Hölder estimates for solutions of integro differential equations like the fractional laplace

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1 Hölder estimates for solutions of integro differential equations like the fractional laplace Luis Silvestre September 4, 2005 bstract We provide a purely analytical proof of Hölder continuity for harmonic functions respect to a class of integro differential equations like the ones associated with purely jump processes. The assumptions on the operator are more flexible than in previous works. Our assumptions include the case of an operator with variable order, without any continuity assumption in that order. Introduction Several regularity results in the nonlinear theory of elliptic differential equations are based on Hölder estimates for linear equations with rough coefficients. These results are closely related with Harnack inequalities as it is standard in the subject. In [0], E. De Giorgi proved a C α estimate for second order uniformly elliptic equations in divergence form with measurable coefficients, and the regularity of minimizers for nonlinear convex functionals followed from there. Other proofs were given by J. Nash [4] and J. Moser [3]. In the non-divergent case, the corresponding result was obtained by Krylov and Safonov [], and it is an essential tool in proving C,α regularity for fully nonlinear elliptic equations (See [9]). The regularity of harmonic functions with respect to nonlocal operators was studied in several recent papers like [7] and [8], however their point of view is probabilistic. We obtain a similar result from a purely analytic point of view. Moreover, our assumptions imply many of the previous Hölder estimates in the nonlocal case, and also include the important case of variable order without assuming any continuity in the order, that was not possible with the previous approach (See [7], Example 2). For nonlocal operators, Harnack inequality does not imply a Hölder estimate like in differential equations, due to the fact that the harmonic function u is assumed to be nonnegative in the whole space. In [6] and [6] a Harnack inequality is obtained, but no Hölder estimate follows. We expect this estimates to be important in a forthcoming nonlinear theory of nonlocal operators. We will study operators T defined by an integral. T u(x) = (u(x) u(x + y) + χ B (y) u(x) y) K(x, y) dy (.) Where K is a nonnegative function that satisfies sup x ( y 2 )K(x, y) dy < +.

2 With some additional hypothesis we will prove that if u is a bounded function so that T u(x) = 0 for every x B 2r, then for α small enough, u(x) u(y) sup x,y B r x y α C r α u We are specially interested in the symmetric case where K(x, y) = K(x, y). However our method works in a more general setting. If K(x, y) dy is replaced by a measure K(x, dy) with a singular component, our result will apply as long as we satisfy (2.2) where K (rx + x 0, ry) r n dy would stand for the pull back of K(x, dy) by the application x rx + x 0. Moreover, our result could be extended to operators that do not necessarily have an integral representation like (.) as remark 4.4 shows. In the symmetric case, when K(x, y) = K(x, y) then the term χ B (y) u(x) y will not have any influence in (.) besides making the integral convergent. In this case χ B (y) u(x) y could be replaced by χ BR (y) u(x) y for any R > 0 without changing the operator. Moreover T u(x) = PV (u(x) u(x + y)) K(x, y) dy It would be a little cleaner to write down all the paper only in the symmetric case. I have seriously considered doing so. But in order to keep enough generality to include most previous results, symmetry is not assumed in this work. The regularity result is based in lemma 4.. It will be proved through point estimates assuming that u is smooth, however the estimates do not depend on the norm of any derivative or modulus of continuity of u. Therefore the result should extend to nonsmooth functions that are merely locally bounded. The problem is that for the time being we do not know any good way to make sense of the concept of harmonic or subharmonic function of an operator like (.) when u is not smooth. We believe that when there is a good way to define a meaning for 4., then it should be possible to apply this result, either adapting the proof or by an approximation process. The paper is organized as follows. In Section 2, we set up the technical assumptions that we require for the operator T in order for our estimates to apply. s our assumptions may not seem simple or natural, Section 3 is devoted to give several common particular cases where our theory applies. In Section 4 a lemma is proved that is the non-local analogous to De Giorgi s lemma for subsolutions of divergence form elliptic PDEs. From this lemma, the main result is derived in Section 5. 2 Notation and special assumptions We will first define the auxiliary bump function b(x) = β( x ), for β(x) = ( x 2 ) 2. The only important thing about b is that it is C 2, radially symmetric, its support is B and decreasing in any ray from the origin. ny function with these characteristics would work for us, but we must stick to the same one for the whole paper. Given a δ > 0, we will make the following assumption for T : there are two positive numbers κ < /4 and η such that for every x κt b(x) + 2 ( 8y η )K(x, y) dy < 2 inf K(x, y) dy (2.) B 2 B 4 >δ 2

