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\hfill{Dec 10, 2001}
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\topmatter
\title{On the equation $\text{ \bf div } \bold Y \pmb = \bold f$ and application to control of
phases}\endtitle
\author{Jean Bourgain$^{(1)}$ and Ha\"im Brezis$^{(2), (3)}$}\endauthor
\thanks{Acknowledgment:
The first author (J.B.) is partially supported by NSF Grant
DMS-9801013.  The second author (H.B.) is partially sponsored by a
European Grant ERB FMRX CT98 0201.  He is also a member of the
Institut Universitaire de France.  The authors thank C. Fefferman, P.~Lax,
P. Mironescu, L. Nirenberg, T. Rivi\`ere, M.~Vogelius and D. Ye for useful comments.}
\endthanks
\endtopmatter
\NoRunningHeads
\document
\bigskip
\noindent
{\bf 1.  Introduction.}
\medskip
The purpose of this paper is to present new results concerning the equation
$$
\text{ div } Y = f \quad \text{ on } \Bbb T^d,  \tag 1.1
$$
\noindent
i.e., we work on $\Bbb R^d$ with $2\pi$ - periodic functions in all
variables.  In what follows we will always assume that $d \ge 2 $ and
that 
$$
\int_Q f = 0  \tag 1.2
$$
\noindent
where $Q = (0,2\pi)^d$.  The notations $L^p,W^{1,p}$, etc... refer to
$L^p(\Bbb T^d), W^{1,p}(\Bbb T^d)$, etc... or to $2\pi$--periodic functions
in $L^p_{loc} (\Bbb R^d), W^{1,p}_{loc}(\Bbb R^d)$, etc...
We denote by $L^p_{\#}$ the space of functions in $L^p$ satisfying (1.2).

\medskip
Clearly, (1.1) is an underdetermined problem which admits many
solutions.  A standard way of tackling (1.1) is to look for a
vector-field $Y$ satisfying the {\it additional} condition
$$
\text{ curl } Y = 0,
$$
i.e., one looks for a {\it special} $Y$ of the form
$$
Y = \text{ grad\,}u.
$$
Equation (1.1) then becomes
$$
\Delta u = f \tag 1.3
$$
\noindent
and the standard $L^p$-regularity theory yields a solution $u \in
W^{2,p}$ when $f \in L^p_{\#}, 1 < p < \infty$.  Consequently (1.1) has a
solution $Y \in W^{1,p}$ for every $f \in L^p_{\#}, 1< p < \infty$.  More
precisely, the operator div :  $W^{1,p} \to L^p_{\#}$ admits a right-inverse
which is a bounded linear operator $K : L^p_{\#} \to W^{1,p}$.  Strictly
speaking, we should write $Y \in (W^{1,p})^{d} (=d$ -fold copy of
$W^{1,p}$), div : $(W^{1,p})^d \to L^p$, etc....  But we will often
omit the superscript $d$ to alleviate notation.
\medskip
Three {\it limiting} cases are of interest:  
\medskip
\noindent
{\bf Case 1: $\bold p \, {\pmb =} \, \bold 1$}.  It is well-known that when $f \in L^1$ equation
(1.3) does not necessarily admit a solution $u \in W^{2,1}$.  However,
one might still hope to have some solution $Y$ of (1.1) in $W^{1,1}$ or
at least in $BV$.  This is not true: for some $f$'s in $L^1$,
equation (1.1) has no solution in $BV$ and not even in $L^{d/(d-1)}$;
see Section 2.1.
\medskip
\noindent
{\bf Case 2: $\bold p\pmb =\pmb\infty$.}
It is well-known that when $f\in L^\infty$ equation (1.3) does not 
necessarily admit a solution $u \in
W^{2,\infty}$.  However, one might hope to find a solution $Y$ of (1.1)
in $W^{1,\infty}$.  This is not true: McMullen [1] has shown that
for some $f$'s in $L^\infty$ (even $f$ continuous) equation (1.1) has no
solution in $W^{1,\infty}$.  This is proved using a duality argument
and a ``non-estimate'' of Ornstein [1]; see Section 2.2.
\medskip

\noindent
{\bf Case 3: $\bold p \, {\pmb =} \, \bold d$}.  This is the heart of our
work.  
For every $f \in L^d_{\#}$, equation (1.3) admits a solution $u\in W^{2,d}$ and  thus
equation (1) admits a solution $Y$ = \text{ \, grad\,}$u \in W^{1,d}$.  Since
$W^{1,d}$ is {\it not } contained in $L^\infty$ (this is a limiting
case for the Sobolev imbedding) we {\it cannot} assert that this $Y$
belongs to $L^\infty$.  In fact, we give in Section 3 (Remark 7) an explicit $f \in L^d$
such that the corresponding $Y$ = grad $u$ does {\it not} belong to
$L^\infty$.  However one might still hope that given any $f \in L^d_{\#}$
there is {\it some} $Y\in L^\infty$ solving (1.1).  This is indeed true:
\medskip
\noindent
{\bf Proposition 1}.  {\it Given any $f \in L^d_{\#}$ 
there exists some $Y \in L^\infty$ solving (1.1) (in the sense of distributions) with} 
$$
\Vert Y\Vert_{L^\infty} \le C(d) \Vert f\Vert_{L^d}.  \tag 1.4
$$
\medskip
\noindent
{\bf Remark 1}.  A more precise statement established in the course of
the proof says that there exists $Y \in C^0$ satisfying (1.1) and
(1.4).
\medskip
The proof of Proposition 1 is quite elementary; see Section 3.  It relies on
the Sobolev - Nirenberg imbedding $W^{1,1} \subset L^{d/(d-1)}$ (and even
$BV \subset L^{d/(d-1)}$) combined with duality, i.e., Hahn-Banach.  As a
consequence, the argument is {\it not constructive}, and $Y$ is not obtained as above via a bounded
linear operator acting on $f$.
In fact, surprisingly, the
operator div has no bounded right-inverse in this setting:
\medskip
\noindent
{\bf Proposition 2}.  {\it There exists no bounded linear operator $K$:
$L^d_{\#} \to L^\infty$ such that} div $Kf = f\quad \forall  f \in L^d_{\#}$
{\it (in the sense of distributions)}.
\medskip
\noindent
{\bf Remark 2}.  Another way of formulating Proposition 2 is to say
that the subspace $\{ Y \in L^\infty; \text{ div } Y =0\}$ admits no
complement in the space $\{Y \in L^\infty;\text{ div } Y \in L^d\}$
equipped with its natural norm. Alternatively, the closed subspace
$\{\text{ grad } u ; u \in W^{1,1}\}$ has no complement in $L^1$; see
Section 3.
\medskip
To summarize: for every $f \in L^d_{\#}$, equation (1.1) admits 
\medskip
\noindent
a)  a solution \ $Y_1 \in W^{1,d}$,
\smallskip
\noindent
b)  a solution \ $Y_2 \in L^\infty$.
\medskip
\noindent
A natural question is whether there exists a solution $Y$ of (1.1) in
$L^\infty\cap W^{1, d}$.  This is indeed one of our main results.
\medskip
\noindent 
{\bf Theorem 1}. {\it For every $f \in L^d_{\#}$ there
exists a solution $Y \in L^\infty \cap W^{1,d}$ of (1.1) satisfying}
$$
\Vert Y\Vert_{L^\infty} + \Vert Y\Vert_{W^{1,d}} \le C (d) 
\Vert f\Vert_{L^d}.  \tag 1.5
$$
\medskip

Despite the simplicity of this statement the argument is rather involved and a simpler proof would be desirable.

We will present two techniques to tackle Theorem 1.
\medskip
\noindent
{\bf First proof of Theorem 1 when ${\bold d \, \pmb= \,\bold 2}$} (see Section 4).  It relies on
Hahn-Banach (via duality) and thus it is {\it not} constructive.  But
it is rather elementary; the main ingredient is the new estimate (1.6)
which is established by $L^2$-Fourier methods.
\medskip
\noindent
{\bf Lemma 1}.  {\it On $\Bbb T^2$ we have}
$$
\Vert u-\notint 
u\Vert _{L^2} \le C \Vert\text{\,grad\,} u\Vert_{L^1+H^{-1}}, \quad \forall u
\in L^2,  \tag 1.6
$$
{\it for some absolute constant $C$}.
\medskip
The main difficulty, in proving (1.6), stems from the fact that if we decompose 
$$
\text{\,grad\,} u = h_1 + h_2
$$
\noindent
with $h_1 \in L^1$ and $h_2 \in H^{-1}$, then $h_1$ and $h_2$ need
{\it not} be gradients themselves; it is only their sum which is a
gradient.
\medskip
The analogue of Lemma 1 for $d > 2$ is the estimate on $\Bbb T^d$,
$$
\Vert u - \notint u \Vert_{L^{d/(d-1)}} \le C (d) \Vert\text{\,grad\,} u
\Vert_{L^1+W^{-1,d/(d-1)}}.  \tag 1.7
$$
\medskip
We have no direct proof of (1.7).  But it can be deduced by duality from
the statement of Theorem 1 (and thus from the second proof presented in Section 7).
\medskip
\noindent
{\bf Second proof Theorem 1, valid for all $\bold d \pmb\ge \bold 2$} (see Sections 5 and 6).  We exhibit via
a {\it constructive} (nonlinear) argument some explicit $Y \in W^{1,d}
\cap L^\infty$ satisfying (1.1) and (1.5).
The argument for $d=2$ is simpler and we start with this case for expository reasons.

\medskip
One should observe a certain analogy with the Fefferman-Stein [1
]decomposition of BMO - functions and Uchiyama's [1] constructive proof
.  Indeed, returning to equation (1.1) and defining $F$ by $|\xi|
\hat{F}(\xi) = \hat{f}(\xi)$, we obtain that $F \in W^{1,d} \subset BMO$ and
(1.1) becomes
$$
F = \sum^{d}_{j=1} R_j\,Y_j \tag 1.8
$$
with $R_j = j^{th} \text{ Riesz  transform }$ $(\widehat{R_j\psi}(\xi)
= \hat{\psi} (\xi) \frac{\xi_j}{|\xi|}), \ Y = (Y_1,..., Y_d)$.  
\medskip
The statement of Theorem 1 is that (1.8) has a solution $Y \in L^\infty
\cap W^{1,d}$.  Recall that according to Fefferman-Stein [1] {\it any} $F \in BMO$
has a decomposition of the form
$$ F = Y_0 + \sum^d_{j=1} R_j Y_j \,\,\,\text{ with } Y_0,
Y_1,...,Y_d \in L^\infty. \tag 1.9
$$
The proof of this decomposition is again by duality and
nonconstructive.  The later constructive approach from Uchiyama [1] gives a
different proof of (1.9).  If we assume moreover that $F \in W^{1,d}$,
Uchiyama's argument gives that (1.9) has a solution $Y_0,
Y_1,...,Y_d \in L^\infty \cap W^{1,d}$.  The new result in this paper
shows that, in fact, for $F \in W^{1,d}$, the $Y_0$ - component is
unnecessary and (1.8) holds for some $Y_1,..,Y_d \in L^\infty \cap W^{1,d}$.
\medskip
It should be mentioned that to achieve our decomposition we do use
significantly different methods from Uchiyama .  This raises the
question what are the function spaces $X$, $W^{1,d} \subset X \subset
BMO$, such that every $F \in X$ has a decomposition
$$
F = \sum^d_{j=1} R_j Y_j \tag 1.10
$$
where $Y_j \in L^\infty$ or (assuming the Riesz transforms bounded on
$X$) the stronger property $Y_j \in L^\infty \cap X$.
\medskip
\noindent
{\bf Remark 3}.  
Using Theorem 1 we will prove (in Sections 4 and 6) that a slightly stronger conclusion holds:

\proclaim
{Theorem 1$'$} For every $f\in L^d_{\#}$ there exists a solution $Y\in C^0\cap W^{1, d}$ of (1.1) satisfying (1.5).
\endproclaim
\medskip
The original motivation for studying (1.1) comes from the following
question about lifting discussed in Bourgain-Brezis-Mironescu [1],[2],[3].  Consider the equation   
$$
g = e^{i\varphi}\quad \text{ on } \Bbb T^d
$$
where $\varphi$ is a smooth real-valued function.
\medskip
\noindent
{\bf Question}:  Assuming $g$ is controlled in $H^{1/2}$, what kind of
estimate can we deduce for $\varphi$?
\medskip
Here is a first easy consequence of Theorem 1.
\medskip
\noindent
{\bf Corollary 1}.  {\it We have}
$$
\Vert\varphi -\notint \varphi\Vert_{L^{d/(d-1)}} \le C(d) (1+\Vert g
\Vert_{H^{1/2}})\Vert g\Vert_{H^{1/2}}.
\tag 1.11
$$
{\bf Proof}.  Write 
$$ 
\text{\,grad\,} g = i e^{i\varphi} \text{\, grad\,} \varphi
$$
and thus
$$
\text{\,grad\,} \varphi = -i \-\bar g (\text{\,grad\,} g).  \tag 1.12
$$
Multiplying by $Y$ gives
$$
\int_Q \varphi \text{ div } Y = \,\, \int_Q i \-\bar g Y
\cdot \text{\,grad\,} g.  \tag 1.13
$$
Given $f \in L^d$ we obtain from Theorem 1 some $Y$ satisfying (1.1)
(with $f$ replaced by $f-\notint f$)and (1.5).  Thus we have
$$
|\int(\varphi - \notint \varphi) f| \le \Vert g\Vert_{H^{1/2}} (\Vert
\-\bar g Y\Vert_{H^{1/2}}). \tag 1.14
$$
But 
$$
\aligned
\Vert \-\bar g Y\Vert_{H^{1/2}} &\le \Vert\-g\Vert_{H^{1/2}} \Vert Y
\Vert_{L^\infty} + \Vert g\Vert_{L^\infty} \Vert Y\Vert_{H^{1/2}}\\
{}\\
(\text{ by } (1.5))  &\le C (\Vert g\Vert_{H^{1/2}} \Vert f\Vert_{L^d} +
\Vert f\Vert_{L^d}) 
\endaligned
\tag 1.15
$$
where we have use the obvious fact that $||Y||_{H^{1/2}} \le C
||Y||_{W^{1,d}}$.  Combining (1.14) and (1.15) yields (1.11).
\medskip
\noindent
{\bf Remark 4}.  Estimate (1.11) cannot be improved, replacing the norm
$||\,\,||_{L^{d/(d-1)}}$ by $||\,\,||_{L^p}, p > d/(d-1)$.  This may be
seen by choosing $g=e^{i\varphi}$ with $\varphi(x) = (|x|^2 +
\varep^2)^{-\alpha/2}$ with $\alpha < d-1, \alpha$ close to $(d-1)$ and
$\varep$ close to $0$ (the same example has already been used in Bourgain-Brezis-Mironescu [1], 
Lemma 5).
There is however a better estimate than (1.11), namely

\proclaim
{Theorem 4}
Let $\vp$ be a smooth real-valued function on $\Bbb T^d$ and set $g=e^{i\vp}$, then
$$
\Vert\vp\Vert_{H^{1/2}+ W^{1,1}} \leq C(d) (1+\Vert g\Vert_{H^{1/2}})\Vert g\Vert_{H^{1/2}}.
$$
\endproclaim

Theorem 4 has been announced in Bourgain-Brezis-Mironescu [2] (Theorem 3) and is proved in Section 8.
Our proof of Theorem 4 is a direct estimate based on paraproducts.
In view of the preceding argument one may wonder whether Theorem 4 can be proved by solving a
divergence equation.
After duality the required statement would be
$$
\Vert u-\notint u\Vert_{H^{1/2}+W^{1, 1}}\leq C\Vert \grad u\Vert_{H^{-1/2}+L^1}\tag 1.16
$$
but we do not know whether (1.16) holds.

