LECTURE22-23:WEYL'SLAW1.FUNCTIONAL CALCULUS OF PSEUDODIFFERENTIAL OPERATORSInsections1and2wewillalwaysassume. m ≥ 1 is an order function,.pS(m)isa real-valued symbol.p+iis elliptic in S(m).Under these assumptions we know thatP =pW : Hn(m) C L?(R") -→ L?(Rn)is a (densely-defined) self-adjoint operator, and moreover, P±i-Id is invertible forh E (O,ho)and the inverse is apseudodifferential operator with symbol in S(1/m).Helffer-Sjostrand formula.InLecture8 wementioned thatforaself-adjoint operatorP onaHilbert spaceH and a Borel measurable function f on R, one can define a new self-adjoint linearoperator f(P)on H using the spectral theorem as follows:By spectral theorem(multiplication form) there is a measurable space (X,μ), a measurable real-valuedfunction h on X and a unitary isomorphism V : H → L?(X, μ) so thatVoPoV*= Mhon L?(X,μ). Then the operator f(P) is defined to bef(P) = V*o Mr(h(r) o V.We notice that by definition,(1)If IfI ≤C, then IIf(P)Ilc≤C.We want to answer the following natural question:Question:Is f(P)a semiclassical pseudodifferential operator if P isa semiclassical pseudodifferential operator and f is a (nice) function?Unfortunately the construction of f(P) above is a bit too abstract to work withHowever, if f E is a Schwartzfunction, then by using the so-called almost analyticertensionf off,Helffer-Sjostrand gaveamore concreteformula for f(P),namely(P) =-元 [5. ()(2- P)-L(da)(2)using which we will prove that f(P) is a semiclassical pseudodifferential operator,and calculate its symbol expansion. Recall that an almost analytic extension f E1
LECTURE 22-23: WEYL’S LAW 1. Functional calculus of pseudodifferential operators In sections 1 and 2 we will always assume • m ≥ 1 is an order function, • p ∈ S(m) is a real-valued symbol, • p + i is elliptic in S(m). Under these assumptions we know that P = pb W : H~(m) ⊂ L 2 (R n ) → L 2 (R n ) is a (densely-defined) self-adjoint operator, and moreover, P ± i · Id is invertible for ~ ∈ (0, ~0) and the inverse is a pseudodifferential operator with symbol in S(1/m) . ¶Helffer-Sj¨ostrand formula. In Lecture 8 we mentioned that for a self-adjoint operator P on a Hilbert space H and a Borel measurable function f on R, one can define a new self-adjoint linear operator f(P) on H using the spectral theorem as follows: By spectral theorem (multiplication form) there is a measurable space (X, µ), a measurable real-valued function h on X and a unitary isomorphism V : H → L 2 (X, µ) so that V ◦ P ◦ V ∗ = Mh on L 2 (X, µ). Then the operator f(P) is defined to be f(P) = V ∗ ◦ Mf(h(x)) ◦ V. We notice that by definition, (1) If |f| ≤ C, then kf(P)kL ≤ C. We want to answer the following natural question: Question: Is f(P) a semiclassical pseudodifferential operator if P is a semiclassical pseudodifferential operator and f is a (nice) function? Unfortunately the construction of f(P) above is a bit too abstract to work with. However, if f ∈ S is a Schwartz function, then by using the so-called almost analytic extension ˜f of f, Helffer-Sj¨ostrand gave a more concrete formula for f(P), namely (2) f(P) = − 1 π Z C ¯∂z ˜f(z)(z − P) −1L(dz) using which we will prove that f(P) is a semiclassical pseudodifferential operator, and calculate its symbol expansion. Recall that an almost analytic extension ˜f ∈ 1
