> [!definition] > > Let $E$ be a [[Vector Space|vector space]] over $K \in \bracs{\real, \complex}$, $\seqi{F}$ be a family of [[Topological Vector Space|topological vector spaces]] over $K$, and $\seqi{T}$ be a family of [[Linear Transformation|linear maps]] with $T_i: E \to F_i$, then there exists a unique [[Topological Space|topology]] $\mathcal T$ on $E$ such that > 1. $(E, \mathcal T)$ is a [[Topological Vector Space|topological vector space]]. > 2. $T_i: E \to F_i$ is [[Continuity|continuous]] for all $i \in I$. > 3. Any other topology on $E$ satisfying $(1)$ and $(2)$ contains $\mathcal T$. > > In other words, $\mathcal T$ is the smallest topology on $E$ making $\seqi{T}$ continuous, known as the **initial topology** on $E$. > > *Remark: The key revelation here is that the topology pulled back into $E$ by each individual map makes $E$ a topological vector space. Continuous linear maps do not simply preserve the topological and linear structures, but also their compatibility.* > > *Proof*. Let $\mathcal T$ be the [[Weak Topology|weak topology]] generated by $\seqi{T}$, then it's sufficient to show $(1)$. > > Let $x, y \in E$ and $U \in \cn^o(x + y)$ be an [[Open Set|open]] [[Neighbourhood|neighbourhood]] of $x + y$. Since $\mathcal T$ is generated by $\seqi{T}$, there exists $\seqf{i_k} \subset I$ and $\seqf{U_k}$ such that $U_k \in \cn^o(T_{i_k}(x + y))$ for each $1 \le k \le n$, and > $ > x + y \in \bigcap_{k = 1}^n T_{i_k}^{-1}(U_{k}) \subset U > $ > Since each $F_i$ is a topological vector space, for each $1 \le k \le n$, there exists $V_{k} \in \cn^o(T_{i_k}x)$ and $W_k \in \cn^o(T_{i_k}y)$ such that $V_k + W_k \subset U_k$. From here, let > $ > V = \bigcap_{k = 1}^n T_{i_k}^{-1}(V_k) \quad W = \bigcap_{k = 1}^n T^{-1}_{i_k}(W_k) > $ > then $V \in \cn^o(x)$ and $W \in \cn^o(y)$ where > $ > V + W \subset \bigcap_{k = 1}^nT_{i_k}^{-1}(V_k) + \bigcap_{k = 1}^n T^{-1}_{i_k}(W_k) \subset \bigcap_{k = 1}^n T^{-1}_{i_k}(U_k) > $ > so addition is continuous in $E$. > > Let $x \in E$, $\lambda \in K$, and $U \in \cn^o(\lambda x)$ be an open neighbourhood of $\lambda x$. Similarly, there exists $\seqf{i_k} \subset I$ and $\seqf{U_k}$ such that $U_k \in \cn^o(T_{i_k}(\lambda x))$ for each $1 \le k \le n$, and > $ > \lambda x \in \bigcap_{k = 1}^n T^{-1}_{i_k}(U_k) \subset U > $ > As each $F_i$ is a topological vector space, for each $1 \le k \le n$, there exists $V_k \in \cn^o(T_{i_k}x)$ and $W_k \in \cn^o(\lambda)$ such that $W_kV_k \subset U_k$. Let > $ > V = \bigcap_{k = 1}^n T^{-1}_{i_k}(V_k) \quad W = \bigcap_{k = 1}^n W_k > $ > then $V \in \cn^o(x)$ and $W \in \cn^o(\lambda)$ with $VW \subset \bigcap_{k = 1}^n T_{i_k}^{-1}(U_k)$, and scalar multiplication is continuous in $E$. > > [!definition] > > Let $E$ be a vector space over $K \in \bracs{\real, \complex}$, $\seqi{F}$ be a family of [[Locally Convex Topological Vector Space|locally convex topological vector spaces]], and $\seqi{T}$ be a family of linear maps with $T_i: E \to F_i$, then there exists a unique topology $\mathcal T$ on $E$ such that > 1. $(E, \mathcal T)$ is a locally convex topological vector space. > 2. $T_i: E \to F$ is continuous for each $i \in I$. > 3. Any other topology on $E$ satisfying $(1)$ and $(2)$ satisfies $\mathcal T$. > > In other words, $\mathcal T$ is the smallest locally convex topology on $E$ such that each $T_i$ is continuous, known as the **initial topology** on $E$. > > *Proof*. Let $\mathcal T$ be the topology generated by $\seqi{T}$. It's sufficient to show that $(E, \mathcal T)$ is locally convex. > > Let $U \in \cn^o(0)$, then there exists $\seqf{i_k} \subset I$ and $\seqf{U_k}$ such that $U_k \in \cn^o(0)$ with respect to $F_{i_k}$ and > $ > 0 \in \bigcap_{k = 1}^nT^{-1}_{i_k}(U_k) \subset U > $ > Since each $F_i$ is locally convex, there exists $\seqf{V_k}$ such that $V_k \in \cn^o(0)$ with respect to $F_{i_k}$, $V_k \subset U_k$, and each $V_k$ is [[Convexity|convex]] , [[Balanced Set|balanced]], and [[Absorbing|absorbing]]. This yields > $ > 0 \in \bigcap_{k = 1}^n T^{-1}_{i_k}(V_k) \subset \bigcap_{k = 1}^n T^{-1}_{i_k}(U_k) \subset U > $ > As preimages of convex, balanced, and absorbing sets by linear maps also have the same properties, $\mathcal T$ is a locally convex topology on $E$.