\input amstex \NoBlackBoxes \parskip=\medskipamount \magnification=\magstep 1 \documentstyle{amsppt} \pageheight{8.5truein} \pagewidth{6.1truein} \def\varep{\varepsilon} \def\notint{\hbox{${\lower 1pt\hbox{$^{_{\slash}}$}\kern -8pt \int}$}} \def\Mint{\diagup\hskip-.50truecm\int} \def\mint{\diagup\hskip-.39truecm\int} \def\ve{\varepsilon} \def\grad{\text {\rm \,grad\,}} \def\vp{\varphi} \hfill{Dec 10, 2001} \bigskip \bigskip \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 $I \circ D$ ~ \input pictex \font\thinlinefont=cmr5 $$ \hbox{\beginpicture \setcoordinatesystem units <.80000cm,.80000cm> \setshadesymbol ({\thinlinefont .}) \setlinear % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\plot 8.191 25.241 5.747 22.892 / % % arrow head % \plot 5.886 23.114 5.747 22.892 5.974 23.022 / % \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\plot 2.695 25.284 5.044 22.839 / % % arrow head % \plot 4.822 22.978 5.044 22.839 4.914 23.066 / % \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 3.080 25.590 to 7.779 25.590 % % arrow head % \plot 7.525 25.527 7.779 25.590 7.525 25.654 / % \special{ps: grestore}% % Fig TEXT object % \put{$\scriptstyle I\circ D$} [lB] at 5.017 25.813 % % Fig TEXT object % \put{$L^2$} [lB] at 7.938 25.432 % % Fig TEXT object % \put{$L^2$} [lB] at 2.540 25.432 % % Fig TEXT object % \put{$L^1$} [lB] at 5.157 22.574 % % Fig TEXT object % \put{$\scriptstyle B$} [lB] at 7.176 23.876 % % Fig TEXT object % \put{$\scriptstyle A$} [lB] at 3.461 23.844 \linethickness=0pt \putrectangle corners at 2.540 26.099 and 8.217 22.574 \endpicture} $$ \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 \font\thinlinefont=cmr5 $$ \hbox{\beginpicture \setcoordinatesystem units <.60000cm,.60000cm> \setshadesymbol ({\thinlinefont .}) \setlinear % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 6.985 23.971 to 15.240 23.971 % % arrow head % \plot 14.986 23.908 15.240 23.971 14.986 24.035 / % \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 11.113 22.225 to 11.113 25.718 % % arrow head % \plot 11.176 25.463 11.113 25.718 11.049 25.463 / % \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\special{ps: gsave 0 0 0 setrgbcolor} \putrectangle corners at 12.224 25.082 and 13.970 23.019 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\special{ps: gsave 0 0 0 setrgbcolor} \putrectangle corners at 8.255 25.051 and 10.001 22.987 \special{ps: grestore}% % Fig TEXT object % \put{$_{\Lambda_j^1\subset [|n_1|\sim 2^j, |n_2|\leq 2^j]}$} [lB] at 1.587 24.606 % % Fig TEXT object % \put{$_{\Lambda_j^2\subset <[|n_2|\sim 2^j, |n_1|\leq 2^j]}$} [lB] at 1.587 23.495 % % Fig TEXT object % \put{$_{n_2}$} [lB] at 10.954 26.035 % % Fig TEXT object % \put{$_{\Lambda_j^1}$} [lB] at 10.001 22.543 % % Fig TEXT object % \put{$_{\Lambda_j^1}$}[lB] at 13.970 22.543 % % Fig TEXT object % \put{$_{n_1}$} [lB] at 15.458 23.971 % % Fig TEXT object % \put{$_{2^j}$} [lB] at 12.986 24.295 \put{$_{-2^j}$} [lB] at 8.800 24.295 \linethickness=0pt \putrectangle corners at 1.587 26.321 and 15.558 22.200 \endpicture} $$ 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 $$ \Lambda_j=\bigcup_{r<\frac 1\varepsilon +1} \Lambda_{j, r} $$ \font\thinlinefont=cmr5 $$ \hbox{\beginpicture \setcoordinatesystem units <.70000cm,.70000cm> \linethickness=1pt \setshadesymbol ({\thinlinefont .}) \setlinear % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 1.270 23.495 to 10.160 23.495 % % arrow head % \plot 9.906 23.431 10.160 23.495 9.906 23.559 / % \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 2.540 26.194 to 2.540 20.796 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 2.540 26.194 to 7.620 26.194 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 2.540 20.796 to 7.620 20.796 \special{ps: grestore}% % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 7.620 26.194 to 7.620 20.796 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 3.651 26.194 to 3.651 20.796 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 4.731 26.226 to 4.731 20.828 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 5.906 26.226 to 5.906 20.796 \putrule from 5.906 20.796 to 5.906 20.860 \special{ps: grestore}% % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \special{ps: gsave 0 0 0 setrgbcolor}\putrule from 5.302 21.273 to 5.302 20.161 % % arrow head % \plot 5.239 20.415 5.302 20.161 5.366 20.415 / % \special{ps: grestore}% % Fig TEXT object % \put{$\Lambda_j$} [lB] at 6.413 25.718 % % Fig TEXT object % \put{$n_1$} [lB] at 10.255 23.368 % % Fig TEXT object % \put{$\Lambda_{j,r}$} [lB] at 5.017 19.717 \linethickness=0pt \putrectangle corners at 1.245 26.251 and 10.255 19.602 \endpicture} $$ \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 $$ K_N(y) =\sum_{|n|j} (1-G_k).