Let $(E,\|\cdot\|)$ be a separable Banach space. Let $E'$ be the topological dual of $E$ equipped with the weak* topology $w^*$.
I read that a certain linear operator $J:(E,\|\cdot\|)\to (E',w^*)$ is continuous because $J(u_n)\xrightarrow{w^*}J(u)$ provided that $u_n\xrightarrow{\|\cdot\|}u$. So, the proof consists in proving that the operator is sequentially continuous.
I'd like to justify that, in this case, sequential continuity is indeed equivalent to continuity. I do not have a good background in topology, but I found (here) the following results which can be applied to the said case.
Every normed space is "bornological". (p. 445)
If $X$ is "bornological" and $Y$ is locally convex, then any sequentially continuous linear operator $A:X\to Y$ is continuous. (p. 452)
As I'm not familiar with the concept of "bornologicity", I'd like to know if there is a more elementary argument for the following question.
In the said case, why sequential continuity implies continuity?
Edit (in response to the comments)
Edit: Daniel Fischers comment makes it easier, I was not aware of this fact.
Since the weak$^\ast$-topology on $E'$ is induced by evaluation mappings $ev_x$ for each $x\in E$, $J$ is continuous if and only if $ev_x\circ J$ is continuous for each $x\in E$ but $ev_x\circ J:E\to \mathbb F$ is a mapping between normed spaces, so it suffices to check sequential continuity for $ev_x\circ J$.
Now if $u_n\to u$ in $E$, then $J(u_n) \xrightarrow{w^*}J(u)$, which is by definition that $J(u_n)[x]\to J(u)[x]$ or $ev_x\circ J(u_n)\to ev_x\circ J(u)$ for each $x\in E$. This is exactly the sequential continuity of $ev_x\circ J$ and hence its continuity.