# 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.

• This is Woronowicz's "z transform", right? What source are you following for this material? Feb 4, 2021 at 9:47
• @MatthewDaws I'm looking at Lance's book, which assumes that $D(T)$ is closed under the action of $A$. I think the operator $T(1+T^*T)^{-1/2}$ is the $z$-transform of $T$. I guess what I was speculating in the comment to the question was that even if $T$ is only $A_0$-linear, then it has an $A$-linear $z$-transform, so its functional calculus can be developed. Feb 4, 2021 at 10:10

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.

• Thanks for the answer! I should have included that $t$ is symmetric in the hypotheses. Let me ponder this a little more, but I think you are right that knowing $D(t)$ is only an $A_0$-submodule won't help show $1+t^*t$ has dense range. However, if one happened to know that $1+t^*t$ has dense range, then this should be sufficient to define functional calculus of $t$, even in the $A_0$-submodule case. Here one won't expect to recover $t$ from its $z$-transform like in the usual case. I think the functional calculus formed in this way should just be the functional calculus of $t^{**}$. Feb 4, 2021 at 11:29
• Be careful with the difference between "symmetric" and "self-adjoint". We definitely need self-adjoint here. Feb 4, 2021 at 11:50
• Right, I should have said closed and symmetric, which together with the density-of-range assumption implies self-adjoint. On the other hand, all symmetric operators are closable. Feb 4, 2021 at 12:11