C OMPOSITIO M ATHEMATICA
S TEPHEN J. G ARDINER
A fine limit property of functions superharmonic outside a manifold
Compositio Mathematica, tome 83, n
o2 (1992), p. 239-249
<http://www.numdam.org/item?id=CM_1992__83_2_239_0>
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A fine limit property of functions superharmonic outside
amanifold
STEPHEN J. GARDINER
Department of Mathematics, University College, Dublin, Ireland Received 7 November 1990; accepted 22 August 1991
Abstract. Let (X’, X") denote a typical point of Rn = RI x Rn-k, where n > 3 and 1 k n - 2. Also, let
E={ |X"| f( IX’I)},
where f : [0, co) - [0, oo) is increasing. A necessary and sufficient condition is given for E to be thin at the origin. This, in turn, is used to study the behaviour of functions u whichare superharmonic on the complement of a C2 k-dimensional manifold S. In particular it is shown that, if u - does not grow too quickly near S, then IX - Yin -
2U(X)
has a finite non-negative fine limitas X ~ Y, for any Y E S.
© 1992 Kluwer Academic Publishers. Printed in the Netherlands.
1. Main results
A set E in Euclidean space R" is said to be thin at a
point
Y if there is asuperharmonic
function u on aneighbourhood
of Y such thatThe classical criterion of Wiener
[7,
Theorem10.21]
characterizes thinness at Y in terms of the convergence of a seriesinvolving
the Newtonian(outer) capacity
of the sets
E n { 2-j-i~ |X - Y| ~ 2 - j}, where jE N. (Here 1 XI
denotes the Euclidean norm ofX.)
The notion of thinness isimportant
in thestudy
of theDirichlet
problem:
aboundary point
Y isregular
for the Dirichletproblem
on(an
openset)
Q if andonly
ifRnB03A9
is not thin at Y In this context a classicalexample
of a set which is thin at theorigin
inR3
is the"Lebesgue spine"
definedby {(x,
y,z):
x > 0 andy2+Z2~ e-clx},
where c > 0(see [7, p. 175]).
Our firstresult
gives
asimple geometric
characterization ofspine-like
sets which are thinat the
origin
O. Let X= (X’, X")
denote atypical point
ofR" = Rk x Rn - k,
wheren > 3 and
k E {l, 2, ... , n - 2}.
THEOREM 1. Let
E = {X: IX"I f( IX’I)}, where f: [0, ~) ~ [0, oo) is
increas-ing.
Then E is thin at 0if
andonly if
The
axially symmetric
case(k =1)
of Theorem 1 has beengiven by
severalauthors under the
stronger hypothesis
thatf (t)/t
isincreasing.
In this form it appears in the recent bookby Hayman [6,
Theorem7.15],
where it is attributedto Câmera
[3]. However,
it can also be found inArmitage [1]
and Port and Stone[8, Chap. 3, Prop. 3.5].
The case k = n - 3 wasrecently
establishedby Burdzy [2,
Theorem2.4] using probabilistic
methods. The case k = n - 1 does not appear in Theorem 1 because a set of the form{(X’,xn)ERn-l
x R :lx.1 f(|X’|)}
is thin at 0 if andonly
if the(increasing) function f
is valued 0 on[0, e)
for some e > 0.(Hyperplanes
are non-thin at each constituentpoint.)
InSection 2 we deduce Theorem 1 from Wiener’s criterion and estimates of the
capacity
of certainellipsoids given
inHayman [6].
The
fine topology
on R" is the coarsesttopology
for which allsuperharmonic
functions are continuous. Thus a
superharmonic
function u on an open set has afine limit at every interior
point.
It also has thefollowing property,
which we shall labelby (P): IX - Yin - 2U(X)
hasa finite non-negatiue fine
limit as X ~Y, for
every interior
point
Y(In fact,
this limit isequal
tou( {Y}), where y
is the Rieszmeasure associated with u : see
[5, 1.XI.4].)
The connection between thin sets and the finetopology
isgiven by
the fact that a set E in R" is thin at apoint
Y if andonly
if Y is not a fine limitpoint
of E.We will use Theorem 1 to establish a fine limit
property
ofsuperharmonic
functions defined on the
complement
of a k-dimensional manifold. Let E be arelatively
closedpolar
subset ofB(1),
whereB(X, r)= {Y: IY -XI r}
andB(r)
=B(O, r).
