8.8 Complete Metric TVSs
Proposition 8.8.1. Let $E$ be a metric TVS with its topology induced by the pseudonorm $\rho: E \to [0, \infty)$, then the following are equivalent:
$E$ is complete.
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 $E$.
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.$\square$
Theorem 8.8.2 (Successive Approximations [Section III.2, SW99]). 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:
$\eta(y - Tx) \le \gamma \eta(y)$.
$\rho(x) \le C \eta(y)$.
then for any $y \in F$, there exists $\seq{x_n}\subset E$ such that:
$\sum_{n \in \natp}\rho(x_{n}) \le C\eta(y)/(1 - \gamma)$.
$y = \limv{N}\sum_{n = 1}^{N} Tx_{n}$.
In particular,
Proof. 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:
$\sum_{n = 1}^{N}\rho(x_{n}) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
$\eta\paren{y - \sum_{n = 1}^N Tx_n}\le \eta(y)\gamma^{N}$.
By assumption, there exists $x_{N+1}\in E$ such that:
$\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n}\le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n}\le \gamma^{N+1}$.
$\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n}\le C\eta(y)\gamma^{N}$.
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), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n}= \limv{N}\eta(y)\gamma^{N} = 0$.$\square$
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 $Tx = y$.$\square$
Proposition 8.8.4. 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 $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
$E$ is complete.
then $T(E)$ is closed.
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 method of successive approximations,
As this holds for all $r > 0$, $T(E) \supset \overline{T(E)}$.$\square$
Theorem 8.8.5 (Open Mapping Theorem). 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:
$T(E)$ is meagre.
$T$ is open.
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 absorbing,
By the Baire Category Theorem, 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,
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 Proposition 8.8.3, $B_{F}(0, t) \subset T(B_{E}(0, r)) \in \cn_{F}(0)$ 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 $E$.$\square$