We now turn to the question of coupling equation (1.1) with the Dirichlet condition
$$
Y=0 \quad \text { on } \partial Q.\tag 1.17
$$
This question was addressed (in various forms) by a few authors; see e.g. Arnold--Scott--Vogelius [1],
Duvaut--Lions [1] (Theorem 3.2), X.~Wang [1], Temam [1] (Proposition 1.2(ii) and Lemma 2.4) and the 
references therein to Magenes--Stampacchia [1] and Ne\v cas [1].
Our aim is to establish the analogue of Theorem 1$'$ under Dirichlet condition.
We start with the following known fact (see e.g. Arnold--Scott--Vogelius [1] for $d=2$).

\proclaim
{Theorem 2}
Given $f\in L^p_{\#}(Q), 1< p<\infty$, there exists some $Y\in W_0^{1, p}(Q)$ satisfying (1.1) with
$$
\Vert Y\Vert_{W^{1, p}} \leq C(p)\Vert f\Vert_{L^p}.\tag 1.18
$$

Moreover $Y$ can be chosen, depending linearly on $f$.
\endproclaim

The operator and the estimate do not depend on $p$ assuming we stay away from the end points.

For the convenience of the reader we include a new proof; our technique is extremely elementary and can be adapted
to establish, for the limiting case $p=d$,

\proclaim
{Theorem 3} Given $f\in L^d_{\#} (Q)$ there exists some $Y\in C^0(\bar Q)\cap W^{1, d}_0(Q)$ satisfying (1.1) with 
$$
\Vert Y\Vert_{L^\infty} +\Vert Y\Vert_{W^{1, d}}\leq C\Vert f\Vert_{L^d}.
$$
\endproclaim

Theorem 3 is stronger than Theorem 1$'$. However it will be deduced from Theorem 1$'$.
There are variants of Theorems 2 and 3 when $Q$ is replaced by a Lipschitz domain in $\Bbb R^d$ (see Section 7.2).

The plan of the paper is the following:

\itemitem {1.} Introduction.

\itemitem {2.} The cases $f\in L^p$ with $p=1$ and $p=\infty$.

\itemitem{3.} Proofs of Proposition 1 and 2 and related questions.

\itemitem{4.} Proof of Theorem 1 when $d=2$ via duality.

\itemitem{5.} Proof of Theorem 1 when $d=2$ (explicit construction).

\itemitem{6.} Proof of Theorem 1 when $d>2$ (explicit construction).

\itemitem{7.} The equation div $Y=f$ with Dirichlet condition. Proof of Theorems 2 and 3.

\itemitem{8.} Estimation of the phase in $H^{1/2}+W^{1, 1}$.
Proof of Theorem 4.
\bigskip

\noindent
{\bf 2.  The cases $\bold f \pmb\in \bold L^{\bold p}$ with $\bold p\pmb =\bold 1$ and $\bold p\pmb=
\pmb\infty$}.
\medskip
We consider here equation (1.1) with $f \in L^p_{\#}$ and ask whether there
exists a solution $Y \in W^{1,p}$ of (1.1) when $p=1$ and $p=\infty$.
As we have already mentioned in the Introduction the answer is
negative.  Here is the proof.
\bigskip
\noindent
{\bf 2.1.  The case $\bold p \pmb = \bold 1$}.
\medskip
Assume by contradiction that for every $f \in L^1_{\#}$
there is some $Y \in W^{1,1}$ satisfying (1.1).  It follows that the
linear operator 
$$
T u = \text{ div } u\, \text {  from }E = W^{1,1} \text{ into } F = L^1_{\#}
$$
is bounded and surjective.  By the open mapping principle there is a
constant $C$ such that for every $f \in F$ there exists a solution $Y
\in E$ of (1.1) satisfying 
$$ 
||Y||_{W^{1,1}} \le C ||f||_{L^1}.
$$
\medskip
We now use a duality argument which occurs frequently in the rest of
the paper.  We will deduce that $W^{1,d} \subset L^\infty$  with
continous injection and since this is false we infer that for some
$f$'s in $F$ there is no $Y \in W^{1,1}$ satisfying (1.1).
\medskip
Let $u \in W^{1,d}$ and set
$$
\text{\,grad\,} u = h \in L^d.  \tag 2.1
$$
\noindent
Given any $f \in L^1$, let $Y \in W^{1,1}$ be such that 
$$
\text{ div } Y = f - \notint f
$$
\noindent
and
$$
||Y||_{W^{1,1}} \le C ||f - \notint f ||_{L^1}.
$$
\noindent
Taking the scalar product of (2.1) with $Y$ and integrating yields
$$
\int_Q (u - \notint_Q u) f = - \int_Q h Y.
$$
\noindent
Consequently
$$
|\int_Q (u - \notint_Q u) f | \le ||h||_{L^d} ||Y||_{L^{d/(d-1)}}.  \tag
2.2
$$
\noindent
By the Sobolev-Nirenberg imbedding we have $W^{1,1} \subset
L^{d/(d-1)}$ and thus 
$$
||Y||_{L^{d/(d-1)}} \le C ||Y||_{W^{1,1}} \le C ||f||_{L^1}. \tag 2.3
$$
\noindent
Combining (2.2) and (2.3) we deduce that $(u - \notint_Q u) \in L^\infty$
with
$$
|| u - \notint_Q u ||_{L^\infty} \le C ||\text{\,grad\,} u||_{L^d}.
$$
\noindent
Impossible.
\medskip
\noindent
{\bf Remark 5}.  The same argument shows that equation (1.1) with $f
\in L^1_{\#}$ need not have a solution $Y$ in the sense
of distributions with $Y \in L^{d/(d-1)}$ (Note, however, that the
solution $Y$ given via (1.3) belongs to $L^p, \, \forall
p < d/(d-1)$ and even to weak - $L^{d/(d-1)})$.  It suffices to follow
the above argument with $E = W^{1,1}$ replaced by
$$
\widetilde E = \{ Y \in L^{d/(d-1)} ; \text{ div } Y \in L^1\}
$$
\noindent
equipped with its natural norm.
\bigskip
\smallskip
\noindent
{\bf 2.2.  The case $\bold p \, \pmb = \, \pmb \infty$}.
\medskip
This case has been settled negatively by C.T. McMullen [1] (the
interest in this kind of problem grew out of the study of the equation
det $(\nabla \varphi) = f $ with $\varphi$ bi-lipschitz and also from
a question of M. Gromov [1] on separated nets;  see
B. Dacorogna-J. Moser [1], D. Ye [1], T. Rivi\`ere-D. Ye [1],[2],
D. Burago-B. Kleiner [1]).
\medskip
For the convenience of the reader we sketch a proof when $d=2$, which
is essentially similar to the one of Mc Mullen [1].  We argue by
contradiction as above. Then, for every $f \in L^\infty$ there is a $Y
\in W^{1,\infty}$ satisfying
$$
\text{ div } Y = f - \notint f 
$$
\noindent
and
$$
||Y||_{W^{1,\infty}} \le C ||f||_{L^\infty}.
$$
Let $\psi$ be a smooth function on $\Bbb T^2$ and set $g =
\psi_{x_1x_2}$.  Write
$$
\int g_{x_1} Y_1 + g_{x_2} Y_2 =- \int g f = -\int \psi_{x_1x_1}
Y_{1x_2} + \psi_{x_2x_2} Y_{2x_1}.
$$
\noindent
Consequently
$$
\bigg|\int g f\bigg| \le C (||\psi_{x_1x_1}||_{L^1} +
||\psi_{x_2x_2}||_{L^1})||f||_{L^\infty} 
$$
and thus
$$
||g||_{L^1} = ||\psi_{x_1x_2}||_{L^1} \le C (||\psi_{x_1x_1}||_{L^1} +
||\psi_{x_2x_2}||_{L^1}).
$$
\noindent
This contradicts a celebrated ``non-inequality'' of Ornstein [1] and
completes the proof.
\medskip
\noindent
{\bf Remark 6}.  The same argument shows that equation (1.1) with $f \in
C^0$ and $\int f = 0$ need not have a solution $Y \in W^{1,\infty}$. 
\bigskip

\noindent
{\bf 3.  Proofs of Proposition 1 and 2 and related questions}.
\medskip
\noindent
{\bf Proof of Proposition 1}.  Recall the Sobolev-Nirenberg imbedding
$W^{1,1} \subset L^{d/(d-1)}$ and, more generally, $ BV \subset
L^{d/(d-1)}$ with
$$
||u -\notint u||_{L^{d/(d-1)}} \le C(d)|| \text{\,grad\,} u||_{\Cal M} \quad
\forall u \in BV. \tag 3.1
$$
\noindent
where $\Cal M$ denotes the space of measures.
\noindent
Set
$$
E = C^0,\quad  F =  L^d_{\#} 
$$
and consider the unbounded linear operator $A = D(A) \subset E \to F$, defined
by 
$$
D(A) = \{Y \in E; \text{ div } Y \in L^d\}, \quad A Y = \text{ div } Y,
$$
so that $A$ is densely defined and has closed graph.  Clearly  we
have 
$$
E^*=\Cal M, \quad F^* =  L_{\#}^{d/(d-1)},
$$
$$
D(A^*) = F^* \cap B V, \ A^* u = \text{\,grad\,} u. 
$$
By (3.1) we have 
$$
||u||_{F^*} \le C(d) ||A^* u||_{E^*} \quad \forall u \in D(A^*).
$$
It follows from the closed-range theorem (see e.g. Brezis [1], Section
II.7), that $A$ is surjective.  More precisely, we claim that for any
$f \in F$ there is some $Y \in E$ satisfying (1.1) and 
$$
\Vert Y\Vert _{L^\infty} \le 2 C(d) \Vert f\Vert_{L^d},
$$
\noindent
where $C(d)$ is the constant in (3.1).
\medskip
Indeed, let $f \in F$ with $||f||_{L^d} = 1$ and consider the two
convex sets
$$
B = \{Y \in E;\,\, ||Y||_E < 2C(d)\}
$$
\noindent
and
$$
L = \{Y \in E; \text{ div } Y = f\}.
$$
\noindent
We have to prove that $B \cap L \ne \emptyset$.  Suppose not, that $B \cap L =
\emptyset$.  Then, by Hahn-Banach there exists $\mu \in E^*, \mu \ne 0$, and
$\alpha \in \Bbb R$ such that 
$$
\langle \mu, Y \rangle \le \alpha \quad \forall Y \in B  \tag 3.2
$$
and
$$ 
\langle \mu, Y \rangle \ge \alpha \quad \forall Y \in L.  \tag 3.3
$$
\noindent
>From (3.2) we have $||\mu|| \le \alpha/2 C(d)$ and from (3.3) we
deduce, in particular, that $\langle \mu, Z \rangle = 0 \quad \forall
Z \in N(A)$.  It follows that $\mu \in N(A)^\perp = R(A^*)$.  Hence
there exists some $u \in F^* \cap B V$ such that grad $u = \mu$.
Applying (3.1) we see that 
$$
||u||_{L^{d/(d-1)}} \le C(d) ||\mu|| \le \alpha/2.  \tag 3.4
$$
\noindent
On the other hand, by (3.3), $\forall Y \in L$, 
$$
\alpha \le \langle \mu,Y\rangle = \langle \text{\,grad\,} u, Y\rangle =
- \int u \text{ div } Y = - \int u f \le ||u||_{L^{d/(d-1)}} \le \alpha/2.
$$
\noindent
This is impossible since $\alpha > 0$ (because $\mu \ne 0$).
\medskip
\noindent
{\bf Remark 7}.  The special solution of (1.1) given by $Y = \text{
\,grad\,} u$, where $u$ is the solution of (1.3) belongs to $W^{1,d}$
when $f \in L^d$; however, in general, it does {\it not} belong to
$L^\infty$.  Here is an example due to L. Nirenberg.
Using $(x_1,x_2,...,x_d)$ as coordinates in $\Bbb R^d$ consider the
function 
$$
u =  x_1 |\text{log }r|^\alpha \zeta
$$
\noindent
where $\zeta$ is a smooth cut-off function with support near $0$ and
$0 < \alpha < (d-1)/d$.  Note that $Y = \text{\,grad\,} u $ does not
belong to $L^\infty$ while 
$$
|\Delta u| \le \frac{C}{r} |\text{ log }r|^{\alpha -1},  
$$
so that $\Delta u \in L^d$.
\medskip

