Characterization of Convergence in Locally Convex Space

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \mathcal P}$ be a locally convex space over $\Bbb F$ with standard topology $\tau$.

Let $x \in X$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.

Then:

$x_n \to x$ in $\struct {X, \tau}$

if and only if:

$\map p {x_n - x} \to 0$ as $n \to \infty$ as a real sequence, for each $p \in \mathcal P$.

Proof

Necessary Condition

Suppose that:

$x_n \to x$ in $\struct {X, \tau}$

Let $p \in \mathcal P$.

Let $\epsilon > 0$.

We aim to show that there exists $N \in \N$ such that:

$\map p {x_n - x} < \epsilon$ for $n \ge N$

showing that:

$\map p {x_n - x} \to 0$ as $n \to \infty$.

From the definition of a convergent sequence in a topological space, we have:

for each $U \in \tau$ such that $x \in U$, there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$.

From the definition of the standard topology $\tau$, we have:

$U = \set {y \in X : \map p {y - x} < \epsilon}$ is open.

So there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$.

That is:

$\map p {x_n - x} < \epsilon$ for $n \ge N$.

Since $\epsilon$ was arbitrary we have:

$\map p {x_n - x} \to 0$ as $n \to \infty$.

$\Box$

Sufficient Condition

Suppose that:

$\map p {x_n - x} \to 0$ as $n \to \infty$ as a real sequence, for each $p \in \mathcal P$.

Let $U \in \tau$ be such that $x \in U$.

We aim to show that there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$.

From Open Sets in Standard Topology of Locally Convex Space, there exists $\epsilon > 0$ and $p_1, p_2, \ldots, p_m \in \mathcal P$ such that:

$\set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } k \in \set {1, 2, \ldots, m} } \subseteq U$

Since:

$\map {p_k} {x_n - x} \to 0$ as $n \to \infty$

for each $k \in \set {1, 2, \ldots, m}$, there exists $N_k \in \N$ such that:

$\map {p_k} {x_n - x} < \epsilon$ for $n \ge N_k$.

Set:

$N = \max \set {N_1, N_2, \ldots, N_k}$

Then, for $n \ge N$, we have:

$\map {p_k} {x_n - x} < \epsilon$ for all $k \in \set {1, 2, \ldots, m}$.

Then:

$x_n \in U$ for $n \ge N$.

Since $U$ was an arbitrary open set containing $x$, we have:

$x_n \to x$ in $\struct {X, \tau}$.

$\blacksquare$