From the previous section we obtain two candidates for convergence type to impose on the \(A_n\)’s in (1): Strong graph convergence and strong convergence of graph projections. That these are actually both natural choices is illuminated by Theorem 8 which states that for sequences of self-adjoint operators each of them is equivalent to strong resolvent convergence – exactly the convergence type we seek! Throughout this section we use the following conventions: Let \(\_^\infty \), \(\_^\infty \), \(\_^\infty \), A, B and U be as in the beginning of Sect. 1. Let \(\Gamma _\infty \) be the strong limit of \(\}(A_n)\}_^\infty \) and \(\Gamma _\infty ^*\) the strong limit of \(\}(A_n^*)\}_^\infty \). Denote by \(P_n\) and P the orthogonal projections in \(}\oplus }\) onto \(}(A_n)\) and \(}(A)\) respectively.

The answer to the question closing Sect. 2 is straightforward and given below in Corollary 9, and in applications it can be useful even in this raw form.

Corollary 9

(to Theorem 8). Consider operators \(A_n\), A, \(B_n\) and B as described in Sect. 1 and suppose \(}(A)\subseteq \Gamma _\infty \). If moreover for every pair (or, equivalently, for k linearly independent pairs) \((\phi ,B\phi )\) from the orthogonal complement of \(}(A)\) inside the Hilbert space \(}(B)\) there exists a sequence \(\_^\infty \subseteq }\) so that \(\phi _n\in D(B_n)\) for all n and \((\phi _n,B_n\phi _n)\rightarrow (\phi ,B\phi )\) then \(B_n\rightarrow B\) in the strong resolvent sense.

Proof

Denoting the strong limit of \(\}(B_n)\}_^\infty \) by \(\Gamma _\infty ^B\) it is basically the assumption above that the orthogonal complement of \(}(A)\) inside \(}(B)\) is contained in \(\Gamma _\infty ^B\). Moreover, \(}(A)\subseteq \Gamma _\infty \subseteq \Gamma _\infty ^B\) since all the \(B_n\)’s are extensions of the \(A_n\)’s. This concludes the proof. For the fact that it suffices to consider k linearly independent pairs, see the first couple of lines of the proof of Theorem 12. \(\square \)

For the remaining part of this section we formulate and prove results like (1) with the different notions of convergence of the \(A_n\)’s introduced above. For this it will be essential to have at our disposal the following characterization of strong convergence of the \(U_n\)’s defining the self-adjoint extensions of the \(A_n\)’s.

Lemma 10

Consider the \(U_n\)’s and the U described in Sect. 1. We have \(U_n\rightarrow U\) strongly if and only if the following statement is true:

$$\begin \text \text \phi _+\in }_+ \text \text \text \text \_^\infty \subseteq } \text \text \nonumber \\ \phi _+^n\in }_+^n \text \text n \text (\phi _+^n,U_n\phi _+^n)\rightarrow (\phi _+,U\phi _+). \end$$

(3)

Note that the condition (3) actually says that the strong limit of the graphs of the \(U_n\)’s considered as operators only on \(}_+^n\) contains the corresponding graph of U.

Proof

(of Lemma 10). Observe firstly that if \(\psi _n\rightarrow \psi \) then the inequalities

$$\begin \Vert U_n\psi _n-U\psi \Vert \le \Vert \psi _n-\psi \Vert +\Vert U_n\psi -U\psi \Vert \le 2\Vert \psi _n-\psi \Vert +\Vert U_n\psi _n-U\psi \Vert \end$$

show that

$$\begin U_n\psi _n\rightarrow U\psi \;\;\; \text \;\;\; U_n\psi \rightarrow U\psi . \end$$

(4)

