Question

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(T\) maps bounded closed sets in \(X\) onto closed sets in \(Y\), then \(T(X)\) is closed in \(Y\). Hint: Assume \(T\left(x_{n}\right) \rightarrow y \notin T(X)\). Put \(M=\operatorname{Ker}(T)\), set \(d_{n}=\operatorname{dist}\left(x_{n}, M\right)\) and find \(w_{n} \in M\) such that \(d_{n} \leq\left\|x_{n}-w_{n}\right\| \leq 2 d_{n} .\) If \(\left\\{x_{n}-w_{n}\right\\}\) is bounded, then \(T\left(x_{n}-w_{n}\right) \rightarrow y \in T(X)\), since the closure of \(\left\\{x_{n}-w_{n}\right\\}\) is mapped onto a closed set containing \(y\). If \(\left\|x_{n}-w_{n}\right\| \rightarrow \infty\), then since \(T\left(x_{n}-w_{n}\right) \rightarrow y\), we have \(T\left(\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right) \rightarrow 0 .\) By the hypothesis, \(M\) must contain a point \(w\) from the closure of \(\left\\{\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right\\}\) since 0 lies in the closure of the image of this sequence. Fix \(n\) so that \(\left\|\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}-w\right\|<1 / 3\). Then \(\left\|x_{n}-w_{n}-\right\| x_{n}-w_{n}\|w\| \leq \frac{1}{3}\left\|x_{n}-w_{n}\right\|<(2 / 3) d_{n}\) and \(w_{n}+\left\|x_{n}-w_{n}\right\| w \in M\) a contradiction.

Short answer

If \(T\) maps bounded closed sets in \(X\) onto closed sets in \(Y\), then \(T(X)\) is closed in \(Y\).

Step-by-step solution

Understanding the problem

Given two Banach spaces, \(X\) and \(Y\), and a linear bounded operator \(T \in \mathcal{B}(X, Y)\), the task is to show that if \(T\) maps bounded closed sets in \(X\) onto closed sets in \(Y\), then \(T(X)\) is closed in \(Y\).

Assume contrary and derive contradiction

Assume, for contradiction, that \(T(X)\) is not closed in \(Y\). This means there exists a sequence \(\{x_n\} \subset X\) such that \(T(x_n) \to y otin T(X)\).

Define kernel and distances

Let \(M = \operatorname{Ker}(T)\) be the kernel of \(T\). Define \(d_n = \operatorname{dist}(x_n, M)\), which is the distance from \(x_n\) to \(M\). For each \(n\), find \(w_n \in M\) such that \(d_n \leq \|x_n - w_n\| \leq 2d_n\).

Examine boundedness of sequence

Consider the sequence \(\{x_n - w_n\}\). We need to consider two cases: either \(\{x_n - w_n\}\) is bounded or it is unbounded.

Bounded case

If \(\{x_n - w_n\}\) is bounded, then \(T(x_n - w_n) \to y\). Since \(T\) maps bounded closed sets to closed sets, the closure of \(\{x_n - w_n\}\) is mapped onto a closed set containing \(y\), which contradicts \(y otin T(X)\).

Unbounded case

If \(\|x_n - w_n\| \to \infty\), then consider the normalized sequence \(\{\frac{x_n - w_n}{\|x_n - w_n\|}\}\). Since \(T(x_n - w_n) \to y\), we have \(T\left(\frac{x_n - w_n}{\|x_n - w_n\|}\right) \to 0\).

Leverage the hypothesis

By the hypothesis, the kernel \(M\) must contain a point \(w\) in the closure of the sequence \(\{\frac{x_n - w_n}{\|x_n - w_n\|}\}\), since \(0\) lies in the closure of this sequence's image.

Derive contradiction

Fix \(n\) such that \(\left\|\frac{x_n - w_n}{\|x_n - w_n\|} - w\right\| < \frac{1}{3}\). Then we get \(\left\|x_n - w_n - \|x_n - w_n\| w\right\| \leq \frac{1}{3} \|x_n - w_n\| < \frac{2}{3} d_n\). The point \(w_n + \|x_n - w_n\| w\) is in \(M\), leading to a contradiction.

Conclusion

The assumption that \(T(X)\) is not closed leads to contradictions in both bounded and unbounded cases. Therefore, \(T(X)\) must be closed in \(Y\).

Key concepts

Banach Spaces

Understanding Banach spaces is key to grasping the problem at hand. A Banach space is a complete normed vector space, meaning that every Cauchy sequence converges within the space. This property ensures that all the limits of sequences we work with in Banach spaces are already part of the space, allowing for thorough analysis.

Bounded Linear Operators

A bounded linear operator is a linear transformation between two normed vector spaces that maps bounded sets to bounded sets. These operators are crucial because they allow us to control the behavior of sequences under transformation. For example, given a bounded sequence in a Banach space X and a bounded linear operator T, the image of this sequence under T remains bounded in Y.

Closed Sets

In the context of functional analysis, a closed set is one that contains all its limit points. If a set contains a sequence that converges within a space, the limit of that sequence is also in the set. This concept is important because one of the conditions in the exercise is that T maps bounded closed sets in X to closed sets in Y. Understanding what it means for a set to be closed helps us understand why the mapping property is significant.

Kernel of a Linear Operator

The kernel of a linear operator T, denoted as Ker(T), is the set of elements in X that T maps to the zero vector in Y. The kernel is always a closed subspace of X. In this exercise, defining the kernel helps us analyze the behavior of sequences and distances to specific subspaces, aiding our contradiction argument.

Distance to a Subspace

The distance from a point x to a subspace M is defined as the infimum of the norms of the differences between x and elements of M. This notion helps in measuring how far elements are from a given subspace, and it plays a critical role in forming the sequences used for our contradiction arguments. Understanding distances to subspaces allows us to break down complex problems into solvable subproblems.