The Green Functions Of Two Dimensional Random Walks Killed On A Line And Their Higher Dimensional Analogues

We obtain asymptotic estimates of the Green functions of random walks on the two-dimensional integer lattice that are killed on the horizontal axis. A basic asymptotic formula whose leading term is virtually the same as the explicit formula for the corresponding Green function of Brownian motion is established under the existence of second moments only. Some reﬁnement of it is given under a slightly stronger moment condition. The extension of the results to random walks on the higher dimensional lattice is also given.


Introduction and Results
Let S x n = x + ξ 1 + · · · + ξ n be a random walk on Z 2 (the two dimensional integer lattice) starting at S x 0 = x ∈ Z 2 . Here ξ j , j = 1, 2, . . ., are independent and identically distributed Z 2 -valued random variables defined on a probability space (Ω, , P). The walk S 0 n is supposed irreducible and having zero mean and finite variances. Let [x 1 , x 2 ] stand for a point of R 2 with components x 1 and x 2 and put L = {[s, 0] : s ∈ R} (the first coordinate axis).
In this paper we obtain asymptotic estimates of the Green function G L of the walk killed on L: where p n L (x, y) = P[S x n = y, S x j / ∈ L for j = 1, . . . , n−1], the n-step transition probability of the killed walk; in particular p 0 L (x, y) = δ x, y (Kronecker's symbol). Note that p n L (x, y) (with n ≥ 1), hence G L (x, y), may be positive even if x, y ∈ L. Our definition, thus different from usual one (but only if |x| · | y| = 0), is convenient when duality relations are considered (see Remark 4 below). Let X and Y be the first and second component of ξ 1 = S 0 1 , respectively, put σ 2 1 = EX 2 , σ 12 = EX Y, σ 2 2 = EY 2 , σ j j = σ 2 j ( j = 1, 2), Q = (σ i j ) (the covariance matrix of S 0 1 ) and σ = | det Q| 1/4 , and define the norm x , x ∈ R 2 by x 2 = σ 2 Q −1 (x), where Q −1 and Q −1 (x) stand for the inverse matrix of Q and its quadratic form , respectively.
For a, b ∈ R, a ∨ b and a ∧ b denote, respectively, the maximum and the minimum of a and b. The function t log t is understood continuously extended to t = 0.   2 . Then the formula of Theorem 1.1 is understood to be in accordance with the one for Brownian motion, say g • (a, b), that is explicit owing to reflection principle (see Appendix (B)). Donsker's invariance principle implies that if two points a, b are taken from the upper half plane R × (0, ∞), > 0 and α x = |a|/|x| (x ∈ Z 2 ), then α 2 x { y:|α x y−b|< } G L (x, y) converges to |u−b|< g • (a, u)du as |x| → ∞ in such a way that lim α x x = a, but not much more. REMARK 2. For n < 0, the formula of Theorem 1.3 hold true as n[(−k) ∨ 1]/ [s, k − n] 2 → 0 by duality. In the case when nk < 0 and |k| ∧ |n| → ∞ we have σ 4 2 a(n)a(−k) + nk = o(nk), so that any proper asymptotic form of G L is not given by it; the determination of it requires more detailed analysis than that carried out in this paper and will be made in a separate paper ( [11]). REMARK 3. It follows from (1.1) that under the constraint s 2 +(n−k) 2 > nk > 0 (for some > 0) the convergence to zero of B s,n,k stated in Theorem 1.1 holds true without assuming E[|S 0 1 | 2 log |S 0 1 |] < ∞, which however cannot be removed in general (see also Theorem 3.2). Similarly the moment condition E[X 2   Because of this duality we may suppose that |k| ≤ n for the proof of Theorems above. In view of (1.4) (that also follows from the duality) the results on H [0,n] (s) obtained in [9] provide the formula of Theorem 1.3 in the case nk = 0. Thus we may further suppose that |k| > 0. This paper is in a sense a continuation of [9] in which the hitting distribution of L for the walk starting at [0, n] is evaluated. As in [9] put and for t = 0, Let H x (s) denote the probability that the first entrance after time 0 of the walk S x n into L takes place at [s, 0]: H x (s) = P[ for some τ ≥ 1, S x τ = [s, 0] and S x n / ∈ L for 0 < n < τ]. The evaluation of H x (s) in [9] is based on the Fourier representation formula while our starting point for evaluation of G L is Let p m (x, y) = P[S x m = y] and G(x, y) := ∞ m=0 p m (x, y), the Green function of the walk S x n . In the next theorem we give results corresponding to Theorems 1.1 and 1.2 under the following moment condition (to be a minimal one for the obtained estimates):  If the walk satisfies a certain condition concerning symmetry and continuity in the vertical direction these identities reduce to the reflection principle. Otherwise, however, direct computation using them together with the estimates of H [0,n] and G (as found e.g. in [9], [8]) does not give any correct asymptotic form of G L unless either the first term of the decomposition is dominant or we have nice estimates of the second order terms of H [0,n] and G; in any case the results obtained in such a way are in general not sharp.
