Boundedness commutator between Sobolev spaces
By Sadek Gala
Universit\'e d'Evry Val d'Essonne
D\'epartement de math\'ematiques
Bd F. Mitterrand. 91025 Evry Cedex. France
In this paper,we will study the boundedness properties of commutator \[ C_{f}=\left[ f,\Delta\right] \] acting from $\overset{.}{H}^{1}\left( \mathbb{R}^{d}\right) $ to $\overset{.}{H}^{-1}\left( \mathbb{R}^{d}\right) .$ %\end{frontmatter} %\begin{frontmatter} %\selectlanguage{francais} \vskip 0.5\baselineskip \noindent% Text of abstract in French %\noindent{R\'esum\'e}\\ %\begin{abstract} %\selectlanguage{francais} % Texte du r�sum� en fran�ais %{R\'esum\'e.} %Dans cet article, on \'{e}tudiera la propri\'{e}t\'{e} de %continuit\'{e} du commutateur agissant de l'espace de Sobolev $\overset{.}% %{H}^{1}\left( \mathbb{R}^{d}\right) $ sur son dual $\overset{.}{H}% %^{-1}\left( \mathbb{R}^{d}\right) .$
math0505169.tex
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\begin{document}
%\begin{frontmatter}
\selectlanguage{english}
\vskip 0.5\baselineskip
\title{Boundedness commutator between Sobolev spaces}
\author{By Sadek Gala\\Universit\'e d'Evry Val d'Essonne\\
  D\'epartement de math\'ematiques\\ Bd F. Mitterrand. 91025 Evry
  Cedex. France\\ Sadek.Gala@maths.univ-evry.fr }
\maketitle
%\ead{Sadek.Gala@maths.univ-evry.fr}
%\address{Universit\'e d'Evry Val d'Essonne\\ D\'epartement de math\'ematiques\\ Bd F. Mitterrand. 91025 Evry Cedex. France}

\begin{abstract} In this paper,we will study the boundedness properties of commutator
\[
C_{f}=\left[  f,\Delta\right]
\]
acting from $\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)  $ to
$\overset{.}{H}^{-1}\left(  \mathbb{R}^{d}\right)  .$
%\end{frontmatter}
%\begin{frontmatter}
%\selectlanguage{francais}
\vskip 0.5\baselineskip
\noindent% Text of abstract in French

%\noindent{\bf R\'esum\'e}\\
%\begin{abstract}

%\selectlanguage{francais}
% Texte du r�sum� en fran�ais
%{\bf R\'esum\'e.}
%Dans cet article, on \'{e}tudiera la propri\'{e}t\'{e} de
%continuit\'{e} du commutateur agissant de l'espace de Sobolev $\overset{.}%
%{H}^{1}\left(  \mathbb{R}^{d}\right)  $ sur son dual $\overset{.}{H}%
%^{-1}\left(  \mathbb{R}^{d}\right)  .$
\end{abstract}
%\end{frontmatter}

%\section{Introduction}
Throughout this paper, $d$ will denote a fixed positive integer greater than 2
and $\Omega$ will denote an open, non-empty subset of $\mathbb{R}^{d}$. We
wish to study linear elliptic differential operator $\Delta$ in divergence
form defined on $C^{2}$ functions by
\[
\Delta f(x)=\frac{1}{2}\sum_{i,j=1}^{d}\partial_{i}\left(  \partial
_{j}f\right)  (x)
\]

Recall the divergence theorem. Suppose $\Omega$ is a nice region, $F$ is a
smooth vector field, $\nu(x)$ is the outward normal vector at $x\in
\partial\Omega$, and $\sigma$ is surface measure on $\partial\Omega$. The
divergence theorem then says that%
\[
\int_{\partial\Omega}F(y).\nu(y)\sigma(dy)=\int_{\Omega}divF(x)dx.
\]

\bigskip A twice continuously differentiable, complex-valued function $v$
defined on $\Omega$ is harmonic on $\Omega$ if
\[
\Delta v=0.
\]