We now turn to the proof of Proposition 2, i.e., the non-existence of
a bounded right-inverse $K : L^d_{\#} \to L^\infty$ for the operator div.
We present two proofs.  The first one is the simplest:  after a
standard averaging trick we obtain a bounded multiplier $L^d \to
L^\infty$ and we reach a contradiction by a direct summability
consideration.  The second proof is related to Remark 2:  the
existence of $K$ would yield a factorization of the identity map $I$:
$W^{1,1} \to L^{d/(d-1)}$ through the Banach space $L^1$; however no
such factorization exists by a general argument from the geometry of
Banach spaces.
\medskip
\noindent
{\bf First proof of Proposition 2}.  Assume $K : L^d_{\#} \to
L^\infty$ is a bounded operator satisfying div $K = I \text{
on } L^d_{\#}$.  Then the averaged operator
$$
\widetilde K =\int \limits_{\Bbb T^d} \tau_{-x} K \tau_x dx,
$$
\noindent
where $\tau_x f(y) = f(y+x)$ still satisfies
$$
\text { div } \widetilde K = I \quad \text{ on } L^d.  \tag 3.5
$$
\noindent
On the other hand $\widetilde K$ is clearly a multiplier
$$
\widetilde K(e^{in \cdot x}) =
(\lambda_1(n),\lambda_2(n),...,\lambda_d(n))e^{in \cdot x}
$$
which is bounded from $L^d$ into $L^\infty$ and hence from
$L^1$ into $L^{d^{\prime}}$ where $d^{\prime} = d/(d-1)$.
\medskip
By (3.5) we have 
$$
\sum_{j=1}^{d} n_j \lambda_j(n) =1 \quad \forall n \in \Bbb Z^d
$$
\noindent
so that
$$
|\lambda(n)|^2 = \sum_{j=1}^{d} |\lambda_j(n)|^2 \ge 1/|n|^2 \quad
\forall n.  \tag 3.6
$$
\noindent
Consider the multiplier 
$$
M (e^{in \cdot x}) = \frac{1}{|n|^{\frac{d}{2}-1}} e^{in \cdot x} ,\quad n \ne 0.
$$
\noindent
Then $M$ is bounded from $L^{d^{\prime}}$ into $L^2$.  Hence $M\widetilde
K$ is a bounded multiplier from $L^1$ into $L^2$.  Thus 
$$
\sum \Sb n \in \Bbb Z^d\\ n \ne 0 \endSb
\frac{|\lambda_j(n)|^2}{|n|^{d-2}} < \infty, \quad \forall j.
$$
\noindent
Summing over $j=1,2, ..., d$, and using (3.6) we deduce
$$
\sum \Sb n \in \Bbb Z^d\\ n \ne 0 \endSb \frac{1}{|n|^d} < \infty.
$$
\noindent
A contradiction.
\medskip
\noindent
{\bf Second proof of Proposition 2}.  Assuming the existence of $K :
L^d_{\#} \to L^\infty$ we obtain a factorization of the identity
map $I : W^{1,1} \to L^{d^{\prime}}$ as
$$
I = K^* \circ \text{\,grad\,}
$$
\noindent
which in, particular, gives a factorization of $I$ through the Banach
space $L^1$.  We claim that there in no such factorization, as a
consequence of Grothendieck's theorem on absolutely summing operators.
Both the result and the method are well-known and we briefly recall
it (see Wojtaszczyk [1] for details).  Take first $d=2$.  Then $I : W^{1,1} \to
L^2$ and we consider the operator $I \circ D$ where $D : L^2 \to
W^{1,1}$ is defined by 
$$
D(e^{in \cdot x}) = \frac{1}{\sqrt{1+|n|^2}} e^{in\cdot x}
$$
\noindent
Thus $D$ is clearly bounded as an operator into $H^1$, hence into
$W^{1,1}$.  Since $I$ is assumed to factor through $L^1$, so does $I
\circ D$
~
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\medskip
Next, recall Grothendieck's theorem that any bounded operator $B : L^1
\to L^2$ is 1- summing, i.e.,
$$
\pi_1(B) \equiv \sup \bigg\{ \sum ||Bx_i||; (x_i) \subset L^1 \text{ and
} \max\limits_{x^* \in L^\infty, ||x^*|| \le 1} \sum |\langle
x_i,x^*\rangle| \le 1\bigg\} \le K_G ||B||,
$$  
where $K_G$ is Grothendieck's constant.
\medskip

>From the usual ideal properties, we obtain
$$
\spreadlines {1.5\jot}
\aligned
\bigg(\sum\limits_{n \in \Bbb Z^2} \frac{1}{1 + |n|^2}\bigg)^{1/2} = ||
I\circ D||_{HS} = \pi_2(I\circ D) \le \pi_1(I\circ D)\\ 
= \pi_1(B\circ A) \le ||A|| \pi_1(B) \le K_G ||A|| ||B|| < \infty, \endaligned
$$
\noindent
which in an obvious contradiction.
\medskip
For $d>2$, we have $I$ :$W^{1,1} \to L^{d^{\prime}}$ and we consider the
multiplier operator $M : L^{d^{\prime}} \to L^2$ given by $M (e^{in\cdot
x}) = (1 + |n|)^{1-\frac{d}{2}} e^{in\cdot x}$.  Hence, considering now
$M\circ I\circ D : L^2 \to L^2$ factoring through $L^1$, we obtain again
a contradiction:
$$
\bigg(\sum \frac{1}{(1 + |n|)^{d-2}(1 + |n|^2)}\bigg)^{1/2} = ||M\circ
I\circ D||_{HS} = \pi_2(M\circ I\circ D) \le \pi_1(M\circ I\circ D) < \infty.
$$
\medskip
\noindent
{\bf Proof of Remark 2}.  Consider the Banach space
$$
E = \{Y \in L^\infty; \text{ div } Y \in L^d\}
$$
equipped with its natural norm $||Y||_{L^\infty} + ||\text{ div }
Y||_{L^d}$.
\noindent
Then
$$
N = \{Y \in L^\infty; \text{ div } Y = 0\}
$$
\noindent
is a closed subspace of $E$ which admits no complement in $E$.  Indeed
set
$$ 
F = L^d_{\#}
$$
and consider the bounded linear operator $T : E \to F$ defined by $TY
=\text{ div } Y$.  By Proposition 1, $T$ is surjective.  If $N =
N(T)$ admits a complement in $E$, then $T$ has a bounded right-inverse,
i.e., an operator $S : F \to E$ such that 
$$
\text{ div } (Sf) =f \quad \forall f \in F
$$
(see e.g. Brezis [1], Th\'eor\`eme II.10).  But this is impossible by
Proposition 2.
\medskip
Similarly, the subspace
$$
R = \{\text{\,grad\,} u;\, u \in W^{1,1}\}
$$
\noindent
of $L^1$ is closed and admits not complement in $L^1$.
Indeed, consider the spaces $E = \{ u \in W^{1,1} ; \int u = 0\}, F =
L^1$  and the operator $T$ = grad as a bounded linear
injective operator from $E$ into $F$.  
If $R = R (T)$ admits a
complement in $F$, then $T$ has a bounded left-inverse $S:F \to E$
(see e.g. Brezis [1], Th\'eor\`eme II.11).  In particular, $S : F
\to L^{d/(d-1)}_{\#}$ satisfies
$$
S (\text{\,grad\,} u) = u, \quad \forall u \in W^{1,1} \text { with} \int u=0.
$$
Then $S^* : L^d_{\#} \to L^\infty$ satisfies
$$
\text{ div } (S^* f) =f, \quad \forall f \in L^d_{\#}
 $$
and this is again impossible by Proposition 2.
\bigskip

\noindent
{\bf 4. Proof of Theorem 1 when $\bold d\, \pmb =\, \bold 2$ via duality.}

We now return to the periodic setting and we will prove the slightly stronger form of Theorem 1,

\proclaim
{Theorem 1$'$ (for $\bold d\, \pmb =\, \bold 2$)}
For every $f\in L^2_{\#}$ there exists a solution $Y\in
C^0\cap H^1$ of (1.1) with
$$
\Vert Y\Vert_{L^\infty}+ \Vert Y\Vert_{H^1}\leq C\Vert f\Vert_{L^2}\tag
4.1
$$
for some absolute constant $C$.
\endproclaim

Theorem 1$'$ is proved by duality from

\proclaim
{Lemma 2}
On $\Bbb T^2$ we have 
$$
\Vert u-\notint u\Vert_{L^2}\leq C\Vert\text{\,grad\,}
u\Vert_{L^1+H^{-1}}, \forall u\in L^2 \tag 4.2
$$
where $C$ is an absolute constant.
\endproclaim

Assuming the lemma we turn to the 

\noindent
{\bf Proof of Theorem 1$'$.}
First observe that
$$
L^1+H^{-1} \subset \Cal M+H^{-1}
$$
and that
$$
\Vert\qquad \Vert_{L^1+H^{-1}}=\Vert \qquad\Vert _{\Cal M+H^{-1}} \text { on
}
L^1+H^{-1}\tag 4.3
$$
(this may be easily seen using regularization by convolution).

Let $E=C^0\cap H^1, F=L^2_{\#}$ and consider the bounded
operator $T:E\to F$ defined by $TY=\text{ div }Y$.
Clearly, $T^*:F^*= F\to E^*=\Cal M +H^{-1}$ is given by $T^*u= \text{\,grad\,}
 u$.
By Lemma 2 we have
$$
\Vert u\Vert_{F^*} \leq C\Vert T^*u\Vert_{E^*} \quad \forall u\in F^*
$$
and therefore $T$ is surjective from $E$ onto $F$.
Estimate (4.1) follows from the open mapping principle or one could
argue directly  using (4.2) and Hahn-Banach as in the proof of
Proposition 1.

\noindent
{\bf Proof of Lemma 2.}
Assume
$$
u\in L^2_{\#}, \tag 4.4
$$
$$
\partial_xu=F_1+h_1, \ \partial_yu=F_2+h_2\tag 4.5
$$
and
$$
\Vert F_1\Vert_{L^1}+\Vert F_2\Vert_{L^1}+\Vert h_1\Vert_{H^{-1}}+\Vert
h_2\Vert_{H^{-1}}\leq 1.\tag 4.6
$$
We have to prove that
$$
\Vert u\Vert_{L^2} \leq C.\tag 4.7
$$

The main ingredient is

\proclaim
{Lemma 3}
Under assumptions (4.4)-(4.6) we have
$$
\sum_{n_1, n_2 \in\Bbb Z} \ \frac{n_1^2n_2^2}{(n_1^2+n^2_2)^2} \ |\hat
u(n_1, n_2)|^2 \leq C(\Vert u\Vert_{L^2}+1).\tag 4.8
$$
\endproclaim

Assuming Lemma 3 we may now complete the proof of Lemma 2.
Define
$$
u'(x', y')=u(x'+y', x'-y')=\sum_{n_1, n_2} \hat u (n_1, n_2) \,
e^{i[(n_1+n_2)x'+(n_1-n_2)y']}\tag 4.9
$$
so that
$$
\widehat {u'}(n_1+n_2, n_1-n_2)=\hat u(n_1, n_2)\tag 4.10
$$
and
$$
\align
\partial_{x'} u'(x', y')&=\partial_x u(x'+y', x'-y')+\partial_y
u(x'+y', x'-y')\\
&=(F_1+F_2)(x'+y', x'-y')+(h_1+h_2)(x'+y', x'-y')\\
&\in L^1+H^{-1}
\endalign
$$
and similarly for $\partial_{y'} u'$.

>From (4.8) and (4.10) we obtain
$$
\spreadlines {1.5\jot}
\align
&\sum_{n_1, n_2} \ \frac{(n_1+n_2)^2(n_1-n_2)^2}{4(n_1^2+n_2^2)^2}
|\hat u(n_1, n_2)|^2 =\sum_{n_1', n_2'} \ \frac
{(n_1')^2(n_2')^2}{\big((n_1')^2+(n_2')^2\big)^2} |\widehat{u'}(n_1',
n_2')|^2\tag 4.11\\
&\leq C(\Vert u'\Vert_{L^2}+1) =C(\Vert u\Vert_{L^2}+1).
\endalign
$$
Addition of (4.8) and (4.11) implies that
$$
\Vert u\Vert^2_{L^2} =\sum_{n_1, n_2}|\hat u(n_1, n_2)|^2 \leq C(\Vert
u\Vert_{L^2}+1)
$$
and the desired estimate (4.7) follows.

\bigskip
We now turn to the

\noindent
{\bf Proof of Lemma 3.}
We have
$$
\sum_{n\not=0} \ \frac{n_1^2n_2^2}{(n_1^2+n_2^2)^2} |\hat u(n)|^2 =\frac 1i
\sum \ \frac{n_1n_2^2}{(n_1^2+n_2^2)^2} \ \widehat{\partial_x u}
(n)\hat u (-n)
$$
$$
\spreadlines{1.5\jot}
\align
&\overset{\text by \  (4.5) }\to = \frac 1i \sum
\frac{n_1n_2^2}{(n_1^2+n_2^2)^2} \ \hat F_1(n) \hat u(-n)+\frac 1i\sum
\frac{n_1n_2^2}{(n_1^2+n_2^2)^2} \hat h_1(n) \hat u(-n)\\
&\quad = (4.12)+(4.13).
\endalign
$$
Estimate
$$
|(4.13)|\leq \sum_{n_1, n_2}\frac{|\hat
h_1(n)|}{\sqrt{n_1^2+n_2^2}} \ |\hat u(-n)| \leq \Vert
h_1\Vert_{H^{-1}}\Vert u\Vert_{L^2}.
\tag 4.14
$$
Write
$$
\spreadlines{1.5 \jot}
\align
(4.12)\ &=\sum \ \frac{n_1n_2}{(n_1^2+n_2^2)^2} \ \hat F_1(n)
\widehat{\partial_yu}(-n)\\
&=\sum \ \frac{n_1n_2}{(n_1^2+n_2^2)^2} \hat F_1(n)  \hat F_2(-n)+\sum
\ \frac{n_1n_2}{(n_1^2+n_2^2)^2} \hat F_1(n)\hat h_2 (-n)\\
& = (4.15)+(4.16).
\endalign
$$
Estimate
$$
\spreadlines {1.5\jot}
\align
\noindent
|{(4.16)}| &\leq \sum \ \frac{|n_1| \, |n_2|}{(n_1^2+n_2^2)^2}
(|\widehat{\partial _x u}(n)|+|\hat h_1(n)|)\,  |\hat h_2(-n)|\\ 
&\leq \sum \ \frac{n_1^2|n_2|}{(n^2_1+n_2^2)^2}\, |\hat u(n)| \ |\hat
h_2(-n)|+\sum\frac{|\hat h_1(n)|}{\sqrt{n_1^2+n_2^2}} \ \frac{|\hat
h_2(-n)|}{\sqrt{n_1^2+n_2^2}}\tag 4.17\\
&\leq \Vert f\Vert_{L^2} \ \Vert h_2\Vert_{H^{-1}}+\Vert
h_1\Vert_{H^{-1}} \ \Vert h_2\Vert_{H^{-1}}.
\qquad\qquad\qquad
\endalign
$$

\noindent
{\bf Estimation of (4.15).}

This is the key point.
Since $\Vert F_1\Vert_{L^1} \leq 1, \Vert F_2\Vert_{L^1}\leq 1$, it suffices
(by convexity) to replace $\widehat{F_i}(n)$ by
$$
\widehat{F_1} (n) =e^{in\cdot a} \qquad \widehat{F_2}(n)=e^{in\cdot b}\tag 4.18
$$
for some $a, b\in \Bbb T^2$ (this amounts to replace $F_1, F_2$ by
resp. the Dirac measures $\delta_a, \delta_b$).

