“…Similarly to(3.23),σ k • ∇ξ s H β−2+ε e k ξ s H β−1+ε and thus k |k| −2α σ k • ∇ξ s 2 H β−2+ε k,l |k| −2α |l| −2(1−β−ε) | ξ s , e k−l |Note that the right-hand side can be written as a, b * c 2 , wherea k = |k| −2α , b k = |k| −2(1−β−ε) and c k = | ξ s , e k | 2 , k ∈ Z 2 0 ; c ∈ 1 (Z 2 0 ) with c 1 = ξ s and for any p ∈ (1/α, 1/(β +ε)), one has a ∈ p (Z 2 0 ), b ∈ p (Z 2 0 ), where p is the conjugate in particular, v ∈ L ∞ t C δ x for any ∈ [0, 1 − γ and v ∈ L 2 t C δ x for any δ < 2 − γ . Remark A 7. By time reversal, similar results hold if we considered the backward equation (∂ t + )h = b • ∇h + c h with given terminal condition h T ∈ L ∞ .…”