We begin with the following simple observation :

\begin{proposition}
\label{prop1}
Let $g$ be a $C^{\infty}$ function with compact support and $f$ a bounded
$C^{\infty}$ function. Then
\[
\int_{\mathbb{R}^{d}}g(x)\Delta f(x)dx=-\frac{1}{2}\int_{\mathbb{R}^{d}%
}\left(  \sum_{i,j=1}^{d}\partial_{i}g(x)\partial_{j}f(x)\right)  dx.
\]
The integrand on the right could be written $\nabla g.\nabla f.$
\end{proposition}

\begin{proof}
We apply the divergence theorem. Let $B$ be a ball large enough to contain the
support of $g$ and let $F(x)$ be the vector field whose $i-$th component is
\[
\frac{1}{2}g(x)\sum_{j=1}^{d}\partial_{j}f(x).
\]
Since $g$ is $0$ on $\partial B$, then $F.\nu=0$ on $\partial B$, and also,
\begin{align*}
divF(x) &  =\frac{1}{2}\sum_{i=1}^{d}\partial_{i}\left(  g(x)\sum_{j=1}%
^{d}\partial_{j}f(x)\right) =\frac{1}{2}\sum_{i,j=1}^{d}\partial_{i}g(x)\partial_{j}f(x)+g(x)\Delta
f(x).
\end{align*}
\end{proof}

\begin{proposition}
\label{prop2}
Suppose $v$ is positive and harmonic in $Q\left(  4\right)  $.
There exists $c_{1}$ independent of $v$ such that if $w=\log v$, then
\[
\int_{Q}\left|  \nabla w(x)\right|  ^{2}dx\leq c_{1}h^{d-2}%
\]
for all cubes $Q$ of side length $h$ contained in $Q(2)$.
\end{proposition}

\begin{proof}
Let $Q^{\ast}$ be the cube with the same center as $Q$ but side length twice
as long. Note $Q^{\ast}\subset Q\left(  4\right)  $. Let $\varphi$ be
$C^{\infty}$ with values in $\left[  0,1\right]  $, equal to $1$ on $Q$,
supported in $Q^{\ast}$, and such that $\left\|  \nabla\varphi\right\|
_{\infty}\leq c_{2}h^{-1}$. Since
\[
\nabla w=\frac{\nabla v}{v}%
\]
and $v$ is harmonic in $Q\left(  4\right)  $,%
\begin{align*}
0  =2\int\frac{\varphi^{2}}{v}\Delta v=-\int\nabla\left(  \frac{\varphi
^{2}}{v}\right)  .\nabla v =-\int2\frac{\varphi\nabla\varphi}{v}.\nabla v+\int\frac{\varphi^{2}}%
{v^{2}}\nabla v.\nabla v \\=-2\int\varphi\nabla\varphi.\nabla w+\int\varphi^{2}\nabla w.\nabla w.
\end{align*}
So by the Cauchy-Schwarz inequality and proposition \ref{prop1},%
\begin{align*}
\int_{Q^{\ast}}\varphi^{2}\left|  \nabla w\right|  ^{2}dx  &  \leq c_{3}%
\int_{Q^{\ast}}\varphi^{2}\nabla w.\nabla w=c_{4}\int_{Q^{\ast}}\nabla
\varphi.\varphi\nabla w\\
&  \leq c_{5}\left(  \int_{Q^{\ast}}\left|  \nabla\varphi\right|
^{2}dx\right)  ^{\frac{1}{2}}\left(  \int_{Q^{\ast}}\varphi^{2}\left|  \nabla
w\right|  ^{2}dx\right)  ^{\frac{1}{2}}.
\end{align*}
Dividing by the second factor on the right, squaring and using the bound on
$\left|  \nabla\varphi\right|  $,%
\[
\int_{Q}\left|  \nabla w(x)\right|  ^{2}dx\leq\left(  \int_{Q^{\ast}}%
\varphi^{2}\left|  \nabla w\right|  ^{2}dx\right)  \leq c_{5}^{2}\left|
Q^{\ast}\right|  c_{2}^{2}h^{-2},
\]
which implies our result.
\end{proof}

