# The Weak Topology Induced by W ⊆ X♯

Recall from The Weak Topology Induced by F page that if $X$ is a nonempty set and if $\mathcal F$ is a collection of functions defined on $X$ then for each $\epsilon > 0$, $F \subseteq \mathcal F$ finite, and $x \in X$ we defined:

(1)We then proved that the collection $\{ N_{\epsilon, F}(x) : \epsilon > 0, \: F \subseteq \mathcal F \: \mathrm{finite}, \: x \in X \}$ is a base for some topology $\tau$ on $X$ and we defined that topology to be the weak topology induced by $\mathcal F$.

Now let $X$ be a linear space. Recall that $X^{\sharp}$ is the collection of all linear operators on $X$. If $W \subseteq X^{\sharp}$ then we can consider the $W$-weak topology on $X$.

The following proposition tells us that the only $W$-weakly continuous functions are those functions in $W$.

Theorem 1: Let $X$ be a linear space and let $W \subseteq X^{\sharp}$. Then a linear functional $\varphi \in X^{\sharp}$ 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. $\blacksquare$