= Solution
A map $h:X\to X$ is a contraction if there is a constant $q<1$ such that
$$
d(hx,hy)\leq qd(x,y)
$$
for every $x,y\in X$.
The <contraction mapping theorem> states that a contraction of a nonempty complete metric space has a unique fixed point, and that the iterates from every starting point converge to it. To prove this, choose $x_0\in X$ and put $x_{n+1}=h(x_n)$. Then
$$
d(x_{n+1},x_n)\leq q^nd(x_1,x_0),
$$
so, for $m>n$,
$$
d(x_m,x_n)\leq\frac{q^n}{1-q}d(x_1,x_0).
$$
Thus $(x_n)$ is Cauchy and converges, by completeness, to some $x_*$. A contraction is continuous, so
$$
h(x_*)=\lim_nh(x_n)=\lim_nx_{n+1}=x_*.
$$
If $y_*$ is another fixed point, then
$$
d(x_*,y_*)\leq qd(x_*,y_*),
$$
forcing $x_*=y_*$.
For the Newton map
$$
g(x)=x-\frac{f(x)}{f'(x)},
$$
one has $g(r)=r$ and
$$
g'(x)=\frac{f(x)f''(x)}{f'(x)^2},
\qquad
\boxed{g'(r)=0}.
$$
On the given neighbourhood,
$$
|g'(x)|\leq\frac{M|f(x)|}{\delta^2}.
$$
Since $f(r)=0$, choose a closed interval $U'$ centred at $r$ and contained in $U$ so small that $M|f(x)|/\delta^2\leq1/2$ there. Then
$$
|g(x)-r|\leq\frac12|x-r|,
$$
so $g(U')\subseteq U'$ and $g$ is a contraction on the complete interval $U'$. The <local contraction proof for Newton iteration> therefore shows that $r$ is the unique fixed point of $g$ on $U'$.
Back to article page
