On the regularity of the free boundary in the $p$-Laplacian obstacle problem

0.42 MB·

by  Alessio Figalli

ON THE REGULARITY OF THE FREE BOUNDARY
IN THE p-LAPLACIAN OBSTACLE PROBLEM
ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
7
1
Abstract. We study the regularity of the free boundary in the obstacle for the
0
2 p-Laplacian, min(cid:8)−∆pu, u−ϕ(cid:9) = 0 in Ω ⊂ Rn. Here, ∆pu = div(cid:0)|∇u|p−2∇u(cid:1),
and p∈(1,2)∪(2,∞).
n
Near those free boundary points where ∇ϕ (cid:54)= 0, the operator ∆ is uniformly
a p
J elliptic and smooth, and hence the free boundary is well understood. However,

9 when ∇ϕ = 0 then ∆p is singular or degenerate, and nothing was known about
1 the regularity of the free boundary at those points.
Here we study the regularity of the free boundary where ∇ϕ = 0. On the
]
P one hand, for every p (cid:54)= 2 we construct explicit global 2-homogeneous solutions
A to the p-Laplacian obstacle problem whose free boundaries have a corner at the
origin. In particular, we show that the free boundary is in general not C1 at
.
h points where ∇ϕ = 0. On the other hand, under the “concavity” assumption
t
a |∇ϕ|2−p∆ ϕ<0, we show the free boundary is countably (n−1)-rectifiable and
p
m we prove a nondegeneracy property for u at all free boundary points.
[

1
v
2 1. Introduction
6
2 In this paper we study the obstacle problem
5
0 min(cid:8)−∆ u, u−ϕ(cid:9) = 0 in Ω ⊂ Rn (1.1)
. p
1
0 for the p-Laplacian operator
7
1 ∆ u = div(cid:0)|∇u|p−2∇u(cid:1), 1 < p < ∞.
: p
v
i The problem appears for example when considering minimizers of the constrained
X
p-Dirichlet energy
r
a
(cid:26)(cid:90) (cid:27)
inf |∇v|p : v ∈ W1,p(Ω), v ≥ ϕ in Ω, v = g on ∂Ω ,
Ω
where ϕ and g are given smooth functions and Ω is a bounded smooth domain.
2010 Mathematics Subject Classification. 35R35.
Key words and phrases. Obstacle problem; p-Laplacian; free boundary.
AF and BK were supported by NSF-FRG grant DMS-1361122. XR was supported by NSF
grant DMS-1565186 and MINECO grant MTM2014-52402-C3-1-P (Spain).
1
2 ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
The regularity of solutions to (1.1) was recently studied by Andersson, Lindgren,
and Shahgholian in [ALS15]. Their main result establishes that if ϕ ∈ C1,1 then
sup (u−ϕ) ≤ Cr2 for all r ∈ (0,1)
Br(x0)
at any free boundary point x ∈ ∂{u > ϕ}. Thus, solutions u leave the obstacle ϕ
0
in a C1,1 fashion at free boundary points x .
0
Notice that, near any free boundary point x ∈ ∂{u > ϕ} at which ∇ϕ(x ) (cid:54)= 0,
0 0
thesolutionuwillsatisfy∇u (cid:54)= 0aswellandhencetheoperator∆ uisuniformlyel-
p
liptic in a neighborhood of x . Therefore, by classical results [Caf77, Caf98, PSU12],
0
the solution u is C1,1 near x , and the structure and regularity of the free boundary
0
is well understood.
Thus, the main challenge in problem (1.1) is to understand the regularity of
solutions and free boundaries near those free boundary points x ∈ ∂{u > ϕ} at
0
which ∇ϕ(x ) = 0. Our first main result is the following.
0
Theorem 1.1. Let p ∈ (1,2)∪(2,∞), and let ϕ(x) = −|x|2 in R2. There exists a
2-homogeneous function u : R2 → R satisfying (1.1) in all of R2, and such that the
set {u > ϕ} is a cone with angle
(cid:18) (cid:114) (cid:19)
p−1
θ = 2π 1− (cid:54)= π.
0
2p
In particular, the free boundary has a corner at the origin.
