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reducing subspace, quantum dynamics, band
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\newtheorem{Theorem}{Theorem}[section]
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\newcommand{\Reals}{{\mathbb R}}
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\begin{document}
\title{Reducing subspaces}
\author{Rowan Killip$^1$
and Christian Remling$^2$}
\date{January 16, 2001}
\maketitle
%\begin{center}
%(to appear in {\it })
%\end{center}
%\vspace{0.5cm}

\noindent
1. University of Pennsylvania, 209 South 33rd Street, Philadelphia PA 19104-6395, USA.\\
On leave at: The Institute for Advanced Study, Einstein Drive, Princeton NJ 08540, USA.\\
E-mail: killip@math.upenn.edu\\[0.1cm]
2.\ Universit\"at Osnabr\"uck, Fachbereich Mathematik/Informatik, 49069 Osnabr\"uck, GERMANY\\
E-mail: cremling@mathematik.uni-osnabrueck.de

\vskip0.3cm

\noindent
2000 AMS Subject Classification: primary 47B25, 81Q10, secondary 42A38 \\[0.3cm]
Key words: self-adjoint operator, reducing subspace, quantum dynamics, band \\[0.3cm]

\begin{abstract}
Let $T$ be a self-adjoint operator acting in a separable Hilbert space
$\h$. We establish a correspondence between the reducing
subspaces of $T$ that
come from a spectral projection and the convex, norm-closed
bands in the set of finite Borel measures on $\mathbb R$.

If $\h$ is not separable, we still obtain a reducing subspace corresponding
to each convex norm-closed band.

These observations lead to a unified treatment of various reducing subspaces;
moreover, they also settle some open questions and suggest new decompositions.
\end{abstract}

\section{Reducing subspaces and bands}

Throughout this paper, we fix a self-adjoint operator $T$ acting in Hilbert space $\h$.
As $T$ is self-adjoint, it admits the representation $T=\int_{\mathbb R}
\lambda\, dE(\lambda)$ where $E(\cdot)$ is a projection-operator-valued measure.
Also, to each $\psi\in\h$, we associate its spectral measure, $\rho_\psi(M)=\|E(M)\psi\|^2$.

Consider a set $\g$ of those $\psi\in\h$ whose spectral measures, $\rho_\psi$, have
certain prescribed properties.  The question we wish to address is: is the set $\g$ the
range of a spectral projection?  That is, is there a Borel set $M\subset \Reals$ so that
$\g=E(M)\h$?  More generally, is $\g$ even a reducing subspace?

%This question arises when considering the dynamics of a quantum mechanical system
%with Hamiltonian $T$.  In a crude sense, the regularity of the spectral measure determines
%the high-frequency behaviour of its Fourier transform, $\hat\rho_\psi$.  Quantum mechanically,
%this corresponds to the long-term behaviour of the probability of finding the system in original
%state.

While our answer, Theorem~\ref{T1.1}, is rather straightforward, we believe that it is
a useful way of thinking about reducing subspaces.  It provides a
unified treatment for the spectral decompositions arising in quantum dynamics.
More significantly, it suggests
further refinements and settles some open questions. We will discuss
these applications in the second section.

We need some notation. Let $\mathcal M$ denote the Banach space of finite Borel
measures on $\Reals$  (the norm of a measure is equal to its total variation).
Further, we write $\M$ to denote the subset of positive measures with the induced
(norm) topology.
A subset $B\subset \M$ is called a
band if it is closed with respect to absolute continuity.  That is, if
$\nu\in B$, $\mu\in\M$ and $\mu\ll\nu$ then $\mu\in B$.
For a subset $B\subset\M$, we define
$\h_B = \{ \psi\in\h : \rho_{\psi}\in B\}$.

\begin{Theorem}
\label{T1.1}
If $B\subset\M$ is a convex {\rm(}norm-{\rm)}closed band then $\h_B$ is a reducing subspace.
Moreover, if $\h$ is separable, $\h_B$ is the range of a spectral projection.
\end{Theorem}

Of course, not all reducing subspaces for $T$ are the range of a spectral projection.
(Consider, for example, the direct sum of an operator with itself.)
We prepare for the proof with two simple lemmas.  We will write $f\mu$
for the measure $(f\mu)(S)=\int_S f(x)\, d\mu(x)$.