We need to distinguish the class of vector fields $\overrightarrow{F}$ such
that the commutator inequality
\[
\left|  \int_{\mathbb{R}^{d}}\overrightarrow{F}.\left(  \overline{u}\nabla
v-v\nabla\overline{u}\right)  dx\right|  \leq C\left\|  u\right\|
_{\overset{.}{H}^{1}}\left\|  v\right\|  _{\overset{.}{H}^{1}}%
\]
for all $u$, $v\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  $. In the
important case where $\overrightarrow{F}=\nabla f$, the preceding inequality
is equivalent to the boundedness of the commutator
\[
C_{f}=\left[  f,\Delta\right]
\]
acting from $\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)  $ to
$\overset{.}{H}^{-1}\left(  \mathbb{R}^{d}\right)  .$

Another useful point in the sequel is the symmetry of the Laplace operator,
\[
\left\langle \Delta\varphi,\psi\right\rangle =-\left\langle D\varphi
,D\psi\right\rangle =\left\langle \varphi,\Delta\psi\right\rangle
\]
for all $\varphi$, $\psi\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  $.

Let us present a proof of the 

\begin{lemma}
For any twice continuously differentiable functions $u:A\rightarrow\mathbb{R}$
and $v:A\rightarrow\mathbb{R}$, where $A\subset\mathbb{R}^{d}$, we have%

\[
div\left(  u\nabla v-v\nabla u\right)  =u\Delta v-v\Delta u.
\]
\end{lemma}

\begin{proof}
Note that
\begin{align*}
&  div\left(  u\nabla v-v\nabla u\right) =div\left(  u\left(  \frac{\partial v}{\partial x_{1}},...,\frac{\partial
v}{\partial x_{d}}\right)  -v\left(  \frac{\partial u}{\partial x_{1}%
},...,\frac{\partial u}{\partial x_{d}}\right)  \right) \\
&  =div\left(  u\frac{\partial v}{\partial x_{1}}-v\frac{\partial u}{\partial
x_{1}},...,u\frac{\partial v}{\partial x_{d}}-v\frac{\partial u}{\partial
x_{d}}\right) \\
&  =\frac{\partial}{\partial x_{1}}\left(  u\frac{\partial v}{\partial x_{1}%
}-v\frac{\partial u}{\partial x_{1}}\right)  +...+\frac{\partial}{\partial
x_{d}}\left(  u\frac{\partial v}{\partial x_{d}}-v\frac{\partial u}{\partial
x_{d}}\right) \\
&  =\left(  u\frac{\partial^{2}v}{\partial x_{1}^{2}}+\frac{\partial
u}{\partial x_{1}}\frac{\partial v}{\partial x_{1}}-\frac{\partial v}{\partial
x_{1}}\frac{\partial u}{\partial x_{1}}-v\frac{\partial^{2}u}{\partial
x_{1}^{2}}\right)  +...+\left(  u\frac{\partial^{2}v}{\partial x_{d}^{2}%
}+\frac{\partial u}{\partial x_{d}}\frac{\partial v}{\partial x_{d}}%
-\frac{\partial v}{\partial x_{d}}\frac{\partial u}{\partial x_{d}}%
-v\frac{\partial^{2}u}{\partial x_{d}^{2}}\right) \\
&  =u\left(  \frac{\partial^{2}v}{\partial x_{1}^{2}}+...+\frac{\partial^{2}%
v}{\partial x_{d}^{2}}\right)  -v\left(  \frac{\partial^{2}u}{\partial
x_{1}^{2}}+...+\frac{\partial^{2}u}{\partial x_{d}^{2}}\right) =u\Delta v-v\Delta u.
\end{align*}
\end{proof}

\begin{lemma}
Let $f\in\mathcal{D}^{\prime}\left(  \mathbb{R}^{d}\right)  $. Then for all
$u$, $v\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  $, we have
\[
\left|  \left\langle C_{f}u,v\right\rangle\right|  =\left|  \int
_{\mathbb{R}^{d}}\overrightarrow{F}.\left(  \overline{u}\nabla v-v\nabla
\overline{u}\right)  dx\right|  .
\]
\end{lemma}

