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I have recently started learning weak topology from Folland and here's how he defines it in section $4.2$:

If $X$ is any set and $\{f_\alpha:X \to Y_\alpha \}_{\alpha \in A}$ is a family of maps from $X$ into some topological spaces $Y_\alpha$, the weak topology on $X$ generated by $\{f_\alpha \}$ is the weakest topology $\tau$ on $X$ that makes all the $f_\alpha$ continuous, namely $\tau$ is generated by the sets of the form $f_\alpha^{-1}(U_\alpha)$, where $U_\alpha \subset Y_\alpha$ is open.

Now in section $5.4$, Folland talks about weak topology and weak$^*$ topology again as follows:

If $X$ is a normed vector space, the weak topology on $X$ is the weak topology generated by $\{f\,|\, f\in X^{*}\}$.

I am slightly confused with the terminology here. The weak topology generated by $\{f\,|\, f\in X^{*}\}$ by definition is the weakest topology such that all such $f$ are continuous. But $f \in X^*$ is continuous by definition. Isn't this redundant?

Finally, he asserts the following which I was able to prove:

If $\langle x_\alpha \rangle$ is a net in $X$, then in the weak topology $\langle x_\alpha \rangle \to x \iff f(x_\alpha) \to f(x) \,\, \forall \, f \in X^*$.

Proof: $\implies$ Let $U$ be any ngh of $f(x)$. Then $V:=f^{-1}(U)$ is a ngh of $x$. Thus $\exists \, \alpha_0$ such that $x_\alpha \in V \, \, \forall \, \alpha \geq \alpha_0$. Hence $f(x_\alpha) \in U \, \, \forall \, \alpha \geq \alpha_0$

$\Leftarrow$ Take $f(x)=x$. Is this correct?

approximation
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    Each $f \in X^{*}$ is continuous w.r.t. the original topology of $X$ but there are smaller topologies that makes each $f$ continuous. The weak topology is the smallest of these. – geetha290krm Feb 17 '23 at 06:18

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The weak topology generated by $\{f\,|\, f\in X^{*}\}$ by definition is the weakest topology such that all such $f$ are continuous. But $f \in X^*$ is continuous by definition. Isn't this redundant?

Not quite, elements in $X^\ast$ are continuous when $X$ is endowed with the norm topology; but maybe some of the open sets are redundant for all linear functionals to be continuous. In general for topologies "more rigid" = "stronger" = "finer" = "larger" = "more expensive" = "higher resolution" = "harder convergence", and "more flexible" = "weaker" = "coarser" = "smaller" = "cheaper" = "lower resolution" = "easier convergence". Also see Hirsch's Differential Topology vs Rudin Functional analysis definition of weak and strong topology. for an application of this.

Proof: $\implies$ Let $U$ be any ngh of $f(x)$. Then $V:=f^{-1}(U)$ is a ngh of $x$. Thus $\exists \, \alpha_0$ such that $x_\alpha \in V \, \, \forall \, \alpha \geq \alpha_0$. Hence $f(x_\alpha) \in U \, \, \forall \, \alpha \geq \alpha_0$

$\Leftarrow$ Take $f(x)=x$. Is this correct?

Hint: Yes, when $X$ is one dimensional. What if $X$ has at least two dimensions? Recall that the weak topology is defined in terms of an optimality condition.

Alp Uzman
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  • Could you please point out how is dimension relevant here? I don't think I am assuming $X$ is one-dimensional anyway. – approximation Feb 17 '23 at 15:50
  • @approximation If $f(x)=x$ is to be a linear functional, then $x$ would need to be in the base field to begin with. – Alp Uzman Feb 17 '23 at 15:57
  • @approximation I mean by a linear functional on $X$ a linear function $X\to\mathbb{K}$, where $\mathbb{K}$ is the field over which $X$ is a normed vector space, and by $X^\ast$ the collection of all such linear functions that are also bounded. – Alp Uzman Feb 17 '23 at 16:07
  • I see your point. For $x \in X$, $x$ need not be in $K$, and thus $f(x)=x$ is not a linear functional. How should I remedy this? – approximation Feb 17 '23 at 16:09