# Existence of fixed points for this Markov operator.

Perhaps math overflow is a better place to put this but I'm looking for some mathematical results that I might be able to apply to see if an operator I'm considering has a fixed point.

In particular consider, $$Ag(x) = \Big\{ \xi(x) + K(g(x))^\frac{1}{\theta} \Big\}^\theta$$ where $$g$$ is some $$L^1$$ or $$C^1$$ (if easier) function, $$K$$ is a linear operator and $$\xi$$ is compact valued and positive. I know for a fact that if the spectral radius condition $$r(K)^\frac{1}{\theta}<1$$ holds then $$K$$ will have a fixed point and so if $$\xi(x) = C$$ were a constant then $$A$$ would have a fixed point.

I'm wondering how to extend it to the case where $$\xi(x)$$ is a compact valued function.

One idea I had would be to note that by compactness $$\xi(x)$$ has an upper and lower bound and for these 'upper' and 'lower' versions of $$A$$ we just consider $$A_0g(x) = \Big\{ C + Kg(x)^\frac{1}{\theta} \Big\}^\theta \leq \Big\{ \xi(x) + Kg(x)^\frac{1}{\theta} \Big\}^\theta \leq \Big\{ C' + Kg(x)^\frac{1}{\theta} \Big\}^\theta = A_1g(x)$$

Since $$A_1$$ and $$A_0$$ have fixed points is there some kind of result that will get $$A$$ to have a fixed point?