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## Hahn-Banach (complex version)

Let $X$ be a complex vector space, $X_0$ one of its subspaces, $p: X \to \mathbb{R}_+$ such that $p(\lambda x)=|\lambda| p(x),\ \forall \lambda \in \mathbb{C}, x \in X$ and $p(x+y) \leq p(x)+p(y),\ \forall x,y \in X$, 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 $|F(x)| \leq p(x),\ \forall x \in X$.
Solution: Since $f$ is linear, it follows that $Re\ f :X_0 \to \mathbb{R}$ is linear and $Re\ f(x) \leq |f(x)| \leq p(x),\ \forall x \in X_0$. By the Hahn-Banach Theorem (real version) there exists $g: X \to \mathbb{R}$ a linear functional such that $g$ is an extension for $Re\ f$ 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 $F(x)=g(x)-ig(ix),\ \forall x \in X$. This is obviously linear and if $x \in X_0$ we have $F(x)=g(x)-ig(ix)=Re\ f(x)- i Re\ i f(x)=$ $Re\ f(x)+i Im\ f(x)=f(x),\ \forall x \in X_0$.
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. Further more, we have $g(e^{i\theta}x) \leq p(e^{i\theta} x)=p(x)$. Combinig the two above, we get $|F(x)|\leq p(x),\ \forall x \in X$, which solves the theorem. $\Box$