A REMARK ON INVARIANT SUBSPACES OF POSITIVE OPERATORS

VLADIMIR G. TROITSKY

Abstract. If S, T, R, and K are non-zero positive operators on a Banach lattice such that S ↔ T ↔ R ⩽ K, where “↔” stands for the commutation relation, T is non-scalar, and K is compact, then S has an invariant subspace.

Throughout this note, X is a (real or complex) Banach lattice. For two operators S and T on X, the notation S ↔ T means that S and T commute. A (norm closed) subspace Y of X is said to be invariant under an operator T in L(X) if {0} ≠ Y ≠ X and T Y ⊆ Y. We follow the notations and terminology of [AA02].

There have been many extensions of Lomonosov’s theorem [Lom73] to positive operators; see Chapter 10 of [AA02] for a review of the subject. In particular, if T ↔ R ⩾ K for some positive non-zero operators T, R, and K with T quasinilpotent and K compact, then T has an invariant subspace (even an invariant closed ideal). The condition T ↔ R ⩾ K can be replaced with T ↔ R ⩽ K or, even more generally, with T ↔ R ⩾ C ⩽ K for some non-zero positive operator C; in the latter case, T is said to be compact friendly. There have been several more recent similar extensions of Lomonosov’s theorem to positive quasinilpotent operators: [Drn01, IM04, AT05, ÇE07, FTT08, PT09, Ges09, FV09, DK11]. In this note we do not require that T be quasinilpotent. Our result was motivated by Theorem 3.5 of [ÇM11], where quasinilpotence is not required either.

Theorem 1. Suppose that S, T, R, and K are non-zero positive operators on a Banach lattice such that S ↔ T ↔ R ⩽ K, T is non-scalar, and K is compact. Then S has an invariant subspace.

Proof. Suppose that S has no invariant subspaces. Let S̃ = ∑_{n=0}^{∞} t^{n}S^{n} where t is a positive real such that series converges. Then S̃ ⩾ I, S̃ ⩾ tS, and S̃ commutes with T.

Date: February 19, 2012.

2010 Mathematics Subject Classification. Primary: 47B65. Secondary: 47A15.

Key words and phrases. positive operators, invariant subspaces.

The author was supported by NSERC.

2 V. G. TROITSKY

Claim: for every x > 0, the vector S̃x is quasi-interior, that is, the order ideal J generated by S̃x is dense in X. Indeed, S̃x ⩾ x > 0, so that J ≠ {0}. Note that J is invariant under S because for every z ∈ J we have |z| ⩽ λS̃x for some λ > 0, so that

|Sz| ⩽ S|z| ⩽ λSS̃x = λ ∑_{n=0}^{∞} t^{n}S^{n+1}x = \frac{λ}{t} ∑_{n=0}^{∞} t^{n+1}S^{n+1}x ⩽ \frac{λ}{t} S̃x.

Since S has no invariant subspaces, J has to be dense in X. This proves the claim.

Since R ≠ 0, there exists x_{0} > 0 such that Rx_{0} > 0. By the claim, S̃Rx_{0} is quasi-interior. Since R is positive and non-zero, it cannot vanish on a quasi-interior vector, hence RS̃Rx_{0} > 0. Iterating this step, we get RS̃RS̃Rx_{0} > 0. It follows that RS̃RS̃R ≠ 0. Since S̃R ⩽ S̃K and the latter operator is compact, RS̃RS̃R is compact by Aliprantis-Burkinshaw’s Cube Theorem [AA02, Theorem 2.34]. Hence, T commutes with a non-zero compact operator. Therefore, T has a hyperinvariant subspace: in case of a complex Banach lattice this follows from Lomonosov’s Theorem, while in the case of a real Banach lattice we use Corollary 2.4 of [Sir05]. □

Remark 2. We have, actually, proved more than stated: we proved that either S has an invariant closed ideal or T commutes with a non-zero compact operator and, therefore, has a hyperinvariant subspace. We would also like to point out that the assumption that T is positive is not really needed.

To put Theorem 1 in perspective, note that, under the assumptions of the theorem, the following facts are well known.

• If both X and X^{∗} have order continuous norm, then R is compact by Dodds-Fremlin Theorem [AA02, Theorem 2.38], so that T has a hyperinvariant subspace by Lomonosov’s Theorem.

• Note that R^{3} is always compact by the Cube Theorem, and T ↔ R^{3}. Thus, if R^{3} ≠ 0 then it follows immediately from Lomonosov’s Theorem that T has a hyperinvariant subspace. On the other hand, if R^{3} = 0 then ker R is a nontrivial subspace invariant under T. Hence, in any case, T has an invariant subspace.

• Note that T is compact-friendly. Therefore, if T is quasinilpotent at a positive vector then Theorem 10.55 of [AA02] guarantees that S has an invariant closed ideal. The following result is an analogue of Theorem 10.55 in our setting.

INVARIANT SUBSPACES OF POSITIVE OPERATORS 3

Theorem 3. Suppose that T, R, and K are non-zero positive operators on a Banach lattice X such that T ↔ R ⩽ K, T is non-scalar, and K is compact. If (S_{n}) is a sequence of positive operators commuting with T then there is a subspace invariant under T, R, and all S_{n}’s.

Proof. Let S = T + R + ∑_{n=1}^{∞} a_{n}S_{n}, where (a_{n}) is a sequence of positive reals such that the series converges. Observe that S is a positive operator commuting with T. If S has an invariant closed ideal then this ideal remains invariant under T, R, and each S_{n} because these operators are dominated by S. However, if S has no invariant closed subspaces, then T has a hyperinvariant subspace by Remark 2. □

Example 4. 0 ⩽ R ⩽ K, K is compact, R is not compact, and R^{2} = 0.

