Topological Vector Spaces over the Field of Real or Complex Numbers

# Topological Vector Spaces over the Field of Real or Complex Numbers

 Definition: A Topological Vector Space is a vector space $E$ with a topology $\tau$ such that the operations of vector addition $+ : E \times E \to E$ and scalar multiplication $\cdot : E \times \mathbf{F} \to E$ are continuous with respect to the topology $\tau$.

When we require $+ : E \times E \to E$ and $\cdot : E \times \mathbf{F} \to E$ to be continuous, we mean that $+$ is continuous on $E \times E$ where $E \times E$ is equipped with the product topology, and similarly, we mean that $\cdot$ is continuous on $E \times \mathbf{F}$ where $E \times \mathbf{F}$ is equipped with the product topology (where $\mathbf{F}$ is equipped with the standard Euclidean topology on $\mathbb{R}$ or $\mathbb{C}$).

If $E$ is a vector space, then a topology $\tau$ is said to be Compatible with $E$ if the operations $+$ and $\cdot$ are continuous with respect to the topology $\tau$. Thus, a topological vector space is a vector space $E$ with a compatible topology $\tau$.

 Proposition 1: Let $E$ be a topological vector space. Then: (1) For all $a \in E$, the function $f_a : E \to E$ defined for all $x \in E$ by $f_a(x) := x + a$ is a homeomorphism of $E$ onto $E$. (2) For all $\alpha \in \mathbf{F}$, $\alpha \neq 0$, the function $f_{\alpha} : E \to E$ defined for all $x \in E$ by $f_{\alpha}(x) := \alpha x$ is a homeomorphism of $E$ onto $E$.
• Proof of (1): Let $a \in E$. Then $f_a : E \to E$ is bijective, with inverse $f^{-1}_a = f_{-a}$ since for all $x \in E$:
(1)
\begin{align} \quad f_a(f_{-a}(x)) = f_a(x - a) = (x - a) + a = x \quad \mathrm{and} \quad f_{-a}(f_a(x)) = f_{-a}(x + a) = (x + a) - a = x \end{align}
• We now show that $f_a$ is continuous. Let $x \in E$. Then $f(x) = x + a$. Let $V$ be a neighbourhood of $f(x)$. Then $V - a$ is a neighbourhood of $x$ such that $f(V - a) = V$. So $f$ is continuous at $x$ and thus continuous on $E$. $\blacksquare$
• Proof of (2): Let $\lambda \in \mathbf{F}$ with $\lambda \neq 0$. Then $f_{\lambda} : E \to E$ is bijective, with inverse $f^{-1}_{\lambda} = f_{1/\lambda}$ since for all $x \in E$:
(2)
\begin{align} \quad f_{\lambda} (f_{1/\lambda}(x)) = f_{\lambda} \left ( \frac{1}{\lambda} x \right ) = \lambda \cdot \frac{1}{\lambda} x = x \quad \mathrm{and} \quad f_{1/\lambda}(f_{\lambda}(x)) = f_{1/\lambda}(\lambda x) = \frac{1}{\lambda} \cdot \lambda x = x \end{align}
• We now show that $f_{\lambda}$ is continuous. Let $x \in E$. Then $f(x) = \lambda x$. Let $V$ be a neighbourhood of $f(x)$. Then $\displaystyle{\frac{1}{\lambda} V}$ is a neighbourhood of $x$ such that $f \left ( \frac{1}{\lambda} V \right ) = V$. So $f$ is continuous at $x$ and thus continuous on $E$. $\blacksquare$