Definition 11.2.6 (Complexification of Normed Spaces).label Let $E$ be a normed vector space over $\real$, then
\[\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_{E}\]
is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric.
Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))}= \norm{T}_{L(E; F)}$.
Proof. For any $\phi \in [0, 2\pi]$ and $x, y \in E$,
\begin{align*}\normn{e^{i \phi}(x, y)}_{\complex(E)}&= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)}\\&= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_{E} \\&= \norm{(x, y)}_{\complex(E)}\end{align*}
so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$,
\[\norm{\iota x}_{\complex(E)}= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_{E} = \norm{x}_{E} \\\]
Therefore $\iota: E \to \complex(E)$ is isometric.
Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
\begin{align*}\norm{\complex(T)}_{L(\complex(E); \complex(F))}&= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1}\\&\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1}= \norm{T}_{L(E; F)}\end{align*}
On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
\begin{align*}\normn{\complex(T)(x, y)}_{\complex(F)}&= \normn{(Tx, Ty)}_{\complex(F)}= \norm{\cos(\theta)Tx + \sin(\theta)Ty}_{F} \\&\le \norm{T}_{L(E; F)}\cdot \norm{\cos(\theta)x + \sin(\theta)y}_{E} \\&\le \norm{T}_{L(E; F)}\cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_{E} \\&= \norm{T}_{L(E; F)}\cdot \norm{(x, y)}_{E}\end{align*}
$\square$
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