# 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.

Let \(x = \lim_{x \rightarrow \infty} A^{kn} x_0\) and \(Ax = \lim_{k \rightarrow \infty} A A^{kn} x_0\).

**Theorem 2** (Picard)

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 \[\phi_1(x_0) = y_{01}, \ldots, \phi_n(x_0) = y_{0n}\]

The *Fredholm equation (of the second kind)* and *Volterra equation* are
defined in the book.

## Problems

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.

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 $λ = \frac{1}{K_1 + K_2} since \[\lambda F^\prime(x) \leq K_2 = \frac{K_2}{K_1 + K_2} < 1\] 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 \(F(x) = 0\).

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.**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.

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.

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

Easy.

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.

Beats me.

**TODO.**