S ÉBASTIEN B LACHÈRE
Harmonic functions on annuli of graphs
Annales mathématiques Blaise Pascal, tome 8, n
o2 (2001), p. 47-59
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Harmonic functions
onannuli of graphs
Sébastien Blachère
Abstract
In this paper, we prove the "relative connectedness" of
graphs
which
satisfy
apolynomial
volumegrowth
and aPoincare-type
in-equality
on balls.By
"relativeconnectedness",
we mean that everytwo vertices at distance R from a vertex x can be
joined by
apath
within an annulus
A(x, aR)
Weapply
this result first to con-trol the behavior of harmonic functions outside a ball and
then,
in thecase of
Cayley graph
of groupshaving
polynomial volumegrowth,
toobtain a
Poincare-type inequality
on the annuli.1
Introduction
Let r be an infinite undirected connected
graph
and note that we call r both thegraph
and itsset
of vertices when there is noambiguity. Let
two verticesx
and y
beneighbors (denoted y)
when r has anedge
between them.We also suppose the
graph
to belocally uniformly finite,
which means:~K > 0, ~x ~ 0393 #{y ~ 0393 :
y ~ x} ~ K .Let
d(x, y)
be the natural distance onr,
that is the minimal number ofedges
between x and y. Then we denote
S(x, R)
={y
E r :d(x, y)
=jR}
andB(x, R)
={ y
E r :d(x, y) R~
thesphere
and the ball of radiusR,
centered at x.
We say that the
spheres
of r(with
respect tod)
arerelatively
con-nected if there exist constants
Ro
> 0 and a > 1 such that: for any R >Ro
and any vertex x E
r,
every two vertices in thesphere S(x, R)
can bejoined by
apath
within the annulusa-I R, aR)
=B(x, aR) 1 B (x,
.We suppose that F has
polynomial
volumegrowth
of exponent D:C-1RD #B(x, R) CRD.
.(1.1)
Our aim is first to prove the relative connectedness of the
spheres
of r(Propo-
sition
2.1)
when it satisfies(1.1)
and aD~Poincar~-type inequality
onballs: there is a constant
C(D)
such that for any function u onr,
~
>yEB(x,R) yeB(x,2R)
(1.2)
where UB =
#B 03A3y~B u(y)
for any setR,
and= ~
.y~x
The method is
adapted
from[5, Prop. 4.5]
to our discretesetting.
We willgive
the wholeproof
for the sake ofcompleteness.
Notethat,
forinstance,
theabove
assumptions
are satisfied forCayley graphs
of groups withpolynomial growth
of exponent D > 2[7,
Th.4.1].
Then,
we extend to annuli(Theorem 3.1)
theelliptic
Harnackinequality
on balls obtained
by
Delmotte[2]
under(1.2)
and thedoubling
of the volume(implied by (1.1)).
From thisinequality,
we deduce a control on the behavior of harmonic functions outside a finite set(Theorem 3.2), by comparison
withthe behavior of the Green function.
Finally,
when r is theCayley graph
of agroup
having polynomial
volumegrowth,
we deducePoincare-type inequali-
ties on annuli
(Theorem 3.3)..
2
Relative connectedness of the spheres
Proposition
2.1 Let r be aninfinite, locally uniformly finite,
undirectedconnected
graph
whichsatisfies (I.1)
and(1.~~. Then,
thespheres of
T arerelatively
connected.PROOF:
First,
note that the definition(1.2)
of theD-Poincare-type inequal- ity
differs from the one in[5],
whichis,
under the sameconditions,
/ Bi/D
1 #B(x,R) 03A3 |u(y)
- u B(x,R)I
-C(D) ) ( £ |~u(y)|D ) .
.(2.3)
#B~~~
yEB(x,2R) /
Indeed, (2.3)
is a consequence of(1.2)
and Holderinequality,
and will be usedbelow.
Actually, (1.2)
and(2.3)
areequivalent (see [5]).
