A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
L UCIO D AMASCELLI F ILOMENA P ACELLA
Monotonicity and symmetry of solutions of p-Laplace equations, 1 < p < 2, via the moving plane method
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 26, n
o4 (1998), p. 689-707
<http://www.numdam.org/item?id=ASNSP_1998_4_26_4_689_0>
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Monotonicity and Symmetry of Solutions of
p-Laplace Equations, 1 p 2,
via the Moving Plane Method
LUCIO DAMASCELLI - FILOMENA PACELLA (1998),
Abstract. In this paper we prove some
monotonicity
and symmetryproperties
of
positive
solutions of theequation -
div Du) =f (u) satisfying
anhomogenuous
Dirichletboundary
condition in a bounded domain Q. We assume1 p 2 and
f locally Lipschitz
continuous and we do notrequire
anyhypoth-
esis on the critical set of the solution. In
particular
we get that if Q is a ball then the solutions areradially symmetric
andstrictly radially decreasing.
Mathematics
Subject
Classification (1991): 35B50(primary),
35J60, 35J70(secondary).
1. - Introduction
In this paper we consider the
problem
where
0p
denotes thep-Laplacian operator Apu
= divDu),
p >1,
Q is a bounded domain inJRN,
N >2,
andf
is alocally Lipschitz
continuousfunction.
We are interested in
studying monotonicity
andsymmetry properties
ofsolutions of
(1.1)
independence
of thegeometry
of the domain Q.In the case p - 2 several results have been obtained
starting
with thefamous paper
[GNN] by Gidas,
Ni andNirenberg where,
among otherthings,
it is
proved that,
if Q is a ball and p =2,
solutions of( 1.1 )
areradially symmetric
andstrictly radially decreasing.
This paper had abig impact
notonly
in virtue of the severalmonotonicity
andsymmetry
results that itcontains,
Pervenuto alla Redazione il 21 ottobre 1997.
but also because it
brought
to attention themoving plane
methodwhich,
sincethen,
has beenlargely
used in many differentproblems.
Thismethod,
whichis
essentially
based on maximumprinciples,
goes back to Alexandrov[H]
andwas first used
by
Serrin in[S]. Quite recently
themoving plane
method hasbeen
improved
andsimplified by Berestycki
andNirenberg
in[BN]
with theaid of the maximum
principle
in small domains.Very
little is known about themonotonicity
andsymmetry
of solutions of( 1.1 )
when 2. In this case the solutions canonly
be considered ina weak sense
since, generally, they belong
to the space(see [Di]
and
[T]). Anyway
this is not adifficulty
because themoving plane
method canbe
adapted
to weak solutions ofstrictly elliptic problems
indivergence
form(see [D]
and[Dal]).
The real
difficulty
withproblem (1.1),
for2,
is that thep-Laplacian
operator
isdegenerate
in the criticalpoints
of thesolutions,
so thatcomparison principles (which
could substitute the maximumprinciples
in order to use themoving plane
method when theoperator
is notlinear)
are not available in thesame form as for p = 2.
Actually counterexamples
both to thevalidity
of com-parison principles
and to thesymmetry
results are available(see [GKPR], [Br])
for any p with differentdegrees
ofregularity
off.
Before
stating
our main theorems let us recall some known results about(1.1).
When Q is a ball in
[BaNa]
thesymmetry
of the solutions of( 1.1 )
isobtained
assuming
that theirgradient
vanishesonly
at the center.In
[GKPR] by
a suitableapproximation procedure
is shown that isolated solutions with nonzeroindex,
in suitable function spaces, aresymmetric.
A different
approach
is used in[KP] where, using symmetrization
tech-niques,
isproved
thatif p
=N,
S2 is a ball andf
isonly continuous,
butf (s)
> 0 for s >0,
then u isradially symmetric
andstrictly radially decreasing.
While we were
completing
this paper F. Brock told us that in[Br]
hegets
the
symmetry
result in the ball in the case1 p 2 or p > 2
butf
monotone.For other
symmetric
domains he shows that solutions are"locally symmetric"
ina suitable sense defined in
[Br].
His method does not usecomparison principles
but the so called "continuous Steiner
symmetrization".
