Problem 11 Let $\left\\{A_{n}\right\\}$ b... [FREE SOLUTION]

Chapter 13: Problem 11

Let $\left\\{A_{n}\right\\}$ be a sequence of contraction maps on a complete metric space $(X, d)$ such that $$ d\left(A_{n}(x), A_{n}(y)\right) \leq \alpha d(x, y) $$ for all $x, y \in X$ and $n \in \mathbb{N}$ and some $\alpha<1$. Suppose that $\left\\{A_{n}(x)\right\\}$ converges to $A(x)$ for each $x \in X .$ Show that $A$ is a contraction mapping on $X$ and that its fixed point can be computed as $$ a=\lim _{n \rightarrow \infty} a_{n} \text { , } $$ where $a_{n}$ is the fixed point of the contraction map $A_{n}$.

Short Answer

Expert verified

The map $A$ is a contraction, and its fixed point can be found as $a = \lim_{n \to \infty} a_n$, where each $a_n$ is a fixed point of $A_n$.

Step by step solution

Understand What a Contraction Map Is

A contraction mapping is a function between two metric spaces where the distance between any two points in the image is always less than the distance between their pre-images, by some fixed ratio $ \alpha < 1 $. This implies that the map brings points closer together.

Apply the Definition to Each $A_n$

Given that each $A_n$ is a contraction map, by definition $d\left(A_{n}(x), A_{n}(y)\right) \leq \alpha d(x, y)$. This relationship holds for all $x, y \in X$ and indicates that $A_n$ makes every pair of points in its input space get proportionally closer by at least a factor of $\alpha < 1$.

Consider Convergence of $A_n(x)$ to $A(x)$

We know $\{A_n(x)\}$ converges to $A(x)$ for each $x \in X$. Therefore, $A$ should also map input points, bringing them closer together like each $A_n$ does.

Prove $A$ is a Contraction

Since $A_n(x)$ converges to $A(x)$, for any given $\epsilon > 0$, there exists a large enough $N$ such that for all $n > N$, $d(A_n(x), A(x)) < \epsilon/2$. Similarly, $d(A_n(y), A(y)) < \epsilon/2$. By the triangle inequality, $d(A(x), A(y)) \leq d(A(x), A_n(x)) + d(A_n(x), A_n(y)) + d(A_n(y), A(y))$. As each $A_n$ is a contraction, $d(A_n(x), A_n(y)) \leq \alpha d(x, y)$. Therefore, $A$ also is a contraction mapping because as $n \rightarrow \infty$, $d(A(x), A(y)) \leq (\alpha + \epsilon)d(x, y)$ and $\epsilon $ is arbitrarily small.

Show Existence of Fixed Point for $A$

Since $A$ is a contraction, by Banach's fixed-point theorem, $A$ has a unique fixed point $a$.

Fixed Point Convergence to $a$

For each $a_n$, the fixed point of $A_n$, $A_n(a_n) = a_n$ holds. Since $A_n(x)$ converges to $A(x)$, $A_n(a_n) \to A(a)$. This implies $a_n \to a$ because, by contraction properties, the points converge to the fixed point. Hence $a = \lim_{n \to \infty} a_n$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Banach Fixed-Point Theorem

The Banach fixed-point theorem is a critical concept in analysis and topology. It states that in a complete metric space, every contraction mapping has exactly one fixed point. This is invaluable because it assures us that under certain conditions, the process of iteratively applying a map will always lead to a single, stable solution.

A contraction mapping moves any two points closer together. For a map to be a contraction, there must exist a constant $ \alpha $, where $ 0 \leq \alpha < 1 $, such that the distance between any two points in the image is reduced by at least the factor $ \alpha $. This brings the points in a sequence close enough to each other so that they converge to a single point.

The Banach fixed-point theorem has vast applications, including proving the existence and uniqueness of solutions to differential equations and various problems in numerical analysis.

Metric Space

A metric space is a foundational concept in mathematics that provides a framework for discussing distances and geometric properties of sets. It consists of a set $X$ and a distance function (or metric) $d$ which defines the distance between any two elements in the set. This function must satisfy four key properties:

Non-negativity: $ d(x, y) \geq 0 $, and $ d(x, y) = 0 $ if and only if $ x = y $.
Symmetry: $ d(x, y) = d(y, x) $.
Triangle inequality: $ d(x, z) \leq d(x, y) + d(y, z) $.
All distances are positive unless the two points coincide.

In the context of contraction mappings, working within a metric space allows us to rigorously define and deal with the concepts of proximity and convergence. Knowing that the space is complete ensures that every Cauchy sequence has a limit in the set, paving the way for applying tools like the Banach fixed-point theorem.

Sequence Convergence

Sequence convergence is about how sequences of numbers or objects approach a particular point, called the limit. A sequence $ \{x_n\} $ converges to a limit $ L $ in a metric space if, for every positive number $ \epsilon $, there exists an index $ N $ such that for all $ n > N $, the distance $ d(x_n, L) < \epsilon $. This means the sequence "settles" towards $ L $ as it progresses.

Contraction mappings often lead to convergence. In a sequence generated by successive applications of a contraction $ A_n $, the elements get closer to the fixed point. This is advantageous because convergence implies predictability and stability in many mathematical and real-world problems. For our specific problem, the sequence of fixed points $ \{a_n\} $ of maps $ A_n $ converges to a fixed point $ a $ of the map $ A $. This ability of sequences to zero in on an exact point is a cornerstone of many mathematical principles and applications.

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Short Answer

Step by step solution

Understand What a Contraction Map Is

Apply the Definition to Each \(A_n\)

Consider Convergence of \(A_n(x)\) to \(A(x)\)

Prove \(A\) is a Contraction

Show Existence of Fixed Point for \(A\)

Fixed Point Convergence to \(a\)

Key Concepts

Banach Fixed-Point Theorem

Metric Space

Sequence Convergence

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