Luis Escauriaza Universidad del País Vasco

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1 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY Luis Escauriaza Universidad del País Vasco Abstract. It is shown that the square of a nonconstant harmonic function u which either vanishes continuously on an open subset V contained in the boundary of a Dini domain or whose normal derivative vanishes on an open subset V in the boundary of a C 1,1 domain in R d, satisfies the doubling property with respect to balls centered at points Q V. Under any of the above conditions, the module of the gradient of u is a B 2 (dσ)-weight when restricted to V, and the Hausdorff dimension of the set of points { Q V : u(q) = 0 } is less or equal than d 2. These results are generalized to solutions to elliptic operators with Lipschitz second order coefficients and bounded coefficients in the lower order terms. Introduction Bourgain and Wolff [4] have constructed a counterexample of a C 1 -harmonic function u in R+ d = { X R d : X d 0 }, d 3, such that u and its gradient vanish simultaneously on a set of positive measure on R+. d This counterexample has been generalized to any C 1,α domain D in R d, for 0 < α < 1 and d 3 [13]. On the other hand, the following conjecture still remains an open question: If u is a harmonic function on a connected Lipschitz domain D in R d, d 3, vanishing continuously on an open set V contained in the boundary of D, D, and whose normal derivative vanishes on a subset of V with positive surface measure, then u is identically zero on D. When u is non-negative, the comparison principle for nonnegative harmonic functions vanishing continuously on an open subset of D, [5], implies that the normal derivative of u is pointwise comparable almost everywhere to the density of harmonic measure with respect to surface measure dσ on any compact subset K contained in V, and it is well known that harmonic measure is mutually absolutely continuous with respect to dσ. Therefore, the answer to the conjecture in this case is positive. The conjecture is well known to hold when the dimension of the ambient space is 2, for in this case the function f(z) = u x iu y is analytic in D and from the subharmonicity of log f on D log f(x) log f(q) dω X (Q) for all X in D, D The first author is partially supported by a grant from the Royal Swedish Academy of Sciences and the second author is supported by Universidad del País Vasco grant UPV EA-099/94 1 Typeset by AMS-TEX

2 2 VILHELM ADOLFSSON where dω X denotes the harmonic measure for D at X. But if the gradient of u is zero on a set of positive surface measure contained in D, the mutual absolute continuity of harmonic measure and surface measure implies that f is identically zero on a subset of D with positive harmonic measure. Thus, the above integral is identically equal to, and f is identically zero in D. We recall that a non-negative function w defined on D is a B 2 (dσ)-weight provided that there is a constant C such that for all Q D and r > 0 the following holds ( 1 σ( r (Q)) r(q) w 2 dσ ) C σ( r (Q)) r(q) where r (Q) = D B r (Q). In [2], the following reduction of the conjectured was obtained. w dσ, Theorem 1. Let D be a Lipschitz domain in R d, d 2, Q 0 D, and u be a harmonic function in D vanishing continuously on 6 (Q 0 ). Assume that there exists a constant M, possibly depending on u, such that for all Q 3 (Q 0 ) and 0 < r < 1 we have (0.1) u 2 dx M u 2 dx Γ 2r(Q) Γ r(q) Then, there exists a constant C depending on M, the Lipschitz character of D and d, such that for all Q 2 (Q 0 ) and 0 < r < 1 (0.2) ( 1 σ( r (Q)) r(q) u N 2 dσ ) C σ( r (Q)) r(q) u N dσ. In particular, the absolute value of the normal derivative of u is a B 2 (dσ)-weight when restricted to 2 (Q 0 ). Also, from the arguments used in [2] to prove the above theorem and standard boundary estimates for harmonic functions, one can obtain the following. Lemma 1. Let D be a Lipschitz domain in R d, d 2, and u be a harmonic function in D vanishing continuously on an open set V contained in D and whose normal derivative vanishes on a subset E of V with positive surface measure. Then, if Q V is a density point of E, u vanishes to infinite order at Q; i.e., for all n > 0 there exits C n > 0 such that u dx C n r n for all 0 < r < 1, Γ r(q) where Γ r (Q) = D B r (Q). It is well known that the doubling condition (0.1) implies that a non zero harmonic function u as in theorem 1 cannot vanish to infinity order at points Q 2 (Q 0 ), and that the B 2 (dσ) condition (0.2) on the normal derivative of u over 2 (Q 0 ) implies that the set { Q 2 (Q 0 ) : u(q) = 0 } has zero surface measure. Therefore, as it was already pointed out in [2], to obtain a positive answer

3 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 3 to the conjecture suffices to show that the doubling property (0.1) holds, which is already a classical method in problems related to unique continuation [6], [10]. In [2], [8] it was shown that the conjecture holds when the domain D is convex, showing that (0.1) holds with a constant M depending on d and u. Also, F. H. Lin [11] has proved that (0.1) holds when the domain D is C 1,1 and with a constant M depending on u, d, and the C 1,1 character of D. In fact in the last work the author shows using the doubling property and Federer s dimension reduction argument [12], that when D is a C 1,1 domain the Hausdorff dimension of the set { Q 2 (Q 0 ) : u(q) = 0 } is less or equal than d 2. In the first version of this paper the results in theorems 2 and 3 did not include the full generality of all the Dini domains because of a minor technicality. After this paper was sent for publication, I. Kukavica [9] extended theorem 2 in the original version to all Dini domains. Since the changes needed in the first version to make this extension were minimal, we have included them in this new version. In [8] the reader will find some applications to control theory of a positive answer to this conjecture. Definition. Let θ be a nonnegative and nondecreasing function defined on [0, + ) verifying 4 θ(r) dr <. r 0 A connected domain D in R d is a Dini domain if for each point Q on the boundary of D there is a coordinate system (x, x d ), x R d 1, x d R, so that with respect to this coordinate system Q = (0, 0), and a ball B centered at Q and a Lipschitz function ϕ : R d 1 R verifying (1) ϕ L (R d 1 ) m for some m > 0. (2) ϕ(x) ϕ(y) θ( x y ) for all x, y R d 1. (3) B D = B { (x, x d ) : x d > ϕ(x) },B D = B { (x, x d ) : x d = ϕ(x) }. In this work, the following results are proved. Theorem 2. Let D be a Dini domain in R d. Then, if Q 0 D and u is a nonconstant harmonic function in D vanishing continuously on 6 (Q 0 ), there exist r 0 > 0 depending on d and the module of continuity of θ at r = 0, and a constant M depending on d, the value of the above integral, and u such that the doubling property (0.1) holds for all Q 3 (Q 0 ) and 0 < r < r 0. Theorem 3. Let D be a Dini domain in R d, Q 0 D, and u be a nonconstant harmonic function in D vanishing continuously on 6 (Q 0 ). Then, the absolute value of the normal derivative of u is a B 2 (dσ)-weight when restricted to 2 (Q 0 ), and the Hausdorff dimension of the set { Q 2 (Q 0 ) : u(q) = 0 } is less or equal than d 2. The proof of the doubling property follows from a suitable modification of the arguments used in [6] to prove that a solution to a divergence form elliptic operator with Lipschitz coefficients in the unit ball of R d satisfies the doubling property in the interior of the ball. The B 2 (dσ) condition follows from theorem 1 and the last statement from the doubling property and the methods developed in [11]. In this paper and for completeness we have included the proof given in [11] of this last statement.

