Fixed gap in maximum modulus strip.

src/dg/complex/zero.tex

@@ -96,6 +96,35 @@
 \end{proof}
 
 
+
+\begin{lemma}
+\label{lemma:maximum-modulus-strip}
+    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
+    \[
+    \norm{f}_{u} = \sup_{z \in \partial S}\norm{f(z)}_E
+    \]
+\end{lemma}
+\begin{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 \hyperref[Maximum Modulus Theorem]{theorem: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
+    \[
+    \norm{f}_u = \sup_{\eps > 0} \norm{\phi_\eps f}_u = \sup_{\eps > 0} \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_E = \sup_{z \in \partial S}\norm{f(z)}_E
+    
+    \]
+\end{proof}
+
+
 \begin{lemma}[Hadamard's Three Lines Lemma]
 \label{lemma:three-lines}
     Let $S = \bracs{z \in \complex| \text{Re}(z) \in [0, 1]}$, $E$ be a Banach space over $\complex$, and $f \in H(S; E) \cap BC(\ol{S}; E)$. For each $s \in [0, 1]$, let
@@ -121,7 +150,7 @@
     h: \ol S \to E \quad z \mapsto \frac{f(z)}{g(z)}
     \]
 
-    then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_E \le 1$ for all $z \in \partial S$. By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{h(z)}_E \le 1$ for all $z \in S$. Thus for every $z \in S$,
+    then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_E \le 1$ for all $z \in \partial S$. By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip}, $\norm{h(z)}_E \le 1$ for all $z \in S$. Thus for every $z \in S$,
     \[
     f(z) \le M(0)^{\text{Re}(z)} M(1)^{1-\text{Re}(z)}
     \]