On Abstract Strichartz Estimates and the Strauss Conjecture for Nontrapping Obstacles
Kunio Hidano, Jason Metcalfe, Hart F. Smith, Christopher D. Sogge, Yi Zhou
aa r X i v : . [ m a t h . A P ] F e b ON ABSTRACT STRICHARTZ ESTIMATES AND THE STRAUSSCONJECTURE FOR NONTRAPPING OBSTACLES
KUNIO HIDANO, JASON METCALFE, HART F. SMITH, CHRISTOPHER D. SOGGE,AND YI ZHOU Introduction.
The purpose of this paper is to show how local energy decay estimates for certain linearwave equations involving compact perturbations of the standard Laplacian lead to optimalglobal existence theorems for the corresponding small amplitude nonlinear wave equationswith power nonlinearities. To achieve this goal, at least for spatial dimensions n = 3 and4, we shall show how the aforementioned linear decay estimates can be combined with“abstract Strichart” estimates for the free wave equation to prove corresponding estimatesfor the perturbed wave equation when n ≥
3. As we shall see, we are only partiallysuccessful in the latter endeavor when the dimension is equal to two, and therefore, atpresent, our applications to nonlinear wave equations in this case are limited.Let us start by describing the local energy decay assumption that we shall makethroughout. We shall consider wave equations on the exterior domain Ω ⊂ R n of acompact obstacle:(1.1) ( ∂ t − ∆ g ) u ( t, x ) = F ( t, x ) , ( t, x ) ∈ R + × Ω u (0 , · ) = f∂ t u (0 , · ) = g ( Bu )( t, x ) = 0 , on R + × ∂ Ω , where for simplicity we take B to either be the identity operator (Dirichlet-wave equation)or the inward pointing normal derivative ∂ ν (Neumann-wave equation). We shall alsoassume throughout that the spatial dimension satisfies n ≥ g is the Laplace-Beltrami operator associated with a smooth, timeindependent Riemannian metric g jk ( x ) which we assume equals the Euclidean metric δ jk for | x | ≥ R , some R . The set Ω is assumed to be either all of R n , or else Ω = R n \K where K is a compact subset of | x | < R with smooth boundary.We can now state the main assumption that we shall make. The first author was supported in part by the Grant-in-Aid for Young Scientists (B) (No.18740069),The Ministry of Education, Culture, Sports, Science and Technology, Japan, and he would like to thankthe Department of Mathematics at the Johns Hopkins University for the hospitality and financial supportduring his visit where part of this research was carried out. The second, third and fourth authors weresupported by the National Science Foundation. The fifth author was supported by project 10728101of NSFC and the “111” project and Doctoral Programme Foundation of the Ministry of Education ofChina.
Hypothesis B.
Fix the boundary operator B and the exterior domain Ω ⊂ R n as above.We then assume that given R > Z ∞ (cid:16) k u ( t, · ) k H ( | x | 0, then with u ′ = ( ∂ t u, ∇ x u ), k u ′ ( t, · ) k L ( | x | 3. For n = 2, if ∂ Ω is assumed to be nonempty one has α ( t ) = O ((log(2 + t )) − (1 + t ) − ) for the Dirichlet-wave equation. Here we have usedthat, due to the Dirichlet boundary conditions and the fundamental theorem of calculus,the local L norm can be controlled by the local L norm of the gradient. Since thesebounds yield α ( t ) ∈ L ( R + ), we conclude that Hypothesis B is valid in these cases. Weremark that when Ω = R and ∆ g = ∆, then α ≈ t − for large t (see [29]), and so, inthis case, α / ∈ L ( R + ). Proofs of these results for n ≥ n = 2 follows fromVainberg [39] (see § For the case where ∆ g is assumed to be a time-independent variable coefficient com-pact perturbation of ∆ and Ω is assumed to be nontrapping with respect to the metricassociated with ∆ g , one also has that (1.2) is valid for the Dirichlet-wave equation for all n ≥ n = 2 if ∂ Ω = ∅ . See Taylor [38] and Burq [4].Having described the main assumption about the linear problem, let us now describethe nonlinear equations that we shall consider. They are of the form(1.3) ( ∂ t − ∆ g ) u ( t, x ) = F p (cid:0) u ( t, x ) (cid:1) , ( t, x ) ∈ R + × Ω Bu = 0 , on R + × ∂ Ω u (0 , x ) = f ( x ) , ∂ t u (0 , x ) = g ( x ) , x ∈ Ω , with B as above. We shall assume that the nonlinear term behaves like | u | p when u issmall, and so we assume that(1.4) X ≤ j ≤ | u | j (cid:12)(cid:12) ∂ ju F p ( u ) (cid:12)(cid:12) . | u | p We are very grateful to Jim Ralston for patiently explaining these results and their history to us. BSTRACT STRICHARTZ ESTIMATES 3 when u is small.We shall be assuming that the data (and some of its derivatives) are small in certainSobolev norms that we now describe.As in the earlier works that proved global Strichartz estimates ([4], [25], [33]), we shallrestrict ourselves to the case where the Sobolev index γ is smaller than n/ 2. One reasonfor this is that the Strichartz estimates that seem to arise in applications always have γ ≤ 1. Another reason is that when | γ | < n/ β ∈ C ∞ ( R n ) is continuous from ˙ H γ ( R n ) to H γ ( R n ) and the two norms are equivalent onfunctions with fixed compact support. Recall that ˙ H γ ( R n ) is the homogeneous Sobolevspace with norm given by k f k H γ ( R n ) = (cid:13)(cid:13) ( √− ∆) γ f (cid:13)(cid:13) L ( R n ) = (2 π ) − n Z R n (cid:12)(cid:12) | ξ | γ ˆ f ( ξ ) (cid:12)(cid:12) dξ, while the inhomogeneous Sobolev space H γ ( R n ) has norm defined by k f k H γ ( R n ) = (cid:13)(cid:13) (1 − ∆) γ/ f (cid:13)(cid:13) L ( R n ) = (2 π ) − n Z R n (cid:12)(cid:12) (1 + | ξ | ) γ/ ˆ f ( ξ ) (cid:12)(cid:12) dξ, with ˆ f denoting the Fourier transform and ∆ denoting the standard Laplacian.Let us now describe the Sobolev spaces on Ω that we shall consider. Let β be a smoothcutoff on R n with β and 1 − β respectively supported where | x | < R and | x | > R . LetΩ ′ be the embedding of Ω ∩ {| x | < R } into the torus obtained by periodic extension ofΩ ∩ [ − R, R ] n , so that ∂ Ω ′ = ∂ Ω. We define k f k H γB (Ω) = k βf k H γB (Ω ′ ) + k (1 − β ) f k H γ ( R n ) k f k ˙ H γB (Ω) = k βf k H γB (Ω ′ ) + k (1 − β ) f k ˙ H γ ( R n ) , | γ | < n/ . The spaces H γB (Ω ′ ) are defined by a spectral decomposition of ∆ g | Ω ′ subject to theboundary condition B . In the homogeneous spaces ˙ H γB (Ω) it is assumed that (1 − β ) f belongs to ˙ H γ ( R n ), so that the Sobolev embedding ˙ H γB (Ω) ֒ → L p (Ω) holds with p =2 n/ ( n − γ ). From this, it is verified that the Sobolev spaces on Ω are independent of thechoice of β and R , and thus the ˙ H γB (Ω) and H γB (Ω) norms are equivalent on functions offixed bounded support. We note that H − γB (Ω) is the dual of H γB (Ω), and ˙ H − γB (Ω) is dualto ˙ H γB (Ω) for | γ | < n/ 2. Also, for γ a nonnegative integer, k f k H γB (Ω) ≈ X | α |≤ γ k ∂ αx f k L (Ω) k f k H γB (Ω) ≈ X | α | = γ k ∂ αx f k L (Ω) . The Sobolev spaces as defined are verified to be an analytic interpolation scale of spaces.The above definition then agrees, for nonnegative integer γ , with the subspace of H γ (Ω)such that B (∆ j g f ) = 0 for all j for which the trace is well defined, and for general γ byduality and interpolation. Finally, for every γ the set of functions f ∈ C ∞ (Ω) such that B (∆ j g f ) = 0 for all j ≥ γ ∈ (0 , ), whilethe ones in the abstract Strichartz estimates to follow only involve certain γ ≤ ( n − / K. HIDANO, J. METCALFE, H. F. SMITH, C. D. SOGGE, AND Y. ZHOU In practice the useful Strichartz-type estimates always involve γ ∈ (0 , R n , ∆ = ∆ g , as well as for the mixed-norm estimates for (1.1) that we shall state.The data ( f, g ) in Theorem 1.1 below will have second derivatives belonging to ˙ H γB (Ω) × ˙ H γ − B (Ω), where γ ∈ (0 , ), thus will locally belong to H γ (Ω) × H γ (Ω). The boundarycondition for ( f, g ) to locally belong to H γB (Ω) × H γB (Ω) for γ ∈ (0 , ) is the same asfor H B (Ω) × H B (Ω), which for the Dirichlet case is f | ∂ Ω = g | ∂ Ω = 0, and for Neumannis ∂ ν f | ∂ Ω = 0. These are the assumptions placed on the data ( f, g ) in Theorem 1.1.If we let { Z } = { ∂ l , x j ∂ k − x k ∂ j : 1 ≤ l ≤ n, ≤ j < k ≤ n } then we can now state our existence theorem for (1.3). Theorem 1.1. Let n = 3 or , and fix Ω ⊂ R n and boundary operator B as above.Assume further that Hypothesis B is valid.Let p = p c be the positive root of (1.5) ( n − p − ( n + 1) p − , and fix p c < p < ( n + 3) / ( n − . Then if (1.6) γ = n − p − , there is an ε > depending on Ω , B and p so that (1.3) has a global solution satisfying ( Z α u ( t, · ) , ∂ t Z α u ( t, · )) ∈ ˙ H γB × ˙ H γ − B , | α | ≤ , t ∈ R + , whenever the initial data satisfiesthe boundary conditions of order , and (1.7) X | α |≤ (cid:16) k Z α f k ˙ H γB (Ω) + k Z α g k ˙ H γ − B (Ω) (cid:17) < ε with < ε < ε . In the case where Ω = R n and ∆ g = ∆ it is known that p > p c is necessary for globalexistence (see John [16], Glassey [12] and Sideris [31]). In this case under a somewhatmore restrictive smallness condition global existence was established by John [16] for thecase where n = 3, then Glassey [12] for n = 2, Zhou [40] for n = 4, Lindblad and Sogge[22] for n ≤ n (see also Tataru [37]).For obstacles, when n = 4, ∆ g = ∆ the results in Theorem 1.1 for the Dirichlet-waveequation outside of nontrapping obstacles under a somewhat more restrictive smallnessassumption was obtained in [8].Also, when Ω = R , ∆ g = ∆, it was shown in Sogge [35] that, for the sphericallysymmetric case, the variant of the condition (1.7) saying that the ˙ H γ ( R ) × ˙ H γ − ( R )norm of the data be small with γ as in (1.6) is sharp. Further work in this direction (forthe non-obstacle case) was done by Hidano [13], [14] and Fang and Wang [10].It is not difficult to see that the condition (1.7) is sharp in the sense that there are noglobal existence results for γ > n − p − . To do this we use well known results concerningblowup solutions for ( ∂ t − ∆) v = | v | p , p > 0, in R + × R n (see Levine [19]). Specifically,we shall use the fact that given δ > C ∞ data ( v , v ) vanishing for | x | < R so that the solution of ( ∂ t − ∆) v = | v | p , v (0 , · ) = v , ∂ t v (0 , · ) = v blows up within BSTRACT STRICHARTZ ESTIMATES 5 time δ . Next, let us assume that the above global existence results for (Ω , B, ∆ g ) heldfor this nonlinearity and some γ > n − p − in (1.7). Then, if λ is sufficiently large, the˙ H γB × ˙ H γ − B norm of ( λ − / ( p − v ( · /λ ) , λ − − / ( p − v ( · /λ )) would be bounded by its˙ H γ ( R n ) × ˙ H γ − ( R n ) norm, which equals λ n/ − / ( p − − γ k ( v , v ) k ˙ H γ ( R n ) × ˙ H γ − ( R n ) . Sincethis goes to zero as λ → ∞ for γ > n − p − , we conclude that if the above existence resultsheld for this value of γ then we would obtain a global solution of ( ∂ t − ∆ g ) u λ = | u λ | p , u λ ( t, x ) = 0 , ( t, x ) ∈ R + × ∂ Ω with initial data ( λ − / ( p − v ( · /λ ) , λ − − / ( p − v ( · /λ )).Since v and v vanish for | x | < R , by finite propagation speed, if