Introduction...2011/03/24 · Volume 24, Number 3, July 2011, Pages 849–869 S 0894-0347(2011)00698-X Article electronically published on March 24, 2011 REGULARITY THEORY FOR PARABOLIC

JOURNAL OF THEAMERICAN MATHEMATICAL SOCIETYVolume 24, Number 3, July 2011, Pages 849–869S 0894-0347(2011)00698-XArticle electronically published on March 24, 2011

REGULARITY THEORY

FOR PARABOLIC NONLINEAR INTEGRAL OPERATORS

LUIS CAFFARELLI, CHI HIN CHAN, AND ALEXIS VASSEUR

1. Introduction

The purpose of this work is to develop a regularity theory for nonlocal evolutionequations of variational type with “measurable” kernels. More precisely, we considersolutions of the evolution equations of the type

(1.1) wt(t, x) =

∫[w(t, y)− w(t, x)]K(t, x, y) dy,

where all that is required of the kernel K is that there exists 0 < s < 2 and 0 < Λ,such that

(1.2)

symmetry in x, y : K(t, x, y) = K(t, y, x) for any x �= y,

1{|x−y|≤3}1

Λ|x− y|−(N+s) ≤ K(t, x, y) ≤ Λ|x− y|−(N+s).

The symmetry of the kernel K makes of the operator∫[w(y)− w(x)]K(x, y) dy

the Euler-Lagrange equation of the energy integral

E(w) =

∫ ∫[w(x)− w(y)]2K(x, y) dx dy.

It suggests a mathematical treatment based on the De Giorgi-Nash-Moser ideas[12, 20] from the calculus of variations. In fact, one of the immediate applicationsof our result is to nonlinear variational integrals

Eφ(w) =

∫ ∫φ(w(x)− w(y))K(x− y) dx dy,

for φ a C2 strictly convex functional. Indeed, the fact that K(x, y) has the specialform K(x−y) makes the equation translation invariant, and as in the second-ordercase, this implies that first derivatives of w satisfy an equation of the type (1.1). Our

Received by the editors March 8, 2010 and, in revised form, August 2, 2010, October 26, 2010,and December 17, 2010.

2010 Mathematics Subject Classification. Primary 35B65, 45G05, 47G10.Key words and phrases. Nonlinear partial differential equation, nonlocal operators, integral

variational problems, De Giorgi methods, image and signal processing.The first author was partially supported by the NSF.The third author was partially supported by both the NSF and the EPSRC Science and

Innovation award to the Oxford Centre for Nonlinear PDE (EP/E035027/1).

c©2011 American Mathematical SocietyReverts to public domain 28 years from publication

849

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850 LUIS CAFFARELLI, CHI HIN CHAN, AND ALEXIS VASSEUR

results are basically that solutions with initial data in L2 become instantaneouslybounded and Holder continuous.

Kassmann showed previously a similar result for the stationary case in a veryinteresting paper where he developed the corresponding Moser scheme [15]. Ina later paper [1], Barlow, Bass, Chen, and Kassmann considered a similar time-dependent equation with less restrictive assumptions on the kernel. Under theirassumptions, they constructed an example where Holder continuity does not hold,due to the fact that their equations are of variable order and their assumptions arenot scale invariant. Note, however, that they do prove a Harnack inequality undertheir general assumptions. For the time-dependent case, a version of the result wasproved by Komatsu [17], using the corresponding Nash scheme (see also Chen [10]).

Other results have been obtained for the nondivergence case. Along these lines,see Bass and Kassmann [3, 2] (see also [4]). There are also recent works of Silvestre(see [21], [7], and the references therein).

We were motivated by our work on Navier-Stokes [23] and the quasi-geostrophicequations [8]. In this work, the full regularity of the solutions to the surface quasi-geostrophic equation is shown in the critical case. It was followed by several workson the same subject in the super-critical case (see for instance [11]). Note also thatthe result was obtained, using completely different techniques by Kiselev, Nazarovand Volberg [16]. Our approach led to some progress in the supercritical case (see[22, 9]). It follows pretty much the lines of De Giorgi’s work [12] with a differentlocalization scheme. Nonlinear equations of this form appear extensively in thephase transition literature (see Giacomin, Lebowitz, and Presutti [13]) and morerecently on issues of image processing (see Gilboa and Osher [14]).

2. Presentation of the results

Consider the variational integral

V (θ) =

∫RN

∫RN

φ(θ(y)− θ(x))K(y − x)dydx,

for φ : R → [0,∞) an even convex function of class C2(R) satisfying the conditions

(2.1)φ(0) = 0,

Λ−1/2 ≤ φ′′(x) ≤ Λ1/2, x ∈ R,

for a given constant 1 < Λ < ∞.The kernel K : RN −{0} → (0,∞) is supposed to satisfy the following conditions

for 0 < s < 2:

(2.2)

K(−x) = K(x), for any x ∈ RN − {0},

1{|x≤3}Λ−1/2

|x|N+s≤ K(x) ≤ Λ1/2

|x|N+s, for any x ∈ R

N − {0}.

With the above setting, the corresponding Euler-Lagrange equation for the varia-tional integral

∫RN

∫RN φ(θ(y)− θ(x))K(y − x)dydx is given by

−∫RN

φ′(θ(y)− θ(x))K(y − x)dy = 0.

We are considering in this paper the associated time-dependent problem:

(2.3) ∂tθ(t, x)−∫RN

φ′(θ(t, y)− θ(t, x))K(y − x)dy = 0.


REGULARITY THEORY FOR PARABOLIC NONLINEAR INTEGRAL OPERATORS 851

The main goal of this paper is to address the regularity problem for solutions tothe above parabolic-type equation and establish the following main theorem.

Theorem 2.1. Consider an even convex function φ verifying Hypothesis (2.1)and a kernel K verifying Hypothesis (2.2) for 0 < s < 2. Then, for any initialdatum θ0 ∈ H1(RN ), there exists a global classical solution to equation (2.3) withθ(0, ·) = θ0 in the sense that ‖θ(t, ·) − θ0‖L2(RN ) → 0, as t → 0. Moreover ∇xθ ∈Cα((t0,∞)× R

N ) for any t0 > 0.

The existence of weak solutions with nonincreasing energy can be constructedfollowing [5]. To address the regularity problem for solutions to equation (2.3),we follow the classical idea of De Giorgi and look at the first derivative Dθ of asolution θ to equation (2.3). First, we use the change of variable y = x + z torewrite equation (2.3) as follows:

(2.4) ∂tθ −∫RN

φ′(θ(x+ z)− θ(x))K(z)dz = 0.

