Cauchy problem for operators with triple effectively hyperbolic characteristics-Ivrii's conjecture-
aa r X i v : . [ m a t h . A P ] F e b Cauchy problem for operators with tripleeffectively hyperbolic characteristics– Ivrii’s conjecture –
Tatsuo Nishitani ∗ Abstract
Ivrii’s conjecture asserts that the Cauchy problem is C ∞ well-posedfor any lower order term if every critical point of the principal symbol iseffectively hyperbolic. Effectively hyperbolic critical point is at most triplecharacteristic. If every characteristic is at most double this conjecturehas been proved in 1980’. In this paper we prove the conjecture for theremaining cases, that is for operators with triple effectively hyperboliccharacteristics. Keywords: Ivrii’s conjecture, B´ezout matrix, triple characteristic, Tricomi type, effectivelyhyperbolic, Cauchy problem, Weyl calculus.Mathematics Subject Classification 2010: Primary 35L30, Secondary 35G10
This paper is devoted to the Cauchy problem(1.1)
P u = D mt u + P m − j =0 P | α | + j ≤ m a j,α ( t, x ) D αx D jt u = 0 ,D jt u (0 , x ) = u j ( x ) , j = 0 , . . . , m − t ≥ x ∈ R d and the coefficients a j,α ( t, x ) are real valued C ∞ functionsin a neighborhood of the origin of R d and D x = ( D x , . . . , D x d ), D x j =(1 /i )( ∂/∂x j ) and D t = (1 /i )( ∂/∂t ). The Cauchy problem (1.1) is C ∞ well-posed at the origin for t ≥ δ > U ofthe origin of R d such that (1.1) has a unique solution u ∈ C ∞ ([0 , δ ) × U ) forany u j ( x ) ∈ C ∞ ( R d ). We assume that the principal symbol of Pp ( t, x, τ, ξ ) = τ m + m − X j =0 X | α | + j = m a j,α ( t, x ) ξ α τ j ∗ Department of Mathematics, Osaka University: [email protected]
1s hyperbolic for t ≥
0, that is there exist δ ′ > U ′ of theorigin such that(1.2) p = 0 has only real roots in τ for ( t, x ) ∈ [0 , δ ′ ) × U ′ and ξ ∈ R d which is indeed necessary in order that the Cauchy problem (1.1) is C ∞ well-posed near the origin for t ≥ C ∞ wellposed for any lower order term then the Hamilton map F p has a pair of non-zero real eigenvalues at every critical point ([7, Theorem 3]). With X = ( t, x ),Ξ = ( τ, ξ ) the Hamilton map F p is defined by F p ( X, Ξ) = ∂ p∂X∂ Ξ ∂ p∂ Ξ ∂ Ξ − ∂ p∂X∂X − ∂ p∂ Ξ ∂X and a critical point ( X, Ξ) is a point where ∂p/∂X = ∂p/∂ Ξ = 0. Note that p ( X, Ξ) = 0 at critical points by the homogeneity in Ξ so that ( X, Ξ) is amultiple characteristic and τ is a multiple characteristic root of p . A criticalpoint where the Hamilton map F p has a pair of non-zero real eigenvalues iscalled effectively hyperbolic ([4], [10]). In [8], Ivrii has proved that if everycritical point is effectively hyperbolic, and p admits a decomposition p = q q with real smooth symbols q i near the critical point, then the Cauchy problemis C ∞ well-posed for every lower order term. In this case the critical point iseffectively hyperbolic if and only if the Poisson bracket { q , q } does not vanishthere. He has conjectured that the assertion would hold without any additionalcondition.If a critical point ( X, Ξ) is effectively hyperbolic then τ is a characteristicroot of multiplicity at most 3 ([7, Lemma 8.1]). If every multiple characteristicroot is at most double, the conjecture has been proved in [8], [20], [11, 12, 13],[21, 23, 22]. When there exists an effectively hyperbolic critical point ( X, Ξ)such that τ is a triple characteristic root where p cannot be factorized, severalpartial results are obtained in [2], [27], [28], [26]. Note that if there is a triplecharacteristic which is not effectively hyperbolic the Cauchy problem is notwell-posed in the Gevrey class of order s > Theorem 1.1.
Assume (1.2) . If every critical point (0 , , τ, ξ ) , ξ = 0 is effec-tively hyperbolic then for any a j,α ( t, x ) with j + | α | ≤ m − , which are C ∞ ina neighborhood of (0 , , there exist δ > , a neighborhood U of the origin and n > such that for any s ∈ R and any f with t − n +1 / h D i s f ∈ L ((0 , δ ) × R d ) there exists u with t − n h D i − n − s + m − j D jt u ∈ L ((0 , δ ) × R d ) , j = 0 , , . . . , m − satisfying P u = f in (0 , δ ) × U. h D i stands for p | D x | . For some more detailed information aboutthe constant n , see (10.1) below. Theorem 1.2.
Under the same assumption as in Theorem 1.1, for any a j,α ( t, x ) with j + | α | ≤ m − , which are C ∞ in a neighborhood of (0 , , there exist δ > and a neighborhood U of the origin such that for any u j ( x ) ∈ C ∞ ( R d ) , j = 0 , , . . . , m − , there exists u ( t, x ) ∈ C ∞ ([0 , δ ) × U ) satisfying (1.1) in [0 , δ ) × U . If u ( t, x ) ∈ C ∞ ([0 , δ ) × U ) with ∂ jt u (0 , x ) = 0 , j = 0 , , . . . , m − ,satisfies P u = 0 in [0 , δ ) × U then u = 0 in a neighborhood of (0 , .Proof. Compute u j ( x ) = D jt u (0 , x ) for j = m, m + 1 , . . . from u j ( x ), j =0 , , . . . , m − P u = 0. By a Borel’s lemma there is w ( t, x ) ∈ C ∞ ( R d ) such that D jt w (0 , x ) = u j ( x ) for all j ∈ N . Since ( D jt P w )(0 , x ) = 0for all j ∈ N it is clear that t − n +1 / h D i s P w ∈ L ((0 , δ ) × R d ) for any s . Thanksto Theorem 1.1 there exists v with t − n h D i − n − s + m − j D jt v ∈ L ((0 , δ ) × R d ), j = 0 , , . . . , m − P v = − P w in (0 , δ ) × U . Since we may assume n ≥ D jt v (0 , x ) = 0, j = 0 , , . . . , m − u = v + w is a desired solution. Local uniqueness follows from Theorem 13.4 because ∂ kt u (0 , x ) = 0 for any k ∈ N by P u = 0.
Remark 1.1.
Under the assumption of Theorem 1.1 we see that p has necessar-ily non-real characteristic roots in the t < , , ξ ) if p (0 , , τ, ξ ) = 0has a triple characteristic root. Therefore P would be a Tricomi type operatorin this case. In fact from [7, Lemma 8.1] it follows that if τ is a triple character-istic root at (0 , , ξ ) and all characteristic roots are real in a full neighborhoodof (0 , , ξ ) then F p (0 , , τ, ξ ) = O . Remark 1.2.
For any characteristic root τ of multiplicity r ≥ t, x, ξ ) with t ≥ t, x, τ, ξ ) is a critical point, where F p ( t, x, τ, ξ ) = O unless r = 3and t = 0 ([7, Lemma 8.1]). For any double characteristic root τ at ( t, x, ξ ) with t ≥ t, x, τ, ξ ) is a critical point if t > t = 0. Here is a simple example P = ( D t − t ℓ D x )( D t + c D x ) , ℓ ∈ N , x ∈ R , t ≥ c ∈ R . Let c = 0 then it is clear that τ = 0 is a double characteristic rootat (0 , , ℓ = 1 then ∂ t p (0 , , ,
1) = − c = 0 and hence (0 , , ,
1) is not acritical point. If ℓ ≥ , , ,
1) is a critical point and F p has non-zeroreal eigenvalues there if and only if ℓ = 2. Let c = 0 then τ = 0 is a triplecharacteristic root at (0 , ,
1) hence (0 , , ,
1) is a critical point. At (0 , , , F p has non-zero real eigenvalues if and only if ℓ = 1. As noted in Introduction, if a critical point ( X, Ξ) is effectively hyperbolic then τ is a characteristic root of multiplicity at most 3. This implies that it is essential3o study operators P of third order(2.1) P = D t + X j =1 a j ( t, x, D ) h D i j D − jt which is differential operator in t with coefficients a j ∈ S , classical pseudodif-ferential operator of order 0, where h D i = op((1 + | ξ | ) / ). One can reduce P to the case with a ( t, x, D ) = 0 and hence the principal symbol is(2.2) p ( t, x, τ, ξ ) = τ − a ( t, x, ξ ) | ξ | τ − b ( t, x, ξ ) | ξ | . All characteristic roots are real for t ≥ a ( t, x, ξ ) − b ( t, x, ξ ) ≥ , ( t, x, ξ ) ∈ [0 , T ) × U × R d . Assume that p (0 , , τ, ¯ ξ ) = 0 has a triple characteristic root ¯ τ , which is neces-sarily ¯ τ = 0. The critical point (0 , , ¯ τ , ¯ ξ ) is effectively hyperbolic if and onlyif(2.4) ∂ t a (0 , , ¯ ξ ) = 0 . So we can assume that a = e ( t + α ( x, ξ )) with e >
0. Add to P a second orderterm M e h D i D t with a large parameter M > a ( t, x, ξ ) changes to e ( t + α + M h ξ i − ) which we still denote by the same a .With U = t ( D t u, h D i D t u, h D i u ) the equation P u = f is reduced to(2.5) D t U = A ( t, x, D ) h D i U + B ( t, x, D ) U + F where A, B ∈ S , F = t ( f, ,
0) and A ( t, x, ξ ) = a b . Let S be the B´ezout matrix of p and ∂p/∂τ , that is S ( t, x, ξ ) = − a a b − a b a then S is nonnegative definite and symmetrizes A , that is SA is symmetricwhich is easily examined directly, though this is a special case of a generalfact (see [14], [25]). We now diagonalize S by an orthogonal matrix T so that T − ST = Λ = diag ( λ , λ , λ ) where 0 < λ < λ < λ are the eigenvalues of S which satisfies (cid:12)(cid:12) ∂ αx ∂ βξ λ j (cid:12)(cid:12) - a − j −| α + β | / h ξ i −| β | , j = 1 , , . × V = T − U ; roughly(2.6) D t V = A T h D i V + B T V, A T = T − AT where Λ symmetrizes A T . A significant feature of λ j is that∆ a - λ - a , λ ≃ a, λ ≃ . From the conditions (2.3) and (2.4) the discriminant ∆ is essentially a thirdorder polynomial in t and we can find a smooth ψ ( x, ξ ) and c > a ≥ c min (cid:8) t , ( t − ψ ) + M ρ h ξ i − (cid:9) where ρ = α + M h ξ i − and ψ satisfies (cid:12)(cid:12) ∂ αx ∂ βξ ψ (cid:12)(cid:12) - ρ −| α + β | / h ξ i −| β | . Since (2.6) is a symmetrizable system with a diagonal symmetrizer Λ, anatural energy will be (cid:0) op(Λ)
V, V (cid:1) = X j =1 (cid:0) op( λ j ) V j , V j (cid:1) and (2.7) suggests that a weighted energy with a scalar weight op( t − n φ − n ) φ = ω + t − ψ, ω = p ( t − ψ ) + M ρ h ξ i − would work, which is essentially the same weight as the weight employed forstudying double effectively hyperbolic characteristics in [23] (see also [24]), there M h ξ i − was used in place of M ρ h ξ i − . A main feature of the weight function tφ is ∂ t ( tφ ) = κ ( tφ ) , κ = 1 t + 1 ω . Our task is now to show the weighted energy Re e − θt (op(Λ)op( t − n φ − n ) V, op( t − n φ − n ) V )works well and can control any lower order term, yielding weighted energy esti-mates for P . In doing so it is crucial that λ j , ω , ρ and φ are admissible weightsfor the metric g = M − (cid:0) h ξ i| dx | + h ξ i − | dξ | (cid:1) and λ j ∈ S ( λ j , g ), φ ∈ S ( φ, g ) so on. This fact enables us to apply the Weylcalculus to op( λ j ), op( φ − n ) and so on. One of main points to derive energyestimates is the following inequalities Re (op(Λ)op( ∂ t ( t − n φ − n )) V, W ) − θ (op(Λ) W, W ) ≤ − n (1 − CM − ) k op( κ / Λ / ) W k − c θ (cid:0) k op(Λ / ) W k + X j =1 M − j kh D i − (3 − j ) / W j k (cid:1) W = op( t − n φ − n ) V , while for B = ( b ij ) with b ij ∈ S (1 , g ) we see (cid:12)(cid:12) (op(Λ)op( B ) W, W ) (cid:12)(cid:12) ≤ C k op( κ / Λ / ) W k + C (cid:0) k op(Λ / ) W k + X j =1 M − j kh D i − (3 − j ) / W j k (cid:1) which are proved applying the Weyl calculus of pseudodifferential operators. Study third order operators P of the form (2.1) with a ( t, x, D ) = 0, hence theprincipal symbol has the form (2.2) where a ( t, x, ξ ) and b ( t, x, ξ ) are homoge-neous of degree 0 in ξ and assumed to satisfy (2.3) with some T > U of the origin of R d . Assume that p ( t, x, τ, ξ ) has a triple char-acteristic root ¯ τ at (0 , , ¯ ξ ), | ¯ ξ | = 1 and (0 , , ¯ τ , ¯ ξ ) is effectively hyperbolic. It isclear that ¯ τ = 0 , a (0 , , ¯ ξ ) = 0 , b (0 , , ¯ ξ ) = 0 . Since ∂ αx ∂ βξ a (0 , , ¯ ξ ) = 0 for | α + β | = 1 and ∂ αx ∂ βξ b (0 , , ¯ ξ ) = 0 for | α + β | ≤ (cid:0) λ − F p (0 , , , ¯ ξ ) (cid:1) = λ d (cid:0) λ − { ∂ t a (0 , , ¯ ξ ) } (cid:1) hence (0 , , , ¯ ξ ) is effectively hyperbolic if and only if ∂ t a (0 , , ¯ ξ ) = 0 . Since a (0 , , ¯ ξ ) = 0 and ∂ t a (0 , , ¯ ξ ) = 0 there is a neighborhood U of (0 , , ¯ ξ ) inwhich one can write a ( t, x, ξ ) = e ( t, x, ξ )( t + α ( x, ξ ))where e > U . Note that α ( x, ξ ) ≥ ξ because a ( t, x, ξ ) ≥ , T ) × U × R d . Introducing a small parameter ǫ we consider τ − e ( t, x, ξ )( t + α ( x, ξ ) + ǫ ) | ξ | τ − b ( t, x, ξ ) | ξ | = τ − a ( t, x, ξ, ǫ ) | ξ | − b ( t, x, ξ ) | ξ | . (3.2)From now on we write b ( X ) or a ( X, ǫ ) and so on to make clearer that thesesymbols are defined in some neighborhood of ¯ X = (0 , ¯ ξ ) or ( ¯ X, t, X, ǫ ) = 4 a ( t, X, ǫ ) − b ( t, X ) . emma 3.1. One can write ∆ = ˜ e ( t, X, ǫ ) (cid:0) t + a ( X, ǫ ) t + a ( X, ǫ ) t + a ( X, ǫ ) (cid:1) in a neighborhood of (0 , ¯ X, where a j ( ¯ X,
0) = 0 , j = 1 , , and ˜ e > .Proof. It is clear that ∂ kt a (0 , ¯ X,
0) = 0 for k = 0 , , ∂ t a (0 , ¯ X, = 0.Show ∂ t b (0 , ¯ X,
0) = 0. Suppose the contrary and hence b ( t, ¯ X,
0) = t (cid:0) b + tb ( t ) (cid:1) with b = 0. Since a ( t, ¯ X,
0) = c t with c > t, ¯ X,
0) = 4 c t − b ( t, ¯ X, ≥ ∂ kt ∆(0 , ¯ X,
0) = 0 for k = 0 , , ∂ t ∆(0 , ¯ X, = 0. Then from the Malgrange preparation theorem (e.g. [5,Theorem 7.5.5]) one can conclude the assertion.Introducing(3.3) ρ ( X, ǫ ) = α ( X ) + ǫ one can also write∆ = 4 e ( t + ρ ) − b = 4 e n ( t + ρ ) − e b o = 4 e (cid:8) ( t + ρ ) − ˆ b (cid:9) with ˆ b = 3 √ b/ e / . Denotingˆ b ( t, X ) = X j =0 ˆ b j ( X ) t j + ˆ b ( t, X ) t where ˆ b ( ¯ X ) = ˆ b ( ¯ X ) = 0 which follows from the proof of Lemma 3.1, one canwrite ∆ / ˜ e = ¯∆ = t + a ( X, ǫ ) t + a ( X, ǫ ) t + a ( X, ǫ )= E (cid:8) ( t + ρ ) − (cid:0) X j =0 ˆ b j ( X ) t j + ˆ b ( t, X ) t (cid:1) (cid:9) (3.4)with E ( t, X, ǫ ) = 4 e / ˜ e . Here note that E (0 , ¯ X,
0) = 1 since e (0 , ¯ X,
0) = ∂ t a (0 , ¯ X,
0) and ˜ e (0 , ¯ X,
0) = 4 ∂ t a (0 , ¯ X, . Lemma 3.2.
There is a neighborhood V of ¯ X such that (cid:12)(cid:12) ˆ b ( X ) (cid:12)(cid:12) ≤ α / ( X ) ( X ∈ V ) . Proof.
It is clear that | ˆ b ( X ) | ≤ α / ( X ). If α ( X ) = 0 then the assertion isobvious. Assume α ( X ) = 0. Since(3.5) ( t + α ( X )) ≥ (cid:0) X j =0 ˆ b j ( X ) t j + ˆ b ( t, X ) t (cid:1) (0 ≤ t ≤ T )7hoosing t = 3 α ( X ) ≤ T , writing α = α ( X ), it follows from (3.5) that8 α / ≥ (cid:12)(cid:12) ˆ b ( X ) + 3ˆ b ( X ) α (cid:12)(cid:12) − Cα ≥ | ˆ b ( X ) | α − Cα − α / hence the assertion is clear because α ( ¯ X ) = 0. Lemma 3.3.
In a neighborhood of ( ¯ X, we have a j ( X, ǫ ) = O (cid:0) ρ ( X, ǫ ) j (cid:1) for j = 1 , , . More precisely a ( X, ǫ ) = E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − ˆ b ( X ) (cid:1) + O ( ρ / ) ,a ( X, ǫ ) = E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − b ( X )ˆ b ( X ) (cid:1) + O ( ρ / ) ,a ( X, ǫ ) = E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − ˆ b ( X ) (cid:1) . Proof.
Since ¯∆(0 , X, ǫ ) ≥ a ( X, ǫ ) = E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − ˆ b ( X ) (cid:1) ≥ b = O ( ρ / ) and consequently a ( X, ǫ ) = O ( ρ ). Since ∂ t ¯∆ (cid:12)(cid:12)(cid:12) t =0 = a ( X, ǫ ) = ∂ t E (0 , X, ǫ ) a ( X, ǫ )+ E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − b ( X )ˆ b ( X ) (cid:1) it follows that ˆ b ( X )ˆ b ( X ) = O ( ρ ) by Lemma 3.2 and hence the above equalityshows the assertion for a ( X, ǫ ). Finally from ∂ t ¯∆ (cid:12)(cid:12)(cid:12) t =0 = 2 a ( X, ǫ ) = ∂ t E (0 , X, ǫ ) a ( X, ǫ )+2 ∂ t E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − b ( X )ˆ b ( X ) (cid:1) +2 E (0 , X, ǫ ) (cid:0) ρ ( X, ǫ ) − ˆ b ( X ) − b ( X )ˆ b ( X ) (cid:1) and Lemma 3.2 one concludes the assertion for a ( X, ǫ ). ψ ( x, ξ ) Denote(3.6) ν ( X, ǫ ) = inf { t | ¯∆( t, X, ǫ ) > } and hence ¯∆( ν, X, ǫ ) = 0. First check that ν ( X, ǫ ) ≤
0. Suppose the contrary.Since ¯∆( t, X, ǫ ) ≥ t ≥ ν ( X, ǫ ) is a double root, that is onecan write ¯∆( t ) = ( t − ν ) ( t − ˜ ν ) with a real ˜ ν . It is clear that ˜ ν = ν and ˜ ν ≤ t ) ≥ t ≥
0. Therefore we have ˜ ν < ν and ¯∆( t ) > ν < t < ν which is incompatible with the definition of ν . Write¯∆( t, X, ǫ ) = ( t − ν ( X, ǫ )) (cid:0) t + A ( X, ǫ ) t + A ( X, ǫ ) (cid:1) where A = ν + a . Here we prepare following lemma.8 emma 3.4. One can find a neighborhood U of ( ¯ X, such that for any ( X, ǫ ) ∈U there is j ∈ { , , } such that | ν j ( X, ǫ ) | ≥ ρ ( X, ǫ ) / where ¯∆( t, X, ǫ ) = Q j =1 ( t − ν j ( X, ǫ )) .Proof. First show that there is 1 / < δ < / n | ρ − ˆ b | / , | ρ − b ˆ b / | / , | ρ − ˆ b / | o ≥ δ ρ. In fact denoting f ( δ ) = 2(1 − δ ) / (1 − δ ) / / √ − − δ it is easy to checkthat f (1 / > f (1 / <
0. Take 1 / < δ < / f ( δ ) = 0.If | ρ − ˆ b | / < δ ρ and | ρ − ˆ b / | < δ ρ then | ˆ b | ≥ (1 − δ ) / ρ / and | ˆ b | ≥ √ − δ ) / ρ / hence | ρ − b ˆ b / | ≥ | ˆ b ˆ b | / − ρ ≥ (cid:0) f ( δ ) + δ (cid:1) ρ = δ ρ which shows that | ρ − b ˆ b / | / ≥ δ ρ . Thus (3.7) is proved. Thanks toLemma 3.3, taking E (0 , ¯ X,
0) = 1 and 1 / < δ , one can find a neighborhood U of ( ¯ X,
0) such that | a ( X, ǫ ) | ≥ ρ/ , | a ( X, ǫ ) | ≥ ρ / , | a ( X, ǫ ) | ≥ ρ / , ( X, ǫ ) ∈ U . Then the assertion follows from the relations between { ν i } and { a i } . Lemma 3.5.
Denote ν defined in (3.6) by ν and by ν j , j = 2 , the other rootsof ¯∆ = 0 in t . Then one can find a neighborhood U of ( ¯ X, and c i > suchthat (3.8) if ν + a < c ρ, ( X, ǫ ) ∈ U then | ν − ν j | ≥ c ρ, j = 2 , . In particular ν ( X, ǫ ) is smooth in ( X, ǫ ) ∈ U ∩ { ν + a < c ρ } .Proof. Set δ = 1 / c < δ/
4. First note that if Re ν j ≥ c δ , j = 2 , | ν − ν j | ≥ | ν − Re ν j | ≥ Re ν j ≥ c δ because ν ≤ Re ν j < c δ, j = 2 , . Write ¯∆( t ) = Y j =1 ( t − ν j ) = ( t − ν ) (cid:0) ( t + A / − D (cid:1) and recall ν + a = A . Consider the case that both ν , ν are real so that D ≥ ν , ν = − A / ± √ D . If D = 0 and hence − c δ < Re ν j = − A / < c δ
9n view of (3.8) and (3.9). Then we see | ν | ≥ δ ρ thanks to Lemma 3.4 andhence | ν − ν j | ≥ | ν | − | ν j | ≥ ( δ − c ) ρ ≥ δρ/ . If D > − A / √ D ≤
0. Otherwise ¯∆( t ) would be negative forsome t > − A / √ D which is a contradiction. Thus √ D ≤ A / ≤ c δ which shows that | ν | , | ν | ≤ | A | / √ D ≤ c ρ and hence | ν | ≥ δ ρ by Lemma 3.4 again. Therefore | ν − ν j | ≥ | ν | − | ν j | ≥ ( δ − c ) ρ ≥ δρ/ . Turn to the case
D < ν , ν = − A / ± i p | D | . As observed aboveone may assume | Re ν j | = | A / | < c δ . Thanks to Lemma 3.4 either | ν | ≥ δρ or | ν | = | ν | ≥ δρ . If | ν | ≥ δρ then it follows that | ν − ν j | ≥ | ν + A / | ≥ | ν | − | A | / ≥ ( δ − c ) ρ ≥ δρ/ . If | ν | = | ν | ≥ δρ so that | A | / p | D | ≥ δρ hence p | D | ≥ δρ − | A | / ≥ ( δ − c ) ρ which proves | ν − ν j | ≥ p | D | ≥ ( δ − c ) ρ ≥ δρ/ . Thus ν ( X, ǫ ) is a simple root and hence smooth provided that ν + a < c ρ .Now define ψ ( X, ǫ ) which plays a crucial role in our arguments derivingweighted energy estimates. Choose χ ( s ) ∈ C ∞ ( R ) such that 0 ≤ χ ( s ) ≤ χ ( s ) = 1 if s ≤ χ ( s ) = 0 for s ≥
1. Define ψ ( X, ǫ ) = − χ (cid:16) ν + a c ρ (cid:17) ν + a ǫ = 0) . We now prove
Proposition 3.1.
