My question
I have the following lemma and proof in my textbook, to apply the Hahn-Banach theorem on complex-linear functionals.
I don't understand the last equality with the question mark, because I don't know how to apply the sub-linearity of $p$ here.
Lemma
Let $p: X \rightarrow \mathbb{R}$ be sub-linear and $l: X \rightarrow \mathbb{C}$ be a complex-linear functional. Show that
$$ \forall x \in X : |l(x)| \leq p(x) \Leftrightarrow \forall x \in X : |\operatorname{Re} l(x)| \leq p(x) $$
Proof
$\Leftarrow$: Is obvious with $|\operatorname{Re} l| \leq |l|$ .
$\Rightarrow$: Let $\forall x \in X : |\operatorname{Re} (l(x))| \leq p(x) $. Let $x \in X$ and write $l(x) = |l(x)| \exp(\text{i} \alpha)$ for some $\alpha \in \mathbb{R}$. Then:
$$ |l(x)| = \operatorname{Re}(|l(x)|) = \operatorname{Re} (l(x) \exp(- \text{i} \alpha)) = \operatorname{Re} (l(\exp(- \text{i} \alpha) x )) \leq p(\exp(- \text{i} \alpha) x ) \stackrel{?}{=} p(x) $$
From http://mathonline.wikidot.com/the-hahn-banach-theorem-complex-version sublinearity of $p$ implies for complex $\lambda$ that $p(\lambda x)=|\lambda|p(x)$. Since when $\alpha$ is real, $|exp(-i\alpha)|=1$, the result follows.