δ > λ if we extend u λ to be zero on Ω c then the resulting functionwould agree with the solution of the Minkowski space wave equation ( ∂ t − ∆) v λ = | v λ | p on [0 , δλ ] × R n with data ( λ − / ( p − v ( · /λ ) , λ − − / ( p − v ( · /λ )). By scaling v ( t, x ) = λ / ( p − v λ ( λt, λx ) would then solve the Minkowski space equation ( ∂ t − ∆) v = | v | p on[0 , δ ] × R n with initial data ( v ( x ) , v ( x )). As we noted before, we can always choose( v , v ) so that this is impossible for a given δ > 0, which allows us to conclude thatthe above existence results do not hold if the Sobolev exponent γ in (1.7) is larger than n − p − .As a final remark, we point out that we have restricted ourselves to the case where p < ( n + 3) / ( n − 1) because of the techniques that we shall employ. However, sincethe solutions obtained are small, the above existence theorem leads to small-data globalexistence of (1.3) when p is larger than or equal to the conformal power ( n + 3) / ( n − L qt L rx estimates on R + × Ω for solutions of (1.1). For certain applications, suchas obtaining the Strauss conjecture in various settings, it is convenient to replace the L rx norm with a more general one. To this end, we consider pairs of normed function spaces X ( R n ) and X (Ω). The spaces are localizable, in that k f k X ≈ k βf k X + k (1 − β ) f k X for smooth, compactly supported β , with β = 1 on a neighborhood of R n \ Ω in case X = X (Ω). Finally, we assume that(1.8) k (1 − β ) f k X (Ω) ≈ k (1 − β ) f k X ( R n ) for such β . Weighted mixed L p spaces, as well as (cid:0) ˙ H γ ( R n ) , ˙ H γB (Ω) (cid:1) , are the examplesused in the proof of Theorem 1.1.We shall let k · k X ′ denote the dual norm (respectively over R n and Ω) so that k u k X = sup k v k X ′ =1 (cid:12)(cid:12)(cid:12) Z u v dx (cid:12)(cid:12)(cid:12) . An important example for us is when k u k X = k | x | α u k L p , for a given 1 ≤ p ≤ ∞ and | α | < n/p , in which case the dual norm is k v k X ′ = k | x | − α v k L p ′ , with p ′ denoting the conjugate exponent. K. HIDANO, J. METCALFE, H. F. SMITH, C. D. SOGGE, AND Y. ZHOU We shall consider time Lebesgue exponents q ≥ k v k L qt X ( R × R n ) . k v (0 , · ) k ˙ H γ ( R n ) + k ∂ t v (0 , · ) k ˙ H γ − ( R n ) , assuming that(1.10) ( ∂ t − ∆) v = 0 in R × R n . Here k v k L qt X ( I × R n ) = (cid:16) Z I k v ( t, · ) k qX dt (cid:17) /q , I ⊂ R . We shall also consider analogous norms on I × Ω, I ⊂ R , k u k L qt X ( I × Ω) = (cid:16) Z I k u ( t, · ) k qX (Ω) dt (cid:17) /q . In addition to Hypothesis B and (1.9), we shall assume that we have the local abstractStrichartz estimates for Ω:(1.11) k u k L qt X ([0 , × Ω) . k f k ˙ H γB (Ω) + k g k ˙ H γ − B (Ω) , assuming that u solves (1.1) with vanishing forcing term, i.e.,(1.12) ( ∂ t − ∆ g ) u = 0 in [0 , × Ω . Definition 1.2. When (1.9) and (1.11) hold we say that ( X, γ, q ) is an admissible triple. We can now state our main estimate. Theorem 1.3. Let n ≥ and assume that ( X, γ, q ) is an admissible triple with (1.13) q > and γ ∈ [ − n − , n − ] . Then if Hypothesis B is valid and if u solves (1.1) with ( ∂ t − ∆ g ) u ≡ , we have theglobal abstract Strichartz estimates (1.14) k u k L qt X ( R × Ω) . k f k ˙ H γB (Ω) + k g k ˙ H γ − B (Ω) . The condition on γ in (1.13) is the one to ensure that γ and 1 − γ are both ≤ ( n − / n = 2, this forces γ to beequal to 1 / 2, while a larger range of γ ∈ (0 , 1) is what certain applications require. Forthis reason, we are unable at present to show that the Strauss conjecture for obstaclesholds when n = 2. See the end of the next section for further discussion. Corollary 1.4. Assume that ( X, γ, q ) and ( Y, − γ, r ) are admissible triples and thatHypothesis B is valid. Also assume that (1.14) holds for ( X, γ, q ) and ( Y, − γ, r ) , andthat ≤ γ ≤ . Then we have the following global abstract Strichartz estimates for thesolution of (1.1)(1.15) k u k L qt X ( R + × Ω) . k f k ˙ H γB (Ω) + k g k ˙ H γ − B (Ω) + k F k L r ′ t Y ′ ( R + × Ω) , where r ′ denotes the conjugate exponent to r and k · k Y ′ is the dual norm to k · k Y . BSTRACT STRICHARTZ ESTIMATES 7 For simplicity, in the corollary we have limited ourselves to the case where 0 ≤ γ ≤ P = p − ∆ g is the square root of minus the Laplacian (with the boundaryconditions B ), then we need show (cid:13)(cid:13)(cid:13) Z t sin (cid:0) ( t − s ) P (cid:1) P − F ( s, · ) ds (cid:13)(cid:13)(cid:13) L qt X ( R + × Ω) . k F k L r ′ t Y ′ ( R + × Ω) . Since q > r ′ , an application of the Christ-Kiselev lemma (cf. [7], [33], [36, chapter 4])shows that it suffices to prove the estimate (cid:13)(cid:13)(cid:13) Z ∞ sin (cid:0) ( t − s ) P (cid:1) P − F ( s, · ) ds (cid:13)(cid:13)(cid:13) L qt X ( R + × Ω) . k F k L r ′ t Y ′ ( R + × Ω) . After factorization of the sin function, it suffices by (1.14) to show that (cid:13)(cid:13)(cid:13) Z ∞ cos( sP ) F ( s, · ) ds (cid:13)(cid:13)(cid:13) ˙ H γ − (Ω) + (cid:13)(cid:13)(cid:13) Z ∞ P − sin( sP ) F ( s, · ) ds (cid:13)(cid:13)(cid:13) ˙ H γ (Ω) . k F k L r ′ t Y ′ ( R + × Ω) . This, however, is the dual version of (1.14) for ( Y, − γ, r ). (cid:3) As a special case of (1.15) when the spaces X and Y are the standard Lebesgue spaces,we have the following Corollary 1.5. Suppose that n ≥ and that Hypothesis B is valid. Suppose that q, ˜ q > , r, ˜ r ≥ and that q + nr = n − γ = 1˜ q ′ + n ˜ r ′ − and q + n − r , q + n − r ≤ n − . Then if the local Strichartz estimate (1.11) holds respectively for the triples (cid:0) L r (Ω) , γ, q (cid:1) and (cid:0) L ˜ r (Ω) , − γ, ˜ q (cid:1) , it follows that when u solves (1.1) k u k L qt L rx ( R + × Ω) . k f k ˙ H γB (Ω) + k g k ˙ H γ − B (Ω) + k F k L ˜ q ′ t L ˜ r ′ x ( R + × Ω) . These results also hold for n = 2 under the above assumption, provided that γ = 1 / . These estimates of course are the obstacle versions of the mixed-norm estimates for R n and ∆ g = ∆. When n ≥ q or ˜ q is 2. For the Dirichlet-wave operator( B = Id ) these results were proved in odd dimensions by Smith and Sogge [33] and thenby Burq [4] and Metcalfe [25] for even dimensions. The Neumann case was not treated,but it follows from the same proof. Unfortunately, the known techniques seem to onlyapply to the case of γ = 1 / n = 2, and Hypothesis B seems also to require B = Id and ∂ Ω = ∅ in this case. The restriction that γ = 1 / n = 2 comes from thesecond part of (1.13), while for n ≥ γ in Corollary 1.5 always satisfy 0 ≤ γ ≤ 1. Also, at present, the knowledge ofthe local Strichartz estimates (1.11) when X = L r (Ω) is limited. When Ω is the exterior K. HIDANO, J. METCALFE, H. F. SMITH, C. D. SOGGE, AND Y. ZHOU of a geodesically convex obstacle, they were obtained by Smith and Sogge [32]. Recently,there has been work on proving local Strichartz estimates when X = L r (Ω) for moregeneral exterior domains ([5], [6], [3], [34]), but only partial results for a more restrictiverange of exponents than the ones described in Corollary 1.5 have been obtained.2. Proof of Abstract Strichartz Estimates. As mentioned before, we shall prove (1.14) by adapting the arguments from [4], [25]and [33]. We shall assume that (1.2) is valid for (Ω , ∆ g ) throughout. A key step in theproof of Theorem 1.3 will be to establish the following result that is implicit in [4]. Proposition 2.1. Let w solve the inhomogeneous wave equation in Minkowski space ( ( ∂ t − ∆) w = F on R + × R n w | t =0 = ∂ t w | t =0 = 0 . Assume as above that (1.9) is valid whenever v is a solution of the homogeneous waveequation (1.10) . Assume further that q > and γ ≥ − n − . Then, if F ( t, x ) = 0 if | x | > R , we have k w k L qt X ( R + × R n ) . k F k L t H γ − ( R + × R n ) . At the end of this section we shall show that when n = 2 the assumption that γ ≥ / n = 2 is necessary even in the model case where X = L r ( R n ) with 2 /q + 1 /r = 1 / /q + 2 /r = 1 − γ .To prove Proposition 2.1, we shall use our free space hypothesis (1.9) and the followingresult from [33]. Lemma 2.2. Fix β ∈ C ∞ ( R n ) and assume that γ ≤ n − . Then Z ∞−∞ (cid:13)(cid:13)(cid:13) β ( · ) (cid:0) e it | D | f (cid:1) ( t, · ) (cid:13)(cid:13)(cid:13) H γ ( R n ) dt . k f k H γ ( R n ) , if | D | = √− ∆ . As was shown in [33], this lemma just follows from an application of Plancherel’stheorem and the Schwarz inequality. The assumption that γ ≤ ( n − / q > 2, by theChrist-Kiselev lemma [7], it suffices to show that(2.1) (cid:13)(cid:13)(cid:13) Z ∞ e i ( t − s ) | D | | D | − β ( · ) G ( s, · ) ds (cid:13)(cid:13)(cid:13) L qt X ( R + × R n ) . k G k L t H γ − ( R n ) , assuming that β ∈ C ∞ ( R n ). If we apply (1.9), we conclude that the left side of thisinequality is majorized by (cid:13)(cid:13)(cid:13) Z ∞ e − is | D | | D | − γ β ( · ) G ( s, · ) ds (cid:13)(cid:13)(cid:13) L ( R n ) . BSTRACT STRICHARTZ ESTIMATES 9 Since k (1 − ∆) ( γ − / G ( s, · ) k = k G ( s, · ) k H γ − , it suffices to see that (cid:13)(cid:13)(cid:13) Z ∞ e − is | D | | D | − γ β ( · )(1 − ∆) (1 − γ ) / H ( s, · ) ds (cid:13)(cid:13)(cid:13) L ( R n ) . k H k L ( R + × R n ) . By duality, this is equivalent to the statement that(2.2) (cid:13)(cid:13) (1 − ∆) (1 − γ ) / β ( · ) e is | D | | D | − γ h (cid:13)(cid:13) L ( R + × R n ) . k h k L ( R n ) . Since we are assuming that γ ≥ − n − , we have that 1 − γ ≤ n − . Therefore, (2.2) followsfrom Lemma 2.2, completing the proof of Proposition 2.1. (cid:3) To prove Theorem 1.3 we also need a similar result for solutions of the wave equation(1.1) for (Ω , B, ∆ g ). Proposition 2.3. Let u solve (1.1) and assume that (2.3) f ( x ) = g ( x ) = F ( t, x ) = 0 , when | x | > R. Then if ( X, γ, q ) is an admissible triple with q > and γ ≥ − n − we have (2.4) k u k L qt X ( R + × Ω) . k f k H γB + k g k H γ − B + k F k L t H γ − B . The key ingredients in the proof are Proposition 2.1 and the following variant of(1.2), which holds for all γ ∈ R , provided (2.3) holds, and β ∈ C ∞ c ( R n ) equals 1 on aneighborhood of R n \ Ω:(2.5) k βu k L ∞ t H γB + k β∂ t u k L ∞ t H γ − B + k βu k L t H γB + k β∂ t u k L t H γ − B . k f k H γB + k g k H γ − B + k F k L t H γ − B . The L t estimates in (2.5) on u follow from (1.2) and elliptic regularity arguments for γ ∈ Z , and by interpolation for the remaining γ ∈ R . The L ∞ t estimates then follow fromenergy estimates, duality, and elliptic regularity.To prove (2.4), let us fix β ∈ C ∞ ( R n ) satisfying β ( x ) = 1, | x | ≤ R and write u = v + w, where v = βu, w = (1 − β ) u. Then w solves the free wave equation ( ( ∂ t − ∆) w = [ β, ∆] uw | t =0 = ∂ t w | t =0 = 0 . An application of Proposition 2.1 shows that k w k L qt X is dominated by k ρu k L t H γB if ρ equals one on the support of β . Therefore, by (2.5), k w k L qt X is dominated by the rightside of (2.4).As a result, we are left with showing that if v = βu