MATH 825 FALL 2014 ANALYSIS AND GEOMETRY IN CARNOT-CARATHÉODORY SPACES

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1 MATH 825 FALL 2014 ANALYSIS AND GEOMETRY IN CARNOT-CARATHÉODORY SPACES Contents 1. Introduction Vector fields and flows Metrics Commutators Consequences of Hörmander s condition Sums of Squares and Variants Four Classical Examples 5 2. The Laplace operator: fundamental solutions and hypoellipticity Invariance properties of A fundamental solution: the Newtonian potential Solutions of the Poisson equation u = g Hypoellipticity of Regularity of the Laplace operator Spaces of homogeneous type Calderón-Zygmund kernels of positive order Hardy-Littlewood-Sobolev estimates Lipschitz Estimates Sobolev estimates L 1 -estimates 25 Appendix A. Test functions and distributions 27 A.1. Some spaces of smooth functions 27 A.2. The space D (Ω) of distributions on Ω 28 A.3. Other spaces of distributions 29 A.4. Convolutions of distributions and test functions 30 A.5. Convolution of two distributions 30 References Introduction Despite its official title 1, this course will focus on the geometry and analysis of Carnot- Carathéodory metrics, sometimes called control metrics, which are constructed from appropriate collections of first order partial differential operators or vector fields. Such metrics arise in many examples, and before dealing with the general case, we will study a number of classical examples. This will involve the development of a number of tools of classical harmonic analysis, Compiled Monday 8 th September, 2014, 2:30pm. 1 Topics in Functional Analysis LectureNotes-1/Math825/Desktop.

2 2 MATH 825 such as maximal operators, and fractional and singular integral operators. We will then discuss applications of the general theory to some more recent problems arising in complex analysis in several variables. However, to introduce the subject, let us quickly introduce some general definitions, and state some theorems we will prove Vector fields and flows. We will work in an open subset Ω R N. A smooth vector field on Ω is a first order partial differential operator N N X = a j (x) = a j (x) xj (1.1) x j j=1 where each a j C (Ω). This is an analytic definition, but we also think geometrically, so that at each point x Ω, X defines a vector X(x) = (a 1 (x),..., a N (x)) having direction and length. For each point y Ω, the flow along X through y is a smooth curve Φ : ( ɛ, +ɛ) R N so that Φ(0) = y and Φ (t) = X(Φ(t)) for t < ɛ; thus the velocity vector of the curve Φ is given by the vector field X. To see that the flow along X always exists for ɛ sufficiently small, write y = (y 1,..., y N ) and Φ(t) = (ϕ 1 (t),..., ϕ N (t)). Then Φ (t) = (ϕ 1(t),..., ϕ N (t)), and so we need to solve the following system of first order ordinary differential equations with initial conditions: ϕ ( 1(t) = a 1 ϕ1 (t),..., ϕ N (t) ), ϕ 1 (0) = y 1,. j=1 ϕ N(t) = a N ( ϕ1 (t),..., ϕ N (t) ), ϕ N (0) = y N. It follows from the basic theorem on ordinary differential equations that if a 1,..., a N C (Ω), there exists ɛ > 0 so that the system (1.2) has a unique solution on an interval ( ɛ, +ɛ) R. Moreover, if E Ω is compact, the same ɛ works for all y E, and the solution depends smoothly on y Metrics. Let X = {X 1,..., X p } be a finite set of smooth vector fields on Ω. At each point y Ω, these vectors span a subspace X y R N y, where R N y is the set of all (tangent) vectors at y. (Thus X y is a subset of the tangent space to R N at y.) We consider a set C(X ) of mappings Φ : [0, 1] Ω defined as follows: Φ C(X ) if and only if: a) There exist points t 0 = 0 < t 1 < < t M 1 < t M = 1 so that Φ is of class C on each closed subinterval [t j, t j+1 ], 0 j M 1. Thus Φ is piecewise of class C, and at the points t j the curve may have two distinct tangents. b) For each 0 j M 1 and each t [t j, t j+1 ], Φ (t) X ( Φ(t) ). Thus the tangent vector to the curve Φ(t) is restricted to lie in the specified subspaces. Now define ρ X : Ω Ω [0, + ] as follows. Let y 0, y 1 Ω, and put. (1.2) C(X )(y 0, y 1 ) = { Φ C(X ) : Φ(0) = y 0, Φ(1) = y 1 }. (1.3) a) If C(X )(y 0, y 1 ) =, set ρ X (y 1, y 1 ) = +. b) If If C(X )(y 0, y 1 ), set ρ X (y 0, y 1 ) equal to the infimum of the set of δ > 0 such that there exists Φ C(X )(y 0, y 1 ) so that for all t [0, 1], there exist (t 1,..., t p ) R p with p p Φ (t) = c j X j (Φ(t)) and c j 2 < δ 2. (1.4) j=1 j=1

3 MATH The function ρ X is most interesting when C(X )(y 0, y 1 ) for all y 0, y 1 Ω. In this case, it is not too hard to show that ρ X is actually a metric: ρ X (y 0, y 1 ) 0, ρ X (y 0, y 1 ) = 0 y 0 = y 1, ρ X (y 0, y 1 ) = ρ X (y 1, y 0 ), ρ X (y 0, y 2 ) ρ X (y 0, y 1 ) + ρ X (y 1, y 2 ). (1.5) Also, it is fairly clear that if X y = R N for every y Ω, then C(X )(y 0, y 1 ) for all y 0, y 1. However, there are interesting and important examples where ρ X is a metric even if X y is a proper subspace of R N. Exercise 1: In R N let X j = metric. x j, and let X = {X 1,..., X N }. Show that ρ X is the Euclidean Exercise 2: Using coordinates (x, y) R 2 let X = x, Y = x y, and let X = {X, Y }. Then X (x,y) = R 2 if and only if x 0. Show that C(X ) ( (0, 0), (0, 1) ). Exercise 3: Using coordinates (x, y, t) in R 3, let X = x y 2 t, Y = y + x 2 and let X = {X, Y }. In this case X (x,y,t) is a two dimensional subspace of R 3 at every point. Nevertheless, ρ X is a metric Commutators. If X, Y are two smooth vector fields, then XY and Y X are second order partial differential operators. However, the equality of mixed derivatives shows that all the second order terms in the difference XY Y X cancel, and the result is again a first order operator or vector field, called the commutator of X and Y, and written [X, Y ]. Explicitly, if X = N j=1 a j(x) xj and Y = N k=1 b k(x) xk then t, [X, Y ] = = N j,k=1 k=1 a j (x) xj b k (x) x k j=1 N j,k=1 b k (x) xk a j (x) x j [ N N [ a j (x) b k (x) b j (x) a ] ] k (x). x j x j x k (1.6) The space of all smooth vector fields on Ω R N is a module over the algebra of smooth functions C (Ω). With the commutator product, this module becomes a Lie algebra. Thus for any vector fields X, Y, Z, 0 = [X, Y ] + [Y, X], (anti-commutativity); 0 = [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]], (Jacobi identity). (1.7)