One can find a neighborhood U of ( ¯ X, such that (3.10) ¯∆( t, X, ǫ ) ≥ υ min (cid:8) t , ( t − ψ ( X, ǫ )) (cid:9) ( t + ρ ( X, ǫ )) holds for ( X, ǫ ) ∈ U , ǫ = 0 and t ∈ [0 , T ] where υ = (cid:0) √ (cid:1) − .Proof. Set δ = 1 / c ≥ υ such that(3.11) ¯∆( t, X, ǫ ) ≥ c t ( t + ρ ) if A = ν + a ≥ . Write ¯∆( t ) = ( t − ν ) (cid:0) ( t + A / − D (cid:1) . Consider the case D = 0. From Lemma3.4 either | ν | ≥ δρ or | A / | = A / ≥ δρ . If | ν | ≥ δρ then t − ν = t + | ν | ≥ t + δρ hence δ − ( t − ν ) ≥ t + ρ . Since ( t + A / ≥ t it is clear that (3.11)holds with c = δ . If A / ≥ δρ then t + A / ≥ t + δρ and t + A / ≥ t ,10 − ν = t + | ν | ≥ t gives (3.11) with c = δ . Next consider the case D > t ) ≥ t ≥ − A / √ D ≤
0. Write¯∆( t ) = ( t − ν )( t − ν )( t − ν )where ν , ν = − A / ± √ D ≤
0. If | ν | ≥ δρ then δ − ( t − ν ) ≥ t + ρ asabove and t − ν i = t + | ν i | ≥ t then (3.11) with c = δ . Consider the case D < ν , ν = − A / ± i p | D | . If | ν | ≥ δρ then (3.11) holds because | t − ν i | ≥ | t + A / | ≥ t . If | ν | = | ν | ≥ δρ then A / p | D | ≥ δρ . Since( t − ν )( t − ν ) = ( t + A / + | D | ≥ (cid:0) t + A / p | D | (cid:1) / ≥ ( t + δρ ) / ≥ δ t ( t + ρ ) / c = δ/ A <
0. In this case, using ψ = − ( ν + a ) / >
0, one canwrite ¯∆( t ) = ( t − ν ) (cid:0) ( t − ψ ) − D (cid:1) . Consider the case | ν | ≥ δρ . Note that D ≤ ψ + √ D > t ) and a contradiction. Then( t − ψ ) − D = ( t − ψ ) + | D | ≥ ( t − ψ ) . Recalling t − ν = t + | ν | ≥ δ ( t + ρ ) we get(3.12) ¯∆( t, X, ǫ ) ≥ c ( t − ψ ) ( t + ρ )with c = δ . Consider the case | ν | = | ν | = (cid:12)(cid:12) ψ ± i p | D | (cid:12)(cid:12) = p ψ + | D | ≥ δρ sothat ( t − ν )( t − ν ) = ( t − ψ ) + | D | ≥ (cid:0) | t − ψ | + p | D | (cid:1) / . Assume ψ ≥ p | D | so that √ ψ ≥ δρ . For 0 ≤ t ≤ ψ/ t ≤ | t − ψ | and ψ/ ≤ | t − ψ | one has | t − ψ | = (1 − γ ) | t − ψ | + γ | t − ψ | ≥ (1 − γ ) t + γψ/ ≥ (1 − γ ) t + ( γδ/ √ ρ ≥ δ (2 √ δ ) − ( t + ρ )with γ = 2 √ / (2 √ δ ). Since | t − ψ | + p | D | ≥ | t − ψ | ≥ t and | t − ν | = t + | ν | ≥ t it is clear that (3.11) holds with c = δ/ √ δ ). For ψ/ ≤ t suchthat | t − ψ | ≤ t one sees t − ν ≥ t = (1 − γ ) t + γt ≥ (1 − γ ) t + γψ/ ≥ δ (2 √ δ ) − ( t + ρ )and hence ( t − ν ) (cid:0) ( t − ψ ) + | D | (cid:1) ≥ c ( t + ρ )( t − ψ ) which is (3.12) with c = δ/ √ δ ). Next assume p | D | ≥ ψ so that √ p | D | ≥ δρ . For 0 ≤ t ≤ ψ/ | t − ψ | ≥ t and hence | t − ψ | + p | D | ≥ t + δρ/ √ ≥ ( δ/ √ t + ρ ) . | t − ν | = t + | ν | ≥ t it is clear that (3.11) holds with c = δ/ √
2. For ψ/ ≤ t we see that | t − ψ | + p | D | ≥ t − | ψ | + p | D | ≥ t, | t − ψ | + p | D | ≥ p | D | ≥ δρ/ √ | t − ψ | + p | D | ≥ δ ( √ δ ) − ( t + ρ ). Recalling | t − ν | = t + | ν | ≥ t again one has (3.11) with c = δ/ √ δ ). Thus the proof iscompleted. Lemma 3.6.
One can find a neighborhood U of ( ¯ X, and ǫ > , C ∗ > suchthat (3.13) | ∂ t ∆( t, X, ǫ ) | ∆( t, X, ǫ ) ≤ C ∗ (cid:16) t + 1 | t − ψ | + √ a ǫ (cid:17) , ( X, ǫ ) ∈ U holds for t ∈ (0 , T ] and < ǫ ≤ ǫ .Proof. It will suffice to show (3.13) for ∆( t, X, √ ǫ ) which we denote by ˜∆( t, X, ǫ ).It is clear that˜∆ = ∆ + 4 e (cid:0) t + ρ ) ǫ + 3( t + ρ ) ǫ + ǫ (cid:1) = ∆ + ∆ r . Writing ˜∆ = ˜ e (cid:0) ¯∆ + ¯∆ r (cid:1) it suffices to show the assertion for ¯∆ + ¯∆ r instead of˜∆. Note that(3.14) | ∂ t ¯∆ r | ¯∆ r ≤ C (cid:0) t + ρ (cid:1) ≤ C ′ t always holds. Write ¯∆ = ( t − ν )( t − ν )( t − ν ) and note that ∂ t ¯∆¯∆ = X j =1 t − ν j . Checking the proof of Proposition 3.1 it is easy to see that (cid:12)(cid:12) ∂ t ¯∆ / ¯∆ (cid:12)(cid:12) ≤ C/t when A ≥
0. Therefore | ∂ t ˜∆ | ˜∆ ≤ | ∂ t ¯∆ | ¯∆ + ¯∆ r + | ∂ t ¯∆ r | ¯∆ + ¯∆ r ≤ | ∂ t ¯∆ | ¯∆ + | ∂ t ¯∆ r | ¯∆ r proves the assertion. Study the case that A <
0. From the proof of Proposition3.1 one can write ¯∆ = ( t − ν ) (cid:0) ( t − ψ ) − D (cid:1) where ψ > D ≤
0. If | D | ≥ aǫ then | t − ψ | ( t − ψ ) + | D | ≤ t − ψ ) + aǫ ) / ≤ √ | t − ψ | + √ a ǫ | t − ν | = t + | ν | ≥ t . Similarly if | t − ψ | ≥ √ a ǫ one has | t − ψ | ( t − ψ ) + | D | ≤ | t − ψ | + √ a ǫ . If | D | < aǫ and | t − ψ | < √ a ǫ it follows that | ∂ t ¯∆ | ≤ ( t − ψ ) + | D | + 2 | t − ν || t − ψ | ≤ aǫ + Ca / ǫ because | t − ν | ≤ Ca . In view of C ¯∆ r ≥ a ǫ one concludes that | ∂ t ¯∆ | ¯∆ + ¯∆ r ≤ | ∂ t ¯∆ | ¯∆ r ≤ C aǫ + Ca / ǫa ǫ ≤ C (cid:16) a + 1 √ a ǫ (cid:17) ≤ C ′ (cid:16) t + 1 | t − ψ | + √ a ǫ (cid:17) which together with (3.14) proves the assertion. In the preceding Sections 3.1 and 3.2 all symbols we have studied are definedin some conic (in ξ ) neighborhood of ( X, ǫ ) = ( ¯ X,
0) or X = ¯ X . In this sectionwe extend such symbols to those on R d × R d following [23] (also [24]). Let ¯ X = (0 , ¯ ξ ) with | ¯ ξ | = 1. Let χ ( s ) ∈ C ∞ ( R ) be equal to 1 in | s | ≤
1, vanishesin | s | ≥ ≤ χ ( s ) ≤
1. Define y ( x ) and η ( ξ ) by y j ( x ) = χ ( M x j ) x j , η j ( ξ ) = χ ( M ( ξ j h ξ i − γ − ¯ ξ j ))( ξ j − ¯ ξ j h ξ i γ ) + ¯ ξ j h ξ i γ for j = 1 , , . . . , d with h ξ i γ = ( γ + | ξ | ) / where M and γ are large positive parameters constrained(4.1) γ ≥ M . It is easy to see that (1 − CM − ) h ξ i γ ≤ | η | ≤ (1 + CM − ) h ξ i γ and(4.2) | y | ≤ CM − , | η/ | η | − ¯ ξ | ≤ CM − with some C > y, η ) is contained in a conic neighborhood of (0 , ¯ ξ ),shrinking with M . Note that ( y, η ) = ( x, ξ ) on the conic neighborhood of (0 , ¯ ξ )(4.3) W M = (cid:8) ( x, ξ ) | | x | ≤ M − , | ξ j / | ξ | − ¯ ξ j | ≤ M − / , | ξ | ≥ γM (cid:9) (cid:12)(cid:12)(cid:12) ξ j h ξ i γ − ¯ ξ j (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) ξ j h ξ i γ − ξ j | ξ | (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ξ j | ξ | − ¯ ξ j (cid:12)(cid:12)(cid:12) ≤ M − |h ξ i γ − | ξ ||h ξ i γ ≤ M − γ h ξ i γ ( h ξ i γ + | ξ | ) ≤ M − if ( x, ξ ) ∈ W M where δ ij is the Kronecker’s delta. Define extensions α ( x, ξ ), a ( t, x, ξ ), b ( t, x, ξ ), ∆( t, x, ξ ), ¯∆( t, x, ξ ), . . . of α ( X ), a ( t, X, ǫ ), b ( t, X ), ∆( t, X, ǫ ),¯∆( t, X, ǫ ), . . . by α ( x, ξ ) = α ( y ( x ) , η ( ξ )) , a ( t, x, ξ ) = a ( t, y ( x ) , η ( ξ ) , ǫ ( ξ )) ,b ( t, x, ξ ) = b ( t, y ( x ) , η ( ξ )) , ∆( t, x, ξ ) = ∆( t, y ( x ) , η ( ξ ) , ǫ ( ξ )) , ¯∆( t, x, ξ ) = ¯∆( t, y ( x ) , η ( ξ ) , ǫ ( ξ ))so on with(4.4) ǫ ( ξ ) = M / h ξ i − / γ . In view of (4.1) and (4.2) such extended symbols are defined in R d × R d . Let G = M ( | dx | + h ξ i − γ | dξ | ) . Then it is easy to see(4.5) y j ∈ S ( M − , G ) , η j − ¯ ξ j h ξ i γ ∈ S ( M − h ξ i γ , G ) , ǫ ( ξ ) ∈ S ( M − , G )for j = 1 , . . . , d . To avoid confusions we denote h η ( ξ ) i γ by [ ξ ] hence(4.6) [ ξ ] ∈ S ( h ξ i γ , G ) , [ ξ ] h ξ i − γ − ∈ S ( M − , G ) . Lemma 4.1.
Let f ( X, ǫ ) be a symbol defined in a conic neighborhood of ( ¯ X, which is homogeneous of degree in ξ . If ∂ αx ∂ βξ ∂ kǫ f ( ¯ X,
0) = 0 for ≤ | α + β | + k < r then f ( x, ξ ) = f ( y ( x ) , η ( ξ ) , ǫ ( ξ )) ∈ S ( M − r , G ) . Let h ( X ) be a symboldefined in a conic neighborhood of ¯ X which is homogeneous of degree in ξ .Then h ( x, ξ ) − h (0 , ¯ ξ ) ∈ S ( M − , G ) . Proof.
We prove the first assertion. By the Taylor formula one can write f ( y, η, ǫ ) = X | α + β | + k = r α ! β ! k ! y α ( η − ¯ ξ h ξ i γ ) β ǫ k ∂ αx ∂ βξ ∂ kǫ f (0 , ¯ ξ h ξ i γ , r + 1) X | α + β | + k = r +1 h α ! β ! k ! y α ( η − ¯ ξ h ξ i γ ) β ǫ k × Z (1 − θ ) r ∂ αx ∂ βξ ∂ kǫ f ( θy, θ ( η − ¯ ξ h ξ i γ ) + ¯ ξ h ξ i γ , θǫ ) dθ i .
14t is clear that y α ( η − ¯ ξ h ξ i γ ) β ǫ k ∂ αx ∂ βξ ∂ kǫ f (0 , ¯ ξ, h ξ i −| β | γ ∈ S ( M − r , G )for | α + β | + k = r in view of (4.5). Since h ξ i γ /C ≤ | θ ( η − ¯ ξ h ξ i γ )+ ¯ ξ h ξ i γ | ≤ C h ξ i γ the integral belongs to S ( h ξ i −| β | γ , G ) hence the second term on the right-handside is in S ( M − r − , G ) thus the assertion. From now on it is assumed that all constants are independent of M and γ ifotherwise stated. We write A - B if A is bounded by constant, independent of M and γ , times B . Recall ρ ( X, ǫ ) = α ( X ) + ǫ so that(4.7) ρ ( x, ξ ) = α ( x, ξ ) + M h ξ i − γ . From Lemma 4.1 we see ρ ∈ S ( M − , G ) hence | ∂ αx ∂ βξ ρ | - h ξ i −| β | γ for | α + β | = 2.Since ρ ≥ (cid:12)(cid:12) ∂ αx ∂ βξ ρ (cid:12)(cid:12) - √ ρ h ξ i −| β | γ ( | α + β | = 1) . Lemma 4.2.
Assume | a ( X, ǫ ) | ≤ Cρ ( X, ǫ ) n with some n > in a conic neigh-borhood of ( ¯ X, and a ( X, ǫ ) is of homogeneous of degree in ξ . Then thereexists C αβ > such that (4.9) (cid:12)(cid:12) ∂ αx ∂ βξ a ( x, ξ ) (cid:12)(cid:12) ≤ C αβ ρ ( x, ξ ) n −| α + β | / h ξ i −| β | γ . Proof.
From the assumption it follows that ∂ αx ∂ βξ ∂ kǫ a (0 , ¯ ξ,
0) = 0 for | α + β | + k < n and hence Lemma 4.1 shows that a ( x, ξ ) ∈ S ( M − n , G ). Therefore for | α + β | ≥ n one sees (cid:12)(cid:12) h ξ i | β | γ ∂ αx ∂ βξ a ( x, ξ ) (cid:12)(cid:12) ≤ CM | α + β |− n ≤ C (cid:0) C ρ − (cid:1) | α + β | / − n = CC | α + β | / − n ρ n −| α + β | / because M ≤ C ρ − . Hence (4.9) holds for | α + β | ≥ n . The case | α + β | ≤ n − X = ( x, ξ ), Y = ( y, η h ξ i γ ) and applyingthe Taylor formula to obtain (cid:12)(cid:12) a ( X + sY ) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) n − X j =0 s j j ! d j a ( X ; Y ) + s n (2 n )! d n a ( X + sθY ; Y ) (cid:12)(cid:12)(cid:12) ≤ C (cid:16) n − X j =0 s j j ! d j ρ ( X ; Y ) + s n (2 n )! d n ρ ( X + sθ ′ Y ; Y ) (cid:17) n (4.10)with some 0 < θ, θ ′ < d j a ( X ; Y ) = X | α + β | = j j ! α ! β ! ∂ αx ∂ βξ a ( x, ξ ) y α η β h ξ i | β | γ . ρ ( x, ξ ) = 0 then ∂ αx ∂ βξ ρ ( x, ξ ) = 0 for | α + β | = 1 because ρ ≥ ∂ αx ∂ βξ a ( x, ξ ) = 0 for | α + β | ≤ n − s > ρ ( x, ξ ) = 0. If ρ ( x, ξ ) ≥ s then onehas (cid:12)(cid:12) ∂ αx ∂ βξ a ( x, ξ ) h ξ i | β | γ (cid:12)(cid:12) ≤ C αβ M − n +2 | α + β | ≤ C αβ ≤ C αβ s − n + | α + β | / s n −| α + β | / ≤ C αβ s − n + | α + β | / ρ n −| α + β | / which proves (4.9). Assume 0 < ρ ( x, ξ ) < s . Note that (cid:12)(cid:12) d n a ( X + sθY ; Y ) (cid:12)(cid:12) ≤ C, d n ρ ( X + sθ ′ Y ; Y ) ≤ Cρ ( X ) − n for any | ( y, η ) | ≤ /
2. Indeed the first one is clear from a ( x, ξ ) ∈ S ( M − n , G ).To check the second inequality it is enough to note that for | α + β | = 2 n (cid:12)(cid:12) ∂ αx ∂ βξ ρ ( X + θ ′′ Y ) (cid:12)(cid:12) ≤ CM − n h ξ + θ ′′ h ξ i γ η i −| β | γ ≤ C ′ ( C ρ ( X ) − ) n − h ξ i −| β | γ since √ h ξ + θ h ξ i γ η i γ ≥ h ξ i γ / | η | ≤ / | θ | <
1. Take s = ρ ( X ) / in(4.10) to get (cid:12)(cid:12)(cid:12) n − X j =0 j ! d j a ( X ; Y ) ρ ( X ) j/ (cid:12)(cid:12)(cid:12) ≤ C (cid:16) n − X j =0 j ! d j ρ ( X ; Y ) ρ ( X ) j/ (cid:17) n + Cρ ( X ) n which is bounded by Cρ ( X ) n because | dρ ( X ; Y ) | ≤ C ′′ ρ ( X ) / in view of (4.8)and | d j ρ ( X ; Y ) | ≤ CM − j ≤ C ( C ρ − ( X )) j/ − for j ≥
3. This gives (cid:12)(cid:12)(cid:12) n − X j =1 j ! d j a ( X ; Y ) ρ ( X ) j/ ρ ( X ) n (cid:12)(cid:12)(cid:12) ≤ C . Replacing ( y, η ) by s ( y, η ), | ( y, η ) | = 1 /
2, 0 < | s | < (cid:12)(cid:12)(cid:12) n − X j =1 s j j ! d j a ( X ; Y ) ρ ( X ) j/ ρ ( X ) n (cid:12)(cid:12)(cid:12) ≤ C . Since two norms sup | s |≤ | p ( s ) | and max {| c j |} on the vector space consistingof all polynomials p ( s ) = P n − j =0 c j s j are equivalent one obtains | d j a ( X ; Y ) | ≤ B ′ ρ ( X ) n − j/ . Since | ( y, η ) | = 1 / Corollary 4.1.
On has (cid:12)(cid:12) ∂ αx ∂ βξ ρ ( x, ξ ) (cid:12)(cid:12) - ρ ( x, ξ ) −| α + β | / h ξ i −| β | γ . Lemma 4.3.
Let s ∈ R . Then (cid:12)(cid:12) ∂ αx ∂ βξ ρ s (cid:12)(cid:12) - ρ s −| α + β | / h ξ i −| β | γ . roof. Since ∂ αx ∂ βξ ρ s = X C α ( j ) β ( j ) ρ s (cid:16) ∂ α (1) x ∂ β (1) ξ ρρ (cid:17) · · · (cid:16) ∂ α ( k ) x ∂ β ( k ) ξ ρρ (cid:17) the assertion follows from Corollary 4.1. Lemma 4.4.
Let a j ( x, ξ ) = a j ( y ( x ) , η ( ξ ) , ǫ ( ξ )) . Then (cid:12)(cid:12) ∂ αx ∂ βξ a j ( x, ξ ) (cid:12)(cid:12) - ρ ( x, ξ ) j −| α + β | / h ξ i −| β | γ , j = 1 , , . Proof.
The assertion follows from Lemmas 3.3 and 4.2.For the extension ψ ( x, ξ ) = ψ ( y ( x ) , η ( ξ ) , ǫ ( ξ )) of ψ ( X, ǫ ) we have
Lemma 4.5.
One has (cid:12)(cid:12) ∂ αx ∂ βξ ψ ( x, ξ ) (cid:12)(cid:12) ≤ C αβ ρ ( x, ξ ) −| α + β | / h ξ i −| β | γ .Proof. Since Lemma 4.2 is not available for ψ ( X, ǫ ) because it is not defined for ǫ = 0 then we show the assertion directly. Let ν ( x, ξ ), a ( x, ξ ) and ¯∆( t, x, ξ ) beextensions of ν ( X, ǫ ), a ( X, ǫ ) and ¯∆( t, X, ǫ ) and hence one has ¯∆( ν ( x, ξ ) , x, ξ ) =0. Note that | ∂ t ¯∆( ν , x, ξ ) | ≥ c ρ ( x, ξ ) if ν ( x, ξ ) + a ( x, ξ ) < c ρ ( x, ξ )thanks to Lemma 3.5. Starting with ∂ t ¯∆( ν , x, ξ ) ∂ αx ∂ βξ ν + ∂ αx ∂ βξ ¯∆( ν , x, ξ ) = 0 ( | α + β | = 1)a repetition of the same argument proving the estimates for λ in Lemma 6.3below together with Lemma 4.4 one obtains(4.11) (cid:12)(cid:12) ∂ αx ∂ βξ ν (cid:12)(cid:12) - ρ −| α + β | / h ξ i −| β | γ , ν + a < c ρ. Here we have used | ν | - ρ which also follows from Lemma 4.4. Using (4.11)and Lemmas 4.3 and 4.4 the assertion follows easily. (3.10) In this subsection we work near (0 , ¯ ξ ) and ( x, ξ ) varies in a neighborhood of(0 , ¯ ξ ). First note that p has no triple characteristic root in t > t + α ( x, ξ ) > t >
0. Define¯ ψ ( x, ξ ) = − ( ν ( x, ξ,
0) + a ( x, ξ, / C ∆( t, x, ξ, ≥ min (cid:8) t , ( t − ¯ ψ ( x, ξ )) (cid:9) ( t + α ( x, ξ )) ( t ≥ . Assume that p has a double characteristic root at ( t, x, ξ ) with t >
0. Denotingby µ ( t, x, ξ ) the other characteristic root of p , which is simple and hence smoothin ( t, x, ξ ) near the reference point, one can write p ( t, x, τ, ξ ) = τ − a ( t, x, ξ ) | ξ | τ − b ( t, x, ξ ) | ξ | = ( τ − µ ( t, x, ξ )) (cid:0) τ + c ( t, x, ξ ) τ + c ( t, x, ξ ) (cid:1) . t, x, ξ, | ξ | = (cid:0) a − b (cid:1) | ξ | = (2 c + c ) (cid:0) c − c (cid:1) where ∆ = ( c − c ) / τ + c τ + c . Since µ + c µ + c = 0hence 2 c + c = 0 it follows from (4.12) that { ( t, x, ξ ) | ∆ = 0 , t > } ⊂ { ( t, x, ξ ) | t = ¯ ψ ( x, ξ ) > } . Note that ¯ ψ ( x, ξ ) > ν ( x, ξ, a ( x, ξ, < ≤ c α ( x, ξ ) hence¯ ψ ( x, ξ ) is smooth there and(4.13) (cid:12)(cid:12) ν ( x, ξ, − ν j ( x, ξ, (cid:12)(cid:12) ≥ c α ( x, ξ ) , j = 2 , Lemma 4.6.
