On the definition of weak and weak-* topologies

Question

I have been studying topological vector spaces, and despite going over numerous resources, the definitions of weak and weak-* topologies have been causing me some confusion. I am having trouble visualizing and understanding these topologies.

Suppose $X$ is a normed vector space.

Then the weak topology on $X$ is the topology generated by $X^*$, in other words the weakest topology making $x \mapsto f(x)$ continuous for all $f \in X^*$.

Similarly, the weak-* topology is the weakest topology making the maps $x \mapsto f(x)$ continuous for all $x \in X$.

I see the big difference here is that one is generated by the dual and the other by the original vector space. However, I have three points of confusion. I suspect part of my difficulty may be due to not properly visualizing topologies generated by a collection of seminorms.

1. The dual space $X^*$ is defined as the set of bounded linear functionals from $X$ to the underlying field. However, I recall reading that the boundedness of a linear map is equivalent to the map being continuous, so I fail to see what sets we are excluding in this new, weaker topology.

2. What do the open sets (or more simply, the basis sets) look like in these two topologies?

3. The resources I am learning this from often note a relation between the double dual $X^{**}$ and the weak-* topology, what is the relation between these spaces exactly?

Answer 1

For any $X$, the weak topology on $X$ is defined to be the coarsest topology that makes $x\mapsto f(x)$ continuous for each $f\in X^*$, the dual space of $X$. Equivalently it is the topology induced by the seminorms $x\mapsto|f(x)|$, and an open neighborhood of origin looks like $\{x\in X: |f_i(x)|\leq\epsilon,\forall\ 1\leq i\leq n \}$. More generally, weak topology makes sense whenever we have a pairing $(X,Y)$ of spaces, see the wiki article.

Now this definition works for any normed space, so in particular it also works for $X^*$: the weak topology on $X^*$ is the coarsest topology that makes $f\mapsto \chi(f)$ continuous for each $\chi\in X^{**}$, the dual space of $X^*$. However there is another natural topology on $X^*$, where we only consider those $\chi$ that comes from elements in $X$, namely we have an embedding $i:X\rightarrow X^{**},x\mapsto\chi_x$ where $\chi_x(f):=f(x)$. This is the weak* topology. In general the weak* topology is weaker than the weak topology on $X^*$, but if $i(X)=X^{**}$ then obviously they are the same (the converse is also true, see here)

The weak/weak* topology is much weaker than the norm topology. Every weak open neighborhood of origin is norm-unbounded. The closed unit ball of $X^*$ is compact in weak* topology (Alaoglu Theorem), while it is far from compact under the norm topology, unless $X$ is finite dimensional.

Hilbert space is an important special case. It is reflexive by Riesz representation theorem, so no need to distinguish weak and weak* topology. Say $(e_i)_{i=1}^\infty$ is an orthonormal basis for the Hilbert space $H$, so each $x\in H$ can be expanded as $x=\sum x_ie_i$, also denoted as $(x_1,x_2,x_3,...)$. Then on any norm-bounded set, weak topology is exactly componentwise convergence, i.e., a sequence $x^{(n)}$ converges to $y$ iff $x^{(n)}_i$ converges to $y_i$ for every $i$. This is a good exercise, you can also find it in this note. There is also an operator version.