$$\def\R{\mathbb{R}} \def\N{\mathbb{N}} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\eps{\varepsilon} \def\epsilon{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant}$$

# Chapter 2 - Metric Spaces

## 2.8 Contraction Mappings

 Main: Index Previous: 2.7 Complete Metric Spaces Next: 3.9 Basic Concepts

### Results

Contraction Mapping
Let $A$ be a mapping of a metric space $R$ into itself. Then $x$ is called a fixed point of $A$ if $Ax = x$, i.e. if $A$ carries $x$ into itself. Suppose there is some number $\alpha < 1$ such that $$\rho(Ax, Ay) \leq \alpha\cdot \rho(x, y)$$ for every pair of points $x,y\in R$. Then $A$ is said to be a contraction mapping.

Theorem 1 (Fixed Point Theorem)

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

Theorem 1'

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

Theorem 2 (Picard)

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, $$|f(x,y) - f(x,\overset{\sim}{y})| \leq M|y - \overset{\sim}{y}|$$ 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 condition $$\phi(x_0) = y_0$$

### Problem 1

Let $A$ be a mapping of a metric space $R$ into itself. Prove that the condition $$\rho(Ax, Ay) < \rho(x, y), \quad(x\not= y)$$ is insufficient for the existence of a fixed point of $A$.

Proof.
Theorem 1 only works in the cases where the metric space is complete. To construct a counterexample, consider the metric space where $X = (0,1)$ and $\rho(x,y) = |x - y|$, and the mapping is given by the function $$Ax = \frac{x}{2}.$$ Then clearly, $$\rho(Ax, Ay) = \left|\frac{x}{2} - \frac{y}{2}\right| = \frac{1}{2}|x - y| < |x - y| = \rho(x,y)$$ The fixed point is the solution of $Ax = x$: $$Ax = x \;\Longrightarrow\; \frac{x}{2} = x \;\Longrightarrow\; x = 0$$ but $0\not\in(0,1)$, so the fixed point does not exist in the metric space, and the condition is insufficient by itself.

### Problem 2

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

Proof.
Define the function $$f(x) = x - \lambda F(x)$$ where $\lambda$ is some parameter. Since $F$ is continuous, we have $F\in C_{[a,b]}$, which is a complete metric space, and it follows that $f\in C_{[a,b]}$. Note that, $$f'(x) = 1 - \lambda F'(x).$$ To ensure that this is a contraction mapping, we select $\lambda$ so the condition $|f'(x)| < 1$ is satisfied. $$|f'(x)| = |1 - \lambda F'(x)| < 1,$$ whenever $\lambda F'(x) < 1$, since $F'(x)$ is strictly positive. This can be achieved by setting, $$\lambda = \frac{1}{K_1 + K_2},$$ since $$\lambda F'(x) \leq \lambda K_2 = \frac{K_2}{K_1 + K_2} < 1.$$ By Theorem 1, $f$ has a fixed point $x_0\in[a,b]$ such that $f(x_0) = x_0$, which means that $\lambda F(x_0) = 0$, and hence $x_0$ is the unique root of $F(x) = 0$.

### Problem 3

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

Skipped... This theorem is not stated in the text.

### Problem 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 $|a_{ij}| < 1/n$ (for all $i$ and $j$), but not if $|a_{ij}| = 1/n$.

Proof.
As shown in Example 2, there are two conditions depending on what metric is used. The contraction condition is satisfied if, $$\sum_{j=1}^n|a_{ij}| \leq \alpha < 1,$$ $$\sum_{i=1}^n\sum_{j=1}^n a^2_{ij} \leq \alpha < 1.$$ Suppose $|a_{ij}| < 1/n$. Then in the first case: $$\sum_{j=1}^n|a_{ij}| < \sum_{j=1}^n\frac{1}{n} = \frac{1}{n}\sum_{j=1}^n 1 = 1,$$ and in the second case, $$\sum_{i=1}^n\sum_{j=1}^n a^2_{ij} < \sum_{i=1}^n\sum_{j=1}^n \frac{1}{n^2} = \frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n 1 = \frac{1}{n}\sum_{i=1}^n 1 = 1.$$ Showing that the contraction condition is satisfied for all three metrics. If we have equality, then each of these sums become equal to 1, and the contraction condition is not satisfied.

### Problem 5

Prove that the condition: $$\sum_{j=1}^n|a_{ij}| \leq \alpha < 1 \tag{6}$$ is necessary for the mapping: $$y_i = \sum_{j=1}^n a_{ij}x_j + b_i \tag{5}$$ to be a contraction mapping in the space $R_0^n$.

Proof.
The metric for $R_0^n$ is given by: $$\rho(x,y) = \max_{1\leq i\leq n}|x_i - y_i|.$$ We assume that the mapping $A$ is a contraction mapping, and show that (6) must follow (i.e. that it is a necessary condition). By definition of a contraction mapping, then for some $\alpha < 1$, $$\rho(y, \overset{\sim}{y}) \leq \alpha\rho(x, \overset{\sim}{x}).$$ This is pretty much just following the result in the book. \begin{align} \rho(y, \overset{\sim}{y}) &= \max_{1\leq i\leq n}|y_i - \overset{\sim}{y}_i| \\ &= \max_{1\leq i\leq n}\left|\left(\sum_{j=1}^n a_{ij}x_j + b_i\right) - \left(\sum_{j=1}^n a_{ij}\overset{\sim}{x}_j + b_i\right)\right| \\ &= \max_{1\leq i\leq n}\left|\left(\sum_{j=1}^n a_{ij}x_j + b_i - a_{ij}\overset{\sim}{x}_j - b_i\right)\right| \\ &= \max_{1\leq i\leq n}\left|\sum_{j=1}^n a_{ij}(x_j - \overset{\sim}{x}_j)\right|\\ &\leq \max_{1\leq i\leq n}\sum_{j=1}^n |a_{ij}||x_j - \overset{\sim}{x}_j| \\ &\leq \max_{1\leq i\leq n}\sum_{j=1}^n |a_{ij}|\max_{1\leq j\leq n}|x_j - \overset{\sim}{x}_j|\\ &= \max_{1\leq i\leq n}\sum_{j=1}^n |a_{ij}|\cdot\rho(x, \overset{\sim}{x}). \end{align} Which shows that the previous condition is satisfied when $\alpha = \max_{1\leq i\leq n}\sum_{j=1}^n |a_{ij}| < 1$, which is also satisfied when: $$\sum_{j=1}^n|a_{ij}| \leq \alpha < 1$$ for $i = 1,2,\ldots,n$.

### Problem 6

Prove that any of the $n$-dimensional contraction mapping conditions imply that: $$\begin{vmatrix} a_{11}-1 & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22}-1 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}-1 \\ \end{vmatrix} \not= 0.$$

Proof.
Whenever one of the conditions mentioned are satisfied, this means we have an $n$-dimensional contraction mapping, this means we solve the equation of the form $Ax = x$. Which means: $$Ax - x = 0 \;\Longrightarrow\; (A-I)x = 0.$$ By assumption, this homogeneous linear system has a solution, which means $A-I$ has rank $n$, which is equivalent to saying that it is invertible and therefore has a nonzero determinant: $\det(A-I)\not= 0$.

Skipped...