Thus we obtain
$$
\align
\sum_{n_1, n_2\in\Bbb Z} \ \frac{n_1n_2}{(n_1^2+n_2^2)^2}
\hat{F_1}(n) \hat F_2 (-n)&= \sum \ \frac{n_1n_2}{(n_1^2+n_2^2)^2} 
e^{i[n_1(a_1-b_1)+n_2(a_2-b_2)]}\\
&=-\sum\frac{n_1n_2}{(n_1^2+n_2)^2} \sin n_1(a_1-b_1) \sin
n_2(a_2-b_2)\tag 4.19
\endalign
$$
by parity considerations.

\proclaim
{Claim}
For all $\theta_1, \theta_2\in \Bbb T$
$$
\bigg|\sum_{n_1, n_2} \ \frac{n_1n_2}{(n_1^2+n_2^2)^2} \sin n_1\theta_1
\sin n_2\theta_2\bigg|\leq C.\tag 4.20
$$
\endproclaim

>From the claim, we conclude that $|(4.15)|, |(4.19)| \leq C$
and, recalling also (4.14), (4.17), inequality (4.8) follows.


\noindent
{\bf Proof of the Claim.}

Splitting $\Bbb Z$ in dyadic intervals, we obtain
$$
\sum_{k_1, k_2\geq 0} \bigg| \sum_{n_1\sim 2^{k_1}, n_2\sim 2^{k_2}}
\ \frac{n_1n_1}{(n_1^2+n_2^2)^2} \sin n_1\theta_1\sin
n_2\theta_2\bigg|.\tag 4.21
$$

Recall the inequality
$$
\bigg|\sum_{n\in I} \sin n\theta\bigg| \lesssim 4^k|\theta| \wedge
\frac 1{|\theta|}\tag 4.22
$$
if $\theta\in \Bbb T$ and $I\subset[2^{k-1}, 2^k]$ is an interval
(where $\wedge $ denotes min).

>From (4.22), assuming $k_1\geq k_2$, we have
$$
\align
&\bigg|\sum_{n_1\sim 2^{k_1}, n_2\sim 2^{k_2}} \
\frac{n_1n_2}{(n_1^2+n_2^2)^2} \sin n_1\theta_1 \sin
n_2\theta_2\bigg|\leq\\
&\left(4^{k_1}|\theta_1|\wedge \frac 1{|\theta_1|}\right)
\left(4^{k_2} |\theta_2|\wedge\frac 1{|\theta_2|}\right)
\bigg\Vert\bigg\{ \frac{n_1n_2}{(n_1^2+n_2^2)^2}\bigg\}\bigg\Vert_
{\ell^\infty(n_1\sim 2^{k_1})\hat\otimes \ell^\infty (n_2\sim
2^{k_2})}\tag 4.23
\endalign
$$
where $\ell^\infty(I)\hat\otimes \ell^\infty(J)$ denotes the usual
projective tensor product.
Thus the last factor in (4.23) may be bounded by
$$
\Vert\partial^2_{n_1n_2} \ \frac{n_1n_2}{(n_1^2+n_2^2)^2}
\Vert_{\ell^1(n_1\sim 2^{k_1}, n_2\sim 2^{k_2})} \leq C\bigg\Vert\frac
1{(n_1^2+n_2^2)^2}\bigg\Vert _{\ell^1(n_1\sim 2^{k_1}, n_2\sim
2^{k_2})}\leq C\frac{2^{k_2}}{8^{k_1}}.\tag 4.24
$$
Substitution of (4.23), (4.24) in (4.21) gives the bound
$$
\align
(4.20), (4.21) &\leq C\sum_{k_1\geq k_2\geq 0}
4^{k_2-k_1}\bigg(2^{k_1}|\theta_1| \wedge \frac 1{2^{k_1}|\theta_1|}
\bigg) \bigg(2^{k_2}|\theta_2|\wedge \frac
1{2^{k_2}|\theta_2|}\bigg)\\
&\lesssim C\prod^2_{i=1} \bigg[\sum_{k\in\Bbb
Z_+}\bigg(2^k|\theta_i|\wedge\frac 1{2^k|\theta_i|}\bigg)\bigg]\leq C.
\endalign
$$
This completes the proof of the Claim and of Theorem 1$'$ for $d=2$.

\bigskip
\noindent
{\bf 5. Proof of Theorem 1 when $\bold d \pmb =\bold 2$ (explicit construction).}

Our aim is to construct $Y\in L^\infty \cap H^1$ such that
$$
\text{div\,} Y = f\in L^2_{\#}(\Bbb T^2).\tag 5.1
$$
Write
$$
\Bbb Z^2=\bigcup_{j\geq 0} (\Lambda_j^1 \cup \Lambda_j^2)
$$
where
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$$
Let
$$
\Lambda^\alpha =\bigcup_j\Lambda_j^\alpha\qquad (\alpha =1, 2).
$$
Decompose
$$
f=f^1+f^2 \text { where } f^\alpha =P_{\Lambda^\alpha} f\equiv
\sum_{n\in\Lambda^\alpha} \hat f(n) e^{in.x}.
$$

\noindent
{\bf Claim.}

{\it Let $\delta>0$ be small enough and $\Vert f\Vert_2\leq \delta$.
Then there are $Y_1, Y_2$ such that
$$
\Vert Y_\alpha\Vert_{L^\infty \cap H^1} \leq 1\tag 5.2
$$
and
$$
\Vert \partial_\alpha Y_\alpha -f^\alpha\Vert_2 \leq\delta^{4/3} \qquad
(\alpha =1, 2).\tag 5.3
$$}

Thus if $\Vert f\Vert_2 =\delta$, then
$$
\Vert f-\partial_1Y_1 -\partial_2Y_2\Vert_2 \leq \delta^{1/3} \Vert
f\Vert_2
$$
and iteration of this gives (5.1).

The construction of $Y_1, Y_2$ is explicit but {\it non-linear} (see Proposition 2).
\bigskip

Take $\alpha =1$ and denote $f^1$ by $f, \Lambda_j^1$ by
$\Lambda_j$.

Define
$$
\align
f_j& = P_{\Lambda_j} f\\
c_j&=\Vert f_j\Vert_2\\
F_j&=D^{-1}_{x_1}f_j \equiv \sum \frac 1{n_1} \hat f_j (n) e^{in.x}.
\endalign
$$
Hence
$$
\bigg(\sum c_j^2\bigg)^{1/2} =\Vert f\Vert_2
$$
$$
\Vert F_j\Vert_\infty \leq \sum_{n\in\Lambda_j} \frac 1{|n_1|}
|\hat f(n)|\lesssim 2^{-j} |\Lambda_j|^{1/2} \Vert f_j\Vert_2 \lesssim c_j.\tag 5.4
$$
Fix $\varepsilon >0$ a small constant and partition
$$
\Lambda_j=\bigcup_{r<\frac 1\varepsilon +1} \Lambda_{j, r}
$$
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$$
\noindent
in stripes $\Lambda_{j, r} $ such that
$$
|\text{Proj}_{n_1}\Lambda_{j, r}|\sim\ve2^j.\tag 5.5
$$
Define first
$$
\tilde F_j(x)=\sum_r \bigg|\sum_{n\in\Lambda_{j, r}}\frac 1{n_1} \hat
f_j(n) e^{in.x}\bigg|.\tag 5.6
$$
Thus
$$
|F_j(x)|\leq |\tilde F_j(x)|\lesssim c_j.\tag 5.7
$$
>From Cauchy-Schwarz
$$
\Vert\tilde F_j\Vert_2 \leq \ve^{-1/2}\Vert F_j\Vert_2\lesssim
\ve^{-1/2} 2^{-j} c_j.\tag 5.8
$$
Observe that if $\text{Proj}_{n_1} \Lambda_{j, r}=[a_r, b_r],
b_r-a_r\sim\ve2^j$, then
$$
|\partial_1\tilde F_j|\leq \sum_r\bigg|\sum_{n\in\Lambda_{j, r}}
\frac{n_1-a_r}{n_1} \hat f_j(n) e^{in\cdot x}\bigg|
$$
where
$$
\left|\frac{n_1-a_r}{n_1}\right|< \varepsilon.
$$
Therefore
$$
\Vert\partial_1 \tilde F_j\Vert_2 \lesssim \sum_r \varepsilon\Vert
P_{\Lambda_{j, r}}f\Vert_2 \lesssim \varepsilon^{1/2} \Vert
P_{\Lambda_j} f\Vert_2 =\varepsilon^{1/2} c_j\tag 5.9
$$
(this is the purpose of the construction of $\tilde F_j$).

We also need to make an appropriate localization of the Fourier
transform of $\tilde F_j$.
Denote
$$
K_N(y) =\sum_{|n|<N} \frac {N-|n|}{N}\, e^{iny}
$$
the usual F\'ejer kernel on $\Bbb T$.
It is easy to see that if
$$
P(y) =\sum_{|n|<N} \hat P(n) e^{iny}
$$
is a trigonometric polynomial, then
$$
|P|\leq 3(|P|*K_N).\tag 5.10
$$
Using this fact in the variables $x_1, x_2$, we see that
$$
|F_j| \leq \tilde F_j\leq G_j\tag 5.11
$$
denoting
$$
G_j=9\tilde F_j*(K_{N_1} \otimes K_{N_2})\tag 5.12
$$
where each $\Delta_{j, r}$ is an $N_1\times N_2$ rectangle, $N_1\sim
\varepsilon 2^j, N_2\sim 2^j$.

Thus, by construction
$$
\text{supp\,} \hat G_{j}\subset [-N_1, N_1]\times[-N_2, N_2]\subset[|n|\leq
2^j]\tag 5.13
$$
and inequalities (5.7), (5.8), (5.9) remain preserved.

Thus
$$
\align
\Vert G_j\Vert_\infty& \leq 9\Vert\tilde F_j\Vert_\infty \lesssim c_j\qquad
(0<\delta <1)\tag 5.14\\
\Vert G_j\Vert_2&\lesssim \ve^{-1/2} 2^{-j} c_j\tag 5.15\\
\Vert\partial_1 G_j\Vert_2&\lesssim \ve^{1/2} c_j\tag 5.16\\
\Vert\nabla G_j\Vert_2&\lesssim \ve^{-1/2} c_j.\tag 5.17
\endalign
$$

Assume $\{f_j| j\leq K\}$ a finite sequence (which is no
restriction).

Define
$$
\align
Y_1&=F_K+F_{K-1} (1-G_K)\\
&+F_{K-2}(1-G_{K-1})(1-G_K)+\cdots\\
&=\sum_{j\leq K} F_j \prod_{k>j} (1-G_k).\tag 5.18
\endalign
$$
Thus from (5.11)
$$
\align
|Y_1|\leq |F_K|&+(1-|F_K|) |F_{K-1}|\\
&+(1-|F_K|)(1-|F_{K-1}|)|F_{K-2}|+\cdots \leq 1.
\endalign
$$
One may also rewrite (5.18) as
$$
Y_1=\sum F_j-\sum G_jH_j\tag 5.19
$$
with
$$
\align
H_j&=F_{j-1} +F_{j-2} (1-G_{j-1})\\
&+F_{j-3}(1-G_{j-2})(1-G_{j-1})+\cdots\\
&=\sum_{k<j} F_k \prod_{k<k'<j}(1-G_{k'}).\tag 5.20
\endalign
$$
Clearly
$$
|H_j|<1.
$$
By construction
$$
\partial_1Y_1=\sum f_j-\sum \partial_1(G_jH_j).\tag 5.21
$$
Next, we estimate the second term in (5.21) that will appear as an
error term.

Observe that since $\text{supp\,}\hat F_j\subset [|n|\sim 2^j]$
and (5.13), also
$$
\text{supp\,}\hat H_j\subset [|n|\lesssim 2^j].\tag 5.22
$$
Denote $P_k$ Fourier projection operators on $[|n|\sim 2^k]$ such that $Id
=\sum_{k\geq 0} P_k$.

>From the preceding, we may thus ensure that
$$
G_jH_j=\sum_{k\leq j} P_k (G_jH_j).\tag 5.23
$$
Estimate then
$$
\bigg\Vert \sum_j\partial_1 (G_jH_j)\bigg\Vert_2 \leq \sum_{s\geq 0}\bigg
(\sum_j\Vert\partial_1P_{j-s}(G_jH_j)\Vert^2_2\bigg)^{1/2}\tag 5.24
$$
(since for fixed $s$, the $P_{j-s}$ have disjoint range).