then(2.6) k v k L qt X ( R + × Ω) . k f k H γB + k g k H γ − B + k F k L t H γ − B , assuming, as above, that (2.3) holds. To do this, fix ϕ ∈ C ∞ (( − , P ∞ j = −∞ ϕ ( t − j ) = 1. For a given j ∈ N , let v j = ϕ ( t − j ) v . Then v j solves ( ∂ t − ∆ g ) v j = − ϕ ( t − j )[∆ , β ] u + [ ∂ t , ϕ ( t − j )] βu + ϕ ( t − j ) FBv j ( t, x ) = 0 , x ∈ ∂ Ω v j (0 , · ) = ∂ t v j (0 , · ) = 0 , while v = v − P ∞ j =1 v j solves ( ∂ t − ∆ g ) v = − ˜ ϕ [∆ , β ] u + [ ∂ t , ˜ ϕ ] βu + ˜ ϕFBv ( t, x ) = 0 , x ∈ ∂ Ω v | t =0 = f, ∂ t v | t =0 = g, if ˜ ϕ = 1 − P ∞ j =1 ϕ ( t − j ) if t ≥ G j = ( ∂ t − ∆ g ) v j bethe forcing term for v j , j = 0 , , , . . . , then, by (2.5), we have that ∞ X j =0 k G j k L t H γ − B ( R + × Ω) . k f k H γB + k g k H γ − B + k F k L t H γ − B . By the local Strichartz estimates (1.11) and Duhamel, we get for j = 1 , , . . . k v j k L qt X ( R + × Ω) . Z ∞ k G j ( s, · ) k H γ − B ds . k G j k L t H γ − B , using Schwarz’s inequality and the support properties of the G j in the last step. Similarly, k v k L qt X ( R + × Ω) . k f k H γB + k g k H γ − B + k G k L t H γ − B . Since q > 2, we have k v k L qt X ( R + × Ω) . ∞ X j =0 k v j k L qt X ( R + × Ω) and so we get k v k L qt X . k f k H γB + k g k H γ − B + k F k L t H γ − as desired, which finishes the proof of Proposition 2.3. (cid:3) End of Proof of Theorem 1.3: Recall that we are assuming that ( ∂ t − ∆ g ) u = 0. ByProposition 2.3 we may also assume that the initial data for u vanishes when | x | < R/ β ∈ C ∞ ( R n ) satisfying β ( x ) = 1, | x | ≤ R and β ( x ) = 0, | x | > R/ u = u − v = (1 − β ) u + ( βu − v ) , where u solves the Cauchy problem for the Minkowski space wave equation with initialdata defined to be ( f, g ) if x ∈ Ω and 0 otherwise. By the free estimate (1.9) and (1.8),we can restrict our attention to ˜ u = βu − v . But( ∂ t − ∆ g )˜ u = − [∆ , β ] u ≡ G is supported in R < | x | < R , and satisfies(2.7) Z ∞ k G ( t, · ) k H γ − B dt . k f k H γB + k g k H γ − B by Lemma 2.2 and the fact that G vanishes on a neighborhood of ∂ Ω. Note also that ˜ u has vanishing initial data. Therefore, since Proposition 2.3 tells us that k ˜ u k L qt X ( R + × R n ) is dominated by the left side of (2.7), the proof is complete. (cid:3) For future reference, we note that the preceeding steps establish the following gener-alization of (2.5), assuming that γ ∈ [ − n − , n − ], and that F ( x ) = 0 for | x | > R : BSTRACT STRICHARTZ ESTIMATES 11 (2.8) k u k L ∞ t ˙ H γB + k ∂ t u k L ∞ t ˙ H γ − B + k βu k L t H γB + k β∂ t u k L t H γ − B . k f k ˙ H γB + k g k ˙ H γ − B + k F k L t H γ − B . In particular, f and g have no support restrictions. To see that (2.8) holds, first considerbounding the terms k β∂ jt u k L t H γ − jB for j = 0 , 1. For these terms, it suffices by (2.5) toconsider F = 0 and f, g = 0 near ∂ Ω. Decomposing u = (1 − β ) u + ˜ u as above, we mayuse (2.5) and (2.7) to deduce the L t bounds in (2.8) for u . These bounds now yield k ( ∂ t − ∆ g )(1 − β ) u k L t H γ − B + k ( ∂ t − ∆ g ) βu k L t H γ − B . k f k ˙ H γB + k g k ˙ H γ − B + k F k L t H γ − B . The L ∞ t bounds on βu now follow from (2.5). Finally, (1 − β ) u satisfies the Minkowskiwave equation on R × R n , with initial data in ˙ H γ × ˙ H γ − , and driving force ˜ F ∈ L t ˙ H γ − which vanishes for | x | ≥ R . The contribution to u from its initial data satisfies the L ∞ t bounds as a result of homogeneous Sobolev bounds for the Minkowski wave group. Thecontribution from ˜ F is bounded using Lemma 2.2 and duality.Let us conclude this section by showing that when n = 2 the restriction in Proposi-tion 2.1 that γ ≥ / X = L r ( R ). In this case, by thestandard mixed-norm Strichartz estimates (see e.g. [18]), the hypotheses of the Proposi-tion are satisfied when 0 ≤ γ < / 4, 1 /q + 2 /r = 1 − γ and 2 /q + 1 /r ≤ / γ and X = L r ( R ) as above, then the L qt L rx ( R + × R ) norm of W F ( t, x ) = Z t −∞ Z R e ix · ξ sin( t − s ) | ξ || ξ | ˆ F ( s, ξ ) dξds would have to be bounded by the L t H γ − norm of F if F ( t, x ) = 0 when | x | > 1. Weshall take F to be a product h T ( s ) β ( x ) where β ∈ C ∞ ( R ) vanishes for | x | > β (0) = 1, while h T is an odd function supported in [ − T, T ]. For this choice of F we have W F ( t, x ) = − i Z R e ix · ξ cos( t | ξ | )ˆ h T ( | ξ | ) ˆ β ( ξ ) dξ/ | ξ | , if t > T. Fix a nonzero function ρ ∈ C ∞ ( R ) supported in (1 / , h T to be the oddfunction which equals T − / ρ ( s/T ) for positive s , then since h T has a non-zero L normwhich is independent of T , if Proposition 2.1 were valid for an L qt L rx space as above, thenit would follow that W F ( t, x ) = − iT / Z R e ix · ξ cos( t | ξ | )ˆ h ( T | ξ | ) ˆ β ( ξ ) dξ/ | ξ | = − iT − / Z R e i xT · ξ cos( tT | ξ | ) ˆ h ( | ξ | ) ˆ β ( ξ/T ) dξ/ | ξ | would belong to L qt L rx ([ T, ∞ ) × R ) with a bound independent of T . An easy calculationshows that this norm equals T − / /q +2 /r (cid:13)(cid:13)(cid:13)Z R e ix · ξ cos( t | ξ | )ˆ h ( | ξ | ) ˆ β ( ξ/T ) dξ | ξ | (cid:13)(cid:13)(cid:13) L qt L rx ([1 , ∞ ) × R ) . Since our assumption that ˆ β (0) = 1 implies that the last factor on the right tends toa positive constant, we conclude that if the conclusion of Proposition 2.1 were valid for X = L r ( R ), then we would need that12 ≥ q + 2 r = 1 − γ. This means that when n = 2, the assumption that γ ≥ / The Strauss conjecture for nontrapping obstacles