Jan 10, 2005 - If 1 ¡ p 5 â, it follows from Theorem 1.2 (b) that Lp-CapP(F; ) = 0. On ..... K. Assuming that Lp-CapP(K; U) = 0 we may find a sequence gk â Câ.
Invent. math. 126, 589 – 623 (1996)
On removable singularities of locally solvable dierential operators Jorge Hounie1 , Joaquim Tavares 2 1 Departamento 2 Departamento
de Matematica, UFSCar, 13.565-905, S˜ao Carlos, Brasil de Matematica, UFPE, 50.740-540, Recife, Brasil
Oblatum 2-IX-1995 & 14-V-1996 Dedicated to F: Treves in his 65th birthday
Abstract. We study removable sets of solutions of locally solvable partial differential operators with smooth coecients and characterize them in terms of capacity in the sense of Harvey and Polking. For operators of order one we are able to give a geometric characterization of removable sets in terms of orbits in the sense of Sussmann. 0 Introduction A classical theorem of Riemann states that a bounded holomorphic function, de ned in an open set from which a point has been omitted, can be holomorphically extended to all of ; in other words, a point can be removed, it is a removable singularity for any bounded holomorphic functions. Almost fty ago, Ahlfors [A] considered the problem of determining which sets of the plane could also be removed as well, and this led him to introduce the notion of analytic capacity (further developed by Garabedian [Gara]); the analytic capacity of a compact set K of the complex plane is
(K) = sup{|f0 (∞)|;
f ∈ H ∞ (C\K); kfk∞ 5 1} ;
where H ∞ (C\K) denotes the space of bounded holomorphic function de ned on C\K (or de ned on the complement of K in the Riemann sphere C ∪ {∞}\K, because {∞} is removable). It turns out that a compact set K is removable if and only if (K) = 0. If one views the holomorphic functions as solutions of a homogeneous partial linear dierential equation, it is then natural to consider removable sets with respect to bounded solutions P(x; D)u = 0 (or solutions belonging to a certain given functional space) for some xed linear partial dierential operator P(x; D). ∗ Both
authors were supported in part by CNPq.
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The work of Harvey and Polking [HaP2] (see also [HaP1]) introduced a notion of capacity of a set associated to any dierential operator P(x; D) de ned on an open set and to any Banach space B ⊂ D0 ( ) that becomes a multiple of the analytic capacity of the set if one takes B = L∞ (C) and P(x; D) = @. They used this notion to give necessary and sucient conditions for the removability of sets with respect to a wide class of constant coecient operators, including elliptic and parabolic operators; as for the Banach space B they could handle B = Lp ( ), 1 ¡ p 5 ∞, C( ) ∩ L∞ ( ), and Lip ( ), 0 ¡ 5 1, for any bounded . In spite of its success, this notion of capacity is mainly focused on operators with constant coecients and does not re ect, in general, the local behavior of a removable set. It is not a local property and the capacity of a set K may be aected by perturbations of away from K. In this work we present a localized version of the notion of capacity of Harvey and Polking that we call total capacity, more suitable for the study of removable sets of local solutions of operators with variable coecient. In Sect. 1 we prove, adapting a reasoning of [HaP2], that for linear operators of any order with smooth coecients which are locally solvable in Lp with loss of one derivative, a closed set F is everywhere removable if and only if its total capacity vanishes (see De nition 1.5, De nition 2.2 and Theorem 1.6 for precise statements). In Sect. 2 we consider operators of any order that satisfy the celebrated condition (P) introduced by Nirenberg and Treves ([NT1], [NT2]) in their study of local solvability, here we may apply the Beals–Feerman L2 local solvability result [BF] to characterize the removable sets in spaces of L2 -solutions in terms of capacity. Similarly, for operators of order one, property (P) is enough to guarantee Lp solvability, 1 ¡ p ¡ ∞, ([HP], [P]), and we may characterize removable sets also in this case. This method cannot be applied, in general, to a rst order operator that satis es (P) in the important case p = ∞, because such operator may not be solvable in L∞ (an example is given in Sect. A). Characterizing removable sets in terms of capacity is of reduced value unless one has criteria to identify sets with zero capacity; fortunately they exist for some operators. For instance, although no characterization of subsets having zero analytic capacity is known, a good deal of information on those sets is available: they are totally disconnected and have Hausdor dimension less than or equal to one; if contained in a Lipschitz graph they have onedimensional Hausdor measure zero ([M], [Ch]) but there exist sets of zero analytic capacity and positive one-dimensional Hausdor measure [Garn]. Here we give a geometric characterization of everywhere removable sets for L∞ solutions of operators L = L(x; D) of order one satisfying (P); this is the main result of this work (Theorem 7.2); for p ¡ ∞ the geometric conditions are sucient for removability (Theorem 7.3). The characterization is made in terms of the interaction of the removable set with the orbits of L (in the sense of Sussmann [Su]) which due to property (P) have dimensions 1 or 2. On orbits of dimension one L behaves as a multiple of a real vector
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eld (one-dimensional behavior); on orbits of dimension two L has twob ⊂ but on the complement \, b which dimensional behavior on a subset is a union of curves with dierent endpoints tangent to Re L and Im L, L behaves again as a multiple of a real vector eld. Very roughly speaking, a set F is everywhere removable if and only if F does not disconnect almost every curve on which L has one-dimensional behavior (that includes both orbits of b is made for any orbit of dimension 1 and the curves out of which \ dimension 2) and furthermore, the intersection of F with almost every reduced orbit of dimension two has zero analytic capacity for the canonical holomorphic structure of reduced orbits (see Sect. 7 for precise statements). Thus, our geometric characterization is, of course, established up to the characterization of sets of zero analytic capacity. The geometric results of Sect. 7 are a byproduct of the proof of Theorem 4.2, where the connection between removability and zero capacity is studied for p = ∞, its long proof is carried out in Sect. 4, Sect. 5 and Sect. 6. One main basic tool in this proof is the Baouendi–Treves approximation theorem ([BT1], [BT2]) and its extension to Lp solutions due to Berhanu and Chanillo [BC], which we describe in Sect. B. The paper is organized as follows: Section Section Section Section Section Section Section Section Section
1. 2. 3. 4. 5. 6. 7. A. B.
Capacity and removable singularities Condition (P) and removable singularities Two basic examples Removing singularities of bounded solutions Beginning of the proof of Lemma 4.6 End of the proof of Lemma 4.6 The geometry of removable sets Inexistence of bounded local solutions The approximation formula in Lp
1 Capacity and removable singularities Consider a linear partial dierential operator with C ∞ coecients P(x; D) de ned in a paracompact manifold (most of the time will be just an open subset of RN ). Suppose that is endowed with a xed positive C ∞ density dx (that will be the Lebesgue measure in the case of open subsets of RN ). This allows us to de ne the transpose operator PRt (x; D) by hP(x; D)u; vi = hu; P t (x; D)vi for all u, v ∈ Cc∞ ( ) where hf; gi = fg dx. If f is a distribution, g is a smooth function and one of them has compact support, hf; gi will also denote the usual duality bracket between distributions and test functions. For any 1 5 p 5 ∞ we will consider the spaces Lp ( ). Let us present the de nition of capacity introduced in [HaP2]: De nition 1.1. For each compact set K ⊂ ; its Lp -capacity with respect to P(x; D) in ; denoted by Lp -CapP (K; ); is sup{|hP(x; D)f; 1)i| : f ∈ Lp ( ); kfkp 5 1; supp P(x; D)f ⊂ K}
(1.1)
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The Lp -capacity of any subset X of is defined by Lp -CapP (X; ) = sup {Lp -CapP (K; )} where the supremum is taken over all compact sets K of ; contained by X . If X is compact the latter definition is consistent with the previous one and the supremum is attained for K = X . A part of which is established in [HaP2] is Theorem 1.2. Let K be a compact subset of ; 1 ¡ p 5 ∞; 1=p + 1=q = 1. (a) There is a function f ∈ Lp ( ) with kfkp 5 1 and P(x; D)f = 0 in \K such that hP(x; D)f; 1i = Lp -CapP (K; ) (b) Lp -CapP (K; ) = inf {kP t (x; D)gkq : g ∈ C0∞ ( ) and g ≡ 1 in a neighborhood of K}. De nition 1.3. Let P(x; D) be a differential operator with smooth coefficients defined in . We say that a relatively closed set F ⊂ is Lp -removable with respect to P(x; D) in if any u ∈ Lp ( ) which satisfies P(x; D)u = 0
in \F
also satisfies P(x; D)u = 0
in :
If F is a closed subset of and Lp -CapP (F; ) ¿ 0 it follows that there is a function u ∈ Lp ( ) such that Pu = 0 in \F but Pu-0 in and F is not Lp -removable in with respect to P(x; D). On the other hand, Harvey and Polking established the converse for a wide class of operators with constant coecients, including elliptic and parabolic operators. However, Lp -CapP (F; ) is a global object – in contrast with remotion of singularities that is a local matter – and re ects, in general, not only the local structure of F but also the geometry of . Example 1. Let be the punctured disk of radius 1 in R2 centered at the origin and consider the operator P(x; D) = x1
@ @ − x2 @x2 @x1
and let F = {(a; 0) : 0 ¡ |a| ¡ 1}. If K is any compact subset of F we can nd a radial function g(x), compactly supported in and equal to 1 in a neighborhood of K, which obviously will satisfy P t (x; D)g = −P(x; D)g = 0. If 1 ¡ p 5 ∞, it follows from Theorem 1.2 (b) that Lp -CapP (F; ) = 0. On the other hand, the function u(x) equal to 1 for x2 ¿ 0 and equal to −1 for x2 ¡ 0 belongs to Lp ( ) and satis es P(x; D)u = 0 in \F but P(x; D)u is a nonvanishing measure concentrated in F, so F is not removable. Also, if G = {(a; 0) : 0 ¡ a ¡ 1}, Lp -CapP (G; ) = 0 and this time G is in fact removable in , but if we delete from a radius dierent from G and call the resulting open set 0 it follows that Lp -CapP (G; 0 ) ¿ 0. Thus, changes in away from G may aect the capacity of G.