3 Moreover, we will need (2.) to hold at every scale. That is, if r > 0 and x 0, let u r,x0 (x) = u(rx + x 0 ). By a change of variables we can obtain an operator T r,x0 such that (T r,x0 u r,x0 )(x) = [T u] r,x0 = T u(rx + x 0 ). This operator is given by the formula: T r,x0 v(x) = (v(x) v(x + y) + χ B (ry) v(x) y)k (rx + x 0, ry) r n dy Notice that when K is symmetric, the terms χ B (ry) v(x) y and χ B (y) v(x) y have the same effect in the integral. If for example T = ( ) s (see 3.), T r,x0 = r 2s T. To obtain the desired regularity in harmonic functions we will require (2.) to hold for every T r,x0. In other words, we will require that there exist η and κ < /4 such that for every r > 0 and x 0 : κt r,x0 b(x) + 2 B 4 ( 8y η )K (rx + x 0, ry) r n dy < 2 inf B 2 >δ K (rx + x 0, ry) r n dy (2.2) Normally, we will consider classes of operators with a similar behavior at every scale. So in practice it will be enough to check only (2.) for that class, and then (2.2) will hold. The formula for (2.2) is not important, what matters is that (2.2) means that (2.) holds for T r,x0 for any value of r and x 0. Usually, we are able to find a lower bound for the right hand side of (2.), and then we have to prove that we can choose η and κ to make the two terms of the left hand side as small as desired. The second term controls the long distance behavior in the proof of lemma 4. (probabilists would say the big jumps ). This is needed since the operators are not local in nature. To control this term (at least at unit scale) it is enough to show that for some η 0 η y 0 K(x, y) dy < + B 4 and then use dominated convergence. The first term of the left hand side of (2.) simply applies T to a fixed test function b. When T b is bounded (which happens for every practical situation), this term can easily be controlled. However, when K is very non-symmetric it seems really difficult to have a uniform estimate of this term at every scale in (2.2). In remark 4.4, we will state a nonlinear analog of these assumptions. 3 Particular cases where our assumptions apply Having (2.2) as an assumption may seem a little obscure and awkward to check whether it holds. This section is devoted to see a bunch of examples to illustrate when the theory applies and when it does not. n observation that can come handy is that the left hand side of (2.) (and also (2.2)) is linear in K and the right hand side is super linear in the sense that for a pair K, K 2 and a, a 2 > 0: inf a K (x, y) + a 2 K 2 (x, y) dy inf a K (x, y) dy + inf a K (x, y) dy B 2 >δ B 2 >δ 3 B 2 >δ

4 This immediately implies the next proposition: Proposition 3.. If K i satisfies (2.) (or (2.2)) for k =... N, and a i is a N-uple of positive real numbers, then a i K i also satisfies (2.) (or (2.2)). Proof. We just have to choose the least value of κ and η from all the ones that we have for each K i. 3. Fractional powers of the Laplacian For s (0, ) the operator ( ) s can be obtained by its singular integral representation: ( ) s u(x) u(y) u(x) = PV n+2s dy x y R n = (u(x) u(x + y) + χ B (y) u(x) y) y n+2s dy If we take T = ( ) s, let us check that T satisfies (2.). Let us see first that T b is a bounded function. The auxiliary function b is C 2 and bounded. Recall that b is a fixed function defined at the beginning of section 2 T b(x) = (b(x) b(x + y) + χ B (y) b(x) y) n+2s dy y R (3.) n = I + I 2 where we split the domain of integration into the unit ball and its complement, I = (b(x) b(x + y) + χ B (y) b(x) y) n+2s dy (3.2) y B I 2 = (b(x) b(x + y) + χ B (y) b(x) y) n+2s dy (3.3) y \B For estimating I, we use that b is a fixed C 2 function to get the upper bound b(x) b(x + y) + χ B (y) b(x) y C y 2 for a universal constant C. I B 2 C y y n+2s C 2 2s for a constant C depending only on dimension. For estimating I 2, we use that b is bounded, to get the estimate (3.4) b(x) b(x + y) C = 2 b L 4

