When do positive operators have eigenvalues?

Question

Let $B$ be a complex Banach lattice and let $T : B \to B$ be a positive operator. Are there any conditions that ensure that $T$ has an eigenvalue? I am interested in particular in non-compact operators.

I'm interested in the following setting for instance: Denote by $S^1$ the unit circle and we consider operators $T: L^2(S^1) \to L^2(S^1)$. We know that $T$ maps positive functions to positive functions and we also know that $T$ is selfadjoint and satisfies $\langle T f,f \rangle \geq 0$ for all $f\in L^2(S^1)$. However, my operator is not compact. Can we impose some further conditions on $T$ such that we can conclude that $T$ has an eigenvalue? I would be even more interested in an eigenfunction with eigenvalue $||T||$.

Edit after further comments: Even more concretely, I am interested in the following operator. Consider the $SL_2(\mathbb{R})$ action on $S^1$, where we view $S^1$ as the boundary of hyperbolic space and denote by $(\rho, L^2(S^1))$ the quasiregular representation associated to this action. Let $\mu$ be a finitely supported probability measure on $SL_2(\mathbb{R})$. Then the operator I care about is $\rho(\mu) : L^2(S^1) \to L^2(S^1)$.

Answer 1

Here is one result that could, sometimes, be helpful in the setting of the question:

Theorem. Let $(\Omega,\mu)$ be a finite measure space and let $0 \not= T: L^2(\Omega,\mu) \to L^2(\Omega,\mu)$ be a positive (in the sense of Banach lattices) and self-adjoint linear operator. Assume moreover that $T$ is hyperbounded, i.e., that there exists a number $q \in (2,\infty)$ such that the range of $T$ is contained in $L^q(\Omega,\mu)$. Then the norm of $T$ (which is equal to the spectral radius of $T$) is an isolated spectral value spectral value of $T$ and an eigenvalue of finite mulitplicity.

This is a consequence of (a more general result in) Corollary 2.4 of my paper "Spectral gaps for hyperbounded operators (2020)" [Link to journal, Link to arXiv] - but the main idea of this result goes back to much earlier papers of Martínez and Lotz (please see the references in the linked paper for details).

Note, however, that the conclusion of the theorem is stronger than what the question is asking for: the theorem does not only yield an eigenvalue, but also a spectral gap (i.e., the spectral radius is isolated in the spectrum). Hence, the assumptions of theorem are so strong that the theorem can, sometimes, not be applied, even if $\|T\|$ is an eigenvalue of $T$.

Further versions of the theorem:

• In [op. cit., Sections 2 and 4.1] you can find more results of this type, along with several references.

• In [op. cit., Section 4.2] you can find results where positivity of the operator is replaced with certain contractivity assumptions.