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Lemma A.2. Let D ⊂ R d and 0 < s < t < 1. If b ∈ C t (D) and v ∈ H s (D), then bv ∈ H s (D) and‖bv‖ H s (D) |b| C t (D) ‖v‖ L 2 (D) + |b| C 0 (D) ‖v‖ H s (D).The hidden constant depends only on t, s and d.Proof. This is a classical result, but we require again the exact dependence of the bound on b.First note that trivially ‖bv‖ L 2 (D) ≤ b max ‖v‖ L 2 (D) where b max := |b| C 0 (D). Now, for any x, y ∈ D,we have|b(x)v(x) − b(y)v(y)| 2 ≤ 2(b(x) 2 |v(x) − v(y)| 2 + v(y) 2 |b(x) − b(y)| 2 ).Continuing v by 0 on R d and denoting this extension by ṽ, this implies|b(x)v(x) − b(y)v(y)|∫∫D 2∫∫2 ‖x − y‖ d+2s dx dy ≤ 2b 2 max‖v‖ 2 H s (D) + 2 v(y) 2 |b(x) − b(y)| 2D 2 ‖x − y‖ d+2s∫∫≤ 2b 2 max‖v‖ 2 H s (D) + 8b 2 v(y) 2max‖x − y‖ d+2s + v(y) 22|b|2 dx dyC t (D)‖x − y‖ d+2(s−t)x,y∈D‖x−y‖≥1∥ ∥ )≤ 2b 2 max‖v‖ 2 H s (D)(8b + 2 1 ‖z‖≥1 ∥∥∥Lmax ∥‖z‖ d+2s + 2|b| 2 1 ‖z‖≤1 ∥∥∥LC t (D) ∥1 (R d ) ‖z‖ d+2(s−t) ‖ṽ‖ 2 L 2 (R d )1 (R d ) b 2 max‖v‖ 2 H s (D) + |b|2 C t (D) ‖v‖2 L 2 (D) .Proof of Lemma 3.3. First we continue the solution w by 0 on R d \ R d + and denote the extension˜w ∈ H 1 (R d ). Take 1 ≤ i ≤ d − 1. Similarly to the previous section, we define for u, v ∈ H 1 (R d )∫∫d(u, v) := A∇u∇Rh i (v) dx − A∇(Rh i )∗ u∇v dxand deduce again thatR d +R d +|d(u, v)| |A| C t (R d + ,S d(R)) |u| H 1 (R d + ) |v| H 1 (R d + ) .We now note that, since i ≠ d, (R i h )∗ w ∈ H 1 0 (Rd +) and (R i h )∗ ˜w ∈ H 1 (R d ) is equal to the continuationby 0 on R d \ R d + of (R i h )∗ w. We deduce, similarly to the proof in Section A.1 using (A.1), that‖(R i h )∗ ˜w‖ H 1 (R d )1()|A|A C t (R d min + ,S d(R)) |w| H 1 (R d + ) + ‖F ‖ H s−1 (R d + ) + ‖w‖ H 1 (R d + ) =: B(w).(Note that we added |w| H 1 (R d + ) to the bound to simplify the notation later.) Hence, by the sametoken as in the previous section, we get∫(1 + |ξ| 2 )(|ξ 1 | 2 + ... + |ξ d−1 | 2 ) s |̂˜w(ξ)| 2 dξ B(w) 2 .(A.2)R dIn particular, this implies that, for 1 ≤ i ≤ d and 1 ≤ j ≤ d − 1, we have∫ ∣ ∣∣∣∣ ̂∂2 ˜w(ξ)R d ∂x i ∂x j ∣2(1 + |ξ| 2 ) s−1 dξ =∫R d |ξ i | 2 |ξ j | 2 |̂˜w(ξ)| 2 (1 + |ξ| 2 ) s−1 dξ B(w) 2 ,23
