Proposition: The Fixed Point Theorem

Let $E$ be a nonempty complete metric space, $F : E \to E$ a function. Suppose that there exists a real number $k$ less than one such that for all $p,q \in E$ we have $$d(F(p),F(q)) \leq kd(p,q)$$ We call $F$ a contraction map. Then there exists a unique point $P \in E$ such that $F(P) = P$. Furthermore if $p_0$ is any point of $E$ and $p_1 = F(p_0)$, $p_2 = F(p_1)$, $p_3 = F(p_2)$, etc., then $$lim_{n \to \infty} p_n = P$$
Let $a,b \in \mathbb{R}, a < b$, and let $F: [a,b] \to [a,b]$ be a continuous function. Suppose that $F$ is differentiable on $(a,b)$ and that there exists a real number $k < 1$ such that $|F'(x)| \leq k$ for all $x \in (a,b)$. Then $F$ is a contraction map, so that the fixed point theorem is applicable to $[a,b]$ and $F$.

Proofs:

First, lets start with the sequence by letting $p_0$ be an arbitrary point of $E$ and letting $p_1,p_2,p_3,...$ be given by $$p_{n+1} = F(p_n), n = 0,1,2,...$$ For any integer $n > 0$ we have $$d(p_n,p_{n+1}) = d(F(p_{n-1}),F(p_n)) \leq k d(p_{n-1},p_n)$$ Repeated application of this gives $$d(p_n,p_{n+1}) \leq k^n d(p_0,p_1)$$ We can then check that the sequence $p$ is a Cauchy sequence. $E$ is complete, so this sequence converges to a limit, say $P$. The inequality $d(F(p),F(q)) \leq d(p,q)$ shows that $F$ is uniformly continuous, hence continuous. Thus $$F(P) = \lim_{n \to \infty} F(p_n) = \lim_{n \to \infty} p_{n+1} = P$$ To show that $P$ is the only point with the property that $F(P) = P$, suppose that $Q \in E, F(Q) = Q$. Then $$d(P,Q) = d(F(P),F(Q)) \leq kd(P,Q)$$ Since $k < 1$ this implies that $d(P,Q) = 0$, so $P = Q$.
Deep. So we need to show that for all $p,q \in [a,b]$ we have $|F(p) - F(q)| \leq k|p-q|$. This is clear if $p = q$, whereas if $p \neq q$ the mean value theorem gives us the existence of some $\zeta$ between $p$ and $q$ such that $$F(p) - F(q) = F'(\zeta) (p - q)$$ $$|F(p) - F(q)| = |F'(\zeta)||p - q| \leq k|p-q|$$

Proposition: The Simplest Case of the Implicit Function Theorem

Let $f$ be a continuous real-valued function on an open subset of $E^2$ that contains the point $(a,b)$, with $f(a,b) = 0$. Suppose that $\partial f / \partial y$ exists and is continuous on the given open subset and that $\frac{\partial f}{\partial y} (a,b) \neq 0$. Then there exist open intervals $U, V \subset \mathbb{R}$, with $a \in U$ and $b \in V$, such that there exists a unique function $\phi: U \to V$ such that $f(x,\phi(x)) = 0$ for all $x \in U$, and such that this function $\phi$ is continuous.

Proofs:

This is kind of like saying that given any reasonably behaved function $f(x,y) = 0$, you can usually find $y$ as a function of $x$ within some region. We begin by defining another real-valued function $F$ on the same open subset of $E^2$ on which $f$ is defined by $$F(x,y) = y - \frac{f(x,y)}{\frac{\partial f}{\partial y} (a,b)}$$ This $F$ has as basic properties that both it and it's partial derivative w.r.t. $y$ are continuous, $F(a,b) = b$, $\frac{\partial F}{\partial y}(a,b) = 0$, and for any $(x,y)$ the equation $f(x,y) = 0$ holds if and only if $F(x,y) = y$. Hang on, for constant $x$ we are looking for fixed points of $F(x,y)$, because those correspond to zeroes of $f$.

