Basic Properties of Locally Convex Topological Vector Spaces

Basic Properties of Locally Convex Topological Vector Spaces

Theorem 1: Let $X$ be a locally convex 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 locally convex 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. $\blacksquare$
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