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Example 2. Let K be a compact subset of the complex plane and 1 ¡ p ¡ 2. If P(x; D) = (1=2)(@=@x1 + i@=@x2 ) is the Cauchy–Riemann operator and ∈ Cc∞ (C) is equal to 1 in a disk containing K it is easily veri ed that (x) = ( x) also has this property for 0 ¡ ¡ 1 and kP(x; D) kq → 0 as → 0, if q ¿ 2 (see the proof of Lemma 3.2 for the case q = 2). Hence, Lp -CapP (K; C) = 0 for any compact set. On the other hand, if is a bounded subset of the plane Lp -CapP (K; ) = 0 if and only if K is empty. Notice that f(x) = (x1 + ix2 )−1 satis es P(x; D)f = 0 outside {0}, is locally in Lp and cannot be holomorphically continued across the origin; f fails to belong to Lp (C) because of its growth at in nity. The examples above suggest that one should consider a notion of capacity that only depends on the behavior of P(x; D) in a neighborhood of K, as it is obvious that the possibility of removing a set K for a solution f will only depend on the behavior of f near K De nition 1.4. The total Lp -capacity of a subset X ⊂ with respect to P(x; D) in ; denoted by P (X; ; p); is
P (X; ; p) = sup{Lp -CapP (K; U )}
(1.2)
where the supremum is taken over all pairs (K; U ) with K ⊂ X compact; U ⊂ open and K ⊂ U . The quantity P (K; ; p) may be in nite for K compact, for instance, this is the case if = R, P = d=dx and K is the standard Cantor set. De nition 1.5. Let P(x; D) be a differential operator with smooth coefficients defined in . We say that a relatively closed set F ⊂ is everywhere Lp -removable with respect to P(x; D) in if for any open U ⊂ each u ∈ Lp (U ) which satisfies P(x; D)u = 0
in U \F
also satisfies P(x; D)u = 0
in U :
If a closed set F ⊂ is everywhere Lp -removable with respect to P(x; D) in , it is clear that P (F; ; p) = 0. We will now establish the converse for a class of locally solvable operators adapting an argument of [HaP2]. For 1 5 p 5 ∞ and real s we shall denote by Lps the Sobolev space (1 − )−s=2 Lp (RN ) and by E0 (U ) the space of distributions with compact support contained in the open set U . Theorem 1.6. (a) Let P(x; D) be a differential operator of order m with smooth coefficients in ⊂ RN having the following property for some p ∈ (1; ∞): every point of is contained in a neighborhood U such that for any g ∈ Lp1−m ∩ E0 (U ) there exist w ∈ Lp (U ) satisfying P(x; D)w = g in U . Then;
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a relatively closed set F ⊂ is everywhere Lp -removable with respect to P(x; D) if and only if there is an open covering of F by sets U ⊂ such that P (F ∩ U ; U ; p) = 0. (b) If P(x; D) has order 1 the same conclusion is valid for p ∈ [1; ∞]. Proof. We will prove the nontrivial implication. Let u ∈ Lp (U ), U ⊂ , satisfy P(x; D)u = 0 in U \F. We wish to prove that P(x; D)u = 0 in U , i.e., hu; P t (x; D)i = 0 for any ∈ Cc∞ (U ). Since this is a local property we may assume that U is contained in U for some and U possesses the solvability in the hypothesis. Let v = P(x; D)u = P(x; D)(u) − P property assumed ||5m−1 a D u = P(x; D)(u) + g. The coecients a are smooth and their support is contained in the support of . Notice that g ∈ E0 (U ) and g ∈ Lp1−m because (1 − )(1−m)=2 a D , || 5 m − 1, is bounded in Lp (being a global pseudo-dierential operator of nonpositive order and type (1,0) in RN , cf. [AH]). Also, when m = 1, g = −uP0 (x; D), where P0 (x; D) is the principal part of P(x; D) and g ∈ Lp ∩ E0 (U ) for p = 1; ∞ as well if u does. Thus, we may solve the equation P(x; D)w = g in U with w ∈ Lp (U ). Now, hP(x; D)(u + w); 1i = hv; 1i = hP(x; D)u; 1i = hP(x; D)u; i and the left hand side is zero because P (F ∩ U ; U ; p) = 0. Indeed, u + w ∈ Lp (U ) and P(x; D)(u + w) = P(x; D)u is compactly supported in F ∩ U . This proves the theorem. The following consequence of Theorem 1.6 will be referred to in the sequel. Lemma 1.7. Let P(x; D) be a differential operator of order 1 satisfying the hypotheses of Theorem 1.6 for some 1 5 p 5 ∞ and let P0 (x; D) denote its principal part. Let U ⊂ be open; bounded and have the following properties: i) if f ∈ C ∞ (U ) the equation P0 (x; D)u = f can be solved with u ∈ C 1 (U ); ii) if f ∈ Lp (U ) ∩ E0 (U ) the equation P(x; D)u = f can be solved with U ∈ Lp (U ). Given A; b ∈ C ∞ (U ); |A| ¿ 0; set Q = AP + b and consider a compact subset K ⊂ U . Then; Lp -CapP (K; U ) = 0 if and only if Lp -CapQ (K; U ) = 0. Proof. Assume that Lp -CapP (K; U ) ¿ 0. Then, there exist u ∈ Lp (U ) such that Pu = 0 in U \K and Pu-0. Setting u = e v this implies that (P + P0 ( ))v = 0 in U \K and (P + P0 ( ))v-0. Solving the equation P0 ( ) = bA−1 in U we nd v ∈ Lp (U ) such that Qv = 0 in U \K and Qv-0. Similarly, we obtain that the equation Q(x; D)w = f also can be solved in Lp (U ) when the right hand side is in Lp ∩ E0 (U ) because this is so for the equation P(x; D)w = f. The proof of Theorem 1.6 now shows that Lp -CapQ (K; U ) cannot be zero, otherwise Qv = 0 in U \K would imply that Qv is zero, a contradiction. The argument may be reversed to obtain the other implication.
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2 Condition (P) and removable singularities Let P(x; D) be a dierential operator of order m = 1 with complex smooth coecients in . The principal symbol of P, de ned on the cotangent bundle T ∗ ( ), will be denoted by p(x; ). P(x; D) is of principal type if p(x; ) = 0;
∈ Rn \0
implies 3p(x; )-0 :
In this work, only operators of principal type will be considered. De nition 2.1. An operator P(x; D) of principal type in is said to satisfy condition (P) if there exist no smooth complex valued function z ∈ T ∗ ( )\0 such that Im zp takes both positive and negative values on a null bicharacteristic of Re zp where z-0. We recall that a bicharacteristic of Re zp is an integral curve of the Hamiltonian vector eld Re Hzp . It turns out that Re zp is constant on the bicharacteristics; when the constant is zero the bicharacteristic is said to be a null bicharacteristic. Beals and Feerman proved in [BF] that (P) implies local solvability in L2 with loss of one derivative for a general operator of principal type, in the sense of the following de nition: De nition 2.2. A partial differential operator of order m P(x; D) defined in
⊂ RN is locally solvable in Lp with loss of one derivative if for every point x0 ∈ and every s ∈ R there is a neighborhood U ⊂ of x0 ; such that for every f ∈ Lps ∩ E0 (U ) the equation P(x; D)u = f can be solved in U with u ∈ Lps+m−1 . Here; Lps = (I − )−s=2 Lp (RN ) . Combining the Beals Feerman theorem with Theorem 1.6 we obtain Theorem 2.3. Let P(x; D) be a differential operator of order m with smooth coefficients in ⊂ RN satisfying condition (P). Then; a relatively closed set F ⊂ is everywhere L2 -removable with respect to P(x; D) if and only if there is an open covering of F by sets U ⊂ such that P (F ∩ U ; U ; 2) = 0. It is also known that operators of order one satisfying (P) are locally solvable in Lp in the sense of De nition 2.2 for 1 ¡ p ¡ ∞ ([P], [HP], [HM]). Thus, another application of Theorem 1.6 yields Theorem 2.4. Let P(x; D) be a differential operator with smooth coefficients in ⊂ RN ; of order 1; satisfying condition (P) and let 1 ¡ p ¡ ∞. Then; a relatively closed set F ⊂ is everywhere Lp -removable with respect to P(x; D) if and only if there is an open covering of F by sets U ⊂ such that P (F ∩ U ; U ; p) = 0.
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In Sect. 4 we shall give a version of Theorem 2.4 that holds for p = ∞ involving a more stringent notion of capacity. This involves methods that do not rely on Theorem 1.6. The use of a dierent approach is essential because, in general, condition (P) does not guarantee local solvability in L∞ as the example in Appendix A shows. The new method will also give a precise geometric description of which sets F satisfy P (F; ; p) = 0, 1 ¡ p 5 ∞, for rst-order operators satisfying (P) which is of interest in the applications of Theorem 2.4 to speci c situations. As a preparation for a geometric approach to removable sets we dedicate the next section to the study of two simple but important examples.
3 Two basic examples Suppose that P(x; D) is a smooth real nonvanishing vector eld in , U is a small open subset of , K ⊂ U is compact and we wish to determine when Lp -CapP (K; U ) = 0. If : U → U 0 is a dieomorphism that takes P(x; D) into P 0 (x; D) and K into K 0 , it is clear that Lp -CapP (K; U ) = 0 if and only if Lp -CapP0 (K 0 ; U 0 ) = 0. In this sense, the condition Lp -CapP (K; U ) = 0 is invariant under change of variables. Hence, by the tubular neighborhood lemma, we may assume that P(x; D) = @=@xN and U is given by |xj | ¡ 1, j = 1; : : : ; N . If N = 1 it is easy to check that Lp -CapP (K; U ) ¿ 0 unless K = ∅. If N ¿ 1 we write N = n + 1, x = (x1 ; : : : ; xn ), t = xN , P = @=@t. The following lemma will be useful. Lemma 3.1. Let P(x; y; Dx ) be a differential operator defined on the set U given by |x| ¡ a; |y| ¡ b; with coefficients depending on x and y but containing derivatives just with respect to x; let K ⊂ U be compact and 1 ¡ p 5 ∞. Then; Lp -CapP (K; U ) = 0 if and only if the Lp -capacity of K ∩ {y = y0 } with respect to P(x; y0 ; Dx ) in {|x| ¡ a} is zero for almost every y0 ; |y0 | ¡ b. Proof. We know from Theorem 1.2 (a) that there is function f with kfkp 5 1 and supp Pf ⊂ K such that Lp -CapP (K; U ) = hP(x; y; Dx )f; 1i = hP(x; y; Dx )f; gi = hf; P t (x; y; Dx )gi for any g ∈ C0∞ (U ) such that g ≡ 1 in a neighborhood of K. Now, if Lp -CapP (K; U ) ¿ 0, Fubini’s theorem implies that R
f(x; y0 )P t (x; y0 ; D)g(x; y0 )dx-0
(3.1)
for all y0 belonging to set of positive measure contained in {|y| ¡ b}. Since f(x; y0 ) is a solution of P(x; y0 ; Dx ) in {|x| ¡ a} × {y0 }\(K ∩ {y = y0 }) for almost every y0 , we reach one of the implications. Next, to prove the remaining implication, we recall that by Theorem 1.2 (b) we have Lp -CapP (K; U ) = inf kP t gkq
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where the in mum is taken over g ∈ C0∞ (U ) and g ≡ 1 in a neighborhood of K. Assuming that Lp -CapP (K; U ) = 0 we may nd a sequence gk ∈ C0∞ (U ) such that g ≡ 1 in a neighborhood of K and lim kP t gk kq → 0 as k → ∞. Passing through a subsequence we may assume that R Gk (y0 ) = |P t (x; y0 ; Dx )gk (x; y0 )|q dx → 0 a:e: y0 because Fubini’s theorem shows that Gk → 0 in L1 . This shows that the Lp capacity of K ∩ {y = y0 } with respect to P(x; y0 ; Dx ) in {|x| ¡ a} is zero for almost every y0 , as required. Returning to the vector eld P = @=@t on U = {(x; t) : |x| ¡ 1; |t| ¡ 1} we conclude that Lp -CapP (K; U ) = 0 for a compact set K ⊂ U if and only if for almost every |x| ¡ 1 the integral curve of P passing through x does not meet K. Suppose now that P(x; D) is a smooth real nonvanishing vector eld in and F is a relatively closed subset of such that P (F; ; p) = 0. If K is a compact part of F contained in a tubular neighborhood U of P(x; D) we have that Lp -CapP (K; U ) = 0 and this implies that almost every integral curve of P in U does not meet K. Since a countable number of K’s exhaust U we see that almost every integral curve of P in U does not meet F and covering
with a countable number of tubular neighborhoods we can give a meaning to the statement “almost every integral curve of P in does not meet F”. It is easy to check that this is independent of the covering. Conversely, if almost every integral curve of P in does not meet F, U is any open subset of
and u ∈ Lp (U ) satis es P(x; D)u = 0 in U \F, we may modify u setting it equal to zero on the orbits that meet F. It is clear that this does not change the class of u in Lp and P(x; D)u = 0 in U because it is constant in all the orbits. We have proved Lemma 3.2. Let P(x; D) be a smooth real nonvanishing vector field in ; let F be a relatively closed subset of and 1 ¡ p 5 ∞. Then P (F; ; p) = 0 if and only if almost every integral curve of P in does not meet F. Assume now that P(x; D) is a complex vector eld in , i.e., P(x; D) = X (x; D) + iY (x; D) and assume that the real vector elds X and Y are linearly independent everywhere and the Frobenius conditions is satis ed: the Lie bracket [X; Y ] is everywhere a linear combination of X and Y . Thus, is foliated by two-dimensional leaves whose tangent plane is generated at every point by X and Y . In the neighborhood of a given point we may nd a local dieomorphism that takes the leaves into 2-planes parallel to the rst 2 coordinates axes and takes P(x; D) into a multiple of the Cauchy-Riemann operator. Thus, in view of Lemma 1.7, we may assume that 1 @ @ P(x; D) = @ = +i 2 @x1 @x2 and U = {(z; y) : |z| ¡ 1; |y| ¡ 1} with the notation z = (x1 ; x2 ), y = (x3 ; : : : ; xN ). It follows from Lemma 3.1 that a compact subset K ⊂ U satis es Lp -CapP (K; U ) = 0 if and only if for almost every y0 the set K ∩ {y = y0 }
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has capacity equal to zero with respect to @ in = {|z| ¡ 1}. If 1 5 p ¡ 2 Lp -Cap @ (K; ) = 0 if and only K is empty because (z − a)−1 ∈ Lp () for 1 5 p ¡ 2. If 2 5 p 5 ∞ points have capacity zero, i.e., Lp -Cap @ ({a}; ) = 0, a ∈ . Indeed, for 0 ¡ ¡ 1, consider the function 1 if 0 5 |z| 5 e−1= (z) = | ln |z|| if e−1= 5 |z| 5 1 : One checks that k@ kq → 0 as → 0 for 1 5 q 5 2 and approximating conveniently by test functions we see, invoking Theorem 1.2 (b), that Lp -Cap @ ({0}; ) = 0 for p = 2 and it is easy to extend the argument to any point a ∈ . Furthermore, L∞ -Cap @ (K; ) = 0 if and only if K has analytic capacity equal to zero in the sense of Ahlfors ([A], [AB]). The latter holds if and only if K is L∞ -removable with respect to @ in ; although no characterization of those sets having zero analytic capacity is known, this has strong topological and metric implications on K, for instance, K is totally disconnected and has Hausdor dimension 5 1 (we refer to [Ch] for details and more information). Since L∞ (\K) ⊂ Lp (\K) it follows for any p that Lp -Cap @ (K; ) = 0 implies that K is totally disconnected. It follows that every point of K has a basis of neighborhoods whose boundaries do not meet K, in particular, K has compact intersection with each of these neighborhoods. Lemma 3.3. Let F be a subset of C; 1 ¡ p 5 ∞. Then; @ (F; C; p) = 0 if and only there exists an open covering {U } of F such that @ (F ∩ U ; U ; p) = 0. Proof. Only the “if” part needs to be proved. Assume there is a covering {U } with the required properties. Passing through a re nement we may assume that each U is a disk. If 1 ¡ p 5 2 we conclude that F is empty. If 2 ¡ p 5 ∞ and K is any compact set of the plane contained in F, we know that K ∩ U is totally disconnected for any U . Thus, we may cover K with a nite number of open sets V1 ; : : : ; Vk such that each Vj is contained in some U , @Vj ∩ K = ∅ for any j and Vk ∩ Vj = ∅ for j-k (we refer the reader to the classical book [HW] for topological facts on totally disconnected subsets of the plane). Writing Kj = K ∩ Vj , it follows that Lp -Cap @ (Kj ; Vj ) = 0 and this implies that Lp -Cap @ (K; C) 5 Lp -Cap @ (K; V1 ∪ · · · ∪ Vk ) = 0 using that the Vj ’s are disjoint. This implies that K is Lp -removable with respect to @ in any open set that contains K and it follows that @ (F; C; p) = 0. We also have an analogue of Lemma 3.2. Let P(x; D) = X (x; D) + iY (x; D) be a smooth complex vector eld in and assume that the real vector elds X and Y are linearly independent everywhere and the Frobenius condition is satis ed. Let U ⊂ be an open subset such that there exists a dieomorphism that takes each U onto the product of a 2-dimensional disk and an (N − 2)dimensional ball B so that each leaf in U of the foliation induced by P is taken onto × {b} for some b ∈ B.