Two relevant matters on some closely related random walks are briefly discussed in the rest of this section.
Walks Killed on a half horizontal axis. Put L ± = {[s, 0] : ±s ≥ 0} ⊂ Z 2 (the non-negative and nonpositive parts of the horizontal axis) and let G L − (x, y) be the Green function of the two dimensional walk killed on L − that is defined analogously to G L (x, y). Then it follows that for y / ∈ L − , where g (−∞,0] (s, s 1 ) denotes the Green function of one dimensional walk which is the trace on n ) killed on L − . Consider the last identity for the time-reversed walk. Indicating the dual objects by putting ∼ over the notation as in Remark 4, we then find the identities G L − (x, y) =G L − ( y, x),H y (s 1 ) = H − y (−s 1 ) andg (−∞,0] (s 1 , s) = g (−∞,0] (s, s 1 ) and substitute these into (1.10) to obtain that where H + x (s) denotes the entrance distribution into L + (defined analogously to H x (s)). Certain asymptotic estimates of H + x (s) are obtained in [10] (under the present setting), [4] (for simple walk) and [1] (for a class of random walks with a finite range of jump by an algebraic method). Computation made in Section 4 of [10] would be helpful for evaluation of the double sum in (1.11).
Walks Killed on a half plane. Let D = {[s, k] ∈ Z 2 : k ≤ 0}, the lower half plane. The evaluation of G L is intimately related to that of G D , the Green function of the walk killed on D. First it is pointed out that if the walk is either upwards or downwards continuous, namely either P[Y ≥ 2] = 0 or P[Y ≤ −2] = 0, then G L and G D agree on Z 2 \ D. Let f + (n) (resp. f − (n)) (n = 1, 2, . . .) be the positive harmonic function that is asymptotic to n of the one dimensional walk killed on {n ≤ 0} whose increment has the same law as Y (resp. −Y ): f ± (n) = E[ f ± (n ± Y ); n ± Y > 0] (n ≥ 1) and lim n→∞ f ± (n)/n = 1, which each exists uniquely ( [7]:p.212). If P[Y ≥ 2] = 0, then for n ≥ 1, f + (n) = 1 2 (σ 2 2 a(n) + n) and f − (n) = σ 2 2 a(−n) = n, and the formula of Theorem 1.3 may be written as . In a separate paper [11] we prove (1.12) to be true in general under E[X 2 log |X |] < ∞, where the proof rests on Theorem 1.3 of the present paper.
The rest of the paper is organized as follows. In Section 2 some preliminary lemmas are established.

Preliminary Lemmas
Throuout the rest of the paper we suppose that σ 12 = 0 unless otherwise stated explicitly, so that the quadratic form Q(θ ) = Q(θ 1 , θ 2 ) = σ i j θ i θ j (θ = (θ 1 , θ 2 ) ∈ R 2 ) is of the form The case σ 12 = 0 can be similarly treated and necessary modifications will be given in Appendix (C).
Let n > 0. As an ideal substitute for π −n (t), we bring in where, as in Remark 1, On writing m = n − k the Fourier representation (1.3) is decomposed as We evaluate these two integrals separately in Propositions 2.1 and 2.2 given below.
