# When the adjoint of an unbounded operator on a Hilbert space coincides with the formal adjoint on its natural domain?

I am trying to work with infinite matrices in a Hilbert space. I want to consider these as unbounded operators, but I have some troubles understanding how the domain of the adjoint operator is defined in this case.

Namely, suppose we have a closed and densely defined operator $$A$$ with a domain $$D(A)$$ which is a subspace of a Hilbert space $$\mathcal{H}$$. Let $$\mathcal{H}$$ have an orthonormal basis $$\{e_n\}_{n=1}^\infty$$. Suppose $$\{e_n\} \in D(A)$$. Then for the operator $$A$$ there exists an infinite matrix $$A_{ij} = \{\langle Ae_j, e_i\rangle\}_{ij}$$.

We know that there is a usual procedure to define $$A^*$$ with its domain $$D(A^*)$$. Suppose $$\{e_n\} \in D(A^*)$$. Now consider the formal adjoint operator $$A_* = \{\overline{A_{ji}}\}$$ with the domain $$D(A_*)$$ consisting of those $$\zeta$$ such that $$\eta_j = \sum {\overline{A_{ji}}} \zeta_i$$ is in $$\ell^2$$. Are there some simple conditions on $$A$$ and on $$D(A)$$ for these domains to coincide: $$D(A^*) = D(A_*)$$?

What can be said on this matter if $$A_{ij}$$ is a finite-band matrix? Or when $$A$$ is formally self-adjoint ($$A_{ij} = \overline{A_{ji}})$$?

• What is the domain of $A_*$? The domain of $A^*$ is defined to be the set of all $v$ for which the map $x\mapsto \langle Ax,v\rangle$ is continuous on the domain of $A$. So what is the Definition of the domain of $A_*$?
– user130903
Jan 8, 2021 at 9:33
• One would guess that $A_*$ is those $\xi = \sum_n \xi_n e_n \in H$ such that the matrix $(\overline{A_{ji}})$ acts on $\xi$, that is, $\sum_i \overline{A_{ji}} \xi_i$ converges for each $j$, and the resulting vector $\eta_j = \sum_i \overline{A_{ji}} \xi_i$ is in $\ell^2$. However, it would be good if the questioner could clarify this in the question. Jan 8, 2021 at 11:53
• One general comment: surely we need $e_n$ to be in the domain of $A$, for each $n$. As $A$ is densely defined, we can always find an orthonormal basis in the domain; but this is an additional assumption. Jan 8, 2021 at 12:54
• While the OP is asking about the dual operator, it might be worthwhile to point out that even the operator $A$ itself need, in general, not coincinde with the operator given by the matrix $(A_{ij})$ on its maximal domain (even if $A$ is closed). Thus, the claim that "$A$ can be viewed as an infinite matrix" is a bit too strong in this generality, even if all $e_n$ are in the domain of $A$ (which is, as mentioned by @MatthewDaws, needed for the matrix $(A_{ij})$ to well-defined) Jan 8, 2021 at 14:12
• Is it clear that the domain $D(A_*)$ is well defined independent of the choice of basis? Jan 10, 2021 at 1:30

Even after addressing the issues raised in the comments, the matrix coefficients $$A_{ij}$$ don't give you enough information to find $$D(A^*)$$. For example, consider $$A_j=-d^2/dx^2$$ on $$L^2(0,1)$$, $$j=1,2$$. More precisely, I take $$A_1$$ as the closure of $$-d^2/dx^2$$ on $$C_0^{\infty}$$ and $$A_2$$ as the self-adjoint operator on $$H^2$$ with Dirichlet boundary conditions. Then $$A_1$$ is only symmetric, not self-adjoint, so $$A_1^*\supsetneqq A_2^*$$.
However, both operators have the same matrix with respect to any ONB $$e_n\in C_0^{\infty}$$.
• This example is useful, thanks. I understand that there is a freedom in defining the unbounded operator even when we fix its values on some dense subset. Accordingly, the domain of adjoint varies even for the same matrix. What I am trying to find out is whether there are some simple conditions when $A$ is defined on its natural domain and its formal adjoint $A_*$ is also defined on its natural domain. I wonder when these two are adjoint (maybe it is known for simpler cases of finite-band $A_{ij}$). Probably, the original statement of the question is not the most accurate. Jan 10, 2021 at 0:44
• @apyshkin: Finite band matrices is the same thing, you can build the same type of example with a limit circle Jacobi matrix (= tridiagonal). I'm not completely sure if I understand your other comments correctly, but one remark is that there need not be an obvious "natural domain" for given matrix elements. Of course, if we do have the domain, then everything is determined and we can (in principle, at least) find $A^*$. Jan 10, 2021 at 19:35