9.5 Inductive Limits
Definition 9.5.1 (Inductive Locally Convex Topology). Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $\seqi{T}$ such that $T_{i} \in \hom(E_{i}; E)$ for all $i \in I$, and $E$ be a vector space over $K$, then there exists a topology $\topo$ on $E$ such that:
$(E, \topo)$ is a locally convex space over $K$.
For each $i \in I$, $T_{i} \in L(E_{i}; E)$.
For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S}\subset T$.
For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T_{i} \in L(E_{i}; F)$ for all $i \in I$.
The family
\[\mathcal{B}= \bracs{U \subset E|U \text{ convex, radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}\]is a fundamental system of neighbourhoods for $E$ at $0$.
The topology $\topo$ is the inductive locally convex topology on $E$ induced by $\seqi{T}$.
Proof. (1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply Proposition 8.1.12.
Every set in $\mathcal{B}$ is radial and circled by definition.
For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
Let $\topo$ be the vector space topology such that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, then $(E, \topo)$ is a locally convex space.
(2): For each $i \in I$ and $U \in \mathcal{B}$, $T_{i}^{-1}(U) \in \cn_{E_i}(0)$, so $T_{i} \in L(E_{i}; E)$.
(U): Let $U \in \cn_{(E, \mathcal{S})}(0)$ be convex, circled, and radial. By (2), $T_{i}^{-1}(U) \in \cn_{E_i}(0)$ for all $i \in I$. Thus the convex, circled, and radial neighbourhoods of $0$ in $(E, \mathcal{S})$ is a subset of $\mathcal{B}$.
(4): Let $i \in I$ and $U \in \cn_{F}(0)$ be convex, circled, and radial. Since $T \circ E_{i} \in L(E_{i}; F)$, $T_{i}^{-1}(T^{-1}(U)) \in \cn_{E_i}(0)$, so $T^{-1}(U) \in \mathcal{B}\subset \cn_{E}(0)$.$\square$
Definition 9.5.2 (Inductive Limit). Let $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of locally convex spaces over $K \in \RC$, then there exists $(E, \bracsn{T^i_E}_{i \in I})$ such that:
$E$ is a locally convex space over $K$.
For each $i \in I$, $T^{i}_{E} \in L({E_i, E})$.
For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
\[\xymatrix{ E_i \ar@{->}[rd]_{T^i_E} \ar@{->}[r]^{T^i_j} & E_j \ar@{->}[d]^{T^j_E} \\ & E }\]For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes
\[\xymatrix{ E_i \ar@{->}[d]_{T^i_E} \ar@{->}[rd]^{S^i_F} & \\ E \ar@{->}[r]_{S} & F }\]for all $i \in I$.
For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^{i}_{E} \in L(E_{i}; F)$ for all $i \in I$.
The family
\[\mathcal{B}= \bracs{U \subset E|U \text{ convex, radial, circled}, (T^i_E)^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}\]is a fundamental system of neighbourhoods for $E$ at $0$.
The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the inductive limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
Proof. Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (Proposition 1.2.9). Equip $E$ with the inductive topology induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
(U): By (U) of Proposition 1.2.9, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of Definition 9.5.1, $S \in L(E; F)$.
(5): By (5) of Definition 9.5.1.$\square$
Remark 9.5.3. The projective topology behaves well across the constraints of topological vector spaces and locally convex spaces: the preimage of a vector space/locally convex topology is also a vector space/locally convex topology.
On the inductive side, the story is not as simple: In principle, the locally convex inductive topology is smaller than the vector space inductive topology, which is smaller than the inductive topology. As such, the same construction must be performed three separate times, each time restricting to a smaller collection of sets.
In addition to the neighbourhood construction given above, the inductive topology may also be constructed as the weak topology generated by all topologies satisfying certain properties. While this more non-constructive method is simpler, it does not directly provide an explicit fundamental system of neighbourhoods at $0$.
Lemma 9.5.4. Let $E$ be a locally convex space over $K$, $M \subset E$ be a subspace, $U \in \cn_{M}(0)$ be convex and circled, then
There exists $V \in \cn_{E}(0)$ circled and radial such that $U = M \cap V$.
For any $x \in E \setminus \ol M$, there exists $V \in \cn_{E}(0)$ circled and radial such that $U = M \cap V$ and $x \not\in U$.
Proof. (1): Let $W \in \cn_{E}(0)$ be circled and radial with $W \cap M \subset U$, and $V = \text{Conv}(W \cup U)$.
For any $u \in U$, $w \in W$, and $t \in [0, 1]$, there exists $\alpha \in (0, 1)$ such that $x = \alpha^{-1}w \in W$. In which case,
\[(1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V\]so $V \in \cn(0)$.
For any $\lambda \in K$ with $\abs{\lambda}\le 1$, $u \in U$, $w \in W$, and $t \in [0, 1]$,
\[\lambda (1 - t)u + \lambda tw = (1 - t)\lambda u + t \lambda w \in V\]as $U$ and $W$ are both circled.
so $V \in \cn_{E}(0)$ is convex and circled.
For any $u \in U$, $w \in W$, and $t \in [0, 1]$, if $(1 - t)u + tw \in M$, then $u \in M \cap U \subset M \cap W$, so $(1 - t)u + tw \in W$.