4 4 VILHELM ADOLFSSON In the case that the domain D is of class C 1 the methods we use yield the following partial answer to the conjecture. Theorem 4. Let D be a connected C 1 domain in R d and u be a harmonic function in D vanishing continuously on an open subset V of D. Assume that for some point Q in V, u satisfies (0.3) Γ r(q) ( u 2 dx C exp 1 ) r β for all 0 < r < 1 and for some absolute constants β and C > 0, then u must be identically zero on D. A natural question arising from the above conjecture is the following: If u is a harmonic function on a connected Lipschitz domain D in R d whose normal derivative vanishes almost everywhere on an open subset V of D, and whose gradient vanishes on a subset of V with positive surface measure, then u must be identically constant on D. In this case we have the following analog of theorem 1. Theorem 5. Let D be a Lipschitz domain in R d, d 2, Q 0 D, and u be a harmonic function in D whose normal derivative vanishes almost everywhere on 6 (Q 0 ). Assume that there exists a constant M, possibly depending on u, such that for all Q 3 (Q 0 ), l R and 0 < r < 1 we have (0.4) (u l) 2 dx M (u l) 2 dx Γ 2r(Q) Γ r(q) Then, there exists a constant C depending on M, the Lipschitz character of D and d, such that for all Q 2 (Q 0 ) and 0 < r < 1 (0.5) ( 1 σ( r (Q)) r(q) u 2 dσ ) C σ( r (Q)) r(q) u dσ. In particular, u is a B 2 (dσ)-weight when restricted to 2 (Q 0 ). Hence, as in the previous case it suffices to establish the doubling property at the boundary to derive that a harmonic function u as above must be constant on D. In this case, we are only able to show that the above conjecture holds in the case of C 1,1 domains, obtaining the following analog of theorems 2 and 3. Theorem 6. Let D be a C 1,1 domain in R d, Q 0 D and u be a nonconstant harmonic function in D whose normal derivative vanishes continuously on 6 (Q 0 ). Then, there exist r 0 depending on the C 1,1 character of D and d, and a constant M depending on d, the C 1,1 character of D and u, such that the doubling property (0.4) holds for all Q 2 (Q 0 ), l R and 0 < r < r 0.

5 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 5 Theorem 7. Let D, Q 0 D and u be as in the previous theorem. Then, the length of the gradient of u is a B 2 (dσ)-weight when restricted to 2 (Q 0 ) and the Hausdorff dimension of the set { Q 2 (Q 0 ) : u(q) = 0 } is less or equal than d 2. In fact, in the remarks after the proofs of the theorems we will show that theorems 1, 2, 3 and 4 hold when we replace the Laplace operator for an elliptic operator in divergence form with Lipschitz coefficients and bounded coefficients in the lower order terms. We will also show that theorems 5, 6 and 7 hold when we replace the Laplace operator for an elliptic operator in divergence form with Lipschitz coefficients, bounded coefficients in the first order term and with a zero coefficient in the term of zero order. In section 1 we show how the methods developed in [2] yield lemma 1 and prove two auxiliary theorems that we need to prove theorems 2 and 6. In section 2 we prove theorems 2, 3 and 4, in section 3 we prove theorems 5, 6 and 7 and in section 4 some of these results are extended to parabolic operators. Acknowledgments:. The authors wish to thank Carlos E. Kenig for all the conversations we had related to this work, and for having introduced us to these problems. 1. Proof of the auxiliary theorems Proof of Lemma 1. Without loss of generality we may assume that V = 6 (Q 0 ). We let u be a harmonic function as in theorem 1 and Q 2 (Q 0 ) denote a density point of the set E = { Q 2 (Q 0 ) : u(q) = 0 }; i.e., σ( r (Q) E) (1.1) lim = 1. r 0 σ( r (Q)) In [2] (See also lemma 3 in this paper) it was shown that for all n > 0 there exists C n depending on the Lipschitz character of D, d and n such that for 0 < r < 1 the following holds (1.2) Γ r(q) u dx C n r 2 2r(Q) u (Q) dσ + 2 n u dx. N Γ 2r(Q) On the other hand, the first integral on the right hand side of the last inequality can be bounded by ( σ( 2r (Q) \ E) 1/2 u 2r(Q) N 2 dσ ) 1/2. Let β denote a vector field supported in Γ 4r (Q) with β r 1, β N C 1 on 2r (Q) for some constant C depending on the Lipschitz character of D and β N 0 on 4r (Q), where N denotes the exterior unit normal to D at points of D. Integrating the Rellich-Necas identity div(β u 2 ) = 2div((β. u) u) + O( β u 2 )

6 6 VILHELM ADOLFSSON over Γ 2r (Q) and since u = u N N almost everywhere in 5(Q 0 ), we obtain 2r(Q) u N 2 dσ Cr 1 Γ 4r(Q) u 2 dx, and from Cacciopoli s inequality and standard estimates for harmonic functions vanishing on the boundary [7] we have for some constant C depending on the Lipschitz character of D ( 2r(Q) u N 2 dσ From the above chain of inequalities Γ r(q) u dx [ ) 1/2 Cr d+3 2 C n ( σ( 2r (Q) \ E) σ( 2r (Q) Γ 6r(Q) ) 1/2 + 2 n ] u dx. u dx. Γ 6r(Q) Then, from (1.1) and choosing n large enough we find that for all ε > 0 there exists r(ε) > 0 such that Γ r(q) u dx ε u dx Γ 6r(Q) and this is well known to imply that which proves the lemma. Γ r(q) for 0 < r < r(ε), u dx = O(r n ) for all n > 0, In the next theorems r and Γ r will denote respectively r (0) and Γ r (0), and we will consider an operator L in divergence form (1.3) Lu = d D i (a ij (X)D j u) + i,j=1 d b i (X)D i u + c(x)u, where the coefficient matrix A(X) = (a ij (X)) is symmetric and the coefficients verify the following properties: i=1 (i) (ii) d λ ξ 2 a ij (X)ξ i ξ j λ 1 ξ 2 for all ξ and X in R d. i,j=1 d b i (X) + c(x) K for all X in R d. i=1

7 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 7 Theorem 8. Let Ω be a Lipschitz domain in R d with 0 Ω and u be a nonconstant solution to Lu = 0 in Γ 1 vanishing continuously on 1, and where the operator L is as in (1.3) and satisfies the following assumptions: (i) A(0) = I and A(Q)Q.N(Q) 0 for almost every Q 1, where N(Q) denotes the exterior unit normal vector to Ω. (ii) There is a nondecreasing function f : (0, + ) (0, + ) such that for all X in R d A(X) f( X )/ X and A(X) A(0) f( X ), and 2 0 f(r) r dr < Then, there exist r 0 depending on λ, d, K and the Lipschitz character of Ω, and a constant C depending on λ, d, K, the Lipschitz character of Ω, the value of the above integral and u, such that the following holds u 2 dx C Γ 2r u 2 dx Γ r for all 0 < r < r 0. We will need the following well known inequalities. Lemma 2. Let Ω be a Lipschitz domain with 0 Ω. Then, there exists a constant C depending on d and the Lipschitz character of Ω such that the following inequalities hold for all functions w W 1,2 (Γ r ) w 2 dx C [r 2 w 2 dx + r Γ r Γ [ r w w dx C r w 2 dx + Γ r Γ r ] w 2 dσ ] w 2 dσ Proof. We may assume that Ω = { X R d : x d > ϕ(x) } where ϕ is a Lipschitz function on R d 1 with ϕ(0) = 0 and consider the vector field β(x) = (x 1,..., x d 1, x d ϕ(x) + x ϕ(x)). Then, div(β) = d and β.n(q) = 0 on Ω and the inequalities follow from Hölder s inequality and the identity Γ r div(β)w 2 dx = 1 r β(q).q w 2 dσ 2 wβ. w dx. Γ r Proof of Theorem 8. In this proof we will consider the function µ and vector field β defined as µ(x) = A(X)X.X/ X 2 and β(x) = A(X)X/µ(X)

