Hahn-Banach Theorem/Real Vector Space/Lemma 1

== Lemma ==
Let $X$ be a vector space over $\R$. 
Let $p : X \to \R$ be a sublinear functional.
Let $G$ be a proper linear subspace of $X$. 
Let $g_0 : G \to \R$ be a linear functional such that: 
:$\map {g_0} x \le \map p x$ for each $x \in G$.
Then there exists a linear subspace $G'$ of $X$ and a linear functional $g : G' \to \R$ such that: 
:$g$ extends $g_0$ with $\map g x \le \map p x$ for each $x \in G'$. 
== Proof ==
Since $G \ne X$, $X \setminus G$ is non-empty. 
Let $z \in X \setminus G$. 
Set: 
:$G' = \map \span {G \cup \set z}$
From Linear Span is Linear Subspace, we have: 
:$G'$ is a linear subspace of $X$.
Then, for all $x \in G'$, there exists unique $x_0 \in G$ and $t \in \R$ such that: 
:$x = x_0 + t z$ 
To show uniqueness, suppose that $x_0' \in G$ and $t' \in \R$ had:
:$x = x_0' + t' z$
then: 
:$0 = \paren {x_0' - x_0} + \paren {t' - t} z$
So that: 
:$x_0 - x_0' = \paren {t' - t} z$
$t' \ne t$, then we would have: 
:$\ds z = \frac 1 {t' - t} \paren {x_0 - x_0'}$ 
Since $G$ is a linear subspace, we would then have: 
:$\ds \frac 1 {t' - t} \paren {x_0 - x_0'} \in G$
and so $z \in G$, so we have a contradiction and so $t' = t$.
We then obtain $x_0 = x_0'$ hence the desired uniqueness.
We look to extend $f_0$ to a linear functional $f : G' \to \R$. 
Let: 
:$\map f x = \map {f_0} x$ 
for $x \in G$.
It remains to pick a value of $\map f x$ for $x \in G' \setminus G$.
If $f : G' \to \R$ is a linear functional extending $f_0$ we must have: 
:$\map f {x_0 + t z} = \map f x + t \map f z$ 
that is: 
:$\map f {x_0 + t z} = \map {f_0} x + t \map f z$
So any such extension is entirely determined by the choice of $\map f z$.
We will therefore determine a suitable choice of $\map f z$ and verify that the thus obtained map is linear.
For brevity, write:
:$\map f z = C$
We require: 
:$\map {f_0} {x_0} + t C \le \map p {x_0 + t z}$
for each $x_0 \in G$ and $t \in \R$.
If $t = 0$, the inequality follows from the hypothesis.
If $t > 0$, then:
:$\map {f_0} {x_0} + t C \le \map p {x_0 + t z}$
is equivalent to: 
:$\ds C \le \frac 1 t \paren {\map p {x_0 + t z} - \map {f_0} {x_0} }$
From linearity, we have: 
:$\ds C \le \frac 1 t \map p {x_0 + t z} - \map {f_0} {\frac {x_0} t}$
Since $G$ is a linear subspace of $X$, this is equivalent to requiring: 
:$\ds C \le \frac 1 t \map p {t x_0 + t z} - \map {f_0} {x_0}$
for each $x_0 \in G$ and $t > 0$.
Now suppose $t < 0$. 
Write $t = -s$ for $s > 0$. 
Then: 
:$\map {f_0} {x_0} - s C \le \map p {x_0 - s z}$ 
is equivalent to: 
:$s C \ge \map {f_0} {x_0} - \map p {x_0 - s z}$ 
From linearity, this is equivalent to: 
:$\ds C \le \map {f_0} {\frac {x_0} s} - \frac 1 s \map p {s x_0 - s z}$ 
Since $G$ is a linear subspace of $X$, this is equivalent to requiring:
:$C \le \map {f_0} {x_0} - \dfrac 1 s \map p {s x_0 - s z}$
for each $x_0 \in G$ and $s > 0$.
Let:
:$\ds a = \sup_{s > 0, \, x_0 \in X} \paren {\map {f_0} {x_0} - \frac 1 s \map p {s x_0 - s z} }$
and:
:$\ds b = \inf_{t > 0, \, x_0 \in X} \paren {\frac 1 t \map p {t x_0 + t z} - \map {f_0} {x_0} }$
We show that $a \le b$. 
Then, for any $C \in \closedint a b$, say: 
:$\ds C = \frac {a + b} 2$ 
we have:
:$C \le \map {f_0} {x_0} - \dfrac 1 s \map p {x_0 - s z}$
for each $x_0 \in G$ and $s > 0$, and: 
:$\ds C \le \frac 1 t \map p {t x_0 + t z} - \map {f_0} {x_0} \le C$
for each $x_0 \in G$ and $t > 0$.
That is, we will obtain: 
:$\map {f_0} {x_0} + t C \le \map p {x_0 + t z}$
for all $x_0 \in G$ and $t \ne 0$.
We will show that for each $t, s > 0$ and $x, y \in G$, we have: 
:$\ds \map {f_0} x - \frac 1 s \map p {s x - s z} \le \frac 1 t \map p {t y + t z} - \map {f_0} y$
This is equivalent to: 
:$\ds \map {f_0} x + \map {f_0} y \le \frac 1 t \map p {t y + t z} + \frac 1 s \map p {s x - s z}$
We have: 
as required.
So we obtain the required $a \le b < \infty$. 
Now set: 
:$\ds C = \frac {a + b} 2$ 
and define $f : G' \to \R$ by: 
:$\map f x = \map {f_0} {x_0} + t C$ 
for each $x \in G'$, where $x_0 \in G$ and $t \in \R$ are such that: 
:$x = x_0 + t z$
We have shown that this decomposition is unique, so $f$ is well-defined.
If $x \in G$, then $t = 0$ and we obtain: 
:$\map f x = \map {f_0} {x_0}$
so $f$ certainly extends $f_0$.
We verify that $f$ is a linear functional.
Let $x, y \in G'$ and $\lambda, \mu \in \R$. 
Decompose: 
:$x = x_0 + t z$
for $x_0 \in G$ and $t \in \R$, and: 
:$y = y_0 + s z$
for $y_0 \in G$ and $s \in \R$. 
Then: 
We therefore have that $f$ is the desired extension.
Category:Hahn-Banach Theorem