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Lemma 3.4. Let P(x; D) = X (x; D) + iY (x; D) and U ⊂ be as above. Let F be a relatively closed subset of U and 1 ¡ p 5 ∞. Then P (F; ; p) = 0 if and only if for almost every leaf O of the foliation induced by X and Y; O ∩ F has Lp -capacity equal to zero in O (with respect to the analytic structure induced on O by P). For 1 ¡ p ¡ 2 this holds if and only if almost all leaves do not meet F.
4 Removing singularities of bounded solutions Let P(x; D) be a dierential operator of order m in ⊂ RN , U ⊂ , U open. If ∈ C ∞ ( ) and we set P1 (x; D) = e P(x; D) it is clear that the equation P(x; D)u = 0 is equivalent to the equation P1 (x; D)u = 0 for any U . Therefore, a closed set F ⊂ U will be Lp -removable with respect to P(x; D) in U if and only if it is Lp -removable with respect to P1 (x; D) in U . On the other hand, although in some cases Lp -CapP (K; U ) = 0 is equivalent to Lp -CapP1 (K; U ) = 0 (cf. Lemma 1.7) it is not a priori true that these two conditions should be equivalent. This suggest the introduction of the following De nition 4.1. The full Lp -capacity of a subset X ⊂ with respect to P(x; D) in ; denoted by 0P (X; ; p); is
0P (X; ; p) = sup{Lp -CapP1 (K; U )} where the supremum is taken over all pairs (K; U ); K ⊂ X compact; U ⊂
open; K ⊂ U and all operators P1 = e P; with ∈ C ∞ ( ) and sup |D | 5 1 for || 5 m. Thus, 0P (X; ; p) = 0 means that Lp -CapP (K; U ) = 0 for all proper pairs (K; U ) and, in addition, that the same holds for all nonvanishing multiples of P, which is clearly a necessary condition in order that a set X be everywhere Lp -removable with respect to P. The notion of full capacity is also adequate for the study of removable solutions of a system of vector elds, where the important object is the bundle generated by the vectors rather than speci c generators. We prove now an analogue of Theorem 2.4 valid for p = ∞. Theorem 4.2. Let L(x; D) be a differential operator of order 1 with smooth coefficients in ⊂ RN satisfying condition (P). Then; a relatively closed set F ⊂ is everywhere L∞ -removable with respect to L(x; D) if and only if there is an open covering of F by sets U ⊂ such that 0L (F ∩ U ; U ; ∞) = 0. We assume in what follows that L(x; D) is a dierential operator of order 1 with smooth coecients in ⊂ RN satisfying condition (P). We need to recall some backgound on the geometry of condition (P). Let us write X = Re L, Y = Im L. The orbits of the pair of real vector elds X and Y in the sense of Sussmann [Su] are called the orbits of L. Two points belong to same orbit if and only if they can be joined by a continuous piecewise dierentiable curve,
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such that each piece is an integral curve of ±X or ±Y . The orbits are connected submanifolds of , tangent to X and Y . They can be used to characterize the vector elds that satisfy (P) [H1, Thoerem 3.1]. In particular, if L satis es (P), the orbits of L are submanifols of dimension 1 or 2. In general, for a complex vector eld without zeros, a ner qualitative description than the simple decomposition of in orbits may be given. This is done by considering the equivalence relation ∼ de ned by De nition 4.3. p1 ∼ p2 if and only if there exist a smooth curve : [a; b] →
such that (a) = p1 ; (b) = p2 and Re L; Im L are parallel to 0 (s) along (s). The equivalent class of p will be denoted by [p]. It is known ([H1]) that Proposition 4.4. Let m ∈ and O denote the orbit of L through m. Then; [p] is a subset of O homeomorphic to one of the following spaces: {0};
R;
R+ = [0; ∞);
I = [0; 1];
S 1 = {ei } :
The inclusion [p] ∈ O is an imbedding if we give [p] its natural differential structure of integral curve. If dim O = 2; O does not contain any class [p] homeomorphic to R or S 1 and for any p ∈ O there is a curve : [0; 1] → O such that (0) = p and Re L; Im L are linearly independent at (1). Let us assume that F ⊂ is a relatively closed set, that u ∈ L∞ ( ) is a weak solution of Lu = 0 in \F and that there is an open covering of F by sets U ⊂ such that 0L (F ∩ U ; U ; ∞) = 0. We must prove that every point of is contained in a neighborhood throughout which the equation Lu = 0 is satis ed. Initially, we will assume that L is a vector eld, i.e., it has no term of order zero. This restriction will be removed later. After localization, choice of appropriate coordinates and multiplication of L by a non-vanishing factor we may write L=
n P @ @ −i bj (x; t) ; @t @x j j=1
|x| 5 1; |t| 5 1 ;
(4.1)
where the coecients bj (x; t) are real smooth functions. We may also assume that the open unit cylinder Q = B × (−1; 1), B = {|x| ¡ 1}, is contained in ˜ one of the U ’s, which implies that Pn L (F ∩ Q; Q; ∞) = 0. We denote by b(x; t) the vector eld in Q given by j=1 bj (x; t)@=@xj , (x; t) ∈ Q. It is well known that the fact that L veri es (P) implies that there exists a unit vector eld ˜v(x) de ned on Q such that ˜b(x; t) = |˜b(x; t)|˜v(x);
(x; t) ∈ Q :
Set N = {x ∈ Rn : |x| ¡ 1; and ˜b(x; t) = 0; ∀ |t| ¡ 1}
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and (x) = sup |˜b(x; t)|; |t|¡1
x ∈ Rn ; |x| ¡ 1 ;
so that N is precisely the set where vanishes. The function (x) is Lipschitz and k3kL∞ 5 k3x ˜bkL∞ . Assuming, without loss of generality, that k3 ˜bkL∞ 5 1 we obtain the following estimate that we note for later reference |(x) − (x0 )| 5 |x − x0 |;
x; x0 ∈ Rn ; |x|; |x0 | ¡ 1 :
(4.2)
Another important feature is the local integrability of L: this means that in the neighborhood of every point p there exist n complex functions Z1 ; : : : ; Zn with locally independent dierentials that satisfy the equations LZj = 0, j = 1; : : : ; n. This property is also a consequence of (P) ([T2]). It is a consequence of the Baouendi–Treves theorem ([BT1], [BT2], [T2], [Co]) that if u is a continuous solution of Lu = 0 it can be uniformly approximated in a suitable neighborhood of p by polynomials in Z1 ; : : : ; Zn . This fact was extended by Berhanu and Chanillo [BC] to Lp solutions, 1 ¡ p ¡ ∞, and in this case the convergence holds in Lp (see Sect. B). In the next lemma Lt indicates the transpose of L. Lemma 4.5. Let be the characteristic function of N and let ∈ Cc∞ (Q). Assume that the bounded function u satisfies Lu = 0 in Q\F. Then R R (4.3) u(x; t)(x)Lt dx dt = 0 : t Proof. P We observe that L = −@=@t + c(x; t) if x ∈ N, where c(x; t) = i( k @bk =@xk ). We may write N = N1 ∪ N2 ∪ N3 with
N1 = {x ∈ N : ∃t ∈ (−1; 1); 3x ˜b(x; t)-0} ; N2 = (N\N1 ) ∩ (F) and N3 = N\(N1 ∪ (F)), where : Q → B is the projection of the cylinder onto its base. Observe that by the implicit function theorem N1 is a negligible set; furthermore, Lt = −t on (N\N1 ) × (−1; 1). Take a point (x0 ; t0 ) ∈ N3 × (−1; 1); it will be contained in a small cylinder Q0 = B0 × (; ) that does not meet F, so Lu = 0 on Q0 . We may assume that there is a sequence of polynomials Pj such that vj (x; t) = Pj (Z1 (x; t); : : : ; Zn (x; t)) converges almost everywhere to u in Q0 (see Theorem B.1). Since the functions vj are smooth and satisfy Lvj = 0, their restrictions to N3 × (−1; 1), where L = @t , are functions of x alone and passing to the limit we conclude that, after modi cation on a set of measure zero, the restriction of u to Q0 ∩ (N3 × (−1; 1)) is independent of t. Let −1 ¡ a ¡ b ¡ 1. Covering {x0 } × [a; b] with a nite number of cylinders as described we nd that there is a ball B00 3 x0 such that u is, a.e., a function of x alone on Q00 = (B00 ∩ N3 ) × [a; b]. We reach the conclusion that, after modifying u on a negligible set, u is independent of t on N3 × [a; b]. If we choose a and b so that the support of
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is contained in B × [a; b], we get R
R N3 ×(−1;1)
u(x; t)L dt dx = − t
R
N3
Ã
Rb @ dt u(x) a @t
! dx = 0 :
To end the proof we need only show that R R u(x; t)Lt dt dx = 0 : N2 ×(−1;1)
It is at this point that we use the hypotheses on the capacity of F, in fact, they imply that N2 has measure zero. Let us shrink Q a little to obtain a subcylinder Q0 = B0 × (a; b), so that K = Q0 ∩ F is compact and set N20 = (N\N1 ) ∩ (K). Consider a function ∈ Cc∞ (Q) be equal to 1 in a neighborhood of K. Then, R R t |Lt | dx dt |L | dx dt = N02 ×(−1;1)
=
R N02
Ã
R1
−1
| t | dt
! dx = 2 m(N02 ) :
The last inequality follows from the observation that for any x ∈ N20 the function t 7→ (x; t) ∈ Cc∞ (−1; 1) and assumes the value 1 at some point. The in mum over all such ’s is zero because of Theorem 1.2 (b) and the assumption that L∞ - CapL (K; Q) = 0. Thus, the measure of N20 is zero and letting B0 × (a; b) grow montonically and countably to Q we see that m(N2 ) = 0 and the lemma is proved. Our next step is to prove that R R u(x; t)(1 − (x))Lt dx dt = 0;
∈ Cc∞ (Q) :
(4.4)
Let B be the open ball |x| ¡ 1 in Rn . Consider a sequence of functions k (x) ∈ Cc∞ (B\N) such that k (x) = 1 if dist (x; N) ¿ 1=k and |3k (x)| 5 Ck where C ¿ 0 is independent of k. This sequence is easily constructed. Lemma 4.6. For k = 1; 2; : : : R R u(x; t)Lt (k ) dx dt = 0;
∈ Cc∞ (Q) :
(4.5)
We postpone the proof of Lemma 4.6 and continue our reasoning. Letting k → ∞ in (4.5) we observe that the integrand converges pointwisely and boundedly to u(x; t)(1 − (x))Lt . This is so because the derivatives with respect to xj of k (x), which are bounded by Ck, appear as factors of the coecients bj which are majorized by (x) and since the support of 3k is contained in a 1=k- neighborhood of N it follows from (4.2) that 5 1=k on the support 3k . Hence, the dominate convergence theorem applies and (4.5) implies (4.4) when k → ∞. Finally, adding (4.3) and (4.4) we obtain that Lu = 0 in the sense of distributions in Q.
