Banach’s Fixed Point Theorem

Theorem: The theorem also known as the contraction mapping theorem, states that if a function $\tau$ is a contraction mapping on a complete metric space $(X,d)$, then $\tau$ has a unique fixed point in $X$.

We approach the proof by first defining the contraction mapping, and then using an iterative method to show the existence of a fixed point.

Some Preliminaries

Contraction: A function $\tau: X \mapsto X$ is a contraction mapping on a metric space $(X,d)$ if there exists a constant $0 < k < 1$ such that $\forall$ $x,y \in X$:

\[ d(\tau(x),\tau(y)) \leq kd(x,y) \]

Complete Metric Space: A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to a point in $X$.

Cauchy Sequence: A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence if $\forall$ $\epsilon > 0$, $\exists$ $N \in \mathbb{N}$ such that $\forall$ $m,n \geq N$:

\[ d(x_m,x_n) < \epsilon \]

Proof

Let $(X,d)$ be a complete metric space and $\tau: X \mapsto X$ be a contraction mapping on $X$. We will show that $\tau$ has a unique fixed point in $X$ ie. $\exists$ $x^* \in X$ such that $\tau(x^*) = x^*$.

We can define an iterative process as follows:

$ x^{(0)} \in X \ x^{(n+1)} = \tau(x^{(n)}) \quad \forall n \in \mathbb{N}\ $

Hence we obtain asequence {$x^{(n)}$} in $X$. We need to show that this sequence converges. Since the sequence is in a complete metric space, it is enough to show that this sequence is Cauchy.

\[ \begin{aligned} d(x^{n}, x^{n+1}) &= d(\tau(x^{(n-1)}), \tau(x^{(n)}))\\ &\leq kd(x^{(n-1)}, x^{(n)})\\ &\leq k^2d(x^{(n-2)}, x^{(n-1)})\\ &\vdots\\ &\leq k^nd(x^{(0)}, x^{(1)})\\ \end{aligned} \]

Now by the triangular inequality we can see that:

\[ \begin{aligned} d(x^{(m)}, x^{(n)}) &\leq d(x^{(m)}, x^{(m+1)}) + d(x^{(m+1)}, x^{(m+2)}) + \dots + d(x^{(n-1)}, x^{(n)})\\ &\leq k^md(x^{(0)}, x^{(1)}) + k^{m+1}d(x^{(0)}, x^{(1)}) + \dots + k^{n-1}d(x^{(0)}, x^{(1)})\\ &= k^md(x^{(0)}, x^{(1)})\sum_{i=0}^{n-m-1}k^i\\ &= k^md(x^{(0)}, x^{(1)})\frac{1-k^{n-m}}{1-k}\\ &\leq \frac{k^md(x^{(0)}, x^{(1)})}{1-k}\\ \end{aligned} \]

Since $0 < K < 1$ we can choose $m$ sufficiently large such that $d(x^{m}, x^{n})$ is small, and hence the sequence is Cauchy.

Finally, since the sequence is Cauchy, it converges to a point $x \in X$, ie. $x^{(n)} \xrightarrow{} x$

We now need to show that this point is a fixed point.

\[ \begin{aligned} d(x,\tau(x)) &\leq d(x,x^{(n)}) + d(x^{(n)}, \tau(x))\\ &= d(x,x^{(n)}) + d(\tau(x^{(n-1)}), \tau(x))\\ &\leq d(x,x^{(n)}) + Kd(x^{(n-1)}, x)\\ \end{aligned} \]

Since the sequence converges to $x$, $d(x,\tau(x)) \leq 0$ and hence $x = \tau(x)$.

Finally, we need to show that this fixed point is unique. Suppose there exists another fixed point $y \in X$ such that $y = \tau(y)$. Then:

\[ \begin{aligned} d(x,y) &= d(\tau(x), \tau(y))\\ &\leq Kd(x,y)\\ \end{aligned} \]

This implies $d(x,y) = 0$ and hence $x = y$ since $0 < K < 1$

Hence we have shown that the sequence of approximations converges to a point that is the fixed point, and that this fixed point is unique.

This completes the proof.