# The W-Weak Topology on a Normed Linear Space

Recall from The F-Weak Topology on a Normed Linear Space page that if $X$ is a normed linear space and $\mathcal F$ is a collection of complex-valued functions on $X$ then the $\mathcal F$-weak topology on $X$ is the weakest topology which makes every function in $\mathcal F$ continuous. For $\epsilon > 0$, $F \subseteq \mathcal F$, and $x \in X$ we denoted:

(1)We saw that a base for the $\mathcal F$-weak topology on $X$ is given by:

(2)And that for each fixed $x \in X$ a local base of $x$ is given by:

(3)Now suppose $X$ is any normed linear space and $W \subseteq X^{\#}$. Then $W$ is a collection of linear functionals on $X$ and so we may consider the $W$-weak topology on $X$. (Later, we will look at the special case when $W = X^* \subseteq X^{\#}$)

Theorem 1: Let $X$ be a normed linear space and let $W \subseteq X^{\#}$. Then a linear functional $\varphi \in X^{\#}$ is continuous with respect to the $W$-weak topology on $X$ if and only if $\varphi \in W$. |

**Proof:**$\Rightarrow$ Let $\varphi : X \to \mathbb{C}$ be a linear functional that is continuous with respect to the $W$-weak topology on $X$. Consider the following set:

- Then the above set is the inverse image of the unit open ball in $\mathbb{C}$. Since $\varphi$ is continuous (with respect to the $W$-weak topology on $X$), the above set is open and also contains $0$. So there exists an element of the base that contains $0$ and is contained in this set. That is, there exists an $\epsilon > 0$ and a finite set $F \subseteq W$ for which:

- Therefore, if $| \psi (x) | < \epsilon$ for all $\psi \in F$ we have that $| \varphi (x) | < 1$.

- Now let $x \in \bigcap_{\psi \in F} \ker \psi$. Then $\psi (x) = 0$ for every $\psi \in F$. Hence:

- Hence $|\varphi (tx)| < 1$ for every $t > 0$ and:

- This shows that $| \varphi(x) | = 0$, so $\varphi (x) = 0$. Hence $x \in \ker \varphi$. So:

- From the theorem on the Expressing a Linear Functional as a Linear Combination of Other Linear Functionals page, we must have that $\varphi$ is a linear combination of the linear functionals in $F$. So $\varphi \in W$.

- $\Leftarrow$ Suppose that $\varphi \in W$. Since the $W$-weak topology is the weakest topology which makes all of the functions in $W$ continuous, we have that $\varphi$ is continuous with respect to the $W$-weak topology. $\blacksquare$

The theorem above is significant. It tells us that if $X$ is a normed linear space and $W \subset X^{\#}$, then the only continuous functions with respect to the $W$-weak topology are those functions in $W$ itself.