\begin{proof}
Let $u$, $v\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  .$ We observe that%
\begin{align*}
\left\langle C_{f}u,v\right\rangle  &  =\left\langle \left[  f,\Delta\right]
u,v\right\rangle =\left\langle f\Delta u-\Delta\left(  fu\right)
,v\right\rangle \\
&  =\left\langle f\Delta u,v\right\rangle -\left\langle \Delta\left(
fu\right)  ,v\right\rangle =\left\langle f,v\Delta\overline{u}\right\rangle
-\left\langle fu,\Delta v\right\rangle \\
&  =\left\langle f,v\Delta\overline{u}\right\rangle -\left\langle
f,\overline{u}\Delta v\right\rangle =\left\langle f,v\Delta\overline
{u}-\overline{u}\Delta v\right\rangle \\
&  =\left\langle f,div\left(  v\nabla\overline{u}-\overline{u}\nabla v\right)
\right\rangle \\
&  =-\left\langle \nabla f,v\nabla\overline{u}-\overline{u}\nabla
v\right\rangle =-\left\langle \overrightarrow{F},v\nabla\overline{u}%
-\overline{u}\nabla v\right\rangle
\end{align*}
Therefore%
\[
\left|  \left\langle C_{f}u,v\right\rangle \right|  =\left|  \int
_{\mathbb{R}^{d}}\overrightarrow{F}.\left(  \overline{u}\nabla v-v\nabla
\overline{u}\right)  dx\right|  .
\]
\end{proof}

We will let $P$ be the Newton potential of $\mu$ defined by%
\[
P\mu(x)=P(x)=C(d)\int_{\mathbb{R}^{d}}\frac{d\mu(y)}{\left|  x-y\right|
^{d-2}},\text{ \ \ }x\in\mathbb{R}^{d}\text{, \ \ }d\geq3.
\]
Then, we have the following lemma :

\begin{lemma}
\label{lemma 2.7}Let $\mu$ be a positive Borel measure on $\mathbb{R}^{d}$
such that $P(x)=I_{2}\mu(x)\neq\infty$. Then the following inequality hold :%

\begin{equation}
\int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla P(x)\right|  ^{2}}{P^{2}%
(x)}dx\leq4\left\|  \nabla u\right\|  _{L^{2}}^{2}\text{, \ \ }u\in
\mathcal{D}\left(  \mathbb{R}^{d}\right)  .\label{eq2}%
\end{equation}
\end{lemma}

\begin{proof}
Suppose $u\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  .$ Then $K=$supp $u$ is
a compact set, and obviously $\underset{x\in K}{\inf}P(x)>0.$ Without loss of
generality, we assume that $\nabla P\in L_{loc}^{2}\left(  \mathbb{R}%
^{d}\right)  $, and hence the left-hand side of (\ref{eq2}) is finite.

Using integration by parts, together with the properties $-\Delta P=\mu$
(understood in the distributional sense) and applying the Schwarz inequality,
we get :%
\begin{align*}
2\int_{\mathbb{R}^{d}}\nabla u(x).\nabla P(x)\frac{u(x)}{P(x)}dx  &
=\int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla P(x)\right|  ^{2}}%
{P^{2}(x)}dx+\int_{\mathbb{R}^{d}}u^{2}(x)\frac{d\mu(x)}{P(x)}dx\\
&  \leq2\left(  \int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla P(x)\right|
^{2}}{P^{2}(x)}dx\right)  ^{\frac{1}{2}}\left(  \int_{\mathbb{R}^{d}}\left|
\nabla u(x)\right|  ^{2}dx\right)  ^{\frac{1}{2}},
\end{align*}
for all $u\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  .$ Using this
inequality, we have
\[
\int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla P(x)\right|  ^{2}}{P^{2}%
(x)}dx\leq2\left(  \int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla
P(x)\right|  ^{2}}{P^{2}(x)}dx\right)  ^{\frac{1}{2}}\left(  \int
_{\mathbb{R}^{d}}\left|  \nabla u(x)\right|  ^{2}dx\right)  ^{\frac{1}{2}}.
\]
Consequently, we obtain%
\[
\int_{\mathbb{R}^{d}}u^{2}(x)\frac{\left|  \nabla P(x)\right|  ^{2}}{P^{2}%
(x)}dx\leq4\int_{\mathbb{R}^{d}}\left|  \nabla u(x)\right|  ^{2}dx.
\]
\end{proof}