Remark 1.2. Let u be a solution to (1.1) with ϕ(x) = −|x|2 as in Theorem 1.1. For
each a ∈ R and b > 0, a−bu is a solution to (1.1) in R2 with ϕ(x) = a−b|x|2 for
which the contact set is a cone with angle θ (cid:54)= π.
0
Remark 1.3. Notice that for p ∈ (2,∞), θ > π and thus u is not convex. This
0
is in contrast with the classical result of Caffarelli on the classifications of global
solutions to the obstable problem for the Laplacian [Caf98].
Remark 1.4. In the process of constructing the solutions u of Theorem 1.1, for p = 9
we will construct a global solution u to ∆ u = 0 in all of R2.
p
Inviewoftheaboveresult, noC1 regularitycanbeexpectedforthefreeboundary
at points at which ∇ϕ = 0. Also, the lack of convexity of possible blow-up profiles
seemstobeamajorobstacleofunderstandingthefinestructureofthefreeboundary
at these points.
Still,aninterestingquestionistodecidewhetherthefreeboundaryhasfiniteHn−1
measure near points at which the gradient of the obstacle vanishes. A standard first
step in this direction is to prove a nondegeneracy result stating that u−ϕ cannot
decay faster than quadratic at free boundary points. [ALS15] previously proved a
similarnondegeneracyresultatthefreeboundarypointsundertheassumptionsthat
ϕ ∈ C2, p > 2, and ∆ ϕ ≤ −c < 0. However, if ϕ ∈ C2 satisfies ∇ϕ(x ) = 0 then
p 0 0
∆ ϕ(x ) = 0, and thus the result in [ALS15] can not be applied to free boundary
p 0
points on {∇ϕ = 0}. We show the following.
FREE BOUNDARY REGULARITY IN THE p-LAPLACIAN OBSTACLE PROBLEM 3
Theorem 1.5. Let p ∈ (1,∞), ϕ ∈ C2(B ), and u be a solution of (1.1) in B .
1 1
Assume that ϕ satisfies
(cid:0) (cid:1)
|∇ϕ|2−pdiv |∇ϕ|p−2∇ϕ ≤ −c < 0 in {∇ϕ (cid:54)= 0}. (1.2)
0
Then for any free boundary point x ∈ ∂{u > ϕ}∩B there exists c > 0 such that
0 1/2 1
sup (u−ϕ) ≥ c r2 for r ∈ (0,1/2),
1
Br(x0)
where the constant c depends only on the modulus of continuity of D2ϕ and on the
1
constant c in (1.2).
0
Remark 1.6. The hypothesis (1.2) is nontrivial in the sense that (1.2) implies that
either ϕ is identically constant on B or {∇ϕ (cid:54)= 0} is an open dense subset of B ,
1 1
see Lemma 3.1 below. In the case that ϕ is identically constant on B , by the Hopf
1
boundary point lemma [Va´z84, Theorem 5] either u ≡ ϕ in B or ∆ u = 0 and
1 p
u > ϕ in B and in particular the free boundary is an empty set.
1
As a consequence of Theorem 1.5 we can deduce that, under the hypotheses of
Theorem 1.5, the free boundary is porous: i.e., there exists a δ > 0 such that for
every B (x ) ⊆ B , there exists B (x) ⊂ B (x )\∂{u > ϕ}. The proof is standard
r 0 1 δr r 0
and follows from combining the optimal regularity of solutions in [ALS15] with
Theorem 1.5 above. Porosity of the free boundary implies that the free boundary
has zero Lebesgue measure. We in fact prove the stronger result that, under the
hypotheses of Theorem 1.5, the free boundary ∂{u > ϕ} is an (n−1)-dimensional
rectifiable set.
Definition 1.7. Let 0 ≤ k ≤ n be an integer. We say a set S ⊆ Rn is countably
k-rectifiableifthereexistsasetE ⊂ Rn withHk(E ) = 0andacountablecollection
0 0
of Lipschitz maps f : Rk → Rn such that
j
∞
(cid:91)
S ⊆ E ∪ f (Rk).
0 j
j=1
Theorem 1.8. Let p ∈ (1,∞), ϕ ∈ C2(B ), and u be a solution of (1.1) in B .
