Proof of Theorem 2.1b

Next, by 3.63 we have W 2 T ≤ Z R d Z t 2 t 1 s −γα Z t 2 −t 1 p u du ∗ e ϕ T y 1+ β dsd y. The Young inequality, substitution u ′ = ut 2 − t 1 , and self-similarity imply W 2 T ≤ Ct 1− γα 2 − t 1− γα t 2 − t 1 1+ β−dαβ || f || 1+ β 1+ β ||ϕ|| 1+ β 1 ≤ C 1 t 1− γα 2 − t 1− γα 1 2+ β−dαβ , 3.67 by 3.41. Combining 3.66, 3.67, 3.61 and using 3.59, we obtain 3.57. It remains to prove 3.56. Applying the usual substitutions to AT given by 3.58 we obtain AT = H T T 2−d α−γα F 2 T Z 1 Z 1 s Z R 3d |x| −γ p s x − y e ϕ T yp u−s y − z e ϕ T zχsχud x dzd y duds, hence, by 3.60, 3.55 and the Hölder inequality, AT ≤ C H T T 2−d α−γα F 2 T Z t 2 t 1 Z t 2 s s −γα || e ϕ T || 1+ β β ||p u−s ∗ e ϕ T || 1+ β duds. We use 3.65, 3.45 and 2.5, obtaining AT ≤ C 1 t 1− γα 2 − t 1−1 α 1 t 2 − t 1 1−d αβ1+β , which implies 3.56 by 3.59. This completes the proof of tightness. ƒ

3.4 Proof of Theorem 2.1b

We prove the theorem for γ = d 2 + β 1 + β α 3.68 see Remark 2.2c. Recall that in this case kT occurring in 2.8 and 2.9 is log T . According to the discussion in Section 3.1 it suffices to prove 3.12, 3.13 and 3.15. By the form of the limit process see 2.3, 3.12 is equivalent to lim T →∞ I 1 T = σS d−1 α Z R d Z 1 p s y Z 1 s p u−s yχudu 1+ β dsd y ‚Z R d ϕzdz Œ 1+ β , 3.69 where σS d−1 is the measure of the unit sphere in R d = 2 if d = 1. 1350 By 3.9, 3.2, 3.6, using similar substitutions as in the previous section, we obtain I 1 T = 1 log T Z R d Z 1 Z R d p s x T −1α − y Z 1 s Z R d p u−s y − zχu e ϕ T zdzdu 1+ β 1 1 + |x| d d y dsd x, 3.70 where e ϕ T is given by 3.34. We write I 1 T = I ′ 1 T + I ′′ 1 T + I ′′′ 1 T , 3.71 where I ′ 1 T = 1 log T Z 1 |x|T 1 α Z 1 Z R d ... 3.72 I ′′ 1 T = 1 log T Z |x|≥T 1 α Z 1 Z R d ... 3.73 I ′′′ 1 T = 1 log T Z |x|≤1 Z 1 Z R d .... 3.74 Passing to polar coordinates in the integral with respect to x we have I ′ 1 T = 1 log T Z T 1 α 1 Z S d−1 Z 1 Z R d p s w r T −1α − yg s ∗ e ϕ T y 1+ β r d−1 1 + r d d y ds σd wd r, where g is defined by 3.35. The crucial step is the substitution r ′ = log r log T , 3.75 which gives I ′ 1 T = Z 1 α Z S d−1 Z 1 Z R d p s w T r−1 α − yg s ∗ e ϕ T y 1+ β T r d 1 + T r d d y ds σd wd r. It is now clear that if one could pass to the limit under the integrals as T → ∞, then I ′ 1 T would converge to the right hand side of 3.69. This procedure is indeed justified by the fact that for f defined by 3.22 we have f ∈ L 2+ β R d , 3.76 which follows from 3.68, 3.23 and 3.24. We omit the details, which are similar to the argument in [BGT6] see 3.51 therein. Next we show that I ′′ 1 T and I ′′′ 1 T tend to zero. In I ′′ 1 T see 3.73 we substitute x ′ = x T −1α and we use 3.36, obtaining I ′′ 1 T ≤ C log T Z |x|1 Z 1 Z R d p s x − y f ∗ e ϕ T y 1+ β T d α 1 + |x| d T d α d y dsd x ≤ C 1 log T || f ∗ e ϕ T || 1+ β 1+ β ≤ C 1 log T || f || 1+ β 1+ β ||ϕ|| 1+ β 1 → 0. 1351 I ′′′ 1 T see 3.74 is estimated as follows: I ′′′ 1 T ≤ C log T Z |x|≤1 f ∗ f ∗ e ϕ T 1+ β x T −1α d x ≤ C 2 log T || f ∗ f ∗ e ϕ T 1+ β || ∞ ≤ C 2 log T || f || 2+ β || f ∗ e ϕ T 1+ β || 2+β1+β ≤ C 2 log T || f || 2+ β 2+ β ||ϕ|| 1+ β 1 → 0, by 3.76. This and 3.71 prove 3.69. To prove 3.13 we use 3.14 and easily obtain I 2 T ≤ C H T T 2−2d α F 2 T Z R 2 f ∗ e ϕ T f ∗ e ϕ T x T −1α 1 1 + |x| d d x. We write the right-hand side as the sum of integrals over {|x| ≤ T 1 α } and {|x| T 1 α }. To estimate the integral over {|x| ≤ T 1 α } we use sup T 2 1 log T Z |x|≤T 1 α 1 1 + |x| d d x ∞, 3.77 and in the second integral we apply 1 1 + |x| d ≤ T −dα . For each of the integrals we use ap- propriately the Hölder inequality, properties of the convolution and 3.45, obtaining the estimates C 1 T 2d α12+β−11+β and C 2 T dα12+β−21+β , respectively the factors involving negative powers of H T and log T have been estimated by constants. These bounds tend to zero as T → ∞ by 3.68. We omit details. This proves 3.13. To prove 3.16 and 3.17 we use 3.18 and 3.19. Again, we consider separately the integrals over {|x| ≤ T 1 α } and {|x| T 1 α }, and apply the same tricks as for I 2 T . For J 1 T we obtain the estimate J 1 T ≤ C T dα1+β2+β−1 → 0 log T and H T appear with negative powers only, whereas J 2 T ≤ C 1 T H β T log T β + C 2 T H β T log T 1+ β → 0 by assumption 2.9. The proof of Theorem 2.1 is complete ƒ

3.5 Proof of Proposition 2.3