Prove that the real part of a complex linear functional, $\phi$, is a bounded linear functional, $\phi_r$, and $||\phi_r|| = ||\phi||$. I have been given the following exercise:
V is a complex vector space and $\phi: V \rightarrow \mathbb{C}$ is a bounded linear functional. Show that if V is considered as a real vector space, then the real part of $\phi$ is a bounded linear functional $\phi_r: V\rightarrow \mathbb{R}$ and $||\phi_r|| = ||\phi||$.
I don't have a problem with showing that $\phi_r$ is a complex-linear functional. My problem is calculating the operator norm: $||\phi_r|| = ||\phi||$.
I done the following so far:
With $\phi(v) = \phi_r(v) -i\cdot\phi_r(iv)$ , (and $\phi_r(v) \text{ is real}$), then $$|\phi_r(v)| \leq |\phi_r(v)-i\cdot\phi_r(iv)| = |\phi(v)|. $$
This implies that $\phi_r(v)\text{ is bounded and  that } ||\phi_r(v)|| \leq ||\phi(v)||$.
But I can't seem to prove the other inequality, i.e. $||\phi(v)|| \leq ||\phi_r(v)||.$ Can anyone help?

Let $w_n$ be a sequence of unit vectors for which $||\phi||=\lim_{n \to \infty}|\phi(w_n)|$ and pick $|\alpha_n|=1$ for which $|\phi(w_n)|=\alpha_n\phi(w_n)$
Then since $\phi$ is complex linear (crucial here) we have that $|\phi(w_n)|=\alpha_n\phi(w_n)=\phi(\alpha_nw_n)$ so if we take $v_n=\alpha_nw_n, ||v_n||=1$ and $\phi(v_n ) \ge 0$ so $\phi(v_n ) =\phi_r(v_n)$  and since $\phi_r(v_n)=\phi(v_n ) \to ||\phi||, ||v_n||=1$ it follows that $||\phi|| \le ||\phi_r||$ so we are done using the easy reverse inequality in the post.
