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e-mail address: umeda@sci.himeji-tech.ac.jp
AMS-Code 35E99, 35J10, 35S99
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relativistic Hamiltonians, relativistic Schr\"odinger operators, distributions
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\documentstyle{amsppt}
\parindent20pt
\addto\tenpoint{\normalbaselineskip16pt\normalbaselines}
\NoRunningHeads
\NoBlackBoxes
\loadeusm
\magnification=\magstep1
\pagewidth{130mm}
\pageheight{204mm}
\font\sm=eusm10
\topmatter
\title Action of $\sqrt{ - \Delta \,}$ on Distributions
\endtitle
\author Tomio Umeda
\endauthor
\affil Department of Mathematics \\
Himeji Institute of Technology \\
Himeji 671-2201 \\ Japan
\endaffil
\email umeda@sci.himeji-tech.ac.jp
\endemail
\dedicatory Dedicated to Professor Yoshimi Satio
on the occasion of his sixtieth birthday
\enddedicatory
\thanks Research on this paper is supported by Grant-in-Aid for
Scientific Research (Basic Research(C), No. 09640212), Japan Society for the
Promotion of Science.
\endthanks
\keywords relativistic Hamiltonians, relativistic Schr\"odinger
operators, \break distributions
\endkeywords
\subjclass
35E99, 35J10, 35S99
\endsubjclass
\abstract
Action of $\sqrt{-\Delta}$ on distributions is examined in the context
of Sobolev spaces, weighted $L^2$ spaces and
weighted Sobolev spaces, respectively.
The results obtained are as follows: Let $k$ be a real
number. (1) If $f$ is in
$H^{k}({\bold R}^n)$,
then $\sqrt{-\Delta}f$ is in
$H^{k-1}({\bold R}^n)$; \ (2) If $f$ is in $\Cal S({\bold R}^n)$, the space
of rapidly decreasing functions, then $f$ is in $L^{2, \, s}({\bold R}^n)$
for any $s < n/2 +1$; \ (3) If $f$ is in
$H^{k,\, s}({\bold R}^n)$ for
some $s > -n / 2 -1$, then $\sqrt{-\Delta}f$ is in
${\Cal S}^{\prime}({\bold R}^n)$, the space of
tempered distributions. Also, it is shown in the
one dimensional case that there exists a $\varphi_0$ in
${\Cal S}(\bold R)$ such that $\sqrt{-\Delta} \varphi_0$
does not belong to $L^{2, \, s}(\bold R)$ for any $s \ge 3/2$.
\endabstract
\endtopmatter
\newpage
\document
\noindent
{\bf \S 1 Introduction.}
%\vskip 5pt
Mathematically rigorous study of relativistic Schr\"odinger operators was initiated
by Weder [8] in 1974. Since then, the relativistic Schr\"odinger operators have
received a considerable attention, and there has been a
substantial amount of literature for the relativistic Schr\"odinger operators
from various view points. See a survey article [2] by
Lieb and Umeda [7] and references therein.
The kinetic energy part of the relativistic Schr\"odinger operator
is given by
%
$$
\sqrt{-\Delta + m^2 \,},
$$
%
where $m$ is a nonnegative constant. If $m$ is positive,
then the operator
$\sqrt{-\Delta + m^2}$ is a pseudodifferential operator
of order 1 with smooth symbol. Actually, the symbol $\sqrt{|\xi|^2 + m^2 \,}$
belongs to the class $S^1_{1,0}$. (See Section 2 for the definition
of the symbol class.)
It is straightforward that $\sqrt{-\Delta + m^2}$ maps $\Cal S({\bold R}^n)$,
the space of rapidly
decreasing functions, continuously into itself, hence by duality,
$\sqrt{-\Delta + m^2}$ maps $\Cal S^{\prime}({\bold R}^n)$,
the space of
tempered distributions, into itself.
The situation, however, changes dramatically for $m = 0$.
The symbol of the operator $\sqrt{-\Delta}$ is given by
$|\xi|$, which is singular at the origin.
Therefore, $\sqrt{-\Delta} \,\varphi$ may not
belong to $\Cal S({\bold R}^n)$
even if $\varphi$ belongs to $\Cal D({\bold R}^n)$,
the space of test functions,
and the definition
%
$$
\big< \sqrt{-\Delta} f, \, \varphi \big>
:= \big< f , \,
\overline{ \sqrt{-\Delta} \, \overline{\varphi} } \, \big>
\qquad
( \, \varphi \in {\Cal D}({\bold R}^n) \, )
$$
%
does not make sense for a generic $f \in {\Cal D}^{\prime}({\bold R}^n)$,
the space of distributions.
A remedy for this difficulty
is found in Lieb and Loss [3, p. 172]: \ {\it If
$f \in \text{H}^{\,1/2}({\bold R}^n)$, the Sobolev space of
order $1/2$,
then $\sqrt{-\Delta} f \in {\Cal D}^{\prime}({\bold R}^n)$}.
A few questions naturally arise.
{\it
\noindent
Question 1. \
Does $\sqrt{-\Delta} f$ belong to $\text{H}^{\,-1/2}({\bold R}^n)$
when
$f \in \text{H}^{\,1/2}({\bold R}^n)$?
\noindent
Question 2. \
Is it possible to define $\sqrt{-\Delta} f$ for $f$ in a wider
subspace of distributions?
\noindent
Question 3.
Can one find subspaces of distributions
which are closed under the operation $\sqrt{-\Delta}$?
}
The aims of this paper is twofold, one of which
is to give answers to these questions.
The answer to Question 1 is affirmative. Furthermore, we shall
show that $\sqrt{-\Delta} f$ makes sense for any
$f$ in $H^{-\infty}({\bold R}^n) =
\cup_{k \in {\bold R}} H^k ({\bold R}^n)$, and that
$\sqrt{-\Delta} f$ belongs
$ H^{k-1} ({\bold R}^n)$ if $f$ belongs to $H^k ({\bold R}^n)$.
This gives answers to Questions 2 and 3; $H^{-\infty}({\bold R}^n)$
is closed under the operation $\sqrt{-\Delta}$.
Therefore, the singularity of the symbol $|\xi|$ causes
$\sqrt{-\Delta}$ to exhibit no difference
in the context of Sobolev spaces
from pseudodifferential operators with symbols in the class
$S^{\mu}_{\rho, \, \delta}$ (e.g. Kumano-go [1, Theorem 2.7, p.124]).