4 4 MATH 825 Now if {X 1,..., X p } are vector fields on Ω, let L(X 1,..., X p ) denote the Lie sub-algebra they generate. Thus L(X 1,..., X p ) is spanned over the space of smooth functions by the vector fields X 1, X 2,..., X p ; [X 1, X 2 ], [X 1, X 3 ],..., [X p 1, X p ]; [X 1, [X 1, X 2 ]],..., etc. Let L(X ) y be the span of all vectors Y (y) where Y L(X ). Then X = {X 1,..., X p } satisfies the Hörmander condition at y if L(X ) y = R N y Consequences of Hörmander s condition. The result in Exercise 3 is a consequence of what is called Chow s Theorem (see [2]). Theorem 1.1 (Chow). Let X = {X 1,..., X p } be smooth vector fields on Ω R N. Suppose that X satisfies Hörmander s condition at y for all y Ω. Then C(X )(y 0, y 1 ) for every pair of points y 0, y 1 Ω, and so the function ρ X is a metric on Ω, the Carnot-Carathéodory metric. We will also study the properties of the Carnot-Carathéodory balls B X (x, δ) = { y Ω : ρ X (x, y) < δ }, (1.8) and in particular, find estimates for their volumes. doubling property : We will establish the following critical Theorem 1.2. Let X = {X 1,..., X p } be smooth vector fields on Ω R N, and suppose that for each y Ω, { X(y) : X L(X 1,..., X n ) } = R N y. Let K Ω be a compact subset. Then there exist constants η > 0 and C > 0 so that for all x K and all δ < η, BX (x, 2δ) C BX (x, δ) Sums of Squares and Variants. Let X = { X 1,..., X p } be smooth vector fields on Ω, and consider the second order partial differential operator L = X X 2 p. (1.9) Motivated by results of J.J. Kohn in complex analysis ([7], [8]), Hörmander [5] established the following fundamental regularity result. Theorem 1.3 (Hörmander). Suppose that the vector fields X = {X 1,..., X p } on Ω Ω satisfy the Hörmander condition at each point of Ω. Then the sum of squares operator L is hypo-elliptic: if u is a distribution on Ω and if Lu C (Ω) where U Ω is open, the u C (Ω). Hörmander established the same result for a variant of L. This time let X = { X 0, X 1,..., X p } be smooth vector fields and consider the second order partial differential operator H = X 0 + X X 2 p. (1.10) Theorem 1.4 (Hörmander). Suppose that the vector fields X = {X 0, X 1,..., X p } on Ω Ω satisfy the Hörmander condition at each point of Ω. Then the operator H is hypo-elliptic.

5 MATH Going beyond Hörmander s results, we will establish more refined regularity theorems for the operators L and H by constructing parametrices for these operators. (A parametrix for L is an operator K so that the operators I KL and I LK are either zero, in which case we have found an inverse, or are appropriately smoothing.) To describe these parametrices, one needs to use the corresponding Carnot-Carathéodory metrics. In the case of the vector fields X = {X 1,..., X p } satisfying Hörmander s condition, the operator K will be given as an integral operator Kf(x) = K(x, y)f(y) dy (1.11) Ω where the kernel K C (Ω Ω \ diagonal) satisfies differential inequalities X α x X β y K(x, y) ρx (x, y) 2 α β BX (x, ρ X (x, y) 1 (1.12) for x, y E Ω with x y. Here if α = (α 1,..., α p ) is a multi-index, X α = X α1 1 X αp p. There are similar results for the operator H, but in that case one needs to modify the definition of the Carnot-Carathéodory metric with the operator X 0 treated like a second order operator. With the parametrix in hand, we will obtain regularity results for the equations Lu = g or Hu = g when g belongs to an L p -Sobolev or Lipschitz space Four Classical Examples. Before starting on the general theory, we will look at several classical examples. 1) The Laplace operator on. If X j = xj, then = 2 x x 2 = n n Xj 2. j=1 2) The heat operator H on +1. If X j = xj for 1 j n and X 0 = t, then the inhomogeneous heat equation u t (x, t) = 2 u (x, t) u x 2 (x, t) g(x, t) n x 2 1 can be written X 0 u + n j=1 X2 j u = g. 3) An example of Grushin on R 2. If X = x and Y = x y, consider the equation Gu = 2 u x 2 + x2 2 u y 2 = (X2 + Y 2 )u = g. The operator G is elliptic except where x = 0. However we will also be able to obtain sharp estimates when x = 0. 4) Complex analysis and the Heisenberg group. We use coordinates (x 1,..., x n, y 1,..., y n, t) on R = R 2n+1. Set X j = Y j = T = t. 2y j x j t, + 2x j y j t,

6 6 MATH 825 It is easy to check that [X j, X k ] = [Y j, Y k ] = [X j, T ] = [Y j, T ] = 0 for 1 j, k n and that [X j, Y k ] = 4δ j,k T. The Kohn Laplacian for the Siegal upper half-space in C n+1 is the operator b = n j=1 ( X 2 j + Y 2 j ). 2. The Laplace operator: fundamental solutions and hypoellipticity 2.1. Invariance properties of. The Laplace operator has invariance properties related to the geometry of that can be expressed by the commutation properties of the operator with three kinds of mappings: translations, dilations, and rotations. Let y, let λ > 0, and let U O(n) (the orthogonal group) so that U is a linear transformation on satisfying UU = I. If f L 1 loc (Rn ), set τ y f(x) = f(x y) D λ f(x) = f(λx) R U f(x) = f(ux) (translation by y); (dilation by λ); (rotation by U). (2.1) If f L 1 loc (Rn ) and ϕ C0 ( ) then τ y f(x)ϕ(x) dx = f(x y)ϕ(x) dx = f(x)ϕ(x + y) dy = f(x)τ y ϕ(x) dx, R n D λ f(x)ϕ(x) dx = f(λx)ϕ(x) dx = f(x)λ n ϕ(λ 1 x) dx = f(x)λ n D λ 1ϕ(x) dx, R n R U f(x)ϕ(x) dx = f(ux)ϕ(x) dx = f(x)ϕ(u x) dx = f(x)r U ϕ(x) dx. The operators τ y, D λ, and R U extend mappings of the space D ( ) of distributions to itself 2 : if u D ( ) and ϕ C 0 ( ), set τy u, ϕ = u, τ y ϕ ; Dλ u, ϕ = u, λ n D λ 1ϕ ; RU u, ϕ = u, R U ϕ. (2.2) The three operators τ y, D λ, R U and the Laplacian thus all map the space D ( ) to itself. We have the following commutation properties. Proposition 2.1. Let y, λ > 0, and U O(n). Then τ y = τ y D λ = λ 2 D λ R U = R U. Proof. These identities are easy computations. The first identity holds for any constant coefficient differential operator of any order, and the second holds for any constant coefficient differential operator of order exactly two. 2 An excellent reference for material on distributions is [6]. A brief summary of needed material is presented in Appendix A below.

7 MATH The third is a special case of a more general commutation formula. Let A = {a j,k } and M = {m j,k } be real n n matrices and for ϕ C0 ( ) set P A [ϕ](x) = n j,k=1 a j,k x 2 j,x k ϕ(x) and R M [ϕ](x) = ϕ(mx). The chain rule shows that n n [ n a j,k x 2 j,x k R M [ϕ](x) = m r,j a j,k m s,k ] x 2 r,x s ϕ(mx) j,k=1 r,s=1 so that P A R M = R M P MAM where M is the transpose (or adjoint) of M. Now if A is the identity matrix, the operator P A = is the Laplace operator, and if M = U O(n), then UU = I. It follows that R U [ϕ] = R U [ϕ] for every test function ϕ. If u E ( ) and ϕ C0 ( ), it follows from the third equality in (2.2) that RU u, ϕ = R U u, ϕ = u, R U ϕ Thus R U = R U. j,k=1 = u, R U ϕ = u, R U ϕ = R U u, ϕ 2.2. A fundamental solution: the Newtonian potential. Let δ x D ( ) be the Dirac delta-function at x given by δ x, ϕ = ϕ(x). Note that δ x = τ x δ 0. We look for a fundamental solution for the Laplace operator at x; i.e. a tempered distribution N x S ( ) such that N x = δ x. Observe that the translation invariance of means that we only need to find a fundamental solution at 0. Thus if N = δ 0, it follows that δ x = τ x δ 0 = τ x N = τ x N so that N x τ x N is a fundamental solution at x. The symmetries of suggest the form of a fundamental solution. Note that D λ δ 0 = λ n δ 0 and R U δ 0 = δ 0 R U. If N is a fundamental solution for, it follows from Proposition 2.1 that [ λ n 2 D λ N ] = λ n 2 D λ N = λ n D λ N = λ n D λ δ 0 = δ 0, [ R U N ] = R U N = R U δ 0 = δ 0. Thus λ n 2 D λ N and R U N would be fundamental solutions for as well. This suggests that we look a fundamental solution such that N = R U N and N = λ n 2 D λ N. The first identity means that N(x) depends only on x, and the second identity means that N(λx) = λ 2 n N(x). This suggest that we might consider N(x) = c x 2 n for some constant c. When n > 2, we see in Lemma 2.3 below that this heuristic reasoning is correct. 3 Definition 2.2. Let ( ω n = 2π n n 2 Γ 2 be the surface measure of the unit sphere in. For n 2 the Newtonian potential on is the function ω2 1 log( x ) when n = 2; N(x) = ωn 1 (2 n) 1 x 2 n when n > 2. Using polar coordinates, it is easy to check that the function N L p loc (Rn ) provided that p < n n 2, and in particluar, N is always locally integrable. We let N S ( ) be the tempered distribution given by integration against N. 3 Note that when n = 2 this formula cannot be correct. ) 1

