## 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

Inserted Note: The second line results fromiteratingthe 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 fromiteratingthe 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 theuniform continuityof \(T\).

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.