∈ X | dist(x, W) ⩽ ε}. The continuity of the map x ↦ dist(x, W) from X to ℝ implies that W_{ε} is a closed set.

Lemma 2.10. Let X be a Banach space and W a bounded circled closed convex subset of X with void interior. Then for any ε > 0 the set W_{ε} contains no ball of radius larger than ε.

Proof. Let Y = span W = ⋃_{n=1}^{∞} nW. Then Y equipped with the norm given by the Minkowski functional of W is a normed space. Let T: Y → X be the inclusion operator. Then T is bounded, one-to-one, and T (B_{Y}) = W. Hence, T is a semiembedding, so by Proposition 2.9 Y is complete.

Clearly, W_{ε} is also circled and convex. Suppose that W_{ε} contains a ball of radius r > ε centered at x_{0}. Since W_{ε} is circled, it also contains the ball of radius r centered at −x_{0}, and since W_{ε} is convex, it contains the ball of radius r centered at the origin. Indeed, if ∥z∥ ⩽ r then x_{0} +z and −x_{0} +z belong to W_{ε}, so that z = \frac{(x_{0}+z)+(−x_{0}+z)}{2} ∈ W_{ε}. Now let ε < ε′ < r. Then rB_{X} ⊆ W_{ε} ⊆ W + ε′B_{X}, thus, B_{X} ⊆ \frac{1}{r}W + \frac{ε′}{r} B_{X}. Since \frac{ε′}{r} < 1, we conclude by by Lemma 2.1 that T is surjective. This yields X = Y = ⋃_{n=1}^{∞} nW, which contradicts the Baire Category Theorem since W has void interior. □

3. The structure of strictly semi-transitive algebras

A subalgebra 𝔄 of L(X) is called strictly semi-transitive if for all non-zero x, y ∈ X there exists an operator A ∈ 𝔄 so that Ax = y or Ay = x.

Proposition 3.1. Let 𝔄 be a unital subalgebra of L(X). Then the following are equivalent.

(i) 𝔄 is strictly semi-transitive; (ii) All the 𝔄-invariant linear subspaces of X are totally ordered by inclusion; (iii) All the orbits are totally ordered by inclusion.

Proof. (i)⇒(ii). Let Y and Z be two 𝔄-invariant linear subspaces of X, and suppose that Z ⊈ Y. Show that Y ⊆ Z. Choose z ∈ Z \ Y and y ∈ Y. There is no T ∈ 𝔄 with T y = z, for otherwise z ∈ Y since Y is 𝔄-invariant. Hence there must be a T ∈ 𝔄 with T z = y, and so y ∈ Z since Z is 𝔄-invariant.

(ii)⇒(iii) is trivial.

(iii)⇒(i). Let x, y ∈ X. Then 𝔄x ⊆ 𝔄y or 𝔄y ⊆ 𝔄x. Since 𝔄 is unital, it follows that x ∈ 𝔄y or y ∈ 𝔄x, hence 𝔄 is strictly semi-transitive. □

Remark 3.2. The implications (i)⇒(ii)⇒(iii) are still valid for non-unital algebras.

The next result is fundamental for our development. Recall that a set is residual if its complement is of first category; residual subsets of X are dense by Baire Category Theorem.

Lemma 3.3. Let (W_{n}) be a sequence of circled bounded closed convex sets, each with void interior, such that W_{n} ⊆ W_{n+1} and ⋃_{n=1}^{∞} W_{n} is dense in X, and let (ε_{n}) and (δ_{n}) be sequences of positive reals tending to zero. Let

G_{1} = {x ∈ X | for all λ > 0, dist(λx, W_{n}) > ε_{n} for infinitely many n},
G_{2} = {x ∈ X | for all λ > 0, dist(λx, W_{n}) < δ_{n} for infinitely many n}.

Then G_{1} ∩ G_{2} is residual.

Proof. Note that x ∈ G_{1} if and only if

(3.1)  ∀ λ > 0 ∀ m ∈ ℕ ∃ n ⩾ m such that dist(λx, W_{n}) > ε_{n}.

In particular,

(3.2)  ∀ k ∈ ℕ ∀ m ∈ ℕ ∃ n ⩾ m such that dist(\frac{1}{k}x, W_{n}) > ε_{n}.

