Section 7.2 The Brouwer and Schauder fixed-point theorems
The Brouwer fixed point theorem is a central piece in algebraic topology in finite dimensions \(\RR^n\text{.}\) It is equivalent to the negative retraction principle, and this is where we start our arguments. The Brouwer fixed point theorem is a rather general result for finite dimensional spaces. We lift it to the infinite dimensional setting in Schauders fixed-point theorem, where we employ some sense of compactness.Definition 7.2.1.
Let \(X\) be a topological space and \(T:X\to M\) a continuous map with \(M\subset X\text{.}\) \(T\) is called a retraction on \(M\) if \(T(x)=x\) for all \(x\in M\text{.}\) In this case \(M\) is called a retract of \(X\text{.}\)Example 7.2.2.
\begin{equation}
T(x) =\left\{\begin{array}{cc} x \amp \qquad \mbox{if} \quad x\in B_R(0), \\
R\frac{x}{|x|} \amp\qquad \mbox{if} \quad x\not\in B_R(0), \end{array}\right.\tag{7.2.1}
\end{equation}
is a continuous retraction of \(X\) to \(M\text{.}\)Theorem 7.2.3.
In \(\RR^n\) there is no continuous map \(T:\bar B_1(0) \to \partial B_1(0)\text{,}\) which leaves the boundary points fixed, i.e.
\begin{equation*}
T(x) =x, \qquad \mbox{for all }\quad \quad x\in \partial B_1(0).
\end{equation*}
Proof.
Theorem 7.2.4.
Each continuous map \(T:\bar B_1(0)\to \bar B_1(0) \) in \(\RR^n\) has a fixed point.Proof.
Corollary 7.2.5.
If \(M\subset \RR^n\) is homeomorphic to \(\bar B_1(0)\text{,}\) then each continuous map \(T:M\to M\) has a fixed point.Proof.
\begin{equation*}
f:=\phi\circ T \circ \phi^{-1}: \bar B_1(0) \to \bar B_1(0).
\end{equation*}
Example 7.2.6.
Proposition 7.2.7.
Let \(K\Subset X\) be a compact subset and let \(\ep>0\) be given. Let \(A\) denote a finite set \(A\subset K\) such that
\begin{equation*}
K\subset \bigcup_{a\in A} B_\ep(a)
\end{equation*}
is a finite open covering of \(K\text{.}\) We define a convex combination of the elements of \(A\) as a map \(\phi_A:K\to K\) as
\begin{equation}
\phi_A(x) = \frac{\sum_{a\in A} m_a(x) \; a}{\sum_{a\in A} m_a(x)}, \quad \mbox{where}\quad m_a(x) =\left\{\begin{array}{ll} 0 \amp \quad \mbox{for } \|x-a\|\geq \ep \\
\ep - \|x-a\| \amp \quad \mbox{for } \|x-a\|<\ep\end{array}\right.. \tag{7.2.2}
\end{equation}
The function \(\phi_A(x)\) denotes a weighted average of the anchor points \(a_i\in A\text{,}\) weighted by the distance to \(x\text{.}\) Then \(\phi_A(x)\) is continuous on \(K\) and for each \(x\in K\) we have
\begin{equation*}
\|\phi_A(x) - x\|<\ep.
\end{equation*}
Proof.
\begin{equation*}
\phi_A(x) - x = \frac{\sum_{a\in A} m_a(x) (a-x)}{\sum_{a\in A} m_a(x) },
\end{equation*}
and for those \(m_a\) with \(m_a(x)>0\) we have \(\|x-a\|<\ep\text{.}\) Hence
\begin{equation*}
\|\phi_A(x) - x\|\leq \frac{\sum_{a\in A} m_a(x) \; \ep }{\sum_{a\in A} m_a(x)} = \ep.
\end{equation*}
Theorem 7.2.8.
Let \(X\) be a Banach space and \(M\subset X\) a nonempty, bounded, and convex subset. Consider a continuous map \(T:M\to M\text{.}\)- If \(M\) is compact, then \(T\) has a fixed point.
- If \(T\) is compact, then \(T\) has a fixed point.
Proof.
\begin{equation*}
K \subset \bigcup_{a\in A_n} B_{\frac{1}{n}} (a),
\end{equation*}
where \(A_n\) is a finite set. We use the above function \(\phi_A\) to define \(\phi_{A_n}\text{.}\) The element \(\phi_{A_n}(x)\in X\) is a convex combination of elements \(a\in A_n\text{,}\) hence
\begin{equation*}
\phi_n(K)\subset \mbox{convex hull}(K) \subset M,
\end{equation*}
since \(T:M\to M\) and \(M\) is convex. Then we define maps \(T_n = \phi_{A_n} \circ T: M\to M\) with
\begin{equation*}
\|T_n(x) - T(x) \|\leq \frac{1}{n},
\end{equation*}
by Proposition Proposition 7.2.7. We define \(M_n := M \cap \mbox{span}(A_n) \text{.}\) Since \(\mbox{span}(A_n)\) is a finite dimensional subspace, the sets \(M_n\) are a bounded, closed and convex subsets of a finite dimensional subspace, and \(T_n:M_n \to M_n \) is continuous. By the Corollary to Brouwers fixed point theorem (Corollary Corollary 7.2.5) we have a fixed point \(x_n\) for each \(n\text{:}\) \(T_n(x_n)=x_n\text{.}\) - If \(M\) is compact, then the bounded sequence \(\{x_n\}_n\) has a convergent subsequence\begin{equation*} x_{n_j}\to x, \qquad \mbox{for}\quad j\to \infty, \end{equation*}and since \(T\) is continuous we have \(T(x)=x\text{.}\)
- If \(T\) is compact, then \(\{T(x_n)\}_n\) has a convergent subsequence, and the result is the same as under item 1.