If u is apositive superharmonic
function onB(1)BE,
then u has apositive superharmonic
extension toB(1) (see [7,
Theorem7.7])
and soproperty (P)
holds. Thepositivity requirement
on u can be relaxed here: if u issuperharmonic
onB(1)BE
and there is anegative
subharmonic function s onB(1)BE
such that u > sthere,
then u can berepresented (outside
apolar set)
asthe difference of two
positive superharmonic
functions onB(1)
and(P)
continuesto hold. Now suppose that E takes the form
{(X’, 0"): X’c- R’l.
If we writeu- =max{O, -u},
then the abovereasoning
shows that anysuperharmonic
function u on
B(1)BE
which satisfieswill have
property (P).
The next result shows that(P)
remains true undersignificantly
weakerassumptions
on thegrowth
of u -, where the Rieszdecomposition
theorem does notapply
in an obvious way. LetS={XEB(1):03A6(X)=O"}, where 03A6:B(1)~Rn-k
is aC2
function whose de- rivative matrix has full rankthroughout B(l),
and letdist(X, S) =
inf {|X - YI: Y E S}.
THEOREM 2. Let g:
(0, 1]
~(0, oo)
be adecreasing continuous function
such thatIf u
is asuperharmonic function
onB(1)BS satisfying u-(X)~ g(dist(X, S)),
thenIX - Y|n- 2U(X) has a finite non-negative fine limit u*(Y) as
X ~Y for any Y E B(1).
Further,
theset {Y E B(r): u*(Y)
>el
isfinite for
each r E(0, 1)
and each e > 0.Theorem 2 is the main result of the paper. It can be
regarded
as an interior fine limitanalogue
of a resultof Rippon [10,
Theorem3]
on minimal fine behaviour of subharmonic functions. We will prove it in Section 3using
Theorem1,
ideasfrom
[9, 10],
and estimates of thebalayage
of the functionX ~ |X|2 - n
relative tocertain sets which are thin at the
origin.
Proof of Theorem 1
2.1. For
a, b, c
> 0 we define the setsLet
W(A)
denote the Newtoniancapacity
of anarbitrary
Borel(and
hencecapacitable)
set A~ R". We refer to Helms[7, Chapters 7, 10]
for basic resultson
capacity.
In addition werequire
thefollowing
result fromHayman [6,
p.432]
concerning
thecapacity
of theellipsoid Ek(a, b).
LEMMA A. As
bla ~ 0,
thefollowing quantities
tendto finite positive
limits cn,k(depending only
on n andk):
2.2. We
begin
withthe if part
of Theorem 1. So letf: [0, ~ )~ [0,~)
beincreasing,
let E ={X: IX"I f (|X’|)},
and assume that(1)
holds. It follows thatthe series
converge. In
particular, f(2-j)/2-j ~
0. SinceK, (al,,1’2, b/#2) ~ Ek(a, b),
wehave
Thus,
for allsufficiently large j,
by
Lemma A. It now follows from Wiener’s criterion that E is thin at O.2.3. It remains to prove the
only if part
of Theorem 1. So let the functionf : [0, oo)
~[0, oc)
beincreasing
and assume that the set E ={X: |X"| f( |X’|)}
is thin at O.
Let 03B4e(0,1)
be chosen smallenough
sothat,
for 0b/a 03B4,
thedisplayed quantities
in Lemma A lie in the interval[2cn,k/3, 4cn,k/3].
Leth(t)
=min{f(t), 03B4t}
on[0, oo)
andEh
={X: IX"J h(|X’|)}.
SinceE, g E,
it follows thatEh
is also thin at O.Hence, by
Wiener’scriterion,
we havewhere d =
22+nI2,
and soUsing
thesubadditivity property
ofcapacity
and LemmaA,
we obtainprovided 03B4~ 1/d.
Hence the seriesconverge, and it follows that
The convergence of these
integrals
and themonotonicity
of himply
thath(t)lt
~ 0 as t ~ 0 +. Henceh(t)
=f (t)
for allsufficiently
small t,establishing (1).
The
proof
of Theorem 1 is nowcomplete.
3. Proof of Theorem 2
3.1.
Let g
be as in the statement of Theorem 2.By adding
a suitable function if necessary, we can assumethat g
isstrictly decreasing
and unbounded on(0, 1].
