# Is the operator $Af = \int_0^1 k(x,y) f(x) dx$ irreducible provided that $Af = f$ has unique solution such that $\int_0^1 f = 1$

Let $k \in L^1((0,1) \times (0,1))$ be non-negative and such that $$\int_0^1 k(x,y) dy = 1, \quad x \in (0,1).$$ Let $A\colon L^1(0,1) \to L^1(0,1)$ be defined by $$Af(y) = \int_0^1 k(x,y) f(x) dx, \quad y \in (0,1).$$ Assume that there exists a unique $f \in L^1(0,1)$ such that

• $f$ is positive a.e.,
• $\int_0^1 f(x) dx = 1$,
• $Af = f$.

Is it true that $A$ is an irreducible operator?

By irreducible I mean

For every measurable $\Omega \subset (0,1)$ with measure $0 < \mu(\Omega) < 1$ ($\mu$ - Lebesgue measure), the set $$I = \{f \in L^1(0,1)\colon f_{|\Omega} = 0 \}$$ i not $A$-invariant, i.e., $$A(I) \not\subset I$$ holds.

Edit: It is clear the the uniqueness of $f$ is crucial. If $k$ is $2$ times the characteristic function of $$(0,1/2] \times (0,1/2) \cup (1/2,1) \times (1/2,1),$$ then $A$ is redducible, however there are also many $f$'s such that $Af = f$.

• I don't know the answer to your question but this notion of irreducibility looks very interesting to me. Do you maybe have a reference for it? – Surb Oct 7 '16 at 9:05
• @Surb In general, $A$ is an irreducible operator in a Banach lattice if there is no $A$-invariant closed ideal different than $\{ 0 \}$ and the whole space. In particular, in $L^1$ every closed ideal is of the form $\{ f\colon f = 0 \text{ on } S \}$ for some $S$, hence I just provided this spacial case characterization. – xen Oct 7 '16 at 9:23
• @Surb These are nice books on the subject: Shaefer, Banach Lattices and Positive Operators, and Meyer-Nieberg, Banach Lattices. – xen Oct 7 '16 at 9:25
• Thank you very much, I'll have a look at that. I'm working a lot with the notion of irreducibility for nonlinear operators in $\Bbb R^n_+$. There, the strong notion of irreducibility have some striking resemblance with your definition (and I guess is exactly the same). – Surb Oct 7 '16 at 18:08
• And actually, you might be interested to know that there are nonnegative matrices that have a unique positive eigenvector but are not irreducible. For an example consider the matrix $$\begin{pmatrix} 0& 0& 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix}$$ I guess that you should be able to transfer this counter-example to your setting – Surb Oct 7 '16 at 18:16