LECTURE13:L?-THEORYOFSEMICLASSICALPsDOs:HILBERT-SCHMIDTANDTRACECLASSOPERATORSIn today's lecture we start with some abstract properties of Hilbert-Schmidtoperators and trace class operators. Then we will investigate the following naturalquestions: for which class of symbols, the quantized operator Opt(a) is a Hilbert-Schmidt or trace class operator?How to compute its trace?1.HILBERT-SCHMIDT AND TRACE CLASS OPERATORS: ABSTRACT THEORY Definitions via singular values.As we havementioned last time, we are aiming at developing the spectral theoryfor semiclassical pseudodifferential operators. For a E S(m), where m is a decayingsymbol, we showed thatOpt(a)is acompact operator acting on L?(Rn).As aconsequence, either Opt(a) has only finitely many nonzero eigenvalues, or it hasinfinitely nonzero eigenvalues which can be arranged to a sequence whose normsconverges to 0.However, in general the eigenvalues of a compact operator A are non-real. Avery simple way to get real eigenvalues is to consider the operator A*A,which isa compact self-adjoint linear operator acting on L?(IRn).Thus the eigenvalues 1 ofA*A can be list? in decreasing order asThe numbers Si (which will always be taken to be the positive one) are called thesingularvalues of A.Definition 1.1. We say a compact operator3 A is a Hilbert-Schmidt operator if(1)IIAlHS<+8The number lAllHs is called the Hilbert-Schmidt norm (or Schatten 2-norm) of A.Another very important class of operators for us is the trace class operators.1Note that by definition, the eigenvalues of A*A are nonnegative.2We always repeat an eigenvalue according to its multiplicity.3The “official" definition of Hilbert-Schmidt operator assumes only the boundedness of A.How-ever, it is easy to prove that if A is Hilbert-Schmidt operator, then it can be written as the limit(withrespectto theoperator normtopology)of a sequence offiniterank operators,and thusmustbe a compact operator.1
LECTURE 13: L 2 -THEORY OF SEMICLASSICAL PSDOS: HILBERT-SCHMIDT AND TRACE CLASS OPERATORS In today’s lecture we start with some abstract properties of Hilbert-Schmidt operators and trace class operators. Then we will investigate the following natural questions: for which class of symbols, the quantized operator Opt ~ (a) is a HilbertSchmidt or trace class operator? How to compute its trace? 1. Hilbert-Schmidt and Trace class operators: Abstract theory ¶ Definitions via singular values. As we have mentioned last time, we are aiming at developing the spectral theory for semiclassical pseudodifferential operators. For a ∈ S(m), where m is a decaying symbol, we showed that Opt ~ (a) is a compact operator acting on L 2 (R n ). As a consequence, either Opt ~ (a) has only finitely many nonzero eigenvalues, or it has infinitely nonzero eigenvalues which can be arranged to a sequence whose norms converges to 0. However, in general the eigenvalues of a compact operator A are non-real. A very simple way to get real eigenvalues is to consider the operator A∗A, which is a compact self-adjoint linear operator acting on L 2 (R n ). Thus the eigenvalues 1 of A∗A can be list2 in decreasing order as s 2 1 ≥ s 2 2 ≥ s 2 3 ≥ · · · . The numbers sj (which will always be taken to be the positive one) are called the singular values of A. Definition 1.1. We say a compact operator3 A is a Hilbert-Schmidt operator if (1) kAkHS := X j s 2 j !1/2 < +∞. The number kAkHS is called the Hilbert-Schmidt norm (or Schatten 2-norm) of A. Another very important class of operators for us is the trace class operators. 1Note that by definition, the eigenvalues of A∗A are nonnegative. 2We always repeat an eigenvalue according to its multiplicity. 3The “official” definition of Hilbert-Schmidt operator assumes only the boundedness of A. However, it is easy to prove that if A is Hilbert-Schmidt operator, then it can be written as the limit (with respect to the operator norm topology) of a sequence of finite rank operators, and thus must be a compact operator. 1
