Fundamental problem about weak topology (remain a continuous function)

Question

I know the definition of weak topology:

Let $\mathcal{B}$ be a infinite dimensional Banach space. The weak topology in $\mathcal{B}$, is the topology generated by $\Sigma = \lbrace \varphi^{-1}(U); \varphi \in \mathcal{B}^{*} \rbrace $, where $U \subset \mathbb{F} $ is open. It is, the coarsest topology in $\mathcal{B}$ such that each element of $X^{*}$ remains a continuous function.

My question is what is "remains a continuous function"?

So if I have the topology worse than the weak topology, some elements $\phi$ of $\mathcal{B}^*$ are not continuous. What do these $\phi$ look like? Moreover, how to know if the topology is the weak topology; it seems difficult to come up with a function in $\mathcal{B}^*$ which is not continuous.

It is difficult for me to understand this. Is there any concrete example to explain this?

Answer 1

Here is my way of understanding. We can define various topologies over a vector space to make it a topological vector space. If $\tau_2$ is weaker than $\tau_1$ ($\tau_1\supset \tau_2$) , then $(X,\tau_1)$ has more continuous linear functional than $(X,\tau_2)$. For example, let $X$ equipped with a trivial topology, say $\tau=\{\emptyset, X\}$, then the only conitnuous linear functional over it is $f=0$. If $\tau=2^X$, then any linear functional over $X$ is continuous. The coarsest topology is the weakest topology to make sure that the number of continuous linear functional doesn't decrease. In other words, the element of $X'$ "remains a continuous function"

In the space $B$, you have two topologies: the weak one and the strong one $\tau_w,\tau_s$. It is obivous that $\tau_w\subset \tau_s$. i.e., it is weak than strong topology, that is why it is called weak. We recall that a functional over $(X,\tau)$ is continuous if and only if $f^{-1}(U)\in \tau$ for any open set in $\mathbb{F}$.

Then you can introduce two type of continuities over $B$. If $f\in X'$, then the defintion of weak topology shows that $f$ is a also a continuous linear function over $(X,\tau_w)$ On the other hand, any functional $f$ that is weak continuous, on can show that $f\in X'$.

Give an example: Let us set $X=L^2(\Omega)$ and $Y=C_0^\infty(\Omega)$ and set $F_h(g)=\int_\Omega h(x) g(x) dx$. Then it is a (strong)-continuous linear functional over $X$. If we define coarsest topology based on $\{F_f\}_{f\in Y}$ instead of $X'$. This topology $\tau_Y$ is even weaker than the one induced by $X'$. There are some continuous linear functions in $X'$ fail to be continuous over $(\tau, Y_\tau)$. Actually, a functional $\Lambda$ is continuous over $(\tau, Y_\tau)$ if and only if $\Lambda=F_{f}$ for some $f\in Y$. Meanwhile, $\Lambda\in X'$ if and only if $\Lambda=F_{f}$ for some $f\in X$.

Answer 2

$\mathcal{B}^\ast$ is the set of continuous linear functions from $\mathcal{B}$ to $\mathbb{F}$, where $\mathcal{B}$ has the norm topology and $\mathbb{F}$ its usual topology (be it the reals or the complex numbers).

So all $\phi$ already were "strong"-continuous (as it's called). It turns out that if we define the weak topology in this way a linear function $\phi: \mathcal{B} \to \mathbb{F}$ is strong-continuous iff it is weak-continuous. So we get the exact same continuous functionals on the Banach space under both topologies, only the weak topology is more directly related to their continuity, as it were.