For the evaluation of the first integral above we observe that Then I ≤ C/(s 2 +1)(m 2 ∧ n 2 +1) since the function  As an immediate corollary of Proposition 2.1 we obtain where the limit value is understood to be 0 or ∞ according as |α| = ∞ or α = β = 0. For each M > 1 the convergence is uniform for s and m such that |m| < M n and |m| ∨ |s| > n/M .
Proof of Proposition 2.1. Let J = J s,n,m be the function given by (2.4). Then According to Lemma 2.1 J → 0 as |s| + (|n| ∧ |m|) → ∞. The repeated integral on the right side is decomposed into the sum of the integral and a similar one (with m in place of n). As |s|+|n| → ∞, the first integral converges to a finite limit n,m remains bounded, and otherwise it diverges to +∞ as |n| → ∞ with s, m being fixed. This completes the proof of (a).
For the proof of (b) we may suppose |m| ≤ n.
To this end we break it into three parts by dividing the inner integral at l = ±1/n. The part corresponding to the interval |l| < 1/n is easy to evaluate to be o(k/n). For the evaluation of the remaining parts we put It remains to prove K + + K − = o(k/m). Since the factor e −ist does not come into play at all, let s = 0 for simplicity. Writing e iml − e inl = (e −ikl − 1)e inl we integrate by parts to have The first term on the right side is dominated by kn −2 π 0 |D(t, 1/n)|d t + r = o(k/n) + r, where r is the boundary term corresponding to π and cancels out with that arising from K − . Integrating by parts once more and using the relations one can easily deduce that the second and third terms are o(k/m) and o(k/n), respectively.
We continue the argument made above for the proof of (c). Obviously K ± = o(log n) (uniformly in s, k), hence one may suppose |m| ∧ |s| → ∞. First consider the case |m| > |s|. We must prove that K ± = o(log(n/|m|)) as n/|m| → ∞. The contribution to K ± of the part involving e inl tends to zero as n → 0 uniformly in s, namely With this in mind we see that as in view of (2.6). Now let s|s| ≥ |m|. In view of (2.7), with symmetric roles of t and l in D(t, l) taken into account, it suffices to prove that uniformly in m, Changing the variable of integration we see that for any > 0 and N > 0 Proof. Write ρ • (t) for 1/π • 0 (t) = σ 2 |t| and make decomposition On using (2.2) the Fourier transform of the first term equals hence gives the principal term of the formula of the lemma.
For the remaining terms we use certain estimates of Fourier type integrals, which are collected in Appendix (D). From Lemma 6.1 (i) there it follows that This, together with (2.9), shows that the middle term on the right side of (2.9) is e −λn|t| (|k|×o(|t|)+ |t| × o(k)), hence its contribution is o(k/n) as above (but this time not only n but |k| must also be made large indefinitely: otherwise o(k/n) must be replaced by O(k/n)).
The asymptotic estimates of a {d=2} ([s, n]) given in [3] or [8] provide better estimates of C s,n,m than in Proposition 2.1 but under certain stronger moment conditions.

Proof of Theorems 1.1 and 1.2
For simplicity we let 0 < |k| ≤ n unless contrary is stated, which gives rise to no loss of generality as being pointed out in Remark 4. We continue to suppose that Q is diagonal.
It follows that [s, n] 2 = σ 2 (s 2 * + n 2 * ). For simple random walk the reflection principle may apply and it immediately follows from an asymptotic expansion of the potential function a {d=2} as found in [3], [5] We are to find a reasonable estimate of r in general case.
We first consider the case when n → ∞ under the constraint This condition is equivalent (or understood to be so) to the condition that there exists a constant , 0 < < 1 such that k > n; |s| < n/ ; and either k < (1 − )n or |s| > n.
