Bath Institute For Complex Systems - ENS de Cachan - Antenne de ...

therefore we getA min ‖(R i h )∗ w‖ 2 H 1 (R d ) |A| C t (R d ,S d (R)) |w| H 1 (R d ) |(R i h )∗ w| H 1 (R d ) +and finally, using (A.1) once more,‖(R i h )∗ w‖ H 1 (R d )‖F ‖ H s−1 (R d ) ‖(R i h )∗ w‖ H 1 (R d ) + A min ‖(R i h )∗ w‖ 2 L 2 (R d ) |A| C t (R d ,S d (R)) |w| H 1 (R d ) |(R i h )∗ w| H 1 (R d ) +‖F ‖ H s−1 (R d ) ‖(R i h )∗ w‖ H 1 (R d ) + A min ‖(R i h )∗ w‖ H −1 (R d )‖(R i h )∗ w‖ H 1 (R d ),1()|A|A C t (R d ,S d (R)) |w| H 1 (R d ) + ‖F ‖ H s−1 (R d ) + ‖w‖ L 2 (R d ).minFor any 1 ≥ h > 0, since |1 − e −h−iξ ih | 2 ≥ |Im(1 − e −h−iξ ih )| 2 = e −2h sin(ξ i h) 2 ≥ e −2 sin(ξ i h) 2 , andsince sin 2 (ξh) ≥ ( 2π ξh) 2 , for all |ξ| ≤ 1/h, we have conversely thatd∑‖(Rh i )∗ w‖ 2 H 1 (R d )i=1≥=≥∫∫|ξ|≤1/h|ξ|≤1/he −2 ∫∫|ξ|≤1/h(1 + |ξ| 2 )d∑i=1| (R ̂h i )∗ w(ξ)| 2 dξd∑(1 + |ξ| 2 )1 − e −h−iξ ih∣ h ∣i=1d∑(1 + |ξ| 2 )sin(ξ i h)∣ h ∣|ξ|≤1/hi=1(1 + |ξ| 2 )|ξ| 2s |ŵ(ξ)| 2 dξ.2s2s|ŵ(ξ)| 2 dξ|ŵ(ξ)| 2 dξHence, for any 0 < h ≤ 1, we obtain‖w‖ 2 H 1+s (R d )≤∫(1 + |ξ| 2 )|ξ| 2s |ŵ(ξ)| 2 dξ +R d ∫(1 + |ξ| 2 )|ŵ(ξ)| 2 dξR dd∑≤ ‖(Rh i )∗ w‖ 2 H 1 (R d ) + ‖w‖2 H 1 (R d )and soi=1‖w‖ H 1+s (R d ) 1 ()|A|A C t (R d ,S d (R)) |w| H 1 (R d ) + ‖F ‖ H s−1 (R d ) + ‖w‖ H 1 (R d ) < +∞.minA.2 Step 2 – The Case D = R d +In order to proof Lemma 3.3 we will need the following two lemmas.Lemma A.1 ([18, Lemma 9.1.12]). Let s > 0, s ≠ 1/2, then(∥ d∑|||v||| s := ‖v‖ 2 L 2 (R d + ) + ∂v ∥∥∥2∥∂x idefines a norm on H s (R d +) that is equivalent to ‖v‖ H s (R d + ) .i=1H s−1 (R d + ) ) 1/222