Question about Rudin's Functional Analysis Closed Graph Theorem In page 51 of Rudin's Functional Analysis, the closed graph theorem is proven, which says that if you have a linear map between two F-spaces whose graph is closed in the product space, then the map is continuous (2.15).
Then in a remark, Rudin says that in applications the closedness of the graph is verified by verifying that the map is sequentially closed. However, why would you even need to use the closed graph theorem if the map is sequentially closed, as for metric spaces (of which F-spaces are one example), sequential continuity anyway implies continuity, and so you wouldn't need to use the closed graph theorem to begin with!
What am I missing?
 A: Let $X$, $Y$ be Banach spaces and $A: X \to Y$ be linear.
Sequential closedness of the graph of $A$ means that if $x_n \to x$ and $Ax_n \to y$, then $Ax =y$. But not any convergent sequence $(x_n)_n$ has to be mapped to a convergent image sequence $(Ax_n)_n$.
Sequential continuity of $A$ means, that if $x_n \to x$, then $Ax_n$ converges as well and $Ax_n \to x$.
For example let
$X = (C^1[a,b], \Vert \cdot \Vert_\infty)$ and $Y= (C^0[a,b], \Vert \cdot \Vert_\infty)$ and $Af = f'$ be the differentiation map. A classical result from analysis tells you that if both $f_n$ and $f_n'$ are uniformly convergent (i.e. convergent in respect to $\Vert \cdot \Vert_\infty$) then $\lim f_n' = (\lim f_n)'$. This means that the graph of $A$ is closed. However $A$ is not continuous as you can see by $f_n (x) = \frac{1}{n} \sin(n^2 x)$: Then $f_n \to 0$ but not $Af_n \to 0$. This is because $Af_n (x) = n \cos(n^2x)$ so $Af_n$ does not converge uniformly at all.
The reason why the closed graph theorem doesn't work here is that $X = (C^1[a,b], \Vert \cdot \Vert_\infty)$ is not complete.
Continuity of $A$ implies closedness of the graph of $A$, but the inverse statement is the closed graph theorem and requires comleteness of $X$ and $Y$.
