I first saw a reference to the core of an unbounded linear operator in the context of Hilbert space and subsequently in the more general context of Banach space.
The definition here https://math.stackexchange.com/q/716486 .....
"For $D \subseteq D(T)$ to be a core for $T$, what you want is $\{(x,Tx): x \in D\}$ to be dense in the graph $\{(x,Tx): x \in D(T)\}$ of $T$ (as subsets of ${\mathscr H} \times {\mathscr H}$, with the norm topology)." .....
is in the context of Hilbert space, but only refers to the norm, so presumably applies equally in Banach space.
Question 1: In some references I can find it is required that $D$ is a subspace while others just identify it as a subset ?
Question 2: I think I could adapt the definition given to a purely topological context and say "For first-countable Hausdorff topological spaces $(X, \mathscr T)$ and $(Y, \mathscr S)$ and a partially defined function $f: D(f) \subset X \to Y$ then $D \subset D(f)$ is a core for $f$ if $G(f|_D) = \{x, f(x): x \in D\}$ is dense in $G(f) = \{x, f(x): x \in D(f)\}$ with the product topology on $(X \times Y)$"
I think this reduces to the Banach space definition since the norm topology on ${\mathscr H} \times {\mathscr H}$ is just the product space topology, but I can't find any reference to confirm or contradict this and would appreciate feedback.
Question 3: taking the Banach space definition, it seems that if $D$ is a core for $T$ then $D$ is a dense in $D(T)$ but the converse doesn't necessarily hold - correct ?
Question 4: I think I get the same result $D$ is a core for $f \implies D$ is a dense in $D(f)$ with my topological definition - correct ?
Proof:
Let $x \in D(T)$ so that $(x, f(x)) \in G(f)$
If $O_X$ is any open set containing $x$ and $O_Y$ is any open set containing $f(x)$ then $O_X \times O_Y $ is an open set in the product topology and contains some point $(x_D, f(x_D))$ with $x_D \in O_X \implies D$ is dense in $D(f)$.
Question 5: I think with the Banach space definition it follows that a closed operator is uniquely defined by its values on a core.
Question 6: I think with my topological space definition it follows that a closed operator is uniquely defined by its values on a core provided that the spaces are first countable and Hausdorff.
Proof:
Relies on the following results
- A convergent sequence in Hausdorff space has a unique limit.
- If a space is first countable, $C$ is closed and $D$ is dense in $C$ then any point in $C$ is the limit of a convergent sequence of points in $D$.
- The product topology of two first-countable spaces is first-countable
Firstly assert that if $f :D(f) \subset X \to Y$ and $G $ is a dense subset of $G(f) = \{(x, f(x)): x \in D(f)\}$ in the product topology then $\overline G = G(f)$. This follows since $X \times Y$ is first countable so any $(x, f(x))$ is the limit of a sequence $(x_n, f(x_n))$ in $G$ and being Hausdorff this limit is unique.
Then let $f,g$ be two closed functions $f:D(f) \subset X \to Y$ and $g:D(g) \subset X \to Y$ with $D$ a core for both and $f = g$ on $D$. Then $G = \{x, f(x): x \in D\} = \{x, g(x): x \in D\}$ is dense in $G(f)$ and $G(g)$
So $G(f) = G(g) = \overline G \implies f = g$ on $D(f) = D(g)$.