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Show that if \(Y\) is a subspace of a Banach space \(X\) and \(X^{*}\) is separable, then so is \(Y^{*}\). Hint: \(Y^{*}\) is isomorphic to the separable space \(X^{*} / Y^{\perp}\).

Short Answer

Expert verified
\(Y^{*}\) is separable because it is isomorphic to the separable space \(X^{*} / Y^{\bot}\), and isomorphisms preserve separability.

Step by step solution

01

- Understand the Given Information

We are given that we have a Banach space, denoted by \(X\), and a subspace of it, denoted by \(Y\). Also, \(X^{*}\), the dual space of \(X\), is separable. We need to show that \(Y^{*}\), the dual space of \(Y\), is also separable.
02

- Use the Hint

The hint tells us that \(Y^{*}\) is isomorphic to \(X^{*} / Y^{\bot}\), where \(Y^{\bot}\) is the annihilator of \(Y\) in \(X^{*}\). This means that for every continuous linear functional \(f\) in \(Y^{*}\), there is a unique corresponding coset in the quotient space \(X^{*} / Y^{\bot}\).
03

- Understand the Concept of Annihilator

The annihilator of \(Y\), \(Y^{\bot}\), consists of all continuous linear functionals in \(X^{*}\) that vanish on \(Y\). Formally, \(Y^{\bot} = \{ f \in X^{*} : f(y) = 0 \text{ for all } y \in Y \}\).
04

- Use Separability

Since \(X^{*}\) is separable, it has a countable dense subset, say \(\{ f_{n} \}\). We need to demonstrate that \(X^{*} / Y^{\bot}\), and hence \(Y^{*}\), is separable. The quotient space \(X^{*} / Y^{\bot}\) inherits separability from \(X^{*}\). This is because the image of the countable dense subset under the quotient map will form a countable dense subset in the quotient space.
05

- Conclude Separability of \(Y^{*}\)

Since \(Y^{*}\) is isomorphic to the separable space \(X^{*} / Y^{\bot}\), and isomorphism preserves separability, we can conclude that \(Y^{*}\) is separable.

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Key Concepts

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

Dual Space
Imagine you have a Banach space, which is just a special type of complete normed vector space. The dual space of this Banach space, denoted as \(X^*\), is the set of all continuous linear functionals on \(X\). In simpler terms, it's a collection of maps that take elements from \(X\) and return real or complex numbers in a linear and continuous way.
The dual space itself is also a Banach space. This comes in handy for many important properties and theorems in functional analysis. To visualize, if you think of a vector as a point in space, a linear functional is like a plane that 'measures' points in that space.
Understanding the dual space is crucial for tackling many problems in functional analysis, as seen in this exercise where you end up showing properties about \(Y^*\) based on information about \(X^*\).
Separability
A space is called separable if it contains a countable dense subset. Think of a dense subset as a smaller set of points that come arbitrarily close to every other point in your space.
For example, the set of rational numbers \(\textbf{Q}\) is dense in the real numbers \(\textbf{R}\), because no matter how close you get to any real number, you can always find a rational number even closer.
In this exercise, we know that the dual space \(X^*\) is separable. This means it has a countable dense subset. We use this information to show that \(Y^*\) is separable. This works because the quotient space \(X^* / Y^\bot\) inherits the separability property from \(X^*\), and since \(Y^*\) is isomorphic to \(X^* / Y^\bot\), it too must be separable.
Annihilator
The annihilator of a subspace \(Y\) in a Banach space \(X\) is a set of functionals in the dual space \(X^*\) that 'kill' or vanish on every element of \(Y\). Formally, it is denoted as \(Y^\perp\) and defined by \(Y^\perp = \{ f \in X^* : f(y) = 0 \text{ for all } y \in Y \}\).
In other words, if you apply any functional from \(Y^\perp\) to any element of \(Y\), the result is zero.
The annihilator plays a key role in this exercise. It lets us use the isomorphism between \(Y^*\) and the quotient space \(X^* / Y^\perp\). The separability of \(X^*\) translates to the separability of the quotient space, and therefore to \(Y^*\). This concept simplifies understanding deep properties of subspaces and their duals.

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Most popular questions from this chapter

Let \(L\) be a closed subset of a compact space \(K\). Show that \(C(L)\) is isomorphic to a quotient of \(C(K)\). Hint: Let \(Q: C(K) \rightarrow C(L)\) be defined for \(f \in C(K)\) by \(Q(f)=\left.f\right|_{L}\). Then \(Q\) is onto by Tietze's theorem; then use Corollary \(2.25\).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If \(T\) is one-to- one, is \(T^{*}\) necessarily onto? Hint: No, consider the identity map from \(\ell_{1}\) into \(\ell_{2}\).

We have \(\|T\|=\left\|T^{*}\right\|\) for a bounded linear operator on a Banach space, so if for a sequence of operators \(T_{n}\) we have \(\left\|T_{n}\right\| \rightarrow 0\), then \(\left\|T_{n}^{*}\right\| \rightarrow\) 0\. Find an example of a sequence of operators \(T_{n}\) on a Banach space \(X\) such that \(\left\|T_{n}(x)\right\| \rightarrow 0\) for every \(x \in X\) but it is not true that \(\left\|T_{n}^{*}\left(x^{*}\right)\right\| \rightarrow 0\) for every \(x^{*} \in X^{*}\). Hint: Let \(T_{n}(x)=\left(x_{n}, x_{n+1}, \ldots\right)\) in \(\ell_{2}\). Then \(T_{n}^{*}(x)=\left(0, \ldots, 0, x_{1}, x_{2}, \ldots\right)\), where \(x_{1}\) is on the \(n\) -th place.

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(Y\) is separable and \(T\) is onto \(Y\), then there is a separable closed subspace \(Z\) of \(X\) such that \(T(Z)=Y\).

Let \(X\) be a Banach space. Show that: (i) \(\overline{\operatorname{span}}(A)=\left(A^{\perp}\right)_{\perp}\) for \(A \subset X\). (ii) \(\operatorname{span}(B) \subset\left(B_{\perp}\right)^{\perp}\) for \(B \subset X^{*}\). Note that in general we cannot put equality. (iii) \(A^{\perp}=\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp}\) for \(A \subset X\) and \(B_{\perp}=\left(\left(B_{\perp}\right)^{\perp}\right)_{\perp}\) for \(B \subset X^{*}\). Hint: (i): Using the definition, show that \(A \subset\left(A^{\perp}\right)_{\perp}\). Then use that \(B_{\perp}\) is a closed subspace for any \(B \subset X^{*}\), proving that \(\frac{\overline{\operatorname{span}}}(A) \subset\left(A^{\perp}\right)_{\perp}\). Take any \(x \notin \overline{\operatorname{span}}(A)\). Since \(\overline{\operatorname{span}}(A)\) is a closed subspace, by the separation \(\left.f\right|_{A}=0\), hence \(f \in A^{\perp}\); also \(f(x)>0\), so \(x \notin\left(A^{\perp}\right)_{\perp}\) (ii): Similar to (i). (iii): Applying (i) to \(A^{\perp}\), we get \(A^{\perp} \subset\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} .\) On the other hand, using \(A \subset\left(A^{\perp}\right)_{\perp}\) and the previous exercise, we get \(\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} \subset A^{\perp} .\) The dual statement is proved in the same way.

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