The F-Weak Topology on a Normed Linear Space

# The F-Weak Topology on a Normed Linear Space

 Definition: Let $X$ be a normed linear space and let $\mathcal F$ be a collection of complex-valued functions on $X$. Then the $\mathcal F$-Weak Topology on $X$ is the weakest topology on $X$ which makes all of the functions in $\mathcal F$ continuous

The $\mathcal F$-weak topology is pretty vague, so we would like to develop a basis for this topology. For every $\epsilon > 0$, $F \subseteq \mathcal F$, and $x \in X$, define:

(1)
\begin{align} \quad V_{\epsilon, F, x} = \{ x' \in X : |f(x) - f(x')| < \epsilon, \: \forall f \in F \} \end{align}

Then a basis for the $\mathcal F$-weak topology on $X$ is:

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

To verify that $\mathcal B$ is indeed a basis of the $\mathcal F$-weak topology on $X$ we must show that $X$ is the union of all elements in $\mathcal B$ and that the intersection of any two basis elements is also a basis element.

For the first statement, it is clear that:

(3)
\begin{align} \quad X = \bigcup_{\epsilon, F, x} V_{\epsilon, F, x} \end{align}

For the second statement, let $\epsilon_1, \epsilon_2 > 0$, $F_1, F_2 \subseteq \mathcal F$ be finite, and let $x_1, x_2 \in X$. Suppose that:

(4)
\begin{align} \quad x \in V_{\epsilon_1, F_1, x_1} \cap V_{\epsilon_2, F_2, x_2} \end{align}

Then by definition $x \in V_{\epsilon_1, F_1, x_1}$ and $x \in V_{\epsilon_2, F_2, x_2}$ and so:

(5)

Let $\delta > 0$ be such that:

(6)
\begin{align} \quad 0 < \delta < \min_{f \in F_1, g \in F_2} \{ \epsilon_1 - |f(x) - f(x_1)|, \epsilon_2 - |g(x) - g(x_2)| \} \end{align}

Now we have that $\delta > 0$ and $F_1 \cup F_2$ is finite (since $F_1$ and $F_2$ are both finite sets). So if $x' \in V_{\delta, F_1 \cup F_2, x}$ we have that for all $f \in F_1$:

(7)
\begin{align} \quad | f(x') - f(x_1) | &\leq | f(x') - f(x) | + | f(x) - f(x_1) | \\ & < \delta + |f(x) - f(x_1)| \\ & < (\epsilon_1 - |f(x) - f(x_1)|) + |f(x) - f(x_1)| \\ & < \epsilon_1 \end{align}

And we have that for all $g \in F_2$ that:

(8)
\begin{align} \quad | g(x') - g(x_2) | & \leq | g(x') - g(x)| + |g(x) - g(x_2)| \\ & < \delta + |g(x) - g(x_2)| \\ & < (\epsilon_2 - |g(x) - g(x_2)|) + |g(x) - g(x_2)| \\ & < \epsilon_2 \end{align}

Therefore $x' \in V_{\epsilon_1, F_1, x_1} \cap V_{\epsilon_2, F_2, x_2}$. Thus:

(9)
\begin{align} \quad V_{\delta, F_1 \cup F_2, x} \subseteq V_{\epsilon_1, F_1, x_1} \cap V_{\epsilon_2, F_2, x_2} \end{align}

So indeed, $\mathcal B$ is a basis of the $\mathcal F$-weak topology.

Now for each fixed $x \in X$, a local basis of $x$ is given by:

(10)
\begin{align} \quad \mathcal B_x = \{ V_{\epsilon, F, x} : \epsilon > 0, F \subseteq \mathcal F \: \mathrm{is \: finite} \} \end{align}

## F-Weak Convergence of Sequences

By definition, if $X$ has the $\mathcal F$-Weak topology, then a sequence of points $(x_n)_{n=1}^{\infty}$ in $X$ converges to $x \in X$ if for every $f \in \mathcal F$ we have that:

(11)
\begin{align} \quad \lim_{n \to \infty} f(x_n) = f(x) \end{align}