2LECTURE22-23:WEYL'SLAWCo(C) of a Schwartz function f E (R) is by definition a smooth function on Csuch thatFlr= f, suppf c (z: [Im(z)|≤1)and such that as |Im(z)l→ 0,0,f(2) = O(Im(z)/~),where as usual, = (Or + iou)/2 for z = r + iy, and L(dz) denotes the Lebesguemeasure on C (wedon't use drdy since rhas different meaningbelow.)To proveformula (2) it is enough to notice the following relation between f and f (c.f.PSet1):f(t) = -元 [ 5. (2)(2 -t)-(d)爪Jand use the spectral theorem (multiplication form) for both (z -P)-1 and f(P).In literature there areat least two different ways to construct an almost analyticextension: thefirst construction is due to Hormander (who proposed the conceptionofalmostanalyticextensionin1968)whoadaptedtheconstructioninBorel'sLemmabyputting(+ i) -一((i) (A),k!where A is a sequence of real numbers that is chosen so that they tends to +oosufficiently fast, and x is a cut-off function. The other way is due to Mather whomake use of the Fourier transform (PSet 1):/ x(yE)f(E)eis(r+i) dE. (+ i) =2x() /Of course the almost analytic extension is not unique in general. Also note thatif f is compactly supported on R,then we can take f to be compactly-supportedin C[For thefirst construction this is obvious, for the second construction wemaymultiply theformula bya cut-off function which is identically one on supp(f))Symboloftheresolvent(z-P)-1.We want to prove that f(P) is a pseudodifferential operator if f is Schwartzfunction on R and P is a semiclassical pseudodifferential operator, and calculate itssymbol.In view of the Helffer-Sjostrant formula (2),we start with the resolventoperator (z-P)-1.Lemma 1.1. Under the previous assumptions, for any zE C R, the operator z-Pisinvertible, andthereeristsr.ES(1/m)suchthatrw= (z-P)-1Proof. This is just a consequence of ellipticity: for any fixed z = a +bi (b o), wehaveinfle-t> 0.tERIt+i
2 LECTURE 22-23: WEYL’S LAW C ∞(C) of a Schwartz function f ∈ S (R) is by definition a smooth function on C such that ˜f|R = f, supp ˜f ⊂ {z : |Im(z)| ≤ 1} and such that as |Im(z)| → 0, ¯∂z ˜f(z) = O(|Im(z)| ∞), where as usual, ¯∂z = (∂x + i∂y)/2 for z = x + iy, and L(dz) denotes the Lebesgue measure on C (we don’t use dxdy since x has different meaning below.) To prove formula (2) it is enough to notice the following relation between f and ˜f (c.f. PSet 1): f(t) = − 1 π Z C ¯∂z ˜f(z)(z − t) −1L(dz) and use the spectral theorem (multiplication form) for both (z − P) −1 and f(P). In literature there are at least two different ways to construct an almost analytic extension: the first construction is due to H¨ormander (who proposed the conception of almost analytic extension in 1968) who adapted the construction in Borel’s Lemma by putting ˜f(x + iy) = X k f (k) (x) k! (iy) kχ(λky), where λk is a sequence of real numbers that is chosen so that they tends to +∞ sufficiently fast, and χ is a cut-off function. The other way is due to Mather who make use of the Fourier transform (PSet 1): ˜f(x + iy) := 1 2π χ(y) Z R χ(yξ) ˆf(ξ)e iξ(x+iy) dξ. Of course the almost analytic extension is not unique in general. Also note that if f is compactly supported on R, then we can take ˜f to be compactly-supported in C [For the first construction this is obvious, for the second construction we may multiply the formula by a cut-off function which is identically one on supp(f)]. ¶Symbol of the resolvent (z − P) −1 . We want to prove that f(P) is