Proof of Theorem 1.2 getdocf0f4. 363KB Jun 04 2011 12:05:10 AM

z n , i.e. the net current flowing in the network when a unit voltage is applied across 0 and z n . Since S , ϕ is recurrent we know that cS 0,n → 0, n → ∞. Recall that cS 0,n satisfies cS 0,n = 1 2 X x, y ∈S ϕ |x − y|ψ n x − ψ n y 2 , 2.5 where ψ n is the electric potential, i.e. the unique function on S that is harmonic in S 0,n , takes the value 1 at 0 and vanishes outside of S 0,n . Given S ∈ Ω, set e S n = ∪ x ∈S 0,n V ′ x . Note that e S n is an increasing sequence of finite sets, covering all S. Collapse all sites in e S n c into a single site ez n and call ce S n the effective conductance between 0 and ez n by construction 0 ∈ e S n . From the Dirichlet principle 1.6 we have ce S n 6 1 2 X u,v ∈e S ϕ|u − v|gu − gv 2 , for any g : e S → [0, 1] such that g0 = 1 and g = 0 on e S n c . Choosing gu = ψ n φ ′ u we obtain ce S n 6 1 2 X x, y ∈S ψ n x − ψ n y 2 X u ∈V ′ x X v ∈V ′ y ϕ|u − v| . From the assumption 2.4 and the recurrence of S , ϕ implying that 2.5 goes to zero, we deduce that E[ce S n ] → 0, n → ∞. Since ce S n is monotone decreasing we deduce that ce S n → 0, P –a.s. This implies the P–a.s. recurrence of e S, ϕ and the claim follows.