A first
step
towardsextending
themoving plane
method to solutions ofproblems involving
thep-Laplacian operator
has been done in[Da2].
In thispaper the author
mainly
proves some weak andstrong comparison principles
for solutions of differential
inequalities involving
thep-Laplacian. Using
theseprinciples
headapts
themoving plane
method to solutions of(1.1) getting
some
monotonicity
andsymmetry
results in the case 1 p 2.Although
thecomparison principles
of[Da2]
arequite powerful
for 1 p2,
thesymmetry
result is notcomplete
and relies on theassumption
that the set of the criticalpoints
of u does not disconnect the caps which are constructedby
themoving plane
method.In this paper we use the results of
[Da2]
toget monotonicity
andsymmetry
for solutions u of( 1.1 )
in smooth domains in the case 1 p 2 without extra-assumptions
on u.To state our results we need some notations.
Let v be a direction in
R ,
i.e. v Eand v ~ =
1. For a real number Àwe define
(i.e. R~
is the reflectionthrough
thehyperplane 7~)
If h >
a (v)
thenS2~
isnonempty,
thus we setFollowing [GNN]
we observe that if Q is smooth and ~, >a (v),
with~, - a (v) small,
then the reflected cap~52~ ) ~
is contained in Q and will remain init,
at least until one of thefollowing
occurs:(i) 52 )
becomesinternally tangent
to a S2 at somepoint
not onT:’
(ii) 7
isorthogonal
to 8Q at somepoint.
Let be the set of those k >
a (v)
such that for each A E(a (v), À)
none of the conditions
(i)
and(ii)
holds and defineThe main result of the paper is the
following.
THEOREM 1.1. Let S2 be a bounded smooth domain in
JRN,
N >2,
and u E C1 (S2)
a weak solutionof ( 1.1 )
with 1 p 2. For any direction v andfor
 inthe interval
(a (v) , h (v)]
we haveMoreover
where Z =
{x
E Q :Du (x)
=0}.
Easy
consequences of Theorem 1.1 are thefollowing.
COROLLARY
1. l. If, for a
direction v, thesymmetric
withrespect
to thehyperplane {x
E v =0}
and~.1 (v) - ~,1 (-v) - 0,
then u issymmetric,
i.e.u (x) -
any xE Q,
anddecreasing in the
v directionin
S2o. Moreover av
> 0 inQõ B
Z.COROLLARY 1.2.
Suppose
that Q is the ballJRN
with center at theorigin
and radius R. Then u isradially symmetric
0for
0 r R.Note that the
previous
theoremimplies
also aregularity
result since from 0 inby
standardregularity results,
we deduce that ubelongs
to
C~(0)B{0}).
The
proof
of Theorem 1.1 islong
andtechnically quite complicated,
there-fore we would like to illustrate the main ideas
beyond it,
so toclarify
also therole of the smoothness of
The
starting point
is Theorem 1.5 of[Da2]
which ispresented
and extendedin Section 3
(Theorem 3 .1 ).
This theorem assertsthat,
once we start themoving plane procedure,
we must reach the maximalpossible position (see
the Definition 3.1 of~,2 (v)
in Section3)
unless the set Z of the criticalpoints
of ucreates a connected
component
C of the set where0,
which issymmetric
with
respect
to a certainhyperplane
is defined in(3.5))
and whereu coincides with the
symmetric
functionu~o~"~ .
Therefore all thesubsequent
efforts are in the direction of
proving
that such a set C cannot exist.A first
result,
deduced fromProposition 3.1,
is that if u is constant on a connected subset of criticalpoints
of 8C whoseprojection
on thehyper- plane
contains an open subset of7~)
then such a set C cannot exist.This is
proved by
a careful use of theHopfs
lemma andgives
aproperty
of the critical set of a solution u of( 1.1 )
which isinteresting
in itself(see Proposition 3.1 ).
As
explained
in Remark 3.1 thehypothesis
that u is constant on a connectedset of critical
points (which
could appearobviously
satisfiedby
anyC
Ifunction)
is not unnecessary, since could not hold if the critical set of u is very
singular.
