Harmonic functions on annuli of graphs

In this paper, we prove the "relative connectedness" of graphs which satisfy a polynomial volume growth and a Poincare-type inequality on balls. By "relative connectedness", we mean that every two vertices at distance R from a vertex x can be joined by a path within an annulus A(x, aR) We apply this result first to control the behavior of harmonic functions outside a ball and then, in the case of Cayley graph of groups having polynomial volume growth, to obtain a Poincare-type inequality on the annuli.


Introduction
Let r be an infinite undirected connected graph and note that we call r both the graph and its set of vertices when there is no ambiguity. Let two vertices x and y be neighbors (denoted y) when r has an edge between them.
We also suppose the graph to be locally uniformly finite, which means: K > 0, ~x ~ 0393 #{y ~ 0393 : y ~ x } ~ K . Let d(x, y) be the natural distance on r, that is the minimal number of edges between x and y. Then we denote S(x, R) = {y E r : d(x, y) = jR} and B(x, R) = { y E r : d(x, y) R~ the sphere and the ball of radius R, centered at x.
We say that the spheres of r (with respect to d) are relatively connected 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 the sphere S(x, R) can be joined by a path within the annulus a-I R, aR) = B(x, aR) 1 B (x, . We suppose that F has polynomial volume growth 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 (Proposition 2.1) when it satisfies (1.1) and a D~Poincar~-type inequality on balls: there is a constant C(D) such that for any function u on r, > yEB(x,R) yeB(x,2R) (1.2) where UB = #B 03A3y~B u(y) for any set R, and The method is adapted from [5,Prop. 4.5] to our discrete setting. We will give the whole proof for the sake of completeness. Note that, for instance, the above assumptions are satisfied for Cayley graphs of groups with polynomial growth of exponent D > 2 [7, Th. 4.1]. Then, we extend to annuli (Theorem 3.1) the elliptic Harnack inequality on balls obtained by Delmotte [2] under (1.2) and the doubling of the volume (implied by (1.1)). From this inequality, we deduce a control on the behavior of harmonic functions outside a finite set (Theorem 3.2), by comparison with the behavior of the Green function. Finally, when r is the Cayley graph of a group having polynomial volume growth, we deduce Poincare-type inequalities on annuli (Theorem 3.3).. 2 Relative connectedness of the spheres Proposition 2.1 Let r be an infinite, locally uniformly finite, undirected connected graph which satisfies (I.1) and (1.~~. Then, the spheres of T are relatively connected. PROOF: First, note that the definition (1.2) of the D-Poincare-type inequality differs from the one in [5], which is, under the same conditions, 3) is a consequence of (1.2) and Holder inequality, and will be used below. Actually, (1.2) and (2.3) are equivalent (see [5]). Let x E r and R E N* be large enough. Let zi, x2 be two vertices on S{x, R) and take a > 21. If d(x1, x2) then they can be joined within A(x, a-I R, aR). So we only need to consider x2) > Let Fi (resp. F2) be a path from zi (resp. 2:2) to S(x, R/2) of length R/2. We suppose there is no path between Fi and F2 within A(x, aR), and will prove that this becomes impossible for large a.
Let u be a function on aR) such that u == 0 on Fl and u(y) > For any z E A(x, aR) and z' ~ z, we easily see that where K is a uniform bound for the number of neighbors.
Then, to prove that (2.4) leads to a contradiction it is suflicient to obtain sup ~ g(z)D ---~ 0.
(2.5) R z~B(x,5R) 03B1~T aking the supremum over R gives the uniformity in R of the constant a. Using the definition of g, we obtain The last inequality uses # A ( x , 2 i 0 3 B 1 -1 R , 2 i + 1 0 3 B 1 -1 R ) C(2i+103B1-1R)D. Since g = 0 on B(x, 03B1-1 R) and D > 1, we obtain (2.5) and then the result. p Remark: In the hypothesis of Proposition 2.1, if we replace D by p > D in the Poincare-type inequality, the result fails. Indeed, take r the graph made of two copies of the two-dimensional lattice joined by a single edge. This graph, with polynomial volume growth of exponent 2, satisfies the above assumptions, but not a 2-Poincare-type inequality, whereas the relative connectedness clearly fails. Conversely, the p-Poincare-type inequality with 1 p D is stronger than the D-Poincare-type inequality.