8 8 MATH 825 Lemma 2.3. The distribution N is a fundamental solution for. Proof. We must show that N = δ 0 as distributions. Since for every ϕ C0 ( ), N, ϕ = N, ϕ so we must show that δ0, ϕ = N, ϕ, i.e. ϕ(0) = N(y) ϕ(y) dy. Choose R > 0 so large that the compact support of ϕ is contained in the open Euclidean ball centered at the origin of radius R. For any 0 < ɛ < R, the function x N(x) has no singularities in a neighborhood of the closure of the spherical shell B(ɛ, R) = {x R n ɛ < x < R}, so Green s theorem gives [ ] N(x) ϕ(x) ϕ(x) N(x) dx = B(ɛ,R) B(ɛ,R) [ ] N(ζ) n ϕ(ζ) ϕ(ζ) n ϕ(ζ) dσ(ζ). Here n denotes the outward unit normal derivative on the boundary B(ɛ, R). The function N is infinitely differentiable away from the origin, and a direct calculation shows that [N](x) = 0 for x 0. Thus, since ϕ and [ϕ] have compact support inside the ball of radius R, the left-hand side of equation (2.3) reduces to N(x) ϕ(x) dx = N(x) ϕ(x) dx. B(ɛ,R) To deal with the right-hand side of equation (2.3), note that the boundary of B(ɛ, R) has two connected components: the set S R where x = R and the set S ɛ where x = ɛ. The function ϕ is identically zero in a neighborhood of S R, so this part of the boundary gives no contribution. Thus equation (2.3) reduces to N(x) ϕ(x) dx = x >ɛ ζ =ɛ x >ɛ [ ] N(ζ) n ϕ(ζ) ϕ(ζ) n ϕ(ζ) dσ(ζ). On S ɛ the outward unit normal derivative is n = r where r = x is the distance from the origin. It follows from the explicit formula for N(x) that when ζ = ɛ we have ω2 1 log(ɛ) if n = 2, N(ζ) = ωn 1 (2 n) 1 ɛ 2 n if n > 2, and N (ζ) = ω 1 n ɛ 1 n for n 2. n Since ω n is the surface measure of the unit sphere, ϕ(0) = ωn 1 ɛ 1 n ζ =ɛ and subtracting ϕ(0), we have N(x) ϕ(x) dx x ɛ (2.3) ϕ(0) dσ(ζ), so adding = ϕ(0) + ωn 1 ɛ 1 n [ ] ω2 1 log(ɛ) S ɛ r ϕ(ζ) dσ(ζ) if n = 2 ϕ(ζ) ϕ(0) dσ(ζ). ζ =ɛ ωn 1 ɛ 2 n S ɛ r ϕ(ζ) dσ(ζ) if n > 2 Now let ɛ 0. The Lebesgue dominated convergence theorem shows that lim N(x) [ϕ](x) dx = N(x) ϕ(x) dx. ɛ 0 x ɛ

9 MATH Also, as ɛ 0, ωn 1 ωn 1 ɛ 1 n S ɛ [ ] ϕ(ζ) ϕ(0) dσ(ζ) ɛ sup ϕ(x) 0; x ω2 1 ϕ log(ɛ) S ɛ n (ζ) dζ ɛ log(ɛ) sup ϕ(x) 0; x R 2 (2 n) 1 ɛ 2 n S ɛ ϕ n (ζ) dζ ɛ (2 n) 1 sup x ϕ(x) 0. Thus the limit as ɛ 0 of the right-hand side of equation (2.3) is ϕ(0), and so ϕ(0) = N(y) ϕ(y) dy. This completes the proof Solutions of the Poisson equation u = g. We first consider the case when the right hand side in this equation is a test function. Since N L 1 loc (Rn ), if ϕ C0 ( ) the convolution ϕ N is given by the absolutely convergent integral ϕ N(x) = ϕ(x y)n(y) dy, and differentiating under the integral sign, it easily follows that ϕ N C ( ) with α [ϕ N](x) = α ϕ(x y)n(y) dy = [ α ϕ] N(x). If ϕ C0 ( ), set ϕ(y) = ϕ( y) and observe that ϕ(y) = ϕ( y). Then [ϕ N](x) = ϕ(x y)n(y) dy = ϕ(y x)n(y) dy = τ x ϕ(y)n(y) dy = N, τ x ϕ = N, τ x ϕ = N, τ x ϕ = δ 0, τ x ϕ = τ x ϕ(0) = ϕ(x). There is a similar result if the right hand side in the Poisson equation is a distribution u E ( ) with compact support. We want to define u N. Note that formally u N(x)ϕ(x) dx = u(y)n(x y)ϕ(x) dy dx = u(y) Ñ(y x)ϕ(x) dx dy R 2n R n = u(y) ϕ Ñ(y) dy. Since Ñ = N, and N ϕ C ( ) if ϕ C 0 ( ), we can rigorously define the distribution u N by the formula u N, ϕ = u, ϕ N. (2.4) Then and [u N], ϕ = u N, ϕ = u, ϕ N = u, ϕ u N, ϕ = u, ϕ N = u, [ϕ N] = u, ϕ so Thus we have proved [u N] = u N = u.

10 10 MATH 825 Lemma 2.4. a) If ϕ C 0 ( ), then ϕ N C ( ), and [ϕ N] = ϕ N = ϕ. b) If u E ( ) then u N D ( ), and [u N] = u N = u. We denote by N the operator Nf = N f when the convolution is defined Hypoellipticity of. A differential operator P [f](x) = α m a α(x) α f(x) on an open set Ω is hypoelliptic if whenever u D (Ω) is a distribution such that P [u] C (Ω), it follows that u C (Ω). Theorem 2.5 (Weyl s Lemma). The Laplace operator is hypoelliptic. Proof. Let V U be any relatively compact open subset. Choose χ C 0 (Ω) with χ(x) 1 in an open neighborhood of V. The distribution χu then has compact support, and [χu] is given on V by integration against a C -function. But χu = N [χu] and hence Singular support (χu) Singular support (N) + Singular support (χu) = {0} + Singular support (χu) = Singular support (χu) \ V. Thus χu is given on V by integration against a C -function. 3. Regularity of the Laplace operator Suppose that u and g are distributions on a domain Ω, and that [u] = g. We can use the Newtonian potential to obtain information on the regularity of the solution u in terms of regularity of the given data g. Thus let χ C 0 (Ω), and suppose χ(x) 1 for x Ω 1 Ω. Then χg is a distribution with compact support, and according to part (2) of Lemma 2.3, we have [ N[χg] ] = χg. Hence [ u N[χg] ] = (1 χ)g, which is identically zero on Ω 1. It follows that the distribution u N[χg] is given by integration against an infinitely differentiable (actually real-analytic) function. Thus modulo smooth functions, u and N[χg] have the same regularity. We will discuss classical regularity results in three scales of function spaces. A. We study the behavior of the Newtonian potential N on the scale of L p spaces on. This operator maps L p ( ) to L q ( ) with q > p, and it is this sense that the operator is improving or smoothing. Our main result is the Hardy-Littlewood Sobolev theorem on fractional integration. Since we are dealing with an operator which is smoothing of positive order, cancellation conditions are not needed.