\begin{Lemma}
\label{L1.1}
Let $B$ be a convex closed band.  For every $\mu\in\M$, there is a
Borel set $M$ such that\\
{\rm\phantom{i}(i)} $\chi_M\mu\in B$, and\\
{\rm(ii)} if $\nu\in B$ and $\nu\ll \mu$, then $\nu\ll \chi_M\mu$.
\end{Lemma}

\medskip

\noindent{\it Proof.} Let
\[
c=\sup \{ \mu(S): S\subset\mathbb R \text{ Borel set}, \chi_S\mu \in B \} .
\]
We claim that the supremum is
attained for some Borel set $M$. To see this, pick Borel sets
$M'_n$ so that $\chi_{M'_n}\mu\in B$ and $\mu(M'_n)\ge c-1/n$.
Define $M_1=M'_1$ and $M_n=M'_n\cup M_{n-1}$ for each $n\ge 2$.
As $\chi_{M_n}\mu\ll \frac{1}{2}(\chi_{M'_n}\mu +\chi_{M_{n-1}}
\mu )$, induction on $n$ shows that $\chi_{M_n}\mu \in B$.
Of course, we still have $\mu(M_n)\ge c-1/n$. Now
$M=\bigcup_{n\in\mathbb N} M_n$ has the desired properties:
$\mu(M)=c$ and $\|\chi_M\mu-\chi_{M_n}\mu \|=
\mu(M\setminus M_n)\to 0$ by monotone convergence.  Thus 
$\chi_M\mu\in B$ because $B$ is closed.

It remains to check property (ii).  As $\nu\ll
\mu$, there are Borel sets $S,T$ with $S\subset M$,
$T\cap M=\emptyset$, so that $\nu$ is equivalent to
$\chi_{S\cup T} \mu$.
It follows that $\chi_T\mu\in B$, but then also $\chi_{M\cup T}\mu\in B$
because this measure is absolutely continuous with respect to
$\frac{1}{2}(\chi_M\mu + \chi_T\mu)$. The definitions of $c$ and $M$
now imply that $\mu(T)=0$, so $\nu\ll \chi_M\mu$ and (ii) holds. \hfill $\square$

\medskip

The decomposition $\mu=\chi_M\mu + \chi_{\mathbb R\setminus M}
\,\mu$ performed in the Lemma is unique. That is, the set
$M$ is uniquely determined up to sets of $\mu$ measure zero.
To see this, notice that if the supremum $c$ were achieved for distinct
sets, $M$ and $N$, then it must also happen that $\mu(M\cup N)=c$.
This shows that $\mu(N)=\mu(M\cup N)=\mu(M)$ so $M$ and $N$ differ by a set
of zero $\mu$ measure.

\begin{Lemma}
\label{L1.2}
If $\psi_n\in\h$, $\psi_n\to\psi$, then $\rho_{\psi_n}\to\rho_{\psi}$.
\end{Lemma}

\medskip

\noindent
{\it Proof.}
For $\varphi,\psi\in \h$,
the definition of the total variation of a measure gives
\[
\| \rho_{\varphi}-\rho_{\psi}\| = \sup \sum_n
\Big| \|E(M_n)\varphi\|^2- \|E(M_n)\psi\|^2 \Big|,
\]
where the supremum is taken over all countable
partitions of $\mathbb R$ into disjoint Borel sets $M_n$. As
\[
\Big| \|E(M)\varphi\|^2- \|E(M)\psi\|^2 \Big|
  \le \|E(M)(\varphi-\psi)\|\,\cdot\,\Big( \|E(M)\varphi\| + \|E(M)\psi\|\Big),
\]
we obtain
\begin{align*}
    & \| \rho_{\varphi}-\rho_{\psi}\| \\
\le & \sqrt{2}\, \sup \Big\{
	\sum_n  \|E(M_n)(\varphi-\psi)\|^2\,
	\sum_n \left( \|E(M_n)\varphi\|^2 +\|E(M_n)\psi\|^2 \right)
	\Big\}^{1/2}  \\
=   & \sqrt{2}\, \|\varphi-\psi\| \left( \|\varphi\|^2 + \|\psi\|^2 \right)^{1/2}.
\end{align*}
Now the assertion is obvious. \hfill $\square$

\medskip

\noindent
{\it Proof of Theorem \ref{T1.1}}.
We begin with the case when $\h$ is separable.
Let $\{\psi_n:n\in\Nats\}$ be a basis for $\h$ and define a measure
$\Lambda=\sum_n 2^{-n} \rho_{\psi_n}$.  Notice that for every $\psi\in\h$,
$\rho_\psi\ll\Lambda$.

