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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 canonical norm of \(\ell_{1}\) is nowhere Fréchet differentiable and is Gâteaux differentiable at \(x=\left(x_{i}\right)\) iff \(x_{i} \neq 0\) for every \(i\). If \(\Gamma\) is uncountable, show that the canonical norm of \(\ell_{1}(\Gamma)\) is not Gâteaux differentiable at any point. Hint: Let \(x \in S_{X} .\) Given \(\varepsilon>0\), find \(i\) such that \(\left|x_{i}\right|<\varepsilon / 2\) and consider \(h=\varepsilon e_{i} .\) Show that \(\|x \pm h\| \geq 1+\varepsilon / 2\) and use Lemma \(8.3\). Note that every vector in \(\ell_{1}(\Gamma)\) has a countable support, choose a standard unit vector outside this support, and use Lemma \(8.3 .\)

Show that the norm \(\|\cdot\|\) on a Banach space \(X\) is LUR if and only if \(\lim \left\|x_{n}-x\right\|=0\) whenever \(x_{n}, x \in S_{X}\) satisfy \(\lim \left\|x_{n}+x\right\|=2\)

Show that \(c_{0}\) contains no two-dimensional subspace on which the standard norm is Gâteaux differentiable. Hint: Then some quotient \(Q\) of \(\ell_{1}\) would have uncountably many extreme points since the dual to a two-dimensional smooth space is strictly convex. Every point of the sphere of \(Q\) that is identified with the restriction to the two-dimensional subspace in question extends to an extremal point of the sphere in \(\ell_{1}\) by the Krein-Milman theorem, considering the face of all the extensions. Thus, there are uncountably many such extreme points of the ball of the standard norm in \(\ell_{1}\), which is a contradiction because the extreme points of the ball of \(\ell_{1}\) are exactly \(\pm e_{i}\).

Let \(X\) be a separable Banach space and assume that \(X^{*}\) has the \(w^{*}\) Kadec-Klee property. Show that if \(C\) is a \(w^{*}\) -compact convex set in \(X^{*}\), then \(C\) has a point where the identity map of \(C\) into \(C\) is \(w^{*}\) -to-norm continuous.

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.

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