Proof. Here again the expression on the right side of (3.
in particular, each of e n (t)/n, te n (t)/n and t 2 e n (t)/n tends to zero as nt → 0 and is uniformly bounded. (Here η ( j) denotes the derivative of the j-th order.) The contribution to G L of the first term on the right side is given in Proposition 2.1 (b) and that of the second is given by Theorem 1.1 of [9] (in view of (1.2)). The sum of these two may be written as 1 as n → ∞ under (i). As for the third term we write Having the bound 2) and Lemma 4.1 (b), we infer that the first integral on the right side is o(k/n). That the second one admits the same estimate follows from Lemma 6.2 with the help of the bound |e −k (t)| ≤ C|k|. Thus we obtain the relation of Theorem 4.1.
Case (ii) n ≤ |s|; |k| = o(|s|). Here we make the decomposition It suffices to compute the Fourier inversions for the first two terms of the right-most member, the other terms being dealt with in [9] (see (4.17) and (4.18) given at the end of the present proof). The first and second of the two is dealt with in Lemmas 4.2 and 4.3 below, respectively.
where w(t) is a smooth cutoff function such that w ≥ 0; w = 1 for |t| < 1/2; w = 0 for |t| > 1. As before we have that for any N as |s| → ∞. As for the error term r, perform integration by parts twice (to have the factor s −2 ) and decompose (e inl − 1)(e −ikl − 1) = sin nl sin kl + A n,k (l), where Observe so that (−π,π] 2 |F (t, l)| × d t l 3 → 0 as l → 0, and then deduce that (cf. Lemma 11.1 of [9]). In order to deal with the contribution of sin nl sin kl we need the moment condition E[X 2 log |X |] < ∞, with which we proceed as in [9]: Lemma 6.3 (estimation of Θ I I ). We decompose F (t, l) = V (t, l) + R(t, l), where and R is the rest. On the one hand R admits once more differentiation with ∂ t R = o((|t| + |l|) −3 ), and the integration by parts shows that its contribution is o(nk). On the other hand V satisfies so that the Riemann-Lebesgue lemma apply. These together show that (Here the outer integral (for the R part) must in general be understood improper and for the integral on |t| < 1/|s| one should integrate by parts back to have the result above.) Consequently r(s, k, n) = The next lemma is subtler than the preceding one. Proof. Although one can proceed by extending the lines of the proof of Theorem 1.2 of [9] (given in Sections 3 and 6), where evaluation of the integral ρe n e −ist d t is carried out, we proceed somewhat differently in a way the proof works better in the higher dimensions.
From Lemma 4.1 we have the expression e n (t) = f n (t) + η n (t) with the estimates where f n (t) = σ −2 d |n| f (λnt). In addition to the fact that both ρ and e n do not necessarily admit the differentiation of the third order, the difference of estimates between the derivatives of f n and those of η n as given above causes the complication of arguments. To make the proof conceptually clear we replace (ρe n e −k )(t) by σ 2 (|t| f n f −k )(t)w(t) and compute the corresponding integral of the latter and that of the difference between the two, separately. First we consider the difference, for which we need to find a way round the lack of differentiability. Write ρ • (t) for σ 2 |t| (as in (2.9)) and put We may suppose that 0 < k ≤ n ≤ s. After integrating by parts once we split the range of integration at t = ±1/|s|. From Lemma 4.1 it follows that under the constraint of the variables s, n, s of the lemma sup which entails the same relation for g in place of (ρe n e −k ) . We then deduce that the Fourier integral of g on |t| < 1/|s| is o(kn/s 2 ), and one more integration by parts gives For the latter we express g as the telescopic sum (ρ − ρ • ) f n f k + (e n − f n )ρ f k + ρe n (e −k − f k ) and deal with them separately. Among them we consider only the last term and verify that the integrals for the other two being evaluated in the same way in view of (4.7).
