s s s k Us s

Since u 1 r = u 1 r − , we have b 1 u 1 r ≤ r r − b 1 u 1 r + 1, thus r − r − ∆b 1 u 1 r + 1 ≤ r − r − ∆b 1 u 1 r + 1 α . Now v i 2 − b i 2 u i 2 v i 2 ∆b i 1 u i 1 v i 1 + 1 . . . v i l − b i l u i l v i l ∆b i l u i l v i l + 1 1 and |∆ 1 u 1 r+1 ∆ i 2 u i2 v i2 +1 . . . ∆ i l u il v il +1 S n Uv| = |∆ 1 u 1 r+1 X i∈I ǫ i S n i| ≤ X i∈I |∆ 1 u 1 r+1 S n i|, 43 where ǫ i = ±1 and I is some appropriate subset of [0, n] ∩ N d with 2 l−1 elements. Denote for convenience Z n = max 1≤k≤k n |∆ 1 k 1 S n k| ∆b 1 k 1 α . 44 Now noting that r − r − = 2 − j and ∆b 1 k 1 depends only on k 1 , we obtain for l ≥ 2 |T l u ′ − T l u| ≤ 2 − jα d − 1 l − 1 2 l−1 Z n . Thus |Ξ n v − Ξ n v − | ≤ d X l=1 2 − jα d − 1 l − 1 2 l−1 Z n = 3 d−1 2 − jα Z n 45 Case 2. u 1 r = u 1 r − + 1. In this case we have b 1 u 1 r − ≤ r − b 1 u 1 r ≤ r. Using previous definitions we can write |Ξ n v − Ξ n v − | ≤ |Ξ n v − Ξ n b 1 u 1

r, s

ℓ | + |Ξ n b 1 u 1

r, s

ℓ − Ξ n v − |. Now r − b 1 u 1 r ∆b 1 u 1 r + 1 ≤ r − b 1 u 1 r ∆b 1 u 1 r + 1 α ≤ 2 − jα ∆b 1 u 1 r + 1 α and similarly b 1 u 1 r − r − ∆b 1 u 1 r − + 1 ≤ 2 − jα ∆b 1 u 1 r − + 1 α . Combining these inequalities with 41 and 42 we get as in45 |Ξ n v − Ξ n v − | ≤ 2 · 3 d−1 2 − jα Z n . 2274 Case 3. u 1 r u 1 r − + 1. Put u = b 1 u 1

r, s

ℓ , u − = b 1 u 1 r − + 1, s ℓ and ψ n r, r − = max k 2:d ≤k n ,2:d ¯ ¯ ¯ ¯ u 1 r X i=u 1 r − +2 ∆ 1 i S n

i, k

2:d ¯ ¯ ¯ ¯ . 46 Then |Ξ n v − Ξ n v − | ≤ |Ξ n v − Ξ n u| + |Ξ n u − Ξ n u − | + |Ξ n u − − Ξ n v − | Recall notation 5 and note that U u 2:d = Uu − 2:d = Uv 2:d . We have Ξ n u = S n Uu+ d−1 X l=1 X 2≤i 1 i 2 ···i l ≤d l Y k=1 v i k − b i k u i k v i k ∆b i k u i k v i k + 1 l Y k=1 ∆ i k u ik v ik +1 S n Uu and similar representation holds for Ξ n u − . Since S n Uu − S n Uu − = u 1 r X i=u 1 r − +2 ∆ 1 i S n

i, Us

ℓ , similar to 43 and 45 we get |Ξ n u − Ξ n u − | ≤ ψ n r, r − d−1 X l=0 2 l ≤ 3 d−1 ψ n r, r − . We can bound |Ξ n v − Ξ n u| and |Ξ n u − − Ξ n v| as in case 2. Thus we get |Ξ n

