Section 7.1 The Banach Fixed-Point Theorem
The main idea is to solve a fixed-point problem \(x=Tx\) iteratively, i.e., start at some \(x_0\) and build a sequence
\begin{equation*}
x_0\in X, \qquad x_{n+1}=T x_n
\end{equation*}
and show that \(x_n\) converges to \(x\) in some appropriate sense. Definition 7.1.1.
Let \((X,d)\) be a metric space and \(M\subset X\text{.}\) A (nonlinear) map \(T:M\to X\) is called-
\(k\)-contractive if and only if\begin{equation*} d(T(x), T(y)) \leq k \; d(x,y), \qquad \mbox{for all }\quad x,y\in M, \qquad 0<k<1, \end{equation*}
-
contractive if and only if\begin{equation*} d(T(x) , T(y)) < d(x,y), \qquad \mbox{for all }\quad x,y\in M, \qquad x\neq y. \end{equation*}
Theorem 7.1.2.
Let \(M\subset X\) be a non-empty, closed subset of a complete metric space \(X\text{,}\) and let \(T:M\to M\) be \(k\)-contractive with \(k<1\text{.}\) Then- \(T\) has exactly one fixed point in \(M\text{.}\)
- The sequence \(x_{n+1}=Tx_n\) converges to the fixed point \(x\) for each initial point \(x_0\in M\text{.}\)
Proof.
\begin{equation*}
d(x_{n+1}, x_n) = d(T(x_n), T(x_{n-1})) \leq k\; d(x_n, x_{n-1}) \leq \cdots\leq k^n d(x_1,x_0).
\end{equation*}
and then
\begin{align*}
d(x_{n+m}, x_n) \amp\leq \amp d(x_{n+m}, x_{n+m-1}) + d(x_{n+m-1}, x_{n+m-2}) + \cdots + d(x_{n+1}, x_n)\\
\amp\leq \amp \left(k^{n+m-1} + k^{n+m-2} + \cdots + k^n\right)d(x_1, x_0)\\
\amp\leq\amp k^n \sum_{j=0}^{m-1} k^j \; d(x_1, x_0)\\
\amp\leq\amp \frac{k^n}{1-k} d(x_1, x_0)\\
\amp\to\amp 0, \qquad \mbox{for}\qquad n\to\infty.
\end{align*}
Hence \(\{x_n\}_n \) is a Cauchy sequence, and since \(X\) is complete, it converges to a point \(x\in X\text{.}\) Moreover, \(T:M\to M\) and \(M\) is closed, hence the limit \(x\in M\text{.}\) While \(T\) is a \(k\)-contraction, it is also continuous, hence \(T(x)=x\) and \(x\) is a fixed point. To show uniqueness, we assume there is another fixed point \(y\in M\text{.}\) Then
\begin{equation*}
d(x,y) = d(T(x), T(y) ) \leq k\; d(x,y)
\end{equation*}
and \(k<1\text{,}\) hence \(d(x,y)=0\text{,}\) and we find \(x=y\text{.}\)