Definition of weak-$*$ topology

Question

Though I learned some basic theory about topological vector spaces, I always confused about the definition of weak-$*$ topology.

Given $x\in X$, let $\phi_x: X^*\to \Bbb R$ denote the evaluation map $u\to u(x)$ at $x$.

The weak-$*$ topology on $X^*$ is the initial topology associated with the family of all evaluation mpas $\phi_x: X^*\to \Bbb R$. Thus, the weak-$*$ topology is the smallest topology on $X^*$ for which all evaluation maps $\phi_x$ are continuous.

1. What is the definition of initial topology, how to understand the first statement?

2. Why the weak-$*$ topology is the smallest topology on $X^*$ for which all evaluation maps $\phi_x$ are continuous?

3.There is also a conclusion: every subset of $X^*$ which is open for the weak-$*$ topology is also open for the strong topology. (1)

I want to show that $(X^*,SOT)\rightarrow (X^*, \|\cdot \|)$ is continuous, but the book I refered mention that "since all evaluation maps are continuous for the strong topology", why can we prove (1) by the reason $"\cdots"$.

Answer 1

I explained the basic theory of initial topologies in this answer.

Given a family of functions defined on a set, where the codomain have topologies, we give the set the smallest topology that makes all these functions continuous wrt the already given topologies on the codomains. I prove existence and unicity in that answer.

In your case we have the set $X^\ast$ and all point-evaluation functions $\phi_x$, where $x$ ranges over $X$, and so the common codomain is the field $\Bbb R$ (or $\Bbb C$ in the case of complex vector spaces), which has its standard topology.

If we take all continuous linear maps from $X^\ast$ (in the norm topology) to $\Bbb R$) we get the weak topology (instead of the weak-star topology) on $X^\ast$; the evaluations are but a small subset of these. We then get a topology in general between (as subsets) the weak-star topology and the norm topology.

This is all general topology theory, but in the case where all $\phi_x$ are linear (as is the case here) and the domain is a vector space, we get a vector space topology on the domain, so that $+$ and scalar multiplication are still continuous on it.

So as to your question 2.: it's true by definition. And we choose it because it's useful, e.g. because the closed unit ball is compact in this topology (which can have useful consequences). And as the norm topology is one of the topologies that makes all $\phi_x$ continuous by minimality $\mathcal{T}_{w\ast} \subseteq \mathcal{T}_{\textrm{strong}}$ and so the inclusion is continuous. QED.

From the general theory plus some reasoning it follows that the following sets form a base for the open sets of the weak-star topology:

$$B(f; \{x_1, x_2, \ldots,x _n\}, r):= \{g \in X^\ast \mid \forall 1 \le i \le n: |f(x_i) - g(x_i) | < r\}$$

where $f \in X^\ast$, $\{x_1, \ldots, x_n\}$ is a finite subset of $X$ and $r>0$.

This makes it quite concrete as a topology.

Answer 2

By initial topology the author means the weakest (coarsest) topology that makes your $\phi_x$ continuous.

To see its the weakest, you just have to write down a topology $\sigma$ that makes all the $\phi_x$ continuous, and realize that the weak topology is contained in this one as well. If you want to be a bit more set theoretical, you can show that the lattice of all topologies on a set $X$ is always complete (i.e all the subsets are bounded above and bellow) under the relation $\tau_2$ is finer (stronger) than $\tau_1$ if $\tau_1 \subset \tau_2$. Then, the set off all topologies that make a family $\{ f_\alpha :X\to Y \} $ continuous always has a weakest ( coarsest) topology associated with it.

For 3, consider that the norm topology makes all $\phi_x$ continuous. So the topology it induces is stronger (finer) than the weak* topology by definition of the later. Hence, if a set U is on the weak* topology, it must also be on the strong topology.

Showing the inclusion map is continuous is then just a consequence of the definitions.