The singularity, however, does cause $\sqrt{-\Delta}$ to exhibit
something different in the
context of
weighted $L^2$ spaces. In fact, we shall show
that there exists a function $\varphi_0$ in
$\Cal S(\bold R)$ such that
$\sqrt{-\Delta} \varphi_0$ belongs to $L^{2, \, s}(\bold R)$
for every $s < 3/2$, but does not for any $s \ge 3/2$. More precisely,
we shall prove that the $k$th derivative of $\sqrt{-\Delta} \varphi_0$
belongs to $L^{2, \, s}(\bold R)$ for all $s < k + 3/2$, but
does not for any $s \ge k + 3/2$. In other words,
$\sqrt{-\Delta} \varphi_0$ together with its all derivatives
decays polynomially
at infinity. The fact we can prove in general is that
if $f$ is in $\Cal S({\bold R}^n)$, then $\sqrt{-\Delta} f$ is
in $L^{2, \, s} ({\bold R}^n)$ for all $s < n/2 + 1$.
Surprisingly enough, we shall find that there
exists a function $f_0$, decaying polynomially at infinity
and belonging to $H^{\infty}({\bold R}^n) =
\cap_{k \in {\bold R}} H^k ({\bold R}^n)$,
such that
$\sqrt{-\Delta} f_0$ together with its all derivatives have
exactly the same asymptotic behaviors at infinity, apart from
multiplicative constants, as
$\sqrt{-\Delta} \varphi_0$ with $\varphi_0$ mentioned above.
The other aim of this paper is to investigate action of
$\sqrt{-\Delta}$ on weighted Sobolev spaces. There are
two reasons. One reason is that
action of $\sqrt{-\Delta}$ on weighted Sobolev spaces
seems to be much more interesting than
that of $\sqrt{-\Delta}$ on Sobolev spaces, as
suggested above. The other reason is that
it natural to seek wider subspaces of distributions
than $H^{-\infty}({\bold R}^n)$, and
that $\Cal S^{\prime}({\bold R}^n)$ is expressed as a
union of weighted Sobolev spaces. (This is an
immediate consequence of Treves[6, Theorem 25.4, p. 272].)
The plan of the paper is as follows. In Section 2, we fix notation
and prove a boundedness result of pseudodifferential operators
on weighted Sobolev spaces, which will be used later in this paper.
In Section 3, we discuss action of $\sqrt{-\Delta}$ on Sobolev
spaces. In Section 4, we consider the one dimensional
case and make explicit computations of $\sqrt{-\Delta}f$
for some specific functions $f$'s.
We examine action of $\sqrt{-\Delta}$ on weighted
$L^2$ spaces in Section 5. The results in Section 5 will be
applied in Section 6, where we discuss action of
$\sqrt{-\Delta}$ in weighted Sobolev spaces.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The end of Section 1 %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 30pt
\noindent
{\bf \S 2 Preliminaries.}
In this section, we introduce the notation which will be used
throughout the paper, and prove a boundedness result
on pseudodifferential operators in
weighted Sobolev spaces, the boundedness
result which we shall use in Section 6.
For $x \in {\bold R}^n$, $|x|$ denotes the Euclidean norm of
$x$, and $\langle x \rangle = \sqrt{1 + |x|^2}$. $D$ stands
for $-i \partial / \partial x$, and
%
$$
\langle D \rangle = \sqrt{1 + |D|^2} = \sqrt{1- \Delta}.
$$
%
The seminorms of $\Cal S({\bold R}^n )$ are defined by
%
$$
| \varphi|_{\ell, \, \Cal S} =
\max_{k + | \alpha| \le \ell} \;
\sup_x \; \{ \, \langle x \rangle^k \, | D^{\alpha}
\varphi(x) | \, \} \qquad ( \, \ell = 0, \, 1, \, 2, \, \cdots ).
$$
%
For $k, \;\;s \in \bold R$, the
weighted Sobolev space $H^{k, \,s}( {\bold R}^n)$ is defined by
%
$$
H^{k, \,s}( {\bold R}^n) = \{ f \in \Cal S^{\prime}({\bold R}^n) \; | \,
{\langle x \rangle}^s {\langle D \rangle}^k f \in L^2({\bold R}^n) \}
$$
%
with the associated norm
%
$$
\Vert f \Vert_{k,s} =
\Vert {\langle x \rangle}^s {\langle D \rangle}^k f {\Vert}_{L^2}.
$$
%
The weighted Sobolev space $H^{0, \,s}( {\bold R}^n)$ turns out to
be the weighted $L^2$ space
%
$$
L^{2, \,s}( {\bold R}^n) = \{ f \in \Cal S^{\prime}({\bold R}^n) \; | \,
{\langle x \rangle}^s f \in L^2({\bold R}^n) \}
$$
%
with the associated norm
%
$$
\Vert f \Vert_{s} =
\Vert {\langle x \rangle}^s f {\Vert}_{L^2},
$$
%
and the weighted Sobolev space $H^{k, \,0}( {\bold R}^n)$ turns out to be
the Sobolev space
%
$$
H^k({\bold R}^n) = \{ f \in \Cal S^{\prime}({\bold R}^n) \; | \,
{\langle D \rangle}^k f \in L^2({\bold R}^n) \}
$$
%
with the associated norm
%
$$
\Vert f \Vert_{k, \, 0} =
\Vert {\langle D \rangle}^k f {\Vert}_{L^2}.
$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 20pt
\noindent
{Definition.} \ {\it A $C^{\infty}$-function $p(x, \, \xi)$ on
${\bold R}^n \times{\bold R}^n$ is said to be
in the class $S^{\mu}_{\rho, \delta}$ \
$( \mu \in {\bold R}, \;\, 0 \le \delta \le \rho \le 1, \;\, \delta < 1)$ if
for any pair $\alpha$ and $\beta$ of
multi-indices there exists a constant $C_{\alpha\beta}$
such that}
%
$$
\Big| \Big( {\partial \over \partial \xi} \Big)^{\alpha}
\Big( {\partial \over \partial x} \Big)^{\beta}
p(x, \, \xi) \Big|
\le
C_{\alpha\beta}
{\langle \xi \rangle}^{\mu + \delta |\beta| -\rho |\alpha|}.
$$
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 20pt
It is well-known that if $p(x, \, \xi) $ belongs to $S^{\mu}_{\rho, \delta}$,
then the
pseudodifferential operator defined by
%
$$
p(x, \, D) \varphi (x) =
(2\pi)^{-n/2} \! \! \int e^{ix \cdot \xi} \,
p(x, \, \xi) \hat{\varphi} (\xi) \, d\xi
$$
%
maps $\Cal S({\bold R}^n)$ continuously into itself,
and by duality, maps $\Cal S^{\prime}({\bold R}^n)$ into itself
(e.g. Kumano-go [1, Theorem 1.3, p. 58] and [1, Definition 1.1,
p. 116]).