Near ¯ X = (0 , ¯ ξ ) the doubly characteristic set of p with t > iscontained in { ( t, x, ξ ) | t = ¯ ψ ( x, ξ ) > } and t − ¯ ψ ( x, ξ ) is a time function for p . It remains to show that t − ¯ ψ is a time function (see e.g. [24]) for p . Let q = τ + c τ + c then F p = c F q at a double characteristic with t > c = 0 then it is enough to prove that t − ¯ ψ is a time function for q . Write q = (cid:0) τ + c / (cid:1) − ∆ and recall [24, Lemma 2.1.3] that t − ¯ ψ is a time function for q if and only if(4.14) { τ + c / , t − ¯ ψ } > , { ∆ , t − ¯ ψ } ≤ c { τ + c / , t − ¯ ψ } ∆ with some 0 < c <
1. Since ∆ ≥ |{ ∆ , t − ¯ ψ }| = |{ ∆ , ¯ ψ }| ≤ C √ ∆ |∇ ¯ ψ | . Taking (4.13) into account, a repetition of the proof of Lemma 4.5shows |∇ ¯ ψ | ≤ C ′ √ α and hence |{ ∆ , t − ¯ ψ }| ≤ Cα ∆ . On the other hand onehas { τ + c / , t − ¯ ψ } = 1 − { c , ¯ ψ } / ≥ − C |∇ ¯ ψ | ≥ − C ′′ √ α then (4.14)holds because α can be assumed to be small there. Recall that α ( x, ξ ), a ( t, x, ξ ), b ( t, x, ξ ), e ( t, x, ξ ), ∆( t, x, ξ ), . . . are extensions of α ( X ), a ( t, X, ǫ ), b ( t, X ), e ( t, X, ǫ ), ∆( t, X, ǫ ), . . . defined in Section 4.2 so that p = τ − a ( t, x, ξ ) | ξ | τ + b ( t, x, ξ ) | ξ | , a = e ( t, x, ξ ) (cid:0) t + α ( x, ξ ) (cid:1) is now defined in R d × R d and coincides with the original p in a conic neighbor-hood W M of (0 , ¯ ξ ). We add a term 2 M e ( t, x, ξ ) h ξ i − γ to p and consider τ − e ( t + α + 2 M h ξ i − γ ) | ξ | τ − b | ξ | . Denoting(4.15) a M ( t, x, ξ ) = e ( t, x, ξ )( t + α ( x, ξ ) + 2 M h ξ i − γ )18onsider the discriminant∆ M ( t, x, ξ ) = 4 e (cid:0) t + α + 2 M h ξ i − γ (cid:1) − b = 4 e (cid:0) t + α + M h ξ i − γ ) − b + ∆ r ( t, x, ξ )(4.16)where, recalling α + M h ξ i − γ = ρ , we have∆ r = 4 e (cid:0) t + ρ ) M h ξ i − γ + 3( t + ρ ) M h ξ i − γ + M h ξ i − γ (cid:1) = 12 e (cid:0) c ( x, ξ ) t + c ( x, ξ ) t + c ( x, ξ ) (cid:1) ≥ e M ( t + ρ ) h ξ i − γ where it is clear that c j ( x, ξ ) verifies (cid:12)(cid:12) ∂ αx ∂ βξ c j (cid:12)(cid:12) - ρ j −| α + β | / h ξ i −| β | γ . Thanks toProposition 3.1 one has¯∆( t, x, ξ ) ≥ υ min { t , ( t − ψ ) } ( t + ρ ) . Since ∆( t, x, ξ ) = ˜ e ¯∆ then∆( t, x, ξ ) = ˜ e ¯∆ ≥ ˜ e υ min { t , ( t − ψ ) } ( t + ρ ) ≥ (cid:0) ˜ e/e (cid:1) υ min { t , ( t − ψ ) } e ( t + ρ ) . (4.17)Therefore choosing a constant ¯ ν > e ≥ (cid:0) ˜ e/e (cid:1) υ ¯ ν one obtains from(4.16), (4.17) that∆ M ≥ (cid:0) ˜ e/e (cid:1) υ min (cid:8) t , ( t − ψ ) (cid:9) e ( t + ρ ) + 12 e ( t + ρ ) M h ξ i − γ ≥ (cid:0) ˜ e/e (cid:1) υ (cid:0) min (cid:8) t , ( t − ψ ) (cid:9) + ¯ ν ( t + ρ ) M h ξ i − γ (cid:1) e ( t + ρ ) ≥ (cid:0) ˜ e/e (cid:1) υ min (cid:8) t , ( t − ψ ) + ¯ νM ρ h ξ i − γ (cid:9) e ( t + ρ ) ( t ≥ . (4.18) Proposition 4.1.
One can write ∆ M = e (cid:0) t + a ( x, ξ ) t + a ( x, ξ ) t + a ( x, ξ ) (cid:1) where < e ∈ S (1 , G ) uniformly in t and a j satisfies (4.19) (cid:12)(cid:12) ∂ αx ∂ βξ a j (cid:12)(cid:12) - ρ j −| α + β | / h ξ i −| β | γ . Moreover there exist ¯ ν > and c > such that (4.20) ∆ M a M ≥ ˜ e e υ min (cid:8) t , ( t − ψ ) + ¯ νM ρ h ξ i − γ (cid:9) , ∆ M a M ≥ c M h ξ i − γ a M for ≤ t ≤ T where ψ and ρ satisfy (4.21) (cid:12)(cid:12) ∂ αx ∂ βξ ψ (cid:12)(cid:12) , (cid:12)(cid:12) ∂ αx ∂ βξ ρ (cid:12)(cid:12) - ρ −| α + β | / h ξ i −| β | γ . Proof.
Choosing ǫ = √ M / h ξ i − / γ in (4.4) and applying Lemma 3.1 onecan write ∆ M as a third order polynomial in t , up to non-zero factor and canestimate the coefficients thanks to Lemmas 3.3 and 4.2 in terms of α + 2 M h ξ i − γ .Noting ρ ( x, ξ ) ≤ α ( x, ξ ) + 2 M h ξ i − γ ≤ ρ ( x, ξ ) we have (4.19). The assertion(4.20) follows from (4.18) for a M ≤ e ( t + ρ ). The estimates (4.21) are nothingbut Corollary 4.1 and Lemma 4.5 with the choice ǫ = M / h ξ i − / γ .19e estimate the ratio of ∂ t b to √ a M for later use. Lemma 4.7.
We have (cid:12)(cid:12) ∂ t b (cid:12)(cid:12) ≤ (1 + CM − ) (cid:0) p / (cid:1) | e (0 , , ¯ ξ ) |√ a M (0 ≤ t ≤ M − ) . Proof.
Write b = β ( x, ξ ) + tβ ( x, ξ ) + t β ( t, x, ξ ). From 27 b ≤ a for 0 ≤ t ≤ T it is clear that | β | ≤ (2 / √ e / α / . We first check that(4.22) | β | ≤ (1 + CM − )(2 / √ e / √ α. If α ( x, ξ ) = 0 then β ( x, ξ ) = 0 by 27 b ≤ a hence (4.22) is clear. When α ( x, ξ ) > t = 3 α it follows from 27 b ≤ a that3 α | β | ≤ / e / / √ α / + | β | + Cα ≤ (6 / √ e / α / + Cα ≤ (6 / √ CM − ) e / α / because α ≤ CM − which proves (4.22). Since | ∂ t b | ≤ | β | + Ct we see that | ∂ t b | ≤ (1 + CM − )(2 / √ e / √ α + CM − √ t ≤ (1 + CM − )(2 / √ e / (cid:0) √ α + √ t (cid:1) from which the proof is immediate. Remark 4.1.
Here we make a remark on e (0 , , ¯ ξ ) = ∂ t a (0 , , ¯ ξ ). In viewof (3.1) it is clear that e (0 , , ¯ ξ ) is the nonzero positive real eigenvalue of F p (0 , , , ¯ ξ ). Since ˜ e/e = 4 e (0 , , ¯ ξ ) (cid:0) O ( M − ) (cid:1) the coefficienet of the right-hand side of (4.20) is, essentially, constant times the square of the nonzeropositive real eigenvalue of the Hamilton map.In what follows we denote ¯ e = e (0 , , ¯ ξ ). g and estimates of ω and φ Introduce the metric g = g ( x,ξ ) ( dx, dξ ) = M − (cid:0) h ξ i γ | dx | + h ξ i − γ | dξ | (cid:1) which is a basic metric with which we work in this paper. Note that S ( M s , G ) ⊂ S ( M s , g )because M s +2 | α + β | h ξ i −| β | γ ≤ M s M −| α + β | / h ξ i ( | α |−| β | ) / γ in view of h ξ i γ ≥ γ ≥ M . The metric g is slowly varying and σ temperate (see [6, Chapter 18.5], inwhat follows we omit “ σ ” because we use only the Weyl calculus in this paper) uniformly in γ ≥ M ≥ emma 5.1. For | α + β | ≥ one has (cid:12)(cid:12) ∂ αx ∂ βξ ψ (cid:12)(cid:12) - M / ρ / h ξ i − / γ M −| α + β | / h ξ i ( | α |−| β | ) / γ . Proof.
It is enough to remark (cid:12)(cid:12) ∂ αx ∂ βξ ψ (cid:12)(cid:12) - ρ −| α + β | / h ξ i −| β | γ - ρ / ρ − ( | α + β |− / h ξ i −| β | γ - ρ / ( M − h ξ i γ ) ( | α + β |− / h ξ i −| βγ = M / ρ / h ξ i − / γ M −| α + β | / h ξ i ( | α |−| β | ) / γ which proves the assertion. Corollary 5.1.
For | α + β | ≥ one has ∂ αx ∂ βξ ψ ∈ S ( M − ( | α + β |− / ρ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ , g ) . ω by metric g Taking Proposition 4.1 into account we introduce a preliminary weight ω ( t, x, ξ ) = q ( t − ψ ( x, ξ )) + ¯ νM ρ h ξ i − γ . Since the exact value of ¯ ν > ν = 1 from now on. In what follows we work with symbols dependingon t . We assume that t varies in some fixed interval [0 , T ] and it is assumedthat all constants are independent of t ∈ [0 , T ] and γ , M if otherwise stated.Now A - B implies that A is bounded by constant, independent of t , M and γ ,times B . Lemma 5.2.
Let s ∈ R . For | α + β | ≥ we have (cid:12)(cid:12) ∂ αx ∂ βξ ω s (cid:12)(cid:12) - ω s ( ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ . Proof.
Recall ω = ( t − ψ ) + M ρ h ξ i − γ . Note that for | α + β | ≥ (cid:12)(cid:12) ∂ αx ∂ βξ ( t − ψ ) (cid:12)(cid:12) - ω | ∂ αx ∂ βξ ψ | + X | ∂ α ′ x ∂ β ′ ξ ψ || ∂ α ′′ ∂ β ′′ ξ ψ | - ωρ −| α + β | / h ξ i −| β | γ + ρ −| α + β | / h ξ i −| β | γ - ω (cid:8) ω − ρ / ρ − ( | α + β |− / + ω − ρρ − ( | α + β |− / (cid:9) h ξ i −| β | γ - ω (cid:8) ω − ρ / ( M − h ξ i γ ) ( | α + β |− / + ω − ρ ( M − h ξ i γ ) ( | α + β |− / (cid:9) h ξ i −| β | γ - ω ( ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ since ω ≥ √ M ρ / h ξ i − / γ . When | α + β | = 1 there is no second term and hence (cid:12)(cid:12) ∂ αx ∂ βξ ( t − ψ ) (cid:12)(cid:12) - ωρ / − ( | α + β |− / h ξ i −| β | γ - ω ( ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ . | α + β | ≥ (cid:12)(cid:12) ∂ αx ∂ βξ ( M ρ h ξ i − γ ) (cid:12)(cid:12) - M ρ h ξ i − γ ρ −| α + β | / h ξ i −| β | γ - M ρ / h ξ i − γ ρ − ( | α + β |− / h ξ i −| β | γ - ω (cid:0) M ω − ρ / h ξ i − γ (cid:1)(cid:0) M − h ξ i γ (cid:1) ( | α + β |− / h ξ i −| β | γ - ω ( ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ because ω ≥ √ M ρ / h ξ i − / γ ≥ M h ξ i − γ . Therefore one concludes that (cid:12)(cid:12) ∂ αx ∂ βξ ω (cid:12)(cid:12) - ω ( ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ which proves the assertion for s = 2. For general s noting (cid:12)(cid:12) ∂ αx ∂ βξ ( ω ) s/ (cid:12)(cid:12) - X (cid:12)(cid:12) ( ω ) s/ − l (cid:0) ∂ α x ∂ β ξ ω (cid:1) · · · (cid:0) ∂ α l x ∂ β l ξ ω (cid:1)(cid:12)(cid:12) - X ω s (cid:0) ω − ρ / h ξ i − / γ ) l M − ( | α + β |− l ) / h ξ i ( | α |−| β | ) / γ - ω s (cid:0) ω − ρ / h ξ i − / γ ) M − ( | α + β |− / h ξ i ( | α |−| β | ) / γ for ω − ρ / h ξ i − / γ ≤ M − / ≤ Corollary 5.2.
We have ω s ∈ S ( ω s , g ) for s ∈ R . Corollary 5.3.
For | α + β | ≥ one has ∂ αx ∂ βξ ω s ∈ S (cid:0) M − ( | α + β |− / ω s ω − ρ / h ξ i − / | α |− β | ) / γ , g (cid:1) . φ by metric g Introduce a wight which plays a crucial role in deriving energy estimates φ ( t, x, ξ ) = ω ( t, x, ξ ) + t − ψ ( x, ξ ) . Start with remarking
Lemma 5.3.
There is
C > such that φ ( t, x, ξ ) ≥ M h ξ i − γ /C .Proof. When t − ψ ( x, ξ ) ≥ φ ≥ ω hence φ ≥ ω ≥ M / ρ / h ξ i − / γ ≥ M h ξ i − γ is obvious for ρ ≥ M h ξ i − γ . Assume t − ψ ( x, ξ ) < ≤ t < ψ ( x, ξ ) ≤ δρ ( x, ξ ) with some δ > | t − ψ ( x, ξ ) | = ψ ( x, ξ ) − t ≤ δρ ( x, ξ ) we have ω ( t, x, ξ ) = ( t − ψ ( x, ξ )) + M ρ ( x, ξ ) h ξ i − γ ≤ δ ρ + M ρ h ξ i − γ ≤ δ ρ + ρ = ( δ + 1) ρ . φ ( t, x, ξ ) ≥ M ρ h ξ i − γ ω + | t − ψ | ≥ M ρ h ξ i − γ ω the proof is immediate.Next show Lemma 5.4.
We have (cid:12)(cid:12) ∂ αx ∂ βξ φ (cid:12)(cid:12) - φM −| α + β | / h ξ i ( | α |−| β | ) / γ .Proof. Let | α + β | = 1 and write(5.2) ∂ αx ∂ βξ φ = − ∂ αx ∂ βξ ψω φ + ∂ αx ∂ βξ ( M ρ h ξ i − γ )2 ω = φ αβ φ + ψ αβ . From Corollaries 5.2 and 4.1 it follows that (cid:12)(cid:12) ∂ µx ∂ νξ (cid:0) ψ αβ (cid:1)(cid:12)(cid:12) - ω − M ρ h ξ i − γ M −| α + β + µ + ν | / h ξ i ( | α + µ |−| β + ν | ) / γ . Noting (5.1) one obtains (cid:12)(cid:12) ∂ µx ∂ νξ (cid:0) ψ αβ (cid:1)(cid:12)(cid:12) - φM −| α + β + µ + ν | / h ξ i ( | α + µ |−| β + ν | ) / γ . On the other hand thanks to Corollaries 5.1 and 5.2 one sees | ∂ µx ∂ νξ φ αβ | - M −| α + β + µ + ν | / h ξ i ( | α + µ |−| β + ν | ) / γ . Hence using (5.2) the assertion is proved by induction on | α + β | .We refine this lemma. Lemma 5.5.
Let | α + β | ≥ then ∂ αx ∂ βξ φ ∈ S ( φ M − ( | α + β |− / ω − ρ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ , g ) . Proof.
From Corollary 5.1 one has ∂ αx ∂ βξ ψ ∈ S ( ρ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ , g ) for | α + β | = 1 hence φ αβ ∈ S ( ω − ρ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ , g ) for | α + β | = 1 byCorollary 5.2. From Corollary 5.3 it follows that (cid:12)(cid:12) ∂ µx ∂ νξ (cid:0) ψ αβ (cid:1)(cid:12)(cid:12) - ω − ρ / M h ξ i − −| β | γ M −| µ + ν | / h ξ i ( | µ |−| ν | ) / γ for | α + β | = 1 because ∂ αx ∂ βξ ( M ρ h ξ i − γ ) ∈ S ( M ρ / h ξ i − −| β | γ , g ). Thanks toLemma 5.3 one sees M h ξ i − γ ≤ Cφ ( t, x, ξ ) and hence ψ αβ ∈ S ( ω − ρ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ φ, g ) , | α + β | = 1 . Since φ ∈ S ( φ, g ) by Lemma 5.4 we conclude the assertion from (5.2).23 B´ezout matrix as symmetrizer
Add − M op( e ( t, x, ξ ) h ξ i − γ )[ D ] D t to the principal part and subtract the sameone from the lower order part so that the operator is left to be invariant;ˆ P = D t − a M ( t, x, D )[ D ] D t − b ( t, x, D ) [ D ] + b ( t, x, D ) D t (cid:0) b ( t, x, D ) + d M ( t, x, D ))[ D ] D t + b ( t, x, D )[ D ] where b j ( t, x, ξ ) ∈ S (1 , G ) and d M ( t, x, ξ ) = 2 M ( e h ξ i − γ ) ξ ] ∈ S ( M, G ). Herenote that(6.1) d M ( t, x, ξ ) − M e ( t, x, ξ ) ∈ S ( M − , g )which follows from (4.6). With U = t ( D t u, [ D ] D t u, [ D ] u ) the equation ˆ P u = f is transformed to(6.2) D t U = A ( t, x, D )[ D ] U + B ( t, x, D ) U + F where F = t ( f, ,
0) and A ( t, x, ξ ) = a M b , B ( t, x, ξ ) = b b + d M b . Let S be the B´ezout matrix of p and ∂p/∂τ , that is S ( t, x, ξ ) = − a M a M b − a M b a M then S is nonnegative definite and symmetrizes A , that is SA is symmetric. Consider the principal symbol τ − a M ( t, x, ξ )[ ξ ] τ − b ( t, x, ξ )[ ξ ] of ˆ P . Denote σ ( t, x, ξ ) = t + α ( x, ξ ) + 2 M h ξ i − γ = t + ρ ( x, ξ ) + M h ξ i − γ hence a M ( t, x, ξ ) = e ( t, x, ξ ) σ ( t, x, ξ ) and (1 − CM − )¯ e σ ≤ a M ≤ (1+ CM − )¯ e σ .In what follows we assume that t varies in the interval0 ≤ t ≤ M − . Since ρ ∈ S ( M − , G ) it is clear that σ ( t, x, ξ ) ∈ S ( M − , G ). Lemma 6.1.
We have (cid:12)(cid:12) ∂ αx ∂ βξ σ (cid:12)(cid:12) - σ −| α + β | / h ξ i −| β | γ . In particular σ ∈ S ( σ, g ) .Proof. It is clear from (4.8) that (cid:12)(cid:12) ∂ αx ∂ βξ σ (cid:12)(cid:12) - √ σ h ξ i −| β | γ for | α + β | = 1. For | α + β | ≥ ρ ∈ S ( M − , G ) that (cid:12)(cid:12) ∂ αx ∂ βξ σ (cid:12)(cid:12) - M | α + β |− h ξ i −| β | γ - σ −| α + β | / h ξ i −| β | γ since Cσ − ≥ M . The second assertion is clear from σ − ≤ M − h ξ i γ .24 orollary 6.1. Let s ∈ R . Then (cid:12)(cid:12) ∂ αx ∂ βξ σ s (cid:12)(cid:12) - σ s −| α + β | / h ξ i −| β | γ . In particular σ s ∈ S ( σ s , g ) . Definition 6.1.
To simplify notations we denote by C ( σ s ) the set of symbols r ( t, x, ξ ) satisfying (cid:12)(cid:12) ∂ αx ∂ βξ r (cid:12)(cid:12) - σ s −| α + β | / h ξ i −| β | γ . It is clear that C ( σ s ) ⊂ S ( σ s , g ) because σ −| α + β | / ≤ M −| α + β | / h ξ i | α + β | / γ .It is also clear that if p ∈ C ( σ s ) with s > p ) − − ∈ C ( σ s ). Lemma 6.2.
One has a sM ∈ C ( σ s ) , ( s ∈ R ) , b ∈ C ( σ / ) , ∂ t a M ∈ C (1) , ∂ t b ∈ C ( √ σ ) . Proof.
The first assertion is clear from Corollary 6.1 because a M = eσ and e ∈ S (1 , G ), 1 /C ≤ e ≤ C . To show the second assertion, recalling b ( t, x, ξ ) isthe extension of b ( t, X ), write b ( t, x, ξ ) = b (0 , y ( x ) , η ( ξ )) + ∂ t b (0 , y ( x ) , η ( ξ )) t + Z (1 − θ ) ∂ t b ( θt, y ( x ) , η ( ξ )) dθ · t . (6.3)Since ∂ αx ∂ βξ b (0 , , ¯ ξ ) = 0 for | α + β | ≤ ∂ t b (0 , , ¯ ξ ) = 0 then thanks to Lemma4.1 one has b (0 , y ( x ) , η ( ξ )) ∈ S ( M − , G ) and ∂ t b (0 , y ( x ) , η ( ξ )) ∈ S ( M − , G ).Since 0 ≤ t ≤ M − we conclude that b ( t, x, ξ ) ∈ S ( M − , G ). Since | b | ≤ Cσ / and σ ∈ S ( M − , G ) a repetition of the same arguments proving Lemma 4.2shows the second assertion. The third assertion is clear because ∂ t a M = e +( ∂ t e ) σ . As for the last assertion, recall Lemma 4.7 that | ∂ t b | ≤ Ca / M ≤ C ′ σ / .Noting ∂ t b ∈ S ( M − , G ) which results from (6.3) one sees |h ξ i | β | γ ∂ αx ∂ βξ ∂ t b | - M | α + β |− - σ / −| α + β | / for | α + β | ≥ ≤ λ ( t, x, ξ ) ≤ λ ( t, x, ξ ) ≤ λ ( t, x, ξ )be the eigenvalues of S ( t, x, ξ ). Recall [26, Proposition 2.1] Proposition 6.1.
There exist M and K > such that ∆ M / (6 a M + 2 a M + 2 a M ) ≤ λ ≤ (cid:0) / Ka M (cid:1) a M , (2 − Ka M ) a M ≤ λ ≤ (2 + Ka M ) a M , ≤ λ ≤ Ka M provided that M ≥ M .Proof. Since a M = e σ and σ ∈ S ( M − , G ) then for any ¯ ǫ > M suchthat e M − ≤ ¯ ǫ . Then the assertion follows from [26, Proposition 2.1]. Corollary 6.2.
The eigenvalues λ i ( t, x, ξ ) are smooth in (0 , M − ] × R d × R d . .2 Estimates of eigenvalues First we prove
Lemma 6.3.
One has λ j ∈ C ( σ − j ) for j = 1 , , . Denote q ( λ ) = det ( λI − S ) so that(6.4) q ( λ ) = λ − (3 + 2 a M + a M ) λ + (6 a M + 2 a M + 2 a M − b ) λ − ∆ M . Note that(6.5) ∂ λ q ( λ i ) ∂ αx ∂ βξ λ i + ∂ αx ∂ βξ q ( λ i ) = 0 , | α + β | = 1 . Let us write ∂ αx ∂ βξ = ∂ α,βx,ξ for simplicity. We show by induction on | α + β | that ∂ λ q ( λ i ) ∂ α,βx,ξ λ i = X | µ + ν | + s ≥ C µ,ν,γ ( j ) ,δ ( j ) ,s ∂ µ,νx,ξ ∂ sλ q ( λ i ) × (cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) (6.6)where µ + P γ ( i ) = α , ν + P δ ( j ) = β and | γ ( i ) + δ ( j ) | ≥
1. The assertionholds for | α + β | = 1 by (6.5). Suppose that (6.6) holds for | α + β | = m . With | e + f | = 1 operating ∂ e,fx,ξ to (6.6) the resulting left-hand side is ∂ λ q ( λ i ) ∂ α + e,β + fx,ξ λ i + ∂ λ q ( λ i )( ∂ α,βx,ξ λ i )( ∂ e,fx,ξ λ i ) + ∂ e,fx,ξ ∂ λ q ( λ i ) ∂ α,βx,ξ λ i = ∂ λ q ( λ i ) ∂ α + e,β + fx,ξ λ i − X | µ + ν | + s ≥ C µ,ν,γ ( j ) ,δ ( j ) ,s ∂ µ,νx,ξ ∂ sλ q ( λ i ) (cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) . On the other hand, the resulting right-hand side is X C ... ∂ µ + e,ν + fx,ξ ∂ sλ q ( λ i ) (cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) + X C ... ∂ µ,νx,ξ ∂ s +1 λ q ( λ i ) (cid:0) ∂ e,fx,ξ λ i (cid:1)(cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) + s X j =1 X C ... ∂ µ,νx,ξ ∂ sλ q ( λ i ) (cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( j ) + e,δ ( j ) + fx,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) which can be written as X | µ + ν | + s ≥ C µ,ν,γ ( j ) ,δ ( j ) ,s ∂ µ,νx,ξ ∂ sλ q ( λ i ) (cid:0) ∂ γ (1) ,δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) ,δ ( s ) x,ξ λ i (cid:1) where µ + P γ ( i ) = α + e , ν + P δ ( i ) = β + f and | γ ( j ) + δ ( j ) | ≥
1. Therefore weconclude (6.6). In order to estimate ∂ α,βx,ξ λ i one needs to estimate ∂ µ,νx,ξ ∂ sλ q ( λ i ).26 emma 6.4. For any s ∈ N and α , β it holds that | ∂ α,βx,ξ ∂ sλ q ( λ j ) | - σ − j − (3 − j ) s −| α + β | / h ξ i −| β | γ , j = 1 , | ∂ α,βx,ξ ∂ sλ q ( λ ) | - σ −| α + β | / h ξ i −| β | γ . Proof.