5 The term χ B (y) b(x) y vanishes for y >, thus we have I 2 C y n+2s = C 2 2s \B (3.5) for a constant C 2 depending only on dimension. Therefore, T b = I + I 2 C 2 2s + C2 2s is a bounded function, with an upper bound depending only on dimension and s. So, choosing κ small, we are able to make κt b have an upper bound as small as we wish. Now, let us estimate the second term in (2.), that in this case reads 2 ( 8y η ) n+2s dy y B 4 We notice that the integral is finite if η < 2s. Moreover, the integrand decreases when η decreases, and tends to zero as η 0. We can apply dominated convergence theorem to see that the integral goes to zero as η 0. Thus, this term can be as small as we wish if we choose η small. For a given ε > 0, the value of η for which this term is less than ε depends on dimension and on s. For the right hand side of (2.), we observe that 2 y n+2s dy 2 δ 2 n+2s then, 2 inf B 2 K(x, y) dy is strictly positive. Therefore, if we pick κ and η small enough, the >δ left hand side if (2.) will be smaller than the right hand side. Notice that the values of κ and η for which this happens depend only on s and dimension. Since T r,x0 = r 2s T and every term in (2.) is linear in T, then every T r,x0 satisfies (2.). Thus T also satisfies (2.2). 3.2 Sum of powers of the Laplacian If we consider T = i=...n a i ( ) si for a i > 0 and s i (0, ). Then T satisfies (2.2) because of proposition Kernels comparable to those of ( ) s In [8] they consider operators like (.) with kernels satisfying for 0 < a and s (0, ). a n+2s K(x, y) y y n+2s (3.6) K(x, y) = K(x, y) (3.7) 5

6 If we follow closely the estimates for the fractional laplacian, we see that the only thing we are using is their growth estimates at zero and infinity. The same proof, line by line, works for these operators too. Now, if T is an operator whose kernel satisfies (3.6) and (3.7), then so is r 2s T r,x0. We do not have to worry about the term χ B (ry) v(x) y in T r,x0 since K is symmetric. Since (2.) is linear in T, then T r,x0 satisfies (2.) uniformly for every r > 0 and x 0. It is important to notice that the choice of η and κ goes to zero as s goes either to 0 or to. This means that this method would not allow us to obtain a similar estimate for second order operators. This suggests that the estimates are not completely sharp, since second order uniformly elliptic equations in nondivergence form can be obtained as a limit of operators like the ones considered here. But from the result of Krylov and Safonov [] we know that the Hölder estimates do hold for second order equations. 3.4 Operators of variable order Taking a close look to (2.) we can notice that the condition must be satisfied for each single value of x, but there is not interaction with the neighboring points of x. Therefore, if we have a family of kernels K α so that they satisfy the assumption (2.2) with a uniform choice of κ and η, then a K such that K(x, y) = K α(x) (x, y) would also satisfy the hypothesis (2.2). This kernel K would produce an operator T that applies a different operator T α (corresponding to each K α ) depending on the point where it is evaluated. We can apply this observation to kernels satisfying (3.6) and (3.7) as long as we keep uniform bounds on κ and η. But no continuity whatsoever is required in x. Therefore we can consider kernels satisfying the following conditions: a K(x, y) (3.8) n+2s(x) y y n+2s(x) K(x, y) = K(x, y) (3.9) for 0 < a and s (0, ) as long as 0 < s s(x) s 2 < so that we can keep a uniform choice for κ and η. We could alternatively prove this case by a direct computation like we did for the fractional Laplace operators. To obtain an upper bound for T b(x), we split the integral like in 3.: T b(x) = I + I 2. This time we have the inequalities I for a constant C depending only on dimension. I 2 C B 2 C y y C (3.0) n+2s2 2 2s 2 \B for a constant C 2 depending only on dimension. y n+2s = C 2 2s (3.) 6