By applying Lemma~\ref{L1.1} to $\Lambda$, we obtain a Borel set $M$.  We will
now show that $\h_B=E(M)\h$:  If $\psi\in \h_B$ then $\rho_\psi\in B$.  But
$\rho_\psi\ll\Lambda$, so, by Lemma~\ref{L1.1}, $\rho_\psi\ll\chi_M\Lambda$.
Thus $\psi\in E(M)\h$.  Conversely, if $\psi\in E(M)\h$ then $\rho_\psi\ll\chi_M\Lambda$
and so $\rho_\psi\in B$, or equivalently, $\psi\in\h_B$.

Consider now, the case that $\h$ is not separable.  Because $B$ is a convex band and 
$\rho_{\psi_1+\psi_2} \ll \frac{1}{2}(\rho_{\psi_1}+\rho_{\psi_2})$, $\h_B$ is a subspace of $\h$.
As $B$ is closed, Lemma~\ref{L1.2} shows that $\h_B$ is closed. Now for any $\psi\in\h$,
the cyclic subspace generated by $\psi$ and $T$ is separable.  Thus, by our earlier treatment
of the separable case, for any bounded measurable function $f$,
$\psi\in\h_B \Rightarrow f(T)\psi\in\h_B$ and $\psi\in\h_B^\perp \Rightarrow f(T)\psi\in\h_B^\perp$.
This proves that $\h_B$ is a reducing subspace. \hfill $\square$

\medskip

For separable
spaces, the converse of Theorem~\ref{T1.1} is both true and easily proved: $E(M)\h$
is equal to $\h_B$ when $B=\{\mu:\mu(\Reals\setminus M)=0\}$.
This gives a correspondence between convex closed bands in $\M$ and
the ranges of spectral projections.
For a fixed operator, this correspondence is not one-to-one: different bands can generate
the same subspace. However, if $\h_B=\h_{B'}$ for all self-adjoint
operators on an infinite-dimensional Hilbert space $\h$,
then $B=B'$. To prove this observation, assume that
there is a $\mu\in B\setminus B'$
and consider a self-adjoint operator
whose spectral representation is multiplication by the independent variable in
the space $L_2(\mathbb R, \mu)$.

The final general remark on Theorem \ref{T1.1} concerns the possibility
of ``completing'' reducing subspaces. Namely, one can show that if $\h_0$
is a reducing subspace then
$B=\{ \rho_{\psi} : \psi\in\h_0 \}$ is a convex, closed band.
Thus one can form $\h'_0 =\h_B$, which,
by its definition, is the smallest reducing subspace containing $\h_0$ that
is generated by a convex, closed band.  In the separable case, Theorem~\ref{T1.1} and its
converse show that it is also the smallest reducing subspace containing $\h_0$ that is
the range of a spectral projection.




\section{Some applications}

First of all, let us point out that the usual decompositions
can also easily be obtained with the
aid of Theorem \ref{T1.1}. For example,
\begin{gather*}
\{ \mu\in\M : \mu \text{ is absolutely continuous} \},\\
\{ \mu\in\M : \mu(\{ x\} )=0 \text{ for all }x\in\mathbb R \}
\end{gather*}
are convex, closed bands. These give rise to the 
absolutely continuous and continuous subspaces, respectively.
In this way, we obtain the well known decomposition of an operator into absolutely
continuous, singular continuous, and point parts.  The
refined decompositions of the singular continuous subspace with respect
to Hausdorff measures, that were introduced by Last \cite{L}, can be obtained
in the same fashion.  For example, Last's $\alpha$-continuous subspace, $\h_{\alpha c}$,
corresponds to $\h_B$ for
$$
B=\{ \mu\in\M : \mu(S)=0 \text{ for all sets $S$ of zero $\alpha$-Hausdorff measure} \}.
$$
Finally, there is the transient/recurrent decomposition
of Avron and Simon \cite{AS}. This will be discussed (and refined)
shortly.  Let us first
note another consequence of Theorem \ref{T1.1}.

