# Proof Attempt of Brouwer (via Separating Hyperplane Theorem)

In part motivated by the discussion here, I have been playing with trying to prove Brouwer's theorem appealing as minimally as possible to topology.

In the 1-dimensional case I believe one can construct a proof based on separation of convex bodies. For a continuous function $f$, $f$ has a fixed point if and only if the convex hull of the epigraph of $f$ and the reflection of this set about the 45-degree line are disjoint. Assuming they are, for contradiction, implies the existence of a separating line; hence, for any line orthogonal to this separating line, the projection of these convex sets onto this orthogonal line must be disjoint. But they cannot be because the diameter of each of these projections down is weakly larger than the projections of the half-spaces formed by the separating line.

My question is whether this logic can be generalized to arbitrary finite dimension. Below is an attempted proof, though may not be correct. Any hole poking or discussion is greatly appreciated!

Theorem: Let $S\subset \mathbb{R}^n$ be a compact, convex set. Let $f: S \to S$ continuous. Then $f$ has a fixed point.

Proof Attempt: Let $\Delta = \{(s,s) \in S \times S : s \in S\}$. Let $\Gamma = \{(s,s') \in S \times S : s' \in f(s)\}$. Let $A =$ epi$f=\{(s,t)\in S \times S : t \succeq f(s)\}$. Let $B=\{(t,s) \in S \times S : (s,t) \in A\}$.

As $f$ is continuous, the graph of $f$, $f^{-1}$ are closed in $S \times S$, hence $A$, $B$ are compact. As linear functions are the minimal convex structure, $f$ has a fixed point (i.e. $\Gamma \cap \Delta \neq \emptyset$) if and only if $conv(A) \cap \Delta \neq \emptyset$ (where $conv(X)$ denotes the convex hull of the set $X$). This in turn is equivalent to $conv(A) \cap conv(B) \neq \emptyset$.

Assume, for contradiction, that $f$ does not have a fixed point. Then $conv(A) \cap conv(B) = \emptyset$. Let $\bar{\Delta}$ denote the collection of all separating hyperplanes containing $\Delta$. Since $conv(A)$ and $conv(B)$ are disjoint, compact, convex sets, we may find a strictly separating hyperplane. Moreover, it may be always chosen to include the diagonal, thus $\bar{\Delta}$ is non-empty.

Let $\bar{\delta} \in \bar{\Delta}$, let $\bar{\delta}'$ be an orthogonal hyperplane, and let $\pi_{\bar{\delta}'}$ denote the projection from $S \times S$ onto $\bar{\delta}'$. Then $\bar{\delta}$ partitions $S\times S$ into two compact, convex half-spaces, $U$ and $L$ ('upper' and 'lower'). Let $D=\min\{\text{diam}(\pi_{\bar{\delta}'}(U), \text{diam}(\pi_{\bar{\delta}'}(L)\}$. But $$\min\{\text{diam}(\pi_{\bar{\delta}'}(conv(A)), \text{diam}(\pi_{\bar{\delta}'}(conv(B))\} \ge D$$ Therefore $\bar{\delta}$ is not a strictly separating hyperplane, contradiction. Hence $conv(A) \cap conv(B) \neq \emptyset$ and thus $conv(A) \cap \Delta \neq \emptyset$ and $f$ has a fixed point.

• I don't know too much about convex analysis, but: 1. You write "$f^{-1}$ denotes the function whose graph ...". In general, $f^{-1}$ will not be a function, if $f$ is not injective. 2. (Maybe related): Are you considering $f$ to be set-valued? Otherwise, why do you write $s' \in f(s)$? 3. What do you mean by $t \succeq f(s)$? Are you comparing coordinate-wise? 4. You write "Let $\overline{\Delta}$ denote the collection of all hyperplanes containing $\Delta$. Do you mean all separating hyperplanes containing $\Delta$? – PhoemueX Jul 20 '15 at 10:45
• 1. Yes, it should be set valued. In fact I think the way I've written it is unneccessary; instead $B$ should simply be $\{(t,s) \in S \times S : (s,t) \in A\}$. 2. I mean less-than in the product order (see ccrma.stanford.edu/~dattorro/gcf.pdf p. 23) 3. Yes that is what I meant. I will edit now, thanks! – see.blue Jul 20 '15 at 10:47