I have a general question about how can we characterize the weak and weak$^\ast$ topologies and continuous maps between then in normed spaces:
Let's start with a normed space $X$. The weak topology on $X$ is the smallest topology such that every $x^\ast\in X^\ast$ is a continuous linear functional. Here $X^\ast$ is the topological dual of $X$, therefore previously equipped with the operator norm topology. Similarly, the weak$^\ast$ topology is the weakest topology on $X^\ast$ such that every linear functional $J_X(x):X^\ast\rightarrow\mathbb{C}$ is continuous, where $J_X:X\rightarrow X^{\ast\ast}$ is the canonical embedding. Both these definitions are a bit unpractical when dealing with claims about continuity, so can we characterize continuity by sequential continuity, i.e., a map $T:X\rightarrow Y$ is weakly continuous (respectively, weak$^\ast$ continuous) iff $Tx_j\xrightarrow{w} Tx$ (respectively, $Tx_j\xrightarrow{w^\ast}Tx$) for every weakly convergent (respectively, weak$^\ast$ convergent) sequence $x_j\xrightarrow{w}x$? Of course this would be true for nets, but usually sequences are easier to work with.
I know generally sequences are not enough to characterize a topology, however I've seen this notion of continuity being used around and it got me wondering if that's the notion of weak and weak$^\ast$ topology that is being referred to in this context.
Sequences are not enough to characterize the weak topology or weak continuity.
Indeed, recall that the normed space $\ell^1$ has this peculiar property that weak convergence of sequences is equivalent to strong convergence of sequences. However, weak topology is not equal to the topology of the norm (in infinite dimension it never is since the open unit ball is not weakly open).
Similarly, the function $\|\cdot\|_1 : \ell^1 \to \Bbb{R}$ is strongly continuous, weakly sequentially continuous but not weakly continuous. Indeed, otherwise for any weakly convergent net $x_j \xrightarrow{w} x$ we would have $$\|x_j - x\| \to 0 \implies x_j \xrightarrow{s} x$$ which would imply that the topologies are equal, but they aren't.