Therefore,
if the critical set Z of u is not very bad the assertion of Theorem 1.1 is a consequence of Theorem 3.1 andProposition 3.1,
withoutexploiting
any smoothnessassumption
on aQ(see
Remark3.1 ).
But,
of course, one cannot ingeneral
have "apriori"
informations on the critical set Z.Hence,
to prove that u is constant on a connected subset ofcritical points
of 8C whoseprojection
on thehyperplane
contains anopen subset of the
hyperplane,
some extra work is needed.To do that a new
argument
ispresented
in Section 4 and consists inmoving hyperplanes orthogonal
to directions close to v in order to prove that the "bad"set C is also
symmetric
withrespect
tonearby hyperplanes
and hence(see Step
3 in theproof
of Theorem 1.1 in Section4)
on itsboundary
there is atleast one connected
piece
where u is constant, Du =0,
and whoseprojection
on the
hyperplane Ty
(v) contains an open subset of thehyperplane.
This
procedure of moving nearby hyperplanes
to beefficient,
needs a certaincontinuity
of the minimal and maximalpositions
of thehyperplanes 7~
withrespect
to v. It is this verycontinuity property
that is ensured whenever is smooth and fails forsimple
domains with comers, as forexample
the square in theplane.
We think that this method oflooking
at directions close to a fixed direction v could be useful also in otherproblems.
The paper is
organized
as follows. In Section 2 we state the maximumand
comparison principles
needed in thesequel.
In Section 3 we recall andextend Theorem 1.5 of
[Da2]
andpresent
some results forgeneral
domains. InSection 4 we
give
theproof
of Theorem 1.1 and its corollaries.2. - Preliminaries
In this section we recall some known results about solutions of
equations involving
thep-Laplacian operator.
Webegin
with a version of thestrong
maximumprinciple
and of theHopf’s
lemma for thep-Laplacian.
It is aparticular
case of a resultproved
in[V].
THEOREM 2.1
(Strong
MaximumPrinciple
andHopf’s Lemma).
Let Q be adomain in
JRN
and suppose that u1 (Q),
u > 0 inQ, weakly
solveswith 1 p oc, q > p -
1,
c > 0 and g ELoo
0 then u > 0 in Q.Moreover
for
anypoint
Xo E a Q where the interiorsphere
condition issatisfied,
and such that u
1 (S2
Ulxo 1)
and u(xo)
= 0 we have thatas
> 0for
any inward directional derivative(this
means thatif y approaches
xo in a ballB C
S2 that has xoon its
boundary
thenlim u(y)-u(xo)
>0).
on ltS oun ary t en ly ol > .
Next we recall some weak and
strong comparison principles,
whoseproofs
can be found in
[Da2] (see
Theorem1.2,
Theorem 1.4 therein and the remarksthat
follow).
_Suppose
that Q is a bounded domain in and that u, v EC (Q) weakly
solve
with
f :
R - Rlocally Lipschitz
continuous. For any setA c
Q we defineand denote
by I A I
itsLebesgue
measure.THEOREM 2.2
(Weak Comparison Principle). Suppose
that 1 p2,
then there exist a, M >0, depending
on p,( S2 I, M~
and the L 00 normsof
u and v suchthat:
if
an open setQ’ C
S2satisfies
SZ’ = A1 U A2, A 1 n A2 ~ [
=0, I
a,MA2
M then u v on a Q’implies
u v in Q’.THEOREM 2.3
(Strong Comparison Principle). Suppose
that 1 p oo anddefine Zu - {x
E S2 :Du(x)
==Dv(x) =0}.
and there existsxo E S2 B Zu
withu (xo)
=v (xo),
then u == v in the connected componentZ~
containing
xo.REMARK 2.1. A function
f :
R - 1R islocally Lipschitz
continuous if andonly
if for each R > 0 there existC2 (R) >
0 such that(i) fl (s)
=f (s) - Cis
isnonincreasing
in[-R, R]
(ii) f2(s)
=f(s)
+C2s
isnondecreasing
in[ - R , R ] .
In the
proof
of Theorem 2.2only (i)
isused,
while theproof
of Theorem 2.3only exploits (ii).
