《半经典微局部分析 Semiclassical Microlocal Analysis》课程教学资源（英文讲义）Lec15 THEORY OF SEMICLASSICAL PSDOS：POSITIVITY

LECTURE15:L?-THEORYOF SEMICLASSICALPsDOs:POSITIVITY1.THEWEAK GARDING INEQUALITYPositivity: a counterexample.The last “quantitative aspect" of the quantization procedure a aw we wantto study is the positivity:If a is positive (and thus real-valued),what kinds of“positivity"does the operator aw admit? Is it always a positive operator? Herewe only use the Weyl quantization, since other t-quantization will not convert real-valued functions to self-adjoint operators in general.Recall that a densely defined symmetric A is called positive if(Au,u) ≥ 0, Vu ED(A).For example, for any densely defined closed operator A, the operator A* A is always apositive operator.As usual we will use the notation A ≥0 if A is a positive operatorand use the notation A >B if A.B aredensely-defined symmetric operators andA-B is a positive operator.It is not hard to prove A is positive if and only if A=B2for somepositive self-adjoint operator B.Equivalently,by the spectral theorem,aself-adjoint linear operator A is positive if and only if Spec(A) C [0, +oo).To study therelationbetween positivity of operatorsand of corresponding sym-bols,let's startwithtwo simpleexamples,whichtells us that thepositiveoperatorneed not have nonnegative symbol, and conversely, the Weyl quantization of a non-negative function need not be a positive operator:1Similarly we can prove that it is possible that a is not nonnegative while aw ispositive:The first example is simple: Consider a(r, $) = r? +2-h. ThenaW = Q? + p2 - h.According to our result for the quantum harmonic oscillator (Lecture 3), aw is apositive operator, but its symbol a is not a non-negative function.lIn general, we call a quantization procedure a positive quantization if it sends nonnegative-valued functions to positive operators.So the Weyl quantization (and in fact anyt-quantization)isnot a positive quantization. The so-called anti-Wick quantization (also known as Berezin-Toeplitzquantization)is apositive quantization.1

LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY 1. The weak G˚Arding inequality ¶ Positivity: a counterexample. The last “quantitative aspect” of the quantization procedure a ba W we want to study is the positivity: If a is positive (and thus real-valued), what kinds of “positivity” does the operator ba W admit? Is it always a positive operator? Here we only use the Weyl quantization, since other t-quantization will not convert real￾valued functions to self-adjoint operators in general. Recall that a densely defined symmetric A is called positive if hAu, ui ≥ 0, ∀u ∈ D(A). For example, for any densely defined closed operator A, the operator A∗A is always a positive operator. As usual we will use the notation A ≥ 0 if A is a positive operator, and use the notation A ≥ B if A, B are densely-defined symmetric operators and A−B is a positive operator. It is not hard to prove A is positive if and only if A = B2 for some positive self-adjoint operator B. Equivalently, by the spectral theorem, a self-adjoint linear operator A is positive if and only if Spec(A) ⊂ [0, +∞). To study the relation between positivity of operators and of corresponding sym￾bols, let’s start with two simple examples, which tells us that the positive operator need not have nonnegative symbol, and conversely, the Weyl quantization of a non￾negative function need not be a positive operator:1 Similarly we can prove that it is possible that a is not nonnegative while ba W is positive: The first example is simple: Consider a(x, ξ) = x 2 + ξ 2 − ~. Then ba W = Q 2 + P 2 − ~. According to our result for the quantum harmonic oscillator (Lecture 3), ba W is a positive operator, but its symbol a is not a non-negative function. 1 In general, we call a quantization procedure a positive quantization if it sends nonnegative￾valued functions to positive operators. So the Weyl quantization (and in fact any t-quantization) is not a positive quantization. The so-called anti-Wick quantization (also known as Berezin-Toeplitz quantization) is a positive quantization. 1