2HILBERT-SCHMIDTANDTRACECLASSPSDOSDefinition 1.2. We say a compact operator A is a trace class operator if(2)Il/Alltr := sj<+00.2ThenumberLAtris called thetracenorm (or theSchatten 1-norm)of A. Abstract definitions via Hilbert basis.In general the singular values of an operator are very hard to compute. Fortu-nately, we have an alternative characterization of Hilbert-Schmidt norm (and thusHilbert-Schmidt operators) via Hilbert bases, which is easier to use. Let H be aseparableHilbert space, and A E C(H) is a bounded linear operator.If [e),(f)are two orthonormal bases of H, then by the Parseval's identity,llAeil?=-ZKAei,f)P=ZKA*fi,ei)P-IA'fil2i.1As a consequence, the quantityIAel?=IAfi2=lAe,lis independent of the choice of the orthonormal basis fe;].Moreover, one can provethatiftheabovequantityisfinite,thenAmustbeacompactoperator.Inparticular.if we take en to be the orthonormal basis consisting of eigenvectors of A*A (whoseexistence is guaranteed by the spectral theory of self-adjoint compact operators),thenllAeil? = (Aej, Aej) = (A*Aej,ei) = s)So we get the following alternative characterization of Hilbert-Schmidt norm:1/2(3)I/Ae;,12IIA|HS :where fel is any basis of H.Similarly we can define trace class operators by using a basis of H: For A eX(H). Then one can define a positive self-adjoint operator |A| E C(H) by[A|? = A*A.Applying the previous computation to [Al, we see that the quantity<Alej,ei)=EI/A/2ell?is independent of the choice of (en].Ifwe take (en) to be the orthonormal basisconsistingofeigenvectorsof A*A,thenE(IAlej,e;) =Zsj.9
2 HILBERT-SCHMIDT AND TRACE CLASS PSDOS Definition 1.2. We say a compact operator A is a trace class operator if (2) kAktr := X j sj < +∞. The number kAktr is called the trace norm (or the Schatten 1-norm) of A. ¶ Abstract definitions via Hilbert basis. In general the singular values of an operator are very hard to compute. Fortunately, we have an alternative characterization of Hilbert-Schmidt norm (and thus Hilbert-Schmidt operators) via Hilbert bases, which is easier to use. Let H be a separable Hilbert space, and A ∈ L(H) is a bounded linear operator. If {ei}, {fi} are two orthonormal bases of H, then by the Parseval’s identity, X i kAeik 2 = X i X j |hAei , fj i|2 = X i,j |hA ∗ fj , eii|2 = X j kA ∗ fjk 2 . As a consequence, the quantity X j kAejk 2 = X j kA ∗ fjk 2 = X j kAe˜jk 2 is independent of the choice of the orthonormal basis {ei}. Moreover, one can prove that if the above quantity is finite, then A must be a compact operator. In particular, if we take en to be the orthonormal basis consisting of eigenvectors of A∗A (whose existence is guaranteed by the spectral theory of self-adjoint compact operators), then X j kAejk 2 = X j hAej , Aej i = X j hA ∗Aej , ej i = X j s 2 j . So we get the following alternative characterization of Hilbert-Schmidt norm: (3) kAkHS = X j kAejk 2 !1/2 , where {ei} is any basis of H. Similarly we can define trace class operators by using a basis of H: For A ∈ K(H). Then one can define a positive self-adjoint operator |A| ∈ L(H) by |A| 2 = A ∗A. Applying the previous computation to |A|, we see that the quantity X j h|A|ej , ej i = X j k|A| 1/2 ejk 2 is independent of the choice of {en}. If we take {en} to be the orthonormal basis consisting of eigenvectors of A∗A, then X j h|A|ej , ej i = X j sj
3HILBERT-SCHMIDTANDTRACECLASSPSDOSIn other words, we get the following alternative expression of the trace norm:(4)IAlltr =《IA|ej,ei)5 Spaces of operators v.s. spaces of sequences.Wewill denotethe space of Hilbert-Schmidt operators on H byC2(H),anddenote the space of trace class operators on H by Ci(H).It turns out that they canbe viewed as a“"non-commutative analogues/generalizations" of the spaces [2 and, and have very similar properties. Here is an interesting table comparing spacesof sequences and spaces of operators on a separable Hilbert space H:space of sequencesspace of operatorssubspaceCoo (eventually zero sequences)F(H) (finiterank operators)Banach Spaces[1 (summable sequences)Ci(H) (trace class operators)Hilbert Spaces? (square summable sequences)C2(H) (Hilbert-Schmidt operators)Banach Spacesco (sequences converge to 0)K(H) (compact operators)[ (bounded sequences)C(H)(bounded operators)Banach SpacesCoo C Il c I? C co C 10inclusionF(H)CLi(H)CC2(H)CK(H)CC(H)denseCoo is dense in (,I Il), inF(H) is dense in (C(H), Il - Iltr), in(2,I I/2) and in (1,I -I/)(C2(H), JL- JHs) and in (K(H), - Jle)I - I/10e ≤ I1 - I/2 ≤ I - I/1II -I/c≤II - I/Hs≤II -Itrnorm()* = [0Duality(C(H))* = L(H)(co)* = [1Duality(K(H))* = Li(H)ideal[0]1, 11[00 c 11C(H)C(H), C(H)Ci(H) C Li(H)[0012, 12100 C 12idealC2(H)C(H), C(H)C2(H) C L2(H)[1 = [212decompositionCi(H) = C2(H)C2(H)decompositionI(anbn)/≤ Il(an)I/2 /l(bn)l/2ABIt≤IIAHsBHSFor proofs of these properties, we refer to Reed-Simon, Volume 1, g6.6
HILBERT-SCHMIDT AND TRACE CLASS PSDOS 3 In other words, we get the following alternative expression of the trace norm: (4) kAktr = X j h|A|ej , ej i. ¶ Spaces of operators v.s. spaces of sequences. We will denote the space of Hilbert-Schmidt operators on H by L2(H), and denote the space of trace class operators on H by L1(H). It turns out that they can be viewed as a “non-commutative analogues/generalizations” of the spaces l 2 and l 1 , and have very similar properties. Here is an interesting table comparing spaces of sequences and spaces of operators on a separable Hilbert space H: space of sequences space of operators subspace c00 (eventually zero sequences) F(H) (finite rank operators) Banach Spaces l 1 (summable sequences) L1(H) (trace class operators) Hilbert Spaces l 2 (square summable sequences) L2(H) (Hilbert-Schmidt operators) Banach Spaces c0 (sequences converge to 0) K(H) (compact operators) Banach Spaces l ∞ (bounded sequences) L(H) (bounded operators) inclusion c00 ⊂ l 1 ⊂ l 2 ⊂ c0 ⊂ l ∞ F(H)⊂L1(H)⊂L2(H)⊂K(H)⊂L(H) dense c00 is dense in (l 1 , k · kl 1 ), in F(H) is dense in (L1(H), k · ktr), in (l 2 , k · kl 2 ) and in (l ∞, k · kl∞) (L2(H), k · kHS) and in (K(H), k · kL) norm k · kl∞ ≤ k · kl 2 ≤ k · kl 1 k · kL ≤ k · kHS ≤ k · ktr Duality (l 1 ) ∗ = l ∞ (L1(H))∗ = L(H) Duality (c0) ∗ = l 1 (K(H))∗ = L1(H) ideal l ∞l 1 , l1 l ∞ ⊂ l 1 L1(H)L(H),L(H)L1(H) ⊂ L1(H) ideal l ∞l 2 , l2 l ∞ ⊂ l 2 L2(H)L(H),L(H)L2(H) ⊂ L2(H) decomposition l 1 = l 2 l 2 L1(H) = L2(H)L2(H) decomposition k(anbn)kl 1 ≤ k(an)kl 2 k(bn)kl 2 kABktr ≤ kAkHSkBkHS For proofs of these properties, we refer to Reed-Simon, Volume 1, §6.6
4HILBERT-SCHMIDTANDTRACECLASSPSDOST The trace of trace class operators.For trace class operators, one can define a linear functional called the trace. Itwill play the same role as the trace for matrices that we learned in linear algebra.In fact, one of our ultimate goals is to study the traces of certain semiclassicalpseudodifferential operators.On the space of trace class operators, we can define a trace functional viaTr : Ci(H) → C, A Tr(A) =(Aej,ej).JAlthough it is not that obvious, the quantity above is independent of the choices of[e]:to see this one start with the“polar decomposition"A=U|A]of A, whereUis an isometric when restricted to the closed subspace Ker(U)+ (For the existenceofpolardecomposition.c.f.Reed-Simon,$6.4).Then(Aej,ei) =(U|A|ej,ei) =(IA]/2ej;|A|/2*e;)=《/A/2ej,fi)《IA/2U*ej,f)=《A|1/2fi,e,)(U|AP/2fi,es)=(U|AP/2 fi,IA1/2 fi)and the conclusion follows. Note that by definition, for any A e C2(H), we haveAlHs = Tr(A*A).We list without proof a couple basis properties of the trace functional:. The trace functional is continuous (with respect to the trace norm):[Tr(A)I≤ IIAlltr-.By definition, if A is trace class and self-adjoint, thenTr(A) = ^jwhere As are eigenvalues of A (counting multiplicity). It turns out that theformula holds for any trace class operator (but the proof is more involved):Theorem 1.3 (Lidskii). Suppose A is of trace class, andSpec(A) =[,],[1/ ≥ [2/ ≥...→ 0,then Tr(A) = , >j.. For any A E Ci(H) and B C(H), we have Tr(AB) = Tr(BA)
