Closure/not closure of the derivative operator on $C^1[0, 1]$ Consider the space $C[0, 1]$ with the usual $\|\cdot\|_{\infty}$.
Now define the operator $D: C^1[0, 1] \rightarrow C[0, 1]$ (the domain inherits the same norm) to be the derivative operator $D(f) = \frac{\mathrm{d}f}{\mathrm{d}x}$
Then is it true that $D$ is closed? Followup: suppose $D$ acts on $C^2[0, 1]$; is it closed then?
Here's my idea:
Consider $f_n(x) = \frac{1}{n}\sin(n\pi x)$. Then $f_n \rightarrow 0$ clearly. But then $D(f_n) = \pi\cos(n\pi x)$ does not converge to $D(0) = 0$.
This suggests to me that $D$ is not closed even on $C^1[0, 1]$.
But this would also imply lack of closure on $C^2[0, 1]$.
The followup question seems to make me wonder if the answer I was supposed to get was yes to the first question then no/yes to the second since the second one is too easy otherwise.
So basically, what I'm asking is- is my argument correct ?
 A: Yes, the operator you have defined is closed. Let $(f_n)_{n \in \mathbb{N}} \subset C^1([0,1])$ be a sequence converging to some $f$ in $\lVert\cdot \rVert_{\infty}$ such that $f_n'$ converges to some $g \in C([0,1])$ in the same norm. Then, we have
\begin{align*}
\sup_{x\in [0,1]}\left|\int_0^x (f_n' - g) \,\mathrm{d}y\right| \leq \lVert f_n' -g\rVert_{\infty} \, ,
\end{align*}
which implies that $f_n$ converges to $f(0)+ \int_0^x g \, \mathrm{d}y$. Thus, we have
\begin{align}
    f(x)=  f(0)+ \int_0^x g \, \mathrm{d}y \, ,
\end{align}
which by the fundamental theorem of calculus implies that $D$ is closed.
The problem with your example (as @MaoWao mentioned) is that the image sequence does not converge to anything in $\lVert\cdot \rVert_\infty$. See https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators for the definition of closedness.
A: Similar to the answer before, but might be helpful,
To show that the graph $\Gamma(D)$ is closed one has to show that every sequence $(f_n,Df_n)_n \in \Gamma(D)$ which converges has its limit $(f,g)\in \Gamma(D)$, i.e. $g = Df$, where $f_n \to f$ and $Df_n \to g$.
But the convergence of the continuous $Df_n$ is uniform and so is the limit $g$ continuous. The continuity of $g$ implies, by the fundamental theorem of calculus, that $g$ has an antiderivative $h=\int g dx \in C^1[0,1]$. Also we have that the limit $f$ is continuous. The uniqueness of this limit $f$ and the following computation show that $f$ is an antiderivative of $g$, which we choose such that $h(0) = f(0)$.
$$
\| f_n - (f(0) + \int_0^x g(t) dt) \|_\infty = \sup_{0 \le x \le 1} \left| f_n(x) - (f(0) + \int_0^x g(t) dt) \right| \\
= \sup_{0 \le x \le 1} \left| f_n(0) - f(0) + \int_0^x(f_n'(t) - g(t)) dt \right| \le |f_n(0) - f(0)| + \| f_n' - g \|_\infty \to 0,
$$
since by assumption we have the sequence with $Df_n = f_n' \to g$. So the graph muss be closed.