3. - Results for
general
domainsHere we prove some
monotonicity
andsymmetry
results for bounded do- mains which do not need to be smooth. From now on, p will be a fixed number in the interval(1,2), Q
a bounded domain inlRN, N > 2,
andf :
1R --+ R alocally Lipschitz
continuous function. For any direction v leta (v), Q~, ~52~~
be as defined in Section 1. Next we define
and,
ifA 2 ( 1)) i-
0We observe that if Q is smooth then
where A
I (v)
is defined as in Section 1.If
a(v) ~, ~,2 (v),
X E U we setwhere x(
is as in(1.4),
Finally
we defineIf
Ao (v) ~ ~
we setObviously
we haveÂ2(v).
Let us now state a first result which is a different formulation and an
extension of Theorem 1.5 in
[Da2].
Wepresent
the details of theproof,
sincewe need them in the
proof
of Theorem 1.1.THEOREM 3.1. Let u E
be a weak solution of ( 1.1 ),
with 1 p 2. For any direction v such that1~2(v) ~ ~
we have thatAO(v) =,4 0 and,
then there exists at least one connected
component
CVof S2~o~v~ B zvo(v)
such thatu = in C".
u =
Xo (v)
In .For any such
component
CV we alsoget
Moreover for
any ~, witha (v)
Àko (v)
we haveand finally
PROOF. Since the
proof
isquite long
we divide it in threesteps.
STEP 1. Let v be a direction such that
A2(V) :A 0.
Fora ( v ) À :S h2(v)
we compare the functions u and
uv
in the open setQX using
Theorem 2.2 andTheorem
2.3,
sinceu ~
satisfies inQX
the same Inparticular,
sincewe can fix a, M >
0, indipendent
fromÀ,
so that Theorem 2.2applies
inQf
to u and v =
If ~, >
a (v)
and h -a (v)
is small thenI
is small. Moreover we haveon the
inequality uv,
since 0 =uv
on aQ n while u =uv
on n
7~ by
definition.Therefore, by
Theorem 2.2 we have thatuf
inS2~
for >~(p), ~ 2013 a (v) small,
so that 0.STEP 2. In the cap
by continuity,
theinequality
holds.Moreover,by
Theorem 2.3 we have that if Cv is a connectedcomponent
of then either u in C~ or u - in C".Suppose
now thath2(v)
and assume,arguing by contradiction,
thatu in
QXo(v) B
Let us choose an open setA,
with c A csuch that
+ M (this
ispossible
sinceDu = = 0 in
Zvo(v) ).
We also fix acompact
set K c such thatK ( 2 (a
and Mbeing
the numbers fixed inStep 1).
IfK B A ~ 0, by
ourassumption
the function u ispositive there,
andsince K B
A iscompact
we have thatminKBA u)
= m > 0.By continuity
there existsE > 0 such that
Ào(v) + E h2(v)
and such that forÀo(v)
hho(v)
+ c we haveand,
if I~B A ~ 0,
(in particular uX -
u > 0 ona (I~ B A) ).
Moreover for such values of h wehave that u
uv
on a(S2~ B (K B A))
because if xo is apoint
on thatboundary
either xo E
aQv
where uuv
asalready
observed inStep 1,
or xo Ea (I B A)
where
u ( -
u ispositive.
Since is the
disjoint
union of =A
I(with
smallmeasure)
and K n A =
A2 (where
thegradients
aresmall),
from Theorem 2.2 weget
that in
Since,
if we have thatuv-u
> 0 inK BA,
we obtain thatuv
inQÀ
forho(v)
h~,o(v) -~-E.
This contradicts the definition of and shows that ifh2(v)
then it is notpossible
that u in
B
so that there exists at least one connectedcomponent
Cv ofB
such that u - in Cv.If Cv is such a connected
component, by definition + [
> 0 inCv,
but since u =uvo(v)
we also have thatDu =1=
0 inCv,
i.e.(3.6).
Finally, by
the very definition ofCv,
we have(3.7)
since 9C" C U 8Q U Z(Z
is defined in(3.4)).