8 8 VILHELM ADOLFSSON and from the hypothesis in theorem 8 we have with X = r (1.4) (1.5) λ µ(x) λ 1 β(x) = O(r) µ(x) = O(f(r)/r) D i β j = δ ij + O(f(r)), and where the constants depend on λ and d. For u as in theorem 8 and 0 < r < 1 we define the following functions H(r) = 1 r d 1 µu 2 dσ, A(r) = 1 r d 2 A u. u dx, Γ r (1.6) D(r) = 1 r d 2 Differentiating H(r) we have from (1.4) Γ r A u. u + udiv(a u) dx, and N(r) = D(r) H(r). H (r) = 2 r d 1 µu u dσ + O(f(r)/r)H(r). r We recall that the conormal derivative associated to the operator L on B r is given by u ν = A u. Q Q. Now, since α = A(X)X/ X µ(x)x/ X is a tangential vector field on B r with α(x) = O(f(r)/r) and u = 0 on 1, we obtain from the divergence theorem and (1.4) the following identity Thus, r d 2 D(r) = = = u u ν dσ = µu u dσ 1/2 r µu u r dσ + u 2 divα dσ µu u r dσ + O(f(r)/r)rd 1 H(r). u u.α dσ (1.7) (1.8) H (r) = 2 D(r) + O(f(r)/r)H(r) r r d (log H(r)) = 2N(r) + O(f(r)). dr Proceeding in the standard way, we will show that there exists r 0 > 0 depending on λ, d, K, and the Lipschitz character of Ω and C > 0 depending on the same parameters and on f(1) such that the frequency function Z(r) = N(r) + 1 verifies (1.9) Z (r) C(1 + f(r)/r)z(r) for all 0 < r < r 0, and this is well known to imply theorem 8. Integrating over Γ r the Rellich-Necas identity div(βa u. u) = 2 div(β. ua u) + div(β)a u. u 2D i β k a ij D j ud k u 2β. u div(a u) + β k D k a ij D i ud j u.

9 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 9 we obtain from (1.5) and the identities β. Q Q = r, β. ua u. Q Q = r µ β.na u. u = β. ua u.n = AQ.NAN.N µ the following formula (1.10) A u. u dσ = 2 +( d 2 r Then, setting B(r) = 2 r d 2 I 1 (r) = 1 r d 2 ( ) 2 u on B r ν ( ) 2 u on r N ( ) 2 1 u dσ + 1 µ ν r r + O(f(r)/r)) A u. u dx 2 β. u div(a u) dx. Γ r r Γ r ( ) 2 1 u dσ + 1 µ ν r d 1 r u div(a u) dx, I 2 (r) = 2 Γ r I 3 (r) = 1 r d 3 ( ) 2 AQ.NAN.N u dσ µ N AQ.NAN.N µ r d 2 u div(a u) dσ ( ) 2 u dσ, N Γ r β. u div(a u) dx, and differentiating D(r) = A(r) + I 1 (r) we get from (1.10) the identity (1.11) D (r) = B(r) 1 r [(d 2)I 1(r) + I 2 (r) + I 3 (r) + O(f(r))A(r)]. From lemma 2, (1.4), (1.5) and the equation satisfied by u we have (1.12) I 1 (r) + I 2 (r) C 1 r[a(r) + H(r)] for all 0 < r < 1, where C 1 depends on λ, d, K and the Lipschitz character of Ω. Also, from the identity (1.10), lemma 2 and Hölder s inequality we obtain 1 (1.13) r d 3 u 2 dσ C 2 [rb(r) + A(r) + H(r)] for all 0 < r < 1, where C 2 depends on λ, d, K, f(1) and the Lipschitz character of Ω. Finally, (1.13) and Hölder s inequality imply (1.14) I 3 (r) C 3 r[h(r) + A(r) + rb(r)h(r)] for all 0 < r < 1. It follows from (1.12) and the identity D(r) = A(r) + I 1 (r), that we can find r 0 > 0 depending on C 1 such that for all 0 < r < r 0 the following holds (1.15) A(r) 2D(r) + H(r) D(r) 2[A(r) + H(r)].

10 10 VILHELM ADOLFSSON and from (1.11), (1.12) and (1.14) we find that with constants depending on λ, d, K,f(1) and the Lipschitz character of Ω (1.16) B(r) 2D (r) + O( 1 )(H(r) + A(r)) for all 0 < r < 1. r From (1.15), the function Z(r) = N(r) + 1 is nonnegative for 0 < r < r 0. Therefore, if Z (r) 0, (1.9) holds and there is nothing to prove. Otherwise, we have D (r)h(r) H (r)d(r), and multiplying (1.16) by rh(r) we get that for 0 < r < 1 rb(r)h(r) 2rD (r)h(r) + C[H(r) + A(r)] 2 2rH (r)d(r) + C(H(r) + A(r)) 2. Replacing the identity (1.7) on the right hand side of the last inequality we get from the second inequality in (1.15) rb(r)h(r) C[A(r) + H(r)] for all 0 < r < r0, where C depends on λ, d, K, f(1) and the Lipschitz character of Ω.. This inequality and (1.14) imply I 3 (r) Cr[A(r) + H(r)], and from the last inequality, (1.12) and (1.11) we obtain D (r) B(r) C(1 + f(r)/r)[a(r) + H(r)]. Then, the above inequality, (1.7) and (1.15) imply (1.17) Z (r) rb(r)h(r) 2D(r)2 rh(r) 2 C(1 + f(r)/r)z(r) for all 0 < r < r 0. On the other hand, by definition we have r 2d 3 [rb(r)h(r) 2D(r) 2 ] = { +2r 1 µ ( u ν ) 2 dσ ( ) 2 AQ.NAN.N u dσ µu 2 dσ r µ N ( µu 2 dσ u u ) 2 } ν dσ and from Hölder s inequality and the assumption that A(Q)Q.N(Q) 0 on 1 it follows that the first term on the right hand side of (1.17) is nonnegative, which proves the theorem. Remark. The above proof when the boundary of Ω is empty is somehow a new proof of the main result in [6], where they proof the doubling property for solutions to divergence form operators with Lipschitz coefficients, and which avoids introducing normal coordinates to obtain the identity (1.10), (in this case the term in (1.10) which involves integration over r does nor appear) and replacing them by a suitable choice of the vector field β in the Rellich-Necas identity.

11 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 11 Theorem 9. Let Ω be a Lipschitz domain in R d with 0 Ω and u be a non zero solution to Lu = 0 in Γ 1 whose conormal derivative vanishes almost everywhere on 1, and where the operator L is as in (1.3) and satisfies the following assumptions: (i) A(0) = I and A(Q)Q.N(Q) = 0 for almost every Q 1. (ii) There is a nondecreasing function f : (0, + ) (0, + ) such that for all X in R d A(X) f( X )/ X and A(X) A(0) f( X ), and 2 0 f(r) r dr < Then, there exist r 0 depending on λ, d, K and the Lipschitz character of Ω and a constant C depending on λ, d, K, the value of the above integral, the Lipschitz character of Ω and u, such that the following holds u 2 dx C Γ 2r u 2 dx Γ r for all 0 < r < r 0. Moreover, if the coefficient c of the zero order term of the operator L is identically zero, both u and u l satisfy the doubling property with a constant C independent of l R. Proof of Theorem 9. In this proof we let µ and β be the function and vector field defined in the proof of theorem 8. For u and L as in theorem 8 we define the functions H(r), A(r), D(r) and N(r) as in theorem 8. Now, since A(Q)Q.N(Q) = 0 on 1 we obtain from the divergence theorem the following identity ( H(r) = r Γ 1 d div A(X)X X u 2) dx r and H (r) = d 1 H(r) + 2r 1 d u u dσ + r1 d r ν div ( ) A(X)X X u 2 dσ. Since Lu = 0 in Γ 1 it follows from the above identity, the divergence theorem and the assumptions on the coefficients matrix A that (1.18) (1.19) H (r) = 2 D(r) + O(f(r)/r)H(r) r r d (log H(r)) = 2N(r) + O(f(r)). dr As in the proof of theorem 8 we integrate over Γ r the Rellich-Necas identity div(βa u. u) = 2 div(β. ua u) + div(β)a u. u 2D i β k a ij D j ud k u 2β. u div(a u) + β k D k a ij D i ud j u