4 HILBERT-SCHMIDT AND TRACE CLASS PSDOS ¶ The trace of trace class operators. For trace class operators, one can define a linear functional called the trace. It will play the same role as the trace for matrices that we learned in linear algebra. In fact, one of our ultimate goals is to study the traces of certain semiclassical pseudodifferential operators. On the space of trace class operators, we can define a trace functional via Tr : L1(H) → C, A 7→ Tr(A) = X j hAej , ej i. Although it is not that obvious, the quantity above is independent of the choices of {ej}: to see this one start with the “polar decomposition” A = U|A| of A, where U is an isometric when restricted to the closed subspace Ker(U) ⊥ (For the existence of polar decomposition, c.f. Reed-Simon, §6.4). Then X j hAej , ej i = X j hU|A|ej , ej i = X j h|A| 1/2 ej , |A| 1/2U ∗ ej i = X j h|A| 1/2 ej , fj ih|A| 1/2U∗ej , fj i = X j h|A| 1/2fj , ej ihU|A| 1/2 fj , ej i = X j hU|A| 1/2 fj , |A| 1/2 fj i and the conclusion follows. Note that by definition, for any A ∈ L2(H), we have kAk 2 HS = Tr(A∗A). We list without proof a couple basis properties of the trace functional: • The trace functional is continuous (with respect to the trace norm): |Tr(A)| ≤ kAktr. • By definition, if A is trace class and self-adjoint, then Tr(A) = X j λj , where λj s are eigenvalues of A (counting multiplicity). It turns out that the formula holds for any trace class operator (but the proof is more involved): Theorem 1.3 (Lidskii). Suppose A is of trace class, and Spec(A) = {λj}, |λ1| ≥ |λ2| ≥ · · · → 0, then Tr(A) = P j λj . • For any A ∈ L1(H) and B ∈ L(H), we have Tr(AB) = Tr(BA)
HILBERT-SCHMIDTANDTRACECLASSPSDOS52.HILBERT-SCHMIDTANDTRACECLASSINTEGRALOPERATORSTHilbert-Schmidt integral operators.A very important class of Hilbert-Schmidt operators are Hilbert-Schmidt inteqraloperators,which arebydefinitionHilbert-Schmidt operators onL?spaces of theformK(r,y)p(y)dy.A= Ak : P-→[AkP](r) = (Of course in the definition of Hilbert-Schmidt integral operators, onemay replaceRnbyanymeasurespace.)Let K = K(r, y) be a measurable function defined on Rn × Rn. We want tofind out conditions so thattheintegral operatorAkis aHilbert-Schmidtoperatoron L?(Rn). For this purpose we fix r E IRn and denote Kr(y) :=K(r, y). Then bydefinition, for any E L?(Rn),AkP(r) = (Kr,P)L2(R).So, if we want Ak(r) to be a well-defined measurable function, we need to requireK, E L?(Rn). Of course we want more: we want Ak E L?(R"). For this purposewe calculate via Cauchy-Schwarz inequality:[ I[Ar](r)Pdr= // / K(r, )u(y)dy daIIAkl/2(R") =((/ [K(r,g)Pdy) (/ [u(y)Pdy) = IKI2(R2m) Il/ll2(Rm)-So the condition we need is K(r, y) e L?(Rn × Rn). Note that if this is satisfied,then by Fubini's theorem, K,(y) := K(r,y) e L?(Rn) for a.e. r e Rn, so our firstrequirement is satisfied automatically. In summary, we have shown:Fact 1.For Ak to be a linear map from L?(IRn) to L?(Rn), we needto require K =K(,y) L?(Rn × Rn).Moreover, under this as-sumption, Ak is in fact a bounded linear map with(5)IIAkllc(L2(R") ≤ IIKIIL2(R2n).To get more information of the operator Ak,we may try to approximate Ak bysimpler operators. By definition, to approximate Ak, it is enough to approximatethe kernel function K. So we let (gibjen be an orthonormal basis of L?(R"). Then[;(r)k(y))j,ken is an orthonormal basis of L?(R" × IRn) and we can decomposeK(r, y) =cikp;(r)pk(y).j.k
HILBERT-SCHMIDT AND TRACE CLASS PSDOS 5 2. Hilbert-Schmidt and trace class integral operators ¶ Hilbert-Schmidt integral operators. A very important class of Hilbert-Schmidt operators are Hilbert-Schmidt integral operators, which are by definition Hilbert-Schmidt operators on L 2 spaces of the form A = AK : ϕ 7→ [AKϕ](x) = Z Rn K(x, y)ϕ(y)dy. (Of course in the definition of Hilbert-Schmidt integral operators, one may replace R n by any measure space.) Let K = K(x, y) be a measurable function defined on R n x × R n y . We want to find out conditions so that the integral operator AK is a Hilbert-Schmidt operator on L 2 (R n ). For this