1 1
Assume that ϕ satisfies (1.2). Then, the free boundary ∂{u > ϕ} is countably
(n−1)-rectifiable.
Relatedobstacle-typeproblemsforthep-Laplacianhavebeenstudiedin[KKPS00,
LS03,CLRT14,CLR12]. Inthoseworks, however, theystudiedthedifferentproblem
∆ u = f(x)χ in Ω ⊂ Rn. (1.3)
p {u>0}
It is important to notice that, when p (cid:54)= 2, the obstacle problems (1.1) and (1.3) are
of quite different nature. For example, when f ≡ 1 solutions to (1.3) are not C1,1
but C1,p−11 near all free boundary points.
The paper is organized as follows. In Section 2 we prove Theorem 1.1. Then, in
Section 3 we prove Theorems 1.5 and 1.8.
4 ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
2. Homogeneous degree-two solutions
We construct here the homogeneous solutions of Theorem 1.1.
Proof of Theorem 1.1. Let 1 < p < ∞ and p (cid:54)= 2, and let ϕ(x) = −|x|2 in R2. We
will show that there exists a global solution u(x ,x ) to (1.1) which is homogeneous
1 2
of degree 2 and such that the free boundary consists of two rays meeting at an angle
θ (cid:54)= π.
0
We use polar coordinates reiθ on R2 where r > 0 and θ ∈ [0,2π]. We want to
construct u ∈ C1(R2) such that
∆ u(reiθ) = 0, u(reiθ) ≥ −r2 for θ ∈ (0,θ ),
p 0
u(reiθ) = −r2 for θ ∈ [θ ,2π], (2.1)
0
∂u ∂u
(r) = (reiθ0) = 0.
∂θ ∂θ
Assumethatu(reiθ) = r2v(θ)forsome2π-periodicfunctionv ∈ C1(R)∩C∞([0,θ ])∩
0
C∞([θ ,2π]). We want to express ∆ u(reiθ) = 0 for θ ∈ (0,θ ) as a ordinary
0 p 0
differential equation of v. We compute
∂ ∂
∇u = 2rv(θ) +v(cid:48)(θ)
∂r ∂θ
and thus
(cid:18) (cid:19) (cid:18) (cid:19)
1 ∂ ∂u 1 ∂ ∂u
∆ u = r|∇u|p−2 + |∇u|p−2
p r ∂r ∂r r2 ∂θ ∂θ
1 ∂ ∂
(cid:0) (cid:1) (cid:0) (cid:1)
= rp(4v2 +(v(cid:48))2)(p−2)/2 ·2v +rp−2 (4v2 +(v(cid:48))2)(p−2)/2v(cid:48)
r ∂r ∂θ
(cid:18) 4v(v(cid:48))2 +(v(cid:48))2v(cid:48)(cid:48)(cid:19)
= rp−2(4v2 +(v(cid:48))2)(p−2)/2 2pv +v(cid:48)(cid:48) +(p−2) .
4v2 +(v(cid:48))2
Thus we can rewrite ∆ u = 0 as
p
4v(v(cid:48))2 +(v(cid:48))2v(cid:48)(cid:48)
2pv +v(cid:48)(cid:48) +(p−2) = 0.
4v2 +(v(cid:48))2
Solving for v(cid:48)(cid:48),
8pv2 +(6p−8)(v(cid:48))2
v(cid:48)(cid:48) = −v . (2.2)
4v2 +(p−1)(v(cid:48))2
Notice that (2.1) is equivalent to v satisfying (2.2) for θ ∈ (0,θ ) and
0
v(θ) ≥ −1 for θ ∈ (0,θ ),
0
v(θ) = −1 for θ ∈ [θ ,2π], (2.3)
0
v(cid:48)(0) = v(cid:48)(θ ) = 0.