2LECTURE15:L?-THEORYOFSEMICLASSICALPSDOS:POSITIVITYThe other direction is a bit more complicated. Let a(r, $) = r2g2. Then byMcCoy'sformula(Lecture7),aw=(P2Q?+ 2QP2Q + Q?P2),If we let b(r,$)= r, then 6w =(QP +PQ), which impliesGW)?=I(PQPQ+QPQP+QP?Q+PQ?P)(P2Q?+iPQ+Q?p2=ihQP+QP2Q+(QP-it)(PQ+i))1h2=aw4Ontheotherhand,itcanbeproventhatSpec((W)?)=[0,+oo).(Formallybysolving ODE one can find thefunction r-1/2 which satisfies[(-)-2 /6W(α-1/2)==0?+2Of course the function r-1/2 is far from being in our space.For example, -1/2 is onlydefined on (o,+oo).So it isnatural to consider thediffeomorphism R-→(o,+oo)which sends r to e.The diffeomorphism induces a natural unitary operator fromL?(R)to L?(O,+oo)), sending f() to er/2 f(e). One can show that this unitaryoperator will conjugate bw (restricted to L?(0,+oo))) to the operator Pon L?(R)Itiswell-knownthatthe spectrum of themomentum operatorPis R.Itfollowsthat Spec(6W) = R and thus Spec((6W)2) = [0, +oo). It follows thatSpec(aW) = [-1h2,+8)and thusaw isnotapositiveoperator. Semiclassical Garding inequality: a simple version.In the remaining of this lecture, we will show that if a is non-negative, then awis"almost positive".We start with a simple version which claims that if a ≥ O, thenaw ≥-e.Id:Theorem 1.1 (Weak Garding inequality). Suppose a E S(1) is nonnegative, i.e.a≥0. Then for any >0, there ecists ho>0 such that for any he(0,ho),(aWu, u) ≥ -llull22(R"),VuE L?(R").Proof.By definition, for any < -e, a- is elliptic in S(1).So according towhat we learned last time, there exists ho = ho() such that for any h e (o, ho), theoperator aw - . Id is invertible and the inverse is a bounded linear operator. Inwhatfollowswewill show:

2 LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY The other direction is a bit more complicated. Let a(x, ξ) = x 2 ξ 2 . Then by McCoy’s formula (Lecture 7), ba W = 1 4 (P 2Q 2 + 2QP2Q + Q 2P 2 ). If we let b(x, ξ) = xξ, then bb W = 1 2 (QP + P Q), which implies (bb W ) 2 = 1 4 (P QP Q + QP QP + QP2Q + P Q2P) = 1 4 (P 2Q 2 + i~P Q + Q 2P 2 − i~QP + QP2Q + (QP − i~)(P Q + i~)) = ba W + 1 4 ~ 2 . On the other hand, it can be proven that Spec((bb W ) 2 ) = [0, +∞). (Formally by solving ODE one can find the function x −1/2 which satisfies bb W (x −1/2 ) = 1 2  x(− 1 2 )x −3/2 + 1 2 x −1/2  = 0. Of course the function x −1/2 is far from being in our space. For example, x −1/2 is only defined on (0, +∞). So it is natural to consider the diffeomorphism R → (0, +∞) which sends x to e x . The diffeomorphism induces a natural unitary operator from L 2 (R) to L 2 ((0, +∞)), sending f(x) to e x/2 f(e x ). One can show that this unitary operator will conjugate bb W (restricted to L 2 ((0, +∞))) to the operator P on L 2 (R). It is well-known that the spectrum of the momentum operator P is R. It follows that Spec(bb W ) = R and thus Spec((bb W ) 2 ) = [0, +∞). It follows that Spec(ba W ) = [− 1 4 ~ 2 , +∞) and thus ba W is not a positive operator. ¶ Semiclassical G˚arding inequality: a simple version. In the remaining of this lecture, we will show that if a is non-negative, then ba W is “almost positive”. We start with a simple version which claims that if a ≥ 0, then ba W ≥ −ε · Id: Theorem 1.1 (Weak G˚arding inequality). Suppose a ∈ S(1) is nonnegative, i.e. a ≥ 0. Then for any ε > 0, there exists ~0 > 0 such that for any ~ ∈ (0, ~0), hba W u, ui ≥ −εkuk 2 L2(Rn) , ∀u ∈ L 2 (R n ). Proof. By definition, for any δ < −ε, a − δ is elliptic in S(1). So according to what we learned last time, there exists ~0 = ~0(δ) such that for any ~ ∈ (0, ~0), the operator ba W − δ · Id is invertible and the inverse is a bounded linear operator. In what follows we will show:

3LECTURE15:L?-THEORYOFSEMICLASSICALPSDOS:POSITIVITYClaim: Fixing a and e, the constant ho = ho() can be chosen to beindependentofthechoiceof2It follows that for any ,la(-)Lis uniformly bounded forall S<-e.So thereexistsaconstantC,uniformlyforall<-e, suchthatII/WIlc(L2(R) ≤Ch2口and the conclusion follows.2More general, for any a(r,$) and any one-variable function f, one can easily check[a(r,), f(a(r,))) = 0

LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY 3 Claim: Fixing a and ε, the constant ~0 = ~0(δ) can be chosen to be independent of the choice of δ < −ε. As a consequence, the interval (−∞, −) lies in the resolvent set of ba W for any ~ ∈ (0, ~0), and thus the spectrum Spec(ba W ) ⊂ [−, +∞) uniformly for all ~ ∈ (0, ~0). By spectral theorem for bounded linear operators, we get hba W u, ui ≥ −εkuk 2 L2(Rn) for any ~ ∈ (0, ~0). So it remains to prove the claim. For this we have to go back to the proof of the invertibility. According to the remark after Corollary 1.4 in Lecture 14, we it is enough to find ~0 so that krb W kL(L2(Rn)) < 1/2, ∀~ ∈ (0, ~0) holds for all δ < −ε, where r = 1 − (a−δ)? 1 a−δ . It is easy to check2 {a − δ, 1 a − δ } = 0, we have 1 − (a − δ) ? 1 a − δ = O(~ 2 ) According to Calderon-Vaillancourt theorem, we need a uniform estimate of k∂ α rkL∞ for all δ < −ε. So we only need to find uniform bound for k∂ α (a − δ)kL∞(Rn) and k∂ α ( 1 a − δ )kL∞(Rn) Since a ∈ S(1), all derivatives of a − δ is uniformly bounded. Note: a − δ itself is NOT bounded. However, we will not need it in computing r. We only need derivatives of a − δ. On the other hand, as in last lecture, by induction one can prove (1) ∂ α ( 1 a − δ ) = 1 a − δ X β1+···+βk=α,|βj |≥1 Cβ1···βk Y j ( 1 a − δ ∂ βja) It follows that for any α, k∂ α ( 1 a−δ )kL∞ is uniformly bounded for all δ < −ε. So there exists a constant C, uniformly for all δ < −ε, such that krb W kL(L2(Rn)) ≤ C~ 2 and the conclusion follows. 2More general, for any a(x, ξ) and any one-variable function f, one can easily check {a(x, ξ), f(a(x, ξ))} = 0.

LECTURE 15: L?-THEORY OF SEMICLASSICAL PSDOS: POSITIVITYN2. THE SHARP GARDING INEQUALITY Sharp (semiclassical)Garding inequality.The weak Garding inequality is “weak"because the lower bound we get is not"semiclassical small",thusash→Othelowerbound could berelativelylarge.Ourmain goal in this lecture is to prove the following sharp Garding inequality,whichclaims that the Weyl quantization of a nonnegative symbol is “semiclassically closeto be positive":Theorem 2.1 (Sharp Garding inequality). Suppose a E S(1) and a > 0. Thenthere erists constant C≥0 and ho> 0 such that for all hE(0,ho),(αWu, u) ≥ -Cillul/2(Rn),VuE L?(R")Remark.With more work, one can prove a stronger inequality:the (semiclassical)Fefferman-Phong inequality:Vu E L?(R").(aWu,u) ≥ Ch?llul22(R"),In view of the examples at the beginning of this lecture, this inequality is sharp.In Lecture 1l, as a consequence of the Calderon-Vaillancourt theorem, we get/aWIl(L2(R) ≤ C sup lal + O(h1/2).R.Nowwecan removetheconstant Cby usingthe SharpGarding inequality:Corollary 2.2. Suppose a E S(1). ThenIa"Wllc(L2(R") ≤ sup [al + O(t),.Proof.For any u EL?with Julz2=1,we haveIawullz= (aw)awu, u)2 = (awawu, u)L2.Sinceaaw=ap"+0(h)weget from the sharp Garding inequalityIaWull2 = (laj2" u, ) +O() ≥ (sup lal)? +O(n),口Similarly one can prove: If a E S(1) is real-valued, then there exists C and hosuch thatinfa-Ch<aw≤supa+Ch

