I am teaching myself Operator Theory, and I found this question in a book I am reading.
We have these facts:
$\bullet$ Every normed linear vectorial space admits a completion to be a Banach space.
$\bullet$ A linear operator $A: D(A) \subset X \rightarrow X$ on a Banach Space $X$ is closed iff $(D(A), \|\cdot \|_G)$ is a Banach sapce while $(\|\cdot\|_G)$ on $D(A)$ is the graph norm defined by:
$$ \|u\|_G=\|u\|_X+\|Au\|_X, \quad \forall u \in D(A)$$
$\bullet$ Not every linear operator is closable.
The question I want to address is: Show that there is no contradiction?
One may think that for every linear operator we can associate the graph norm to its domain and complete it to get a Banach space and by then a closed operator, the answer to this is that we can complete it to be a Banach space, but never to be a Graph of an operator, suppose for example that $A: D(A) \subset X \rightarrow X$ is a non-closable operator which means:
There exists sequence $(x_n)_{x \geq 0}$ in $D(A)$ such that $x_n \rightarrow 0$ and $ Ax_n \rightarrow \alpha \in X$ such that $\alpha \neq 0$ $(*)$
The completion of $(D(A), \|\cdot \|_G)$ will never solve this issue but it will add limits of some sequences which didn't exist in $D(A)$
I find this question very educative and that this kind of question helps a lot to understand the topic whatever it is. My question is do you know any questions of this kind? In Operator theory or any other topics in mathematic? I would appreciate also it if you comment on my example above. Thank you in advance.