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

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:

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:

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:


 * $\tau = \tau'$