The Hahn-Banach Theorem for Vector Spaces Part 2

The Hahn-Banach Theorem for Vector Spaces Part 2

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 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 2: Let $E$ be a vector space over $\mathbb{C}$. If $H$ is a real hyperplane of $E$ then $H \cap (iH)$ is a complex hyperplane of $E$.
  • Proof: Let $H$ be a real hyperplane. Then $H$ is a proper real subspace of $E$. Since the intersection of two subspaces is a subspace, we see that $H \cap (iH)$ is a real subspaces of $E$, that is proper since $H \cap (iH) \subseteq H$ and $H$ is proper. Moreover, it is a complex subspace of $E$. Since if $x \in H \cap (iH)$, then $x \in iH$ implies that $ix \in -H = H$. Similarly, $x \in iH$ implies that $\frac{x}{i} \in H$. Thus, for all $\lambda + \mu i \in \mathbb{C}$ and for all $x \in H \cap (iH)$ we have that:
\begin{align} (\lambda + \mu i) x = \lambda x + \mu i x \in H + H = H \quad \mathrm{and} \quad (\lambda + \mu i)x = \lambda x + \mu i x = \lambda i \frac{x}{i} + \mu i x \in iH + iH = iH \end{align}
  • So $(\lambda + \mu i) x \in H \cap (iH)$, and thus $H \cap (iH)$ is closed under scalar multiplication of the complex numbers.
  • Let $a \not \in H \cap (iH)$ with $a \not \in H$. We will show that $\mathrm{span}_{\mathbb{C}} (H \cap (iH) \cup \{ a \}) = E$ to conclude that $H \cap (iH)$ is a complex hyperplane of $E$.
  • Since $a \not \in H$ we have that $ia \not \in iH$. But $iH$ is a real hyperplane, and so $\mathrm{span}_{\mathbb{R}} (iH \cup \{ ia \}) = E$. So $a$ can be written in the form:
\begin{align} \quad a = \alpha ia + b \end{align}
  • Where $\alpha \in \mathbb{R}$ and $b \in iH$. So $b = a - \alpha i a$ or equivalently, $b = (1 - \alpha i)a$. Multiplying both sides of this equation by $1 + \alpha i$ yields:
\begin{align} \quad (1 + \alpha i)b &= (1 + \alpha i)(1 - \alpha i)a \\ &= (1 + \alpha^2)a \end{align}
  • But since $H$ is a real hyperplane (and hence a subspace) for which $a \not \in H$, we have that $(1 + \alpha^2)a \not \in H$, and consequently, $(1 + \alpha i)b \not \in H$. This implies that $b \not \in H$. (Note that if instead $b \in H$, then since $b \in iH$ already, we have that $\frac{b}{i} \in H$ too, and so:
\begin{align} \quad (1 + \alpha i) b = b + \alpha i b = b + \alpha i \cdot i \cdot \frac{b}{i} = b - \alpha \frac{b}{i} \in H - H = H \end{align}
  • contradicting the fact that $(1 + \alpha i) b \not \in H$).
  • Since $H$ is a real hyperplane and since $b \not \in H$, we have that $\mathrm{span}_{\mathbb{R}} (H \cup \{ b \}) = E$. So for each $x \in E$, there exists $\beta \in \mathbb{R}$ and $y \in H$ such that:
\begin{align} \quad x = \beta b + y \end{align}
  • Since $y \in H \subseteq E$ and since $iH$ is a real hyperplane, $b \not \in H$, so $ib \not \in H$ and we have that $\mathrm{span}_{\mathbb{R}} (iH \cup \{ ib \}) = E$, and thus there exists $\gamma \in \mathbb{R }$ and $z \in iH$ such that:
\begin{align} \quad y = \gamma ib + z \end{align}
  • Also we have that $z \in H$. (This is because $y \in H$, and $b \in iH$, so $\gamma b \in iH$, and $\gamma ib \in i^2 H = -H = H$).
  • Hence, $x \in E$ can be written as:
\begin{align} \quad x &= \beta b + y \\ &= \beta b + \gamma ib + z \\ &= (\beta + \gamma i)b + z \\ &= (\beta + \gamma i)(a - \alpha i a) + z \\ &= \beta a - \alpha \beta i a + \gamma i a - \alpha \gamma i a + z \\ &= (\beta - \alpha \beta + [\gamma - \alpha \gamma]i)a + z \end{align}
  • Let $\lambda := \beta - \alpha \beta$ and let $\mu := \gamma - \alpha \gamma$. Then $\lambda, \mu \in \mathbb{R}$, and moreover, $x = (\lambda + \mu i ) a + z$, with $z \in H \cap (iH)$. So $\mathrm{span}_{\mathbb{C}} (H \cap (iH) \cup \{ a \}) = E$, and thus $H \cap (iH)$ is a complex hyperplane. $\blacksquare$
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