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We have proved Theorem 4.2 assuming that L has no zero-order term. For the general case, write L = L0 + c(x; t), with L0 a vector eld satisfying (P). The transpose operators of L0 of L may be written as Lt0 = −L0 + c0 (x; t) and Lt = −L0 + c1 (x; t) respectively. In a convenient small neighborhood U of a given point we may solve the equation L0 = c1 − c0 with ∈ C ∞ (U ) ([Hor]). By hypothesis, if U is small enough, the L∞ -capacity of any compact set K ⊂ F ∩ U in U with respect to e L is zero. Applying Theorem 1.2. b) we obtain a sequence of test functions gj ∈ Cc∞ (U ) such that gj ≡ 1 in a neighborhood of K and k(e L)t gj k1 → 0. But (e L)t gj = e (−L0 − L0 ( ) + c1 )gj = e (−L0 + c0 )gj = −e Lt0 gj and it follows that kLt0 gj k1 → 0 which implies that the L∞ -capacity of K in U with respect to L0 is zero. This shows that we may cover F with sets U0 ⊂ such that 0L0 (F ∩ U ; U ; ∞) = 0. Thus, by the rst part of the theorem, F is everywhere L∞ -removable with respect to L0 in
. Consider now an arbitrary open set V ⊂ and a function u ∈ L∞ (V ) that satis es Lu = 0 in V \F. Any point of V ∩ F is contained in a neighborhood U , chosen so that the equation L0 = c can be solved with ∈ C ∞ (U ). Setting u = e− v in U , it follows that v ∈ L∞ (U ) and L0 v = e Lu = 0 in U \F. Using the fact that F is everywhere L∞ -removable with respect to L0 , we conclude that L0 v = 0 holds throughout U , which implies that Lu = 0 in U . This proves that Lu = 0 in V , which implies that F is everywhere L∞ -removable with respect to L in . The proof of Theorem 4.2 is now complete except for the proof of Lemma 4.6, a task that will be carried out in the next sections.
5 Beginning of the proof of Lemma 4.6 We start making the following observation. Let P(x; y; Dx ) be a dierential operator with coecients depending on x ∈ Rn and y ∈ Rm containing derivatives just with respect to x. If u(x; y) ∈ Lploc , 1 5 p 5 ∞, is a weak solution of P(x; y; Dx )u = 0 in Rn+m , then v(x) = u(x; y0 ) is a weak solution of P(x; y0 ; Dx )v = 0 in Rn for almost every y0 . For the proof it is enough to consider test functions of the form (x) (y), observe that P t ( ) = P t () and then use Fubini’s theorem and a density argument to conclude that R v(x)P t () dx is de ned and vanishes for a.e. y0 . In particular, assume that L is a vector eld de ned on an open set U ⊂ Rn and that L is tangent to the leaves of a regular foliation of U . If a bounded measurable function u satis es Lu = 0 in U in the sense of distributions and is a generic leaf of the foliation, then the restriction v of u to satis es Lv = 0 in for almost every (we are identifying L and its pull-back to ). This observation, together with Lemma 3.1, allows us to work leaf by leaf when removing a set of capacity zero. We begin the proof of Lemma 4.6. Since k ∈ Cc∞ (Q\N × (−1; 1)), (4.5) will follow if we prove that u is a weak solution of Lu = 0 in Q∗ = Q\N × (−1; 1) under the assumption that u ∈ L∞ (Q∗ ) and Lu = 0 in Q∗ \F. Every point of Q∗ is contained in a two-dimensional orbit of L in Q because all one-dimensional orbits are contained in N × (−1; 1) and L has only orbits
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of dimension one and two ([H1], [T2]). In a neighborhood of any such point of Q∗ we may perform a change of variables involving only the x coordinates so that the two-dimensional orbits are given by xj = constant, j = 2; : : : ; n. This is easily achieved by (locally) rectifying the ow of ˜v(x) which is smooth outside N. In such coordinates, L assumes the form @ @ − ib(x; t) ; @t @x1
|x| 5 a; |t| ¡ 1 ;
with b(x; t) = 0. Furthermore, For all x ∈ [−a; a] the function (−1; 1) 3 t → b(x; t) does not vanish identically : Considering x0 = (x2 ; : : : ; xn ) as a parameter we may now deal with a vector eld in two dimensions. The function u0 (t; x1 ) = u(t; x1 ; x00 ) is measurable, bounded and satis es the equation Lu0 = 0 in {x0 = x00 }\F for almost every x00 (see the observation at the begining of the section). Furthermore, the capacity of {x0 = x00 } ∩ F with respect to L(x1 ; x00 ; t; D) is zero for almost every x00 , by Lemma 3.1. To simplify the notation we write x instead of x1 and consider a vector eld @ @ − ib(x; t) ; (x; t) ∈ R2 ; L= @t @x with b(x; t) = 0 de ned in a neighborhood of = [−a; a] × [−T; T ]. Solving the equation LU = ibx in a neighborhood of we may construct a smooth function Z such that LZ = 0;
Zx -0;
(x; t) ∈ :
This is the property of local integrability. Using (ReZ; t) as new coordinates in a neighborhood of the origin we may assume, after rede ning a and T , that we are in the following situation (cf. [HT, p. 833]): i) Z(x; t) = x + i(x; t) with (x; t) real valued, |x| 5 a, |t| 5 T , ii) |x (x; t)| ¡ 1=2, |x| 5 a, |t| 5 T , iii) the function t → (x; t) is monotone nondecreasing and not constant for any x, iv) L = @=@t − B(x; t)@=@x with B(x; t) = Zt =Zx = it =(1 + ix ), v) F is relatively closed in = (−a; a) × (−T; T ) and 0L (F; ; ∞) = 0, vi) u ∈ L∞ ( ) satis es Lu = 0 in \F. If we can prove that Lu = 0 in , that is, if we can remove F, this will prove Lemma 4.6. Consider the bers of Z, i.e., the sets Z −1 (z), for z = x + iy ∈ Z( ). It is a consequence of iii) that they belong to one of the three following types: 1) Z −1 (z) = {x} × (−T; sx ]; 2) Z −1 (z) = {x} × [rx ; T ) or 3) Z −1 (z) = {x} × [r; s] (the last type includes single points when r = s). Let : → (−a; a) denote the projection (x; t) 7→ x and consider the union P (resp. R) of all the bers of type 1) (resp. of type 2)). It is clear that P and R are closed sets.
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Lemma 5.1. The sets (P∩F) and (R∩F) have one-dimensional Lebesgue measure equal to zero. In particular; F ∩ (P ∪ R) is a negligible subset of . Proof. The arguments are similar to those of Lemma 4.5. On the bers of P we have that Lt = −L = −@t (it is clear that Bx is zero on bers of types 1 and 2 because t = 0 so B vanishes of second order when it does). If K is a compact subset of P ∩ F, ∈ Cc∞ ( ) is equal to 1 in a neighborhood of K and = {x} × R(−T; sx ] is one of the bers that meet F, the contribution of to the integral |Lt | dx dt is R
| t | dt =
Rsx −T
| t | dt = 1 ;
R which integration in x gives |Lt | dx dt = m((K)). Since the in mum R after of |Lt | dx dt over all ∈ Cc∞ ( ) which are equal to 1 in a neighborhood of K is zero because of v), it follows that (K) is negligible and so is (P ∩ F). The proof for R is analogous. Let us write ∗ = \(P ∪ R). The set Z( ∗ ) is open and described by Z( ∗ ) = {x + iy : −a ¡ x ¡ a; (−T; x) ¡ y ¡ (T; x)} ; furthermore, the map Z : ∗ −→ Z( ∗ ) is proper. Lemma 5.2. Let K ⊂ ∗ be compact and set K [ = Z −1 (Z(K)). Then; K [ ⊂
∗ is compact and L∞ -CapL (K; ∗ ) = L∞ -CapL (K [ ; ∗ ) :
(5.1)
We will assume Lemma 5.2 and leave its proof for the end of the next section. Lemma 5.3. The set F˜ = Z(F ∩ ∗ ) has capacity equal to zero with respect to the Cauchy–Riemann operator; i.e.; ˜ Z( ∗ )) = 0 : L∞ − Cap @ (F;
(5.2)
Proof. We are assuming that 0L (F; ; ∞) = 0, which implies in a trivial way that also 0L (F ∩ ∗ ; ∗ ; ∞) = 0. Thus, for any compact subset K ⊂ F ∩ ∗ we have, in view of (5.1), that L∞ -CapL (K [ ; ∗ ) = 0. It follows that F [ = Z −1 (Z(F ∩ ∗ )) ⊃ F is a relatively closed subset of ∗ , Z(F ∩ ∗ ) = Z(F [ ) and 0L (F [ ; ∗ ; ∞) = 0. Thus, replacing F ∩ ∗ by F [ we may assume from the beginning that F has the following property: If p ∈ ∗ ∩ F, the ber Z −1 (z) of z = Z(p) is contained in F :
(5.3)
If p ∈ ∗ ∩ F, the ber Z −1 (z) of z = Z(p) is necessarily of type 3) and identical with the equivalence class [p] (cf. De niton 4.3 and Proposition 4.4). Thus, either Z −1 (z) is a single point or it is homeomorphic to the unit interval and, in any case, [p] ⊂ F. It follows that ˜ : F ∩ ∗ = Z −1 (F)
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Let K˜ ⊂ F˜ be compact. We must show that any v ∈ L∞ (Z( ∗ )) that satis es @v = 0 in Z( ∗ )\K˜ must also satisfy h@v; 1i = 0. Let ∈ Cc∞ (Z( ∗ )) and for ¿ 0 set @ @ Z (x; t) = Z(x; t) + t; ; L1 = Zx − Zt @t @x g = ◦ Z ; w = v ◦ Z : The map (x; t) → Z (x; t) = (x; y) is a dieomorphism for any ¿ 0 and we have R R v @ dx dy = v ◦ Z (x; t) (@ ) ◦ Z (x; t)(t + ) dx dt R = −2i w (x; t)L1 g (x; t) dx dt : Letting → 0 and writing w = v ◦ Z and L1 = Zx @t − Zt @x = Zx L we get h@v; i = −hv; @ i = 2ihw; L1 ( ◦ Z)i :
(5.4)
Observing that Lt1 = −L1 , and choosing ≡ 1 in a neighborhood of K˜ we see that (5.4) implies that h@v; 1i = −2ihL1 w; 1i. This is so because ◦ Z ≡ 1 in a ˜ and L1 w is supported in Z −1 (K). ˜ Indeed, the function neighborhood of Z −1 (K) −1 ˜ w = v ◦ Z is a holomorphic function of Z o Z (K) and L1 w vanishes there. ˜ ⊂ Z −1 (F) ˜ is a compact subset of F ∩ ∗ , thus But w ∈ L∞ ( ∗ ) and Z −1 (K) 0 0 hL1 w; 1i = 0 because L1 (F; ; ∞) = L (F; ; ∞) = 0. Hence, h@v; 1i = 0 and (5.2) is proved. Lemma 5.4. The set F ∩ ∗ has Lebesgue measure equal to zero. Proof. We wish to show that any compact set K ⊂ F ∩ ∗ has measure zero. By Fubini’s theorem it is enough to show that for every xed t ∈ (−T; T ) the set Kt = {x ∈ (−a; a) : (x; t) ∈ K} has one-dimensional Lebesgue measure zero. For xed t, the map x 7→ Z(x; t) = (x; (x; t)) maps dieomorphically the interval [−a; a] onto the smooth curve : [−a; a] 3 x 7→ Z(x; t) and the measure of Kt will be zero if and only if the arc-length vanishes. But Z(Kt ) ⊂ F˜ must have analytic capacity equal of Z(Kt ) ⊂ to zero by Lemma 5.3. It is known (cf. [Ch, Theorem VII.7], [M], [Ca]) that a compact subset E of a recti able curve of the plane that has positive one-dimensional Hausdor measure 1 (E) ¿ 0 must have positive analytic capacity (this result is refered to as the Denjoy conjecture). For subsets E of a smooth curve , 1 (E) coincides R with the arc-length measure of E. Thus, 1 (Z(Kt )) = 0 implies that Kt dx = 0 and the lemma is proved. 6 End of the proof of Lemma 4.6 We keep the notation of Sect. 5; let us return to the bounded function u described in vi) that satis es Lu = 0 o F and study its restriction to ∗ \F. If p ∈ ∗ \F the equivalence class [p] = {xp } × [rp ; sp ], −T ¡ rp 5 sp ¡ T ,
