LECTURE12:L?-THEORYOFSEMICLASSICALPsDOs:COMPACTNESS1.COMPACTNESSOFSEMICLASSICALPsDOS Compactness operators.As we have explained at the very beginning of this course, in the quantum partof the theory, the quantum energies should be discrete.For this purpose, we needto find a criteria for a semiclassical pseudodifferential operator aw to have discretespectrum. One very useful way to prove discreteness of spectrum that we learnedin functional analysis is through compact operators.Recall:A bounded linear operator AE L(Hi,H2) is called a compact operatorA maps bounded sets in Hi into pre-compact subsets in H2 For every bounded sequence [rk] in Hi, the image [Ark] contains aconvergent subsequence.For example, if A E C(Hi, H2) has a finite dimensional range, then A is compact.Such operators are called finite rank operators. Compact operators are very usefulin spectral theory because.For any compact operator A, the spectrum o(A) consists of eigenvalues withfinite multiplicities, and the only possible accumulation point is 0.. For a self-adjoint operator A to have discrete spectrum, it is enough to provethe compactness of the resolvent (A - )-1)We list several important facts for compact operators:. The set of compact operators, K(Hi, H2), is a closed vector subspace ofC(Hi, H2) (with respect to the operator norm topology)..Any compact operator can be approximated bya sequence of finiterankoperators.. (Schauder) An operator Ae C(Hi,H2) is compact if and only if A* is compact.CompactnessforSchwartz symbols.We would like to show that if the symbol a is a Schwartz function, then aw is acompact operator on L?(IRn).During the proof we will need to control the Lo-normof various functions of the form raoaWu(r). So we first prove
LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS 1. Compactness of semiclassical PsDOs ¶ Compactness operators. As we have explained at the very beginning of this course, in the quantum part of the theory, the quantum energies should be discrete. For this purpose, we need to find a criteria for a semiclassical pseudodifferential operator ba W to have discrete spectrum. One very useful way to prove discreteness of spectrum that we learned in functional analysis is through compact operators. Recall: A bounded linear operator A ∈ L(H1, H2) is called a compact operator ⇐⇒A maps bounded sets in H1 into pre-compact subsets in H2 ⇐⇒ For every bounded sequence {xk} in H1, the image {Axk} contains a convergent subsequence. For example, if A ∈ L(H1, H2) has a finite dimensional range, then A is compact. Such operators are called finite rank operators. Compact operators are very useful in spectral theory because • For any compact operator A, the spectrum σ(A) consists of eigenvalues with finite multiplicities, and the only possible accumulation point is 0. • For a self-adjoint operator A to have discrete spectrum, it is enough to prove the compactness of the resolvent (A − i) −1 . We list several important facts for compact operators: • The set of compact operators, K(H1, H2), is a closed vector subspace of L(H1, H2) (with respect to the operator norm topology). • Any compact operator can be approximated by a sequence of finite rank operators. • (Schauder) An operator A∈L(H1,H2) is compact if and only if A∗ is compact. ¶ Compactness for Schwartz symbols. We would like to show that if the symbol a is a Schwartz function, then ba W is a compact operator on L 2 (R n ). During the proof we will need to control the L ∞-norm of various functions of the form x α∂ β x ba W u(x). So we first prove 1
2LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESSLemma 1.1. If a E (R2n), then for any multi-indices α and β, there eristsconstant Ca,B,n so that for any ue L2(IR"),(1)sup [raa awul≤Ca,B,nllull2.TER"Proof.The Schwartz kernel of aw is(r+u,E)dEet(r-y)-SaK(r,9) = (2h)n J2Since a E, we have Ke.So for any multi-index a and β,sup [rao awul and definebk(c) = (r)NaWpr(r).SinceawkawpllLll)-NiLaIkllLit is enough to find k so that the sequence k converges in L(Rn).To find such a sequence,we use the standard diagonalization trick.We denoteQ" = {ri, 2, ... ]. We first observe that, according to (1), the family[b(r) := (r)NaWo(r) / E B)is uniformly bounded. So we can choose a sequence () C B so that (r)Nawpconverges at i, and in general we can inductively choose a subsequence (om] C[pm-1) so that (r)NaW pm converges at the point rk. If we set Pk := pk, then thesequence bk converges at all r E Qn.Finally we prove the sequence is a Cauchy sequence in L(R"), and thusconverges in L(Rn). By (1), there exists a constant M so that for all k,/V()/≤M/3, O and chooseRlargeenough sothatM/R<e.Choosepoints (y1,".,yp) CQnsuchthatB(0, R) C UP- B(up, 一)
2 LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS Lemma 1.1. If a ∈ S (R 2n ), then for any multi-indices α and β, there exists constant Cα,β,n so that for any u ∈ L 2 (R n ), (1) sup x∈Rn |x α ∂ β x ba W u| ≤ Cα,β,nkukL2 . Proof. The Schwartz kernel of ba W is K(x, y) = 1 (2π~) n Z Rn e i ~ (x−y)·ξ a( x + y 2 , ξ)dξ. Since a ∈ S , we have K ∈ S . So for any multi-index α and β, sup x∈Rn |x α ∂ β x ba W u| ≤ sup (x,y)∈R2n |x α ∂ β x hyi nK(x, y)| Z Rn hyi −n |u(y)|dy ≤ Cα,β,nkukL2 . Now we prove the compactness of ba W for a Schwartz symbol a: Theorem 1.2. Suppose a = a(x, ξ) ∈ S (R 2n ). Then ba W : L 2 (R n ) → L 2 (R n ) is a compact operator. Proof. Let B ⊂ L 2 (R n ) be a bounded set. We need to find a sequence {ϕk} ⊂ B so that ba W ϕk converges in L 2 . We fix N > n 2 and define ψk(x) = hxi Nba W ϕk(x). Since kba W ϕk − ba W ϕlkL2 ≤ khxi −N kL2 · kψk − ψlkL∞, it is enough to find ϕk so that the sequence ψk converges in L ∞(R n ). To find such a sequence, we use the standard diagonalization trick. We denote Qn = {x1, x2, · · · }. We first observe that, according to (1), the family {ψ(x) := hxi Nba W ϕ(x) | ϕ ∈ B} is uniformly bounded. So we can choose a sequence {ϕ 1 k } ⊂ B so that hxi Nba W ϕ 1 k converges at x1, and in general we can inductively choose a subsequence {ϕ m k } ⊂ {ϕ m−1 k } so that hxi Nba W ϕ m k converges at the point xk. If we set ϕk := ϕ k k , then the sequence ψk converges at all xl ∈ Qn . Finally we prove the sequence ψk is a Cauchy sequence in L ∞(R n ), and thus converges in L ∞(R n ). By (1), there exists a constant M so that for all k, |∇ψk(x)| ≤ M/3, hxi|ψk(x)| ≤ M/2. Fix ε > 0 and choose R large enough so that M/R < ε. Choose points {y1, · · · , yP } ⊂ Qn such that B(0, R) ⊂ ∪P p=1B(yp, ε M ).
