AN ENDPOINT LITTLEWOOD-PALEY INEQUALITY FOR BVP ASSOCIATED WITH THE LAPLACIAN ON LIPSCHITZ DOMAINS

We prove a commutator inequality of Littlewood-Paley type between partial derivatives and functions of the Laplacian on a Lipschitz domain which gives interior energy estimates for some BVP. It can be seen as an endpoint inequality for a family of energy estimates.


Introduction
Let Ω be a Lipschitz connected domain in R n .Consider the heat equation Simple integration by parts gives us the following energy estimate where X = (x, t) ∈ Q.If f happens to be smoother then similar weighted estimates of Littlewood-Paley type can be obtained via functional calculus where η(X) = t is the distance of X = (x, t) to the bottom boundary of Q and f s = (−∆) s/2 f 2 which is roughly a semi-norm on H s (Ω).This holds for 0 ≤ s < 1.A proof will be given later for convenience.This proof shows that c s blows up as s ↑ 1.In fact, this inequality does fail for s = 1.The question is whether a suitable correction can be made on u to obtain an endpoint estimate.In this case, one imposes f to be in H 1 (Ω), and at first sight, it seems quite natural to compare ∇ x u with ∇f at least for small time but the integral 1 0 Ω |e t∆ ∇f (x) − ∇f (x)| 2 dxdt t diverges in general.A better correction is to take instead of ∇f the solution v of the heat equation with initial data ∇f .It turns out as we will show that Using semigroup notations, we have for X =(x, t) ∈ Q, u(X) = (e t∆ f )(x) and v(X) = (e t∆ ∇f )(x).This inequality amounts to the commutator inequality (1) where [A, B] = AB − BA.
A similar phenomenon appears with those harmonic functions on Q solutions of with Dirichlet or Neumann condition on the lateral boundary of Q and u going to 0 at ∞.One has for s ∈ [0, 1), and for s = 1, correcting u with the harmonic function v taking values ∇f on Ω we will obtain (2 This amounts to studying another commutator: [∇ x , e −t(−∆) 1/2 ].
Both inequalities (1) and ( 2) can be seen from different viewpoints: they are either endpoint estimates in a family of Littlewood-Paley estimates and or a measure of the defect of the semigroups of not being convolution operators.It should be observed that, if Ω = R n , both semigroups are convolution operators and the commutators vanish.We also suspect some connections with the Hodge theory for the Dirichlet or Neumann Laplacian on Ω.Such estimates arose in connection with our work [2] on the square roots of second order elliptic divergence operators on Lipschitz domains for which we had to understand this defect precisely.
This paper is organised as follows.In Section 2, we state the main theorem.We explain the result in the case of the upper half-space in Section 3. Then we prove it for special Lipschitz domains in Section 4 and for bounded Lipschitz domains in Section 5. We conclude with other commutators in Section 6.
We want to thank S. Hofmann and A. McIntosh with whom we have discussed these topics.

Statement of the main result
As we shall see in Section 6, the commutators introduced above have the same nature as the one defined in terms of the resolvent family R t = (1 − t 2 ∆) −1 for t > 0. Specific features of potential theory forces us to use resolvents.Recall that for given is obtained by letting v = u.Here and thereafter, ∇ is the n-tuple of partial derivatives ∂ ∂xj , 1 ≤ j ≤ n, defined on V .Also, V = H 1 0 (Ω) in the case of the Dirichlet Laplacian, and V = H 1 (Ω) in the case of the Neumann Laplacian.
Before moving on, let us recall some basic facts for the square root of −∆.By a straightforward integration by parts, we have where the last equality comes from Green's formula.By the boundedness of ∂ ∂xj (−∆) −1/2 and density Thus, (−∆) −1/2 ∂ ∂xj also extends to a bounded operator on L 2 (Ω).Next, introduce the commutator between the partial derivatives and the resolvent of ∆, that is The cancellation in the commutator brings a better result.
Theorem 1.We have Here and from now on V = H 1 0 (Ω) for a Dirichlet boundary condition and V = H 1 (Ω) for a Neumann boundary condition.

Remark.
The constant c depends only on the Lipschitz character of Ω when Ω is a special Lipschitz domain (see Section 4).When Ω is a bounded Lipschitz domain, the proof gives a right hand side of the form c( where d has the homogeneity of a length and c depends only on the Lipschitz character of Ω.On gets rid of Ω |f (x)| 2 dx by Poincaré's inequality, but this makes the constant in (7) depend on other geometrical parameters of the domain.Still, it remains invariant under rigid motion between domains.

