Real Hahn-Banach Theorem
Suppose $X$ is a vector space over $\mathbb{R}$, $p: X \to \mathbb{R}$ has the following properties:
- $p(X) = \lambda p(x)$, $\forall x \in X$, $\lambda \in \mathbb{R}_+$ and $p(x + y) \leq p(x) + p(y)$, $\forall x, y \in X$.
- Let $X_0$ be a subspace of $X$ and $u: X_0 \to \mathbb{R}$ a linear functional such that $u(x) \leq p(x)$, $\forall x \in X_0$.
Then we can find $f: X \to \mathbb{R}$ a linear functional such that $f|_{X_0} = u$ and $f(x) \leq u(x)$, $\forall x \in X$.
Proof: Let $Y$ is a subspace of $X$, $g: Y \to \mathbb{R}$ is a linear functional which extends $u$ and $g \leq p$ on $Y$
Consider the set $M = { (Y, g) }$. Define an order relation on $M$ like this $(Y_1, g_1) \leq (Y_2, g_2)$ if $Y_1 \subset Y_2$ and $g_2$ is an extension for $g_1$.
We show that in $M$ every chain has an upper bound. Suppose $M_0$ is a totally ordered subset of $M$. Then define $Y_0 = \bigcup_{(Y,g) \in M_0} Y$ and $g: Y_0 \to \mathbb{R}$, $g(y) = g_0(y)$ if $y \in Y_0$ and $(Y_0, g) \in M_0$. This function is well defined, and $Y_0$ is a subspace of $X$ because the set $M_0$ is totally ordered.
Furthermore, from the definition for $g_0$, we have that $g_0 \leq p$. Therefore $(Y_0, g_0) \in M$, and is obviously an upper bound for $M_0$. By Zorn’s Lemma, we find that $M$ has at least one maximal element $(Z, h)$.
Suppose $X \neq Z$. Then we can find $x_0 \in X \setminus Z$. Define $W = \text{Span}{Z, x_0} = \mathbb{R} \cdot x_0 \oplus Z$. Therefore, $W$ is a linear subspace in $X$. Let $y, z \in Z$. Then $$ h(y) + h(z) = h(y + z) \leq p(y + z) = p(y - x_0 + x_0 + z) \leq p(y - x_0) + p(x_0 + z) $$ Therefore, we have $$ h(z) - p(-x _0 + z) + h(y) - p(y - x _0) \leq - h(y) + p(x _0 + y), \quad\forall y, z \in Z $$
Therefore, we can say $$ a = \sup_{z \in Z} (h(z) - p(-x_0 + z)) \leq - \inf_{y \in Z} (-h(y) + p(x_0 + y)) $$ Pick one $c \in [a, b]$ and define $h_1(z) = \lambda c + h(y)$, where $z = \lambda x_0 + y$ (unique representation), $h_1$ is linear, and extends $h_1$ on $W$, which means that it extends $u$ on $X_0$.
We can check that $(W, h_1) \in M$ and the maximal element $h_1$ is the requested functional element, which is a contradiction.
Therefore $Z = X$, and the maximal element $h_1$ is the requested functional.