Let $H$ be a (separable complex) Hilbert space and let $A$ and $B$ be two densely-defined, maximally-defined linear operators on $H$ with domains $D(A)$ and $D(B)$ respectively. (By maximall-defined, I mean that $A$ and $B$ admit no extensions). Then, we can define the operator $A+B$ on $D(A)\cap D(B)$, however, in general, the operator $\left( D(A)\cap D(B),A+B\right)$ will not be maximally-defined. The question is: does this operator admit a unique maximal extension?
My conjecture is that the answer is no, but I would absolutely love for the answer to be yes.
Any ideas?
Thanks again!
There are no maximally-defined operators except those defined everywhere. That is, if $D(A)$ is any proper subspace of $H$, you can take any $u \notin D(A)$ and define an extension $\tilde{A}$ on $D(\tilde{A}) = \text{span}(D(A),u)$ by $\tilde{A}(x+cu) = Ax$ for $x \in D(A)$ and scalars $c$.
Perhaps you might be interested in operators related to self-adjoint operators. There I have a little result that you might find interesting. Let $A$ be any self-adjoint unbounded linear operator with purely discrete spectrum. Consider $T = U A$ and $T^* = A U^*$ where $U$ is a unitary operator. Then the set of $U$ for which ${\cal D}(T) \cap {\cal D}(T^*) = \{0\}$ is a dense $G_\delta$ in the unitary operators on $H$.