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:
- $\tuple {G, g} \preceq \tuple {H, h}$ if and only if:
- $(1): \quad$ $G \subseteq H$
- $(2): \quad$ $h$ 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:
\(\ds \lambda \map g x + \mu \map g y\) | \(=\) | \(\ds \lambda \map {g_\gamma} x + \mu \map {g_\gamma} y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {g_\gamma} {\lambda x + \mu y}\) | since $g_\gamma$ is a linear functional | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g {\lambda x + \mu y}\) | since $g$ extends $g_\gamma$ |
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.
$\blacksquare$