real part of bounded linear functional Let $X$ be a normed space over the complex numbers and let $f \in X^*$. Now, apparently we have $\|f\| = \|\Re(f)\|$. Why is this the case? I'm sure this is trivial I just can't figure it out at the moment.
 A: You have
$$\lvert f(x)\rvert = \sup_{\vartheta\in [0,2\pi]} \lvert\Re f(e^{i\vartheta}x)\rvert,$$
since for an appropriate choice of $\vartheta$, $f(e^{i\vartheta}x)$ is real and non-negative. Thus
$$\lVert f\rVert = \sup_{\lVert x\rVert = 1} \lvert f(x)\rvert = \sup_{\lVert x\rVert = 1}\sup_\vartheta \lvert\Re f(e^{i\vartheta}x)\rvert = \sup_{\lVert x\rVert = 1} \lvert\Re f(x)\rvert.$$
A: Remember that
(a) For all $z\in\mathbb{C}$, we have $|\Re(z)|\leq|z|$;
(b) For all $z\in \mathbb{C}$, there exists $\lambda_z\in\mathbb{C}$ with $|\lambda_z|=1$ such that $z=|z|\lambda_z$.
From (a), we get
$|\Re(f)(x)|=|\Re(f(x))|\leq|f(x)|$ for all $x\in X$ and thus
$$\sup_{x\in S}|\Re(f)(x)|\leq \sup_{x\in S} |f(x)|,\tag{1}$$
where $S=\{x\in X;\;\|x\|=1\}$.
From (b), for all $x\in X$ there exists $\beta_x\in \mathbb{C}$ with $|\beta_x|=1$ such that $|f(x)|=\beta_xf(x)$. So,
$$\Re(f(\beta_xx))=\Re(\beta_xf(x))=\Re(|f(x)|)=|f(x)|$$
for all $x\in X$ and thus
$$\sup_{x\in S} |f(x)|=\sup_{x\in S} |\Re(f(\beta_xx))|=\sup_{y\in S'} |\Re(f(y))|\leq\sup_{y\in S} |\Re(f(y))|=\sup_{x\in S} |\Re(f)(x)|,\tag{2}$$
where $S'=\{\beta_xx\in X;\;x\in S\}$ (the last inequality is valid because $S'\subset S$).
It follows from $(1)$ and $(2)$ that $f$ is bounded iff $\Re(f)$ is bounded and, in the affirmative case, 
$$\|f\|=\sup_{x\in S}|f(x)|=\sup_{x\in S}|\Re(f)|=\|\Re(f)\|.$$