The case of the upper half-space
It is interesting to examine in some detail the case of a flat boundary (Ω = R n + ) since the reflection principle yields a kernel representation of the commutator.
Let E(x) be the fundamental solution of 1 − ∆ on R n , which vanishes at ∞. Hence E t (x) = 1 t n E( x t ) is the fundamental solution of 1 − t 2 ∆ and the kernel of R t is given by where the − sign is for Dirichlet boundary condition and the kernel is called G t (x, y) and the + sign is for Neumann boundary condition and the kernel is called N t (x, y).Here, y * = (y 1 , . . ., y n−1 , −y n ) denotes the reflection of y across the boundary in the vertical direction.
Since R t is of convolution type in the first n − 1 variables, it commutes with ∂ ∂xj for 1 ≤ j ≤ n−1.When j = n, easy computations give a kernel representation of C t f in terms of ∂f ∂xn .More precisely, for Note that (8a) would not hold if f had a non vanishing trace on the boundary.
Hence (7) in both cases follows from an inequality of the type As we shall see in Section 4.3, this is basically a consequence of Hardy inequality together with the classical estimates for E which we now recall (see [7]).
In the case of a general Lipschitz domain, kernel representation will also be our basic tool together with Green formula.

Special Lipschitz domains
Assume that Ω is the open set above a Lipschitz graph.Let us introduce some notations.We are given a function φ : is its vertical projection on ∂Ω and x * = (x , 2φ(x ) − x n ) is its vertical reflection across ∂Ω.We shall consistently use the notation x to denote a point on ∂Ω.It is worth noticing that x → x * is a bilipschitz transformation from Ω onto c Ω with jacobian determinant equal to 1. Finally, σ denotes surface measure on ∂Ω and N (x) the exterior unit normal.
We first assume that Ω is smooth, that is with a C ∞ boundary, in order to make the computations rigourous.The limiting argument is done in Section 4.4.Of course, we only use quantitatively the Lipschitz character of Ω.