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does not intersect F in view of (5.3). We may nd a small rectangle Up containing {xp } × [rp ; sp ] and contained in ∗ \F such that Zt -0 on the upper and lower horizontal sides of Up . It follows ([H2, Theorem 3.1]) that L is globally hypoelliptic in U . Since this holds for any p ∈ ∗ \F we conclude that L is globally hypoelliptic in ∗ \F and u is in fact smooth in ∗ \F. Applying the Baouendi–Treves theorem to u in each rectangle Up we nd a function vp de ned and holomorphic on the open set U˜p = Z(Up ) such that u = vp ◦ Z in Up . The sets U˜p cover Z( ∗ )\F˜ as p varies over ∗ \F and also vp = vq in U˜p ∩ U˜q , p; q ∈ ∗ \F. This determines a bounded holomorphic function v de ned in Z( ∗ )\F˜ such that u = v ◦ Z in ∗ \F. The set F˜ can be removed by Lemma 5.3 and v admits a holomorphic extension to ∗ that we denote by v] . Set u] (x; t) = v] ◦ Z(x; t) for (x; t) ∈ ∗ . Then, Lu] = 0 in ∗ and u] = u a.e. in ∗ by Lemma 5.4. Thus, F ∩ ∗ can be removed and dropping the ] from the notation we may assume that u satis es Lu = 0 in ∗ = \(P ∪ Q). It remains to remove F ∩ (P ∪ Q). Observe that P is mapped by Z into a subset of the smooth curve − given by (−a; a) 3 x 7→ (x; (x; −T )) and the image of P ∩ F by Z has Lebesgue one-dimensional measure equal to zero by Lemma 5.1. An analogous remark holds for Q and the corresponding curve + . By Fatou’s lemma, the nontangential limit of v exist a.e. at the boundary of Z( ∗ ) (in particular, at + ∪ − ) and we will denote v0 the extension of v to Z( ∗ ). If G is the Riemann map that sends Z( ∗ ) to the unit disk , G can be extended as a homeomorphism of Z( ∗ ) onto (that we still denote by G), furthermore, G and G −1 are smooth up to the boundary, except at the 4 corners of @(Z( ∗ )), i.e., the points (±a; (±a; ±T )). Consider the bounded holomorphic function V () = v ◦ G −1 (), ∈ . Let 0 ¡ rn ¡ 1, n = 1; 2; : : : ; be an increasing sequence converging to 1 and set Vn () = V (rn ) for || ¡ rn−1 and vn = Vn ◦ G −1 . It follows that vn is holomorphic in a neighborhood of Z( ∗ ) and vn converges to v0 , pointwise in Z( ∗ ) and a.e. in + ∪ − . Hence, un = vn ◦ Z satis es Lun = 0 in a neighborhood of Z −1 (Z( ∗ )) ⊃ and un converges a.e. and dominatedly to u0 = v0 ◦ Z in , in particular, Lu0 = 0 in and u0 = u in ∗ . Let’s write P 0 = P ∩ F (that is the union of the bers of type 1) that intersect F) and, similarly, R0 = R ∩ F. We wish to show that u = u0 in \(P 0 ∪ R0 ). Given p = (x0 ; t0 ) ∈ P\P 0 we may nd a thin vertical strip that contains p, does not intersect F and contains a point q = (x0 ; t1 ) ∈ ∗ . Now, u = u0 in a neighborhood of q and L(u − u0 ) = 0 in , thus, by uniqueness in the Cauchy problem ([CH], [T1]), u and u0 coincide on a neighborhood of the vertical segment {(x0 ; t) ∈ }, in particular, u = u0 in a neighborhood of p. Similarly, u = u0 in the neighborhood of any point of R\R0 . Thus, u0 = u a.e. in by Lemma 5.1 and, since we know that Lu0 = 0, the set F has been completely removed. Finally, we must prove Lemma 5.2. Of course, since K ⊂ K [ , (5.1) will follow if we prove the inequality L∞ -CapL (K; ∗ ) = L∞ -CapL (K [ ; ∗ ) :
(6.1)
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By the de nition of capacity, (6.1) is a consequence of the following extension lemma: Lemma 6.1. With the notation of Lemma 5.2; let K ⊂ ∗ be compact; K [ = Z −1 (Z(K)) and consider a function f ∈ L∞ ( ∗ ) such that Lf = 0 in ∗ \K [ . Then; there exists a function f∗ ∈ L∞ ( ∗ ) such that i) ii) iii) iv)
f∗ = f on ( ∗ \K [ ) ∪ K; Lf∗ = 0 on ∗ \K; hLf∗ ; 1i = hLf; 1i. kf∗ k∞ = kfk∞ ;
Proof. Since f∗ is to verify i), this determines it on the complement on K [ \K and we must de ne f∗ on K [ \K. It should be pointed out that every point of K [ \K is contained in a ber of Z that is homeomorphic to [0; 1]. Consider a point p = (px ; pt ) ∈ K [ \K and its ber [p] = {px } × [; ]. The value of f∗ (p) will depend on the location of p in [p]\K. The latter set is an at most countable union of vertical segments {px } × Ij and Ij is either an open interval (j ; j ), ¡ j ¡ j ¡ , or a semi-open interval [; j ) or a semi-open interval (j ; ]. The latter cases may happen at most twice so we reserve the values j = 0; 1 for them if they occur, i.e., I0 = [; 0 ), I1 = (1 ; ], and the other segments are indexed by integers j = 2. Then, if p belongs to one of the segments {px } × Ij with j = 2 we set f∗ (p) = c, with c = kfk∞ . If p belongs to {px } × I0 ∪ {px } × I1 the de niton of f∗ (p) requires more elaboration. Let’s say that p ∈ {px } × I1 , so 1 ¡ pt ¡ . We may nd a cube Q = (px − ; px + ) × (pt − ; pt + ) that does not meet K and such that L is elliptic in a neighborhood of the top side (px − ; px + ) × {pt + } (use that the coecient B = Zt =Zx vanishes at p but B(px ; pt + )-0 for all ¿ 0 suciently small). In particular, Q\K [ -∅. Set Q∗ = Q\K [ . The set Q∗ is obtained from Q removing the classes [q] that possess some point in K. Since Q ∩ K = ∅ and no class that is homeomorphic to [0; 1] can intersect the top side of Q we conclude that Q∗ is obtained by removing from Q closed vertical segments that stem from the bottom side of Q and end at an interior point of Q. Reasoning as at the beginning of this section (where a holomorphic function v – de ned in Z( ∗ )\F˜ and satisfying u = v ◦ Z on ∗ \F – was constructed), we may nd a holomorphic function g de ned on the open set Z(Q∗ ) such that f = v ◦ Z in Q∗ . Next, using the Riemann map and Fatou’s lemma technique (as in the construction of u0 out of u in the end of the proof of Lemma 4.8) we may nd a function f0 ∈ L∞ (Q) that coincides with f on Q∗ , kf0 k∞ 5 kfk∞ and Lf0 = 0 in Q. We then set f∗ (p) = f0 (p). To make sense of this de nition (f0 is only de ned up to a negligible set) we may cover {px } × (1 ; ] with (a countable number of) cubes like the cube Q described above and one checks that by uniqueness in the Cauchy problem the corresponding functions f0Q agree a.e. in the intersection of dierent cubes. Thus, we obtain an open set Up , equal to the union of those cubes, and a function f0 ∈ L∞ (Up ) such that Up ∩ [p] = {px } × I1 , Lf0 = 0 in Up and f = f0 in Up \K [ . If we take the union of the opens sets Up as p varies over
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all points of K [ \K (Up = ∅ if the segment {p} × I1 (p) does not occur) we get an open set U contained in the complement of K and a function fU ∈ L∞ (U ) such that U meets K [ exactly at the segments {px } × I1 (p), LfU = 0 on U , kfU k∞ 5 kfk∞ and f = fU on U \K [ . Similarly, there is an open set V contained in the complement of K and a function fV that do a similar job for the segments of the form {px } × I0 (p). The construction of U and V shows that their intersection does not contain points of K [ : if QU and QV are, respectively, cubes used in the construction of U and V , the coecient B = Zt =Zx does not vanish on the upper side of QU , nor on the lower side of QV , so any ber [p] with p ∈ QU ∩ QV must be con ned inside QU ∪ QV ⊂ ∗ \K. Hence, U ∩ V ⊂ ∗ \K [ and fU , fV and f coincide on U ∩ V . Summing up, the function f1 de ned by f, fU and fV on the measurable set ( ∗ \K [ ) ∪ K ∪ U ∪ V satis es Lf1 = 0 outside K and extends f without increasing its essential supremum. As said before, we de ne f∗ by f1 on the set where f1 is de ned and by the constant value c on the rest of K [ \K. It is clear that f∗ is measurable, extends f and its norm is not larger than that of f (so they are equal) and properties i) and iv) are satis ed. It remains to be proved that f∗ satis es the equation Lf∗ = 0 in a convenient neighborhood of any arbitrary point of K [ \(K ∪ U ∪ V ). Let p ∈ K [ \(K ∪ U ∪ V ) and consider a cube Q = (px − ; px + ) × (pt − ; pt + ) that does not meet K. Let [q] be a class homeomorphic to [0; 1] that meets Q. If [q] does not meet K, that means that [q] does not meet K [ either. If [q] ∩ K-∅ the intersection necessarily occurs outside Q and [q] ∩ Q = {qx } × (pt − ; pt + ) is contained in a vertical segment {qx } × Ij (q), for some j = 0; 1; 2 : : :. If j = 0; 1 [q] ∩ Q ⊂ U ∪ V and we already know that Lf∗ = 0 on U ∪ V . If j = 2 we have that f∗ = c throughout {qx } × Ij (q) ⊃ [q] ∩ Q. Then, either Lf∗ = 0 in a neighborhood of [q] ∩ Q or f∗ = c on [q] ∩ Q and [q] ∩ Q = {qx } × (pt − ; pt + ). This shows that Q ∩ (K [ \(U ∪ V )) = H × (pt − ; pt + ) for some relatively closed set H ⊂ (px − ; px + ) and that Lf∗ = 0 on Q\(H × (pt − ; pt + )). Observe that the coecient B = Zt =Zx vanishes of second order at H × (pt − ; pt + ) (recall that Zt (x; t) = it (x; t) and t = 0 because t → (x; t) is monotone nondecreasing). Let ∈ Cc∞ (Q) and let’s show that hf∗ ; Lt i = 0. Consider a sequence of functions k (x) ∈ C ∞ ((px − ; px + )), 0 5 k 5 1, equal to one in a neighborhood of H , that converges decreasingly to the characteristic function H of H and such that their derivatives |k0 (x)| are bounded by a constant (independent of k) times the distance from H to the complement of supp k . It follows that B(x; t)k0 (x) → 0 pointwise and |B(x; t)k0 (x)| 5 C. Then hf∗ ; Lt i = hf∗ ; Lt (k )i = lim hf∗ ; Lt (k )i k→∞ R = − f∗ (x; t)H (x) t (x; t) dx dt = 0 :
(6.2)
We have used that f∗ (x; t) = c when x ∈ H and integrated rst in t to obtain the last equality in (6.2). This shows that, Lf∗ = 0 in Q. Thus, Lf∗ = 0 is satis ed outside K and this proves ii).