We claim that (3.1) and (3.2) are, in fact, equivalent. Indeed, suppose that (3.2) holds. Given λ > 0 and m ∈ ℕ, find a positive integer k such that \frac{1}{k} < λ, then there exists n ⩾ m such that dist(\frac{1}{k}x, W_{n}) > ε_{n}. Since W_{n} is convex and circled, it follows that dist(λx, W_{n}) ⩾ dist(\frac{1}{k}x, W_{n}) > ε_{n}.

It follows from (3.2) that x ∉ G_{1} if and only if there exist k, m ∈ ℕ such that \frac{1}{k}x ∈ (W_{n})_{ε_{n}} for all n ⩾ m. Therefore,

∼G_{1} = ⋃_{k,m=1}^{∞} ⋂_{n=m}^{∞} k(W_{n})_{ε_{n}} = ⋃_{k,m=1}^{∞} k ⋂_{n=m}^{∞}(W_{n})_{ε_{n}}.

It follows from Lemma 2.10 that (W_{n})_{ε_{n}} contains no balls of radius larger than ε_{n}. Therefore, for every m the set ⋂_{n=m}^{∞}(W_{n})_{ε_{n}} has void interior, so that ∼ G_{1} is of first category.

Similarly, x ∈ G_{2} if and only if

(3.3)  ∀ k ∈ ℕ ∀ m ∈ ℕ ∃ n ⩾ m such that dist(kx, W_{n}) < δ_{n}.

Indeed, if x ∈ G_{2} then (3.3) is satisfied trivially. Conversely, suppose that x satisfies (3.3), let λ > 0 and m ∈ ℕ. Let k ∈ ℕ be such that λ ⩽ k. By (3.3) there exists n ⩾ m such that dist(kx, W_{n}) < δ_{n}. But since W_{n} is convex and circled we have dist(λx, W_{n}) ⩽ dist(kx, W_{n}) < δ_{n}. Note that dist(kx, W_{n}) < δ_{n} is equivalent to x ∈ \frac{1}{k}(W_{n} + δ_{n}B_{X}^{∘}), where B_{X}^{∘} stands for the open unit ball of X. Thus,

G_{2} = ⋂_{k,m=1}^{∞} \frac{1}{k} ⋃_{n=m}^{∞}(W_{n} + δ_{n}B_{X}^{∘}).

Since ⋃_{n=m}^{∞}(W_{n} + δ_{n}B_{X}^{∘}) is open and dense in X for every m, then by Baire Theorem G_{2} is residual. Since G_{1} and G_{2} are both residual, so is their intersection. □

The next result is our main lemma. It shows that if W_{n}’s are as in the previous lemma, then there are always u and v in X which “see” the W_{n}’s very differently.

Lemma 3.4. Let (W_{n}) be a nested increasing sequence of closed convex bounded circled sets, each with void interior, so that Y = ⋃_{n=1}^{∞} W_{n} is dense in X. Then given any u ∈ X \ Y, there is a vector v ∈ X \ Y so that

(3.4)  lim sup_{n→∞} \frac{dist(v, W_{n})}{dist(u, W_{n})} = ∞ and lim inf_{n→∞} \frac{dist(v, W_{n})}{dist(u, W_{n})} = 0.

Proof. Note that Y is a linear subspace of X of first category, so Y ≠ X by the Baire Category Theorem. Let τ_{n} = dist(u, W_{n}), then the sequence (τ_{n}) is decreasing and lim_{n→∞} τ_{n} = 0. Now let G_{1} and G_{2} be as in Lemma 3.3 with ε_{n} = \sqrt{τ_{n}} and δ_{n} = τ_{n}/n, then G_{1} ∩ G_{2} is residual, hence non-empty. Let v ∈ G_{1} ∩ G_{2}, then

\frac{dist(v, W_{n})}{dist(u, W_{n})} > \frac{ε_{n}}{τ_{n}} = \frac{1}{\sqrt{τ_{n}}}

for infinitely many n, proving the first equality in (3.4), and

\frac{dist(v, W_{n})}{dist(u, W_{n})} < \frac{δ_{n}}{τ_{n}} = \frac{1}{n}

for infinitely many n, proving the second equality. □

Remark 3.5. Notice that every v in G_{1}∩G_{2}, which is a residual set in X, satisfies (3.4). Furthermore, by definition the sets G_{1} and G_{2} are positively homogeneous, that is, with every vector they contain the entire ray passing through that vector, so that G_{1} ∩ G_{2} is also positively homogeneous.

