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03e7899904
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@@ -96,6 +96,35 @@
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\end{proof}
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\end{proof}
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\begin{lemma}
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\label{lemma:maximum-modulus-strip}
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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
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\[
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\norm{f}_{u} = \sup_{z \in \partial S}\norm{f(z)}_E
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\]
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\end{lemma}
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\begin{proof}
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Let $\eps > 0$ and $\phi_\eps(z) = e^{\eps(z^2 - 1)}$, then $\phi f \in H(S; E) \cap BC(\ol S; E)$.
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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},
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\[
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\norm{\phi_\eps f}_u = \lim_{R \to \infty} \sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_E
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\]
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However, since $\phi_\eps f(z) \to 0$ as $|\text{Im}(z)| \to \infty$,
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\[
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\norm{\phi_\eps f}_u = \lim_{R \to \infty} \sup_{z \in \partial S_R}\norm{\phi_\eps f(z)}_E = \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_E
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\]
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Therefore
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\[
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\norm{f}_u = \sup_{\eps > 0} \norm{\phi_\eps f}_u = \sup_{\eps > 0} \sup_{z \in \partial S}\norm{\phi_\eps f(z)}_E = \sup_{z \in \partial S}\norm{f(z)}_E
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\]
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\end{proof}
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\begin{lemma}[Hadamard's Three Lines Lemma]
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\begin{lemma}[Hadamard's Three Lines Lemma]
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\label{lemma:three-lines}
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\label{lemma:three-lines}
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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
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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
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@@ -121,7 +150,7 @@
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h: \ol S \to E \quad z \mapsto \frac{f(z)}{g(z)}
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h: \ol S \to E \quad z \mapsto \frac{f(z)}{g(z)}
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\]
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\]
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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 $z \in S$,
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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 $z \in S$,
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\[
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\[
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f(z) \le M(0)^{\text{Re}(z)} M(1)^{1-\text{Re}(z)}
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f(z) \le M(0)^{\text{Re}(z)} M(1)^{1-\text{Re}(z)}
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\]
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\]
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@@ -3,5 +3,6 @@
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\input{./definitions.tex}
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\input{./definitions.tex}
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\input{./polar.tex}
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\input{./polar.tex}
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\input{./mackey.tex}
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@@ -6,7 +6,7 @@
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Let $S = \bracs{z \in \complex| \text{Re}(z) \in (0, 1)}$ and $(E_0, E_1)$ be a compatible couple of Banach spaces over $\complex$, then the \textbf{Calderón space} $\cf(E_0, E_1)$ is the Banach space of functions $f: \ol S \to E_0 + E_1$ such that:
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Let $S = \bracs{z \in \complex| \text{Re}(z) \in (0, 1)}$ and $(E_0, E_1)$ be a compatible couple of Banach spaces over $\complex$, then the \textbf{Calderón space} $\cf(E_0, E_1)$ is the Banach space of functions $f: \ol S \to E_0 + E_1$ such that:
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\begin{enumerate}
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\begin{enumerate}
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\item $f$ is holomorphic on $S$.
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\item $f$ is holomorphic on $S$.
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\item $f$ is continuous on $\ol S$.
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\item $f$ is bounded and continuous on $\ol S$.
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\item For each $t \in \real$, $f(it) \in E_0$, and $\lim_{|t| \to \infty}\norm{f(it)}_{E_0} = 0$.
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\item For each $t \in \real$, $f(it) \in E_0$, and $\lim_{|t| \to \infty}\norm{f(it)}_{E_0} = 0$.
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\item For each $t \in \real$, $f(1 + it) \in E_1$, and $\lim_{|t| \to \infty}\norm{f(1 + it)}_{E_1} = 0$.
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\item For each $t \in \real$, $f(1 + it) \in E_1$, and $\lim_{|t| \to \infty}\norm{f(1 + it)}_{E_1} = 0$.
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\end{enumerate}
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\end{enumerate}
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@@ -17,9 +17,9 @@
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\]
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\]
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\end{definition}
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\end{definition}
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\begin{proof}
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\begin{proof}
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By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem} applied to $f$ as a function in $H(S; E_0 + E_1)$, $\norm{\cdot}_{\cf(E_0, E_1)}$ is a norm.
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By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip} applied to $f$ as a function in $H(S; E_0 + E_1)$, $\norm{\cdot}_{\cf(E_0, E_1)}$ is a norm.
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By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, \autoref{proposition:holomorphic-complete}, and \autoref{proposition:uniform-limit-continuous}, $\cf(E_0, E_1)$ is complete.
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By the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip}, \autoref{proposition:holomorphic-complete}, and \autoref{proposition:uniform-limit-continuous}, $\cf(E_0, E_1)$ is complete.
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\end{proof}
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\end{proof}
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\begin{definition}[The Complex Interpolation Method]
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\begin{definition}[The Complex Interpolation Method]
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@@ -52,7 +52,7 @@
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As the above holds for all $\delta > 0$, $E_0 \cap E_1$ is continuously embedded in $[E_0, E_1]_\theta$.
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As the above holds for all $\delta > 0$, $E_0 \cap E_1$ is continuously embedded in $[E_0, E_1]_\theta$.
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Let $x \in [E_0, E_1]_\theta$ and $f \in \cf(E_0, E_1)$ with $f(\theta) = x$, then by the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem},
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Let $x \in [E_0, E_1]_\theta$ and $f \in \cf(E_0, E_1)$ with $f(\theta) = x$, then by the \hyperref[Maximum Modulus Theorem]{lemma:maximum-modulus-strip},
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\begin{align*}
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\begin{align*}
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\norm{x}_{E_0 + E_1} &= \norm{f(\theta)}_{E_0 + E_1} \\
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\norm{x}_{E_0 + E_1} &= \norm{f(\theta)}_{E_0 + E_1} \\
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&\le \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_0 + E_1}, \sup_{t \in \real}\norm{f(1 + it)}_{E_0 + E_1}} \\
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&\le \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_0 + E_1}, \sup_{t \in \real}\norm{f(1 + it)}_{E_0 + E_1}} \\
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