Let $S :=F^E$ be the set of all maps from a set $E$ to a t.v.s. $F$. Then $S$ is also a t.v.s.

Question

I'm trying to prove below result about topological vector space.

Let $E$ be a set and $F$ a t.v.s. Let $S :=F^E$ be the set of all maps from $E$ to $F$. Let $F_x := F$ for all $x\in E$. Then we can write $$S = \prod_{x\in E} F_x.$$ In this way, we endow $S$ with the product topology. On the other hand, $S$ is also a vector space with pointwise addition and pointwise scalar multiplication. Then $S$ is a t.v.s.

Could you have a check on my attempt?

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. If other people post answers, I'm happy to accept theirs.

Answer 1

Consider $T: S \times S \to S, (f, g) \mapsto f+g$ and $L: \mathbb R \times S \to S, (\lambda, f) \mapsto \lambda f$. Our goal is to show that $T,L$ are continuous. For $x\in E$, let $\pi_x: S \to F_x, f \mapsto f(x)$ be the canonical projection. It suffices to show that $\pi_x \circ T$ and $\pi_x \circ L$ are continuous for all $x\in E$. The claim then follows from

• the diagrams $$ {S \times S \atop (f, g)} \, { \longrightarrow \atop \longmapsto } \, {F_x \times F_x \atop (f(x),f(x))} \, { \longrightarrow \atop \longmapsto } \, { F_x \atop f(x)+f(x)} \quad \text{and} \quad {\mathbb R \times S \atop (\lambda, f)} \, { \longrightarrow \atop \longmapsto } \, {\mathbb R \times F_x \atop (\lambda, f(x))} \, { \longrightarrow \atop \longmapsto } \, { \mathbb R \atop \lambda f(x)}. $$

• product topology is the same as topology of pointwise convergence.

• $F_x$ is a topological vector space and $\pi_x$ is continuous.