0
FREE BOUNDARY REGULARITY IN THE p-LAPLACIAN OBSTACLE PROBLEM 5
Moreover, by integration by parts, (2.2), and the homogeneity of u, for all ζ ∈
C1(R2 \{0}) it holds
c
(cid:90)
− |∇u|p−2∇u·∇ζ
B1
(cid:90) θ0(cid:90) ∞ (cid:90) 2π(cid:90) ∞
= − |∇u|p−2∇u·∇ζrdrdθ− |∇u|p−2∇u·∇ζrdrdθ
0 0 θ0 0
(cid:90) θ0(cid:90) ∞ (cid:90) 2π(cid:90) ∞ (cid:90) 2π
= ∆ uζrdrdθ+ ∆ uζrdrdθ+lim |∇u|p−2D uζrdθ
p p r
r↓0
0 0 θ0 0 0
(cid:90) 2π(cid:90) ∞
= ∆ (−r2)ζrdrdθ.
p
θ0 0
Hence, ∆ u = ∆ (−r2)χ ≤ 0 weakly in R2 \{0}. This together with (2.1)
p p {θ0<θ<2π}
implies that u is a solution to (1.1).
Now let us solve (2.2). Set
X = v(θ) and Y = v(cid:48)(θ)
so that we transform (2.2) into the first order system
X(cid:48) = Y,

8pX2 +(6p−8)Y2 (2.4)
Y(cid:48) = −X on (0,θ ).
 4X2 +(p−1)Y2 0
Now, (2.3) implies that
X(θ) ≥ −1 for θ ∈ (0,θ ),
0
(X(0),Y(0)) = (X(θ ),Y(θ )) = (−1,0). (2.5)
0 0
Notice that (2.4) states that X(cid:48) = Y and Y(cid:48) equals a homogeneous degree one
function of (X,Y). Thus it is convenient to set
X = ρ(θ) cos(ψ(θ)) and Y = ρ(θ) sin(ψ(θ))
for some functions ρ and ψ, so that (2.4) is equivalent to
 ρ(cid:48)
 cos(ψ)−ψ(cid:48) sin(ψ) = sin(ψ),

 ρ
ρ(cid:48) 8p cos2(ψ)+(6p−8) sin2(ψ)
 sin(ψ)+ψ(cid:48) cos(ψ) = −cos(ψ)

ρ 4 cos2(ψ)+(p−1) sin2(ψ)
for all θ ∈ (0,θ ). Let
0
8p cos2(ψ)+(6p−8) sin2(ψ) (8p−4) cos2(ψ)+(5p−7) sin2(ψ)
F (ψ) := −1 =
p 4 cos2(ψ)+(p−1) sin2(ψ) 4 cos2(ψ)+(p−1) sin2(ψ)
6 ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
so that
 ρ(cid:48)
 cos(ψ)−ψ(cid:48) sin(ψ) = sin(ψ),

 ρ
(2.6)
ρ(cid:48)
 sin(ψ)+ψ(cid:48) cos(ψ) = −cos(ψ)−cos(ψ)F(ψ)

ρ
for all θ ∈ (0,θ ). Note that (2.6) can be rewritten as
0
ρ(cid:48)
= −cos(ψ) sin(ψ)F (ψ), (2.7)
p
ρ
ψ(cid:48) = −1−cos2(ψ)F (ψ), (2.8)
p
and this system can be solved by first solving (2.8) to find ψ, and then integrating
(2.7) to find ρ. We compute that
∂ ∂ (cid:18)(8p−4) cos2(ψ)+(5p−7) sin2(ψ)(cid:19)
F (ψ) =
∂p p ∂p 4 cos2(ψ)+(p−1) sin2(ψ)
∂ (cid:18)(3p+3) cos2(ψ)+5p−7(cid:19)
=
∂p (−p+5) cos2(ψ)+p−1
(3 cos2(ψ)+5)(5 cos2(ψ)−1)−(3 cos2(ψ)−7)(−cos2(ψ)+1)
=
((−p+5) cos2(ψ)+p−1)2
18 cos4(ψ)+12 cos2(ψ)+2
= > 0
((−p+5) cos2(ψ)+p−1)2
for all ψ ∈ [0,2π], so
1 1 3
1+cos2(ψ)F (ψ) ≥ 1+cos2(ψ)F (ψ) = 1+cos2(ψ)− sin2(ψ) = + cos2(ψ) > 0
p 1
2 2 2
for all ψ ∈ [0,2π]\{π/2,3π/2}. Note that, when ψ = π/2,3π/2, F (ψ) degenerates
p
as p ↓ 1, but 1+cos2(ψ)F (ψ) = 1 for all p > 1. Thus, solving (2.8), we find that
p
ψ(θ) = Θ−1(θ) for all θ ∈ [0,θ ] where Θ : R → R is the strictly decreasing function
0
defined by
(cid:90) ψ dσ
Θ(ψ) := − (2.9)
1+cos2(σ)F (σ)
ψ(0) p
for ψ(0) to be determined. Integrating (2.7) over [0,θ], we obtain
(cid:18) (cid:90) θ (cid:19)
ρ(θ) = ρ(0) exp − cos(ψ(τ)) sin(ψ(τ))F (ψ(τ))dτ (2.10)
p
0
for all θ ∈ [0,θ ] and for ρ(0) to be determined. Notice that ψ,ρ ∈ C∞([0,θ ]) and
0 0
thus v = ρ cos(ψ) ∈ C∞([0,θ ]).
