Help following an outline of Markov–Kakutani Fixed Point Theorem proof.

Question

I'm trying to prove Markov–Kakutani Fixed Point Theorem by following the outline on page 173 here: https://people.math.ethz.ch/~salamon/PREPRINTS/funcana-ams.pdf

Let X be a locally convex Hausdorff topological vector space and let $A$ be a collection of pairwise commuting continuous linear operators A : X → X. Let C ⊂ X be a nonempty A-invariant compact convex subset of X, so that:

A(C) ⊂ C for all A ∈ A.

a) Let $$A_k(c)=\sum_0^{k-1}A^i(c)$$

Then $A_k(C)$ is a nonempty compact convex subset of $C$.

no problems here, I could use some advice showing this set is non-empty since mine seems clumsy

b) Show $A_k(B_l(C))\subset A_k(C)\cap B_l(C)$

no problems here either. This follows from the commodity relation

Use this to show $\bigcap_{k\in {1,2..}}\bigcap_{A\in\A}A_k(C)$ is not empty.

Having a very hard time here. If I can understand this better I think I can use Hahn-Banach to finish the proof I'm having trouble extending the pairwise relationship to the family intersection. I think if I can show the inner intersection is compact for any k then nested interval theorem should allow the conclusion.

c) Prove that every element x ∈ F is a fixed point of A. Hint: Fix an element A ∈ A. If $Ax-x\not=0$ find a continuous linear functional Λ : X → R such that Λ(x − Ax) = 1. Prove that, for every k ∈ N, there exists an element y ∈ C such that $$A_k(y)=x$$.

I think the last part will come from the definition of $F$. I'm unsure how to produce this functional. I'm only familiar with Hahn Banach on normed vector spaces. Here is it trivial to produce such a functional. Can this be used here in some way?

Answer 1

At page 173 there is nothing to do with this theorem, searching the document also does not give any returns for "markov", "kakutani", or "fixed" although that may be a problem with my reader. Are you sure this is the right document?

Anyway, you have a typo in the definition of $A_k$, you want $$A_k(c)=\frac1k\sum_{i=0}^{k-1}A^i(c),$$ otherwise you have no guarantee that $A_k(C)\subseteq C$ (indeed for $A=\Bbb1$ and $k=2$ you would get $\Bbb1_2(C)=2C\not\subseteq C$ with your definition). Further, the reason $A_k(C)$ is not empty is because $A_k = \sum_{i=0}^{k-1} A^i$ is a well defined (even linear) map, hence $A_k(C)$ is the image of a non-empty set under some function, as such it must be non-empty.

You will repeat this argument for b), you have seen that $A_k(C)\cap B_l(C) = A_k(B_l(C))$. Hence $A_k(C)\cap B_l(C)$ is never empty for any $A, B\in \mathfrak A$ and $k,l\in\Bbb N$ (since it is the image of the non-empty $B_l(C)$ under the map $A_k$). You can extend this to finite intersections $A_k(C)\cap B_l(C)\cap ... \cap Z_m(C)$ to get that no finite intersection of such sets is empty.

Now when you take the intersection over all $A_k(C)$ for $A\in\mathfrak A$ and $k\in\Bbb N$ we have just seen that no finite intersection of these sets is empty. Further these sets are all closed, because they are the image of the compact $C$ under the continuous $A_k$. Since they are closed, contained in the compact $C$, and have the property that no finite intersection is empty, you find that the entire intersection $\bigcap_{A\in\mathfrak A, k\in\Bbb N} A_k(C)$ is non-empty by compactness of $C$ (this is the finite intersection property of compact sets, which is "complementary" to the finite covering property).

For c) I assume $F$ is the big intersection. You get the functional by Hahn-Banach, in particular the separation version (take $A=\{x-Ax\}, B= \{0\}$ to get a functional that is non-zero on $x$ and then multiply it with an appropriate constant).

If you want to use this functional to get the contradiction then you should note:

$$x-Ax = A_ky-AA_ky=\frac1k(\sum_{i=0}^{k-1}A^iy-A^{i+1}y) = \frac1k(y-A^ky)$$ Since $C$ is invariant under $A$ you find that $A^ky\in C$, hence $\Lambda(y-A^ky) \subseteq \Lambda(C)+\Lambda(-C)$ which is bounded in $\Bbb C$, since $C$ is compact and $\Lambda$ continuous. But: $\Lambda(y-A^ky) = \Lambda( k(x-Ax))=k$ is unbounded.