Recall that a measure $\mu$ is called Ra\kern-0.1exjchman if $\lim_{|t|\to
\infty} \widehat{\mu}(t)=0$; here, $\widehat{\mu}$ denotes the
Fourier transform $\widehat{\mu}(t)=\int_{\mathbb R} e^{-itx}\,
d\mu(x)$. An absolutely continuous measure is Ra\kern-0.1exjchman by the
Riemann-Lebesgue Lemma, while Wiener's Theorem shows that
a point measure is never Ra\kern-0.1exjchman.
A singular continuous measure may or may not be Ra\kern-0.1exjchman.

\begin{Corollary} The set
\label{C2.1}
$\h_{\Raj}= \{ \psi\in \h : \rho_{\psi}\text{ is Rajchman} \}$
is a reducing subspace.
\end{Corollary}

\noindent{\it Proof.} The set of Ra\kern-0.1exjchman measures is obviously convex
and closed.  That it is also a band is known as the Milicer-Gru\u zewska Theorem
\cite[Chapter XII, Theorem 10.9]{Zyg}.\hfill $\square$

\medskip

The question ``is $\h_\Raj$ a reducing subspace?'' appeared in \cite{Amr,L} and was
the original motivation for the present work.  Corollary~\ref{C2.1}
also follows from Lyons's characterization
of the Ra\kern-0.1exjchman measures as those which
give zero weight to all Weyl sets \cite{Ly}.
In this context, a Theorem of
Mokobodzki is of interest; it gives criteria for a band
to consist of exactly those measures which annihilate certain sets.
See \cite[Chapter IX]{KecL} for further information on this.

We are grateful to B.~Simon for pointing out to us that an earlier proof
(using the Lyons/Mokobodzki results) was unnecessarily complicated.

If $T$ is the Hamiltonian of a quantum mechanical system and if
the initial state $\psi$ is normalized
(i.e.\ $\|\psi\|=1$), then $|\widehat{\rho}_{\psi}(t)|^2$ is
the probability of finding the system in the state $\psi$ at time
$t$. Thus it is interesting to study other decompositions of $T$ which
carry information on the asymptotics of the Fourier transform of
the spectral measures. A large class of such decompositions can
be obtained using the following Proposition, which comes in two
variants.
\begin{Proposition}
\label{P2.1}
Let $P$ be a convex subset of $C_b(\mathbb R)$, and suppose
that $C_c^{\infty}(\mathbb R) * P \subset P$. Define
\[
B_1  = \overline{ \{ \mu\in \M : \widehat{\mu} \in P \} }, \quad\quad
B_2  = \overline{ \{ \rho_{\psi}:
\psi\in\h, \widehat{\rho}_{\psi} \in P \} }.
\]
Then $B_1$ and $B_2$ are convex, closed bands.
\end{Proposition}
Here, $C_b(\mathbb R)$ is the Banach space of bounded continuous
functions on $\mathbb R$, $C_c^\infty$ is the space of infinitely differentiable functions
of compact support and the star denotes convolution.

\medskip

\noindent{\it Proof.} $B_1$ and $B_2$ are obviously closed and it is also
clear that $B_1$ is convex. We will
now show that
$B_1$ is a band. So suppose that $\nu\in B_1$ and $\mu\in\M$ with
$\mu\ll \nu$. By the definition of $B_1$, there are $\nu_n\in \M$, so that
$\nu_n\to\nu$ and $\widehat{\nu}_n\in P$. By the Radon-Nikodym Theorem,
we have that $\mu=f\nu$ for some $f\in L_1(\mathbb R, \nu)$,
$f\ge 0$. If
$\epsilon>0$ is given, we determine a function $g\ge 0$
with $\widehat{g}\in C_c^{\infty}(\mathbb R)$, so that
$\|f-g\|_{L_1(\mathbb R, \nu)} < \epsilon$. To see that this can be done,
pick $h\in C_c^{\infty}(\mathbb R)$, $h\ge 0$
with $\|f-h\|_{L_1(\mathbb R, \nu)} < \epsilon/2$. Next, take any
real valued $\theta\not\equiv 0$ with $\widehat{\theta}\in
C_c^{\infty}(\mathbb R)$, and let $\varphi(x)=\theta^2(x)/\int \theta^2$.
Then $\varphi\ge 0$, $\widehat{\varphi}\in C_c^{\infty}(\mathbb R)$,
and $\int\varphi(x)\, dx =1$. Let $\varphi_{\delta}(x)=(1/\delta)
\varphi(x/\delta)$; then $g=\varphi_{\delta}*h$ with a sufficiently
small $\delta>0$ has the desired properties.

