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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!

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Well, $D(A) \cap D(B)$ can be almost empty, no? What do you want to do with it? –  Jonas Teuwen Jul 25 '11 at 22:45
    
Well, that has been the route I've been taking to try to find a counterexample, that is, try to make $D(A)\cap D(B)$ as small as possible, but no luck yet. What exactly did you have in mind? –  Jonathan Gleason Jul 25 '11 at 22:58
    
In case you're wondering why I'm interested, I am trying to turn the collection of all densely-defined linear operators on $H$ into a $^*$-algebra, where two operators are considered equal iff they have the same domain and agree on that domain. I then added the condition of being maximally-defined so that we have equality with $A^{**}=A$. But then, I just realized, it is not obvious (and probably not true) that addition is well-defined, hence the question. –  Jonathan Gleason Jul 25 '11 at 23:02
    
I will think about an example in the morning if nobody else has given one, I'm too tired now. –  Jonas Teuwen Jul 25 '11 at 23:11
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If you add the requirement of self-adjointness (so a symmetric operator must have the same domain as its adjoint to be "self-adjoint"), you can avoid the particular issue of having no maximally-defined operators, as in Robert I's answer below. In general, there is not a unique self-adjoint extension of a symmetric operator, of course, and compositions of self-adjoint are not necessarily self-adjoint in any case... but this notion may clarify some technicalities for you. –  paul garrett Jul 25 '11 at 23:46
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1 Answer 1

up vote 5 down vote accepted

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 $H$.

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Thank you. That was rather stupid of me, haha. –  Jonathan Gleason Jul 25 '11 at 23:36
    
@Robert: do you have a reference to your result? Thanks in advance. –  Willie Wong Jul 25 '11 at 23:48
    
Actually I should have said, contains a dense $G_\delta$. See "Some Generic Results in Mathematical Physics", Markov Processes and Related Fields 10 (2004), 517-521. –  Robert Israel Jul 26 '11 at 4:45
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