12 1. THE LOCAL THEORY OF THE CAUCHY PROBLEM If so, we would have (1.22) R3 0 t 0 sin ( (t s) −Δ ) −Δ u5(x, s)ds h(x, t)dxdt = R3 0 v(x, t)h(x, t)dxdt. Since h is arbitrary in C0 (R3 × I+), we deduce, in view of the definition of v, the desired integral formula. Let us now show the limits above. We will prove the first one, the second one being similar. Using that (v, ∂tv)|t=0 = (0, 0), (v, ∂tv) C I ˙ 1 × L2 , we deduce that ∀t I, v(t) L2(R3) and limt→0 1 t v(t) L2 = 0. Let ε 0 and choose a0 such that v(t) L2 x εt for t (0,a0]. Then, using that ϕ ( t a ) = 0 for t a or t 0, we have, for a a0, R4 v(x, t) 1 a2 ϕ t a H(x, t)dxdt C a 0 εt a2 dt Cε, which concludes the proof. In the rest of the chapter we will be studying properties of Lorentz transforma- tions of solutions, using results in [62] and [35]. We start out with a preliminary lemma from [62] on solutions of (LW). Lemma 1.23. Let w be a solution to (LW), with (w0,w1) ˙ 1 × L2 and h Lt 1 Lx. 2 Then, for |l| 1, we have sup t ∇xw x lt √1 1 l2 , x , t lx1 1 l2 L2(dxdx ) + sup t ∂tw x lt √1 1 l2 , x , t lx1 1 l2 L2(dxdx ) C w0 ˙ 1 (R3) + w1 L2(R3) + h L1L2 t x . Proof. Let v(x, t) = U(t)f be given by v(ξ, t) = eit(ξ)f(ξ),f L2. We will show that sup t v x lt √1 1 l2 , x , t lx1 1 l2 L2(dxdx ) C f L2 , which easily implies the desired estimate. But, v(x, t) = R3 eix−ξeit|ξ|f(ξ)dξ = eix1ξ1eit|ξ|eix ξ f(ξ)dξ1dξ = R3 eix1ξ1eit ξ2+|ξ 1 |2 eix ·ξ f(ξ1,ξ )dξ1dξ ,
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