4 LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY 2. The sharp G˚Arding inequality ¶ Sharp (semiclassical) G˚arding inequality. The weak G˚arding inequality is “weak” because the lower bound we get is not “semiclassical small”, thus as ~ → 0 the lower bound could be relatively large. Our main goal in this lecture is to prove the following sharp G˚arding inequality, which claims that the Weyl quantization of a nonnegative symbol is “semiclassically close to be positive”: Theorem 2.1 (Sharp G˚arding inequality). Suppose a ∈ S(1) and a ≥ 0. Then there exists constant C ≥ 0 and ~0 > 0 such that for all ~ ∈ (0, ~0), hba W u, ui ≥ −C~kuk 2 L2(Rn) , ∀u ∈ L 2 (R n ). Remark. With more work, one can prove a stronger inequality: the (semiclassical) Fefferman-Phong inequality: hba W u, ui ≥ C~ 2 ku| 2 L2(Rn) , ∀u ∈ L 2 (R n ). In view of the examples at the beginning of this lecture, this inequality is sharp. In Lecture 11, as a consequence of the Calderon-Vaillancourt theorem, we get kba W kL(L2(Rn)) ≤ C sup Rn |a| + O(~ 1/2 ). Now we can remove the constant C by using the Sharp G˚arding inequality: Corollary 2.2. Suppose a ∈ S(1). Then kba W kL(L2(Rn)) ≤ sup Rn |a| + O(~). Proof. For any u ∈ L 2 with kukL2 = 1, we have kba W uk 2 L2 = h(ba W ) ∗ba W u, uiL2 = hba W ba W u, uiL2 . Since ba W ba W = |da| 2 W + O(~), we get from the sharp G˚arding inequality kba W uk 2 = h|da| 2 W u, ui + O(~) ≥ (sup |a|) 2 + O(~). Similarly one can prove: If a ∈ S(1) is real-valued, then there exists C and ~0 such that inf a − C~ ≤ ba W ≤ sup a + C~.

LECTURE15:L?-THEORYOFSEMICLASSICALPSDOS:POSITIVITY5TProof of Sharp Gardinginequality:First thoughts.Now we try toprove the Sharp Garding inequality.Again, it is enough to provethat for -1now.Again we wantto control the L-norms of partial derivatives of a- S and .-. Again the L-normof eachderivativeof a-isuniformlycontrolled.So the onlyproblem is togeta uniform control of the L-norm of each partial derivative of -, which is quitenon-trivial now.We still need to apply the formula (1). Since a ≥ 0 and Fortunately, there is a critical symbol class, Si/2(1), which is not as good asS(1)=So(1)since we will not have those nice asymptotic expansions as we haveexplained in Lecturelo, but it is alsonot that bad in the sense that we can still getmany useful results after hard working, e.g. we still have a nice explicit formula fortheMoyalproduct.Tofinish theproofof theSharpGarding inequality,wefirstnoticethatifβ≥2,thenwehave (Note:[|<1.):allL≤CoJ1;1/2So in view of (1), it is possible to improve the symbol class from Si(1) to Si/2(1) aslong as we can prove the same inequality for|B=1. For this purpose we will needto use the following elementary gradient estimate:

LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY 5 ¶ Proof of Sharp G˚arding inequality: First thoughts. Now we try to prove the Sharp G˚arding inequality. Again, it is enough to prove that for δ −1 now. Again we want to control the L ∞-norms of partial derivatives of a−δ and 1 a−δ . Again the L ∞-norm of each derivative of a − δ is uniformly controlled. So the only problem is to get a uniform control of the L ∞-norm of each partial derivative of 1 a−δ , which is quite non-trivial now. We still need to apply the formula (1). Since a ≥ 0 and δ 1 2 . Fortunately, there is a critical symbol class, S1/2(1), which is not as good as S(1) = S0(1) since we will not have those nice asymptotic expansions as we have explained in Lecture 10, but it is also not that bad in the sense that we can still get many useful results after hard working, e.g. we still have a nice explicit formula for the Moyal product. To finish the proof of the Sharp G˚arding inequality, we first notice that if |β| ≥ 2, then we have (Note: |δ| < 1.) k 1 a − δ ∂ βjakL∞ ≤ Cβj 1 |δ| ≤ C 1 |δ| |βj |/2 . So in view of (1), it is possible to improve the symbol class from S1(1) to S1/2(1) as long as we can prove the same inequality for |β| = 1. For this purpose we will need to use the following elementary gradient estimate:

6LECTURE15:L?-THEORYOFSEMICLASSICALPSDOS:POSITIVITYDETOUR: A gradient estimate for positive functions.Beforeweprovethe theorem,we need a couplepreparation.Supposef eC2(R)and f>0, then by Taylor's expansion, there exists E (,r +t) such thatt2f"()0≤f(α+t)=f(r)+tf'()+Thus if If"l≤ A and if we take t =-f'/A we concludeIf'≤V2Af.This simple fact has the following multi-variable generalization,Lemma 2.3 (Gradient estimate).Suppose f :IRn →R is non-negative, C?, and[ofl≤ A (meaning rT(f) ≤A|r2 for any r E Rn). Then/VfI≤V2Af.Proof. The proof is almost the same as the one variable case. The only difference isthat we use the following Taylor's expansion with integral remainder:(1 - s)(02 f(r + st)t, t)ds.0≤ f(α+t)=f(r)+(Vf,t)+Taking t=-Vf(r), we get/VfI?≤Af +A2(1 - s)ds . (t,t) = Af -/VfIP/2,口fromwhich the conclusionfollows.DETOUR: The critical symbol class S1/2(m)We also list some results for Si/2(m) that we need. Recall that a E Ss(m) if foranya,10al ≤Cah-dlalm.We need the following results for the critical symbol class Si/2(m):W(I) For any symmetric non-singular 2n × 2n matrix Q, the operator esTQsmapsSi/2(m)toSi/2(m).(Butwewillnothavetheasymptoticexpansionas in the case of <(II) If aE S1/2(m1),be Si/2(m2), then a*b e S1/2(mim2) andab(r,S) = [e(DeDe-De-D)(a(r,)b(y,n)(Il1) (Calderon-Vaillancourt for critical symbol) For a E Si/2(1), aW is boundedon L?(Rn) andIIaWllc(L2(R") ≤c Z hla/2[0allL (Rn),Ja<Mn