Let x E r and R E N* be
large enough.
Let zi, x2 be two vertices onS{x, R)
and take a > 21. Ifd(x1, x2)
thenthey
can bejoined
withinA(x, a-I R, aR).
So weonly
need to considerx2)
> LetFi (resp.
F2)
be apath
from zi(resp. 2:2)
toS(x, R/2)
oflength R/2.
We supposethere is no
path
betweenFi
andF2
withinA(x, aR),
and will provethat this becomes
impossible
forlarge
a.Let u be a function on
aR)
such that u == 0 onFl
andu(y)
> 1for y E
F2. Suppose
uB(x,R)~ 1/2
and let y EF2,
then> 1/2.
Let
Bi
=B(y, 22~~R)
and C be a constant whose value maychange
fromone line to another. Then
by (1.1),
asB{x, R)
CB1,
1 uB1| I + u(y) I
oo
UBI + 03A3 |uBi+1 -
uBiI
~
2C 03A31 #Bi 03A3|u(z) -
i=l~
’ xEB~uBi|
.So, using (2.3),
we obtainoc / B1/D
|uB(x,R)-u(y)| ~ C03A3( 03A3 |~u(z)|D )
lID
CR1/D
supr 1 03A3 |~u(z)|D ) 1/D 2(2-i)/D.
r4R
zEB(y,r)~
i= 0So,
there is a constant Cdepending
on D and the constants from(1.1)
and(1.2),
such that for each y EF2,
there is a radiusr(y)
4R withE Cr{y)I R
.z~B(y,r(y))
By
thecovering
Lemma[5,
Th.14.12],
among all theB(y, r(y)),
there isa sub-collection of balls
r(yi)) (yi
EF2), pointwise disjoint,
such thatF2
C As a consequence,by
definition ofF2,
we must have~=
>R/20. So,
E |~u(z)|D
> C’z~B(x,5R) i z~B(yi,r(yui))
1/2,
thentaking Fl
instead ofF2
leads to the same result.Finally
E
>C’
.(2.4)
zEB(x,5R)
As C’ does not
depend
on a, the aim is to choose u such that the left hand side of(2.4)
tends to 0 as a goes toinfinity.
Let
g(z)
=(In
when z EA(x, aR)
and 0 elsewhere.We define the function .-
.
u(z)
= infL g(y)
>where the
03B3z’s
are thepaths
from z toFl.
Then u m 0on F1 and, assuming
that
Fi
andF2
cannot bejoined
withinA(x, aR),
anypath
betweenthem must leave the annulus. If it does so
through
its exteriorboundary,
the
path
should go fromS(x, R)
toS(x, aR)
and come back.Likewise,
ifthe
path
exits the annulusthrough
its interiorboundary,
it should go fromS(x, R/2)
toS(x,
and come back.Then,
for y EF2,
we obtainu(y) ~ min {(ln 03B1)-12k-1, (ln 03B1)-12 k-1} ~ 1
.For any z E
A(x, aR)
and z’ ~ z, weeasily
see thatu(z’)~ ~ ~9(z) ~
.So,
where K is a uniform bound for the number ofneighbors.
Then,
to prove that(2.4)
leads to a contradiction it is suflicient to obtainsup
R~ g(z)D
---~ 0.(2.5)
z~B(x,5R)
03B1~~
Taking
the supremum over Rgives
theuniformity
in R of the constant a.Using
the definition of g, we obtainE (ln a) D E d(x, z)-D
(In a)-D (~21og2 a]
+1) C2D
.The last
inequality
uses#A(x,2i03B1-1R,2i+103B1-1R) C(2i+103B1-1R)D.
Sinceg = 0 on
B(x, 03B1-1 R)
and D >1,
we obtain(2.5)
and then the result. p Remark: In thehypothesis
ofProposition 2.1,
if wereplace
Dby
p > D in thePoincare-type inequality,
the result fails.Indeed,
take r thegraph
made of two
copies
of the two-dimensional latticejoined by
asingle edge.
