Hahn-Banach Theorem/Real Vector Space/Lemma 3

== 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 ordering $\preceq$ on $P$ by:
:$\tuple {G, g} \preceq \tuple {H, h}$ :
::$(1): \quad$ $G \subseteq H$
::$(2): \quad$ $h$ extends $g$. 
Then: 
:every non-empty chain in $\struct {P, \preceq}$ has an upper bound.
== Proof ==
Let:
:$C = \set {\tuple {G_\alpha, g_\alpha} : \alpha \in A}$ 
be a non-empty chain in $\struct {P, \preceq}$. 
Let: 
:$\ds G = \bigcup_{\alpha \in A} G_\alpha$
We show that $G$ is a linear subspace of $X$. 
Clearly we have $G \subseteq X$. 
Let $x, y \in G$ and $\lambda \in \R$.
From One-Step Vector Subspace Test, it suffices to show $x + \lambda y \in G$. 
Then $x \in G_\alpha$ for some $\alpha \in A$, and $y \in G_\beta$ for some $\beta \in A$.
Since $C$ is a chain and $\tuple {G_\alpha, g_\alpha}, \tuple {G_\beta, g_\beta} \in C$, we have that:
:$\tuple {G_\alpha, g_\alpha}$ and $\tuple {G_\beta, g_\beta}$ are comparable.
So either $G_\alpha \subseteq G_\beta$ or $G_\beta \subseteq G_\alpha$.
So either $x, y \in G_\alpha$ or $x, y \in G_\beta$.
Suppose that $x, y \in G_\gamma$ for $\gamma \in A$.
Since $G_\gamma$ is a vector subspace of $X$, we have: 
:$x + \lambda y \in G_\gamma$
so:
:$x + \lambda y \in G$
So, by the One-Step Vector Subspace Test, we have that $G$ is a vector subspace of $X$. 
Since:
:$X_0 \subseteq G_\alpha$
we have:
:$X_0 \subseteq G$
from Set is Subset of Union, so we have $(1)$ for $G$. 
We construct a linear functional $g : G \to \R$ that extends each $g_\alpha$. 
Let $x \in G$. 
Then $x \in G_\alpha$ for some $\alpha \in A$. 
If $x \in G_\alpha$ for exactly one $\alpha \in A$, we can safely pick $\map g x = \map {g_\alpha} x$. 
However, if $x \in G_\alpha$ for more than one $\alpha \in A$, the choice is initially unclear.
Suppose $x \in G_\alpha$ and $x \in G_\beta$ for $\alpha, \beta \in A$ with $\alpha \ne \beta$.
Since $C$ is a chain, $\tuple {G_\alpha, g_\alpha}$ and $\tuple {G_\beta, g_\beta}$ are comparable, and we either have: 
:$\tuple {G_\alpha, g_\alpha} \preceq \tuple {G_\beta, g_\beta}$
or:
:$\tuple {G_\beta, g_\beta} \preceq \tuple {G_\alpha, g_\alpha}$
If:
:$\tuple {G_\alpha, g_\alpha} \preceq \tuple {G_\beta, g_\beta}$
then $g_\beta$ extends $g_\alpha$.
Then: 
:$\map {g_\alpha} t = \map {g_\beta} t$ for all $t \in G_\alpha$.
In particular: 
:$\map {g_\alpha} x = \map {g_\beta} x$
Similarly, if: 
:$\tuple {G_\beta, g_\beta} \preceq \tuple {G_\alpha, g_\alpha}$
then $g_\alpha$ extends $g_\beta$.
Then: 
:$\map {g_\alpha} t = \map {g_\beta} t$ for all $t \in G_\beta$.
In particular: 
:$\map {g_\alpha} x = \map {g_\beta} x$
So, if $x \in G_\alpha \cap G_\beta$, then $\map {g_\alpha} x = \map {g_\beta} x$. 
So, for each $x \in G$ we safely choose $\map g x = \map {g_\alpha} x$ for any $\alpha \in A$ with $x \in G_\alpha$.
We now verify that $g$ is a linear functional $g : G \to \R$ that extends each $g_\alpha$.
Let $\alpha \in A$. 
Note that, from Set is Subset of Union, we have: 
:$G_\alpha \subseteq G$
From construction, we have: 
:$\map g x = \map {g_\alpha} x$
for any $x \in G_\alpha$.
So $g$ extends $g_\alpha$ for each $\alpha \in A$. 
We show that $g$ is a linear functional.
Let $\lambda, \mu \in \R$ and $x, y \in G$. 
Then $x \in G_\alpha$ for some $\alpha \in A$, and $y \in G_\beta$ for some $\beta \in A$.
Since $C$ is a chain and $\tuple {G_\alpha, g_\alpha}, \tuple {G_\beta, g_\beta} \in C$, we have that:
:$\tuple {G_\alpha, g_\alpha}$ and $\tuple {G_\beta, g_\beta}$ are comparable.
So either $G_\alpha \subseteq G_\beta$ or $G_\beta \subseteq G_\alpha$.
So either $x, y \in G_\alpha$ or $x, y \in G_\beta$.
Suppose that $x, y \in G_\gamma$ for $\gamma \in A$.
Then, we have: 
So:
:$g$ is a linear functional. 
We now show that $\tuple {G, g} \in P$. 
We have already shown that: 
:$G$ is a linear subspace of $X$ with $X_0 \subseteq G$.
It remains to show that: 
:$g : G \to \R$ is a linear functional extending $f_0$
and:
:$\map g x \le \map p x$ for each $x \in G$.
Let $x \in X_0$, then $x \in G$. 
So $x \in G_\alpha$ for some $\alpha \in A$. 
We then have: 
:$\map g x = \map {g_\alpha} x$ 
Since $g_\alpha$ extends $f_0$, we have: 
:$\map g x = \map {f_0} x$ 
for each $x \in X_0$.
So $g$ extends $f_0$. 
We also have: 
:$\map {g_\alpha} x \le \map p x$
so:
:$\map g x \le \map p x$
for each $x \in X_0$. 
So we have $\tuple {G, g} \in P$.
Since $G_\alpha \subseteq G$ for each $\alpha \in A$, and $g$ extends $g_\alpha$ we have that: 
:$\tuple {G, g}$ is an upper bound for $C$ in $\tuple {P, \preceq}$.
So:
:$C$ has an upper bound.
Since $C$ was an arbitrary non-empty chain, we have: 
:every non-empty chain in $\struct {P, \preceq}$ has an upper bound.
Category:Hahn-Banach Theorem