For each n denote by \(P_n\) the orthogonal projection onto \(}_+^n\) and by P the orthogonal projection onto \(}_+\). Assume \(U_n\rightarrow U\) strongly. Then, for any \(\phi _+\in }_+\) and \(\psi \in }\), we have

$$\begin \langle P_n\phi _+,\psi \rangle =\langle U_n^*U_n\phi _+,\psi \rangle =\langle U_n\phi _+,U_n\psi \rangle \longrightarrow \langle U\phi _+,U\psi \rangle =\langle P\phi _+,\psi \rangle =\langle \phi _+,\psi \rangle \end$$

so that \(P_n\phi _+\rightarrow \phi _+\) weakly in \(}\). As further

$$\begin \Vert \phi _+ \Vert \le \liminf _\Vert P_n\phi _+ \Vert \le \limsup _\Vert P_n\phi _+ \Vert \le \Vert \phi _+ \Vert \end$$

by lower semi-continuity of the norm it is apparent that additionally \(\Vert P_n\phi _+ \Vert \rightarrow \Vert \phi _+ \Vert \) and consequently \(P_n\phi _+\rightarrow \phi _+\) with respect to the Hilbert space norm. We claim that letting \(\_^\infty :=\_^\infty \) for each \(\phi _+\in }_+\) verifies (3). Indeed, since then \(\phi _+^n\rightarrow \phi _+\), the strong convergence \(U_n\rightarrow U\) and (4) yield the desired conclusion.

Suppose now that (3) is satisfied. For any \(\phi _+\in }_+\) we can choose the sequence from this condition and (4) implies that \(U_n\phi _+\rightarrow U\phi _+\). For proving convergence of \(U_n\psi \) for \(\psi \in [}_+]^\perp \) fix such \(\psi \) and consider an orthonormal basis \(\,\dots ,\phi _\}\) of \(}_+\). By (3) there exist sequences \(\^n\}_^\infty ,\dots ,\^n\}_^\infty \subseteq }\) with \(\phi _^n\in }_+^n\) for all n and \(\ell =1,\dots ,k\) and \(\phi _^n\rightarrow \phi _\) for all \(\ell \). Now by applying the Gram-Schmidt process to \(\^n,\dots ,\phi _^n\}\) for each n we obtain new sequences \(\_^n\}_^\infty ,\dots ,\_^n\}_^\infty \subseteq }\) with \(\_^n,\dots ,\widetilde_^n\}\) an orthonormal basis of \(}_+^n\) for sufficiently large n. Induction in \(\ell \) shows that also \(\widetilde_^n\rightarrow \phi _\) for all \(\ell \). Consequently,

$$\begin[}_+^n]^\perp \ni \psi _n:=\psi -\sum _^\langle \widetilde_^n\;,\psi \rangle \widetilde_^n\longrightarrow \psi \end$$

and, since \(U_n\psi _n=0=U\psi \) for large n, a final application of (4) proves that \(U_n\psi \rightarrow U\psi \). \(\square \)

Remark 11

Lemma 10 is actually the main reason why we assume that the deficiency indices of the \(A_n\)’s are finite, since then we can simply restate the condition (3) as strong convergence of the \(U_n\)’s – which is exactly the kind of formulation we seek. If one, in the case of infinite deficiency indices, replaces “\(U_n\rightarrow U\) strongly” with (3) then the remaining results of this note in Theorem 12 and Corollaries 14 and 15 indeed remain valid. One can realize that these conditions are truly different in the infinite case by taking the \(U_n\)’s and the U to be projections and recalling the content of Remark 4.

While this description of strong convergence of the \(U_n\)’s not at first sight simplifies things, the fact that it is so closely related to the definition of strong graph convergence will help us apply our theory from Sect. 2 via Theorem 8. With this, we are now in a position to state and prove the main theoretical statement of this note.

Theorem 12

Let \(\_^\infty \), \(\_^\infty \), \(\_^\infty \), A, B and U be as in the beginning of Sect. 1. Let \(\Gamma _\infty \) be the strong limit of \(\}(A_n)\}_^\infty \), and denote by \(P_n\) and P the orthogonal projections in \(}\oplus }\) onto \(}(A_n)\) and \(}(A)\) respectively. Then the following holds:

(a)

If \(}(A)\subseteq \Gamma _\infty \) and \(U_n\rightarrow U\) strongly then \(B_n\rightarrow B\) in the strong resolvent sense and \(P_n\rightarrow P\) strongly.

(b)

If \(B_n\rightarrow B\) in the strong resolvent sense and \(P_n\rightarrow P\) strongly then \(U_n\rightarrow U\) strongly.