0
It remains to determine θ , ψ(0), and ρ(0), and to verify that (2.5) holds true.
0
To this aim, we observe that (2.5) is equivalent to
ρ(0) = ρ(θ ) = 1, ψ(0) = π, ψ(θ ) = −(2k −1)π (2.11)
0 0
FREE BOUNDARY REGULARITY IN THE p-LAPLACIAN OBSTACLE PROBLEM 7
for some integer k ≥ 1. Then, in view of the fact that X(cid:48) = Y (and so X(θ) attains
its minimum value when ψ(θ) = −(2j −1)π for some integer j), we see that
ρ(θ) ≤ 1 whenever ψ(θ) = −(2j −1)π, for j = 1,2,...,k −1. (2.12)
Hence, by (2.11), we should choose ρ(0) = 1 and ψ(0) = π. To choose θ observe
0
that, by (2.9), ψ(θ ) = −(2k −1)π if and only if
0
(cid:90) π dσ (cid:90) π/2 dσ
θ = = 4k ,
0 1+cos2(σ)F (σ) 1+cos2(σ)F (σ)
−(2k−1)π p 0 p
where the last step follows by symmetry. We compute that
(cid:90) π/2 dσ
1+cos2(σ)F (σ)
0 p
(cid:90) π/2 4 cos2(σ)+(p−1) sin2(σ)
= dσ
4 cos2(σ)+(p−1) sin2(σ)+(8p−4) cos4(σ)+(5p−7) cos2(σ) sin2(σ)
0
(cid:90) π/2 (−p+5) cos2(σ)+p−1
= dσ
(3p+3) cos4(σ)+(4p−2) cos2(σ)+p−1
0
(cid:90) π/2 (−p+5) cos2(σ)+p−1
= dσ
(3 cos2(σ)+1)((p+1) cos2(σ)+p−1)
0
(cid:90) π/2(cid:18) 2 p−1 (cid:19)
= − dσ
3 cos2(σ)+1 (p+1) cos2(σ)+p−1
0
(cid:90) π/2(cid:18) 2 p−1 (cid:19)
= − sec2(σ)dσ
tan2(σ)+4 (p−1) tan2(σ)+2p
0
(cid:90) ∞(cid:18) 2 p−1 (cid:19)
= − dt
t2 +4 (p−1)t2 +2p
0
(cid:20) (cid:18) (cid:19) (cid:114) (cid:18)(cid:114) (cid:19)(cid:21)∞
t p−1 p−1
= arctan − arctan t
2 2p 2p
0
(cid:18) (cid:114) (cid:19)
π p−1
= 1− ,
2 2p
where we let t = tan(σ). Thus, we need to choose
(cid:18) (cid:114) (cid:19)
p−1
θ = 2kπ 1− (2.13)
0
2p
for some integer k ≥ 1. Since
(cid:18) (cid:114) (cid:19)
kπ p−1
< 2kπ 1− = θ < 2π,
0
2 2p
we deduce that k ≤ 3.
8 ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
1 < p < 2 p = 2 2 < p < ∞
θ θ θ
0 0 0
u = ϕ
u = ϕ u = ϕ
Figure 1. The angle θ and the contact set {u = ϕ} of the homoge-
0
neous solution for 1 < p < 2, p = 2, and 2 < p < ∞, respectively.