0 (v) STEP 3. To prove
(3.8)
isenough
to show thatIn fact if
(3.8)
is false andu(xo) = uf(xo)
for apoint
xo EQX B Z~ ,
thenu =
uX
in thecomponent
ofQ’ B Z’
to which xobelongs,
and thisimplies
thatboth IDu(xo)1 and IDuXo(v)(xo)1 I
are not zero, i.e. xo EQX B
Z so that(3.10)
does not hold.
Let us now prove
(3.10)
and assume, forsimplicity
ofnotations,
thatv = e 1 =
(1, 0,..., 0).
We write coordinates in asx = ( y, z) with y
ER,
.z E
R N-
1 and we omit thesuperscript v
= el inuv,
etc.Suppose, by contradiction,
that there exists it, witha (el )
ILand xo
= (yo, zo) E Q, B
Z such thatu (xo)
=By
Theorem 2.3 we have that u * u, in thecomponent
C ofQ, B Z,
to which xobelongs.
If h > it and À - IL is small we have that(xo)x
=(x),,
where x =(y, zo)
is apoint
ofC with y yo, so that
u(x)
= =u(xo),
since for A > it,~. - A
small(more precisely
forh s Ào)
theinequality
ux holds inQx.
Soif y yo, with yo - y small we
get
theinequality u(y, zo) a u(yo, zo),
whichimplies
thatu(y, zo)
=u(yo, zo)
because u isnondecreasing
in theel-direction
in
QÀo’
Therefore the setand
is
nonempty.
Let us now defineWe claim that x 1 =
(y 1, zo)
E a SZ . In fact suppose that x 1 E Q and setÀl
=By continuity u(xi )
=u(xo)
and Xi 1 E since(xi),xl
= xoand 0. Moreover from Theorem 2.3 we obtain that u - in the
component
ofQx, B
to which x 1belongs,
which in turnimplies
that0,
as before.Repeating
theprevious arguments, with it
and xosubstituted
by ~,1
and xl, we obtain thatu((y, zo)) = u((yo, zo))
for y yl, yl - ysmall,
and this contradicts the definition of yl. So xl E and 0 =u (xl )
=u(xo)
> 0. This contradiction proves(3.10)
and hence(3.8).
Finally (3.9)
is a consequence of(3.8)
and the usualHopf’s
lemma forstrictly elliptic operators.
In fact let x =(À, z)
EZ,
I,e, hÀo
andD u,(x ) ~ 0.
In a ball B =Br (x )
we havethat
c >0,
so thatI Du 1, 1 Dux I
> E > 0 in B f1Qx.
Thisimplies by
standard results that u EC2(B)
andthat
the difference u satisfies a linearstrictly elliptic equation L (u~, - u) = 0 (see [S]
and also[BaNa]).
On the other hand wehave, by (3.8),
that u > 0 in B n while
u (x) - ux (x)
because xbelongs
toTx.
Hence, by
the usualHopf’s
lemma weget
0 > 1-2aau (x)
1 i.e.(3.9)
holds. 0
Now we prove a
proposition
whichgives
a useful information on how the set Z of the criticalpoints
of u can intersect the capPROPOSITION 3.1.
Suppose
that u E is a weak solutionof ( 1.1 ),
with1 p 2. For any direction v the cap does not contain any subset r
of
Zon which u is constant and whose
projection
on thehyperplane
contains anopen subset
(relatively
to the inducedtopology).
PROOF. For
simplicity
we take v as thejci-direction
and denote apoint
xin
JRN as x = (y, z)
withZER N-1 .
As usual we omit thesuperscript
v = el in
Qv, uv,
etc.Arguing by
contradiction we assume that containsa set r with the
properties:
(i)
there exist y > 0 and zo ER N-1
suchthat,
for eachpoint (ho, z) E T~,o
with z - zo I
y there exists yho
with(y, z)
E r(ii) Du(x)
= 0 for all x E r(iii) u(x)
= m > 0 for all x E r.Note that
r
satisfies the sameproperties
as r and thatby (iii)
Let (o = wy be the
(N - 1)
dimensional ball centered at zo with radius y. We consider thecylinder
R x cv and denoteby E
the intersection of thiscylinder
with the cap Now we
distinguish
two cases.In this case we consider the "left"
part El
of E withrespect
to the setr, i.e.
where
=
inf ~ y
such that(y, z)
Ea2 (z)
=sup {y
such that(y ", z) ~ r
for anyy’ y} .
