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Chapter 1: Problem 30

Show that a Banach space \(X\) is separable if and only if \(S_{X}\) is separable. Hint: If \(\left\\{x_{n}\right\\}\) is dense in \(S_{X}\), consider \(\left\\{r_{k} x_{n}\right\\}_{k, n}\) for some dense sequence \(\left\\{r_{k}\right\\}\) in \(\mathbf{K}\). If \(\mathcal{S}\) is countable and dense in \(X\), consider \(\left\\{\frac{x}{\|x\|} ; x \in \mathcal{S}\right\\}\).

Short Answer

Expert verified
A Banach space \(X\) is separable if and only if its unit sphere \(S_X\) is separable.

Step by step solution

01

Define Separability

A space is separable if it contains a countable, dense subset. Therefore, need a countable subset whose closure is the whole space.
02

Define the Unit Sphere

In a Banach space \(X\), the unit sphere \(S_X\) is defined as the set \(S_X = \{x \in X : \|x\| = 1\}\).
03

Assume \(S_X\) is Separabile

Assume that the unit sphere \(S_X\) in \(X\) is separable. Then there exists a countable dense subset \(\{x_n\}\) in \(S_X\).
04

Construct a Dense Set in \(X\)

Consider a dense sequence \(\{r_k\}\) in the scalar field \(\mathbf{K}\). If \(X\) is over the reals, \(\mathbf{K} = \mathbb{R}\), use \(\mathbb{Q}\). If \(X\) is complex, \(\mathbf{K} = \mathbb{C}\), use \(\mathbb{Q}(i)\). Construct the set \(\{r_k x_n\}_{k, n}\), where \(x_n \in S_X\).
05

Verify Density in \(X\)

The linear combinations of elements in \(\{r_k x_n\}\) will be dense in \(X\). Therefore, \(X\) will be separable because a countable dense subset was constructed in \(X\).
06

Assume \(X\) is Separable

Now assume \(X\) is separable, so there exists a countable dense subset \(\mathcal{S} = \{x_n \} \subset X\).
07

Normalize Dense Subset

Consider the set \(\left\{ \frac{x}{\|x\|} ; x \in \mathcal{S} \right\}\). This forms a countable subset of \(S_X\).
08

Check for Density in \(S_X\)

Since \(\mathcal{S}\) is dense in \(X\), the closure of \(\left\{ \frac{x}{\|x\|} ; x \in \mathcal{S} \right\}\) is dense in \(S_X\). Hence, \(S_X\) is separable.
09

Conclude the Proof

Since both implications have been shown, it has been proven that a Banach space \(X\) is separable if and only if \(S_X\) is separable.

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Key Concepts

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

Dense Subset
In mathematical spaces, the concept of a dense subset is crucial. A subset \(\beta\) of a space \(\tau\) is called dense in \(\tau\) if every point in \(\tau\) can be approximated arbitrarily closely by points in \(\beta\).

To understand this better, imagine a space like the number line. A subset would be dense if, no matter how close you look at any point on the number line, you will always find members of this subset near that point.

Formally, a subset is dense if its closure is the whole space. The closure of a set is all the limit points of that set. In simpler terms, it means that you can't isolate a part of the space without finding a piece of the dense subset.

For example, the rational numbers \( \mathbb{Q} \ \) are dense in the real numbers \( \mathbb{R} \) because between any two real numbers, no matter how close, there is always a rational number.
Unit Sphere
The unit sphere in a Banach space is an interesting concept. In Banach space \( X \), the unit sphere (denoted by \( S_X \)) is defined as the set of all elements \(x \in X \) such that their norm (which is a measure of their 'size') is exactly 1.

Formally, we write this as \(\big\{ x \in X : \| x \| = 1 \big\}\). The unit sphere is pivotal because it helps us in understanding the structure and properties of the space.

A Banach space may contain infinitely many elements. However, the idea of normalizing elements (transforming each element to have a norm of 1 without changing their direction) lets us simplify and study the space more effectively. It tells us how every element in this Banach space can be related back to the unit sphere.
Scalar Field
In the context of Banach spaces, the scalar field \$ \mathbf{K} \$ plays a key role. The scalar field can be either the real numbers \(\boldsymbol{\textbf{R}}\) or the complex numbers \(\boldsymbol{\textbf{C}}\).

The scalar field is essentially the set of numbers that we use to scale elements in our Banach space. Imagine you have a vector; by multiplying it by a scalar, you change its length but not its direction.

For instance, if \( X \) is a Banach space over the real numbers, we use \(\boldsymbol{\textbf{R}}\), and if it is over the complex numbers, we use \(\boldsymbol{\textbf{C}}\).

In practical terms, when we talk about dense sequences in \(\boldsymbol{\textbf{K}}\) for constructing dense sets in \( X \), we're essentially saying that we can approximate any member of our Banach space by a combination of elements from the dense subset scaled by elements from a dense subset in the scalar field (like rational numbers in \( \boldsymbol{\textbf{R}}\) or rational numbers with imaginary parts in \( \boldsymbol{\textbf{C}} \)).

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Most popular questions from this chapter

Show that \(C\) is a convex set in a vector space if and only if \(\sum \lambda_{i} x_{i} \in C\) whenever \(x_{1}, \ldots, x_{n} \in C\) and \(\lambda_{1}, \ldots, \lambda_{n} \geq 0\) satisfy \(\sum \lambda_{i}=1\) Hint: (a) \(\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}=\left(\lambda_{1}+\lambda_{2}\right)\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}} x_{1}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}} x_{2}\right)+\lambda_{3} x_{3}\) and induction.

