LCTVS over the Field of Real or Complex Numbers

Table of Contents

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. $\blacksquare$

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