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Let \(X\) be a separable Banach space. Show that if \(X^{*}\) has the \(w^{*}\) Kadec-Klee property, then \(X^{*}\) is separable.

Short Answer

Expert verified
If \(X\) is a separable Banach space, then \(X^{*}\) is separable because the \(w^{*}\) Kadec-Klee property ensures norm convergence in \(X^{*}\), implying it is separable.

Step by step solution

01

- Review Definitions

Understand the definitions: A Banach space is a complete normed vector space. A separable Banach space has a countable dense subset. The dual space, denoted by \(X^{*}\), consists of all continuous linear functionals on \(X\). The \(w^{*}\) Kadec-Klee (KK) property states that if a sequence \((f_n)\) in \(X^{*}\) converges weakly* to \(f\) and also converges in norm to \(f\), then \(f_n\) converges to \(f\) strongly in the norm of \(X^{*}\).
02

- Understand Given Properties

Since \(X\) is separable, there is a countable dense subset \(D \subseteq X\). Each \(x \in D\) defines a mapping in \(X^{*}\). The weak* topology \(w^{*}\) on \(X^{*}\) is the weakest topology such that every functional evaluation \(f \mapsto f(x)\) is continuous for every \(x \in X\).
03

- Apply the \(w^{*}\) Kadec-Klee Property

We need to use the fact that \(X^{*}\) having the \(w^{*}\) Kadec-Klee property implies some strong convergence properties. Specifically, we know that \(f_n \to f\) in the weak* topology implies norm convergence due to the Kadec-Klee property.
04

- Leverage Separable Dual Space

Because \(X^{*}\) has the \(w^{*}\) Kadec-Klee property and \(X\) is separable, the dual ball \(B_{X^{*}}\) is metrizable in the weak* topology. The metrizability implies that \(B_{X^{*}}\) is separable in the weak* topology because every metrizable space is separable.
05

- Conclusion

Therefore, since the weak* topology on \(X^{*}\) makes \(B_{X^{*}}\) separable and the norm topology follows from the \(w^{*}\) Kadec-Klee property, \(X^{*}\) is indeed separable.

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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 complete normed vector space. This means that it is a vector space equipped with a norm, which is a function that assigns a positive length to each vector. Furthermore, the space is complete, meaning that every Cauchy sequence (a sequence where the elements get arbitrarily close to each other) converges to an element within the space. This is a fundamental concept in functional analysis because it ensures that limits of sequences within the space stay within the space.
weak* Kadec-Klee property
The weak* Kadec-Klee (KK) property for a Banach space's dual space, denoted as \(X^{*}\), describes a specific compatibility between the weak* topology and the norm topology. Specifically, it states that if a sequence \((f_n)\) in \(X^{*}\) converges to some function \(f\) in the weak* topology and also in the norm topology, then \(f_n\) converges to \(f\) strongly, meaning in the norm of \(X^{*}\). This property is critical because it provides a bridge between two different types of convergence, making some forms of analysis simpler.
dual space
The dual space of a Banach space \(X\), denoted by \(X^{*}\), is the set of all continuous linear functionals on \(X\). A linear functional is a function from \(X\) to the real (or complex) numbers that is linear. 'Continuous' means that small changes in the input result in small changes in the output. The dual space itself is a Banach space when equipped with the operator norm. In the context of the given problem, the elements of the dual space are used to study the properties of the original space \(X\).
separability
A Banach space \(X\) is separable if there exists a countable dense subset \(D\) within \(X\). This means that every element in \(X\) can be approximated arbitrarily closely by elements from \(D\). In simpler terms, it means that we can find a countable set of vectors such that any vector in \(X\) is as close as desired to some vector in this countable set. In essence, separability simplifies many problems because it allows the use of sequences (which are countable) rather than dealing with uncountable collections.
weak* topology
The weak* topology on a dual space \(X^{*}\) is the weakest topology that makes all evaluation maps continuous. An evaluation map takes a functional \(f\) in the dual space \(X^{*}\) and maps it to a scalar by evaluating it at a point \(x\) in \(X\). Formally, a sequence \((f_n)\) in \(X^{*}\) converges to \(f\) in the weak* topology if and only if \(f_n(x)\) converges to \(f(x)\) for each \(x\) in \(X\). This topology is coarser than the norm topology, meaning there are fewer open sets, making it easier for sequences to converge.

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Most popular questions from this chapter

Show that the norm of \(C[0,1]\) is nowhere Fréchet differentiable. Show that the norm of \(C[0,1]\) is Gâteaux 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 Šmulian lemma, \(x\) is not a point of Fréchet 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).

Let \(X\) be a Banach space and let \(f\) be a continuous convex function on \(X^{*}\) that is \(w^{*}\) -lower semicontinuous. Show that if \(f\) is Fréchet differentiable at \(x^{*} \in X^{*}\), then \(f^{\prime}\left(x^{*}\right) \in X\). Hint: The derivative, as a uniform limit of quotients in \(B_{X^{*}}\), is also \(w^{*}\) -lower semicontinuous. Then use its linearity to see that \(f^{\prime}\left(x^{*}\right)\) is a functional that is \(w^{*}\) -continuous on \(B_{X *}\) and apply Theorem \(4.44\).

Show that every standard unit vector \(e_{i}\) is a strongly exposed point of \(B_{\ell_{1}}\).

Let \(\|\cdot\|_{\infty}\) denote the canonical of \(\ell_{\infty}\) and set \(p(x)=\limsup \left|x_{i}\right|\). Define \(\|x\|=\|x\|_{\infty}+p(x)\) for \(x \in \ell_{\infty} .\) Show that \(\|\cdot\|\) is nowhere Gâteaux differentiable. Hint: It is enough to show that \(p\) is nowhere differentiable. If \(x=\left(x_{i}\right) \in \ell_{\infty}\) and \(x_{n_{k}} \rightarrow 1=p(x)\), consider the direction \(h=\sum(-1)^{k} e_{n_{k}}\).

We say that a subset \(A\) of a Banach space \(X\) has property \((*)\) if \(A\) is a nonempty, closed, convex, and bounded subset of \(X\) and cvery point \(a \in A\) is a proper support point; that is, given \(a \in A\), there exists \(a^{*}\) in \(X^{*}\) such that \(a^{*}(a)=\sup _{A}\left(a^{*}\right)\) and there is \(x \in A\) such that \(a^{*}(x)<\sup _{A}\left(a^{*}\right)\)

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