LECTURE11:L?-THEORY OF SEMICLASSICALPsDOs:BOUNDEDNESSIn the previous several lectures, we have studied the definition and basic prop-erties of semiclassical pseudodifferential operators, but mainly as an operator actingon(Rn).However, as wehave seen, in quantum part (=the spectral part) of thestory the natural space should be a Hilbert space:(Rn) is not.In the next sev-erallecturesweshall studyproperties of semiclassicalpseudodifferentialoperatorsas linear operators acting on L?(Rn), or in cases we need more regularity, acting onthe Sobolev spaces H"(R").1.L?-BOUNDEDNESS OFOP(a)FOR SCHWARTZSYMBOLSSuppose a = a(r, ) E J(R2n) is a Schwartz function. Then as we have seen,the operator aw, or more generally, the operator Opi(a) for any t e [0,1], maps'(Rn) into(R"). In particular, these operators are linear maps from L?(R")intoL?(IRn).It turns out thatfor a Schwartz symbol, the operator Opt(a)is alwaysa bounded linear operator (and as we will prove next time, is a compact operator)on L?(R"). In what follows we will provide two different proofs of this fact.I Schur's test.To prove the L2-boundedness of linear operators like Opt(a) which are definedby Schwartz kernels, a very useful criterion is the following Schur's test:Lemma1.1 (Schur's Test).Let K :RnxRn→Cbe a continuous function satisfying[K(r, y)]dy< +o0Ci = supandC2= supIK(r, y)]dr < +00,Rand let A be the linear operator with Schwartz kernel K:Au(r) =K(r,y)u(y)dyThenAisa boundedlinear operatorfromL?(Rn)toL?(Rn)with(1)IIAllc(L2(R") ≤(CIC2)Proof. For any u E L?(IR") the Cauchy-Schwartz inequality gives[Au(r)?≤ / [K(a,g)]dy./ [K(r,9)]lu(g)P’dy ≤C1 /[K(r, )]lu(g)'dy
LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS In the previous several lectures, we have studied the definition and basic properties of semiclassical pseudodifferential operators, but mainly as an operator acting on S (R n ). However, as we have seen, in quantum part (=the spectral part) of the story the natural space should be a Hilbert space: S (R n ) is not. In the next several lectures we shall study properties of semiclassical pseudodifferential operators as linear operators acting on L 2 (R n ), or in cases we need more regularity, acting on the Sobolev spaces Hs (R n ). 1. L 2 -boundedness of Opt ~ (a) for Schwartz symbols Suppose a = a(x, ξ) ∈ S (R 2n ) is a Schwartz function. Then as we have seen, the operator ba W , or more generally, the operator Opt ~ (a) for any t ∈ [0, 1], maps S 0 (R n ) into S (R n ). In particular, these operators are linear maps from L 2 (R n ) into L 2 (R n ). It turns out that for a Schwartz symbol, the operator Opt ~ (a) is always a bounded linear operator (and as we will prove next time, is a compact operator) on L 2 (R n ). In what follows we will provide two different proofs of this fact. ¶ Schur’s test. To prove the L 2 -boundedness of linear operators like Opt ~ (a) which are defined by Schwartz kernels, a very useful criterion is the following