| course |
course_year |
question_number |
tags |
title |
year |
Analysis II |
IB |
19 |
|
1.II.11F |
2008 |
State and prove the Contraction Mapping Theorem.
Let $(X, d)$ be a nonempty complete metric space and $f: X \rightarrow X$ a mapping such that, for some $k>0$, the $k$ th iterate $f^{k}$ of $f$ (that is, $f$ composed with itself $k$ times) is a contraction mapping. Show that $f$ has a unique fixed point.
Now let $X$ be the space of all continuous real-valued functions on $[0,1]$, equipped with the uniform norm $|h|_{\infty}=\sup {|h(t)|: t \in[0,1]}$, and let $\phi: \mathbb{R} \times[0,1] \rightarrow \mathbb{R}$ be a continuous function satisfying the Lipschitz condition
$$|\phi(x, t)-\phi(y, t)| \leqslant M|x-y|$$
for all $t \in[0,1]$ and all $x, y \in \mathbb{R}$, where $M$ is a constant. Let $F: X \rightarrow X$ be defined by
$$F(h)(t)=g(t)+\int_{0}^{t} \phi(h(s), s) d s$$
where $g$ is a fixed continuous function on $[0,1]$. Show by induction on $n$ that
$$\left|F^{n}(h)(t)-F^{n}(k)(t)\right| \leqslant \frac{M^{n} t^{n}}{n !}|h-k|_{\infty}$$
for all $h, k \in X$ and all $t \in[0,1]$. Deduce that the integral equation
$$f(t)=g(t)+\int_{0}^{t} \phi(f(s), s) d s$$
has a unique continuous solution $f$ on $[0,1]$.