Contraction Mapping Definition
Let \((X, d)\) be a metric space, A mapping \(T : X \mapsto X\) is a contraction mapping, or contraction, if \(\exists c, 0 \leq c < 1\) s.t. \(\forall x,y \in X\), we have \[d(T(x), T(y)) \leq c d(x, y).\]
Note 1: The contraction literally means shortening the distance between points.
Note 2: The point \(x \in X\) s.t. \(T(x) = x\) is called a fixed point of \(T\).
Note 3: The contraction mapping is uniformly continuous (not showing the proof in this post).
Contraction Mapping Theorem
If \(T: X \mapsto X\) is a contraction mapping on a complete metric space \((X, d)\), then \(\exists x \in X\) be fixed point.
Note 1: A metric space \((X, d)\)is said to be complete if every Cauchy sequence in \(X\) converges to a point in \(X\).
Proof. The proof uses a constructive method by creating a sequence converging to the fixed point. Let \(x_0\) be any point in \(X\). We define a sequence \((x_n)\) in \(X\) by \(x_{n+1} = Tx_n\) for \(n \in \mathbb{N}\). A direct consequence is that \(x_n = T^nx_0\).
First, we show that \((x_n)\) is a Cauchy sequence. If \(n \geq m \geq 1\), then from the definition of contractions and the triangle inequality, we have
\[\begin{aligned} d\left(x_{n}, x_{m}\right) &=d\left(T^{n} x_{0}, T^{m} x_{0}\right) \\ & \leq c^{m} d\left(T^{n-m} x_{0}, x_{0}\right) \\ & \leq c^{m}\left[d\left(T^{n-m} x_{0}, T^{n-m-1} x_{0}\right)+d\left(T^{n-m-1} x_{0}, T^{n-m-2} x_{0}\right)\right.\\ & \leq c^{m}\left[\sum_{k=0}^{n-m-1} c^{k}\right] d\left(x_{1}, x_{0}\right) \\ & \leq c^{m}\left[\sum_{k=0}^{\infty} c^{k}\right] d\left(x_{1}, x_{0}\right) \\ & \leq\left(\frac{c^{m}}{1-c}\right) d\left(x_{1}, x_{0}\right) \end{aligned}\]Inserted Note: The second line results from iterating the definition of contractions for \(m\) times, the first iteration would be \[d(T^n x_0, T^m x_0 \leq cd(T^{n-1}x_0, T^{m-1}x_0).\] The third line results from iterating the triangle inequality for \(n-m-1\) times, the first iteration would be \[c^{m} d\left(T^{n-m} x_{0}, x_{0}\right) \leq c^m [d(T^{n-m}x_0, T^{n-m-1}x_0) + d(T^{n-m-1}x_0, x_0)].\] The fourth line, again, results from applying the defintion of contractions on each term separately. The last line gives an upper bound and since \(0 \leq c < 1\), the bound converges to \(0\).
Hence, \((x_n)\) is Cauchy, and it converges to \(x \in X\) since \(X\) is complete. This further implies that \[Tx = T \lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} x_{n+1} = x.\]
Inserted Note: The limit sign can be pulled out because of the uniform continuity of \(T\).
Finally, suppose \(x\) and \(y\) are two fixed points, then we have \[ 0 \leq d(x, y) = d(Tx, Ty) \leq cd(x, y),\] which implies that \(d(x,y) = 0\) otherwise it contradicts to the definition of contractions.
References
The source of the proof is available here.