11 MATH B. We study the behavior of the Newtonian potential N on the scale of Lipschitz spaces Λ k α( ). These are spaces of functions which are k-times continuously differentiable, and all k th -order derivatives satisfy a Lipschitz or Hölder condition of order α. (This is defined precisely in Section 3.4 below). We show that for 0 < α < 1, the operator N gains two derivatives: if f Λ k α( ) then N[f] Λ k+2 α ( ). Such results are often known as Schauder estimates. C. Instead of measuring smoothness with the scale of spaces {Λ k α( )}, one can use Sobolev spaces L p k (Rn ) of functions which have k-derivatives in L p ( ). The precise definition will be given in Section 3.5, and the result is that if f L p k (Rn then N f L p k+2 (Rn ). There are many good references for a detailed discussion of these matters 4, and it is not our objective to give an exhaustive account of this material. Rather, we want to indicate that many of these results do not depend on the explicit formula for N given in Definition 2.2, but are true for any operator K[f](x) = K(x, y)f(y) dy whose Schwartz kernel K(x, y) satisfies appropriate differential inequalities and cancellation conditions. Our purpose here is to formulate one possible definition of such a more general class of operators. It is convenient to formulate this theory in terms of the notion of spaces of homogeneous type Spaces of homogeneous type. Let Ω be open, let ρ be a metric on Ω, and let dx be Lebesgue measure. We say that the triple (Ω, ρ, dx) is a space of homogeneous type if there is a constant A so that for all x Ω and all δ > 0, B(x, 2δ) A B(x, δ) where E is the Lebesgue measure of a subset E Ω, and B(x, δ) = {y Ω : ρ(x, y) < δ}. Note that the space or any open subset with the standard Euclidean metric ( n d E (x, y) = x y = (x j y j ) 2) 1 2 is a space of homogeneous type since the volume of the Euclidean ball is j=1 BE (x, δ) = ω n n δn. When n 3, there is an explicit connection between the Newtonian potential N and the Euclidean metric d E since we can write 1 N(x y) = n(n 2) d E (x, y) ( 2 B E x, de (x, y) ). At this stage, it may seem strange to introduce the inverse of the volume of the ball of radius d E (x, y) rather than simply write a constant times d E (x, y) 2 n. However, in later examples, we will see that the volume of the ball is not always equal to a power of the radius, but estimates of fundamental solutions still have this form. Using just the Euclidean distance and volume, we formulate the essential estimates for derivatives of the Newtonian potential as follows. 4 See, for example, books on elliptic partial differential equations such as [1] or [4], or books on pseudodifferential operators such as [11]. 5 In the late 1960 s and early 1970 s there were a number of authors who introduced concepts like this, but the term may have been first used in [3].

12 12 MATH 825 Lemma 3.1. Suppose n 3 and let N(x, y) = N(x y) = c n x y n+2. For any multi-indices α, β there is a constant C α,β > 0 so that for all x, y α x y β N(x, y) d E (x, y) 2 α β Cα,β ( B E x, de (x, y) ). The same inequality holds when n = 2 provided that α + β > 0. Proof. The function N is homogeneous of degree n + 2, and x α N is homogeneous of degree n α + 2. Thus the function x x n+α 2 x α N(x) is continuous and homogeneous of degree zero on the unit sphere, and hence bounded there. This gives the required estimate Calderón-Zygmund kernels of positive order. Let Ω be a connected open set and let d : Ω Ω [0, ) be a metric. For x Ω, let B(x, δ) = { y Ω : d(x, y) < δ }, and assume that B(x, δ) is an open subset of Ω for all x and δ. Assume also that there is a constant A > 0 so that for all x Ω and all δ > 0, B(x, 2δ) A B(x, δ). Let D Ω = {(x, y) Ω Ω : x = y} be the diagonal. A Calderón-Zygmund kernel of positive order m > 0 on Ω is a smooth function K C (Ω Ω \ D Ω ) such that for all multi-indices α and β there is a constant C α,β such that α x β yk(x, y) Cα,β d(x, y) m α β B ( x, d(x, y) ) 1. Lemma 3.2. If K is a Calderón-Zygmund kernel on Ω of order m > 0, the functions x K(x, y) and y K(x, y) are locally integrable. There is a constant C(A, m) > 0 so that for all x, y Ω, K(x, t) dt C(A, m) and K(t, y) dt C(A, m). t B(x,1) t B1,y) Proof. Since K is smooth away from the diagonal, it is clearly locally integrable away from the diagonal. Thus it suffices to prove the two estimates above. If B j (x) = B(x, 2 j )\B(x, 2 j 1 ), K(x, t) dt = K(x, t) dt d(x, t) m ( ) B x, d(x, t) 1 dt t B(x,1) j=0 B j(x) j=0 j=0 B j(x) 2 mj B(x, 2 j 1 ) B 1 dt 2 mj B(x, 2 j 1 ) 1 j j=0 j=0 B(x,2 j ) = 2 mj B(x, 2 j 1 ) 1B(x, 2 j ) A [2 m ] j = A[1 2 m ] 1. The second estimate is proved in the same way Hardy-Littlewood-Sobolev estimates. We study the regularity of the Newtonian potential or more general Calderón-Zygmund operators of positive order on the scale of L p -spaces. For example, if f L p ( ), we investigate for which q is it true that N[f] L q ( ) or N[f] L q loc (Rn ) where j=0 N[f](x) = N f(x) = c n x y n+2 f(y) dy. dt

13 MATH We start by observing that invariance properties of the function N easily provide a necessary condition for the existence of a constant C p,q such that for all f L p ( ), N[f] Lq ( ) C p,q f Lp ( ). (3.1) For f L p ( ) and λ > 0 set D λ f(x) = f(λx). Then Dλ f p = λ n/p f p. On the other hand, N[D λ f](x) = c n x y n+2 f(λy) dy = c n λ n x λ 1 y n+2 f(y) dy = λ 2 c n λx y n+2 f(y) dy = λ 2 D λ N[f](x). Thus if the inequality in equation (3.1) holds for all f L p, we can apply it to the functions D λ [f] and obtain λ n q N[f] q = D λ N[f] q = λ 2 ND λ [f] q C p,q λ 2 D λ [f] p = C p,q λ 2 n p f p. Thus a consequence of inequality (3.1) is the estimate λ n q + n p +2 N[f] q C p,q f p which holds for all λ > 0. If the exponent n q n p 2 0, we could let λ tend to zero or infinity, and conclude that f p = +. Thus a necessary condition for the boundedness of the operator N asserted in (3.1) is that q 1 = p 1 2n 1. This relationship between p, q, and n is also sufficient. In fact, we shall show the following more general result. Let Ω be open, let d be a metric on Ω, and let B(x, δ) = {y Ω : d(x, y) < δ}. Suppose that d satisfies i) there is a constant A > 0 so that for all x Ω and all δ > 0, B(x, 2δ) A B(x, δ). ii) there is a constant B > 0 so that for all x Ω and all δ > 0, B(x, δ) B δ ν. (For Lebesgue measure, we have ν = n.) Let m > 0 and let K : Ω Ω C be a measurable function such that for x, y Ω, Finally, let K(x, y) C d(x, y) mb ( x, d(x, y) ) 1. K[f](x) = Ω K(x, y)f(y) dy. Theorem 3.3 (Hardy-Littlewood-Sobolev). With the above notation, let 1 p < ν m. (a) If f L p (Ω) with 1 p < ν m then the integral defining K[f] converges absolutely for almost every x Ω. 1 (b) If p > 1 and if q = 1 p m ν > 0, there is a constant C p,m,ν so that for every f L p (Ω) we have K[f] Lq ( ) C p,m,µ f Lp ( ).

