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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.

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

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