We are now ready for our first main structural result for strictly semi-transitive algebras.

Theorem 3.6. Let 𝔄 be norm-closed and strictly semi-transitive. Then 𝔄x is closed for every x ∈ X.

STRICTLY SEMI-TRANSITIVE OPERATOR ALGEBRAS 9

Proof. Let X′ = \overline{𝔄x}. Define T: 𝔄 → X′ by T (A) = Ax for A ∈ 𝔄, and let

W_{n} = n\overline{T (B_{𝔄})} = \overline{\{Ax | A ∈ 𝔄, ∥A∥ ⩽ n\}}

for all n. Of course, W_{n} = nW_{1} and W_{n} ⊂ X′ for every n. It suffices to prove that W_{1} has non-void relative interior in X′ because in this case Remark 2.2 would imply that X′ = T (𝔄) = 𝔄x. Suppose not. Then, of course, each W_{n} is a closed convex bounded circled set, with void relative interior in X′. Pick any u in X′ \ ⋃_{n=1}^{∞} W_{n}. It follows from Lemma 3.4 that we may choose v in X′ \ ⋃_{n=1}^{∞} W_{n} satisfying (3.4). Since 𝔄 is strictly semi-transitive, there is an operator A ∈ 𝔄 so that Au = v or Av = u. It follows from Remark 3.5 that by scaling v and A we can also assume without loss of generality that ∥A∥ = 1. Then A(W_{n}) ⊆ W_{n} for every n. If Au = v, then

dist(v, W_{n}) ⩽ dist(Au, AW_{n}) ⩽ dist(u, W_{n}),

for every n, but this contradicts the first equality in (3.4). Similarly, the second equality in (3.4) is violated if Av = u. This contradiction completes the proof. □

Theorem 3.7. Let 𝔄 be a strictly semi-transitive operator algebra on a complex Banach space X. If 𝔄 is transitive, then it is WOT-dense in L(X).

Proof. We may assume without loss of generality that 𝔄 is norm-closed. But then 𝔄 is strictly transitive by Theorem 3.6, hence WOT-dense in L(X). □

We now arrive at a second basic structural result.

Theorem 3.8. Let 𝔄 is strictly semi-transitive. Then Lat 𝔄 is well-ordered by reverse inclusion. That is, every non-empty subset of Lat 𝔄 has a maximal element.

Proof. We may without loss of generality assume that 𝔄 is norm closed. Suppose that W is a non-empty subset of Lat 𝔄 with no maximal element. Then we can find an infinite sequence X_{1} ⫋ X_{2} ⫋ X_{3} ⫋ … in W. Let Y = ⋃_{n=1}^{∞} X_{n}. Then Y is not closed; indeed each X_{n} is nowhere dense in the induced topology of \overline{Y}, so that Y is of first category in \overline{Y}. Furthermore, \overline{Y} is 𝔄-invariant (i.e., \overline{Y} ∈ Lat 𝔄), and 𝔄 reduced to \overline{Y} is again a strictly semi-transitive algebra. Set W_{n} = nB_{X_{n}} for all n. Again by Lemma 3.4 we may choose u, v ∈ \overline{Y} \ Y satisfying (3.4). Again, choose A ∈ 𝔄 of norm one so that Au = v or Av = u. Then A(W_{n}) ⊆ W_{n} because X_{n} is an invariant subspace for 𝔄. Now the rest of the argument for Theorem 3.6 yields a contradiction. □

The next result yields the surprising fact that every 𝔄-invariant linear subspace is in Lat 𝔄.

Corollary 3.9. Let 𝔄 be unital, norm-closed, and strictly semi-transitive. Then every 𝔄-invariant linear subspace Y of X is closed. Furthermore, Y = 𝔄x for some x ∈ X.

