Question

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

Short answer

The subspace \(Y\) is closed because it is the image of a bounded linear operator from a Banach space.

Step-by-step solution

Understanding Definitions

First, recall the definitions: A subspace is finite-codimensional if the dimension of its quotient space is finite. A space is closed if it contains all its limit points.

Contextualize the Problem

Given: A Banach space \(X\) and its finite-codimensional subspace \(Y\). Also given: A bounded linear operator \(T\) from another Banach space \(Z\) onto \(Y\). We need to show that \(Y\) is closed in \(X\).

Use the Bounded Operator

Since \(T\) is a bounded linear operator from the Banach space \(Z\) onto \(Y\), it implies \(T(Z) = Y\). This establishes a continuous and surjective operator mapping.

Consider the Closed Image of a Bounded Operator

Using the Closed Graph Theorem, which states that if a linear operator between Banach spaces has a closed graph, then the operator is bounded. Since \(Y\) is the image of a bounded operator, it must be closed.

Conclusion

Since \(T(Z) = Y\) and \(T\) is a bounded linear operator, by the properties of Banach spaces and bounded operators, \(Y\) is closed in \(X\).

Key concepts

Banach Space

A Banach space is a complete normed vector space. This means that every Cauchy sequence (a sequence where elements get arbitrarily close to each other) in the space converges to an element within the space. Consider spaces like \( \text{L}^p(\Omega) \) or \( \text{C}(K) \) — these are classic examples of Banach spaces.
A Banach space is important because it allows us to work with infinite-dimensional spaces in a structured, manageable way. The rules of convergence and continuity still apply, which is essential for functional analysis and many applications in fields such as engineering and physics.

Bounded Linear Operator

A bounded linear operator is a function between two Banach spaces that is both linear and has a bounded (or limited) norm. In simpler terms, if you apply this operator to a vector, the result won't be arbitrarily large unless the original vector is.
Mathematically, an operator \( T \) is bounded if there exists a constant \( C \) such that \[ ||T(x)|| \leq C \cdot ||x|| \] for all vectors \( x \) in the domain of \( T \).
This is crucial because a bounded operator ensures stability and predictability, meaning the transformation it performs won't lead to wild, unmanageable results.

Finite-Codimensional Subspace

A subspace \(Y \) is finite-codimensional in a Banach space \(X \) if the quotient space \(X / Y \) has finite dimension. Essentially, this means you can 'factor out' \(Y \) from \(X \) and what's left is finite-dimensional.

Understanding finite-codimensional subspaces is significant because they often retain many properties of the larger space while reducing complexity. For example, problems in infinite dimensions can sometimes be simplified by considering their finite-codimensional subspaces.

Closed Graph Theorem

The Closed Graph Theorem is a key result in functional analysis, stating that if a linear operator between Banach spaces has a closed graph, the operator must be bounded. A graph being closed means that if a sequence \( \{ (x_n, T(x_n)) \} \) in the graph of \( T \) converges to \( (x, y) \), then \( y \) must be equal to \( T(x) \).

This theorem unifies the concepts of boundedness and closedness, providing a robust tool for analysts to determine when an operator behaves well. When applying the Closed Graph Theorem, it simplifies the verification that an operator is well-behaved simply by checking that its graph is closed, which often is easier to establish than directly proving boundedness.