Proof of Theorems 1.2 and 1.4

Theorem 2.1. For any Borel set A ⊆ ∂ D, P x Bτ D ∈ A = 1 2 Z A – ∂ ∂ n y G D x, y ™ σd y, x ∈ D, where ∂ ∂ n y is the inward normal derivative at y ∈ ∂ D and σd y is the surface measure on ∂ D induced by the usual Riemannian structure on R d . ƒ The unboundedness of D complicates the proof of Theorem 2.1. We will break up the proof into several pieces, but before that, we now use it to prove Theorems 1.2 and 1.4.

2.1 Proof of Theorems 1.2 and 1.4

For d ≥ 3, let Ω d −1 be the unit ball in R d −1 and set ν = d −3 2 . As pointed out earlier, the first Dirichlet eigenvalue of ∆ R d −1 on Ω d −1 is j 2 ν , where j ν is the first positive zero of the Bessel function J ν . Furthermore, the corresponding eigenfunction is w ˜ x = C |˜x| −ν J ν j ν |˜x|, ˜ x ∈ Ω d −1 , where C is chosen so that w0 = 1. The next theorem is due to Cranston and Li 1997—see the two paragraphs just after the proof of their Theorem 2.1. Note that although they make the blanket assumption at → 0 as t → ∞, this is not used to prove the version of their theorem that we use. Theorem 2.2. Let a : [M , ∞ → 0, ∞ be continuous and, for sufficiently large t, twice differentiable with lim t →∞ [|a ′ t| + at|a ′′ t|] = 0. Set D M = {x 1 , ˜ x ∈ R × R d −1 : x 1 M , |˜x| ax 1 }, and suppose H is bounded and Hölder continuous on D M with lim t →∞ at 2 sup |z|at |Ht, z| = 0. If u satisfies u on D M ∆ R d + Hu = 0 on D M u = 0 on {x 1 M } ∩ ∂ D M lim x 1 →∞ ux 1 , ˜ x = 0, then for each δ ∈ 0, j ν there exist M 1 M and C 0 such that C −1 w ˜ x ax 1 exp ‚ − j ν + δ Z x 1 M d t at Œ ≤ ux 1 , ˜ x ≤ C w ˜ x ax 1 exp ‚ − j ν − δ Z x 1 M d t at Œ for all x = x 1 , ˜ x ∈ D M 1 . ƒ 2663 For y ∈ ∂ D, recall n y is the inward unit normal to ∂ D at y. Lemma 2.3. Let y = y 1 , ˜ y ∈ ∂ D with y 1 1 2 . Then for z = z 1 , ˜ z = y + hn y , lim h →0 + w˜ z az 1 h = −C j ν J ′ ν j ν p 1 + [a ′ y 1 ] 2 a y 1 . Remark 2.4. It is known that J ′ ν j ν 0. Proof of Lemma 2.3. For y = y 1 , ˜ y ∈ ∂ D with y 1 1 2 , it is a simple matter to show that n y = [1 + [a ′ y 1 ] 2 ] −12 a ′ y 1 , − ˜ y a y 1 . Since | ˜y| = a y 1 , as h → 0 + we get ˜ z az 1 = ˜ y • 1 − h a y 1 p 1+[a ′ y 1 ] 2 ˜ a y 1 + ha ′ y 1 p 1 + [a ′ y 1 ] 2 = a y 1 a y 1 + ha ′ y 1 p 1 + [a ′ y 1 ] 2 1 − h a y 1 p 1 + [a ′ y 1 ] 2 7 → 1. Thus lim h →0 + w˜ z az 1 h = lim h →0 + C h ˜ z az 1 −ν J ν ‚ j ν ˜ z az 1 Œ = C lim h →0 + 1 h J ν ‚ j ν ˜ z az 1 Œ = C lim h →0 + J ′ ν ‚ j ν ˜ z az 1 Œ ∂ ∂ h ˜ z az 1 j ν = −C j ν J ′ ν j ν p 1 + [a ′ y 1 ] 2 a y 1 . Proof of Theorem 1.2. Define x 1 N to be the first coordinate of the intersection of the circle ρ 2 + z 2 = N 2 with the curve z = a ρ in the ρz-plane: x 1 N 2 + ax 1 N 2 = N 2 . 