a pseudodifferential operator if f is Schwartz function on R and P is a semiclassical pseudodifferential operator, and calculate its symbol. In view of the Helffer-Sj¨ostrant formula (2), we start with the resolvent operator (z − P) −1 . Lemma 1.1. Under the previous assumptions, for any z ∈ C\R, the operator z−P is invertible, and there exists rz ∈ S(1/m) such that rbz W = (z − P) −1 . Proof. This is just a consequence of ellipticity: for any fixed z = a + bi (b 6= 0), we have inf t∈R |z − t| |t + i| > 0
3LECTURE22-23:WEYL'SLAWWe need a more explicit lower bound below.So let's try tofind aconstantC>0 suchthat[2 - t]≥bC,(3)VtER.[t+iThis is equivalent to(a-t)?+b2-(bC)(t?+1)≥0,VtER.We will take C small enough so that bC < 1.By calculating thediscriminant and simplifying it, we get the condition on C:1-C2(1 - C2b2) ≥ a2.C2As a consequence, if we assume |zl < Co, then we can find a constantC so that (3) holds for all t E R.It follows that[z-pl≥Clp+i≥cm口and thus z- p is elliptic in S(m).To apply Helffer-Sjostrand formula, we need to study the dependence of rz onz.For this purpose, we introduce the following variation of Beals's theorem.Recall that in Lecture 14 we have seen that any continuous linearoperatorA:y→gcanbewrittenasA=awforsomea:a(r,s,h) e i, and Beals's theorem tells us a e S(1) if and only ifIladaw o o adw Allc(L2) =O(N)for any N and any linear functions li,..:,l on IR2n. Of course themainpart in the proof of Beals'stheorem isto provethe conditionabove implies a E S(1), namely, to prove oa.al≤ Co.Proposition 1.2 (Beals's Estimate with Parameter). Suppose a =a(r,E,z;h), i.e. a depends on a parameter z. Let s = S(z) be afunction valued in (o,1] such thatIladgw o ... 0 adwAllc(L2) = O(8-N)holds for any N and any linear functions li,..:,ln on R2n. Thenthere erists a universal constant M such that for any Q,[oe,ea(r,5,z; h)/ ≤ Camax(1, Vh/8)Ms-lal.We will leave the proof as an exercise. [c.f. Helffer-Sjostrand,Spectral Asymptotics in the Semi-Classical Limit, Prop. 8.4.]Using this result, we will prove the following resolvent symbol estimate:
LECTURE 22-23: WEYL’S LAW 3 We need a more explicit lower bound below. So let’s try to find a constant C > 0 such that (3) |z − t| |t + i| ≥ bC, ∀t ∈ R. This is equivalent to (a − t) 2 + b 2 − (bC) 2 (t 2 + 1) ≥ 0, ∀t ∈ R. We will take C small enough so that bC < 1. By calculating the discriminant and simplifying it, we get the condition on C: 1 − C 2 C2 (1 − C 2 b 2 ) ≥ a 2 . As a consequence, if we assume |z| < C0, then we can find a constant C so that (3) holds for all t ∈ R. It follows that |z − p| ≥ C|p + i| ≥ cm and thus z − p is elliptic in S(m). To apply Helffer-Sj¨ostrand formula, we need to study the dependence of rz on z. For this purpose, we introduce the following variation of Beals’s theorem. Recall that in Lecture 14 we have seen that any continuous linear operator A : S → S 0 can be written as A = ba W for some a = a(x, ξ, ~) ∈ S 0 , and Beals’s theorem tells us a ∈ S(1) if and only if kadlb1 W ◦ · · · ◦ ad lcN W AkL(L2) = O(~ N ) for any N and any linear functions l1, · · · , lN on R 2n . Of course the main part in the proof of Beals’s theorem is to prove the condition above implies a ∈ S(1), namely, to prove |∂ α x,ξa| ≤ Cα. Proposition 1.2 (Beals’s Estimate with Parameter). Suppose a = a(x, ξ, z; ~), i.e. a depends on a parameter z. Let δ = δ(z) be a function valued in (0, 1] such that kadlb1 W ◦ · · · ◦ ad lcN W AkL(L2) = O(δ −N ~ N ) holds for any N and any linear functions l1, · · · , lN on R 2n . Then there exists a universal constant M such that for any α, |∂ α x,ξa(x, ξ, z; ~)| ≤ Cα max(1, √ ~/δ) Mδ −|α| . We will leave the proof as an exercise. [c.f. Helffer-Sj¨ostrand, Spectral Asymptotics in the Semi-Classical Limit, Prop. 8.4.] Using this result, we will prove the following resolvent symbol estimate:
4LECTURE22-23:WEYL'SLAWTheorem 1.3. Fir Co > 0. For any z E CR with |zl <Co, we have[0,er-/ ≤ Ca max(1, /2|m(2)]-1)[Im(2)]-1-lal,where r E S(1/m) is the symbol of the resolvent of P, namely (z - P)-1 = rWProof.ByProposition1.2,we only need toprove(4)ladew o... 0 adw(z -P)-illc(L2) = O(Im(z)-1-NN)Using the formulaeada(B-1) = -B-1(adAB)B-1andadA(BC) = (ad^B)C + Bad^Cwe can write ad.ww o ... o adw(z - P)-1 as a summation of terms of the form±(z- P)-lada(P)(z-P)-lad(P) ... (z- P)-lad(P)(z-P)-1,where aj = [aj1,*-,ain,} such that (aj, / Vj, ] = [1,.*, N]. Note thatpE S(m) = [l,p) E S(m) = ad(P)=pW for some p, EniS(m),Thus in view of the fact P +i has symbol in S(1/m), we get thatIlad(P)(z - P)-illc(L2) ≤ Iad(P)(P + i)-Ilc(L2) Il(P +i)(z - P)-illc(L2)≤O(ni[Im(z)/-1),where in the last step we used Calderon-Vaillancourt theorem for the first term, andII(P+i)(z - P)-1llc(L2(Rn) ≤C|Im(z)-1for the second term, which is a consequence of.The fact () at the beginning of this lecture, which is a consequence of thespectraltheorem,.The argument in theproof of Lemma 1.1,i.e.if weassume[zl< Co,thenthere exists a universal constant C that is independent of z such that (3)holds. In other words长二≤CIm(2)]-1, Vte R.[t+iSimilarly we haveI(z - P)-1llc(L2(R") ≤ [Im(z)]-1,口so the estimate (4) holds,which completes the proof
4 LECTURE 22-23: WEYL’S LAW Theorem 1.3. Fix C0 > 0. For any z ∈ C \ R with |z| < C0, we have |∂ α x,ξrz| ≤ Cα max(1, ~ 1/2 |Im(z)| −1 ) M|Im(z)| −1−|α| . where rz ∈ S(1/m) is the symbol of the resolvent of P, namely (z − P) −1 = rbz W . Proof. By Proposition 1.2, we only need to prove (4) kadlb1 W ◦ · · · ◦ ad lcN W (z − P) −1 kL(L2) = O(|Im(z)| −1−N ~ N ). Using the formulae adA(B −1 ) = −B −1 (adAB)B −1 and adA(BC) = (adAB)C + BadAC we can write adlb1 W ◦ · · · ◦ ad lcN W (z − P) −1 as a summation of terms of the form ±(z − P) −1 adα1 blW (P)(z − P) −1 adα2 blW (P)· · ·(z − P) −1 adαk blW (P)(z − P) −1 , where αj = {αj,1, · · · , αj,nj } such that {αj,l | ∀j, l} = {1, · · · , N}. Note that p ∈ S(m) =⇒ {l, p} ∈ S(m) =⇒ adαj blW (P) = pbj W for some pj ∈ ~ njS(m). Thus in view of the fact P + i has symbol in S(1/m), we get that kadαj blW (P)(z − P) −1 kL(L2) ≤ kadαj blW (P)(P + i) −1 kL(L2) · k(P + i)(z − P) −1 kL(L2) ≤ O(~ nj |Im(z)| −1 ), where in the last step we used Calderon-Vaillancourt theorem for the first term, and k(P + i)(z − P) −1 kL(L2(Rn)) ≤ C|Im(z)| −1 for the second term, which is a consequence of • The fact (1) at the beginning of this lecture, which is a consequence of the spectral theorem, • The argument in the proof of Lemma 1.1, i.e. if we assume |z| < C0, then there exists a universal constant C that is independent of z such that (3) holds. In other words, |z − t| |t + i| ≤ C|Im(z)| −1 , ∀t ∈ R. Similarly we have k(z − P) −1 kL(L2(Rn)) ≤ |Im(z)| −1 , so the estimate (4) holds, which completes the proof.
