C OMPOSITIO M ATHEMATICA
R OBERT K AUFMAN
J ANG -M EI W U
Parabolic measure on domains of class Lip
12Compositio Mathematica, tome 65, n
o2 (1988), p. 201-207
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Parabolic
measure ondomains of class Lip 1/2
ROBERT KAUFMAN & JANG-MEI WU
Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA
Received 5 February 1987; accepted in revised form 12 August 1987 Compositio
© Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands
1. We shall construct a domain f2 in
R2x,t’
whoseboundary
isgiven by
thegraph of a Lip 2 function x
=F(t),
so that on an theparabolic
measure 03C9 and theadjoint parabolic
measure co* are concentrated on twodisjoint
sets, whoseprojections
onto the t-axis haveHausdorff
dimensionsstrictly
less than 1.We do not known how small the dimensions can be made in the
example.
But the dimensions must be at
least -1, by
a theoremof Taylor
and Watson[8;
p.337],
which states that a set E on aLip 2
curve x =F(t)
has heatcapacity
zero if andonly
if itsprojection
onto the t-axis has zero2 -capacity.
In a
previous
paper[3],
we constructed aLip 2
domain{x
>F(t)l
satisfying
the weakerproperty
that theprojections
ofsupports
of (O and cv*have zero 1-dimensional Hausdorff measure. There are two technical
improvements
made here: anexplicit
constructionof F(t)
isgiven
and shownto
satisfy
anexplicit inequality
in classLip -L,
and a more careful estimation ofparabolic
measure isnecessary. F is
alacunary
sum but the gaps are not toolarge
to obtain an estimate of the dimension.thé
size of the gaps is critical inobtaining
anexplicit
estimate of F inLip 1 2.
Because the
only diffeomorphisms
that preserve the solutions of the heatequation ((~2/~x2) - (~/~t))u
= 0 are{(x, t)
-(ax
+b, a2 t
+c)l, [2],
domains S2 =
{x
>f(t)},
withLip 2 boundary x
=f(t)
are very natural forstudying
solutions of the heatequation.
It follows from theorems of Pet-rowsky [7]
that these domains are Dirichletregular
for the solutions of the heatequation.
For a Borel set E c
~03A9,
we denoteby 03C9(x,t) (E) (or 03C9*(x,t (E))
theparabolic
measure
(or
theadjoint parabolic measure)
of E withrespect
toS2, i.e.,
the solution of the heat
equation (or
theadjoint
heatequation)
on 03A9with
boundary
value 1 onE,
0 on~03A9BE,
in the Brelot-Perron-Wienersense.
We say that co is
supported
on a Borel set scanprovided
that03C9(x,t)(~03A9BS) =
0 for every(x, t)
EQ;
andsimilarly
for 03C9*.202
We say that u is a
parabolic (or adjoint parabolic)
functionprovided
thatIn the case of
Laplace’s equation,
on theboundary
of a Jordan domainin
R2,
the harmonic measure is concentrated on a set of Hausdorff dimen- sion1 ;
and any set of Hausdorff dimension less than 1 has zero harmonicmeasure
[6].
For the heatequation,
weconjecture
that theparabolic
measureon the
boundary
of a Jordan domain issupported
on a set of"parabolic
dimension" at most 2. We refer to
[8]
for the definitionof parabolic dimension,
and recall that the
parabolic
dimension of the line{t
=01
is1,
of the line{x
= at +bl
is2,
and ofR2
is 3.The
parabolic
measure for{x
>01
issupported only by
sets of full linear measure, thereforeof parabolic
dimensionexactly
2. For theexample
to beconstructed,
theparabolic
measure issupported by
a subset of thegraph {x
=F(t)}
ofparabolic
dimension 2.2. A. function of class
Lip 2
Proof.
To prove(a)
we observe thatWe obtain each term inside the minimum
by using
one of theinequalities
onh. The numbers n =
1, 2, 3,
... are divided into three groups.(i)
N-n03B5|t
- s|1/2. Using
N-n in theminimum,
and N4,
we see thatthe contribution from this group is at most
281t - sI1/2N(N - 1)-1
303B5|t -
S|1/2.
(ii) Nn03B5-1/2|t - s| 81 t
-s|1/2.
The same estimateapplies,
if we usethe third term in the minimum.
(iii) 03B5|t - s|1/2
N-"03B5-3/2|t - s|1/2.
