Equivalence of Definitions of Weak Convergence on Topological Vector Space

Theorem

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

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

Let $X^\ast$ be the topological dual space of $X$.

Suppose that:

for each $x, y \in X$ with $x \ne y$, there exists $f \in X^\ast$ such that $\map f x \ne \map f y$.

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

Let $x \in X$.

The following definitions of the concept of weak convergence on topological vector space are equivalent:

Definition 1

Let $w$ be the weak topology on $X$.

We say that $\sequence {x_n}_{n \mathop \in \N}$ converges weakly if and only if it converges in $\struct {X, w}$.

We write:

$x_n \weakconv x$

Definition 2

We say that $\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$ if and only if:

for each $f \in X^\ast$ we have $\map f {x_n} \to \map f x$ in $\GF$.

Proof

Definition 1 implies Definition 2

Suppose that $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {X, w}$.

Let $f \in X^\ast$ and $\epsilon > 0$.

We want to show that:

$\map f {x_n} \to \map f x$ in $\GF$.

Let $\map {B_r} {\lambda, \GF}$ be the open ball in $\GF$ of radius $r$ and Centre $\lambda$.

From Characterization of Continuity of Linear Functional in Weak Topology, we have that $f$ is continuous in $\struct {X, w}$.

From the definition of the initial topology, we then have:

$f^{-1} \sqbrk {\map {B_r} {\map f x, \GF} }$ is weakly open for each $r > 0$ and $x \in X$.

So by the definition of convergence in $\struct {X, w}$, there exists $N \in \N$ such that:

$x_n \in f^{-1} \sqbrk {\map {B_r} {\map f x, \GF} }$ for $n \ge N$.

Then:

$\map f {x_n} \in \map {B_r} {\map f x, \GF}$ for $n \ge N$

so that:

$\cmod {\map f {x_n} - \map f x} < \epsilon$

Since $\epsilon$ was arbitrary, we have:

$\map f {x_n} \to \map f x$ in $\GF$.

$\Box$

Definition 2 implies Definition 1

Suppose that:

for each $f \in X^\ast$ we have $\map f {x_n} \to \map f x$ in $\GF$.

Let $U$ be a weak open neighborhood of $x$.

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

From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, the initial topology $w$ is induced as a locally convex space by:

$\set {p_f : f \in X^\ast}$

where $p_f : X \to \hointr 0 \infty$ is defined by:

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

for each $x \in X$.

From Open Sets in Standard Topology of Locally Convex Space, there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that:

$\set {y \in X : \cmod {\map {f_i} y - \map {f_i} x} < \epsilon \text { for each } 1 \le i \le n} \subseteq U$

Since $\map {f_i} {x_n} \to \map {f_i} x$ for each $1 \le i \le n$, for each $1 \le i \le n$ there exists $N_i \in \N$ such that:

$\cmod {\map {f_i} {x_n} - \map {f_i} x} < \epsilon$

Set:

$\ds N = \max_{1 \le i \le n} N_i$

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

$\cmod {\map {f_i} {x_n} - \map {f_i} x} < \epsilon$ for each $1 \le i \le n$.

That is, for $n \ge N$ we have:

$x_n \in \set {y \in X : \cmod {\map {f_i} y - \map {f_i} x} < \epsilon \text { for each } 1 \le i \le n}$

and so $x_n \in U$, as required.

$\blacksquare$