The Weak Topology (X*-Weak Topology) on a Normed Linear Space

Recall from The W-Weak Topology on a Normed Linear Space page that if $X$ is a normed linear space and $W \subseteq X^{\#}$ then a linear functional $\varphi \in X^{\#}$ is continuous with respect to the $W$-weak topology if and only if $\varphi \in W$. That is, the $W$-weak topology on $X$ makes every linear functional in $W$ continuous and ONLY the linear functionals in $W$ continuous.

We now look at a particular $W$-weak topology on a normed linear space $X$ - the $X^*$-weak topology.

Definition: Let $X$ be a normed linear space. Then the Weak Topology on $X$ is the $X^*$-weak topology.

Observe that the Weak topology on $X^*$ is the weakest topology which makes every function in $X^*$ continuous. The norm topology on $X^*$ already makes every function in $X^*$ continuous. Therefore, the weak topology on $X$ is weaker than the norm topology on $X$. Furthermore, if $W \subseteq X^*$ then the $W$-weak topology is weaker than the weak topology on $X$. So:

(1)

\begin{align} \quad W-\mathrm{Weak \: Topology} \subseteq \mathrm{Weak \: Topology} \subseteq \mathrm{Norm \: Topology} \end{align}

Observe that a base for the weak topology on $X$ is:

(2)

\begin{align} \quad \mathcal B = \{ V_{\epsilon, F, x} : \epsilon > 0, F \subseteq X^* \: \mathrm{is \: finite}, x \in X \} \end{align}

And for each $x \in X$, a local base of $x$ is:

(3)

\begin{align} \quad \mathcal B_x = \{ V_{\epsilon, F, x} : \epsilon > 0, F \subseteq X^* \: \mathrm{is \: finite} \} \end{align}

Weak Convergence of Points in the Weak Topology

Definition: Let $X$ be a normed linear space. Then a sequence of points $(x_n)_{n=1}^{\infty}$ in $X$ Weak Converges to $x \in X$ if for every $\varphi \in X^*$ we have that $(\varphi(x_n))_{n=1}^{\infty}$ converges to $\varphi (x)$.