From Proposition 6.1 and (6.4) one sees that | q ( λ i ) | - | λ i | + | a M || λ i | + | a M | , | ∂ α,βx,ξ q ( λ i ) | - (cid:0) | ∂ α,βx,ξ a M | + | ∂ α,βx,ξ b | (cid:1) | λ i | + | ∂ α,βx,ξ a M | + | ∂ α,βx,ξ b | ( | α + β | ≥ | ∆ M | - a M and | b | - a / M . Therefore thanks to Proposition 6.1 andLemma 6.2 one obtains the assertions for the case s = 0. Since | ∂ λ q ( λ i ) | - | λ i | + | a M | , | ∂ sλ q ( λ i ) | - , s ≥ , | ∂ α,βx,ξ ∂ λ q ( λ i ) | - | ∂ α,βx,ξ a M || λ i | + | ∂ α,βx,ξ a M | + | ∂ α,βx,ξ b | ( | α + β | ≥ , | ∂ α,βx,ξ ∂ λ q ( λ i ) | - | ∂ α,βx,ξ a M | , ∂ α,βx,ξ ∂ sλ q ( λ i ) = 0 , s ≥ | α + β | ≥ s ≥ ∂ λ q ( λ i ) = Q k = i ( λ i − λ k ) it follows from Proposition6.1 that(6.7) 6 a M (1 − Ca M ) ≤ (cid:12)(cid:12) ∂ λ q ( λ i ) (cid:12)(cid:12) ≤ a M (1 + Ca M ) , i = 1 , , ∂ λ q ( λ ) ≃ . Then for | α + β | = 1 one has (cid:12)(cid:12) ∂ α,βx,ξ λ j (cid:12)(cid:12) - (cid:12)(cid:12) ∂ α,βx,ξ q ( λ j ) /∂ λ q ( λ j ) (cid:12)(cid:12) - σ − j − / h ξ i −| β | γ , j = 1 , , s = 0. Assume that (cid:12)(cid:12) ∂ α,βx,ξ λ j (cid:12)(cid:12) - σ − j −| α + β | / h ξ i −| β | γ , j = 1 , , | α + β | ≤ m . Lemma 6.4 and (6.6) show that (cid:12)(cid:12) ∂ λ q ( λ ) ∂ α,βx,ξ λ (cid:12)(cid:12) - X σ − s −| µ + ν | / σ −| γ (1) + δ (1) | / · · · σ −| γ ( s ) + δ ( s ) | / h ξ i −| β | γ - X σ −| µ + ν | / σ −| γ (1) + δ (1) | / · · · σ −| γ ( s ) + δ ( s ) | / h ξ i −| β | γ - σ −| α + β | / h ξ i −| β | γ . This together with (6.7) proves the estimate for λ . The same arguments showthe assertion for λ . The estimate for λ is clear from (6.6) because of (6.7).Thus we have the assertion for | α + β | = m + 1 and the proof is completed byinduction on | α + β | .Turn to estimate ∂ t λ i . Lemma 6.5.
One has ∂ t λ ∈ C ( σ ) , ∂ t λ ∈ C (1) and ∂ t λ ∈ C (1) . roof. First examine that ∂ λ q ( λ i ) ∂ α,βx,ξ ∂ t λ i can be written as X | α ′ + β ′ | < | α + β | C ... ∂ µ,νx,ξ ∂ s +1 λ q ( λ i ) (cid:0) ∂ α ′ ,β ′ x,ξ ∂ t λ i (cid:1)(cid:0) ∂ γ (1) + δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) + δ ( s ) x,ξ λ i (cid:1) + X C ... ∂ µ,νx,ξ ∂ sλ ∂ t q ( λ i ) (cid:0) ∂ γ (1) + δ (1) x,ξ λ i (cid:1) · · · (cid:0) ∂ γ ( s ) + δ ( s ) x,ξ λ i (cid:1) (6.8)where α ′ + µ + P γ ( i ) = α , β ′ + ν + P δ ( i ) = β and | γ ( i ) + δ ( i ) | ≥
1. Indeed(6.8) is clear when | α + β | = 0 from(6.9) ∂ λ q ( λ i ) ∂ t λ i + ∂ t q ( λ i ) = 0 . Differentiating (6.9) by ∂ e,fx,ξ and repeating the same arguments proving (6.6)one obtains (6.8) by induction. To prove Lemma 6.5 first check that(6.10) | ∂ α,βx,ξ ∂ sλ ∂ t q ( λ j ) | - σ − j − (3 − j ) s −| α + β | / h ξ i −| β | γ , j = 1 , , . In fact from(6.11) ∂ t q ( λ ) = − ∂ t (2 a M + a M ) λ + ∂ t (6 a M + 2 a M + 2 a M − b ) λ − ∂ t ∆ M it follows that | ∂ t q ( λ i ) | - λ i + σ and | ∂ α,βx,ξ ∂ t q ( λ i ) | - ( λ i + σ ) σ −| α + β | / h ξ i −| β | γ for | α + β | ≥ s = 0. Since | ∂ α,βx,ξ ∂ sλ ∂ t q ( λ i ) | - σ −| α + β | / h ξ i −| β | γ for s ≥ λ by induction on | α + β | . Assume(6.12) (cid:12)(cid:12) ∂ α,βx,ξ ∂ t λ (cid:12)(cid:12) - σ −| α + β | / h ξ i −| β | γ . It is clear from (6.9), (6.7) and (6.10) that (6.12) holds for | α + β | = 0. Assumethat (6.12) holds for | α + β | ≤ m . For | α + β | = m + 1, thanks to the inductiveassumption, Lemma 6.4 and Lemma 6.3 it follows that X | α ′ + β ′ | < | α + β | (cid:12)(cid:12) ∂ µ,νx,ξ ∂ s +1 λ q ( λ ) (cid:0) ∂ α ′ ,β ′ x,ξ ∂ t λ (cid:1)(cid:0) ∂ γ (1) + δ (1) x,ξ λ (cid:1) · · · (cid:0) ∂ γ ( s ) + δ ( s ) x,ξ λ (cid:1)(cid:12)(cid:12) - X σ − s +1) −| µ + ν | / σ −| α ′ + β ′ | / σ −| γ (1) + δ (1) | / · · · σ −| γ ( s ) + δ ( s ) | / h ξ i −| β | γ - σ −| α + β | / h ξ i −| β | γ . On the other hand one sees
X (cid:12)(cid:12) ∂ µ,νx,ξ ∂ sλ ∂ t q ( λ ) (cid:0) ∂ γ (1) + δ (1) x,ξ λ (cid:1) · · · (cid:0) ∂ γ ( s ) + δ ( s ) x,ξ λ (cid:1)(cid:12)(cid:12) (cid:22) X σ − s −| µ + ν | / σ −| γ (1) + δ (1) | / · · · σ −| γ ( s ) + δ ( s ) | / h ξ i −| β | γ - σ −| α + β | / h ξ i −| β | γ in view of (6.10) and Lemma 6.3. This proves that (6.12) holds for | α + β | = m +1and hence for all α , β . As for λ , λ the proof is similar.28 .3 Eigenvectors of B´ezout matrix We sometimes denote by C ( σ s ) a function belonging to C ( σ s ). If we write n ij forthe ( i, j )-cofactor of λ k I − S then t ( n j , n j , n j ) is, if non-trivial, an eigenvectorof S corresponding to λ k . We take k = 1, j = 3 and hence a M (2 a M − λ )3 b ( λ − λ − λ − a M ) = ℓ ℓ ℓ is an eigenvector of S corresponding to λ and therefore t = t t t = 1 d ℓ ℓ ℓ , d = q ℓ + ℓ + ℓ is a unit eigenvector of S corresponding to λ . Thanks to Proposition 6.1 andrecalling b ∈ C ( σ / ) it is clear that d = q a M + C ( σ ) = 6 a M p C ( σ ) = 6 a M (1 + C ( σ )) . Therefore since ℓ = C ( σ ), ℓ = C ( σ / ) and ℓ = 6 a + C ( σ ) we have t = t t t = a M / C ( σ ) − b/ (2 a M ) + C ( σ )1 + C ( σ ) . Similarly choosing k = 2 , j = 2 and k = 3 , j = 1 − a M b ( λ − λ − a M ) − a M b ( λ − = ℓ ℓ ℓ , ( λ − a M )( λ − a M ) − b − a M b − a M ( λ − a M ) = ℓ ℓ ℓ are eigenvectors of S corresponding to λ and λ respectively and t j = t j t j t j = 1 d j ℓ j ℓ j ℓ j , d j = q ℓ j + ℓ j + ℓ j are unit eigenvectors of S corresponding to λ j , j = 2 ,
3. Thanks to Proposition6.1 it is easy to check d = 3 λ (1 + C ( σ )) , d = λ (1 + C ( σ )) . Then repeating the same arguments one concludes t t t = C ( σ / ) − C ( σ ) − b/λ + C ( σ ) , t t t = C ( σ ) C ( σ / ) − a M /λ + C ( σ ) . T = ( t , t , t ) = ( t ij ) is an orthogonal matrix which diagonalizes S ;Λ = T − ST = t T ST = λ λ
00 0 λ . Note that Λ symmetrizes A T = T − AT ; t (Λ A T ) = t ( t T SAT ) = t T t ( SA ) T = t T SAT = Λ A T . Summarize what we have proved above in
Lemma 6.6.
Let T be defined as above. Then there is M such that T has theform T = a M / C ( σ ) C ( σ / ) 1 + C ( σ ) − b/ (2 a M ) + C ( σ ) − C ( σ ) C ( σ / )1 + C ( σ ) − b/λ + C ( σ ) − a M /λ + C ( σ ) = C ( σ ) C ( σ / ) 1 + C ( σ ) C ( σ / ) − C ( σ ) C ( σ / )1 + C ( σ ) C ( σ / ) C ( σ ) for M ≥ M . In particular T, T − ∈ S (1 , g ) . Lemma 6.7.
We have ∂ t T = ∂ t ( a M /
3) + C ( σ ) C ( σ / ) C (1) − ∂ t (3 b/ a M ) + C (1) C (1) C ( σ / ) C (1) − ∂ t (3 b/λ ) + C (1) − ∂ t ( a M /λ ) + C ( σ ) = C (1) C ( σ / ) C (1) C ( σ − / ) C (1) C ( σ / ) C (1) C ( σ − / ) C (1) . Proof.
Note that every entry of T is a function in a M , b and λ j . Then theassertion is clear from Lemmas 6.2 and 6.5.From Lemma 6.6 it follows that(6.13) h ξ i | β | γ ∂ αx ∂ βξ T = C ( √ σ ) C ( σ ) C ( √ σ ) C (1) C ( √ σ ) C ( σ ) C ( √ σ ) C (1) C ( √ σ ) , | α + β | = 1 . Lemma 6.8.
There is M such that A T = T − AT has the form A T = C ( √ σ ) − C ( σ ) C ( √ σ ) C ( σ ) C ( √ σ ) − C ( σ ) C ( σ / ) C ( σ ) C ( σ / ) , M ≥ M . roof. Writing A T = (˜ a ij ) it is clear that˜ a ij = t i a M t j + t i b t j + t i t j + t i t j . Then the assertion follows from Lemma 6.6.
Corollary 6.3.
Let A T = (˜ a ij ) . Then ˜ a = λ C ( √ σ ) , ˜ a = λ C (1) , ˜ a = λ C ( σ − ) . Proof.
Lemma 6.8 givesΛ A T = λ C ( √ σ ) λ ( − C ( σ )) λ C ( √ σ ) λ ˜ a λ ˜ a λ ( − C ( σ )) λ ˜ a λ ˜ a λ ˜ a . Since Λ A T is symmetric it follows immediately˜ a = λ C ( √ σ ) λ , ˜ a = λ ( − C ( σ )) λ , ˜ a = λ ( − C ( √ σ )) λ . This proves the assertion because 1 /λ ∈ C (1) and 1 /λ ∈ C ( σ − ).From Corollary 6.3 one can improve Lemma 6.8 such that ˜ a = C ( σ / ) for λ ∈ C ( σ ). Corollary 6.4.
We have h ξ i | β | γ ∂ αx ∂ βξ A T = C (1) C ( √ σ ) C (1) C ( √ σ ) C (1) C ( √ σ ) C ( σ ) C ( √ σ ) C ( σ ) , | α + β | = 1 . Proof.
The proof is clear since h ξ i | β | γ ∂ αx ∂ βξ ( − C ( σ )) = C ( √ σ ).Before closing the section we consider T − ( ∂ t T ). Note that( ∂ t T − ) T = ( ∂ t ( t T )) T = ( h ∂ t t i , t j i )and h ∂ t t i , t j i = −h t i , ∂ t t j i = −h ∂ t t j , t i i so that ( ∂ t T − ) T is antisymmetric.From Lemmas 6.6 and 6.7 one has(6.14) T − ( ∂ t T ) = − ∂ t (3 b/ a M ) + C (1) C (1) ∂ t (3 b/ a M ) + C (1) 0 C ( √ σ ) ∂ t ( a M /
3) + C ( σ ) C ( √ σ ) 0 . For later use we estimate (2 , , T − ( ∂ t T ). Recalling a M = e ( t + α + 2 M h ξ i − γ ) and 0 ≤ t ≤ M − it is clear(6.15) ∂ t a M − ¯ e ∈ S ( M − , g ) . Taking | b /a M | ≤ /
27 into account, thanks to Lemma 4.7 it follows that (cid:12)(cid:12) √ a M ∂ t (3 b/ a M ) (cid:12)(cid:12) ≤ (cid:0)(cid:12)(cid:12) ∂ t b/ √ a M (cid:12)(cid:12) + (cid:12)(cid:12) b/a / M (cid:12)(cid:12) | ∂ t a M | (cid:1) / ≤ (1 + CM − ) (cid:0) (1 + 3 √ / √ (cid:1) ¯ e. (6.16) 31 φ and λ j are admissible weights for g Write z = ( x, ξ ) and w = ( y, η ). It is clear that g σz ( dx, dξ ) = M (cid:0) h ξ i γ | dx | + h ξ i − γ | dξ | (cid:1) = M g z ( dx, dξ ) . Note that | ξ − η | ≤ c h ξ i γ with 0 < c < − c ) h ξ i γ / √ ≤ h η i γ ≤ √ c ) h ξ i γ . If g z ( w ) < c then | ξ − η | < c M h ξ i γ = c M h ξ i − γ h ξ i γ ≤ c h ξ i γ then g z ( dx, dξ ) /C ≤ g w ( dx, dξ ) ≤ Cg z ( dx, dξ )with C independent of γ ≥ M ≥ g z is slowly varying uniformly in γ ≥ M ≥
1. Similarly noting that | ξ − η | ≥ ( γ + | ξ | ) / ≥ h ξ i γ / h η i γ ≤ h ξ i γ / √ | ξ − η | ≥ ( γ + | η | ) / ≥ h η i γ / h η i γ ≥ √ h ξ i γ it is clear that(7.1) h ξ i γ h η i γ + h η i γ h ξ i γ ≤ C (cid:0) h η i − γ | ξ − η | (cid:1) ≤ C (1 + g σw ( z − w ) (cid:1) that is g is temperate uniformly in γ ≥ M ≥ g is an admissible metric . It is clear from (7.1) that(7.2) g σz ( z − w ) ≤ C (cid:0) g σw ( z − w ) (cid:1) . ρ and σ are admissible weights for g We adapt the same convention as in Sections 5, 6 even to weights for g so thatwe omit to say uniformly in t ∈ [0 , M − ]. Lemma 7.1. ρ is an admissible weight for g .Proof. First study ρ / . Assume g z ( w ) = M − h ξ i γ (cid:0) | y | + h ξ i − γ | η | (cid:1) < c ( < / M − h ξ i − γ | η | < c hence | η | < c h ξ i γ for M h ξ i − γ ≤
1. Thus h ξ + sη i − γ ≤ C h ξ i − γ ( | s | <
1) and Lemma 4.3 shows | ρ / ( z + w ) − ρ / ( z ) | ≤ C (cid:0) | y | + h ξ + sη i − γ | η | (cid:1) ≤ CM / h ξ i − / γ g / z ( w ) . Since ρ ( z ) ≥ M h ξ i − γ this yields(7.3) | ρ / ( z + w ) − ρ / ( z ) | ≤ Cρ / ( z ) g / z ( w ) . Choosing c such that C c < / (cid:12)(cid:12) ρ ( z + w ) /ρ ( z ) − (cid:12)(cid:12) < / ρ / ( z + w ) / ≤ ρ / ( z ) ≤ ρ / ( z + w ) / ρ / is g continuous hence so is ρ . Note that(7.4) M h ξ i − γ ≤ ρ ( z ) ≤ CM − ≤ C. If | η | ≥ c h ξ i γ / g σz ( w ) ≥ M c h ξ i γ / g σz ( w ) ≥ M c | η | / ρ ( z + w ) ≤ C ≤ C h ξ i γ ρ ( z ) ≤ C ′ ρ ( z )(1 + g σz ( w )) . If | η | ≤ c h ξ i γ then (7.3) gives(7.5) ρ / ( z + w ) ≤ Cρ / ( z )(1 + g z ( w )) / ≤ Cρ / ( z )(1 + g σz ( w )) / so that(7.6) σ ( t, z + w ) ≤ C σ ( t, z )(1 + g σz ( w ))hence ρ is g temperate in view of (7.2). Thus ρ is an admissible weight. Lemma 7.2. σ is an admissible weight for g and σ ∈ S ( σ, g ) .Proof. Since σ ( t, z ) = t + ρ ( z ) + M h ξ i − γ and ρ ( z ) + M h ξ i − γ is admissible for g by Lemma 7.1 it is clear that σ is admissible for g . The second assertion isclear from (cid:12)(cid:12) ∂ αx ∂ βξ σ (cid:12)(cid:12) - σ −| α + β | / h ξ i −| β | γ - σ ( M − h ξ i γ ) | α + β | / h ξ i −| β | γ - σ M −| α + β | / h ξ i ( | α |−| β | ) / γ . for σ ≥ M h ξ i − γ . ω and φ are admissible weights for g We start with showing
Lemma 7.3. ω and φ are g continuous.Proof. Denote f = t − ψ and h = M / ρ / h ξ i − / γ so that ω = f + h . Notethat | ω ( z + w ) − ω ( z ) | = | ω ( z + w ) − ω ( z ) || ω ( z + w ) + ω ( z ) |≤ | f ( z + w ) − f ( z ) | + 2 | h ( z + w ) − h ( z ) | (7.7)because | f ( z + w ) + f ( z ) || ω ( z + w ) + ω ( z ) | ≤ , | h ( z + w ) + h ( z ) || ω ( z + w ) + ω ( z ) | ≤ . Assume g z ( w ) < c ( ≤ /
2) which implies | η | < √ c h ξ i γ for M h ξ i − γ ≤ h ξ + sη i γ /C ≤ h ξ i γ ≤ C h ξ + sη i γ C is independent of | s | ≤
1. It is assumed that constants C may changefrom line to line but independent of γ ≥ M ≥
1. Noting | f ( z + w ) − f ( z ) | = | ψ ( z + w ) − ψ ( z ) | it follows from Lemma 5.1 that | f ( z + w ) − f ( z ) | ≤ Cρ / ( z + sw ) (cid:0) | y | + h ξ + sη i − γ | η | (cid:1) ≤ Cρ / ( z + sw ) (cid:0) | y | + h ξ i − γ | η | (cid:1) ≤ CM / ρ / ( z ) h ξ i − / γ g / z ( w )(7.9)since ρ is g continuous. Noting that ω ( z ) ≥ M / ρ / ( z ) h ξ i − / γ it results(7.10) | f ( z + w ) − f ( z ) | ≤ Cω ( z ) g / z ( w ) . Similar arguments shows that | h ( z + w ) − h ( z ) | ≤ CM / h ξ i − γ g / z ( w ). Taking ω ( z ) ≥ M / ρ / ( z ) h ξ i − / γ ≥ M h ξ i − γ into account we have | h ( z + w ) − h ( z ) | ≤ CM − / ω ( z ) g / z ( w ) . Therefore from (7.7) one has | ω ( z + w ) − ω ( z ) | ≤ Cω ( z ) g / z ( w ). Choosing c such that C c < / ω is g continuous.Next consider φ . Since φ = ω + f one can write φ ( z + w ) − φ ( z )= ( f ( z + w ) − f ( z ))( φ ( z + w ) + φ ( z )) + h ( z + w ) − h ( z ) ω ( z + w ) + ω ( z ) . (7.11)Since ω is g continuous, decreasing c > ω ( z + w ) /C ≤ ω ( z ) ≤ C ω ( z + w )which together with (7.10) gives | f ( z + w ) − f ( z ) | / ( ω ( z + w ) + ω ( z )) ≤ Cg / z ( w ) . Recalling h ( z ) = M ρ ( z ) h ξ i − γ and repeating similar arguments as above onesees | h ( z + w ) − h ( z ) | ≤ CM ρ / ( z ) h ξ i − / γ g / z ( w ) ≤ CM / ρ ( z ) h ξ i − γ g / z ( w )(7.12)for ρ / ( z ) ≥ M / h ξ i − / γ . Taking (5.1) into account it follows from (7.12) that | h ( z + w ) − h ( z ) | / ( ω ( z + w ) + ω ( z )) ≤ Cφ ( z ) g / z ( w ) . Combining these estimates we obtain from (7.11) that (cid:12)(cid:12)(cid:12) φ ( z + w ) φ ( z ) − (cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12) φ ( z + w ) φ ( z ) + 1 (cid:12)(cid:12)(cid:12) g / z ( w ) + Cg / z ( w )which proves φ ( z ) /C ≤ φ ( z + w ) ≤ C φ ( z ) choosing c > φ is g continuous. 34 emma 7.4. ω and φ are admissible weights for g and ω ∈ S ( ω, g ) , φ ∈ S ( φ, g ) .Proof. Note that(7.13) h ξ i − γ ≤ M h ξ i − γ ≤ √ M √ ρ h ξ i − / γ ≤ ω ≤ CM − ≤ C. Assume | η | ≥ c h ξ i γ hence g σz ( w ) ≥ M c h ξ i γ ≥ c h ξ i γ . Therefore(7.14) ω ( z + w ) ≤ C ≤ C h ξ i γ ω ( z ) ≤ C ′ ω ( z )(1 + g σz ( w )) . Assume | η | ≤ c h ξ i γ and note that (7.5) holds provided | η | ≤ c h ξ i γ . Thenchecking the proof of Lemma 7.3 we see that | f ( z + w ) − f ( z ) | ≤ Cω ( z )(1+ g σz ( w ))and | h ( z + w ) − h ( z ) | ≤ Cω ( z )(1 + g σz ( w )) / . Then (7.14) follows from (7.7)which proves that ω is g temperate hence admissible for g .Turn to φ . From (5.1) and (7.4), (7.13) it follows that h ξ i − γ /C ≤ M h ξ i − γ /C ≤ φ ( z ) = ω ( z ) + f ( z ) ≤ CM − ≤ C. If | η | ≥ h ξ i γ / g σz ( w ) ≥ M h ξ i γ / ≥ h ξ i γ / φ ( z + w ) ≤ C ≤ C h ξ i γ φ ( z ) ≤ Cφ ( z )(1 + g σz ( w )) . Assume | η | ≤ h ξ i γ / | f ( z + w ) − f ( z ) | ≤ Cρ / ( z ) h ξ i − / γ (1 + g σz ( w )) . Recalling (7.5) and M g z ( w ) = g σz ( w ) the same arguments obtaining (7.12)shows that | h ( z + w ) − h ( z ) | ≤ Cρ / ( z ) h ξ i − / γ (1 + g σz ( w )) . Taking these into account (7.11) yeilds | φ ( z + w ) − φ ( z ) | ≤ C (cid:16) ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) (cid:0) φ ( z + w ) + φ ( z ) (cid:1) + ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) (cid:17) (1 + g σz ( w )) . (7.15)Applying Lemma 5.3 to (7.15) to obtain | φ ( z + w ) − φ ( z ) | ≤ C (cid:16) ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) (cid:0) φ ( z + w ) + φ ( z ) (cid:1) + ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) φ ( z ) (cid:17) (1 + g σz ( w ))= C (cid:0) φ ( z + w ) + 2 φ ( z ) (cid:1) ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) (1 + g σz ( w )) . ρ / ( z ) h ξ i − / γ (1 + g σz ( w )) / ( ω ( z + w ) + ω ( z )) < ε then it follows (cid:12)(cid:12)(cid:12) φ ( z + w ) φ ( z ) − (cid:12)(cid:12)(cid:12) ≤ ε (cid:16) φ ( z + w ) φ ( z ) + 2 (cid:17) from which we have φ ( z + w ) /C ≤ φ ( z ) ≤ C φ ( z + w ). If ρ / ( z ) h ξ i − / γ ω ( z + w ) + ω ( z ) (1 + g σz ( w )) ≥ ε we have ε − (1 + g σz ( w )) ≥ h ξ i γ ρ ( z ) ω ( z + w ) ω ( z ) ≥ φ ( z + w ) 12 φ ( z )by (5.1) and an obvious inequality φ ( z + w ) ≤ ω ( z + w ). Thus we concludethat φ is g temperate hence φ is an admissible weight for g . λ j are admissible weights for g Lemma 7.5.
