Question

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.

Short answer

We have \(\overline{\operatorname{span}}(A) = \left(A^{\perp}\right)_{\perp}\). Also, for subsets \(B \subset X^{*}\), \(\operatorname{span}(B) \subset (B_{\perp})^{\perp}\). Finally, \(A^{\perp} = \left(\left(A^{\perp}\right)_{\perp}\right)^{\perp}\) and \(B_{\perp} = \left(\left(B_{\perp}\right)^{\perp}\right)_{\perp}\).

Step-by-step solution

Show Inclusion

First, show that for any subset \(A \subset X\), we have \(A \subset \left(A^{\perp}\right)_{\perp}\). By definition, \(A^{\perp}\) consists of all functionals in \(X^{*}\) that vanish on \(A\). Therefore, any element \(x \in A\) must satisfy \(f(x) = 0\) for all \(f \in A^{\perp}\), thus belonging to the annihilator of \(A^{\perp}\), denoted as \(\left(A^{\perp}\right)_{\perp}\).

Span Closure Subset

Since \(\left(A^{\perp}\right)_{\perp}\) is a closed subspace and contains \(A\), it must also contain the closure of the span of \(A\), that is, \(\overline{\operatorname{span}}(A) \subset \left(A^{\perp}\right)_{\perp}\).

Show Equality

Assume there exists an element \(x otin \overline{\operatorname{span}}(A)\). Since \(\overline{\operatorname{span}}(A)\) is a closed subspace, the Hahn-Banach theorem allows for a functional \(f \in X^{*}\) such that \(f(\overline{\operatorname{span}}(A)) = 0\) but \(f(x) > 0\). This implies \(f \in A^{\perp}\) and \(x otin \left(A^{\perp}\right)_{\perp}\), showing that \(\overline{\operatorname{span}}(A) = \left(A^{\perp}\right)_{\perp}\).

Span and Annihilator Relationship

For any subset \(B \subset X^{*}\), any element in \(\operatorname{span}(B)\) by definition is a finite linear combination of elements of \(B\). Functionals in \(B_{\perp}\) annihilate all elements of \(B\), so they will also annihilate any finite linear combination of them. Hence, \(\operatorname{span}(B) \subset (B_{\perp})^{\perp}\).

Lack of General Equality

Since \(B_{\perp}\) is closed while \(\operatorname{span}(B)\) might not be, equality \(\operatorname{span}(B) = (B_{\perp})^{\perp}\) does not generally hold unless span is closed.

Applying Inclusions to Perpendicularity

From Step 1, we know \(A \subset \left(A^{\perp}\right)_{\perp}\), and applying results from (i) gives \(A^{\perp} \subset \left(\left(A^{\perp}\right)_{\perp}\right)^{\perp}\). Conversely, since \(\left(A^{\perp}\right)_{\perp} = \overline{\operatorname{span}}(A)\), it shows \(\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} \subset A^{\perp}\), concluding equality.

Dual Statement

By symmetry, applying the above to \(B \subset X^{*}\), results duality similarly provides \(B_{\perp} = \left(\left(B_{\perp}\right)^{\perp}\right)_{\perp}\).

Key concepts

Closed Subspaces

In the context of Banach spaces, a subspace is any subset that is also a vector space. When we talk about closed subspaces, we mean subspaces that are closed under the limit operation. This means if a sequence of elements in the subspace converges to some limit, then that limit must also be within the subspace. The idea of closure is crucial because it ensures completeness, a key property of Banach spaces.

When dealing with linear operators, the closure of subspaces becomes particularly important. For example, if you have a set A within a Banach space X, the closed subspace generated by A is denoted by \(\overline{\operatorname{span}}\left(A\right)\)). This is the smallest closed subspace containing all linear combinations of elements in A. This concept helps us generalize and extend the properties of finite-dimensional spaces to infinite-dimensional ones, allowing for more robust and far-reaching mathematical theories.

Understanding closed subspaces helps in analyzing the structure of Banach spaces and is essential for the application of significant theorems, such as the Hahn-Banach theorem.

Hahn-Banach Theorem

The Hahn-Banach theorem is a fundamental result in functional analysis, concerning the extension of linear functionals. Simply put, it states that given a sublinear function p and a linear functional f defined on a subspace, f can be extended to the whole space without increasing its norm, while still remaining bounded by p.

Mathematically, if X is a real vector space, p: X → ℝ is sublinear, and f is a linear functional defined on a subspace Y of X such that for all y in Y, f(y) ≤ p(y), then there exists a linear extension F: X → ℝ such that for all x in X, F(x) ≤ p(x) and F agrees with f on Y.

This theorem has several powerful implications. It is often used to prove other important results in functional analysis. For example, when working with closed subspaces, the Hahn-Banach theorem allows us to separate points outside a closed subspace from the subspace itself using continuous linear functionals. This separation property is employed in our problem to show the equality \( \overline{\operatorname{span}}(A)=\left(A^{\perp}\right)_{\perp} \).

The Hahn-Banach theorem is widely used for dual spaces, optimization, and in proving other theorems. It ensures that the rich structure of functional analysis can be applied in many practical and theoretical scenarios.

Annihilator

The concept of an annihilator is pivotal in the study of dual spaces. Given a subset A of a Banach space X, the annihilator of A, denoted as \(A^{\perp}\), is the set of all continuous linear functionals in the dual space X* that vanish on A. In simpler terms, it's the collection of all functionals that evaluate to zero for every element in A.

Mathematically, \(A^{\perp} = \{ f \in X^{*} \mid f(a) = 0, \forall a \in A \}\). This concept is used to understand and decompose the structure of subspaces. For example, if A is a subset of X, the annihilator of A's annihilator, denoted as \((A^{\perp})_{\perp}\), gives us the largest closed subspace of X, in which A is dense. Hence, \(\overline{\operatorname{span}}(A) = \left(A^{\perp} \right)_{\perp}\).

Similarly, for a subset B of the dual space X*, the annihilator of B, denoted as \((B_{\perp})\), consists of all elements in X that are annihilated by every element of B. This interplay between spaces and their duals is pivotal in understanding linear operators and dual spaces in functional analysis.

Understanding the annihilator helps to bridge the connection between spaces and their corresponding dual spaces, offering a profound insight into the nature of Banach spaces and their subspaces.