Hahn-Banach Theorem/Real Vector Space

Theorem
Let $X$ be a vector space over $\R$.

Let $p : X \to \R$ be a sublinear functional.

Let $X_0$ be a linear subspace of $X$.

Let $f_0 : X_0 \to \R$ be a linear functional such that:


 * $\map {f_0} x \le \map p x$ for each $x \in X_0$.

Then there exists a linear functional $f$ defined on the whole space $X$ which extends $f_0$ and satisfies:


 * $\map f x \le \map p x$ for each $x \in X$.

That is, there exists a linear functional $f : X \to \R$ such that:


 * $\map f x \le \map p x$ for each $x \in X$

and:


 * $\map f x = \map {f_0} x$ for each $x \in X_0$.

Proof
We first prove a lemma:

Lemma 1
Let $P$ be the set of pairs $\tuple {G, g}$ such that:


 * $(1): \quad$ $G$ is a linear subspace of $X$ with $X_0 \subseteq G$
 * $(2): \quad$ $g : G \to \R$ is a linear functional extending $f_0$
 * $(3): \quad$ $\map g x \le \map p x$ for each $x \in G$.

Define the relation $\preceq$ on $P$ by:


 * $\tuple {G, g} \preceq \tuple {H, h}$ :


 * $(1): \quad$ $G \subseteq H$
 * $(2): \quad$ $h$ extends $g$.

Lemma 2
We show that every chain in $\struct {P, \preceq}$ has an upper bound.

We will then invoke Zorn's Lemma.

Lemma 3
From Zorn's Lemma, we have that $P$ has a maximal element.

Let $\struct {G, g}$ be a maximal element of $P$.

Then:


 * $G$ is a linear subspace of $X$ with $X_0 \subseteq G$

and:


 * $g : G \to \R$ is a linear functional extending $f_0$ with:


 * $\map g x \le \map p x$ for each $x \in G$.

suppose that $G \ne X$.

Then, by Lemma 1, there exists a linear subspace of $X$ with $G'$ and a linear functional $g^* : G' \to \R$ such that:


 * $(1): \quad$ $G'$ is a linear subspace of $X$ and is a proper superset of $G$
 * $(2): \quad$ $g^\ast$ extends $f_0$ with $\map {g^\ast} x \le \map p x$ for each $x \in G'$.

So:


 * $\tuple {G, g} \preceq \tuple {G', g^\ast}$

Since $\tuple {G, g}$ is maximal, we have:


 * $\tuple {G, g} = \tuple {G', g^\ast}$

So $G = G'$.

But $G'$ is a proper superset of $G$, which implies $G \ne G'$.

So we have a contradiction and so $G = X$.

Setting $f = g$, we have therefore found a linear functional $f : X \to \R$ extending $f_0$ with:


 * $\map f x \le \map p x$ for each $x \in X$.