Schur’s test: Lemma 1.1 (Schur’s Test). Let K : R n×R n → C be a continuous function satisfying C1 = sup x Z Rn |K(x, y)|dy < +∞ and C2 = sup y Z Rn |K(x, y)|dx < +∞, and let A be the linear operator with Schwartz kernel K: Au(x) = Z Rn K(x, y)u(y)dy. Then A is a bounded linear operator from L 2 (R n ) to L 2 (R n ) with (1) kAkL(L2(Rn)) ≤ (C1C2) 1 2 Proof. For any u ∈ L 2 (R n ) the Cauchy-Schwartz inequality gives |Au(x)| 2 ≤ Z Rn |K(x, y)|dy · Z Rn |K(x, y)||u(y)| 2 dy ≤ C1 Z Rn |K(x, y)||u(y)| 2 dy. 1
2LECTURE 11:L?-THEORY OF SEMICLASSICAL PSDOS:BOUNDEDNESSIntegrating withrespectto ,weget(C1- / K(r,y)llu(g)Pdy) dr≤CiCall/2IIAul/≤/口L?-boundedness for PsDOs with Schwartz symbolsAs an immediate consequence,Theorem 1.2. If a = a(r, E) is a Schwartz function, then for any t e [0, 1],Opt(a) : L?(Rn) → L?(Rn)is a bounded linear operator withI(a)lc(L2(R")≤sup sup, (ga)(,)/(R)lal<≤n+1Proof. Recall that the Schwartz kernel of the operator Opt(a) iset(a-)-Sa(tr +(1 -t)y,s)d =b(tr+(1 -t)yka(r,y)(2元h)n(2元h)*节where b is the“partial Fourier transform" of a given bye-it"a(r,s)deb(r, z) = Fe-→2[a(r,E)] =Since a is a Schwartz function, b is also a Schwartz function (Check this!).Thus1[6(tr + (1 - t)y, "二-[kg(r,y)]dr =/da(2元h)n方1[b(y - thz,z)[dz(2元)n1e)-n-1dz . sup Kz)n+1b(y,z2)l(2元)n.2e1(2)-n-1dz·sup,sup, l[Fe2al](y,z)l~(Rn)≤(2元）ny[al<n)-n-1dz·supsup1[Fg-(oga)](y,z)/L(Rn)(2元)nyJal<n1(z)-n-1dz·sup sup Il(oga)(y,E)llzi(R)≤Ci :(2元)nyJal<n+and similarly1(2)-n-dz·sup sup I(oga)(r,E)li(R)[ka(r, y)]dy ≤ C2(2元)nJa<n+口Now the conclusion follows from Schur's test
2 LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS Integrating with respect to x, we get kAuk 2 L2 ≤ Z Rn C1 · Z Rn |K(x, y)||u(y)| 2 dy dx ≤ C1C2kuk 2 L2 . ¶ L 2 -boundedness for PsDOs with Schwartz symbols. As an immediate consequence, Theorem 1.2. If a = a(x, ξ) is a Schwartz function, then for any t ∈ [0, 1], Opt ~ (a) : L 2 (R n ) → L 2 (R n ) is a bounded linear operator with kOpt ~ (a)kL(L2(Rn)) ≤ sup x sup |α|≤n+1 k(∂ α ξ a)(x, ξ)kL1(Rn ξ ) . Proof. Recall that the Schwartz kernel of the operator Opt ~ (a) is k a t (x, y) = 1 (2π~) n Z Rn e i ~ (x−y)·ξ a(tx + (1 − t)y, ξ)dξ = 1 (2π~) n b(tx + (1 − t)y, y − x ~ ), where b is the “partial Fourier transform” of a given by b(x, z) = Fξ→z[a(x, ξ)] = Z Rn e −iξ·z a(x, ξ)dξ. Since a is a Schwartz function, b is also a Schwartz function (Check this!). Thus Z Rn |k a t (x, y)|dx = 1 (2π~) n Z Rn |b(tx + (1 − t)y, y − x ~ )|dx = 1 (2π) n Z Rn |b(y − t~z, z)|dz ≤ 1 (2π) n Z Rn hzi −n−1 dz · sup y,z∈Rn |hzi n+1b(y, z)| ≤ 1 (2π) n Z Rn hzi −n−1 dz · sup y sup |α|≤n+1 kz α [Fξ→za](y, z)kL∞(Rn z ) ≤ 1 (2π) n Z Rn hzi −n−1 dz · sup y sup |α|≤n+1 k[Fξ→z(∂ α ξ a)](y, z)kL∞(Rn z ) ≤ C1 := 1 (2π) n Z Rn hzi −n−1 dz · sup y sup |α|≤n+1 k(∂ α ξ a)(y, ξ)kL1(Rn ξ ) and similarly Z Rn |k a t (x, y)|dy ≤ C2 := 1 (2π) n Z Rn hzi −n−1 dz · sup x sup |α|≤n+1 k(∂ α ξ a)(x, ξ)kL1(Rn ξ ) . Now the conclusion follows from Schur’s test.
