Closable operator - Knowlpedia

Let $T:\operatorname{Dom}(T)\subseteq H\to K$ be a linear operator between Hilbert spaces; its domain need not be all of $H$ . The operator $T$ is closable if the closure of its graph in $H\oplus K$ is the graph of an operator. That operator is denoted $\overline T$ , called the closure of $T$ , and is the smallest closed operator extending $T$ . Equivalently, whenever $x_n\in\operatorname{Dom}(T)$ , $x_n\to0$ , and $Tx_n\to y$ , one must have $y=0$ . This criterion prevents the graph closure from assigning two different values to the same input.

Characterizations by extensions

An operator is closable exactly when it has at least one closed extension. If $S$ is any closed extension of $T$ , then $\overline T\subseteq S$ , including containment of domains. An operator is already closed precisely when $T=\overline T$ . These statements concern graph closure, not merely closure of $\operatorname{Dom}(T)$ in $H$ .

Adjoint criterion

If $T$ is densely defined, its Hilbert-space adjoint $T^*$ is closed, and

T\text{ is closable}\quad\Longleftrightarrow\quad \operatorname{Dom}(T^*)\text{ is dense in }K.

In that case $\overline T=T^{**}$ . Density of the original domain is needed to define $T^*$ as an operator, but it is not needed for the graph-based definition of closability itself.

Examples and a non-example

The derivative $d/dx$ on $C_c^\infty(0,1)\subset L^2(0,1)$ is closable; its closure has domain $H_0^1(0,1)$ . By contrast, let $T:c_{00}\subset \ell^2\to\ell^2$ satisfy $Te_n=n e_1$ . Then $x_n=e_n/n\to0$ but $Tx_n=e_1$ , so $T$ is not closable. This example shows that a densely defined linear operator need not be closable.

References

Tosio Kato, Perturbation Theory for Linear Operators, 2nd ed., corrected reprint, Springer, 1995. DOI record. Relevant: Chapter III, §5 on closed and closable operators.
Konrad Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Springer, 2012. DOI record. Relevant: Chapter 1 on graphs, closures, and adjoints.