12 12 VILHELM ADOLFSSON and from (1.5) and the identities β. Q Q = r, β. ua u. Q Q = r µ ( ) 2 u on B r ν β.na u. u = AQ.N µ A u. u = 0 and β. ua u.n = 0 on r we obtain the following formula (1.20) ( ) 2 1 u A u. u dσ = 2 dσ µ ν +( d 2 + O(f(r)/r)) A u. u dx 2 β. u div(a u) dx. r Γ r r Γ r Then, setting B(r) = 2 ( ) 2 1 u r d 2 dσ I 1 (r) = 1 µ ν r d 2 u div(a u) dx Γ r I 2 (r) = 2 r d 2 β. u div(a u) dx. I 3 (r) = 1 Γ r r d 3 u div(a u) dσ and differentiating D(r) = A(r) + I 1 (r) we get from (1.20) the identity (1.21) D (r) = B(r) 1 r [(d 2)I 1(r) + I 2 (r) + I 3 (r) + O(f(r))A(r)]. At this point, from (1.18) and (1.21) the argument proceeds as in the proof of theorem 8, obtaining that the function Z(r) = N(r)+1 is nonnegative for 0 < r < r 0 and it either satisfies Z (r) 0 or (1.22) Z (r) On the other hand, by definition { 2r 1 µ rb(r)h(r) 2D(r)2 rh(r) 2 ( u ν C(1 + f(r)/r)z(r). r 2d 3 [rb(r)h(r) 2D(r) 2 ] = ) 2 ( dσ µu 2 dσ u u ν dσ ) 2 } and from Hölder s inequality it follows that the first term on the right hand side of (1.22) is nonnegative. Thus, in any case we obtain Z (r) C(1 + f(r)/r)z(r) for all 0 < r < r 0, which proves the first claim in theorem 9. When the coefficient c of the zero order term in the operator L is identically zero, both u and u l are solutions to Lw = 0 whose conormal derivative vanishes on 1 for all l R, and if H l (r), D l (r) and N l (r) denote the corresponding functions defined in (1.6) and associated to u l, we observe that to prove the second claim in theorem 9 suffices to show that sup l R N l (r 0 ) is finite. This clearly holds when u is constant, otherwise, from (1.15) we have N l (r 0 ) A(r 0) H l (r 0 ) for all l R, and since the operator L has the interior unique continuation property [3], it follows that inf l R H l (r 0 ) > 0, which proves the theorem.

13 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY Zero Dirichlet boundary condition Proof of theorem 2. Let D and u be as in theorem 2. We fix a point Q 3 (Q 0 ). Then, after a translation we may assume Q = 0, and that u is harmonic on Γ 1, where Γ 1 is the set of points Y in the unit ball of R d, Y = (y, y d ), such that y d > ϕ(y), where ϕ is a Lipschitz function in R d 1 verifying ϕ(0) = 0, ϕ(y) ϕ(0) θ( y ) for all y R d 1, and where θ is a Dini function. From the mean value theorem we get and in particular y ϕ(y) ϕ(y) 2 y θ( y ) for all y R d 1, (2.1) y ϕ(y) ϕ(y) 2 y θ( y ) for all y R d 1. We define θ(r) = 1 2r 1 2t θ(s) log 2 2 r t t s dsdt and consider the change of variables Y = Ψ(X) = (x, x d + 3 X θ( X )). calculus we have y i = δ ij for all i = 1,..., d 1 and j = 1,..., d x j (2.2) y n = δ jd + x jα( X ) for all j = 1,..., d, x j X From and where α(r) = 3 d dr (r θ(r)). If JΨ denotes the Jacobian of the transformation Ψ we have det JΨ(X) = y d = 1 + x dα( X ) 1 α( X ), x d X and since α(r) 3θ(4r) and θ(r) tends to zero as r tends to zero, the mapping Ψ defines a C 1 -diffeomorphism from a neighborhood of zero in the X-space onto a neighborhood of zero in the Y -space. In particular there exist r 1 and r 0 depending on θ such that Ψ : B r1 B r0 defines a C 1 -diffeomorphism, and 1/2 det JΨ(X) 2 for all X B r1. Defining w(x) = u Ψ(X) and Ω = { X R d : x d > ϕ(x) 3 X θ( X ) }, we have that w is a W 1,2 -solution to Lw = div(a w) = 0 on Γ r1 = Ω B r1 which vanishes continuously on the associated set r1. Moreover, the coefficients matrix A(X) is given by (2.3) A(X) = det (JΨ(X))JΨ t (X)JΨ 1 (X). From (2.2) and simple algebra we obtain writing A(X) = (a ij (X)) a ij (X) = δ ij [1 + x dα( X ) ] X for i, j = 1,..., d 1. a jd (X) = a dj (X) = x jα( X ) X for j = 1,..., d 1. a dd (X) = 1 + x 2 α( X ) 2 X x dα( X ) X

14 14 VILHELM ADOLFSSON We will show that at scale r 1, the function w, the domain Ω and the operator L, satisfy the conditions in theorem 8, which implies that w satisfies the doubling property. Then, undoing the change of variables defined by Ψ, we will obtain that our original harmonic function u satisfies the corresponding doubling property. From (2.3) we have that the matrix A verifies the following properties: First, A is symmetric, A(0) = I, and taking r 1 sufficiently small we have that A satisfies the ellipticity condition 1/2 ξ 2 a ij (X)ξ i ξ j 2 ξ 2 for all X B r1 and all ξ R d. i,j=1 and for some constant C depending on d A(X) Cθ(4 X )/ X, A(X) A(0) Cθ(4 X ) for all X R d Moreover, we claim that A(Q)Q.N(Q) 0 on the bottom boundary of Ω. To see this, we observe that from our assumptions on θ and the implicit function theorem, Ω near 0 Ω is the region above a C 1 -graph defined implicitly by a function φ satisfying the equation (2.4) { φ(x) = ϕ(x) 3 Q θ( Q ) φ(0) = 0 where Q = (x, φ(x)). Differentiating (2.4) implicitly we have ϕ(x) α( Q ) x Q (2.5) φ(x) = 1 + φ(x)α( Q ) Q for x < r 1. Now, the exterior unit normal to Ω is given almost everywhere by N(Q) = ( φ(x), 1) 1 + φ(x) 2, and (2.3), (2.5) and elementary algebra show that for Q Ω we have x ϕ(x) φ(x) (2.6) A(Q)Q.N(Q) =. det JΨ(Q) 1 + φ(x) 2 But from (2.4), (2.1) and observing that θ θ we obtain (2.7) x ϕ(x) φ(x) = x ϕ(x) ϕ(x)+3 Q θ( Q ) Q θ( Q ) 0 for x < r 1. Therefore, A(Q)Q.N(Q) 0 for all Q r1, which proves the claim and theorem 2. Remarks.

