The Hahn-Banach Theorem for Vector Spaces Part 1

# The Hahn-Banach Theorem for Vector Spaces Part 1

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 $x \in E$. |

## Preliminary Results

Lemma 1: Let $E$ be a locally convex topological vector space over $\mathbf{R}$. If $A \subset E$ is an open and convex subset of $E$, and if $H \subset E$ is a subspace of $E$ such that $A \cap H = \emptyset$, then either $H$ is a hyperplane of $E$ or there exists a point $x \not \in H$ such that $A \cap \mathrm{span} (H \cup \{ x \}) = \emptyset$. |

*That is, in a real locally convex topological vector space, if $A$ is an open convex set and $H$ is a subspace for which $A$ and $H$ do not intersect, then $H$ is a hyperplane of $E$ or there exists a point $x$ outside of $H$ for which $A$ and the subspace spanned by $H$ and $x$ do not intersect.*

**Proof:**Let $A \subset E$ be open and convex, and let $H$ be a subspace of $E$. Let:

\begin{align} \quad C := H + \bigcup_{\lambda > 0} \lambda A \end{align}

- Since $A$ is open we have that $\lambda A$ is open for each $\lambda > 0$, and so $\displaystyle{\bigcup_{\lambda > 0} \lambda A}$ is open as it is a union of open sets. Then, for each $h \in H$, $\displaystyle{h + \bigcup_{\lambda > 0} \lambda A}$ is open as it is a translation of an open set. Thus, $\displaystyle{C = \bigcup_{h \in H} \left [ h + \bigcup_{\lambda > 0} \lambda A \right ]}$ is open.

- Observe that:

\begin{align} \quad -C = H + \bigcup_{\lambda < 0} \lambda A \end{align}

- Since $\displaystyle{-C = -H - \bigcup_{\lambda > 0} \lambda A = H + \bigcup_{\lambda > 0} -\lambda A = H + \bigcup_{\lambda < 0} \lambda A}$ (where $H = -H$ since $H$ is a subspace of $E$). Also observe that:

\begin{align} \quad C \cap (-C) = \emptyset \end{align}

- This is because if $x \in C \cap (-C)$ then $x = h_1 + \lambda_1 a_1$ and $x = h_2 - \lambda_2 a_2$ for some $h_1, h_2 \in H$, $\lambda_1, \lambda_2 > 0$, and $a_1, a_2 \in A$. Then $h_1 - h_2 = \lambda_1 a_1 + \lambda_2 a_2$. By setting $y : = h_1 - h_2 = \lambda_1a_1 + \lambda_2 a_2$, we see that $y \in H - H = H$ (since $H$ is a subspace of $E$. Furthermore, we see that $y = \lambda_1 a_1 + \lambda_2 a_2 \in \lambda_1 A + \lambda_2A \subseteq (\lambda_1 + \lambda_2) A$ by the convexity of $A$. Thus $\frac{y}{\lambda_1 + \lambda_2} \in A$. But $\frac{y}{\lambda_1 + \lambda_2} \in H$ too since $H$ is a subspace, and thus $\frac{y}{\lambda_1 + \lambda_2} \in A \cap H$, which is a contradiction since $A \cap H = \emptyset$.

- There are now two cases to consider.

**Case 1:**Let $H \cup C \cup (-C) \neq E$. We will show that for every point $x \not \in H$ and $x \not \in C \cup (-C)$, that $A \cap \mathrm{span} (H \cup \{ x \}) = \emptyset$. So let $x \in E$ be such that $x \not \in H$, $x \not \in C \cup (-C)$.

- Suppose instead that $A \cap \mathrm{span} (H \cup \{ x \}) \neq \emptyset$. Then, there exists a point $y$ such that $y \in A$ and $y \in \mathrm{span} (H \cup \{ x \})$. But then $y \in H + \lambda x$ where $\lambda \neq 0$ (note that $\lambda \neq 0$ since otherwise this would imply that $y \in H$. But then $y \in A \cap H$ contradicting the assumption that $A \cap H = \emptyset$). So $-\lambda x \in -y + H$, and dividing by $-\lambda \neq 0$ yields:

\begin{align} \quad x \in \frac{-1}{\lambda}y + \left ( \frac{-1}{\lambda} \right ) H \end{align}

- Setting $\mu := -\frac{1}{\lambda}$ and observe that $\mu H = H$ (since $H$ is a subspace of $E$), we have that:

\begin{align} \quad x \in H + \mu y \in H + \mu A \subseteq (C) \cup (-C) \end{align}

- This is a contradiction though, since in this particular case, $x \not \in C \cup (-C)$. Hence, we must have that $A \cap \mathrm{span} (H \cup \{ x \}) = \emptyset$ in this case.

**Case 2:**Suppose that $H \cup C \cup (-C) = E$. We will show that in this case, $H$ is necessarily a hyperplane of $E$.

- Suppose instead that $H$ is not a hyperplane of $E$. Then there exists a point $a \in E \setminus H \subseteq C \cup (-C)$ for which $\mathrm{span} (H \cup \{ a \}) \neq E$. We may take $a \in C$ (for if $a \in -C$, then $-a \in C$ will work.). Since $E \neq \mathrm{span} (H \cup \{ a \})$ and since $E \cup C \subseteq \mathrm{span} \(H \cup \{ a \})$, there must also exist a poinr $b \in -C$ such that $b \not \in \mathrm{span} (H \cup \{ a \})$.

- Let $f : [0, 1] \to E$ be defined for all $\lambda \in [0, 1]$ by:

\begin{align} \quad f(\lambda) = (1 - \lambda)a + \lambda b \end{align}

- Then $f$ is a continuous function on $[0, 1]$. Since $C$ and $-C$ are open sets in $E$, we have by the continuity of $f$ that $f^{-1}(C)$ and $f^{-1}(-C)$ are open subsets of $[0, 1]$ such that $0 \in f^{-1}(C)$ and $1 \in f^{-1}(-C)$. Furthermore, $f^{-1}(C) \cap f^{-1}(-C) =f^{-1} (C \cap (-C)) = f^{-1}(\emptyset) = \emptyset$.

- Let:

\begin{align} \quad \alpha := \sup \{ \lambda : \lambda \in f^{-1}(C) \} \end{align}

- Then observe that:

\begin{align} \quad \alpha \in \overline{ f^{-1}(C)} \cap \overline{[0, 1] \setminus f^{-1}(C)} \subseteq \overline{[0, 1] \setminus f^{-1}(-C)} \cap \overline{[0, 1] \cap f^{-1}(C)} = [0, 1] \setminus f^{-1}(-C) \cap [0, 1] \setminus f^{-1}(C) \end{align}

- Thus $a \not \in f^{-1}(C) \cup f^{-1}(-C) = f^{-1} (C \cup -C)$, and so $f(\alpha) \not \in C \cup (-C)$. But since $E = H \cup C \cup (-C)$ we must have that $f(\alpha) = (1 - \alpha) a + \alpha b \in H$. So $\alpha b \in H - (1 - \alpha)a$, showing that $b \in \mathrm{span} \{H \cup \{ a \})$, which is a contradiction by choice of $b$.

- Thus $H$ must be a hyperplane of $E$. $\blacksquare$