7 Therefore, T b(x) = I + I 2 C 2 2s 2 + C 2 2s is bounded independently of x, with an upper bound depending only on s, s 2, and dimension. So, as before, we can chose κ small, in order to make κt b(x) as small as we wish. For the second term in (2.), 2 ( 8y η )K(x, y) dy 2 ( 8y η ) 42s 2 2s y n+2s dy B 4 B 4 nd, as before, we can make it as small as we wish as η goes to zero. For the right hand side of (2.), we observe that K(x, y) dy 2 2 δ a 2 n+2s 2 then, 2 inf B 2 K(x, y) dy is strictly positive. Therefore, as in 3., if we pick κ and η small >δ enough, the left hand side of (2.) will be smaller than the right hand side. The values of η and κ for which this happens depend on s, s 2, a,, and dimension. It is very important to observe that the operator r 2sr,x 0 (x) T r,x0 satisfies also conditions (3.8) and (3.9) for s r,x0 (x) = s(rx + x 0 ) instead of s(x). This implies that it also satisfies (2.) for κ and η depending only on δ, a,, s and s 2. But we see that assumption (2.) remains invariant if we multiply T by a function of x. Then T r,x0 satisfies (2.) for any r and x 0 with a uniform choice of κ and η, or in other words, T satisfies (2.2). This lack of continuity in x can not be achieved with the hypothesis of [7], as Example 2 in that paper points out. Since it is known ([]) that for purely diffusive operators the solutions are Hölder continuous, then it would not be unreasonable to expect that to be true for operators like the above but without the condition s 2 <. t the present time we are unable to prove or disprove this. 3.5 The assumptions of Bass and Kassmann In [7], they study the operator (.) having a structure with a (not necessarily absolutely continuous) measure K(x, dy) (They actually use the negative of (.), but that does not make any difference). For the assumptions on K they define: S(x, r) = K(x, dy) (3.2) y r L(x, r) = S(x, r) + r yk(x, dy) + r 2 y 2 K(x, dy) (3.3) y r y <r { N(x, r) = inf K(x, x) : B(x, 2r), } B(x, r) (3.4) 3 2d nd then they assume: sup L(x, ) < x 7

8 and also the following. (a) There exist κ > 0 and σ > 0 such that S(x, λr) S(x, r) κ λ σ, x R d, < λ < /r, r < (3.5) (b) There exist κ 2 > 0 such that if x R d, r <, r/2 s 2r, and x x 2 2r, then N(x, r) κ 2 L(x 2, s) (3.6) These assumptions are designed to apply to Hölder estimates in balls of radius less than, and they are sharpenned in that respect. ssumption (a) would be the same as the bound in the second term of (2.2) if we did not have λ < /r. The assumption (b) is almost the same as our bound for the first term of (2.2) as L is used in [7] to estimate the operator in a smooth test function. However our estimate uses the same point in both sides of the inequality; the fact that we have to consider two different points x and x 2 in (b) prevents from applying the result of [7] to very irregular cases like in Minimizers of functionals in H s Given s (0, ), one of the possible norms for the Sobolev space H s ( ) is given by u 2 H = s u(x) 2 2 u(x) u(y) dx + x y n+2s dx dy Given g H s ( ) and λ h(x, y) Λ, a boundary value problem can be stated in this space as considering the minimizer of a functional like h(x, y) J(u) = x y n+2s u(x) u(y) 2 dx dy (3.7) over all u H s ( ) such that u(x) = g(x) for x \ B. The Euler-Lagrange equation for J gives an equation for the minimizer u. Let us compute the Frechet derivative of J. h(x, y) J(u + tv) = x y n+2s u(x) + tv(x) u(y) tv(y) 2 dx dy h(x, y) = J(u) + 2t x y n+2s (v(x) v(y)) (u(x) u(y)) dx dy + O(t2 ) Then DJ(u), v = 2t h(x, y) n+2s (v(x) v(y)) (u(x) u(y)) dx dy x y 8

