The Hahn-Banach Theorem for Vector Spaces Part 4

Table of Contents

We are about to prove the famous Hahn-Banach theorem for vector spaces stated below. We first need to prove a few preliminary results though. These results can be found on the following pages:

The Hahn-Banach Theorem for Vector Spaces Part 1
The Hahn-Banach Theorem for Vector Spaces Part 2
The Hahn-Banach Theorem for Vector Spaces Part 3
The Hahn-Banach Theorem for Vector Spaces Part 4

We are now ready to prove the following theorem:

Theorem 1 (The Hahn-Banach Extension Theorem): Let

$E$

be a vector space and let

M \subseteq E

be a subspace of

$E$

. If

$f$

is a linear form on

$M$

and if

$p$

is a seminorm on

$E$

with the property that

|f(x)| \leq p(x)

for all

x \in M

, then there exists a linear form

$f_1$

$E$

which extends

$f$

and such that

|f_1(x)| \leq p(x)

for all

x \in E

Proof: If $$f$$ is the zero linear form on $$M$$ , then clearly, by setting $$f_1$$ to be the zero linear form on $$E$$ , we see that $$f_1$$ extends $$f$$ and is such that $|f_1(x)| = 0 \leq p(x)$ for all $x \in E$ trivially.

So assume that $$f$$ is a nonzero linear form on $$M$$ . Then, there exists a point $a \in M$ such that $$f(a) = 1$$ , and $M = f^{-1}(0) \oplus \mathbf{F}a$ .

Equip $$E$$ with the coarsest topology determined by $\{ p \}$ . Then $$E$$ equipped with topology is a locally convex topological vector space. Since $U := \{ x : p(x) < 1 \}$ is an open absolutely convex neighbourhood of the origin, we see that in particular translation set $$a + U$$ is a convex open neighbourhood of $$a$$ .

Since $$f$$ is a nonzero linear form, we have that $f^{-1}(0)$ is a closed subspace of $$M$$ and hence is a closed subspace of $$E$$ . For each $y \in a + U$ we can write $$y = a + x'$$ with $x' \in U$ . Then, if $x' \in M$ we have that

(1)

\begin{align} \quad f(y) = f(a + x') = f(a) + f(x') = 1 + f(x') \neq 0 \end{align}

(Since $|f(x')| \leq p(x') < 1$ ). On the otherhand, if $x' \not \in M$ then $y \not \in M$ so that $(a + U) \cap M = \emptyset$ . But since $f^{-1}(0)$ is a subspace of $$M$$ we'd have that $(a + U) \cap f^{-1}(0) = \emptyset$ . Thus in either case we see that:

(2)

\begin{align} \quad f^{-1}(0) \cap (a + U) = \emptyset \end{align}

Since $$E$$ equipped with the coarsest topology determined by $\{ p \}$ is a locally convex topological space such that $$a + U$$ is an open convex set and $f^{-1}(0)$ is a closed subspace of $$E$$ such that $f^{-1}(0) \cap (a + U) = \emptyset$ , by the theorem on The Hahn-Banach Theorem for Vector Spaces Part 3 page, there exists a closed hyperplane $$H$$ of $$E$$ such that $f^{-1}(0) \subseteq H$ and $H \cap (a + U) = \emptyset$ .

Let $f_1 : E \to \mathbb{R}$ be the linear form on $$E$$ specified by $f_1^{-1}(0) := H$ , and $$f_1(a) := 1$$ . Then for each $x \in E$ , write $x = y + \lambda a$ for $y \in H$ and $\lambda \in \mathbf{F}$ and define $f_1(x) := \lambda$ .

Observe that if $m \in M = f^{-1}(0) \oplus \mathbf{F} a$ , then $m := y + \lambda a$ where $y \in f^{-1}(0)$ and $\lambda \in \mathbf{F}$ . Then $y \in H$ , and so $f_1(m) = \lambda$ and:

(3)

\begin{align} \quad f(m) = f(y + \lambda a) = f(y) + \lambda f(a) = \lambda = f_1(y + \lambda a) \end{align}

So $$f_1$$ is an extension of $$f$$ to $$E$$ . Furthermore, since $$U$$ is a balanced set we have by the lemma on the Closed Preimage Criterion for a Linear Form to be Continuous in a TVS page that $(a + U) \cap f^{-1}(0) = (a + U) \cap H = \emptyset$ implies that:

(4)

\begin{align} \quad U \subseteq \{ x : |f_1(x)| < 1 \} \end{align}

So, since $$p$$ and $$|f_1|$$ are two seminorms on $$E$$ such that $$p(x) < 1$$ implies that $|f_1(x)| < 1 \leq 1$ , we have by the lemma on the Seminorms and Norms on Vector Spaces page that $|f_1(x)| \leq p(x)$ for all $x \in E$ , which completes the proof. $\blacksquare$

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The Hahn-Banach Theorem for Vector Spaces Part 4