Weak topologies and weak convergence - Looking for feedbacks

Solution 1:

I would say "Yes" to all your questions.

The so-called weak* topology is an example of a weak topology.

Weak convergence of measures can be considered a weak* topology.

To clarify a new example of convergence of this kind, you should master that $\sigma (\mathcal{F}, X)$ notation. (I think it goes back to Mackey, and was popularized by Bourbaki.)

Solution 2:

Dual Pair

Two vector spaces with a bilinear form satisfying: $$\langle\cdot,\cdot\rangle:X\times Y\to\mathbb{C}:\quad X^\perp=(0),\,Y_\perp=(0)$$ (Informally, this way one guarantees that it separates points.)

Weak Topology

The initial topology generated by the dual: $$\sigma(X;Y):=\tau\left(\bigcup_{y\in Y}\varepsilon_y^{-1}(\mathcal{T}_\mathbb{C})\right)\quad(\varepsilon_y:=\langle\cdot,y\rangle)$$ (This way one guatantees that they become continuous.)

Example

Consider a Banach space with its continuous dual: $$\langle\cdot,\cdot\rangle:E\times E'\to\mathbb{C}:\quad\langle x,f\rangle:=f(x)$$ By the initial topology weak convergence boils down to: $$x_n\rightharpoonup x\iff f(x_n)\to f(x)\quad(f\in E')$$ (That is in terms of product topology convergence by components.)

Remark

The difference between the weak and weak* topology lie in the chosen dual: $$\langle x,f\rangle:=f(x):\quad\sigma(E';E)$$ $$\langle F,f\rangle:=F(f):\quad\sigma(E';E'')$$ (The comparison becomes solid on embedding the Banach space into its double dual.)