There is also no loss of
generality
inassuming
thatg(1)
= 1. Letf
denote theinverse of the
increasing
function t H{g(t)} 1/(2 -n).
We aregoing
to show that(1)
follows from
(2).
Let
ô, e
E(0, 1)
andk E { 1, ... , n - 3 } . Using
thedecreasing property of g
and(2)
we have
Hence
as e - 0 +.
If k =
n - 2,
thenIt follows that
log(1/t)
=o(log(1 / f(t))
as t - 0 +.Thus,
forsuitably
small a > 0 and s E(0, a),
we haveas 8 ~ 0 +,
using (2).
It follows from
(2)
that(1)
holds for allk E {l, ... , n - 2}. Hence, by
Theorem1,
the setE = {X: IX"I 2f( |X’|)}
is thin at O.3.2. Let
03A6,
S be as in theparagraph preceding
Theorem2,
let r E(0, 1),
and letZ E S n B(r).
From theimplicit
function theorem we can(using
a suitable newcoordinate
system
centered atZ)
find aC2
function03C8: Rk~Rn-k and
numbers ar > 1 and p, > 0(depending
on r and 4) but not onZ)
such thatand
It can be
arranged that pr
E(0, 1/(4ar)). Further,
sincef (t)/t
~0 as t ~ 0 +(where f
is as defined in Section3.1),
we can choose Pr to besufficiently
small so thatf (t)/t ~ 1/4
fortE(0,2Pr).
Thus p, nowdepends
also on g. We can also find anumber
b,
> 0(depending
on r and (D but not onZ)
such thatThis new coordinate
system
will remain in force in what follows.3.3. Let F : Rn~ R" be defined
by F(X’, X") = (X’,
X" +03C8(X’)),
let E be as definedat the end of Section
3.1,
letE1 = {XeE: 1 X’l p, 1
andE 2 = F( El). Further,
letvl, v, denote the
balayage
of the fundamental functionX ~ |X|2 -n
relative tothe sets
E1, E2 respectively.
LEMMA 1.
(i)
The setE2 is
thin at O.To prove the
lemma, let u
be the measure associated with the Newtonianpotential
v,, and let w be thepotential corresponding
to the measure v definedon Borel sets A
by v(A) = u( {X: F(X)
EAI).
SinceEl is
thin at 0(by
Section3.1)
we have
u({0}) = 0,
and hencev({O})=O
also.If X EE1,
thenand so
|F(X)| ~ 31XI/4. Using (4)
and(5)
we haveIt follows that
E2
is thin at0,
and also thatIt remains to prove
(ii).
IfY E E1 (so that |Y"l ~ 2f (| Y’|) ~ I Y’ I /2 by
ourchoice
of pr
in Section3.2)
andIX"J
>21X’I,
then|X - Y|~ 31 Y115. Using (5)
wehave
and so
Combining (6)
and(7)
we obtain(ii).
The lemma is nowproved.
3.4. Let
E2
be as above and let U ={X: IX’I
p,,,IX"I Pr}BE2
Sincewe have from
(3)
thatdist(X, S) ~ f (|X’| ) ) (X E U). (8)
Now let u be as in the statement of Theorem 2. For X E U we have
dist(X, S) pr#2 2pr
and sof (dist(X, S)) ~ dist(X, S).
Henceusing (3)
and(8).
IfIX" - 03C8(X’)|
~IX"1/2,
then(9)
shows thatu -(X) C (32/ |X|2)n - 2/2.
Otherwise we have|X" -03C8(X’)| IX"I/2,
whenceand
(9)
now shows thatu - (X ) ~ (2/|X|)n-2.
In either case we thus haveWe now know that
|X|n-2U(X)
is bounded below on U.Also,
0 is anirregular boundary point
of the open setU, by
Lemma1(i).
It follows(see
Doob[5, 1.XI.21])
that|X|n-2u(X)
has a finite fine limit 1 as X ~ O. Inparticular (see [4]), rn-2u(rY)~
l as r~ 0 + for allYE8B(1)BA,
where A is somepolar
set.So,
if l 0 and we choose Y EaB(1)BA
such thatY" :0 0",
thenu(rY) (1/2)r2- n
for all
sufficiently
small r > 0.Combining
this with ourhypothesis
on u -, it follows thatg(t)
dominates apositive multiple
oft2 -n
on some interval of theform
(0, n), where 1
> 0.This,
in turn, contradicts(2).