\tag 5.18 \endalign $$ 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 $$ \align H_j&=F_{j-1} +F_{j-2} (1-G_{j-1})\\ &+F_{j-3}(1-G_{j-2})(1-G_{j-1})+\cdots\\ &=\sum_{kFrom the preceding, we may thus ensure that $$ 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_{kFrom 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 $$ \Vert \nabla H_j\Vert_\infty \leq \ve^{\frac 2d-1}\sum_{k \bold 2$.} The argument is somewhat bizarre: one uses duality twice! First, from Theorem 1 we easily deduce the estimate on $\Bbb T^d$ $$ \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 $$ For simplicity we investigate first the case where the domain is a cube and then the case of a Lipschitz bounded domain. \noindent {\bf 7.1. The case of a cube.} Let $Q=(0, 1)^d$. Here is a first result: \proclaim {Theorem 2} Given $f\in L_{\#}^p(Q), 12$. Let $$ Q=\{ (x, y) \in\Bbb R^2; \quad 0From 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 $$ \hat Z=0 \quad \text { on the edge }\{(x, 0); -20$ such that for every point $p\in\partial \Omega$, $\partial \Omega\cap B_\delta(p)$ is the graph of a Lipschitz function (in an appropriate coordinate system varying with $p$). We have the following variants of Theorems 2 and 3. \proclaim {Theorem 2$'$} Given any $f\in L^p_{\#} (\Omega), 1< p<\infty$, there exists some $Y\in W_0^{1, p}(\Omega)$ solving (1.1) with $$ \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')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 $\Bbb R_+$ \font\thinlinefont=cmr5 $$ \hbox{\beginpicture \setcoordinatesystem units <.60000cm,.60000cm> \linethickness=1pt %\setplotsymbol ({\makebox(0,0)[l]{\tencirc\symbol{'160}}}) %\setshadesymbol ({\thinlinefont .}) \setlinear % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \putrule from 2.540 25.400 to 2.540 23.495 % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \putrule from 2.540 23.495 to 10.160 23.495 % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \setdashes < 0.1270cm> \plot 6.032 25.400 6.032 23.495 / % % Fig POLYLINE object % \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) \setsolid \putrule from 2.540 23.463 to 2.540 26.003 \linethickness= 0.500pt \setplotsymbol ({\thinlinefont .}) % % Fig INTERPOLATED PT SPLINE % \plot 2.540 25.400 2.651 25.404 2.759 25.408 2.866 25.412 2.971 25.415 3.074 25.419 3.175 25.422 3.275 25.426 3.372 25.429 3.468 25.432 3.562 25.435 3.654 25.437 3.745 25.440 3.834 25.443 3.921 25.445 4.006 25.447 4.090 25.449 4.172 25.451 4.253 25.453 4.332 25.455 4.410 25.457 4.486 25.458 4.560 25.459 4.634 25.461 4.705 25.462 4.776 25.463 4.845 25.463 4.912 25.464 4.978 25.465 5.043 25.465 5.107 25.466 5.230 25.466 5.349 25.466 5.462 25.465 5.572 25.463 5.677 25.462 5.778 25.459 5.874 25.457 5.968 25.453 6.057 25.449 6.143 25.445 6.226 25.440 6.306 25.435 6.383 25.429 6.458 25.422 6.530 25.415 6.600 25.408 6.668 25.400 6.732 25.390 6.810 25.375 6.897 25.355 6.988 25.333 7.079 25.310 7.165 25.286 7.241 25.262 7.303 25.241 7.375 25.210 7.464 25.166 7.552 25.121 7.620 25.082 7.728 25.017 7.794 24.975 7.862 24.930 7.931 24.885 7.995 24.841 8.096 24.765 8.169 24.698 8.258 24.609 8.345 24.518 8.414 24.448 8.484 24.378 8.572 24.289 8.661 24.200 8.731 24.130 8.798 24.058 8.882 23.966 8.970 23.876 9.049 23.812 9.111 23.777 9.184 23.742 9.271 23.707 9.374 23.671 9.495 23.633 9.564 23.612 9.638 23.591 9.719 23.569 9.806 23.546 9.900 23.521 10.001 23.495 / % % Fig TEXT object % \put{$^{2^{k-2}}$} [lB] at 5.874 22.800 % % Fig TEXT object % \put{$^1$} [lB] at 2.159 25.241 % % Fig TEXT object % \put{$^{2^{k-1}}$} [lB] at 9.684 22.800 \linethickness=0pt \putrectangle corners at 2.159 26.029 and 10.185 22.816 \endpicture} $$ 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 $$ \sum_{k_1\sim k_2} +\sum_{k_1k_2+4} =(8.12)+(8.13)+(8.14). $$ 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_1k_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 \item{}{\bf R.A. Adams [1]}, `` Sobolev spaces'', Acad.~Press, 1975. \item{} {\bf D.N.~Arnold, L.R.~Scott and M.~Vogelius [1]}, {\it Regular inversion of the divergence operator with Dirichlet boundary condition on a polygon}, {Ann.~Sc.~Norm. Pisa, Serie IV, {\bf 15} (1988), 169--192}. \item{} {\bf J. Bourgain, H. Brezis and P. 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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