Closed and Closable Operators

Closedness is the condition that an operator respects simultaneous limits of vectors and their images. Closability asks whether the original graph can be completed without assigning two different outputs to one input. These properties justify passing from convenient test domains to operators stable under approximation; neither property alone establishes self-adjointness.

Helpful background. Dense Domains introduces graph norms and operator cores.

Closed graphs encode stable limits

The graph of $A:D(A)\subseteq\mathcal H\to\mathcal K$ is

$$G(A)=\{(\psi,A\psi):\psi\in D(A)\} \subseteq\mathcal H\oplus\mathcal K.$$

The operator is closed when this is a closed subspace. In sequence language, if $\psi_n\to\psi$ and $A\psi_n\to\eta$, then $\psi\in D(A)$ and $A\psi=\eta$.

This definition requires both limits. It does not claim that every convergent input sequence has a convergent image sequence. Unbounded closed operators are discontinuous in the ambient Hilbert norm but behave continuously when their domain is equipped with the graph norm.

In fact, $A$ is closed exactly when $D(A)$ is complete for

$$\|\psi\|_A^2=\|\psi\|^2+\|A\psi\|^2.$$

The map $\psi\mapsto(\psi,A\psi)$ identifies this normed domain isometrically with its graph. A closed subspace of a Hilbert space is complete; conversely, a complete linear subspace is closed. This proves the equivalence.

The domain itself need not be closed in $\mathcal H$. For example, the domain of position is dense and proper, whereas the graph of position is closed.

The zero-sequence criterion for closability

An operator is closable if it has a closed extension. The useful test is

$$\psi_n\to0,\quad A\psi_n\to\eta \quad\Longrightarrow\quad \eta=0.$$

To see why, any closed extension must take the graph limit $(0,\eta)$ to its value at zero, which is zero. Conversely, if this criterion holds, the closure of $G(A)$ contains no nonzero vertical vector $(0,\eta)$. Two putative outputs at one input would differ by such a vector. The closed graph is therefore the graph of a single-valued linear operator, called $\overline A$.

Its domain consists of limits of sequences for which both $\psi_n$ and $A\psi_n$ converge. The operator $\overline A$ is the smallest closed extension: every closed graph containing $G(A)$ contains its closure. The closure is unique, but other closed extensions can still exist.

For a densely defined operator, a fundamental equivalent criterion is that $D(A^*)$ be dense. In that case $\overline A=A^{**}$. The proof uses orthogonal complements of graphs and is explained alongside Adjoints of Unbounded Operators.

Derivatives illustrate closure and extension

Start with $A_0=-i\,d/dx$ on $C_c^\infty(0,1)$ in $L^2(0,1)$. Its graph norm is the $H^1$ norm. By the definition of $H_0^1$ as the closure of compactly supported smooth functions,

$$D(\overline A_0)=H_0^1(0,1), \qquad \overline A_0f=-if'.$$

The derivative is interpreted weakly. Both endpoint traces vanish in this closure.

The operator with the same action on $H^1(0,1)$ is another closed extension, now with no endpoint restriction. Periodic and quasiperiodic momentum operators are closed extensions as well. Closing the original graph does not arbitrarily choose one of these larger domains.

On the whole real line, the same compactly supported seed closes to the derivative on $H^1(\mathbb R)$, where there are no finite endpoint traces to specify. The contrast shows that the ambient interval is part of the mathematical problem.

A densely defined operator that is not closable

Let $\mathcal H=L^2(0,1)$, take $D(A)=C^\infty([0,1])$, and set

$$Af=f(0)\,g$$

for a fixed nonzero $g\in L^2(0,1)$. Smooth representatives make the boundary evaluation meaningful on this domain, which is dense in $L^2$.

Choose a smooth cutoff $\chi$ with $\chi(0)=1$ and support in $[0,1)$, and define $f_n(x)=\chi(nx)$. Then $\|f_n\|_2\to0$ but $Af_n=g$ for every $n$. The zero-sequence criterion fails. No closed extension of this operator exists on $L^2(0,1)$.

Boundary evaluation becomes continuous on $H^1(0,1)$, so changing the ambient norm changes the conclusion. The example is a warning against using pointwise data as if they were continuous functionals on $L^2$.

Exercises

Prove that maximal multiplication by a finite measurable function $a$ is closed on $L^2$.

Solution

Suppose $f_n\to f$ and $af_n\to h$ in $L^2$. After taking a common subsequence, both convergences hold almost everywhere. Pointwise multiplication then gives $h=af$ almost everywhere. Since $h\in L^2$, the vector $f$ belongs to the maximal multiplication domain and its image is $h$.

Let $A$ be closed and $B$ bounded on $\mathcal H$. Show that $A+B$ on $D(A)$ is closed.

Solution

If $\psi_n\to\psi$ and $(A+B)\psi_n\to\eta$, boundedness gives $B\psi_n\to B\psi$. Hence $A\psi_n\to\eta-B\psi$. Closedness of $A$ yields $\psi\in D(A)$ and $A\psi=\eta-B\psi$, proving the claim. Without boundedness or another suitable relative estimate, a sum of closed operators need not be closed.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1, Chapter 1 — closed graphs and adjoint operators.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — closed symmetric operators and differential examples.