17.2 The Complex Interpolation Method

Definition 17.2.1 (Calderón Space).label 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 Calderón space $\cf(E_{0}, E_{1})$ is the Banach space of functions $f: \ol S \to E_{0} + E_{1}$ such that:

  1. (1)

    $f$ is holomorphic on $S$.

  2. (2)

    $f$ is continuous on $\ol S$.

  3. (3)

    For each $t \in \real$, $f(it) \in E_{0}$, and $\lim_{|t| \to \infty}\norm{f(it)}_{E_0}= 0$.

  4. (4)

    For each $t \in \real$, $f(1 + it) \in E_{1}$, and $\lim_{|t| \to \infty}\norm{f(1 + it)}_{E_1}= 0$.

equipped with the norm

\[\norm{f}_{\cf(E_0, E_1)}= \max\braks{\sup_{t \in \real}\norm{f(it)}_{E_0}, \sup_{t \in \real}\norm{f(1 + it)}_{E_1}}\]

Proof. By the 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 Maximum Modulus Theorem, Proposition 28.4.2, and Proposition 7.3.2, $\cf(E_{0}, E_{1})$ is complete.$\square$

Definition 17.2.2 (The Complex Interpolation Method).label Let $(E_{0}, E_{1})$ be a compatible couple of Banach spaces over $\complex$, $\cf(E_{0}, E_{1})$ be their Calderón 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:

  1. (1)

    $[E_{0}, E_{1}]_{\theta}$ is an intermediate Banach space between $E_{0}$ and $E_{1}$.

  2. (2)

    The mapping $C_{\theta}$ defined by $(E_{0}, E_{1}) \mapsto [E_{0}, E_{1}]_{\theta}$ is an interpolation functor of exact exponent $\theta$.

and functor $C_{\theta}$ is the method of complex interpolation.

Proof, [Theorem 4.1.2, BL76]. (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 Lemma 12.1.3.

For any $x \in E_{0} \cap E_{1}$ and $\delta > 0$, let $f_{\delta}(z) = x_{0} e^{(z - \theta)^2}$, 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 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 $f(\theta) = x$,

\[\norm{Tx}_{[F_0, F_1]_\theta}\le \norm{T}_{L(E_0; F_0)}^{1 - \theta}\norm{T}_{L(E_1; F_1)}^{\theta}\norm{x}_{[E_0, E_1]_{\theta}}\]

$\square$

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