As in title, $A$ is a differential operator with the domain $D(A)=\{f\in C^1[0,1]: \text{there is $\epsilon>0$ such that $f(x)=0$ for $x\in [0,\epsilon)$}\}$
This came up as an example of non-closed operators, but it was not explained why so - it just stated that it is an example. I do think I have a explanation:
Let $\|\cdot\|'$ denote the graph norm in $D(A)$. That is,
$$\|f\|'=\|f\|+\|f'\|$$
where $\|\cdot\|=\|\cdot\|_{\infty}$. It's enough to show that $D(A)$ is not closed with respect to this norm. For $n\ge 2$, consider following sequence of $C^1$ functions:
$$ f_n(x)=\begin{cases} 0 & \text{if $x\in [0,1/n)$} \\ \frac{1}{\left(1-\frac{1}{n}\right)^2} \left(x-\frac{1}{n}\right)^2 & x\in [1/n,1]\end{cases} $$
Clearly $f_n$ are in $D(A)$, and
$$ g_n(x):=x^2-f_n(x)=\begin{cases} x^2 & \text{if $x\in [0,1/n)$} \\ (1-c_n)x^2 +\frac{2c_n}{n}x -\frac{c_n}{n^2}& x\in [1/n,1]\end{cases} $$
where $c_n=\frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n^2}{(n-1)^2}$. Then
$$ g_n'(x)=\begin{cases} 2x & \text{if $x\in [0,1/n)$} \\ 2(1-c_n)x +\frac{2c_n}{n} & x\in [1/n,1]\end{cases} $$
Since $c_n>1$ for all $n\ge 2$, $g_n'$ is decreasing in $[1/n , 1]$, and $g_n'=0$ when
$$x=x_0:=-\frac{c_n}{n(1-c_n)}=\frac{n}{2n-1}=1-\frac{n-1}{2n-1}\in (0.5,1) $$
Hence
$$ \|g_n'\|=\frac{2}{n}\to 0$$
and
$$ \|g_n\|=g_n(x_0)=-\frac{c_n^2}{n^2(1-c_n)}-\frac{c_n}{n^2} $$
since $c_n\to 1$ as $n\to \infty$ and as $n^2(1-c_n)=\frac{n^2-2n^3}{(n-1)^2}\to -\infty$ we also have that $\|g_n\|\to 0$
Thus $g_n\to x^2$ in $\|\cdot\|'$ norm, but $x^2\notin D(A)$ so we are done.
I think what I came up with is correct, but as one can see there is quite a bit of messy computations involved in it. And although it makes intuitive sense that $g_n$ should converge to $x^2$ I wouldn't have completely believed that this example worked without those computations.
Thus I'd like to know if there are simpler examples/ or simpler solution that does not rely on explicit construction of an example that goes wrong somehow.
- Also, although this probably counts as another individual question, nevertheless I'd like to know if there is any connection between the graph norm and original norm (here $\|\|' \text{and} \|\|_{\infty}$), because the way I first thought up of such an example was by recognizing that the set $D(A)$ is in fact not closed with respect to $\|\|_{\infty}$ and then tried to see if the same holds for the graph norm.