Functional Analysis 5 -- Hahn-Banach Theorem Extension

This is course note from Siran Li-Shanghai Jiao Tong University.

There are two problems related to HB-theorem: extension and separation.

1. Extension

Given \(Y \le X\) being subspace of \(X\), and \(f: Y \rightarrow Z\), does there exist \(F: X \rightarrow Z, s.t. F\mid_Y = f?\) If so, can \(F\) keep some natural properties of \(f?\)

2. Separation

Can we separate bounded closed sets by hyperplanes?

Definition: Dual Space

Let \(X\) be a NVS over \(\mathbb{K} = \mathbb{R} \ or \ \mathbb{C}.\) Its algebraic dual is \(X' = \mathcal{L}(X, \mathbb{K}).\) Its NVS-dual/topological dual is \(X^* = B(X, \mathbb{K}).\)

Note: If \(X\) is finite dimension, then \(X' = X^*.\) Elements in \(X'\) are called functional.

Definition: Sublinear Functional

Let \(X\) be a real VS. Say \(p: X\rightarrow \mathbb{R}\) is a sublinear functional if 1. subaddictive \(p(x+y) \le p(x) + p(y)\) 2. positive 1-homogenity \(p(\lambda x) = \lambda x, \forall \lambda \ge 0.\)

Examples:

1. Seminorms are sublinear
2. On \(\mathbb{R}^d,\) consider \(p(x) = \max \{x^d, 0\}\) = dist(\(x, \mathbb{R}^d_{-}\)) is sublinear but not seminorm.

Theorem 5.1: Hahn Banach Theorem

Let \(X\) be a NVS over \(\mathbb{R}\). Let \(Y \le X\) be a subspace. Then \(\forall f \in Y^*,\) there is an extension \(F \in X^*, F: X \rightarrow \mathbb{R}\) such that \(F\mid_Y = f,\) and \(||F||_{X^*} = ||f||_{Y^*}.\)

proof:[[TODO]]

Remark: Extension obtained by HB is not unique in general. If \(Y \le X\) is dense, any \(f \in Y^*\) can be uniquely extended to \(F \in X^*\). Also, ||F|| = ||f||.

Application

1. Let \(X\) be a NVS. Let \(x \in X\backslash\{0\}.\) Then \(\exists f \in X^*, \ s.t.\ f(x) = ||x||\) and \(||f|| = 1\). Here \(f\) is the "norming functional" of \(X.\)

proof:

On \(<x>,\) define \(g(x) = \lambda ||x||, \forall \lambda \in \mathbb{R}.\) Clearly \(||g||_{<x>^*} = 1\) and \(g(x) = ||x||.\) So, by Hahn-Banach theorem, there exists an extension \(f \in X^*, \ s.t. \ ||f|| = 1\) and \((x) = ||x||.\) $

Proposition 5.2

Let \(X\) be any NVS. Then \(X^* = B(X, \mathbb{K})\) separate points.

proof:

Let \(x\neq y\) in \(X\). Pick \(f \in X^*\) to be the norming functional of \(x-y.\) Then \(f(x-y) = ||x - y|| \neq 0.\) \(\Box\)

Lemma 5.3

Let \(X\) be a NVS. \(0 \neq f \in X'.\) Then \(\forall x_0 \notin \ker f.\) We have \(X = span \{x_0 + \ker f\}.\)

Corollary 5.4

\(\forall x\neq y \in X\) being a NVS, \(\exists f \in X^*\) and \(c \in \mathbb{R}, s.t. \ f(x) < c < f(y).\)

Lemma 5.5

For \(X\) being NVS, \(f \in X'\backslash\{0\},\) given any \(x_0 \in X \backslash \ker f.\) One has \(X = span\{x_0 + \ker f\}.\)

Remark: \[\dim(X/\ker f) = 1 \Leftrightarrow \text{codim}(\ker f) = 1\]

Definition: Hyperplane

A codim-1 subspace of a VS is called a hyperplane.

Corollary 5.6

Let \((X, ||\cdot||)\) be a Banach space. Let \(\{0\} \neq Y \le X\) be a closed subspace. Then \(\forall x_0 \in X \backslash Y, \exists f \in X^* \ s.t. f\mid_Y = 0, ||f|| = 1\) and \(f(x_0) = dist(x_0, Y).\)