Definition 11.8.1 (Inductive Locally Convex Topology).label Let $\seqi{E}$ be locally convex spaces over $K \in \RC$, $E$ be a vector space over $K$, and $\seqi{T}$ such that $T_{i} \in \hom(E_{i}; E)$ for all $i \in I$, then there exists a topology $\topo$ on $E$ such that:
- (1)
$(E, \topo)$ is a locally convex space over $K$.
- (2)
For each $i \in I$, $T_{i} \in L(E_{i}; E)$.
- (U)
For any topology $\mathcal{S}$ on $E$ satisfying (1) and (2), $\mathcal{S}\subset T$.
- (4)
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 $i \in I$.
- (5)
The family
\[\mathcal{B}= \bracs{U \subset E|U \text{ convex, circled, radial}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}\]is a fundamental system of neighbourhoods for $E$ at $0$.
- (6)
If $E$ is spanned by $\bigcup_{i \in I}T_{i}(E_{i})$, then
\[\fB = \bracs{\Gamma\paren{\bigcup_{i \in I}T_i(U_i)} \bigg | U_i \in \cn_{E_i}(0)}\]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 10.1.12.
- (TVB1)
Every set in $\mathcal{B}$ is radial and circled by definition.
- (TVB2)
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 $\mathcal{B}$.
(5): 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 $T^{-1}(U) \in \mathcal{B}\subset \cn_{E}(0)$.
(6): If $E$ is spanned by $\bigcup_{i \in I}T_{i}(E_{i})$, then each set in $\fB$ is radial. Hence $\fB$ is a family of neighbourhoods of $E$ at $0$.
Let $U \in \cn_{E}(0)$ be convex, circled, and radial, then for each $i \in I$, $T_{i}^{-1}(U) \in \cn_{E_i}(0)$, so $U \supset \bigcup_{i \in I}T_{i}[T_{i}^{-1}(U)]$. Since $U$ is convex and circled, $U \supset \Gamma\paren{\bigcup_{i \in I}T_i[T_i^{-1}(U)]}\in \fB$. Therefore $\fB$ forms a fundamental system of neighbourhoods for $E$ at $0$.$\square$
Post a Comment