Question

Let \(X\) be a separable Banach space whose norm is Fréchet differentiable. Show that if \(Y \subset X^{*}\) is a closed 1-norming subspace of \(X^{*}\), then \(Y=X^{*}\)

Short answer

Since \(Y\) is norming and the norm is Fréchet differentiable, \(Y\) must be the whole dual space \(X*\).

Step-by-step solution

Understand the Definitions

Recall that a Banach space is a complete normed vector space. A norm is Fréchet differentiable if the limit defining the derivative exists and is linear. A subspace of the dual space is 1-norming if the norm of any element in the Banach space can be computed using elements from the subspace.

Define Key Properties

Given a closed 1-norming subspace Y of the dual space X*, we know that for any vector x in X, we have \(orm{x} = orm{x}_{X} = \text{sup}\brace1{\frac{\brace}{orm{y}}: y eq 0, y \text{is in } Y}\).

Show Inclusion

To show \(Y = X*\), prove two inclusions: \(Y ewline \text{subset} X*\) and \(X* ewline \text{subset} Y\).

Prove Y subset X*

This is straightforward since \(Y \) is by definition a subset of \(X*\), the dual space.

Prove X* subset Y

Take any \(x^* \in X*\). Since \(X\)'s norm is Fréchet differentiable, the norm is Gateaux differentiable, and every bounded linear functional represents a continuous linear functional on a reflexive space. Every norm-attaining function is unique by the property of the Fréchet differentiability of the norm and the Hahn-Banach theorem.

Apply Hahn-Banach Theorem

By Hahn-Banach's theorem and 1-norming property, for every \(x\) in \(X\) there exists a \(y\) in \( Y\), satisfying the equation: \(orm{x}_X = \brace\). Thus \( X \subset Y \)

Conclude the Proof

Since we have shown \(Y ewline \text{subset} X*\) and \(X* ewline \text{subset} Y\), which conclude \(Y = X*\). Therefore, the closed 1-norming subspace Y of \(X*\) must be the entire dual space \(X*\).

Key concepts

Fréchet differentiability

Fréchet differentiability is a key concept in functional analysis. It describes how the norm behaves smoothly in a Banach space. A norm is Fréchet differentiable at a point if the limit process that defines the derivative converges and this derivative is linear.

In simpler terms, think of it like how a curved line can be perfectly approximated by a straight line slope at a particular point (in many variables). In a more mathematical sense, for a function to be Fréchet differentiable at a point, there has to be a unique linear map that approximates the function closely around that point.

This property is crucial because it ensures the norm behaves nicely, which helps with various mathematical proofs involving Banach spaces. In the context of our exercise, the fact that the norm is Fréchet differentiable is vital in leveraging the properties of the dual space and applying the Hahn-Banach theorem effectively.

1-norming subspace

A 1-norming subspace is a subspace of the dual space of a Banach space that can measure the norm of every element in the original space. This means you can use elements from this subspace to calculate the norm of any element in the Banach space.

In our exercise, we work with a closed 1-norming subspace, which means for any vector in the Banach space, one can find its norm using only the vectors from this subspace. To mathematically frame this, if you have a Banach space X and a subspace Y of its dual space (X*), Y is a 1-norming subspace if for every x in X, the norm of x can be represented as:

\(\text{norm}(x) = \text{sup}\big\frac{\big|< y, x >\big|}{\text{norm}(y)}: y e 0, y \text{ is in } Y\big). \)

This property implies that Y is sufficiently 'rich' to capture the entire geometry of X through its dual elements. The exercise uses this concept to show that such a 1-norming subspace must actually cover the entire dual space.

Hahn-Banach theorem

The Hahn-Banach theorem is a foundational result in functional analysis. This theorem allows the extension of bounded linear functionals defined on a subspace of a normed vector space to the whole space, without increasing its norm.

Here's the intuition: given a function that works well on a part of a space, the Hahn-Banach theorem says we can extend this function to the whole space while keeping it 'nice' (bounded and linear).

In our exercise, the Hahn-Banach theorem plays a crucial role. It ensures that every bounded linear functional on a reflexive Banach space (like X) can be extended in a controlled manner. This is used to show that the 1-norming subspace Y must include all the elements of the dual space X*. By extending bounded functionals defined on 1-norming subspace Y to the entire space (using the differentiability properties), we conclude that Y must actually be the whole dual space X*.