9 t this point one is tempted to split the integral in two terms and exchange x with y in one of the terms to obtain ( ) h(x, y) + h(y, x) DJ(u), v = 2t v(x) x y n+2s (u(x) u(y)) dy dx (3.8) However we can not always do that. The inner integral in (3.8) is not absolutely convergent. If further symmetry is assumed, it can be defined as a principal value, and DJ(u) takes a form like in (.) with K(x, y x) = h(x,y)+h(y,x). When this K satisfies K(x, y) = K(x, y), then a simple x y n+2s calculation shows that h(x, y) + h(y, x) DJ(u), v = 2t v(x) lim r 0 x y n+2s (u(x) u(y)) dy dx \B r (x) = 2t v(x) PV (u(x) u(x + y))k(x, y) dy dx = 2t T u, v kernel K obtained by this means would always satisfy K(x, y x) = K(y, x y). In other words, not any functional like (3.7) gives rise to an operator like (.) nor the other way round. This essential difference has to be understood as the difference between divergence and nondivergence second order equations. Our result does not apply in general to minimizers of (3.7). The result in this paper is in this respect more in the flavor of Krylov-Safonov and not De Giorgi-Nash-Moser Harnack inequality. 4 The main lemma The proof of main theorem of this paper is based in the following lemma. The idea of proving a result like this was taken from the paper of De Giorgi [0], where he obtained Hölder regularity for weak solutions to divergence form elliptic equations using a Lemma that looks very similar to this one. Lemmas of this type are sometimes called growth lemmas. They became a common tool in regularity theory for elliptic equations since the work of Landis [2]. To adapt it to nonlocal equations, it was necessary to add the condition (4.3) to control the behavior of u away from the origin. Lemma 4.. Suppose our operator satisfies (2.), and u is a function that satisfies the following assumptions (where δ and η are the same as in (2.2)): T u(x) 0 when x B (4.) u(x) when x B (4.2) u(x) 2 2x η when x \ B (4.3) Then u γ in B /2 for some γ > 0 depending only on κ. δ < {x B : u(x) 0} (4.4) 9

10 Remark 4.2. We are assuming that (2.) holds for triplet of values δ, η and κ. For a given kernel K, the value of κ and η for which the assumption (2.) holds will depend on δ. The value of γ in the last Lemma depends only on κ (γ = κ(β(/2) β(3/4)), where β is the fixed function β(x) = ( x 2 ) 2 ). But since κ depends on δ, then it could be said that γ depends on δ too. The only value of δ for which we are actually going to apply the Lemma is δ = 2 B. Proof. Let γ = κ(β(/2) β(3/4)) (recall b(x) = β( x )). Suppose there is a point x 0 B /2 such that u(x 0 ) > γ = κ β(/2) + κ β(3/4). Then u(x 0 ) + κb(x 0 ) > + κ β(3/4), and for every y B \ B 3/4, u(x 0 ) + κb(x 0 ) > u(y) + κb(y). That means that the supremum of u(x) + κb(x) for x B is greater than and is achieved in an interior point of B 3/4. Let us call that point x. Now we will evaluate T (u + κb)(x ). On one hand, T (u + κb)(x ) = T u(x ) + κt b(x ) κt b(x ). On the other hand we have T (u + κb)(x ) = ((u + κb)(x ) (u + κb)(x + y) + (u + κb)(x ) y)k(x, y) dy Since u + κb has a local maximum at x, (u + κb)(x ) = 0. Besides, for any point z B we know (u + κb)(x ) (u + κb)(z). Let 0 = {y : x + y B u(x + y) = 0}. We use (4.3) and that κ < /4 to obtain the lower bound: T (u + κb)(x ) ((u + κb)(x ) (u + κb)(x + y))k(x, y) dy (x +y) \B + ((u + κb)(x ) (u + κb)(x + y))k(x, y) dy (x +y) B ( ( (2 2 2 y + 3 η )K(x, y) dy + ( 2κ)K(x, y) dy 4)) \B /4 0 (2 2 8y η )K(x, y) dy + (/2)K(x, y) dy \B /4 0 Therefore κt b(x ) 2 ( 8y η )K(x, y) dy + 2 K(x, y) dy \B /4 0 2 ( 8y η )K(x, y) dy + 2 inf K(x, y) dy B 2 \B /4 >δ But this is a contradiction with (2.). 0