Now consider the measures $g\nu_n\in\M$. We have that
$(g\nu_n)\,\widehat{ }= \widehat{g} * \widehat{\nu}_n\in P$. Moreover,
\[
\|g\nu_n - \mu\| \le \|g\|_{\infty} \|\nu_n-\nu\|
+\|g-f\|_{L_1(\mathbb R, \nu)} < 2\epsilon
\]
for all sufficiently large
$n$. Hence $\mu\in B_1$, as desired.

To prove the claim for $B_2$, note that a measure of the form
$f\rho_{\psi}$ with $f\in L_1(\mathbb R,\rho_{\psi})$, $f\ge 0$ is a
spectral measure (i.e.\ $f\rho_{\psi}=\rho_{\varphi}$ for some $\varphi\in\h$).
So the argument
from above can also be used to show that $B_2$ is a band.
Finally, suppose that $\psi_1,\psi_2\in\h$ and $\widehat{\rho}_{\psi_i}
\in P$. It is easy to see, by restricting to the reducing subspace generated
by $\psi_1,\psi_2$ and using spectral representations, that
any convex combination $t\rho_{\psi_1}+(1-t)\rho_{\psi_2}$ is again
a spectral measure. Thus $\{ \rho_{\psi}: \widehat{\rho}_{\psi}\in P \} $
and hence also $B_2$ are convex sets. \hfill$\square$

\medskip

Given a space $P$ satisfying the hypotheses of Proposition \ref{P2.1},
we can form the subspaces $\h_{B_1}$ and $\h_{B_2}$. Of course,
since $B_1\supset B_2$, we also have that
$\h_{B_1}\supset \h_{B_2}$, and the inclusion may be proper, as we
will see in a moment. For most purposes, $\h_{B_2}$ is the more useful space;
it can also be described as follows.

\begin{Theorem}
\label{T2.1}
If $P$ and $B_2$ are as in Proposition \ref{P2.1}, then
\[
\h_{B_2} = \overline{ \{ \psi\in\h : \widehat{\rho}_{\psi}
\in P \} } .
\]
\end{Theorem}

\noindent
{\it Proof.} By definition,
\begin{equation}
\label{HB2}
\h_{B_2} = \{ \psi\in\h : \text{ There are } \psi_n\in\h
\text{ so that } \widehat{\rho}_{\psi_n}\in P\text{ and }\rho_{\psi_n}
\to\rho_{\psi} \}.
\end{equation}
Lemma \ref{L1.2} now shows
that the set $\overline{ \{ \psi\in\h : \widehat{\rho}_{\psi}
\in P \} }$ from the statement of Theorem \ref{T2.1}
is contained in $\h_{B_2}$.

