Hahn-Banach Theorem/Real Vector Space/Lemma 2

Lemma

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 $x \in X_0$.

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:

$\tuple {G, g} \preceq \tuple {H, h}$ if and only if:

$(1): \quad$ $G \subseteq H$

$(2): \quad$ $h$ extends $g$.

Then:

$\struct {P, \preceq}$ is an ordered set.

Proof

We check each of the conditions of an ordering for $\preceq$.

Proof of $(1)$

We show that for each $\tuple {G, g} \in P$, we have:

$\tuple {G, g} \preceq \tuple {G, g}$

We have, from Set is Subset of Itself:

$G \subseteq G$

We also have:

$g$ extends $g$.

So, we have:

$\tuple {G, g} \preceq \tuple {G, g}$

$\Box$

Proof of $(2)$

Suppose that $\tuple {G, g}, \tuple {H, h}, \tuple {I, i} \in P$ have:

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

and:

$\tuple {H, h} \preceq \tuple {I, i}$

We show that:

$\tuple {G, g} \preceq \tuple {I, i}$

We have:

$G \subseteq H$

and:

$H \subseteq I$

From Subset Relation is Transitive, we have:

$G \subseteq I$

Since $h$ extends $g$, we have:

$\map g x = \map h x$ for each $x \in G$.

Since $i$ extends $h$, we have:

$\map h x = \map i x$ for each $x \in H$.

Since $G \subseteq H$, we in particular have:

$\map h x = \map i x$ for each $x \in G$.

So, we have:

$\map g x = \map i x$ for each $x \in G$.

Since $G \subseteq I$, we therefore have:

$i$ extends $g$.

So, we have:

$\tuple {G, g} \preceq \tuple {I, i}$

$\Box$

Proof of $(3)$

Suppose that $\tuple {G, g}$ and $\tuple {H, h}$ have:

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

and:

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

We show that:

$\tuple {G, g} = \tuple {H, h}$

That is:

$G = H$ and $g = h$.

Since:

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

we have that:

$G \subseteq H$

and:

$g$ extends $h$.

Since:

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

we have:

$H \subseteq G$

and:

$h$ extends $g$.

So, from the definition of set equality, we have:

$G = H$

So:

$h$ is actually a linear functional $G \to \R$.

Since $h$ extends $g$, we have:

$\map h x = \map g x$ for each $g \in G$.

That is:

$g = h$

So, we have:

$\tuple {G, g} = \tuple {H, h}$

$\blacksquare$