Dirichlet boundary condition.
Let G t (x, y) be the Green's function of I − t 2 ∆ on Ω.It is defined for (x, y) ∈ Ω × Ω and x = y.Elliptic boundary regularity tells us that G t is C ∞ where it is defined.Moreover, for fixed y ∈ Ω, G t and where E t (x) is the fundamental solution of I − t 2 ∆ on R n as introduced above.In other words, for fixed y ∈ Ω, We now come to the proof of (7).By density, it suffices to take f ∈ C 1 0 (Ω).Denote by E t and H t the operators on C 1 0 (Ω) associated with the kernels above.Since E t is a convolution operator, [∇, E t ] = 0 when acting on C 1 0 (Ω).Thus, we have The difference no longer plays any role and (7) will follow from Pointwise interior estimates on H t (x, y) give us a direct proof of (11a).The proof of (11b) is more involved and use instead boundary representation and Rellich inequalities in the spirit of [5].
We begin with controlling I 1 .Let us make a simple but useful geometric observation whose proof is left to the reader.
The key estimate is the following Lemma 4.There are numbers c > 0 and a ≥ 1 depending only on M such that Proof: Let us observe that 0 ≤ H t (x, y) follows from the minimum principle, since E t (x − y) is non-negative everywhere for x ∈ ∂Ω.
Let α 2 < α 1 be the constants of Lemma 2 and set a = c(M )α 1 /α 2 where c(M ) is the constant in (12).Then, there exists a constant c > 0 such that Now, using (10) we also have The estimate (11a) is now a simple consequence of Lemma 5 whose proof is postponed for the moment.
where c depends only on M , α, β and n.
We now turn to proving (11b).First we reduce things to boundary integrals.
The proof is classical.Integrate by parts 0 Then apply Cauchy-Schwarz inequality and let R tends to ∞.
For f ∈ C 1 0 (Ω) the qualitative properties of H t (x, y) imply that for each t > 0, u = H t f satisfies the hypothesis of Lemma 6.Hence, we have I 2 ≤ √ I 3 I 4 , where and The integral I 3 is the easiest to deal with.Since f vanishes on ∂Ω, we can write where e n = (0, . . ., 0, 1) so that we obtain, using Fubini's theorem, where Next, using Lemma 4 and Lemma 3, we obtain when n ≥ 4 and similar estimates when n = 2 and n = 3.We conclude applying the following lemma to the operator Ht t whose proof is in Section 4.3.
where c depends only on M , α, β and n.
Of course, one cannot do the same thing with ∇ x H t (x, y) as we do not have enough information on the normal component (recall that we do not want to use quantitatively the smoothness assumption on Ω).This is where we use the Rellich identity.Lemma 8. Let u ∈ C 2 (Ω) be a real valued function satisfying (I − t 2 ∆)u = 0 in Ω. Assume furthermore that u and ∇u have rapid decay at ∞.
where ∇ T denotes the tangential gradient at the boundary, and c depends only on M .
Hence, by Stokes theorem (as in Lemma 6 do it on Ω R and let R tend to ∞) we obtain The tangential gradient is defined by ∇ T u = ∇u − (∇u • N )N so that it is orthogonal to the normal derivative.Hence Note that all integrals converge thanks to the assumptions.Now choose e = (0, . . ., 0, −1).Since Ω is a special Lipschitz domain, there exists α > 0 depending on M only such that 1 ≥ e•N ≥ α a.e. on ∂Ω.Using this and Remark.As we were finishing writing this paper, we learned that Ancona proved the Rellich identity on bounded Lipschitz domains for functions in the domain of the Laplacian, i.e. u ∈ H 1 0 (Ω) with ∆u ∈ L 2 (Ω).See [1] for details.The weak version given here is enough for our needs.
Since f has compact support, this lemma applies to u = H t f for each t > 0 thanks to the qualitative properties of H t (x, y).Integrating with respect to t gives us By definition of H t , we have on ∂Ω, so that it suffices to prove where now the gradient is taken over all directions.Using the same technique as for I 3 , we have where It is well-known that |∇E(x)| ≤ c|x| 1−n e −α|x| , so that an easy calculation gives with the usual change if n = 2.We apply again Lemma 7. The proof of Theorem 1 in this special case is complete.
Remark.We owe Steve Hofmann another argument to estimate I 2 which avoids the use of Rellich identity.The idea is to prove the identity, (14 for some number c depending only on M .Thus, (14) implies , which is even more precise than (11b).Let us prove (14).Fix t > 0. We start out with the similar integration by parts: The key observation comes from Green's theorem which gives us that for all y ∈ Ω, Hence, by Fubini's theorem, Once this is done it suffices to replace u t by H t f and to express as before f in terms of ∂f /∂ yn since f vanishes on ∂Ω.Then integrate in t to obtain an explicit but quite messy expression for K(x, y).The control of its size uses Lemma 2, Lemma 3, Lemma 4 and elementary calculations which we leave to the interested reader.

Neumann boundary condition.
We now work with the Neumann Laplacian.We prove by following a similar strategy.This is enough since Define a function F t (x, y) on Ω × Ω by (16) where N t (x, y) is the Neumann function of I − t 2 ∆ on Ω.In other words, for fixed y ∈ Ω, This is a simple reformulation of the definition of N t (x, y).Since Ω is smooth, N t (x, y) extends smoothly to Ω × Ω away from the diagonal and for fixed y ∈ Ω, N t (x, y) and ∇ x N t (x, y) decay rapidly to 0 as |x| → ∞ with x ∈ Ω.The same properties hold for F t (x, y) as well.Let us list further properties of F t (x, y).

