The W-Weak Topology on a Normed Linear Space

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

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

(2)
\begin{align} \quad \mathcal B = \{ V_{\epsilon, F, x} : \epsilon > 0, F \subseteq \mathcal F \: \mathrm{is \: finite}, x \in X \} \end{align}

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

(3)
\begin{align} \quad \mathcal B_x = \{ V_{\epsilon, F, x} : \epsilon > 0, F \subseteq \mathcal F \: \mathrm{is \: finite} \} \end{align}

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:
(4)
\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:
(5)
\begin{align} \quad 0 \in V_{\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:
(6)
\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:
(7)
\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:
(8)
\begin{align} \quad \bigcap_{\psi \in F} \ker \psi \subseteq \ker \varphi \end{align}
  • $\Leftarrow$ 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.

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