LECTURE22-23:WEYL'SLAW5The functional calculus.Now we are ready to prove that the operator f(P)is also a semiclassical pseu-dodifferential operator:Theorem 1.4. If f e J, then f(P) = aw, where a E S(m-k) for any k E N.Moreover,wehave anasymptoticerpansiona(r,) ~hax(r,E),k≥0where ao(r, $)= f(p(r,), and in general,as(r,s) =(2)()*[()(,,)()Proof. Using Helffer-Sjostrand formula, we see f(P) = aW, where[ (0f(2)r2(r,3)L(dz),a(r,s) =-1where is an almost analytic extension of f.Although the (r,)-derivatives of rzis unbounded as Im(z) → 0, the unboundedness is controlled by Theorem 1.3. Soby using the fact ,f(z) = O(|Im(z)[) we see a E S(1). More generally, for anykeN, ifwe applythe above arguments to f(t)=f(t)(t+i),we can provethatfu(P) = (P + i)*f(P) is a pseudodifferential operator with symbol in S(1), whichimplies a e S(m-k).We alsoneed an asymptotic expansionof a.For thispurposewe startwith theasymptotic expansion of rz.Recall that by construction (Lecture14), rz=r*(1-u)for some u E hS(1), and r can be solved form the equation(z-p)*r,- 1=0(h)inductively,which hastheform (exercise)qk(r,s,2)~k(5)(2 - p(r, $)2k+1k=0where qk is a degree 2k polynomial in z (and thus is holomorphic in z):2kqk(r,E,z) =dk,3(r, )2ij=0with qo = 1, q1 = 0 and in general, qkj E S(m2k-j)Again the expansion (5) is an expansion for each fixed z, and is not a goodone as Im(z) → 0. However, we may resolve this problem by fixing a E (0, 1/2
LECTURE 22-23: WEYL’S LAW 5 ¶The functional calculus. Now we are ready to prove that the operator f(P) is also a semiclassical pseudodifferential operator: Theorem 1.4. If f ∈ S , then f(P) = ba W , where a ∈ S(m−k ) for any k ∈ N. Moreover, we have an asymptotic expansion a(x, ξ) ∼ X k≥0 ~ k ak(x, ξ), where a0(x, ξ) = f(p(x, ξ), and in general, ak(x, ξ) = 1 (2k)! (∂t) 2k [f(t)qk(x, ξ, t)] t=p(x,ξ) . Proof. Using Helffer-Sj¨ostrand formula, we see f(P) = ba W , where a(x, ξ) = − 1 π Z C ( ¯∂z ˜f(z))rz(x, ξ)L(dz), where ˜f is an almost analytic extension of f. Although the (x, ξ)-derivatives of rz is unbounded as Im(z) → 0, the unboundedness is controlled by Theorem 1.3. So by using the fact ¯∂z ˜f(z) = O(|Im(z)| ∞) we see a ∈ S(1). More generally, for any k ∈ N, if we apply the above arguments to fk(t) = f(t)(t + i) k , we can prove that fk(P) = (P + i) k f(P) is a pseudodifferential operator with symbol in S(1), which implies a ∈ S(m−k ). We also need an asymptotic expansion of a. For this purpose we start with the asymptotic expansion of rz. Recall that by construction (Lecture 14), rz = rez ?(1−u) for some u ∈ ~ ∞S(1), and rez can be solved form the equation (z − p) ? rez − 1 = O(~ ∞) inductively, which has the form (exercise) (5) rz ∼ X∞ k=0 ~ k qk(x, ξ, z) (z − p(x, ξ))2k+1 where qk is a degree 2k polynomial in z (and thus is holomorphic in z): qk(x, ξ, z) = X 2k j=0 qk,j (x, ξ)z j with q0 = 1, q1 = 0 and in general, qk,j ∈ S(m2k−j ). Again the expansion (5) is an expansion for each fixed z, and is not a good one as Im(z) → 0. However, we may resolve this problem by fixing a δ ∈ (0, 1/2)