2
$\begingroup$

I had asked this question on MSE but did not get any response.


I would like some reference to books that talk about irreducible operators on Banach lattices and its properties. A quick google search did not reap fruitful results. I'm especially interested in properties of the adjoint of an irreducible operator.

$\endgroup$
  • 2
    $\begingroup$ The Banach lattice books of Schaefer (1974) and of Meyer-Nieberg (1991) contain quite some material about irreducible operators, and there are also some useful research and survey articles. Could you be a bit more specific on which properties of irreducible operators you are interested in? In my experience, irreducible operators are most often studied with a focus on spectral theory. $\endgroup$ – Jochen Glueck Feb 27 '18 at 22:38
  • 1
    $\begingroup$ By the way, concerning adjoints: one can prove that a positive operator on a reflexive Banach lattice is irreducible if and only if its adjoint operator is positive. On more general Banach lattices, this is not in general true. But of course, it might still be possible to prove non-trivial results about adjoints of irreducible operators. $\endgroup$ – Jochen Glueck Feb 27 '18 at 22:43
  • 1
    $\begingroup$ I don't have a copy to hand (so might be completely wrong), but I seem to remember there being quite bit about this sort of thing in the book "An invitation to operator theory" by Abramovich and Aliprantis. It's a nice book in any case, $\endgroup$ – DCM Feb 27 '18 at 23:21
  • 1
    $\begingroup$ @JochenGlueck I'll look into both the books. I'm mainly interested in knowing if there are any conditions which guarantee that adjoint of an irreducible operator is irreducible. $\endgroup$ – Mark Feb 28 '18 at 9:44
  • 1
    $\begingroup$ Arrrgh, my comment above contains a really annoying mistake: I intended to write "A positive operator on a reflexive Banach lattice is irreducible if and only if its adjoint is irreducible". What I wrote instead clearly doesn't make any sense at all. $\endgroup$ – Jochen Glueck Feb 28 '18 at 10:25
4
$\begingroup$

Preliminary remarks:

  • In the comments the OP noted that he is primarly interested in the question whether the dual operator of an irreducible operator is irreducible, so I will focus on this aspect here.

  • The OP asked for a reference where this topic is treated. But to the best of my knowledge, all the standard references for Banach lattices and positive operators do not discuss characterisations of irreducible operators in detail. The topic might appear in some research papers or theses, but the only reference I could find is Gao's thesis quoted in Remark 6 below.

  • Below I discuss a few results (with proofs) which might be of interest for the OP. Several similar results are contained in Gao's thesis, but with a focus on band irreducibility rather than irreducibility of the dual operator (note that the term "irreducibility" is usually interpreted as ideal irreducibility unless otherwise specified).

Assumptions: Throughout, let $E$ be a (real or complex) Banach lattice and let $T$ be a positive (and thus bounded) linear operator on $E$. We denote the dual space of $E$ by $E'$ and the dual (or adjoint) operator of $T$ by $T'$; obviously, $T'$ is positive, too.

To rule out any misunderstandings, let us agree on the following terminology:

Definition. The operator $T$ is called irreducible if $0$ and $E$ are the only $T$-invariant closed ideals in $E$.

The following characterisation of irreducibility is quite useful for our purposes:

Proposition 1. The operator $T$ is irreducible if and only if for every $f \in E_+ \setminus \{0\}$ and every $\varphi \in E'_+ \setminus \{0\}$ there exists an integer $n \in \mathbb{N}_0 := \{0,1,2,\dots\}$ such that $\langle \varphi, T^n f\rangle > 0$.

Proof. "$\Rightarrow$" Assume that there exist $f \in E_+ \setminus \{0\}$ and $\varphi \in E'_+ \setminus \{0\}$ such that $\langle \varphi, T^nf\rangle = 0$ for all $n \in \mathbb{N}_0$. We define \begin{align*} I := \{g \in E: \; \langle \varphi, T^n|g|\rangle = 0 \text{ for all } n \in \mathbb{N}_0\}. \end{align*} Then $I$ is a closed and $T$-invariant ideal in $E$. We have $f \in I$, so $I$ is non-zero. Moreover, every positive vector in $I$ is contained in the kernel of $\varphi$. Since $\varphi$ is non-zero, we conclude that $I \not= E$. Hence, $T$ is not irreducible.

"$\Leftarrow$" Assume that $T$ is not irreducible. Then there exists a $T$-invariant closed ideal $I$ in $E$ such that $\{0\} \subsetneq I \subsetneq E$. The quotient space $E/I$ is non-zero, so there exists a non-zero positive functional $\psi \in (E/I)'$. Let $q: E \to E/I$ denote the quotient mapping and define $\varphi = \psi \circ q = q'\psi \in E'$. Then $\varphi \in E'_+ \setminus \{0\}$, and $\varphi$ vanishes on $I$. Since $I$ is non-zero, we can find a vector $0 \le f \in I \setminus \{0\}$. We have $T^nf \in I$ for all $n \in\mathbb{N}_0$ (since $I$ is $T$-invariant) and hence, $\langle \varphi, T^nf\rangle = 0$ for all $n \in \mathbb{N}_0$.

Corollary 2. If $T'$ is irreducible, then so is $T$.

Corollary 3. Assume that $E$ is reflexive. Then $T$ is irreducible if and only if $T'$ is irreducible.