In the sequel, we will always denote a and Ro the constants related to the relative connectedness of the spheres. Now, we give a control of the length of the path between two vertices of a sphere of r, within the annulus defined in Proposition 2.1. . Proposition 2.2 Under the hypothesis of Proposition ~.1, there exists a positive constant a such that, for all R > Ro and all x1, x2 on the sphere S(x, R) of r, there is a path from xl to x2, within the annulus A(x, (2a)-1R, 2aR), of length at most ~R.
PROOF: By Proposition 2.1, we can pick one path between xl and X2 within A(x, 03B1-1R, aR). Let us take a sequence (vi) of vertices on this path oriented from Xl to 2:2? by the following rules: . ~i > . Given vi, is the last vertex along this path at distance 03B1-1R/2 (the lowest greater integer) from v;, . We stop at i = 7 when x2 belongs to the ball jB(f/, 03B1-1R/2). Such a sequence exists and is finite. Note that all the balls a-iR/4) are disjoint by construction. Recall the constant C in (1.1). All these balls are included in B(x, (a+a-1/2)R), whose volume is less than /2)DRD.
Let u be a non-negative function defined on r, harmonic on an annulus A(z, s, t) = B(z, t)BB(z, s). We write A(s, t) for A(z, s, t). Recall the constant a from Proposition 2.1. First we extend to annuli an elliptic Harnack inequality on balls. Theorem 3.1 Assume that s > Ra (R0 > 0 large enough), and t/a > 4sa. Let u be a non-negative function defined on r, harmonic on A(s, t). . Then u satisfies an elliptic Harnack inequality on the annulus A(2sa, t/(2a)), namely: max u ~ c(t/s) min u, where c(t/s) is a constant depending only on t/s and the graph. PROOF: Let x, y be in A(2sa, t/(2a)). By. Proposition 2.1, there is a path between x and y within A(2s, t/2). We first take a sequence of vertices vi (i == 1 to I ) along this path, as in the proof of Proposition 2.2, with ~s/3ĩ nstead 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] has proved that an harmonic function on a ball B(x, 2n) satisfies an elliptic Harnack inequality on B(x, n) whose constant is independent of x and r: We can apply this result for u on all the balls B(vi, s/6), because their doubles B(vi, s/3) are all included in A(s, t). Moreover we can write the Harnack inequality on the union of all these balls, putting the constant to the power I depending only on t/s. The result follows. 0 We have an immediate corollary. Corollary 3.1 Let u be a non-negative function defined on r, harmonic outside a ball B(z, N) (z E rand N > 0). There exists N0 ~ N such that u satisfies an elliptic Harnack inequality on all the dyadic annuli A(z, 2n, 2n+1) (2n > No) with the same constant Ch. Now, we take the hypothesis of the previous corollary, and we study the asymptotic behavior of such a function u. We begin by an analysis made by Moser in [6] about its oscillations on the spheres. Since the center z is fixed, we denote Ixl = d(z, x). We define: M(r) = max u(x) and m(r) = min . |x|=r |x|=r Suppose Ajf(r) has two relative minima, say ri and r2 (r2 > ri). Then, in the annulus r2), M(r) attains his maximum inside the domain, so does u. This contradicts the maximum principle. So we are left with two cases: either M(r) has one relative minimum at f and so M(r) is increasing for r > r. Or M(r) has no relative minimum and so it is decreasing. Likewise we see that m(r) has at most one relative maximum r. Therefore either m(r) is decreasing for r > r, or m(r) is increasing. Finally, for r bigger than some ro, M(r) and m(r) are both monotone, and we have four cases: (i) Case 1 implies that osc(r) tends to infinity at least like a power p of r, and (ii) Cases 2, 3, 4 imply that osc(r) tends to zero. Hence: lim u(x) = u~ exists. |x|~R emark: ; As the four cases cover all possibilities, (i) and (ii) correspond respectively to u unbounded and u bounded.
. For case 2, the proof of Moser gives also that osc(r) tends to zero at most like a power p' = log2((Ch -1)/(Ch + 1)) of r. |u(x) -u~| 2U 2U(C'N/|x|)D-2, and so the result still holds. As u is bounded below, we can look at the behavior of u + U which has the same speed of convergence. Hence we can use the previous analysis for nonnegative functions. If u corresponds to case 4, then -u + U corresponds to case 3 and has the san~e speed of convergence. So we just deal with cases 2, 3.