Proof

(a): Recall that, cf. [1] X.1,

$$\begin }(B)=}(A)\oplus \}_+\}, \end$$

where the sum is orthogonal, from which the k-dimensional orthogonal complement of \(}(A)\) in \(}(B)\) is apparent. Applying Lemma 10 we can for any \(\phi _+\in }_+\) find \(\_^\infty \subseteq }\) so that \(\phi _+^n\in }_+^n\) for all n and

$$\begin (\phi _+^n+U_n\phi _+^n,i\phi _+^n-iU_n\phi _+^n)\longrightarrow (\phi _++U\phi _+,i\phi _+-iU\phi _+). \end$$

Hence, Corollary 9 implies \(B_n\rightarrow B\) in the strong resolvent sense. Likewise we have also (cf. [1] X.1)

$$\begin }(A^*)=}(A)\oplus \}_+\}\oplus \}_+\} \end$$

and a similar application of Lemma 10 tells us that \(}(A^*)\subseteq \Gamma _\infty ^*\) (the strong limit of \(\}(A_n^*)\}_^\infty \)). By invoking Proposition 7 we get thus additionally \(P_n\rightarrow P\) strongly.

(b): We note that by Theorem 8 (and using the notation herein) we have \(Q_n\rightarrow Q\) strongly, and consequently \(Q_n-P_n\rightarrow Q-P\) strongly. Now \(Q_n-P_n\) is the orthogonal projection onto the orthogonal complement of \(}(A_n)\) inside \(}(B_n)\) and similarly for \(Q-P\). But we have just seen in the proof of (a) that these are exactly

$$\begin & \}_+^n\}\\ & \qquad \text \quad \}_+\} \end$$

respectively. Hence, Lemma 3(b) tells us that for each \(\phi _+\in }_+\) there exists a sequence \(\_^\infty \subseteq }\) so that \(\phi _+^n\in }_+^n\) for all n and

$$\begin (\phi _+^n+U_n\phi _+^n,i\phi _+^n-iU_n\phi _+^n)\longrightarrow (\phi _++U\phi _+,i\phi _+-iU\phi _+). \end$$

By taking linear combinations of the entries we see that for this sequence \(\phi _+^n\rightarrow \phi _+\) and \(U_n\phi _+^n\rightarrow U\phi _+\), and to wrap things up Lemma 10 yields the claimed strong convergence of the \(U_n\)’s towards U as needed. \(\square \)

Remark 13

We present here a more transparent way of proving \(B_n\rightarrow B\) in the strong resolvent sense in Theorem 12(a) than the one presented above which in particular avoids the use of Corollary 9 and hence of Theorem 8.

Define the subspace \(V:=\}_+\}\) in \(}\) and write

$$\begin }=R(B+i)=R(A+i)+R(B\vert _V+i). \end$$

Now since we assume \(}(A)\subseteq \Gamma _\infty \) the convergence of \((B_n+i)^\) towards \((B+i)^\) on \(R(A+i)\) is proved as in (iii)\(\Rightarrow \)(i) in Theorem 8. Notice then that

$$\begin (B+i)(\phi _++U\phi _+)=2i\phi _+\qquad \text \qquad (B_n+i)(\phi _+^n+U_n\phi _+^n)=2i\phi _+^n \end$$

for any \(\phi _+\in }_+\) and \(\phi _+^n\in }_+^n\). This proves that \(R(B\vert _V+i)=}_+\), and for each \(\phi _+\in }_+\) we can use Lemma 10 to find a sequence \(\_^\infty \subseteq }\) so that \(\phi _+^n\in }_+^n\) for each n and

$$\begin \Vert (B_n+i)^\phi _+-(B+&i)^\phi _+ \Vert \\&\le \Vert (B_n+i)^\phi _+-(B_n+i)^\phi _+^n \Vert \\&\quad \quad +\Vert (B_n+i)^\phi _+^n-(B+i)^\phi _+ \Vert \\&\le \Vert \phi _+-\phi _+^n \Vert +\frac\Vert (\phi _+^n+U_n\phi _+^n)-(\phi _++U\phi _+) \Vert \longrightarrow 0. \end$$

We can now use Theorem 12 to prove various statements of the form (1). Taking \(A_n\rightarrow A\) to be in terms of strong convergence of the orthogonal projections onto the graphs, i.e. \(P_n\rightarrow P\) strongly, we have also \(}(A)\subseteq \Gamma _\infty \) due to Proposition 7, and thus we get a particularly clean statement.

