> [!definition] > > Let $E$ be a [[Locally Convex Topological Vector Space|locally convex topological vector space]], then $E$ is a **LF-space** if there exists a [[Countable Direct Limit of Locally Convex Topological Vector Spaces|countable inductive system]] of [[Fréchet Space|Fréchet spaces]] $(\seq{E_n}, \bracsn{T^m_n})$ such that $E$ is its [[Direct Limit|direct limit]]. If $(\seq{E_n}, \bracsn{T^m_n})$ is strict, then $E$ is a **strict LF-space**, and denote $T^m_n = \iota^m_n$. > [!theorem] > > Let $E$ be a strict LF-space, $(\seq{E_n}, \bracsn{\iota^m_n})$ be a strict inductive system of Fréchet spaces with $E$ as its limit, and $F$ be a locally convex topological vector space, then > 1. $E$ is [[Hausdorff Space|Hausdorff]]. > 2. Let $\seq{x_n} \subset E$ and $x \in E$, then $x_n \to x$ if and only if there exists $N \in \nat$ such that $x_n \in E_N$ for all $n \in \nat$, $x \in E_N$, and $x_n \to x$ in $E_N$. > 3. Let $T: E \to F$ be a [[Linear Transformation|linear map]], then $T$ is continuous if and only if $T$ is sequentially continuous. > > *Proof.* $(2)$: Since each $E_n$ is complete, $E_n$ is a closed subset of $E_{n + 1}$ for all $n \in \nat$. > > $(3)$: $T$ is continuous if and only if its restriction to each $E_n$ is continuous. Since each $E_n$ is first countable, $T|_{E_n}$ is continuous if and only if it is sequentially continuous on $E_n$. As convergent sequences in $E$ corresponds to convergent sequences in each $E_n$, $T$ is continuous if and only if it is sequentially continuous on $E$. > [!theorem] > > Suppose that $E$ is strict and let $B \subset E$, then $B$ is [[Bounded Set|bounded set]] if and only if there exists $n \in \nat$ such that $B \subset E_n$ and $B$ is bounded in $E_n$. > > *Proof*. It's sufficient to show that there exists $n \in \nat$ such that $B \subset E_n$. Assume without loss of generality that there exists $\seq{x_n}$ such that $x_1 \in E_1$, $x_{n+1} \in E_{n+1} \setminus E_{n}$ for all $n \in \nat$. > > Since each $E_{n}$ is embedded as a closed subspace of $E$, for each $n \in \nat$, there exists a seminorm $[\cdot]_{n}: E \to \real^+$ such that $[x_{n}]_n > 0$ but $[x]_{n} = 0$ for all $x \in E_{n - 1}$. From here, let > $ > [x] = \sum_{n = 1}^\infty n\frac{[x]_n}{[x_n]_n} > $ > then $[\cdot]$ is also a seminorm, with its restriction to each $E_n$ also continuous. Therefore $[\cdot]$ is a continuous seminorm. > > However, for any $k \in \nat$, > $ > [x_k] = \sum_{n = 1}^\infty n\frac{[x]_n}{[x_n]_n} \ge k \frac{[x_k]_k}{[x_k]_k} = k > $ > which means that $[\cdot]$ is not bounded on $B$. Therefore if $B$ is bounded, there must exist $n \in \nat$ such that $B \subset E_n$. > [!theorem] > > Suppose that $E$ is strict, then $E$ is [[Bornologic Space|bornologic]]. > > *Proof*. Let $[\cdot]: E \to \real^+$ be a [[Seminorm|seminorm]] that is bounded on each bounded set. Since each $E_n$ is a [[Fréchet Space|Fréchet space]], it is bornologic. Thus $[\cdot]$ restricted to $E_n$ is continuous for all $n \in \nat$, so $[\cdot]$ is continuous as well.