purpose we fix x ∈ R n x and denote Kx(y) := K(x, y). Then by definition, for any ϕ ∈ L 2 (R n ), AKϕ(x) = hKx, ϕiL2(Rn y ) . So, if we want AKϕ(x) to be a well-defined measurable function, we need to require Kx ∈ L 2 (R n y ). Of course we want more: we want AKϕ ∈ L 2 (R n ). For this purpose we calculate via Cauchy-Schwarz inequality: kAKϕk 2 L2(Rn) = Z Rn |[AKϕ](x)| 2 dx = Z Rn Z Rn K(x, y)u(y)dy 2 dx ≤ Z Rn Z Rn |K(x, y)| 2 dy Z Rn |u(y)| 2 dy dx = kKk 2 L2(R2n) · kϕk 2 L2(Rn) . So the condition we need is K(x, y) ∈ L 2 (R n x × R n y ). Note that if this is satisfied, then by Fubini’s theorem, Kx(y) := K(x, y) ∈ L 2 (R n y ) for a.e. x ∈ R n , so our first requirement is satisfied automatically. In summary, we have shown: Fact 1. For AK to be a linear map from L 2 (R n ) to L 2 (R n ), we need to require K = K(x, y) ∈ L 2 (R n x × R n y ). Moreover, under this assumption, AK is in fact a bounded linear map with (5) kAKkL(L2(Rn)) ≤ kKkL2(R2n) . To get more information of the operator AK, we may try to approximate AK by simpler operators. By definition, to approximate AK, it is enough to approximate the kernel function K. So we let {ϕj}j∈N be an orthonormal basis of L 2 (R n ). Then {ϕj (x)ϕk(y)}j,k∈N is an orthonormal basis of L 2 (R n × R n ) and we can decompose K(x, y) = X j,k cjkϕj (x)ϕk(y)
6HILBERT-SCHMIDTANDTRACECLASSPSDOSNow the kernel function K is naturally approximated by the “truncated" kernelKn(a,y) := cjkp;(r)p(y).j,k<NSince the operatorAkn = Z Ck(,Pk)pjj.k≤Nis a finite rank operator and by (5),IIAk - Ak llc(L2(R) ≤KN - KIL2(R") → 0as N → oo, we immediately deduce that Ak is a compact operator on L?(R") aslong as the kernel K = K(r, y) E L?(Rn × Rn). But of course this is not our finalgoal. We want Ak to be Hilbert-Schmidt operator. For this purpose we calculatethe Hilbert-Schmidt norm of Ak via the basis (p]:IIAkl/s =IAkpi/2(Rn) = Ilcjk;(a)0ki/2(R) =lcil2 =IKI2(R2n).k,li.lSo we conclude:Fact 2 For any K = K(r, y) E L?(R2n), the integral operator Akis a Hilbert-Schmidt operator on L?(Rn) with(6)IAklHS = IIK|L2(R2n)We can conclude more: Consider the mapA: L?(R2n) →C?(L?(R")),K -→AkThen by (6), A is an isometric embedding. In particular, the image of A is closed.On the other hand, any finite rank operator on L?(Rn) is an integral operator withL?-kernel. Proof: If A is a finite rank operator, i.e. Im(A) is finitely dimensional.then after fixing a basis i,., bm of Im(A), we can write A uniquely asAp=ci()1+...+Cm(p)m.The linearity of A implies that each c;() is linear functional on L?(R"). By Rieszrepresentation theorem, there exists P, such that c()=(p,P,). It follows[E(;(ar);() (u)dy,Ap(r)=ie.Ais an integral operatorwithkernel Em=(b;(r);(y))L(R2n).Sincethe set of finite rank operator is dense in C2(L?(Rn)), we conclude that A is alsosurjective. In other words, any Hilbert-Schmidt operator on L?(R") is an integraloperator with L?(R2n)-kernel!In summary, we proved
6 HILBERT-SCHMIDT AND TRACE CLASS PSDOS Now the kernel function K is naturally approximated by the “truncated” kernel KN (x, y) := X j,k≤N cjkϕj (x)ϕk(y). Since the operator AKN = X j,k≤N cjkh·, ϕkiϕj is a finite rank operator and by (5), kAK − AKN kL(L2(Rn)) ≤ kKN − KkL2(Rn) → 0 as N → ∞, we immediately deduce that AK is a compact operator on L 2 (R n ) as long as the kernel K = K(x, y) ∈ L 2 (R n x × R n y ). But of course this is not our final goal. We want AK to be Hilbert-Schmidt operator. For this purpose we calculate the Hilbert-Schmidt norm of AK via the basis {ϕk}: kAKk 2 HS = X l kAKϕlk 2 L2(Rn) = X j,k,l kcjkϕj (x)δklk 2 L2(Rn x) = X j,l |cjl| 2 = kKk 2 L2(R2n) . So we conclude: Fact 2. For any K = K(x, y) ∈ L 2 (R 2n ), the integral operator AK is a Hilbert-Schmidt operator on L 2 (R n ) with (6) kAKkHS = kKkL2(R2n) . We can conclude more: Consider the map A : L 2 (R 2n ) → L2 (L 2 (R n )), K 7→ AK. Then by (6), A is an isometric embedding. In particular, the image of A is closed. On the other hand, any finite rank operator on L 2 (R n ) is