3LECTURE11:L?-THEORYOFSEMICLASSICALPSDOS:BOUNDEDNESSRemark. One can also prove the L?-boundedness of Opt(a) directly as follows: Westart with Weyl's decomposition (c.f.the formula (11)in Lecture 9,page 7)[op(e(r+)] [(Fr)ss)-(w)a(y, )dydn.(Opt(a))(r) =(2元h)2nSince the operator Opi(et(u+n-s) is unitary on L2(R"), the triangle inequality im-pliesI[(Fn)(s,s)(y,n)al(y, n)]dydn =IIOp;(a)l(L2(R") ≤(2元h)2n(2m)2~1/ (g.5)(0,)al/ 1.It remains to estimate F(s.t)-(yn)alli. Note that here we are using the usualFourier transform, not the semiclassical one. So our estimate is uniform w.r.t. h.We state and prove a general result:The L1-norm of the Fourier transform of a Schwartz function can becontrolled by the Ll-norm of its partial derivatives:Lemma 1.3. There erists a constant C = Cn such that for anyaE(Rn),Falli≤Cnsuplaallzi.al<n+1Proof.We haveIIFal/μ = / [Fa()] ds ≤ [ (5)-n-1 d I(s)n+1 Fa()lL≤Cn,supllsaFallL≤Cnsuploallzi.[al<n+1al≤n+1As a consequence, we get:Proposition 1.4. For any a E (R2n), and any t e [0, 1](2)IIOp(a)lc(L2(R")≤Cnsup IIoallLI.lall<2n+2.L?BOUNDEDNESSFORMOREGENERALSYMBOLSBoundedness of symbols v.s. L?-boundedness of operators.We would like to extend the L?-boundedness result we proved above to semi-classical pseudo-differential operators with symbols in more general classes. This isnot always possible. For example,·NeithertheoperatorQj=“multiplication by
LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS 3 Remark. One can also prove the L 2 -boundedness of Opt ~ (a) directly as follows: We start with Weyl’s decomposition (c.f. the formula (11) in Lecture 9, page 7) (Opt ~ (a))(x) = 1 (2π~) 2n Z R2n h Opt ~ (e i ~ (y·x+η·ξ) ) i [(F~)(s,ξ)→(y,η)a](y, η)dydη. Since the operator Opt ~ (e i ~ (y·x+η·ξ) is unitary on L 2 (R n ), the triangle inequality implies kOpt ~ (a)kL(L2(Rn)) ≤ 1 (2π~) 2n Z R2n |[(F~)(s,ξ)→(y,η)a](y, η)|dydη = 1 (2π) 2n kF(s,ξ)→(y,η)akL1 . It remains to estimate kF(s,ξ)→(y,η)akL1 . Note that here we are using the usual Fourier transform, not the semiclassical one. So our estimate is uniform w.r.t. ~. We state and prove a general result: The L 1 -norm of the Fourier transform of a Schwartz function can be controlled by the L 1 -norm of its partial derivatives: Lemma 1.3. There exists a constant C = Cn such that for any a ∈ S (R n ), kFakL1 ≤ Cn sup kαk≤n+1 k∂ α akL1 . Proof. We have kFakL1 = Z Rn |Fa(ξ)| dξ ≤ Z Rn hξi −n−1 dξ · khξi n+1Fa(ξ)kL∞ ≤ Cn sup |α|≤n+1 kξ αFakL∞ ≤ Cn sup kαk≤n+1 k∂ α akL1 . As a consequence, we get: Proposition 1.4. For any a ∈ S (R 2n ), and any t ∈ [0, 1], (2) kOpt ~ (a)kL(L2(Rn)) ≤ Cn sup kαk≤2n+1 k∂ α x,ξakL1 . 2. L 2 boundedness for more general symbols ¶ Boundedness of symbols v.s. L 2 -boundedness of operators. We would like to extend the L 2 -boundedness result we proved above to semiclassical pseudo-differential operators with symbols in more general classes. This is not always possible. For example, • Neither the operator Qj = “multiplication by xj”
4LECTURE11:L?-THEORYOFSEMICLASSICALPSDOS:BOUNDEDNESSnor the operatorhaPj=ioriisbounded onL?(Rn)(check thisstatementbyproviding counterexamples!).This is reasonable since neither the function i (the classical counterpartof Qi) nor the function E, (the classical counterpart of P) are boundedfunctions..On the otherhand,ifa(r)is a bounded continuous function, namely a(r)l≤Cfor all r eRn, then obviously the operatorOpi(a) = Ma(a) =“multiplication by a(r)"is abounded operatoron L?since/ la(r)u(r)’drI/Ma(a) ul 2 =≤Cllulz2.. Similarly if a(s) is a bounded function, i.e. [a($)I ≤ C, thenOp(a) = Fr-1 0 Ma(s) 0 Fhis a bounded operator on L?, since the Plancherel's Theorem (c.f. Lecture4, Prop. 1.7)implies1CI/Oph(a)ullL2 =(2h)/l/Ful/2 = Il/2.(2 h)/lMa()Fu/2 Calderon-Vaillancourt Theorem: Idea of proof.So one may guess that for any bounded symbol a(r, E), the operator Oph(a) isa bounded operator on L?. Unfortunately this is not quite true in general. (I needan example here.) However, we will prove that if a E S(1), namely if we assume thesymbol a itself togetherwith all its derivatives are bounded, then Opt(a)is boundedon L?(Rn):Theorem 2.1 (Calderon-Vaillancourt). If a E S(1), then the operatorOpt(a) : L?(Rn) → L?(Rn)is a bounded linear operator on L?(Rn)with(3)IIOpi(a)lc(L2(R) ≤C Z hl/2 sup|8°alR21lal<Mnfor some universal constant MlIn the rest of this lecture, we prove Calderon-Vaillancourt's theorem.The ideais the following:IWe can take Mn to be 10n + 6