Assume that λ ∈ C ( σ ) and λ ≥ cM σ h ξ i − γ with some c > .Then λ is an admissible weight for g .Proof. Consider √ λ . Assume g z ( w ) < c and hence h ξ + sη i γ ≈ h ξ i γ . Since √ λ ∈ C ( σ ) it follows that (cid:12)(cid:12)p λ ( z + w ) − p λ ( z ) (cid:12)(cid:12) ≤ C p σ ( z + sw ) (cid:0) | y | + h ξ + sη i − γ | η | (cid:1) ≤ C p σ ( z + sw ) h ξ i − / γ g / z ( w )(7.16)which is bounded by C ′ p σ ( z ) h ξ i − / γ g / z ( w ) since σ is g continuous. By as-sumption λ ( z ) ≥ cM σ ( z ) h ξ i − γ one has (cid:12)(cid:12)p λ ( z + w ) − p λ ( z ) (cid:12)(cid:12) ≤ C ′′ M − / p λ ( z ) g / z ( w ) ≤ C ′′ p λ ( z ) g / z ( w ) . Choosing c > C ′′ √ c < p λ ( z ) is g continuous and sois λ ( z ). From cM h ξ i − γ ≤ cM σ h ξ i − γ ≤ λ ≤ C ′ σ ≤ C ′ M − one sees c M h ξ i − γ ≤ c M / σ / h ξ i − / γ ≤ p λ ( z ) ≤ C. If | η | ≥ h ξ i γ / g σz ( w ) ≥ M h ξ i γ / p λ ( z + w ) ≤ C ≤ C ( c M ) − h ξ i γ p λ ( z ) ≤ C ′ p λ ( z ) g σz ( w ) . If | η | ≤ h ξ i γ / (cid:12)(cid:12)p λ ( z + w ) − p λ ( z ) (cid:12)(cid:12) ≤ C p σ ( z ) h ξ i − / γ (1 + g σz ( w )) ≤ C ′ p λ ( z )(1 + g σz ( w ))which proves that √ λ is g temperate and hence so is λ .36 emma 7.6. Assume that λ ∈ C ( σ ) and λ ≥ cM h ξ i − γ with some c > . Then λ is an admissible weight for g . If λ ∈ C (1) and λ ≥ c with some c > then λ is an admissible weight for g .Proof. It is enough to repeat the proof of Lemma 7.5.
Lemma 7.7.
Assume that λ ∈ C ( σ ) and λ ≥ cM σ h ξ i − γ with some c > .Then ∂ αx ∂ βξ λ ∈ S ( √ σ √ λ h ξ i −| β | γ , g ) , | α + β | = 1 . In particular λ ∈ S ( λ, g ) .Proof. From λ ∈ C ( σ ) we have |h ξ i | β | γ ∂ αx ∂ βξ λ | ≤ C σ for | α + β | = 2. Since λ ≥
0, thanks to the Glaeser’s inequality one has (cid:12)(cid:12) ∂ αx ∂ βξ λ (cid:12)(cid:12) ≤ C ′ √ σ √ λ h ξ i −| β | γ , | α + β | = 1 . For | α ′ + β ′ | ≥ (cid:12)(cid:12) ∂ α ′ x ∂ β ′ ξ ′ (cid:0) ∂ αx ∂ βξ λ (cid:1)(cid:12)(cid:12) - σ / −| α ′ + β ′ | / h ξ i −| β |−| β ′ | γ - σ − ( | α ′ + β ′ |− / h ξ i −| β | γ h ξ i −| β ′ | γ - σ ( M − h ξ i γ ) ( | α ′ + β ′ |− / h ξ i −| β ′ | γ h ξ i −| β | γ - σM −| α ′ + β ′ | / M / h ξ i − / γ h ξ i ( | α ′ |−| β ′ | ) / γ h ξ i −| β | γ - √ σM −| α ′ + β ′ | / √ λ h ξ i ( | α ′ |−| β ′ | ) / γ h ξ i −| β | γ because λ ≥ cM σ h ξ i − γ which proves the first assertion. Noting √ σ h ξ i −| β | γ = √ σ h ξ i − / γ h ξ i ( | α |−| β | ) / γ ≤ CM − / √ λ h ξ i ( | α |−| β | ) / γ it is clear that λ ∈ S ( λ, g ). Lemma 7.8.
Assume that λ ∈ C ( σ ) and λ ≥ cM h ξ i − γ with some c > . Then λ ∈ S ( λ, g ) . If λ ∈ C (1) and λ ≥ c with some c > . Then λ ∈ S ( λ, g ) .Proof. It suffices to repeat the proof of Lemma 7.7.
Corollary 7.1.
For s ∈ R we have λ sj ∈ S ( λ sj , g ) , j = 1 , , . Define κ = 1 t + 1 ω = t + ωt ω , ( t > . Lemma 7.9. κ is an admissible weight for g and κ s ∈ S ( κ s , g ) for s ∈ R .Proof. Since ω − is g continuous and g temperate it is clear that κ = t − + ω − is g continuous and g temperate. Noting that ω − ∈ S ( ω − , g ) and ω − ≤ κ itis also clear that (cid:12)(cid:12) ∂ αx ∂ βξ κ (cid:12)(cid:12) = (cid:12)(cid:12) ∂ αx ∂ βξ ω − (cid:12)(cid:12) - M −| α + β | / ω − h ξ i ( | α |−| β | ) / γ - M −| α + β | / κ h ξ i ( | α |−| β | ) / γ , | α + β | ≥ κ ∈ S ( κ, g ). 37 emma 7.10. One has ∂ αx ∂ βξ κ s ∈ S (cid:0) M − ( | α + β |− / κ s ω − ρ / h ξ i − / | α |− β | ) / γ , g (cid:1) , | α + β | ≥ . Proof.
Since ∂ αx ∂ βξ κ s = κ s − ∂ αx ∂ βξ κ it is enough to show the case s = 1. Theproof of the case s = 1 follows easily from Corollary 5.3. Lemma 7.11.
There is
C > such that κλ ≤ e υ (1 + CM − ) κ, σ κ ≤ κ. Proof.
In view of Propositions 4.1 and 6.1 one sees λ ≥ ¯ e υ (1 − CM − ) min (cid:8) t , ω (cid:9) . Denote c = 3 / (¯ e υ (1 − CM − )) = (3 / ¯ e υ )(1 + CM − ). If ω ≥ t and hence λ ≥ t /c then 1 /λ ≤ c/t which shows that1 κλ ≤ cκt = c t ω ( t + ω ) t = c ω ( t + ω ) t ≤ c ( t + ω ) tω = cκ. If t ≥ ω and hence λ ≥ ω /c then 1 /λ ≤ c/ω and hence1 κλ ≤ cκω = c t ω ( t + ω ) ω = c t ( t + ω ) ω ≤ c ( t + ω ) t ω = cκ then the first assertion. To show the second assertion it suffices to note σ ≥ t and then σ ( t + ω ) ≥ t ( t + ω ) ≥ t ω . op( λ i ) Introduce a metric independent of M ¯ g = h ξ i γ | dx | + h ξ i − γ | dξ | so that g = M − ¯ g . We start with Lemma 8.1.
Let m be an admissible weight for g and p ∈ S ( m, g ) satisfy p ≥ c m with some constant c > . Then p − ∈ S ( m − , g ) and there exist k, ˜ k ∈ S ( M − , g ) such that p p − k ) = 1 , (1 + k ) p p − = 1 , p − k ) p = 1 ,p − p k ) = 1 , (1 + ˜ k ) p − p = 1 , p k ) p − = 1 . roof. It is clear that p − ∈ S ( m − , g ). Write p p − = 1 − r where r ∈ S ( M − , g ). Fix any M . Since | r | ( l ) S (1 , ¯ g ) = sup | α + β |≤ l, ( x,ξ ) ∈ R d (cid:12)(cid:12) h ξ i ( | β |−| α | ) / γ ∂ αx ∂ βξ r (cid:12)(cid:12) ≤ C l M − from the Calder´on-Vaillancourt theorem we have k op( r ) k ≤ CM − . Thereforefor large M there exists the inverse (1 − op( r )) − which is given by 1+ P ∞ ℓ =1 r ℓ ∈ S (1 , ¯ g ). (see [1]). Denote k = P ∞ ℓ =1 r ℓ ∈ S (1 , ¯ g ) and prove k ∈ S ( M − , g ). Itis easy to see from the proof (see, e.g. [16], [18]) that there is M such that forany l ∈ N one can find C l > | k | ( l ) S (1 , ¯ g ) ≤ C l holds uniformly in M ≥ M . Note that k satisfies (1 − r ) k ) = 1, that is(8.1) k = r + r k. Since r ∈ S ( M − , g ) it follows from (8.1) that (cid:12)(cid:12) k (cid:12)(cid:12) ( l ) S (1 , ¯ g ) ≤ C l M − uniformly in M ≥ M . Assume that(8.2) sup (cid:12)(cid:12) h ξ i ( | β |−| α | ) / γ ∂ αx ∂ βξ k (cid:12)(cid:12) ≤ C α,β,ν M − − l/ , | α + β | ≥ l for 0 ≤ l ≤ ν . Let | α + β | ≥ ν + 1 and note that ∂ αx ∂ βξ k = ∂ αx ∂ βξ r + X C ··· (cid:0) ∂ α ′′ x ∂ β ′′ ξ r (cid:1) (cid:0) ∂ α ′ x ∂ β ′ ξ k (cid:1) where α ′ + α ′′ = α and β ′ + β ′′ = β . From the assumption (8.2) we have ∂ α ′ x ∂ β ′ ξ k ∈ S ( M − −| α ′ + β ′ | / h ξ i ( | α ′ |−| β ′ | ) / γ , ¯ g ) if | α ′ + β ′ | ≤ ν and ∂ α ′ x ∂ β ′ ξ k ∈ S ( M − − ν/ h ξ i ( | α ′ |−| β ′ | ) / γ , ¯ g ) if | α ′ + β ′ | ≥ ν + 1. Since r ∈ S ( M − , g ) one has (cid:0) ∂ α ′′ x ∂ β ′′ ξ r (cid:1) (cid:0) ∂ α ′ x ∂ β ′ ξ k (cid:1) ∈ S ( M − − ( ν +2) / h ξ i ( | α |−| β | ) / γ , ¯ g )which implies that (8.2) holds for 0 ≤ l ≤ ν + 1 and hence for all ν by inductionon ν . This proves that k ∈ S ( M − , g ). The proof of the assertions for ˜ k issimilar.Here recall [24, Lemmas 3.1.6, 3.1.7]. Lemma 8.2.
Let q ∈ S (1 , g ) satisfy q ≥ c with a constant c independent of M .Then there is C > such that (cid:0) op( q ) u, u ) ≥ ( c − CM − / ) k u k . Proof.
One can assume c = 0. We see q ( x, ξ )+ M − / is an admissible weight for¯ g and ( q + M − / ) / ∈ S (( q + M − / ) / , ¯ g ). Moreover ∂ αx ∂ βξ ( q + M − / ) / ∈ S ( M − / h ξ i ( | α |−| β | ) / γ , ¯ g ) for | α + β | = 1. Therefore q + M − / = ( q + M − / ) / q + M − / ) / + r, r ∈ S ( M − , ¯ g )which proves the assertion. 39 emma 8.3. Let q ∈ S (1 , g ) then there is C > such that k op( q ) u k ≤ (cid:0) sup | q | + CM − / (cid:1) k u k . Lemma 8.4.
Let m > be an admissible weight for g and m ∈ S ( m, g ) . If q ∈ S ( m, g ) then there is C > such that (cid:12)(cid:12) (op( q ) u, u ) (cid:12)(cid:12) ≤ (cid:0) sup (cid:0) | q | /m (cid:1) + CM − / (cid:1) k op( √ m ) u k . Proof.
First note that m ± / are admissible weights and m ± / ∈ S ( m ± / , g ).Write ˜ q = (1 + k ) m − / q m − / k ) ∈ S (1 , g )where m / k ) m − / = 1 and m − / k ) m / = 1 such that m / q m / = q. Since k , ˜ k ∈ S ( M − , g ) one can write ˜ q = qm − + r with r ∈ S ( M − , g ). Thanksto Lemma 8.3 we have k op( qm − ) v k ≤ (sup (cid:0) | q | /m (cid:1) + CM − / ) k v k hence (cid:12)(cid:12) (op( q ) u, u ) (cid:12)(cid:12) ≤ (cid:12)(cid:12) (op( qm − )op( m / ) u, op( m / ) u ) | + CM − k op( m / ) u k proves the assertion. Lemma 8.5.
Let m i > be two admissible weights for g and assume that m i ∈ S ( m i , g ) and m ≤ C m with C > . Then there is C ′ > such that (cid:13)(cid:13) op( m ) u (cid:13)(cid:13) ≤ C ′ (cid:13)(cid:13) op( m ) u (cid:13)(cid:13) . Proof.
Write ˜ m = m m − k ) such that m = ˜ m m with k ∈ S ( M − , g ). Since ˜ m ∈ S (1 , g ) one has k op( m ) u k = k op( ˜ m )op( m ) u k ≤ C ′ k op( m ) u k which proves the assertion. op( λ j ) Lemma 8.6.
There exist
C > and M such that Re (cid:0) op( λ j κ ) u, u (cid:1) ≥ (1 − CM − ) (cid:13)(cid:13) op( κ / λ / j ) u (cid:13)(cid:13) , M ≥ M . Proof.
Since κ ∈ S ( κ, g ) and λ j ∈ S ( λ j , g ) one can write λ j κ = κλ j + r j + r j where r j is pure imaginary and r j ∈ S ( M − κλ j , g ). Thanks to Lemma 8.4 itfollows that Re (op( λ j κ ) u, u ) ≥ (op( κλ j ) u, u ) − CM − k op( λ j / κ / ) u k . κλ j ) u, u ). Since λ / j κ / ∈ S ( λ / j κ / , g ) then( λ / j κ / ) λ / j κ / ) = λ j κ + ˜ r j with ˜ r j ∈ S ( M − λ j κ, g ). Applying Lemma 8.4 to op(˜ r j ) one obtains (cid:0) op( λ j κ ) u, u ) ≥ (1 − CM − ) k op( λ / j κ / ) u k which proves the assertion. Lemma 8.7.
There exist c > and M such that Re (cid:0) op( λ ) u, u (cid:1) ≥ c (cid:13)(cid:13) op( λ / ) u (cid:13)(cid:13) + cM (cid:13)(cid:13) h D i − γ u (cid:13)(cid:13) , M ≥ M . Proof.
From Propositions 4.1 and 6.1 it follows that λ ≥ c ′ M σ h ξ i − γ with some c ′ >
0. Denote ˜ λ = λ / − c M σ h ξ i − γ where c > λ ≥ c M σ h ξ i − γ with c >
0. Note that ˜ λ ∈C ( σ ) since M σ h ξ i − γ ∈ C ( σ ). Thanks to Lemmas 7.5 and 7.7 it follows that˜ λ ∈ S (˜ λ , g ) and ˜ λ is an admissible weight for g . Thus a repetition of theabove arguments shows (cid:0) op(˜ λ ) u, u ) ≥ (1 − CM − ) k op(˜ λ / ) u k where the right-hand side is nonnegative if M ≥ √ C = M . Since( σ / h ξ i − / γ ) σ / h ξ i − / γ ) = σ h ξ i − γ + r with r ∈ S ( M − σ h ξ i − γ , g ) and then (cid:0) op( σ h ξ i − γ ) u, u ) ≥ (1 − CM − ) k op( σ / h ξ i − / γ ) u k . Recalling op( λ /
2) = op(˜ λ ) + c M op( σ h ξ i − γ ) it follows that(8.3) (op( λ / u, u ) ≥ c M (1 − CM − ) k op( σ / h ξ i − / γ ) u k for M ≥ M . Since M h ξ i − γ ≤ M σ h ξ i − γ it follows from Lemma 8.5 that(8.4) M kh D i − γ u k ≤ CM k op( σ / h ξ i − / γ ) u k . Finally writing λ = λ / λ / + r with r ∈ S ( M − λ , g ) one obtains (cid:0) op( λ / u, u (cid:1) ≥ (1 / − CM − ) (cid:13)(cid:13) op( λ / ) u (cid:13)(cid:13) which together with (8.3) and (8.4) proves the assertion. Lemma 8.8.
There exist c > and M such that Re (cid:0) op( λ ) u, u (cid:1) ≥ c (cid:13)(cid:13) op( λ / ) u (cid:13)(cid:13) + c M (cid:13)(cid:13) h D i − / γ u (cid:13)(cid:13) , M ≥ M . roof. A repetition of the same arguments shows that(op( λ ) u, u ) ≥ (1 − CM − ) k op( λ / ) u k . Note that one can find
C > M such that k op( σ / ) u k /C ≤ k op( λ / ) u k ≤ C k op( σ / ) u k for M ≥ M . Noting σ ≥ M h ξ i − γ we conclude the assertion. Lemma 8.9.
There exist c > and M such that (cid:0) op( λ ) u, u (cid:1) ≥ c (cid:13)(cid:13) u (cid:13)(cid:13) , M ≥ M . Summarize what we have proved in
Proposition 8.1.
There exist c > , C > and M such that Re (op(Λ κ ) W, W ) ≥ (1 − CM − ) k op( κ / Λ / ) W k , Re (op(Λ) W, W ) ≥ c X j =1 (cid:0) k op(Λ / ) W k + k op( D ) W k (cid:1) for M ≥ M where D = M h ξ i − γ M / h ξ i − / γ
00 0 1 . Diagonalizing the B´ezout matrix introduced in Section 6 we reduce the system(6.2) to a system with a diagonal symmetrizer.
Lemma 9.1.
Let p ∈ C ( σ k ) then ∂ αx ∂ βξ p ∈ S ( σ k −| α + β | / h ξ i −| β | γ , g ) .Proof. The proof is clear from (cid:12)(cid:12) ∂ α ′ x ∂ β ′ ξ ( ∂ αx ∂ βξ p ) (cid:12)(cid:12) - σ k −| α ′ + β ′ + α + β | / h ξ i −| β ′ + β | γ - σ k −| α + β | / h ξ i −| β | γ σ −| α ′ + β ′ | / h ξ i −| α ′ + β ′ | / γ h ξ i ( | α ′ |−| β ′ | ) / γ - σ k −| α + β | / h ξ i −| β | γ M −| α ′ + β ′ | / h ξ i ( | α ′ |−| β ′ | ) / γ since σ ≥ ρ ≥ M h ξ i − γ . Lemma 9.2.
Let p ∈ C ( σ k ) and q ∈ C ( σ ℓ ) . Then p p − p ∈ S ( σ k − h ξ i − γ , g ) , p q − pq ∈ S ( σ k + ℓ − h ξ i − γ , g ) . roof. The assertions follows from Lemma 9.1 and the Weyl calculus of pseu-dodifferential operators.Since a ∈ C ( σ ), b ∈ C ( σ / ) one sees A ξ ] = A ( t, x, ξ )[ ξ ] + R with R ∈ S ( M − , g ) for ∂ βξ [ ξ ] ∈ S (1 , g ) by (4.6) one can replace A ( t, x, D ) D ] by op( A [ ξ ])in (6.2), moving R to B . Denote L = D t − op( ˜ A ) − op( B ) , ˜ A = a b [ ξ ] . Consider T − T = I − R where R ∈ S ( M − , g ). Thanks to Lemma 8.1 thereis K ∈ S ( M − , g ) such that ( I − R ) I + K ) = I = ( I + K ) I − R ) and hence T − T I + K ) = I, ( I + K ) T − T = I, T I + K ) T − = I. Therefore one can write(9.1) L op( T ) = op( T ) ˜ L where˜ L = D t − op(( I + K ) T − A + B ) T ) + op(( I + K ) T − D t T )) . Lemma 9.3.
One has K ∈ S ( M − h ξ i − γ , g ) .Proof. Write T = ( t ij ) then T − T = ( P k =1 t ki t kj ) and denote X k =1 t ki t kj = δ ij + r ij . Taking Lemma 6.6 into account, we see r ii ∈ S ( σ − h ξ i − γ , g ) ⊂ S ( M − h ξ i − γ , g )and r ij ∈ S ( σ / h ξ i − γ , g ) ⊂ S ( M − h ξ i − γ , g ) for i = j thanks to Lemma 9.2hence R ∈ S ( M − h ξ i − γ , g ). Since K ∈ S ( M − , g ) satisfies K = R + R K weconclude the assertion.Therefore K T − A + B ) T ∈ S ( M − , g ) is clear. Hence˜ L = D t − op( T − A + B ) T − T − D t T )) + op( S ( M − , g )) . To simplify notations sometimes we abbreviate S ( m, g ) to S ( m ) where m isadmissible for g . In view of Lemmas 6.6 and 6.7 it follows from Lemma 9.2 that T − ∂ t T ) = T − ∂ t T + S ( σ − h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( σ − h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( h ξ i − γ ) (9.2) 43ence T − ∂ t T ) = T − ∂ t T + S ( M − , g ) because σ ≥ M h ξ i − γ .Turn to study T − A T . Noting that ∂ αx ∂ βξ a ∈ S ( σ / h ξ i −| β | γ , g ), ∂ αx ∂ βξ b ∈ S ( σ h ξ i −| β | γ , g ) for | α + β | = 1 and ∂ βξ [ ξ ] ∈ S (1 , g ), | β | = 1 we have T − A = T − ˜ A + R, R = S (1) S ( M − ) S ( M − ) S ( M − ) S (1) S ( M − ) S ( M − ) S ( M − ) S ( M − ) . Therefore T − A T = ( T − ˜ A ) T + R with R = R T = S ( M − ) S ( M − ) S (1) S ( M − ) S (1) S ( M − ) S ( M − ) S ( M − ) S ( M − ) . Note that T − ˜ A = C ( σ / ) 1 + C ( σ ) C ( σ / ) − C ( σ ) C ( σ / ) C ( σ ) C ( σ / ) C ( σ ) C ( σ / ) [ ξ ]and hence h ξ i | β | γ ∂ αx ∂ βξ (cid:0) T − ˜ A (cid:1) = S (1) S (1) S ( M − ) S (1) S (1) S ( M − ) S ( M − ) S ( M − ) S ( M − ) for | α + β | = 1. Then thanks to (6.13) one sees( T − ˜ A ) T = T − ˜ AT + R , R = S (1) S ( M − ) S ( M − ) S (1) S ( M − ) S ( M − ) S ( M − ) S ( M − ) S ( M − ) . Thus we obtain T − A T = T − ˜ AT + R + R where R + R = S (1) S ( M − ) S ( M − ) S (1) S ( M − ) S ( M − ) S ( M − ) S ( M − ) S ( M − ) . Recall B = b b + d M b and consider T − B T . Since d M ∈ S ( M, g ) one sees by Lemma 6.6 that T − B = S ( σ ) S ( M σ ) S ( σ ) S ( σ / ) S ( M σ / ) S ( σ / ) b + S ( σ ) b + d M + S ( M σ ) b + S ( σ ) . σ ≤ CM − we conclude that T − B T is written(9.3) S ( σ ) S ( M σ ) S ( σ ) S ( σ / ) S ( M σ / ) S ( σ / ) b + S ( M σ / ) − b − d M + S ( σ / ) b + S ( σ ) . Thus using σ ≤ CM − again T − B T = b − M e + S (1) S (1) + S ( M − , g )where we have used (6.1). We summarize what we have proved in Proposition 9.1.
One can write L · op( T ) = op( T ) · ˜ L where ˜ L = D t − op( A + B − T − D t T ) , A = ( T − AT )[ ξ ] , B = S (1) S (1) S (1) S (1) S (1) S (1) b + S ( M − ) − M e + S (1) S (1) . Note that from Lemma 4.1 it follows that(9.4) b ( t, x, ξ ) − ¯ b ∈ S ( M − , g ) , ¯ b = b (0 , , ¯ ξ ) .
10 Weighted energy estimates
Let w = tφ ( t, x, ξ ) and consider the energy with the scalar weight op( w − n ); E ( V ) = e − θt (cid:0) op(Λ)op( w − n ) V, op( w − n ) V (cid:1) where θ > n is fixed such that(10.1) n > υ − / (cid:16) | b + i ¯ e | ¯ e + 6 + √ (cid:17) + C ∗ + 2where υ − = (cid:0) √ (cid:1) and C ∗ is given by (3.13) and ¯ e is the nonzeropositive real eigenvalue of F p (0 , , , ¯ ξ ) (cf. [24, (7.2.3)]).Note that ∂ t φ = ω − φ and hence ∂ t w − n = − n (cid:16) t + 1 ω (cid:17) w − n = − nκ w − n . Recall that V satisfies(10.2) ∂ t V = op( i A + i B ) V + F, B = B − T − D t T. ∗ = op(Λ) one has ddt E = − θe − θt (cid:0) op(Λ)op( w − n ) V, op( w − n ) V (cid:1) − n Re e − θt (cid:0) op(Λ)op( κw − n ) V, op( w − n ) V )+ e − θt (cid:0) op( ∂ t Λ)op( w − n ) V, op( w − n ) V (cid:1) +2 Re e − θt (cid:0) op(Λ)op( w − n )(op( i A + i B ) V + F ) , op( w − n ) V ) . (10.3)Consider op( w − n )op(Λ)op( κw − n ) = op( w − n κw − n )). Since κ and φ − n are admissible weights for g one can write κ φ − n = κφ − n − r, r ∈ S ( M − κφ − n , g ) . Let ˜ r = r φ n k ) ∈ S ( M − κ, g ) such that r = ˜ r φ − n and hence κφ − n =( κ + ˜ r ) φ − n thus κw − n = ( κ + ˜ r ) w − n . Therefore we have Re (cid:0) op(Λ)op( κw − n ) V, op( w − n ) V (cid:1) ≥ Re (cid:0) op(Λ κ )op( w − n ) V, op( w − n ) V (cid:1) − (cid:12)(cid:12)(cid:0) op(Λ r )op( w − n ) V, op( w − n ) V ) (cid:12)(cid:12) . Since λ j r ∈ S ( M − κλ j , g ) thanks to Lemma 8.4 the second term on the right-hand side is bounded by CM − k op( κ / Λ / )op( w − n ) V k . Applying Proposition 8.1, one can conclude, denoting W j = op( w − n ) V j , that Re (cid:0) op(Λ)op( κw − n ) V, op( w − n ) V (cid:1) ≥ (1 − CM − ) k op( κ / Λ / ) W k . Applying Proposition 8.1 again one obtains Re (cid:0) op(Λ)op( w − n ) V, op( w − n ) V (cid:1) ≥ c (cid:0) k op(Λ / ) W k + k op( D ) W k (cid:1) for M ≥ M . Definition 10.1.
