The Hahn-Banach Theorem for Vector Spaces Part 4

# The Hahn-Banach Theorem for Vector Spaces Part 4

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 $f_1$ on $E$ 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

\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:

\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:

\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:

\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. $\blacksquare$