3LECTURE12:L?-THEORYOF SEMICLASSICALPSDOS:COMPACTNESSSince each sequence [bk(yp)) is a Cauchy sequence, there exists K so thatC[k(yp) - b(yp)l K and all 1 ≤ p ≤ P. Moreover, according to the choice of R,2sup [()-()/≤sup()/b()/RTo estimate suplel K, we haveIlk-llL ≤max(sup [r() -bi(r)l, sup ((r) -bi(r)) R口This completes the prove. Compactness for more general symbolsNow we want to study the compactness of aw for more general symbols a ES(m),where m is some order function.The idea is similar to the proof of Calderon-Vaillancourt's theorem, namely we first use the periodic partition of unity1=xeEZ2nthat we constructed last time to decompose the symbol a into countably manyalmost disjoint compact supported symbolswhich are uniformly controllable:a(r,E) = aa(r,5),where as last time, aa := Xaa. Since each aa is compactly supported, the operatoraw is compact. As a result, any finite sum Zjalm aaw is compact. Since the setof compact operators is closed in the set of bounded linear operators, to prove thataw is compact, it is enough to prove the convergence of .daw.So, again we needthe Cotlar-Stein lemma to add these compact operators:Lemma 1.3 (Cotlar-Stein Lemma).Let Hi,H2 be Hilbert spacesFor jeN let A, E C(Hi,H2)be bounded linear operators satisfyingsup I/A,Al|/2 <Cand supllA,Al/2<C.1The the series Z- A, converges in the strong operator topology toA E L(Hi, H2) with IAll≤C
LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS 3 Since each sequence {ψk(yp)} is a Cauchy sequence, there exists K so that |ψk(yp) − ψl(yp)| K and all 1 ≤ p ≤ P. Moreover, according to the choice of R, sup |x|≥R |ψk(x) − ψl(x)| ≤ 2 R sup |x|≥R hxi|ψl(x)| K, we have kψk − ψlkL∞ ≤ max( sup |x|<R |ψk(x) − ψl(x)|, sup |x|≥R |ψk(x) − ψl(x)|) < ε. This completes the prove. ¶ Compactness for more general symbols. Now we want to study the compactness of ba W for more general symbols a ∈ S(m), where m is some order function. The idea is similar to the proof of CalderonVaillancourt’s theorem, namely we first use the periodic partition of unity 1 = X α∈Z2n χα that we constructed last time to decompose the symbol a into countably many almost disjoint compact supported symbols which are uniformly controllable: a(x, ξ) = X α∈Z2n aα(x, ξ), where as last time, aα := χαa. Since each aα is compactly supported, the operator ba W α is compact. As a result, any finite sum P |α|≤M acα W is compact. Since the set of compact operators is closed in the set of bounded linear operators, to prove that ba W is compact, it is enough to prove the convergence of P α acα W . So, again we need the Cotlar-Stein lemma to add these compact operators: Lemma 1.3 (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 < C and sup j X∞ k=1 kAjA ∗ kk 1/2 ≤ C. The the series P∞ j=1 Aj converges in the strong operator topology to A ∈ L(H1, H2) with kAk ≤ C.
4LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESSHowever, this time the convergence in strong operator norm is not enough for us,since we need .aaw to converge in operator norm topology. Fortunately, theCotlar-Stein Lemma can also beapplied to prove convergence in operator normtopology:ToproveA,convergestoAintheoperatornormtopology9 To prove A, - A= A, converges to Bm with Bmll ≤8iMApplyCotlar-SteintoZCA,forM-→ooandtrytogetboundsi>Mthat tends to zero.aaw to control on the "interactions"So we still need the“almost orthogonality"of aawbaw - (aaW)*oaw. But this time, we need a better control. Fortunately, a betterestimate exists, as long as we use a nicer symbol class S(m) instead of s(1):Again suppose h = 1.Lemma 1.4 ("Almost orthogonality" for S(m)). Suppose a E S(m)Then for each N and each multi-inder , there is a constantZC = C(, N, n)sup[aal,DTJa/≤2N+4n+1+hlsuch that for all z = (r,s) e R2na+β(2)[0%ba(2)/ ≤Cm(α)m(B)(α - B)-N (z 2As a consequence, we get from the explicitformula of the C(L2)-bound of aw withSchwartz symbols (c.f.Proposition 1.4 in Lecture 11)that(3)Ilaβ"llc(L2(R") ≤Cm(a)m(β)(α-β)-N.Let's put the above ideas together. We want(4)Il aaWllc(L2(Rn) →0 as M -00.[a|>MSo instead of applying the Cotlar-Stein Lemma to the whole sum .aw, we willapply it to the sum Zja>M aaw, which gives ussup,Z /aw(aw)/, sup, ZI(aaw)apwZaa≤maxlal>M /B>MJal>M (B>M[a|>Mc(L2)W)*oa.wSince (aba,wehaveN i(a")ya"a C suVmam(a-)-~/2≤ sp, m(a).sup1a|>M (8]>M[a|>M1al>M (8]>M