Proof. Let W be the set of all orbits contained in Y, i.e., W = {𝔄x | 𝔄x ⊆ Y }. Note that W ⊆ Lat 𝔄 by Theorem 3.6. Then Theorem 3.8 yields that W has a maximal element, say 𝔄x. We claim that Y = 𝔄x. Suppose not, then there exists y ∈ Y \𝔄x. Then 𝔄x ⊆ 𝔄y ⊆ Y by Proposition 3.1, and 𝔄x ≠ 𝔄y because y ∈ 𝔄y\𝔄x. But this contradicts the maximality of 𝔄x. Hence Y = 𝔄x, so that Y is closed by Theorem 3.6. □

The following result may alternatively be deduced from [RR73, Theorem 4.4] and Theorem 3.6. We present a simple direct proof of it.

Corollary 3.10. Let 𝔄 be unital, norm-closed, and strictly semi-transitive. Then 𝔄 has (a residual set of) strictly cyclic vectors.

Proof. If 𝔄 is transitive, then every non-zero vector is strictly cyclic by Theorem 3.6. Otherwise, 𝔄 has a maximal proper closed invariant subspace Y by Theorem 3.8. Then X \ Y is residual. But if x ∈ X \ Y, then 𝔄x = X by Theorem 3.6 and the maximality of Y. □

We conclude this section with an explicit description of Lat 𝔄 when 𝔄 a strictly semitransitive algebra acting on a separable space, and a description of all “full” strictly semi-transitive algebras.

Corollary 3.11. Let X be separable and 𝔄 be strictly semi-transitive. Then there exists a countable ordinal η and a family of closed subspaces (X_{α})_{α⩽η} such that

(i) X_{0} = X, X_{η} = {0}, and X_{α+1} ⫋ X_{α} for all α < η;

(ii) Lat 𝔄 = {X_{α} | α ⩽ η}.

Proof. This holds for arbitrary X by Theorem 3.8, except that η need not be countable. However, if X is assumed to be separable, it follows that η must be countable, otherwise X would have an uncountable strictly decreasing chain of closed subspaces, which is impossible. □

Remark 3.12. Assume that η and X_{α}’s are as in the above statement. It follows that if α_{1} < α_{2} < … with α_{n} < η for all n

and α_{n} → α, then X_{α} = ⋂_{n=1}^{∞} X_{α_{n}}. (3.5)

Indeed, let Y = ⋂_{n=1}^{∞} X_{α_{n}}, then Y ∈ Lat 𝔄, hence Y = X_{β} for some β. But then β ⩾ α_{n} for all n, hence β ⩾ α. This yields X_{β} ⊆ X_{α}, but X_{α} ⊆ X_{α_{n}} for all n, so that

⋂_{n=1}^{∞} X_{α_{n}} = X_{β} ⊆ X_{α} ⊆ ⋂_{n=1}^{∞} X_{α_{n}},

hence X_{β} = X_{α}.

The above remark motivates our final result of this section, which is a partial converse to Theorem 3.8.

Proposition 3.13. Let X be a separable Banach space, η a countable ordinal with η > 1, and (X_{α})_{α⩽η} a family of closed subspaces of X satisfying (i) of Corollary 3.11 and (3.5). Let

𝔄 = {T ∈ L(X) | T X_{α} ⊆ X_{α} for all α ⩽ η}.

Then 𝔄 is strictly semi-transitive and Lat 𝔄 = {X_{α} | α ⩽ η}.

Proof. Let x, y ∈ X \ {0}. Put

α = sup{γ ⩽ η | x ∈ X_{γ}}  and  β = sup{γ ⩽ η | y ∈ X_{γ}}

It follows from (3.5) that x ∈ X_{α} and y ∈ X_{β}. Without loss of generality, α ⩽ β and, of course, β < η. Then by the Hahn Banach Theorem, there is an f ∈ X^{∗} with X_{β+1} ⊆ ker f and f(x) = 1. Define T ∈ L(X) by T u = f(u)y for all u ∈ X. Evidently T x = y. If γ ⩾ β + 1 then X_{γ} ⊆ X_{β+1} ⊆ ker f, so T X_{γ} = {0}, and if γ ⩽ β, then X_{β} ⊆ X_{γ} and, of course, T X = [y] ⊆ X_{γ}. In either case, T X_{γ} ⊆ X_{γ}. Thus, T ∈ 𝔄, and we have shown that 𝔄 is strictly semi-transitive.

