\section{The Complex Interpolation Method}
\label{section:complex-interpolation}

\begin{definition}[Calderón Space]
\label{definition:calderon-space}
    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:
    \begin{enumerate}
        \item $f$ is holomorphic on $S$. 
        \item $f$ is continuous on $\ol S$. 
        \item For each $t \in \real$, $f(it) \in E_0$, and $\lim_{|t| \to \infty}\norm{f(it)}_{E_0} = 0$. 
        \item For each $t \in \real$, $f(1 + it) \in E_1$, and $\lim_{|t| \to \infty}\norm{f(1 + it)}_{E_1} = 0$. 
    \end{enumerate}
    
    equipped with the norm
    \[
    \norm{f}_{\cf(E_0, E_1)} = \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_0}, \sup_{t \in \real}\norm{f(1 + it)}_{E_1}}
    \]
\end{definition}
\begin{proof}
    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. 

    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. 
\end{proof}

\begin{definition}[The Complex Interpolation Method]
\label{definition:complex-interpolation-method}
    Let $(E_0, E_1)$ be a compatible couple of Banach spaces over $\complex$, $\cf(E_0, E_1)$ be their \hyperref[Calderón space]{definition:calderon-space}, $\theta \in [0, 1]$, and
    \[
    [E_0, E_1]_\theta = \bracsn{f(\theta)| f \in \cf(E_0, E_1)}
    \]

    with the norm
    \[
    \norm{x}_{[E_0, E_1]_\theta} = \inf\bracsn{\norm{f}_{\cf(E_0, E_1)}| f \in \cf(E_0, E_1), x = f(\theta)}
    \]

    then:
    \begin{enumerate}
        \item $[E_0, E_1]_\theta$ is an intermediate Banach space between $E_0$ and $E_1$.
        \item The mapping $C_\theta$ defined by $(E_0, E_1) \mapsto [E_0, E_1]_\theta$ is an interpolation functor of exact exponent $\theta$. 
    \end{enumerate}

    and the functor $C_\theta$ is the \textbf{method of complex interpolation}. 
\end{definition}
\begin{proof}[Proof, {{\cite[Theorem 4.1.2]{BerghInterpolation}}}. ]
    (1): Let $\seq{x_n} \subset [E_0, E_1]_\theta$ with $\sum_{n \in \natp}\norm{x_n}_{[E_0, E_1]_\theta} < \infty$, then there exists $\seq{f_n} \subset \cf(E_0, E_1)$ such that for each $n \in \natp$, $f_n(\theta) = x_n$ and $\norm{f_n}_{\cf(E_0, E_1)} \le 2\norm{x_n}_{[E_0, E_1]_\theta}$. Since $\cf(E_0, E_1)$ is complete, there exists $f \in \cf(E_0, E_1)$ such that $f = \sum_{n = 1}^\infty f_n$. Let $x = f(\theta)$, then since $\sum_{n = 1}^N f_n \to f$ in $\cf(E_0, E_1)$ as $N \to \infty$, $\sum_{n = N}^\infty x_n \to x$ in $[E_0, E_1]_\theta$ as $N \to \infty$. Therefore $[E_0, E_1]_\theta$ is a Banach space by \autoref{lemma:banach-criterion}. 

    For any $x \in E_0 \cap E_1$ and $\delta > 0$, let $f_\delta(z) = x_0 e^{\delta(z - \theta)^2}$, then $f_\delta \in \cf(E_0, E_1)$ with $\norm{f_\delta}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}$. Thus $x \in [E_0, E_1]_\theta$ with
    \[
    \norm{x}_{[E_0, E_1]_\theta} \le \norm{f}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}
    \]

    As the above holds for all $\delta > 0$, $E_0 \cap E_1$ is continuously embedded in $[E_0, E_1]_\theta$. 

    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}, 
    \begin{align*}
    \norm{x}_{E_0 + E_1} &= \norm{f(\theta)}_{E_0 + E_1}  \\
    &\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}} \\ 
    &\le  \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_1}, \sup_{t \in \real}\norm{f(1 + it)}_{E_1}} = \norm{f}_{\cf(E_0, E_1)}
    \end{align*}

    so $[E_0, E_1]_\theta$ is continuously embedded in $E_0 + E_1$. 
    

    (2): Let $(F_0, F_1)$ be a compatible couple of Banach spaces over $\complex$, and $T \in L(E_0 + E_1; F_0 + F_1)$ such that $T|_{E_0} \in L(E_0; F_0)$ and $T|_{E_1} \in L(E_1; F_1)$. Let $x \in [E_0, E_1]_\theta$ and $f \in \cf(E_0, E_1)$ such that $f(\theta) = x$. For each $z \in \bracs{y \in \complex| \text{Re}(y) \in [0, 1]}$, let
    \[
    g(z) = \norm{T}_{L(E_0; F_0)}^{z - 1} \norm{T}_{L(E_1; F_1)}^{-z} \cdot T \circ f(z)
    \]

    then for each $t \in \real$, 
    \begin{align*}
        \norm{g(it)}_{F_0} &= \norm{T}_{L(E_0; F_0)}^{-1} \cdot \norm{T \circ f(it)}_{F_0} \le \norm{f(it)}_{E_0} \\
        \norm{g(1 + it)}_{F_0} &= \norm{T}_{L(E_1; F_1)}^{-1} \cdot \norm{T \circ f(1 + it)}_{F_1} \le \norm{f(1 + it)}_{E_1}
    \end{align*}
    
    so $g \in \cf(F_0, F_1)$ with $\norm{g}_{\cf(F_0, F_1)} \le \norm{f}_{\cf(E_0, E_1)}$. Thus
    \begin{align*}
    g(\theta) &= \norm{T}_{L(E_0; F_0)}^{\theta - 1} \norm{T}_{L(E_1; F_1)}^{-\theta} \cdot T \circ f(\theta) \\
    &= \norm{T}_{L(E_0; F_0)}^{\theta - 1} \norm{T}_{L(E_1; F_1)}^{-\theta} \cdot Tx
    \end{align*}

    and $Tx \in [F_0, F_1]_\theta$ with
    \begin{align*}
        \norm{Tx}_{[F_0, F_1]_\theta} &\le \norm{T}_{L(E_0; F_0)}^{1 - \theta} \norm{T}_{L(E_1; F_1)}^{\theta}\norm{g}_{\cf(F_0, F_1)} \\
        &\le \norm{T}_{L(E_0; F_0)}^{1 - \theta} \norm{T}_{L(E_1; F_1)}^{\theta} \norm{f}_{\cf(E_0, E_1)}
    \end{align*}

    Since the above holds for all $f \in \cf(E_0, E_1)$ with $f(\theta) = x$, 
    \[
    \norm{Tx}_{[F_0, F_1]_\theta} \le \norm{T}_{L(E_0; F_0)}^{1 - \theta} \norm{T}_{L(E_1; F_1)}^{\theta} \norm{x}_{[E_0, E_1]_{\theta}}
    \]
\end{proof}