Reducing Kakutani’s fixed-point theorem to Brouwer’s using a selection theorem

fixed-point-theoremsproof-verification

Kakutani's fixed-point theorem is quite similar to Brouwer's fixed point theorem – the main difference is that Brouwer speaks about single-valued functions and Brouwer about multi-valued functions.

There is a way to go from multi-valued functions to single-valued ones – it is Michael's selection theorem. So, I thought of the following simple proof to Kakutani's theorem:

  1. Given a multi-valued function $F$ satisfying Kakutani's conditions [$F$ has closed graph, $F(x)$ is nonempty and convex for all $x$], by Michael's selection theorem, there is a continuous single-valued selection-function $f$, such that $f(x)\in F(x)$ for all $x$.

  2. By Brouwer's fixed-point theorem, the continuous function $f$ has a fixed-point $x_0$ such that $f(x_0)=x_0$.

  3. Since $f$ is a selection function, $x_0\in F(x_0)$, so it is a fixed point of $F$ too.

The main thing I am not sure about is all the conditions of the theorems. Particularly, Michael's selection theorem requires that, if $F$ is a function from $X$ to $2^Y$, then:

  • $X$ is a paracompact space;
  • $Y$ is a Banach space;
  • $F$ is lower hemicontinuous;
  • For all $x \in X$, the set $F(x)$ is nonempty, convex and closed.

I read the definitions of these concepts, but I am still not sure whether they fit exactly the requirements of Kakutani's theorem.

MY QUESTION: is the proof I wrote above correct? If not, can it be fixed?

Best Answer

Michael's selection theorem will not generally apply to a function satisfying the conditions of Kakutani's fixed point theorem. In particular, though the other three conditions will always be satisfied, the condition that $F$ is lower hemicontinuous might fail. An example where this fails (adapted from an example on the wikipedia page) is the function $F : [0, 1] \to 2^{[0, 1]}$ for which $$F(x) = \begin{cases} \{1\} & x < 1/2 \\ [0, 1] & x = 1/2 \\ \{0\} & x > 1/2 \end{cases}$$ Kakutani's theorem guarantees that this function has a fixed point, but Michael's selection theorem does not apply -- indeed, you can check that $F$ has no continuous selection function. To me, this suggests there isn't a way to fix your proof.

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