Now, we consider w = Deθ, the derivative in the direction e of θ. Derivating(formally) equation (2.3) in the direction e we find

∂tw −∫RN

φ′′(θ(x+ z)− θ(x)){w(x+ z)− w(x)}K(z)dz = 0.

We then perform a change of variable back to y = x + z to rewrite the aboveequation in the following way:

∂tw −∫RN

φ′′(θ(y)− θ(x)){w(y)− w(x)}K(y − x)dz = 0.

Consider the new kernelK(t, x, y) = φ′′(θ(t, y)−θ(t, x))K(y−x) (with an obviousslight abuse of notation). Since φ is an even function, φ′′ is also an even function,and hence the new kernel K(t, x, y) is symmetric in x and y. Moreover, Hypotheses(2.2) and (2.1) imply that K(t, x, y) satisfies the condition

1{|x−y|≤3}Λ−1

|x− y|N+s≤ K(t, x, y) ≤ Λ

|x− y|N+s.

As a result, the function w = Deθ satisfies equation (1.1) with the kernel K(t, x, y)verifying Hypothesis (1.2). Our goal is then to show that the solutions to equation(1.1) are in Cα.

To make the argument rigorous, we will consider the difference quotientDhe θ(·) =

1h{θ(·+he)− θ(·)}. We use again the version (2.4) of equation (2.3). For any given

η ∈ C∞c (RN ), we use the difference quotient D−h

e η to test against it, and we get∫RN

∂tθ(t, x)D−he η(x)dx−

∫RN

∫RN

φ′(θ(t, x+ z)− θ(t, x))D−he η(x)dxK(z)dz = 0.

Using discrete integration by parts,∫RN f(x)D−h

e g(x)dx=−∫RNDh

e f(x)g(x)dx,we find∫

RN

∂tDhe θ(t, x) · η(x)dx−

∫RN

∫RN

Dhe [φ

′(θ(·+ z)− θ(·))](x) · η(x)dxK(z)dz = 0.

The change of variable y = x+ z leads to∫RN

∂tDhe θ(t, x) ·η(x)dx−

∫RN

∫RN

Dhe [φ

′(θ(·+y−x)−θ(·))](x)·η(x)K(y−x)dxdy=0.



Note that φ is an even function, so φ′ is an odd function and consequently

Dhe [φ

′(θ(·+ y − x)− θ(·))](x) = −Dhe [φ

′(θ(·+ x− y)− θ(·))](y).

Using also the symmetry of K, we can symmetrize the operator to get

∫RN

∂tDhe θ(t, x) · η(x)dx

(2.5)

− 1

2

∫RN

∫RN

Dhe [φ

′(θ(·+ y − x)− θ(·))](x) · [η(x)− η(y)]K(y − x)dxdy = 0.

Setting Y = θ(y + he)− θ(x+ he) and X = θ(y)− θ(x), we get

Dhe [φ

′(θ(·+ y − x)− θ(·))](x) = 1

h{φ′(θ(y + hei)− θ(x+ hei))− φ′(θ(y)− θ(x))}

=Y −X

h

∫ 1

0

φ′′(X + s(Y −X)) ds

= [Dhe θ(y)−Dh

e θ(x)]

∫ 1

0

φ′′((1− s)[θ(t, y)− θ(t, x)]

+ s[θ(t, y + he)− θ(t, x+ he)]) ds.

Hence, w = Dhe θ solves the following equation:∫

RN

∂tw(t, x)η(x)dx+

∫RN

∫RN

Kh(t, x, y)[η(x)− η(y)][w(t, x)− w(t, y)]dydx = 0,

where

Kh(t, x, y) = K(y−x)

∫ 1

0

φ′′((1−s)[θ(t, y)−θ(t, x)]+s[θ(t, y+he)−θ(t, x+he)]) ds.

Note that this new kernel verifies independently on h the properties (1.2) with thesame Λ.

Theorem 2.1 is then a consequence of the following theorem.

Theorem 2.2. Let w be a weak solution of (1.1) with a kernel satisfying the prop-erties (1.2). Then for every t0 > 0, w ∈ Cα((t0,∞) × R

N ). The constant α andthe norm of w depend only on t0, N , ‖w0‖L2 , and Λ.

Passing into the limit h → 0 gives the result of Theorem 2.1. The rest of thepaper is dedicated to the proof of Theorem 2.2.

Remark. 1. With the Hypothesis of Theorem 2.1, if in addition θ0 ∈ H1+s(RN ),then ∂tθ(0, ·) ∈ L2(RN ). Then, applying again Theorem 2.2 on w = ∂tθ showsthat ∂tθ ∈ Cα and so θ ∈ C1,α on (t0,∞) × R

N for any t0 > 0. As an alternativeto the incremental quotient, an approximation can be used to justify this formalargument. We provide such an approximation scheme in the appendix.

2. If s ≤ 1, the theorem shows that the solutions are classical. For 1 < s < 2,if the function φ ∈ C3,s−1+ε, and K(z) = |z|−(N+s), then ∇θ is C1,α in time, andCs,α in x. In this case, the solution is also classical. We provide the proof of thisfact in the appendix.



3. The first De Giorgi’s lemma

In this section and the next section, we focus on the differential equation statedin the sense of weak formulation in (1.1). We rewrite it in the following way:

(3.1)

∫RN

∂tw(t, x) · η(x)dx+B[w(t, ·), η] = 0, ∀η ∈ C∞c (RN ),

B[u, v] =

∫RN

∫RN

K(t, x, y)[u(x)− u(y)] · [v(x)− v(y)]dxdy,

where the kernel K(t, x, y) is assumed to satisfy the Hypothesis (1.2). We firstintroduce the following function ψ:

(3.2) ψ(x) = (|x| s2 − 1)+.

For any L ≥ 0, we define

(3.3) ψL(x) = L+ ψ(x).

With the above setting, the first De Giorgi’s lemma is as follows.

Lemma 3.1. Let Λ be the given constant in condition (1.2). Then, there existsa constant ε0 ∈ (0, 1), depending only on N , s, and Λ, such that for any solutionw : [−2, 0]× R

N → R to (3.1), the following implication for w holds true.If it is verified that

∫ 0

−2

∫RN

[w(t, x)− ψ(x)]2+dxdt ≤ ε0,

then we have

w(t, x) ≤ 1

2+ ψ(x)

for (t, x) ∈ [−1, 0] × RN . (Hence, we have in particular that w ≤ 1/2 on [−1, 0] ×

B(1) .)