Returning to the parameter $0<\ve<1$ introduced earlier, write
$$
\ve =2^{-s_*} \ (s_*>0)\tag 5.25
$$
and estimate (5.24) in the ranges
$$
\align
&s>s_*\tag 5.26\\
&0\leq s\leq s_*.\tag 5.27
\endalign
$$

\noindent
{\bf Contribution of (5.26).}

Since $|H_j|\leq 1$ and (5.15)
$$
\align
\Vert \partial_1 P_{j-s}(G_jH_j)\Vert_2 &\lesssim 2^{j-s} \Vert G_j
H_j\Vert_2\\
& \leq 2^{j-s} \Vert G_j\Vert_2\leq \ve^{-1/2} \,
2^{-s} c_j.\tag 5.28
\endalign
$$
Substitution in (5.24) gives the contribution
$$
\sum_{s\geq s_*} 2^{-s} \ve^{-1/2}\bigg(\sum c_j^2\bigg) ^{1/2}
< 2^{-s_*}\ve^{-1/2} \Vert f\Vert_2 < \ve^{1/2} \Vert f\Vert_2.\tag 5.29
$$

\noindent
{\bf Contribution of (5.27).}

Estimate now
$$
\align
\Vert\partial_1P_{j-s} (G_jH_j)\Vert_2 \leq \Vert\partial_1(G_jH_j)
\Vert_2 &\leq \Vert\partial_1 G_j\Vert_2+\Vert G_j \partial_1H_j\Vert_2\\
&\leq \ve^{1/2} c_j+\Vert G_j\partial_1H_j\Vert_2.\tag 5.30
\endalign
$$
using (5.16).

Recalling definition (5.20) of $H_j$, one easily verifies that
$$
|\nabla H_j|\leq \sum_{k<j} (|\nabla F_k|+ |\nabla G_k|).\tag 5.31
$$
Hence
$$
\Vert\nabla H_j\Vert_\infty \leq \sum_{k<j} 2^k c_k\tag 5.32
$$
and from (5.15)
$$
\Vert G_j\partial_1H_j\Vert_2 \leq \ve^{-1/2} c_j\bigg(\sum_{k<j} 2^
{-(j-k)} c_k\bigg).\tag 5.33
$$
Substitution of (5.30), (5.33) in (5.24) gives the following bound on the contribution of
(5.27)
$$
s_* \ve^{1/2} \bigg(\sum c_j^2\bigg)^{1/2} + s_*\ve^{-1/2} \bigg[
\sum_jc_j^2\bigg(\sum_{k<j}2^{-(j-k)} c_k\bigg)^2\bigg]^{1/2}
$$
$$
\leq\bigg(\log \frac 1\ve\bigg) \ve^{1/2} \Vert f\Vert_2 +
\bigg(\log\frac 1\ve\bigg) \ve^{-1/2} \Vert f\Vert_2^2.\tag 5.34
$$
Consequently, from (5.21), (5.29), (5.34)
$$
\Vert f-\partial_1 Y_1\Vert_2= \bigg\Vert\sum_j\partial_1(G_jH_j)
\bigg\Vert_2 \leq \log\frac 1\ve (\ve^{1/2}
\Vert f\Vert_2+\ve^{-1/2} \Vert f\Vert^2_2).\tag 5.35
$$
Under the assumption $\Vert f\Vert_2 \leq \delta$, letting $\ve=\delta$
in (5.35), we obtain thus
$$
\Vert f-\partial_1Y_1\Vert_2 \leq\delta^{\frac 32 -} \leq\delta^{\frac 43}\tag 5.36
$$
which is (5.3).

It remains to estimate $\Vert Y_1\Vert_{H^1} =\Vert \nabla Y_1\Vert_2$.

By (5.19)
$$
\Vert\nabla Y_1\Vert_2 \leq \bigg\Vert \sum_j\nabla F_j\bigg\Vert_2
 +\bigg\Vert \sum \nabla (G_jH_j)\bigg \Vert_2.\tag 5.37
$$
>From the definition of $F_j$ and since $\text{supp\,} \hat F_j\subset
\Lambda^1_j$,
it follows that
$$
\bigg\Vert\sum_j\nabla F_j\bigg\Vert_2 \sim \bigg(\sum
\Vert f_j\Vert^2_2\bigg)^{1/2}=\Vert f\Vert_2.\tag 5.38
$$
Estimate the second term in (5.37) as in (5.24)
$$
\bigg\Vert \sum_j \nabla(G_jH_j)\bigg\Vert_2 \leq \sum_{s\geq 0} \bigg(\sum_j\Vert
\nabla P_{j-s} (G_jH_j)\Vert_2^2\bigg)^{1/2}\tag 5.39
$$
and
$$
\Vert\nabla P_{j-s} (G_jH_j)\Vert_2 \lesssim 2^{j-s} \Vert G_jH_j\Vert_2\leq \ve^{-1/2}
2^{-s} c_j.\tag 5.40
$$
Thus
$$
(5.39) \leq \ve^{-1/2} \sum_{s\geq 0} 2^{-s} \bigg(\sum_jc_j^2\bigg)^{1/2}
\leq \ve^{-1/2}\Vert f\Vert_2\tag 5.41
$$
and
$$
\Vert \nabla Y_1\Vert_2 \leq \delta^{-1/2} \Vert f\Vert_2\leq \delta^{1/2}.\tag 5.42
$$
Since $\Vert Y_1\Vert_\infty\lesssim 1$, this establishes (5.2).

This proves the Claim and completes the proof of Theorem 1 for $d=2$.
\bigskip

\noindent
{\bf 6. Proof of Theorem 1 when $\bold d\pmb> \bold 2$ (explicit construction).}

Let $f\in L^d_{\#}(\Bbb T^d)$.
Our aim is to construct a solution $Y$ of $\text{div\,}Y=f$ satisfying
$$
\align
\Vert Y\Vert_\infty&\leq C\Vert f\Vert_d\tag 6.1\\
\Vert \nabla Y\Vert_d &\leq C\Vert f\Vert_d.\tag 6.2
\endalign
$$
We do this by standard modification of the previous $L^2$-argument with as main additional
ingredient, the Littlewood-Paley square function theory.
Consider again a partition
$$
\Bbb Z^d =\bigcup_{j\geq 0} (\Lambda_j^1 \cup\cdots\cup\Lambda_j^d)
$$
of disjoint $d$-rectangles $\Lambda_j^\alpha$ of side length $\sim 2^j$.

Formulate the analogue of the Claim with $Y_\alpha$ satisfying bounds (6.1), (6.2).
Letting $\alpha =1, f=f^1$, define again
$$
F_j =D_{x_1}^{-1} f_j\tag 6.3
$$
satisfying
$$
\Vert F_j\Vert_\infty \lesssim (2^{j/d})^d \Vert F_j\Vert_d =2^j\Vert D^{-1}_{x_1} f_j
\Vert_d\sim \Vert f_j\Vert_d \equiv c_j.\tag 6.4
$$
Define $\tilde F_j$ and $G_j$ as in (5.6), (5.12).
Thus (5.11), (5.13) hold.
Also
$$
\align
\Vert G_j\Vert_\infty &\lesssim\Vert\tilde F_j\Vert_\infty\leq
\ve^{-1/d'} \bigg(\sum_{r<\frac 1\ve}\bigg\Vert\sum_
{n\in\Lambda_{j, r}} \frac 1{n_1}
\hat f_j(n) e^{inx}\bigg \Vert^d_\infty\bigg)^{1/d}\\
&\leq \ve^{-1/d'}\bigg(\sum_{r<\frac 1\ve}\big(2^{j\frac {d-1}{d}}
(\ve 2^j)^{\frac 1d}
\bigg\Vert\sum_{n\in\Lambda_{j, r}} \frac 1{n_1} \hat f_j(n)
e^{in\cdot x}\bigg\Vert_d\big)^d\bigg)^{1/d}\\
&\lesssim\ve^{-1/d'+1/d}\bigg(\sum_{r<\frac 1\ve}\bigg\Vert
\sum_{n\in\Lambda_{j, r}} \hat f_j(n) e^{inx}\bigg\Vert^d_d\bigg)^
{\frac 1d}\\
&\lesssim \ve^{\frac2d-1} \Vert f_j\Vert_d =\ve^{\frac 2d-1} c_j\leq
\ve^{\frac 2d-1}\delta.\tag 6.5
\endalign
$$
(We assume $\delta$ small enough compared
 with $\ve$ to ensure in particular that $\ve^{\frac 2d-1}
\delta\ll 1$.)

Repeat the construction from Section 5.
In place of estimate (5.24) we now have
$$
\bigg\Vert\sum_j\partial_1(G_jH_j)\bigg\Vert_d \leq \sum_{s\geq 0}
\bigg\Vert\sum_j |\partial_1 P_{j-s} (G_jH_j)|^2\bigg)^{1/2}
\bigg\Vert_d\tag 6.6
$$
and distinguish the cases (5.26), (5.27).

\noindent
{\bf Contribution of (5.26).}

Estimate
$$
\align
&\bigg\Vert\bigg(\sum_j|\nabla P_{j-s}(G_jH_j)|^2\bigg)^{1/2}\bigg\Vert_d
\lesssim\\
&\bigg\Vert\bigg(\sum_j 4^{j-s}|P_{j-s} (G_jH_j)|^2\bigg)^{1/2}\bigg\Vert_d\lesssim\\
&2^{-s}\bigg\Vert\bigg(\sum_j 4^j|G_jH_j|^2\bigg)^{1/2}\bigg\Vert_d\lesssim\\
&2^{-s} \bigg\Vert \bigg(\sum_j 4^j(\tilde F_j* K_j)^2\bigg)^{1/2}\bigg\Vert_d
\tag 6.7
\endalign
$$
where $K_j$ is a product of F\'ejer kernels
$$
K_{N_1}\otimes K_{N_2}\otimes\cdots\otimes K_{N_d}\qquad N_1\sim\ve2^j
 \quad \text {and } N_2, \ldots,
N_d\sim 2^j.
$$
Again from standard square function inequalities
$$
(6.7) \lesssim 2^{-s} \bigg\Vert\bigg(\sum_j 4^j(\tilde F_j)^2\bigg)^
{1/2}\bigg\Vert_d.
\tag 6.8
$$
Recalling the definition of $\tilde F_j$, estimate
$$
(\tilde F_j)^2 \leq \ve^{-1} \sum_{r\leq\ve^{-1}} \bigg|\sum_{n\in\Lambda^1_{j, r}}
\frac 1{n_1} \hat f(n) e^{inx}\bigg|^2.\tag 6.9
$$
Substituting in (6.8), this gives
$$
\aligned
&\ve^{-1/2}2^{-s} \bigg\Vert\bigg(\sum_j\sum_{r<\ve^{-1}}\bigg|
 \sum_{n\in\Lambda^1_{j, r}}\frac{2^j}{n_1} \hat f(n) e^{inx}|^2\bigg)
^{1/2}\bigg\Vert_d\\
&\lesssim \ve^{-1/2} \, 2^{-s} \bigg\Vert \bigg(\sum_j\sum_{r<\ve^{-1}}\bigg|\sum_{n\in\Lambda^1_{j, r}} \hat f(n) e^{i
nx}\bigg|^2\bigg)^{1/2}\bigg\Vert_d.\\
\endaligned
\tag 6.10
$$
We use here the fact that $|n_1|\sim|n|\sim 2^j$ for $n\in \Lambda_j^1$.

Recall also the definition of $\Lambda_{j, r}$ obtained by partitioning the $n_1$-variable
in intervals of size $\ve 2^j$.

At this stage, we use the following (1-variable) inequality due to Rubio de Francia,
which generalizes the Littlewood-Paley inequality to arbitrary intervals.

\proclaim
{Proposition 3}
Let $\{I_\alpha\}$ be disjoint intervals in $\Bbb Z$ and
$$
P_If=\sum_{n\in I}\hat f(n) e^{inx}
$$
the corresponding Fourier projection.

Then, for $2\leq d<\infty$, there is the (one-sided) inequality
$$
\bigg\Vert\bigg(\sum|P_{I_\alpha} f|^2\bigg)^{1/2}\bigg\Vert_d \leq C\Vert f\Vert_d.\tag 6.11
$$
\endproclaim

Since $\{\text{Proj}_{n_1} \Lambda^1_{jr}\}$ are disjoint intervals in $\Bbb Z$,
application of (6.11) in the $x_1$-variable implies that
$$
(6.6) \lesssim \ve^{-1/2}\, 2^{-s} \Vert f\Vert_d.\tag 6.12
$$
Summation of (6.12) for $s\geq s_*$ gives then
$$
\text{(5.26)-contribution } \leq \ve^{1/2} \Vert f\Vert_d.\tag 6.13
$$

\noindent
{\bf Remark 8.}
We used the general Proposition 3 for convenience; the present case could in fact
be treated by more elementary means.

\noindent
{\bf Contribution of (5.27).}

Estimate
$$
\align
&\bigg\Vert\bigg(\sum_j|\partial_1P_{j-s}(G_jH_j)|^2\bigg)^{1/2}
\bigg\Vert_d \lesssim
\bigg\Vert \bigg(\sum_j|\partial_1(G_jH_j)|^2\bigg)^{1/2}\bigg
\Vert_d\leq\\
&\bigg\Vert\bigg(\sum_j|\partial_1G_j|^2\bigg)^{1/2}\bigg\Vert_d+
\bigg
\Vert \bigg(\sum_j|G_j(\partial_1H_j)|^2\bigg)^{1/2}\bigg\Vert_d=(6.14)+(6.15).
\endalign
$$
Estimate (6.14) by
$$
\bigg\Vert\bigg(\sum_j|\partial_1\tilde F_j|^2\bigg)^{1/2}\bigg\Vert_d.
\tag 6.16
$$
We have that
$$
\align
|\partial_1\tilde F_j|&\leq\sum_{r<\ve^{-1}}\bigg|
\sum_{n\in\Lambda^1_{j, r}} \frac{n_1-a_{j, r}}{n_1} \hat f(n)
e^{inx}\bigg|\\
 &\leq \ve^{-1/2} \bigg(\sum_{r<\ve^{-1}}\bigg|\sum_{n\in\Lambda^
1_{jr}} \frac {n_1-a_{j,r}}{n_1} \hat f(n) e^{inx}\bigg|^2\bigg)^{1/2}
\endalign
$$
where
$ \text{Proj}_{n_1} \Lambda^1_{jr}=[a_{jr}, b_{jr}], b_{jr}-a_{jr}
\sim \ve2^j$.
Thus $\big|\frac{n_1-a_{j, r}}{n_1}\big|\leq\ve$.