6 LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY ¶ DETOUR: A gradient estimate for positive functions. Before we prove the theorem, we need a couple preparation. Suppose f ∈ C 2 (R) and f ≥ 0, then by Taylor’s expansion, there exists ˜x ∈ (x, x + t) such that 0 ≤ f(x + t) = f(x) + tf0 (x) + t 2 2 f 00(˜x). Thus if |f 00| ≤ A and if we take t = −f 0/A we conclude |f 0 | ≤ p 2Af. This simple fact has the following multi-variable generalization, Lemma 2.3 (Gradient estimate). Suppose f : R n → R is non-negative, C 2 , and |∂ 2 f| ≤ A (meaning x T (∂ 2 f)x ≤ A|x| 2 for any x ∈ R n ). Then |∇f| ≤ p 2Af. Proof. The proof is almost the same as the one variable case. The only difference is that we use the following Taylor’s expansion with integral remainder: 0 ≤ f(x + t) = f(x) + h∇f, ti + Z 1 0 (1 − s)h∂ 2 f(x + st)t, tids. Taking t = − 1 A∇f(x), we get |∇f| 2 ≤ Af + A 2 Z 1 0 (1 − s)ds · ht, ti = Af − |∇f| 2 /2, from which the conclusion follows. ¶ DETOUR: The critical symbol class S1/2(m). We also list some results for S1/2(m) that we need. Recall that a ∈ Sδ(m) if for any α, |∂ α a| ≤ Cα~ −δ|α|m. We need the following results for the critical symbol class S1/2(m): (I) For any symmetric non-singular 2n × 2n matrix Q, the operator e\i 2~ ξ T Qξ W maps S1/2(m) to S1/2(m). (But we will not have the asymptotic expansion as in the case of δ < 1 2 .) (II) If a ∈ S1/2(m1), b ∈ S1/2(m2), then a ? b ∈ S1/2(m1m2) and a ? b(x, ξ) = h e i~ 2 (Dξ·Dy−Dx·Dη) (a(x, ξ)b(y, η))i y=x,η=ξ (III) (Calderon-Vaillancourt for critical symbol) For a ∈ S1/2(1), ba W is bounded on L 2 (R n ) and kba W kL(L2(Rn)) ≤ C X |α|≤Mn ~ |α|/2 k∂ α akL∞(Rn) .

LECTURE15:L?-THEORYOFSEMICLASSICALPSDOS:POSITIVITY7We will only prove (I),since theproofs to (HI) and (III) will be similar to earlierproofs.Proof of (1).Were-scalebysettingi=h-1/2w.Then二「 e-Po-1(m)a(z+ w)dwe(TQsa(z) =hne-po-1(m)a(z+h1/2)da=C=C//e-po-1(m)a(z+/2w)x()di +C /e-pQ-1()a(z+h//2)(1-X1())dwhere Xi is a compactly supported cut-off function which equals 1 near the origin.For the first part. one hase-±PQ-1()a(z + h1/2)x1()du|≤Ch-lal/2m(2),a0Vpo-1-hDand for the second part, one use an integration by parts argument via L Vpo-12口as we did in proving non-stationary phase.Proof of SharpGarding inequality:continuedWe have mentioned that for [β| ≥ 2, we havePa≤CCjj13;1/2For B,/=1 we use the gradient estimate to getIVal≤CVa≤ca-Voand thus50%all≤cVCombining these two cases together, we conclude that for any α,[8]-/0/2a0llL≤CaNowwefor e(-1,-Ch),we write&=-h/h.So he (h,).Since≤we can rewrite the above inequality asIlL≤Cah.h-lal/2I1%i.e. hS1/2(1)

LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY 7 We will only prove (I), since the proofs to (II) and (III) will be similar to earlier proofs. Proof of (I). We re-scale by setting we = ~ −1/2w. Then e i 2~ ζ T Qζa(z) = C ~ n Z R2n e − i 2~ pQ−1 (w) a(z + w)dw = C Z R2n e − i 2 pQ−1 (we) a(z + ~ 1/2we)dwe = C Z R2n e − i 2 pQ−1 (we) a(z + ~ 1/2we)χ1( ˜w)dwe + C Z R2n e − i 2 pQ−1 (we) a(z + ~ 1/2we)(1 − χ1(we))dw, e where χ1 is a compactly supported cut-off function which equals 1 near the origin. For the first part, one has     ∂ α z Z R2n e − i 2 pQ−1 ( ˜w) a(z + ~ 1/2w˜)χ1( ˜w)dw˜     ≤ C~ −|α|/2m(z), and for the second part, one use an integration by parts argument via L = ∇pQ−1 ·~D |∇pQ−1 | 2 as we did in proving non-stationary phase. ¶ Proof of Sharp G˚arding inequality: continued. We have mentioned that for |βj | ≥ 2, we have k 1 a − δ ∂ βjakL∞ ≤ Cβj 1 |δ| ≤ C 1 |δ| |βj |/2 . For |βj | = 1 we use the gradient estimate to get |∇a| ≤ C √ a ≤ C a − δ p |δ| and thus k 1 a − δ ∂ βjakL∞ ≤ C 1 p |δ| . Combining these two cases together, we conclude that for any α, k∂ α 1 a − δ kL∞ ≤ Cα 1 a − δ |δ| −|α|/2 . Now we for δ ∈ (−1, −C~), we write δ = −~/~˜. So h˜ ∈ (~, 1 C ). Since 1 a−δ ≤ 1 −δ , we can rewrite the above inequality as k∂ α ~ a − δ kL∞ ≤ Cα~˜ · ~ −|α|/2 , i.e. ~ a−δ ∈ ~˜S1/2(1),