The main difficulty in our approach is due to the nonlocal operator. In [8], alocalization of the problem was performed at the cost of adding one more variableto the problem. This was based on the “Dirichlet to Neuman” map. This approachstill works for any fractional Laplacian (see Caffarelli and Silvestre [6]). Howeverit breaks down for general kernels as (1.2). Instead, we keep track of the far awaybehavior of the solution via the function ψ.

Remark. All the computations on weak solutions in the proof can be justifiedby replacing the variable kernel in a neighborhood of the origin by the fractionalLaplacian through a smooth cutoff, and smoothing out the kernel outside of thisneighborhood (in the same spirit as the approximation scheme provided in the ap-pendix). Then the equation becomes a fractional heat equation with a smoothright-hand side, thus C2 in space. This makes the integrals involved uniformlyconvergent. Once the a priori Holder continuity is proven, we pass to the limit. Ac-tually, such an approximation can be performed for the original nonlinear problemof Theorem 2.1. We provide a more detailed proof of this in the appendix.

Proof. We split the proof into several steps.



First step: Energy estimates. Let w : [−2, 0] × RN → R be a solution to

equation (3.1). For 0 ≤ L ≤ 1, we consider the truncated function [w − ψL]+,where ψL is defined by (3.3). Then, we take the test function η to be [w−ψL]+ inthe weak formulation of equation (3.1), which gives

0=1

2

d

dt

∫RN

[w − ψL]2+dx+B[w, (w − ψL)+]

=1

2

d

dt

∫RN

[w−ψL]2+dx+B[(w − ψL)+, (w − ψL)+]+B[(w − ψL)−, (w − ψL)+]

+B[ψL, (w − ψL)+].

(3.4)

Now, due to the observation that (w− ψL)+ · (w−ψL)− = 0 and the symmetryof K in x, y, we have

B[(w − ψL)−, (w − ψL)+] = 2

∫RN

∫RN

K(t, x, y)(w − ψL)+(x)(w − ψL)neg(y)dxdy,

where we denote (w − ψL)neg = −(w − ψL)− ≥ 0 . In particular,

B[(w − ψL)−, (w − ψL)+] ≥ 0.

This “good term” is not fully exploited in this section. It will be used in a crucialway in the next section. The remainder can be written as:

B[ψL, (w − ψL)+]

=1

2

∫ ∫|x−y|≥1

K(t, x, y)[ψL(x)− ψL(y)] · {(w − ψL)+(x)− (w − ψL)+(y)}dxdy

+1

2

∫ ∫|x−y|<1

K(t, x, y)[ψL(x)− ψL(y)] · {(w − ψL)+(x)− (w − ψL)+(y)}dxdy.

(3.5)

Using the inequality |ψ(x) − ψ(y)| ≤ 2|y − x| s2 , for any x and y with |y − x| ≥ 1,we get the following estimation of the “far-away” contribution:∣∣∣∣∣

∫ ∫|x−y|≥1

K(t, x, y)[ψL(x)− ψL(y)] · [w − ψL]+(x)dxdy

∣∣∣∣∣=

∣∣∣∣∣∫ ∫

|x−y|≥1

K(t, x, y)[ψ(x)− ψ(y)] · [w − ψL]+(x)dxdy

∣∣∣∣∣≤

∫RN

∫|y−x|≥1

2Λ

|x− y|N+s2|y − x| s2 dy · (w − ψL)+(x)dx

= 4Λ|SN−1|∫ ∞

1

r−s2 dr

∫RN

(w − ψL)+(x)dx ≤ C

∫RN

(w − ψL)+(x)dx.

By symmetry we end up with

∣∣∣∣∣∫ ∫

|x−y|≥1

K(t, x, y)[ψL(x)− ψL(y)] · {(w − ψL)+(x)− (w − ψL)+(y)}dxdy∣∣∣∣∣

≤ C

∫RN

(w − ψL)+(x)dx.

(3.6)



The other part of the remainder can be controlled in the following way:

∣∣∣∣∣∫ ∫

|x−y|<1

K(t, x, y)[ψL(x)− ψL(y)] · {(w − ψL)+(x)− (w − ψL)+(y)}dxdy∣∣∣∣∣

≤ 2

∫ ∫|x−y|<1

K(t, x, y)χ{[w−ψL](x)>0}|ψL(x)− ψL(y)|

· |(w − ψL)+(x)− (w − ψL)+(y)|dxdy

(3.7)

where, in the above inequality, we have used the fact that

|(w − ψL)+(x)− (w − ψL)+(y)|≤ {χ{[w−ψL](x)>0} + χ{[w−ψL](y)>0}}|(w − ψL)+(x)− (w − ψL)+(y)|,

and the symmetry in x and y.Now, by Holder’s inequality, and using the elementary inequality |ψ(y)−ψ(x)| <

|y − x| , for any x, y in RN , we can have the following estimation:

2

∫ ∫|x−y|<1

K(t, x, y)χ{[w−ψL](x)>0}|ψL(x)− ψL(y)|

· |(w − ψL)+(x)− (w − ψL)+(y)|dxdy

≤ a ·∫ ∫

|x−y|<1

K(t, x, y){(w − ψL)+(x)− (w − ψL)+(y)}2dydx

+1

a·∫ ∫

|x−y|<1

K(t, x, y)|ψ(x)− ψ(y)|2 · χ{[w−ψL](x)>0}dydx,

(3.8)

in which the arbritary a > 0 will be chosen later. Finally

∫ ∫|x−y|<1

K(t, x, y)|ψ(x)− ψ(y)|2dy · χ{[w−ψL](x)>0}dx

≤∫RN

∫|x−y|<1

2Λ

|x− y|N+s|y − x|2dy · χ{[w−ψL](x)>0}dx = Cs

∫RN

χ{[w−ψL](x)>0}dx.

