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 \(X^{*}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
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.
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.
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:
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.
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:
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.