2
$\begingroup$

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)).

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I've been thinking about the construction of the $A^s$ operators in terms of the spectral measure, and the domains, $D(A^s)$ in this way. What I'm having a mental block on is how to infer that if $\{x_n\}$ is Cauchy in $X_t$, $s<t$ so that the $\|x\|_s\leq C(s,t)\|x\|_t$ giving Cauchyness in $X_s$, I can make the identification of the limits to make sense of the inclusion mapping, and, presumably, make the inclusion injective. $\endgroup$ – user2379888 Aug 29 '19 at 18:05
  • 1
    $\begingroup$ @user2379888, you are correct in worrying about the injectivity! It needs a more delicate argument. $\endgroup$ – paul garrett Aug 29 '19 at 18:58
  • $\begingroup$ I'm still missing something as to why injectivity is an immediate consequence. $\endgroup$ – user2379888 Aug 30 '19 at 0:20
  • 1
    $\begingroup$ Since the function $y$ in my answer never vanishes, all these spaces are naturally embedded into the vector space of measurable functions. Hence, injectivity is not an issue (all in the sense of „almost everywhere“ as is customary in this context). $\endgroup$ – user131781 Aug 30 '19 at 4:00
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's a reasonable argument, but my real question is in the case of negative indices. There are definitely counterexamples in the more general case. $\endgroup$ – user2379888 Aug 30 '19 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.