Pulling this inequality in (3.7) with a = 1/2, and gathering it together with (3.5),(3.6), (3.7), we can rewrite the energy inequality as

(3.9)

d

dt

∫RN

[w − ψL]2+dx+

1

2B[(w − ψL)+, (w − ψL)+]

≤ CN,Λ,s{∫RN

(w − ψL)+(x)dx+

∫RN

χ{[w−ψL](x)>0}dx},

where CN,Λ,s is some universal constant depending only on N and Λ and s. Next,in order to employ the Sobolev embedding theorem, we need to compare



B[(w − ψL)+, (w − ψL)+] with ‖(w − ψL)+‖2H

s2 (RN )

as follows:

‖(w − ψL)+‖2H s2 (RN )

=

∫ ∫|x−y|≤2

{(w − ψL)+(x)− (w − ψL)+(y)}2|x− y|N+s

+

∫ ∫|x−y|>2

{(w − ψL)+(x)− (w − ψL)+(y)}2|x− y|N+s

≤ Λ ·B[(w − ψL)+, (w − ψL)+]

+ 2

∫ ∫|x−y|>2

1

|x− y|N+s{(w − ψL)

2+(x) + (w − ψL)

2+(y)}dxdy

≤ Λ ·B[(w − ψL)+, (w − ψL)+] + C

∫RN

(w − ψL)2+dx.

Hence,

d

dt

∫RN

[w − ψL]2+dx+

1

Λ‖(w − ψL)+‖2H s

2 (RN )

≤ CN,Λ,s{∫RN

(w − ψL)+dx+

∫RN

χ{w−ψL>0}dx+ |∫RN

(w − ψL)2+dx}.

(3.10)

Second step: Nonlinear recurrence. From this energy inequality, we establisha nonlinear recurrence relation to the following sequence of truncated energy:

Uk = supt∈[Tk,0]

∫RN

(w − ψLk)2+(t, x)dx+

∫ 0

Tk

‖(w − ψLk)+(t, ·)‖2H s

2 (RN )dt,

where, in the above expression, Tk = −1 − 12k

and Lk = 12 (1 −

12k). Moreover, we

will use the abbreviation Qk = [Tk, 0]× RN .

Now, let us consider two variables σ, t that satisfy Tk−1 ≤ σ ≤ Tk ≤ t ≤ 0. Bytaking the time integral over [σ, t] in inequality (3.10), we obtain

∫RN

[w − ψLk]2+(t, x)dx+

∫ t

σ

‖(w − ψLk)+‖2H s

2 (RN )ds

≤∫RN

[w − ψLk]2+(σ, x)dx

+ CN,Λ,s{∫ t

σ

∫RN

(w − ψLk)+ + χ{w−ψLk

>0} + (w − ψLk)2+dxds}.

Next, by first taking the average over σ ∈ [Tk−1, Tk], and then taking the sup overt ∈ [Tk, 0] in the above inequality, we deduce from the above inequality that

(3.11) Uk ≤ 2k(1 +CN,s,Λ){∫Qk−1

(w− ψLk)+ + χ{w−ψLk

>0} + (w− ψLk)2+dxds}.

We now use the very classical Sobolev embedding for fractional spaces (see for

instance [18]). The Sobolev embedding theorem Hs2 (RN ) ⊂ L

2NN−s (RN ) and inter-

polation give

‖(w − ψLk)+‖L2(1+ s

N)(Qk)

≤ CNU12

k .



Using the Tchebychev inequality we get∫Qk−1

(w − ψLk)+ ≤

∫Qk−1

(w − ψL)+χ{w−ψLk−1> 1

2k+1 }

≤ (2k+1)1+2sN

∫Qk−1

(w − ψLk−1)2(1+ s

N )+

≤ (2k+1)1+2sN C

2(1+ sN )

N U1+ s

N

k−1 ;∫Qk−1

χ{w−ψLk>0} ≤ (2k+1)2(1+

sN )

∫Qk−1

(w − ψLk−1)2(1+ s

N )+

≤ (2k+1)2(1+sN )C

2(1+ sN )

N U1+ s

N

k−1 ;∫Qk−1

(w − ψLk)2+ ≤

∫Qk−1

(w − ψL)2+χ{w−ψLk−1

> 1

2k+1 }

≤ (2k+1)2sN

∫Qk−1

(w − ψLk−1)2(1+ s

N )+

≤ (2k+1)2sN C

2(1+ sN )

N U1+ s

N

k−1 .

The above three inequalities, together with inequality (3.11), give

(3.12) Uk ≤ {CN,Λ,s}kU1+ s

N

k−1 , ∀k ≥ 0,

for some universal constant CN,Λ,s depending only on N , s, and Λ. Due to thenonlinear recurrence relation (3.12) for Uk, we know there exists some sufficientlysmall universal constant ε0 = ε0(CN,Λ,s), depending only on CN,Λ,s, such that thefollowing implication is valid.

If U1 ≤ ε0, then it follows that limk→∞ Uk = 0.Equation (3.11) with the Tchebychev inequality gives that

U1 ≤ C

∫ 0

−2

∫RN

|w − ψ|2 dx dt,

and Uk converges to 0 implies that

w ≤ ψ +1

2t ∈ [−1, 0]× R

N . �

We have the following corollary of Lemma 3.1. It shows that any solutions areindeed bounded for t > 0.

Corollary 3.2. Any solution to (1.1) with initial value in L2(RN ) is uniformlybounded on (t0,∞)× R

N for any 0 < t0 < 2. Indeed:

supt>t0,x∈RN

|w(t, x)| ≤ ‖w0‖L2

2√ε0(t0/2)(N/s+1)/2

.

Proof. Fixing 0 < t0 < 2 and x0 ∈ RN , for any t > −2, x ∈ R

N , we consider

w(t, x) =(t0/2)

(N/s+1)/2√ε0‖w0‖L2

w(t0 + t(t0/2), x0 + x(t0/2)1/s).

The function w still satisfies equation (3.1) with another kernel verifying Hypothesis(1.2) with the same constant Λ. From the decreasing of energy, w satisfies theassumptions of Lemma 3.1. Hence w(0, 0) ≤ 1/2. Working with −w gives that−w(0, 0) ≤ 1/2 too. �



We define ψ(x) = (|x|s/4 − 1)+. We can rewrite the main lemma of this sectionin the following way. It will be useful for the next section.

Corollary 3.3. Let Λ be the given constant in condition (1.2). Then, there existsa constant δ ∈ (0, 1), depending only on N , s, and Λ, such that for any solutionw : [−2, 0]× R

N → R to (3.1) satisfying

w(t, x) ≤ 1 + ψ(x) on [−2, 0]× RN

and

|{w > 0} ∩ ([−2, 0]×B2)| ≤ δ,

we have

w(t, x) ≤ 1

2, (t, x) ∈ [−1, 0]× B1.