Let f be a non-negative function on r, sub-harmonic outside the ball B(z, N) and vanishing at infinity. We denote Gz(x) the Green's function rooted at z, i.e. the unique solution of ~~c = ~z which vanishes at infinity. 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) For case 3, f = u~ -u is a non-negative function on F, sub-harmonic outside the ball B(z, N) and vanishing at infinity. So, by the above argument, the claim follows. For case 2, the sign of f = u-u~ may change since M(r) is decreasing and m(r) is increasing, both tending to u~ at infinity. We denote 11 = max(0, f ) and f 2 = max(o, -f ). These are non-negative functions and we easily see that they are sub-harmonic outside B(z, N). As they vanish at infinity, we obtain (3.6) for 11 and f 2 and the result is also true.. Here, r is the Cayley graph of a finitely generated group G, associated with a symmetric finite generating set S: its vertices are the elements of G, and there is an edge between x and y when yx-1 E S. We assume the polynomial volume growth (1,1), then (see [1,7]) r satisfies (1.2), and so the relative connectedness of the spheres. On this kind of graph, the proof of the Poincaré-type inequality on balls relies on the construction of a particular path 03B3x,y between each pair of vertices x, y in a ball B(z, R) (z E r, R > 0). The set of these paths should have the property to pass "not too often" through any edge in B(z, 2R). To obtain the same type of inequality on annuli we will need to define the path in a way adapted to our setting. Recall the constants a and .Re from Theorem 2.1. The Poincaré-type inequalities on annuli are the following.  Likewise, we define i(y). The path 'Y intersects a sequence of B/s denoted Bi(k) (k = 0 to K) and we can take i(O) = i(x) and i(K) = i(y) (adding Bi(x) and to the sequence if necessary). We first construct a sequence of vertices (zn) (n = 0 to 2K + 1), from x to y as follows z2k = and (3.7) We denote 3Bi(k) = and 5Bi(k) = 503B1-1s/6). For all k, Z2k and z2k+I belong to and z2k+2 belongs to 3Bi(k). So, for all n, d(zn, zn+1) 2a-IS/3. Let g(z-1nzn+1) be a minimal path from e to z-1nzn+1. We join zn and zn+1 by the translated path zng(z-1n zn+1) which stays within 5Bi([n/2]) (where [.] denotes the integer part). Finally, we obtain a path 03B3x,y from x to y within A' = + 5a-ls/6), whose length is bounded by Cs(t/s)D for some non-negative constant C. Indeed, 2K (length of 4I03B1-1s/3 Cs(t/ s)D . n=0 Now, we prove our Poincare-type inequality using the same technique as the one on balls (see [1,7]). We denote H(a, b) = # {{x, y} : (a, b) E 03B3x,y}, where (a, b) denotes the edge between two neighbors a and b.
To estimate H(a, b), let first write {x, ?/} such that i(x) = i, i(y) = j and (a, b) E Saying that the edge (a, b) belongs to means that (a, b) belongs to one of the paths 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, and likewise, a bound K2 for the number of balls in B that could contain such zn+i . Then, we want a bound K3 for the number of elements h E F such that h = z-1nzn+1. Finally, we need a bound K4 for the number of possible zn. By (3.7), va(E~n+11~2~~, z~, i = i(x) and j = i(y) fully determine the pair {x, y}, so H(a, b) _ I2K1K2K3K4 . (3.8) Since has length less than 2a-ls/3, the vertices a and b should be at distance less than 503B1-1s/6 from the center of any ball in B that contains such zn. By definition of the overlapping bound N, for a fixed edge (a, b), there are at most (N + 1)5 such balls. So, K1 ~ (JV + 1)5 and likewise K2 (iV + 1)5. As E B(e, 03B1-1s/3), there are, at most, #B(e,03B1-1s/3) choices for z-1nzn+1, so K3 CsD. Once z-1nzn+1 is fixed, knowing that (a, b) is one of the edges of leaves 2a-is/3 choices for the starting point zn, so K4 2a~1s~3.
Therefore, the result follows. r-j