4 LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS nor the operator Pj = ~ i ∂ ∂xj is bounded on L 2 (R n ) (check this statement by providing counterexamples!). This is reasonable since neither the function xj (the classical counterpart of Qj ) nor the function ξj (the classical counterpart of Pj ) are bounded functions. • On the other hand, if a(x) is a bounded continuous function, namely |a(x)| ≤ C for all x ∈ R n , then obviously the operator Opt ~ (a) = Ma(x) = “multiplication by a(x)” is a bounded operator on L 2 since kMa(x)ukL2 = Z |a(x)u(x)| 2 dx−1/2 ≤ CkukL2 . • Similarly if a(ξ) is a bounded function, i.e. |a(ξ)| ≤ C, then Opt ~ (a) = F −1 ~ ◦ Ma(ξ) ◦ F~ is a bounded operator on L 2 , since the Plancherel’s Theorem (c.f. Lecture 4, Prop. 1.7) implies kOpt ~ (a)ukL2 = 1 (2π~) n/2 kMa(ξ)F~ukL2 ≤ C (2π~) n/2 kF~ukL2 = CkukL2 . ¶ Calderon-Vaillancourt Theorem: Idea of proof. So one may guess that for any bounded symbol a(x, ξ), the operator Opt ~ (a) is a bounded operator on L 2 . Unfortunately this is not quite true in general. (I need an example here.) However, we will prove that if a ∈ S(1), namely if we assume the symbol a itself together with all its derivatives are bounded, then Opt ~ (a) is bounded on L 2 (R n ): Theorem 2.1 (Calderon-Vaillancourt). If a ∈ S(1), then the operator Opt ~ (a) : L 2 (R n ) → L 2 (R n ) is a bounded linear operator on L 2 (R n ) with (3) kOpt ~ (a)kL(L2(Rn)) ≤ C X |α|≤Mn ~ |α|/2 sup R2n |∂ α a| for some universal constant M. 1 In the rest of this lecture, we prove Calderon-Vaillancourt’s theorem. The idea is the following: 1We can take Mn to be 10n + 6
LECTURE11:L?-THEORYOFSEMICLASSICALPSDOS:BOUNDEDNESS5.Wefirst decompose a into countably many compactly-supported symbolsa=Em am.This can be done by choosing any partition of unity 1 = m Xmsuch that each Xm is compactly supported, and letting am = Xma. ThenformallywehaveOpk(a) =Opk(am),and by compactness of supp(am), each Opi(am) is L?-bounded.? In general, if A=Am, to conclude the boundedness of A from the bound-edness of Am's,a necessary condition weneed is thatthe bound of Am's is uniform for all m.In view of (2),this“uniformlyboundedness"can be fulfilled if we chooseour partition of unity in a“"uniform" way.-The“uniformly boundedness" of components is still not enough, sincethere may be “interactions" between different Am's. Of course the bestdream is that if we can choose these Am's so that there are“no interac-tion",or in other words,if these Am's are"orthogonal"to each other.(namelyifthedecompositionA=mAmisan“"orthogonaldecompo-sition" A =@mAm), then the“uniformly boundedness"of Am's doesimply the boundedness of A. But that's only a dream: we can't chooseourdecompositionmOpt(am)tobeorthogonal.-However, the dream shed a light on the correct direction! The“orthogonality"or the“no-interaction condition"implies that Am o Am=Oform' + m. This is too strong, since it is enough to assume "almost-orthogonality", or in other words, it is enough if we have a nicely-controlled“interaction". Back to our decomposition. Although we can't make Opt(am)oOpi(am) = 0forall m'+m,wecanmakeOpt(am)oOpt(am)=O(h)formostm+m.In fact,according to Corollary 1.4 inLecture9,wehaveOpt(am) 0 Oph(am) = O()if supp(am)nsupp(am)= 0. In other words, wedo have“almost-orthogonality"of Opt(am),if we start with"almost-disjoint"symbols am's.[In other words"almost-orthogonality"is the quantum analogue of the“"almost disjointnessoffunctionsl.In summary, here is how we prove thetheorem:A We first carefully choose a partition of unity 1 = Em Xm, in a uniformway,so that the resulting decomposition a =(xma)decompose a into asummation of almost disjoint compactly-supported symbols am = Xma.B We then control the interaction between different Opi(am)s
LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS 5 • We first decompose a into countably many compactly-supported symbols a = P m am. This can be done by choosing any partition of unity 1 = P m χm such that each χm is compactly supported, and letting am = χma. Then formally we have Opt ~ (a) = X m Opt ~ (am), and by compactness of supp(am), each Opt ~ (am) is L 2 -bounded. • In general, if A = PAm, to conclude the boundedness of A from the boundedness of Am’s, – a necessary condition we need is that the bound of Am’s is uniform for all m. In view of (2), this “uniformly boundedness” can be fulfilled if we choose our partition of unity in a “uniform” way. – The “uniformly boundedness” of components is still not enough, since there may be “interactions” between different Am’s. Of course the best dream is that if we can choose these Am’s so that there are “no interaction”, or in other words, if these Am’s are “orthogonal” to each other, (namely if the decomposition A = P m Am is an “orthogonal decomposition” A = L m Am), then the “uniformly boundedness” of Am’s does imply the boundedness of A. But that’s only a dream: we can’t choose our decomposition P m Opt ~ (am) to be orthogonal. – However, the dream shed a light on the correct direction! The “orthogonality” or the “no-interaction condition” implies that Am ◦Am0 = 0 for m0 6= m. This is too strong, since it is enough to assume “almostorthogonality”, or in other words, it is enough if we have a nicelycontrolled “interaction”. • Back to our decomposition. Although we can’t make Opt ~ (am)◦Opt ~ (am0) = 0 for all m0 6= m, we can make Opt ~ (am) ◦Opt ~ (am0) = O(~ ∞) for most m0 6= m. In fact, according to Corollary 1.4 in Lecture 9, we have Opt ~ (am) ◦ Opt ~ (am0) = O(~ ∞) if supp(am)∩supp(am0) = ∅. In other words, we do have “almost-orthogonality” of Opt ~ (am), if we start with “almost-disjoint” symbols am’s.[In other words, “almost-orthogonality” is the quantum analogue of the “almost disjointness” of functions]. In summary, here is how we prove the theorem: A We first carefully choose a partition of unity 1 = P m χm, in a uniform way, so that the resulting decomposition a = P(χma) decompose a into a summation of almost disjoint compactly-supported symbols am = χma. B We then control the interaction between different Opt ~ (am)s
6LECTURE11:L?-THEORY OFSEMICLASSICALPSDOS:BOUNDEDNESSC Finally we add the“"almost-orthogonal" Opt(am)s to get a bounded operator.Technically,this is done by applying the Cotlar-Stein lemma below (whichtells us how to “add a sequence of almost-orthogonal bounded operators"):Lemma 2.2 (Cotlar-Stein Lemma).Let Hi, H, be Hilbert spaces. For j ENletA,EC(Hi,H2)be bounded linearoperators satisfyingsupIA,A/2Candsupl/A,Al1/?≤C.k=1k=1The the series i= A, converges in the strong operator topology? to A EC(Hi, H2) with All≤C.A proof will be given in next lecture.Decompositionof symbol.As we just explained, we first decompose a into a family of “almost disjoint"compactly supported symbols which are“uniformly controllable".For this purposewe will use integer points Q E Rd as our labels (which are evenly distributed in Rd),and we prove the following"periodic partition of unity":Lemma 2.3. There erists Xo E Co(Rd) so that 0 ≤ Xo ≤ 1 on Rd, supp(xo) B(0, Vd) 3 andonRdxa=1(4)aEZdwhere for any α E Zd, Xa(z) := Xo(z-a).Proof. First choose E Co(IRd) so that ≥ 0 on Rd, = 1 on B(0, Va/2) and= 0 on RdB(0, V@). Letb(2) = (z-α).aeZdThen for z in any bounded set, the sum above is a finite sum. Thus is well-definedand is a smooth function. Moreover, by construction,.for each z one has (z) ≥1 (since for each z one can always find an α E zdso that[zi-Qil ≤ forall i,i.e. z-Q EB(O,Va/2).). for any a e Zd, w(z+a) = (z).2Recall: the operator strong topology on C(Hi, H2) is the topology so thatA, →AA,(r)→A(r) for all EHi.Note that in Cotlar-Stein Lemma, the sum A, does not converge in operator norm topology.3In Zworski, Xo is taken to be supported in B(0,2),which can't be true since after translation,these balls can't cover R2n for n > 4: you need a larger radius to cover points like (1/2, ..-,1/2)
