LEC CAL1 10

The Banach contraction principle

Definition 179.1(Definition).

A point $x\in X$ is called a fixed point of the mapping $T:X\to X$ if $T(x)=x$.

Definition 179.2(Definition).

A mapping $T$ from a metric space $(X,\rho )$ into itself is said to be Lipschitz if there exists $c\ge 0$, called a Lipschitz constant for the mapping, for which

$$\rho (T(u),T(v))\le c\rho (u,v)\forall u,v\in X.$$

It is clear that a Lipschitz mapping is uniformly continuous. If $c<1$, the Lipschitz mapping is called a contraction.

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Theorem 179.3(Banach contraction principle).

Let $X$ be a complete metric space and the mapping $T:X\to X$ be a contraction. Then $T$ has exactly one fixed point.

Proof.

Choose any$x_0\in X$. Let $x_n=T^n{x}_0$. For $n\le m$,

$$\begin{array}{rl}d(x_n,x_m)& \le d(T^n{x}_0,T^m{x}_0)\\ & \le c^{m}d(x_0,T^{m-n}{x}_0)\\ & \le c^m(d(x_0,T{x}_0)+\cdots +d(T^{n-m+1}{x}_0,T^{n-m}{x}_0))\\ & \le c^m\left(\frac{d(x_0,T{x}_0)}{1-c}\right)\to 0\text{ as }n,m\to \infty \end{array}$$

So, $\{T^n{x}_0\}$ is Cauchy. Since $X$ is complete, it converges, say to $x$. Since $T$ is continuous, $T(x)$ must be $x$. Showing uniqueness is trivial.□

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