an integral operator with L 2 -kernel. Proof: If A is a finite rank operator, i.e. Im(A) is finitely dimensional, then after fixing a basis ψ1, · · · , ψm of Im(A), we can write A uniquely as Aϕ = c1(ϕ)ψ1 + · · · + cm(ϕ)ψm. The linearity of A implies that each cj (ϕ) is linear functional on L 2 (R n ). By Riesz representation theorem, there exists ϕj such that cj (ϕ) = hϕ, ϕj i. It follows Aϕ(x) = Z Rn Xm j=1 ψj (x)ϕj (y) ϕ(y)dy, i.e. A is an integral operator with kernel Pm j=1 ψj (x)ϕj (y) ∈ L 2 (R 2n ). Since the set of finite rank operator is dense in L 2 (L 2 (R n )), we conclude that A is also surjective. In other words, any Hilbert-Schmidt operator on L 2 (R n ) is an integral operator with L 2 (R 2n )-kernel! In summary, we proved
7HILBERT-SCHMIDTANDTRACECLASSPSDOSTheorem 2.1. An integral operator Ak with Schwartz kernel K(c,y) is a Hilbert-Schmidt integral operator on L?(IRn) if and only if K(r,y) EL?(Rn ×Rn). Moreover,inthiscasewehaveIAkHS=IKIL2. Semiclassical PsDOs as Hilbert-Schmidt operators.As a consequence, we getTheorem 2.2. The operator Opi(a) is a Hilbert-Schmidt operator if and only ifa = a(r, 5) e L?(R2n), in which case we have1I:()lis 2h) /ll2(am)Proof.Weknowthat thekernel of Opt(a)is thepartial Fourier transformet(r-)-Ea(tr +(1- t)y,S)dEkt(r,y) =(2元h)(2h)[()--2d(ta +(1-t), -),So we get from thePlancherel's theorem (c.f. Lecture 4)1/21kall 2(R2m) =Ja(ta+(1-t)r,y-r)}'drdy(2元h)n/21/21Ja(a,dedy(2元h)n/2(2 h)n/lll/2(R2n)口and theconclusionfollows.In particular, we see that if the order function m E L?(R2n),then for anya E S(m),the semiclassical pseudodifferential operator Opt(a)is a Hilbert-Schmidtoperator whoseHilbert-Schmidtnorm isbounded by O(h-n/2). Trace class integral operators.Now we study trace class integral operators.Unlike Hilbert-Schmidt operatorsfor which we have Theorem 2.1, there is no simple criteria for an integral operator tobea trace class operator. So in what follows wewill only give a sufficient condition(on the kernel) so that an integral operator is in trace class.Since any trace class operator can be written as the composition of two Hilbert-Schmidt operator, we will start with the trace class integral operators which can bewritten asthe composition of twoHilbert-Schmidt integral operators
HILBERT-SCHMIDT AND TRACE CLASS PSDOS 7 Theorem 2.1. An integral operator AK with Schwartz kernel K(x, y) is a HilbertSchmidt integral operator on L 2 (R n ) if and only if K(x, y) ∈ L 2 (R n×R n ). Moreover, in this case we have kAKkHS = kKkL2 . ¶ Semiclassical PsDOs as Hilbert-Schmidt operators. As a consequence, we get Theorem 2.2. The operator Opt ~ (a) is a Hilbert-Schmidt operator if and only if a = a(x, ξ) ∈ L 2 (R 2n ), in which case we have kOpt ~ (a)kHS = 1 (2π~) n/2 kak 2 L2(R2n) . Proof. We know that the kernel of Opt ~ (a) is the partial Fourier transform k t a (x, y) = 1 (2π~) n Z Rn e i ~ (x−y)·ξ a(tx + (1 − t)y, ξ)dξ = 1 (2π~) n [(F~)ξ→y−xa](tx + (1 − t)y, y − x). So we get from the Plancherel’s theorem (c.f. Lecture 4) kk t akL2(R2n) = 1 (2π~) n/2 Z R2n |a(tx + (1 − t)x, y − x)| 2 dxdy1/2 = 1 (2π~) n/2 Z R2n |a(x, e ye)| 2 dxde ye 1/2 = 1 (2π~) n/2 kakL2(R2n) and the conclusion follows. In particular, we see that if the order function m ∈ L 2 (R 2n ), then for any a ∈ S(m), the semiclassical pseudodifferential operator Opt ~ (a) is a Hilbert-Schmidt operator whose Hilbert-Schmidt norm is bounded by O(~ −n/2 ). ¶ Trace class integral operators. Now we study trace class integral operators. Unlike Hilbert-Schmidt operators for which we have Theorem 2.1, there is no simple criteria for an integral operator to be a trace class operator. So in what follows we will only give a sufficient condition (on the kernel) so that an integral operator is in trace class. Since any trace class operator can be written as the composition of two HilbertSchmidt operator, we will start with the trace class integral operators which can be written as the composition of two Hilbert-Schmidt integral operators.
