Question

Let \(T\) be a contraction from a Banach space \(X\) into \(X .\) Show that \(F=I_{X}-T\) is an open one-to-one map from \(X\) onto \(X\).

Short answer

F is one-to-one because it preserves distances scaled by k. It is onto as every y in X can be written F(x). It is open as it maps open sets to open sets.

Step-by-step solution

Understanding Contraction Mapping

Given a contraction mapping, we know that there exists a constant, say, 0 < k < 1, such that for all x, y in X, we have \(orm{T(x) - T(y)} \leq k \orm{x - y}\). This will help us show that F is an open map and one-to-one.

Define the Map F

Define the map F by \(F(x) = x - T(x)\). The goal is to show that F is one-to-one and open, mapping X onto X.

Show F is One-to-One

Assume \(F(x_1) = F(x_2)\). This means \(x_1 - T(x_1) = x_2 - T(x_2)\). Rearranging gives \(x_1 - x_2 = T(x_1) - T(x_2)\). Using the contraction property, we get \(orm{x_1 - x_2} = orm{T(x_1) - T(x_2)} \leq k orm{x_1 - x_2}\). Since 0 < k < 1, this implies \(x_1 = x_2\), thus F is one-to-one.

Show F is Onto (Surjective)

Given any \(y \in X\), we need to find an \(x \in X\) such that \(F(x) = y\). Define a sequence \(x_{n+1} = T(x_n) + y\). Starting with \(x_0\), we show by induction that this sequence converges to a point \(x\). Since \(T\) is a contraction, \(x_n\) is a Cauchy sequence and converges to some \(x\) satisfying \(x - T(x) = y\), proving that F is onto.

Show F is Open

To show F is an open map, take an open set \(U \subseteq X\). For any \(x \in U\), since T is a contraction and continuous, there exists an \(\text{open ball around } x, \ \text{say}, \ x \in B \subseteq U\), such that the image under F is open. Hence, F(U) is open, proving F is an open map.

Key concepts

Banach Space

A Banach space is a complete normed vector space. Essentially, it's a space where you can measure lengths and distances using a norm, and every Cauchy sequence converges within the space.
This completeness ensures that sequences won't 'escape' the space as they converge.
Banach spaces are important in analysis and its applications because they provide a robust framework for various types of problems, including those involving differential equations and functional analysis.

Contraction Mapping

A contraction mapping on a metric space is a function that brings points closer together. Formally, a map T is a contraction if there is a constant k with 0 < k < 1 such that for all points x and y, the inequality \( \| T(x) - T(y) \| \leq k \| x - y \| \) holds.
This property is powerful because of the Banach fixed-point theorem, which states that any contraction mapping on a complete metric space has a unique fixed point. This means that there exists a single point x such that \( T(x) = x \).
Contraction mappings are extremely useful in proving the existence and uniqueness of solutions in various problems.

Surjective Map

A surjective (onto) map is a function that covers every point in the target space. This means that for every point y in the target space, there is at least one point x in the domain such that \( T(x) = y \).
In the context of the task, showing that our function F is surjective implies that for every y in the Banach space, we can find an x such that \( F(x) = y \).
This aspect of a function is crucial in determining that our map is an 'onto' mapping and ensures that no elements in our target space are missed.

One-to-One Mapping

A one-to-one (injective) map is a function where each element in the domain maps to a unique element in the target space. This means that if \( F(x_1) = F(x_2) \), then \( x_1 = x_2 \).
For the function F defined as \( F(x) = x - T(x) \), showing injectivity helps confirm that distinct elements in the domain map to distinct elements in the target space.
This property is essential for proving that F is a bijective map, meaning it is both injective and surjective.

Open Map

An open map is a function that maps open sets to open sets. In other words, if you take an open set in the domain, its image under the function will also be an open set in the target space.
To prove our function F is an open map, we need to show that every open ball in the domain Banach space remains open in the image under F.
This property is significant because it ensures that not only does our map cover the entire target space (surjective) and map distinct points to distinct points (injective), but it also preserves the 'open nature' of sets, which is a crucial aspect in topology and analysis.