Fixed typos in the sequence space duality result.
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@@ -65,7 +65,7 @@
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\begin{theorem}
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\begin{theorem}
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\label{theorem:lp-sum-dual}
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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 $p, q \in [1, \infty]$ be Hölder conjugates. For each $y \in [l^q(I); X_i^*]$, let
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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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\[
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\phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
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\phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i}
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\]
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\]
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@@ -78,31 +78,23 @@
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is an isometric isomorphism.
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is an isometric isomorphism.
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\end{theorem}
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\end{theorem}
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\begin{proof}
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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 $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$.
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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 $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$.
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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 $\dpn{x_i, y_i}{X_i} \ge \alpha \norm{y_i}_{X_i}$. If $q = 1$, then
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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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\begin{align*}
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\alpha \norm{y}_{[l^1(I); X_i^*]} &\le \alpha \sum_{i \in I} \norm{y_i}_{X_i^*} \le
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\sum_{i \in I}\dpn{x_i, y_i}{X_i} \\
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&=
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\dpn{x, \phi_y}{[l^\infty(I); X_i]} \le \norm{\phi}_{[l^\infty(I); X_i]^*}
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\end{align*}
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Otherwise, for each $J \subset I$ finite and $i \in I$, let $F_J(i) = \one_{J} \cdot \norm{y_i}_{X_i^*}^{q - 1}$, then by \autoref{lemma:holder-conjugate-gymnastics},
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\[
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\[
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\norm{F_J x}_{[l^p(I); X_i]}^p = \sum_{j \in J}\norm{y_j}_{X_i^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_i^*}^{q}
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\norm{F_J x}_{[l^p(I); X_i]}^p \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q}
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\]
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\]
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so
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so
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\begin{align*}
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\begin{align*}
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\alpha \sum_{j \in J} \norm{y_i}_{X_i^*}^q & \le
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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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\sum_{i \in I}F_J(i)\dpn{x_i, y_i}{X_i} =
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\dpn{F_J x, \phi_y}{[l^p(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_i^*}^{q}}^{1/p} \\
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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_i}_{X_i^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*}
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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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\end{align*}
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As the above holds for all $\alpha \in (0, 1)$, $y \in [l^p(I); X_i]^*$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi_y}_{[l^p(I); X_i]^*}$.
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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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\end{proof}
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