Here
%
$$
{\hat \varphi}(\xi) = (2\pi)^{-n/2} \! \!
\int e^{-ix \cdot \xi} \,
\varphi(x) \, dx.
$$
%
With an abuse of notation, we write $p(x, \, D) \in S^{\mu}_{\rho, \delta}$
when $p(x, \, \xi) \in S^{\mu}_{\rho, \delta}$.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 2.1.} \ {\it Let $k$ and $s$ be in $\bold R$, and let
$p(\xi)$ be in $S^{\mu}_{0, \, 0}$. Then $p(D)$ is a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k-\mu, \,s}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ To
prove the lemma, it is sufficient to
show that the operator $T$ defined by
%
$$
\eqalign{
T:&= \langle x {\rangle}^s \langle D {\rangle}^{k-\mu}
p(D)
\langle D {\rangle}^{-k} \langle x {\rangle}^{-s} \cr
{} &= \langle x {\rangle}^s
p(D)
\langle D {\rangle}^{-\mu} \langle x {\rangle}^{-s} \cr}
$$
%
is bounded in $L^2({\bold R}^n)$ for $s>0$. (The boundedness
of $T$ for $s<0$ can be reduced to the boundedness of $T^*$.)
Introducing a new symbol $q(x, \, \xi)$
by
%
$$
q(x, \, \xi) = \hbox{Os}-\iint_{{\bold R}^{2n}}
e^{-iy\cdot \eta}
p ( \xi + \eta)
\langle \xi + \eta {\rangle}^{-\mu} \langle x + y {\rangle}^{-s} \,
( 2 \pi )^{-n} \, dy\, d\eta \, , \leqno(2.1)
$$
%
we see that $T = \langle x {\rangle}^{s} q(x, \, D)$
(cf. [1, Theorem 2.1 (1), p. 74]). By differentiation under
the oscillatory integral sign
and integration by parts, we deduce that for any pair of multi-indices
$\alpha$ and $\beta$
%
$$
\Big| \Big( {\partial \over \partial \xi} {\Big)}^{ \!\alpha}
\Big( {\partial \over \partial x} {\Big)}^{ \!\beta}
q(x, \, \xi) \, \Big|
\le C_{s\alpha\beta}
\langle x {\rangle}^{-s} , \leqno(2.2)
$$
%
where $C_{s\alpha\beta}$ is a nonnegative constant.
Therefore, the symbol of the operator $T$,
which is given by $\langle x {\rangle}^{s}\, q(x, \, \xi)$,
belongs to $S^0_{0, \, 0}$. Thus, we can apply
the Calder\'on-Vaillancourt theorem to conclude
that $T$ is a bounded operator
in $L^2({\bold R}^n)$. \ \ \ {$\square$}
\vskip 30pt
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The end of Section 2 %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf \S3 $\sqrt{-\Delta}$ on Sobolev spaces.}
In this section we work in Sobolev spaces.
We start with the formal
definition of the operator $\sqrt{-\Delta} \,$:
%
$$
\sqrt{-\Delta} \,f (x) =
(2 \pi)^{-n/2 } \!\! \int e^{ix \cdot \xi} \, |\xi|
{\hat f}(\xi) \, d\xi.
$$
%
If $f \in \Cal S({\bold R}^n)$ and the support of ${\hat f}$ does not
contain the origin in the
$\xi$-space, then it follows that
$\sqrt{-\Delta} f (x) \in \Cal S({\bold R}^n)$.
This fact leads to a decomposition of
$\sqrt{-\Delta}$ into a singular part and a regular part:
%
$$
\sqrt {-\Delta} = \chi (D) \sqrt {-\Delta} +
\big( 1 - \chi (D) \big) \sqrt {-\Delta}, \leqno(3.1)
$$
%
where $\chi$ is a $C^{\infty}$-function satisfying
%
$$
\chi(\xi) = \cases
1 &\text{ if \ $| \xi | \le 1 $ } \\
0 & \text{ if \ $| \xi | \ge 2 $} .
\endcases \leqno(3.2)
$$
%
This decomposition shall make the discussions in this
section rather simple.
It is evident that $\big(1 - \chi (D) \big) \sqrt{-\Delta}$ maps
$\Cal S({\bold R}^n)$ continuously
into itself. Hence $\big(1 - \chi (D) \big) \sqrt{-\Delta}$ maps
$\Cal S^{\prime}({\bold R}^n)$ into itself. More precisely, we
have the following result.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 3.1.}\ {\it Let $k$ be in $\bold R$.
Then $ \big( 1 - \chi (D) \big) \sqrt {-\Delta}$ is a
bounded operator from $H^k( {\bold R}^n)$
to $H^{k-1}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ Let $f$ be in $H^k({\bold R}^n)$. We then see that
%
$$
\eqalign{
\Vert \big( 1 - &\chi(D) \big) \sqrt{-\Delta}f
{\Vert}_{k-1, \, 0} \cr
\noalign{\vskip 3pt}
{}& =
\Big\{ \int \big| \langle \xi {\rangle}^{k-1}
\big( 1 - \chi (\xi) \big) \, |\xi| \hat f (\xi) \,
{\big|}^2 \, d\xi \,{\Big\} }^{1/2} \cr
\noalign{\vskip 3pt}
{}& \qquad
\le \Big\{ \sup_{\xi} \big( \langle \xi {\rangle}^{-1}
\big( 1 -\chi(\xi) \big)
\, |\xi| \big) \Big\} \,
\Vert f {\Vert}_{k, \, 0}.
}
$$
%
This inequality proves the lemma. \ \ \ $\square$
\vskip 20pt
We next show a boundedness result of
the singular part $\chi (D) \sqrt{-\Delta}$.
Since the symbol $\chi(\xi) |\xi|$ is a bounded function,
we readily see that
$\chi (D) \sqrt{-\Delta}$ defines a bounded operator in
$L^2({\bold R}^n)$. Furthermore, we can prove that
$\chi (D) \sqrt{-\Delta}$ defines a bounded operator from
a Sobolev space of any order to another Sobolev space
of any order.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 3.2.}\ {\it Let $k$ and $\ell$ be in $\bold R$.
Then $\chi (D) \sqrt {-\Delta}$ is
bounded from $H^k( {\bold R}^n)$
to $H^{\ell}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ Let $f$ be in $H^k({\bold R}^n)$.