8 Fix x ∈ D and suppose M |x| and δ ∈ 0, j ν . Combined with the fact that G D x, y goes to 0 as the modulus of y ∈ D goes to infinity, Theorem 2.2 applied to u· = G D x, · and H ≡ 0 on D M shows that we can choose M 1 M and C 0 such that 2664 C −1 w ˜ y a y 1 exp ‚ − j ν + δ Z y 1 M d t at Œ ≤ G D x, y ≤ C w ˜ y a y 1 exp ‚ − j ν − δ Z y 1 M d t at Œ 9 for all y = y 1 , ˜ y ∈ D M 1 . Since G D x, · = 0 on ∂ D, for y = y 1 , ˜ y ∈ ∂ D with y 1 M , ∂ ∂ n y G D x, y = lim h →0 + G D x, y + hn y h . Using this and Lemma 2.3 in 9, we get that for some positive C 1 and C 2 C 1 p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν + δ Z y 1 M d t at Œ ≤ ∂ ∂ n y G D x, y ≤ C 2 p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν − δ Z y 1 M d t at Œ for y = y 1 , ˜ y ∈ ∂ D with y 1 M 1 . Then by Theorem 2.1, also using that P x |Bτ D | N = P x B 1 τ D x 1 N recall 8, we get that for N M 1 , 1 2 C 1 Z y 1 x 1 N p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν + δ Z y 1 M d t at Œ d y ≤ P x |Bτ D | N ≤ 1 2 C 2 Z y 1 x 1 N p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν − δ Z y 1 M d t at Œ d y. The integrands depend only on y 1 , so this reduces to 1 2 C 1 Z ∞ x 1 N p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν + δ Z y 1 M d t at Œ a y 1 d −1 d y 1 ≤ P x |Bτ D | N ≤ 1 2 C 2 Z ∞ x 1 N p 1 + [a ′ y 1 ] 2 a y 1 exp ‚ − j ν − δ Z y 1 M d t at Œ a y 1 d −1 d y 1 . 10 Since a ′ t → 0 as t → ∞, the identity log at = Z t 1 a ′ u au du + log a1 implies –Z t M du au ™ −1 log at → 0 as t → ∞. 11 2665 By l’Hôpital’s rule and the fact that a ′ x → 0 as x → ∞, for any γ 0 that is close to j v , lim N →∞ R ∞ N p 1 + [a ′ y 1 ] 2 exp − γ R y 1 M d t at a y 1 d −2 d y 1 aN d −1 exp − γ R N M d t at = 1 γ . Combining this with 10, we get that for some positive C 3 and C 4 , for large N , C 3 j ν + δ ax 1 N d −1 exp − j ν + δ Z x 1 N M d t at ≤ P x |Bτ D | N ≤ C 4 j ν − δ ax 1 N d −1 exp − j ν − δ Z x 1 N M d t at . 12 Take the natural logarithm, multiply by [ R x 1 N M d t at ] −1 , let N → ∞ and use 11 to get − j ν + δ ≤ lim inf N →∞   Z x 1 N M d t at   −1 log P x |Bτ D | N ≤ lim sup N →∞   Z x 1 N M d t at   −1 log P x |Bτ D | N ≤ − j ν − δ. Let δ → 0 and use the fact that Z x 1 N M d t at ∼ Z x 1 N 1 d t at as N → ∞ to get lim N →∞   Z x 1 N 1 d t at   −1 log P x |Bτ D | N = − j ν . To complete the proof of Theorem 1.2, we show Z x 1 N 1 d t at ∼ Z N 1 d t at as N → ∞. Indeed, recalling the definition of x 1 N from 8 and writing x 1 for x 1 N , we have N = x 1 È 1 + ax 1 x 1 2 . 2666 Then by the mean value theorem, there is ˜ x ∈ x 1 , N such that aN ax 1 − 1 = aN − ax 1 ax 1 = a ′ ˜ xx 1 –Ç 1 + ax 1 x 1 2 − 1 ™ ax 1 = a ′ ˜ xx 1 ax 1 – 1 + 1 2 ax 1 x 1 2 + o ‚ ax 1 x 1 2 Œ − 1 ™ = a ′ ˜ x 1 2 ax 1 x 1 + o1 ax 1 x 1 → 0 as N → ∞, since au u → 0 and a ′ u → 0 as u → ∞. Thus aN ax 1 → 1 as N → ∞. Upon differentiating 8 with respect to N , x ′ 1 = N x 1 1 + ax 1 x 1 a ′ x 1 = Ç 1 + ax 1 x 1 2 1 + ax 1 x 1 a ′ x 1 → 1 as N → ∞. To finish, use these limits and l’Hôpital’s rule to get lim N →∞ R x 1 M d t at R N M d t at = lim N →∞ x ′ 1 aN ax 1 = 1, as desired. Proof of Theorem 1.4. Under the additional hypotheses of Theorem 1.2, the conclusion of Theorem 2.2 holds with δ = 0. Thus the argument leading to 12 goes through with δ = 0 there and we get the conclusion of Theorem 1.4. 2667

2.2 Proof of Theorem 2.1: Preliminaries and a Reduction