Note that
by (i)
and(iv)
the definition ofEl
makes sense andcrl (z) ho.
Moreover we haveand
In fact if
(y, z)
EEl
thenu ((y, z)) u((a2(z), z))
= m since uis nondecreasing
in the xl direction in and the
point (o~2(z), z) belongs
tor (otherwise
itwould have a
positive
distance fromr
and we would have that(y, z) V r
fory >
a2 (z)
and close toa~2 (z)).
Moreover in
El
since u = 0 on aQ and aQ n 0.Since
f (m) 0
we havefor any
A >
0. On the other side sincef
islocally Lipschitz
continuous there existsA >
0depending
on such that isnondecreasing
fors E
[0, (see
Remark2.1 ).
For such a value of A(3.13) gives
Now let us observe that for some
point
x’ on n r the interiorsphere
condition is
satisfied.
In fact let us takeyo G R
with 0152l(zo)
yo ~2(zo).
Sincedist
((yo, zo), r)
> 0 there exists E > 0 such that 0 E y,BE ((yo, zo))
CEl.
For y G R let
B(y)
be the ball centered at(y, zo)
with radius E and let us defineSince if
B(yl)
n r0
then dist(B(yl), r)
>0,
it is clear thatB(yl)
nr ~
ø.Moreover
B(y1 )
n r as it is easy tocheck from
the definition of yi.So c
El
and there existsx’
E88(yi)
n r withu(x’)
= mby (iii).
By (3.12), (3.14)
and theHopf’s
lemma(Theorem 2.1 )
we obtain that 0 for any interior(with respect
toEl )
directionalderivative,
which contradicts(ii).
CASE 2.
f (m )
> 0._
In this case we consider the
"right" part E,
=E B El
of E withrespect
to
r,
i.e.Again ~r
is well defined andby
themonotonicity
of u in we haveand
because if u were flat in
£r
thenf (m)
should be zero,against
ourassumptions.
As
before, exploiting
thelipschitzianity
off
and the fact thatf (m )
ispositive
we
get
for some
A >
0. From(3.15)
and(3.16) applying
theHopf’s
lemma(Theo-
rem
2.1 )
in apoint x’
weget (being u (x’)
= m =minw
> 0 for any interior(with respect
toEr )
directionalderivative,
which contradicts(ii).
C7REMARK 3.1. A first consequence of Theorem 3.1 and
Proposition
3.1 isthe
following. Suppose
that u is a solution of( 1.1 )
whose critical set Z isquite regular
and such that u is constant on any connectedcomponent
of the set Z of its criticalpoints. Then,
for any direction v, we have that~,2 (v) .
In
fact,
if for a direction v ithappens
thatho(v) ~,2 (v) then, by
The-orem 3.1 there exists a connected
component
Cv of such thatu - in C" . Thus acv would contain a set r on which Du = 0
(by definition,
see(3.7)),
u is constant(by assumption)
and whoseprojection
on thehyperplane
contains an open subset of this would beimpossible by Proposition
3.1.Note that in what we have
just
stated we have used theassumption
that uis constant on any connected
component
of the set Z of its criticalpoints.
At a
superficial glance
one could think that thishypothesis
isalways
sat-isfied
by
anyC
1 function: but this is not true andactually
thequestion
offinding
sufficient conditions on a connected set of criticalpoints
of aC
I func-tion g
which ensurethat g
is constantthere,
is verydeep
andcomplicated.
Asa matter of fact there is a famous
counterexample
due toWhitney ([W])
onthis
subject
as well as manysubsequent
research papers(see [N])
which showthat this
question
isstrictly
related to Sard’s lemma and thetheory
of fractalsets.
Roughly speaking
in our case one could say that if u is not constant onthe connected
components
ofZ,
then Z isgeometrically
verycomplicated
andcontains fractal subsets.
Also in view of some
counterexamples ([GKPR], [Br])
it could be reason-able to think that solutions u of
( 1.1 )
do not have such bad sets of criticalpoints,
so that theequality
=k2 (V)
should hold for any solution of( 1.1 )
in the case 1 p
2,
forgeneral
domains.4. - Proof of Theorem 1.1 and its corollaries
We
begin
with asimple topological
lemma that will be used later.LEMMA 4. l. Let
A, B
benonempty
open sets in atopological
space such that An B ~ ø, B Sf
A.If B is
connected then a A nB =1=
0.PROOF.
