The Hahn-Banach Theorem for Vector Spaces Part 3

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

The final proof can be found in part 4.

Theorem (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_(x)| \leq p(x)

for all

x \in E

Theorem 3: Let

$E$

be a locally convex topological vector space. If

A \subseteq E

is an open and convex subset of

$E$

and if

M \subseteq E

is a subspace of

$E$

such that

M \cap A = \emptyset

, then there exists a closed hyperplane

$H$

$E$

such that

M \subseteq H

and

H \cap A = \emptyset

Proof: Let $A \subseteq E$ be an open and convex subset of $$E$$ , and let $M \subseteq E$ be a subspace of $$E$$ such that $M \cap A = \emptyset$ .

Real Case: Let $$E$$ be a real vector space and let $\mathcal S$ be the set of all real vector subspaces $S \subseteq E$ such that $M \subseteq S$ and $S \cap A = \emptyset$ . Let $\mathcal C := \{ M \}$ . Since $\mathcal C$ is a chain (trivially), by the maximal axiom there exists a maximal chain $\mathcal N$ such that:

(1)

\begin{align} \quad \mathcal C \subseteq \mathcal N \subseteq \mathcal S \end{align}

Let $H := \bigcup_{N \in \mathcal N} N$ . Then $$H$$ is a vector subspace of $$E$$ since $\mathcal N$ is a chain of subspaces, and furthermore, since $\mathcal N \subseteq \mathcal S$ we have that $N \cap A = \emptyset$ for all $N \in \mathcal N$ and so $H \cap A = \emptyset$ .

By the lemma on The Hahn-Banach Theorem for Vector Spaces Part 1 page, since $$E$$ is a real, locally convex topological vector space and since $$A$$ is an open and convex subset of $$E$$ for which $H \cap A \neq \emptyset$ , we have that either $$H$$ is a hyperplane of $$E$$ or there exists a point $x \not \in H$ such that $\mathrm{span}(H \cup \{ x \}) \cap A$ . But this second possibility cannot happen by the maximality of the chain $\mathcal N$ as then $\mathrm{span} (H \cup \{ x \})$ could be added to $\mathcal N$ . Furthermore by the proposition on the Hyperplanes of a Vector Space page, $$H$$ is either closed or dense in $$E$$ , but $$H$$ cannot be dense in $$E$$ since $$A$$ is an open set with $H \cap A = \emptyset$ .

Thus $$H$$ is a real closed hyperplane of $$E$$ such that $M \subseteq H$ and $H \cap A = \emptyset$ . $\blacksquare$

Complex Case: Now suppose that $$E$$ is a complex vector space. Then it is also a real vector space by restricting the field of scalars from $\mathbb{C}$ to $\mathbb{R}$ . Thus from above, there exists a real closed hyperplane $$K$$ of $$E$$ such that $M \cap K$ and $K \cap A = \emptyset$ . Let $H := K \cap (iK)$ . Then $$H$$ is closed (because both $$K$$ and $$iK$$ are) and by the lemma on The Hahn-Banach Theorem for Vector Spaces Part 2 page, we have that $$H$$ is a closed complex hyperplane. Furthermore, since $M \subseteq K$ we have that $iM \subseteq K$ . So $M \cap (iM) \subseteq K \cap (iK) = H$ . But $M \cap (iM) = M$ since $$M$$ is a subspace of $$E$$ . Thus $M \subseteq H$ . Also since $A \cap M = \emptyset$ , we see that $A \cap H = \emptyset$ .

Thus $$H$$ is a complex closed hyperplane of $$E$$ such that $M \subseteq H$ and $H \cap A = \emptyset$ . $\blacksquare$

Corollary 2: Let

$E$

be a locally convex topological vector space. If

M \subseteq E

is a closed proper subspace of

$E$

, then

$M$

is the intersection of all closed hyperplanes containing

$M$

Proof: We will show that for all $a \in E \setminus M$ there exists a closed hyperplane $$H_a$$ containing $$M$$ but that does not contain $$a$$ .

So let $a \in E \setminus M$ . Since $$M$$ is closed, $E \setminus M$ is open, and so there exists a convex open neighbourhood $$U_a$$ of $$a$$ such that $U_a \subseteq E \setminus M$ . Then $M \cap U_a = \emptyset$ . By the previous theorem, there then exists a closed hyperplane $$H_a$$ of $$E$$ such that $M \subseteq H_a$ and $H_a \cap U_a = \emptyset$ . But since $$U_a$$ is an open neighbourhood of $$a$$ , this equation implies that $a \not \in H_a$ . This holds true for all such $a \in E \setminus M$ . So if $\mathcal S$ is the collection of all closed hyperplanes containing $$M$$ then:

(2)

\begin{align} \quad \bigcap_{S \in \mathcal S} S \subseteq \bigcap_{a \in E \setminus M} H_a \subseteq M \end{align}

But of course, by definition of $\mathcal S$ , $M \subseteq S$ for all $S \in \mathcal S$ , so that:

(3)

\begin{align} \quad \bigcap_{S \in \mathcal S} S \supseteq M \end{align}

So indeed, $$M$$ is the intersection of all closed hyperplanes of $$E$$ which contain $$M$$ . $\blacksquare$

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The Hahn-Banach Theorem for Vector Spaces Part 3