\begin{lemma}
\bigskip\label{lemma 2}The Sobolev space $\overset{.}{H}^{1}\left(
\mathbb{R}^{d}\right)  $ be invariant under the transformation :%

\begin{align*}
u  & \rightarrow\widetilde{u}=e^{iw}u\text{, \ \ \ }v\rightarrow\widetilde
{v}=e^{iw}v,\\
\left\|  u\right\|  _{\overset{.}{H}^{1}}  & \simeq\left\|  \widetilde
{u}\right\|  _{\overset{.}{H}^{1}},\text{ \ \ \ \ \ \ }\left\|  v\right\|
_{\overset{.}{H}^{1}}\simeq\left\|  \widetilde{v}\right\|  _{\overset{.}%
{H}^{1}}%
\end{align*}
\end{lemma}

\begin{proof}
Clearly, We observe that $v\in\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)
$
\[
\nabla\left(  e^{iw}u\right)  =\left(  iu\nabla w+\nabla u\right)  e^{iw}.
\]
Consequently, for every $u\in\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)
,$%
\[
\left\|  \widetilde{u}\right\|  _{\overset{.}{H}^{1}}=\left\|  e^{iw}%
u\right\|  _{\overset{.}{H}^{1}}\leq\left\|  u\nabla w\right\|  _{L^{2}%
}+\left\|  \nabla u\right\|  _{L^{2}}.
\]
We set 
\[
w=\log P\mu\text{, \ \ }v=P\mu.
\]
Note that
\[
\nabla w=\frac{\nabla P}{P}.
\]
Hence, by lemma \ref{lemma 2.7},
\[
\left\|  u\nabla w\right\|  _{\overset{.}{H}^{1}}=\left\|  u\frac{\nabla P}%
{P}\right\|  _{\overset{.}{H}^{1}}\leq2\left\|  u\right\|  _{\overset{.}%
{H}^{1}}.
\]
From this, it follows
\[
\left\|  \widetilde{u}\right\|  _{\overset{.}{H}^{1}}\leq3\left\|  u\right\|
_{\overset{.}{H}^{1}}.
\]
Using similar estimates for $\widetilde{u}\rightarrow u=e^{-iw}\widetilde{u}$,
we deduce
\[
\frac{1}{3}\left\|  u\right\|  _{\overset{.}{H}^{1}}\leq\left\|  \widetilde
{u}\right\|  _{\overset{.}{H}^{1}}.
\]
This completes the proof of lemma.
\end{proof}

We now establish the main result of this paper.

\begin{theorem}
\label{theorem}
Let $\overrightarrow{F}\in\mathcal{D}^{\prime}\left(  \mathbb{R}^{d}\right)
$, $d\geq3.$ Suppose that, for all $u$, $v\in\mathcal{D}\left(  \mathbb{R}%
^{d}\right)  $%

\begin{equation}
\left|  \left\langle\overrightarrow{F},v\nabla\overline{u}-\overline{u}\nabla
v\right\rangle\right|  \leq C\left\|  u\right\|  _{\overset{.}{H}^{1}%
}\left\|  v\right\|  _{\overset{.}{H}^{1}}.\label{eq4}%
\end{equation}
Then there is a vector-field $\overrightarrow{G}=\nabla\left(  \Delta
^{-1}div\overrightarrow{F}\right)  \in L_{loc}^{2}\left(  \mathbb{R}%
^{d}\right)  $ such that
\[
\int_{\mathbb{R}^{d}}\left|  \overrightarrow{G}(x)\right|  ^{2}\left|
u(x)\right|  ^{2}dx\leq C\left\|  u\right\|  _{\overset{.}{H}^{1}}^{2}%
\]
for all $u\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  $, where the constant
$C$ does not depend on  $u\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  .$
\end{theorem}

