\section{Complete Metric TVSs}
\label{section:tvs-complete-metric}

\begin{proposition}
\label{proposition:tvs-complete-metric}
    Let $E$ be a metric TVS with its topology induced by the pseudonorm $\rho: E \to [0, \infty)$, then the following are equivalent:
    \begin{enumerate}
        \item $E$ is complete.
        \item For any $\seq{x_n} \subset X$ with $\sum_{n \in \natp}\rho(x_n) < \infty$, $\limv{N}\sum_{n = 1}^N x_n$ exists in $E$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    $(2) \Rightarrow (1)$: Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k} \subset \natp$ such that for each $k \in \natp$, $\rho(x_{n_{k+1}} - x_{n_{k}}) < 2^{-k}$.

    Let $x = x_{n_1} + \limv{N}\sum_{k = 1}^N (x_{n_{k+1}} - x_{n_k})$, then $x = \limv{k}x_{n_k} \in E$. Since $\seq{x_n}$ is a Cauchy sequence that admits a convergent subsequence, it is convergent.
\end{proof}

\begin{theorem}[Successive Approximations]
\label{theorem:successive-approximations}
    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorm $\rho$ and $\eta$, respectively. Let $T \in L(E; F)$, $r > 0$, $\gamma \in (0, 1)$, and $C \ge 0$. Suppose that for every $y \in B_F(0, r)$, there exists $x \in E$ such that:
    \begin{enumerate}
        \item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
        \item[(b)] $\rho(x) \le C \eta(y)$.
    \end{enumerate}

    then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
    \begin{enumerate}
        \item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
        \item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
    \end{enumerate}

    In particular,
    \[
    T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
    \]

\end{theorem}
\begin{proof}[Proof {{\cite[Section III.2]{SchaeferWolff}}}. ]
    Let $y_0 = y$ and $x_0 = 0$. Let $N \in \natz$ and suppose inductively that $\seqf[N]{x_n} \subset E$ has been constructed such that:
    \begin{enumerate}
        \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
        \item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
    \end{enumerate}

    By assumption, there exists $x_{N+1} \in E$ such that:
    \begin{enumerate}
        \item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
        \item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
    \end{enumerate}

    Combining (I) and (ii) shows that $\sum_{n = 1}^N \rho(x_n) \le C \eta(y) \sum_{n = 0}^N \gamma^n$. Therefore there exists $\seq{x_n} \subset E$ such that (I) and (II) holds for all $N \in \natp$.

    By (I), $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)\sum_{n \in \natz}\gamma^n = C \eta(y)/(1 - \gamma)$. By (II), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
\end{proof}

\begin{proposition}
\label{proposition:successive-approximation-all}
    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$ respectively, and $T \in L(E; F)$. If
    \begin{enumerate}
        \item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
        \item[(b)] $E$ is complete.
    \end{enumerate}

    then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
\end{proposition}
\begin{proof}
    Let $s > r$ and $\seq{s_n}, \seq{\delta_n} \subset (0, \infty)$ such that
    \begin{enumerate}
        \item[(i)] $s = \sum_{n \in \natp}s_n$.
        \item[(ii)] $s_1 = r$.
        \item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
        \item[(iv)] $\rho_1 = \rho$.
    \end{enumerate}

    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:
    \begin{enumerate}
        \item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
        \item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
    \end{enumerate}

    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
    \[
    x = \limv{N}x_N = \limv{N}\sum_{n = 1}^N(x_n - x_{n-1})
    \]

    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 $Tx = y$.
\end{proof}


\begin{proposition}
\label{proposition:coercive-closed-range}
    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$, respectively, and $T \in L(E; F)$. If
    \begin{enumerate}
        \item[(a)] For any $r > 0$, there exists $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
        \item[(b)] $E$ is complete.
    \end{enumerate}

    then $T(E)$ is closed.
\end{proposition}
\begin{proof}
    Let $r > 0$ and $\gamma \in (0, 1)$. For any $y_0 \in B_F(0, r) \cap \overline{T(E)}$, there exists $y \in B_F(0, r)$ such that $\eta(y) \le \eta(y_0)$ and $\eta(y - y_0) \le \gamma \eta(y_0)$. By assumption (a), there exists $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y) \le C\eta(y_0)$.

    By the \hyperref[method of successive approximations]{theorem:successive-approximations},
    \[
    T(E) \supset T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r) \cap \overline{T(E)}
    \]

    As this holds for all $r > 0$, $T(E) \supset \overline{T(E)}$.
\end{proof}

\begin{theorem}[Open Mapping Theorem]
\label{theorem:open-mapping}
    Let $E, F$ be complete metric TVSs over $K \in \RC$, $T \in L(E; F)$ with $T(E)$ dense, then exactly one of the following holds:
    \begin{enumerate}
        \item $T(E)$ is meagre.
        \item $T$ is open.
    \end{enumerate}
\end{theorem}
\begin{proof}
    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is radial,
    \[
    E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
    \]

    By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $N \in \natp$, $s > 0$, and $y \in nT(B_E(0, r))$ such that $B_F(y, s) \subset \overline{nT(B_E(0, r))}$. In which case,
    \[
    B_F(0, s) = B_F(y, s) - y \subset \overline{nT(B_E(0, r)) + nT(B_E(0, r))} \subset \overline{nT(B_E(0, r_0))}
    \]

    By (TVS2), there exists $t > 0$ such that $n^{-1}B_F(0, s) \supset B_F(0, t)$, so $\overline{T(B_E(0, r_0))} \supset B_F(0, t)$.

    Thus by \autoref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
\end{proof}