Theorem 10.1.3 (Successive Approximation). Let $E, F$ be normed spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
$\norm{x}_{E} \le C\norm{y}_{F}$.
$\norm{y - Tx}_{F} \le \gamma \norm{y}_{F}$.
then for any $y \in F$, there exists $\seq{x_n}\subset E$ such that:
$\sum_{n \in \natp}\norm{x_n}_{E} \le C\norm{y}_{F}/(1 - \gamma)$.
$\sum_{n = 1}^{\infty} Tx_{n} = y$.
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 $Tx = y$.
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 $\norm{y_{n+1}}_{F} \le \gamma \norm{y_n}_{F}$.
For each $n \in \nat$,
Since $\norm{x_n}_{E} \le C\norm{y_n}_{F}$,
In addition,
so $\sum_{n = 1}^{\infty} Tx_{n} = y$.$\square$