To simplify notations we denote E ( V ) = k op( κ / Λ / )op( w − n ) V k = t − n k op( κ / Λ / )op( φ − n ) V k , E ( V ) = k op(Λ / )op( w − n ) V k + k op( D )op( w − n ) V k = t − n k op(Λ / )op( φ − n ) V k + t − n k op( D )op( φ − n ) V k . Now we summarize
Lemma 10.1.
One can find
C > , c > and M such that n Re (cid:0) op(Λ)op( κw − n ) V, op( w − n ) V (cid:1) + θ Re (cid:0) op(Λ)op( w − n ) V, op( w − n ) V (cid:1) ≥ n (1 − CM − ) E ( V ) + c θ E ( V ) , M ≥ M . (op(Λ)op( w − n )op( B ) V, op( w − n ) V ) First recall that λ i ∈ S ( λ i , g ) and λ ≤ Cσλ ≤ Cσ λ with some C >
0. Let b ∈ S ( σ − / , g ) and consider (op( λ j )op( b ) W i , W j ) for i ≥ j . Write r = (1 + k ) κ − / λ − / j ) λ j b ) λ − / i k ) ∈ S ( κ − / σ − / λ / j λ − / i , g ) ⊂ S ( σ ( i − j ) / , g ) ( i ≥ j )for σκ ≥
1, such that ( κ / λ / j ) r λ / i = λ j b . Then we have | (op( λ j )op( b ) W i , W j ) | ≤ M − k op( κ / Λ / ) W k + CM k op(Λ / ) W k for i ≥ j . Turn to study (op( λ i )op( b ) W j , W i ) for 1 ≤ j < i . Let b ∈ S ( M l , g )and denote r = (1 + k ) κ − / λ − / ) λ b ) λ − / k )such that ( κ / λ / ) r λ / = λ b . Since r ∈ S ( κ − / λ / λ − / , g ) hence r ∈ S ( σ / λ − / , g ) ⊂ S (1 , g ) in view of Lemma 7.11 which proves (cid:12)(cid:12) (op( λ )op( b ) W , W ) (cid:12)(cid:12) = (cid:12)(cid:12) (op( r )op( κ / λ / ) W , op( λ ) W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k + CM l k op(Λ / ) W k . We next check (op( λ )op( b ) W , W ) for b ∈ S (1 , g ). Write r = (1 + k ) κ − / λ − / ) λ b ) κ − / λ − / ) k )such that ( κ / λ / ) r κ / λ / ) = λ b . Since k, ˜ k ∈ S ( M − , g ) it is easyto see that r = b λ / λ − / κ − ) + ˜ r with ˜ r ∈ S ( M − / , g ). By Proposition 6.1 and Lemma 7.11 one sees that (cid:12)(cid:12) λ / λ − / κ − (cid:12)(cid:12) ≤ / (¯ e υ / ) + CM − hence k op( λ / λ − / κ − ) u k ≤ / (¯ e υ / ) (1 + C ′ M − / ) k u k . Therefore | (op( λ )op( b ) W , W ) | = | (op( r )op( κ / λ / ) W , op( κ / λ / ) W ) |≤ (cid:0) / (¯ e υ / ) k op( b ) k + CM − / (cid:1) k op( κ / Λ / ) W k . Now consider (op( λ )op( b ) W , W ) for b ∈ S ( σ − / , g ) = S ( λ − / , g ). Denote r = (1 + k ) κ − / λ − / ) λ b ) λ − / κ − / ) k )such that ( κ / λ / ) r λ / κ / ) = λ b . Since one can write r = ( λ / b ) κ − λ − / ) + ˜ r r ∈ S ( M − , g ) because κ − λ − / ∈ S (1 , g ). Thus repeating the samearguments as above one conclude (cid:12)(cid:12) (op( λ )op( b ) W , W ) (cid:12)(cid:12) ≤ (cid:0) √ / (¯ e υ / ) k op( λ / b ) k + CM − / (cid:1) k op( κ / Λ / ) W k . We summarize the above estimates in
Lemma 10.2.
We have (cid:12)(cid:12) (op( λ j )op( b ) W i , W j ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k + CM k op(Λ / ) W k for b ∈ S ( σ − / , g ) and i ≥ j and | (op( λ )op( b ) W , W ) | ≤ CM − k op( κ / Λ / ) W k + CM l k op(Λ / ) W k for b ∈ S ( M l , g ) and | (op( λ )op( b ) W , W ) | ≤ (cid:0) / (¯ e υ / ) k op( b ) k + CM − / (cid:1) k op( κ / Λ / ) W k for b ∈ S (1 , g ) and (cid:12)(cid:12) (op( λ )op( b ) W , W ) (cid:12)(cid:12) ≤ (cid:0) √ / (¯ e υ / ) k op( λ / b ) k + CM − / (cid:1) k op( κ / Λ / ) W k for b ∈ S ( σ − / , g ) . In particular, this lemma implies
Corollary 10.1.
Let B = ( b ij ) ∈ S (1 , g ) . Then (cid:12)(cid:12) (op(Λ)op( B ) W, W ) (cid:12)(cid:12) ≤ (3 / (¯ e υ / ) k op( b ) k + CM − / ) E ( V ) + C E ( V ) . From Proposition 9.1 it results φ − n B − B φ − n ∈ S ( M − φ − n , g ) thenone concludes by Corollary 10.1 that(10.4) (cid:12)(cid:12) (op(Λ)[op( w − n ) , op( B )] V, W ) (cid:12)(cid:12) ≤ CM − E ( V ) + C E ( V )where W = op( w − n ) V again. Write T − ∂ t T = (˜ t ij ) and recall (6.14) and notethat ˜ t = − ˜ t ∈ C ( σ − / ) and ˜ t ∈ S (1 , g ). Then thanks to Lemma 5.5 onehas λ φ − n t − ˜ t φ − n ) φ n ∈ S ( ω − ρ / h ξ i − γ , g ) ⊂ S ( M − p κλ p κλ , g ) ,λ φ − n t − ˜ t φ − n ) φ n ∈ S ( σ − / ω − ρ / h ξ i − γ , g ) ⊂ S ( M − p κλ p κλ , g )because Cλ ≥ M σ h ξ i − γ , Cλ ≥ σ ≥ M h ξ i − γ and ω − ≤ κ . Therefore repeat-ing similar arguments one concludes(10.5) (cid:12)(cid:12) (op(Λ)[op( w − n ) , op( T − ∂ t T )] V, W ) (cid:12)(cid:12) ≤ CM − E ( V ) . B = B − T − D t T it follows from (10.4) and (10.5) that(10.6) (cid:12)(cid:12) (op(Λ)[op( w − n ) , op( B )] V, W ) (cid:12)(cid:12) ≤ CM − E ( V ) + C E ( V ) . With B = ( q ij ) we see that q = i∂ t (3 b/ a M ) + S (1) , q = b + i∂ t a M / S ( M − )and q = − M e + S (1), q ij ∈ S ( σ − / , g ) for j ≥ i . Applying Lemma 10.2 wehave from (6.15), (6.16), (9.4) and Proposition 6.1 that (cid:12)(cid:12) (op(Λ)op( B )op( w − n ) V, op( w − n ) V ) (cid:12)(cid:12) ≤ (cid:0) υ − / ( | b + i ¯ e | / ¯ e + 6 + √
2) + CM − / (cid:1) E ( V ) + CM E ( V ) . (10.7)Combining the estimates (10.7) and (10.6) we obtain Lemma 10.3.
We have (cid:12)(cid:12) (op(Λ)op( w − n )op( B ) V, op( w − n ) V ) (cid:12)(cid:12) ≤ (cid:0) υ − / ( | b + i ¯ e | / ¯ e + 6 + √
2) + CM − / (cid:1) E ( V ) + CM E ( V ) . (op(Λ)op( w − n )op( i A ) V, op( w − n ) V ) Study g − n A − A g − n . Recall Lemma 6.8(10.8) A = [ ξ ] C ( √ σ ) − C ( σ ) C ( √ σ ) C ( σ ) C ( √ σ ) − C ( σ ) C ( σ / ) C ( σ ) C ( σ / ) . Let r ∈ C ( σ s ) then thanks to Lemma 5.5 it follows that φ − n ξ ] r ) − ([ ξ ] r ) φ − n ∈ S ( φ − n σ s − / ω − ρ / , g ) . Denoting φ − n A − A φ − n = ( r ij ), in view of Lemma 5.5 it follows that r ij ∈ S ( φ − n ω − ρ / , g ) ⊂ S ( M − κφ − n , g )for i ≤ j because ω − ≤ κ . Writing ˜ r ij = r ij φ n k ) ∈ S ( M − κ, g ) suchthat r ij = ˜ r ij φ − n one obtains (cid:12)(cid:12) (op( λ i )op( r ij ) V j , W i ) (cid:12)(cid:12) = (cid:12)(cid:12) (op( λ i r ij ) W j , W i ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k since λ i r ij ∈ S ( M − κλ i , g ). It rests to estimate (op( λ i )op( r ij ) V j , W i ) for i > j . From Corollary one sees ˜ a = λ C ( σ − ) hence thanks to Lemmas 5.5and 7.7 r = φ − n a [ ξ ]) − φ − n ˜ a [ ξ ] ∈ S ( σ − / λ / ω − ρ / φ − n , g ) ⊂ S ( λ / κφ − n , g )49ecause ω − ≤ κ hence r = ˜ r φ − n with ˜ r ∈ S ( κλ / , g ). Then noting λ / ≤ CM − we have (cid:12)(cid:12) (op( λ )op( r ) V , W ) (cid:12)(cid:12) = (cid:12)(cid:12) (op( λ r ) W , W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k . Similarly from ˜ a = λ C ( σ / ), ˜ a = λ C (1) and Lemma 7.7 it follows that r ∈ S ( σλ / ω − ρ / φ − n , g ) ⊂ S ( M − λ / κφ − n , g ) ,r ∈ S ( λ / ω − ρ / φ − n , g ) ⊂ S ( M − λ / κφ − n , g ) . Here we have used(10.9) ∂ αx ∂ βξ λ ∈ S ( λ / h ξ i −| β | γ , g ) , | α + β | = 1which follows from λ ∈ C ( σ ) easily. Then one obtains (cid:12)(cid:12) (op( λ )op( r ) V , W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k , (cid:12)(cid:12) (op( λ )op( r ) V , W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k . Therefore (op(Λ)op( w − n )op( A ) V, op( w − n ) V ) − (op(Λ)op( A ) W, W ) is boundedby constant times M − E ( V ).Next study Λ A − Λ A = ( q ij ). From Lemmas 6.8 and 7.7 it follows that λ a j [ ξ ]) − λ ˜ a j [ ξ ] ∈ S ( σ / λ / , g ) ⊂ S ( M − λ κ, g ) ,λ a j [ ξ ]) − λ ˜ a j [ ξ ] ∈ S ( λ / , g ) ⊂ S ( λ κ / , g )because λ / κ ≥ Cλ κ ≥
1. Then (cid:12)(cid:12) (op( q j ) W j , W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k + C k op(Λ / ) W k for j = 1 , , (cid:12)(cid:12) (op( q j ) W j , W ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / ) W k + CM k op(Λ / ) W k for j = 2 ,
3. Repeating similar arguments, applying Lemmas 6.8 and 7.7, onehas λ a [ ξ ]) − λ ˜ a [ ξ ] ∈ S ( σ − / λ / λ / , g ) ⊂ S ( κ / λ / λ / , g ) ,λ a [ ξ ]) − λ ˜ a [ ξ ] ∈ S ( σ / λ / , g ) ,λ a [ ξ ]) − λ ˜ a [ ξ ] ∈ S ( σ − / λ / , g ) ⊂ S ( κ / λ / , g )since κσ ≥
1. Therefore we have | (op( q ) W , W ) | + | (op( q ) W , W ) | + | (op( q ) W , W ) |≤ CM − k op(Λ / κ / ) W k + CM k op(Λ / ) W k . (cid:12)(cid:12) (op(Λ)op( w − n )op( A ) V, op( w − n ) V ) − (op(Λ A ) W, W ) (cid:12)(cid:12) ≤ CM − E ( V ) + CM E ( V ) . (10.10)Since Λ ∗ = Λ a repetition of the same arguments proves that (cid:12)(cid:12) (op(Λ)op( w − n ) V, op( w − n )op( A ) V ) − (op( A Λ ∗ ) W, W ) (cid:12)(cid:12) is also bounded by the right-hand side of (10.10). Recalling that Λ A = A ∗ Λand Λ ∗ = Λ we have Lemma 10.4.
One can find
C > such that (cid:12)(cid:12) Re (cid:0) op(Λ)op( w − n )op( i A ) V, op( w − n ) V ) (cid:12)(cid:12) ≤ CM − E ( V ) + CM E ( V ) . (op( ∂ t Λ)op( w − n ) V, op( w − n ) V ) Start with
Lemma 10.5.
We have ∂ t λ j ∈ S ( κλ j , g ) , j = 1 , .Proof. Note that Lemma 3.6 with ǫ = √ M h ξ i − γ implies | ∂ t ∆ M | ≤ C ∗ (cid:16) t + 1 ω (cid:17) ∆ M = C ∗ κ ∆ M . Recalling ∂ t λ = − ∂ t q ( λ ) /∂ λ q ( λ ) it follows from (6.11) and (6.7) that | ∂ t λ | ≤ (1 + CM − ) (cid:0) | ∂ t a M /a M | λ + | ∂ t ∆ M | / a M (cid:1) . Since (1+ CM − ) λ ≥ ∆ M / a M by Proposition 6.1 and 1 /a M ≤ κ/e by Lemma7.11 one concludes | ∂ t λ | ≤ (1 + CM − )( C ∗ + 1) κλ . Since ∂ t λ ∈ C ( σ ) then | ∂ αx ∂ βξ ∂ t λ | ≤ Cσ −| α + β | / h ξ i −| β | γ ≤ Cσ / h ξ i − / γ h ξ i ( | α |−| β | ) / γ . From Lemma 7.11 and Cλ ≥ M σ h ξ i − γ it follows that κλ ≥ κ p λ M / σ / h ξ i − / γ /C ≥ M / σ / h ξ i − / γ /C which proves | ∂ αx ∂ βξ ∂ t λ | ≤ CM − / κλ h ξ i ( | α |−| β | ) / γ for | α + β | = 1. For | α + β | ≥ | ∂ αx ∂ βξ ∂ t λ | - σ −| α + β | / h ξ i −| β | γ - σ − ( | α + β |− / h ξ i −| β | γ - ( M − h ξ i γ ) ( | α + β |− / h ξ i −| β | γ = M h ξ i − γ M −| α + β | / h ξ i ( | α |−| β | ) / γ ≤ σ − M σ h ξ i − γ M −| α + β | / h ξ i ( | α |−| β | ) / γ ≤ Cκλ M −| α + β | / h ξ i ( | α |−| β | ) / γ because κσ ≥
1. Therefore we conclude ∂ t λ ∈ S ( κλ , g ). On the other hand ∂ t λ j ∈ S ( κλ j , g ), j = 2 , ∂ t λ j ∈ C (1) ⊂ S (1 , g ) ⊂ S ( κλ , g ) for Cλ κ ≥
1. This completes the proof. 51ote that from (6.10), (6.11), (6.7) and Proposition 6.1 we see that (cid:12)(cid:12) ∂ t λ (cid:12)(cid:12) ≤ | ∂ t a M | λ /a M + Ca M ≤ (cid:0) CM − (cid:1) κλ for κσ ≥
1. Now applying Lemma 8.4 one obtains (cid:12)(cid:12) (op( ∂ t λ ) W , W ) (cid:12)(cid:12) ≤ ( C ∗ + 1 + CM − / ) k op( κ / λ / ) W k , (cid:12)(cid:12) (op( ∂ t λ ) W , W ) (cid:12)(cid:12) ≤ (1 + CM − / ) k op( κ / λ / ) W k . Since (cid:12)(cid:12) (op( ∂ t λ ) W , W ) (cid:12)(cid:12) ≤ C k op( λ ) W k is clear summarizing the above es-timates we obtain Lemma 10.6.
We have (cid:12)(cid:12) (op( ∂ t Λ)op( w − n ) V, op( w − n ) V ) (cid:12)(cid:12) ≤ ( C ∗ + 2 + CM − / ) E ( V ) + C E ( V ) . Consider the term Re (cid:0) op(Λ)op( w − n ) F, op( w − n ) V ) where F = t ( F , F , F ).Write R = (1 + K ) κ − / Λ / ) κ / Λ − / ) K )such that Λ = ( κ / Λ / ) R κ − / Λ / ). Since R ∈ S (1 , g ) it follows that (cid:12)(cid:12) (op(Λ)op( w − n ) F, op( w − n ) V ) (cid:12)(cid:12) = (cid:12)(cid:12) (op( R )op( κ − / Λ / )op( w − n ) F, op( κ / Λ / )op( w − n ) V ) (cid:12)(cid:12) ≤ CM − k op( κ / Λ / )op( w − n ) V k + CM k op( κ − / Λ / )op( w − n ) F k . Therefore we have
Lemma 10.7.
There exist
C > , M such that (cid:12)(cid:12) Re (cid:0) op(Λ)op( w − n ) F, op( w − n ) V ) (cid:12)(cid:12) ≤ CM − E ( V )+ CM k op( κ − / Λ / )op( w − n ) F k , M ≥ M . Because of the choice of n it follows from (10.3) and Lemmas 10.1, 10.3,10.4, 10.6, 10.7 one can find c i > M , γ , θ such that ddt E ≤ − c e − θt E − c θe − θt E + CM e − θt k op( κ − / Λ / )op( w − n ) F k for M ≥ M , γ ≥ γ and θ ≥ θ . Recalling w − n = t − n φ − n and integrating theabove differential inequality in t we obtain52 roposition 10.1. There exist c i > and M , γ , θ such that c t − n e − θt (cid:16) k op(Λ / )op( φ − n ) V ( t ) k + k op( D )op( φ − n ) V ( t ) k (cid:17) + c Z t e − θs s − n k op( κ / Λ / )op( φ − n ) V ( s ) k ds + c θ Z t e − θs s − n (cid:16) k op(Λ / )op( φ − n ) V ( s ) k + k op( D )op( φ − n ) V ( s ) k (cid:17) ds ≤ CM Z t e − θs s − n k op( κ − / Λ / )op( φ − n ) F ( s ) k ds for V satisfying lim t → +0 t − n (cid:0) op(Λ)op( φ − n ) V ( t ) , op( φ − n ) V ( t ) (cid:1) = 0 and for ≤ t ≤ M − , M ≥ M , γ ≥ γ and θ ≥ θ . Fix M such that Proposition 10.1 holds. Since φ > κ ≥ t − and h ξ i − / j/ γ ≤ Cλ / j one sees that t − / h ξ i − / j/ γ ≤ Cκ / λ / j φ − n , ≤ j ≤ . Hence it follows from Lemma 8.5 that t − / kh D i − γ V k ≤ t − / k op( D ) V k ≤ C k op( κ / Λ / φ − n ) V k . Writing κ / λ j φ − n = r j φ − n with r j ∈ S ( κ / λ / j , g ) it is clear that(10.11) t − / kh D i − γ V k ≤ t − / k op( D ) V k ≤ C k op( κ / Λ / )op( φ − n ) V k . Similarly we see that k op( κ − / Λ / )op( φ − n ) F k ≤ C k op( κ − / φ − n Λ / ) F k . Thanks to Lemma 5.3 one has κ − / φ − n λ / j ∈ S ( √ t h ξ i nγ , g )hence applying Lemma 8.5 again k op( κ − / Λ / )op( φ − n ) F k ≤ C k op( κ − / φ − n Λ / ) F k≤ C √ t kh D i nγ F k . (10.12)Remarking that (cid:12)(cid:12) (op(Λ)op( φ − n ) V ( t ) , op( φ − n ) V ( t )) (cid:12)(cid:12) ≤ C kh D i nγ V ( t ) k one con-cludes that Corollary 10.2.
We have t − n e − θt k op( D ) V ( t ) k + Z t s − n − e − θs k op( D ) V ( s ) k ds ≤ C Z t s − n +1 e − θs kh D i nγ F ( s ) k ds for V satisfying lim t → +0 t − n kh D i nγ V ( t ) k = 0 . Let s ∈ R and try to obtain estimates for h D i sγ V . In what follows we fix M and γ (actually it is enough to choose γ = M , see (4.1)) such that Proposition 10.1holds, while θ remains to be free. From (10.2) one has ∂ t ( h D i sγ V ) = (cid:0) op( i A + i B ) + i [ h D i sγ , op( A + B )] h D i − sγ (cid:1) h D i sγ V + h D i sγ F. Lemma 11.1.
For any s ∈ R there is C > such that (cid:12)(cid:12) ([ h D i sγ , op( A )] V, op(Λ) h D i sγ V ) (cid:12)(cid:12) ≤ C E ( h D i sγ V ) . Proof.
Denoting T − AT = (˜ a ij ) thanks to Corollary 6.3 and Lemma 7.7 and(10.9) we see that((˜ a [ ξ ]) h ξ i sγ − h ξ i sγ a [ ξ ])) h ξ i − sγ ∈ S ( σ p λ , g ) , ((˜ a [ ξ ]) h ξ i sγ − h ξ i sγ a [ ξ ])) h ξ i − sγ ∈ S ( p λ , g ) , ((˜ a [ ξ ]) h ξ i sγ − h ξ i sγ a [ ξ ])) h ξ i − sγ ∈ S ( σ − / p λ , g )(11.1)where S ( σ √ λ , g ) ⊂ S ( M − √ λ , g ) and S ( σ − / √ λ , g ) = S ( λ − / √ λ , g ).From Lemma 6.8 it is easy to see ((˜ a ij [ ξ ]) h ξ i sγ −h ξ i sγ a ij [ ξ ])) h ξ i − sγ ∈ S (1 , g )for j ≥ i then taking (11.1) into account the assertion is easily proved. Lemma 11.2.
For any s ∈ R and any ǫ > there is C > such that (cid:12)(cid:12) ([ h D i sγ , op( B )] V, op(Λ) h D i sγ V ) (cid:12)(cid:12) ≤ ǫ E ( h D i sγ V ) + C E ( h D i sγ V ) . Proof.
Write B = (˜ b ij ). Since ˜ b ij ∈ S (1 , g ) by (9.3) it suffices to consider ˜ b ij with i > j . Taking b , b ∈ S (1 , G ) and d M ∈ S ( M, G ) (here recall that M being fixed) into account, it follows from (9.3) that λ h ξ i sγ b − ˜ b h ξ i sγ ) h ξ i − sγ ∈ S ( σ / h ξ i − / γ , g ) ⊂ S ( σ / λ / λ / , g ) ,λ h ξ i sγ b − ˜ b h ξ i sγ ) h ξ i − sγ ∈ S ( σ / h ξ i − / γ , g ) ⊂ S ( λ / λ / , g ) ,λ h ξ i sγ b − ˜ b h ξ i sγ ) h ξ i − sγ ∈ S ( σ / h ξ i − / γ , g ) ⊂ S ( h ξ i − / γ λ / λ / , g )since λ ≥ M σ h ξ i − γ . This proves(11.2) (cid:12)(cid:12) ([ h D i sγ , op( B )] V, op(Λ) h D i sγ V ) (cid:12)(cid:12) ≤ C E ( h D i sγ V ) . Next consider T − ∂ t T = (˜ t ij ). Recalling ˜ t ∈ C ( σ − / ) and ˜ t ∈ C (1) we have λ h ξ i sγ t − ˜ t h ξ i sγ ) h ξ i − sγ ∈ S ( h ξ i − γ , g ) ⊂ S ( M − p κλ p λ , g ) ,λ h ξ i sγ t − ˜ t h ξ i sγ ) h ξ i − sγ ∈ S ( σ − / h ξ i − γ , g ) ⊂ S ( M − p κλ p λ , g )since σκ ≥ Cλ ≥ M σ h ξ i − γ and Cλ ≥ σ ≥ M h ξ i − γ . Therefore we have (cid:12)(cid:12) ([ h D i sγ , op( T − ∂ t T )] V, op(Λ) h D i sγ V ) (cid:12)(cid:12) ≤ CM − q E ( h D i sγ V ) q E ( h D i sγ V ) ≤ ǫ E ( h D i sγ V ) + C M − ǫ − E ( h D i sγ V )which together with (11.2) proves the assertion.54hoosing ǫ > c in Proposition 10.1 and choosing θ large weconclude Proposition 11.1.
Let s ∈ R be given. There exist C > , θ > such that t − n e − θt (cid:16) k op(Λ / )op( φ − n ) h D i sγ V ( t ) k + k op( D )op( φ − n ) h D i sγ V ( t ) k (cid:17) + Z t e − θτ τ − n k op( κ / Λ / )op( φ − n ) h D i sγ V ( τ ) k dτ + θ Z t e − θτ τ − n (cid:16) k op(Λ / )op( φ − n ) h D i sγ V ( τ ) k + k op( D )op( φ − n ) h D i sγ V ( τ ) k (cid:17) dτ ≤ C Z t e − θτ τ − n k op( κ − / Λ / )op( φ − n ) h D i sγ F ( τ ) k dτ for ≤ t ≤ M − = δ , θ ≤ θ and V satisfying lim t → +0 t − n (cid:0) op(Λ)op( φ − n ) h D i sγ V ( t ) , op( φ − n ) h D i sγ V ( t ) (cid:1) = 0 . Lemma 11.3.