We get therefore
$$
\align
(6.16) &\leq \ve^{-1/2} .\ve.\bigg\Vert\bigg(\sum_j\sum_{r< \ve^{-1}}
\bigg|\sum_{n\in\Lambda^1_{jr}}\hat f(n)
e^{inx}\bigg|^2\bigg)^{1/2}\bigg
\Vert_d\\
&
\lesssim \ve^{1/2} \Vert f\Vert_d.\tag 6.17
\endalign
$$
To estimate (6.15), use again inequality (5.31), together with (6.4),
(6.5).
Thus
$$
\Vert \nabla H_j\Vert_\infty \leq \ve^{\frac 2d-1}\sum_{k<j} 2^k
c_k<\ve^{\frac 2d-1} \,2^j\Vert f\Vert_d.\tag 6.18
$$
Hence
$$
\align
(6.15)&\leq \ve^{\frac 2d-1} \Vert f\Vert_d \bigg\Vert\sum_j
4^jG_j^2\bigg)^{1/2}\bigg\Vert_d\\
&\leq \ve^{\frac 2d-1} \Vert f\Vert_d \bigg\Vert\bigg(\sum_j(2^j
\tilde F_j)^2\bigg)^{1/2}\bigg\Vert_d\\
&\leq \ve^{\frac 2d-\frac 32}\Vert f\Vert^2_d\tag 6.19
\endalign
$$
applying again the (6.8)-bound using Proposition 3.

Thus the (5.27)-contribution is
$$
\leq \ve^{1/2}\log\frac 1\ve \Vert f\Vert_d+ \ve^{\frac 2d-\frac 32}
\log \frac 1\ve \Vert f\Vert^2_d.\tag 6.20
$$
Collecting estimates (6.13), (6.20), it follows that
$$
\align
\Vert f-\partial_1 Y\Vert_d&=\bigg\Vert\sum_j\partial_1(G_j H_j)
\bigg\Vert_d \\
&\leq \ve^{1/2}\log \frac 1\ve \Vert f\Vert_d+
\ve^{\frac 2d -\frac 32}\log\frac 1\ve \Vert f\Vert^2_d\tag 6.21
\endalign
$$
which is the analogue of (5.35).
Assuming $\Vert f\Vert_d=\delta$, take $\ve=\delta^{1/2}$ to obtain
$$
\Vert f-\partial_1 Y\Vert_d \leq \delta^{1/5} \Vert f\Vert_d.\tag 6.22
$$
Remains to estimate
$$
\Vert\nabla Y\Vert_d \leq\bigg\Vert\sum \nabla F_j\bigg\Vert_d+
\bigg\Vert \sum\nabla(G_jH_j)\bigg\Vert_d= (6.23)+(6.24).
$$
We have
$$
(6.23)\sim\bigg\Vert \bigg(\sum |\nabla F_j|^2\bigg)^{1/2}\bigg\Vert_d
\sim \bigg\Vert\bigg(\sum|f_j|^2)^{1/2}\bigg\Vert_d \lesssim
\Vert f\Vert_d.
$$
Estimate (6.24) as
$$
\bigg\Vert \sum_{s\geq 0}\bigg(\sum_j|\nabla
P_{j-s}(G_jH_j)|^2\bigg)^{1/2}
\bigg\Vert_d \lesssim \ve ^{-1/2}\Vert f\Vert_d\tag 6.25
$$
using (6.7)-(6.12).

This completes the argument.
\bigskip

We conclude this Section with a

\noindent
{\bf Proof of Theorem 1$'$ when $\bold d\pmb> \bold 2$.}
The argument is somewhat bizarre: one uses duality twice!
First, from Theorem 1 we easily deduce the estimate on $\Bbb T^d$
$$
\Vert u- \notint u\Vert_{L^{d/(d-1)}} \leq C(d) \Vert\grad u\Vert_{L^1+W^{-1, d/(d-1)}}, 
\forall u\in L^{d/(d-1)}.\tag
6.26
$$
Next, we argue as in the beginning of Section 4.
Observe that
$$
L^1+W^{-1, d/(d-1)} \subset \Cal M +H^{-1}
$$
and that
$$
\Vert \qquad \Vert_{L^1+W^{-1, d/(d-1)}} =\Vert \qquad \Vert_{\Cal M+W^{-1, d/(d-1)}} \ \text { on } L^1+W^{-1,
d/(d-1)}\tag 6.27
$$
(this may be easily seen using regularization by convolution).

Let $E=C^0\cap W^{1, d}, F=L^d_{\#}$ and consider the bounded operator $T: E\to F$ defined by $TY=\text { div\,} Y$.
Clearly $T^*:F^*\to E^*=\Cal M+W^{-d, d/(d-1)}$ is given by $T^*u=\grad u$.
By (6.26) and (6.27) we obtain
$$
\Vert u\Vert_{F^*} \leq C\Vert T^* u\Vert_{E^*} \quad\forall u\in F^*
$$
and therefore $T$ is surjective from $E$ onto $F$.
Applying the open mapping principle (or use Hahn-Banach as in the proof of Proposition 1).
We see that for every $f\in F$ there is some $Y\in E$ satisfying $TY=f$ and $\Vert Y\Vert_E
\leq C\Vert f\Vert_F$.

\noindent
{\bf Remark 9.}
Alternatively, one may approximate $f\in L^d_{\#} (\Bbb T^d)$ by trigonometric polynomials.
If $f$ is a trigonometric polynomial we may clearly obtain $Y$ as a trigonometric polynomial (after convolution).
A standard limit procedure permits then to complete the argument.

\bigskip
\noindent
 {\bf 7. The equation $\text{div\,} \bold Y \pmb= \bold f$ with Dirichlet condition.
Proof of Theorems 2 and 3.}


So far we have studied  problem (1.1) coupled with a periodic condition.
We consider here problem (1.1) coupled with a Dirichlet condition.
Usually one associates with (1.1) the ``partial'' Dirichlet condition
$$
Y\cdot n =0 \quad \text { on } \partial Q\tag 7.1
$$
($n$ normal to $\partial Q$).
It is quite standard that for every $f \in L_{\#}^p$, $1< p<\infty$, there is some $Y\in W^{1, p}$ satisfying (1.1),
(7.1) and
$$
\Vert Y\Vert_{W^{1, p}}\leq C\Vert f\Vert _{L^p}.
$$
Indeed, one may look for a {\it special} $Y$ of the form $Y= \text {\,grad\,} u$ and one is led to the Neumann
problem
$$
\cases
\Delta u =f \qquad &{\text { in }} Q\\
\frac{\partial u}{\partial n} =0 \qquad &\text { on } \partial Q\endcases\tag 7.2
$$
which admits a solution $u\in W^{2, p}$ such that
$$
\Vert u\Vert_{W^{2, p}} \leq C\Vert f\Vert_{L^p}.
$$

It is also possible to couple problem (1.1) with the {\it full} Dirichlet condition
$$
Y=0 \qquad \text { on } \partial Q.\tag 7.3
$$
For simplicity we investigate first the case where the domain is a cube and then the case of a Lipschitz bounded
domain.

\noindent
{\bf 7.1. The case of a cube.}

Let $Q=(0, 1)^d$.
Here is a first result:


\proclaim
{Theorem 2}
Given $f\in L_{\#}^p(Q), 1<p<\infty$, there exists some $Y\in W^{1, p}_0(Q)$ solving (1.1) with
$$
\Vert Y\Vert _{W^{1, p}} \leq C(p, d)\Vert f\Vert_{L^p},
$$
where we use the standard notation
$$
W^{1, p}_0(Q) =\{Y\in W^{1, p} (Q); Y=0 \text { on } \partial Q\}.
$$
Moreover $Y$ can be chosen, depending linearly on $f$.
\endproclaim

We will make use of the following lemma (which is a special case of Theorem 2).

\proclaim
{Lemma 4}
 Given $f\in W_0^{1, p}(Q), 1<p<\infty$, with $\int f=0$, there exists $Y\in W_0^{1, p}(Q)$,
such that
$$
\text {div\,} Y=f
$$
and
$$
\Vert Y\Vert_{W^{1, p}(Q)} \leq C(d)\Vert f\Vert_{W^{1, p}(Q)}.\tag 7.4
$$
Moreover $Y$ can be chosen, depending linearly on $f$.
\endproclaim

\noindent
{\bf Proof.}
Following a known construction (see Adams [1], p. 58 and Nirenberg [1]), we construct $Y$ by induction on the dimension $d$.
The assertion is obvious for $d=1$.
Assume that it holds in dimension $(d-1)$.
Let $f\in W_0^{1, p}(Q_d)$, where $Q_d=(0, 1)^d$, with $\int_{Q_d} f=0$.

Set
$$
g(x')=\int^{1}_0 f(x', t)dt, \text { where } x'=(x_1, \ldots, x_{d-1})\in Q_{d-1}.
$$
Clearly, $g\in W_0^{1, p}(Q_{d-1})$ with
$$
\Vert g\Vert _{W^{1, p}(Q_{d-1})} \leq C\Vert f\Vert_{W^{1, p}(Q_d)}
$$
and also $\int_{Q_{d-1}} g=0$.
By the induction assumption there is some $Z\in W^{1, p}_0(Q_{d-1})$ such that
$$
\text {div}_{x'} Z =g \quad \text { on } Q_{d-1}\tag 7.5
$$
and
$$
\Vert Z\Vert_{W^{1, p}(Q_{d-1})}\leq C\Vert g\Vert_{W^{1, p}(Q_{d-1})}\leq C\Vert f\Vert_{W^{1, p}(Q_d)}.
$$
Fix a function $\zeta \in C_0^\infty(0, 1)$ such that
$$
\int_0^{1}\zeta (t)dt=1.\tag 7.6
$$
For $x=(x', x_d)\in Q_d$ set
$$
h(x) =\int_0^{x_d} (f(x', t) -\zeta(t) g(x'))dt.
$$
It is easy to see (using (7.6)) that $h\in W_0^{1, p}(Q_d)$ and
$$
\Vert h\Vert_{W^{1, p}(Q_d)} \leq C\Vert f\Vert_{W^{1, p}(Q_d)}.
$$
Moreover
$$
\frac {\partial h}{\partial x_d}(x) =f(x) -\zeta (x_d)g(x').
$$
Combining this with (7.5) yields
$$
f(x) =\text{ div}_{x'} \big(\zeta(x_d)Z(x')\big) +\frac {\partial h}{\partial x_d}
$$
i.e., the conclusion  holds with
$$
Y(x) =\big(\zeta(x_d)Z(x'), h(x)\big).
$$

\noindent
{\bf Proof of Theorem 2.}
For simplicity we assume that $d=2$; the argument is similar for $d>2$.

Let
$$
Q=\{ (x, y) \in\Bbb R^2; \quad 0<x<1, \, 0<y<1\}.
$$
Given $f\in L^p_{\#} (Q), 1<p<\infty$, we will construct a solution $Y\in W_0^{1, p}(Q)$ of (1.1); moreover
$$
\Vert Y\Vert_{W^{1, p}}\leq C_p\Vert f\Vert_{L^p}\tag 7.7
$$
and $Y$ depends linearly on $f$.
This is done in three steps.

\noindent
{\bf Step 1.}
Construct a solution $Y\in W^{1, p}(Q)$ of (1.1) satisfying (7.7) and
$$
Y=0  \ \text { on the edge } \{(x, 0); \quad 0<x<1\}.\tag 7.8
$$

\noindent
{\bf Proof.} Set
$$
\tilde Q =\{(x, y); \quad 0<x<1, -2< y<1\}
$$
and
$$
\tilde f=\cases f  \qquad \text { in } Q\\
0 \qquad \text { in } \tilde Q\backslash Q.
\endcases
\tag 7.9
$$

Let $Z\in W^{1, p}(\tilde Q)$ be the solution of
$$
\text{div\,} Z =\tilde f \quad \text { in } \tilde Q\tag 7.10
$$
obtained via (7.2) (or via periodic conditions on $\tilde Q$).

The heart of the matter is the following construction.
Write $Z=(Z_1, Z_2)$ and define in $Q$, $Y=(Y_1, Y_2)$ where
$$
\aligned
&Y_1(x, y)=Z_1 (x, y)+3Z_1(x, -y)-4Z_1(x, -2y)\\
& Y_2(x, y)= Z_2(x, y)-3Z_2(x, -y)+ 2Z_2(x, -2y).\endaligned
\tag 7.11
$$ 
(This type of ``reflection'' is reminiscent of standard extension techniques in $W^{m, p}, m\geq 2$; see e.g. Adams
[1]).

It is easy to see using (7.9), (7,10) and (7.11) that
$$
\text{div \,} Y= f\quad \text { in } Q
$$
while (7.8) is clear from the definition of $Y$.

It is important (for the next step) to observe that if we had started with the additional information
$$
Z=0 \quad \text {on the edge } \{0, y);\quad -2<y<1\} \text {  of } \tilde Q
$$
then we could infer that $Y$ {\it also vanishes} on the edge $\{(0, y); \quad 0<y<1\}$ of $Q$.

\noindent
{\bf Step 2.} Construct a solution $Y\in W^{1, p}(Q)$ of (1.1) satisfying (7.7) and
$$
Y=0 \text{ on the 2 adjacent edges } \{(x, 0); \ \, 0<x<1\} \text { and } \{(0, y);\ \, 0<y<1\}.\tag 7.12
$$

\noindent
{\bf Proof.} Set
$$
\hat Q =\{(x, y);\quad -2<x<1, \quad 0<y<1\}
$$
and
$$
\hat f =\cases f\quad \text { in } Q\\
0\quad\text { in } \hat Q\backslash Q.\endcases
$$
>From Step 1 applied to $\hat f$ in $\hat Q$ we obtain a solution $\hat Z$ of
$$
\text {div\,} \hat Z =\hat f \quad \text { in } \hat Q
$$
such that
$$
\hat Z=0 \quad \text { on the edge }\{(x, 0); -2<x<1\} \text { of } \hat Q.
$$
Starting with $\hat Z$ (instead of $Z$) we repeat the construction of Step 1 changing the roles of $x$ and $y$.
We thus obtain a $Y\in W^{1, p}(Q)$ satisfying (1.1) in $Q$, (7.7) and (7.12).