r, s

ℓ − Ξ n r − , s ℓ | ≤ 3 d−1 ψ n r, r − + 2 · 3 d−1 2 − jα Z n . 47 Substituting this inequality into 39 we get that Π − J , n; ǫ ≤ Π 1 J , n; ǫ2 · 3 d−1 + Π 2 n; ǫ4 · 3 d−1 where Π 1 J , n; ǫ = P Z n ǫ 48 and Π 2 J , n; ǫ = P sup j≥J 2 − jα max r∈D j ψ n r, r − ǫ . 49 2275 Thus 37 will hold if lim J →∞ lim sup n→∞ Π 1 J , n; ǫ = 0, 50 lim J →∞ lim sup n→∞ Π 2 J , n; ǫ = 0. 51 Proof of 50. Using Markov and Doob inequalities for q 112 − α P Z n ǫ ≤ k 1 n X k=1 P max k 2:d ≤k n ,2:d |∆ 1 k S n k| ǫ∆b 1 k α ≤ k 1 n X k=1 ǫ −q ∆b 1 k −qα E max k 2:d ≤k n ,2:d |∆ 1 k S n k| q ≤ k 1 n X k=1 ǫ −q ∆b 1 k −qα E |∆ 1 k S n k n | q . By applying the Rosenthal inequality we get P Z n ǫ ≤ c k 1 n X k=1 ǫ −q ∆b 1 k −qα ∆b 1 k q2 + k 2 n X k 2 =1 · · · k d n X k d =1 E |X n ,k | q . 52 We have k 1 n X k=1 ∆b 1 k q12−α ≤ max 1≤k≤k 1 n ∆b 1 k q12−α−1 k 1 n X k=1 ∆b 1 k = max 1≤k≤k 1 n ∆b 1 k q12−α−1 → 0, as mn → ∞, due to 9 and the fact that q 12 − α. Also k 1 n X k=1 ∆b 1 k −qα k 2 n X k 2 =1 · · · k d n X k d =1 E |X n ,k | q = X k≤k n ∆b 1 k 1 −qα E |X n ,k | q ≤ X k≤k n σ −2qα n ,k E |X n ,k | q → 0, as mn → ∞, due to 10, since ∆b 1 k 1 −qα ≤ σ −2qα n ,k for all 1 ≤ k ≤ k n . Reporting these estimates to 52 we see that 9 and 10 imply 50. Proof of 51. We have Π 2 J , n, ǫ ≤ X j≥J P2 α j max r∈D j ψ n r, r − ǫ ≤ X j≥J X r∈D j ǫ −q 2 α jq E |ψ n r, r − | q . 2276 Doob inequality together with 34 gives us E ψ n r, r − q ≤ E ¯ ¯ ¯ ¯ X k 2:d ≤k n ,2:d u 1 r X k 1 =u 1 r − +2 X n ,k ¯ ¯ ¯ ¯ q ≤ c u 1 r X k 1 =u 1 r − +2 X k 2:d ≤k n ,2:d σ 2 n ,k q2 + u 1 r X k 1 =u 1 r − +2 X k 2:d ≤k n ,2:d E |X n ,k | q . Due to definition of u 1 r u 1 r X k 1 =u 1 r − +2 X 1≤k 2:d ≤k n ,2:d σ 2 n ,k = u 1 r X k 1 =u 1 r − +2 ∆b 1 k 1 ≤ r − r − = 2 − j , thus Π 1 J , n, ǫ ≤ c ǫ q X j≥J 2 qα+1−q2 j + c ǫ q X j≥J X r∈D j 2 qα j u 1 r X k 1 =u 1 r − +2 X k 2:d ≤k n ,2:d E |X n ,k | q . 53 Denote by IJ , n, q the second sum without the constant cǫ −q . By changing the order of summation we get I J , n, q = X 1≤k≤k n E |X n ,k | q X j≥J 2 αq j X r∈D j 1{u 1 r − + 1 k 1 ≤ u 1 r}. 54 The proof further proceeds as in one dimensional case [7]. Consider for fixed k 1 the condition u 1 r − + 1 k 1 u 1 r. 55 Suppose that there exists r ∈ D j satisfying 55 and take another r ′ ∈ D j . Since u 1 is non decreasing, if r ′ r − we have u 1 r ′ u 1 r − + 1 k, and thus r ′ cannot satisfy 55. If r ′ r, we have that r ′− r, and we have that k ≤ u 1 r ≤ u 1 r ′− u 1 r ′− + 1 and again if follows that r ′ cannot satisfy 55. Thus there will exists at most only one r satisfying 55. If such r exists we have r − ≤ u 1 r − +1 X i=1 ∆b 1 i k 1 X i=1 ∆b 1 i ≤ u 1 r X i=1 ∆b 1 i ≤ r. Thus ∆b 1 k 1 ≤ 2 − j . So ∀k 1 = 1, . . . , k 1 n , X r∈D j 1{u 1 r − + 1 k 1 ≤ u 1 r} ≤ 1{∆b 1 k 1 ≤ 2 − j } so X j≥J 2 αq j X r∈D j 1{u 1 r − + 1 k 1 ≤ u 1 r} ≤ 2 qα 2 qα − 1 ∆b 1 k 1 −αq 56 we can sum only those j, for which ∆b 1 k 1 ≤ 2 − j , because for larger j, r and r − will be closer together and will fall in the same R n ,k . 2277 Reporting estimate 56 to 54 we get I J , n, q ≤ C X k≤k n ∆b 1 k 1 −qα E |X n ,k | q ≤ X k≤k n σ −2qα n ,k E |X n ,k | q and substituting this to inequality 53 we get Π 1 J , n; ǫ ≤ C 1 ǫ −q 2 −J qα+1−q2 + C 2 X k≤k n σ −2qα n ,k E |X n ,k | q . Thus 51 follows from 10, and condition ii holds.

5.2 Proof of the theorem 13