For any s ∈ R there exists C s > such that t − n k op( D ) h D i sγ V ( t ) k + Z t τ − n − k op( D ) h D i sγ V ( τ ) k dτ ≤ C s Z t τ − n +1 kh D i n + sγ ˜ LV ( τ ) k dτ for ≤ t ≤ δ and for V satisfying lim t → +0 t − n kh D i n + sγ V ( t ) k = 0 . Recall (9.1) so that ˜ L = op( I + K )op( T − ) · L · op( T ) with T , T − ∈ S (1 , g )then kh D i n + sγ ˜ LV k ≤ C s kh D i n + sγ L · op( T ) V k . Since kh D i s − γ op( T ) V k ≤ C s kh D i s − γ V k ≤ C s k op( D ) h D i sγ V k it results fromLemma 11.3 that t − n kh D i s − γ op( T ) V ( t ) k + Z s τ − n − kh D i s − γ op( T ) V ( τ ) k dτ ≤ C s Z t τ − n +1 kh D i n + sγ L · op( T ) V ( τ ) k dτ. Replacing op( T ) V by U one obtains Lemma 11.4.
For any s ∈ R there exists C s > such that t − n kh D i s − γ U ( t ) k + Z t τ − n − kh D i s − γ U ( τ ) k dτ ≤ C s Z t τ − n +1 kh D i n + sγ LU ( τ ) k dτ, ≤ t ≤ δ for V satisfying lim t → +0 t − n kh D i n + sγ U ( t ) k = 0 . P . Since U = t ( D t u, h D i γ D t u, h D i γ u ) and LU = t ( ˆ P u, ,
0) wehave t − n X j =0 kh D i s +1 − jγ D jt u ( t ) k + X j =0 Z t τ − n − kh D i s +1 − jγ D jt u ( τ ) k dτ ≤ C s Z t τ − n +1 kh D i n + sγ ˆ P u ( τ ) k dτ, ≤ t ≤ δ. (11.3)Now consider the adjoint operator ˆ P ∗ of ˆ P . Noting a M ∈ C ( σ ), b ∈ C ( σ / )and (4.6) we see thatˆ P ∗ = D t − a M ( t, x, D )[ D ] D t − b ( t, x, D ) [ D ] + b D t + (cid:0) ˜ b + d M )[ D ] D t + ˜ b [ D ] + ˜ c D t + ˜ c [ D ]with ˜ b j ∈ S (1 , g ) and ˜ c j ∈ S ( M , g ) hence ˜ c j [ D ] − ∈ S ( M − , g ) where it is notdifficult to check that ˜ b − (cid:0) b + ie (cid:1) ∈ S ( M − , g ) . Since the power n of the weight φ − n depends only on a , b and b (see (10.1))then we can assume that one can choose the same n for ˆ P ∗ as for ˆ P . Thenemploying the weighted energy E ∗ ( V ) = e θt (cid:0) op(Λ)op( w n ) V, op( w n ) V (cid:1) and repeating the same arguments as before and making the integration − Z δt ddt E ∗ dt we have Proposition 11.2.
There exist c i > and M , γ , θ such that c t n e θt (cid:16) k op(Λ / )op( φ n ) V ( t ) k + k op( D )op( φ n ) V ( t ) k (cid:17) + c Z δt e θτ τ n k op( κ / Λ / )op( φ n ) V ( τ ) k dτ + c θ Z δt e − θτ τ n (cid:16) k op(Λ / )op( φ n ) V ( τ ) k + k op( D )op( φ n ) V ( τ ) k (cid:17) dτ ≤ Cδ n e θδ (op(Λ)op( φ n ) V ( δ ) , op( φ n ) V ( δ ))+ CM Z δt e θτ τ n k op( κ − / Λ / )op( φ n ) F ∗ ( τ ) k dτ, ≤ t ≤ δ = M − for M ≥ M , γ ≥ γ and θ ≥ θ where F ∗ = op( T ) t ( ˆ P ∗ f, , . M such that Proposition 11.2 holds. From Lemma 5.3 and Cλ j ≥ M h ξ i − γ we have t − / h ξ i − n − / j/ γ ∈ S ( κ / λ / j φ n , g ) , κ / λ / j φ n ∈ S ( √ t, g )which shows that t n e θt k op( D ) h D i − nγ V ( t ) k + Z δt τ n − e θτ k op( D ) h D i − nγ V ( τ ) k dτ ≤ Cδ n e θδ k V ( δ ) k + C Z δt τ n +1 e θτ k F ∗ ( s ) k dτ, < t ≤ δ. Therefore repeating the same arguments as before we have
Lemma 11.5.
For any s ∈ R there is C s > such that t n kh D i s − n − γ U ( t ) k + Z δt τ n − kh D i s − n − γ U ( τ ) k dτ ≤ C s (cid:16) δ n kh D i sγ U ( δ ) k + Z δt τ n +1 kh D i sγ F ∗ ( τ ) k dτ (cid:17) , < t ≤ δ. Lemma 11.5 implies that X j =0 (cid:16) t n kh D i s +1 − jγ D jt u ( t ) k + Z δt τ n − kh D i s +1 − jγ D jt u ( τ ) k dτ (cid:17) ≤ C s (cid:16) k δ n X j =0 kh D i n + s +2 − jγ D jt u ( δ ) k + Z δt τ n +1 kh D i n + sγ ˆ P ∗ u ( τ ) k dτ (cid:17) (11.4)for 0 < t ≤ δ . Replacing s by − n − − s then (11.4) gives Z δ t n − kh D i − n − s u ( t ) k dt ≤ C Z δ t n +1 kh D i − − s ˆ P ∗ u ( t ) k dt for u ∈ C ∞ ((0 , δ ) × R d ). This implies (cid:12)(cid:12)(cid:12) Z δ ( f, v ) dt (cid:12)(cid:12)(cid:12) ≤ (cid:16) Z δ t − n +1 kh D i n + s f k dt (cid:17) / (cid:16) Z δ t n − kh D i − n − s v k dt (cid:17) / ≤ C (cid:16) Z δ t − n +1 kh D i n + s f k dt (cid:17) / (cid:16) Z δ t n − kh D i − − s ˆ P ∗ v k dt (cid:17) / for all v ∈ C ∞ ((0 , δ ) × R d ) and f such that R δ t − n +1 kh D i n + s f k dt < ∞ . Usingthe Hahn-Banach theorem to extend the anti-linear form in ˆ P ∗ v ;(11.5) ˆ P ∗ v Z δ ( f, v ) dt
57e conclude that there is some u with R δ t − n +1 kh D i s u k dt < + ∞ such that Z δ ( f, v ) dt = Z δ ( u, ˆ P ∗ v ) dt. This implies that ˆ
P u = f . Since we may assume that 2 n − ≥ h D i s u ∈ L ((0 , δ ) × R d ) it follows from [6, Theorem B.2.9] that h D i s − j D jt u ∈ L ((0 , δ ) × R d ) , j = 0 , , . In view of the estimate (11.3) the following estimate holds for this ut − n X j =0 kh D i − j + s D jt u ( t ) k + X j =0 Z t τ − n − kh D i s − j D jt u ( τ ) k dτ ≤ C Z t τ − n +1 kh D i n + s f ( τ ) k dτ, ≤ t ≤ δ. (11.6) Theorem 11.1.
There exists δ > such that for any s ∈ R and any f with t − n +1 / h D i n + s f ∈ L ((0 , δ ) × R d ) there is a unique u with t − n − / h D i s − j D jt u ∈ L ((0 , δ ) × R d ) , j = 0 , , satisfying ˆ P u = f and (11.6) . Instead of (11.5) considering the anti-linear form in ˆ
P v ;ˆ P v Z δ ( f, v ) dt + X j =0 (cid:0) w − j , D jt v ( δ, · ) (cid:1) + (cid:0) w , ( D t − h D i a ( δ, x, D )) v ( δ, · ) (cid:1) for v ∈ C ∞ ((0 , ∞ ) × R d ) and repeating similar arguments adopting (11.3) weconclude Theorem 11.2.
There exists δ > such that for any s ∈ R and any f with t n +1 / h D i n + s f ∈ L ((0 , δ ) × R d ) and any w j with h D i n + s +2 − jγ w j ∈ L ( R d ) , j = 0 , , , there is a unique u with (11.7) t n − / h D i s − j D jt u ∈ L ((0 , δ ) × R d ) , D jt u ( δ, · ) = w j , j = 0 , , satisfying ˆ P ∗ u = f and (11.4) . Indeed we first see that there is u with t n − / h D i s u ∈ L ((0 , δ ) × R d )satisfying D jt u ( δ ) = w j , j = 0 , , h D i n + s f ∈ L (( ε, δ ) × R d ) and h D i s u ∈ L (( ε, δ ) × R d ) for any ε > h D i s − j D jt u ∈ L (( ε, δ ) × R d ), 0 ≤ j ≤
2. Applying(11.4) with t = ε we conclude (11.7), since ε > Remark 11.1.
It is clear from the proof that for any n ′ ≥ n , Theorems 11.1and 11.2 hold. 58 In Section 11 we have proved an existence result of the Cauchy problem forˆ P , which coincides with the original P only in W M . Following [21], [22], [9](also [24]) we show that the micro support of u ( t, · ), obtained by Theorem 11.1,propagates with a finite speed via estimates of Sobolev norms of Φ u , cut off bya suitable Φ. This fact enables us to solve the Cauchy problem for the original P through that of ˆ P . Let χ ( x ) ∈ C ∞ ( R d ) be equal to 1 near x = 0 and vanish in | x | ≥
1. Set d ǫ ( x, ξ ; y, η ) = (cid:8) χ ( x − y ) | x − y | + | ξ h ξ i − γ − η h η i − γ | + ǫ (cid:9) / ,f ǫ ( t, x, ξ ; y, η ) = t − T + νd ǫ ( x, ξ ; y, η )where ( y, η ) ∈ R d × ( R d \ { } ) and ν is a positive small parameter and T > (cid:12)(cid:12) ∂ αx ∂ βξ d ǫ (cid:12)(cid:12) ≤ C h ξ i −| β | γ , | α + β | = 1where C is independent of ǫ >
0. Define Φ ǫ by(12.2) Φ ǫ ( t, x, ξ ) = ( exp (1 /f ǫ ( t, x, ξ )) if f ǫ <
00 otherwiseand set Φ ǫ = f − ǫ Φ ǫ . Note that Φ ǫ , Φ ǫ ∈ S (1 , g ) for any fixed ǫ > g = | dx | + h ξ i − γ | dξ | and Φ ǫ − f ǫ ǫ ∈ S ( h ξ i − γ , g ) . Since ∂ t Φ ǫ = − Φ ǫ /f ǫ writing ∂ t (op(Φ ǫ ) V ) = − op( f − ǫ Φ ǫ ) V + (op( i A + i B ))op(Φ ǫ ) V +[op(Φ ǫ ) , op( i A + i B )] V + op(Φ ǫ ) F (12.3)we estimate E (op(Φ ǫ ) V ) = e − θt (cid:0) op(Λ)op( w − n )op(Φ ǫ ) V, op( w − n )op(Φ ǫ ) V (cid:1) . SinceΦ ǫ B − B ǫ ∈ S ( c ( M ) h ξ i − / γ , g ) by Proposition 9.1 it is not difficult to seefrom the proof of Corollary 10.1 that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )[op(Φ) , op( B )] V, op( φ − n )op(Φ) V (cid:1)(cid:12)(cid:12) ≤ c ( M, ǫ ) N ( h D i − / γ V )where, to simplify notations, we have set E ( V ) + E ( V ) = t − n N ( V ) . ǫ T − ∂ t T ) − ( T − ∂ t T ) ǫ = ( ϕ ij ) hence ϕ ∈ S ( σ − h ξ i − γ , g ) and ϕ ∈ S ( σ − / h ξ i − γ , g ) in view of (6.14). Then we have λ ϕ ∈ S ( h ξ i − γ , g ) ⊂ S ( h ξ i − / γ p κλ p κλ , g ) ,λ ϕ ∈ S ( σ − / h ξ i − γ , g ) ⊂ S ( h ξ i − / γ p κλ p κλ , g )because Cλ ≥ M σ h ξ i − γ , Cλ ≥ σ and κσ ≥
1. A repetition of similar argu-ments proving (10.5) shows that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )[op(Φ ǫ ) , op( T − ∂ t T )] V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) ≤ c ( M, ǫ ) N ( h D i − / γ V ) . Note that Φ ǫ A − A ǫ can be written X | α + β | =1 ( − | α | (2 i ) | α + β | α ! β ! (cid:16) ∂ αx ∂ βξ Φ ǫ ∂ βx ∂ αξ A − ∂ βx ∂ αξ Φ ǫ ∂ αx ∂ βξ A (cid:17) + R ǫ = H ǫ + R ǫ where it follows from (10.8) that R ǫ = S ( σ − / h ξ i − γ ) S ( M h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( h ξ i − γ ) S ( σ − / h ξ i − γ ) S ( M h ξ i − γ ) S ( σ / h ξ i − γ ) S ( h ξ i − γ ) S ( σ / h ξ i − γ ) ∈ S ( c ( M ) h ξ i − / γ , g )for σ ≥ M h ξ i − γ . It is not difficult to see from the proof of Corollary 10.1 that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )op( R ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) ≤ c ( M, ǫ ) N ( h D i − / γ V ) . Study (op(Λ)op( φ − n )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V ). Note that H ǫ ∈ S (1 , g ) be-cause ∂ αx ∂ βξ A ∈ S ( h ξ i −| β | γ , g ) for | α + β | = 1. WriteΦ ǫ = f ǫ ǫ + r ǫ , r ǫ ∈ S ( h ξ i − γ , g )and note φ − n f ǫ − f ǫ φ − n ∈ S ( ω − ρ / h ξ i − γ φ − n , g ) ⊂ S ( M − / φ − n h ξ i − / γ , g ),then a repetition of similar arguments proves that the difference (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1) − (cid:0) op( f )op(Λ)op( φ − n )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) is bounded by c ( M, ǫ ) N ( h D i − / γ V ). Since λ j ∈ S ( λ j , g ) it follows that f ǫ λ j − λ j f ǫ ∈ S ( M − / λ j h ξ i − / γ , g )60hen applying a similar arguments one can see that the difference (cid:12)(cid:12) (op( f ǫ )op(Λ)op( φ − n )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1) − (cid:0) op(Λ)op( φ − n )op( f ǫ )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) is bounded again by c ( M, ǫ ) N ( h D i − / γ V ). Here look at iH ǫ more carefully.Note that iH ǫ = (cid:16) X | α + β | =1 ∂ αx ∂ βξ (˜ a ij [ ξ ])( ∂ βx ∂ αξ f ǫ ) 1 f ǫ Φ ǫ (cid:17) = (cid:0) h ǫij (cid:1) f ǫ Φ ǫ Taking h ǫij ∈ S (1 , g ) and f − ǫ Φ ǫ , Φ ǫ ∈ S (1 , g ) into account one can write f ǫ iH ǫ ) = (cid:0) h ǫij (cid:1) ǫ + R ǫ where R ǫ ∈ S ( M − / h ξ i − / γ , g ) hence denoting ˜ H ǫ = (cid:0) h ǫij (cid:1) it results that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )op( f ǫ )op( iH ǫ ) V, op( φ − n )op(Φ ǫ ) V ) − (op(Λ)op( φ − n )op( ˜ H ǫ )op(Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) is bounded by c ( M, ǫ ) N ( h D i − / γ V ). From Lemma 6.8 we see that h ǫij ∈ C (1) , j ≥ i, h ǫ , h ǫ ∈ C ( σ / ) , h ǫ ∈ C ( σ )then in view of Lemma 5.5 λ i (cid:0) φ − n h ǫij − h ǫij φ − n (cid:1) ∈ S ( κλ i h ξ i − γ φ − n , g ) , j ≥ i,λ (cid:0) φ − n h ǫ − h ǫ φ − n (cid:1) ∈ S ( M − / κλ λ / h ξ i − / γ φ − n , g ) ,λ (cid:0) φ − n h ǫ − h ǫ φ − n (cid:1) ∈ S ( κλ λ / h ξ i − γ , g ) ,λ (cid:0) φ − n h ǫ − h ǫ φ − n (cid:1) ∈ S ( M − / κλ λ / h ξ i − / γ , g ) . From this it follows that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )op( ˜ H ǫ )op(Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V ) − (op(Λ)op( ˜ H ǫ )op( φ − n )op(Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) is bounded by c ( M, ǫ ) N ( h D i − / γ V ). Lemma 12.1.
One can write h ǫij = X | α + β | =1 k ǫijαβ l ijαβ + r ǫij where k ǫijαβ ∈ S (1 , g ) such that | k ǫijαβ | ≤ Cν with some C > independent of ν and ǫ for any ≤ i, j ≤ . As for l ijαβ and r ǫij one has l ijαβ ∈ S (1 , g ) , r ǫij ∈ S ( σ − / h ξ i − γ , g ) , ( j ≥ i ) ,l αβ ∈ S ( σ − / p λ , g ) , r ǫ ∈ S ( M − / σ − / p λ h ξ i − / γ , g ) ,l αβ ∈ S ( σ p λ , g ) , r ǫ ∈ S ( M − / σ p λ h ξ i − / γ , g ) ,l αβ ∈ S ( p λ , g ) , r ǫ ∈ S ( M − / p λ h ξ i − / γ , g ) . roof. Set k ǫijαβ = h ξ i | α | γ ∂ βx ∂ αξ f ǫ and l ijαβ = h ξ i −| α | γ ∂ αx ∂ βξ (˜ a ij [ ξ ]) then the asser-tion for k ǫijαβ is clear from (12.1). The assertions for l ijαβ follow from Lemma6.8, Corollary 6.3 and Lemma 7.7. To prove the assertions for r ǫij note that ∂ µx ∂ νξ l ijαβ ∈ S ( σ − / h ξ i −| ν | γ , g ), | µ + ν | = 1 for j ≥ i and ∂ µx ∂ νξ l αβ , ∂ µx ∂ νξ l αβ ∈ S ( h ξ i −| ν | γ , g ) , ∂ µx ∂ νξ l αβ ∈ S ( σ / h ξ i −| ν | γ , g )for | µ + ν | = 1 which follows from ˜ a , ˜ a ∈ C ( σ ) and ˜ a ∈ C ( σ / ). Thenremarking that σ ≥ M h ξ i − γ and λ ≥ M σ h ξ i − γ the assertions for r ǫij are checkedimmediately.With R ǫ = ( r ǫij ) and W = op( φ − n )op(Φ ǫ ) V , recalling λ ≤ Cσλ ≤ Cσ λ ,it is easy to see that (cid:12)(cid:12) (op( R ǫ ) W, op(Λ) W ) (cid:12)(cid:12) ≤ c ( M, ǫ ) k op(Λ / ) h D i − / γ W k . Turn to (op( h ǫij ) W j , op( λ i ) W i ). Write λ i = λ / i k i ) λ / i with k i ∈ S ( M − , g ) then thanks to Lemma 12.1 it follows that (cid:12)(cid:12) (op( λ i )op( h ǫij ) W j , op( λ i ) W i ) (cid:12)(cid:12) ≤ (1 + CM − ) X | α + β | =1 (cid:12)(cid:12) (op( λ / i )op( l ijαβ ) W j , op( k ǫijαβ )op( λ / i ) W i ) (cid:12)(cid:12) ≤ C (1 + M − ) k op( λ / i )op( l ijαβ ) W j kk op( k ǫijαβ )op( λ / i ) W i k≤ C ′ (1 + M − ) k op( λ / j ) W j kk op( k ǫijαβ )op( λ / i ) W i k because λ / i l ijαβ ∈ S ( λ / j , g ) in view of Lemma 12.1. On the other hand,taking Lemma 12.1 into account, it follows from the sharp G˚arding inequality(e.g. [6, Theorem 18.1.14]) k op( k ǫijαβ )op( λ / i ) W i k ≤ Cν k op( λ / i ) W i k + C ( M, ν, ǫ ) k op( λ / i ) h D i − / γ W i k . Therefore applying the above obtained estimates one can find
C > ǫ , ν and M such that (cid:12)(cid:12) Re (cid:0) op(Λ)op( ˜ H ǫ ) W, W ) (cid:12)(cid:12) ≤ C ( ν + M − / ) k op(Λ / ) W k + C ( M, ν, ǫ ) k op(Λ / ) h D i − / γ W k ≤ C ( ν + M − / ) k op(Λ / )op( φ − n )op(Φ ǫ ) V k + C ′ ( M, ν, ǫ ) k op(Λ / )op( φ − n ) h D i − / γ V k . Since it follows from the same reasoning that (cid:12)(cid:12)(cid:0) op(Λ)op( φ − n )op( f − ǫ Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1) − (cid:0) op(Λ)op( φ − n )op(Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1)(cid:12)(cid:12) ≤ c ( M, ǫ ) N ( h D i − / γ V )62e obtain finally − Re (cid:0) op(Λ)op( φ − n )op( f − ǫ Φ ǫ ) V, op( φ − n )op(Φ ǫ ) V (cid:1) + Re (cid:0) op(Λ)op( φ − n )[op(Φ ǫ ) , op( i A )] V, op( φ − n )op(Φ ǫ ) V (cid:1) ≤ − (cid:0) − C ( ν + M − / ) (cid:1) k op(Λ / )op( φ − n )op(Φ ǫ ) V k + c ( M, ν, ǫ ) N ( h D i − / γ V ) . (12.4)We fix M and ν such that 1 − C ( ν + M − / ) ≥ γ = M and δ = 1 /M are assumed to be fixed.Applying Proposition 10.1 to op(Φ) V instead of V one obtains, in view of (12.3)and (12.4) that Proposition 12.1.
For any < ν ≤ ν and any ǫ > one can find C > suchthat E (op(Φ ǫ ) V ) + Z t τ − n N (op(Φ ǫ ) V ) dτ ≤ C Z t τ − n k op( κ − / Λ / )op( φ − n )op(Φ ǫ ) ˜ LV k dτ + C Z t τ − n N ( h D i − / γ V ) dτ. Applying h D i sγ to (12.3) and repeating similar arguments proving Proposi-tions 11.1 and 12.1 one obtains Proposition 12.2.
For any s ∈ R , any < ν ≤ ν and any ǫ > one can find C > such that E ( h D i sγ op(Φ ǫ ) V ) + Z t τ − n N ( h D i sγ op(Φ ǫ ) V ) dτ ≤ C Z t τ − n +1 kh D i n + sγ op(Φ ǫ ) ˜ LV k dτ + C Z t τ − n N ( h D i s − / γ V ) dτ for ≤ t ≤ δ . Lemma 12.2.
Assume t − n h D i l γ V ∈ L ((0 , δ ) × R d ) and t − n +1 / h D i l γ ˜ LV ∈ L ((0 , δ ) × R d ) and t − n +1 / h D i n + s γ op(Φ ǫ ) ˜ LV ∈ L ((0 , δ ) × R d ) with some l , l ∈ R and s ∈ R . Then for every < ǫ < ǫ we have t − n h D i sγ Φ ǫ V ∈ L ((0 , δ ) × R d ) for all s ≤ s − / . Moreover Z t τ − n kh D i sγ op(Φ ǫ ) V ( τ ) k dτ ≤ C Z t (cid:16) τ − n kh D i l γ V ( τ ) k + τ − n +1 kh D i l γ ˜ LV ( τ ) k (cid:17) dτ + C Z t τ − n +1 kh D i n + s γ op(Φ ǫ ) ˜ LV ( τ ) k dτ or < t ≤ δ .Proof. We may assume l ≤ s otherwise nothing to be proved. Let J be thelargest integer such that l + J/ ≤ s . Take ǫ j > ǫ < ǫ < · · · <ǫ J = ǫ . We write Φ ǫ j = Φ j and f j = f ǫ j in this proof. Inductively we show that Z t τ − n N ( h D i l + j/ γ op(Φ j ) V ) dτ ≤ C Z t τ − n kh D i l γ V ( τ ) k dτ + C Z t τ − n +1 (cid:8) kh D i l γ ˜ LV ( τ ) k + kh D i l + n + j/ γ op(Φ ) ˜ LV ( τ ) k (cid:9) dτ. (12.5)Note that from (10.12) and (10.11) it follows(12.6) kh D i − γ V k /C ≤ N ( V ) ≤ C kh D i nγ V k . Choose ψ j ( x, ξ ) ∈ S (1 , g ) so that supp ψ j ⊂ { f j < } and { f j +1 < } ⊂ { ψ j =1 } . Noting thatop(Φ j +1 ) ˜ L op( ψ j ) = op(Φ j +1 ψ j ) ˜ L + op(Φ j +1 )[ ˜ L, op( ψ j )]we apply Proposition 12.2 with s = l + ( j + 1) /
4, Φ = Φ j +1 and V = op( ψ j ) V .Since Φ j +1 ψ j − Φ j +1 ∈ S −∞ then kh D i l +( j +1) / nγ op(Φ j +1 ) ˜ L op( ψ j ) V k isbounded by c kh D i l +( j +1) / nγ op(Φ j +1 ) ˜ LV k + C ( j ) kh D i l γ V k and hence by(12.7) C ( j ) (cid:8) kh D i l +( j +1) / nγ op(Φ ǫ ) ˜ LV k + (cid:8) kh D i l γ ˜ LV k + kh D i l γ V k (cid:9) because Φ j +1 − k j ǫ ∈ S −∞ with some k j ∈ S (1 , g ). Since ψ j − ˜ k j j ∈ S −∞ with some ˜ k j ∈ S (1 , g ) it follows that N ( h D i l + j/ γ op(Φ j +1 )op( ψ j ) V ) ≤ C N ( h D i l + j/ γ op(Φ j ) V ) + C kh D i l γ V k . Consider N ( h D i l +( j +1) / γ op(Φ j +1 )op( ψ j ) V ). Noting that Φ j +1 ψ j − Φ j +1 ∈ S −∞ the same reasoning shows that N ( h D i l +( j +1) / γ op(Φ j +1 ) V ) ≤ C N ( h D i l +( j +1) / γ op(Φ j +1 )op( ψ j ) V ) + C kh D i l γ V k . (12.8)Multiply (12.8) and (12.7) by t − n and t − n +1 respectively and integrate it from0 to t we conclude from Proposition 12.2 that (12.5) holds for j + 1 and hencefor j = J . Since l + J/ ≤ s and l + J/ > s − / i ( i = 1 , ,
3) be open conic sets in R d × ( R d \ { } ) with relativelycompact basis such that Γ ⋐ Γ ⋐ Γ . Take h i ( x, ξ ) ∈ S (1 , g ) with supp h ⊂ Γ and supp h ⊂ Γ \ Γ . Consider a solution V with t − n h D i lγ V ∈ L ((0 , δ ) × R d )to the equation˜ LV = op( h ) F, t − n +1 / h D i sγ F ∈ L ((0 , δ ) × R d ) . Proposition 12.3.