Lemma 9.
(i) For some constants c, a > 0 depending only on M , we have (ii) For all x ∈ Ω and t > 0, Ω F t (x, y) dy = 0.
Remark.The mean value property is of crucial importance; it is the reason of our choice for F t (x, y).Proof: To see (i), recall by (12) that |x − y| ∼ |x − y * | for x ∈ ∂Ω and y ∈ Ω.Since Ω has the extension property, we have the estimate (see [3]), where c depends only on M , while α is independent of Ω.Hence, we have for some c > 0 and a > 1.We conclude as in the proof of Lemma 4 applying the minimum principle to the functions cE at (x − y * ) ± F t (x, y).
To prove (ii), it suffices to remark that by change of variable y → y * (18) and that Ω N t (x, y) dy = 1 from the construction of N t (x, y).The lemma is proved.
Lemma 10.For f ∈ C 1 0 (Ω) and x ∈ Ω, we have where Proof: Denote by E t , E t, * and F t the operators defined on C 1 0 (Ω) associated with the kernels in (16).Writing Note that although E t is a convolution operator, [∇, E t ] is not 0 when acting on smooth functions that do not vanish on the boundary of Ω.Indeed, integrating by parts via Green's theorem (with the classical way of taking care of the singularity at x for E t (x − y) and using Now, and since N points into c Ω, we have Using the change of variables y → y * one can see that where J(ȳ) is the jacobian matrix of y → y * , which depends only on ȳ.
A straightforward computation shows that where N (ȳ) is defined in the statement.Thus, Putting altogether these equalities proves the lemma.
We now prove (15).Using Lemma 5 and the interior estimates on E t (x − y * ) and F t (x, y), we obtain good control on both terms Ω E t (x − y * ) ∂f ∂yn (y) N (ȳ) dy and Ω F t (x, y)∇f (y) dy.By Lemma 10, it remains to look at the last term, namely Ω ∇ x F t (x, y)f (y) dy.We use again boundary integrals.By the qualitative properties of F t and the fact that f has compact support, we easily see that u = F t f satisfies the hypotheses of Lemma 6 and we have 2 Thus, it is enough to show that To estimate J 1 , we use the mean value property (ii) in Lemma 9 to write we obtain, using Fubini's theorem, (20) where The term with F t (x, y) is handled similarly as the one with H t (x, y) in the Dirichlet case, by using Lemma 7.
The first integral in (20) is of a new type and can be estimated by using the following result.
Admit this lemma for the moment.Then, it is classical that the integral in the right hand side of ( 21) is equivalent to the square of the norm of f in the homogeneous Sobolev space Ḣ1/2 (∂Ω), and the trace theorem, therefore, implies that See [6].This yields the desired control of J 1 and (19a) is proved.
We now turn to the control of J 2 .The explicit value of the normal derivative of F t (x, y) at the boundary given by ( 17) implies Now, observe that as in ( 18) Hence, we can proceed as before and obtain for (∇F t f )(x) • N (x) a representation similar to (20) where F t (x, y) is replaced by so that both terms can be treated similarly using Lemma 11.Further details are left to the reader.This proves (19b) and with it (15).

Proofs of technical lemmas.
For simplicity, we assume that Ω = R n + .If not the case, we can pull back Ω to R n + via the transformation F : (x , x n ) → (x , φ(x ) + x n ) which is a bilipschitz homeomorphism from R n + onto Ω and with jacobian determinant equal to 1.With Ω = R n + , we have x * = (x , −x n ) when x = (x , x n ) and the boundary is identified with R n−1 via x = (x , 0).

Proof of Lemma 5:
It is easy to obtain from the definition of w that . Hence, Young's inequality and the definition of A t give us Thus, integrating with respect to Next, we expand the square and integrate with respect to dt t to obtain a bound The proof is complete.
Proof of Lemma 7: Using the same setting and notation as in the previous proof, we have Since w ∈ L 1 (R n−1 ), we obtain Now, integrate against dt and finish the argument as in the preceding proof.
Proof of Lemma 11: We write x = (x , 0).By Schwarz inequality, since w ∈ L 1 (R n−1 ) and the conclusion follows immediately by integrating with respect to x .

Limiting argument.
Let us use a particular change of variable introduced by Kenig and Stein to approximate a special Lipschitz domains by smooth ones with a uniform character.
Let g ∈ C ∞ 0 (R n−1 ) be even with g ≥ 0 and g = 1.Let g t (x) = t −n+1 g(x/t).Then, choosing c > 0 large enough (c ≥ 2M g(y)|y| dy works), the mapping We begin with the case of a Dirichlet boundary condition.Let us see that if (7) holds for all Ω k with a constant that depends only on M then it holds for Ω.
Denote by R k t and C k t respectively the resolvent of the Dirichlet Laplacian on Ω k and the associated commutator.Let f ∈ C 1 0 (Ω), then f ∈ C 1 0 (Ω k ) for k large enough and C k t f is well-defined on Ω k ; extend it by 0 outside.

