The question is: Let A be the Operator defined by $A:C^2([-1,1])\subset C([-1,1])\to C(-1,1]),\; x\mapsto x'$. Show that the operator is not closed.
I suppose an example where a derivative sequence converge to a point that is not in the graph of A would suffice, but isn't a property of a sequence of differentiable functions $f_n$, that if the sequence of derivatives $f'_n$ converges uniformly to a function $g$, then the sequence of functions converges to a function $f$ with $f'=g$?
Thank you in advance.
Comment: "Not closed" would be shown by an example where $$ f_n \to f,\qquad f_n' \to g,\qquad f' \ne g \tag1$$
Of course the derivative operator is closed in the $C^2$ metric. But perhaps the way of stating it $$C^2([-1,1])\subset C([-1,1])\to C(-1,1])$$ is meant to mean in the $C$ topology, that is: uniform convergence.
So find an example of $(1)$ where $f_n, f \in C^2$, $f_n',g \in C$, and $\to$ means uniform convergence.