15 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY The reader will realize that we can relax the condition on the domain D to obtain the doubling property. In fact, we could just assume that D is given locally as the graph of Lipschitz functions ϕ satisfying the following lower side condition: for almost every x R d 1 and all y R d 1 (2.8) (x y) ϕ(x) (ϕ(x) ϕ(y)) 2 x y θ( x y ), and where θ is a Dini function. If we set P = (x, ϕ(x)) and Q = (y, ϕ(y)), we can rewrite the above condition as (2.9) (P Q).N(P ) P Q θ( P Q ) for a.e. P D and all Q D, where N(P) denotes the exterior unit normal vector to D at P D. We observe that condition (2.9) is invariant under linear transformations. To see this, we let A denote an invertible linear transformation in R d, and consider the domain A 1 (D) = { X R d : AX D }. Clearly, A 1 (D) is a Lipschitz domain and any two boundary points P and Q can be respectively written as A 1 P and A 1 Q, for some P and Q in D. Moreover, N(P ) = At N(P ) A t N(P ) and from this we have (P Q ).N(P N(P ) ) = (P Q). A t N(P ) P Q θ( P Q ), where θ(r) = Cθ(Cr) for some C depending on the linear transformation A. In this sense, we can generalize the result for convex domains in [2] to M-convex domains, where by an M-convex domain we understand a Lipschitz domain D given locally as the graph of functions ϕ verifying that ϕ + M x 2 is a convex function, since in this case, (2.8) holds with θ(r) = (M/2)r. 2. Theorem 2 also holds under the same hypothesis when u is a solution to a divergence form operator L with Lipschitz coefficients and bounded coefficients in the lower order terms as in (1.3), but in this case, the constants r 0 and C also depend on K, the bound on the L norm of the gradient of the coefficients matrix and on λ. To see this, we could assume in the previous proof that instead of being u harmonic in D, u is a solution to Lu = 0 on D, where L is as in (1.3). After a linear change of coordinates we could assume that A(0) = I, then the function w = u Ψ would be a solution in Ω to Lw = div(ã w) + b w + c w = 0 with w = 0 on r1, and where the coefficient matrix Ã(X) = (ã ij(x)) and the coefficients b = ( b 1,..., b d ) and c are given by Ã(X) = det JΨ(X)JΨ t (X)A Ψ(X)JΨ 1 (X) b(x) = det JΨ(X)JΨ 1 (X)b Ψ(X) c(x) = det JΨ(X)c Ψ(X).

16 16 VILHELM ADOLFSSON Then, it is clear that the operator L satisfies the condition (ii) in theorem 8 with f(r) = C(r + θ(4r)), and where C depends on d, λ and the L norm of the gradient of the matrix A. Moreover, from (2.6) and (2.7), we see that for Q r1 and taking r 1 smaller if necessary, we have Ã(Q)Q.N(Q) 1 2 Q θ( Q ) + det JΨ(Q)JΨ(Q) t (A Ψ(Q) I)JΨ(X) 1 Q.N(Q) = 1 2 Q θ( Q ) + O( Q 2 ), which would remain larger than 0. Thus, w, the operator L and Ω verify the assumptions in theorem 8, implying the doubling property for w and as a consequence for u. 3. From the previous remarks, the results in [2] for convex domains can be extended to solutions to divergence form operators with Lipschitz coefficients and bounded coefficients in the lower order terms, showing that on and M-convex domain D, a nonconstant solution u to one of these operators which vanishes continuously on an open set V contained in D satisfies the doubling property with respect to balls centered at points Q V, and obtaining that it can not vanish to infinite order at those points. 4. To prove theorem 3 we will need a more detailed definition of the frequency function that we have cooked up to prove the doubling property. In fact, if u and D are as in the last proof and Q = (y, ϕ(y)) 3 (Q 0 ), undoing the change of variables Ψ and defining A(Q, r) = r 2 d u 2 dx (2.10) H(Q, r) = r 1 d N(Q, r) = S r(q) D B r(q+3r θ(r)e d ) D A(Q, r) H(Q, r), where e d = (0,..., 0, 1) and S r = 8 and standard arguments 0<s<r (2.11) e β(r) N(Q, r), where β(r) = C ru 2 r + 3(x d ϕ(y) 3r θ(r))(r θ(r)) dσ B s (Q + 3s θ(s)e d ), we have from theorem r 0 θ(4t) t is a nondecreasing function for 0 < r < r 0, and where r 0 is as in theorem 2 and C depends on d, and the value of the integral from 0 to 1 of θ(r)/r. Proof of Theorem 4. Let D, u, V and Q V be as in theorem 4. Without loss of generality we may assume that V = 6 (Q 0 ), Q = 0 and that the boundary of D below Γ 1 is given as the graph of a C 1 function ϕ in R d 1 verifying ϕ(0) = 0. From the continuity of ϕ we find that for all ε > 0 we can find r = r(ε) > 0 such that dt ϕ(y) ϕ(0) ε whenever y r(ε), and in particular (2.12) y ϕ(y) ϕ(y) 2ε y for y r(ε).

17 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 17 Proceeding as in the proof of theorem 2, we consider the change of variables Ψ(X) = (x, x d + 3ε X ) where now θ(r) = ε for all r. The same calculations as before and with α(r) = 3ε for all r give, det JΨ(X) = 1 + x dα( X ) X = 1 + 3εx d X. When 3ε < 1 the transformation Ψ defines a bilipschitz mapping from R d onto itself whose inverse can be explicitly found. Defining as before w = u Ψ and Ω = { X R d : x d > ϕ(x) 3ε X }, we have that w is an W 1,2 -solution to Lw = div(a w) = 0 on Γ r1 = Ω B r1, which vanishes continuously on the corresponding set r1. Moreover, the coefficients matrix A(X) is as in the proof of theorem 2 given by the formula (2.3) and with α(r) = 3ε for all r. It is then easy to check that defining A(0) = I, we have that w and the operator L satisfy the condition (ii) in theorem 8 with f(r) = Cε. Moreover, since y d x d > 0 when X 0 we have that Ω is the region above a Lipschitz graph φ defined implicitly by the equation { φ(x) = ϕ(x) 3ε Q φ(0) = 0 where Q = (x, φ(x)). Proceeding from here as in the proof of theorem 2 we find that for Q Ω we have x ϕ(x) ϕ(x) + 3ε Q A(Q)Q.N(Q) =, det JΨ(Q) 1 + φ(x) 2 and from this identity and (2.12) we obtain that A(Q)Q.N(Q) 0 on r(ε). At this point, the argument in the proof of theorem 8 implies that the function H(r) and the frequency function N(r) associated to w, L and Ω satisfy the differential inequalities N (r) Cε r N(r) r d for log H(r) 2N(r) + Cε dr all r r(ε), where C depends only on d. Integrating these inequalities we obtain ( ) ( Cε r H(r) H(r(ε)) exp 2 ( ) Cε r(ε) N(r(ε))) r(ε) Cε r But assuming that (0.3) in theorem 4 holds for u, implies that ( H(r) Cexp 1 ) r β 2 for all r < r(ε), for all r < r(ε).