4 LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS However, this time the convergence in strong operator norm is not enough for us, since we need P α acα W to converge in operator norm topology. Fortunately, the Cotlar-Stein Lemma can also be applied to prove convergence in operator norm topology: ⇐⇒ To prove P j Aj converges to A in the operator norm topology ⇐⇒ To prove P j≤M Aj − A = P j>M Aj converges to BM with kBMk ≤ ε ⇐⇒ Apply Cotlar-Stein to P j>M Aj for M → ∞ and try to get bounds ⇐⇒ that tends to zero. So we still need the “almost orthogonality” of acα W to control on the “interactions” bcαβ W = (acα W ) ∗ ◦abβ W . But this time, we need a better control. Fortunately, a better estimate exists, as long as we use a nicer symbol class S(m) instead of S(1): Again suppose ~ = 1. Lemma 1.4 (“Almost orthogonality” for S(m)). Suppose a ∈ S(m). Then for each N and each multi-index 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 , (2) |∂ γ bαβ(z)| ≤ Cm(α)m(β)hα − βi −N hz − α + β 2 i −N . As a consequence, we get from the explicit formula of the L(L 2 )-bound of ba W with Schwartz symbols (c.f. Proposition 1.4 in Lecture 11) that (3) kbcαβ W kL(L2(Rn)) ≤ Cm(α)m(β)hα − βi −N . Let’s put the above ideas together. We want (4) k X |α|>M acα W kL(L2(Rn)) → 0 as M → ∞. So instead of applying the Cotlar-Stein Lemma to the whole sum P α ba W , we will apply it to the sum P |α|>M acα W , which gives us X |α|>M acα W L(L2) ≤ max sup |α|>M X |β|>M kacα W (abβ W ) ∗ k 1/2 , sup |α|>M X |β|>M k(acα W ) ∗ abβ W k 1/2 . Since (acα W ) ∗ ◦ abβ W = bcαβ W , we have sup |α|>M X |β|>M k(acα W ) ∗ ◦abβ W k 1/2 L(L2) ≤ C sup |α|>M X |β|>M p m(α)m(β)hα−βi −N/2 ≤ C sup |α|>M m(α)
LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESS5Similarlysup IldaW(agw)*1/2 ≤C sup m(a).Ja|>M [3]>MJa>MSo for (4) to hold, it is enough to assumeas z→ o!m(z) → 0Of course we assumed h =1 in these arguments.However,it is easy to extend thearguments to general h by using the re-scaling trick as we did last time. We can alsouse the change of quantization formula to extend the theorem to other semiclassicalt-quantizations. Thus we getTheorem 1.5. Suppose m = m(r, s) is an order function on R2n such that(5)lim m(z) = 0.-Then for any a E S(m), the operator aW : L?(Rn) L?(Rn) is a compact operator.2.PROOF OF LEMMA 1.4.Note: In this section we take h = 1.Proof.Recall (See Lecture 9,page 5)aa *ag(r, ) = e(De-D-Da-Da)[aa(r, E)ag(y, n)n=1e-(-g--)aa(r+,+E)ag(+,s+n)didedydn(元h)2nWith the notation z = (r,s), wi = (-i,-), w2 = (-j, -i) and (wi, w2) :E.y-i.jone gets1e-2ip(u1,w2)a(z - w1)ag(z - w2)dwidw2.baB(2) = aas(2) =元2nJROur aim is to find an estimation oforbasl,which should be“small"forα-βlarge. Since each ag is supported near the lattice point , it is natural to separatethe integral above into two parts, ... So we choose a cut-off function( = ((w1, w2) E C(R4n)such that 0 ≤(≤ 1 on R4n, ( = 1 on B4n(0, 1) and (= 0 on R4n /B4n(0, 2). Byplugging 1 = S+ (1 - ) in the integrand, we can split the above integral into twoparts:baβ(z)= I1 +I2.It remains to estimateailand10"12].In estimating these terms, we also need to use the following elementary estima-tion (which, of course, is a mathematical way to say that the partition of unity we
LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS 5 Similarly sup |α|>M X |β|>M kacα W (abβ W ) ∗ k 1/2 ≤ C sup |α|>M m(α). So for (4) to hold, it is enough to assume m(z) → 0 as z → ∞! Of course we assumed ~ = 1 in these arguments. However, it is easy to extend the arguments to general ~ by using the re-scaling trick as we did last time. We can also use the change of quantization formula to extend the theorem to other semiclassical t-quantizations. Thus we get Theorem 1.5. Suppose m = m(x, ξ) is an order function on R 2n such that (5) limz→∞ m(z) = 0. Then for any a ∈ S(m), the operator ba W : L 2 (R n ) → L 2 (R n ) is a compact operator. 2. Proof of Lemma 1.4. Note: In this section we take ~ = 1. Proof. Recall (See Lecture 9, page 5) a¯α ? aβ(x, ξ) = e i~ 2 (Dξ·Dy−Dx·Dη) [¯aα(x, ξ)aβ(y, η)] y=x,η=ξ = 1 (π~) 2n Z R4n e − 2i ~ (ξ˜·y˜−x˜·η˜) a¯α(x + ˜x, ξ + ˜ξ)aβ(x + ˜y, ξ + ˜η)dxd˜ ˜ξdyd˜ η. ˜ With the notation z = (x, ξ), w1 = (−x, ˜ −˜ξ), w2 = (−y, ˜ −η˜) and ϕ(w1, w2) := ˜ξ · y˜ − x˜ · η˜ one gets bα,β(z) = ¯aα ? aβ(z) = 1 π 2n Z R4n e −2iϕ(w1,w2) a¯α(z − w1)aβ(z − w2)dw1dw2. Our aim is to find an estimation of |∂ γ bαβ|, which should be “small” for |α − β| large. Since each aα is supported near the lattice point α, it is natural to separate the integral above into two parts, . So we choose a cut-off function ζ = ζ(w1, w2) ∈ C ∞ 0 (R 4n ) such that 0 ≤ ζ ≤ 1 on R 4n , ζ = 1 on B4n (0, 1) and ζ = 0 on R 4n \ B4n (0, 2). By plugging 1 = ζ + (1 − ζ) in the integrand, we can split the above integral into two parts: bαβ(z) = I1 + I2. It remains to estimate |∂ γ I1| and |∂ γ I2|. In estimating these terms, we also need to use the following elementary estimation (which, of course, is a mathematical way to say that the partition of unity we
6LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESSchose is“uniformly controllable"):for any multi-index ,there exists a constant Csuchthat[aa(w)] =[[xo(w -)a(w)]l ≤ Csup[aPxo(w -)|m(w) ≤ Cm(α),where in the last step we used the fact that Xo(w - a) = 0 for [w - al > 2n, andforw-al<V2nwehavem(w) ≤Cm(α)by the definition of an order function m.Estimatei.For the first parte-2ip(w1,w2)c(w1, w2)aa(z - wi)ap(z - w2)dwidw2,I1=72we have[] ≤C[aa(z -w1)]ag(z -w2)]dwidw2B4n(0.2)CXo(z-W1-a)xo(z-w2-)la(z-wi)lla(z-w2)dwidw2/ B4n (0,2)Recall that xo is supported in B(0, V2n). So the integrand vanishes unless[z-1-αl≤V2n and [z-W2-βl≤2nfor some [w|≤ 2. It followsaα+β≤2+V2n.Ja-βl≤4+2V2nand2Since a E S(m) and |z - w1 - αl ≤ V2n, we have1a(z-wi)/≤Cam(z-w1)≤Cm(α)So for each N one could find C = CN,n so that)] ≤ Cn,n(aα =β)-Nm(a)m(B)(≥ - B)2Similarly for each multi-index one could find C= Cn,on so that[a≤CNm.am(a)m(B)(α-B)-N(≥-)-Awhere the dependence of Cn,,n,a on a is clear: C is bounded by a constant multipleof supp≤laPalEstimateI2.Toestimatethe secondpart上e-2ip(wi,w2)(1 - (wi, w2)aa(z -wi)ag(z - w2)dwidw2,I2 =元2n
