|
|
A2-metricspaces.pdf
THEOREM 6.4.2 (Contraction mapping theorem). Let $X$ be a nonempty complete metric space and suppose that $f: X \rightarrow X$ is a contraction. Then $f$ has a unique fixed point, that is, there is a unique $x \in X$ such that $f(x)=x$.
Proof. We begin by showing that there cannot be two fixed points. Suppose that $f\left(x_1\right)=x_1$ and that $f\left(x_2\right)=x_2$. Then we have
\[
d\left(x_1, x_2\right)=d\left(f\left(x_1\right), f\left(x_2\right)\right) \leqslant K d\left(x_1, x_2\right) .
\]
Since $d\left(x_1, x_2\right) \geqslant 0$ and $K<1$, we are forced to conclude that $d\left(x_1, x_2\right)=0$ and hence that $x_1=x_2$.
Now we show that there is a fixed point. The proof is constructive (and may be used in practical situations to find fixed points numerically). The idea is as follows. Pick an arbitrary $x_0 \in X$, and form the sequence of iterates $x_1:=f\left(x_0\right)$, $x_2:=f\left(x_1\right)$, and so on. We claim that (no matter which $x_0$ we started with) the sequence $\left(x_n\right)_{n=1}^{\infty}$ converges to some limit $x$, and that $f(x)=x$
To show that $\left(x_n\right)_{n=1}^{\infty}$ converges, it suffices to show that it is Cauchy, since $X$ is complete. To do this, first observe that by repeated use of the contraction property and the definition of the sequence $\left(x_n\right)_{n=1}^{\infty}$ we have
\[
d\left(x_n, x_{n-1}\right) \leqslant K d\left(x_{n-1}, x_{n-2}\right) \leqslant K^2 d\left(x_{n-2}, x_{n-3}\right) \leqslant \ldots \leqslant K^{n-1} d\left(x_0, x_1\right)
\]
(you could prove this formally by induction if you wanted). Therefore if $n>m$ we have
\[
\begin{aligned}
d\left(x_n, x_m\right) & \leqslant d\left(x_n, x_{n-1}\right)+\cdots+d\left(x_{m+1}, x_m\right) \\
& \leqslant\left(K^{n-1}+K^{n-2}+\cdots+K^m\right) d\left(x_0, x_1\right) \\
& \leqslant K^m\left(1+K+K^2+\ldots\right) d\left(x_0, x_1\right)=C K^m
\end{aligned}
\]
where $C=d\left(x_0, x_1\right) /(1-K)$ (by summing the geometric series).
It follows that if $n, m \geqslant N$ then $d\left(x_m, x_n\right) \leqslant C K^N$.
Since $K<1$, for any $\varepsilon>0$ there is some $N$ such that $C K^N<\varepsilon$, and therefore $\left(x_n\right)_{n=1}^{\infty}$ is indeed a Cauchy sequence.
Since $X$ is complete, $x_n \rightarrow x$ for some $x \in X$. To complete the proof we must show that $f(x)=x$. This is quite straightforward. Indeed, since $f$ is continuous
we have
\[
f(x)=\lim _{n \rightarrow \infty} f\left(x_n\right)=\lim _{n \rightarrow \infty} x_{n+1}=x .
\]
This finishes the proof.
Remarks. Over the years, many people have lost a mark in exam questions for forgetting that $X$ must be non-empty.
Let us conclude by giving a couple of examples to show that the hypotheses of the theorem are necessary. First, we observe that the weaker condition that $d(f(x), f(y))<d(x, y)$ for all $x \neq y$ is not sufficient. For instance, it may be checked that the function $f:[1, \infty) \rightarrow[1, \infty)$ defined by $f(x)=x+1 / x$ has this property, but it obviously has no fixed points.
More obviously, the requirement that $X$ is complete is important. For instance, if we define $f:(0,1) \rightarrow(0,1)$ by $f(x)=x / 2$ then clearly $f$ is a contraction, but $f$ has no fixed points in $(0,1)$. |
|
Czhang271828
发表于 2022-7-22 16:49
hbghlyj
发表于 2022-12-23 00:16
