Extending a pre-Hilbert module operator over a pre-$C^*$-algebra to a Hilbert module over the completion Let $A_0$ be a pre-$C^*$-algebra, i.e. $A_0$ satisfies all the criteria of a $C^*$-algebra except that it need not be complete. Then we can view $A_0$ as an inner product module over itself, i.e. a $\mathbb{C}$-vector space that is a right $A_0$-module together with an $A_0$-valued inner product, $\langle\,,\,\rangle$, defined by
$$\langle a,b\rangle=a^*b$$
for all $a,b\in A_0$. This structure satisfies the following conditions:

*

*$\lambda(ab)=(\lambda a)b=a(\lambda b)$;

*$\langle a,\lambda b+\mu c\rangle=\lambda\langle a,b\rangle+\mu\langle a,c\rangle$;

*$\langle a,bc\rangle=\langle a,b\rangle c$;

*$\langle b,a\rangle=\langle a,b\rangle^*$;

*$\langle a,a\rangle\geq 0$, and if $\langle a,a\rangle=0$ then $a=0$,

for all $\lambda,\mu\in\mathbb{C}$ and $a,b,c\in A_0$.
Now suppose $T$ is a $\mathbb{C}$-linear and $A_0$-linear operator on the inner product module $A_0$. Let $A$ be the $C^*$-algebra completion of $A_0$.
Question: Can $T$ be extended to a densely defined (unbounded) $A$-linear operator on the right Hilbert-$A$-module $A$?
Remark: In general, an unbounded operator $T$ on a Hilbert $A$-module needs to be defined on a dense domain $D(T)$, where $D(T)$ is closed under the action of $A$. In this case, the naive choice $D(T)=A_0$ doesn't work because it is not quite an $A$-module. As a result, it doesn't make sense to say that $T$ is an $A$-linear operator.
I was thinking of getting around this by defining $D(T)=A_0\cdot A^+$, where $A^+$ is the unitization of $A$. This choice of $D(T)$ is dense $A$ and is naturally a right $A$-module, and the action of $T$ on $A_0\cdot A^+$ would then be defined by:
$$T(ab)=\lim_n (Ta)b_n=(Ta)b,$$
where $a\in A_0$ and $b_n$ is a sequence in $A_0$ converging to $b\in A$.
However, it isn't a priori clear that if this is well-defined: one needs to show that if $ab=a'b'$ for some $a'\in A_0$ and $b'\in A$, then $T(ab)=T(a'b')$.
Remark: Perhaps there is standard technique to do this? However, it seems that all of the examples of unbounded operators on Hilbert modules have been of the type where $D(T)$ is already closed under the action of $A$ (e.g. in the setting when the Hilbert module is a Hilbert space and $T$ is an unbounded $\mathbb{C}$-linear operator).
 A: One of the reasons why the theory of bounded operators on Hilbert's space is so advanced is perhaps
because every bounded operator has an adjoint.  The same cannot be said about unbounded operators, and
this is partly the reason why their theory is so hard to develop.  That is, except for self-adjoint
unbounded operators, which, to a large extent, may be studied with a similar level of success as compared
to their bounded cousins.
In other words, the existence of the adjoint is what makes studying operators possible.
Regarding operators on Hilbert modules, even bounded ones sometimes fail to possess an adjoint.  So,
virtually all authors consider only the adjointable ones, namely operators which admit an adjoint.  For
example, if $E$ and $F$ are Hilbert modules over the same C*-algebra, the universally adopted notation
$\mathcal L(E,F)$ refers to the set of all bounded adjointable operators, while there appears to be no
consensus on a notation for the rarely studied (if at all) set of all bounded opertors.
Thus, it is perhaps sensible to assume that the operator $T$ in the question presents a minimum of
adjointness, such as the existence of an opertor $T^*$, defined on a dense subset $D$ (which may or may
not be contained in $A_0$), and such that
$$
  \langle T(a),d\rangle  =   \langle a,T^*(d)\rangle , \quad \forall a\in  A_0,\quad \forall d\in  D.
  $$
With this much,  it is possible to give an affirmative answer,  namely  if $a, a'\in  A_0$,  and $b, b'\in  A$ are
such that $ab=a'b'$,  then $T(a)b = T(a')b'$.
To see why,  choose sequences $\{b_n\}_n$ and $\{b'_n\}_n$ in
$A_0$,  respectively converging to $b$ and $b'$.  Then,  for every $d\in  D$,  one has
$$
  \langle T(a)b,d\rangle  =
  \lim_{n\to\infty } \langle T(a)b_n,d\rangle  =
  \lim_{n\to\infty } \langle T(ab_n),d\rangle  = $$$$ =
  \lim_{n\to\infty } \langle ab_n,T^*(d)\rangle  =
  \langle ab,T^*(d)\rangle  =
  \langle a'b',T^*(d)\rangle  = $$$$ =
  \lim_{n\to\infty } \langle a'b'_n,T^*(d)\rangle  =
  \lim_{n\to\infty } \langle T(a'b'_n),d\rangle  = $$$$ =
  \lim_{n\to\infty } \langle T(a')b'_n,d\rangle  =
  \langle T(a')b',d\rangle .
  $$
Since $D$  is dense, it follows that $T(a)b=T(a')b'$.