14 14 MATH 825 Proof. For x Ω and λ > 0 K[f](x) K(x, y) f(y) dy = Ω =: I λ (x) + II λ (x). d(x,y) λ K(x, y) f(y) dy + d(x,y)>λ We estimate the two terms separately. To estiamte I λ (x) let R j = {y Ω 2 j 1 λ < d(x, y) 2 j λ}. Then I λ (x) = C j=0 K(x, y) f(y) dy R j 2 mj λ m B(x, 2 j 1 λ) 1 j=0 = Cλ m Cλ m j=0 j=0 2 mj B(x, 2 j λ) B(x, 2 j 1 λ) B(x,2 j λ) 1 B(x, 2 j λ) 2 mj B(x, 2 j λ) B(x, 2 j 1 λ) M[f](x) f(y) dy B(x,2 j λ) K(x, y) f(y) dy f(y) dy where { 1 M[f](x) = sup B(y, δ) B(y,δ) f(t) dt : x B(y, δ) and δ > 0 } is the Hardy-Littlewood maximal operator. 6 Using the doubling property for the volumes of balls in spaces of homogeneous type, it follows that B(x, 2 j λ) A B(x, 2 j 1 λ). Thus since m > 0, [ ] I λ (x) ACλ m 2 mj M[f](x) = AC[1 2 m ] 2 λ m M[f](x) =: C 1 λ m M[f](x). (a) j=0 To estimate II λ (x), we use Hölder s inequality: II λ (x) = K(x, y) f(y) dy f [ Lp (Ω) ] 1 K(x, y) p p dy d(x,y)>λ d(x,y)>λ 6 We will use the following result. Theorem 3.4 (Hardy, Littlewood). (1) There is a constant A 1 > 0 so that if f L 1 (Ω), { x Ω : M[f](x) > λ} A1 λ 1 f L 1 (Ω). (2) For 1 < p there is a constant A p > 0 so that if f L p (Ω) then M[f] L p (Ω) and M[f] L p (Ω Ap f L p (Ω).

15 where 1 p + 1 p = 1. Let S j = {y Ω 2 j λ < d(x, y) 2 j+1 λ}, then K(x, y) p dy MATH d(x,y)>λ = j=0 S j K(x, y) p AC 2 mp λ mp j=0 = ABC 2 mp λ mp +ν(1 p ) dy Cλ mp j=0 2 mp (j+1) B(x, 2 j λ) p B(x, 2 j+1 λ) 2 mp j B(x, 2 j λ) 1 p ABC 2 mp λ mp 2 mp j (2 j λ) ν(1 p ) j=0 2 [mp +ν(1 p )]j = ABC 2 mp [ 1 2 mp +ν(1 p ) ] 1 λ mp +ν(1 p ) =: C p 2 λmp +ν(1 p ) j=0 since p = p p 1 and so mp + ν(1 p ) = mp ν p 1 < 0. Since m + ν 1 p p = m ν p, it follows that II λ (x) C 2 f Lp (Ω) λm ν p It now follows from equations (a) and (b) that there is a constant C depending only on ν, p, and m so that [ K[f](x) C λ m M[f](x) + λ m ν p ] f L. (c) p (Ω) Since M[f](x) < for almost all x Ω, it follows that the integral defining K[f](x) converges absolutely for almost all x. Choose λ so that the two terms on the right-hand side of (c) are equal; that is, let λ = f p/ν L p (Ω) M[f](x) p/ν. It follows that K[f](x) C mp f mp ν L p M[f](x)1 ν = C (Ω) f 1 p q L p (Ω) M[f](x) p q. Since M[f] p C p f p for 1 < p we have K[f] L q C f L p. (b) We make several remarks about the theorem and its proof. 1. If p = 1, if f L 1 (Ω), and if 1 q = 1 m ν, it does not follow that K[f] Lq (Ω). The proof breaks down because the Hardy-Littlewood maximal operator is not bounded on L 1. In the case of the Newtonian potential, if we take f to be the characteristic function of the unit ball, then for large x one can check that N[f](x) c x n+2, and when raised to the power 2 n 2 this just fails to be integrable. 2. Using the fact that the operator M is weak-type (1,1), it does follow that if f L 1 ( ), then K[f](x) < for almost every x, and if 1 q = 1 m ν, { } x f q K[f](x) > λ C 1 λ q. (3.2) 3. If p = ν m and f Lp ( ), it does not follow that K[f] L ( ). The proof breaks down because d(x,y) 1 K(x, y) p dy is now infinite. A discussion of the relevant examples which show that the conclusion is false can be found, for example, in [9], pages If f L p ( ), then K[f] L q loc (Rn ) if 1 q 1 p m n.

16 16 MATH Lipschitz Estimates. We next turn to the study of regularity measured on the scale of Lipschitz spaces. For the standard Euclidean metric, these are defined as follows. For 0 < α < 1, Λ α ( ) denotes the space of complex-valued functions f on for which f Λα = sup f(x) + x sup x 1 x 2 f(x 2 ) f(x 1 ) x2 x 1 α <. We say that f Λ α satisfies a Lipschitz or Hölder condition of order α. If k is a positive integer and 0 < α < 1, Λ k α( ) is the space of k-times continuously differentiable complexvalued functions f on such that every partial derivative of f of order k satisfies a Lipschitz condition of order α. We put f Λ k = sup α f(x) + β f α x Λα. 0 α <k Our objective is to show that the operators like the Newtonian potential N increases smoothness by two orders in this scale of spaces: if 0 < α < 1 and k is a non-negative integer, there is a constant C k,α so that if f Λ k α( ), then N[f] Λ k+2 α ( ) and N[f] Λ k+2 C k,α f α Λ k. (3.3) α To establish estimates of this sort, we must use more information about the Schwartz kernel N(x, y) = c n x y n+2 of the operator N than the simple size estimates used in the proof of the Hardy-Littlewood-Sobolev Theorem. There are two main points. 1. N[ϕ] = N(y)ϕ(x y) dy is given by a convolution, and hence it commutes with differentiation: α N[ϕ](x) = N(y) α ϕ(x y) dy = N[ α ϕ](x) for every test function ϕ. 2. Two derivatives of the kernel N(x) = c n x n+2 has vanishing mean value over any sphere centered at the origin. One computes 2 N x j x k (ζ) = c n n(n 2)ζ j ζ k ζ n 2 β =k c n n(n 2)ζj 2 ζ n 2 (n 2) ζ n if j k if j = k. Then in all cases, if R > 0, ζ =R 2 N x j x k (ζ) dσ(ζ) = 0. This follows if j k since the integrand is odd in both ζ j and ζ k. On the other hand, for any 1 j, k n, symmetry of the sphere shows that ζ =R ζ2 j dσ(ζ) = ζ =R ζ2 k dσ(ζ), and hence n n(n 2)ζj 2 ζ n 2 dσ(ζ) = (n 2)ζk ζ 2 n 2 dσ(ζ) ζ =R k=1 = ζ =R ζ =R (n 2) ζ n dσ(ζ).