8LECTURE15:L?-THEORY OFSEMICLASSICALPSDOS:POSITIVITYSince a - E S(1) C S1/2(1) and - e hSi/2(1), the Moyal star product(a -) a-, is given by the above formula. To get an explicit expression, we noticethat for any one-variable smooth function f, one hasf(1) = f(0) + f'(0) + / (1 - s) f"(s)dsApply this toF(t) = [e (DeD,-D-D)(a(r, ) - ) aa(y,n)-and noticeh(De-Dy-DrDn)(a(r,s)-0)f'(0) =Ra三a(y,n)-we get[e学(De-Du-Da-Dn)F(De· Dy- Dar· Dn)(1-s)(a(r,E)b(y, n)ds(a-8)=1+a-d10r,n=Finally as in the proof of the weak Garding inequality, we denote1r(c,)=1-(a-)*a-sWe want to prove I/Wllc(L2(Rn) < uniformly for h <ho and d e (-e, -1)Observe that - hS1/2(1) implies h?aa(-) e hS/2 for any [Q| = 2. Thistogether with (I) implies r hSi/2(1). So by (111),I/W lc(L2(R") ≤Cih.So if we take C large enough, we can get Ci/C < 1/2. The conclusion follows

8 LECTURE 15: L 2 -THEORY OF SEMICLASSICAL PSDOS: POSITIVITY Since a − δ ∈ S(1) ⊂ S1/2(1) and ~ a−δ ∈ ~˜S1/2(1), the Moyal star product (a − δ) ? 1 a−δ is given by the above formula. To get an explicit expression, we notice that for any one-variable smooth function f, one has f(1) = f(0) + f 0 (0) + Z 1 0 (1 − s)f 00(s)ds Apply this to f(t) =  e it~ 2 (Dξ·Dy−Dx·Dη) ((a(x, ξ) − δ) 1 a(y, η) − δ )  y=x,η=ξ and notice f 0 (0) = i~ 2 (Dξ ·Dy−Dx·Dη)  (a(x, ξ) − δ) 1 a(y, η) − δ  y=x,η=ξ = i~ 2 {a−δ, 1 a − δ } = 0 we get (a−δ)? 1 a − δ = 1+Z 1 0 (1−s) " e is~ 2 (Dξ·Dy−Dx·Dη)  i~ 2 (Dξ · Dy − Dx · Dη) 2 (a(x, ξ)b(y, η))# y=x,η=ξ ds Finally as in the proof of the weak G˚arding inequality, we denote r(x, ξ) = 1 − (a − δ) ? 1 a − δ . We want to prove krb W kL(L2(Rn)) < 1 2 uniformly for ~ < ~0 and δ ∈ (−ε, −1). Observe that ~ a−δ ∈ ~˜S1/2(1) implies ~ 2∂ α ( 1 a−δ ) ∈ ~˜S1/2 for any |α| = 2. This together with (I) implies r ∈ ~˜S1/2(1). So by (III), kre W kL(L2(Rn)) ≤ C1~˜. So if we take C large enough, we can get C1/C < 1/2. The conclusion follows