It follows by definition that {X_{α} | α ⩽ η} ⊆ Lat 𝔄, so we must prove that, conversely, if Y ∈ Lat 𝔄, then Y = X_{α} for some α. Assume Y ≠ {0} or X, and let α be the greatest such that Y ⊆ X_{α}. We claim that Y = X_{α}. Suppose this were false. Now since Y ⫋ X_{α}, it follows from Proposition 3.1 and the definition of α that X_{α+1} ⫋ Y. Now we may choose z ∈ X_{α} \ Y and y ∈ Y \ X_{α+1}. But then by our initial discussion, since neither z nor y are in X_{α+1}, yet both belong to X_{α}, there is a T ∈ 𝔄 with T y = z. But then T Y ⊈ Y, contradicting he assumption that Y ∈ Lat 𝔄. □

Example 3.14. Let X = ℓ_{p} for 1 ⩽ p < +∞ and let (e_{i}) be the standard basis of ℓ_{p}. Let 𝔄 be the set of all the bounded operators A on ℓ_{p} such that

(i) the matrix of A is lower triangular, that is, Ae_{n} ∈ [e_{i}]_{i=n}^{∞} for every n, and (ii) the matrix of A contains only finitely many non-zero columns or, equivalently,

there exists a positive integer n such that Ae_{i} = 0 for all i ⩾ n.

12 H. P. ROSENTHAL, V. G. TROITSKY

It is easy to see that 𝔄 is an algebra. Show that 𝔄 is strictly semi-transitive. Given x = (x_{i}) and y = (y_{i}) in ℓ_{p}, let x_{k} be the first non-zero component of x and y_{m} the first non-zero component of y. Without loss of generality, k ⩽ m. Define an operator A as follows: Ae_{k} = \frac{1}{x_{k}} y and Ae_{i} = 0 for all i ≠ k. Then A ∈ 𝔄 and Ax = y.

Notice that this algebra is not norm closed. However, it follows that the algebra of all lower-triangular compact operators and the algebra of all lower-triangular bounded operators are strictly semi-transitive.

On the other hand, consider the algebra of all upper-triangular bounded operators. The linear subspace of all sequences with finitely many non-zero entries is invariant under the algebra, but it is not closed in ℓ_{p}. It follows from Corollary 3.9 that the upper-triangular algebra is not strictly semi-transitive. However, it is easy to see that this algebra is unicellular, hence semi-transitive.

4. Operator ranges in Banach spaces

A subspace Y of a Hilbert space H is termed an operator range by Foiaş [Foi72] if there exists an operator T ∈ L(H) with Y = T (H). We generalize this to arbitrary Banach spaces X as follows.

Definition 4.1. Let X be a Banach space and Y a linear subspace of X. Then Y is called an operator range if there is a closed subspace M ⊆ X^{n} for some n ⩾ 1 and a bounded linear operator T⃗: M → X with Y = Range T⃗. In this case, Y is called an operator range of order n. Finally, Y is called an injective operator range if T⃗ may be chosen one-to-one.

Of course, every closed subspace of X is an operator range of order one. It is proved in [Foi72] (see also [RR73, Theorem 8.9]) that if 𝔄 is a WOT-closed algebra of operators on a Hilbert space with no non-trivial invariant operator ranges, then 𝔄 = L(H). The main result of this section generalizes this fact.

Theorem 4.2. Let 𝔄 be a subalgebra of L(X) for a complex Banach space X, and n ⩾ 1. Then either 𝔄 has a non-trivial invariant injective operator range of order n, or 𝔄 is n-transitive.

Note that 𝔄 is WOT-dense in L(X) if and only if 𝔄 is n-transitive for all n, c.f. [RR73, Theorem 7.1]. This yields the following consequence.

Corollary 4.3. Let 𝔄 be a subalgebra of L(X) for a complex Banach space X such that 𝔄 has no non-trivial invariant injective operator ranges. Then 𝔄 is WOT-dense in L(X).

STRICTLY SEMI-TRANSITIVE OPERATOR ALGEBRAS 13

Remark 4.4. It is easily seen that every operator range in a Hilbert space is injective of order one. It seems we need the concept of operator range of order n, however, to obtain a general result. Also, we don’t know if every operator range in a Banach space is also an injective operator range. This question was posed by P. Rosenthal.

Before proving the theorem, we need some preliminary results, which also hold in real Banach spaces. We start with the following simple consequences of the definition of an operator range, given in [Foi72] in a Hilbert space case.