Notice that, for each k = 1,2,3, θ given by (2.13) is decreasing as a function of
0
p. In particular, when k = 1, θ = 2π for p = 1, π < θ < 2π for 1 < p < 2, θ = π
0 0 0
for p = 2, and 0 < θ < π for 2 < p < ∞, see Figure 1.
0
Hence if p ∈ (1,2)∪(2,∞], by setting k = 1 we can construct u for which
(cid:18) (cid:114) (cid:19)
p−1
θ = 2π 1− ∈ (0,π)∪(π,2π).
0
2p
When k = 2, for all 1 < p < 2 we have θ > 2π and consequently we do not
0
obtain a solution u, for p = 2 we have θ = 2π, and for all 2 < p < ∞ we obtain a
0
solution u with π < θ < 2π. Similarly, when k = 3, we do not obtain a solution u
0
for 1 < p < 9, θ = 2π for p = 9, and we obtain a solution u with π < θ < 2π for
0 0
9 < p < ∞.
To conclude, we need to verify ρ(θ ) = 1. For this, suppose that θ is such that
0
ψ(θ) = −(2j −1)π for an integer j ≥ 1. Observe that by (2.9) and symmetry,
(cid:90) θ (cid:90) π cos(σ) sin(σ)F (σ)
p
cos(ψ(τ)) sin(ψ(τ))F (ψ(τ))dτ = dσ = 0,
p 1+cos2(σ)F (σ)
0 −(2j−1)π p
where σ = ψ(τ). Therefore by (2.10) ρ(θ) = ρ(0) = 1. In particular, when j = k,
we get ρ(θ ) = ρ(0) = 1.
0
Notice that for k = 1 the contact set {u > ϕ} is precisely {θ ≤ θ ≤ 2π}, whereas
0
for k = 2,3 the contact set {u > ϕ} is the union of {θ ≤ θ ≤ 2π} and the rays
0
(cid:16) (cid:113) (cid:17)
ψ(θ) = −(2j −1)π/2, i.e. θ = 2πj 1− p−1 , for j = 1,...,k −1. (cid:3)
2p
Remark 2.1. Observe that when k = 1 and p = 2, the above argument produces a
solution to u to (2.1) with θ = π. In other words, the contact set {u = ϕ} is a
0
half-space. On the other hand, when k = p = 2, or when k = 3 and p = 9, the
FREE BOUNDARY REGULARITY IN THE p-LAPLACIAN OBSTACLE PROBLEM 9
above argument produces solutions ρ and ψ to (2.6) with θ = 2π so that
0
ρ(0) = ρ(2π), ψ(2π)−ψ(0) = −2kπ,
and we thereby obtain u ∈ C1(R2) such that ∆ u = 0 in all of R2. Note that
p
ρ(0) > 0 and ψ(0) are arbitrary and this corresponds to the invariance of ∆ u = 0
p
in R2 under scaling and rotations.
While this solution for p = 9 is new (at least to our knowledge), these solutions
for p = 2 are well-known. Indeed, when p = 2, (2.2) reduces to
v(cid:48)(cid:48) = −4v,
which obviously has the solution
v(θ) = Acos(2θ)+Bsin(2θ) (2.14)
for constants A,B ∈ R. Assuming that v is given by (2.14) for all θ ∈ [0,θ ] and
0
v satisfies the boundary conditions v(0) = v(θ ) = −1 and v(cid:48)(0) = v(cid:48)(θ ) = 0, we
0 0
obtain θ = π, A = −1, and B = 0 so that
0
u(x) = r2v(reiθ) = −|x|2 +2(x )2
2 +
so that w = u − ϕ is the well-known global solution w = 2(x )2 to the obstacle
2 +
problem min{∆w,w} = 0 in R2. If instead we assume that v is given by (2.14) for
all θ ∈ [0,2π], then
u(x) = r2v(reiθ) = A(x2 −x2)+2Bx x ,
1 2 1 2
giving us the usual homogeneous degree two harmonic polynomials.
