# Existence of a fixed point for this operator

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 $$L^1(X)$$ or $$C^1(X)$$ (if easier) function, $$X$$ is a compact set, $$K$$ 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 compact valued.

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? I tried taking an interpolation argument by defining for each $$x \in X$$ a constant $$\xi(x) = \xi_x$$ such that $$A$$ has a fixed point $$g_x$$ and then defining $$g^*(x) = g_x (x)$$. The problem would then reduce to showing that $$g^* \in C^1(X)$$.

Any help would be appreciated.

• What does "compact valued" mean here? – Nik Weaver Apr 1 '19 at 13:49
• That the image is compact - (since the domain is compact) – Debreu Apr 1 '19 at 20:25
• So, $\xi$ is a map from $X$ into what? – Nik Weaver Apr 1 '19 at 21:39
• ... maybe you just mean that $\xi$ is a bounded real-valued function? – Nik Weaver Apr 1 '19 at 21:54