For the proof of (4.10) we claim that Since |ρe n | ≤ C|nt| and η −k (t) = k × o(t −2 ), the integrand is nk × o(1/t). Let F and V be the functions given in (4.5) and (4.6), respectively. We may write with r n (t), a nice function that is negligible for the present purpose, and then and τ k is the rest. Then, as in the preceding proof, we see that the function τ k (t) is differentiable for t = 0, τ k (t)/k = o(1/t 3 ) and its contribution is o(nk) and that if E[X 2 log |X |] < ∞, the Riemann-Lebesgue lemma yields That the last estimate is uniform in n and k requires proof. Since, by dominated convergence, π −π |t v k (t)|d t/k → 0 as |k| → ∞, it suffices for the proof to show that n −1 times the integral restricted on |t| > tends to zero uniformly in n for each > 0 and k. This follows from the fact that e n = π −n −π 0 +a(n) and sup <|t|<π |π −n (t)−π 0 (t)| = o(n). Thus the claim (4.11) has been verified.
For the proof of (4.10) we must evaluate the integrals of other terms of ρe n [e −k − f k ] (t), e.g., ρ (t)e n (t)[e −k − f k ](t), but their evaluations are quite similar, hence omitted.

It remains to prove
It is not hard at all to verify this as in a similar way to the above, but we take up another way. We are concerned with the Fourier integral that has an explicit form if w is removed and we shall seek out it. To this end the following decomposition of ρ • f n f k is convenient: Remember the identity π n (t) = e −λ|nt| /σ 2 |t| given in (2.2) and observe that ρ • π • n π • k = π • n+k , so that the first term on the right side of (4.14) equals π • n+k − π • n − π • k + π • 0 , whose Fourier transform, already computed in the preceding lemma, equals  Observe that under the constraint of (ii), (2π) −1 times the right side of the formula in Lemma 4.2 may be written as n * k * /π [s, n] 2 + o(nk/s 2 ) and that the sum of (4.17) and (4.18) may be written as σ 2 2 a * (n)a * (−k)/π [s, n] 2 +o(nk/s 2 ) since lim a(n)/n = 1/σ 2 2 . Combined with (4.3) and Lemma 4.3, these observations yield the estimate of Theorem 4.1.
The proof of Theorem 4.1, hence that of Theorem 1.3 is complete.

The Walks in Dimensions d ≥ 3
This section consists of five subsections and some preliminary discussions given preceding them. Here our primary purpose is to prove Theorems 1. Let the random walk S 0 n on Z d be irreducible and has zero mean and finite variances. Throughout this section the dimension d is supposed to be greater than or equal to three. For x ∈ Z d−1 and n ∈ Z we denote by [x, n] the d-dimensional point (x 1 , . . . , x n−1 , n) ∈ Z d and by L the hyper plane {[x, 0] : x ∈ Z d−1 }. The random variables X , Y , the norm [x, n] , the Green function G L , the functions ρ, π n etc. are all understood to be analogously defined. The Euclidian norm is denoted by | · | and we often write θ 2 for |θ | 2 , θ ∈ R d−1 .
An obvious analogue of the Fourier representation formula (1.3) (as well as that of (1.2)) is valid in the dimensions d ≥ 3. The leading term in the asymptotic formula of Theorem 1.4 comes from the explicit expression of the Fourier integral that intrinsically arises when 1 − ψ is replaced by 1 2 Q in the representation formula of G L . The problem is to estimate the error term that is caused by the replacement; for the estimation we need some moment condition, of which the condition (1.5) (resp. E|X | d < ∞) is appropriate under the constraint |x| < M (|n| ∨ |k|) (resp. |x| ≥ M −1 |n| ∨ |k|).
Let T = [−π, π) d−1 be the range of the variable θ ; w(θ ) denotes a smooth cut off function as before (w equals 1 about the origin and vanishes for |θ | > 1) . For every N > 0, as s → ∞ Indeed, together with the change of variables of integration, Green's formula transforms the in-  Proof. The proof is similar to that of Proposition 2.2. We have the same decomposition of ρ[π 0 − π k ]π −n as in (2.9) and the same estimate for each terms of it but with |θ | in place of |t|. Thus for the first term, whose Fourier coefficient (i.e. integral on T ) therefore agrees with the principal term of the formula of the lemma up to o(n −N ) for any N . The middle term is of the form e −λn|θ | (|k| × o(|θ |) + |θ | × o(k)) and it is readily inferred that its Fourier coefficientl is o(k/n d−1 ) under (1.6). For the last term, which is ρ[π 0 − π k ](π −n − π • −n ), we first observe that (see Lemma 6.1 (i) for the case |kθ | < 1) and then infer from (i) and (ii) of Lemma 6.2 that its contribution is o(n −d+2 ) under (1.6) and o(k/n d−1 ) under (1.5), respectively.