Lemma 12. With the above notation, for all
Admitting this lemma, an application of Fatou lemma yields (7) on Ω for such an f and a density argument concludes the proof.
The proof of Lemma 12 follows from a classical fact which we recall for convenience.Let g ∈ L 2 (Ω) and define u k = R k t g and u = R t g.This means that u ∈ H 1 0 (Ω) is the unique solution of and Extend u k to be 0 outside of Ω k then (u k ) is bounded in H 1 0 (Ω), so it has a weakly convergent subsequence, which, by taking weak limits in the variational formulation (with v having compact support in Ω) must converge to u.Thus (u k ) converges weakly to u in H 1 0 (Ω).Now expanding the squares and using the equations, one finds and the above fact applied with g = f and g = ∇f proves the claim.
We now turn to the case of a Neumann boundary condition.This time we approximate Ω from outside (applying the same construction to c Ω) so that Ω k ↓ Ω. Recall that H 1 (Ω) is the space of restrictions to Ω of functions in H 1 (R n ) and similarly for Ω k replacing Ω.Let f ∈ H 1 (R n ).Thus with evident notations C k t f and C t f are well-defined on Ω k and Ω respectively.
Lemma 13.With the above notation, for all t > 0, Admitting this lemma, let f ∈ H 1 (Ω) be extended in such a way that it belongs to 2 .By (7) on Ω k we have and (7) on Ω follows from Fatou lemma and the above lemma.
To prove Lemma 13, it is enough to consider resolvents.Let g ∈ L 2 (R n ), then u = R t g is the unique solution in H 1 (Ω) of Recall also that and this tends to 0 by dominated convergence.Thus u k converges weakly to u in H 1 (Ω).
Strong convergence now follows from the inequality obtained from the defining equations for u k and u, the last integral being controlled by which tends to 0 as k → ∞ by dominated convergence.

Bounded Lipschitz domains
We now assume that Ω is a bounded and connected Lipschitz domain with Lipschitz character M .Before going into details, let us remark that it is enough to obtain (7) with a right hand side equal to c Ω (|f Ω f yields the desired result.See the remark in Section 2 for the behavior of the constants. Since Ω is a bounded Lipschitz domain, there exists a finite number of C ∞ 0 (R n ) functions, χ 1 , . . ., χ s , with the following properties: 1. 1≤k≤s χ k (x) = 1, x ∈ Ω; 2. For each k, there exists Ω k , image of a special Lipschitz domain under an orthogonal transformation in R n such that Supp The Lipschitz character M of Ω is, by definition, the infimum of the numbers sup M k , where M k is the Lipschitz character of Ω k , taken over all decompositions of Ω in such a way.From now on the letter c denotes constants that depend only on M .
Since there are a finite number of sets, there is This distance d depends on the chosen partition: the largest possible value has an intrinsic geometrical meaning, that is not related to the Lipschitz character.
For large scales (ie t > 1) we have Proof: This follows immediatly from ∇R t f 2 ≤ t −1 f 2 and From Lemma 14, it is enough to prove that Let us discuss first the case of a Dirichlet boundary condition.
where C k t is the commutator defined on Ω k using the resolvent which we denote by R k t .
Analysis of III.By (7) on Ω k (this inequality is clearly invariant under orthogonal transformations) we have for each Analysis of II.We use a Cacciopoli inequality for each term in the sum.We remark that For g = f k , we obtain Now, by the maximum principle, the kernels of R t and R k t are bounded by at −2 |x − y| 2−n e −(α|x−y|/t) for some a < ∞ and α > 0 depending only on dimension (usual change if n = 2).Since |x − y| ≥ d for x ∈ Supp ∇η k and y ∈ Supp f k and t ≤ d, is easy to deduce a bound of the form for other values of a and α.Similarly, for g = ∇f k , one obtains It is then easy to get Analysis of I.If g ∈ L 2 with Supp g ⊂ Supp χ k and u = R t g ∈ H 1 0 (Ω) we have Inserting v = u(1 − η k ) 2 yields a null right hand side because η k = 1 on the support of g.We conclude using the same argument as before and the fact that d(Supp g, Supp(1 − η k )) ≥ d.Further details are left to the reader.
A similar analysis can be done under Neumann boundary condition (Here, the kernels of R t and R k t have estimates with constants that depend on M ).The proof of Theorem 1 is complete.
Remark.Minor modifications of the argument shows that the result in Theorem 1 is valid (with an extra term c f 2  2 on the right hand side of ( 7)) on all strongly Lipschitz domains, which are those connected open sets in R n whose boundary is covered by finitely many parts of (rotated) Lipschitz graphs, possibly one of those parts being infinite.

Other commutators
Using functional calculus we can consider other commutators such as the ones in (1) and ( 2).The argument is identitical for the Dirichlet and Neumann Laplacians.Hence, (22) follows from (7).
Let us go back to the case 0 ≤ s < 1 mentioned in the introduction and finish by proving ∂u ∂t − ∆u = 0 in Q = Ω × (0, +∞), u(0) = fon Ω, where ∆ denotes the Laplacian in Ω with Dirichlet or Neumann boundary condition and u vanishes as t goes to ∞.That is u satisfies either a Dirichlet or Neumann boundary condition on the lateral boundary of Q.