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)

\begin{align} \quad N_{\epsilon, F}(x) = \{ y \in X : |f(x) - f(y)| < \epsilon, \: \forall f \in F \} \end{align}

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:

(2)

\begin{align} \quad \{ x \in X : |\varphi(x)| < 1 \} \end{align}

• 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:

(3)

\begin{align} \quad 0 \in N_{\epsilon, F} (0) = \{ x \in X : | \psi (x) - \psi(0) | < \epsilon, \forall \psi \in F \} = \{ x \in X : | \psi (x) | < \epsilon, \forall \psi \in F \} \subseteq \{ x \in X : |\varphi (x)| < 1 \} \end{align}

• 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:

(4)

\begin{align} \quad | \psi (tx) | = 0 < \epsilon, \quad \forall t > 0, \forall \psi \in F \end{align}

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

(5)

\begin{align} \quad | \varphi (x) | < \frac{1}{t}, \quad \forall t > 0 \end{align}

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

(6)

\begin{align} \quad \bigcap_{\psi \in F} \ker \psi \subseteq \ker \varphi \end{align}

• 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. $\blacksquare$