6 LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS chose is “uniformly controllable”): for any multi-index γ, there exists a constant Cγ such that |∂ γ aα(w)| = |∂ γ [χ0(w − α)a(w)]| ≤ Cγ sup ρ≤γ |∂ ρχ0(w − α)|m(w) ≤ Cγm(α), where in the last step we used the fact that χ0(w − a) = 0 for |w − a| > √ 2n, and for |w − a| < √ 2n we have m(w) ≤ Cm(α) by the definition of an order function m. Estimate I1. For the first part I1 = 1 π 2n Z R4n e −2iϕ(w1,w2) ζ(w1, w2)¯aα(z − w1)aβ(z − w2)dw1dw2, we have |I1| ≤ C Z B4n(0,2) |a¯α(z − w1)||aβ(z − w2)|dw1dw2 = C Z B4n(0,2) χ0(z − w1 − α)χ0(z − w2 − β)|a(z − w1)||a(z − w2)|dw1dw2. Recall that χ0 is supported in B(0, √ 2n). So the integrand vanishes unless |z − w1 − α| ≤ √ 2n and |z − w2 − β| ≤ √ 2n for some |w| ≤ 2. It follows |α − β| ≤ 4 + 2√ 2n and z − α + β 2 ≤ 2 + √ 2n. Since a ∈ S(m) and |z − w1 − α| ≤ √ 2n, we have |∂ α a(z − w1)| ≤ Cαm(z − w1) ≤ Cm(α). So for each N one could find C = CN,n so that |I1| ≤ CN,nhα − βi −Nm(α)m(β)hz − α + β 2 i −N . Similarly for each multi-index γ one could find C = CN,γ,n so that |∂ γ I1| ≤ CN,γ,n,am(α)m(β)hα − βi −N hz − α + β 2 i −N , where the dependence of CN,γ,n,a on a is clear: C is bounded by a constant multiple of supρ≤γ |∂ ρa| Estimate I2. To estimate the second part I2 = 1 π 2n Z R4n e −2iϕ(w1,w2) (1 − ζ(w1, w2))¯aα(z − w1)aβ(z − w2)dw1dw2
LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESS>we use integration by parts trick. As in Lecture 5 we introduce the operatorL = 4 +(Vb,Du),where w = (w1, w2) and (w) := -2(wi, w2) = -2(·g--n). Since /Vl = 2|wl,we have14(ujg (et) = et.Since the integrant of I2 vanishes for [w 1, we can integrate by parts via L manytimestoget(w)-Mta,M(z -w1)cp,m(z-w2)dwidw2[12] ≤CM[w|>for some smooth Ca,M supported in B(α, V2n) and cg.M supported in B(B, V2n).Soagaintheintegrand vanishes unless[z-W1-al≤V2n and [z-w2-βl≤2nforsomew≥1,whichimplies[Q-β|≤2V2n+|w1/ +[w2/ ≤(2V2n+2)(w)anda+β≤V2m+l(V2n+ 1)(),22So if we take M ≥ 2N + 4n + 1,,(u)2N-M(α - β)-~ (≥ - B)-~ du[2] ≤CM≤Cm(α-β)-N(z-a+BSimilarly if we estimate the derivative JoI2l, we will need to take no more than口M+lderivatives ofa and the conclusion follows.3.APPENDIX:PROOFOFTHECOTLAR-STEINLEMMALet Hi,H, be Hilbert spaces and A C(Hi, H2) a bounded linear operator.Recall that the norm of A is by definition(6)I/All = sup [Ar].[r|=1If we let A* E C(H2, Hi) be the adjoint of A, then one has (exercise!)(7)andIAI? = A*ALIA| = I/A*|I In the case Hi = H2 = H so that Am C(H, H) for all m E N, one hasIAmIl≤IIAIIm
LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS 7 we use integration by parts trick. As in Lecture 5 we introduce the operator L = 4 + h∇ψ, Dwi, where w = (w1, w2) and ψ(w) := −2ϕ(w1, w2) = −2(˜ξ · y˜−x˜ · η˜). Since |∇ψ| = 2|ω|, we have 1 4hwi 2 L(e iψ) = e iψ . Since the integrant of I2 vanishes for |w| ≤ 1, we can integrate by parts via L many times to get |I2| ≤ CM Z |w|≥1 hωi −Mc¯α,M(z − ω1)cβ,M (z − ω2)dω1dω2 for some smooth cα,M supported in B(α, √ 2n) and cβ,M supported in B(β, √ 2n). So again the integrand vanishes unless |z − w1 − α| ≤ √ 2n and |z − w2 − β| ≤ √ 2n for some |w| ≥ 1, which implies |α − β| ≤ 2 √ 2n + |w1| + |w2| ≤ (2√ 2n + 2)hwi and z − α + β 2 ≤ √ 2n + |w1| + |w2| 2 ≤ ( √ 2n + 1)hwi. So if we take M ≥ 2N + 4n + 1, |I2| ≤ CM Z |w|≥1 hwi 2N−Mhα − βi −N hz − α + β 2 i −N dw ≤ CMhα − βi −N hz − α + β 2 i −N . Similarly if we estimate the derivative |∂ γ I2|, we will need to take no more than M + |γ| derivatives of a and the conclusion follows. 3. Appendix: Proof of the Cotlar-Stein Lemma Let H1, H2 be Hilbert spaces and A ∈ L(H1, H2) a bounded linear operator. Recall that the norm of A is by definition (6) kAk = sup |x|=1 |Ax|. If we let A∗ ∈ L(H2, H1) be the adjoint of A, then one has (exercise!) (7) kAk = kA ∗ k and kAk 2 = kA ∗Ak. In the case H1 = H2 = H so that Am ∈ L(H, H) for all m ∈ N, one has kA mk ≤ kAk m.
