LCTVS over the Field of Real or Complex Numbers

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

 Definition: A Locally Convex Topological Vector Space (LCTVS) is a topological vector space $E$ that has a base of convex neighbourhoods of the origin.

Note that we have already proven that a topological vector space has a base of balanced neighbourhoods of the origin.

 Proposition 1: Let $E$ be a locally convex topological vector space. Then there exists a base of neighbourhoods of the origin, $\mathcal U$, with the following properties: (1) For each $U, V \in \mathcal U$ there exists a $W \in \mathcal U$ with $W \subseteq U \cap V$. (2) For each $U \in \mathcal U$ and $\alpha \in \mathbf{F}$ with $\alpha \neq 0$, $\alpha U \in \mathcal U$. (3) Each $U \in \mathcal U$ is absolutely convex and absorbent. Conversely, if $E$ is a vector space and if there exists a nonempty collection $\mathcal U$ of subsets of $E$ that satisfy properties (1)-(3) above for all $U, V \in \mathcal U$, then there exists a topology on $E$ which makes $E$ a locally convex topological vector space with $\mathcal U$ as a base of neighbourhoods of the origin.
• Proof: We will begin this proof by constructing a base of neighbourhoods of the origin, and then show that this base of neighbourhoods satisfies properties (1)-(3).
• Let $E$ be a locally convex topological vector space. Then $E$ has a base of convex neighbourhoods of the origin. Let $U$ be a convex neighbourhood of the origin that is contained in this base. Since $E$ is a topological vector space, by a proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space, the set:
(1)
\begin{align} \quad W := \bigcap_{|\mu| \geq 1} \mu U \end{align}
• is a balanced neighbourhood of the origin such that $W \subseteq U$. But, since $U$ is convex, $\mu U$ is also convex for each $\mu \in \mathbf{F}$ with $|\mu| \geq 1$. Hence $W$ is an intersection of convex sets and is thus convex. So $W$ is a convex and balanced neighbourhood of the origin with $W \subseteq U$.
• So for each neighbourhood $U'$ of the origin, there exists a convex neighbourhood $U$ of the origin with $U \subseteq U'$ (since $U$ is chosen from a base of convex neighbourhoods of the origin), and there exists a convex and balanced neighbourhood $W$ of the origin with $W \subseteq U \subseteq U'$. Thus $E$ has a convex and balanced base of neighbourhoods of the origin.
• Let $\mathcal V$ be this convex and balanced base of neighbourhoods of the origin.
• Observe that for each $\alpha \in \mathbf{F}$ with $\alpha \neq 0$ and for each $V \in \mathcal V$, $\alpha V$ is convex (since each $V$ is), balanced (because since each $V$ is balanced, $\lambda V \subseteq V$ so that $\lambda \alpha V \subseteq \alpha V$ implying $\alpha V$ is balanced), and is a neighbourhood of the origin (because since $V$ is a neighbourhood of the origin there is an open set $V'$ containing the origin with $o \in V' \subseteq V$, and since $E$ is a topological vector space, $\alpha V'$ is open and is such that $o \in \alpha V' \subseteq \alpha V$). So let:
(2)
\begin{align} \quad \mathcal V' := \{ \alpha V : \alpha \in \mathbf{F}, \: \alpha \neq 0, \: \mathrm{and} \: V \in \mathcal V \} \end{align}
• Then as mentioned above, $\mathcal V'$ is a base of convex and balanced neighbourhoods of the origin. We are now ready to prove that $\mathcal V'$ satisfies properties (1)-(3).
• Proof of (1): Let $U, V \subseteq \mathcal V'$. Then, since $\mathcal V'$ is a collection of (convex and balanced) neighbourhoods of the origin, by definition, $U \cap V \in \alpha \mathcal V$. So indeed, there exists a $W \in \mathcal V'$ with $W \subseteq U \cap V$. $\blacksquare$
• Proof of (2): For each $U \in \mathcal V'$ write $U = \beta V$ with $\beta \in \mathbf{F}$, $\alpha \neq 0$, and $V \in \mathcal V$. Then for all $\alpha \in \mathbf{F}$ with $\alpha \neq 0$, we have that $\alpha \beta \neq 0$, and so $\alpha U = \alpha \beta V \in \mathcal V'$. $\blacksquare$
• Proof of (3): As mentioned early, each $U \in \mathcal V'$ is convex and balanced, and thus by definition, is absolutely convex. Furthermore, by the proposition mentioned earlier, each neighbourhood in a base of neighbourhoods of the origin is absorbent, and so each $U \in \mathcal V'$ is absorbent. $\blacksquare$