Notations being as above. There exists δ ′ = δ ′ (Γ i ) > suchthat for any r ∈ R there is C > such that Z t τ − n kh D i rγ op( h ) V ( τ ) k dτ ≤ C Z t (cid:8) τ − n +1 kh D i sγ F ( τ ) k + τ − n kh D i lγ V ( τ ) k (cid:9) dτ, < t ≤ δ ′ . Proof.
Let f ǫ = t − ν ˆ τ + ν d ǫ ( x, ξ ; y, η ) with a small ˆ τ >
0. It is clear thatthere is ˆ ǫ > { t ≥ } ∩ { f ˆ ǫ ≤ } ∩ (cid:0) R × supp h (cid:1) = ∅ for any ( y, η ) / ∈ Γ . Take ˆ ǫ < ˜ ǫ < ˆ τ . It is also clear that one can find a finitenumber of ( y i , η i ) ∈ Γ \ Γ , i = 1 , . . . , M such that with δ ′ = ν (ˆ τ − ˜ ǫ ) / \ Γ ⋐ (cid:16) M [ i =1 { f ˜ ǫ ( δ ′ , x, ξ ; y i , η i ) ≤ } (cid:17) , { t ≥ } ∩ { f ˜ ǫ ( t, x, ξ ; y i , η i ) ≤ } ∩ (cid:0) R × supp h (cid:1) = ∅ . Now Φ iǫ is defined by (12.2) with f ǫ ( t, x, ξ ; y i , η i ). Then since P Φ i ˜ ǫ > , δ ′ ] × supp h there is k ∈ S (1 , g ) such that h − k P Φ i ˜ ǫ ∈ S −∞ . Noting that t − n +1 / h D i rγ op(Φ i ˆ ǫ )op( h ) F ∈ L ((0 , δ ) × R d ) for any r ∈ R we apply Lemma12.2 with Φ ǫ = Φ ˆ ǫ , Φ ǫ = Φ i ˜ ǫ and s = r + 5 / Z t τ − n kh D i rγ op(Φ i ˜ ǫ ) V ( τ ) k dτ ≤ C Z t τ − n kh D i lγ V ( τ ) k dτ + Z t τ − n +1 (cid:0) kh D i n + r +5 / γ op(Φ i ˆ ǫ )op( h ) F ( τ ) k + kh D i sγ F ( τ ) k (cid:1) dτ for kh D i sγ ˜ LV ( τ ) k ≤ C kh D i sγ F ( τ ) k . Since Φ i ˆ ǫ h ∈ S −∞ summing up the aboveestimates over i = 1 , . . . , M one concludes the desired assertion. Lemma 12.3.
The same assertion as Proposition 12.3 holds for L .Proof. Assume that U satisfies LU = op( h ) F, t − n h D i lγ U ∈ L ((0 , δ ) × R d )where t − n +1 / h D i sγ F ∈ L ((0 , δ ) × R d ). Choose ˜Γ i such that Γ ⋐ ˜Γ ⋐ ˜Γ ⋐ Γ ⋐ Γ ⋐ ˜Γ and ˜ h i ∈ S (1 , g ) such that supp ˜ h ⊂ ˜Γ , supp ˜ h ⊂ ˜Γ \ ˜Γ h i = 1 on the support of h i . Recall that L op( T ) = op( T ) ˜ L . Then with U = op( T ) V one has ˜ LV = ( I + op( K ))op( T − )op( h ) F. Since there is ˜ T ∈ S (1 , g ) such that ( I + K ) T − h − ˜ h ˜ T ∈ S −∞ it followsfrom Proposition 12.3 (or rather its proof) that Z t τ − n kh D i rγ op(˜ h ) V ( τ ) k dτ ≤ C Z t (cid:8) τ − n kh D i lγ V ( τ ) k + τ − n +1 kh D i sγ F ( τ ) k } dτ. Similarly since there is ˜ T ∈ S (1 , g ) such that h T − ˜ h ˜ T ∈ S −∞ repeating thesame arguments we conclude the assertion.Returning to ˆ P we have Proposition 12.4.
Notations being as above. Then there exists δ ′ = δ ′ (Γ i ) > such that for any s , r ∈ R there is C such that for any solution u to ˆ P u = op( h ) f, t − n h D i l +2 − j D jt u ∈ L ((0 , δ ′ ) × R d ) , j = 0 , , with t − n +1 / h D i s f ∈ L ((0 , δ ′ ) × R d ) one has Z t τ − n X j =0 kh D i r +2 − j op( h ) D jt u ( τ ) k dτ ≤ C Z t (cid:8) τ − n +1 kh D i s f ( τ ) k + τ − n X j =0 kh D i l +2 − j D jt u ( τ ) k (cid:9) dτ, < t ≤ δ ′ . Denote by H n,s ((0 , δ ) × R d ) the set of all u such that Z δ τ − n kh D i s f ( τ, · ) k dτ < + ∞ . Thanks to Theorem 11.1 for any f ∈ H − n +1 / ,n + s ((0 , δ ) × R d ) there is a uniquesolution u ∈ H − n,s +1 ((0 , δ ) × R d ) to ˆ P u = f satisfying (11.6). Denote this mapby ˆ G : H − n +1 / ,n + s ((0 , δ ) × R d ) ∋ f u ∈ H − n,s +1 ((0 , δ ) × R d )then it follows from Proposition 12.4 and Theorem 11.1 that Z t τ − n X j =0 kh D i r +2 − j op( h ) D jt ˆ G op( h ) f ( τ ) k dτ ≤ C Z t (cid:8) τ − n +1 kh D i n + s f ( τ ) k + τ − n X j =0 kh D i s +1 − j D jt u ( τ ) k (cid:9) dτ ≤ C Z t τ − n +1 kh D i n + s f ( τ ) k dτ. n + s by s and r + 2 by r we obtain Z t τ − n X j =0 kh D i r − j op( h ) D jt ˆ G op( h ) f ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s f ( τ ) k dτ. Summarizing we conclude
Proposition 12.5.
Notations being as above and let Γ i ( i = 1 , , be openconic sets in R d × ( R d \{ } ) with relatively compact basis such that Γ ⋐ Γ ⋐ Γ and h i ( x, ξ ) ∈ S (1 , g ) with supp h ⊂ Γ and supp h ⊂ Γ \ Γ . Then thereexists δ ′ = δ ′ (Γ i ) > such that for any r , s one can find C > such that Z t τ − n X j =0 kh D i r − j op( h ) D jt ˆ G op( h ) f ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s f ( τ ) k dτ, < t ≤ δ ′ for any f ∈ H − n +1 / ,s ((0 , δ ′ ) × R d ) . Denote by H ∗ n,s ((0 , δ ] × R d ) the set of all f with R ∞ t n kh D i s f k dt < + ∞ such that f = 0 for t ≥ δ . Thanks to Theorem 11.2 for any f ∈H ∗ n +1 / ,n + s ((0 , δ ] × R d ) there is a unique solution u ∈ H ∗ n,s +1 ((0 , δ ] × R d ) toˆ P ∗ u = f satisfying (11.6). Denote this map byˆ G ∗ : H ∗ n +1 / ,n + s ((0 , δ ] × R d ) ∋ f u ∈ H ∗ n,s +1 ((0 , δ ] × R d ) . Repeating similar arguments one obtains
Proposition 12.6.
Notations being as in Proposition 12.5. Then there exists δ ′ = δ ′ (Γ i ) > such that for any r , s one can find C > such that Z δ ′ t τ n X j =0 kh D i r − j op( h ) D jt ˆ G ∗ op( h ) f ( τ ) k dτ ≤ C Z δ ′ t τ n +1 kh D i s f ( τ ) k dτ, < t ≤ δ ′ for any f ∈ H ∗ n +1 / ,s ((0 , δ ′ ] × R d ) . Remark 12.1.
It is clear from the proof that for any n ′ ≥ n , Propositions 12.5and 12.6 hold.
13 Proof of Theorem 1.1
Applying the fact that the micro support of u ( t, · ), solution to ˆ P u = f obtainedby Theorem 11.1, propagates with a finite speed (Proposition 12.5) we proveTheorem 1.1 following [22], [24]. 67 Consider(13.1) P = D mt + m X j =1 a j ( t, x, D ) h D i j D m − jt which is differential operator in t with coefficients a j ∈ S . We say that G isa parametrix for P with finite propagation speed of micro supports (which weabbreviate to “parametrix with fps” from now on) with loss of ( n, l ) derivativesif G satisfies the following conditions:(i) There exists δ > s ∈ R there is C >
P Gf = f and m − X j =0 Z t τ − n kh D i − l + s + m − j D jt Gf ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s f ( τ ) k dτ, f ∈ H − n +1 / ,s ((0 , δ ) × R d ) . (ii) For any h j ( x, ξ ) ∈ S (1 , g ), j = 1 , h ⋐ ( R d × R d ) \ supp h there exists δ ′ > r , s ∈ R there is C > m − X j =0 Z t τ − n kh D i r − j op( h ) D jt G op( h ) f ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s f ( τ ) k dτ, < t ≤ δ ′ (13.2)holds for any f ∈ H − n +1 / ,s ((0 , δ ′ ) × R d ).Let P and P be two operators of the form (13.1). We say P ≡ P at (ˆ x, ˆ ξ )if there exist δ ′ > W of (ˆ x, ˆ ξ ) such that(13.3) P − P = m X j =1 R j ( t, x, D ) h D i j D m − jt with R j ∈ S which are in S −∞ ( W ) uniformly in 0 ≤ t ≤ δ ′ . Theorem 13.1.
Assume that for any (ˆ x, η ) , | η | = 1 one can find P η of theform (13.1) having a parametrix with fps with loss of ( n, ℓ ( η )) derivatives suchthat P ≡ P η at (ˆ x, η ) . Then there exist δ > , ℓ ≥ and a neighborhood U of ˆ x such that for every f ∈ H − n +1 / ,s + ℓ ((0 , δ ) × R d ) there exists u with D jt u ∈ H − n,s + m − j ((0 , δ ) × R d ) , ≤ j ≤ m − , satisfying P u = f in (0 , δ ) × U where ℓ = sup | η | =1 ℓ ( η ) . roof. By assumption P η has a parametrix G η with fps with loss of ( n, ℓ ( η ))derivatives. There are finite open conic neighborhood W i of (ˆ x, η i ) such that ∪ i W i ⊃ Ω × ( R d \ { } ), where Ω is a neighborhood of ˆ x , and P ≡ P η i at(ˆ x, η ) with W = W i in (13.3). Now take another open conic covering { V i } ofΩ × ( R d \ { } ) with V i ⋐ W i , and a partition of unity { α i ( x, ξ ) } subordinate to { V i } so that X i α i ( x, ξ ) = α ( x )where α ( x ) is equal to 1 in a neighborhood of ˆ x . Define G = X i G η i α i . Then denoting P − P η i = R i we have P Gf = X i P G η i α i f = X i P η i G i α i f + X i R i G i α i f = α ( x ) f − Rf where R = P i R i G η i α i . Then Z t τ − n kh D i s + ℓ Rf ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s + ℓ f ( τ ) k dτ for 0 ≤ t ≤ δ ′′ with some δ ′′ > ℓ = max i ℓ ( η i ).Choosing δ > Z t τ − n kh D i s + ℓ Rf ( τ ) k dτ ≤ Z t τ − n kh D i s + ℓ f ( τ ) k dτ, < t ≤ δ for f ∈ H − n,s + ℓ ((0 , δ ) × R d )). With S = P ∞ k =0 R k one has Sf ∈ H − n,s + ℓ ((0 , δ ) × R d ) and Z t τ − n kh D i s + ℓ Sf ( τ ) k dτ ≤ Z t τ − n kh D i s + ℓ f ( τ ) k dτ, < t ≤ δ . Let γ ( x ) ∈ C ∞ ( R d ) be equal to 1 near ˆ x such that supp γ ⋐ { α = 1 } . Since γ ( α − R ) S = γ ( I − R ) S = γ it follows that γ ( x ) P GSf = γ ( x ) f. With u = GSf one has m − X j =0 Z t τ − n kh D i s + m − j D jt u ( τ ) k dτ ≤ C Z t τ − n +1 kh D i s + ℓ Sf ( τ ) k dτ ≤ C ′ Z t τ − n kh D i s + ℓ f ( τ ) k dτ which proves the assertion. 69e define a parametrix with fps for P ∗ with obvious modifications then Theorem 13.2.
Assume that for any (ˆ x, η ) , | η | = 1 one can find P ∗ η of theform (13.1) such that P ∗ ≡ P ∗ η for which parametrix with fps exists. Thenthere exist δ > , ℓ ≥ and a neighborhood U of ˆ x such that for every f ∈H ∗ n +1 / ,s + ℓ ((0 , δ ] × R d ) there exists u with D jt u ∈ H ∗ n,s + m − j ((0 , δ ] × R d ) , ≤ j ≤ m − , satisfying P ∗ u = f in (0 , δ ) × U. First consider a third order operator P of the form (2.1). To reduce P to thecase a ( t, x, D ) = 0 we apply a Fourier integral operator, which is actually thesolution operator S ( t ′ , t ) of the Cauchy problem D t u + a ( t, x, D x ) u = 0 , u ( t ′ , x ) = φ ( x )such that S ( t ′ , t ) : φ u ( t ) then it is clear that S ( t, D t + a ) S (0 , t ) = D t .Let S ( t, P S (0 , t ) = ˜ P and assume that ˜ P has a parametrix with fps ˜ G with loss of ( n, ℓ ) derivatives.Then one can show that G = S (0 , t ) ˜ GS ( t,
0) is a parametrix of P with fps withloss of ( n, ℓ ) derivatives.Let | η | = 1 be given. Assume that p has a triple characteristic root ¯ τ at(0 , , η ) and (0 , , ¯ τ, η ) is effectively hyperbolic. Theorem 11.1 and Proposition12.4 imply that ˆ P , which coincides with the original P in W M , given by (4.3),has a parametrix with fps with loss of ( n, n + 2) derivatives. Now assume that p has a double characteristic root ¯ τ at (0 , , η ) such that (0 , , ¯ τ , η ) is effectivelyhyperbolic characteristic if it is a critical point. Note that one can write p ( t, x, τ, ξ ) = (cid:0) τ + b ( t, x, ξ ) (cid:1)(cid:0) τ + a ( t, x, ξ ) τ + a ( t, x, ξ ) (cid:1) = p p in a conic neighborhood of (0 , , η ) where p (0 , , ¯ τ, η ) = 0. Note that thereexist ˆ P i such that P ≡ ˆ P · ˆ P at (0 , η )where the principal symbol of ˆ P j coincides with p j in a conic neighborhoodof (0 , , η ). Note that if ˆ P i has a parametrix with fps G i with loss of ( n, ℓ i )derivatives then one can see that G G is a parametrix with fps for ˆ P · ˆ P withloss of ( n, ℓ + ℓ ) derivatives.First assume that (0 , , ¯ τ, η ) is a critical point. Then it is easy to see that F p (0 , , ¯ τ , η ) = cF p (0 , , ¯ τ , η )with some c = 0 and hence (0 , , ¯ τ , η ) is effectively hyperbolic characteristic of p . Then following [22, 24] there is a parametrix with fps for ˆ P . Since ˆ P is afirst order operator with real principal symbol p it is easy to see that ˆ P has a70arametrix with fps. Therefore P has a parametrix with fps. Turn to the casethat (0 , , ¯ τ, η ) is not a critical point. Writing p as p ( t, x, τ, ξ ) = τ − a ( t, x, ξ ) | ξ | it is easily seen that (0 , , ¯ τ , η ) is not a critical point implies that ∂ t a (0 , , η ) > P is a hyperbolic operator of principal type and somedetailed discussion is found in [6, Chapter 23.4]. It is easily proved that ˆ P hasa parametrix with fps, because it suffices to employ the weight t − n ( φ − n is nowabsent) in order to obtain weighted energy estimates.Turn to the general case. Let | η | = 1 be arbitrarily fixed. Write p (0 , , τ, η ) = Q rj =1 (cid:0) τ − τ j ) m j where P m j = m and τ j are real and distinct from each other,where m j ≤ δ > U of (0 , η ) such that one can write p ( t, x, τ, ξ ) = r Y j =1 p ( j ) ( t, x, τ, ξ ) ,p ( j ) ( t, x, τ, ξ ) = τ m j + a j, ( t, x, ξ ) τ m j − + · · · + a j,m j ( t, x, ξ )for ( t, x, ξ ) ∈ ( − δ, δ ) × U where a j,k ( t, x, ξ ) are real valued, homogeneous ofdegree k in ξ and p ( j ) (0 , , τ, η ) = ( τ − τ j ) m j and p ( j ) ( t, x, τ, ξ ) = 0 has onlyreal roots in τ for ( t, x, ξ ) ∈ [0 , δ ) × U . If (0 , , τ j , η ) is a critical point of p , andnecessarily m j ≥
2, then (0 , , τ j , η ) is a critical point of p ( j ) and it is easy tosee F p (0 , , τ j , η ) = c j F p ( j ) (0 , , τ j , η )with some c j = 0 and hence F p ( j ) (0 , , τ j , η ) has non-zero real eigenvalues if F p (0 , , τ j , η ) does and vice versa. It is well known that one can find P ( j ) suchthat P ≡ P (1) P (2) · · · P ( r ) at (0 , η )where P ( j ) are operators of the form (13.1) with m = m j whose principal symbolcoincides with p ( j ) in some conic neighborhood of (0 , , η ). Since each P ( j ) hasa parametrix with fps thanks to Theorem 11.1 and Proposition 12.5 hence sodoes P . Therefore Theorem 1.1 results from Theorem 13.1 noting Remark 12.1.Repeating a parallel arguments to the existence proof for P above we obtain Theorem 13.3.
Under the same assumption as in Theorem 1.1 there exist δ > , a neighborhood U of the origin and n > such that for any s ∈ R and any f ∈ H ∗ n +1 / ,s ((0 , δ ] × R d ) there exists u with D jt u ∈ H ∗ n, − n − s + m − j ((0 , δ ] × R d ) , j = 0 , , . . . , m − satisfying P ∗ u = f in (0 , δ ) × U. Now we prove a local uniqueness result for the Cauchy problem for P ap-plying Theorem 13.3. From the assumption one can find a neighborhood W of the origin of R d and T > p on71 t, x, ξ ) ∈ (0 , T ) × W is at most double and double characteristic is effectivelyhyperbolic. Let f ∈ C ∞ ((0 , δ ′ ) × {| x | < ε } ) ( δ ′ ≤ T ) and let v be a solution to P ∗ v = f vanishing in t ≥ δ ′ . Then thanks to [15, Main Theorem] there existsˆ c > x v ( t, · ) ⊂ {| x | ≤ ε + ˆ c δ ′ } , < t ≤ δ ′ . Now assume that u satisfies P u = 0 in (0 , δ ) × U and ∂ kt u (0 , x ) = 0 for all k .Choose ε > δ ′ > {| x | ≤ ε + ˆ c δ ′ } ⊂ U , δ ′ ≤ δ . Then we see0 = Z δ ′ (cid:0) P u, v (cid:1) dt = Z δ ′ (cid:0) u, P ∗ v (cid:1) dt = Z δ ′ (cid:0) u, f (cid:1) dt. Since f ∈ C ∞ ((0 , δ ′ ) × {| x | < ε } ) is arbitrary, we conclude that u ( t, x ) = 0 , ( t, x ) ∈ (0 , δ ′ ) × {| x | ≤ ε } . Theorem 13.4. If u ( t, x ) ∈ C ∞ ([0 , δ ) × U ) satisfies P u = 0 in [0 , δ ) × U and ∂ kt u (0 , x ) = 0 for all k then u = 0 in a neighborhood of (0 , . References [1] R. Beals: Characterization of pseudodifferential operators and applications,Duke Math. J., (1977), 45-57.[2] E.Bernardi, A.Bove, V.Petkov: Cauchy problem for effectively hyperbolicoperators with triple characteristics of variable multiplicity , Journal Hyper.Differ. Equ., (2015), 535-579.[3] E.Bernardi, T.Nishitani: Counterexamples to C ∞ well posedness for somehyperbolic operators with triple characteristics , Proc. Japan Acad., , Ser.A (2015), 19-24.[4] L.H¨ormander: The Cauchy problem for differential equations with doublecharacteristics , J. Anal. Math., (1977), 118-196.[5] L.H¨ormander: The Analysis of Linear Partial Differential Operators, I,Springer, Berlin, 1990.[6] L.H¨ormander: The Analysis of Linear Partial Differential Operators, III,Springer, Berlin, 1985.[7] V.Ivrii and V.Petkov: Necessary conditions for the Cauchy problem fornon-strictly hyperbolic equations to be well posed , Uspekhi Mat. Nauk, (1974), 3-70, English translation: Russ. Math. Surv., (1974), 1-70.[8] V.Ivrii: Sufficient conditions for regular and completely regular hyperbolic-ity , Tr. Mosk. Mat. Obs., (1975), 3-65 (in Russian), English translation:Trans. Mosc. Math. Soc., (1978), 1-65.729] V.Ivrii: The well-posed Cauchy problem for non-strictly hyperbolic opera-tors, III: The energy integral , Tr. Mosk. Mat. Obs., (1977), 151-170 (inRussian), English translation: Trans. Mosc. Math. Soc., (1978), 149-168.[10] V.Ivrii: Linear Hyperbolic Equations, In: Encyclopaedia of Mathemati-cal Science, Partial Differential Equations IV, Yu.V.Egorov, M.A.Shubin(Eds.) pp. 149-235 (1993).[11] N.Iwasaki: The Cauchy problem for effectively hyperbolic equations (a spe-cial case) , J. Math. Kyoto Univ., (1983), 503-562.[12] N.Iwasaki: The Cauchy problem for effectively hyperbolic equations (a stan-dard type) , Publ. Res. Inst. Math. Sci., (1984), 551-592.[13] N.Iwasaki: The Cauchy problem for effectively hyperbolic equations (generalcase) , J. Math. Kyoto Univ., (1985), 727-743.[14] E.Jannelli: The hyperbolic symmetrizer: theory and applications , in: Ad-vances in Phase Space Analysis of Partial Differential Equations, in: Progr.Nonlinear Differential Equations Appl., vol. , Birkh¨auser Boston, 2009,pp. 113-139.[15] K.Kajitani, S.Wakabayashi, T.Nishitani; The Cauchy problem for hyper-bolic operators of strong type , Duke Math. J. , No. 2 (1994), 353-408.[16] H.Kumano-go: Pseudo-Differential Operators, The MIT Press, Cambridge,Massachusetts, and London, 1974.[17] P.D.Lax: Asymptotic solutions of oscillatory initial value problem , DukeMath. J., (1957), 627-646.[18] N.Lerner, Metrics on the Phase Space and Non-selfadjoint Pseudo-Differential Operators, Birkh¨auser, Basel, 2010.[19] S.Mizohata: The Theory of Partial Differential Equations, Cambridge Uni-versity Press, Cambridge, 1973.[20] R.Melrose: The Cauchy problem for effectively hyperbolic operators ,Hokkaido Math. J., (1983), 371-391.[21] T.Nishitani: On the finite propagation speed of wave front sets for effectivelyhyperbolic operators , Sci. Rep. College Gen. Ed. Osaka Univ., (1983),1-7.[22] T.Nishitani: The Cauchy problem for effectively hyperbolic operators , InNonlinear variational problems, A.Marino, L.Modica, S.Spagnolo, M. De-Giovanni eds., Rec. Notes in Math. , Pitman, London, 1985, pp. 9-23.[23] T.Nishitani:
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