Question

Let \(f\) be a linear functional on a Banach space \(X\). Show that \(f\) is continuous if and only if \(f^{-1}(0)\) is closed. Show that if \(f\) is not continuous, then \(f^{-1}(0)\) is dense in \(X\).

Short answer

A functional \(f\) is continuous if and only if \(f^{-1}(0)\) is closed. If \(f\) is not continuous, then \(f^{-1}(0)\) is dense in \(X\).

Step-by-step solution

Definition of Continuity

A functional \(f\) on a Banach space \(X\) is continuous if for every sequence \((x_n)\) in \(X\) that converges to \(x\in X\), the sequence \(f(x_n)\) converges to \(f(x)\).

Definition of Inverse Image

The set \(f^{-1}(0)\) is defined as \(\{x \in X : f(x) = 0\}\). We need to prove that \(f\) is continuous if and only if \(f^{-1}(0)\) is closed.

Prove 'If' Direction – Continuity Implies Closed

Assume \(f\) is continuous. Consider a converging sequence \((x_n) \to x \) where \(x_n \in f^{-1}(0)\). Since \(f(x_n) = 0\) for all \(n\) and \(f\) is continuous, we have \(f(x) = \lim f(x_n) = 0\). Thus, \(x \in f^{-1}(0)\). Hence, \(f^{-1}(0)\) is closed.

Prove 'Only If' Direction – Closed Implies Continuity

Assume \(f^{-1}(0)\) is closed. Suppose \(f\) is not continuous. Then there exists a bounded sequence \((x_n)\) converging to \(x \in X\) such that \(f(x_n - x) \to c eq 0\). Set \(y_n = x_n - x\). Since \((y_n) \to 0\) and \((y_n + \frac{c}{\|f\|^2} x) \in f^{-1}(0)\), this shows \(0 \in f^{-1}(0)\) is not closed, a contradiction.

If f is not Continuous

Assume \(f\) is not continuous. Using the Baire Category Theorem, if \(f^{-1}(0)\) is not closed, then \(f^{-1}(0)\) has a dense complement. This implies \(f^{-1}(0)\) is dense in \(X\). Hence, non-continuity of \(f\) forces \(f^{-1}(0)\) to be dense.

Key concepts

Banach space

Understanding Banach spaces is essential when working with functional analysis. A Banach space is a complete normed vector space. This means every Cauchy sequence in the space converges within the space. Examples include spaces like \(L^p\) spaces for \((1 \leq p \leq \infty)\). To grasp the concept deeply, remember:

• \

inverse image

In mathematics, the inverse image (or preimage) of a set under a function is critical to understand. For any set \(B\) in the target space, the inverse image \(f^{-1}(B)\) is the set of all points in the domain that map to \(B\). For the linear functional \(f\) in our exercise, \(f^{-1}(0)\) is a special inverse image since it represents the kernel or null space of \(f\). This means it's the collection of all points in the Banach space \(X\) that map to zero.

• Kernel represents solutions to \(f(x) = 0\)
• The nature of the inverse image, e.g., being closed, tells us about the continuity of \(f\).

Baire Category Theorem

The Baire Category Theorem is a fundamental result in topology with far-reaching implications. It states that any complete metric space (like Banach spaces) is of second category, meaning it cannot be expressed as a countable union of nowhere-dense sets. This theorem helps us understand the structure of Banach spaces better and applies significantly to functional analysis.
In the exercise, we use the Baire Category Theorem to infer that \(f^{-1}(0)\) is dense in \((X)\) if the functional \(f\) is not continuous. This deep result connects topological properties of spaces with the behavior of functions defined on them.
Understanding how these connect:

• If \(f^{-1}(0)\) is not closed, its complement is dense in \(X\).
• Thus, the non-continuity of \((f)\) effectively forces \((f^{-1}(0))\) to be dense in \(X\).