18 18 VILHELM ADOLFSSON and comparing the last two inequalities, we see that taking ε > 0 such that Cε < β 2 implies that H(r(ε)) must be equal to zero. Thus, u must be identically zero on an open neighborhood of zero, which proves the theorem. Remark. With the obvious changes in the above proof, theorem 4 holds when we replace the Laplace operator for an elliptic selfadjoint operator with Lipschitz coefficients and bounded coefficients in the lower order terms. Proof of Theorem 3. The first claim in theorem 3 follows from the doubling property and theorem 1. To prove the second claim we fix m > 0 and a Dini function θ. We let F denote the set of all mappings ϕ : R d 1 R verifying the following conditions (1) ϕ L (R d 1 ) m and ϕ(0) = 0. (2) ϕ(x) ϕ(y) θ( x y ) for all x, y R d 1. For each ϕ F we define D(ϕ) = { X R d : x < 3 and ϕ(x) < x d < ϕ(x) + 3m + 3 }. Now, if ϕ F and u is a harmonic function on D(ϕ) vanishing continuously on the bottom boundary of D(ϕ), the change of coordinates (2.13) { yi = x i y d = x d + ϕ(x) transforms u into a function ũ defined on the set Ω = { X R d : x < 3, 0 < x d < 3m + 3 } and such that if A ϕ (X) = (a ij (X)) is the matrix given by (2.14) a ij (X) = δ ij for i, j = 1,..., d 1 a jd (X) = a dj (X) = D j ϕ(x) for j = 1,..., d 1 a dd (X) = 1 + ϕ(x) 2 and L ϕ denotes the divergence form operator L ϕ = div(a ϕ.), then ũ is a solution to L ϕ w = 0 on Ω vanishing continuously on the bottom boundary of Ω. Next, we let L denote the set formed by all the subsets E of the open unit ball B 1 in R d 1 verifying that there exists a function ϕ F and a nonconstant harmonic function u defined on D(ϕ) vanishing continuously on the bottom boundary of D(ϕ), such that E = { x B 1 : ũ(x, 0) = 0 } = { x B 1 : u(x, ϕ(x)) = 0 }. From [14], it follows that if ϕ F and u is harmonic function on D ϕ vanishing continuously on the bottom boundary of D ϕ, there is a constant C depending on d, m and the value of the integral from 0 to 1 of θ(r)/r, such that (2.15) ( ) 1/2 ũ L (B + 2 (0)) C ũ 2 dx X X ( ) 1/2 ũ(x) ũ(x θ(r) ) C dr ũ 2 dx for all X, Y B r (0). Ω From this, it is clear that for E L, E is a relatively closed subset of B 1. We say E i E for a sequence of {E i } L and E L, if for each ε > 0 there is a i(ε) such that B 1 ε E i { x B 1 : d(e, x) < ε } for all i i(ε). Ω

19 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 19 Now we want to verify the following properties of L. (P 1 ) (Closure under appropriate scaling and translation). If y 1 λ, 0 < λ < 1, and if E L, then E y,λ B 1 L. Here, E y,λ = λ 1 (E y). This is because the fact that if u is a nonconstant harmonic function on a domain D(ϕ) vanishing on the bottom boundary of D(ϕ), setting Y = (y, ϕ(y)), u y,λ (X) = u(λx + Y ) and ϕ y,λ (x) = (ϕ(λx + y) ϕ(y)) /λ, then ϕ y,λ F and u y,λ is a nonconstant harmonic function on D(ϕ y,λ ) vanishing on the bottom boundary of D(ϕ y,λ ) for y 1 λ and 0 < λ < 1. Then, it is easy to see that E y,λ B 1 = { x B 1 : u y,λ (x, ϕ y,λ (x)) = 0 }. Thus, E y,λ B 1 L. (P 2 ) (Existence of homogeneous degree zero tangent set). If y < 1, if {λ k }, 0 < λ k < 1, decreases to zero, and if E L, then there is a subsequence {λ k } and F L such that E y,λk B 1 F and F 0,λ B 1 = F for each 0 < λ < 1. To show (P 2 ), we let ϕ F and u be a nonconstant harmonic function on D(ϕ) vanishing continuously on the bottom boundary of D(ϕ), and such that E = { x B 1 : u(x, ϕ(x)) = 0 }. For y < 1, and a decreasing sequence {λ k } converging to zero, we set as before Y = (y, ϕ(y)), u y,λk (X) = u(λ k X + Y ), ϕ y,λk (x) = (ϕ(λ k x + y) ϕ(y)) /λ. Without loss of generality, we may assume u(y ) = 0, for otherwise, by the continuity of u (or ũ) up to the bottom boundary of D(ϕ), it is clear that E y,λk B 1 = for all k sufficiently large. Then, we simply take F =, which is associated to the harmonic function u(x) = x d and the domain D(0) = { X R d : x d > 0 }, which shows that (P 2 ) holds in this case. Since u is nonconstant, we have from (2.11) (2.16) e β(r) N(Y, r) e β(r0) N(Y, r 0 ) < + for all 0 < r < r 0, where N(Y, r) is the frequency function defined in (2.10) and associated to u, D(ϕ), θ and Q = Y. Now, for each λ k (r 0 (1 y ))/2, we define v k = ( u y,λk B 1(0) D(ϕ y,λk ) u 2 y,λ k dσ ) 1/2. We observe that if N k (r) denotes the frequency function defined in (2.10) and associated to v k, D(ϕ y,λk ), Q = 0 and θ k (r) = θ(λ k r) we have (2.17) β k (r) = β(λ k r), N k (r) = N(Y, λ k r) for all 0 < r < 2. This implies that the squares of the harmonic functions v k satisfy on the corresponding domains of definition the doubling property with respect to balls centered at zero and with a constant independent of k. Through the change of coordinates (2.13) associated to ϕ y,λk and the uniform doubling property, we obtain that there is a constant C depending on d, m, the value of the integral from 0 to 3 of θ(r)/r and u, such that for k large enough C 1 ṽk 2 dx C Ω

20 20 VILHELM ADOLFSSON Since L ϕy,λk ṽ k = 0 on Ω, A ϕy,λk (X) converges to A ψ (X), where ψ(x) = ϕ(y).x, the above inequalities and (2.15), we can find a subsequence of k s such that ṽ k converges in C 1 (Ω) to a function ṽ. Then L ψ ṽ = 0 on Ω, ṽ is nonconstant and vanishes continuously on the bottom boundary of Ω. Clearly, the pull back of ṽ under the change of variables (2.13) corresponding to ψ is a harmonic function on D(ψ) and setting F = { x B 1 : v(x, ψ(x)) = 0 } L we have, E y,λk B 1 F. Moreover, one observes from the formulae (2.10) that defining we have D 0 (r) = r 2 d B r D(ψ) v 2 dx, N 0 (r) = D 0(r) H 0 (r) lim N k (r) = N 0(r) for all 0 < r < 1 and k H 0 (r) = r 1 d B r D(ψ) v 2 dx K = lim r 0 e β(r) N(Y, r) e β(λ kr) N(Y, λ k r) = e β k(r) N k (r) e β(λ k) N(Y, λ k ) for all 0 < r < 1. Hence the frequency function N 0 (r) for v satisfies, N 0 (r) = K for all 0 < r < 1. This implies that v(λx) = λ K v(x) for all 0 < λ < 1. Hence, v is a homogeneous harmonic function of degree K and F 0,λ B 1 = F for 0 < λ < 1, proving (P 2 ). Now, by [12, Appendix A] we have the following Theorem. Let L be a collection of relatively closed proper subsets of the unit open ball in R d 1 which satisfies (P 1 ) and (P 2 ). Then, the Hausdorff dimension of each of the elements in L is less or equal than d 2. This finishes the proof of theorem 3. Remarks. 1. Theorem 3 also holds when we replace the Laplace operator for an operator L as in (1.3) whose coefficient matrix is Lipschitz. In this case, the second claim in theorem 3 follows from the doubling property and the obvious changes in the last proof. The first claim follows from the doubling property, the arguments used in [2] to prove theorem 1 and the following analog of the inequality (1.2). Lemma 3. Let D be a Lipschitz domain in R d, L an operator as in (1.3) whose coefficient matrix A is Lipschitz and satisfies A L (R d ) m for some m > 0. Then, for each ε > 0 there exists a constant C(ε) depending on the Lipschitz character of D, m, λ, K and ε, such that if Q D, 0 < r < 1 and u is a solution to Lu = 0 on Γ 2r (Q) vanishing continuously on 2r (Q) the following holds u dx C(ε)r 2 u N dσ + ε u dx. Γ 2r(Q) Γ r(q) 2r(Q) Proof. We let let F denote the set of all Lipschitz maps ϕ : R d 1 R satisfying ϕ(0) = 0 and ϕ L (R d 1 ) m for some m > 0, and L denote the set of

