Hilbert Scale Inclusions

Question

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)).

Answer 1

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.

Answer 2

The injectivity of $H^t\to H^s$ for $s\le t$ seems a bit special to Hilbert spaces. Without mentioning a spectral theorem, and for $0\le s\le t\in \mathbb Z$, we can give a more elementary argument for the injectivity.

Let $D$ be the (dense) domain of a symmetric (I think self-adjointness is not quite necessary, though for invoking a spectral theorem it would be) operator $T$ on a Hilbert space $V$ with $T\ge c\cdot 1$ for $c>0$. Define Sobolev-like norms $|v|_s=\langle T^s v,v\rangle$ for non-negative integer $s$, and let $H^s$ be the completion of $D$ with respect to $|\cdot|_s$. We claim that $H^{s+1}\to H^s$ is injective. It suffices to treat $s=1$.

Let $j:H^1\to H^0$ be the natural continuous linear map. Let $0\not=v\in H^1$. By density of $D$ in $H^1$, there is $w\in D$ such that $\langle v,w\rangle_1\not=0$. (Here we use the fact that there is an inner product $\langle,\rangle_1$ giving the norm $|\cdot|_1$.) Then $$ 0\not= \langle v,w\rangle_1 \;=\; \langle jv,Tw\rangle $$ Thus, $jv\not=0$.

Also, I seem to recall that it is possible to easily make artificial but elementary counter-examples to the injectivity in the case of Banach spaces. But perhaps I mis-recall. Anyway, in practice, of course, we'd have other info to use to prove the injectivity on natural function spaces.