\noindent
{\bf Step 3.} Proof of Theorem 2 completed.

Consider a smooth partition of unity $(\theta_i), i= 1, 2, 3, 4$, 
subordinate to the covering of $Q$ consisting of the 4
discs of radius 1 centered at the 4 vertices.
Let $Y_i\in W^{1, p}(Q)$ be the solution constructed in Step 2 relative to each vertex.

Set
$$
Z=\sum^4_{i=1}  \theta_i Y_i.
$$
It is easy to see from this construction that $\theta_i Y_i\in W_0^{1, p}(Q)$, $\forall i$ and thus $Z\in W_0^{1,
p}(Q)$.
Moreover
$$
\text {div\,} Z= f+\sum_i \nabla \theta_i\cdot Y_i
$$
and $\sum_i\nabla \theta_i\cdot Y_i\in W_0^{1, p}(Q)$.
By Lemma 4 we may construct $X\in W_0^{1, p} (Q)$ satisfying
$$
\text {div\,} X=\sum_i\nabla \theta_i\cdot Y_i
$$
and $Y=Z-X$ has all the desired properties in Theorem 2.
\bigskip


Next we have a variant of Theorem 1$'$ for the full Dirichlet condition.


\proclaim
{Theorem 3}
Given $f\in L^d_{\#}(Q)$ there exists some $Y\in C^0(\bar Q)\cap W_0^{1, d}(Q)$ 
satisfying  (1.1) with
$$
\Vert Y\Vert_{L^\infty}+ \Vert Y\Vert_{W^{1, d}} \leq C\Vert f\Vert_{L^d}.
$$
\endproclaim
\bigskip

\noindent
{\bf Remark 10.}
Clearly, Theorem 3 implies Theorem 1$'$ since the function $Y$ extended by periodicity belongs to $C^0(\Bbb T^d)\cap
W^{1, d}(\Bbb T^d)$ and satisfies (1.1) on $\Bbb T^d$.
However its proof relies heavily on Theorem 1$'$.

\bigskip 
\noindent
{\bf Proof of Theorem 3.}
Follow the same strategy as in the proof of Theorem 2.
The only difference is that in Step 1 use Theorem 1$'$ to obtain $Z$ (instead of taking the special $Z$ in the form
of a gradient).
Of course the dependence of $Y$ on $f$ is not linear anymore.

In Step 3 rely on the following variant of Lemma 4 (with an identical proof).

\proclaim
{Lemma 4$'$}
Given $f\in C^0 (\bar Q)\cap W_0^{1, p} (Q), 1<p<\infty$, with $\int f=0$, there exists $Y\in C^0(\bar Q)\cap
W_0^{1, p}(Q)$ such that
$$
\text {\rm div\,} Y=f
$$
and
$$
\Vert Y\Vert_{L^\infty}+\Vert Y\Vert_{W^{1, p}} \leq C(\Vert f\Vert_{L^\infty} +\Vert f\Vert_{W^{1, p}}).
$$
\endproclaim
\bigskip

\noindent
{\bf 7.2. The case of Lipschitz domains.}

Let $\Omega$ be a Lipschitz, connected, bounded domain in $\Bbb R^d$.
Recall that $\Omega $ is Lipschitz if there is a $\delta>0$ such that for every point $p\in\partial \Omega$,
$\partial \Omega\cap B_\delta(p)$ is the graph of a Lipschitz function (in an appropriate coordinate system varying
with $p$).

We have the following variants of Theorems 2 and 3.

\proclaim
{Theorem 2$'$} Given any $f\in L^p_{\#} (\Omega), 1< p<\infty$, there exists some $Y\in W_0^{1, p}(\Omega)$ solving
(1.1) with
$$
\Vert Y\Vert _{W^{1, p}}\leq C(p, \Omega) \Vert f\Vert_{L^p}.\tag 7.13
$$
Moreover $Y$ can be chosen, depending linearly on $f$.
\endproclaim

\proclaim
{Theorem 3$'$}
For every $f \in L^d_{\#} (\Omega)$ there exists some $Y
\in C^0(\bar \Omega)\cap W_0^{1, d}(\Omega)$ solving (1.1) with
$$
\Vert Y \Vert_{L^\infty}+ \Vert Y\Vert_{W^{1, d}}\leq C(p, \Omega)\Vert f\Vert_{L^d}.\tag 7.14
$$
\endproclaim

The heart of the argument (for both theorems) is the following.

\proclaim
{Lemma 5} There is a bounded operator $S:L^p(\Omega) \rightarrow W_0^{1, p}(\Omega)$ such that
$$
f- \text{\rm \,div\,} Sf\in W_0^{1, p} \qquad \forall f\in L^p
$$
and
$$
\Vert f -\text{\rm \,div\,} Sf\Vert_{W^{1, p}}\leq C\Vert f\Vert_{L^p}.\tag 7.15
$$
\endproclaim

The variant needed for the proof of Theorem 3$'$ is 

\proclaim
{Lemma 5$'$} There is a nonlinear map
$S:L^d(\Omega)\to C^0(\bar \Omega)\cap W_0^{1, d}(\Omega)$ such that
$$
\Vert Sf\Vert_{L^\infty}+ \Vert Sf\Vert_{W^{1, d}}\leq C\Vert f\Vert_{L^d}\tag 7.16
$$
and
$$
\Vert f-\text{\rm \,div\,} Sf\Vert_{W^{1, d}}\leq C\Vert f\Vert_{L^d}.\tag 7.17
$$
\endproclaim

The proof of Lemma 5 relies on the following construction.
Let $Q'$ be a cube of side $\delta$ in $\Bbb R^{d-1}$ and set
$$
U=\{(x', y)\in Q'\times\Bbb R; \quad \psi(x')<y<\psi(x')+\delta\}
$$
where $\psi \in {\text {\,Lip\,}}(Q')$.

\proclaim
{Lemma 6}
Assume
$$
\Vert\nabla\psi\Vert_{L^\infty(Q')}\leq \ve_0(d) \text{ sufficiently small (depending only on $d$)}
\tag 7.18
$$ 
Then, given any $g\in L^p(U)$ there is some $Z\in W^{1, p}(U)$ satisfying
$$
\text{div\,} Z= g\quad \text { in } U,\tag 7.19
$$
{\rm (7.20)} $\qquad Z=0 \ \text {on } \{y=\psi(x'); \  x'\in Q'\}
\text { and on the lateral boundary of $U$,}$

\noindent
with
$$
\Vert Z\Vert_{W^{1, p}(U)}\leq C(p, d)\Vert g\Vert_{L^{p}(U)}.
$$
Moreover $Z$ can be chosen to depend linearly on $g$.
\endproclaim

\noindent
{\bf Proof.} For $x'\in Q'$ and $0<y<\delta$ set
$$
\tilde g(x', y)= g\big(x',  y+\psi(x')\big).
$$
Note that
$$
\Vert\tilde g\Vert_{L^p(Q)} =\Vert g\Vert_{L^p(U)}
$$
where $Q=Q'\times (0, \delta)$.

By Theorem 2 there exists $\tilde Z \in W^{1, p}(Q)$ such that
$$
\cases
\text{div\,} \tilde Z&= \tilde g \qquad \text { in } Q,\\
\quad \ \, \tilde Z&= 0 \qquad \text { on } \{(x', 0); \  x' \in Q'\}\bigcup\big(\partial Q'\times (0, \delta)\big)
\endcases
$$
with
$$
\Vert\tilde Z\Vert_{W^{1, p}(Q)}\leq C(d)\Vert\tilde g\Vert_{L^p(Q)}.\tag 7.21
$$
\quad Note that here $\int\tilde g=0$ is not required since 
we may consider in $\hat Q=Q' \times (0, 2\delta)$ the function
$$
\hat g(x', y)=\cases \tilde g(x', y)\quad &\text { for } x'\in Q' \text { and } 0<y<\delta\\
-\tilde g(x', y-\delta) &\text { for } x'\in Q \, \text { and } \delta<y<2\delta,
\endcases
$$
and then solve (using Theorem 2).
$$
\align
\text {div\,} \hat Z&= \hat g \qquad \text{ in } \hat Q\\
\hat Z&= Q \qquad \text{on } \partial \hat Q,
\endalign
$$
with
$$
\Vert\hat Z\Vert_{W^{1, p}(\hat Q)} \leq C(d) \Vert\tilde g\Vert_{L^p(Q)}.
$$
The restriction $\tilde Z$ of $\hat Z$ to $Q'\times (0, \delta)$ satisfies the desired properties.


Also, it is clear by scaling that the constant in (7.21) is independent of $\delta$.

Returning to $(x', y)\in U$, set
$$
Z(x', y)= \tilde Z\big(x', y-\psi(x')\big);
$$
it is easy to see, using (7.18) and (7.21) that
$$
\Vert\text {div\,} Z-g\Vert_{L^p(U)} \leq C(d)\ve_0\Vert g\Vert_{L^p(U)}
$$
and
$$
\Vert Z\Vert_{W^{1, p}(U)}\leq C(d)(1+\ve_0)\Vert g\Vert_{L^p(U)}.
$$
Choosing $\ve_0$ such that
$C(d)\ve_0<1$ and iterating this construction yields the lemma.
\bigskip

The variant necessary for Theorem 3$'$ is

\proclaim
{Lemma 6$'$}
Assume (7.18).
Then given $g\in L^d(U)$ there is some $Z\in C^0(\bar U)\cap W^{1, p}(U)$ satisfying (7.19), (7.20) and
$$
\Vert Z\Vert_{L^\infty (U)} +\Vert Z\Vert_{W^{1, d}(U)}\leq C(d)\Vert g\Vert_{L^d(U)}.
$$
\endproclaim

Next, we remove the smallness condition (7.18) on the Lipschitz constant of $\psi$.

\proclaim
{Lemma 7} With the same notation as in Lemma 6, assume only that $\psi\in\text{\rm\,Lip\,}(Q')$.

Then, given any $g\in L^p(U)$, there is some $Z\in W^{1, p}(U)$ satisfying (7.19), (7.20) and
$$
\Vert Z\Vert_{W^{1, p}(U)} \leq C(p, d, \Vert\nabla\psi\Vert_{L^\infty(Q')})\Vert g\Vert_{L^p(U)}.
$$
Moreover $Z$ can be chosen to depend linearly on $g$.
\endproclaim

\noindent
{\bf Proof.} Consider the dilation $x' \mapsto \tilde x'=Nx'$ (only in $x'$, not in the full $x$-variable).
Set $\tilde Q' =NQ'$ and define on $\tilde Q'$ the function
$$
\tilde\psi(\tilde x') =\psi (\tilde x'/N).
$$
Fix an integer $N$ sufficiently large so that
$$
\Vert \nabla \tilde \psi\Vert_{L^\infty(\tilde Q')} =\frac 1N\Vert\nabla \psi\Vert_{L^\infty(Q')} \leq \ve _0(d)
$$
where $\ve_0(d)$ comes from (7.18).

Set
$$
\tilde g(\tilde x', y) =g\bigg(\frac {\tilde x'}N, y\bigg).
$$
Divide the cube $\tilde Q'$ (of side $N\delta$) into $N^{d-1}$ cubes of side $\delta$ and apply, in each one of
them, Lemma 6 to $\tilde\psi$ and $\tilde g$.
By gluing the corresponding solutions (this is possible because all these solutions vanish on the lateral boundaries
of their domains), we obtain some $\tilde Z(\tilde x', y)\in W^{1, p}(\tilde U)$ satisfying
$$
\cases
\text{div\,}_{\tilde x', y}\tilde Z&= \tilde g \qquad \text { in } \tilde U=\{(\tilde x', y)\in \tilde Q' \times 
\Bbb R;  \ \tilde\psi(\tilde x')<y<\tilde \psi(\tilde x')+\delta\}\\
\qquad  \quad  \tilde Z&=0 \qquad \text { on } \{y=\tilde\psi (\tilde x'); \tilde x' \in\tilde Q'\}
\endcases
$$
and the corresponding $W^{1, p}$-estimate for $\tilde Z$

We now return to the variables $(x', y)\in U$.
Write the components of $\tilde Z$ as
$$
\tilde Z =(\tilde Z', \tilde Z_d)
$$
and set
$$
Z(x', y)=\bigg(\frac 1N\tilde Z'(Nx', y), \tilde Z_d(Nx', y)\bigg).
$$
It is easy to check that $Z$ satisfies all the required properties.
\bigskip

The variant necessary for Theorem 3$'$ is 

\proclaim
{ Lemma 7$'$}
With the same notation as in Lemma 6, assume only that $\psi\in\text{\,\rm Lip\,}(Q')$.

Then, given any $g\in L^d(U)$, there is some
$Z\in C^0(\bar U)\cap W^{1, p}(U)$ satisfying (7.19), (7.20) and
$$
\Vert Z\Vert_{L^\infty(U)}+\Vert Z\Vert_{W^{1, p}(U)} \leq C(d, \Vert\nabla\psi\Vert_{L^\infty(Q')}) \Vert
g\Vert_{L^p(U)}.
$$
\endproclaim

We now return to the

\noindent
{\bf Proof of Lemma 5.} Consider a finite covering of $\partial\Omega$ by a collection of cubes $Q_i$ $i=1, \ldots,
k$, of side $\delta$ such that in each $Q_i$, $\partial\Omega\cap Q_i$ admits a 
Lipschitz parametrization $\psi_i$.
To this covering we associate functions $\theta_0, \theta_1, \ldots, \theta_k$ such that
$$
\align
&\theta_0+\sum^k_{i=1} \theta_i = 1 \qquad {\text { on }} \Omega,\\
&\theta_0\in C_0^\infty(\Omega) \text { and }
\theta_i\in C_0^\infty(Q_i)\text { for $i=1, \ldots, k$.}
\endalign
$$
Given $g\in L^p(\Omega)$ solve, using Lemma 7, for $i=1, 2, \ldots, k$
$$
\cases
\ \text{div}\!\!\!\!\!&Z_i = g\qquad\text { in } U_i\\
&Z_i= 0\qquad \text{ on } \partial \Omega\cap Q_i.
\endcases
$$
Next solve
$$
\text{\rm div\,} Z_0 = g \qquad \text { in } \  \Omega
$$
for example $Z_0=\grad (\Delta)^{-1}$ where $\Delta^{-1}$ is used with zero Dirichlet condition on $\partial\Omega$.