Corollary 14

Consider the set-up in Theorem 12 and suppose \(P_n\rightarrow P\) strongly. Then \(B_n\rightarrow B\) in the strong resolvent sense if and only if \(U_n\rightarrow U\) strongly.\(\square \)

The downside of Corollary 14 is, however, that the condition \(P_n\rightarrow P\) strongly is often not easy to verify in concrete cases. Another approach is to assume the convergence of the \(A_n\)’s only in the sense that \(}(A)\subseteq \Gamma _\infty \). We note that this is a strictly weaker notion of convergence than strong convergence of the graph projections, so one cannot expect the implications of this assumption to be as strong as the equivalence between strong convergence of the \(B_n\)’s and of the \(U_n\)’s in Corollary 14. Another application of Proposition 7 yields:

Corollary 15

Consider the set-up in Theorem 12 and suppose \(}(A)\subseteq \Gamma _\infty \). Then \(U_n\rightarrow U\) strongly if and only if both \(B_n\rightarrow B\) in the strong resolvent sense and \(}(A^*)\subseteq \Gamma _\infty ^*\).\(\square \)

An obvious question now arises: Is this the best we can do? In particular we can in the light of Corollary 14 ask whether the condition \(}(A^*)\subseteq \Gamma _\infty ^*\) in Corollary 15 is actually needed. As a matter of fact it is by the following observations.

Remark 16

We do not in general have the result “Suppose \(}(A)\subseteq \Gamma _\infty \). Then \(U_n\rightarrow U\) strongly if and only if \(B_n\rightarrow B\) in the strong resolvent sense.” as the example below shows. Even changing \(}(A)\subseteq \Gamma _\infty \) to \(A=}A_n\) does not make the statement true. The backbone of the example is the extension theory for a well-studied class of operators on \(L^2(}^3)=}\). This is treated in for example [3] I.1.1 to which we refer for the details.

Let \(\_^\infty \subseteq }^3\) be a sequence yet to be specified and define for each n the operator \(A_n\) to be the closure of \(-\Delta \) on \(C_c^\infty (}^3\backslash \)\). One can now find the deficiency subspaces

$$\begin }_^n=}\phi _\pm ^n,\qquad \phi _\pm ^n(x)=\frac|x-y_n |}} \end$$

where \(\Im \sqrt>0\). Moreover, if one defines a self-adjoint extension \(B_n\) of \(A_n\) by the unitary map \(U_n:}_+^n\ni \phi _+^n\mapsto -\phi _-^n\in }_-^n\) as in Sect. 1 then \(B_n=B\) is actually the free Laplacian \(-\Delta \) defined on the Sobolev space \(H^2(}^3)\) independently of n. Now we have the orthogonal decomposition

$$\begin }(B)=}(A_n)\oplus }(\phi _+^n-\phi _-^n,i\phi _+^n+i\phi _-^n)=:}(A_n)\oplus }v_n, \end$$

and consequently \(}(A_n)\) is the orthogonal complement of \(}v_n\) in \(}(B)\) for each n. Notice now that the \(v_n\)’s depend only on the \(y_n\)’s. Choosing \(y_n\) so that \(|y_n |\rightarrow \infty \) it is not difficult to realize that the sequences \(\_^\infty \) converge weakly towards 0 in \(L^2(}^3)\): This follows from the fact that they are translations of a fixed \(L^2\)-function. With such sequence of \(y_n\)’s we get thus

$$\begin \langle (\phi ,\psi ),v_n \rangle =\langle \phi ,\phi _+^n \rangle -\langle \phi ,\phi _-^n \rangle +i\langle \psi ,\phi _+^n \rangle +i\langle \psi ,\phi _-^n \rangle \longrightarrow 0 \end$$

for all \((\phi ,\psi )\in }\oplus }\), i.e. \(v_n\rightarrow 0\) weakly in \(}\oplus }\) and hence in \(}(B)\).

We observe from the above facts that by choosing a sequence of \(y_n\)’s which is a fixed \(y_n=y_0\) for n odd and with \(\\}_^\infty \) unbounded we can make the sequence \(\}(A_n)\}_^\infty \) of subspaces of the Hilbert space \(}(B)\) into a sequence like \(\_^\infty \) in Remark 4. Consequently, the operator \(A=A_1\) is the strong graph limit of the \(A_n\)’s (and of course \(B_n\rightarrow B\)), but the orthogonal projections onto the graphs \(}(A_n)\) do not converge strongly towards the orthogonal projection onto \(}(A)\), and hence Theorem 12(a) tells us that we cannot have \(U_n\rightarrow U\) strongly. Alternatively this can be checked more directly by using Lemma 10.