We even have the following result which is a bit more general than Corollary 3:

Corollary 4 Assume that both $E$ and $E'$ have order continuous norm. Then $T$ is irreducible if and only if $T'$ is irreducible.

Proof. We only have to show "$\Rightarrow$". Assume that $T$ is irreducible, let $\varphi \in E'_+ \setminus \{0\}$ and $\psi \in E''_+ \setminus \{0\}$. Since $E$ has order continuous norm, the band in $E''$ generated by $E$ coincides with the space of all order-continuous functionals on $E'$ (see [Schaefer: Banach Lattices and Positive Operators, 1974, Corollary 1 to Theorem II.5.10 on page 90]). However, $E'$ has also order continuous norm, so every element of $E''$ is an order-continuous functional on $E'$. Hence, the band generated in $E''$ by $E$ equals $E''$. Moreover, $E$ is an ideal in $E''$ since $E$ has order continuous norm. Thus, there exists a vector $f \in E_+ \setminus \{0\}$ such that $f \le \psi$. Since $T$ is irreducible, there exists (by Proposition 1) an integer $n \in \mathbb{N}_0$ such that $\langle \varphi, T^nf \rangle > 0$. Hence, $\langle \psi, (T')^n\varphi\rangle \ge \langle \varphi, T^nf\rangle > 0$, so $T'$ is irreducible (again by Proposition 1).

Remark 5: The classical example where Corollary 4 can be applied, while Corollary 3 cannot, is the space $E = c_0$ of null sequences whose dual space is given by $E' = \ell^1$.

Remark 6: One can prove a bit more general results if one also considers the notion of band irreducibility (in particular on the dual space $E'$). Details can for instance be found in the PhD thesis of Niushan Gao, in particular in Lemma 0.31 on page 20.

Finally, it is certainly worthwhile to point out a few negative results:

Proposition 7. Assume that $E$ is not reflexive, but has order continuous norm. Then the bi-dual operator $T''$ on $E''$ is never irreducible (no matter whether $T$ and $T'$ are irreducible).

Proof. As $E$ has order continuous norm, it is an ideal in $E''$. However, we have $E \not= \{0\}$ and $E \not= E''$ since $E$ is not reflexive. Hence, $E$ is a non-trivial closed $T$-invariant ideal in $E''$.

Proposition 8. Assume that $E_+$ does not contain a quasi-interior point. Then $T$ cannot be irreducible.

Proof. Let $\lambda$ be a real number which is larger than the spectral radius of (the complex extension of) $T$. Choose an arbitray vector $f \in E_+ \setminus \{0\}$ and set $g := (\lambda - T)^{-1}f = \sum_{n=0}^\infty \frac{T^n}{\lambda^{n+1}}$. Then it is easy to see that the closed ideal generated by $g$ is $T$-invariant. This ideal is not equal to $E$ since $g$ is, by assumption, not a quasi-interior point of $E_+$; moreover, the ideal is non-zero since it contains $g$. Hence, $T$ is not irreducible.

Corollary 9. Let $K$ be an uncountable compact Hausdorff space and let $E = C(K)$ by the space of all scalar-valued continuous functions on $K$ (endowed with the supremum norm). Then $T'$ is not irreducible (not matter whether $T$ is irreducible or not).

Proof. According to [Schaefer, op. cit., Example 4 on page 99], the dual space $E'$ does not contain quasi-interior points. Hence, the assertion follows from Proposition 8.

Remark 10. If consider Corollary 9 and Remark 6, it is worthwhile to point out that the dual operator $T'$ in Corollary 9 is not even band irreducible. Indeed, the dual space $E' = C(K)'$ is an AL-space, so it has order continuous norm, and therefore the notions irreducible and band irreducible coincide on $E'$.

$\endgroup$
  • $\begingroup$ Regarding your Remark 5, I am a little bit confused. In the book "Positive linear systems by M. A. Krasnoselskij et al", there is an example in Page 114 which says Reflexivity in corollary 3 is necessary. $\endgroup$ – La Rias Jun 3 '18 at 13:56
  • $\begingroup$ @LaRias: Thanks for your comment. That's strange, so let's try to find the problem. Unfortunately, I currently have difficulties to find the book you mentioned. Could you please provide more specific bibliographic data, such as "names of all authors", "publisher", "year of publication", "language of the book (Russian or English)"? [By the way, I just corrected a typo in the proof of Corollary 4]. $\endgroup$ – Jochen Glueck Jun 3 '18 at 14:17
  • $\begingroup$ M. A. Krasnosel’skij, J. A. Lifshits, and A. V. Sobolev (1989). Positive linear systems: The method of positive operators; Transl. fromthe Russian by Jurgen Appell. Heldermann, Berlin. $\endgroup$ – La Rias Jun 3 '18 at 15:51
  • 1
    $\begingroup$ @LaRias: Thank you for the detailed reference. The point is that the authors of this books consider not only Banach lattices but more general ordered Banach spaces. They show by a counterexample that the assumption of reflexivity cannot be dropped in their Theorem 11.3, in general, and their counterexample is indeed constructed on the space $c_0$. Yet, in this counterexample the space $c_0$ is endowed with a particular cone that does not render the space a Banach lattice (while I consider the standard cone on $c_0$). This is why their counterexample does not contradict Remark 5 above. $\endgroup$ – Jochen Glueck Jun 4 '18 at 16:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.