2.1 Proof of Theorem 1.2

We first prove part i of the theorem, by applying the general statement derived in Lemma 2.1 in the case S = Z d and ϕ = ϕ p, α . Since S , ϕ p, α is transient whenever d 3, or d = 1, 2 and α d the classical proof of these facts through harmonic analysis is recalled in Appendix B, we only need to verify condition 2.1. For the moment we only suppose that ψt 6 C ′ t −γ for some γ 0. Let us fix p, q 1 s.t. 1p + 1q = 1. Using Hölder’s inequality and then Schwarz’ inequality or simply using Hölder inequality with the triple 1 2q, 12q, 1p, we obtain E N φxN φ y rφx, φ y 2.6 6 E ” N φx 2q — 1 2q E ” N φ y 2q — 1 2q E r φx, φ y p 1 p for any x 6= y in Z d . By assumption 1.11 we know that r φx, φ y p 6 c r p, α φx, φ y p := c 1 ∨ |φx − φ y| pd+ α . 2.7 We shall use c 1 , c 2 , . . . to denote constants independent of x and y below. From Jensen’s inequality |φx − φ y| pd+ α 6 c 1 € |φx − x| pd+ α + |x − y| pd+ α + |φ y − y| pd+ α Š . 2591 From 2.7 and the fact that |x − y| 1 we derive that E r φx, φ y p 6 c 2 sup z ∈Z d E ” |φz − z| pd+ α — + c 2 |x − y| pd+ α . 2.8 Now we observe that |φz − z| t if and only if B z,t ∩ S = ;. Hence we can estimate E ” |φz − z| pd+ α — 6 1 + Z ∞ 1 ψ  t 1 pd+ α ‹ d t 6 1 + C Z ∞ 1 t − γ pd+ α d t 6 c 3 , provided that γ pd + α. Therefore, using |x − y| 1, from 2.8 we see that for any x 6= y in Z d : E r φx, φ y p 1 p 6 c 4 r p, α x, y . 2.9 Next, we estimate the expectation E ” N φx 2q — from above, uniformly in x ∈ Z d . To this end we shall need the following simple geometric lemma. Lemma 2.3. Let Ex, t be the event that S ∩ Bx, t 6= ; and S ∩ Bx ± 3 p d t e i , t 6= ;, where {e i : 1 6 i 6 d } is the canonical basis of R d . Then, on the event Ex, t we have φx ∈ Bx, t, i.e. |φx − x| t, and z 6∈ V φx for all z ∈ R d such that |z − x| 9d p d t Assuming for a moment the validity of Lemma 2.3 the proof continues as follows. From Lemma 2.3 we see that, for a suitable constant c 5 , the event N φx c 5 t d implies that at least one of the 2d +1 balls Bx, t, Bx ±3 p d t e i , t must have empty intersection with S. Since Bx ±⌊3 p d t ⌋e i , t − 1 ⊂ Bx ± 3 p d t e i , t for t 1, we conclude that P ” N φx c 5 t d — 6 2d + 1ψt − 1 , t 1. Taking c 6 such that c − 1 d 5 c 1 2qd 6 = 2, it follows that E ” N φx 2q — = Z ∞ P N φx 2q td t 6 c 6 + 2d + 1 Z ∞ c 6 ψ c − 1 d 5 t 1 2qd − 1 d t 6 c 6 + c 7 Z ∞ 1 t − γ 2qd d t 6 c 8 , 2.10 as soon as γ 2qd. Due to 2.6, 2.9 and 2.10, the hypothesis 2.1 of Lemma 2.1 is satisfied when ψt 6 C t −γ for all t 1, where γ is a constant satisfying γ pd + α , γ 2qd = 2pd p − 1 . We observe that the functions 1, ∞ ∋ p 7→ pd + α and 1, ∞ ∋ p 7→ 2pd p −1 are respectively increasing and decreasing and intersect in only one point p ∗ = 3d+ α d+ α . Hence, optimizing over p, it is enough to require that γ inf p 1 max {pd + α, 2pd p − 1 } = p ∗ d + α = 3d + α . 2592 This concludes the proof of Theorem 1.2 i. Proof of lemma 2.3. The first claim is trivial since S ∩ Bx, t 6= ; implies φx ∈ Bx, t. In order to prove the second one we proceed as follows. For simplicity of notation we set m := 3 p d and k := 9d. Let us take z ∈ R d with |z − x| k p d t. Without loss of generality, we can suppose that x = 0, z 1 0 and z 1 |z i | for all i = 2, 3, . . . , d. Note that this implies that k p d t |z| 6 p dz 1 , hence z 1 kt. Since min u ∈B0,t |z − u| = |z| − t max u ∈Bmte 1 ,t |z − u| 6 |z − mte 1 | + t , if we prove that |z| − |z − mte 1 | 2t , 2.11 we are sure that the distance from z to each point in S ∩ B0, t is larger than the distance from z to each point of S ∩ Bmte 1 , t. Hence it cannot be that z ∈ V φ0 . In order to prove 2.11, we first observe that the map 0, ∞ ∋ x 7→ p x + a − p x + b ∈ 0, ∞ is decreasing for a b. Hence we obtain that |z| − |z − mte 1 | p dz 1 − Æ z 1 − mt 2 + d − 1z 2 1 . Therefore, setting x := z 1 t, we only need to prove that p d x − p x − m 2 + d − 1x 2 2 , ∀x k . By the mean value theorem applied to the function f x = p x p d x − p x − m 2 + d − 1x 2 1 2 p d x € d x 2 − x − m 2 − d − 1x 2 Š = 2x m − m 2 2 p d x m p d − m 2 k = 2 . This completes the proof of 2.11. Proof of Theorem 1.2 Part ii. We use the same approach as in Part i above. We start again our estimate from 2.6. Moreover, as in the proof of 2.10 it is clear that hypothesis 1.13 implies E [N φx 2q ] ∞ for any q 1, uniformly in x ∈ Z d . Therefore it remains to check that r x, y := E r φx, φ y p 1 p , 2.12 defines a transient resistor network on Z d , for any d 3, under the assumption that r φx, φ y 6 C e δ|φx−φ y| β . For any β 0 we can find a constant c 1 = c 1 β such that r φx, φ y 6 C exp € δ c 1 ” |φx − x| β + |x − y| β + |φ y − y| β —Š . Therefore, using Schwarz’ inequality we have E r φx, φ y p 1 p 6 c 2 exp € δ c 2 |x − y| ⠊ E ” exp € δ c 2 |φx − x| ⠊— 1 2 E ” exp € δ c 2 |φ y − y| ⠊— 1 2 . 2593 For γ 0 E ” exp € γ|φx − x| ⠊— 6 1 + Z ∞ 1 ψ 1 γ log t 1 β d t = 1 + γβ Z ∞ ψs e γs β s β−1 ds , where, using 1.13, the last integral is finite for γ a. Taking γ = δ c 2 and δ sufficiently smal we arrive at the conclusion that uniformly in x, E r φx, φ y p 1 p 6 c 3 exp € c 3 |x − y| ⠊ =: er x, y . Clearly, er x, y defines a transient resistor network on Z d and the same claim for r x, y follows from monotonicity. This ends the proof of Part ii of Theorem 1.2. Proof of Theorem 1.2 Part iii. Here we use the criterion given in Lemma 2.2 with S = Z d . With this choice of S we have that V ′ x ⊂ {x} ∪ S ∩ Q x,1 , x ∈ Z d . Recalling definition 1.15 we see that for all x 6= y in Z d , E    X u ∈V ′ x X v ∈V ′ y ϕ|u − v|    6 c 1 ϕ x, y E ” 1 + SQ x,1 1 + SQ y,1 — . Using Schwarz’ inequality and condition 1.14 the last expression is bounded above by c 2 ϕ x, y. This implies condition 2.4 and therefore the a.s. recurrence of S, ϕ.

2.2 Proof of Corollary 1.3