Proof of Theorem 1.5. Under (1.6) it holds that for each M > 1, uniformly for |x| < M n, as n → ∞ (see [8] for the identification of the leading term and Lemma 6.2 for the error estimate). Combining this with Lemma 5.1 (its first case) we find the formula of Theorem 1.5.
The following lemma that corresponds to (b) of Proposition 2.1 will be applied in the proof of Theorem 1.4.

Lemma 5.2.
If (1.5) holds, s 2 + (n 2 ∧ m 2 ) = 0 and C x,n,m is defined via the equation Proof. We have an analogue of (2.5). By making a suitable truncation argument by means of w(l) the term J in it is evaluated to be negligible. As for the double integral in it, writing e iml − e inl = (e −ikl − 1)e inl (k = n − m), we first integrate its inner integral by parts d − 2 times successively, and then proceed as in the proof of (b) of Proposition 2.1 (with the help of Lemma 6.2 of Appendix (D)) to obtain the required estimate.
REMARK 7. The case |n/m| → ∞ is excluded in Lemma 5.2, since in its subcase |m|/s → 0 we need to impose an additional moment condition on X for identifying the asymptotic form of
which may also be written as (5.11) The function f comes up from the corresponding integral of the second term: where the formula a 2 ∞ 0 [π(a 2 + l 2 )l] −1 sin y l d l = 1 2 (1 − e −|a| y ) (valid for y ≥ 0) is used for the last equality. It accordingly follows that 2πη n (θ ) = π −π W (θ , l)(e inl − 1)d l+ a negligible term. The evaluation of the last integral is made in the same way as for the two dimensional case for here the integration by θ is not involved. This gives (5.9). Since ∇W = ∇D, we have for k ≥ 1, and the formula (5.10) is obtained as in the proof of Lemma 5.3.
; moreover the estimates of Lemmas 5.3 and 5.4 are improved as follows: for k = 0, . . . , d, respectively, and the use of these considerably simplifies the argumenys in the next section.

Estimation of H [0,n] (x) and Proof of Theorem 1.4
As has been noticed in the two-dimensional case Theorem 1.4 includes, as a special case of k = 0, the asymptotic estimate of the hitting distribution H [0,n] (x). For the proof of Theorem 1.4 this special case is essential and first dealt with. (ii) Suppose that E|X | d < ∞. Then uniformly for |n| < M |x|, as |x| → ∞ Proof. First we prove (i). Let n > 0. The leading term should be given by (see Section 10 of [9]). Our task is to prove that uniformly for |x| < M n, Now suppose (1.5) to hold. We decompose ρπ −n − ρ • π • −n as follows: In view of (5.1) (i.e., π • −n (θ ) = e −nλ|θ | R /ρ • (θ )) and the estimate ρ We shall see in Appendix (D) (Lemma 6.2 (ii)) that the Fourier coefficient of ρ[π −n − π • −n ] admits the same estimate as above. Thus the proof the part (i) is complete.
Evaluation of I 1 is made in a similar way to the proof of Proposition 5.1. We employ Lemma 5.4. After integrating by parts d − 1 times we split the range of integration along the (d − 2)-dimensional sphere of radius 1/s. The integral inside the sphere is easy to evaluate. For the integral on its outside we integrate by parts once more. The typical term that then arises is a constant multiple of plus the boundary term, which is o(1). For 1 ≤ j < d, further performing integration by parts once we see that the last integral is also o (1). For the case j = 0 we use the second formula of Lemmas 5.4 to obtain the same estimate. As for the case j = d, separating the non-differentiable term from ∇ d ω ρ according to (5.6), we have only to make the same argument. Thus we have I 1 = o(1) as s → ∞.
