Proof that Esn getdoc18fd. 410KB Jun 04 2011 12:04:08 AM

where c 1 is as in the statement of the separation lemma. We also let A n = {S[1, σ n ] ∩ b S 4n [0, b σ n ] = ;}, z = b S 4n b σ n , and D n = n −1 min {distSσ n , b S 4n [0, b σ n ], distz , S[0, σ n ]}. By the strong Markov property for random walk, b E ” P ¦ S[1, σ 4n ] ∩ b S 4n [0, b σ 4n ] = ; ©— ≥ cbE ” 1 {b S 4n [ b σ n , b σ 4n ] ⊂ W z }P A n ; D n ≥ c 1 — . By Lemma 2.4 and Corollary 3.7, b E ” 1 {b S 4n [ b σ n , b σ 4n ] ⊂ W z }P A n ; D n ≥ c 1 — ≥ cbE P A n ; D n ≥ c 1 . Finally, by the separation lemma, b E P A n ; D n ≥ c 1 ≥ cbE P A n , and therefore, b E ” P ¦ S[1, σ 4n ] ∩ b S 4n [0, b σ 4n ] = ; ©— ≥ cbE ” P ¦ S[1, σ n ] ∩ b S 4n [0, b σ n ] = ; ©— .

5.2 Proof that Esn

≍ Esm Esm, n Proposition 5.2. There exists C ∞ such that for all m and n with m ≤ n, Esn ≤ C Esm Esm, n. Proof. Let l = ⌊m4⌋ and fix η 1 = η 1 l and η 2 = η 2 m,n . For any path η in Ω n , P S[1, σ n ] ∩ η = ; ≤ P ¦ S[1, σ l ] ∩ η 1 η = ;; S[1, σ n ] ∩ η 2 η = ; © = X z ∈∂ B l P ¦ S[1, σ l ] ∩ η 1 η = ;; Sσ l = z © P z ¦ S[1, σ n ] ∩ η 2 η = ; © . However, η 2 η ⊂ Λ \ B m , and thus by the discrete Harnack principle, for any z, z ′ ∈ ∂ B l , P z ¦ S[1, σ n ] ∩ η 2 η = ; © ≍ P z ′ ¦ S[1, σ n ] ∩ η 2 η = ; © . Therefore, P S[1, σ n ] ∩ η = ; ≤ CP ¦ S[1, σ l ] ∩ η 1 η = ; © P ¦ S[1, σ n ] ∩ η 2 η = ; © . 1060 We now let η = b S n [0, b σ n ]. By Proposition 4.6, for any ω ∈ Ω l , λ ∈ e Ω m,n , P ¦ η 1 b S n [0, b σ n ] = ω; η 2 b S n [0, b σ n ] = λ © ≍ P ¦ η 1 b S n [0, b σ n ] = ω © P ¦ η 2 b S n [0, b σ n ] = λ © . Therefore, Esn = b E ” P ¦ S[1, σ n ] ∩ b S n [0, b σ n ] = ; ©— ≤ CbE ” P ¦ S[1, σ l ] ∩ η 1 b S n [0, b σ n ] = ; © P ¦ S[1, σ n ] ∩ η 2 b S n [0, b σ n ] = ; ©— ≤ CbE ” P ¦ S[1, σ l ] ∩ b S n [0, b σ l ] = ; ©— b E ” P ¦ S[1, σ n ] ∩ η 2 b S n [0, b σ n ] = ; ©— = C b E ” P ¦ S[1, σ l ] ∩ b S n [0, b σ l ] = ; ©— Esm, n. By corollary 4.5, since 4l ≤ n, b E ” P ¦ S[1, σ l ] ∩ b S n [0, b σ l ] = ; ©— ≍ bE ” P ¦ S[1, σ l ] ∩ b S[0, b σ l ] = ; ©— = e Esl. Finally, by Lemma 5.1, e Esl ≍ Esm, which finishes the proof of the proposition. Proposition 5.3. There exists c 0 such that for all m and n with m ≤ n2, Esn ≥ c Esm