17 MATH To formulate a result for an operator K[f](x) = K(x, y)f(y) dy which generalizes the estimate in equation (3.3) we need to make assumptions about the kernel K(x, y) which reflect the two facts listed above for the special kernel N(x, y) = c n x y n+2. Since Lipschitz spaces are defined using the standard Euclidean metric, we will restrict attention to spaces of homogeneous type where the the metric is given by d(x, y) = x y. Definition 3.5. Let m be a non-negative integer. A function K C ( \ D ) belongs to the class CZ(, m) if A) (Differential inequalities) There is a positive integer m and for any muti-indices α, β there is a constant C α,β so that α x β yk(x, y) Cα,β x y m α β B(x, x y ) 1. Here B(x, δ) is the Euclidean ball of radius δ so B(x, x y ) = cn x y n. B) (Cancellation condition) For any 0 < R <, if α + β = m x α yk(x, β ζ) dσ(ζ) = x α yk(ζ, β y) dσ(ζ) = 0. x ζ =R y ζ =R C) (Commutation property) For any multi-index α there is a family of functions {K β (x, y) : 0 β α } so that for any test function ϕ, [ ] x α K(x, y)ϕ(y) dy = K β (x, y) β ϕ(y) dy β α and each function K β C ( \ D ) also satisfies the differential inequalities and cancellation conditions in A) and B). Remark 3.6. If K CZ(, m) and if α, β are multi-indices with α + β m then α x β yk CZ(, m α β ). Theorem 3.7. Let K CZ(, m), and put K[f](x) = K(x, y)f(y) dy. Let 0 < α < 1 and R > 0. There is a constant C α,r so that if f Λ α has compact support in the ball of radius R then K[f] is m-times continuously differentiable, and K[f] Λ m α C α,r f Λα. Moreover, there is a constant C α independent of R so that if β = m then β f(x 2 ) β f(x 1 ) sup x 1 x 2 x 2 x 1 α C α f Λα. The proof of this result proceeds in several steps. 1. We first show that for β m 1, we can differentiate under the integral sign to obtain β K[f](x) = xk(x, β y)f(y) dy. This shows that K[f] is (m 1)-times continuously differentiable. 2. To show that K[f] is m-times continuously differentiable, we replace β xk(x, y) by L(x, y) CM(, 1). It then suffices to show that if L[f](x) = L(x, y)f(y) dy and if f Λ α has compact support, then for 1 j n it follows that j L[f] is continuous. In general,

18 18 MATH 825 however, we cannot just differentiate one more time under the integral sign. Rather, we show that j L[f](x) = x(x, j y) [ f(y) f(x) ] dy + x(x, j y)f(y) dy + A j (x) f(x) x y 1 x y >1 where A j is a smooth function. The function A j f is then belongs to Λ α, while the two integrals are absolutely convergent and define continuous functions. 3. Finally to show that j L[f] satisfies a Lipschitz condition of order α Proposition 3.8. Let f be continuous with compact support. Then K[f] is (m 1)-times continuously differentiable and if 0 β m 1 then β K[f](x) = xk(x, β y)f(y) dy. Moreover, if f is supported in a ball of radius R, then sup β K[f](x) CR m β sup f(x). x x Proof. Choose χ C (R) so that 0 χ(t) 1 for all t and 0 if t 1, χ(t) = 1 if t 2. For x and ɛ > 0 put ϕ ɛ (x) = χ ( ɛ 1 x ). Then ϕ ɛ is supported where x ɛ and if γ 1, γ ϕ ɛ is supported where ɛ < x < 2ɛ and and β ϕ ɛ (x) C β ɛ β 2 β C β x β. Put K ɛ [f](x) = K(x, y) ϕ ɛ (x y) f(y) dy. Then K[f](x) K ɛ [f](x) = K(x, y) [ 1 ϕ ɛ (x y) ] f(y) dy C n sup f(x) K(x, y) dy x x y <2ɛ C n sup f(x) x y m n dy x C n sup x f(x) ɛ m. x y <2ɛ Thus K ɛ [f] K[f] uniformly on as ɛ 0. On the other hand, the function x K ɛ (x, y) =: K(x, y) ϕ ɛ (x y) is infinitely differentiable. Thus K ɛ [f] is infinitely differentiable, and differentiating under the integral sign, for any β < m we have β K ɛ [f](x) = xk β ɛ (x, y) f(y) dy. Put K β [f](x) = xk(x, β y) f(y) dy.

19 MATH Then Kβ [f](x) β K ɛ (x) = C n C n β x [ K(x, y)(1 ϕɛ )(x y) ] f(y) dy sup f(x) x x y <2ɛ sup f(x) ɛ m β. x ( x y n+m β + ɛ β x y n+m) dy Thus β K ɛ [f] K β [f] uniformly on as ɛ 0. Combined with our first observation, this shows that K[f] is (m 1)-times continuously differentiable and β K[f](x) = K β [f](x) for β < m. The upper bound for β K[f](x) follows easily, completing the proof. To show that K[f] is m-times continuously differentiable when f Λ α, we have already observed that we cannot simply differentiate under the integral sign since if β = m, the kernel β xk(x, y) CZ(, 0) is not locally integrable. However if f Λ α, we do have β x K(x, y) [ f(y) f(x) ] Cn f Λα y x α n and so β xk(x, y) [ f(y) f(x) ] is locally integrable as a function of y. We use this observation in the next result. Lemma 3.9. Let L(x, y) CZ(, 1) and let f Λ α have compact support. Then L[f](x) = L(x, y)f(y) dy is continuously differentiable, and for 1 j n j L[f](x) = xl(x, j y) [ f(x) f(y) ] dy + xl(x, j y)f(y) dy + A j (x)f(x) where x y <1 A j (x) = x ζ =1 x y 1 j xl(x, ζ)(x j ζ j ) dσ(ζ). Proof. With ϕ ɛ defined as in Proposition 3.8, put L ɛ [f](x) = L(x, y)ϕ ɛ (x y)f(y) dy. Then for γ = m 1, by the same argument as in the last proposition, L[f](x) L ɛ [f](x) C ɛ sup x f(x) and so L ɛ [f] L[f] uniformly on as ɛ 0. The function L ɛ [f] is infinitely differentiable. We will show that j L ɛ [f] converges uniformly as ɛ 0 and hence L[f] is continuously differentiable. Since the kernel of L ɛ has no singularities, we have j L ɛ [f](x) = x[l(x, j y)ϕ ɛ (x y)]f(y) dy R n = x[l(x, j y)ϕ ɛ (x y)] [ f(y) f(x) ] dy x y <1 + f(x) + x y <1 x y 1 = I ɛ (x) + II ɛ (x) + III ɛ (x). j x[l(x, y)ϕ ɛ (x y)] dy j x[l(x, y)ϕ ɛ (x y)]f(y) dy

20 20 MATH 825 Now j x[l(x, y)ϕ ɛ (x y)] [ f(y) f(x) ] [ x y n + x y n+1 ɛ 1] x y α and since 1 ϕ ɛ (x y) is supported where x y < 2ɛ, it follows that I ɛ (x) xl(x, j y) [ f(y) f(x) ] dy x y <1 uniformly as ɛ 0. Also ϕ ɛ (x y) 1 if x y > 2ɛ, so if ɛ < 1 2, III ɛ (x) = xl(x, j y)f(y) dy. x y 1 It remains to study the term II ɛ (x). We have [ L(x, y)ϕ ɛ (x y) ] = xl(x, j y)ϕ ɛ (x y) + L(x, y) xϕ j ɛ (x y) j x = j xl(x, y)ϕ ɛ (x y) L(x, y) j yϕ ɛ (x y) = xl(x, j y)ϕ ɛ (x y) + yl(x, j y)ϕ ɛ (x y) y j [ L(x, y)ϕɛ (x y) ] Now ϕ ɛ (x y) is constant on the set of y where x y = r, and it follows from the cancellation hypothesis on the kernel L that xl(x, j y)ϕ ɛ (x y) dy = yl(x, j y)ϕ ɛ (x y) dy = 0. x y <1 Using the divergence theorem, we have for ɛ < 1 2 y[ j L(x, y)ϕɛ (x y) ] dy = x y <1 = x y <1 x ζ =1 x ζ =1 L(x, ζ)ϕ ɛ (x ζ) (ζ j x j ) dσ(ζ) L(x, ζ)(ζ j x j ) dσ(ζ). Hence II ɛ (x) A j (x)f(x) as ɛ 0, uniformly on. It now follows that L[f] is continuously differentiable and j L[f](x) is as given in the statement of the Lemma. To complete the proof of Theorem 3.7, it now suffices to show that if f Λ α has compact support, and if F (x) = xl(x, j y) [ f(y) f(x) ] dy + xl(x, j y) f(y) dy x y <1 x y 1 then F Λ α and F Λα C f Λα for some constant C independent of f and its support. In fact, this holds if the kernel j xl(x, y) is replaced by any kernel K satisfying (1) For all x y, K(x, y) C 0 x y n. (2) For all x y, x K(x, y) + y K(x, y) C 1 x y n 1. (3) For all 0 < R 1 < R 2 we have K(x, y) dx = R 1< x y <R 2 K(x, y) dy = 0. R 1< x y <R 2 If f Λ α has compact support, we can define K[f](x) = K(x, y) [ f(y) f(x) ] dy + x y <1 and both integrals converge absolutely. x y 1 K(x, y) f(y) dy.