Conversely, suppose that $\psi\in\h_{B_2}$.
By \eqref{HB2},
there are $\varphi^{(n)}
\in\h$ so that $\widehat{\rho}_{\varphi^{(n)}}\in P$ and
$\rho_{\varphi^{(n)}}\to \rho_{\psi}$. We will now work
in the reducing subspace of $T$ generated by $\psi$ and
the $\varphi^{(n)}$; clearly, this space (call it $\h_0$) is separable.
We may pass to a spectral representation of this part of $T$
and thus assume that $TP_{\h_0}$
is multiplication by the variable in the space
\[
\h_0 = \bigoplus_{i=1}^N L_2(\mathbb R, f_i\rho) .
\]
Here, $\rho\in\M$,
$N\in\mathbb N \cup \{\infty \}$, and $f_i\ge 0$, $f_i\in
L_1(\mathbb R, \rho)$.
Write $\psi=(\psi_i)_{i=1}^N$, $\varphi^{(n)}=(\varphi_i^{(n)}
)_{i=1}^N$,
and let
\[
g(x)= \sum_{i=1}^N \left| \psi_i(x)\right|^2 f_i(x), \quad\quad
g_n(x)= \sum_{i=1}^N \left| \varphi_i^{(n)}(x)\right|^2 f_i(x).
\]
Then $g,g_n\in L_1(\mathbb R, \rho)$ and $\rho_{\psi}=g\rho$,
$\rho_{\varphi^{(n)}}=g_n\rho$; in particular,
\begin{equation}
\label{gn}
\|g_n-g\|_{L_1(\mathbb R, \rho)} = \|\rho_{\varphi^{(n)}}
-\rho_{\psi}\| \to 0 .
\end{equation}
Now define $\psi^{(n)}\in\h_0$ by $\psi_i^{(n)}(x)=
\psi_i(x) \sqrt{ g_n(x)/g(x)}$ if $g(x)\not= 0$ and
$\psi_i^{(n)}(x)= 0$ if $g(x)=0$. Then $\rho_{\psi^{(n)}}=
\rho_{\varphi^{(n)}}$, and a brief computation shows that
\[
\|\psi^{(n)}-\psi\|^2 = \int \left| \sqrt{g_n(x)} - \sqrt{g(x)}
\right|^2 \, d\rho(x) .
\]
This tends to zero by \eqref{gn} and the elementary inequality
$\left( \sqrt{a}-\sqrt{b} \right)^2 \le |a-b|$ ($a,b\ge 0$). $\square$

There are many possible choices for $P$. With $P=C_0(\mathbb R)$,
the continuous functions vanishing at infinity, one recovers
$\h_{\Raj}$ (note that since $C_0$ is a closed subspace of $C_b$,
the closure in the definition of $B_1, B_2$ is superfluous).
Next, $P=L_p\cap C_b$ also satisfies the hypotheses of Proposition
\ref{P2.1}. So Theorem \ref{T2.1} shows that the spaces
\[
\h_p = \overline{ \{ \psi\in\h : \widehat{\rho}_{\psi} \in L_p(\mathbb R)
\} }
\]
are reducing spaces for $1\le p<\infty$.
This answers a question of
Avron and Simon \cite[pg.\ 9]{AS}. It is known that $\h_2=\h_{ac}$
\cite{RS3}, so for $1\le p \le 2$, the spaces $\h_p$ are subspaces
of $\h_{ac}$ (since $L_p\cap C_b \subset L_2\cap C_b$ for these $p$). Also,
$\h_1=\h_{tac}$, the transient subspace introduced
in \cite{AS}. For our purposes, we may take this as the definition
of $\h_{tac}$, so
\[
\h_{tac} = \overline{ \{ \psi\in\h : \widehat{\rho}_{\psi} \in L_1(\mathbb R)
\} } .
\]
The space $\mathcal S$ of infinitely differentiable functions which together
with their derivatives decay faster than any polynomial is also convex
and closed under convolution with $C_c^{\infty}$ functions, so using
Proposition \ref{P2.1} and Theorem \ref{T2.1}, we deduce that
\[
\h_{\mathcal S} = \overline{
\{ \psi\in\h : \widehat{\rho}_{\psi} \in \mathcal S \} }
\]
also is a reducing subspace. Since $\widehat{\mathcal S} = \mathcal
S$, there is the alternate description
\[
\h_{\mathcal S} = \overline{
\{ \psi\in\h : {\rho}_{\psi}=g\, dx \text{ with } g\in \mathcal S \} } .
\]
From the results of \cite{AS}, we have that $\h_{\mathcal S}=\h_{tac}$.
Indeed, it is obvious that
$\h_{\mathcal S}\subset \h_{tac}$, and conversely, if $\widehat{\rho}_{\psi}
\in L_1$, then $\rho_{\psi}= f(x)\, dx$ with some continuous density
$f$. In particular,
the set $\Omega=\{ x: f(x)>0 \}$ is open, and we can approximate
$\rho_{\psi}$ by measures of the form $g(x)\, dx$, with $g\in \mathcal S$
and $g$ supported by $\Omega$. Now an argument similar to
the one used in the proof of Theorem \ref{T2.1} shows that $\psi$ itself can be
approximated by vectors whose spectral measures are absolutely continuous
with densities in $\mathcal S$. Thus $\h_{\mathcal S}\supset \h_{tac}$,
as claimed.

