Topological Vector Spaces (TVS)

# Topological Vector Spaces (TVS)

 Definition: A normed linear space $X$ over $\mathbb{R}$ (or $\mathbb{C}$) is said to be a Topological Vector Space (abbreviated TVS) if $X$ is also equipped with a Hausdorff topology $\tau$ and such that: 1) The operation of addition $+ : X \times X \to X$ defined by $(x, y) \to x + y$ is continuous on $X \times X$. 2) The operator of scalar multiplication $\cdot : \mathbb{R} \times X \to X$ defined by $(\lambda, x) \to \lambda x$ is continuous on $\mathbb{R} \times X$.

For example, let $X$ be a normed linear space and let $\tau$ be the norm topology. Note that $X$ with the norm topology is a metric space and every metric space is Hausdorff. Thus, $\tau$ is Hausdorff. Furthermore, $+$ and $\cdot$ are continuous on $X$ and so $(X, \tau)$ is a topological vector space.

We will now look at some basic properties of topological vector spaces.

 Proposition 1: Let $X$ be a topological vector space. Then a subset $U$ of $X$ is open if and only if $U + x_0$ is open for every $x_0 \in X$.
• Proof: Let $x_0 \in X$ and let $T_{x_0} : X \to X$ be the map defined for all $x \in X$ by:
(1)
\begin{align} \quad T_{x_0}(x) = x + x_0 \end{align}
• Then $T_{x_0}$ is clearly bijective and is continuous since vector addition is continuous. So $T_{-x_0}$ is also continuous and is such that:
(2)
\begin{align} \quad T_{x_0} \circ T_{-x_0} = \mathrm{id} = T_{-x_0} \circ T_{x_0} \end{align}
• Therefore $T_{x_0}$ is a homomorphism. Let $U \subseteq X$. Then $U$ is open if and only if $T_{x_0}(U) = U + x_0$ is open. $\blacksquare$
 Theorem 2: Let $X$ be a topological vector space. Then a subset $U$ of $X$ is open if and only if $\lambda U$ is open for every $\lambda \in \mathbb{C}$, $\lambda \neq 0$.
• Proof: Let $\lambda \in \mathbb{C}$, $\lambda \neq 0$ and let $T_{\lambda} : X \to X$ be the map defined for all $x \in X$ by:
(3)
\begin{align} \quad T_{\lambda}(x) = \lambda x \end{align}
• Then $T_{\lambda}$ is clearly bijective and is continuous since scalar multiplication is continuous. So $T_{1/\lambda}$ is continuous and is such that:
(4)
\begin{align} \quad T_{\lambda} \circ T_{1/\lambda} = \mathrm{id} = T_{1/\lambda} \circ T_{\lambda} \end{align}
• Hence $T_{\lambda}$ is a homeomorphism. Let $U \subseteq X$. Then $U$ is open if and only if $T_{\lambda}(U) = \lambda U$ is open. $\blacksquare$

The propositions above give a nice characterization of open neighbourhoods of points. Note that if $U$ is an open neighbourhood of $0$ then $U + x$ is an open neighbourhood of $x$. If we have a base at $0$ then we can obtain a base at every $x \in X$. Thus, a base at $0$ completely determines the topology on a topological vector space.