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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.