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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)

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    It's all much easier. As a normed space, $E$ is first countable. On first countable spaces, sequential continuity and continuity are equivalent. – 2017-01-15
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    @DanielFischer Isn't the codomain important? The result that I found says that "if $X$ and $Y$ are first countable then any sequentially continuous map from $X$ to $Y$ is continuous". As $E'$ [is not first countable](http://math.stackexchange.com/q/623642/70305) (in the infinite dimensional case), I thought that this result couldn't be applied. Can we in general drop the first countability of $Y$? – 2017-01-15
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    The topology on the codomain is irrelevant. Let $X$ be first countable, and $Y$ a topological space. Suppose $f\colon X \to Y$ is not continuous at $x_0\in X$. Then there is a neighbourhood $W$ of $f(x_0)$ such that $f^{-1}(W)$ is not a neighbourhood of $x_0$. Choose a neighbourhood basis $U_1 \supset U_2 \supset \dotsc$ of $x_0$, and $x_k \in U_k \setminus f^{-1}(W)$. Then $x_k \to x_0$ (since the $U_k$ are a neighbourhood basis) and $f(x_k) \not\to f(x_0)$ (since $f(x_k) \notin W$), so $f$ is not sequentially continuous at $x_0$. – 2017-01-15
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    @DanielFischer Thanks (is the [Willard's book](https://books.google.com.br/books/about/General_Topology.html?id=-o8xJQ7Ag2cC&redir_esc=y) that assumes $Y$ first countable, Corollary 10.5). – 2017-01-15
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    That's odd. I wouldn't expect that a book on topology includes unnecessary hypotheses in such a theorem. What exactly is the statement of the theorem in Willard's book? – 2017-01-15
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    @DanielFischer See my edit. – 2017-01-15
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    I have to repeat myself. That's odd. I see no reason why he assumed $Y$ first countable too. – 2017-01-15

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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.

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    Thanks, it's a good solution (probably, the most elementary possible). – 2017-01-15
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    LeBtz, inspired by your answer I obtained [this solution](http://math.stackexchange.com/a/2099312/70305) for other question of mine. Do you agree with that argument? Can you see any flaw? – 2017-01-15
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    I think it works that way, yes. @Pedro – 2017-01-15