(2): Since $x \not\in \ol M$, there exists $W \in \cn_{E}(0)$ circled and radial with $x + W \cap M = \emptyset$. Let $V$ as in (1), then $x \not\in W + M \supset W + U = V$.$\square$
Proposition 9.5.5. Let $(\seq{E_n}, \seq{\iota_n^m|m, n \in \natp, m \le n})$ be a strict inductive system of locally convex spaces over $K \in \RC$ and $(E, \seq{\iota_n})$ be its direct limit, then:
For each $n \in \natp$, the topology of $E_{n}$ is induced by $\iota_{n}$, which allows the identification of $E_{n}$ as a subspace of $E$.
If $E_{n}$ is separated for all $n \in \natp$, then $E$ is also separated.
If $\iota^{m}_{n}(E_{m}) \subset E_{n}$ is closed for all $m, n \in \natp$ with $m \le n$, then the following are equivalent for any $B \subset E$:
$B$ is bounded.
There exists $n \in \natp$ such that $B \subset E_{n}$ is bounded.
If $E_{n}$ is complete for each $n \in \natp$, then $E$ is also complete.
Proof. (1): Let $U \in \cn_{E_n}(0)$. By Lemma 9.5.4, there exists $\bracs{U_m| m \in \natp, m \ge n}\subset 2^{E}$ such that $U_{n} = U$, $U_{m} \in \cn_{E_m}(0)$ and $U_{m}= U_{m + 1}\cap E_{m}$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_{m}$, then $V \cap E_{m} = U_{m}$ for all $m \ge n$. In particular, $V \cap E_{n} = U_{n} = U$.
(2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_{n}$. Since $E_{n}$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By Lemma 9.5.4 and (1), there exists $V \in \cn_{E}(0)$ such that $V \cap E_{n} = U$, so $x \not\in V$.
(3), $\neg (b) \Rightarrow \neg (a)$: If $B \not\subset E_{n}$ for all $n \in \natp$, then there exists a subsequence $\bracsn{n_k}_{0}^{\infty} \subset \natp$ and $\seq{x_k}\subset B$ such that $x_{k} \in E_{n_{k}}\setminus E_{n_{k - 1}}$ for all $k \in \natp$.
Since $E_{n_k}\subset E_{n_{k+1}}$ is closed for all $k \in \natp$, there exists $\seq{U_k}\subset 2^{E}$ such that:
For each $k \in \natp$, $U_{k} \in \cn_{E_{n_k}}(0)$.
For each $k \in \natp$, $U_{k} = U_{k+1}\cap E_{n_k}$.
For each $k \in \natp$, $n^{-1}x_{k} \not\in U_{k}$.
then $V = \bigcup_{k \in \natp}U_{k} \in \cn_{E}(0)$ with $V \cap E_{n_k}= U_{k}$ for all $k \in \natp$. For any $n \in \natp$, $x_{k} \not\in nU_{k} = nV \cap E_{n_k}$. Therefore $B$ is not bounded.
(3), $(b) \Rightarrow (a)$: Let $U \in \cn_{E}(0)$, then $U \cap E_{n} \in \cn_{E_n}(0)$, so there exists $\lambda \in K$ with $\lambda (U \cap E_{n}) \supset B$.
(4): Let $\fF \subset 2^{E}$ be a Cauchy filter and
then $\fU$ is also a Cauchy filter, which converges if and only if $\fF$ does.
Since each $E_{n}$ is complete, it is sufficient to show that there exists $n \in \natp$ such that $F + U \cap E_{n} \ne \emptyset$ for all $F \in \fF$ and $U \in \cn_{E}(0)$.
Suppose for contradiction that for every $n \in \natp$, there exists $F_{n} \in \fF$ and $U_{n} \in \cn_{E}(0)$ such that $E_{n} \cap F_{n} + U_{n} = \emptyset$. Assume without loss of generality that for every $n \in \natp$, $U_{n}$ is convex and circled with $U_{n} \supset U_{n+1}$. Let
then since each $U_{n}$ is circled, so is $U$. Thus $U \cap E_{n} \supset U_{n} \cap E_{n} \in \cn_{E_n}(0)$, and $U \in \cn_{E}(0)$.
Now, suppose that $(F_{n} + U) \cap E_{n} \ne \emptyset$. Let $y \in (F_{n} + U) \cap E_{n}$, then there exists $N \in \natp$, $\bracs{x_k}_{1}^{N} \subset E$, $\bracs{\lambda_k}_{1}^{N} \subset [0, 1]$, and $z \in F_{n}$ such that
For each $1 \le k \le N$, $x_{k} \in U_{k} \cap E_{k}$.
$\sum_{k = 1}^{N} \lambda_{k} = 1$.
$y = z + \sum_{k = 1}^{N} \lambda_{k} x_{k}$.
In which case, since $U_{k}\supset U_{k+1}$ for all $k \in \natp$,
which is impossible. Therefore $(F_{n} + U) \cap E_{n} = \emptyset$ for all $n \in \natp$.
Finally, since $\fF$ is a Cauchy filter, there exists $F \in \fF$ such that $F - F \subset U$. Let $z \in F$, then there exists $n \in \natp$ such that $z \in E_{n}$. In which case, for any $y \in F \cap F_{n}$,
which contradicts the fact that $(F_{n} + U) \cap E_{n} = \emptyset$.$\square$