6 LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS C Finally we add the “almost-orthogonal” Opt ~ (am)s to get a bounded operator. Technically, this is done by applying the Cotlar-Stein lemma below (which tells us how to “add a sequence of almost-orthogonal bounded operators”): Lemma 2.2 (Cotlar-Stein Lemma). Let H1, H2 be Hilbert spaces. For j ∈ N let Aj ∈ L(H1, H2) be bounded linear operators satisfying sup j X∞ k=1 kA ∗ jAkk 1/2 4: you need a larger radius to cover points like (1/2, · · · , 1/2)
7LECTURE11:L?-THEORY OFSEMICLASSICALPSDOS:BOUNDEDNESSIt is easy to see that the functionXo(2) = p(z)/(z)口is what we want.As a consequence,ifwe fix sucha function x =Xo E Co(Rn × IRn),and for anya = a(r, ) E S(1) if we denoteaa(c,s) = Xa(c,s)a(c,E),then we geta(r,s) = aa(r,s).QEZ2Moreover,thesea(c,)form a countable family of“almost disjoint"compactly-supported symbols such that for any β e N2n, there exists Cg (which comes fromthe bounds of finitely many derivatives of a E S(1)togetherwith the bounds offinitely many derivatives of xo) such thatJoaal≤Cp,VαEz2n.As a consequence, there exists C (which depends on a) such that for all α E Z2nIIOph(aa)I/c(L2(R") ≤ C.“Almost orthogonality"In view of the Cotlar-Stein lemma, we need to control the operator norm ofOph(aa)*oOp(ag) and Op(aa) Opt(ag)*. For simplicity we only consider thecase of t =1/2,namelythecase of Weyl quantization.Thegeneral case can beargued either by a similarproof, orby using the change of quantizationformulaFor any a = a(r,s) e S(1), we denote aα=Xaa as above, and letbaB=aaap,where * is the Moyal star product. The crucial estimate we need isAssumeh=1.]Lemma 2.4. Suppose a E S(1). Then for each N and each multi-inder,thereis a constantC = C(%, N,n)supaal,Jo/≤2N+4n+1+hl R"such that for all z = (r,s) e R2nJ0%bap(2)/ ≤C(α-B)-N(z- +B)(5)Wewill not prove this theorem now.Instead, we will prove astronger version next time: instead of assume a e S(1), we will onlyassume a E S(m) (but the upper bound will also be m-dependent)
LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS 7 It is easy to see that the function χ0(z) = ϕ(z)/ψ(z) is what we want. As a consequence, if we fix such a function χ = χ0 ∈ C ∞ 0 (R n × R n ), and for any a = a(x, ξ) ∈ S(1) if we denote aα(x, ξ) = χα(x, ξ)a(x, ξ), then we get a(x, ξ) = X α∈Z2n aα(x, ξ). Moreover, these aα(x, ξ) form a countable family of “almost disjoint” compactlysupported symbols such that for any β ∈ N 2n , there exists Cβ (which comes from the bounds of finitely many derivatives of a ∈ S(1) together with the bounds of finitely many derivatives of χ0) such that |∂ β aα| ≤ Cβ, ∀α ∈ Z 2n . As a consequence, there exists C (which depends on a) such that for all α ∈ Z 2n , kOpt ~ (aα)kL(L2(Rn)) ≤ C. ¶ “Almost orthogonality”. In view of the Cotlar-Stein lemma, we need to control the operator norm of Opt ~ (aα) ∗ ◦ Opt ~ (aβ) and Opt ~ (aα) ◦ Opt ~ (aβ) ∗ . For simplicity we only consider the case of t = 1/2, namely the case of Weyl quantization. The general case can be argued either by a similar proof, or by using the change of quantization formula. For any a = a(x, ξ) ∈ S(1), we denote aα = χαa as above, and let bαβ = ¯aα ? aβ, where ? is the Moyal star product. The crucial estimate we need is Assume ~ = 1. Lemma 2.4. Suppose a ∈ S(1). Then for each N and each multiindex γ, there is a constant C = C(γ, N, n) X |α|≤2N+4n+1+|γ| sup Rn |∂ α a|, such that for all z = (x, ξ) ∈ R 2n , (5) |∂ γ bαβ(z)| ≤ Chα − βi −N hz − α + β 2 i −N . We will not prove this theorem now. Instead, we will prove a stronger version next time: instead of assume a ∈ S(1), we will only assume a ∈ S(m) (but the upper bound will also be m-dependent).