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Finally, we prove iii). Since Lf∗ = 0 outside K and Lf = 0 outside K [ , given a function ∈ Cc∞ ( ∗ ) that is ≡ 1 in a neighborhood of K [ ⊃ K we have that hLf∗ ; 1i = hf∗ ; Lt i and hLf; 1i = hf; Lt i. We must show that R R (f∗ − f)Lt dx dt = (f∗ − f)Lt dx dt = 0 : (6.3)
∗
K [ \K
We have already noticed that the coecient B of L = @t − B@x vanishes of order 2 on K [ \K and this implies that Lt = − t + B x + Bx reduces to − t = 0 on K [ \K, showing that (6.3) holds. [ Remark. We have seen in the proof S of Lemma 6.1 that if p ∈ K \K the complement [p]\K is of the form j {px } × Ij . We distinguish two situations: either Ij = ∅ for all j = 2 (i.e., all the Ij ’s that occur are semi-open and touch the endpoints of [p]) or at least one open interval Ij occurs (necessarily with j = 2 by notation). In the rst case we say that K does not separate [p] from its endpoints and in the second case we say that [p] is separated by K from its endpoints.
Corollary 6.2. Let F ⊂ = (−a; a) × (−T; T ) be relatively closed and assume that (6.4) L∞ -CapL (F ∩ ∗ ; ∗ ) = 0 : Let S ⊂ (−a; a) be the set of points x such that for some = x + iy ∈ Z( ) the fiber Z −1 () is of type 3) and is separated by F from its endpoints; i.e.; the set Z −1 ()\F contains open vertical segments. Then; S has one-dimensional Lebesgue measure equal to zero. Proof. As remarked right before Lemma 5.1, there exist three types of bers and Lemma 5.1 establishes that bers of type 1) and 2) that intersect F project into a negligible set if the capacity of F in is zero, therefore this corollary is an analogue result for bers of type 3). With the notation of Lemma 5.2, we consider the bers contained in ∗ that are separated by F from its endpoints. Expressing F ∩ ∗ as a countable increasing union of compact sets of ∗ we see that it is enough to prove the corollary assuming that F ∩ ∗ = K is compact. Set K [ = Z −1 (Z(K)) and consider the bers [p] homeomorphic to [0; 1] such that [p]\K contains a component {px } × Ij , j = 2, (cf. the proof of previous lemma). De ne a function u ∈ L∞ ( ) by taking the constant value c on all the components {px } × Ij , j = 2, and equal to Z on the rest of . It follows that Z = u outside K [ . By hypothesis K has capacity zero, by Lemma 6.1 K [ has capacity zero and by the proof of Theorem 4.2 K [ is removable (see the arguments used at the beginning of the section to remove F ∩ ∗ ). It follows that Lu = 0 and by uniqueness in the Cauchy problem Z = u almost everywhere. Since Z is bounded we may assume that c was chosen outside the range of Z. It follows that the union of all components {px } × Ij , j = 2, as px ∈ S has two-dimensional measure zero, which implies by Fubini’s theorem that S has one-dimensional measure zero.
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7 The geometry of removable sets Although the proof of Theorem 4.2 is considerably more dicult and intrincate than that of Theorem 2.4 it has the advantage of yielding precious information about the geometry of removable sets, something that the abstract method of proof of Theorem 2.4 does not furnish. Let L(x; D) be a rst order dierential operator with smooth coecients in ⊂ RN satisfying condition (P) and let F ⊂ be an everywhere L∞ -removable relatively closed subset of . The proof of Theorem 4.2 (cf. the proof of Lemma 4.5) then shows that the local behavior of F with respect to one-dimensional orbits in a neighborhood U of a given point is as follows: almost every one-dimensional orbit of L in U does not meet F. Covering with a countable number of such neighborhoods it is easy to conclude that (i) Almost all one-dimensional orbits of L in do not meet F. This was precisely the behavior of removable sets of real vector elds described in Sect. 3. To examine the behavior with respect to two-dimensional orbits we need to recall some facts. Let be a two-dimensional orbit, then the classes [q], q ∈ , as described by De nition 4.3 and Proposition 4.4, may be homeomorphic to one of the following spaces: {0}, R+ = [0; ∞), I = [0; 1]. If we excise from all classes homeomorphic to R+ we obtain the excised orbit ∗ . The equivalence classes contained in ∗ are homeomorphic to {0} or to I . The quotient space 0 = ∗ =∼ equipped with the quotient topology is called the reduced orbit and has a structure of Riemann surface; in a relative neighborhood of any class [q] ⊂ ∗ there exists a function Z such that LZ = 0 and dZ-0. The function Z is constant on [q] and if we have two of them, Z1 and Z2 , they related by a biholomorphism H , Z1 = H ◦ Z2 ; these two facts are standard consequences of the Baouendi–Treves approximation theorem. The functions Z just described endow 0 with a structure of Riemann surface. The local behavior of F with respect to two-dimensional orbits in a small neighborhood U is that almost all two-dimensional orbits in U present the following characteristics: a) almost all classes [q] ⊂ homeomorphic to R+ do not meet F (cf. Lemma 5.1); b) almost all classes [q] ⊂ ∗ homeomorphic to [0; 1] are not separated by F from its endpoints (cf. Corollary 6.2); c) [F ∩ ∗ ] = (F ∩ ∗ )=∼ ⊂ 0 has full L∞ -capacity zero with respect to the holomorphic structure of Riemann surface of 0 (cf. Lemma 5.3). Observe that for 1 5 p 5 ∞ the de nitions of total capacity (De nition 1.4) and full capacity (De nition 4.1) are locally equivalent for the Cauchy–Riemann operator by Lemma 1.7 and may be interchanged. We may cover once again with a countable a number of such small neighborhoods to reach an analogous conclusion for the global orbits in . We get (ii) Almost all two-dimensional orbits of L in exhibit the following behavior: a) almost all classes [q] ⊂ homeomorphic to R+ do not meet F; b) almost all classes [q] ⊂ ∗ homeomorphic to [0; 1] are not separated by F from its endpoints; c) [F ∩ ∗ ] has full L∞ -capacity zero with respect to the holomorphic structure of Riemann surface of 0 .
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We now wish to show that, conversely, a relatively closed set that ful lls properties i) and ii) is also everywhere L∞ -removable with respect to L(x; D). This will give a geometric characterization of everywhere removable sets which, in a sense, is a combination of the behaviors of the two basic examples described in Sect. 3. It will be enough to reason with a vector eld and we shall do so. Let F ⊂ be a relatively closed set such that i) and ii) hold and consider a function u ∈ L∞ ( ) such that Lu = 0 in \F. We must show that Lu = 0 in
. For starters, we consider a special case in two variables that was already described in Sect. 5. We assume that = (−a; a) × (−T; T ) and L is a vector eld with a rst integral Z such that i) Z(x; t) = x + i(x; t) with (x; t) real valued, |x| 5 a, |t| 5 T , ii) |x (x; t)| ¡ 1=2, |x| 5 a, |t| 5 T , iii) the function t → (x; t) is monotone nondecreasing and not constant for any x, iv) L = @=@t − B(x; t)@=@x with B(x; t) = Zt =Zx = it =(1 + ix ), Consider the bers of Z, i.e., the sets Z −1 (z), for z = x + iy ∈ Z( ). We know that they belong to one of the three following types: 1) Z −1 (z) = {x} × (−T; sx ]; 2) Z −1 (z) = {x} × [rx ; T ) or 3) Z −1 (z) = {x} × [r; s]. Letting :
→ (−a; a) be the projection (x; t) 7→ x we denote by P (resp. R) the closed set equal to the union of all the bers of type 1) (resp. of type 2)). We write
∗ = \(P ∪ R), recall that the set Z( ∗ ) is open and described by Z( ∗ ) = {x + iy : −a ¡ x ¡ a; (−T; x) ¡ y ¡ (T; x)} ˜ Hence, Z −1 (z) ∩ F = ∅ and there exists an and set F˜ = Z(F ∩ ∗ ). Let z ∈| F. −1 open rectangle Uz ⊇ Z (z) such that Uz ∩ F = ∅, U˜ z = Z(Uz ) is open, and there exist a holomorphic function vz de ned on U˜ z such that u = vz ◦ Z in Uz . The subsets U˜ z cover Z( ∗ )\F˜ as z varies over ∗ \F and vz = vz0 in Uz ∩ Uz0 , z; z 0 ∈ ∗ \F. This determines a bounded holomorphic function v in Z( ∗ )\F˜ such that u = v ◦ Z in ∗ \F [ where F [ = Z −1 (Z(F ∩ ∗ )). By c) of property ii) the set F˜ has zero analytic capacity and v can be extended to a bounded holomorphic function v] de ned throughout Z( ∗ ). Reasoning as in Sect. 6 (using Fatou’s lemma and the Riemann mapping theorem) we further extend v] ◦ Z from ∗ to a bounded function u0 that satis es Lu0 = 0 in . By construction, we know that u = u0 on ∗ \F [ and we want to know that u = u0 on \F almost everywhere. Let’s take a point p ∈ \F and prove that u = u0 a.e. in a neighborhood of p = (px ; pt ). If {px } × (−T; T ) does not meet F there exist a rectangle containing p and disjoint from F that has one of its horizontal sides contained in ∗ \F [ , where u = u0 . By uniqueness in the Cauchy problem, u = u0 in the whole rectangle. Next, assume that ({px } × (−T; T )) ∩ F-∅. By a) of property ii) the projection (F ∩ (P ∪ Q)) has onedimensional Lebesgue measure zero and there is no restriction in assuming that px ∈| (F ∩ (P ∪ Q)). Hence, p ∈ ∗ . If p ∈| F [ we already know that u = u0 nearby p, so let’s assume that p ∈ F [ and consider the class [p] = {px } ×
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[; ]. Then, [p]\F = {px } × (I0 ∪ I1 ∪ · · ·). Here, we are using the notation of the proof of Lemma 6.1, reserving the indices j = 2 for open intervals and setting I0 = [; 0 ), I1 = (1 ; ] for the semi-open ones. By b) of property ii) we may assume, after removing a negligible set, that no Ij with j = 2 actually occurs. Thus, p ∈ {px } × (I0 ∪ I1 ), say p ∈ {px } × I0 , and it follows that {px } × [; pt ] does not meet F. We may enlarge this segment to a thin rectangle disjoint from F with its lower horizontal side passing slightly below (px ; ) contained in ∗ \F [ . Once again, uniqueness in the Cauchy problem shows that u = u0 a.e. in a neighborhood of p. We have shown that u0 is an extension of u to and Lu0 = 0. By the arguments of Lemma 5.4, the fact that F˜ has null analytic capacity implies that F ∩ ∗ has measure zero and since F ∩ (P ∪ Q) is negligible as well we conclude that u = u0 a.e. and F has been removed in the special case. We now consider the general case. We must show that the support S(Lu) is empty. The union of the two-dimensional orbits is open so if p belongs to a two-dimensional orbit, there is neighborhood U of p foliated by 2-orbits. If the U is small and the coordinates are chosen conveniently, we will be in the situation of the special case for almost every orbit in U (cf. Sect. 5) and we conclude that S(Lu) ∩ U = ∅. This proves that S(Lu) is contained in the union of the one-dimensional orbits and we must then show that Lu vanishes in a neighborhood of an arbitrary point of a one-dimensional orbit. Since this is a local property we may reason on an open cube Q = B × (−1; 1) and assume that in local coordinates de ned in a neighborhood of Q the vector eld L(x; D) is given by (4.1). We also de ne ˜b(x; t), ˜v(x) and N as in the paragraph of section 4 that precedes Lemma 4.5. Lemma 7.1. Let be the characteristic function of N and let ∈ Cc∞ (Q). Assume that u ∈ L∞ (Q) satisfies Lu = 0 in Q\F. Then RR (7.1) u(x; t)(x)Lt dx dt = 0 : Proof. Following the proof of Lemma 4.5 we consider the decomposition N = N1 ∪ N2 ∪ N3 ; reasoning as before we may conclude that N1 has measure zero, and that R R u(x; t)Lt dt dx = 0 : N3 ×(−1; 1)
