From 845e616453c4fb75cdec990adefea6ecd5aa121f Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 16 Jun 2026 12:35:36 -0400 Subject: [PATCH] Fixed typos in the sequence space duality result. --- src/fa/lp/seq.tex | 26 +++++++++----------------- 1 file changed, 9 insertions(+), 17 deletions(-) diff --git a/src/fa/lp/seq.tex b/src/fa/lp/seq.tex index a16fe29..f238eaf 100644 --- a/src/fa/lp/seq.tex +++ b/src/fa/lp/seq.tex @@ -65,7 +65,7 @@ \begin{theorem} \label{theorem:lp-sum-dual} - 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 + 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 \[ \phi_y: [l^p(I); X_i] \to K \quad x \mapsto \sum_{i \in I}\dpn{x_i, y_i}{X_i} \] @@ -78,31 +78,23 @@ is an isometric isomorphism. \end{theorem} \begin{proof} - 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 $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. + 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 $x \in X_i$, $\dpn{x_i \cdot \one_{\bracs{i}}, \phi}{[l^p(I); X_i]} = \dpn{x_i, y_i}{X_i}$. - 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 - \begin{align*} - \alpha \norm{y}_{[l^1(I); X_i^*]} &\le \alpha \sum_{i \in I} \norm{y_i}_{X_i^*} \le - \sum_{i \in I}\dpn{x_i, y_i}{X_i} \\ - &= - \dpn{x, \phi_y}{[l^\infty(I); X_i]} \le \norm{\phi}_{[l^\infty(I); X_i]^*} - \end{align*} - - 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}, + 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}, \[ - \norm{F_J x}_{[l^p(I); X_i]}^p = \sum_{j \in J}\norm{y_j}_{X_i^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_i^*}^{q} + \norm{F_J x}_{[l^p(I); X_i]}^p \le \sum_{j \in J}\norm{y_j}_{X_j^*}^{p(q - 1)}= \sum_{j \in J}\norm{y_j}_{X_j^*}^{q} \] so \begin{align*} - \alpha \sum_{j \in J} \norm{y_i}_{X_i^*}^q & \le + \alpha \sum_{j \in J} \norm{y_j}_{X_j^*}^q & \le \sum_{i \in I}F_J(i)\dpn{x_i, y_i}{X_i} = - \dpn{F_J x, \phi_y}{[l^p(I); X_i]} \\ - &\le \norm{\phi}_{[l^p(I); X_i]^*} \cdot \braks{\sum_{j \in J}\norm{y_j}_{X_i^*}^{q}}^{1/p} \\ - \alpha \braks{\sum_{j \in J}\norm{y_i}_{X_i^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*} + \dpn{F_J x, \phi}{[l^p(I); X_i]} \\ + &\le \norm{\phi}_{[l^p(I); X_i]^*} \cdot \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^{q}}^{1/p} \\ + \alpha \braks{\sum_{j \in J}\norm{y_j}_{X_j^*}^q}^{1/q} &\le \norm{\phi}_{[l^p(I); X_i]^*} \end{align*} - As the above holds for all $\alpha \in (0, 1)$, $y \in [l^p(I); X_i]^*$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi_y}_{[l^p(I); X_i]^*}$. + As the above holds for all $\alpha \in (0, 1)$ and $J \subset I$ finite, $y \in [l^q(I); X_i^*]$ with $\norm{y}_{[l^q(I); X_i^*]} = \norm{\phi}_{[l^p(I); X_i]^*}$. \end{proof}