By
thehypothesis
A n B is anonempty
proper open subset of B.Arguing by
contradiction suppose that a A n B = 0. Since B is open wealso have that a B n B =
0,
andbeing a (A
n a A U a B we obtain that__
This
implies
thatB B (A n B) = B B (A n B)
is open,nonempty
anddisjoint
from A n
B, contradicting
theassumption
that B is connected. D COROLLARY 4.1. LetA, B
open connected sets in atopological
space andassume that A n
B =A ø, A fl
B. Then(aA
nB)
U(aB
nA) =,4
0.Now we prove Theorem 1.1 and its corollaries.
So,
from nowon, Q
will be a bounded smooth domain in
N >
2. Let v be a direction and~,2 (v),
as defined in Section 1 and Section 3. Since Q is smooth we have that0 # Ai (v) C n2(v),
so thatAo(~) 7~ 0 by
Theorem3.1,
and
a (v) X2 (V) -
In view of the definition of and of Theorem3.1,
Theorem 1.1 will beproved
if we show thatSince the
proof
istechnically quite complicated
we would like tohelp
thereader, spending
a few words about it. The idea of theproof
is to show thatif then there exists a small set r of critical
points
of u in thecap on which u is constant and whose
projection
on thehyperplane
contains an open subset of Once we show this we reach a contradiction with the statement of
Proposition
3.1. As observed in Remark3.1,
if thecritical set Z of u is not very "bad" then
ho(v) = ~,2(v).
From this would followh i (v)
and Theorem 1.1 would beproved. But,
since apriori
the critical set Z of u could be sonasty
that u is not constant on its connectedcomponents (see
Remark3.1 ),
to prove the existence of the set r we use Theorem 3.1 anda new method which consists in
moving hyperplanes orthogonal
to directions"close to v" and which
requires
the smoothness of aQ as observed in the introduction.As usual let v be a direction and define
0v
as the collection of the connectedcomponents
C" of such that u mu v
inCv,
0 inC",
Du = 0 on
aCv B
If
Ào(v) Àl (v)
thenho(v) h2(v)
so thatby Theorem
3.1.If this is the case and Cv E we also have u - in
CV,
so that= 0 since u = 0 on
a S2,
while > 0 inCVBT;oev),
becauseby
the definition of(see
Section1)
we have thatB
C Q.Hence there are
two
alternatives: eitherDu(x) = 0
for all x E inwhich case we define
ev = C,
or there arepoints x
E n suchthat 0. In this latter case we define
ev =
Cv UCl
UC2
whereCf
=C2 = {x 0).
It is easy to check thatev
is open andconnected,
with 0 inC~,
Du = 0 on9C".
Let us
finally
define the collection0v
_{ C" :
C v EREMARK 4.1. At this
point
a crucial remark for thesequel is the following:
if v 1, v2 are directions and C"1 E CV2 E then either
nev2
= 0or
--__
C’2. In fact if c,lnC’2 # 0
andC’2,
thenby Corollary
4.1either
n CV2
or isnonempty,
and this is notpossible
since0 in
CB
Du = 0 on i =1,
2.PROOF OF THEOREM 1.1. If v is a direction and 6 > 0 let us denote
by
the set
As
already
observed the theorem will beproved
if we show that~,1 (v)
for any direction v.
As in the
proof
of Theorem 3.1 we can fix a, M > 0 so that Theorem 2.2applies,
i.e. for any direction v,any k
E(a(v), ~,1 (v))
and any open subset Q’of the
inequality u X
on aS2’implies
theinequality u X
in Q’provided
Q’ =A 1 U A2 with IAll
a,MA2
=supA2 + IDuXI)
M.Suppose
now that vo is a direction such thatÀo(vo)
Sinceit follows that so that from Theorem 3.1
we
get Fo 0
0. SinceJRN
is aseparable
metric space and everycomponent
is open, contains at most
countably
manycomponents
ofS2~o~v°~ ~
so
=== {C~ , ~
i EI c
In case I isinfinite,
since thecomponents
aredisjoints,
we have that£Zi lCv 0 I
oo, so that we can choose no > 1 for whichIf I is finite let no be its
cardinality.