There can be at most
oriè
solution n to thispair
ofinequalities:
ifn1 n2
and both aresolutions, then (n2 - n1) log N log 03B5-5/2, so n2 -
n, 1 or n2 = n1. The contribution from(iii)
is thus at most21 t
- s|1/2.
Adding
up for(i), (ii), (iii) gives (a).
In view of
(a), inequality (b)
is evident.To prove
(c),
with s = 1:, we observe that n = kbelongs
to(iii).
However,
a moreprecise
estimate can begiven: N -k h(N 2k t) - N-kh(N2k03C4) = N -k h(N 2kt) = (t
-03C4)1/2.
The total contribution from(i)
and(ii)
is at most603B5(t - r)1/2.
Since 6e1/2, (c)
follows.3. Estimâtes of
parabolic
measureSuppose
that x =F(t)
isLip -1
on(-00, oo) with |F(t) - F(s)|
Mit - SI1/2.
For a >0, we denote by 0394(t, a) = {(F(s), s): Is - tl | al
andA(t, a) = (F(t)
+2MJa,
t +2a).
Lemma 1.4 in[5]
can be restated asfollows.
LEMMA 2. There is a
positive
constantC(M) depending
on Monly,
so thatif
u is anonegative parabolic function
on Q ~{x
>F(t)l, vanishing
on{(F(s), S): |s - t|
>al641,
thenWe may view Lemma 2 as a
quantitative
version of the Harnackinequality.
Given 0 e
10-4
and N03B5-4,
letf(t)
be thefunction defined
inLemma
1, F(t) = 22f(t)
and Q ={x
>F(t)l.
It follows from(a)
inLemma 1
that |F(t) - F(s) 1 9|
t -s |1/2
for all s, t. We fix a referencepoint (100, 100),
and denoteW(lOO,lOO) by
w.LEMMA 3. There is an absolute constant Co >
0,
so that whenever 03C4 = mN-2k
with m =0, 1,
... ,N2k - 1,
we have
03C9(Ek)
c.e’/2 w(lk)’
204
Proof
For a fixed L =m N-2k,
we let A =(F(t)
+5N-k ,
L +1 8N-2k),
Jk
={(F(t), t):
0 t - 03C41 4 N-2k}, and Lk
={(F(t), t):|t - 03C4| 1 16N-2k}.
Applying
Lemma 2 with M =10, a = /6 N-2k
and u =03C9(Ek),
we obtainwhere
C,
is an absolute constant.To estimate
wA(Ek)’
we define 03A6 to be the transformation:(x, t)
~(20J£
+(x - F(03C4)Nk,
e +(t
-!)N2k).
Hence03A6(A) = (20J£
+5, e
+1 8);
andbecause of
(b)
and(c),
Since 03A6 preserves
parabolic functions, u(Q) - 03C903A6-1(Q)(Ek)
is theparabolic
measure
of 03A6(Ek)
withrespect
to the domain03A6(03A9),
andu(03A6(A)) =
COA(Ek ).
Because u = 0 on
~03A6(03A9)
n{t 03B5},
we have u = 0 on03A6(03A9)
n{t
=03B5}.
Letwhich is
parabolic
for t >0,
and ispositive
when x >2t.
When(x, t)
e03A6(Ek), K(x, t) c203B5-3/2
for some absoluteconstant c2
> 0.Applying
themaximum principle to c-12 03B53/2 K and u over the region 03A6(03A9 ~ {03B5 t 03B5 + 1 4},
we have
where
C3
is an absolute constant. Since03C9A(Ek) = u(03A6(A)),
we conclude the lemma.REMARK. If we choose x =
Bf(t),
with B >22,
in the constructionofQ,
the domain has abigger Lip 2
constant. We need then to usehigher partials (ê"lôr) W(x, t; 0,, 0)
as thecomparison
functions in estimation and obtain03C9(Ek) CBf/1Bw(lk)
in the lemma withCB
> 0 and QB >3/2 depending
on B.4. Conclusion
LEMMA 4.
Suppose
that E is thereciprocal of a positive
eveninteger
withE
min{10-4, (2c0)-2},
Co as in Lemma3,
and that N =E-4.
Then there exist sets T, T* ~(-
~,~) of dimension strictly
less than1,
so that 03C9 and03C9* are
supported on {x
=F(t): t
ET}
and{x = F(t): t
ET*} respectively.
Moreover T and T* can be chosen to be
disjoint.
Proof
BecauseF(t) = F( - t),
the conclusion for 03C9* follows from that forcv
by symmetry.