Next we want to see that R
R
u(x; t)Lt dt dx = 0 :
(7.2)
N2 ×(−1; 1)
Consider a point q = (x0 ; t0 ) with x0 ∈ N2 . Since L = @t on {x0 } × (−1; 1), it follows that the class [q] is homeomorphic to R, S 1 , [0; 1] or R+ . We further split N2 into N21 and N22 so that x0 ∈ N21 if q = (x0 ; 0) belongs to a one-dimensional orbit (equivalently, [q] is homeomorphic to R or S 1 ) and x0 ∈ N22 if q = (x0 ; 0) belongs to a two-dimensional orbit (equivalently, [q] is homeomorphic to R+ or [0; 1]). Since i) holds, the measure of N21 is zero and
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it will be enough to prove (7.2) with N2 replaced by N22 . Also, almost every class [q], q ∈ N22 is homeomorphic to [0; 1] by a) of property ii), so we may assume that all of them are. The connected components of the intersection of a two-dimensional orbit with Q may be written as = × (−1; 1) where
is a maximal integral curve of ˜v(x) in B and a standard argument shows that ∩ Q has at most a countable number of connected components. Let x1 be an arbitray point in N22 ; since we are assuming that the class [q1 ] of q1 = (x1 ; 0) is homeomorphic to [0; 1] we may nd a dieomorphism that recti es the ow of Re L and is de ned in a neighborhood of the compact interval of [q1 ] that lies between x1 and an endpoint of [q1 ]. In a small ball B1 centered at x1 we may perform a change of the x-variables that recti es the
ow of ˜v(x) so that the expression in B1 × (−T; T ) of the principal part of L is L0 = @=@t − ib(x; t)@=@x1 ; with b(x; t) = 0 and b(x; T ) ¿ 0 (cf. Sect. 5). Thus, assuming without loss of generality that is supported in 1 = B1 × (−T; T ), we must show that R R @ dt dx = 0 : (7.3) u(x; t) @t N ×(−T; T ) 22
Introducing the notation x0 = (x2 ; : : : ; xn ), we observe that in 1 the twodimensional orbits are given by x0 = {(x1 ; x0 ; t) ∈ 1 : x0 = const:} and the classes homeomorphic to [0; 1] are given by x0 = const:, x1 = const:, 5 t 5 . Furthermore, the function b(x; t) vanishes identically on such classes. We integrate rst in the variables x1 and t in (7.3), keeping x0 xed. Since the integration in the variables (x1 ; t) takes place within a two-dimensional orbit, where we know that u can be extended to a solution, we conclude that u is almost everywhere constant on each class x1 = const:, 5 t 5 (this follows from the a.e. approximation of the Baouendi–Treves formula). Hence, for almost every x = (x1 ; x0 ) the function t 7→ u(x; t) is constant for |t| ¡ T and integral in t in (7.3) vanishes for a.e. x ∈ N22 . This proves the lemma. To conclude that Lu = 0 in Q we have to prove that RR u(x; t)(1 − (x))Lt dx dt = 0 : This easily done approximating (1−) by functions k like those of Lemma 4.6. R We will have that uLt (k ) dx dt = 0 because k is supported in the union of the two-dimensional orbits and we already know that S(Lu) is contained in the union of the one-dimensional orbits. This ends the proof that F is L∞ -removable. Combining this with the results of Theorem 4.2 we obtain Theorem 7.2. Let L(x; D) be a differential operator of order 1 with smooth coefficients in ⊂ RN satisfying condition (P) and let F ⊂ be a relatively closed set. The following properties are equivalent: 1) F is everywhere L∞ -removable with respect to L(x; D); 2) F satisfies the geometric conditions i) and ii); 3) there is an open covering of F by sets U ⊂ such that 0L (F ∩ U ; U ; ∞) = 0.
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Remark. Theorem 2.4 states that the analogues for Lp , 1 ¡ p ¡ ∞, of conditions of 1) and 3) of Theorem 7.2 are equivalent and it is natural to ask if this is still the case for the analogue of the geometric conditions i) and ii) of 2). Inspection of the proof of the case p = ∞ shows that 2) implies 1) for 1 ¡ p ¡ ∞, that is, the Lp analogues of the geometric conditions i) and ii) are still sucient for Lp -removability. Concerning the necessity, i) as well as a) and b) of ii) are satis ed by any Lp -removable set, but the analogue of c) of property ii) for p ¡ ∞ does not seem to be necessary, although we do not have a counterexample. The diculty comes from the following simple fact: the connection between holomorphic removability on the two-dimensional orbits and removability for the given vector eld L is given by the observation that v 7→ u = v ◦ Z takes holomorphic functions v to solutions u of Lu = 0, where Z is a rst integral of L. In this correspondence, u is bounded if and only if v is bounded, while for 1 ¡ p ¡ ∞, only the implication u ∈ Lp ⇒ v ∈ Lp holds (locally) and the converse is false in general. We have Theorem 7.3. Let L(x; D) be a differential operator of order 1 with smooth coefficients in ⊂ RN satisfying condition (P) and let F ⊂ be a relatively closed set; 1 ¡ p 5 ∞. Assume that F satisfies the following conditions: i) Almost all one-dimensional orbits of L in do not meet F. ii) Almost all two-dimensional orbits of L in exhibit the following behavior: a) almost all classes [q] ⊂ homeomorphic to R+ do not meet F; b) almost all classes [q] ⊂ ∗ homeomorphic to [0; 1] are not separated by F from its endpoints; c) [F ∩ ∗ ] has full Lp capacity zero with respect to the holomorphic structure of Riemann surface of 0 . Then F is everywhere Lp -removable with respect to L(x; D) We nish this section with some examples. Example 7.3. Let be the punctured disk of radius 1 in R2 centered at the origin, 1 ¡ p ¡ ∞, and consider the operator P(x; D) = x1
@ @ − x2 : @x2 @x1
A relatively closed set F ⊂ is everywhere Lp -removable in if and only if the set of radii of circles that intersect F is a set of one-dimensional Lebesgue measure equal to zero Example 7.4. Consider the function exp(−1=t); B(t) = − exp(1=t); with derivative b(t) = B0 (t) =
if t = 0 ; if t 5 0 ;
1 exp(−1=|t|) ; t2
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and de ne the dierential operator on R2 L=
@ @ − ib(t) : @t @x
This operator satis es condition (P) but is not solvable in L∞ as shown in Sect. A. There is only one orbit equal to R2 . Write Z(x; t) = x + iB(t). A set closed F is everywhere L∞ -removable with respect to L if and only if Z(F) is everywhere L∞ -removable with respect to @, this in turn is equivalent with the fact that the analytic capacity of Z(F) is zero. Example 7.5. Let = (0; 1) × (0; 1) and consider a Cantor set C ⊂ (0; 1) of positive measure. Let G ⊂ be a closed set such that a) the {x : (x; t) ∈ G} = C and b) for any x ∈ C the intersection {x} × (0; 1) ∩ G is a closed interval Ix that does not reduce to a point. Let b(x; t) = 0 a smooth function that vanishes exactly at G. Let F ⊂ G be a closed set de ned by ({x} × (0; 1)) ∩ F = {cx }, where cx is an interior point of Ix for each x ∈ C. Then, F is not L∞ -removable by Corollary 6.2 (the same holds for any p ∈ (1; ∞]).
A. Inexistence of bounded local solutions We give now an example of an operator in two variables satisfying condition (P) which is not locally solvable in L∞ . Consider the smooth function of one variable exp(−1=t); if t = 0 ; B(t) = − exp(1=t); if t 5 0 ; its derivative b(t) = B0 (t) =
1 exp(−1=|t|) ; t2
and de ne the dierential operator on R2 L=
@ @ − ib(t) : @t @x
It is easily veri ed that L satis es condition (P) and Lt = −L. The function B(t) is strictly increasing for −∞ ¡ t ¡ ∞ and has an inverse (s) : (−1; 1) → (−∞; ∞) given by (±|s|) = ±1=| ln |s|| . There is a homeomorphism (x; s) = (x; (s)) : R × (−1; 1) → R × (−∞; ∞) which is a dieomorphism for 0 ¡ |s| ¡ 1. Let u ∈ L∞ (R2 ) and f ∈ L∞ (R2 ) be such that Lu = f
(A.1)
in the sense of distributions and set v(x; s) = u(x; (s));
g(x; s) =
f(x; (s)) : s ln2 s
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Lemma A.1. Let L; u; f; v; g as above. Then; v ∈ L∞ ; g ∈ L1loc and 1 @ v= g @z 2
for − 1 ¡ s ¡ 1 ;
(A.2)
in the sense of distributions; in particular; if w is any solution of @ w=g; @z in a neighborhood of the origin; w must be essentially bounded in a neighborhood of the origin. Proof. If U is an open subset of R2 and V = −1 U , then V is an open subset of R × (−1; 1) and its Lebesgue measure is given by m(U ) =
R V
1 dx ds : |s| ln2 |s|
It follows that the Borel measure (X ) = m( (X )) is absolutely continuous with respect to the Lebesgue measure on R × (−1; 1), since |s|−1 ln−2 |s| is locally integrable in R × (−1; 1). Thus, v(x; s) and g(x; s) are measurable and the identity 2
R
v(x; s)
R @(x; s) dx ds = g(x; s)(x; s) dx ds; @z
∈ Cc∞ (R × (−1; 1)) ; (A.3)
follows if we introduce the change of variables (x; s) = (x; B(t)) on both integrals. Indeed, (x; t) = (x; B(t)) is a test function and (A.3) becomes hu; Lt i = hf; i which is precisely (A.1). Furthermore, if w=2 is a local solution of (A.3) it follows that v − w=2 is holomorphic in a neighborhood of the origin. By the lemma, we will have our example if we show that for an appropriate choice of f ∈ L∞ , equation (A.2) has a solution which is not locally bounded in any neighborhood of the origin. We choose f so that F = f ◦ is the characteristic function of the sector K described in polar coordinates by 0 5 r 5 1=2, 4=3 5 5 5=3. Hence, g = 0 ∈ L1c (R2 ) and a solution w(x; s) of (A.2) is obtained convolving g with the standard fundamental solution of the Cauchy–Riemann operator. Thus, Re w(x; s) =
1 R x − x0 1 dx0 ds0 : 0 2 0 2 0 2 K |x − x | + |s − s | s ln2 s0
We see that for (x; s) = (0; 0) the integral above is given by C
1=2 R 5=3 R 0 4=3
1=2 R − cos 1 0 dr d = C dr = ∞ 2 2 r sin ln (r sin ) 0 r ln r
for some C; C 0 ¿ 0 and it is easy to conclude that Re w cannot be essentially bounded in any neighborhood of the origin.