So to do this, we need to prove that within some region, $F$ acts like a contraction map. Which, if you stare at $F$ and think about $f$ hard enough, intuitively makes sense. The details go a little something like this. Pick an $r > 0$ such that the open ball in $E^2$ of center $(a,b)$ with radius $r$ is entirely contained in the open set on which $f$ is defined. Since $\partial F / \partial y$ is continuous and its partial derivative with respect to $y$ at $(a,b)$ is 0, we may assume that $r$ is taken so small that $|\partial F / \partial y| < 1/2$ at each point in the ball. Choose $k$ such that $0 < k < r$ and $h$ such that $0 < h < \sqrt{r^2 - k^2}$ and such that $|F(x,b) - b| < k / 2$ whenever $|x - a| < h$, this last demand being justifiable by the continuity of $F$. We shall prove the theorem with $U = (a-h, a+h)$, $V = (b-k, b + k)$. Consider any fixed $x \in U$. For any $y \in \mathbb{R}$ such that $|y - b| \leq k$ we have $$d((x,y),(a,b)) = \sqrt{(x - a)^2 + (y - b)^2} < \sqrt{h^2 + k^2} < r$$ so that $(x,y)$ is in our open ball of radius $r$. If also $y' \in \mathbb{R}$, $|y' - b| \leq k$, then by the mean value theorem we have $$F(x,y) - F(x,y') = \frac{\partial F}{\partial y} (x,y'')(y - y')$$ for some $y''$ between $y$ and $y'$ (or, if $y = y'$, for $y'' = y = y'$). The point $(x,y'')$ is also in our ball of radius $r$, so we deduce $$|F(x,y) - F(x, y')| \leq \frac{1}{2}|y - y'|$$ Also $$|F(x,y) - b| \leq |F(x,y) - F(x,b)|+|F(x,b) - b|$$ $$< \frac{1}{2}|y - b| + \frac{k}{2} \leq \frac{k}{2} + \frac{k}{2} = k$$ Thus the fixed point theorem applies, and for each $x$ we can find a unique $y*$ such that $F(x,y*) = y*$, or that $f(x,y*) = 0$. Hence our function is $\phi(x) = y*$.

Proposition: Existence and Uniqueness Theorems for Ordinary Differential Equations

Let $f$ be a continuous real-valued function on an open subset of $E^2$ that contains the point $(a,b)$. Suppose that there exists $M \in \mathbb{R}$ such that $$|f(x,y) - f(x,z)| \leq M |y - z|$$ whenever $(x,y)$ and $(x,z)$ are in the given open set. Then there exists $h \in \mathbb{R}$, $h > 0$, such that there exists one and only one real-valued function $\phi$ on $(a-h,a+h)$ such that $\phi'(x) = f(x,\phi(x))$ on this interval and $\phi(a) = b$.
Let $f_1, ... , f_n$ be continuous real-valued functions on an open subset of $E^{n+1}$ that contains the point $(a,b_1,...,b_n)$. Suppose that $f_1, ..., f_n$ satisfy Lipschitz conditions, that is there exists $M \in \mathbb{R}$ such that $$|f_i(x,y_1,...,y_n) - f_i(x,z_1,...,z_n)| \leq M ((y_1 - z_1)^2 + ... + (y_n - z_n)^2)^{1/2}$$ for $i = 1, ... , n$ whenever $(x,y_1,...,y_n)$ and $(x,z_1,...,z_n)$ are in the given open set. Then there exists $h \in \mathbb{R}, h>0$, such that there exists one and only one $n$-tuple of real-valued functions $(\phi_1,...,\phi_n)$ on $(a-h,a+h)$ such that for $i = 1,...,n$ we have $\phi_i'(x) = f_i(x,\phi_1(x),...,\phi_n(x))$ on this interval and $\phi_i(a) = b_i$.

Proofs:

Same __ different __. You fill in the gaps. Finding a $\phi(x)$ such that $\phi' (x) = f(x,\phi (x))$ and $\phi(a) = b$ is equivalent to finding a $\phi(x)$ to solve the "integral equation" $$\phi(x) = \int_a^x f(t,\phi (t))dt + b$$ If to a function $\psi$ we associate another function $F(\psi)$ whose value at any $x$ is $(F(\psi))(x) = \int_a^x f(t,\psi (t))dt + b$, we see that (illuminati) solving the integral equation is the same as finding a function $\phi$ such that $F(\phi) = \phi$, that is, solving a kind of fixed point problem; this is the basic idea. To fill in the details, we want to be working in the metric space of all continuous real-valued functions on the compact metric space $[a-h,a+h]$ such that the distance between any two functions is defined as the maximum between any two points of the images of the functions given the same input (end of chapter IV).