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Me when I forget to commit.

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Bokuan Li
2026-01-17 23:11:19 -05:00
parent af924f6225
commit 307f23ad57
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10
src/fa/norm/normed.tex
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@@ -11,13 +11,13 @@
\end{enumerate}
then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
\begin{enumerate}
\item $\sum_{n \in \natp}\norm{x_n}_E < C\norm{y}/(1 - \gamma)$.
\item $\sum_{n \in \natp}\norm{x_n}_E \le C\norm{y}_F/(1 - \gamma)$.
\item $\sum_{n = 1}^\infty Tx_n = y$.
\end{enumerate}
In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}/(1 - \gamma)$ and $Tx = y$.
In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
\end{theorem}
\begin{proof}
Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}} \le \gamma \norm{y_n}_F$.
Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$.
For each $n \in \nat$,
\[
@@ -25,11 +25,11 @@
\]
Since $\norm{x_n}_E \le C\norm{y_n}_F$,
\[
\sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}}{1 - \gamma}
\sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma}
\]
In addition,
\[
\norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}} \le \gamma^n \norm{y}_F
\norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F
\]
so $\sum_{n = 1}^\infty Tx_n = y$.
\end{proof}