To make things simplier, let $X$ be a normed space and $f\colon X\to \mathbb{R}$ be a linear functional. What does that mean that $f$ is weakly continuous? Continuity is defined by preimages, so we demand $f^{-1}(B)$ to be open in $X$ for every $B$ open in $\mathbb{R}$ endowed with natural topology. On $X$ we may introduce new topology – weak topology. So my intuition is to use the preimages once again. Can we say that $f$ is weakly continuous if $f^{-1}(B)$ is weakly open, i.e., $f^{-1}(B)$ belongs to weak topology for every $B$ being open in $\mathbb{R}$?
Since weak topology is weaker than the normed one, i.e., every weakly open set is strongly open, then the above definition may have sense – weak continuity is something more genreal then continuity. Am I correct?
There is one more observation: by the definition we obtain that $\{f_i\}\subset X^*$ which induce weak topology are continuous. But also, every $f_j^{-1}(B_j)$ belongs to the base of the weak topology, which automatically makes them weakly continuous.
Best Answer
Formally, the weak topology on some locally convex space $X$ can be defined as the Initial Topology with respect to the topological dual $X^*$, i.e. the weakest topology that makes all $f\in X^*$ continuous. In this sense, your first statement
and your third one
are correct. Your second statement however
is false, assuming that "more general" means that the class of weakly continuous functions is contained in the continuous functions - it is exactly the opposite way around: Every weakly continuous function is continuous, but there are continuous functions that are not weakly continuous. A classic example is the function $x\mapsto\|x\| $ on infinite-dimensional Hilbert Spaces - "normal" continuity is trivial, but this function is not weakly continuous since an orthonormal set weakly converges to zero, but has constant $1$ as image. I like to think of this in the sense of sequential continuity: In the weak topology, there are "more" converging sequences than in the standard topology, hence the requirement $$ \lim_n f(x_n)=f(\lim_n x_n) $$ for all convergent sequences $(x_n)_{n}$ is harder to fulfil. In general, I find that thinking about the different topologies with respect to the sequences rather than the open sets is often very helpful, although you should be careful with that when it comes to proofs: The weak topology is in general not first-countable.