Since the support of $\chi$ is compact, it follows that
$\langle \xi {\rangle}^{\ell}\chi (\xi)|\xi| \hat f$ belongs to
$L^2({\bold R}^n_{\xi})$, and that
%
$$
\Vert \langle D {\rangle}^{\ell}\chi (D) \sqrt{-\Delta} f \Vert_0
\le
\Big\{ \sup_{\xi} \big( \big|
\langle \xi {\rangle}^{\ell} \chi(\xi) \, |\xi| \, \langle \xi {\rangle}^{-k}
\big| \big) \Big\} \Vert f \Vert_{k, \, 0}.
$$
%
This implies the lemma. \ \ \ {$\square$}
\vskip 15pt
Combining Lemmas 3.1 and 3.2, we obtain one of the main results.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Theorem 3.3.}\ {\it Let $k$ be in $\bold R$. Then
$\sqrt{-\Delta}$ is bounded from $H^k( {\bold R}^n)$
to $H^{k-1}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The end of Section 3 %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 30pt
\noindent
{\S 4 \bf Examples.}
\vskip 10pt
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section, we deal with $\sqrt{-\Delta}$ on the line, i.e.
$\sqrt{-d^2 / dx^2}$, and make explicit computations of
$\sqrt{-d^2 / dx^2}f$
for
$f(x) = \varphi_0(x):=\exp (-x^2 \!/2)$ and
$f(x) =f_0(x):= 1 /(1+x^2)$. The results obtained below
show that
$\sqrt{-d^2 / dx^2}\, \varphi_0$ and
$\sqrt{-d^2 / dx^2} \, f_0$, together with
their all derivatives, have exactly the same asymptotic behaviors,
apart from a multiplicative constant, as $|x| \to \infty$. This fact
reveals that the operator $\sqrt{-\Delta}$ possesses a
surprisingly interesting property, which neither differential operators
nor pseudodifferential operators with smooth symbols possess.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Example 4.1.} \ Consider the Gaussian function
%
$$
\varphi_0 (x) = \exp ( -x^2 /2 ),
$$
%
which obviously belongs to $\Cal S ( {\bold R})$. Since the Fourier
transform $\hat\varphi_0$ is given by $\exp ( -\xi^2 /2 )$
(e.g. Strichartz [5, p. 41]) which
is an even function, we see that
%
$$
\sqrt{-d^2 / dx^2}\, \varphi_0 (x) =
\sqrt{\frac{2}{\pi}} \int_0^{\infty} \xi \, \hat \varphi_0 (\xi)
\cos (x \xi) \, d\xi . \leqno (4.1)
$$
%
In view of the fact that
$(d /d\xi) \hat \varphi_0 = - \xi \hat \varphi_0$, we can integrate the
right hand side of (4.1), and obtain
%
$$
\sqrt{-\frac{d^2}{dx^2} }\, \varphi_0 (x) =
\sqrt{\frac{2}{\pi}} \,
\Big\{ 1 - x \int_0^{\infty}
\exp \big( \! - \frac{\xi^2}{2} \big)
\sin (x \xi) \, d\xi \, \Big\} . \leqno (4.2)
$$
%
To investigate the
asymptotic behavior of $\sqrt{-d^2 / dx^2}\, \varphi_0$
as $| x | \to \infty$, we put
%
$$
g(x) = \int_0^{\infty}
\exp \big( \! - \frac{\xi^2}{2} \big)
\sin (x \xi) \, d\xi . \leqno (4.3)
$$
%
Then it is straightforward to verify that $g$ is a $C^{\infty}$-function,
and satisfies the ordinary differential
equation with an initial condition:
%
$$
\left\{
\aligned
{}&\frac{dg}{dx} + x g = 1 \\
{}&g(0) = 0.
\endaligned \right. \leqno (4.4)
$$
%
(The solution to the equation (4.4) is given by
%
$$
g(x) = \exp \big( - \frac{x^2}{2} \big)
\int_0^x \exp \big( \frac{t^2}{2} \big) \, dt,
$$
%
which is not an elementary function, and not a suitable expression
for our purpose. ) It then follows from
(4.2) -- (4.4) that
%
$$
\sqrt{-\frac{d^2}{dx^2} }\, \varphi_0 (x) =
\sqrt{\frac{2}{\pi}} \,\frac{dg}{dx}(x) .
$$
%
Differentiating the right hand side of (4.3) under
the integral sign, and integrating by parts twice
yield
%
$$
x^2 \frac{dg}{dx}(x) =
-1 - \int_0^{\infty} \frac{d^2}{d\xi^2} \Big\{
\xi \exp \big( \! - \frac{\xi^2}{2} \big) \Big\}
\cos (x \xi) \, d\xi .
$$
%
The Riemann-Lebesgue theorem now implies that
%
$$
x^2 \frac{dg}{dx}(x) \to -1
\quad \text{as } \; |x| \to \infty.
$$
%
Furthermore, we find, by induction, that
%
$$
\aligned
x^{k+1} \frac{d^k g}{dx^k}(x) &=
( -1 )^k \, k! \\
{}& \;\; + \int_0^{\infty} \frac{d^{k+1}}{d\xi^{k+1}}
\Big\{
\xi^{k} \exp \big( \! - \frac{\xi^2}{2} \big) \Big\}
\sin (x \xi + \frac{2k+1}{2}\pi) \, d\xi \\
{}& \qquad \qquad \qquad ( k = 1, \, 2, \, 3, \cdots ),
\endaligned
$$
%
from which it follows that
%
$$
x^{k+2} \Big(
\frac{d^k }{dx^k}\sqrt{-\frac{d^2}{dx^2} }\, \varphi_0 \Big)
(x) \to (-1)^{k+1}
\sqrt{\frac2{\pi}} \, (k+1)! \leqno (4.5)
$$
%
as $|x| \to \infty$. Here we have used the Riemann-Lebesgue theorem
again. Thus we have shown the following
facts for all $k = 0, \, 1, \, 2, \cdots$:
\vskip 5pt
\item{(i)} $ \dfrac{d^k }{dx^k}\sqrt{-\dfrac{d^2}{dx^2} }\, \varphi_0 $ belongs
to $L^{2, \, s}(\bold R)$ for all $s < k + \dfrac3{2}$;
\vskip 5pt
\item{(ii)} $ \dfrac{d^k }{dx^k}\sqrt{-\dfrac{d^2}{dx^2} }\, \varphi_0 $
does not belong
to $L^{2, \, s}(\bold R)$ for any $s \ge k + \dfrac3{2}$.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Example 4.2.} Let $f_0$ be a function defined by
%
$$
f_0(x)=\frac1{x^2 +1}.
