Fixed point theorem and its uses

The fixed point theorem is a useful tool for proving existence of solutions and also for convergence of iterative schemes. In fact the latter is used to prove the former. If the problem is to find the solution of

$$L(u) = 0$$

then we convert it to a problem of finding a fixed point

$$u = F(u)$$

Statement

Let $X$ be a complete metric space with distance $d(\cdot,\cdot)$ and let $F: X \to X$ be contractive, i.e., for some $0 < \alpha < 1$,

$$d( F(x), F(y) ) \le \alpha d(x,y), \forall x, y \in X$$ (5)

Then there exists a unique fixed point $u \in X$ such that

$$u = F(u)$$ (6)

Proof

Choose any $x_o \in X$ and define the sequence $(x_n)$ by the following iterative scheme

$$x_{n+1} = F(x_n)$$ (7)

If we can show that the sequence $(x_n)$ has a unique limit independent of the starting value then we would have proved the existence of the fixed point. The first step is to show that $(x_n)$ is a Cauchy sequence.

$$\begin{eqnarray*} d(x_{m+1}, x_m) &=& d(F(x_m), F(x_{m-1}))\\ &\le& \alpha d( x... ..._{m-1}, x_{m-2} ) \\ &.& \\ &.& \\ &\le& \alpha^m d(x_1, x_o) \end{eqnarray*}$$

Now using the triangle inequality we get, for $n > m$

$$\begin{eqnarray*} d(x_m, x_n) &\le& d(x_m, x_{m+1}) + d(x_{m+1}, x_{m+2}) + ....... ... d(x_o, x_1) \\ &\le& \frac{ \alpha^m }{1 - \alpha} d(x_o, x_1) \end{eqnarray*}$$

where the last inequality follows because $1 - \alpha^{n-m} < 1$. Now since $\alpha^m \to 0$ as $m \to \infty$ we see that

$$d(x_m, x_n) \to 0, \textrm{ as } m,n \to \infty$$

which proves that $(x_n)$ is a Cauchy sequence. Since $X$ is complete, this sequence converges to a unique limit $u$. It is now left to show that the limit is a fixed point.

$$\begin{eqnarray*} d(u, F(u)) &\le& d(u, x_m) + d(x_m, F(u)) \\ &\le& d(u, x_m) + \alpha d(x_{m-1}, u) \end{eqnarray*}$$

and since $u$ is the limit of the sequence we see that the right hand side goes to zero so that

$$d(u, F(u)) = 0 \Longrightarrow u = F(u)$$

The uniqueness of the fixed point follows easily. Let $u$ and $\bar{u}$ be two fixed points. Then

$$d(u, \bar{u} ) = d(F(u), F(\bar{u})) \le \alpha d(u, \bar{u})$$

and since $0 < \alpha < 1$ we conclude that

$$d(u, \bar{u}) = 0 \Longrightarrow u = \bar{u}$$