Let \(f \in L_{p_{0}}[0,1]\) for some \(p_{0}>1 .\) Show that \(\lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}} .\) If \(f \in L_{\infty}[0,1]\), then \(\lim _{p \rightarrow \infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}\) Let \(x \in \ell_{q}\) for some \(q \geq 1\). Show that \(\lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}\).

Let \(X\) be a normed space, \(M \subset X\) and \(\varepsilon>0 .\) We say that \(A \subset M\) is an \(\varepsilon=n e t\) in \(M\) if for every \(x \in M\) there is \(y \in A\) such that \(\|x-y\|<\varepsilon\). We say that \(A \subset M\) is \(\varepsilon\) -separated in \(M\) if \(\|x-y\| \geq \varepsilon\) for all \(x \neq y \in A\). Clearly, a maximal \(\varepsilon\) -separated subset of \(M\) in the sense of inclusion is an \(\varepsilon\) -net in \(M\). Let \(X\) be an \(n\) -dimensional real normed space. Show that if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\), then \(N \geq \varepsilon^{-n} .\) On the other hand, there is an \(\varepsilon\) -net \(\left\\{x_{j}\right\\}_{j=1}^{N}\) for \(B_{X}\) with \(N \leq\left(\frac{2}{c}+1\right)^{n}\). Hint: Fix a linear isomorphism \(T\) of \(X\) onto \(\mathbf{R}^{n}\). For a compact subset \(C\) of \(X\), we define the volume of \(C\) by \(\operatorname{vol}(C)=\lambda(T(C))\), where \(\lambda\) is the Lebesgue measure on \(\mathbf{R}^{n}\). Let \(B(x, r)\) be the closed ball centered at \(x\) with radius \(r\). We have \(B_{X} \subset \bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\), so $$ \begin{aligned} \operatorname{vol}\left(B_{X}\right) & \leq \operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\right) \leq \sum_{j=1}^{N} \operatorname{vol}\left(B\left(x_{j}, \varepsilon\right)\right) \\ &=N \cdot \operatorname{vol}\left(\varepsilon B_{X}\right)=N \cdot \varepsilon^{n} \cdot \operatorname{vol}\left(B_{X}\right) \end{aligned} $$ Thus \(N \geq \varepsilon^{-n}\). On the other hand, if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is a maximal \(\varepsilon\) -separated set in \(B_{X}\), then \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\). Since \(B\left(x_{j}, \varepsilon / 2\right) \cap B\left(x_{i}, \varepsilon / 2\right)=\emptyset\) if \(i \neq j\) and \(B\left(x_{j}, \varepsilon / 2\right) \subset(1+\varepsilon / 2) B_{X}\), we have $$ \begin{aligned} N \cdot(\varepsilon / 2)^{n} \cdot \operatorname{vol}\left(B_{X}\right) &=\operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon / 2\right)\right) \\ & \leq \operatorname{vol}\left((1+\varepsilon / 2) B_{X}\right)=(1+\varepsilon / 2)^{n} \operatorname{vol}\left(B_{X}\right) . \end{aligned} $$ Thus \(N \leq\left(\frac{1+\varepsilon / 2}{\varepsilon / 2}\right)^{n}\).

Let \(X\) be an infinite-dimensional Banach space. Show that there is no translation-invariant Borel measure \(\mu\) on \(X\) such that \(\mu(U)>0\) for every open set \(U\) and such that \(\mu\left(U_{1}\right)<\infty\) for some open \(U_{1}\). Hint: Every open ball contains an infinite number of disjoint open balls of equal radii (see Lemma \(1.23\) ).

Let \(\sum x_{i}\) be unconditionally convergent. Show that: (i) \(S_{1}=\left\\{\sum_{i=1}^{\infty} \varepsilon_{i} x_{i} ; \varepsilon_{i}=\pm 1\right\\}\) is a compact set in \(X\), (ii) \(S_{2}=\left\\{\sum_{i=1}^{n} \varepsilon_{i} x_{i} ; n \in \mathbf{N}, \varepsilon_{i}=\pm 1\right\\}\) is a relatively compact set in \(X\). Hint: (i): The space \(\\{-1,1\\}^{\mathrm{N}}\) is compact in the pointwise topology. We claim that the map \(\left\\{\varepsilon_{i}\right\\} \mapsto \sum \varepsilon_{i} x_{i}\) is continuous. Indeed, from the unconditional Cauchy condition we have that if \(\sigma\) is a finite set with \(\min (\sigma)\) sufficiently large, then \(\left\|\sum_{i \in \sigma} x_{i}\right\|<\varepsilon\), which gives that if \(n_{0}\) is sufficiently large, then \(\left\|\sum_{i=n}^{m} \varepsilon_{i} x_{i}\right\|<\varepsilon\) for every possible \(\varepsilon_{i}\). (ii): Use (i) and write \(2 \sum_{i=1}^{n} \varepsilon_{i} x_{i}=\sum_{i=1}^{\infty} \varepsilon_{i} x_{i}+\sum_{i=1}^{\infty} \varepsilon_{i}^{\prime} x_{i}\), where \(\varepsilon_{i}^{\prime}=\varepsilon_{i}\) for \(i=1, \ldots, n\) and \(\varepsilon_{i}^{\prime}=-\varepsilon_{i}\) for \(i=n+1, n+2, \ldots\)
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