11 Remark 4.3. It is to be noticed that the condition T u(x) 0 is used only at one (carefully chosen) point x. nother important observation is that the condition T u(x) 0 could be replaced by T u(x) ε for small enough ε (depending also on κ). If we take ε < κ 2 sup T b, then we get the result with γ = κ 2 (β(/2) β(3/4)). Remark 4.4. The integral representation of the operator T has little to do with the proof. Instead, the proof is based on the behavior of T with respect to a few test functions and some sort of ellipticity. Linearity is only used for the inequality T (u + κb) T u + κt b. If we could get this inequality in some way, then the lemma would apply to any operator T that applies to C 2 functions u with the property that u(x) ( + x 2 ) η0 (0 < η 0 < ), such that the following holds: (ellipticity) If for a pair of functions u and v Then T u(x 0 ) T v(x 0 ). u(x 0 ) = v(x 0 ) u(x) v(x) for every x There is η > 0 and /4 > κ > 0 such that: { } κt b(x) inf T u(x) : u(y) ( + x y 2 ) η {y B 2 (x) : u(y) = 0} δ The property is therefore nonlinear, in the sense that lemma 4. holds for any operator satisfying these assumptions. The minimum or the maximum of two operators satisfying these assumptions will also satisfy them. It seems interesting to try to find an explicit (probably nonlinear) operator that is a maximal for a class of operators satisfying these hypothesis. This operator would play the role of the maximal Pucci operator in the theory of uniformly elliptic equations. This maximal operator takes a particularly simple form for operators of constant order like in 3.3. For s (0, ) and 0 < a, we can define ( u(x) u(x+y)+u(x y) M + 2 u(x) = y n+2s ) + a ( ) u(x) u(x+y)+u(x y) 2 y n+2s dy (4.5) where x + = max(x, 0) and x = max( x, 0). It is simple to check that M + (a, ) is the supremum of all the operators like in 3.3. Similarly, the infimum can be computed by ( ) + ( ) u(x) u(x+y)+u(x y) M 2 u(x) u(x+y)+u(x y) 2 u(x) = a y n+2s y n+2s dy (4.6) Notice that these operators satisfy M h(x) M (u + h)(x) M u(x) M + h(x) M h(x) M + (u + h)(x) M + u(x) M + h(x)

12 By scaling Lemma 4., we can obtain the following extension: Corollary 4.5. Suppose our operator satisfies (2.2), and u is a function that satisfies the following assumptions in a ball B r (x 0 ), for a given δ > 0: T u(x) 0 when x B r (x 0 ) (4.7) u(x) when x B r (x 0 ) (4.8) ( u(x) 2 2x x η ) 0 r when x \ B r (x 0 ) (4.9) δ < {x B r(x 0 ) : u(x) 0} r n (4.0) Then u ( γ) in B r/2 (x 0 ) for some γ > 0 depending on κ, η and δ. Proof. We consider v(x) = u(rx + x 0). By definition of T r,x0, T r,x0 v(x) = [T u] (rx + x 0). Then v satisfies T r,x0 v(x) 0 when x B (4.) v(x) when x B (4.2) v(x) 2 2x η when x \ B (4.3) δ < {x B : v(x) 0} (4.4) Since T satisfies (2.2), then T r,x0 satisfies (2.). So we can apply Lemma 4. to obtain v γ in B /2. Recalling that u(x) = v( r (x x 0)), we obtain the result of the Corollary. 5 Hölder regularity Theorem 5.. Let u be a bounded function so that T u(x) = 0 for every x B 2r. Suppose that K satisfies (2.2) for δ = B /2. Then for α small enough, u C α (B r ) and u(x) u(y) sup x,y B r x y α C r α u where the value of α and C depend only on the constants κ and η of (2.2). Proof. We can reduce the problem to the case r = and osc u = by considering the rescaled function u(x) = u(rx) 2 u. So, we will assume this case. Let x 0 B. We want to show that there exist a C > 0 so that u(x 0 ) u(y) C x 0 y α (5.) for any y. Let α = min(η, log( γ/2) log 2 ), where γ is the constant of lemma 4. and η is the one from (2.2). We chose C = 2 α. Since α depends only on η and the γ from Lemma 4. and C is computed from α, then α and C depend only on the constants κ and η of (2.2). 2

