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)).