8LECTURE12:L?-THEORY OFSEMICLASSICALPSDOS:COMPACTNESS(One could have strictly inequality above, for example, when A is nilpotent.) How-ever,if we assumeA is self-adjoint,i.e.A=A*,then for all m EN.IAm|l =IIAm.Proof of Cotlar-Stein Lemma. Let's first assume that A, = 0 for all j > N so thatA is obviously a well-defined bounded linear operator. Since A*A is self-adjoint,IIA|2m =IA*Al" =I(A*A)"=ZA,Aj2. Ajm-1Ag2where the summation is over J = (i,..., j2m) e [1,..., Mj2m.According to theCauchy-Schwarz inequalityI/A,Aj2. Af2m-1Aj2mll ≤I/A,Ajll-.. |Af2m--Aj2mllandI/A,AgA2m-Aj2m l≤I/A,I/AAll-..I/Ajm l!Together with the facts l/Aill ≤C and IIAj2mll ≤ C, one getsIAfA2*. Aj2m-1A2ml≤C/AA,l1/2AgAI/2..- /Af2m-1Ag2mI//2ItfollowsII12m ≤CIIA, A4g /IAjAI1/ IIA4f2m-1 Aj2m 1/2J=CI/Ag,Aal/2IAA,I/2...I/Agam-Ak2ml/2j1.j2j3j2n≤NC2mand thus (since N is fixed and m is arbitrary)II/AIl ≤ C.To prove the general case, we first take E Hi so that = Aty for some k andsome y E H2. ThenZA=ZAAIAA/2AA/ ll≤CllIt follows that the series , Ajr converges for allrEU :=span(AyI yEH2,k≥1]Since we have already proved II Ei= Aill ≤ C uniformly for all N, one could com-mute the summation with limit and conclude that the series E,Ajr converges forall r U, and IIE, Ajrl ≤ Clrll.Finally if e U, then E ker(Ak) for all k, so we trivially have Ajr = 0
8 LECTURE 12: L 2 -THEORY OF SEMICLASSICAL PSDOS: COMPACTNESS (One could have strictly inequality above, for example, when A is nilpotent.) However, if we assume A is self-adjoint, i.e. A = A∗ , then for all m ∈ N, kA mk = kAk m. Proof of Cotlar-Stein Lemma. Let’s first assume that Aj = 0 for all j > N so that A is obviously a well-defined bounded linear operator. Since A∗A is self-adjoint, kAk 2m = kA ∗Ak m = k(A ∗A) mk = X J A ∗ j1Aj2 · · · A ∗ j2m−1Aj2m , where the summation is over J = (j1, · · · , j2m) ∈ [1, · · · , N] 2m. According to the Cauchy-Schwarz inequality, kA ∗ j1Aj2 · · · A ∗ j2m−1Aj2mk ≤ kA ∗ j1Aj2 k · · · kA ∗ j2m−1Aj2mk and kA ∗ j1Aj2 · · · A ∗ j2m−1Aj2mk ≤ kA ∗ j1 kkAj2A ∗ j3 k · · · kAj2mk. Together with the facts kA1k ≤ C and kAj2mk ≤ C, one gets kA ∗ j1Aj2 · · · A ∗ j2m−1Aj2mk ≤ CkA ∗ j1Aj2 k 1/2 kAj2A ∗ j3 k 1/2 · · · kA ∗ j2m−1Aj2mk 1/2 It follows kAk 2m ≤ C X J kA ∗ j1Aj2 k 1/2 kAj2A ∗ j3 k 1/2 · · · kA ∗ j2m−1Aj2mk 1/2 = C X j1,j2 kA ∗ j1Aj2 k 1/2X j3 kAj2A ∗ j3 k 1/2 · · ·X j2m kA ∗ j2m−1Aj2mk 1/2 ≤ NC2m and thus (since N is fixed and m is arbitrary) kAk ≤ C. To prove the general case, we first take x ∈ H1 so that x = A∗ k y for some k and some y ∈ H2. Then X∞ j=1 Ajx = X∞ j=1 AjA ∗ k y ≤ X∞ j=1 kAjA ∗ kk 1/2 kAjA ∗ kk 1/2 kyk ≤ C 2 kyk. It follows that the series P j Ajx converges for all x ∈ U := span{A ∗ k y | y ∈ H2, k ≥ 1}. Since we have already proved k PN j=1 Ajk ≤ C uniformly for all N, one could commute the summation with limit and conclude that the series P j Ajx converges for all x ∈ U, and k P j Ajxk ≤ Ckxk. Finally if x ∈ U ⊥ , then x ∈ ker(Ak) for all k, so we trivially have PAjx = 0.