Proposition 4.5. Let Y_{1} and Y_{2} be operator ranges in X. Then Y_{1} + Y_{2} and Y_{1} ∩ Y_{2} are operator ranges.

Proof. For i = 1, 2 choose n_{i}, M_{i} a closed subspace of X^{n_{i}}, and T_{i}: M_{i} → X bounded linear with Y_{i} = Range T_{i}. Now if

M = M_{1} ⊕ M_{2} ⊆ X^{n_{1}} ⊕ X^{n_{2}} = X^{n_{1}+n_{2}},

and T = T_{1} ⊕ T_{2}, then T (M) = Y_{1} + Y_{2}, hence Y_{1} + Y_{2} is an operator range. The argument for Y_{1} ∩ Y_{2} is not quite obvious. Let

W = {(w_{1}, w_{2}) ∈ M | T_{1}w_{1} = T_{2}w_{2}}.

Evidently, W is a linear subspace of M. If (u_{n}, v_{n}) → (u, v) with (u_{n}, v_{n}) ∈ W for all n, then T_{1}u = lim_{n→∞} T_{1}u_{n} = lim_{n→∞} T_{2}v_{n} = T_{2}v, so that (u, v) ∈ W, hence W is closed. Show that T (W) = Y_{1} ∩ Y_{2}. Indeed, if (w_{1}, w_{2}) ∈ W then T (w_{1}, w_{2}) = T_{1}w_{1} + T_{2}w_{2} = 2T_{1}w_{1} = 2T_{2}w_{2} ∈ Y_{1} ∩ Y_{2}. Conversely, if y ∈ Y_{1} ∩ Y_{2}, then there exist w_{i} ∈ M_{i} with T_{i}w_{i} = y for i = 1, 2. Then (\frac{w_{1}}{2} , \frac{w_{2}}{2} ) ∈ W and T (\frac{w_{1}}{2} , \frac{w_{2}}{2} ) = y. □

The following theorems refine some of the results in [Arv67], we will use them in the proof of Theorem 4.2.

Theorem 4.6. Let 𝔄 be (n − 1)-transitive for some n > 1 and M = \overline{𝔄^{(n)}x⃗} for some linearly independent n-tuple x⃗ = (x_{1}, …, x_{n}) in X^{n}. Then either M = X^{n} or M consists only of linearly independent n-tuples and zero.

Proof. First, we show that if N is a closed 𝔄^{(n)}-invariant subspace of X^{n} such that M ⊆ N, then N satisfies the following two properties.

(i) If N contains an n-tuple of the form (u_{1}, …, u_{n−1}, 0) where u_{1}, …, u_{n−1} are linearly independent, then N = X^{n}.

(ii) If N contains both linearly independent and non-zero linearly dependent ntuples, then N = X^{n}.

14 H. P. ROSENTHAL, V. G. TROITSKY

Note that if 𝔄 is unital, then x⃗ ∈ M; in this case applying (ii) with N = M would immediately yield the conclusion of the theorem.

To prove (i), notice that since 𝔄 is (n − 1)-transitive then X^{n−1} ⊕ {0} ⊆ N. Since 𝔄 is transitive, then Ax_{n} ≠ 0 for some A ∈ 𝔄. It follows that there exists (w_{1}, …, w_{n}) in M with w_{n} ≠ 0. But X^{n−1} ⊕ {0} ⊆ N implies (w_{1}, …, w_{n−1}, 0) ∈ N, so that (0, …, 0, w_{n}) ∈ N, which yields {0}^{n−1} ⊕ X ⊆ N. Hence, N = X^{n}.

To prove (ii) suppose that N contains a linearly independent n-tuple v⃗ = (v_{1}, …, v_{n}) and a non-zero linearly dependent n-tuple y⃗ = (y_{1}, …, y_{n}). Without loss of generality, y_{1}, …, y_{k} are linearly independent for some k < n, and y_{i} = ∑_{j=1}^{k} α_{ij}y_{j} as i = k + 1, …, n. Since 𝔄 is k-transitive, there exists a sequence (A_{m}) in 𝔄 such that A_{m}y_{1} → v_{1} and A_{m}y_{i} → 0 as i = 2, …, k. Then for k + 1 ⩽ i ⩽ n we have A_{m}y_{i} → α_{i,1}v_{1}. It follows that

A_{m}^{(n)}y⃗ → (v_{1}, 0, …, 0, α_{k+1,1}v_{1}, … α_{n,1}v_{1}),

so that the latter n-tuple belongs to N. Subtracting it form v⃗ we see that N contains an element of the form (0, z_{2}, …, z_{n}), where z_{2}, …, z_{n} are linearly independent. Now N = X^{n} by (i).

