I'm looking at properties of the scale of Hilbert spaces $(X\_s)\_{s\in \mathbb{R}}$, which are constructed as follows. Starting with $A:D(A)\subset H\to H$, $A$ a densely defined, strictly positive ($A \geq \gamma I$, with $\gamma>0$), self-adjoint operator on $H$, let
$$
M = \bigcap\_{s\in \mathbb{R}}D(A^s),
$$
where the operator $A^s$ is defined via the spectral measure. It can be shown this space is dense in $H$.


The spaces $X\_s$ are then defined as the closures of $M$ with respect to the inner product (and norm)
$$
\langle f,g\rangle\_s=\langle Af,Ag\rangle.
$$


Based on the reference I am using (and intuition), I would expect that for $s<t$, $X\_t \subset X\_s$, and is even dense in it. Here is my question: I can see how to obtain this inclusion for $0\leq s<t$ by using the closedness of the operators $A^s$ and $A^t$. 


The argument is that since 
$$\|x\|\_s\leq C(s,t)\|x\|\_t$$
for $x \in M$, Cauchy sequences in $X\_t$ are Cauchy in $X\_s$ **and** $H$. Being Cauchy in $X\_t$, with limit $x$, means that $A^t x\_n$ is Cauchy in $H$, thus, we have a closed operator $A^t:D(A^t)\to H$, with a Cauchy sequence $x\_n$ in $H$ for which $A^t x\_n$ is also Cauchy. Thus the limit, $x'\in D(A^t)\subset H$, satisfies $A^t x' = y\in H$ . Consequently, we can identify the limit in the $X\_t$ norm, $x$, with the limit in the $H$ norm $x$, and the limit in the $X\_s$ norm, $x''$, with the same limit in the $H$ norm.


It's then fairly clear how to handle the case $s<0\leq t$.


**Question:** Now, once we switch to negative index spaces, $s<t<0$, how can I handle this? I can no longer rely on the underlying $H$ norm to control things. I suspect this can be done by duality, but I wonder if there is a simpler answer. The reason I wonder if there is a simpler answer is that the reference I am working from (Engl, Hanke, and Neubauer), gives almost no detail for this result (Proposition 8.19(i)).



Hilbert Scale Inclusions
The easiest (if, perhaps, not most elementary) way to do this, is to use the spectral theorem in the form that every such operator is representable as multiplication by a positive (unbounded) measurable function on an $L^2$-spaces. The $X\_s$ are then weighted $L^2$-spaces and the results become quite transparent. In many applications, the operator has discrete spectrum and so is represented by multiplication by a sequence—-in this case, one gets weighted $\ell^2$-spaces, even more transparent.


Edit: in view of the comments below, I would like to flesh out my answer. In order to simplify it, I will assume that the Hilbert space is separable but this is not necessary. One version of the spectral theorem states that the operator is unitarily eqivalent to a multiplication operator $x\mapsto yx$ on some $L^2(\mu)$-space where $y$ is measurable with $y\geq \gamma$. The domain of $A$ is the space of functions $x$ with $\int y^2|x|^2$ finite, i.e., $L^2(y^2 \mu)$, a weighted $L^2$-space as I mentioned above. Then the scale of Hilbert spaces is just the family $\{L^2(y^{2\alpha}\mu)\}$ for $\alpha$ real and the density and injectivity properties you mention follow immediately.


