Symmetry of the derivative (normed/frechet).
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Bokuan Li
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src/fa/tvs/complete-metric.tex
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src/fa/tvs/complete-metric.tex
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\section{Complete Metric TVSs}
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\label{section:tvs-complete-metric}
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\begin{proposition}
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\label{proposition:tvs-complete-metric}
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Let $E$ be a metric TVS with its topology induced by the pseudonorm $\rho: E \to [0, \infty)$, then the following are equivalent:
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\begin{enumerate}
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\item $E$ is complete.
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\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 $E$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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$(2) \Rightarrow (1)$: Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k} \subset \natp$ such that for each $k \in \natp$, $\rho(x_{n_{k+1}} - x_{n_{k}}) < 2^{-k}$.
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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.
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\end{proof}
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\begin{theorem}[Successive Approximations {{\cite[Section III.2]{SchaeferWolff}}}]
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\label{theorem:successive-approximations}
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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:
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\begin{enumerate}
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\item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
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\item[(b)] $\rho(x) \le C \eta(y)$.
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\end{enumerate}
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then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
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\begin{enumerate}
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\item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
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\item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
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\end{enumerate}
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In particular,
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\[
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T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
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\]
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\end{theorem}
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\begin{proof}
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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:
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\begin{enumerate}
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\item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
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\item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
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\end{enumerate}
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By assumption, there exists $x_{N+1} \in E$ such that:
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\begin{enumerate}
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\item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
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\item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
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\end{enumerate}
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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 $N \in \natp$.
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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), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
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\end{proof}
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\begin{proposition}
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\label{proposition:successive-approximation-all}
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Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$ respectively, and $T \in L(E; F)$. If
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\begin{enumerate}
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\item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
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\item[(b)] $E$ is complete.
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\end{enumerate}
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then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
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\end{proposition}
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\begin{proof}
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Let $s > r$ and $\seq{s_n}, \seq{\delta_n} \subset (0, \infty)$ such that
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\begin{enumerate}
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\item[(i)] $s = \sum_{n \in \natp}s_n$.
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\item[(ii)] $s_1 = r$.
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\item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
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\item[(iv)] $\rho_1 = \rho$.
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\end{enumerate}
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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:
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\begin{enumerate}
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\item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
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\item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
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\end{enumerate}
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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 $\eta(Tx_{N+1} - y) \le \rho_{N+2}$.
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By (I), $\seq{x_N}$ is a Cauchy sequence, so
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\[
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x = \limv{N}x_N = \limv{N}\sum_{n = 1}^N(x_n - x_{n-1})
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\]
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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 $Tx = y$.
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\end{proof}
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\begin{proposition}
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\label{proposition:coercive-closed-range}
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Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$, respectively, and $T \in L(E; F)$. If
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\begin{enumerate}
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\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 $\rho(x) \le C\eta(y)$.
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\item[(b)] $E$ is complete.
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\end{enumerate}
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then $T(E)$ is closed.
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\end{proposition}
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\begin{proof}
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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 $\rho(x) \le C\eta(y) \le C\eta(y_0)$.
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By the method of successive approximations (\ref{theorem:successive-approximations}),
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\[
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T(E) \supset T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r) \cap \overline{T(E)}
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\]
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As this holds for all $r > 0$, $T(E) \supset \overline{T(E)}$.
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\end{proof}
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\begin{theorem}[Open Mapping Theorem]
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\label{theorem:open-mapping}
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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:
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\begin{enumerate}
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\item $T(E)$ is meagre.
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\item $T$ is open.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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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,
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\[
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E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
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\]
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By the Baire Category Theorem (\ref{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,
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\[
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B_F(0, s) = B_F(y, s) - y \subset \overline{nT(B_E(0, r)) + nT(B_E(0, r))} \subset \overline{nT(B_E(0, r_0))}
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\]
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By (TVS2), there exists $t > 0$ such that $n^{-1}B_F(0, s) \supset B_F(0, t)$, so $\overline{T(B_E(0, r_0))} \supset B_F(0, t)$.
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Thus by \ref{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 $E$.
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\end{proof}
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Let $U \in \cn_F(0)$, then $T^{-1}(U) \in \cn_E(0)$, so there exists $\lambda \in K$ such that $\lambda T^{-1}(U) = T^{-1}(\lambda U) \supset B$ and $\lambda U \supset T(B)$.
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\end{proof}
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\begin{definition}[Initial Uniformity]
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\label{definition:tvs-initial}
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Let $E$ be a vector space over $K \in \RC$, $\seqi{F}$ be TVSs, and $\seqi{T}$ where $T_i \in \hom(E; F_i)$ for all $i \in I$, then there exists a uniformity $\fU$ on $E$ such that:
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\begin{enumerate}
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\item For each $i \in I$, $T_i \in L(E; F_i)$.
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\item[(U)] If $\mathfrak{V}$ is a uniformity on $E$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
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\end{enumerate}
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Moreover,
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\begin{enumerate}
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\item[(3)] $\fU$ is translation-invariant.
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\item[(4)] $E$ equipped with the topology induced by $\fU$ is a topological vector space.
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\item[(5)] For any TVS $F$ over $K$ and linear map $T \in \hom(F; E)$, $T \in L(F; E)$ if and only if $T_i \circ T \in L(F; F_i)$ for all $i \in I$.
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\end{enumerate}
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The uniformity and its induced topology are the \textbf{initial uniformity/topology} induced by $\seqi{T}$.
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\end{definition}
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\begin{proof}
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(1), (U): By \ref{definition:initial-uniformity}.
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Let $U \in \fU$, then there exists $J \subset I$ finite and translation-invariant entourages $\seqj{U}$ such that
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\[
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U \subset V = \bigcap_{j \in J}(T_j \times T_j)^{-1}(U_j)
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\]
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(3): For each $j \in J$, $(x, y) \in (T_j \times T_j)^{-1}(U_j)$, and $z \in E$,
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\[
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(T_j \times T_j)(x + z, y + z) = (T_jx + T_jz, T_jy + T_jz) \in U_j
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\]
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so $(T_j \times T_j)^{-1}(U_j)$ is translation-invariant, and so is $V$.
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(4): By (TVS1) and (TVS2), for each $j \in J$, there exists an entourage $V_j$ of $F_j$ and $\eps_j > 0$ such that for any $(x, x'), (y, y') \in V_j$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \eps_j$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in U_j$.
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Therefore, for any $(x, x'), (y, y') \in \bigcap_{j \in J} T_j^{-1}(V_j)$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \min_{j \in J}\eps$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in V$.
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(5): By \ref{definition:continuous-linear} and (4) of \ref{definition:initial-uniformity}.
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\end{proof}
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\begin{definition}[Product Topology]
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\label{definition:tvs-product}
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Let $\seqi{E}$ be TVSs over $K \in \RC$ and $E = \prod_{i \in I}E_i$ be their product as a vector space, and $\fU$ be the initial uniformity generated by the projection maps, then
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@@ -8,6 +8,7 @@
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\input{./src/fa/tvs/continuous.tex}
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\input{./src/fa/tvs/quotient.tex}
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\input{./src/fa/tvs/completion.tex}
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\input{./src/fa/tvs/complete-metric.tex}
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\input{./src/fa/tvs/projective.tex}
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\input{./src/fa/tvs/inductive.tex}
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\input{./src/fa/tvs/spaces-of-linear.tex}
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