Closedness and continuity in infinite dimensional spaces I cannot understand why the operator $A=d/dx: D(A)(\subset C[a,b])\to C[a,b]$ is closed when the domain $D(A)$ is chosen to be $C^1[a,b]$ while we know that we can converge to a non-differentiable function by a sequence of differentiable functions under sup-norm. Your explanation will definetly help. Thanks
 A: "Closed" is different from "continuous". This example is perhaps the prototypical one of a non-continuous closed operator. By definition, $(A, D(A))$ is closed if and only if $D(A)$ is a Banach space when equipped with the graph norm 
$$
\|f\|_{D(A)}= \|f\|_{C([0,1])} + \|Af\|_{C([0,1])}.$$ 
In this case the graph norm is exactly the usual norm on $C^1([0,1])$, which is Banach. So the operator is closed. 
To get a better grasp on this, let us examine a different proof. Note that closedness of $(A, D(A))$ is equivalent to the following property: 
$$
\text{ for every sequence }f_n\in D(A)\ \text{such that }f_n\to f \text{ and }Af_n\to \text{something,}$$
$$
\text{one has that }f\in D(A)\text{ and }Af_n\to Af.$$
In the present case, consider a sequence $f_n \in C^1([0,1])$ such that 
$$
f_n\to f\quad f'_n\to g,$$
where $f, g\in C([0,1])$. We claim that $f\in C^1$ and that $g=f'$. To prove this claim, we start from the identity 
$$
f_n(t)=f_n(0)+\int_0^t f'_n(s)\, ds, $$
which becomes, in the limit as $n\to \infty$ (note that uniform convergence commutes with integrals), 
$$
f(t)=f(0)+\int_0^t g(s)\, ds,$$
and since $g$ is continuous, $f\in C^1$. Differentiating, we see that $f'=g$. And this concludes the proof that $(A, D(A))$ is closed.