6HILBERT-SCHMIDTANDTRACECLASSPSDOSProposition 2.3.If Ki,K2 EL?(R2n),thenAk,Ak,is a traceclass operator withAK,Ak lltr ≤IKillL2(R") -|K2l/L2(R") andKi(r,y)K2(y,r)dedytr(AKiAK2)=Proof. The first half of the proposition is obvious. To calculate the trace we firstobserve bydefinition thatAk(r,) = Ak(y,a)ItfollowsTr(Ak,Ak2)=E(Ak,AkPj,pi)2E(AkPj,Akp)=E(AK2pj,Pk)-(AkP,Pk)=E(K2(r, y), p;(y)pk(a)L2(R2n) (Ki(y,a), p;(y)pk(a)L2(R2n)j.kKi(y,r)K2(r, y)drdyKi(r, y)K2(y, r)drdy口 Semiclassical PsDO with Schwartz symbols as trace class operators.Finally we study the trace property of semiclassical PsDOs. We start with acorollary of Proposition 2.3:4Corollary 2.4. Suppose there erist b,c e L2(Rn × Rr) such that aKN =6KN ocKN.Then akN is a trace class operator with1Tr(aKN)a(r, E)drde(2元h)nProof. The Schwartz kernel of 6KN is (2)[(Fr)s-y-ab)(a, y - a). Thus1Tr(6KN0CKN)[(Fn)e→y-rb](r,y -r)[(Fh)s→y-rc(y, r- y)dady(2元h)2n4The result holds for other quantizations. We stated it via Kohn-Nirenberg quantization onlybecause in this case it is easier to do the computations
8 HILBERT-SCHMIDT AND TRACE CLASS PSDOS Proposition 2.3. If K1, K2 ∈ L 2 (R 2n ), then AK1AK2 is a trace class operator with kAK1AK2 ktr ≤ kK1kL2(Rn) · kK2kL2(Rn) and tr(AK1AK2 ) = Z R2n K1(x, y)K2(y, x)dxdy. Proof. The first half of the proposition is obvious. To calculate the trace we first observe by definition that A ∗ K(x,y) = AK(y,x) . It follows Tr(AK1AK2 ) = X j hAK1AK2ϕj , ϕj i = X j hAK2ϕj , A∗ K1 ϕj i = X j X k hAK2ϕj , ϕki · hA∗ K1 ϕj , ϕki = X j,k hK2(x, y), ϕj (y)ϕk(x)iL2(R2n) · hK1(y, x), ϕj (y)ϕk(x)iL2(R2n) = Z R2n K1(y, x)K2(x, y)dxdy = Z R2n K1(x, y)K2(y, x)dxdy. ¶ Semiclassical PsDO with Schwartz symbols as trace class operators. Finally we study the trace property of semiclassical PsDOs. We start with a corollary of Proposition 2.3:4 Corollary 2.4. Suppose there exist b, c ∈ L 2 (R n x × R n ξ ) such that ba KN = bb KN ◦ bc KN . Then ba KN is a trace class operator with Tr(ba KN ) = 1 (2π~) n Z R2n a(x, ξ)dxdξ. Proof. The Schwartz kernel of bb KN is 1 (2π~)n [(F~)ξ→y−xb](x, y − x). Thus Tr(bb KN ◦ bc KN ) = 1 (2π~) 2n Z R2n [(F~)ξ→y−xb](x, y − x)[(F~)ξ→y−xc](y, x − y)dxdy. 4The result holds for other quantizations. We stated it via Kohn-Nirenberg quantization only because in this case it is easier to do the computations.