This
graph,
withpolynomial
volumegrowth
of exponent2,
satisfies the aboveassumptions,
but not a2-Poincare-type inequality,
whereas the rel- ative connectednessclearly
fails.Conversely,
thep-Poincare-type inequality
with 1 p D is
stronger
than theD-Poincare-type inequality.
In the
sequel,
we willalways
denote a andRo
the constants related to therelative connectedness of the
spheres. Now,
wegive
a control of thelength
of the
path
between two vertices of asphere
ofr,
within the annulus defined inProposition
2.1. .Proposition
2.2 Under thehypothesis of Proposition ~.1,
there exists a po- sitive constant a suchthat, for
all R >Ro
and all x1, x2 on thesphere S(x, R) of r,
there is apath from
xl to x2, within the annulusA(x, (2a)-1R, 2aR), of length
at most ~R.PROOF:
By Proposition 2.1,
we canpick
onepath
between xl and X2 withinA(x, 03B1-1R, aR).
Let us take a sequence(vi)
of vertices on thispath
orientedfrom Xl to 2:2?
by
thefollowing
rules:. ~i >
. Given vi, is the last vertex
along
thispath
at distance03B1-1R/2 (the
lowestgreater integer)
from v;,. We stop at i = 7 when x2
belongs
to the balljB(f/, 03B1-1R/2).
Such a sequence exists and is finite. Note that all the balls
a-iR/4)
aredisjoint by
construction. Recall the constant C in(1.1).
All these balls areincluded in
B(x, (a+a-1/2)R),
whose volume is less than/2)DRD.
On the other hand
Hence we must have I +
03B1-1/2)D(403B1)D.
Note also that all theB(vi,
are included inThen,
within this an-nulus,
there arepaths
from Vi to oflength ~a-1R/2~ .
.So,
the annulusA(x, (2a)"1R, 2aR)
contains apath
between x and y oflength
less than ARwith A =
4C2(a
+ 03
Behavior of harmonic functions outside
aball
We
always
suppose r to be aninfinite, locally uniformly finite,
undirectedconnected
graph
which satisfies(1.1)
and(1.2).
A functionf
on r is calledharmonic on a set of vertices E if
for all x E
E,
=~( f (x)
-fey))
= 0 .y~x
Likewise we call a function sub-harmonic
(resp. super-harmonic)
0
(resp. 0 f (x)
>0).
Let u be a
non-negative
function defined onr,
harmonic on an annulusA(z,
s,t)
=B(z, t)BB(z, s).
We writeA(s, t)
forA(z,
s,t).
Recall the con-stant a from
Proposition
2.1. First we extend to annuli anelliptic
Harnackinequality
on balls.Theorem 3.1 Assume that s >
Ra (R0
> 0large enough),
andt/a
>4sa. Let u be a
non-negative function defined
onr,
harmonic onA(s, t).
.Then u
satisfies
anelliptic
Harnackinequality
on the annulusA(2sa, t/(2a)), namely:
max u ~
c(t/s)
min u,A(2s03B1,t/(203B1)) A(2s03B1,t/(203B1))
where
c(t/s)
is a constantdepending only
ont/s
and thegraph.
PROOF: Let x, y be in
A(2sa, t/(2a)). By. Proposition 2.1,
there is apath
between x
and y
withinA(2s, t/2).
We first take a sequence of vertices vi(i
== 1 toI ) along
thispath,
as in theproof
ofProposition 2.2,
with~s/3~
instead of for the distance between two successive vertices.
Then,
we obtain that I
c(t/s)D
where D is the constant in(1.1).
With
(1.1)
and(1.2),
Delmotte[2]
hasproved
that an harmonic functionon a ball
B(x, 2n)
satisfies anelliptic
Harnackinequality
onB(x, n)
whoseconstant is
independent
of x and r:We can
apply
this result for u on all the ballsB(vi, s/6),
because their doublesB(vi, s/3)
are all included inA(s, t).