8LECTURE11:L?-THEORY OFSEMICLASSICALPSDOS:BOUNDEDNESSAs a consequence, we get from (2) the following control of Il(baβ")h=llc(L2(R") (inwhich we take N = 2n+1 so that (z- )N is in Ll, and use all with /l ≤ 2n+1so that we can apply (2)):Corollary 2.5. For any N there is a constantC=C(n) sup[8°alla<8n+4Rmso thatI1(baβ")n=llc(L2(R") ≤ C(α - β)-2n-1(6)Proof of Calderon-Vailancourt Theorem.Finally wefinish theproof of Calderon-Vailancourt TheoremProof.Step 1 We first prove a special case: the bound for Weyl quantization with h = 1:(7)II(aW)=llc(L2(R)≤C sup|8°allal/≤8n+4R2Set Aα = (aW)h=1, then (6w)h=1 = AAβ. By the previous corollaryII6pllc(L2) ≤C(α - B)-2n-1.Itfollowssup II/AαAal1/2 ≤C(α - β)-(2n+1)/2 ≤C.nBβBythesameway one hassup I/A Aal/2 ≤C.Since(aw)h=1=.Aa,theconclusionfollowsfromtheCotlar-Steinlemma.Step 2We then prove the theorem for Weyl quantization with general h, i.e.[aW llc(L2(R")≤C hlal/2 sup [8°alR2[α/≤8n+4This can be done by a re-scaling technique. First we notice that (7) is uniform forall h. In particular, it holds for h = 1.Next we introduce the following re-scaling:=h-1/2z，g=h-1/2y，=h-1/2and define() := u(r) = u(1/2), a(,) := a(,) =a(/2,/2)
8 LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS As a consequence, we get from (2) the following control of k(dbαβ W )~=1kL(L2(Rn)) (in which we take N = 2n+1 so that hz− α+β 2 i N is in L 1 , and use all γ with |γ| ≤ 2n+1 so that we can apply (2)): Corollary 2.5. For any N there is a constant C = C(n) X |α|≤8n+4 sup Rn |∂ α a| so that (6) k(bcαβ W )~=1kL(L2(Rn)) ≤ Chα − βi −2n−1 . ¶ Proof of Calderon-Vailancourt Theorem. Finally we finish the proof of Calderon-Vailancourt Theorem. Proof. Step 1 We first prove a special case: the bound for Weyl quantization with ~ = 1: (7) k(ba W )~=1kL(L2(Rn)) ≤ C X |α|≤8n+4 sup R2n |∂ α a| Set Aα = (ba W )~=1, then (bb W αβ)~=1 = A∗ αAβ. By the previous corollary, kbb W αβkL(L2) ≤ Chα − βi −2n−1 . It follows sup α X β kAαA ∗ βk 1/2 ≤ C X β hα − βi −(2n+1)/2 ≤ C. By the same way one has sup α X β kA ∗ αAβk 1/2 ≤ C. Since (ba W )~=1 = P α Aα, the conclusion follows from the Cotlar-Stein lemma. Step 2 We then prove the theorem for Weyl quantization with general ~, i.e. kba W kL(L2(Rn)) ≤ C X |α|≤8n+4 ~ |α|/2 sup R2n |∂ α a| This can be done by a re-scaling technique. First we notice that (7) is uniform for all ~. In particular, it holds for ~ = 1. Next we introduce the following re-scaling: x˜ = ~ −1/2x, y˜ = ~ −1/2 y, ˜ξ = ~ −1/2 ξ and define u˜(˜x) := u(x) = u(~ 1/2x˜), a˜(˜x, ˜ξ) := a(x, ξ) = a(~ 1/2x, ˜ ~ 1/2 ˜ξ)