21 MATH Theorem Let 0 < α < 1. There is a constant C α depending only on C 0 and C 1 so that for all f Λ α with compact support we have K[f](x 2 ) K[f(x 1 ) sup x 1 x 2 x 2 x 1 α C α f Λ α. Proof. Let x 1 x 2 and put δ = x 2 x 1. Since f has compact support, there exists R > 3δ (depending on f) so that x j y R implies f(y) = 0. Using the fact that the function y K(x, y) has mean value zero on the set 1 x y R, it follows that K[f](x j ) = K(x j, y) [ f(y) f(x j ) ] dy. x j y <R Let ψ C0 ( ) be a radial function such that ψ(x) = 1 if x R and 0 ψ(x) 1 and ψ(x) x 1 for all x. Then we can write K[f](x j ) = K(x j, y) ψ(x j y) [ f(y) f(x j ) ] dy. Then K[f](x 2 ) K[f](x 1 ) = (K ψ)(x 2, y) ( f(y) f(x 2 ) ) (K ψ)(x 1, y) ( f(y) f(x 1 ) ) dy R n = K(x 2, y) ( f(y) f(x 2 ) ) K(x 1, y) ( f(y) f(x 1 ) ) dy B(x 2,2δ) + = I + II. (K ψ)(x 2, y) ( f(y) f(x 2 ) ) (K ψ)(x 1, y) ( f(y) f(x 1 ) ) dy B(x 2,2δ) c Now B(x 2, 2δ) B(x 1, 3δ). Thus using the size estimates (1) for K we have I C0 f Λα y x 2 α K(x 2, y) dy y x 2 <2δ + C 0 f Λα y x 1 <3δ y x 1 α K(x 1, y) dy C 0ω n [ f Λα (2δ) α + (3δ) α] = C(n, α) f Λα δ α. α To deal with II, we rewrite the integrand as ( f(y) f(x1 ) )[ (K ψ)(x 2, y) (K ψ)(x 1, y) ] + [ f(x 1 ) f(x 2 ) ] (K ψ)(x 2, y). Since ψ is radial, we can use the cancellation condition (3) to conclude that B(x 22δ) c [ f(x1 ) f(x 2 ) ] (K ψ)(x 2, y) dy = 0. Thus II f Λ α y x 1 α (K ψ)(x 2, y) (K ψ)(x 1, y) dy y x 2 >2δ If y x 2 > 2δ, it follows that y x 1 y x 2 + x 2 x 1 = y x 2 + δ < y x y x 2, and so y x 1 < 3 2 y x 2.

22 22 MATH 825 Also, using the mean value theorem, it follows that (K ψ)(x 2, y) (K ψ)(x 1, y) = x 2 x 1 x (K ψ)(λx 2 + (1 λ)x 1, y) 3 2 (C 0 + C 1 ) δ y x 2 n 1. Thus ( ) 1+α II 3 (C 0 + C 1 ) f Λα 2 δ = This completes the proof. ( ) 1+α 3 (C 0 + C 1 ) 2 y x 2 n 1+α y x 2 >2δ ω n f Λα δ α = C(n, α) f Λα δ α. 1 α 3.5. Sobolev estimates. Instead of measuring the smoothness of a k th -derivatrive with a Lipschitz norm, we can also measure it with an L p -norm. Let X be a space of functions on contained in L 1 loc (Rn ). Then if f L 1 loc (Rn ) and α Z n is a multi-index, we say that x α f X in the sense of distributions if there exists a function g α X so that for every ϕ D( ) we have f(x) x α ϕ(x) dx = ( 1) α g(x)ϕ(x) dx. (3.4) If g is a locally integrable function and if g(x)ϕ(x) dx = 0 for every ϕ C 0 ( ), then g(x) = 0 for almost every x. Thus if α x f X exists, it is unique. Also, if f is smooth, then a derivative in the sense of distributions equals the classical derivative since equation (3.4) is then a consequence of integration by parts. Definition If k is a non-negative integer and 1 p, { } L p k (Rn ) = f L p ( ) x α f L p ( ) in the sense of distributions for all α k. If f L p k (Rn ), set f L p k = 0 α k α x f Lp ( ). We shall use the following properties of these Sobolev spaces. Lemma (1) The space L p k (Rn ) is complete with the norm introduced in Definition 3.11, and so is a Banach space. L 2 k (Rn ) is a Hilbert space with the inner product (f, g) = ( x α f, x α g) L 2 ( ). 0 α k (2) A function f L p k (Rn ) if and only if there is a sequence of functions f n L p ( ) C ( ) such that for every α k, the sequence { α x f n } is Cauchy in L p ( ). (3) A function f L 2 k (Rn ) if and only if f(ξ) 2 (1 + ξ 2 ) k dξ < + where f is the Fourier transform of f. (4) If f L p k (Rn ) and if ψ C 0 ( ), then ψf L p k (Rn ). Thus membership in L p k (Rn ) depends on the local behavior of f.

23 MATH ] We have seen that the operator which takes a function f to j,k[ 2 N[f] can be written as an operator of the form K[ϕ](x) = K(y) [ ϕ(x y) ϕ(x) ] dy + K(y) ϕ(x y) dy y <R y R where K C 1 ( {0}) and we assume that (1) For all x 0, K(x) C 0 x n. (2) For all x 0, K(x) C 1 x n 1. (3) For all 0 < R 1 < R 2 we have K(x) dx = 0. R 1< x <R 2 We want to show that the operator K, defined for ϕ C0 ( ), has a bounded extension to L 2 ( ). Since this operator is given by convolution with a distribution K, it is natural to use the Fourier transform and the Plancherel theorem to reduce the problem to showing that the Fourier transform of the distribution is uniformly bounded. This can certainly be done, but since we will not have the Fourier transform available in later examples, we prefer to use a method with wider application. This is based on an almost orthogonality argument, and the key result is the following. Theorem 3.13 (Cotlar-Stein). Let {T j }, j Z, be bounded operators on a Hilbert space H. Assume there are constants C and ɛ > 0 so that for all j, k Z we have (1) T j C. (2) Tj T k C 2 ɛ j k. (3) T j Tk C 2 ɛ j k. There is a constant A so that for all N N j= N T j A. This is proved, for example, in [10], pages , so we do not reproduce the argument here. Now let χ C 0 (R) satisfy (i) 0 χ(t) 1 for all t R; (ii) We have 0 if t 1 8, χ(t) = 1 if 1 4 t 1 2, 0 if 1 t. Put χ j (t) = χ(2 j t). Then each t > 0 is in the support of at most 4 of the functions {χ j }. Put Ψ j (t) = [ χ k (t) ] 1 χj (t). Then Ψ j is supported on 2 j 3 t 2 j, and (iii) Ψ j (t) 1 for all t > 0. (iv) j= k N Ψ j is supported on 2 N 3 t 2 N, and is identically 1 for 2 N t 2 N 3. j= N