We now also see that the band $B_1$ from Proposition \ref{P2.1} can
lead to a space larger than the one from Theorem \ref{T2.1}. Namely,
if $P=\mathcal S$, then $B_1$ is the set of all absolutely
continuous measures from $\M$ (again by an approximation argument), so
$\h_{B_1}=\h_{ac}$, which, of course, can be strictly larger than
$\h_{B_2}=\h_{tac}$.

We have already mentioned the fact that Theorem \ref{T2.1} suggests
refined decompositions. We conclude this paper with a discussion of
one such example. Let
\[
P_{\alpha} = L_2(\mathbb R, (1+x^2)^{\alpha/2}\, dx) \cap C_b(\mathbb R).
\]
Then $P_{\alpha}$ satisfies the assumptions of Proposition \ref{P2.1}
for every $\alpha\in\mathbb R$, and if we denote the corresponding space
from Theorem \ref{T2.1} by $\h_{\alpha}$, then
\[
\h_{\alpha} = \overline{ \Big\{ \psi\in\h :
\int \left| \widehat{\rho}_{\psi}(t)\right|^2
{(1+{t}^2)}^{\alpha}\, dt < \infty \Big\} } .
\]
For $\alpha\ge 0$, the scale of these reducing subspaces
gives a refinement of the transient/recurrent
decomposition of Avron and Simon.

\begin{Theorem}
\label{T2.2}
a) If $\alpha\ge\beta$ then $\h_{\alpha}\subset \h_{\beta}$.\\
b) $\h_0=\h_{ac}$ and if $\alpha> 1/2$, $\h_{\alpha}=\h_{tac}$.\\
c) For any $\alpha>0$, it may happen that $\h_{\alpha}\not=\h_{ac}$.\\
d) There are operators with $\h_{1/2}\not=\h_{tac}$.
\end{Theorem}

\noindent
{\it Proof.} a) This is immediate from $P_{\alpha}\subset P_{\beta}$
($\alpha\ge\beta$).

b) As $P_0=L_2\cap C_b$, the identification
$\h_0=\h_{ac}$ is one of the facts
mentioned above. The Cauchy-Schwarz inequality
shows that $P_{\alpha}\subset L_1$ for $\alpha>1/2$, thus
$\h_{\alpha}\subset\h_{tac}$ for these $\alpha$. On the other hand,
the $P=\mathcal S$ characterization of $\h_{tac}$ implies that
$\h_{tac}\subset \h_{\alpha}$ for all $\alpha$.

c) Given $\alpha>0$,
we will construct a (Cantor type) set $C$ of positive (and finite)
Lebesgue
measure, so that for all non-zero $f\in L_1(C)$, $\widehat{f}\notin P_{\alpha}$.
Then the operator of multiplication by the variable in
$L_2(C)$ has $\h_0=\h_{ac}=\h$, but $\h_{\alpha}=\{ 0\}$, so this
construction will prove the claim.