LECTURE11:L?-THEORYOFSEMICLASSICALPSDOS:BOUNDEDNESS9One can check:aWu=(a)h=1uSince the change of variable will create an h-n/4-factor for L2-norms, namelyIlal/2= h-n/4|ul/2, and /aW ull/2= h-n/41(a)=1illz2the conclusion follows from step 1 andsup [0x,a| = hlal/2 sup [0°al.RnRnStep 3 Finally we prove the theorem for any t-quantization. We apply the changeof quantization formula (Theorem 3.1 in Lecture 9).Namely,if we set b(r,)ei(t-)DrDea(r,E), then we have Opi(a) = W. It followsIIOp;(a)c(L2(R) = [16Wllc(L2(R") ≤C Z l/2 sup [8°b]R2m[α|<8n+4. We notice that by applying Proposition 1.1 in Lecture 7 to the matrix Q =wewill get-e-t(t-1)u-na(r + y,$ +n)dydnb(r, ) = ei(t-s)Da-Dea(r, E) =(2元h)nBy inserting n + 1 times before the term e-(t-)y- the operator(Dy,)2L=1+(n,)-2+(-t)2jwhich satisfies1)=e-(t-)yn(0 n)2 (e-(-1)) we will get[0]<cJaPalsupIpl≤/+2n+2and thus we concludeIIOpi(a)lc(L2(R") ≤Cnl/2 sup |0%al.R2nla/<10n+6口Remark. More generally if a e Ss(1) for some 0 ≤ < , one hasIIOp(a)l(L2) ≤C Z hl/2 sup|8°al.Rnlal<MnIn particular, for sucha one hasIIOp;(a)lc(L2) ≤ C sup [a(r, 3)I + O(1/2-5)R21
LECTURE 11: L 2 -THEORY OF SEMICLASSICAL PSDOS: BOUNDEDNESS 9 One can check: ba W u = (ba˜ W )~=1u. ˜ Since the change of variable will create an ~ −n/4 -factor for L 2 -norms, namely ku˜kL2 = ~ −n/4 kukL2 , and kba W ukL2 = ~ −n/4 k(ba˜ W )~=1u˜kL2 the conclusion follows from step 1 and sup Rn |∂x, ˜ ξ˜a˜| = ~ |α|/2 sup Rn |∂ α a|. Step 3 Finally we prove the theorem for any t-quantization. We apply the change of quantization formula (Theorem 3.1 in Lecture 9). Namely, if we set b(x, ξ) = e i(t− 1 2 )~Dx·Dξ a(x, ξ), then we have Opt ~ (a) = bb W . It follows kOpt ~ (a)kL(L2(Rn)) = kbb W kL(L2(Rn)) ≤ C X |α|≤8n+4 ~ |α|/2 sup R2n |∂ α b| . We notice that by applying Proposition 1.1 in Lecture 7 to the matrix Q = 0 I I 0 , we will get b(x, ξ) = e i(t−s)~Dx·Dξ a(x, ξ) = 1 (2π~) n Z Rn Z Rn e − i ~ (t− 1 2 )y·η a(x + y, ξ + η)dydη. By inserting n + 1 times before the term e − i ~ (t− 1 2 )y·η the operator L = 1 +X j (Dηj ) 2 ( 1 2 − t) 2 + X j (Dyj ) 2 ( 1 2 − t) 2 which satisfies 1 hy, ηi 2 L(e − i ~ (t− 1 2 )y·η ) = e − i ~ (t− 1 2 )y·η , we will get |∂ γ b| ≤ C sup |ρ|≤|γ|+2n+2 |∂ ρ a| and thus we conclude kOpt ~ (a)kL(L2(Rn)) ≤ C X |α|≤10n+6 ~ |α|/2 sup R2n |∂ α a|. Remark. More generally if a ∈ Sδ(1) for some 0 ≤ δ < 1 2 , one has kOpt ~ (a)kL(L2) ≤ C X |α|≤Mn ~ |α|/2 sup Rn |∂ α a|. In particular, for such a one has kOpt ~ (a)kL(L2) ≤ C sup R2n |a(x, ξ)| + O(~ 1/2−δ ).