Proof. Consider R ≥ 21s such that 1 + ψ(|y| + 1) ≤ ψ(|y|), for any |y| ≥ R. Note

that R depends only on s. For any (t0, x0) ∈ [−1, 0]×B1 we introduce wR definedon (−2, 0)× R

N by

wR(s, y) = w(t0 +s

Rs, x0 +

y

R).

Note that wR satisfies equation (3.1) with another kernel KR(t, x, y) given by

KR(t, x, y) =1

RN+sK(t0 +

t

Rs, x0 +

x

R, x0 +

y

R).

The point here is that the kernel KR(t, x, y) satisfies the following constraint:

1

Λ1{|x−y|≤3R}

1

|x− y|N+s≤ KR(t, x, y) ≤

1

Λ

1

|x− y|N+s,

which is stronger than the one in hypothesis (1.2) (since R ≥ 21s > 1). Hence, the

conclusion of Lemma 3.1 can be applied to wR.Next, we show that ψ(x0 + ·

R ) ≤ ψ(x0 + ·). Since |x0| ≤ 1, we have that

{y ∈ RN : ψ(x0 + y) = 0} = {y ∈ R

N : |y + x0| ≤ 1} contains the origin O of RN .This in turn ensures that {y ∈ R

N : ψ(x0 +yR ) = 0} = {y ∈ R

N : |y + Rx0| ≤ R},which is the dilation of the disc {y ∈ R

N : |y + x0| ≤ 1} about the origin O with afactor of R, must contain the disc {y ∈ R

N : |y + x0| ≤ 1} itself. That is, we have(3.13){y ∈ R

N : ψ(x0 + y) = 0} = {y ∈ RN : |y + x0| ≤ 1} ⊂ {y ∈ R

N : |y +Rx0| ≤ R}.

Then, we observe that the graph of ψ(x0 +·R ) is obtained by dilating (flattering)

the graph of ψ(x0 + ·) spatially by a factor of R ≥ 21s . This indicates that the

pointwise slope of ψ(x0+·R ) (outside of the disc {y ∈ R

N : |y+Rx0| ≤ R}) shouldbe less than that of ψ(x0 + ·), which together with the relation in (3.13) impliesthat the graph of ψ(x0 +

·R ) must be strictly below the graph of ψ(x0 + ·). That

is, we must have the desired relation ψ(x0 +·R ) ≤ ψ(x0 + ·).

Since ψ increases with respect to |x|, for |x| > 1 we have(3.14)

wR(s, y) = w(t0 +s

Rs, x0 +

y

R) ≤ 1 + ψ(x0 +

y

R) ≤ 1 + ψ(x0 + y) ≤ 1 + ψ(|y|+ 1).



So, from the definition of R, for |y| ≥ R we have wR(s, y) ≤ ψ(y) (recall that1 + ψ(|y|+ 1) ≤ ψ(|y|), for any |y| ≥ R). Hence, from the hypothesis we have∫ 0

−2

∫RN

[wR(s, y)− ψ(y)]2+ =

∫ 0

−2

∫|y|≤R

[wR(s, y)− ψ(y)]2+dy ds

≤∫ 0

−2

∫|y|≤R

[w(t0 +s

Rs, x0 +

y

R)]2+dy ds

≤ RN+s

∫ t0

t0− 2Rs

∫{x0}+B(1)

[w(s, y)]2+dy ds

≤ RN+s

∫ 0

−2

∫B(2)

[w(s, y)]2+dy ds

≤ RN+s(1 + ψ(2))2δ.

(3.15)

Notice that in the above estimation, we have used the facts that {x0}+B(1) ⊂B(2) (since |x0| ≤ 1) and that [t0 − 2

Rs , t0] ⊂ [−2, 0] (since −1 ≤ t0 ≤ 0 and

R ≥ 21s ).

So, choosing δ = R−(N+s)(1+ ψ(2))−2ε0 gives that w(t0, x0) ≤ 1/2 for (t0, x0) ∈(−1, 0)×B1. �

4. The second De Giorgi’s lemma

This section is dedicated to a lemma of local decrease of the oscillation of asolution to equation (3.1). We define the following function:

(4.1) F (x) = sup(−1, inf(0, |x|2 − 9)).

Note that F is Lipschitz, compactly supported in B3, and equal to −1 in B2.For λ < 1/3, we define

ψλ(x) = 0 if |x| ≤ 1

λ4/s

= ((|x| − 1/λ4/s)s/4 − 1)+ if |x| ≥ 1

λ4/s.

The normalized lemma will involve three consecutive cutoffs:

ϕ0 = 1 + ψλ + F,

ϕ1 = 1 + ψλ + λF,

ϕ2 = 1 + ψλ + λ2F.

We prove the following lemma:

Lemma 4.1. Let Λ be the given constant in condition (1.2) and δ the constantdefined in Corollary 3.3. Then, there exists μ > 0, γ > 0, and λ ∈ (0, 1), dependingonly on N , Λ, and s, such that for any solution w : [−3, 0] × R

N → R to (3.1)satisfying

w(t, x) ≤ 1 + ψλ(x) on[−3, 0]× RN ,

|{w < ϕ0} ∩ ((−3,−2)×B1)| ≥ μ,

then we have either

|{w > ϕ2} ∩ ((−2, 0)× RN )| ≤ δ



or

|{ϕ0 < w < ϕ2} ∩ ((−3, 0)× RN )| ≥ γ.

The lemma says that in going from the ϕ0 cutoff to the ϕ2 cutoff, i.e., fromthe set {w > ϕ0} to {w > ϕ2} “some mass” is lost; i.e., if |{w > ϕ2}| is not yetsubcritical (i.e., ≤ δ), then

|{w > ϕ2}| ≤ |{w > ϕ0}| − γ.

Proof. In all the proof, we denote by C constants which depend only on s, N andΛ, but which can change from one line to another. We may fix any 0 < μ < 1/8.We will fix δ smaller than the one in Corollary 3.3 and such that the term Cδ in(4.8) is smaller than 1/4. The task consists now in showing that for 0 < λ < 1/3small enough, there exists a γ > 0 for which the lemma holds. The constraints onλ are (4.3), (4.5), (4.7), and (4.9). We split the proof into several steps.