3. Structure of the free boundary
In this section we prove Theorem 1.5 and Theorem 1.8. First we will use the
implicit function theorem to show that (1.2) implies that either ϕ is a constant
function or {∇ϕ = 0} is countably (n−1)-rectifiable. One immediate consequence
is that {∇ϕ (cid:54)= 0} is either empty or an open dense subset, which we use to prove
Theorem 1.5. Another immediate consequence is Theorem 1.8.
Lemma 3.1. Let p ∈ (1,∞) and ϕ ∈ C2(B ) such that (1.2) holds true. Then
1
either ϕ is identically constant on B or {∇ϕ = 0} is countably (n−1)-rectifiable.
1
Proof. First we will show that (1.2) implies that either ϕ is identically constant on
B or
1
c
|D2ϕ| ≥ 0 in B , (3.1)
1
n+p−2
where |D2ϕ(x)| denotes the operator norm of the matrix D2ϕ(x).
By (1.2),
(cid:12) (cid:12)
(cid:12) (cid:104)∇ϕ,D2ϕ∇ϕ(cid:105)(cid:12)
c ≤ (cid:12)∆ϕ+(p−2) (cid:12) ≤ (n+p−2)|D2ϕ| in {∇ϕ (cid:54)= 0}.
0 (cid:12) |∇ϕ|2 (cid:12)
10 ALESSIO FIGALLI, BRIAN KRUMMEL, AND XAVIER ROS-OTON
Hence, noting that ϕ ∈ C2(B ), we can express B as the union of the disjoint sets
1 1
(cid:110) c (cid:111)
|D2ϕ| ≥ 0 and int{∇ϕ = 0},
n+p−2
which are both relatively open and closed in B , and use the connectedness of B
1 1
to reach our desired conclusion.
Now, suppose (3.1) holds true. Let x ∈ B ∩{∇ϕ = 0}. By (3.1), D2ϕ(x ) has
0 1 0
rank k ≥ 1. Hence after an orthogonal change of variables, we may assume that
(cid:18) (cid:19)
A 0
D2ϕ(x ) =
0 0 0
for some diagonal k×k matrix A with full rank. By the implicit function theorem,
there is an open neighborhood of x in which M = {D ϕ = 0 for i = 1,2,...,k} is
0 i
a C1 (n−k)-dimensional submanifold and {∇ϕ = 0} ⊆ M. Therefore {∇ϕ = 0} is
countably (n−1)-rectifiable. (cid:3)
Next we will prove Theorem 1.5. For this, we will need the following Lemma.
Lemma 3.2. Let ϕ ∈ C2(B ) be a function satisfying (1.2). Let x ∈ B be such
1 0 1/2
that ∇ϕ(x ) = 0. Then, there exists ε > 0 and δ > 0 such that
0
(cid:18) |x|2(cid:19)
∆ ϕ(x)+ε ≤ 0 in B (x ).
p δ 0
2
The constants ε and δ depend only on the modulus of continuity of D2ϕ and on the
constant c in (1.2).
0
Proof. We may assume x = 0. Let us denote
0
(cid:0) (cid:1)
∆(cid:101) w := |∇w|2−p∆ w = |∇w|2−pdiv |∇w|p−2∇w
p p
(cid:104)∇w,D2w∇w(cid:105)
= ∆w+(p−2)
|∇w|2
= ∆w+(p−2)∆ w
∞
wherever ∇w (cid:54)= 0. We know that by (1.2)
∆(cid:101) ϕ ≤ −c < 0 in B ∩{∇ϕ (cid:54)= 0}, (3.2)
p 0 1
and we want to show that ∆(cid:101) (ϕ+ 1ε|x|2) ≤ 0 in B ∩{∇ϕ+εx (cid:54)= 0}.
p 2 δ
Let
λ (x) = λ (x) ≤ λ (x) ≤ ··· ≤ λ (x) = λ (x)
min 1 2 n max
denote the eigenvalues of D2ϕ(x) and λ = λ (0), λ = λ (0), and λ =
i i min min max
λ (0). By continuity of D2ϕ, we have that λ (x) are continuous in x.
max i
Case 1. Assume first that λ ≤ 0, i.e., λ ≤ 0 for all i = 1,...,n.
max i
Noting that
λ (x) ≤ ∆ ϕ(x) ≤ λ (x)
min ∞ max