Continuity of Linear Functionals in Initial Topology on Vector Space Generated by Linear Functionals

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 $F$.

Let $g : X \to \GF$ be a linear functional.

Then $g$ is $\tau$-continuous if and only if $g \in F$.

Proof

Sufficient Condition

From the definition of the initial topology, $\tau$ is the coarsest topology making all $g \in F$ continuous.

So if $g \in F$ then $g$ is $\tau$-continuous.

$\Box$

Necessary Condition

Suppose that $g$ is $\tau$-continuous.

For each $x \in \GF$ and $\epsilon > 0$, let:

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

Then since $g$ is $\tau$-continuous, we have:

$g^{-1} \sqbrk {\map {B_1} {0, \GF} } \in \tau$

and so $g^{-1} \sqbrk {\map {B_1} {0, \GF} }$ is an open neighborhood of ${\mathbf 0}_X$.

From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $\tau$ is generated by the seminorms:

$\set {p_f : f \in F}$

where we define $p_f : X \to \R_{\ge 0}$ by:

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

for each $x \in X$.

So, from Open Sets in Standard Topology of Locally Convex Space, there exists $g_1, \ldots, g_n \in F$ and $\epsilon > 0$ such that:

$\set {x \in X : \cmod {\map {g_k} x} < \epsilon \text { for each } 1 \le k \le n} \subseteq g^{-1} \sqbrk {\map {B_1} {0, \GF} }$

We will deduce that:

$g \in \span \set {g_1, \ldots, g_n}$

so that $g \in F$.

Let:

$\ds x \in \bigcap_{k \mathop = 1}^n \ker g_k$

Then $\map {g_k} {\lambda x} = 0$ for each $\lambda \in \GF$ and $1 \le k \le n$.

In particular, $\cmod {\map {g_k} {\lambda x} } < \epsilon$ for each $\lambda \in \GF$ and $1 \le k \le n$.

So $\lambda x \in g^{-1} \sqbrk {\map {B_1} {0, \GF} }$ for each $\lambda \in \GF$.

Then:

$\cmod {\map g {\lambda x} } < 1$

for each $\lambda \in \GF$.

So, by the linearity of $g$ we have:

$\cmod \lambda \cmod {\map g x} < 1$

for each $\lambda \in \GF$.

If $\cmod {\map g x} > 0$, we could take:

$\ds \cmod \lambda > \frac 1 {\cmod {\map g x} }$

to get:

$\cmod \lambda \cmod {\map g x} > 1$

Hence we must have $\map g x = 0$, so that $g \in \ker f$.

So, we have:

$\ds \bigcap_{k \mathop = 1}^n \ker g_k \subseteq \ker g$

So from Condition for Linear Dependence of Linear Functionals in terms of Kernel, we have:

$g \in \span \set {g_1, \ldots, g_n}$

and so $g \in F$ as desired.

$\blacksquare$