Closure of an operator

Remarks

Explicitly, $x \in D(\overline{T})$ if and only if there is a sequence $(x_n)$ in $D(T)$ with $x_n \to x$ such that $(T x_n)$ converges, and then $\overline{T} x = \lim T x_n$. The operator $\overline{T}$ is closed, it extends $T$, and it is contained in every closed extension of $T$.

Example: on $C[0, 1]$ with the supremum norm, let $T$ be differentiation on polynomials. Every $f \in C^1[0, 1]$ is a limit of polynomials $p_n$ with $p_n' \to f'$ uniformly: take polynomials $q_n \to f'$ uniformly and set $p_n(x) = f(0) + \int_0^x q_n$. Combined with the closedness of differentiation on $C^1[0, 1]$, this gives $D(\overline{T}) = C^1[0, 1]$ and $\overline{T} f = f'$.

Non-example: an operator that is not closable has no closure, because the closure of its graph contains a pair $(0, y)$ with $y \ne 0$ and is not a graph.

A closed operator is its own closure. A subspace $D \subseteq D(\overline{T})$ with $\overline{T|_D} = \overline{T}$ is called a core of $\overline{T}$.