∥∂which means that2 ˜w∂x i ∂x j∈ H s−1 (R d ) and ∥ ∂2 ˜w ∥∥H∂x i ∂x j B(w). In particular, for all (i, j) ≠s−1 (R d ) ∥∂(d, d), this further implies that2 w∂x i ∂x j∈ H s−1 (R d +) and that ∥ ∂2 w ∥∥H∂x i ∂x j B(w). Usings−1 (R d + )Lemma A.1 we deduce that ∂w∂x j∈ H s (R d +) and that∥ ∂w ∥∥∥H∥ B(w), for all 1 ≤ j ≤ d − 1. (A.3)∂x j s (R d + )∥ It remains to bound ∥ ∂w ∥∥H∂x d, which is rather technical. To achieve it we will use the PDE (3.1),s (R d + )Lemma A.2 and the following result.Lemma A.3. For almost all x d ∈ R, we have ∂ ˜w∂x d(., x d ) ∈ H s (R d−1 ) and∫∥ ∂ ˜w∫(., x d ) ∥ 2 ∣ R ∂x dx d H s (R d−1 d = (1 + |ξ ′ | 2 ) s |ξ d | 2 ∣∣̂˜w(ξ) ∣ 2 dξ B(w) 2 .) R dProof. This follows from Fubini’s theorem and Plancherel’s formula, together with (A.2).From this we deduce that ∂w∂x d(., x d ) ∈ H s (R d−1 ), for almost all x d ∈ R + , and that∫ ∥ ∥∥ ∂w(., x d ) ∥ 2R +∂x dx d H s (R d−1 d B(w) 2 .)∂wLet 1 ≤ i ≤ d − 1. Using Lemma A.2 we deduce that A id ∂x d(., x d ) ∈ H s (R d−1 ), for almost allx d ∈ R + , and that( )∂w∥ A id (., x d )∂x d∥ |A id (., x d )| C ∂wt (R d−1 ) ∥ (., x d )H s (R d−1 )∂x d∥L 2 (R d−1 )+ ‖A id (., x d )‖ C ∂w0 (R d−1 ) ∥ (., x d ) ∥ .∂x d∥H s (R d−1 )Therefore, since by definition ‖A id ‖ C 0 (R d ) ≤ A max , we get∫ ∥ ∥∥Aid ∂w∥ ∥∥2R +∂x dx d H s (R d−1 d |A id | 2) C t (R d + ) |w|2 H 1 (R d + ) + A2 max B(w) 2 A 2 max B(w) 2 .∂∂x iSince is linear continuous from H 1−s (R d−1 ) to H −s (R d−1 ) (cf. [18, Remark 6.3.14(b)]) we can( )∂ ∂wdeduce from this that∂x iA id ∂x d∈ H s−1 (R d +) and that( )∥ ∂ ∂w ∥∥∥H∥ A id A max B(w), for all 1 ≤ i ≤ d − 1. (A.4)∂x i ∂x d s−1 (R d + )To see this take ϕ ∈ D(R d +). Then〈 ( ) 〉∣ ∂ ∂w∣∣∣∣ A∣id , ϕ=∂x i ∂x d D ′ (R d + ),D(Rd + )≤≤∥ A ∂wid (x ′ , x d ) ∂ϕ(x ′ , x d ) ∥∂x d ∂x i ∥ ∥ A ∂w ∥∥∥L ∂ϕid∂x d∥ (x ′ , x d ) ∥2 (R + ,H s (R d−1 ))∂x i ∥ ∥ A ∂w ∥∥∥Lid∂x d 2 (R + ,H s (R d−1 ))∥L 2 (R + ,H s (R d−1 ))∥L 2 (R + ,H −s (R d−1 ))‖ϕ‖ L 2 (R + ,H 1−s (R d−1 )).24
Page 1 and 2: BICSBath Institute for Complex Syst
Page 3 and 4: only Hölder continuous with expone
Page 5 and 6: for almost all ω ∈ Ω. The diffe
Page 7 and 8: 2.4 Truncated Karhunen-Loève Expan
Page 9 and 10: 3.1 Regularity of the SolutionPropo
Page 11 and 12: Theorem 3.4. Let Assumptions A1-A3
Page 13 and 14: Corollary 3.10. Let Assumptions A1-
Page 15 and 16: 3.4 Truncated FieldsCombining the r
Page 17 and 18: Since all the expectations E[Y l ]
Page 19 and 20: Figure 1: Let d = 1 and σ 2 = 1 in
Page 21 and 22: Figure 4: Left: Number of samples N
Page 23: therefore we getA min ‖(R i h )
Page 27 and 28: Proof. Since supp(u 0 ) ⊂ D 0 , w
Page 29 and 30: We now extend A i to R d +. Since w

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