To complete the proof, assume that M contains a linearly dependent non-zero ntuple and M ≠ X^{n}. Notice that M + [x⃗] is 𝔄-invariant, so that M + [x⃗] = X^{n} by (ii). In particular, there exists w⃗ ∈ M and a scalar λ such that w⃗ + λx⃗ = (x_{1}, 0, …, 0). Then

w⃗ = ((1 − λ)x_{1}, −λx_{2}, …, −λx_{n})).

If λ = 1 then w⃗ = (0, −x_{2}, …, −x_{n}), so that M = X^{n} by (i), contradiction. If λ ≠ 1 and λ ≠ 0 then all the components of w⃗ are linearly independent, hence M = X^{n} by (ii) which, again, would contradict our assumptions. It follows that λ = 0, and so (x_{1}, 0, …, 0) ∈ M. Similarly, (0, …, 0, x_{i}, 0, …, 0) ∈ M for every i = 1, …, n. It follows that x⃗ ∈ M, so that M = M + [x⃗] = X^{n}, contradiction. □

Remark 4.7. Assuming that 𝔄 is unital and (n − 1)-transitive, it can be shown that any M ∈ Lat 𝔄^{(n)} is either in Lat(L(X))^{(n)} (that is, consists of n-tuples satisfying a fixed set of linear dependence relations), or consists only of linearly independent n-tuples and zero.

Corollary 4.8. Let 𝔄 be transitive and assume that for any linearly independent x_{1} and x_{2} in X there exists an operator A ∈ 𝔄 with (Ax_{1}, Ax_{2}) non-zero and linearly dependent. Then 𝔄 is WOT-dense in L(X).

STRICTLY SEMI-TRANSITIVE OPERATOR ALGEBRAS 15

Proof. We will show by induction that 𝔄 is n-transitive for all n ⩾ 1, then \overline{𝔄}^{WOT}= L(X) by [RR73, Theorem 7.1]. It is given that 𝔄 is 1-transitive. Suppose that 𝔄 is (n − 1)- transitive for some n > 1, and let x⃗ = (x_{1}, …, x_{n}) be a linearly independent n-tuple. Choose A ∈ 𝔄 so that (Ax_{1}, Ax_{2}) is non-zero and linearly dependent. Then x⃗ is cyclic for 𝔄^{(n)} by Theorem 4.6. Hence, 𝔄 is n-transitive. □

Our next result is a refinement of Corollary 2.5 in [Arv67] (see also Lemma 8.8 in [RR73]).

Theorem 4.9 (Graph Theorem). Suppose that 𝔄 is n-transitive. Then 𝔄 is not (n+1)- transitive if and only if there exists a closed operator T⃗: D ⊆ X → X^{n} where D is a dense 𝔄-invariant linear subspace of X and T⃗ = T_{1} ⊕ ⋯ ⊕ T_{n}, each T_{i} commutes with 𝔄, and (x, T_{1}x, …, T_{n}x) is linearly independent for each non-zero x ∈ D. In particular, each T_{i} is one-to-one and non-scalar.

Proof. Suppose that 𝔄 is n-transitive. If 𝔄 is not (n+1)-transitive then by Theorem 4.6 there exists a closed 𝔄^{(n+1)}-invariant subspace M of X^{n+1} such that every non-zero element of M is a linearly independent (n + 1)-tuple. We claim that M is the graph of a closed operator T⃗: D ⊆ X → X^{n} satisfying the required conditions. Indeed, let D = {x_{0} | (x_{0}, x_{1}, …, x_{n}) ∈ M}. For (x_{0}, x_{1}, …, x_{n}) ∈ M define T_{i}x_{0} = x_{i} for i = 1, …, n, and put T⃗ = T_{1} ⊕ ⋯ ⊕ T_{n}. Notice that T⃗ is well-defined: suppose that (x_{0}, x_{1}, …, x_{n}) and (x_{0}, x′_{1}, …, x′_{n}) are both in M, then (0, x_{1} − x′_{1}, …, x_{n} − x′_{n}) ∈ M, but this (n + 1)-tuple is linearly dependent, hence equals zero, so that x_{i} = x′_{i} for all 1 ⩽ i ⩽ n. Clearly, T⃗ is closed because M is closed.