Hence 1 ~ 0.We have shown that
|X - Z|n-2u(X)
has a finitenon-negative
fine limit asX~ Z for any
Z E Sn B(r).
Sincer E (0, 1)
wasarbitrary,
and sinceproperty (P)
holds
automatically
onB(1)BS,
the first assertion of Theorem 2 is nowestablished.
Before
proving
the final sentence of Theorem2,
we make some furtherobservations. We claim that
To see
this,
we denote the left-hand sideof ( 11 ) by - s,
so that s is subharmonicon U.
Further,
Hence the function
s+
is subharmonic onRnB{o},
if weassign
it the value 0 outside U.Also,
the thinnessof E2
at 0implies (see [5, 1.Xl.3])
thatIX ln - 2V2(X),
and hence
IXln-2s+(X),
has fine limit 0 at 0. Thus there is an open setE3
cU,
thin at0,
such thatSince
s(X) ~ (8n-2+l)|X|2-n
onE3,
and since the surface area measure ofE3noB(R)
tends to 0 asR ~ 0,
we now haveRn-2L(s+,R)-+0
asR ~ 0,
whereL(s +, R)
denotes the mean ofs +
overêB(O, R).
It follows from easy estimates of the Poisson kernel forR"BB(R)
thats +
= 0 onRnB{o}, proving (11).
From Lemma
1(ii)
we now havefor X E U
satisfying IX"I
>2|X’|. By
the thinness ofEl
at 0 it follows thatfor some
suitably small ô,
> 0(depending
on r, but not onZ).
3.5. We are now in a
position
to establish the final assertion of Theorem 2.Suppose that,
forgiven rE(O, 1)
and 8 >0,
theset {YEB(r): u*( Y)
>el
is infinite.Then we can find a
convergent
sequence( Y)
ofpoints
in this set with some limitZ. Because the Riesz measure associated with u is
locally
finite inB(1)BS,
we canconclude that
Z E S n B(r).
We choose a new coordinatesystem
centered at Z as in Section 3.2 for thefollowing
discussion.There are three cases to consider. The first is where
By selecting
a suitablesubsequence
of(Yj)
we can find ilE (o,1 )
such thatB(Y, 3n|Yj|Yj)
isdisjoint
from S forall j
E N.Applying
the minimumprinciple
onthe set
B(Yj, 2n |Yj|),
it follows thatSince
(1 - n) | Yj| ~ |X| ~ (1 + n])| Yj| in Bj = B(Jj, 1]1 Jjl),
we now havefor
X E Bj.
Since we haveSince
UjBj
isclearly
non-thin at 0and n E (0, 1)
can bearbitrarily small,
weobtain a contradiction to the fact that
u*(O)
is finite.The second case is where
infinitely
many of the(Yj)
are in S.By taking
asuitable
subsequence,
we can assume thatY E S
forall j.
Let il E(0, 1/4)
and letIt follows from
(12) that,
for allsufficiently large j,
Since
|X| ~ (1- 4n)/|Yj| for X E Bj,
we now haveThe set
Uj Bj
is non-thin at0,
so theright-hand
sideof (13) is a
lower bound foru*(O). But n > 0
can bearbitrarily
small. Henceu*(O) =
oo, which is acontradiction.
The final case is where
dist(Yj, S)l | Y|~
0as j
- oo, butonly finitely
many ofthe
points Y
are in S. These fewpoints
can beignored.
Werequire
a suitablemodification
of (11),
as follows. Let Y be suchthat |Y| pr/2 and | Y"|
>21 Y’|,
and let 1 be the fine limit of
|X - Yin - 2U(X)
as X - YReasoning
as in theproof
of
(11),
it can be seen thatfor X E
U,
where vy denotes thebalayage
of the functionX~ |X - y| 2 - n
relativeto the set
E2.
It follows from estimates in Section 3.3 thatIX - Y| ~ 7|X|/25
forX E
E2.
Hencevy ~ (7/25)2-nv2
on R".Also, IX - Y|2 -n~ (2|X|)2-n
for|X| > 1 YI.
Combining
these observations andusing
Lemmal(i),
we obtainfor some
suitably small dr
> 0(independent
of Y andZ).
Theremaining argument
is now similar to that for the second case, with(14) replacing (12).
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