Lemma 31.2.5.label Let $S = \bracs{z \in \complex| \text{Re}(z) \in (0, 1)}$, $E$ be a Banach space over $\complex$, and $f \in H(S; E) \cap BC(\ol S; E)$, then
\[\norm{f}_{u}= \sup_{z \in \partial S}\norm{f(z)}_{E}\]
Proof. Let $\eps > 0$ and $\phi_{\eps}(z) = e^{\eps(z^2 - 1)}$, then $\phi f \in H(S; E) \cap BC(\ol S; E)$.
For each $R > 0$, let $S_{R} = \bracs{z \in \complex| \text{Re}(z) \in (0, 1), |\text{Im}(z)| < R}$, then by the Maximum Modulus Theorem,
\[\norm{\phi_\eps f}_{u} = \lim_{R \to \infty}\sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_{E}\]
However, since $\phi_{\eps} f(z) \to 0$ as $|\text{Im}(z)| \to \infty$,
\[\norm{\phi_\eps f}_{u} = \lim_{R \to \infty}\sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_{E} = \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_{E}\]
Therefore
\[\norm{f}_{u} = \sup_{\eps > 0}\norm{\phi_\eps f}_{u} = \sup_{\eps > 0}\sup_{z \in \partial S}\norm{\phi_\eps f(z)}_{E} = \sup_{z \in \partial S}\norm{f(z)}_{E}\]
$\square$
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