Let $(X,\tau)$ be a topological vector space with continuous dual $X^*$. "The weak topology" on $X$ is the weakest/coarsest topology under which every element of $X^*$ remains continuous on $X$.
I want to check if I'm understanding this right. For $f\in X^*$ to be continuous wrt a topology, does this mean: for every open $\epsilon$-ball $B_\epsilon(x)$ in that topology, $\epsilon \downarrow 0$ implies $f(x) – f(y) \downarrow 0$ for every $y\in B_\epsilon(x)$ and for every $f\in X^*$?
I am also confused because weaker/coarser topologies have fewer balls than a stronger topology. Which is why convergence in the weak topology is a weaker statement than convergence in a strong topology. But wouldn't this mean that continuity of $f\in X^*$ is a weaker statement in the weak topology? (I.e. why should functions be “less likely" to be continuous in the weak topology?)
Similarly, if a set is compact in a stronger topology, does this imply it is compact in a weaker topology or vice versa?
Best Answer
The weak topology on $X$ is the weakest topology such that every $\varphi\in X^*$ is continuous. That is, it's the topology whose open sets (other than $\emptyset$) are unions of sets of the form$$\varphi_1^{-1}(A_1)\cap\varphi_2^{-1}(A_2)\cap\ldots\cap\varphi_n^{-1}(A_n),$$with each $\varphi_k$ in $X^*$ and where each $A_k$ is an open subset of $\Bbb R$. So, it has the minimum amount of open sets such that each $\varphi\in X^*$ is continuous. Since this topology is weaker than $\tau$, every continuous function defined on $X$ which is continuous with respect to the weak topology is also continuous with respect to $\tau$, but this is not necessarily true in the reverse direction.