13 We will show by induction, that for any integer number k, osc u B 2 k (x 2 kα (5.2) 0) More precisely, we will construct a nondecreasing sequence b k and a nonincreasing sequence a k such that b k u(x) a k when x B 2 k(x 0 ) and a k b k = 2 kα (5.3) Suppose that 2 k x 0 y < 2 k+, then by (5.2) we would conclude u(y) u(x 0 ) 2 ( k+)α 2 α x 0 y α obtaining (5.). Let us construct a k and b k by induction. For k 0, we can take b k = inf u and a k = b k + 2 kα because osc u =. Let us assume we already have a j and b j for any j k, we have to find suitable a k+ and b k+. Let m = a k + b k 2 then, by (5.3), u m 2 2 kα in B 2 k(x 0 ). Consider ū(x) = 2.2 αk (u(2 k x + x 0 ) m). Then ū(x) for x B and T 2 k,x 0 ū = 0 in B. Let us suppose that x B : ū(x) 0 2 B. When y >, let j 0 such that 2 j y < 2 j+, then 2 k+j 2 k y < 2 k+j+, then by the inductive hypothesis ū(y) = 2.2 αk (u(2 k y + x 0 ) m) 2.2 αk (a k j m) 2.2 αk (a k j b k j + b k m) 2.2 αk (2 (k j )α 2 2 kα ) 2.2 (j+)α 2 2y α The function ū is harmonic in B respect to the operator T 2 k,x 0 that also satisfies (2.), since K satisfies (2.2). We have all what we need to apply lemma 4. and obtain ū(x) γ for every x B /2. We then scale back to u to see that u b k + 2 γ 2 2 kα in B 2 k. Then we can define b k+ = b k and a k+ = 2 ( k )α + b k, and we still satisfy u a k+ in B 2 k because α was chosen so that 2 γ 2 2 α. In the case x B : ū(x) 0 < 2 B, we do the same reasoning with ū instead of ū to obtain that u a k 2 γ 2 2 kα. So we define a k+ = a k and b k+ = a k 2 ( k )α. Remark 5.2. Theorem 5. holds for any operator such that if T u = 0 then T (u + C) = 0, and T r,x satisfies the assumptions of remark 4.4 for every r > 0 and x. If we were interested only in the constant order case, we could state the theorem 5. in the following fashion 3

14 Theorem 5.3. Given s (0, ) and 0 < a, let M + and M be the operators defined in (4.5) and (4.6). Let u be a bounded function so that M + u 0 and M u 0 in B 2r. Then for a small α, u C α (B r ) and u(x) u(y) sup x,y B r x y α C r α u Moreover, if u is merely continuous, the conditions M + u 0 and M u 0 in B 2r could be taken in the viscosity sense (for more on viscosity solutions to nonlocal equations see [], [2], [5], [5], [3] and [4]). By this we mean that for every smooth function ϕ touching u from above at a point x 0, i.e. ϕ(x 0 ) = u(x 0 ) ϕ(x) u(x) for every x then M ϕ(x 0 ) 0. nd if ϕ touches u from below at x 0, then M ϕ(x 0 ) 0. If the operator T has a structure like in section 3.4, not only is Theorem 5. valid, but we can also have a right hand side using Remark 4.3 and the scaling properties of the operators of 3.4. Theorem 5.4. Let u be a bounded function so that T u(x) = f(x) for every x B 2r. Where f is a bounded function and T is of the form (.) with K satisfying. a K(x, y) (5.4) n+2s(x) y y n+2s(x) K(x, y) = K(x, y) (5.5) for 0 < a and s (0, ) and 0 < inf s(x) sup s(x) <. Then for α small enough, u C α (B r ), and we have the estimate u(x) u(y) x y α C r α ( u + max(r inf 2s, r sup 2s ) f ) for any x B r, where α and C depend only on a,, inf s and sup s. Proof. First of all, we normalize u so that r =, osc u and f ε, where ε is the constant of Remark 4.3. To achieve this, we consider the following ū instead of u: so that r 2s(rx) T r,0 ū(x) = ū(x) = u(rx) 2 u + ε max(rinf 2s, r sup 2s ) f (5.6) r 2s(rx) u + ε max(r2 inf s, r 2 sup s ) f f(rx) =: f(x) ε and the operators r 2s(rx) T r,0 satisfy (5.4) and (5.5) with s(rx) instead of s(x). Then we can continue in the same way as in the proof of Theorem 5. but using the observation in Remark 4.3 instead of Lemma 4.. We are able to obtain an improvement of the oscillation in the first iteration step because we are considering a small enough right hand side. 4