21 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 21 all the operators L as in (1.3) whose coefficient matrix A satisfies A(0) = I and A L (R d ) m. For each ϕ F we define D(ϕ) = { X Rd : x d > ϕ(x) }. To prove this lemma suffices to show the following: For each ε > 0 there exists a constant C(ε) depending on λ, ε, K and m, such that if ϕ F, L L and u is a solution to Lu = 0 on B 2 D(ϕ) vanishing continuously on B 2 D(ϕ) the following holds B 1 D(ϕ) u dx C(ε) B 2 D(ϕ) To prove it we proceed by contradiction. sequences {ϕ n } in F, {L n } in L with u N dσ + ε L n w = div(a n w) + B n. w + C n w, B 2 D(ϕ) u dx. Otherwise, there exist ε > 0 and and a sequence {u n } of functions verifying L n u n = 0 on B 2 D(ϕ n ),u n = 0 in B 2 D(ϕ n ), (2.18) u n dx = 1 and (2.19) ε B 2 D(ϕ n) B 1 D(ϕ n) u n dx + n B 2 D(ϕ n) u n dσ 1. N For each n we let v n denote the zero extension of u n to the whole ball B 2 outside of the domain D(ϕ n ). Since all the above sequences are compact in the proper topologies, we can find subsequences which we may assume to be the whole sequences, such that ϕ n ϕ L and A n A uniformly over compact sets, B n B and C n C weakly in L 2 (B 3/2 ), where the operator Lw = div(a w) + B. w + Cw lies in L, and v n v uniformly on B 3/2, weakly in W 1,2 (B 3/2 ), and v n converges uniformly to v over compact sets contained in B 3/2 \ D(ϕ), [7]. Moreover, the divergence theorem and (2.19) imply that there is a constant C depending on λ, such that for all ψ C0 (B 3/2 ) and n 1 we have (2.20) A n v n ψ B n. v n ψ C n v n ψ dx C n ψ L (B 3/2 ). Passing to the limit in (2.18) and (2.20) we find a function v W 1,2 (B 3/2 ) satisfying Lv = 0 on B 3/2, (2.21) v dx = 1, B 1 and v is identically zero on B 1 \ D(ϕ). The Lipschitz character of ϕ implies that we can find a nonempty open set U contained in B 1 \ D(ϕ) where v is identically zero, but an operator L L has the interior unique continuation property [3], thus v = 0 on B 3/2.This is in contradiction with (2.21) and proves the lemma. 2. The previous lemma can also be used to show that lemma 1 holds when replace the Laplace operator for an operator L as in (1.3) whose coefficients matrix is Lipschitz.

22 22 VILHELM ADOLFSSON 3. Zero Neumann Boundary Condition Proof of Theorem 5. To prove theorem 5 we need the following analog of the inequality (1.2) and lemma 3 Lemma 4. Let D be a Lipschitz domain in R d. Then, for each ε > 0 there exists a constant C(ε) depending on ε, d and the Lipschitz character of D, such that if Q D and u is a harmonic function on Γ 2r (Q) whose normal derivative vanishes almost everywhere on 2r (Q) and u dσ = 0 the following holds Γ r(q) u dx C(ε)r 2 r(q) 2r(Q) u dσ + ε u dx. Γ 2r(Q) Proof. As in lemma 3 we let F denote the set of Lipschitz mappings ϕ on R d 1 verifying ϕ(0) = 0 and ϕ L (R d 1 ) m for some m > 0. Then, if lemma 4 were false we would find ε > 0 and sequences {ϕ n } in F and {u n } verifying that u n is harmonic on B 2 D(ϕ n ), un N = 0 on B 2 D(ϕ n ), (3.1) u n dσ = 0, u n dx = 1 and (3.2) ε B 1 D(ϕ n) B 2 D(ϕ n) B 1 D(ϕ n) u n dx + n u dσ 1. B 2 D(ϕ n) For each n we let v n and f n = (fn, 1..., fn) d denote the zero extensions to the whole ball B 2 outside of the domain D(ϕ n ) of u n and u n respectively. From (3.2) and standard interior estimates [7], there is a constant C depending on m, d and ε such that v n L2 (B 3/2 ) + f n L2 (B 3/2 ) C for all n 1. Since the above sequences are compact in the proper topologies, we can find subsequences which we may assume to be the whole sequences, such that ϕ n ϕ F uniformly over compact sets, v n v, f n f = (f 1,..., f d ) weakly in L 2 (B 3/2 ) and uniformly over compact sets contained in B 3/2 \ D(ϕ). From (3.2), the divergence theorem and Poincare inequality on 2 we find that there is a constant C depending on m and d such that for all ψ C0 (B 3/2 ) and 1 j d (3.3) v n D j ψ + fnψ j dx + f n. ψ dx C n ψ L (R d ). Taking limits in (3.1) and (3.3) we find that the limit v satisfies: v W 1,2 (B 3/2 ), v is harmonic on B 3/2, (3.4) v dx = 1, B 1 D(ϕ)

23 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 23 and v vanishes on B 1 \ D(ϕ). As in lemma 3 these imply that v is identically zero on B 3/2, which contradicts (3.4) and proves the lemma. Now, if D and u are as in theorem 5, r > 0 and Q D we obtain from the doubling property (0.4) and choosing ε > 0 sufficiently small in lemma 4 that there is a constant C depending on M, d and the Lipschitz character of D verifying the following (3.5) u l dx Cr 2 u dσ Γ r/2 (Q) r(q) where l denotes the average of u over r/2 (Q). Now, from the Rellich identity, the doubling property (0.4) and standard estimates for harmonic functions whose normal derivative vanishes almost everywhere on the boundary [7] we obtain (3.6) ( ) 1/2 u 2 dσ Cr d+3 2 r(q) Γ r/2 (Q) u l dx, and (3.5), (3.6) imply the inequality (0.5), proving the theorem. Proof of theorem 6. Let D and u be as in theorem 6. We fix a point Q 3 (Q 0 ). Then after a translation we may assume Q = 0, and that u is harmonic in Γ 1, where Γ 1 is the set of points Y = (y, y d ) in the unit ball of R d such that y d > ϕ(y), where ϕ is a C 1,1 function in R d 1 verifying ϕ(0) = 0. For j = 1,..., d 1 we consider the functions u j (x, x d ) = x j x d Θ xd D j ϕ(x) and define { uj (x, x d ) for x d 0 ψ j (x, x d ) = 4u j (x, x d /2) 3u j (x, x d ) for x d < 0, where Θ is a smooth compactly supported regularization of the identity in R d 1 and denotes convolution. Next, we consider the C 1,1 -function Ψ(X) = (ψ 1 (x, x d ),..., ψ d 1 (x, x d ), x d + ϕ(x)) whose Jacobian at points (x, 0), x R d 1, is given by the matrix JΨ(x, 0) = D 1 ϕ D 2 ϕ D 3 ϕ D d 1 ϕ D 1 ϕ D 2 ϕ D 3 ϕ... D d 1 ϕ 1 and det JΨ(x, 0) = 1 + ϕ(x) 2. Setting JΨ 1 (x, 0) = (a ij (x)) we have a dd (X) = ϕ 2