$$
%
It is well-known (cf. Stricharts [5, p.49]) that
%
$$
\hat f_0 (\xi) = \sqrt{ \frac{\pi}{2} } e^{-|\xi|} .
$$
%
By a similar computation as in Example 5.1, we see that
%
$$
h(x):= \sqrt{-\frac{d^2}{dx^2}}f_0 (x)
= \int_0^{\infty} \xi \, e^{-\xi}\cos (x\xi) \, d\xi.
$$
%
We then find, by integration by parts, that
%
$$
x^2 h(x) = -1 - h(x) + 2
\int_0^{\infty} \, e^{-\xi}\cos (x\xi) \, d\xi,
$$
%
which implies that
%
$$
h(x) = - \frac{1}{x^2 + 1} + \frac{2}{(x^2 + 1)^2}. \leqno (4.6)
$$
%
It follows from (4.6) that for $k=0, \, 1, \, 2, \cdots $
%
$$
x^{k+2} \Big(
\frac{d^k }{dx^k}\sqrt{-\frac{d^2}{dx^2} } \, f_0 \, \Big)
(x) \to (-1)^{k+1} \, (k+1)!
$$
%
as $|x| \to \infty$. This is exactly the same as (4.5), apart from a
multiplicative constant.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%The end of section 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 30pt
\noindent
{\S 5 \bf $\sqrt{-\Delta}$ in weighted $L^2$-spaces.}
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The examples in the previous section suggest that the operator
$\sqrt{-\Delta}$ should be considered in weighted $L^2$-spaces and
weighted Sobolev spaces. In this section, we consider
action of $\sqrt{-\Delta}$ in weighted $L^2$-spaces.
As was mentioned in Section 3, the regular part $\big(1 -\chi (D) \big)
\sqrt{-\Delta}$ maps $\Cal S({\bold R}^n)$ continuously into
itself, so it maps $\Cal S({\bold R}^n)$ continuously into
$L^{2, \, s}({\bold R}^n)$ for any $s$ in $\bold R$. So the main task
in this
section is to examine whether the singular part
$\chi (D) \sqrt{-\Delta}$ maps $\Cal S({\bold R}^n)$
into $L^{2,\,s}({\bold R}^n)$ for some $s>0$. To this end, we first write
$\chi (D) \sqrt{-\Delta}$
as an integral operator:
%
$$
\eqalign{
\chi(D) \sqrt{-\Delta} f (x) &=
(2\pi)^{-n/2} \!\!
\int e^{ix \cdot \xi} \, \chi(\xi) \, |\xi|
\, \Big\{ (2\pi)^{-n/2} \!\!
\int e^{- iy \cdot \xi} f(y) \, dy \Big\} \, d\xi \cr
\noalign{\vskip 3pt}
{}&= \int K(x-y) f(y) \, dy, \cr
{} & = K *f (x)
} \leqno(5.1)
$$
%
where
%
$$
K(y) = (2\pi)^{-n} \!\!
\int e^{iy \cdot \xi} \, \chi(\xi) \, |\xi| \, d\xi .
$$
%
We note that the integrals in (5.1) are finite
at least for $f \in \Cal S({\bold R}^n)$.
Also, by inspection, we see that $K$ is
a $C^{\infty}$-function, and that $K$ together with
its all derivatives is bounded on ${\bold R}^n$.
More generally, we introduce an integral kernel
%
$$
K_{\alpha}(y) = (2\pi)^{-n} \!\!
\int e^{iy \cdot \xi} \, \chi(\xi) \,
|\xi| \,{\xi}^{\alpha} \, d\xi
$$
%
for each multi-index $\alpha$.
It is true again that $K_{\alpha}$ is a $C^{\infty}$-function
bounded together with its all derivatives.
Note that
%
$$
\chi(D) \sqrt{-\Delta} D^{\alpha} f = K_{\alpha} * f. \leqno(5.2)
$$
%
Also, note that $K_0 = K$.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 5.1.} \ {\it For each $\, 0 < \sigma <1 $,
there exists a positive constant $C_{\sigma}$, depending
also on $\alpha$ and $n$, such that
%
$$
| K_{\alpha} (y) | \le C_{\sigma} \langle y
{\rangle}^{\sigma -n-1 -|\alpha|}. \leqno (5.3)
$$
%
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.} (Following the idea of Nagase and Umeda [4, Lemma 2.2].)
\ Since $K_{\alpha}$ is a bounded function, it
is sufficient to show that for each $\, 0 < \sigma <1 $,
there exists a positive constant $C_{\sigma}$ such that
%
$$
| K_{\alpha}(y) | \le C_{\sigma} |y|^{\sigma -n-1 -|\alpha|}. \leqno (5.4)
$$
%
To show (5.4), we need to introduce a cut-off function at the
origin in the $\xi$-space:
Let $\psi$ be a $C^{\infty}$-function such that
%
$$
\psi(\xi) = \cases 0 \;\; &\text{if $| \xi | \le 1$} \\
1 \;\; &\text{if $|\xi| \ge 2$} . \endcases
$$
%
We then put
%
$$
K_{\alpha}(y; \varepsilon) = (2\pi)^{-n} \!\! \int e^{iy\cdot \xi} \,
\psi({\varepsilon}^{-1} \xi) \,
\,\chi(\xi) \,
|\xi| \, {\xi}^{\alpha} \, d\xi \leqno (5.5)
$$
%
for every $\varepsilon > 0$.
The Lebesgue dominated convergence theorem implies
that $K_{\alpha}(y; \varepsilon)$ converges pointwise
to
$K_{\alpha}(y)$ as $\varepsilon \to 0$.
Hence, in order to obtain (5.4), it is sufficient to show that for each
$ \,0 < \sigma <1$,
there exists a positive constant $C_{\sigma}$, independent of $\varepsilon >0$,
such that
%
$$
| K_{\alpha} (y; \varepsilon) | \le C_{\sigma} | y |^{\sigma -n-1-|\alpha|}.
\leqno (5.6)
$$
%
Since the integrand of the right hand side of (5.5)
is a $C^{\infty}$-function of $\xi$ with compact support, we see,
by repeated use of integration by parts, that
%
$$
y^{\beta}K_{\alpha}(y; \varepsilon) = (2\pi)^{-n} \!\!