\begin{proof}
Suppose that
\[
\left|  \left\langle \overrightarrow{F},v\nabla\overline{u}-\overline{u}\nabla
v\right\rangle \right|  \leq C\left\|  u\right\|  _{\overset{.}{H}^{1}%
}\left\|  v\right\|  _{\overset{.}{H}^{1}}.
\]
holds. By the continuity the bilinear form on the left-hand side can be
extended to all $u$, $v\in\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)  $.
Let $v$ be a nonnegative function such that $w=\log$ $v$ and by the
proposition \ref{prop1}
\[
\nabla w=\frac{\nabla v}{v}\in L_{loc}^{2}\left(  \mathbb{R}^{d}\right)  .
\]
We need $w$ to be chosen so that the Sobolev space $\overset{.}{H}^{1}\left(
\mathbb{R}^{d}\right)  $ be invariant under the transformation :%
\begin{align*}
u &  \rightarrow\widetilde{u}=e^{iw}u\text{, \ \ \ }v\rightarrow\widetilde
{v}=e^{iw}v,\\
\left\|  u\right\|  _{\overset{.}{H}^{1}} &  \simeq\left\|  \widetilde
{u}\right\|  _{\overset{.}{H}^{1}},\text{ \ \ \ \ \ \ }\left\|  v\right\|
_{\overset{.}{H}^{1}}\simeq\left\|  \widetilde{v}\right\|  _{\overset{.}%
{H}^{1}}%
\end{align*}
Applying lemma \ref{lemma 2} and (\ref{eq4}), with $\widetilde{u}$ and
$\widetilde{v}$ in place of $u$ and $v$, we get%
\begin{align*}
\left|  \left\langle \overrightarrow{F},\widetilde{v}\nabla\overline
{\widetilde{u}}-\overline{\widetilde{u}}\nabla\widetilde{v}\right\rangle
\right|   &  \leq C\left\|  \widetilde{u}\right\|  _{\overset{.}{H}^{1}%
}\left\|  \widetilde{v}\right\|  _{\overset{.}{H}^{1}}\\
&  \leq9C\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}}.
\end{align*}
Notice that
\[
\widetilde{v}\nabla\overline{\widetilde{u}}-\overline{\widetilde{u}}%
\nabla\widetilde{v}=v\nabla\overline{u}-\overline{u}\nabla v-2i\overline
{u}v\nabla w.
\]
Combining the preceding estimates, we obtain%
\begin{align*}
2\left|  \left\langle \overrightarrow{F},\overline{u}v\nabla w\right\rangle
\right|   &  \leq\left|  \left\langle \overrightarrow{F},v\nabla\overline
{u}-\overline{u}\nabla v\right\rangle \right|  +\left|  \left\langle
\overrightarrow{F},\widetilde{v}\nabla\overline{\widetilde{u}}-\overline
{\widetilde{u}}\nabla\widetilde{v}\right\rangle \right|  \\
&  \leq10C\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}}.
\end{align*}
Observe that
\[
v\nabla w=\nabla v.
\]
Thus, we arrive at the inequality%
\[
\left|  \left\langle \overrightarrow{F},\overline{u}\nabla v\right\rangle
\right|  \leq C^{\prime}\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|
v\right\|  _{\overset{.}{H}^{1}}.
\]
From the preceding estimate and (\ref{eq4}), we deduce%
\begin{align*}
\left|  \left\langle \overrightarrow{F},\nabla\left(  \overline{u}v\right)
\right\rangle \right|   &  =\left|  \left\langle \overrightarrow{F}%
,v\nabla\overline{u}+\overline{u}\nabla v\right\rangle \right|  \leq\left|
\left\langle \overrightarrow{F},v\nabla\overline{u}\right\rangle \right|
+\left|  \left\langle \overrightarrow{F},\overline{u}\nabla v\right\rangle
\right|  \\
&  \leq C"\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}}.
\end{align*}
This yields
\[
\left|  \left\langle \left(  div\overrightarrow{F}\right)  u,v\right\rangle
\right|  \leq C"\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}},
\]
where $u\in\overset{.}{H}^{1}\left(  \mathbb{R}^{d}\right)  $ and $v=P\mu$. By
theorem 2.2 in [MV],%
\[
div\overrightarrow{F}\in\mathcal{M}\left(  \overset{.}{H}^{1}\rightarrow
\overset{.}{H}^{-1}\right)  .
\]
Hence, there is a vector-field $\overrightarrow{G}\in L_{loc}^{2}\left(
\mathbb{R}^{d}\right)  $ such that
\[
\overrightarrow{G}=\nabla\left(  \Delta^{-1}div\overrightarrow{F}\right)
=\nabla f\in\mathcal{M}\left(  \overset{.}{H}^{1}\rightarrow L^{2}\right)  .
\]
\end{proof}