We conclude by proving a simple requirement for having \(}(A)\subseteq \Gamma _\infty \), thus providing a procedure for checking the assumptions in Corollary 15. Recall that a core for a closed operator A is a subspace of D(A) satisfying that the restriction of A to this has closure A. We obtain now:

Proposition 17

Assume that \(}\) is a common core for A and all \(A_n\)’s. If \(A_n\phi \rightarrow A\phi \) for all \(\phi \in }\) then \(}(A)\subseteq \Gamma _\infty \).

Proof

The assumption tells us that \(\Gamma _}:=\}\}\subseteq \Gamma _\infty \). Thus, if we argue that \(\Gamma _\infty \) is closed, we have also \(}(A)=\overline}}\subseteq \Gamma _\infty \). But closedness is a general property of any strong limit of subspaces by the following argument:

Let \(\_^\infty \) be any sequence of subspaces of a Hilbert space \(}\) and denote as usual its strong limit by \(V_\infty \). If we consider an arbitrary convergent sequence \(\_^\infty \subseteq V_\infty \) with limit \(x_0\) then we need only to find a sequence \(\}_n\}_^\infty \subseteq }\) with \(}_n\in V_n\) for all n such that \(}_n\rightarrow x_0\) in order to obtain \(x_0\in V_\infty \) and hence prove that \(V_\infty \) is closed. We now construct such sequence. Firstly we choose for each k a sequence \(\_^\infty \) with \(x_n^k\in V_n\) for all n and \(x_n^k\rightarrow x_k\), and then we take natural numbers \(N_1<N_2<N_3<\cdots \) so that \(\Vert x_n^k-x_k \Vert <1/k\) for all \(n\ge N_k\). Defining \(}_n:=x_n^1\) for \(n=1,2,\dots ,N_2-1\); \(}_n:=x_n^2\) for \(n=N_2,\dots ,N_3-1\) and generally \(}_n:=x_n^k\) for \(n=N_k,\dots ,N_-1\) one can check using the triangular inequality that this is indeed a sequence with the properties we seek. \(\square \)

Example 18

To make things even more concrete than requiring pointwise convergence of the \(A_n\)’s on a common core, we can ask what this means for differential operators like those in Example 1. To simplify things let us consider a sequence of Schrödinger operators – that is, the \(A_n\)’s are the closures of \(-\Delta +\Phi _n\) defined on \(C_c^\infty (\Omega )\subseteq L^2(\Omega )\) for some open set \(\Omega \subseteq }^d\) and some potentials \(\Phi _n\) (say, real-valued and continuous) on this set. Hence, \(C_c^\infty (\Omega )\) is a common core for the \(A_n\)’s and also for \(A=-\Delta +\Phi \) if we define this in the same manner. Now, for any \(\phi \in C_c^\infty (\Omega )\),

$$\begin \Vert A_n\phi -A\phi \Vert ^2 & =\Vert \Phi _n\phi -\Phi \phi \Vert ^2\nonumber \\ & =\int _\Omega |\phi |^2|\Phi _n-\Phi |^2\,dx\le \Vert \phi \Vert _\infty ^2\int _}\phi }|\Phi _n-\Phi |^2\,dx \end$$

where \(\Vert \,\cdot \, \Vert _\infty \) is the supremum norm. Now if \(\Phi _n\rightarrow \Phi \) in \(L^2_}}(\Omega )\) then we conclude that \(A_n\phi \rightarrow A\phi \) for all \(\phi \in C_c^\infty (\Omega )\). If, on the other hand, we assume the latter, then we see that \(\Phi _n\rightarrow \Phi \) in \(L^2(K)\) for any compact subset \(K\subseteq \Omega \) by choosing \(\phi \equiv 1\) on K, i.e. we get \(\Phi _n\rightarrow \Phi \) in \(L^2_}}(\Omega )\). Being able to consider only local \(L^2\)-convergence is often desirable if one deals for example with potentials with singularities.