\int e^{iy\cdot \xi} \,
\Big( i \frac{\partial}{\partial \xi} {\Big)}^{\!\beta} \,
\! \Big\{
\psi({\varepsilon}^{-1} \xi) \,
\,\chi(\xi) \,
|\xi| \, {\xi}^{\alpha} \Big\} d\xi \leqno (5.7)
$$
%
for any multi-index $\beta$. In particular, if $\beta \not = 0$, taking
$y=0$ in (5.7) gives
%
$$
0= (2\pi)^{-n} \!\! \int
\Big( i \frac{\partial}{\partial \xi} {\Big)}^{\!\beta} \,
\! \Big\{
\psi({\varepsilon}^{-1} \xi) \,
\,\chi(\xi) \,
|\xi| \, {\xi}^{\alpha} \Big\} d\xi . \leqno (5.8)
$$
%
Subtracting (5.8) from (5.7), we get
%
$$
|y^{\beta}K_{\alpha}(y; \varepsilon)| \le (2\pi)^{-n} \!
\int \Big| ( e^{iy\cdot \xi} -1 )
\Big( i \frac{\partial}{\partial \xi} {\Big)}^{\!\beta} \,
\! \big\{
\psi({\varepsilon}^{-1} \xi) \,
\,\chi(\xi) \, |\xi| \, {\xi}^{\alpha}
\big\} \Big| \, d\xi \leqno (5.9)
$$
%
for any $\beta \not= 0$. Here we note that the support of every derivative
of order $\ge 1$ of $\psi({\varepsilon}^{-1} \xi)$ is contained in
the annulus $\varepsilon \le |\xi| \le 2 \varepsilon$.
Using this fact, we deduce that
%
$$
\Big| \Big( i \frac{\partial}{\partial \xi} {\Big)}^{\!\beta} \,
\! \big\{
\psi({\varepsilon}^{-1} \xi) \, \chi(\xi) \,
|\xi| \, {\xi}^{\alpha} \big\} \Big|
\le C_{\alpha\beta}
| \xi |^{1+ |\alpha| -|\beta|} , \leqno (5.10)
$$
%
where $C_{\alpha\beta}$ is a constant
independent of $\varepsilon >0$.
On noting the inequalities
$| e^{iy\cdot \xi} -1 | \le 2 $ and $ |e^{iy\cdot \xi} -1 | \le |y| \,|\xi|$,
we see, by interpolation,
that
%
$$
|e^{iy\cdot \xi} -1 | \le 2 |y|^{\sigma} \,|\xi|^{\sigma} \leqno(5.11)
$$
%
for any $0 < \sigma < 1$.
Combining (5.9) -- (5.11), we now obtain
%
$$
|y^{\beta}K_{\alpha}(y; \varepsilon)|
\le (2\pi)^{-n} \, C_{\alpha\beta} \, 2|y|^{\sigma}
\int_{|\xi| \le 2}
|\xi|^{\sigma + |\alpha| +1 - | \beta |}
\, d\xi \leqno(5.12)
$$
%
for any $\beta \not= 0$ and any $0 < \sigma < 1$,
where $C_{\alpha\beta}$ is the same constant as in (5.10).
The integral on the right hand side of (5.12) is finite for all
$\beta$ satisfying $| \beta | \le |\alpha| + 1+n$.
This fact implies (5.6)
\ \ \ {$\square$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
\noindent
{\bf Remark 5.2.} The inverse Fourier transform of $|\xi|$ is
formally given by\linebreak
$\text{const.} |x|^{-n -1}$, of which
singularity is very severe. This fact
illustrates that it is not
quite appropriate to handle
$\sqrt{-\Delta}$ as an integral operator. As was shown in Lemma 5.1,
the decomposition
of $\sqrt{-\Delta}$ into the singular part and the regular part
enables us to avoid the singularity of the integral
kernel at the cost of an arbitrarily small loss of
the decay rate at infinity.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 5.3.} \ {\it If $s < n/2 + 1$, then $\chi(D) \sqrt{-\Delta}$
maps $\Cal S({\bold R}^n)$ continuously
into $L^{2, \, s}({\bold R}^n)$.
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ Let $\varphi$ be in $\Cal S ({\bold R}^n)$, and
choose, if necessary, $0 < \sigma < 1$ so that
%
$$
\sigma < \frac{n}2 + 1 - s . \leqno (5.13)
$$
%
It follows from (5.1) and Lemma 5.1 with $\alpha = 0$ that
%
$$
| \chi(D) \sqrt{-\Delta} \varphi (x) |
\le C_{\sigma} \int
\langle x - y {\rangle}^{\sigma - n -1}
| \varphi(y) | \, dy. \leqno (5.14)
$$
%
Appealing to the inequality
%
$$
\langle x - y {\rangle}^{-1} \le 2
\langle x {\rangle}^{-1} \langle y \rangle , \leqno (5.15)
$$
%
we get
%
$$\eqalign{
| \chi(D) \sqrt{-\Delta} \varphi (x) |
& \le C^{\prime}_{\sigma}
\langle x {\rangle}^{\sigma -n -1} \int
\langle y {\rangle}^{-\sigma +n +1}
| \varphi(y) | \, dy \cr
\noalign{\vskip 4pt}
{} & \le
C^{\prime\prime}_{\sigma}
\langle x {\rangle}^{\sigma -n -1} \, | \varphi|_{2n +1, \, \Cal S},
}
$$
%
which implies that
%
$$
\Vert \chi(D) \sqrt{-\Delta} \varphi { \Vert}_s^2 \le
{\big( C^{\prime\prime}_{\sigma} | \varphi|_{2n +1, \, \Cal S} \big)}^2
\int \langle x {\rangle}^{2( s + \sigma -n -1)} \,dx.
\leqno(5.16)
$$
%
Since $2( s + \sigma -n -1) < -n$ by (5.13), the assertion of
the lemma follows
from (5.16). \ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Theorem 5.4.} \ {\it If $s < n/2 + 1$, then $\sqrt{-\Delta}$
maps $\Cal S({\bold R}^n)$ continuously
into $L^{2, \, s}({\bold R}^n)$.
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ Lemma 5.3, together with the fact that
the regular part $\big( 1 - \chi (D) \big) \sqrt{-\Delta}$
maps $\Cal S ({\bold R}^n)$ continuously into
$L^{2, \, s}({\bold R}^n)$ for any $s$, gives the lemma.
\ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
%%%%%%%%%%%%%
%%%%%%%%%%%%%
\noindent
{\bf Remark 5.5.} When $n=1$, Theorem 5.4 is optimal in the
following sense: there exists a function
$\varphi_0 \in \Cal S (\bold R)$ such that
$ \sqrt{-d^2 / dx^2} \, \varphi_0$
belongs to $L^{2, \, s}({\bold R})$
for all $s < 3/2$, but
does not belong to $L^{2, \, s}({\bold R})$
for any $s \ge 3/2$ (see Example 4.1 in Section 4).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
Theorem 5.4 allows us to define $\sqrt{-\Delta} f$ as a
tempered distribution for
some $f$ which does not belong to $H^{-\infty}({\bold R}^n)$.
\vskip 15pt
%%%%%%%%%%%%%%
%%%%%%%%%%%%%%
\noindent
{\bf Theorem 5.6.}\ Let $f$ be in $L^{2, \, -s}({\bold R}^n)$ for
some $s < n/2 +1$. Then $\sqrt{-\Delta} f$ is in
${\Cal S}^{\prime}({\bold R}^n)$.
%%%%%%%%%%%%%%
%%%%%%%%%%%%%%
\noindent
{\it Proof.} \ It follows from the definition
%
$$
\big< \sqrt{-\Delta} f, \, \varphi \big>
= \big< f , \,
\overline{ \sqrt{-\Delta} \, \overline{\varphi} } \, \big>
\qquad
( \, \varphi \in {\Cal S}({\bold R}^n) \, )
$$
%
and Theorem 5.4 that
%
$$\eqalign{
\big| \big< \sqrt{-\Delta} f, \, \varphi \big> \big|
& \le \Vert f {\Vert}_{-s}
\Vert \sqrt{-\Delta} \overline{ \varphi} {\Vert}_s \cr
{} & \le C \Vert f {\Vert}_{-s} \,
| \varphi |_{\ell, \, \Cal S} ,
}
$$
%
where $C$ is a positive constant and $\ell$ is a
positive integer. \ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The end of Section 5 %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 30pt
\noindent
{\S 6 \bf $\sqrt{-\Delta}$ on weighted Sobolev spaces.}
\vskip 10pt
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section we work in weighted Sobolev
spaces. We exploit the decomposition
(3.1) again. As to the regular part
$\big( 1 - \chi(D) \big) \sqrt{-\Delta}$, Lemma 2.1
immediately gives a boundedness result in weighted
Sobolev spaces.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.1.} \ {\it Let $k$ and $s$ be in $\bold R$.
Then $\big( 1 - \chi(D) \big) \sqrt{-\Delta}$ is a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k-1, \,s}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
To investigate action of the singular part $\chi(D) \sqrt{-\Delta}$
in weighted Sobolev spaces, we need some preliminary lemmas, all of
which are based on Lemma 5.1.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.2.} \ {\it If $0 \le s < n/2 + 1$ and $t < s - n/2$, then
%
$$
\Vert \chi(D) \sqrt{-\Delta} \varphi {\Vert}_t
\le C \Vert \varphi {\Vert}_s
\quad \qquad
\big( \varphi \in {\Cal S ({\bold R}^n)} \big), \leqno (6.1)
$$
%
where $C$ is a constant depending on $s$ and $t$.
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ Let $\varphi$ be in $\Cal S ({\bold R}^n)$, and
choose $0 < \sigma <1$ so that
%
$$
\sigma < \frac{n}2 +1 -s . \leqno (6.2)
$$
%
Then it follows from (5.1) and Lemma 5.1, with $\alpha=0$, that
%
$$\eqalign{
\Vert \chi(D) \sqrt{-\Delta} \varphi {\Vert}_t ^2
& \le C_{\sigma}^2 \int \langle x {\rangle}^{2t} \,
\Big\{ \int \langle x-y {\rangle}^{\sigma -n -1}
| \varphi (y) | \, dy
{ \Big\} }^2 dx \cr
\noalign{\vskip 4pt}
{}&
\le C_{\sigma}^2 \, \Vert \varphi {\Vert}_s^2
\int \langle x {\rangle}^{2t} \,
\Big\{ \int \langle x-y {\rangle}^{2(\sigma -n -1)}
\langle y {\rangle}^{-2s} \, dy
\Big\} dx ,
}
$$
%
where the Schwarz inequality was used in the second inequality.
In view of the inequality (5.15) and the fact that $s \ge 0$, we have
%
$$\eqalign{
\langle x-y {\rangle}^{2(\sigma -n -1)} &=
\langle x-y {\rangle}^{2(\sigma -n -1+s)}
\langle x-y {\rangle}^{-2s} \cr
{} & \le
2^{2s} \langle x-y {\rangle}^{2(\sigma -n -1+s)}
\langle x {\rangle}^{-2s} \langle y {\rangle}^{2s}.
}
$$
%
Hence we get
%
$$
\Vert \chi(D) \sqrt{-\Delta} \varphi {\Vert}_t ^2
\le C_{\sigma}^2 \, 2^{2s} \, \Vert \varphi {\Vert}_s^2
\int \langle x {\rangle}^{2(t-s)} \, dx \,
\int \langle x-y {\rangle}^{2(\sigma -n -1+ s )} dy.
$$
%
This implies (6.1), since $2(t -s ) < -n$, by assumption of the lemma,
and since $2(\sigma -n -1 +s) < -n$ by (6.2).
\ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.3.} \ {\it If $s \ge n/2 + 1$ and $t < 1$,
then the inequality (6.1) holds.
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \ By assumption it is possible to choose $\tilde s \ge 0$ so
that
%
$$
\frac{n}2 + t < {\tilde s} < \frac{n}2 +1. \leqno (6.3)
$$
%
Then Lemma 6.2 yields
%
$$
\Vert \chi(D) \sqrt{-\Delta} \varphi {\Vert}_t
\le C \Vert \varphi {\Vert}_{\tilde s}
\quad \qquad
\big( \varphi \in \Cal S( {\bold R}^n ) \big) . \leqno (6.4)
$$
%
Since ${\tilde s} < s$ by (6.3) and the
assumption of the lemma, it follows
from (6.4)
that (6.1) holds with the same constant $C$ as in (6.1).
\ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.4.} \ {\it If $-n/2 - 1 -|\alpha| < s < 0 $ and $t < s - n/2$,
then
%
$$
\Vert \chi(D) \sqrt{-\Delta} \, D^{\alpha} \varphi {\Vert}_t
\le C \Vert \varphi {\Vert}_s
\quad \qquad
\big( \varphi \in {\Cal S ({\bold R}^n)} \big),
$$
%
where $C$ is a constant depending on $s$, $t$ and $\alpha$.