\begin{corollary}
Under the assumtions of theorem \ref{theorem}, it follows that if
\[
\left|  \left\langle \overrightarrow{F},v\nabla\overline{u}-\overline{u}\nabla
v\right\rangle \right|  \leq C\left\|  u\right\|  _{\overset{.}{H}^{1}%
}\left\|  v\right\|  _{\overset{.}{H}^{1}},
\]
then
\[
\Delta f\in\mathcal{M}\left(  \overset{.}{H}^{1}\rightarrow\overset{.}{H}%
^{-1}\right)  .
\]
\end{corollary}

\begin{theorem}
Let $f\in\mathcal{D}^{\prime}\left(  \mathbb{R}^{d}\right)  .$ Then the
inequality
\[
\left|  \left\langle \left[  f,\Delta\right]  u,v\right\rangle \right|  \leq
C\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|  _{\overset{.}%
{H}^{1}},
\]
holds for all $u,v\in\mathcal{D}\left(  \mathbb{R}^{d}\right)  $, if and only
if there is a vector-field $\overrightarrow{G}\in L_{loc}^{2}\left(
\mathbb{R}^{d}\right)  $ such that%
\[
\int_{\mathbb{R}^{d}}\left|  \overrightarrow{G}(x)\right|  ^{2}\left|
u(x)\right|  ^{2}dx\leq C\left\|  u\right\|  _{\overset{.}{H}^{1}}^{2},
\]
where $\overrightarrow{G}=\nabla f.$
\end{theorem}

\begin{proof}
Suppose that $\overrightarrow{G}=\nabla\left(  \Delta^{-1}div\overrightarrow
{F}\right)  $ and
\[
\int_{\mathbb{R}^{d}}\left|  \overrightarrow{G}(x)\right|  ^{2}\left|
u(x)\right|  ^{2}dx\leq C\left\|  u\right\|  _{\overset{.}{H}^{1}}^{2}.
\]
Then $\overrightarrow{G}.\nabla$ is form bounded, since by Schwarz's
inequality, we have
\[
\left|  \left\langle \overrightarrow{G}.\nabla u,v\right\rangle \right|
\leq\left\|  u\overrightarrow{G}\right\|  _{L^{2}}\left\|  \nabla v\right\|
_{L^{2}}\leq C\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}}.
\]
Hence, this inequality obviously yields
\begin{align*}
\left|  \left\langle \overrightarrow{F},v\nabla\overline{u}-\overline{u}\nabla
v\right\rangle \right|    & \leq\left|  \left\langle \overrightarrow
{F},v\nabla\overline{u}\right\rangle \right|  +\left|  \left\langle
\overrightarrow{F},\overline{u}\nabla v\right\rangle \right|  \\
& \leq C\left\|  u\right\|  _{\overset{.}{H}^{1}}\left\|  v\right\|
_{\overset{.}{H}^{1}}.
\end{align*}
This completes the proof of theorem.
\end{proof}


\begin{thebibliography}{00}                                                                                       
\bibitem[LL]{LL}Lieb,E.H., and Loss,M.,: Analysis, second Edition,
Amer.math.Soc.Providence, RI, 2001.

\bibitem[MV]{MV}Maz'ya, V., Verbitsky,I. The Schr\"{o}dinger operator on the
enrgy space: Boundness and cmpactness criteria, Acta Mathematica, 188
(2002), 263-302.

\end{thebibliography}


\end{document}



math0505169.extracted_bib
\begin{thebibliography}{00}                                                                                       
\bibitem[LL]{LL}Lieb,E.H., and Loss,M.,: Analysis, second Edition,
Amer.math.Soc.Providence, RI, 2001.

\bibitem[MV]{MV}Maz'ya, V., Verbitsky,I. The Schr\"{o}dinger operator on the
enrgy space: Boundness and cmpactness criteria, Acta Mathematica, 188
(2002), 263-302.

\end{thebibliography}