Let us then choose a
compact Ko
cB
so thatFinally
we take nocompact
setsKi
cC:o,
i =1,...,
no, such thatSo we have
decomposed
the setsKo, ~i,... ,Kno
and in aremaining
part
with measureIf A =
{x
E+ [ ~},
sinceKo B A
iscompact
andu
u vo
inKo B ~ by
Theorem2.3,
there exists m > 0 such thatSince Q is smooth the functions
a (v)
and defined in Section 1 arecontinuous with
respect
to v.By continuity
there exist Eo ,So
> 0 such that ifso
thenwhere
K’
is the reflection~ (Ki)
We now
proceed
in severalsteps
in order to show thatthere
existsi
I E{ 1, ... , n o }
and a direction v 1 E/80 (vo)
such thatC °
E0v
for any direc- tion in a suitableneighbourhood Is(vl)
of vl, and8C(°
contains a set r as inProposition
3.1(with respect
to the directionVI).
1
In what follows we
implicity
assume that E > 0 means 0eo, 6
> 0means 0
8 :S 80.
STEP 1. Here we show that the function is
continuous,
i.e. for eachE > 0 there exists 8 > 0 such that if V E
Is(vo)
thenMoreover for each v E
18(Eo) (vo)
we havePROOF OF STEP 1. Let E > 0
(E eo)
be fixed.By
the definition ofÀo(vo)
there exist h E
(Ào(vo),
-~-E)
and x E such thatu(x)
>By continuity
there exists~1
> 0 such that for every v EIs, (vo) x belongs
toQv
and
u (x )
> Thisimplies
that for all v EIs 1 ( vo)
we haveNext we show that there exists
62
> 0 such thatho(v)
>ho(vo) - e
for anyv E
Is2 (Vo). Suppose
thecontrary,
then there exists a sequence of directionssuch that vn -~ vo and
Up
to asubsequence
we have thatconverges to a
number k - Moreover,
since >a (vn )
and
a (vn )
-~a (vo),
we also have thata (vo) I.
Actually
thisinequality
is strict because the caps have measuregreater
than orequal
to the number a of Theorem 2.2(otherwise
wouldnot be maximal with
respect
to theinequality
usee the proof
ofStep
1in Theorem
3.1 );
thenalso
a, whichimplies
that >a (vo).
Now,
sinceho(vo), by (3.8)
of Theorem3.1,
we haveArguing
as inStep
2 of Theorem 3.1 we can construct an open set A cQvo
and a
compact
set K C such thatand,
if(a
and Mbeing
the usual numbers which come from Theorem2.2).
By continuity
there existr, 3
> 0 such thatif v E
Ir (vo)
and À E(h - 8, 1
+3).
For such values of v and
À, applying
Theorem 2.2exactly
as inStep
2of Theorem
3.1,
weget u ~
in This inparticular
holds for v = vn, h = for any nsufficiently large and il sufficiently small, contradicting
the definition of
Ào(vn).
Hence(i)
isproved.
Observe
that,
since weimplicity
assume that Eo~o, by (i)
and
(4.1 )
we have thatand
(4-2)-(4-5)
hold for v E =ho(v).
Let us now prove
(ii)
and fix a direction v ESuppose
that thereexist i E
{I, ... , no }
and apoint xi E Ki
such thatu(xi)
Since0, by
Theorem 2.3 weget
u = in thecomponent
CV E towhich xi belongs.
We also have that xi EC"
nbecause Ki
CCi 0
and
hence, by
Remark4.1, C" ---
and(ii)
isproved.
The other
possibility
is that for each i E{!,... no}
and every xE Ki
we have
u (x )
If this is the case,by (4.2)-(4.5)
theinequality
u holds in
except
for a set with measure less than a and for theThen, arguing
as in theproof
of(i) (i.e. repeating
theproof
ofStep
2 of Theorem3.1),
weget
theinequality
uuv
inQv
for h >À -