when n = 3 , . Let us start the proof of Theorem 1.1 by going over the Minkowski space results thatwill be used. These will form the assumption (1.9) of Theorem 1.3. Lemma 3.1. Let u solve the Minkowski wave equation ( ∂ t − ∆) u = F, ( t, x ) ∈ R × R n u (0 , · ) = f, ∂ t u (0 , · ) = g. Then, for ≤ p ≤ ∞ , and γ satisfying (3.1) 12 − p < γ < n − p , and < − γ < n , we have the following estimate (3.2) (cid:13)(cid:13)(cid:13) | x | n − n +1 p − γ u (cid:13)(cid:13)(cid:13) L pt L pr L ω ( R + × R n ) . k f k ˙ H γ ( R n ) + k g k ˙ H γ − ( R n ) + (cid:13)(cid:13)(cid:13) | x | − n +1 − γ F (cid:13)(cid:13)(cid:13) L t L r L ω ( R + × R n ) . Here, and in what follows, we are using the mixed-norm notation with respect to thevolume element k h k L qr L pω = (cid:16) Z ∞ (cid:16) Z S n − | h ( rω ) | p dσ ( ω ) (cid:17) q/p r n − dr (cid:17) /q for finite exponents and k h k L ∞ r L pω = sup r> (cid:16) Z S n − | h ( rω ) | p dσ ( ω ) (cid:17) /p . We first note that, by the trace lemma for the unit sphere and scaling, we have(3.3) sup r> r n − s (cid:16) Z S n − | v ( rω ) | dσ ( ω ) (cid:17) / . k v k ˙ H s ( R n ) , < s < n , where dσ denotes the unit measure on S n − . Consequently,sup r> r n − s (cid:16)Z S n − (cid:12)(cid:12) (cid:0) e it | D | ϕ (cid:1) ( rω ) (cid:12)(cid:12) dσ ( ω ) (cid:17) / . k ϕ k ˙ H s ( R n ) , < s < n , which is equivalent to(3.4) k | x | − α e it | D | ϕ k L ∞ r L ω . k ϕ k ˙ H n α ( R n ) , − n − < α < . BSTRACT STRICHARTZ ESTIMATES 13 Note that by applying (3.3) to the Fourier transform of v , we see that it is equivalentto the uniform bounds (cid:16) Z S n − | ˆ v ( λω ) | dσ ( ω ) (cid:17) / . λ − n + s k | x | s v k L ( R n ) , λ > , < s < n , which by duality is equivalent to(3.5) (cid:13)(cid:13)(cid:13) | x | − s Z S n − h ( ω ) e iλx · ω dσ ( ω ) (cid:13)(cid:13)(cid:13) L x ( R n ) . λ s − n k h k L ω ( S n − ) , for λ > / < s < n/ 2. Using this estimate we can obtain(3.6) (cid:13)(cid:13) | x | − s e it | D | ϕ k L ( R + × R n ) . k | D | s − ϕ k L ( R n ) , < s < n , for, by after Plancherel’s theorem with respect to the t -variable, we find that the squareof the left side of (3.6) equals(2 π ) − Z ∞ Z R n (cid:12)(cid:12)(cid:12) | x | − s Z S n − e ix · ρω ρ n − ˆ ϕ ( ρω ) dσ ( ω ) (cid:12)(cid:12)(cid:12) dx dρ . Z ∞ Z S n − ρ n − | ˆ ϕ ( ρω ) | ρ s − n dσ ( ω ) dρ = k | D | s − ϕ k L ( R n ) , using (3.5) in the first step.If we interpolate between (3.4) and (3.6) we conclude that, for 2 ≤ q ≤ ∞ ,(3.7) (cid:13)(cid:13)(cid:13) | x | n − n +1 q − γ e it | D | ϕ (cid:13)(cid:13)(cid:13) L qt L qr L ω ( R + × R n ) . k ϕ k ˙ H γ ( R n ) , − q < γ < n − q . This estimate in turn implies that if v solves the Cauchy problem ( ∂ t − ∆) v = 0 in R + × R n then(3.8) (cid:13)(cid:13)(cid:13) | x | n − n +1 q − γ v (cid:13)(cid:13)(cid:13) L qt L qr L ω ( R + × R n ) . k v (0 , · ) k ˙ H γ ( R n ) + k ∂ t v (0 , · ) k ˙ H γ − ( R n ) , − q < γ < n − q . The estimate dual to (3.3) is(3.9) k ϕ k ˙ H γ − ≤ (cid:13)(cid:13) | x | − n +1 − γ ϕ (cid:13)(cid:13) L r L ω . By the Duhamel formula and (3.8)-(3.9), we then have(3.10) (cid:13)(cid:13)(cid:13) | x | n − n +1 p − γ u (cid:13)(cid:13)(cid:13) L pt L pr L ω ( R + × R n ) . k u (0 , · ) k ˙ H γ ( R n ) + k ∂ t u (0 , · ) k ˙ H γ − ( R n ) + (cid:13)(cid:13)(cid:13) | x | − n +1 − γ ( ∂ t − ∆) u (cid:13)(cid:13)(cid:13) L t L r L ω ( R + × R n ) , provided that γ and 1 − γ satisfy the condition in (3.8) for q equal to p and ∞ , respectively,i.e., (3.1). (cid:3) A calculation shows that if(3.11) γ = n − p − , and p c < p < ( n + 3) / ( n − then (3.1) holds: p > p c is needed for the first part, and p < ( n + 3) / ( n − 1) for thesecond. Additionally, as far as the powers of | x | go in (3.10), we have(3.12) p (cid:16) n − n + 1 p − γ (cid:17) = p (cid:16) ( n + 1) − ( n − pp ( p − (cid:17) = − n − γ , if γ = n − p − . As a result, by the arguments to follow, (3.2) is strong enough to show that for thenon-obstacle, Minkowski space case, i.e. Ω = R n , ∆ g = ∆, if 2 ≤ n ≤ p c < p < ( n + 3) / ( n − n = 3 and 4 we shall use a slightlyweaker inequality for which it will be easy to show that we have the corresponding localStrichartz estimates (1.11) for (Ω , ∆ g ). To this end, if R is chosen so that ∂ Ω is containedin | x | < R and ∆ = ∆ g for | x | ≥ R then we define X = X γ,q ( R n ) to be the space withnorm defined by(3.13) k h k X γ,q = k h k L sγ ( | x | < R ) + (cid:13)(cid:13) | x | n − n +1 q − γ h k L qr L ω ( | x | > R ) , if n (cid:0) − s γ ) = γ. We then prove the following obstacle variant of (3.2). Lemma 3.2. For solutions of (1.1) if n ≥ and p > : (3.14) (cid:13)(cid:13)(cid:13) | x | n − n +1 p − γ u (cid:13)(cid:13)(cid:13) L pt L pr L ω ( R + ×{| x | > R } ) + k u k L pt L sγx ( R + ×{ x ∈ Ω: | x | < R } ) . k f k ˙ H γB + k g k ˙ H γ − B + (cid:13)(cid:13)(cid:13) | x | − n +1 − γ F (cid:13)(cid:13)(cid:13) L t L r L ω ( R + ×{| x | > R } ) + k F k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) provided that (3.1) holds. By (3.8) and Lemma 2.2 we have that the assumption (1.9) of Theorem 1.3 is valid if1 / − /q < γ < n/ − /q and 2 ≤ q ≤ ∞ , i.e.