Complex Hahn-Banach Theorem

Let $X$ be a complex vector space, $X_0$ one of its subspaces, $p: X \to \mathbb{R}_+$ such that

$$ p(\lambda x) = |\lambda| p(x), \quad \forall \lambda \in \mathbb{C}, x \in X \text{ and } p(x + y) \leq p(x) + p(y), \quad \forall x, y \in X, $$

satisfying $|f(x)| \leq p(x)$, $\forall x \in X_0$, where $f: X_0 \to \mathbb{C}$ is linear.

Under these conditions, there exists a linear functional $F: X \to \mathbb{C}$ such that $F|_{X_0} = f$ and

$$ |F(x)| \leq p(x), \quad \forall x \in X. $$

Proof: Since $f$ is linear, it follows that $\text{Re } f: X_0 \to \mathbb{R}$ is linear and $$ \text{Re } f(x) \leq |f(x)| \leq p(x), \quad \forall x \in X_0. $$

By the Real Hahn-Banach Theorem there exists $g: X \to \mathbb{R}$ a linear functional such that $g$ is an extension for $\text{Re } f$ and $g(x) \leq p(x)$, $\forall x \in X$. We also have $g(x) = -g(-x) \geq -p(x)$ so $|g(x)| \leq p(x)$, $\forall x \in X$.

Define now $F(x) = g(x) - i g(ix)$, $\forall x \in X$. This is obviously linear and if $x \in X_0$ we have $$ F(x) = g(x) - i g(ix) = \text{Re } f(x) - i \text{Re } i f(x) = \text{Re } f(x) + i \text{Im } f(x) = f(x), \quad \forall x \in X_0. $$

For the last part we have $|F(x)| = e^{i\theta} F(x) = F(e^{i\theta} x) = g(e^{i\theta} x)$, because this is a real number. Furthermore, we have $g(e^{i\theta} x) \leq p(e^{i\theta} x) = p(x)$. Combining the two above, we get $$ |F(x)| \leq p(x), \quad \forall x \in X, $$ which solves the theorem.