This is the case in Example 5.19 of [AB06]; it is one of the few classical examples showing that Dodds-Fremlin Theorem may fail when X^{∗} is not order continuous. Here is the example. Put X = ℓ_{1} ⊕ L_{2}. Let (e_{i})_{i=1}^{∞} stand for the unit vector basis of ℓ_{1}, (r_{i})_{i=1}^{∞} stand for the sequence of the Rademacher functions in L_{2}, and r_{0} = 𝟙 stand for the constant one function in L_{2}. Recall that the sequence (r_{i})_{i=0}^{∞} is an orthonormal sequence in L_{2}. Note also that r_{i}^{+} = \frac{1}{2}(r_{i} + 𝟙) for all i. We define R_{0}, K_{0} : ℓ_{1} → L_{2} via K_{0}e_{i} = 𝟙 and R_{0}e_{i} = r_{i}^{+} for all i ⩾ 1. It is easy to see that the both operators are bounded, K_{0} is compact, R_{0} is not compact, and 0 ⩽ R_{0} ⩽ K_{0}. Now put R = [ \begin{matrix}0 & 0 \\ R_{0} & 0 \end{matrix} ] and K = [ \begin{matrix}0 & 0 \\ K_{0} & 0 \end{matrix} ]. Then R and K are two operators on X with 0 ⩽ R ⩽ K, K is compact, R is not compact, and R^{2} = 0.

Example 5. With R and K as in Example 4, we will construct T such that T commutes with R but not with K. Put T = [ \begin{matrix}P & 0 \\ 0 & Q \end{matrix} ] where P: ℓ_{1} → ℓ_{1} is the left shift: P e_{i} = e_{i−1} if i > 1 and P e_{1} = 0; and Q: L_{2} → L_{2} is defined as follows. Put Q𝟙 = 𝟙, Qr_{1} = −𝟙, Qr_{i} = r_{i−1} for i > 1 and define Q arbitrarily on the orthogonal complement of the closed span of (r_{i})_{i=0}^{∞} in L_{2}. Using the fact that r_{i}^{+} = \frac{1}{2}(r_{i} + 𝟙) we see that Q acts as a left shift on the sequence (r_{i}^{+})_{i=1}^{∞}. It is easy to see that T commutes with R because for every ∑_{i=1}^{∞} α_{i}e_{i} in ℓ_{1} and every f ∈ L_{2} we have T R(∑_{i=1}^{∞} α_{i}e_{i}, f) = (0, ∑_{i=1}^{∞} α_{i+1}r_{i}^{+}) = RT (∑_{i=1}^{∞} α_{i}e_{i}, f). However, T does not commute with K because T K(e_{1}, 0) = (0, 𝟙) while KT (e_{1}, 0) = (0, 0). Note that T is not positive.

Example 6. We construct three non-zero positive operators T, R, and K such that 0 ⩽ R ⩽ T, K is compact, R is not compact, and T commutes with R but not with

4 V. G. TROITSKY

K. In particular, the operators K, R, and T, together with any positive operator S which commutes with T satisfy the assumptions of Theorem 1.

We construct R and K similarly to Example 4. We again put X = ℓ_{1} ⊕ L_{2}, but this time we consider ℓ_{1} indexed by ℕ ∪ {0}, so that the unit basis now starts with e_{0}. Again, we define R = [ \begin{matrix}0 & 0 \\ R_{0} & 0 \end{matrix} ] and K = [ \begin{matrix}0 & 0 \\ K_{0} & 0 \end{matrix} ] where R_{0}e_{i} = r_{i}^{+} and K_{0}e_{i} = 𝟙 for all i = 0, 1, 2, … (recall that r_{0} = 𝟙). We still have 0 ⩽ R ⩽ K, K is compact, R is not compact, and R^{2} = 0. Put T = [ \begin{matrix}P & 0 \\ 0 & Q \end{matrix} ] where P: ℓ_{1} → ℓ_{1} and Q: L_{2} → L_{2} are defined as follows. Fix a positive real parameter α. For f ∈ L_{2}, put

(Qf)(t) = f(\frac{t}{2}) + 2α ∫ 1 \frac{1}{2} f, t ∈ [0, 1].

It is easy to see that Q𝟙 = (1 + α)𝟙, Qr_{1} = (1 − α)𝟙, and Qr_{i} = r_{i−1} for i > 1. It follows from r_{i}^{+} = \frac{1}{2}(r_{i} + 𝟙) that Qr_{1}^{+} = 𝟙 and Qr_{i}^{+} = r_{i−1}^{+} + \frac{α}{2}𝟙 whenever i > 1. Now we define P so that the action of P on (e_{i})_{i=0}^{∞} matches the action of Q on (r_{i}^{+})_{i=0}^{∞}, namely,

P e_{i} = \begin{cases}(1 + α)e_{0} & i = 0, \\ e_{0} & i = 1, \\ e_{i−1} + \frac{α}{2}e_{0} & i > 1. \end{cases}

Clearly, Q and P are positive, hence so is T. It is easy to verify that T commutes with R. However, T does not commute with K as T K(e_{1}, 0) = T (0, 𝟙) = (0, (1 + α)𝟙), while KT (e_{1}, 0) = K(e_{0}, 0) = (0, 𝟙).

Note that (0, 𝟙) is an eigenvector of T; it follows that T has a hyperinvariant subspace. Also, if α = 1 then T commutes with the compact positive operator C defined by C(x, f) = (0, (∫_{0}^{1} f)𝟙). We do not know whether T commutes with a compact operator when α ≠ 1.

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Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1. Canada

  E-mail address: troitsky@ualberta.ca