Hahn-Banach theorem

I’m pleased to say that I find the derivation of the Hahn-Banach theorem pretty straightforward by now. Let me first state it, for the real case.

Hahn-Banach theorem: Let V be a real vector space. Let p: V \to \mathbb{R} be sublinear. If f : U \to \mathbb{R} be a linear functional on the subspace U \subset V with f(x) \leq p(x) for x \in U, then there exists a linear extension of f to all of V (call it g) such that f(x) \leq g(x) for x \in V with f(x) = g(x) for x \in U and g(x) \leq p(x) for all x \in V.

To show this, start by taking any x_0 \in V \setminus U. We wish to assign some \alpha to x_0 that keeps p as the dominating function in the vector space U + \mathbb{R}x_0. For this to happen, applying the linearity of f and the domination constraint, we can derive

\frac{f(y) - p(y - \lambda x_0)}{\lambda} \leq \alpha \leq \frac{p(y+\lambda x_0) - f(y)}{\lambda}, \quad y \in U, \lambda > 0.

This reduces to

\sup_{y \in U} p(y+x_0) - f(y) \leq \inf_{y \in U} f(y) - p(y-x_0).

Such can be proven via

f(y_1) + f(y_2) = f(y_1 + y_2) \leq p(y_1 + y_2) \leq p(y_1 - x_0) +p(y_2 + x_0), \quad y_1, y_2 \in U.

Now take the space of linear functionals defined on some specific subspace dominated by p. Denote an element of it as (f, U). We introduce a partial order wherein (f, U) \leq (f', U') iff f(x) = f'(x) for x \in U and U \subset U'. We can apply Zorn’s lemma on this, as we can take the union to derive an upper bound for any chain. Any maximal element is necessarily (g, V) as if the domain is not the entire vector space, we can by above construct a larger element.