Moreover we can write the Harnackinequality
on the union of all theseballs, putting
the constant to the power Idepending only
ont/s.
The result follows. 0We have an immediate
corollary.
Corollary
3.1 Let u be anon-negative function defined
onr,
harmonicoutside a ball
B(z, N) (z
E rand N >0).
There existsN0 ~
N such that usatisfies
anelliptic
Harnackinequality
on all thedyadic
annuliA(z, 2n, 2n+1) (2n
>No)
with the same constantCh.
Now,
we take thehypothesis
of theprevious corollary,
and westudy
theasymptotic
behavior of such a function u. Webegin by
ananalysis
madeby
Moser in
[6]
about its oscillations on thespheres.
Since the center z isfixed,
we denote
Ixl
=d(z, x).
We define:M(r)
=max u(x)
andm(r)
= min .|x|=r |x|=r
Suppose Ajf(r)
has two relativeminima,
say ri and r2(r2
>ri). Then,
inthe annulus
r2), M(r)
attains his maximum inside thedomain,
so doesu. This contradicts the maximum
principle.
So we are left with two cases:either
M(r)
has one relative minimum at f and soM(r)
isincreasing
forr > r. Or
M(r)
has no relative minimum and so it isdecreasing.
Likewise we see that
m(r)
has at most one relative maximum r. Thereforeeither
m(r)
isdecreasing
for r > r, orm(r)
isincreasing. Finally,
for rbigger
than some ro,
M(r)
andm(r)
are bothmonotone,
and we have four cases:. Case 1:
M(r) ~
andm(r) B
. Case 2:
M(r) ~
andm(r) ~’
. Case 3:
M(r) j~
andm(r) ~
. Case 4:
M(r) B
andm(r) ~
We denote
osc(r) =1lI (r)
the oscillations of u on thesphere
of radiusr. The
following proposition
mimics[6,
Th4,5].
Proposition
3.1 LetCh
be the constantof Corollary 3..I,
then:(i)
Case 1implies
thatosc(r)
tends toinfinity
at least like a power pof
r, and(ii)
Cases2, 3,
4imply
thatosc(r)
tends to zero. Hence:lim
u(x)
= u~ exists.|x|~~
Remark:
; As the four cases cover all
possibilities, (i)
and(ii) correspond
respec-tively
to u unbounded and u bounded.. For case
2,
theproof
of Mosergives
also thatosc(r)
tends to zero atmost like a power
p’
=log2((Ch - 1)/(Ch
+1))
of r.For cases
2,3,4,
we want to boundu~| by
anexplicit
power of|x|
which
depends only
on the constant D.Theorem 3.2
Suppose
r hasexponent
D > 2. Let u be a bounded(by
sayU) function
onr,
harmonic outside a ballB(z, N) (z
E r and N >0),
then:lim
u (x)
= u~exists,
|x|~~
and
|u(x)
- u~| ~ cND-2U|x|2-D
,where c is a
positive
constantdepending only
on r.PROOF: First note that we can restrict ourselves to the non-constant func- tions. To
apply Corollary 3.1,
we need to consideronly
the x’s such that~x~
>No
with someNo large enough. Actually,
weonly
needNo
= C’Nwith C’ a constant
depending
on F.Then,
when|x| No,
we obtain|u(x)
-u~|
2U2U(C’N/|x|)D-2,
and so the result still holds. Asu is bounded
below,
we can look at the behavior of u + U which has thesame
speed
of convergence. Hence we can use theprevious analysis
for non-negative
functions.If u
corresponds
to case4,
then -u + Ucorresponds
to case 3 and hasthe san~e
speed
of convergence. So wejust
deal with cases2,
3.Let
f
be anon-negative
function onr,
sub-harmonic outside the ballB(z, N)
andvanishing
atinfinity.
We denoteGz(x)
the Green’s function rooted at z, i.e. theunique
solution of ~~c =~z
which vanishes atinfinity.
