The Hahn-Banach Theorem (Complex Version)

The Hahn-Banach Theorem (Complex Version)

Recall from The Hahn-Banach Theorem (Real Version) page that if $X$ is a linear space and $Y \subset X$ is a subspace and if $p : X \to [0, \infty)$ is a positively homogeneous and subadditive function such that and $\varphi : Y \to \mathbb{R}$ is a $\mathbb{R}$-linear functional such that $\varphi (y) \leq p(y)$ for all $y \in Y$ then there exists a $\mathbb{R}$-linear function $\Phi : X \to \mathbb{R}$ such that:

  • $\Phi(y) = \varphi(y)$ for all $y \in Y$.
  • $\Phi(x) \leq p(x)$ for all $x \in X$.

We now give the more general complex version of the Hahn-Banach theorem.

Theorem 1 (The Hahn-Banach Theorem (Complex Version)): Let $X$ be a linear space and let $Y \subset X$ be a subspace. Let $p : X \to [0, \infty)$ be a subadditive function such that $p(\lambda x) = |\lambda| p(x)$ for all $x \in X$ and for all $\lambda \in \mathbb{C}$. If $\varphi : Y \to \mathbb{C}$ is a linear functional such that $|\varphi(y)| \leq p(y)$ for all $y \in Y$ then there exists a linear function $\Phi : X \to \mathbb{C}$ such that:
1) $\Phi(y) = \varphi(y)$ for all $y \in Y$.
2) $|\Phi(x)| \leq p(x)$ for all $x \in X$.
  • Proof: Define a new function $\psi : Y \to \mathbb{R}$ for all $y \in Y$ by:
(1)
\begin{align} \quad \psi(y) = \mathrm{Re} \varphi (y) \end{align}
  • We first show that $\psi$ is an $\mathbb{R}$-linear function. Let $y, y' \in Y$ and let $\lambda \in \mathbb{R}$. Then:
(2)
\begin{align} \quad \psi (y + y') &= \mathrm{Re} \varphi (y + y') \\ &= \mathrm{Re} [\varphi (y) + \varphi (y')] \\ &= \mathrm{Re} \varphi (y) + \mathrm{Re} \varphi (y') \\ &= \psi (y) + \psi (y') \end{align}
(3)
\begin{align} \quad \psi (\lambda y) &= \mathrm{Re} \varphi (\lambda y) \\ &= \mathrm{Re} [\lambda \varphi (y)] \\ &= \lambda \mathrm{Re} \varphi (y) \\ &= \lambda \psi (y) \end{align}
  • So indeed $\psi$ is an $\mathbb{R}$-linear functional on $Y$. Furthermore, we have that:
(4)
\begin{align} \quad \psi (y) \leq | \varphi (y) | \leq p(y), \quad \forall y \in Y \end{align}
  • So by the real version of the Hahn-Banach theorem there is an $\mathbb{R}$-linear functional $\Psi : X \to \mathbb{R}$ such that:
(5)
\begin{align} \quad \Psi(y) = \psi(y), \quad \forall y \in Y \quad (*) \end{align}
(6)
\begin{align} \quad \Psi(x) \leq p(x), \quad \forall x \in X \quad (**) \end{align}
  • Now define $\Phi : X \to \mathbb{C}$ for all $x \in X$ by:
(7)
\begin{align} \quad \Phi(x) = \Psi(x) - i \Psi (ix) \end{align}
  • We first show that $\Phi$ is a linear functional. Let $x, x' \in X$ and let $\lambda = a + bi \in \mathbb{C}$. Then:
(8)
\begin{align} \quad \Phi (x + x') &= \Psi (x + x') - i \Psi(i(x + x')) \\ &= \Psi(x) + \Psi(x') - i\Psi(ix + ix') \\ &= \Psi(x) + \Psi(x') - i\Psi(ix) - i\Psi(ix') \\ &= [\Psi(x) - i\Psi(ix)] + [\Psi(x') - i\Psi(ix')] \\ &= \Phi(x) + \Phi(x') \end{align}
(9)
\begin{align} \quad \Phi(\lambda x) &= \Psi (\lambda x) - i \Psi(i \lambda x) \\ &= \Psi([a + bi]x) - i \Psi (i[a + bi]x) \\ &= \Psi(ax + bix) - i \Psi(aix - bx) \\ &= \Psi(ax) + \Psi(bix) - i\Psi(aix) - i\Psi(-bx) \\ &= a\Psi(x) + b\Psi(ix) - ai\Psi(ix) + bi\Psi(x) \\ &= [a + bi]\Psi(x) + [b - ai]\Psi (ix) \\ &= [a + bi]\Psi(x) - i[a + bi]\Psi(ix) \\ &= \lambda \Psi(x) - i\lambda \Psi(ix) \\ &= \lambda[\Psi(x) - i\Psi(ix)] \\ &= \lambda \Phi(x) \end{align}
  • Now let $y \in Y$. Then:
(10)
\begin{align} \quad \Phi (y) &= \Psi(y) - i\Psi(iy) \\ &=\psi(y) - i\psi(iy) \\ &= \mathrm{Re} \varphi (y) - i \mathrm{Re} (iy) \\ &= \mathrm{Re} \varphi (y) + i \mathrm{Im} \varphi (y) \\ &= \varphi (y) \end{align}
  • Therefore $\Phi (y) = \varphi(y)$ for all $y \in Y$. Lastly, for all $x \in X$ and for $\lambda \in \mathbb{C}$ with $|\lambda| = 1$ we have that:
(11)
\begin{align} \quad |\Phi(\lambda x)| = |\lambda \Phi(x)| = |\lambda|\Phi(x) = \Phi(x) \end{align}
  • Therefore for all $x \in X$ we have that:
(12)
\begin{align} \quad \Phi(x) \leq p(\lambda x) = |\lambda| p(x) = p(x) \quad \blacksquare \end{align}
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