First step: The energy inequality. We start again with the energy inequality(3.4), but use better the “good” term

B((w − ϕ)+, (w − ϕ)−) =

∫∫R2N

(w − ϕ)+(x)K(t, x, y)(w− ϕ)neg(y)) dx dy

that we just neglected before.We have, for ϕ1 the intermediate cutoff (see (3.4)):

∫((w − ϕ1)+)

2 dx∣∣∣T2

T1

+

∫ T2

T1

B((w − ϕ1)+, (w − ϕ1)+) dt

= −∫ T2

T1

B((w − ϕ1)+, ϕ1) dt−∫ T2

T1

B((w − ϕ1)+, (w − ϕ1)−) dt.

The remainder term can be controlled in the following way:

B((w − ϕ1)+, ϕ1) ≤1

2B((w − ϕ1)+, (w − ϕ1)+)

+ 2

∫∫[ϕ1(x)− ϕ1(y)]K(x, y)[ϕ1(x)− ϕ1(y)][χB3

(x)].

The first term 12B((w−ϕ1)+, (w−ϕ1)+) is absorbed on the left. The second term

is smaller than

4λ2

∫∫[F (x)− F (y)]K(x, y)[F (x)− F (y)]

+ 4

∫∫[ψλ(x)− ψλ(y)]K(x, y)[ψλ(x)− ψλ(y)][χB3

(x)],



which is smaller than Cλ2. This is obvious for the first term since F is Lipschitzand compactly supported. Since ψλ(x) = 0 for |x| < 3, the second term is equal to

4

∫∫ψλ(y)

2[χB3(x)]K(t, x, y) dx dy

≤ 4Λ|B3|∫{|y|>1/λ4/s}

(((|y| − 1/λ4/s)s/4 − 1)2+(|y| − 3)N+s

dy

≤ 4Λ|B3|λ2

∫{|z|>1}

(((|z| − 1)s/4 − λ)2+(|z| − 3λ4/s)N+s

dz

≤ 4Λ|B3|λ2

∫{|z|>1}

(((|z| − 1)s/4)2+(|z| − 1/3)N+s

dz

≤ Cλ2,

since λ < 1/3.This leaves us with the inequality∫

(w − ϕ1)2+dx

∣∣∣T2

T1

+1

2

∫ T2

T1

B((w − ϕ1)+, (w − ϕ1)+) dt

+

∫ T2

T1

∫R2N

(w − ϕ1)+(x)K(x, y)(w − ϕ1)neg(y) dx dy dt ≤ C λ2(T2 − T1).

In particular, since the second and third terms are positive, we get that for −3 <T1 < T2 < 0:

H(t) =

∫RN

(w − ϕ1)2+(t, x)dx

satisfies

H ′(t) ≤ C λ2.

Next, observe that, since [w(t, ·) − ϕ]+ ≤ λ1B(3), we have H(t) =∫B(3)

[w(t, ·) −ϕ]2+ ≤ λ2|B(3)|. Hence the following estimation is valid for any −3 < T1 < T2 < 0:

(4.2)

∫ T2

T1

∫R2N

(w − ϕ1)+(x)K(x, y)(w − ϕ1)neg(y) dx dy dt

≤ C λ2[T2 − T1] + |H(T2)−H(T1)|≤ Cλ2.

Note that, up to now, those estimates hold for any 0 < λ < 1/3.

Second step: An estimate on those time slices where the “good” extraterm helps. Remember that μ < 1/8 is fixed from the beginning of the proof.

From our hypothesis

|{w < ϕ0} ∩ ((−3,−2)×B1)| ≥ μ,

the set of times Σ in (−3,−2) for which |{w(·, T ) < ϕ0} ∩ B1| ≥ μ/4 has at leastmeasure μ/(2|B1|).

We estimate now that except for a few of those time slices,

∫RN

(w − ϕ1)2+ dx is

very tiny.



Since inf|x−y|≤3

K(t, x, y) ≥ CΛ−1 we have that

C λ2 ≥∫ −2

−3

B((w − ϕ1)+, (w − ϕ1)−)dt ≥ C Λ−1μ

8

∫Σ

∫RN

(w − ϕ1)+ dx dt

≥ C Λ−1 μ

8λ

∫Σ

∫RN

(w − ϕ1)2+ dx dt

since (w − ϕ1)+ ≤ λ.In other words,∫

Σ

∫RN

[(w − ϕ1)+(x)]2 dx dt ≤ C

λ3

μ≤ λ3−1/8

if λ is small enough such that

(4.3) λ ≤(μ

C

)8

.

In particular, from Tchebychev’s inequality:

(4.4)

∫(w − ϕ1)

2+(t, x) dx ≤ λ3− 1

4

for all t ∈ Σ, except for a very small subset F of t’s of measure smaller than λ1/8.We need it still much smaller than μ ∼ |Σ|. Indeed, if λ is small enough such that

(4.5) λ ≤(

μ

4|B1|

)8

,

then (4.4) holds on a set of t’s in [−3,−2] of measure greater than μ/(4|B1|).

Third step: In search of an intermediate set, where w is between ϕ0 andϕ2. Let us go now to (w − ϕ2)+.

Assume that for at least one time T0 > −2,

|{x | (w − ϕ2)+(T0, x) > 0}| > δ/2,

i.e., goes over critical for the first lemma, and let’s go backwards in time until wereach a slice of time T1 ∈ Σ, where∫

RN

(w − ϕ1)2+(T1, x) dx ≤ λ3− 1

4 .

At T0, for the intermediate cutoff, ϕ1, we have

(4.6)

∫RN

(w − ϕ1)2+(T0, x) dx ≥

∫(ϕ1 − ϕ2)

2χ{(w−ϕ2)+>0}

≥∫(λ− λ2)2F 2(x)χ{(w−ϕ2)+>0} ≥ CF

λ2

4δ3,

where the constant CF depends only on the fixed function F . Indeed we haveλ < 1/2, and F is increasing with respect to |x| and smaller than −C(3 − |x|) for|x| < 3 close to 3. Hence, the integral is minimum when all the mass {(w−ϕ2)+ > 0}is concentrated on 3− Cδ < |x| < 3.

Now, at T1, ∫RN

(w − ϕ1)2+(T1, x) dx ≤ λ3− 1

4 .



