norm of a functional attains in real part

Question

I tried to prove that, when $X$ is a Banach space, it holds that $$\sup Ref(B_X)=\|f\|$$, where $f \in X^*$ is a functional and the norm is defined by $\|f\|=\sup\{ |f(x)|: x \in B_X \}$.

One of the inequalities is straightforward, but I do not see the other one.

Answer 1

If $z=re^{i\theta}$ is any complex number ($r \geq 0, \theta \in \mathbb R$) then we can write $|z|=Re (cz)$ where $c=e^{-i\theta}$. Note that $|c|=1$. Taking $z=f(x)$ where $\|x\| \leq 1$ we see that $|f(x)|= Re f(y)$ for some $y$ with $\|y\|=1$. This proves the other inequality.