Densely defined, closed and closable operators, and cores

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a linear operator on $H$ with domain $D(T)$ and graph $\mathrm{\Gamma }(T)$ (Unbounded linear operators: domain, graph and extension).

$T$ is densely defined when $D(T)$ is a dense subset of $H$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets); closed when $\mathrm{\Gamma }(T)$ is a closed subset of $H\oplus H$; and closable when $T$ has a closed extension. Recall that the graph norm is $\mathrm{\parallel }x{\mathrm{\parallel }}_T=(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Tx{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$.

The graph-norm dictionary. The following three statements are part of the definition's content and are proved in the remarks below rather than assumed:

1. $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$ is a norm on $D(T)$, and $x\mapsto (x,Tx)$ is an isometric isomorphism of $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ onto $\mathrm{\Gamma }(T)$;
2. $T$ is closed if and only if $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space);
3. if $T$ is closed, then $D(T)$ is a Hilbert space for the inner product $⟨x,y{⟩}_T=⟨x,y⟩+⟨Tx,Ty⟩$ whose induced norm is $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$.

A linear subspace $D_0\subseteq D(T)$ is a core for a closed operator $T$ when $D_0$ is dense in $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$, equivalently when $\overline{\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})}=\mathrm{\Gamma }(T)$, where the closure is taken in $H\oplus H$ and $T{\mathrm{\mid }}_{D_0}$ denotes the restriction of $T$ to $D_0$.

Claim 1. $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T=\mathrm{\parallel }\cdot \mathrm{\parallel }$ holds on $\ker T$, and $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x{\mathrm{\parallel }}_T$ always. The map $J:x\mapsto (x,Tx)$ is linear, and $\mathrm{\parallel }x{\mathrm{\parallel }}_T=\mathrm{\parallel }Jx{\mathrm{\parallel }}_{H\oplus H}$. Hence homogeneity and the triangle inequality for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$ are the corresponding norm properties in $H\oplus H$ pulled back along $J$. It is definite because $\mathrm{\parallel }x{\mathrm{\parallel }}_T=0$ forces $\mathrm{\parallel }x\mathrm{\parallel }=0$ and $x=0$. The map $J$ is therefore linear and isometric, and its image is $\mathrm{\Gamma }(T)$ with $\mathrm{\parallel }(x,Tx){\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}_T^2$; a linear isometry is injective, so it is a bijection onto $\mathrm{\Gamma }(T)$.

Claim 2. The space $H\oplus H$ is complete with $\mathrm{\parallel }(x,y)\mathrm{\parallel }$ (Complete metric space: every Cauchy sequence converges in the space, Hilbert space), while $\mathrm{\Gamma }(T)$ carries the subspace metric. If $\mathrm{\Gamma }(T)$ is closed, every Cauchy sequence in it converges in $H\oplus H$ and its limit remains in $\mathrm{\Gamma }(T)$, so the graph is complete. Conversely suppose $\mathrm{\Gamma }(T)$ is complete and $q\in \overline{\mathrm{\Gamma }(T)}$. Countable Choice selects $q_n\in \mathrm{\Gamma }(T)$ with $\mathrm{\parallel }{q}_n-q\mathrm{\parallel }<1\mathrm{/}(n+1)$ for every $n\in \mathbb{N}$. Then $(q_n)$ is Cauchy, so it converges to a point of $\mathrm{\Gamma }(T)$; uniqueness of metric limits makes that point $q$. Thus $\mathrm{\Gamma }(T)$ is closed. Since the map of claim 1 is an isometry onto $\mathrm{\Gamma }(T)$, the space $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ is complete exactly when $\mathrm{\Gamma }(T)$ is closed, that is, exactly when $T$ is closed.

Claim 3. If $T$ is closed then $\mathrm{\Gamma }(T)$ is a closed subspace of the Hilbert space $H\oplus H$, hence a Hilbert space in its own right, and the inner product $⟨(x,Tx),(y,Ty)⟩=⟨x,y⟩+⟨Tx,Ty⟩$ transfers to $D(T)$ through the isometry of claim 1, making $D(T)$ a Hilbert space with inner product $⟨x,y{⟩}_T$ and induced norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T$.

Core. The isometry $J:x\mapsto (x,Tx)$ of claim 1 is a homeomorphism from $(D(T),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_T)$ onto $\mathrm{\Gamma }(T)$ and maps $D_0$ onto $\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})$. Hence it carries the closure of $D_0$ onto the closure of $\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})$ inside $\mathrm{\Gamma }(T)$. Therefore $D_0$ is graph-norm dense exactly when $\overline{\mathrm{\Gamma }(T{\mathrm{\mid }}_{D_0})}=\mathrm{\Gamma }(T)$; this topological argument does not replace density by sequential density.