9HILBERT-SCHMIDTANDTRACECLASSPSDOSRecall from Lecture9that a=bknchas the expression-e-(a)b(,$+)c(r+,s)dda(r,s)(2h)n1e-(y-r):(n-$)b(r, n)c(y, )dydn(2元h)nwhich implies11e-(y-r)(n-5)b(r,n)c(y,s)dydndardea(r,s)drds(2元h)2n(2元h)nJ1[(Fh)e→y-rb](r, y - r)[(Fh)s-→y-rcl(y, a -y)dady(2元h)2nSo we concludeTr(aKN) = Tr(6KN ocKNa(r, E)drde(2元h）n口Now we can work on the case where the symbols are Schwartz functions.Notethat we can easily write a Schwartz function as the product of two L?-functions. Soin view of the function-operator correspondence, we should be able to decomposeOpt(a) (wherea isa Schwartz function)intothecomposition oftwoHilbert-Schmidtoperators, and then use the above theorem to calculate its trace.Proposition 2.5. If a(r, ) E J(R2n), then Opk(a) is a trace class operator withIIopi(a)lt≤Ch-" hl/210°al/a(R2n),lal≤MdwhereM is someconstant,andTr(Op(a)) =a(r,E)dr de(2元h）nSketch of proof. It is enough to work on the case of Kohn-Nirenberg quantization,since then the general caseis a consequence of the special case together with thechangeof quantization formula.(Work outthe detail!)To"decompose"the oper-ator aKN, we let b(r,E) = (r)-n-1(E)-n-1. Then b E L(R2n) and 6KN = FFl。KN(r)-n-1()-n-1 o Fn. So if we let 2KN = (6KN)-1 oaKN = (μ)n+1(E)n+1)OaKNthen the formula for the trace follows immediately since c= (r)n+i($)n+1)*kN a EThe trace norm estimate also follows from the decomposition:Iak Ilr ≤ 1I6k I s 1K lHs ≤Ch-" Z hl/210°ll-(R2n),lal<Md
HILBERT-SCHMIDT AND TRACE CLASS PSDOS 9 Recall from Lecture 9 that a = b ?KN c has the expression a(x, ξ) = 1 (2π~) n Z R2n e − i ~ (˜x·ξ˜) b(x, ξ + ˜ξ)c(x + ˜x, ξ)dxd˜ ˜ξ = 1 (2π~) n Z R2n e − i ~ ((y−x)·(η−ξ))b(x, η)c(y, ξ)dydη which implies 1 (2π~) n Z R2n a(x, ξ)dxdξ = 1 (2π~) 2n Z R4n e − i ~ ((y−x)·(η−ξ))b(x, η)c(y, ξ)dydηdxdξ = 1 (2π~) 2n Z R2n [(F~)ξ→y−xb](x, y − x)[(F~)ξ→y−xc](y, x − y)dxdy. So we conclude Tr(ba KN ) = Tr(bb KN ◦ bc KN ) = 1 (2π~) n Z R2n a(x, ξ)dxdξ. Now we can work on the case where the symbols are Schwartz functions. Note that we can easily write a Schwartz function as the product of two L 2 -functions. So in view of the function-operator correspondence, we should be able to decompose Opt ~ (a) (where a is a Schwartz function) into the composition of two Hilbert-Schmidt operators, and then use the above theorem to calculate its trace. Proposition 2.5. If a(x, ξ) ∈ S (R 2n ), then Opt ~ (a) is a trace class operator with kOpt ~ (a)ktr ≤ C~ −n X |α|≤Md ~ |α|/2 k∂ α akL1(R2n) , where M is some constant, and Tr(Opt ~ (a)) = 1 (2π~) n Z R2n a(x, ξ)dx dξ. Sketch of proof. It is enough to work on the case of Kohn-Nirenberg quantization, since then the general case is a consequence of the special case together with the change of quantization formula. (Work out the detail!) To “decompose” the operator ba KN , we let b(x, ξ) = hxi −n−1 hξi −n−1 . Then b ∈ L 2 (R 2n ) and bb KN = F −1 ~ ◦ hxi −n−1 hξi −n−1 ◦ F~. So if we let bc KN = (bb KN ) −1 ◦ ba KN = (hxi n\+1hξi n+1) KN ◦ ba KN , then the formula for the trace follows immediately since c = (hxi n+1hξi n+1) ?KN a ∈ S . The trace norm estimate also follows from the decomposition: kba KN ktr ≤ kbb KN kHS · kbc KN kHS ≤ C~ −n X |α|≤Md ~ |α|/2 k∂ α akL1(R2n) ,
10HILBERT-SCHMIDTANDTRACECLASSPSDOSwherethelaststep comesfromthefactthat1e-()n+1(s+E)n+la(r+i,s)dedc(r,E) =(2元h)nJR2m口work outthedetails.For the general case we only state the theorem as follows, and leave the proofas an exercise: Again we localize, re-scale, change quantization etc.Theorem 2.6. Suppose a E S(1) satisfiesZ I1allL≤cJal<Mnfor all hE(O,ho),thenthe operator Opt(a)istrace class, withIIOp(a)ltr≤Ch-d hl/2)10°all 1(R2n)[al<Mdfor some constant M, and1Tr(Opt(a) =a(r,s,h)dr d(2元h)dJ/R21
10 HILBERT-SCHMIDT AND TRACE CLASS PSDOS where the last step comes from the fact that c(x, ξ) = 1 (2π~) n Z R2n e − i ~ x˜·ξ˜ hxi n+1hξ + ˜ξi n+1a(x + ˜x, ξ)dxd˜ ˜ξ. work out the details. For the general case we only state the theorem as follows, and leave the proof as an exercise: Again we localize, re-scale, change quantization etc. Theorem 2.6. Suppose a ∈ S(1) satisfies X |α|≤Mn k∂ α akL1 ≤ C for all ~ ∈ (0, ~0), then the operator Opt ~ (a) is trace class, with kOpt ~ (a)ktr ≤ C~ −d X |α|≤Md ~ |α|/2 k∂ α akL1(R2n) for some constant M, and Tr(Opt ~ (a)) = 1 (2π~) d Z R2n a(x, ξ, ~) dx dξ.