Functional calculus for "pre-linear" regular operators on a Hilbert module Let $E$ be a Hilbert module over a $C^*$-algebra $A$. Let $T\colon E\to E$ be a densely defined, unbounded $A$-linear operator. (In particular, the initial domain of $T$ is an $A$-submodule of $E$.) If the operators $T\pm i$ each has dense range, then $T$ is an essentially self-adjoint, regular operator on $E$, and there is a continuous functional calculus
$$F_T\colon C_b(\mathbb{R})\to\mathcal{B}_A(E),$$
where $C_b(\mathbb{R})$ denotes the bounded continuous $\mathbb{C}$-valued functions on $\mathbb{R}$, and $\mathcal{B}_A(E)$ denotes the $C^*$-algebra of bounded adjointable $A$-linear operators on $E$.
Question: Suppose $D(T)$ is not an $A$-submodule of $E$, but only an $A_0$-submodule, where $A_0$ is a (metrically) dense $*$-subalgebra of $A$. Then, still assuming that $T\pm i$ each has dense range, is it possible to construct a continuous functional calculus for $T$?
Comment: It seems to me that even though in this situation $T$ is only $A_0$-linear, the operator $1+T^*T$ is injective, and that $(1+T^*T)^{-1}$ extends to an $A$-linear operator. Similarly, it seems possible to make sense of the operator $T(1+T^*T)^{-1/2}$ as being $A$-linear.
 A: I think you need to be very careful about hypotheses.  Looking in Lance's book, we have

Lemma 9.8: Suppose $t:E\rightarrow E$ is densely-defined and self-adjoint.  Then $t$ is regular if and only if the operators $t\pm i$ are surjective.

(Lance's notation $t:E\rightarrow F$ means that $t$ is defined on $D(t)$ a submodule of $E$.)  Notice here that $t$ needs to be self-adjoint, which you don't mention in your question.  I am afraid I cannot find anything in Lance where $t\pm i$ merely have dense range is sufficient.
Suppose we have $t$ where $D(t)$ is only a dense $A_0$-submodule.  What is $t^\ast$?  Well, the definition is
$$ D(t^\ast) = \{ y\in E : \exists\, z\in E, \ (tx|y) = (x|z) \ (x\in D(t) \} $$
with $t^\ast(y) = z$.  Suppose $y\in D(t^\ast)$ with $t^\ast(y) = z$.  For $a\in A$ we have
$$ (x|za) = (x|z)a = (tx|y)a = (tx|ya) \qquad(x\in D(t), $$
just using $A$-linearity of the inner-product.  So $ya\in D(t^\ast)$ and $t^\ast(ya) = za = t^\ast(y)a$.  So $D(t^\ast)$ is an $A$-submodule, and $t^\ast$ is $A$-linear.
Thus, if $t=t^*$, then $D(t)$ is already an $A$-submodule!  We could go back to the definition of $t$ being "regular", but that requires knowing that $1+t^*t$ has dense range.  I suspect in any application, having the weaker hypothesis that $D(t)$ is only an $A_0$-submodule will not be an aid in showing $1+t^*t$ has dense range.
