Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $F$ be a set of linear functionals on $X$ that form a vector space over $\GF$.

That is, for each $\lambda, \mu \in \GF$ and $f, g \in F$, we have:

$\lambda f + \mu g \in F$

Let $\tau$ be the initial topology on $X$ generated by $F$.

For each $f \in F$, define $p_f : X \to \R_{\ge 0}$ by:

$\map {p_f} x = \cmod {\map f x}$

and let:

$\PP = \set {p_f : f \in F}$


Then the standard topology on the locally convex space $\struct {X, \PP}$ is precisely $\tau$.


Proof

Let $\tau'$ be the standard topology on the locally convex space.

From definition, a sub-basis for $\tau'$ is given by:

$\SS' = \set {\map {B_{p_f} } {\epsilon, x} : f \in F, \, \epsilon > 0, \, x \in X}$

where:

$\map {B_{p_f} } {\epsilon, x} = \set {y \in X : \map {p_f} {y - x} < \epsilon}$

For each $u \in \GF$, let:

$\map {B_\epsilon} {u, \GF} = \set {v \in \GF : \cmod {u - v} < \epsilon}$

From Open Balls form Basis for Open Sets of Metric Space:

$\BB_\GF = \set {\map {B_\epsilon} {u, \GF} : u \in \GF, \, \epsilon > 0}$

forms a basis for $\GF$.

From Sub-Basis for Initial Topology in terms of Sub-Bases of Target Spaces:

$\SS = \set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0}$

is a sub-basis for $\tau$.

We show that $\SS = \SS' \cup \set \O$.

First, take $f = 0$.

Then:

$\ds f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } = \begin{cases}X & 0 \in \map {B_\epsilon} {u, \GF} \\ \O & \text{otherwise}\end{cases} \in \SS' \cup \set \O$

On the other hand, if $f = 0$ and $x \in X$ then:

\(\ds \map {B_{p_f} } {\epsilon, x}\) \(=\) \(\ds \set {y \in X : \map {p_f} {y - x} < \epsilon}\)
\(\ds \) \(=\) \(\ds \set {y \in X : \cmod {\map f {y - x} } < \epsilon}\)
\(\ds \) \(=\) \(\ds \set {y \in X : 0 < \epsilon}\)
\(\ds \) \(=\) \(\ds X\)
\(\ds \) \(\in\) \(\ds \SS\)

Now suppose that $f \ne 0$.

Then from Linear Functional on Vector Space is Zero or Surjective, $f$ is surjective.

So, for each $u \in \GF$ there exists $x \in X$ such that $u = \map f x$.

Then we have:

\(\ds f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} }\) \(=\) \(\ds f^{-1} \sqbrk {\map {B_\epsilon} {\map f x, \GF} }\)
\(\ds \) \(=\) \(\ds \set {y \in X : \map f y \in \map {B_\epsilon} {\map f x, \GF} }\)
\(\ds \) \(=\) \(\ds \set {y \in X : \cmod {\map f y - \map f x} < \epsilon}\)
\(\ds \) \(=\) \(\ds \set {y \in X : \cmod {\map f {y - x} } < \epsilon}\)
\(\ds \) \(=\) \(\ds \set {y \in X : \map {p_f} {y - x} < \epsilon}\)
\(\ds \) \(=\) \(\ds \map {B_{p_f} } {\epsilon, x}\)

So we have:

$\set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0} = \set {\map {B_{p_f} } {\epsilon, x} : f \in F \setminus \set 0, \, \epsilon > 0}$

Combining with our result for $f = 0$, we obtain the desired:

$\SS = \SS' \cup \set \O$

Since $\O$ is contained in any topology on $X$ we have:

$\map \tau \SS = \map \tau {\SS'}$

and hence:

$\tau = \tau'$

$\blacksquare$

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