Here, the Riemann-Lebesgue lemma is applied to the parts involving the functions (O(·) terms in (5.6) and (5.10) ) that are integrable but not necessarily differentiable. That the convergence is uniform in n ≤ s/M for such parts is verified as in the relevant discussion following (4.12) in the proof of Lemma 4.3.
Now we turn to I 2 . We need to derive a bound of ∇ j ω f n (θ ). First observe that Combining this with the estimate of ∇ Case (iii) s ≤ n; (−k) ∨ 1 n. The assertion follows from Theorem 1.5 (under (1.6)).

An Estimate of H
If E[X 2 log |X |] < ∞, the error term in (5.17 Differentiate both sides of (5.7) and observe that −ρ 2 ∇ 2 We do not make truncation by means of w(θ ) at this stage; the boundary terms that arise in the integration by parts (of non-periodic functions) that will be performed once more cancel out one another since ∇ 2 ω ρ is periodic. By (5.2) we obtain that For both Λ 1 and Λ 2 we have a situation similar to one arising in the proofs of Lemmas 6.2 and 6.3 of [9] (but only with the appearance here more complicated) and by virtually the same arguments as are made therein it is shown that T Λ j (θ )e −x·θ dθ = o(s −1 log s) for j = 1, 2, where the error term is replaced by o(s −1 ) plus the boundary term if E[X 2 log |X |] < ∞.
On integrating by parts and applying Fubini's theorem The first term on the right side is o(1/s). Indeed if the expectation above is restricted to the event |X | < K, it is o(1/s) for each K in view of the Riemann-Lebesgue lemma. Changing the variables of the outer integral we see that the same expectation but on |X | ≥ K is bounded in absolute value by a positive multiple of The second term is similarly dealt with. If the expectation involved in it is restricted to |X | ≤ K, the corresponding part is o(1/s). The other part, after integrating by parts once more, is disposed of in the same way. These together verify We are left with ζ 0 . Let 0 < < 1/2. Split the range of integration into two parts according as |X − x| ≥ s or |X − x| < s and call J 1 and J 2 , respectively, their contributions to T 1/s ζ 0 (θ )e −i x·θ dθ .
Since J 2 = (2π) 2 M (x) if < 1/2, it suffices to show that J 1 = o(1). We integrate by parts with respect to θ by factorizing the integrand as e i(X −x)·θ ×(the other) to deduce that for each > 0, J 1 equals   ). It is readily inferred (cf. [9], Appendix A) that for 0 < |t| ≤ π, ∞ m=0 s∈Z p m (0, [s, k])e ist = π k (t) and ∞ τ=0 j∈Z F n (τ, j)e i j t = ρ(t)π −n (t) and with the help of these identities we derive from (6.1) (B) Let B t be the two dimensional standard Brownian motion and consider its linear transform X t = Q 1/2 B t . If g • (x, y) denotes the Green function of X t killed on L and g B (x, y) = − 1 π log | y − x| and g • B (x, y) = g B (x, y) − g B (x , y), where x stands for the mirror image of x relative to the line Q −1/2 L, then, g • (x, y) = g • B (Ax, Ay) det A (x ∈ R, y > 0) where A = Q −1/2 . We may write (Ax) = Ax, so that π −n+k (t) − ρ(t)π −n (t)π k (t) e −i(s+µn−µk)t d t.
The proof of (i) is carried out similarly. For d = 4 we apply integration by parts two times, which results in o(r −4 ) for the integrand, and, the further integration by parts being not allowed under the condition EY 2 < ∞, we impose the logarithmic moment condition on Y to guarantee the integrability. In the case d = 2 we also need to suppose the same moment condition of Y to guarantee the integrability of D w (so that I n = o(1)), but the reason is slightly different: if d = 2 we cannot dispose of the integral about the origin in advance, D w itself being possibly non-integrable.
We must ensure the uniformity of convergence with respect to functions h for the integral that is disposed of by means of the Riemann-Lebesgue lemma. This is readily done by approximating the function of l that results in from integration over |θ | > by a smooth function for each positive , the other integral approaching zero uniformly as → 0.