153 lines
7.1 KiB
TeX
153 lines
7.1 KiB
TeX
\section{Sequence Spaces}
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\label{section:lp-direct-sum}
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\begin{definition}[$c_0$-Direct Sum]
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\label{definition:c0-direct-sum}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For any $x \in \prod_{i \in I}X_i$, $x$ \textbf{vanishes at infinity} if for each $\eps > 0$, $\bracs{i \in I| \norm{x_i}_{X_i} \ge \eps}$ is finite. The space
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\[
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[c_0(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | x \text{ vanishes at infinity}}
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\]
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equipped with the uniform norm
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\[
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\norm{x}_{[c_0(I); X_i]} = \sup_{i \in I}\norm{x_i}_{X_i}
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\]
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is the \textbf{$c_0$-direct sum} of $\seqi{X}$.
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\end{definition}
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\begin{definition}[$l^p$-Direct Sum]
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\label{definition:lp-direct-sum}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the \textbf{$l^p$-direct sum} of $\seqi{X}$ is the space
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\[
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[l^p(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sum_{i \in I}\norm{x_i}_{X_i}^p < \infty}
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\]
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equipped with the norm
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\[
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\norm{x}_{[l^p(I); X_i]} = \braks{\sum_{i \in I}\norm{x_i}_{X_i}^{p}}^{1/p}
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\]
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\end{definition}
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\begin{definition}[$l^\infty$-Direct Product]
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\label{definition:l-infty-direct-product}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$, then the \textbf{$l^\infty$-direct product} of $\seqi{X}$ is the space
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\[
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[l^\infty(I); X_i] = \bracs{x \in \prod_{i \in I}X_i \bigg | \sup_{i \in I}\norm{x_i}_{X_i} < \infty}
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\]
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equipped with the norm
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\[
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\norm{x}_{[l^\infty(I); X_i]} = \sup_{i \in I}\norm{x_i}_{X_i}
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\]
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\end{definition}
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\begin{proposition}
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\label{proposition:lp-direct-sum-gymnastics}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty]$.
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\begin{enumerate}
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\item (\textbf{Hölder's Inequality}) Let $q \in [1, \infty]$ be the Hölder conjugate of $p$, $\seqi{Y}, \seqi{Z}$ be normed spaces, and $\seqi{\lambda}$ such that for each $i \in I$, $\lambda \in L^2(X_i, Y_i; Z)$.
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For each $x \in [l^p(I); X_i]$ and $y \in [l^q(I); Y_i]$, let $\lambda(x, y)_i = \lambda_i(x_i, y_i)$, then
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\[
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\norm{\lambda(x, y)}_{[l^1(I); Z_i]} \le \norm{x}_{[l^p(I); X_i]} \cdot \norm{y}_{[l^q(I); X_i]} \cdot \sup_{i \in I}\norm{\lambda_i}_{L^2(X_i, Y_i; Z_i)}
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\]
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\item (\textbf{Minkowski's Inequality}) For each $x, y \in [l^p(I); X_i]$,
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\[
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\norm{x + y}_{[l^p(I); X_i]} \le \norm{x}_{[l^p(I); X_i]} + \norm{y}_{[l^p(I); X_i]}
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\]
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\item (\textbf{Markov's Inequality}) If $p < \infty$, then for each $\alpha > 0$ and $x \in [l^p(I); X_i]$,
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\[
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|\bracsn{i \in I|\ \norm{x}_{X_i} \ge \alpha}| \le \frac{1}{\alpha^p}\norm{f}_{[l^p(I); X_i]}^p
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\]
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In particular, $\bracs{i \in I|x_i \ne 0}$ is countable.
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\item For any $q \in [p, \infty]$, $[l^p(I); X_i] \subset [l^q(I); X_i]$, where for any $x \in [l^p(I); X_i]$, $\norm{x}_{[l^q(I); X_i]} \le \norm{x}_{[l^p(I); X_i]}$.
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\item If $X_i$ is a Banach space for all $i \in I$, then so is $[l^p(I); X_i]$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1), (2), (3): By the classical \hyperref[Hölder's inequality]{theorem:holder}, \hyperref[Minkowski's inequality]{theorem:minkowski}, and \hyperref[Markov's inequality]{theorem:markov-inequality}.
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(4): For each $i \in I$, $\norm{x_i}_{X_i} \le \norm{x}_{[l^p(I); X_i]}$, so the result holds when $q = \infty$.
