Hahn-Banach Theorem/Real Vector Space/Lemma 1

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

Let $p : X \to \R$ be a convex function. Let $G$ be a 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:


 * $(1): \quad$ if $G \ne X$, then $G'$ is a proper superset of $G$
 * $(2): \quad$ $g$ extends $g_0$ with $\map g x \le \map p x$ for each $x \in G'$.

Proof
If $G = X$, then we can simply take $G' = X$ and $g = g_0$.

Suppose that $G \ne X$.

That is, $X \setminus G$ 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:


 * $\ds \map {f_0} x + \map {f_0} y \le \frac 1 t \map p {t y + t z} + \frac 1 s \map p {s x - s z}$

Set:


 * $\ds \lambda = \frac 1 {\frac 1 s + \frac 1 t} = \frac {s t} {s + t}$

so that:


 * $\ds \frac \lambda t + \frac \lambda s = 1$

Note that $\lambda > 0$.

We have:

Dividing through $\lambda$ we obtain:


 * $\ds \map {f_0} x + \map {f_0} y \le \frac 1 t \map p {t y + t z} + \frac 1 s \map p {s x - s z}$

and so:


 * $\ds \map {f_0} x - \frac 1 s \map p {s x - s z} \le \frac 1 t \map p {t y + t z} - \map {f_0} y$

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.