Hahn-Banach Theorem/Real Vector Space/Lemma 3

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

Let $p : X \to \R$ be a Minkowski 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:


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


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

Then:
 * every chain in $\struct {P, \preceq}$ has an upper bound.

Proof
Let:


 * $C = \set {\tuple {G_\alpha, g_\alpha} : \alpha \in A}$

be a 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 chain, we have:


 * every chain in $\struct {P, \preceq}$ has an upper bound.