24 24 MATH 825 Now put K j (x) = Ψ j ( x ) K(x). Then it is not difficult to check that {K j }, j = 0, ±1, ±2,... is a doubly infinite sequence of continuously differentiable functions on, and there is a constant C so that (1) For all j Z and all x we have K j (x) C 2 nj. (2) For all j Z and all x we have K j (x) C 2 (n+1)j. (3) For all j Z, K j is supported in the ball B E (0; 2 j ). (4) For all j Z, K j (x) dx = 0. Moreover (5) K j (x) = K(x) for x 0. j= (6) K [N] = N j= N K j is supported 2 N 3 x 2 N, and is identically equal to K for 2 N x 2 N 3. Moreover, the kernel K [N] satisfies the same conditions (1), (2), and (3) as K with constants that are independent of N. Set K j [f](x) = K j ϕ(x) = K j (y) f(x y) dy. It follows that if ϕ C0 ( ), we have K[ϕ](x) = K(y) [ ϕ(x y) ϕ(x) ] dy + K(y) ϕ(x y) dy y <R = lim N = lim N j= N N N j= N K j (y) ϕ(x y) dy K j [ϕ](x). Thus if we can show that the operators {K j } satisfy the almost orthogonality conditions of Theorem 3.13, it follows from Fatou s lemma that for ϕ C 0 ( ) we have K[ϕ] 2 = L 2 ( It follows from (1) and (3) that lim N lim sup N +N j= N +N A 2 ϕ 2 L 2 ( ). j= N y 1 K j [ϕ](x) 2 dx K j [ϕ](x) 2 dx Kj L1 ( ) = x <2 j Kj (x) dx C 2 nj BE (0; 2 j ) C n 1 ω n since n 1 ω n is the volume of the Euclidean unit ball in. Now if f, g L 1 ( ), we always have f g L 1 = f g(x) dx f(x y) g(y) dy dx = f L 1 g L 1,

25 MATH and consequently K j K k L 1 (C n 1 ω n ) 2. However, the differential inequality (2) and the cancellation property (4) allow us to get a better estimate when j k. Proposition For all j, k Z we have Kj K k L 1 2 n (C n 1 ω n ) 2 2 j k. Proof. Without loss of generality, assume that j < k. Then K j K k (x) = K k (x y) K j (y) dy R n [ = Kk (x y) K k (x) ] K j (y) dy (using assumption (4)) R n K k (x y) K k (x) K j (y) dy B E (0;2 j ) B E (0;2 j ) y sup z K k (z) K j (y) dy (Mean Value Theorem) C 2 j (n+1)k Kj L 1 (using estimate (3)). On the other hand, if K j K k (x) 0 we must have y 2 j and x y 2 k, so x x y + y 2 2 k. Thus Kj K k L 1 C 2 j (n+1)k Kj L 1 B(0; 2 2 k ) 2 n (C n 1 ω n ) 2 2 j k. This completes the proof. Lemma Let {K j } be functions satisfying conditions (1) through (4), and for each j Z define an operator T j by setting T j [f](x) = K j f(x) = K j (x y) f(y) dy. There exists a constant C so that for any integer N we have N T j [f] C f L. L 2 ( ) 2 ( ) j= N 3.6. L 1 -estimates. It is not true that the operator K from section 3.5, defined on the space C 0 ( ), extends to a bounded operator on L 1 ( ), but we do have the following replacement, which is called a weak type (1,1) estimate. Let us write K [N] = +N j= N K j. We will need the following estimate, which is sometimes called the Calderón-Zygmund estimate. Lemma Let η > A 2, let B = B ρ (x 0 ; δ 0 ), and let B = B ρ (x 0 ; η δ 0 ). Suppose x 1, x 2 B. Then B K [N] (y x 1 ) K [N] (y x 2 ) dy C. Theorem There is a constant A independent of N so that if f L 1 ( ), then { x } K [N] A [f](x) > α f λ. L1 ( ) Proof. We apply the Calderón-Zygmund } decomposition of Theorem to the function f. Let E α = {x M[f](x) > α. Then we can write f = g + j b j = g + b where g and {b j } satisfy

26 26 MATH 825 (1) b j is supported on a ball B E (x j ; δ j ) E α and b j (x) dx = 0 while b j L1 ( ) C α Bj; (2) If x / E α, then x x j 2δ j ; (3) j BE (x j ; δ j ) C Eα C f L α 1 ( ) ; (4) g(x) C α for almost all x. Since K [N] [f](x) K [N] [g](x) + K [N] [b](x), we have { x K [N] [f](x) > α} { x K [N] [g](x) > α }{ x K [N] [b](x) > α }. 2 2 Now K [N] is a bounded operator on L 2 ( ) with norm A independent of N. Thus α 2 4 { x K [N] [g](x) > α } K [N] [g](x) 2 dx 2 A g 2 L 2 A C α f L 1 and so { x K [N] [g](x) > α 2 } 4AC α f L 1. Then Now suppose we can show that E α K [N] [b](x) dx C f L 1. (3.5) { x K [N] [b](x) > α } { 2 E α + x E α K [N] [b](x) > α } 2 C 1 f L K [N] [b](x) dx α α E α C f α L 1, which would complete the proof. Thus the key is to prove the estimate is equation (3.5). Suppose x / E α. Since the integral of b j is zero, we have K [N] [b j ](x) = K R [N] (x y) b j (y) dy n [ = K [N] (x y) K(x x j ) ] b j (y) dy R n K [N] (x y) K [N] (x x j ) bj (y) dy

27 MATH and hence It follows that K [N] [b j ](x) dx E α [ K [N] (x y) K [N] (x x j ) ] dx b j (y) dy E α C b j (y) dy C α B j. K [N] [b](x) dx E α K [N] [b j ](x) dx j E α C α B j j C f L 1. Appendix A. Test functions and distributions A.1. Some spaces of smooth functions. Let Ω be open. If m is a non-negative integer, C m (Ω) denotes the space of m-times continuously differentiable functions on Ω and C (Ω) = m=1 Cm (Ω) the space of infinitely differentiable functions on Ω. Note that C m (Ω) C m+1 (Ω) C (Ω). If K Ω, we put f K,m = sup sup α m x K α ϕ(x). (A.1) Then K,m is a semi-norm on the space C m (Ω) and also on the space C (Ω). Define a topology on C (Ω) so that a sequence {f n } C (Ω) converges to a limit function f 0 C (Ω) if and only if lim n f0 f n K,m = 0 for every compact subset K Ω and for every non-negative integer m. The space C (Ω) equipped with this topology is denoted by E(Ω). If {f n } E(Ω) is a sequence which is Cauchy in every semi-norm K,m, then the sequence converges to a limit f 0 E(Ω). If f C m (Ω), the support of f is the closure of the set of points x Ω such that f(x) 0, and is denoted by suppt(f). Then C 0 (Ω) is the subspace of C (Ω) consisting of functions whose support is a compact subset of Ω. There is a topology on C 0 (Ω) such that a sequence {f n } C 0 (Ω) converges to a limit function f 0 C 0 (Ω) if and only there is a compact set K Ω so that suppt(f n ) K for all n 0 and lim n f 0 f n K,m = 0 for every non-negative integer m. The space C 0 (Ω) with this topology is denoted by D(Ω). 7 The Schwartz space S( ) consists of all ϕ C ( ) such that for all multi-indices α and β, sup x x α β ϕ(x) = ϕ α,β < +. The natural topology on S( ) is given by the collection of these semi-norms. Thus a sequence {ϕ 1, ϕ 2,...} in S( ) converges to ϕ 0 if and only if lim n ϕn ϕ 0 α,β = 0. In case Ω =, the inclusions of spaces D( ) S( ) E( ) 7 Note that this topology is not the same as the topology that C 0 (Ω) inherits as a subspace of E(Ω).