The Hahn-Banach Theorem for Vector Spaces Part 3

# The Hahn-Banach Theorem for Vector Spaces Part 3

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 $f_1$ on $E$ 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 $H$ of $E$ 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:

\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:

\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:

\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 $M$. $\blacksquare$