Closed operator and closed domain. I am a little bit confused about the following remark in the functional analysis textbook by H.Brezis (Functional Analysis, Sobolev Spaces and Partial Differential Equations):

$\bullet$ Remark $11 .$ In order to prove that an operator $A$ is closed, one proceeds in general
as follows. Take a sequence $\left(u_{n}\right)$ in $D(A)$ such that $u_{n} \rightarrow u$ in $E$ and $A u_{n} \rightarrow f$ in
$F .$ Then check two facts:
(a) $u \in D(A)$, (b) $f=A u$

(a) means that $D(A)$ is closed, but there are many examples that $D(A)$ is densely defined and $A$ is closed. So what's wrong with my arguement?
 A: $(a)$ means that $D(A)$ is closed, under the assumption that $Au_n$ is a convergent sequence. 
So, for example if $u_n \in D(A)$, $u_n \to e \in E$ but $Au_n$ is not a convergent sequence in $F$, then it is not necessary that $e \in D(A)$. Thus $D(A)$ need not be closed.
For example, let $C^1[-1,1] \subset C[-1,1]$ be the set of all real valued continuously differentiable functions, under the supremum norm(or uniform convergence). Consider $Au = u'$ the differentiation operator. It is well known that the former subspace is dense in the latter and containment is strict, therefore $C^1[-1,1]$ is not closed.
Let $f_n(x) = \sqrt{x^2 + \frac 1{n^2}}$. These functions $f_n$ converge uniformly to $|x|$ , which is not differentiable. However, each $f_n$ is differentiable, and the derivative is $x(x^2+\frac 1{n^2})^{-\frac 12}$ , which one can see as a sequence of functions doesn't converge uniformly anywhere, because it goes to $-1$ for $x$ negative and $+1$ for $x$ positive, so the uniform limit can't be continuous if it existed, a problem.
So $f_n$ converge uniformly but their limit is not in $C^1$ because $f_n'$ don't converge uniformly.
So $(a)$ and $(b)$ hold under the condition of $Au_n$ converging. If that does not happen then there is no reason for $e \in D(A)$.