Under
(1.2)
and(1.1)
with exponent D >2,
there exists a constant c such that(see [3])
for every x
~
z,Gz(x)
.With
No
= C’N EN,
we denotea = min
Gz(x)
and b = maxf
.S(z,No)
Then
(b/a)Gz
>f
onS(z, No). (b/a)Gz
-f
is asuper-harmonic
functiontherefore we can use the minimum
principle.
Since(bla)Gz - f
vanishes atinfinity
and isnon-negative
onS(z, No),
this function remainsnon-negative
out of
B(z, No).
Otherwise it should have a localminimum,
which would contradict the minimumprinciple.
Hence for all x~ B(z, No),
f(x) (b/a)Gz(x)
.(3.6)
For case
3, f
= u~ - u is anon-negative
function onF,
sub-harmonic outside the ballB(z, N)
andvanishing
atinfinity. So, by
the aboveargument,
the claim follows. For case2,
thesign of f = u-u~
maychange
sinceM(r)
isdecreasing
andm(r)
isincreasing,
bothtending
to u~ atinfinity.
We denote11
=max(0, f )
andf 2
=max(o, - f ).
These arenon-negative
functions andwe
easily
see thatthey
are sub-harmonic outsideB(z, N).
Asthey
vanish atinfinity,
we obtain(3.6)
for11
andf 2
and the result is alsotrue..
Example:
Let be thesimple
random walk on 0393 started at x, and B(z,N) be thehitting
time ofB(z, N). Then,
the functionu(x)
=Px{B(z,N) ~}
is harmonic outside
B(z, N),
boundedby
1 and tends to 0 when~x~
goes toinfinity.
Then Theorem 3.2gives
00} ~
’3.1
Poincare-type inequality
onannuli
Here,
r is theCayley graph
of afinitely generated
groupG,
associated witha
symmetric
finitegenerating
set S: its vertices are the elements ofG,
andthere is an
edge
between x and y whenyx-1
E S. We assume thepoly-
nomial volume
growth (1,1),
then(see [1, 7])
r satisfies(1.2),
and so therelative connectedness of the
spheres.
On this kind ofgraph,
theproof
ofthe
Poincaré-type inequality
on balls relies on the construction of apartic-
ular
path
03B3x,y between eachpair
of vertices x, y in a ballB(z, R) (z
Er,
R >
0).
The set of thesepaths
should have the property to pass "not too often"through
anyedge
inB(z, 2R).
To obtain the sametype
ofinequality
on annuli we will need to define the
path
in a wayadapted
to our set-ting.
Recall the constants a and.Re
from Theorem 2.1. ThePoincaré-type inequalities
on annuli are thefollowing.
Theorem 3.3 Let r be the
Cayley graph of
a group withpolynomial
volumegrowth of exponent
D > I. .Then, for
all p >1,
there exists a constantC(p)
such that
for
all z Er,
s, t ~ N(t
> s > and anyfunction f
onr,
wehave
£
-£
y~A(z,03B1-1s/3,03B1t+203B1-1s/3)
PROOF: As
before,
we omit the reference to the center z of the annuli.We cover the annulus
A(03B1-1s, at) by
a minimal number I of ballsB1
=such that all the
B(vi, a-‘1s~12)
aredisjoint.
We denote ~i =: 1 i
I}
thiscovering. Then,
r
im
which
leads,
as inProposition 2.2,
to Ic(t/s)D.
LetNz
be the number of balls intersectedby Bi.
Let us denote them03B1-1s/6) (k
= 1 toNz).
Remark that
N~
U
CB(03C5i,
k=I
and,
as all the aredisjoint,
Ni
~ (x ls/12) ~ ~~1SI2)
.k=1
So the
Ni’s
are boundedby
sayN,
which does notdepend
on s nor t.Let x, y E
A(s, t).
We construct apath
from x to y in thespirit
of[4,
Ex.2.3]. By
Theorem2.1,
there is apath I
from xto y
withinA(03B1-1s, at).