Thus, for λ small enough such that

(4.7) λ1−1/4 ≤ CFδ3

64,

in going from T0 backwards to T1, H(t) =∫RN (w−ϕ1)

2+(t, x) dx has crossed a range

from CFλ2 δ3

4 to CFλ2 δ3

16 . Since H ′(t) ≤ C λ2, in order to do so the energy H(t) ofthe truncation [w − ϕ1]+(t, ·) has to pass through a range of times D, of at leastlength ∼ δ3, where

D = {t ∈ (T1, T0) : CFλ2

16δ3 < H(t) < CF

λ2

4δ3}.

We want to show that in this range, we pick up an intermediate set, of nontrivialmeasure, where (w − ϕ0)+ > 0 and (w − ϕ2)+ = 0, implying that the measure

A2 = |{(w − ϕ2)+ > 0} ∩ {t ∈ (−3, 0)}|effectively decreases some fixed amount from

A0 = |{w − ϕ0)+ > 0} ∩ {t ∈ (−3, 0)}| .In these ranges of times t ∈ D, given the gap between ϕ1 and ϕ2, we should have

(4.8) |{(w − ϕ2)+(t, ·) > 0}| ≤ δ

2.

Indeed, (4.8) can be shown as follows. Assume towards a contradiction that wehave |{(w − ϕ2)+(τ, ·) > 0}| > δ

2 for some τ ∈ D. Then, at the time slice τ ,we can go through the same estimation in (4.6) to deduce that we would have

H(τ ) ≥ CFλ2

4 δ3, which is in contradiction with τ ∈ D by the definition of the setD.

As said at the beginning of the proof, we may consider a δ such that Cδ < 1/4.Moreover, those times of D for which

|{(w − ϕ0)+ ≤ 0} ∩B2| ≥ μ

are in an exceptional subset F of very small size. Indeed,

Cλ2 ≥∫ 0

−3

∫RN

(w − ϕ1)+K(t, x, y)(w − ϕ1)neg

≥ Cμ

∫F

∫B3

(w − ϕ1)+ dx dt ≥ Cμ

λ

∫F

∫RN

(w − ϕ1)2+ dx dt

≥ Cμ|F|(CFλ2δ3/16)

λ,

where, in the last line of the above estimation, we have used H(t) =∫RN [w −

ϕ1](t, ·) ≥ CFλ2

16 δ3 (for t ∈ F ∈ D) which is ensured by the definition of the time

set D. Hence

|F| ≤ Cλ

μδ3.

Therefore, for λ small enough such that

(4.9) λ ≤ μδ3|D|/(2C),

we have

|F| ≤ |D|2

.



Note that the constraint (4.9) can be expressed depending only on s, N , Λ, δ, and,μ, since |D| < Cδ3. For these times in D that are not in F , we have:

A(t) = |{ϕ0 ≤ w(t, ·) ≤ ϕ2}| ≥ 1/2.

That is,

|{ϕ0 < w < ϕ2} ∩ ((−3, 0)× RN )| ≥

∫ 0

−3

A(t) dt

≥∫D\F

A(t) dt ≥ |D|4

≥ Cδ3. �

5. Proof of the Cαregularity

We are now ready to show the following oscillation lemma. First, for λ as in theprevious section, we define for any ε > 0,

ψε,λ(x) = 0 if |x| ≤ 1

λ4/s

= ((|x| − 1/λ4/s)ε − 1)+ if |x| ≥ 1

λ4/s.

Lemma 5.1. There exists ε > 0 and λ∗ such that for any solution to (3.1) in[−3, 0]× R

N such that

−1− ψε,λ ≤ w ≤ 1 + ψε,λ,

we have

sup[−1,0]×B1

w − inf[−1,0]×B1

w ≤ 2− λ∗.

Proof. We may assume that

|{w < ϕ0} ∩ ((−3,−2)×B1)| > μ.

Otherwise this is verified by −w, and we may work on this function.Consider k0 = |(−3, 0)×B3|/γ. Then we fix ε small enough such that

(|x|ε − 1)+λ2k0

≤ (|x|s/4 − 1)+

for all x. We may take ε = (s/4)λ2k0 for instance. For k ≤ k0, we consider thesequence

wk+1 =1

λ2(wk − (1− λ2)), w0 = w.

By induction, we have that

(wk)+(t, x) ≤ 1 +1

λ2kψε,λ(x), t ∈ (−3, 0), x ∈ R

N .

So, for k ≤ k0 we have wk ≤ 1 + ψλ. By construction |{wk < ϕ0} ∩ (−3,−2)×B1|is increasing, thus greater than μ for any k. Hence, we can apply Lemma 4.1 onwk. As long as |{wk > ϕ2} ∩ ((−2, 0)× R

N )| ≥ δ, we have

|{wk+1 > ϕ0}| = |{wk+1 > ϕ2}|+ |{ϕ0 < wk+1 < ϕ2}|and

|{wk+1 > ϕ2}| ≤ |{wk+1 > ϕ0}| − γ

≤ |{wk > ϕ2}| − γ ≤ |(−3, 0)×B3| − kγ.



This cannot be true up to k0. So there exists k ≤ k0 such that

|{wk > ϕ2} ∩ ((−2, 0)× RN )| ≤ δ.

We can then apply the first De Giorgi lemma on wk+1. Indeed,

wk+1 ≤ 1 + ψλ ≤ 1 + ψ1 on (−3, 0)× RN ,

and

|{wk+1 > 0} ∩ ((−2, 0)×B2)| ≤ |{wk+1 > ϕ0} ∩ ((−2, 0)×B2)|≤ |{wk > ϕ2} ∩ ((−2, 0)× R

N )| ≤ δ.

Hence, from Corollary 3.3, we have

wk+1 ≤ 1/2 on(−1, 0)× B1.

This gives the result with

λ∗ =λ2k0

2. �

The Cα regularity follows in a standard way. More precisely, by using Lemma5.1, we can now complete the proof of Theorem 2.2 as follows.

Proof. For any (t0, x0) ∈ (0,∞)× RN , consider first K0 = inf(1, t0/4)

1/s and

w0(t, x) = w(t0 +Ks0t, x0 +K0x).

This function still satisfies an equation of the type of (3.1) in (−4, 0) × RN with

a kernel K satisfying (1.2). From Corollary 3.2, it is bounded on (−3, 0) × RN .

Consider K < 1 such that

1

1− (λ∗/2)ψλ,ε(Kx) ≤ ψλ,ε(x), for |x| ≥ 1/K.

