Question

Let \(Z\) be a hyperplane of a Banach space \(X .\) Show that if there exists a bounded linear operator \(T\) from some Banach space \(Y\) onto \(Z\), then \(Z\) is closed in \(X\).

Short answer

If \( T \) is a bounded linear operator mapping \( Y \) onto \( Z \), then \( Z \) is closed due to the closedness of the image of a closed set under a continuous mapping.

Step-by-step solution

Understand the Problem Statement

Given a hyperplane \( Z \) in a Banach space \( X \), we need to show that \( Z \) is closed if there exists a bounded linear operator \( T \) from some Banach space \( Y \) onto \( Z \).

Properties of a Bounded Linear Operator

Recall that a bounded linear operator \( T: Y \to Z \) is continuous, and being surjective means it maps \( Y \) onto \( Z \), i.e., for every element in \( Z \) there is a corresponding element in \( Y \).

Using the Definition of Continuity

Since \( T \) is continuous (bounded) by definition, the image of a closed set under \( T \) is also closed. In particular, the image of the whole space \( Y \), which is closed in itself, is closed in \( X \).

Conclude via Image Closedness

Thus, if \( T(Y) = Z \) and \( T \) is continuous, then \( Z \) must be closed in \( X \) because it is the continuous image of a closed space.

Key concepts

hyperplane

In Banach space theory, a hyperplane is a special subspace with an interesting structure. To start, a hyperplane in a Banach space is typically defined as a subspace of codimension one. This means it can be described as the kernel of a bounded linear functional. More formally, if we denote our Banach space by \(X\), a hyperplane \(Z\) can be written as: Z = \{ x \, \in \, X \, | \, f(x) = 0 \} Here, \(f\) is a non-zero bounded linear functional on \(X\). Consequently, hyperplanes play an important role because they help understand and analyze the underlying structure of Banach spaces. They are closely related to the idea of linear separation in functional analysis.

bounded linear operator

A bounded linear operator is a fundamental concept in functional analysis. To put it simply, if you have two Banach spaces, \(Y\) and \(X\), a linear operator \(T: Y \to X\) is termed as bounded if there exists a constant \(M\) such that: \(\|T(y)\| \leq M \cdot \|y\|\) for all \(y \in Y\). Here’s why bounded linear operators are crucial: * Continuity: They preserve the structure and continuity of spaces. This implies that small changes in \(Y\) will result in small changes in \(X\). * Mapping Properties: Bounded linear operators, especially when surjective (mapping \(Y\) onto \(Z\)), ensure that every element in \(Z\) is the image of some element in \(Y\). This feature is especially useful for proving other properties about the space, such as closedness.* Stability: In the context of our exercise, the bounded linear operator from \(Y\) onto the hyperplane \(Z\) guarantees that \(Z\) is a closed set due to its continuity.

closed set

In Banach space theory, understanding closed sets is essential. A set \(Z\) in a Banach space \(X\) is considered closed if it contains all its limit points. In other words, for every convergent sequence \((z_n)\) in \(Z\), its limit \(z\) also belongs to \(Z\).Why is this important?

1. It ensures completeness: Closed sets help maintain the structure of the space, ensuring that operations or functions defined on them behave nicely.
2. Image of continuous functions: A fundamental property of continuous (hence bounded) functions is that they map closed sets to closed sets.

In our exercise, this concept is pivotal. Since we know the bounded linear operator \(T\) is continuous and surjective, the image of the whole space \(Y\) under \(T\), which is \(Z\), must be closed. This ensures that if a sequence in \(Z\) converges, its limit is still within \(Z\). Combining these insights, one can clearly see why the hyperplane \(Z\) is closed when the conditions mentioned hold true.