Wedefine an index
i(x)
associated to x:i(x)
=inf
i :d(v;, x)
= mind(v;, ~) ~
We remark that x E
Likewise,
we definei(y).
The
path ’Y
intersects a sequence ofB/s
denotedBi(k) (k
= 0 toK)
and we can take
i(O)
=i(x)
andi(K)
=i(y) (adding Bi(x)
and to thesequence if
necessary).
We first construct a sequence of vertices(zn) (n
= 0to 2K +
1),
from xto y
as followsz2k = and
(3.7)
We denote
3Bi(k)
= and5Bi(k)
=503B1-1s/6).
For allk,
Z2k and z2k+Ibelong
to and z2k+2belongs
to3Bi(k). So,
for all n,d(zn, zn+1) 2a-IS/3.
Letg(z-1nzn+1)
be a minimalpath
from e toz-1nzn+1.
We
join
zn and zn+1by
the translatedpath zng(z-1n zn+1)
whichstays
within5Bi([n/2]) (where [.]
denotes theinteger part). Finally,
we obtain apath
03B3x,y from x to y within A’ = +5a-ls/6),
whoselength
is boundedby Cs(t/s)D
for somenon-negative
constant C.Indeed,
2K
(length
of ~4I03B1-1s/3 Cs(t/ s)D
.n=0
Now,
we prove ourPoincare-type inequality using
the sametechnique
asthe one on balls
(see [1, 7]).
We denoteH(a, b)
=# {{x, y}
:(a, b)
E03B3x,y},
where
(a, b)
denotes theedge
between twoneighbors
a and b.To estimate
H(a, b),
let first writeI I
b) = # {{x,y} ~ A(s, t)
:i(x) = i, i(y) = j, (a, b) E 03B3x,y}
Now,
suppose iand j
fixed. We want to bound the number ofpair
of vertices{x, ?/}
such thati(x)
=i, i(y) = j
and(a, b)
ESaying
that theedge (a, b) belongs
to means that(a, b) belongs
to one of thepaths zng(z-1nzn+1)
defined above.
We need to obtain a bound
Ki
for the number of balls in B that could contain such zn, andlikewise,
a boundK2
for the number of balls in B that could contain such zn+i .Then,
we want a boundK3
for the number of elements h E F such that h =z-1nzn+1. Finally,
we need a boundK4
forthe number of
possible
zn.By (3.7),
va(E~n+11~2~~, z~, i =i(x)
andj
=i(y) fully
determine thepair {x, y},
soH(a, b)
_I2K1K2K3K4
.(3.8)
Since has
length
less than2a-ls/3,
the vertices a and b should be at distance less than503B1-1s/6
from the center of any ball in B that contains such zn.By
definition of theoverlapping
boundN,
for a fixededge (a, b),
there are at most
(N
+1)5
such balls.So, K1 ~ (JV + 1)5
and likewiseK2 (iV + 1)5.
As EB(e, 03B1-1s/3),
there are, at most,#B(e,03B1-1s/3)
choices for
z-1nzn+1,
soK3 CsD.
Oncez-1nzn+1
isfixed, knowing
that(a, b)
is one of the
edges
of leaves2a-is/3
choices for thestarting point
zn, soK4 2a~1s~3.
Finally, plugging
these bounds and Ic(t/s)D
into(3.8),
we obtainH(a, b) CI2N10sD+1 C(t/s)2DsD+1.
Therefore,
the result follows.r-j
Acknowledgment:
I am verygrateful
to L. Saloff-Coste for his greathelp during
all thepreparation
of thismanuscript.
I also thank T. Coulhon and T. Delmotte for useful remarks.References
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SEBASTIEN BLACHERE
LSP, UNIVERSITÉ PAUL SABATIER
AND DMA,
ECOLE
POLYTECHNIQUE FÉDÉRALE DE LAUSANNEECUBLENS 1015 LAUSANNE SUISSE