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If $q < \infty$, then by \autoref{proposition:lp-intersection-interpolation}, there exists $\lambda \in [p, q]$ such that
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\[
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\norm{x}_{[l^p(I); X_i]} \le \norm{x}_{[l^p(I); X_i]}^{\lambda}\norm{x}_{[l^\infty(I); X_i]}^{1 - \lambda} \le \norm{x}_{[l^p(I); X_i]}
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\]
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\end{proof}
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\begin{theorem}
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\label{theorem:c0-sum-dual}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$. For each $y \in [l^1(I); X_i^*]$, let
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\[
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\phi_y: [c_0(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
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\]
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then the mapping
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\[
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[l^1(I); X_i^*] \to [c_0(I); X_i]^* \quad y \mapsto \phi_y
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\]
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is an isometric isomorphism.
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\end{theorem}
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\begin{proof}
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By \hyperref[Hölder's Inequality]{proposition:lp-direct-sum-gymnastics}, for each $y \in [l^1(I); X_i^*]$, $\norm{\phi_y}_{[c_0(I); X_i]^*} \le \norm{y}_{[l^1(I); X_i^*]}$.
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Let $\phi \in [c_0(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x_i \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[c_0(I); X_i]} = \dpn{x_i, y_i}{X_i}$.
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Let $J \subset I$ be finite and $\alpha \in (0, 1)$, then there exists $x \in [c_0(I); X_i]$ such that
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\begin{enumerate}
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\item $\{i \in I|x_i \ne 0\} \subset J$.
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\item $\norm{x}_{[c_0(I); X_i]} \le 1$.
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\item For each $j \in J$, $\dpn{x_j, y_j}{X_j} \ge \alpha\norm{y_j}_{X_j^*}$.
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\end{enumerate}
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Thus
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\[
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\alpha\sum_{j \in J}\norm{y_j}_{X_j^*} \le \sum_{j \in J}\dpn{x_j, y_j}{X_j} = \dpn{x, \phi}{[c_0(I); X_i]} \le \norm{\phi}_{[c_0(I); X_i]^*}
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\]
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As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^1(I); X_i^*]$ with $\norm{y}_{[l^1(I); X_i^*]} \le \norm{\phi}_{[c_0(I); X_i]^*}$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\dpn{x, \phi_y}{[c_0(I); X_i]} = \dpn{x, \phi}{[c_0(I); X_i]}$ for all $x \in [c_0(I); X_i]$. Hence the map is an isometric isomorphism.
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\end{proof}
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\begin{theorem}
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\label{theorem:lp-sum-dual}
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Let $\seqi{X}$ be normed vector spaces over $K \in \RC$ and $p \in [1, \infty)$ and $q \in (1, \infty]$ be Hölder conjugates. For each $y \in [l^q(I); X_i^*]$, let
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\[
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\phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
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\]
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then the mapping
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\[
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[l^q(I); X_i^*] \to [l^p(I); X_i]^* \quad y \mapsto \phi_y
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\]
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is an isometric isomorphism.
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\end{theorem}
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\begin{proof}
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Let $\phi \in [l^p(I); X_i]^*$, then there exists $y \in [l^\infty(I); X_i^*]$ such that for each $i \in I$ and $x_i \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$.
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Since the $q = \infty$ case has been ruled out, assume that $q \in (1, \infty)$. For each $\alpha \in (0, 1)$, there exists $x \in [l^\infty(I); X_i]$ with $\norm{x_i}_{X_i} \le 1$ and $\dpn{x_i, y_i}{X_i} \ge \alpha \norm{y_i}_{X_i^*}$. For each $J \subset I$ finite and $i \in I$, let $F_J(i) = \one_{J}(i) \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by \autoref{lemma:holder-conjugate-gymnastics},
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\[
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\norm{F_J x}_{[l^p(I); X_i]}^p \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}
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\]
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so
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\begin{align*}
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\alpha \sum_{j \in J} \norm{y_j}_{X_j^*}^q & \le
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\sum_{i \in I}F_J(i)\dpn{x_i, y_i}{X_i} =
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\dpn{F_J x, \phi}{[l^p(I); X_i]} \\
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&\le \norm{\phi}_{[l^p(I); X_i]^*} \cdot \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^{q}}^{1/p} \\
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\alpha \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*}
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\end{align*}
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As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^q(I); X_i^*]$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi}_{[l^p(I); X_i]^*}$.
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\end{proof}
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