Esm, n. Proof. We will use the following abbreviations. Let l = ⌊m4⌋ and let η 1 = η 1 l b S n [0, b σ n ]; η 2 = η 2 m,n b S n [0, b σ n ]; η ∗ = η ∗ l,m,n b S n [0, b σ n ]. Then b S n [0, b σ n ] = η 1 ⊕ η ∗ ⊕ η 2 . We also decompose S[1, σ n ] into S 1 = S[1, σ 2l ] and S 2 = S[σ 2l + 1, σ n ]. Let c 1 be as in the statement of the separation lemma Theorem 4.7. Let W and W ∗ be the half- wedges W = {z : 1 − c 1 4 l ≤ |z| ≤ 1 + c 1 4 m, argz ≤ c 1 4 }; W ∗ = {z : 1 − c 1 4 l ≤ |z| ≤ 1 + c 1 4 m, argz ≤ c 1 2 }. and let A = B l ∪ W . Let K 1 be the set of η 1 such that η 1 ∩ ∂ B l ⊂ {z : argz ∈ − c 1 8 , c 1 8 }, and K 2 be the set of η 2 such that η 2 ∩ ∂ B m ⊂ {z : argz ∈ − c 1 8 , c 1 8 }. 1061 Then, Esn = b E ” P ¦ S[1, σ n ] ∩ b S n [0, b σ n ] = ; ©— = b E ” P ¦ S 1 ∩ η 1 ⊕ η ∗ = ;; S 2 ∩ η 1 ⊕ η ∗ ⊕ η 2 = ; ©— ≥ bE ” 1 {η 1 ∈K 1 } 1 {η 2 ∈K 2 } 1 {η ∗ ⊂W } P ¦ S 1 ∩ η 1 ∪ W ∗ = ;; S 2 ∩ η 2 ∪ A = ; ©— = b E ” 1 {η 1 ∈K 1 } P ¦ S 1 ∩ η 1 ∪ W ∗ = ; © × 1 {η 2 ∈K 2 } P ¦ S 2 ∩ η 2 ∪ A = ; S 1 ∩ η 1 ∪ W ∗ = ; © 1 {η ∗ ⊂W } — Therefore, Esn ≥ bE ” X η 1 Y η 2 1 {η ∗ ⊂W } — , where X η 1 = 1 {η 1 ∈K 1 } P ¦ S 1 ∩ η 1 ∪ W ∗ = ; © , and Y η 2 = 1 {η 2 ∈K 2 } inf z ∈∂ B 2l \W ∗ P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © . By Lemma 2.4 and Corollary 3.8, for any ω 1 ∈ K 1 and ω 2 ∈ K 2 , b P ¦ η ∗ ⊂ W η 1 = ω 1 , η 2 = ω 2 © ≥ c, and therefore Esn ≥ cbE ” X η 1 Y η 2 — . However, by Proposition 4.6, η 1 and η 2 are independent up to constants, and therefore, Esn ≥ cbE ” X η 1 — b E ” Y η 2 — . To prove the Proposition, it therefore suffices to show that b E ” X η 1 — ≥ c Esm, 17 and b E ” Y η 2 — ≥ c Esm, n. 18 To prove 17, note that b E ” X η 1 — = b E ” 1 {η 1 ∈K 1 } P ¦ S 1 ∩ η 1 ∪ W ∗ = ; ©— ≥ cbE ” P ¦ S 1 ∩ η 1 ∪ W ∗ = ; ©— ≥ cbE ” P ¦ S[1, σ l ] ∩ η 1 = ;; distSσ l , η 1 ≥ cl; S[1, σ l ] ∩ W ∗ = ; ©— ≥ cbE ” P ¦ S[1, σ l ] ∩ η 1 = ; ©— , where the last inequality is justified by the separation lemma Theorem 4.7. However, by Corollary 4.5 and Lemma 5.1, b E ” P ¦ S[1, σ l ] ∩ η 1 = ; ©— ≍ bE ” P ¦ S[1, σ l ] ∩ b S[0, b σ l ] = ; ©— = e Esl ≍ Esm. 1062 We now prove 18. Since η 2 ⊂ Λ\B m , by the discrete Harnack inequality, for any z 1 , z 2 ∈ ∂ B 2l \W ∗ , P z 1 ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © ≍ P z 2 ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © . Therefore, fixing a z ∈ ∂ B 2l \ W ∗ , Y η 2 ≥ c1 {η 2 ∈K 2 } P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © . By Lemma 3.1, P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © = Gz; B n \ η 2 ∪ A Gz; B n X y ∈∂ B n P y ¦ ξ z ξ A ∧ ξ η 2 ξ z σ n © P z S σ n = y . For any y ∈ ∂ B n , P y ¦ ξ z ξ A ∧ ξ η 2 ξ z σ n © ≥ X w ∈∂ B m \W ∗ P w ¦ ξ z ξ A ∧ ξ η 2 ξ z σ n © P y ¦ ξ w ξ W ∗ ∧ ξ η 2 ξ z σ n © For any w ∈ ∂ B m \ W ∗ , P w n ξ B cl 4 z ξ A ∧ ξ η 2 ξ z σ n o ≥ c. Furthermore, by Lemma 3.4, for any u ∈ ∂ B cl 4 z, P u ¦ ξ z ξ A ∧ ξ η 2 ξ z σ n © P u ¦ ξ z ξ η 2 ξ z σ n © ≥ P u n ξ z ξ B cl 2 z ξ z ξ η 2 ∧ σ n o ≥ c. It also follows from 8 in Lemma 3.4 that for any two paths η 2 , e η 2 ∈ e Ω m,n and any u ∈ ∂ B cl 4 z, P u ¦ ξ z ξ η 2 ξ z σ n © ≍ P u ¦ ξ z ξ e η 2 ξ z σ n © , and therefore there exists f n, m such that for all η 2 ∈ e Ω m,n and u ∈ ∂ B cl 4 z, P u ¦ ξ z ξ η 2 |ξ z σ n © ≍ f n, m. Thus, P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; © ≥ c f n, m Gz; B n \ η 2 ∪ A Gz; B n × X y ∈∂ B n P y ¦ ξ m ξ η 2 ∧ ξ W ∗ |ξ z σ n © P z S σ n = y . Let r 1 = distz, η 2 ∪ ∂ B n and r 2 = distz, η 2 ∪ A ∪ ∂ B n . Then r 2 c 1 l c 1 r 1 . Therefore, Gz; B n \ η 2 ∪ A ≥ Gz; B r 2 z ≥ cGz; B l z 1063 and by Lemma 3.3 applied to the ball Bz, r 1 , Gz; B l z ≥ cGz; B r 1 z ≥ cGz; B n \ η 2 . Finally, by the reverse separation lemma Theorem 4.10, b E ” P y ¦ ξ m ξ η 2 ∧ ξ W ∗ |ξ z σ n ©— ≥ cbE ” P y ¦ ξ m ξ η 2 |ξ z σ n ©— , and thus b E ” Y η 2 — ≥ cbE ” 1 {η 2 ∈K 2 } P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; ©— ≥ cbE ” P z ¦ S[1, σ n ] ∩ η 2 ∪ A = ; ©— ≥ c Gz; B l Gz; B n f n, mb E    X y ∈∂ B n P y ¦ ξ m ξ η 2 ∧ ξ W ∗ |ξ z σ n © P z S σ n = y    ≥ cbE    Gz; B n \ η 2 Gz; B n f n, m X y ∈∂ B n P y ¦ ξ m ξ η 2 |ξ z σ n © P z S σ n = y    ≥ cbE    Gz; B n \ η 2 Gz; B n X y ∈∂ B n P y ¦ ξ z ξ η 2 |ξ z σ n © P z S σ n = y    . However, by applying Lemma 3.1 again, Gz; B n \ η 2 Gz; B n X y ∈∂ B n P y ¦ ξ z ξ η 2 |ξ z σ n © P z S σ n = y = P z ¦ S[1, σ n ] ∩ η 2 = ; © . and thus b E ” Y η 2 — ≥ c Esm, n.

5.3 Intersection exponents for SLE