10 1. THE LOCAL THEORY OF THE CAUCHY PROBLEM We next give 3 suﬃcient conditions for a function u to be a solution to (NLW). The first one comes from [62], Remark 2.14. Claim 1.19. Let u ∈ L8 ( R3 × I ) be such that (u, ∂tu) ∈ C I ˙ 1 × L2 . Assume that there exists a sequence of solutions {uk} to (NLW) such that sup t∈I (u − uk,∂tu − ∂tuk) ˙ 1 ×L2 k→∞ −→ 0 sup k uk L8L8 I x ∞. Then, u is a solution of (NLW). Proof of Claim 1.19. We need to show that D 1 2 u W (I) ∞ and u is a so- lution of (NLW). This follows by showing that D 1 2 uk W (I) ≤ B, where B is inde- pendent of k. To do this, first find A, N such that supk supt∈I (uk(t),∂tuk(t)) ˙ 1 ×L2 ≤ A, sup k uk L8L8 I x ≤ N. Next, partition I = ∪M j=1 Ij, where Ij is such that uk SI j ≤ δ, where δ = δ(A) is to be chosen. Notice that M = M(N, δ). We then use the integral equation for uk, and the estimate (1.2) D 1 2 F (uk) L 4 3 Ij Lx 4 3 ≤ Cδ4 D 1 2 u W (Ij) . Hence, from the proof of Theorem 1.4, we see that D 1 2 uk W (Ij) ≤ CA + Cδ4 D 1 2 uk W (Ij) . Thus, for δ = δ(A) small, we obtain D 1 2 uk W (Ij) ≤ 2CA. Adding in j we obtain the desired bound. Claim 1.20. Let I be an open interval containing 0, and (u0,u1) ∈ ˙ 1 × L2. Assume that u ∈ L5 I Lx 10 , I ⊂⊂ I satisfies the integral equation. Then u is a solution of (NLW). Proof. By the definition of a solution, it suﬃces to check that u ∈ S(I ),D 1 2 u ∈ W (I ),I ⊂⊂ I. Since by our assumption on u, u5 ∈ L1 I Lx, 2 this follows from the Strichartz estimate (S2). We now prove that a distributional solution of the differential equation, that satisfies an appropriate space-time bound is also a solution of (NLW). Lemma 1.21. Let (u0,u1) ∈ ˙ 1 × L2 be an open interval such that 0 ∈ I, u ∈ LILx 5 10 and (u, ∂tu) ∈ C I ˙ 1 × L2 . Assume furthermore that (u, ∂tu)|t=0 = (u0,u1) and ∂2u t − Δu = u5 in D (R3 × I). Then u is a solution of (NLW) Proof. In view of Claim 1.20, it suﬃces to check that u satisfies the integral equation. We prove the integral equation for t ∈ I+ = I ∩ (0, ∞). The proof for t ∈ I ∩ (−∞, 0) is exactly the same. Let v(t) = u(t) − S(t)(u0,u1).

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