So let $\alpha>0$, and fix $\epsilon\in (0,2\alpha)$ and
$l_0\in (0,1)$. For technical reasons, we also take $l_0$ so small
that $(1+\epsilon)(l_0/2)^{\epsilon}\le 1$.
Put $C_0=[0,l_0]$. To carry out the general step,
assume that $C_{n-1}$ has been constructed and that $C_{n-1}$ is
the union of $2^{n-1}$ closed, disjoint intervals of length $l_{n-1}$.
For each of these intervals, delete an open subinterval in the middle
of the old interval to obtain two smaller intervals of length $l_n$
each, where $l_n$ is determined from the equation
\[
2 l_n (1-l_n^{\epsilon}) = l_{n-1} (1-l_{n-1}^{\epsilon}).
\]
Note that the left-hand side is strictly increasing as a function
of $l_n\in [0,l_{n-1}/2]$, and it is zero at $l_n=0$ and larger than
the right-hand side at $l_n=l_{n-1}/2$.
So $l_n\in (0,l_{n-1}/2)$ is well-defined.
We can now let $C_n$ be the union of
the $2^n$ new intervals obtained from this process, and we put
$C=\bigcap_{n\in\mathbb N} C_n$. The sequence $2^nl_n$ is decreasing,
hence $q=\lim_{n\to\infty} 2^nl_n$ exists. Since $C_n\subset C_{n-1}$,
we also have that $|C|=q$. The recursion defining $l_n$ shows that
the combination $2^nl_n(1-l_n^{\epsilon})$ is independent of $n$. Letting
$n\to\infty$ therefore gives
\[
q=2^nl_n (1-l_n^{\epsilon})\quad\text{for all }n\in\mathbb N_0 ;
\]
in particular, $q>0$. The length of the intervals that are deleted at
step $n$ is equal to $l_{n-1}-2l_n$, hence
\begin{align*}
\left| [0,l_n]\setminus C\right| & = \sum_{k=1}^{\infty}
2^{k-1} (l_{n+k-1}-2l_{n+k}) \\
& = \lim_{N\to\infty} 2^{-n} (2^nl_n - 2^Nl_N) = 2^{-n}(2^nl_n-q)
= l_n^{1+\epsilon}.
\end{align*}
If $x\in C$, then $[x-l_n,x+l_n]\setminus C$ contains
a translate of $[0,l_n]\setminus C$, so
\[
\left| [x-l_n,x+l_n]\setminus C\right| \ge l_n^{1+\epsilon}.
\]

Thus if $f$ is supported by $C$, and if $x\in C$, $f(x)\not= 0$,
then
\begin{align*}
\int_{-1}^1 \left| f(x+t)-f(x)\right|^2\, \frac{dt}{|t|^{1+2\alpha}}
& \ge \int_{-l_n}^{l_n} \left| f(x+t)-f(x)\right|^2\,
\frac{dt}{|t|^{1+2\alpha}}\\
& \ge |f(x)|^2 l_n^{-1-2\alpha}
\left| [x-l_n,x+l_n]\setminus C\right| \\
& \ge |f(x)|^2 l_n^{\epsilon-2\alpha} \to \infty
\quad\quad\quad (n\to\infty).
\end{align*}
But if $\widehat{f}(x)(1+x^2)^{\alpha/2}$ were in $L_2$,
then the integral estimated above would have
to exist for almost every $x\in\mathbb R$
(see, e.g., \cite{St}). Hence the set $C$ does not support non-zero
functions with Fourier transform in $P_{\alpha}$.

d) A special case of results of Polking
\cite{Pol} states that there are nowhere dense sets $C$ that support functions
with Fourier transform in $P_{1/2}$. So if the operator $T$ is again
multiplication by the variable in $L_2(C)$, then $\h_{1/2}\not= \{ 0\}$.
On the other hand, $C$ does not support continuous functions and hence
$\widehat{f}\notin L_1$ for all $f\in L_1(C)$, $f\not\equiv 0$.
Therefore $\h_{tac}=\{ 0 \}$.
This proves the final claim of Theorem \ref{T2.2}. \hfill $\square$

\medskip

The spaces $\h_{\alpha}$ can also be used for $\alpha<0$. One then gets
a decomposition of the continuous subspace $\h_c$,
which is similar to the
decompositions discussed in \cite{L}. Here, the interesting range for
the parameter $\alpha$ is $[-1/2,0]$.
More precisely, one can show that
$\h_{-1/2}\subset \h_c$ (where, in general, equality need not hold) and
$\h_{\alpha}=\h$ if $\alpha< -1/2$.

The decomposition discussed above is based on the usual Sobolev spaces
and so is rather natural. One can, of course, consider
other decompositions which are similar in spirit. For instance,
\[
P_{\beta}= \{ f\in C_b(\mathbb R) : f(x)=o(|x|^{-\beta})
\text{ as } |x|\to\infty \} ,
\]
gives a decomposition of $T$ which, roughly speaking, classifies
vectors according to the Fourier dimension of the support of the
associated spectral measure.

\medskip

{\bf Acknowledgment:} C.R.\ acknowledges financial support by
the Heisenberg program of the Deutsche Forschungsgemeinschaft.
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\end{document}
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