# Metric Spaces: Contraction Mappings

Definition Let $$A$$ be a mapping of a metric space $$R$$ onto itself. Then $$x$$ is a fixed point of $$A$$ if $$Ax = x$$. Suppose there exists $$\alpha < 1$$ such that $$\rho(Ax, Ay)$$ for all $$x, y \in R$$, then $$A$$ is a contraction mapping. Every contraction mapping is automatically continuous, since $$Ax_n \rightarrow Ax$$ whenever $$x_n \rightarrow x$$.

Theorem 1 (Fixed point theorem)

Every contraction mapping $$A$$ defined on a complete metric space $$R$$ has a unique fixed point.

Proof

Keep contracting, i.e. applying $$A$$.

Theorem 1’

Given a continuous mapping of a complete metric space $$R$$ onto itself, suppose $$A^n$$ is a contraction mapping ($$n$$ an integer $$>1$$). Then $$A$$ has a unique fixed point.

Proof

Let $$x = \lim_{x \rightarrow \infty} A^{kn} x_0$$ and $$Ax = \lim_{k \rightarrow \infty} A A^{kn} x_0$$. Since $$A^n$$ is a contraction mapping, then $\rho(Ax, x) = \lim_{k \rightarrow \infty} \rho(AA^{kn}x_0, A^{kn} x_0) = 0$ i.e. $$Ax = x$$ so $$x$$ is a fixed point of $$A$$. Has to be unique, otherwise $$A^n$$ also has multiple fixed points, which is impossible by Theorem 1 since $$A^n$$ is a contraction mapping.

Note The most interesting applications of Theorems 1 and 1’ arise when the space $$R$$ is a function space. Below we find out the conditions for when a solution for initial value problems exist (and is unique).

Theorem 2 (Picard, also known as Picard–Lindelöf)

Given a function $$f(x,y)$$ defined and continuous on a plane domain $$G$$ containing the point $$(x_0, y_0)$$, suppose $$f$$ satisfies a Lipschitz condition of the form $|f(x,y) - f(x,\tilde{y})| \leq M |y - \tilde{y}|$ in the variable $$y$$. Then there is an interval $$|x - x_0| \leq \delta$$ in which the differential equation $\frac{dy}{dx} = f (x, y)$ has a unique solution $y = \phi(x)$ satisfying the initial conditions $\phi(x_0) = y_{0}$

Theorem 2’

Given $$n$$ functions $$f_i (x, y_1, \ldots, y_n)$$ defined and continuous on an $$(n+1)$$​-dimensional domain $$G$$ containing the point $(x_n, y_{01}, \ldots, y_{0n})$ suppose each $$f_i$$ satisfies a Lipschitz condition of the form $|f_i(x, y_1, \ldots, y_n) - f_i(x, \tilde{y}_1, \ldots, \tilde{y}_n)| \leq M \max_{1 \leq i \leq n} |y_i - \tilde{y}_i|$ in the variables $$y_1, \ldots, y_n$$. Then there is an interval $$|x - x_i| \leq \delta$$ in which the system of differential equations $\frac{dy_i}{dx} = f_i (x, y_1, \ldots, y_n) \quad (i = 1, \ldots, n)$ has a unique solution $y_1 = \phi_1(x), \ldots, y_n = \phi_n(x)$ satisfying the initial conditions $\phi_1(x_0) = y_{01}, \ldots, \phi_n(x_0) = y_{0n}$

For additional information Check out libretext or these notes by Denise Gutermuth from the University of Berkeley.

Note Now show the method of successive approximations can be used to prove the existence and uniqueness of solutions of two kinds of integral equations: Fredholm and Volterra.

Example 1 A Fredholm equation (of the second kind) is an integral equation of the form $f(x) = \lambda \int_b^a K(x, y) f(y) dy + \phi(x)$ involving two given functions $$K$$ and $$\phi$$, an unknown function $$f$$ and an arbitrary parameter $$\lambda$$.

Example 2 A Volterra equation $f(x) = \lambda \int_b^x K(x, y) f(y) dy + \phi(x)$ differs from the Fredholm equation by having the variable $$x$$ rather than the fixed number $$b$$ as the upper limit of integration.

## Problems

1. Let $$A$$ be a mapping of a metric space $$R$$ into itself. Prove the condition $\rho(Ax, Ay) < \rho(x, y) \quad (x \neq y)$ is insufficient for the existence of a fixed point of f$$A$$.

Let $$R$$ be $$(0, 1)$$ and $$A = \frac{1}{2}$$, then fixed point would be $$0$$, but $$0 \notin R$$. Need $$R$$ to be complete.

2. Let $$F(x)$$ be a continuously differentiable function on the interval $$[a, b]$$ such that $$F(a) < 0$$, $$F(b)> 0$$ and $0 < K_1 \leq F^\prime(x) \leq K_2 (a \leq x \leq b)$ Use Theorem 1 to find the unique root of the equation $$F(x) = 0$$

Hint. Introduce the auxiliary function $$f(x) = x - \lambda F(x)$$, and choose $$\lambda$$ such that the theorem works for the equivalent equation $$f(x) = x$$.

$f(x) = x - \lambda F(x) \quad f^\prime(x) = 1 - \lambda F^\prime(x)$ Select λ so that $$|f^\prime(x)| = |1 - \lambda F^\prime(x)| < 1$$, whenever $$\lambda F^\prime(x) < 1$$ since $$F^\prime(x)$$ is strictly positive. Can do this by setting $$\lambda = \frac{1}{K_1 + K_2}$$ since $\lambda F^\prime(x) \leq \lambda K_2 = \frac{K_2}{K_1 + K_2} < 1$ By Theorem 1, $$f$$ has a fixed point $$x_0 \in [a, b]$$ such that $$f(x_0) = x_0$$, which means $$\lambda F(x_0) = 0$$, and $$x_0$$ is the unique root of $$F(x) = 0$$.

3. Devise a proof of the implicit function theorem based on the use of the fixed point theorem.

Don’t know what the implicit function theorem is. TODO.

4. Prove that the method of successive approximations can be used to solve the system $x_i = \sum_{j=1}^n a_{ij}x_j + b_i$ if $$|a_{ij}| < \frac{1}{n}$$ (for all $$i$$ and $$j$$), but not if $$|a_{ij}| = \frac{1}{n}$$.

Easy.

5. Prove that the condition $$\sum_{j=1}^n|a_{ij}| \leq \alpha < 1$$ is sufficient for the mapping $$y_i = \sum_{j=1}^n a_{ij}x_j + b_i$$ to be a contraction mapping in the space $$R_0^n$$.

Easy.

6. Prove that any of the $$n$$​-dimensional contraction mapping conditions imply $\det\left(\begin{matrix} a_{11}-1 & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22}-1 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}-1 \\ \end{matrix}\right) \not= 0$

If one of the conditions is satisfied, then $Ax = x \implies Ax - x = 0 \implies (A - I)x = 0$ meaning that $$A - I$$ has rank $$n$$ and is thus invertible and therefore has a nonzero determinant.

7. Consider the nonlinear integral equation $f(x) = \lambda\int_a^b K(x, y; f(y)) dy + \psi(x)$ with continuous $$K$$ and $$\psi$$, where $$K$$ satisfies a Lipschitz condition of the form $|K(x, y; z_1) - K(x, y; z_2)| \leq M|z_1 - z_2|$ in its ‘functional’ argument. Prove that the equation has a unique solution for all $|\lambda| < \dfrac{1}{M(b-a)}$ Write the successive approximations to this solution.

Tweak the method used for Fredholm’s equation.