The Hahn-Banach Lemma

# The Hahn-Banach Lemma

 Lemma 1 (The Hahn-Banach Lemma): Let $X$ be a linear space and let $Y \subset X$ be a subspace. Let $p : X \to [0, \infty)$ be a sublinear functional. If $\varphi : Y \to \mathbb{R}$ is an $\mathbb{R}$-linear functional such that $\varphi(y) \leq p(y)$ for all $y \in Y$ then for each $z \in X \setminus Y$ there is an $\mathbb{R}$-linear functional $\Phi : Y \oplus \mathrm{span} (z) \to \mathbb{R}$ such that: 1) $\Phi (y) = \varphi (y)$ for all $y \in Y$. 2) $\Phi (x) \leq p(X)$ for all $x \in Y \oplus \mathrm{span} (z)$.
• Proof: Let $y, y' \in Y$. Then since $p$ is a subadditive function we have that:
(1)
\begin{align} \quad \varphi (y) + \varphi (y') = \varphi (y + y') \leq p(y + y') = p(y - z + z - y') \leq p(y - z) + p(z - y') \end{align}
• Therefore:
(2)
\begin{align} \quad \varphi(y) - p(y - z) \leq -\varphi(y') + p(z - y') \end{align}
• And hence:
(3)
\begin{align} \quad S = \sup_{y \in Y} \{ \varphi (y) - p(y - z) \} \leq \inf_{y \in Y} \{ -\varphi (y) + p(z - y) \} = I \end{align}
• Let $\varphi (z) \in [S, I]$. Then we have that:
(4)
\begin{align} \quad \varphi(y) - p(y - z) \leq \varphi(z) \leq -\varphi(y) + p(y + z) \quad \forall y \in Y \end{align}
• There are three cases to consider.
• Case 1: Suppose that $\lambda > 0$. Then for all $y \in Y$:
(5)
\begin{align} \quad \varphi(z) &\leq -\varphi \left ( \frac{1}{\lambda} y \right ) + p \left ( \frac{1}{\lambda}y + z \right ) \\ & \leq -\frac{1}{\lambda} \varphi(y) + \frac{1}{\lambda} p(y + \lambda z) \end{align}
• Hence:
(6)
\begin{align} \quad \varphi(y) + \lambda \varphi(z) \leq p(y + \lambda z) \end{align}
• Case 2: Suppose that $\lambda < 0$. Then for all $y \in Y$:
(7)
\begin{align} \quad \varphi (z) &\geq \varphi \left ( -\frac{1}{\lambda} y \right ) - p \left ( -\frac{1}{\lambda} y - z \right ) \\ & \geq -\frac{1}{\lambda} \varphi (y) + \frac{1}{\lambda} p(y + \lambda z) \end{align}
• Hence:
(8)
\begin{align} \quad \varphi (y) + \lambda \varphi (z) \leq p(y + \lambda z) \end{align}
• Case 3: Suppose that $\lambda = 0$. Then for all $y \in Y$ we are given that $\varphi (y) \leq p(y)$.
• So there exists a real number $\varphi (z) \in \mathbb{R}$ such that:
(9)
\begin{align} \quad \varphi (y) + \lambda \varphi (z) \leq p(y + \lambda z) \quad \forall y \in Y, \: \forall \lambda \in \mathbb{R} \end{align}
• So define $\Phi : Y \oplus \mathrm{span} (z) \to \mathbb{R}$ for all $y + \lambda z \in Y \oplus \mathrm{span} (z)$ by:
(10)
\begin{align} \quad \Phi (y + \lambda z) = \varphi (y) + \lambda (z) \end{align}
• Then $\Phi$ is an $\mathbb{R}$-linear functional, $\Phi (y) = \varphi (y)$ for all $y \in Y$, and $\Phi (y + \lambda z) \leq p(y + \lambda z)$ for all $y + \lambda z \in Y \oplus \mathrm{span} (z)$ and for all $\lambda \in \mathbb{R}$. $\blacksquare$