(3.15) k v k L qt X γ,q ( R + × R n ) . k v (0 , · ) k ˙ H γ ( R n ) + k ∂ t v (0 , · ) k ˙ H γ − ( R n ) , if ( ∂ t − ∆) v = 0 in R + × R n , under the additional assumption that γ ≤ ( n − / β ∈ C ∞ ( R n )equals one when | x | ≤ R then Sobolev estimates yield k v ( t, · ) k L sγ ( | x | < R ) . k β ( · ) v ( t, · ) k ˙ H γ ( R n ) . Thus, k v k L qt L sγ ( R + ×{| x | 0, and, as above, 1 / − /q < γ < n/ − /q . Bythe finite propagation speed of the wave equation, it is clear that the contribution of thesecond term in the right side of (3.13) will enjoy this estimate. As before, the first termsatisfies it because of Sobolev estimates. This completes the proof of (3.14). (cid:3) Let us also observe a related estimate(3.16) k u k L ∞ t ˙ H γB ( R + × Ω) + k ∂ t u k L ∞ t ˙ H γ − B ( R + × Ω) + k u k L ∞ t L sγx ( R + × Ω) + k βu k L t H γB ( R + × Ω) . k f k ˙ H γB + k g k ˙ H γ − B + (cid:13)(cid:13)(cid:13) | x | − n +1 − γ F (cid:13)(cid:13)(cid:13) L t L r L ω ( R + ×{| x | > R } ) + k F k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) , assuming that (3.1) holds. Indeed, this is a direct consequence of (2.8) and the Duhamelformula, together with the inclusion ˙ H γB (Ω) ֒ → L s γ (Ω), and the following consequence of(3.9), and the dual estimate to Sobolev embedding ˙ H − γB (Ω) ֒ → L s − γ (Ω),(3.17) k g k ˙ H γ − B . k | x | − n +1 − γ g k L r L ω ( | x | > R ) + k g k L s ′ − γ ( x ∈ Ω: | x | < R ) . To prove Theorem 1.1 we shall require a variation of the last two estimates involvingthe vector fields { Γ } = { ∂ t , Z } where, as before, { Z } are the vector fields { ∂ i , x j ∂ k − x k ∂ j : 1 ≤ i ≤ n, ≤ j < k ≤ n } .Note that all the { Γ } commute with (cid:3) g = ∂ t − ∆ g when | x | > R because ∂ Ω ⊂ { x : | x | < R } and ∆ = ∆ g for | x | > R .The main estimate we require is the following. Lemma 3.3. With p and γ as in Lemma 3.2, u solving (1.1) with n ≥ , and ( f, g, F ) satisfying H B × H B × H B boundary conditions, then X | α |≤ (cid:16) k | x | n − n +1 p − γ Γ α u k L pt L pr L ω ( R + ×{| x | > R } ) + k Γ α u k L pt L sγx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) (3.18) . X | α |≤ (cid:16) k Z α f k ˙ H γB + k Z α g k ˙ H γ − B (cid:17) + X | α |≤ (cid:16) k | x | − n +1 − γ Γ α F k L t L r L ω ( R + ×{| x | > R } ) + k Γ α F k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) . The boundary conditions on ( f, g, F ) imply that ∂ jt u is locally in H γ − j (Ω), j =0 , , , which will be implicitly used in elliptic regularity arguments. We will also use thefact that the Cauchy data for Γ α u is bounded in ˙ H γB × ˙ H γ − B by the right hand side of(3.18) for | α | ≤ 2. This is clear if Γ α is replaced by Z α . On the other hand, the Cauchydata for ∂ t u is ( g, ∆ g f + F (0 , · )). We may control X | α |≤ (cid:16) k Z α g k ˙ H γB + k Z α ∆ g f k ˙ H γ − B (cid:17) ≤ X | α |≤ (cid:16) k Z α f k ˙ H γB + k Z α g k ˙ H γ − B (cid:17) . Recall that γ ∈ (0 , ), so that ˙ H γB (Ω) = ˙ H γ (Ω). To control the term F (0 , · ), we recallthat Γ = { ∂ t , Z } , and use the bound(3.19) X | α |≤ k Γ α F k L ∞ t ˙ H γ − B ( R + × Ω) ≤ X | α |≤ k Γ α F k L t ˙ H γ − B ( R + × Ω) which by (3.17) is seen to be dominated by the right hand side of (3.18). Similar consid-erations apply to the Cauchy data for ∂ t u .Let us now give the argument for (3.18). We first fix β ∈ C ∞ satisfying β = 1 for | x | ≤ R and supp β ⊂ {| x | < R } . Then the first step in the proof of (3.18) will be toshow that X | α |≤ (cid:16) k | x | n − n +1 p − γ (1 − β )Γ α u k L pt L pr L ω ( R + ×{| x | > R } ) + k (1 − β )Γ α u k L pt L sγx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) (3.20) . X | α |≤ (cid:16) k Z α f k ˙ H γB + k Z α g k ˙ H γ − B (cid:17) + X | α |≤ (cid:16) k | x | − n +1 − γ Γ α F k L t L r L ω ( R + ×{| x | > R } ) + k Γ α F k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) . Since the Γ commute with (cid:3) g when | x | ≥ R , we have (cid:3) g (cid:0) (1 − β )Γ α u (cid:1) = (1 − β )Γ α F − [ β , ∆ g ]Γ α u . We can therefore write (1 − β )Γ α u as v + w where (cid:3) g v = (1 − β )Γ α F and v has initialdata (cid:0) (1 − β )Γ α u (0 , · ) , ∂ t (1 − β )Γ α u (0 , · ) (cid:1) , while (cid:3) g w = − [ β , ∆ g ]Γ α u and w hasvanishing initial data. If we do this, it follows by (3.14) that if for | α | ≤ − β )Γ α u by v in the left side of (3.20), then the resulting expressionis dominated by the right side of (3.20). If we use (2.4), we find that if we replace(1 − β )Γ α u by w then the resulting expression is dominated by(3.21) X | α |≤ k [ β , ∆ g ]Γ α u k L t H γ − B . X j ≤ k β ∂ jt u k L t H γ +2 − jB assuming that β equals one on supp( β ) and is supported in | x | < R . As a result, wewould be done with the proof of (3.20) if we could show that the right hand side of (3.21)is dominated by the right side of (3.20). By (3.16) we control k β ∂ t u k L t H γB by the righthand side of (3.20). On the other hand, k β ∂ t u k L t H γ +1 B . k β ∂ t u k L t H γB k β u k L t H γ +2 B so it suffices to dominate k β u k L t H γ +2 B . Since ∆ g u = ∂ t u − F , then if β equals one onsupp( β ) and is supported in the set where | x | < R , we may use elliptic regularity andthe equation to bound k β u k L t H γ +2 B . k β ∆ g u k L t H γB + k β u k L t H γB . k β ∂ t u k L t H γB + k β u k L t H γB + k β F k L t H γB . BSTRACT STRICHARTZ ESTIMATES 17 The first two terms are dominated as above using (3.16). On the other hand, Sobolevembedding and duality yields k β F k L t H γB . X | α |≤ k ∂ αx F k L t L s ′ − γ ( R + ×{ x ∈ Ω: | x |≤ R } ) (3.22) . X | α |≤ k ∂ αt,x F k L t L s ′ − γ ( R + ×{ x ∈ Ω: | x |≤ R } ) . To finish the proof of (3.18), we need to show that the analog of (3.20) is valid when(1 − β ) is replaced by β . Since the coefficients of Γ are bounded on supp( β ), if β equals one on supp( β ) and is supported in | x | < R , then by Sobolev embedding X | α |≤ k β Γ α u k L pt L sγx ( R + × Ω) . X j ≤ k β ∂ jt u k L pt H γ +2 − jB . X j ≤ (cid:16) k β ∂ jt u k L t H γ +2 − jB + k β ∂ jt u k L ∞ t H γ +2 − jB (cid:17) . The terms in L t H γ +2 − j are dominated as above. To control the L ∞ t H γ +2 − j terms, andconclude the proof of (3.18), we establish the following estimate:(3.23) X | α |≤ k Γ α u k L ∞ t ˙ H γB + k ∂ t Γ α u k L ∞ t ˙ H γ − B . X | α |≤ (cid:16) k Z α f k ˙ H γB (Ω) + k Z α g k ˙ H γ − B (Ω) (cid:17) + X | α |≤ (cid:16) k | x | − n +(1 − γ ) Γ α F k L t L r L ω ( R + ×{| x | > R } ) + k Γ α F k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) . The inequality where Γ α u is replaced by (1 − β )Γ α u in (3.23) follows by energyestimates on R n , since the right hand side dominates k (1 − β )Γ α F k L t ˙ H γ − , togetherwith (2.8) using the bound (3.21) to handle the commutator term. If Γ α u is replacedon the left hand side by β Γ α u , the result is dominated by P j ≤ k β ∂ jt u k L ∞ t H γ − jB . Forthe case j = 0 , 1, we write ✷ g ( β u ) = β F − [∆ g , β ] u , and use (2.5) with the Duhamelformula to bound k β u k L ∞ t H γ +2 B + k β ∂ t u k L ∞ t H γ +1 B . k β f k H γ +2 B + k β g k H γ +1 B + k β u k L t H γ +2 B + k β F k L t H γ +1 B . The term on the right involving u is controlled previously; on the other hand, since F satisfies the H γ +1 B boundary conditions, then k β F k L t H γ +1 B . X | α |≤ k ∂ αx F k L t L s ′ − γx . To handle the terms for j = 2 , X j =2 , k β ∂ jt u k L ∞ t H γ − jB ≤ X j =0 , (cid:16) k β ∂ jt ∆ g u k L ∞ t H γ − jB + k β ∂ jt F k L ∞ t H γ − jB (cid:17) . The terms involving ∆ g u are dominated by k β ∂ jt u k L ∞ t H γ +2 − jB with j = 0 , 1. The termsinvolving F are controlled for j = 1 by (3.19), and for j = 0 by observing that (3.22)holds with L t replaced by L ∞ t . This completes the proof of (3.18) and (3.23). (cid:3) We shall now use these estimates to prove Theorem 1.1. Proof of Theorem 1.1: We assume Cauchy data ( f, g ) satsifying the smallness condi-tion (1.7), and let u solve the Cauchy problem (1.1) with F = 0. We iteratively define u k , for k ≥ 1, by solving ( ∂ t − ∆ g ) u k ( t, x ) = F p ( u k − ( t, x )) , ( t, x ) ∈ R + × Ω u (0 , · ) = f∂ t u (0 , · ) = g ( Bu )( t, x ) = 0 , on R + × ∂ Ω . Our aim is to show that if the constant ε > M k = X | α |≤ (cid:16) (cid:13)(cid:13) Γ α u k (cid:13)(cid:13) L ∞ t ˙ H γB ( R + × Ω) + (cid:13)(cid:13) ∂ t Γ α u k (cid:13)(cid:13) L ∞ t ˙ H γ − B ( R + × Ω) + (cid:13)(cid:13) | x | n − n +1 p − γ Γ α u k (cid:13)(cid:13) L pt L pr L ω ( R + ×{| x | > R } ) + k Γ α u k k L pt L sγx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) for every k = 0 , , , . . . For k = 0, it follows by (3.18) and (3.23) that M ≤ C ε , with C a fixed constant.More generally, (3.18) and (3.23) yield that M k ≤ C ε + C X | α |≤ (cid:16) (cid:13)(cid:13) | x | − n +1 − γ Γ α F p ( u k − ) (cid:13)(cid:13) L t L r L ω ( R + ×{| x | > R } ) (3.24) + k Γ α F p ( u k − ) k L t L s ′ − γx ( R + ×{ x ∈ Ω: | x | < R } ) (cid:17) . Note that our assumption (1.4) on the nonlinear term F p implies that for small v X | α |≤ | Γ α F p ( v ) | . | v | p − X | α |≤ | Γ α v | + | v | p − X | α |≤ | Γ α v | . Furthermore, since u k will be locally of regularity H γ +2 B ⊂ L ∞ and F p vanishes at 0, itfollows that F p ( u k ) satisfies the B boundary conditions if u k does.Since the collection Γ contains vectors spanning the tangent space to S n − , by Sobolevembedding for n = 3 , k v ( r · ) k L ∞ ω + X | α |≤ k Γ α v ( r · ) k L ω . X | α |≤ k Γ α v ( r · ) k L ω . Consequently, for fixed t, r > X | α |≤ k Γ α F p ( u k − ( t, r · )) k L ω . X | α |≤ k Γ α u k − ( t, r · ) k pL ω . By (3.12), the first summand in the right side of (3.24) is dominated by C M pk − . We next observe that, since s γ > n ≤ 4, it follows by Sobolev embedding on { Ω ∩ | x | < R } that k v k L ∞ ( x ∈ Ω: | x | < R ) + X | α |≤ k Γ α v k L ( x ∈ Ω: | x | < R ) . X | α |≤ k Γ α v k L sγ ( x ∈ Ω: | x | < R ) . BSTRACT STRICHARTZ ESTIMATES 19 Since s ′ − γ < 2, it holds for each fixed t that X | α |≤ k Γ α F p ( u k − ( t, · )) k L s ′ − γ ( x ∈ Ω: | x | < R ) . X | α |≤ k Γ α u k − ( t, · ) k pL sγ ( x ∈ Ω: | x | < R ) . The second summand in the right side of (3.24) is thus also dominated by C M pk − , andwe conclude that M k ≤ C ε + 2 C C M pk − . For ε sufficiently small, then(3.25) M k ≤ C ε, k = 1 , , , . . . To finish the proof of Theorem 1.1 we need to show that u k converges to a solution ofthe equation (1.3). For this it suffices to show that A k = (cid:13)(cid:13) | x | n − n +1 p − γ ( u k − u k − ) (cid:13)(cid:13) L pt L pr L ω ( R + ×{| x | > R } ) + k u k − u k − k L pt L sγx ( R + ×{ x ∈ Ω: | x | < R } ) tends geometrically to zero as k → ∞ . Since | F p ( v ) − F p ( w ) | . | v − w | ( | v | p − + | w | p − )when v and w are small, the proof of (3.25) can be adapted to show that, for small ε > C so that A k ≤ CA k − ( M k − + M k − ) p − , which, by (3.25), implies that A k ≤ A k − for small ε . Since A is finite, the claimfollows, which finishes the proof of Theorem 1.1. (cid:3) References [1] M. 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