15 For any x 0 B, we construct as in the proof of Theorem 5. a pair of sequences a k and b k bounding ū from above and below respectively in B 2 k(x 0 ), so that a k b k = 2 kα. In each iteration step, we rescale ū by considering v(x) = 2 αk (ū(2 k x + x 0 ) m), where m = a k +b k 2. This function satisfies the equation: (2 k ) 2s 2 k,x 0 (x) T 2 k,x 0 v(x) = (2 k ) α+2s 2 k,x 0 (x) f(2 k x + x 0 ) (5.7) That means that for α < 2 inf s, the right hand side is less or equal to f(2 k x x 0 ) < ε. Moreover, the operators (2 k ) 2s 2 k,x (x) 0 T 2 k,x 0 satisfy (5.4) and (5.5) uniformly. Thus, we can apply the Remark 4.3 to either v or v to get an improvement of the oscillation of ū in B 2 k and all the iteration steps can be carried out like in the proof of Theorem 5.. We obtain the estimate for any y. Thus, replacing ū by (5.6) ū(x 0 ) ū(y) x 0 y α 2 α u(x) u(y) x y α C r α ( u + max(r inf 2s, r sup 2s ) f ) for any x B r. Where C and α depend on κ and η from (2.2) and ε from Remark 4.3, but all those quantities can be computed from a,, inf s and sup s. 6 pplications 6. Liouville property Corollary 6.. Let T be an operator that satisfies (2.2), if u is a bounded global solution of T u(x) = 0, then u is constant. Proof. Given x, y, take any r > 0 so that x, y B r, by theorem 5. u(x) u(y) x y α C r α u Taking r large enough, the right hand side converges to zero. Therefore u(x) = u(y) for any x and y, and u is constant. 6.2 Nonlinear equations s an example we will consider two kernels K and K 2 depending only on y so that the operators: T i u(x) = PV (u(x) u(x + y) + u(x) y)k i (y) dy satisfy (2.2). We consider kernels depending only on y so that the operators T i commute with translations and therefore also with differentiation. 5

16 Now, let F : R 2 R be a function that is strictly increasing in each coordinate in the sense that i F C for i =, 2. The estimate will not depend on its smoothness. We can obtain an interior C,α estimate for bounded solutions u to the nonlinear equation F (T u(x), T 2 u(x)) = 0 (6.) We have all the necessary ingredients to perform a proof almost identical to the one of section 5.3 in [9]. The idea is that the difference between u and a translation of u solves an equation with an operator that satisfies (2.2), then a telescopic sum iterational method shows that u is Lipschitz, and finally the derivative of u also solves an equation for which we can conclude that u C α. The same property could be obtained for a nonlinear operator F that maps C 2 functions u such that u(x) ( + x 2 ) η0 into continuous functions such that there is a uniform choice of η and κ so that for each such function u, the operator: T v = F (u + v) F (u) satisfies the assumptions of remark 5.2. In the constant order case, the right assumption would be M v F (u + v) F (u) M + v where M + and M are the operators defined in (4.5) and (4.6). References [] Olivier lvarez and gnès Tourin. Viscosity solutions of nonlinear integro-differential equations. nn. Inst. H. Poincaré nal. Non Linéaire, 3(3):293 37, 996. [2] nna Lisa madori. Nonlinear integro-differential evolution problems arising in option pricing: a viscosity solutions approach. Differential Integral Equations, 6(7):787 8, [3] Sayah watif. Équations d Hamilton-Jacobi du premier ordre avec termes intégro-différentiels. I. Unicité des solutions de viscosité. Comm. Partial Differential Equations, 6(6-7): , 99. [4] Sayah watif. Équations d Hamilton-Jacobi du premier ordre avec termes intégro-différentiels. II. Existence de solutions de viscosité. Comm. Partial Differential Equations, 6(6-7): , 99. [5] Guy Barles, Rainer Buckdahn, and Etienne Pardoux. Backward stochastic differential equations and integral-partial differential equations. Stochastics Stochastics Rep., 60(-2):57 83, 997. [6] Richard F. Bass and Moritz Kassmann. Harnack inequalities for non-local operators of variable order. Trans. mer. Math. Soc., 357(2): (electronic), [7] Richard F. Bass and Moritz Kassmann. Hölder continuity of harmonic functions with respect to operators of variable order. To appear. [8] Richard F. Bass and David. Levin. Harnack inequalities for jump processes. Potential nal., 7(4): ,

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