The coefficient K depends only on λ, λ∗ and ε. Then we define by induction:

w1(t, x) =w0(t, x)

‖w0‖L∞, (t, x) ∈ (−3, 0)× R

N ,

wk+1(t, x) =1

1− λ∗/4(wk(K

st,Kx)− wk) , (t, x) ∈ (−3, 0)× RN ,

where

wk =1

|B1|

∫ 0

−1

∫B1

wk(t, x) dx dt.

By construction, wk satisfies the hypothesis of Lemma 5.1 for any k. Hence:

sup(t0+(−Kks,0))×(x0+B

Kk )

w − inf(t0+(−Kks,0))×(x0+B

Kk )w ≤ C(1− λ∗/4)k.

So, w is Cα with

α =ln(1− λ∗/4)

ln(Ks). �



Appendix A. Classical solutions if φ is smooth enough

We consider the case 1 < s < 2. We show in this appendix that if φ ∈ C3,s−1+ε

and K(z) = |z|−(N+s), then the solution to (2.3) is, indeed, a classical solution.We linearize the equation of the first derivatives w = ∂xi

θ which is Cα. Since φis even, φ′′′(0) = 0, and(A.1)|φ′′(θ(x)− θ(y))− φ′′(0)| = |R(t, x, y)| ≤ C|θ(x)− θ(y)|s+ε ≤ C inf(|x− y|s+ε, 1).

Hence, w is a solution to

(A.2) ∂tw − φ′′(0)Δs/2w =

∫RN

R(t, x, y)K(x− y)(w(y)− w(x)) dy = F (t, x).

Since w = ∇θ is Cα, the right-hand side term can be shown to be also Cα/2 inspace and time. To show that it is Cα/2 in x, we write

F (x+ z)− F (x)

=

∫R(x, y + x)

|y|N+s(w(x+ z + y)− w(x+ z)− w(x+ y) + w(x)) dy

+

∫[R(x+ z, y + x+ z)−R(x, y + x)]

w(x+ z + y)− w(x+ z)

|y|N+sdy.

From (A.1), the first term can be controlled by C|z|α. For the second term, notethat φ′′ is C1 and θ is C1 in x, so, using (A.1) again, we find

[R(x+ z, y + x+ z)−R(x, y + x)]

≤ C[R(x+ z, y + x+ z)−R(x, y + x)]α/2

× (|R(x+ z, y + x+ z)|1−α/2 + |R(x, y + x)|1−α/2)

≤ C|z|α/2 inf(|y|(1−α/2)(s+ε), 1).

This second term can be controlled by C|z|α/2.Then, from (A.2), the Cα/2 regularity of F , and using Madych-Riviere [19],

∂tw ∈ Cα/2 and w is Cs,α/2 in x. Thus all the terms in the equation are classical.

Appendix B. The approximation scheme for the nonlinear problem

We can approach (a priori) weak solutions to the nonlinear problem of Theorem2.1 by classical solutions of approximated equations using the following approxima-tion scheme. We consider K(z) = |z|−(N+s). For ε > 0, we replace the function φby an approximated function φε such that

φε(X) = |X|2, for |X| < ε,

and we consider θε, the solution to

∂tθε(t, x)−∫RN

φ′ε(θε(t, y)− θε(t, x))K(y − x)dy = 0.

Note, first, that the weak solutions θε are uniformly Lipschitz. This estimate doesnot need approximation (it is the boundedness of wε = ∂xi

θε). It can be done usingthe incremental quotients (as in the second-order case). So the result is also truefor θ. The corresponding operator for ∇xθε is

∂twε =

∫RN

φ′′ε (θε(x)− θε(y))(wε(y)− wε(x))K(x− y) dy.



But φ′′ε = 1 for |θε(x)− θε(y)| < ε. Since θε is Lipschitz, φ′′

ε = 1 for |x − y| ≤ Cε.Thus, the quantity

(B.1)

∫RN

[φ′′ε (θε(x)− θε(y))− 1][wε(x)− wε(y)]K(x− y) dy

is bounded. It is equal to

(B.2) ∂twε −Δs/2wε.

By classical theory, this implies that wε is Cα in space and time. In turn, it impliesthat (B.1) is Cα, and so (B.2) is Cβ (see Appendix A). This implies that wε hasenough regularity to make the equation classical.

We can now show the convergence of θε to θ. We can rewrite the equation on θεas

∂tθε =

∫RN

φ′(θε(x)− θε(y))K(x− y) dy

+

∫RN

[φ′ε(θε(x)− θε(y))− φ′(θε(x)− θε(y))]K(x− y) dy.

Hence, θ − θε verifies the linearized equation

∂t(θ − θε) =

∫RN

K(t, x, y)[(θ − θε)(y)− (θ − θε)(x)] dy

+

∫RN

[φ′ε(θε(x)− θε(y))− φ′(θε(x)− θε(y))]K(x− y) dy,

where

K(t, x, y) =

∫ 1

0

φ′′((1− s)(θ(x)− θ(y)) + s(θε(x)− θε(y)))ds.

The quantity [φ′ε(θε(x)−θε(y))−φ′(θε(x)−θε(y))] is bounded by Cmin(ε, |θε(x)−

θε(y)|). We multiply by (θ − θε) and integrate (note that θ − θε = 0 at t = 0) toobtain

supT

∫RN

(θ − θε)2(T ) dx+

∫ T

0

∫R2N

|(θ − θε)(x)− (θ − θε)(y)|2K(t, x, y) dx dy dt

≤∫ T

0

∫R2N

((θ − θε)(x)− (θ − θε)(y))[φ′ε(θε(x)− θε(y))

− φ′(θε(x)− θε(y))]K(x− y) dy dt.

The quantity φ′′ is bounded from below, so K(x− y) ≤ CK(t, x, y). By Cauchy weget

supT

∫RN

(θ − θε)2(T ) dx+

1

2

∫R2N

|(θ − θε)(x)− (θ − θε)(y)|2K(t, x, y) dx dy

≤ C

∫R2N

[min(ε, θε(x)− θε(y))]2K(x− y) dy

≤ Cδε2−s−δ,

for any δ > 0. Hence θε converges to θ in L∞(0, T ;L2(RN )) ∩ L2(0, T ; Hs/2(RN )).



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Introduction...2011/03/24 · Volume 24, Number 3, July 2011, Pages 849–869 S 0894-0347(2011)00698-X Article electronically published on March 24, 2011 REGULARITY THEORY FOR PARABOLIC