## 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.