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B. The approximation formula in Lp Let = B × D be a neighborhood of the origin in Rxm × Rtn , where the coordinates are denoted x = (x1 ; : : : ; xm ), t = (t1 ; : : : ; tn ), n = 0 and m = 1, and we are assuming that B ⊂ Rm and D ⊂ Rn are open balls centered at the origin. Suppose that m smooth real functions 1 (x; t); : : : ; m (x; t), vanishing at the origin, are given in a neighborhood of . We will assume that the norm of the matrix @j (x; t) x (x; t) = @xk is uniformly small, to be precise, kx (x; t)k ¡ 1=2;
(x; t) ∈ ;
as a linear operator in L(Rm ). Consider the complex functions Zj (x; t) = xj + ij (x; t);
(x; t) ∈ ; j = 1; : : : ; m ;
and the linearly independent smooth vector elds L1 ; : : : ; Ln de ned in by the orthogonality conditions Lk Zj = 0;
Lk tr = kr ;
k; r = 1; : : : ; n; j = 1; : : : ; m :
The Baouendi–Treves theorem ([BT1], [BT2], [T2]) states that if u ∈ D0 ( ) is a distribution in satisfying the overdetermined system of equations Lk u = 0;
k = 1; : : : ; n ;
(B.1)
there is a sequence of polynomials of m variables P‘ (z1 ; : : : ; zm ) such that P‘ ◦ Z(x; t) = P‘ (Z1 (x; t); : : : ; Zm (x; t)) converges in D0 (U ) to u, where U ⊂
is a suitable neighborhood of the origin. Furthermore, if u ∈ C k ( ) the approximation takes place in the topology of C k (U ). Theorem B.1. Under the above hypothesis on Z1 ; : : : ; Zm ; L1 ; : : : Ln ; assume that u ∈ Lploc ( ); 1 5 p 5 ∞; satisfies the equations (B:1). Then there is a neighborhood U ⊂ of the origin and a sequence of polynomials P‘ (z1 ; : : : ; zm ) such that P‘ ◦ Z(x; t) converges a.e. in U to u. If 1 5 p ¡ ∞ the sequence converges in Lp (U ). Let’s choose one for ever a cut-o function h(x) ∈ Cc∞ (B), equal to 1 in a neighborhood of the origin, and consider the operator E u(x; t) =
m=2 R exp{−[Z(x; t) − Z(x0 ; 0)]2 } u(x0 ; 0)h(x0 ) dZ(x0 ; 0) : m R (B.2)
The notation [z]2 for a complex vector z = (z1 ; : : : ; zm ) indicates the number [z]2 = z12 + · · · + zm2 and dZ(x; t) = det(Zx (x; t)) dx. One should also observe that a distribution that satis es (B.1) always has a trace u(x; 0) so h(x)u(x; 0) ∈
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E0 (Rm ) can be paired with any smooth function of x (that is the meaning of the integral if u is not a function). The de nition of E depends on the choice of h, however, the latter is kept xed so there is no need to bother with this. A basic step in the proof of the Baouendi–Treves theorem is to prove that E u(x; t) → u(x; t) uniformly in a neighborhood of the origin as → ∞ when u ∈ C 0 ( ). Similarly, to prove Theorem B.1 we must study the convergence of E u(x; t) when u ∈ Lp ( ). If u ∈ Lp ( ), Fubini’s theorem implies that for almost every t ∈ D the function ut (x) = u(x; t) is de ned, measurable and ∈ Lp (B). If, in addition, u satis es the system (B.1), the wave front set WF(u) ⊂ T ∗ ( ) is contained in {(x; t; ; ) : = 0}, in particular u has a trace Tt u(x) de ned for all t ∈ B, the map D 3 t 7→ Tt u(x) is a smooth function of t with values in D0 (B) and the map D 3 t 7→ h(x)Tt u(x) is a continuous function of t with values in Lp−1 (Rm ) ∩ E0 . We now compare ut and Tt u. Lemma B.2. If u ∈ Lp ( ); 1 5 p 5 ∞; and u satisfies the system (B.1) then Tt u = ut ;
for a:e: t ∈ D ;
in particular Tt u ∈ Lp (B) for a.e. t ∈ D. Proof. Take test functions (x) ∈ Cc∞ (B), (t) ∈ Cc∞ (D). Then, t 7→ hTt u; i ∈ C ∞ (D), t 7→ hut ; i ∈ Lp (D) and R R R R u(x; t)(x) dx (t) dt : (B.3) hTt u; i (t) dt = hut ; i (t) dt = R If we take (t) = j (t − t0 ), j (t) = j n (jt), 0 5 ∈ Cc∞ ({|t| ≤ 1}), dt = 1, and let j → ∞, the left hand side of (B.3) converges for every t ∈ D to hTt u; i while the right hand side converges a.e. to hut ; i. Hence, there is a null set E() ⊂ D such that hTt u; i = hut ; i;
∈ Cc∞ (B); t ∈| E() :
If we apply the last identity to a dense sequence {k } ⊂ Cc∞ (B) and set E = S E(n ) we obtain that Tt u = ut as elements of D0 (B) when t is not in the null set E. Remark. Of course, one cannot expect in general that under the conditions of Lemma B.2 Tt u ∈ Lp for all t. For instance, if = (−1; 1) × (−1; 1) ⊂ R2 , Z = x + it 2 =2, L = @t − it@x is the Mizohata operator and u(x; t) = Z −1 , it is simple to verify that u ∈ Lp ( ) for 1 5 p ¡ 3=2, Lu = 0 in the sense of distributions and Tt u ∈ C ∞ ([−1; 1]) ⊂ L∞ (−1; 1) ⊂ Lp (−1; 1) for t-0 but for t = 0 we have T0 u = pv(1=x) − i(x) ∈| Lp (−1; 1). Let’s now start with the proof of Theorem B.1. Shrinking slightly we may assume that u ∈ Lp ( ) ⊂ L1 ( ) and that it satis es the system (B.1) in a neighborhood of . Consider the operator m=2 R exp{−[Z(x; t) − Z(x0 ; t)]2 } Tt u(x0 )h(x0 ) dZ(x0 ; t) : G u(x; t) = Rm (B.4)
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The error operator R u(x; t) = G u(x; t) − E u(x; t) has a standard expression obtained by application of Stokes’ theorem R u(x; t) =
m R P
rj (x; t; t 0 ; ) dtj0
(B.5)
[0; t] j=1
rj (x; t; t 0 ; ) m=2 R = exp{−[Z(x; t) − Z(x0 ; t 0 )]2 } Tt 0 u(x0 )Lj h(x0 ) dZ(x0 ; t 0 ) : 0 m x ∈R P The path of integration of the smooth 1-form rj dtj0 with coecients depending on (x; t; ) as parameters is the straight segment [0; t] that joins the origin to the point t. Lemma B.3. Let u be a distribution that satisfies the system of equations (B.1) in a neighborhood of . There exists a neighborhood U of the origin such that R u(x; t) → 0 in C ∞ (U ). Since h ≡ 1 in a neighborhood of the origin, Lj h(x0 ) ≡ 0 if |x0 | 5 for some positive . Thus, if we take |x| 5 =2 = 1 in (B.5) it follows that the absolute value of the exponential in the integrand that de nes rj is exp(−(|x − x0 |2 − |(x; t) − (x0 ; t 0 )|2 )) and the exponent is 5 −(|x − x0 |2 − |x − x0 |2 =2 − K|t − t 0 |2 ) 5 −(2 =8 − |(x; t) − (x0 ; t)|). K|t − t 0 |2 ) (we have used √ here that kx k 5 1=2 to estimate 0 Thus, taking |t| 5 =(4 K) = 2 and using that t is in the segment joining the origin to t, so |t − t 0 | 5 |t|, we see that the exponential in the integrand of rj can be estimated by exp(−2 =16) for |x| 5 1 , |t| 5 2 . On the other hand, the information on the wave front set WF(u) ⊂ {(x; t; ; ) : = 0} ⊂ T ∗ ( ) implies that for a certain positive integer k we may write Lj h(x)Zx (x; t)u(x; t) = (1 − x )k (1 − x )−k (Lj h(x)Zx (x; t)u(x; t)) = (1 − x )k vj (x; t) with vj (x; t) ∈ C 0 ( ). Then, rj (x; t; t 0 ; ) = Cm=2
R
5|x0 |5C
(1 − x0 )k [exp{−[Z(x; t) − Z(x0 ; t 0 )]2 }] vj (x0 ; t 0 ) dx0 :
It follows that rj (x; t; t 0 ; ) depends continuously on t 0 and converges to 0 uniformly in |x| 5 1 , |t 0 | 5 |t| 5 2 as → ∞ (notice that the derivatives of the exponential contribute with powers of that are dominated by the exponential decay of the exponential factor). Hence, R u(x; t) → 0 uniformly for
Locally solvable dierential operators
621
|x| 5 1 , |t| 5 2 and it is apparent that the same happens to its derivatives of any order with respect to x and t. Lemma B.3 shows that no matter how irregular the solution u is the error R u converges to zero in the best possible way, so the responsibility for the convergence of E u = G u − R u in a given space with a coarser topology than the C ∞ topology is borne by G u alone. Let’s go back to G and its corresponding maximal operator G ∗ u(x; t) = sup |G u(x; t)| : =1;
Lemma B.4. There exist a positive constant C ¿ 0 such that G ∗ u(x; t) 5 CM (h(x)Tt u(x));
u ∈ L1 ( ) ;
(B.6)
for any t such that Tt u ∈ L1 (B); where M is the Hardy–Littlewood maximal operator acting in the x-variable ∈ Rm Mf(z) = sup r¿0
R 1 |f(y)| dy ; m(B(z; r)) B(z; r)
B(z; r) ⊂ R is the ball of center z ∈ Rm and radius r and m(B(z; r)) its Lebesgue measure. m
Proof. The exponential in the integrand of (B.4) is dominated by exp(−(|x − x0 |2 − |(x; t) − (x0 ; t)|2 )) 5 exp(−3|x − x0 |2 =4) : Thus, G ∗ u(x; t) is dominated uniformly in t ∈√D by the maximal function of the convolution operator f ∗ (hTt u), where = , f (x) = m f(x) and f(x) = 2 Ce−c|x| is a Gaussian (the convolution is performed in the x variable and t is consider as a parameter). Since f is radial decreasing and integrable it is classical [St] that the latter maximal function can be estimated by M (hTt u). We return to the proof of Theorem B.1 and the solution u ∈ Lp ( ). Inequality (B.6), the fact that G u(x; t) converges uniformly to h(x)u(x; t) when u is continuous and well known properties of the Hardy–Littlewood operator imply in a standard fashion [St] that for every t such that Tt u(x) ∈ L1 (B) there exists a null set Et ⊂ B such that G u(x; t) → h(x)u(x; t) for x ∈| Et , → ∞. Hence, G u(x; t) → h(x)u(x; t) a.e. in . Thus, if U is the neighborhood guaranteed by Lemma B.3 and we take (x; t) ∈ U such that Tt u ∈ L1 (B) and x ∈| Et we get lim E u(x; t) = lim G u(x; t) = h(x)u(x; t) = u(x; t) : →∞
→∞
This proves that E u(x; t) → u(x; t) at almost every point of a neighborhood U of the origin. To prove the Lp convergence of G u we may majorize |G u(x; t)| by the convolution with a Gaussian as before, |G u(x; t)| 5 C|(f ∗ (hTt u))(x)|, = √ , and then Young’s inequalities give, for values of t such that ut = Tt u (recall Lemma B.2), kG ( · ; t)kLp (dx) 5 Ckhut kLp (dx) :
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This inequality may be raised to the power p and integration with respect to t yields kG ukp 5 Ckhukp , 1 5 p 5 ∞. Since G u → hu uniformly in
when u is continuous, a density argument that holds for 1 5 p ¡ ∞ implies that G u → hu in Lp ( ) and E u → u in Lp (U ), 1 5 p ¡ ∞. Finally, since E u(x; t) is an entire function of Z(x; t) for each , E u(x; t) can be uniformly approximated on U , for any xed , by polynomials in Z(x; t). This proves Theorem B.1. Acknowledgements. We are greatful to Berhanu and Chanillo for allowing the inclusion of their unpublished results [BC] in Appendix B.
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