The Hahn-Banach Theorem (Real Version)

Theorem 1 (The Hahn-Banach Theorem (Real Version)): Let $X$ be a linear space and let $Y \subset X$ be a subspace. Let $p : X \to \mathbb{R}$ be a sublinear functional. If $\varphi : Y \to \mathbb{R}$ is a $\mathbb{R}$-linear functional such that $\varphi(y) \leq p(y)$ for all $y \in Y$ then there exists an $\mathbb{R}$-linear functional $\Phi : X \to \mathbb{R}$ such that: 1) $\Phi (y) = \varphi (y)$ for all $y \in Y$. 2) $\Phi (x) \leq p(x)$ for all $x \in X$.

In other words, the Hahn-Banach theorem tells us that if $\varphi$ is a real linear functional on the subspace $Y$ of $X$ that is dominated by a sublinear functional $p$ on $Y$ then there exists an extension $\Phi$ that is a real linear functional on the whole space $X$ that is dominated by $p$ on all of $X$.

Note that the Hahn-Banach Theorem is VERY general! We do not require $X$ to even be a normed linear space!

• Proof: Let $\mathcal F$ be the collection of pairs $(Z, \psi)$ where $Y \subseteq Z \subset X$ is a subspace, $\psi |_Y = \varphi$, and $\psi(z) \leq p(z)$ for all $z \in Z$. We define a partial ordering $\prec$ on $\mathcal F$ as follows. If $(Z_1, \psi_1), (Z_2, \psi_2) \in \mathcal F$ then we write:

(1)

\begin{align} \quad (Z_1, \psi_1) \prec (Z_2, \psi_2) \end{align}

• if $Z_1 \subseteq Z_2$ and $\psi_2 |_{Z_1} = \psi_1$. Then $\mathcal F$ is a partially ordered set.

• Let $\mathcal F_0$ be a chain in $\mathcal F$. We want to show that $\mathcal F_0$ has an upper bound in $\mathcal F$. Let:

(2)

\begin{align} \quad Z_0 = \bigcup_{(Z, \psi) \in \mathcal F_0} Z \end{align}

• And define $\psi_0 : Z_0 \to \mathbb{R}$ as follows. For each $z \in Z_0$ by definition, there exists a $(Z, \psi) \in \mathcal F_0$ such that $z \in Z$, so define:

(3)

\begin{align} \quad \psi_0(z) = \psi(z) \end{align}

• We want to show that $\psi_0$ is well-defined. Suppose that $z \in Z_1 \cap Z_2$ where $(Z_1, \psi_1), (Z_2, \psi_2) \in \mathcal F_0$. Since $\mathcal F_0$ is a chain we may assume without loss of generality that $(Z_1, \psi_1) \prec (Z_2, \psi_2)$ so that then $\psi_2 |_{Z_1} = \psi_1$. Therefore $\psi_2(z) = \psi_1(z)$. So indeed, $\psi_0$ is well-defined.

• Furthermore, $\psi_0$ is linear, $\psi_0 |_Y = \varphi$, and $\psi_0(z) \leq p(z)$ for all $z \in Z_0$. Hence $(Z_0, \psi_0) \in \mathcal F$ and is clearly an upper bound of the chain $\mathcal F_0$. By Zorn's Lemma, there exists a maximal element $(Z^*, \psi^*) \in \mathcal F$.

• Now we show that $Z^* = X$. Suppose not. Then there exists a $z \in X \setminus Z^*$. By The Hahn-Banach Lemma there exists a linear function $\Psi^* : Z^* \oplus \mathrm{span} (z) \to \mathbb{R}$ such that $\Psi^* |_Y = \varphi$ and $\Psi^*(x) \leq p(x)$ for all $x \in Z^* \oplus \mathrm{span}(z)$. Therefore $(Z^* \oplus \mathrm{span}(z), \Psi^*) \in \mathcal F$ and:

(4)

\begin{align} \quad (Z^*, \psi^*) \prec (Z^* \oplus \mathrm{span}(z), \Psi^*) \end{align}

• But this contradicts the maximality of $(Z^*, \psi^*) \in \mathcal F$. So we must have that $Z^* = X$.

• Hence by definition, $\psi^* : X \to \mathbb{R}$ is a $\mathbb{R}$-linear functional such that $\psi^*(y) = \varphi(y)$ for all $y \in Y$ and $\psi^*(x) \leq p(x)$ for all $x \in X$. $\blacksquare$