Note that
$$
Z=\sum^k_{i=0} \theta_iZ_i\in W_0^{1, p}
$$
and
$$
\text{div\,} Z= g+\sum^k_{i=0} \nabla\theta_i\cdot Z_i.
$$
All the conclusions of Lemma 5 hold with
$$
Sg=Z.
$$

\noindent
{\bf Proof of Lemma 5$'$.}
We make the same construction as above, using Lemma 7$'$ in place of Lemma 7 and Theorem 2 to solve $\text{div\,}
Z_0=g$ in any large cube containing $\Omega$.
\bigskip

Theorem 2$'$ is an immediate consequence of Lemma 5 and the following general functional analysis argument applied
with $E= W_0^{1, p}, F=L^p_{\#}$ and $T=\text{\,div}$.
(Note that $T^*=\grad$ is injective on $F^*=L^q_{\#}$, since $\Omega$ is connected).

\proclaim
{Lemma 8}
Let $E, F$ be two Banach spaces and let $T$ be a bounded operator from $E$ into $F$.
Assume
$$N(T^*)=\{0\}.\tag 7.22
$$
$$
\cases \text{ There is a bounded operator $S$ from $F$ into $E$ and }\\
\text{  a compact operator $K$ from $F$ into itself such that}\\
\qquad\qquad\qquad T\circ S=I+K.
\endcases\tag 7.23
$$

Then $T$ admits a right--inverse.
\endproclaim

\noindent
{\bf Proof.}
First we note that $T$ is onto.
Indeed in view of (7.22) it suffices to show that $T$ (or equivalently $T^*$) has closed range.
This is an obvious consequence of the inequality
$$
\Vert f\Vert \leq C\Vert T^* f\Vert+\Vert K^* f\Vert\qquad \forall f\in F^*
$$
(which follows from (7.23)).

Next, let $X$ be a complementing subspace for $N(I+K)$ in $F$ and set $Y=R(I+K)$.
Since $u=(I+K)_{|X}$ is an isomorphism onto $Y$, its inverse $u^{-1}:Y\to X\subset F$ satisfies
$$
(I+K)\circ u^{-1} =I \text { on } Y.\tag 7.24
$$

Let $Q$ be a projector from $F$ onto $Y$;
since $R(I-Q)$ is finite dimensional we may choose a base $(e_\alpha)$ of $R(I-Q)$ and write 
$$
f=Qf+\sum_\alpha \langle e_\alpha^*, f\rangle e_\alpha \qquad \forall f\in F,\tag 7.25
$$
for some $e_\alpha^*$'s in $F^*$.

Since we showed that $T$ is onto, one has, for each $\alpha$, some $\bar e_\alpha \in E$ satisfying
$$
T\bar e_\alpha =e_\alpha \qquad \forall \alpha.\tag 7.26
$$
Consider the operator $S_1: F\to E$ defined for every $f\in F$, by
$$
S_1 f=S\circ u^{-1} \circ Qf+\sum_\alpha \langle e_\alpha^*, f\rangle \bar e_\alpha.
$$
Using (7.24), (7.25) and (7.26) we see that
$$
\align
T\circ S_1 f&= (I+K)\circ u^{-1} \circ Qf+\sum_\alpha \langle e^*_\alpha, f\rangle e_\alpha\\
&= Q f+\sum_\alpha \langle e^*_\alpha, f\rangle e_\alpha = f
\endalign
$$
for every $f\in F$.
Thus $S_1$ is a right-inverse for $T$.
\bigskip

\noindent
{\bf Proof of Theorem 3$'$.} Given $f\in L^d$ write, using Lemma 5$'$,
$$
f=\text {\rm div\,} Y_1+R
$$
with $Y_1\in C^0(\bar\Omega)\cap W_0^{1, d}(\Omega)$ and $R\in W_0^{1, d}(\Omega) $ (and the corresponding
estimates).

If $\int f=0$, then $\int R=0$ and we may apply Theorem 2$'$ in any $L^p$ (since $W^{1, d}\subset L^p$, 
$\forall p<\infty$).
In particular if we choose $p>d$ we obtain $Y_2\in W_0^{1, p}(\Omega)$ such that
$$
R=\text{\,div\,} Y_2.
$$
By the Sobolev imbedding, $Y_2\in C^0(\bar\Omega)$ and $Y=Y_1+Y_2$ satisfies all the required properties.


\bigskip
\noindent
{\bf 8. Estimation of the phase in $H^{1/2}+W^{1, 1}$.
Proof of Theorem 4.}

We return in this last Section to the question discussed in the Introduction concerning the control of the phase
$\vp$ in terms of $\Vert e^{i\vp}\Vert_{H^{1/2}}$.

Let $\vp$ be a smooth real-valued function on $\Bbb T^d$ and set $g=e^{i\vp}$.
The main result is the estimate
$$
\Vert\vp\Vert_{H^{1/2}+W^{1, 1}}\leq C(d)(1+\Vert g\Vert_{H^{1/2}}) \Vert g\Vert_{H^{1/2}}.\tag 8.1
$$

Write $g$ as a Fourier series
$$
g=\sum_{\xi\in\Bbb Z^d} \hat g(\xi) e^{ix.\xi}.
$$
The $H^{1/2}$-component in the decomposition of $\varphi$
will be obtained as a paraproduct of $g$ and $\bar g$
$$
P=\sum_k\bigg[\sum_{\xi_2} \lambda_k(|\xi_2|)
\overline{\hat g(\xi_2)} e^{-ix.\xi_2}\bigg]
\bigg[\sum_{2^k\leq|\xi_1|<2^{k+1}}\hat
g(\xi_1)e^{ix.\xi_1}\bigg]\tag 8.2
$$
where for each $k$ we let $0\leq \lambda_k\leq 1$ be a
smooth function on $\Bbb R_+$

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We claim that
$$
\Vert P\Vert_{H^{1/2}} \leq C\Vert g\Vert_\infty
\Vert g\Vert_{H^{1/2}}\tag 8.3
$$
and
$$
\Vert\varphi-\frac 1iP\Vert_{W^{1,1}} \leq C\Vert
g\Vert^2_{H^{1/2}}.\tag 8.4
$$
\noindent
{\bf Proof of (8.3)}

This is totally obvious from the construction
$$
\align
\Vert P\Vert^2_{H^{1/2}} &\sim
\sum_k2^k\bigg\Vert\bigg[\sum_{\xi_2}
\lambda_k(|\xi_2|)\overline{\hat g(\xi_2)}
e^{-ix\xi_2}\bigg] \bigg[\sum_{2^k\leq |\xi_1|<
2^{k+1}}\hat g(\xi_1)
e^{ix\xi_1}\bigg]\bigg\Vert^2_2\\
& \leq \sum_k2^k\bigg\Vert\sum\lambda_k(|\xi|)
\overline{\hat g(\xi)}
e^{-ix\xi}\bigg\Vert^2_\infty\bigg[\sum_{|\xi|\sim
2^k}|\hat g(\xi)|^2\bigg]\\
&\leq C\Vert g\Vert^2_\infty \Vert g\Vert^2_{H^{1/2}}.\tag 8.5
\endalign
$$

\noindent
{\bf Proof of (8.4)}

We estimate for instance
$$
\Vert\partial_1\varphi -\frac 1i \partial_1
P\Vert_{L^1}.\tag 8.6
$$
Thus, letting $\xi =(\xi^1, \ldots, \xi^d)\in \Bbb
Z^d$
$$
\partial_1\varphi =\frac 1i \bar g\partial_1g
=\sum_{\xi_1, \xi_2\in \Bbb Z^d} \xi^1_1 \hat
g(\xi_1)\overline{\hat g(\xi_2)} \,
e^{ix.(\xi_1-\xi_2)}\tag 8.7
$$
\noindent
and by (8.2)
$$
\frac 1i\partial_1P=\sum_k\sum_{2^k\leq
|\xi_1|<2^{k+1},
\xi_2}(\xi_1^1-\xi^1_2)\lambda_k(|\xi_2|)\hat
g(\xi_1)\overline{\hat g(\xi_2)}
e^{ix.(\xi_1-\xi_2)}\tag 8.8
$$
$$
\partial_1\varphi-\frac 1i\partial_1P =\sum_k \sum_{2^
k\leq |\xi_1|<2^{k+1}, \xi_2} m_k
(\xi_1,\xi_2)\hat g(\xi_1)\overline{\hat g(\xi_2)}
e^{ix.(\xi_1-\xi_2)}\tag 8.9
$$
where by definition of $\lambda_k$
$$
m_k(\xi_1, \xi_2)= \xi_1^1
-\lambda_k(|\xi_2|)(\xi_1^1-\xi^1_2)=\cases \xi_2^1
\quad \text { if } |\xi_2|\leq 2^{k-2}\\ \xi^1_1 \quad \text { if
} |\xi_2| \geq 2^{k-1}\endcases.\tag 8.10
$$
Estimate
$$
\Vert\partial_1\varphi-\frac 1i\partial_1 P\Vert_1\leq
\sum_{k_1, k_2}\bigg\Vert\sum_{|\xi_1|\sim 2^{k_1},
|\xi_2|\sim 2^{k_2}} m_{k_1} (\xi_1, \xi_2)\hat
g(\xi_1)\overline{\hat g(\xi_2)}
e^{ix.(\xi_1-\xi_2)}\bigg\Vert_1.\tag 8.11
$$
Distinguish the contributions of 
$$
\sum_{k_1\sim k_2}
+\sum_{k_1<k_2-4}+\sum_{k_1>k_2+4} =(8.12)+(8.13)+(8.14).
$$

Clearly $2^{-k} m_k(\xi_1, \xi_2)$ restricted to
$[|\xi_1|\sim 2^k]\times [|\xi_2|\sim 2^k]$ is a
smooth multiplier satisfying the usual derivative
bounds.
Therefore
$$
(8.12) \leq C\sum_k2^k\bigg\Vert\sum_{|\xi_1|\sim 2^k}
\hat g(\xi_1)e^{ix.\xi_1}\bigg\Vert_2\bigg \Vert
\sum_{|\xi_2|\sim 2^k}\hat
g(\xi_2)e^{ix\xi_2}\bigg\Vert_2 \sim \Vert
g\Vert^2_{H^{1/2}}.\tag 8.15
$$
If $k_1< k_2-4, |\xi_2|> 2^{k_1}$ and $m_{k_1}(\xi_1,
\xi_2)=\xi_1^1$ by (8.10). Therefore
$$
\align
(8.13)&= \sum_{k_1<k_2-4}\bigg\Vert\sum_{|\xi_1|\sim
2^{k_1}, |\xi_2|\sim 2^{k_2}}\xi^1_1 \, \hat
 g(\xi_1)\overline{\hat g(\xi_2)}
e^{ix.(\xi_1-\xi_2)}\bigg\Vert_1
\\
&\leq \sum_{k_1<k_2-4} 2^{k_1}
\bigg\Vert\sum_{|\xi_1|\sim 2^{k_1}}
\hat g(\xi_1)e^{ix.(\xi_1)}
\bigg\Vert_2.\bigg\Vert\sum_{|\xi_2|\sim 2^{k_2}}
\hat g(\xi_2)
e^{ix.\xi_2}\bigg\Vert_2\\
&\leq \sum_{k_1<k_2} 2^{k_1}\bigg(\sum_{|\xi_1|<
2^{k_1}}|\hat g(\xi_1)|^2\bigg)^{1/2}
\bigg(\sum_{|\xi_2|\sim 2^{k_2}}|\hat
g(\xi_2)|^2\bigg)^{1/2} \leq C\Vert
g\Vert^2_{H^{1/2}}.\tag 8.16
\endalign
$$
If $k_1>k_2+4, |\xi_2|< 2^{k_1-2}$ and $m_{k_1}(\xi_1,
\xi_2)= \xi^1_2$ and the bound on (8.14) is similar.
\bigskip
\bigskip
\centerline{\bf Bibliography}
\bigskip
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\item{} {\bf G.~Duvaut and J.L.~Lions [1]}, ``Les in\'equations en m\'ecanique et en physique'', Dunod, 1972;
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M. A. Roller eds), Cambridge Univ. Press, 1993.


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{\it I problemi al contorno per le equazioni differenziali di tipo ellitico},
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40-49.

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l'\'equation \` a forme volume prescrite}, C. R. Acad. Sc. Paris {\bf
319} (1994), 25-28; {\bf [2]}, {\it Resolutions of the prescribed
volume form equations}, {Nonlinear Differential Equations Appl.}
{\bf 3 } (1996), 323-369.

\item{} {\bf R.~Temam [1]}, ``Navier-Stokes equations'', North--Holland, revised edition, 1979.

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Fefferman-Stein decomposition of BMO $(\Bbb R^n)$}, {Acta Math.}
{\bf 148} (1982), 215-241.

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Applic. Anal. {\bf 51} (1993), 35--40.

\item{} {\bf R.~Wojtaszczyk [1]}, ``Banach spaces for analysts'',
Cambridge Univ. Press, 1991.

\item{} {\bf D. Ye [1]}, {\it Prescribing the Jacobian determinant in
Sobolev spaces}, Ann. Inst. H. Poincar\'e, Anal. Nonlin\'eaire
{\bf 11} (1994), 275-296.




\bigskip
\line{\null\hfil{\vtop{
\hbox{(1) Institute for Advanced Study}
\hbox{Princeton, NJ 08540}
\hbox { email: bourgain\@math.ias.edu}
\bigskip
\hbox{(2) Analyse Num\'erique}
\hbox{Universit\'e P. et M.~Curie, B.C. 187}
\hbox{4 Pl. Jussieu}
\hbox{75252 Paris Cedex 05, France}
\hbox { email: brezis\@ccr.jussieu.fr}
\bigskip
\hbox{ and }
\bigskip
\hbox{(3) Department of Mathematics}
\hbox{Rutgers University}
\hbox{Hill Center, Busch Campus}
\hbox{110 Frelinghuysen Rd.}
\hbox{Piscataway, NJ 08854, USA}
\hbox{ email: brezis\@math.rutgers.edu} }\hfill}}
\bigskip



\enddocument