Because a set of the form
{x
=F(t): t ~ E}, with |E|
=0,
can be writtenas the union of two
disjoint
setsEl and E2
with03C9(E1)
= 0 and03C9*(E2) =
0at any
point
in 0[9];
we maymodify
T and T* to becomedisjoint
after weprove their existence.
Because F has
period
one, we needonly
tostudy
thesupport
of 03C9 v on ao~ {0
t1}.
We shall fix the referencepoint (100, 100)
and denoteby 03C9 = 03C9(100,100).
An
increasing
sequenceAo
cAl
~A2
~ ... ofalgebras
of subsets of[0, 1)
is defined as follows.Ak
is thealgebra generated by
the intervals[2pN-2k, (2p
+2)N-2k), where p
is aninteger
and 02p N2k -
2.(Ao
is the trivial
algebra).
Since2pN-2k - 2pN2 N-2k-2,
we haveAk
~Ak+1.
Let 1: =
(2p
+1)N-2k as above; then the interval [03C4, 03C4
+EN-2k ) belongs to
the
algebra Ak+l
if(2p
+1)N-2k
+EN-2k = 2qN-2k-2
defines aninteger q.
Now q = N2(2p
+ 1 +E)
=03B5-8(2p
+ 1 +E),
sothat q
is indeed aninteger.
The interval[03C4,
T +EN-2k ),
calledB( p, k),
is contained in a basic intervalB( p, k)
ofAk
andÀ(B(p, k)) c003B53/2 03BB((p, k)),
where À is the normalizedprojection
of 03C9 on[0, 1) with 03BB([0, 1))
=1. Let Bk be
the unionof the sets
B( p, k), 0 p N2k - 2;
the conditionalprobability 03BB(Bk| Ak) CoE3/2.
Let h
be the characteristic function ofBk. Hence gk - fk - E(fk 1 Ak)
defines an
orthogonal
sequencewith |gk|
1.Using
theorthogonality,
andChebyshev’s inequality
as in[4]
we see that g2 + g4 + ... + g2, =o(r)
03BB-almost
everywhere, or 1;
+h
+ ... +f2r coE3/2r
+o(r) 03BB -
a.e.Thus for /t-almost
all t,
the numbern(r, t)
ofintegers k
r, such thatt E
B2k ,
is at mostCoE3/2 r
+o(r). (The
number Co retains the samevalue.)
Fix£5,
withCoE3/2
£58/2
and letEm = {t
E[0, 1): n(r, t) £5r,
for every rm}. Then 03BB(~~1 Em) =
1.Let t
= 03A3~1 Ck(t)N-2k
be theexpansion
in baseN2 , excepting
the rationalnumbers. Then t E
B2k if
andonly
ifC2k (t)
is odd and 0C’2k+ 1 (t) EN2.
206
For
large
r,Em
is contained in a union of K basic intervals ofA2r,
whereThis may be seen as follows.
Fixing
m, r m, and0 n ôr,
we considerthose t E
Em,
so thatn(r, t)
= n,i.e.,
those t EEm,
such that the event t EB2k
occurs forexactly n
values ofk, 0 k
r. These nplaces
can bechosen in
(rn)
ways. When t EB 2k the
number of choices forC2k
andC2k+l
is
(03B5/2)N4; when t e B2k ,
the number of choices forC2k
andC2k+ is N4 - (03B5/2)A4.
Thisyields
the first estimation for K.To obtain the second
estimation,
we useStirling’s
formula n1 ~nne-n Jiz,
for
large
n. Thelargest
term occurs when n - ôr(since b e/2)
and isapproximately N4r~r,
where il =03B4-03B4(1 - 03B4)-(1-03B4)(03B5/2)03B4(1 - (BI2)Y-b.
Nowq
depends
on03B4,
and increases upto q
= 1 when ô =e/2. Especially q
1when ô
e/2.
From the definition of Hausdorffdimension,
we concludethat
~~1 Em
has dimension at mostlog M/log
N 1.(Apart
from the detailsintroduced to make the
a-algebras
match up, the method is due to Besicovitch[1];
certain choices of successivedigits
in theexpansion
occur with afrequency
différent from - in
fact,
less than - the naturalfrequency.
Thispushes
thedimension below
1.)
Therefore the domain
{x F(t)l
so constructed has all theproperties promised
at thebeginning.
Acknowledgements
The authors
acknowledge
valuable comments from the referee.This work was
partially supported by
the National Science Foundation.References
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