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Chapter 8: Problem 86

Let \(X\) be a Banach space. Let \(\|\cdot\|\) be an equivalent norm on \(X^{*}\) such that the \(w^{*}\) - and norm topologies coincide. Show that then \(\|\cdot\|\) is a dual norm on \(X^{*}\).

Short Answer

Expert verified
\(\|\bullet\|\) is a dual norm on \(X^{*}\) because the weak-* topology and norm topology coincide, implying compactness in the norm topology.

Step by step solution

01

Understand the Problem Statement

Given a Banach space \(X\) and an equivalent norm \(\|\bullet\|\) on \(X^{*}\) where the weak-* topology (\(w^{*}\)) and the norm topology coincide, we need to show that \(\|\bullet\|\) is a dual norm on \(X^{*}\).
02

Define Equivalent Norms

Recall that two norms \(\|\bullet\|_1\) and \(\|\bullet\|_2\) on a space are equivalent if there exist constants \(C_1 > 0\) and \(C_2 > 0\) such that for all \(x \in X\), \(C_1 \|x\|_1 \leq \|x\|_2 \leq C_2 \|x\|_1\).
03

Define Weak-* Topology

The weak-* topology on \(X^{*}\) is defined by the weakest topology for which all the evaluation maps \(\phi_x: X^{*} \rightarrow \mathbb{R}\) given by \(\phi_x(f) = f(x)\) for all \(x \in X\) are continuous.
04

Weak-* Topology and Norm Topology Coincidence

We are given that the weak-* topology and the norm topology on \(X^{*}\) coincide under the norm \(\|\bullet\|\). This is a significant condition that influences the nature of \(\|\bullet\|\).
05

Consequence of Topology Coincidence

Since the weak-* and norm topologies coincide under \(\|\bullet\|\), this implies that the closed unit ball in \(X^{*}\) is compact in the norm topology of \(\|\bullet\|\).
06

Apply Banach-Alaoglu Theorem

By the Banach-Alaoglu theorem, the closed unit ball in \(X^{*}\) is compact in the weak-* topology. Given that the weak-* topology and norm topology coincide, this implies compactness in the norm topology.
07

Identify Dual Norm

The condition of compactness in the norm topology and the equivalence of norms suggests that \(\|\bullet\|_D\) is equivalent to an operator norm. Therefore, \(\|\bullet\|\) must be a dual norm.
08

Conclusion

Given that the weak-* and norm topologies coincide, and the closed unit ball is compact in the norm topology, it follows that \(\|\bullet\|\) is a dual norm on \(X^{*}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Banach Space
A Banach space is a type of vector space that's equipped with a norm and is complete with respect to this norm. The idea of completeness means that every Cauchy sequence in this space will converge to a point within the space. Here's what you need to know about Banach spaces:

  • **Vector Space:** A collection of elements called vectors where you can perform vector addition and scalar multiplication.
  • **Norm:** A function that assigns a non-negative length or size to each vector in the space.
  • **Complete:** Every Cauchy sequence (a sequence where elements get arbitrarily close to each other) has a limit that is also in the space.
Banach spaces are critical in functional analysis as they allow for the generalization and extension of methods from finite-dimensional spaces to those that are infinite-dimensional.
Weak-* Topology
The weak-* topology is a topology on the dual space of a Banach space. This topology is fundamental in the study of Banach spaces because it often allows us to exploit compactness properties we wouldn't have in the norm topology.

  • **Dual Space:** The set of all continuous linear functionals on a Banach space.
  • **Weak-* Topology:** The weakest topology such that all evaluation maps \(\phi_x(f) = f(x)\) for \( x \in X \) are continuous.
  • **Evaluation Map:** Takes an element from the dual space and evaluates it at a fixed vector from the original Banach space.
The weak-* topology on the dual space \( X^{*} \) is less fine than the norm topology, meaning fewer sets are open. This makes many compactness arguments possible with the weak-* topology.
Equivalent Norms
Two norms \( \| \bullet \|_1 \) and \( \| \bullet \|_2 \) on a vector space \( X \) are said to be equivalent if they induce the same topology. That is, they have essentially the same 'size' or 'shape' concept.

To be more precise: there exist constants \( C_1 > 0 \) and \( C_2 > 0 \) such that for all \( x \in X \): \( C_1 \| x \|_1 \leq \| x \|_2 \leq C_2 \| x \|_1\). This implies:
  • **Uniform Behavior:** Vectors in the space behave similarly under both norms.
  • **Topological Equivalence:** Open sets under one norm are also open under the other.
Equivalence ensures that if you prove something under one norm, it also holds for the other.
Banach-Alaoglu Theorem
The Banach-Alaoglu theorem is an essential result in functional analysis that deals with the compactness of the closed unit ball in the dual space of a normed space.

Here's a concise breakdown:

  • **Compactness:** A set is compact if every sequence has a subsequence that converges to a point within the set.
  • **Closed Unit Ball:** The set of all points in the dual space whose norm is less than or equal to 1.
  • **Dual Space Context:** For the dual space \( X^* \), the closed unit ball is compact in the weak-* topology even if it isn't in the norm topology.
The Banach-Alaoglu theorem shows that despite the high dimensionality of these spaces, we can still find compact sets, which is incredibly useful in various analyses.

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Most popular questions from this chapter

Show that the norm of \(C[0,1]\) is nowhere Fr茅chet differentiable. Show that the norm of \(C[0,1]\) is G芒teaux differentiable at \(x \in S_{C[0,1]}\) if and only if \(|x|\) attains its maximum at exactly one point of \([0,1]\). Hint: Note that the distance between two different Dirac measures in \(C[0,1]^{*}\) is two. Given \(x \in S_{C[0,1]}\), choose \(t_{0} \in[0,1]\) such that \(x\left(t_{0}\right)=1\). Then choose \(t_{n} \neq t_{0}\) such that \(x\left(t_{n}\right) \rightarrow 1\). By the 艩mulian lemma, \(x\) is not a point of Fr茅chet differentiability of the supremum norm on \(C[0,1]\). For the second part, assume that \(x \in S_{C[0,1]}\) is such that \(x\left(t_{0}\right)=1\) and \(|x(t)|<1\) for every \(t \neq t_{0} .\) Put \(H=\left\\{f \in C[0,1]^{*} ;\|f\| \leq 1, f(x)=1\right\\} .\) If \(H \cap B_{C[0,1] *} \neq\left\\{\delta_{t_{0}}\right\\}\), then this intersection would have at least two extreme points that would be extreme points of \(B_{C[0,1]^{*}}\). All the extreme points of \(B_{C[0,1]^{*}}\) are \(\pm\) Dirac measures (Lemma 3.42).

Show that every separable Banach space \(X\) can be renormed by an equivalent norm so that every convex set with more than one point has a non-diametral point.

Let \(X\) be an infinite-dimensional separable Banach space. Show that there is a continuous convex function \(f\) on \(X\) such that \(f\) is unbounded on \(B_{X}\)

Assume that \(X\) is a reflexive Banach space whose norm is LUR (resp. Fr茅chet differentiable). Let \(Y\) be a closed subspace of \(X\). Show that the quotient norm of \(X / Y\) is LUR (resp. Fr茅chet differentiable).

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