24 24 VILHELM ADOLFSSON and a ij (x) = δ ij D iϕd j ϕ 1 + ϕ 2 a jd (X) = D jϕ 1 + ϕ 2 a dj (X) = D jϕ 1 + ϕ 2 for i, j = 1,..., d 1. From the inverse function theorem, Ψ defines a C 1 -diffeomorphism from a neighborhood of zero in the X-space onto a neighborhood of zero in the Y -space. In particular, there exist r 1 and r 0 depending on ϕ such that Ψ : B r1 B r0 defines a C 1 -diffeomorphism. Defining w(x) = u Ψ(X), w solves Lw = div(a w) = 0 on B r1 Ω, where A(X) = det JΨ(X)JΨ t (X)JΨ 1 (X). and Ω = { X R d : x d +ϕ(x) ϕ(ψ 1 (x, x d ),..., ψ d 1 (x, x d )) > 0 }. Since Ψ(x, 0) = (x, ϕ(x)) and x d [x d + ϕ(x) ϕ(ψ 1 (x, x d ),..., ψ d 1 (x, x d ))] = 1 + ϕ(x) 2 + O( x d ) we have that for sufficiently small r 1, B r1 Ω = { X B r1 : x d > 0 }. A calculation shows that for x R d 1, A(x, 0) = I, and since we are assuming that u N = 0 on 1 and w = JΨ u Ψ, the conormal derivative of w associated to the operator L on Ω B r1 is given by A(x, 0) w(x, 0).( e d ) = D d w(x, 0) = D j u(x, ϕ(x))d j ϕ(x) D d u(x, ϕ(x)) = 0. These imply that w, L and Ω verify the conditions in theorem 9 with f(r) = Cr, where C depends on d and the C 1,1 -norm of ϕ. Thus, proving theorem 6. Proof of theorem 7. The first claim in theorem 7 follows from the doubling property and theorem 5. The second claim is proved with an argument similar to the one given in the proof of theorem 3. Actually, the same proof we gave there will work just by considering harmonic functions whose normal derivative vanishes at the bottom boundary of the domains D(ϕ) instead of harmonic functions vanishing continuously on the bottom boundary. Remark. Theorems 5, 6, 7 and lemma 4 also hold when we replace the Laplace operator by an operator L as in (1.3) whose coefficient matrix is Lipschitz and where the coefficient c of the zero order term is identically zero, and when we consider solutions u to Lu = 0 whose conormal derivative vanishes on the corresponding boundary portion of the domain. In fact, it is well known that these results are false when the coefficient c of the zero order term is not zero, and that a non zero solution to u + cu = 0 could be constant on a ball. For instance, the function { 1 for X 1 u(x) = ( ) 1 exp 1 X + 1 for X 1 2 satisfies u + cu = 0 with c = u/u L (R d ).

25 C 1,α DOMAINS AND UNIQUE CONTINUATION AT THE BOUNDARY 25 To proof theorem 6 in this general setting, we consider an operator L as in (1.3) with Lu = div(a u) + b. u, A(X) = (a ij (X)) and assume that u is a solution to Lu = 0 on Γ 1 whose conormal derivative vanishes on 1, where Γ 1 is the set of points Y = (y, y d ) in the unit ball of R d such that y d > ϕ(y), ϕ(0) = 0, ϕ C 1,1 (R d 1 ), A(0) = I, and A L (R d ) < +. Then, for each i = 1,..., d we consider the functions and define d 1 β i (x) = a ij (x, ϕ(x))d j ϕ(x) a id (x, ϕ(x)) j=1 u j (x, x d ) = x j x d Θ xd β j (x) for j = 1,..., d 1, u d (x, x d ) = ϕ(x) x d Θ xd β d (x), { uj (x, x d ) for x d 0 ψ j (x, x d ) = 4u j (x, x d /2) 3u j (x, x d ) for x d < 0. for j = 1,..., d, where Θ is a smooth compactly supported regularization of the identity on R d 1. Next, we consider the C 1,1 -function Ψ(X) = (ψ 1 (x, x d ),..., ψ d 1 (x, x d ), ψ d (x, x d )) whose Jacobian at points (x, 0), x R d 1, is given by the matrix D 1 ϕ D 2 ϕ D 3 ϕ JΨ(x, 0) = D d 1 ϕ β 1 β 2 β 3... β d 1 β d and det JΨ(x, 0) = β d (x) + d 1 j=1 β j D j ϕ(x) = 1 + ϕ(x) 2 + O( x ). Setting JΨ 1 (x, 0) = (c ij (x)) and β = (β 1,..., β d 1 ) we have c dd (X) = c ij (x) = δ ij D iϕβ j β. ϕ β d c jd (X) = D jϕ β. ϕ β d β j c dj (X) = β. ϕ β d 1 β. ϕ β d, for i, j = 1,..., d 1. The inverse function theorem implies as before that Ψ is a C 1 -diffeomorphism from a neighborhood of zero in the X-space onto a neighborhood of zero in the Y -space. Defining w(x) = u Ψ(X), w solves Lw = div(ã w) + b. w = 0

26 26 VILHELM ADOLFSSON on B r1 Ω, where Ω = { X R d : x d > 0 } and Ã(X) = det JΨ(X)JΨ t (X)A Ψ(X)JΨ 1 (X) b(x) = det JΨ(X)JΨ 1 (X)b Ψ(X) A calculation shows that the operator L, w and Ω satisfy the conditions in theorem 9 with f(r) = Cr, where C depends on λ, d, the C 1,1 -norm of ϕ and A L (R d ), and these prove this generalization of theorem The parabolic case Here we show that a similar result holds for caloric functions vanishing continuously on the lateral boundary of a cylinder D (0, ), where D is as in theorem 2. In particular we have the following theorem. Theorem 10. Let D be a Dini domain in R d and u(x, t) satisfy u u t = 0 on D (0, ) u(x, 0) = f(x) for X D u(q, t) = 0 for Q D and t > 0 for some f in a suitable class. Assume, that the set E = { (Q, t) D (0, ) : u (Q, t) = 0 } N has positive surface measure on D (0, ). Then, u must be identically zero. Proof. In the first paragraph in [2, section 3] the following was claimed: The reader will observe in the next proof, that in general, the same result can be obtained when D is just a Lipschitz domain and the corresponding unique continuation property holds at points (Q, 0) with Q D, for harmonic functions defined on W D (0, ) and vanishing continuously on W (D (0, )), where W is an open set in R d+1 containing the boundary point (Q, 0). In fact, this is not precise and it should say that the corresponding doubling property should be known to hold at points (Q, τ) W p (D (0, )), where p denotes the parabolic boundary. Hence, from [2, Theorem 4] to prove the above theorem suffices to show that if u(x, y) is harmonic on Ω = D (0, 3) and vanishes continuously on p Ω, there are constants C and r 0 depending on u, d and the Dini character of D, such that for all (Q, τ) p (D (0, 1)) and 0 < r < r 0 the following holds (4.1) u 2 dxdy C u 2 dxdy. B 2r(Q,τ) Ω B r(q,τ) Ω Now, it is an easy exercise to show that when D satisfies (2.8), there exists a constant C depending on the Lipschitz character of D such that for almost every (P, t) p Ω and all (Q, τ) p Ω [(P, t) (Q, τ)].n(p, t) C P Q 2 + (t τ) 2 θ( P Q 2 + (t τ) 2 ),