Now let x ∈ D, then (x, T_{1}x, …, T_{n}x) ∈ M. Since M is 𝔄^{(n+1)}-invariant, if A ∈ 𝔄 then (Ax, AT_{1}x, …, AT_{n}x) ∈ M. It follows that Ax ∈ D, so that D is 𝔄-invariant and, therefore, dense in X. Furthermore, it also follows that AT_{i}x = T_{i}(Ax) for all i = 1, …, n, so that T_{i} commutes with 𝔄. Finally, (x, T_{1}x, …, T_{n}x) is in M for each non-zero x ∈ D, so that this (n + 1)-tuple is linearly independent. It follows that each T_{i} is non-scalar. It also follows that T_{i}x = 0 implies x = 0, so that T_{i} is one-to-one.

To prove the converse, suppose that T⃗ is such an operator, and let M be the graph of T⃗. Then M is closed and every non-zero element of M is linearly independent. It follows that M ≠ X^{n+1}. It is easy to see that M is 𝔄^{(n+1)}-invariant, so that every linearly independent (n + 1)-tuple in M is not cyclic for 𝔄^{(n+1)}, hence 𝔄 is not (n + 1)-transitive. □

Now we are ready to prove Theorem 4.2.

Proof of Theorem 4.2. Suppose that 𝔄 has no non-trivial invariant injective operator ranges of order n for some n. It follows that 𝔄 has no non-trivial invariant injective operator ranges of order k whenever 1 ⩽ k ⩽ n. In particular, 𝔄 is transitive.

Note first, that 𝔄′ = [I]. Indeed, let T ∈ 𝔄′. Pick λ ∈ σ(T), and show that T = λI. Suppose not, then Range(T − λI) ≠ 0. Since Null(T − λI) is an 𝔄-invariant closed subspace then T − λI is one-to-one. Then Range(T − λI) is an 𝔄-invariant injective operator range of order 1, so that Range(T − λI) = X. But this would mean that T − λI is invertible, contradiction.

Suppose that 𝔄 is not n-transitive, and show that this leads to a contradiction. Let k be the greatest such that 𝔄 is k-transitive but not (k +1)-transitive, then 1 ⩽ k < n. Let T⃗: D → X^{k} where T = T_{1}⊕⋯⊕T_{k} is as in Theorem 4.9. Let M be the graph of T⃗. Also, let P: M → X be the projection on the first component. Then D = Range P is an injective operator range of order (k + 1). It follows that D = X, so that T_{i} ∈ L(X) for each i = 1, …, k and, therefore, T_{i} ∈ 𝔄′. But then T_{i} have to be scalar because 𝔄′ = [I], contradiction. □

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[Arv67] William B. Arveson. A density theorem for operator algebras. Duke Math. J., 34:635–647, 1967.

[DHP01] Allan Donsig, Alan Hopenwasser, and David R. Pitts. Automatic closure of invariant linear manifolds for operator algebras. Illinois J. Math., 45(3):787–802, 2001.

[Foi72] Ciprian Foiaş. Invariant para-closed subspaces. Indiana Univ. Math. J., 21:887–906, 1971/72.

[Hop01] Alan Hopenwasser. Invariant linear manifolds for CSL-algebras and nest algebras. Proc. Amer. Math. Soc., 129(2):389–395, 2001.

[Lam71] Alan Lambert. Strictly cyclic operator algebras. Pacific J. Math., 39:717–726, 1971.

[Ric50] C. E. Rickart. The uniqueness of norm problem in Banach algebras. Ann. of Math. (2), 51:615–628, 1950.

[RR73] Heydar Radjavi and Peter Rosenthal. Invariant subspaces. Springer-Verlag, New York, 1973. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 77.

[Yood49] Bertram Yood. Additive groups and linear manifolds of transformations between Banach spaces. Amer. J. Math., 71:663–677, 1949.

Department of Mathematics, University of Texas, Austin, TX 78712. USA

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

  E-mail address: rosenthl@math.utexas.edu, vtroitsky@math.ualberta.ca