Proposition 8.8.3. Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$ respectively, and $T \in L(E; F)$. If
For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))}\supset B_{F}(0, \delta(r))$.
$E$ is complete.
then for every $s > r$, $T(B_{E}(0, s)) \supset B_{F}(0, \delta(r))$.
Proof. Let $s > r$ and $\seq{s_n}, \seq{\delta_n}\subset (0, \infty)$ such that
$s = \sum_{n \in \natp}s_{n}$.
$s_{1} = r$.
For all $n \in \natp$, $\overline{T(B_E(0, s_n))}\supset B_{F}(0, \delta_{n})$.
$\rho_{1} = \rho$.
Let $y_{0} \in B(0, r)$ and $x_{0} = 0$. Let $N \in \natp$ and suppose inductively that $\bracs{x_n}_{1}^{N} \subset E$ has been constructed such that:
For each $0 \le n \le N - 1$, $\rho(x_{n+1}- x_{n}) < s_{n}$.
For each $0 \le n \le N$, $\eta(Tx_{n} - y) \le \rho_{n+1}$.
By density of $T(x_{N} + B_{E}(0, s_{N}))$ in $Tx_{N} + B_{F}(0, \rho_{N})$, there exists $x_{N+1}\in T(x_{N} + B_{E}(0, s_{N}))$ such that $\eta(Tx_{N+1}- y) \le \rho_{N+2}$.
By (I), $\seq{x_N}$ is a Cauchy sequence, so
exists in $E$. In addition, $\rho(x) \le \sum_{n \in \natp}\rho(x_{n} - x_{n-1}) < \sum_{n \in \natp}s_{n} = s$, so $x \in B_{E}(0, s)$. Finally, $\eta\paren{Tx - y}= \limv{N}\rho(Tx_{N} - y) = 0$ and $Tx = y$.$\square$