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.}\ \
Choose $0 < \sigma <1$ so that
%
$$
\sigma < \frac{n}2 +1 + |\alpha| + s . \leqno (6.5)
$$
%
It then follows from (5.2) and Lemma 5.1 that
%
$$\eqalign{
\Vert \chi(D) \sqrt{-\Delta} \, D^{\alpha}\varphi {\Vert}_t ^2
& \le C_{\sigma}^2 \int \langle x {\rangle}^{2t} \,
\Big\{ \int \langle x-y {\rangle}^{\sigma -n -1 -|\alpha|}
| \varphi (y) | \, dy
{ \Big\} }^2 dx \cr
\noalign{\vskip 4pt}
\le C_{\sigma}^2 \,
&\Vert \varphi {\Vert}_s^2
\int \langle x {\rangle}^{2t} \,
\Big\{ \int \langle x-y {\rangle}^{2(\sigma -n -1 -|\alpha| )}
\langle y {\rangle}^{-2s} \, dy
\Big\} dx .
} \leqno (6.6)
$$
%
Because $-2s >0$, the inequality $\langle y {\rangle} \le 2
\langle x - y {\rangle} \langle x {\rangle}$
yields
%
$$
\langle y {\rangle}^{-2s} \le 2^{-2s}
\langle x - y {\rangle}^{-2s} \langle x {\rangle}^{-2s} ,
$$
%
so that (6.6) implies that
%
$$\eqalign{
\Vert \chi(D) \sqrt{-\Delta} \, D^{\alpha}\varphi {\Vert}_t ^2
& \le C_{\sigma}^2 \, 2^{-2s} \,
\Vert \varphi {\Vert}_s^2 \, \times \cr
{} & \qquad \int \langle x {\rangle}^{2(t-s)} \, dx
\int \langle x-y {\rangle}^{2(\sigma -n -1 -|\alpha|-s )}
\, dy .
} \leqno (6.7)
$$
%
The integrals on the right hand side of (6.7) are finite,
since $2(t-s) < -n$ and $2(\sigma -n -1 -|\alpha|-s ) < -n.$
\ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
As a direct consequence of Lemma 6.4, we have the following estimates.
\vskip 15pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.5.} \ {\it If $-n/2 - 1 < s < 0 $ and $t < s - n/2$,
then
%
$$
\Vert \chi(D) \sqrt{-\Delta} \varphi {\Vert}_t
\le C \Vert \varphi {\Vert}_s
\quad \qquad
\big( \varphi \in {\Cal S ({\bold R}^n)} \big).
$$
%
}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
We now notice the following two facts:
\vskip 5pt
\noindent
{\it Fact 1. $\Cal S ({\bold R}^n)$ is dense in
$H^{k, \, s}({\bold R}^n)$ for all $k$ and $s$ in $\bold R$.}
\vskip 5pt
\noindent
{\it Fact 2. If $\varphi$ is in $\Cal S ({\bold R}^n)$, then
$\langle D {\rangle}^k \varphi$ is also in
$\Cal S ({\bold R}^n)$ for all $k$ in $\bold R$ and
$\Vert \langle D {\rangle}^k \varphi{\Vert}_s=
\Vert \varphi{\Vert}_{k, \,s}$ for all $s$ in $\bold R$.}
\vskip 10pt
Combining these two facts with Lemmas 6.2 -- 6.5, we immediately
get
results on the boundedness of
the singular part in weighted Sobolev spaces.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.6.} \ {\it Let $k$ be in $\bold R$.}
\item{(i)} {\it If $s \ge 0$ and $t < \min\{ 1, \, s -n/2\}$,
then $\chi(D) \sqrt{-\Delta}$ can be extended to a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k, \,t}( {\bold R}^n)$.}
\item{(ii)} {\it If $- n/2 - 1 < s < 0$ and $t < s -n/2$,
then the same conclusion as in (i) is valid.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Lemma 6.7.} \ {\it Let $k$ be in $\bold R$.
Under the same assumption as in Lemma 6.4,
$\chi(D) \sqrt{-\Delta} D^{\alpha}$ can be extended to a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k, \,t}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 15pt
Finally we can state the main results on action
of $\sqrt{-\Delta}$ on weighted Sobolev spaces.
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Theorem 6.8.} \ {\it Let $k$ be in $\bold R$.}
\item{(i)} {\it If $s \ge 0$ and $t < \min\{ 1, \, s -n/2\}$,
then $\sqrt{-\Delta}$ can be extended to a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k-1, \,t}( {\bold R}^n)$.}
\item{(ii)} {\it If $- n/2 - 1 < s < 0$ and $t < s -n/2$,
then the same conclusion as in (i) is valid.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.} \ Conclusions (i) and (ii) follow immediately from
Lemmas 6.6 and 6.1. \ \ \ $\square$
\vskip 20pt
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf Theorem 6.9.} \ {\it Let $k$ be in $\bold R$.
If $-n/2 -1 -|\alpha| < s <0$ and
$t < s - n/2$, then
$\sqrt{-\Delta} D^{\alpha}$ can be extended to a
bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k-1 -|\alpha|, \,t}( {\bold R}^n)$.}
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%
\vskip 5pt
\noindent
{\it Proof.} \ It follows from
Lemma 2.1 that $\big( 1 - \chi(D) \big) \sqrt{-\Delta}D^{\alpha}$
is a bounded operator from
$ H^{k, \,s} ( {\bold R}^n)$ to $ H^{k-1 -|\alpha|, \,s}( {\bold R}^n)$.
This fact, together with Lemma 6.7, gives the conclusion of the theorem.
\ \ \ $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The end of Section 6 %%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\yr1990 \vol 22 \jour Bull. Amer. Math. Soc. \endref
\ref \no 3 \by E. H. Lieb and M. Loss \book Analysis \publ American
Mathematical Society
\yr1997 \endref
\ref \no 4 \by M. Nagase and T. Umeda \pages 277--296
\paper Weyl quantized Hamiltonians of
relativistic spinless particles in magnetic fields
\yr1990 \vol 92 \jour J. Funct. Analysis \endref
\ref \no 5 \by R. Strichartz \book A guide to
distribution theory and Fourier transforms \publ CRC Press
\yr1994 \endref
\ref \no 6 \by F. Treves \book Topological vector spaces,
distributions and kernels \publ Academic Press
\yr1967 \endref
\ref \no 7 \by T. Umeda \pages 277--296 \paper Radiation
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\yr1995 \vol 63 \jour Ann. Inst. Henri Poincar\'e, Phys. th\'eor. \endref
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\endRefs
\enddocument
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