LECTURE 10—10/26/2020QUANTIZINGGENERALSYMBOLS1. SEMI-CLASSICAL SYMBOLSToday we would like to extend the theory of semiclassical pseudo-differentialoperators with Schwartz functions as symbols to more general symbols which maydepends on h and more importantly,which may grow as r,→ oo.The key obser-vation is the following: According to the factet(a=)s = (1 + [51)-[1 + (hD,)1(e(r-w)-5)(1)= (1 + [ - y12)-1[1 + (hDe)](et(r-)-5),we may gain factors ~ [sl-m or ~ [r -yl-m after (formally) applying integrationby parts arguments many times. As a result, we may allow our symbol a(r, ) togrow in r or S, as long as thegrowth rateis under control, namely,in an“at mostpolynomially"way. Order functions.Now we introduce a conception that can be used to describe such a polynomialgrowth.Forsimplicitywewillusethenotation1(2) := (1 + [2/2)1/2for zE Rd.Definition 1.1. A measurable function g : Rd → R+ is called an order function ifthere exists constants C, N so that for any z, w e Rd,(2)m(w) ≤C(z- w)Nm(z)Note that if we take z= O, we immediately getm(w)≤C(w)Mfor some constant C and N. So order functions grow at most polynomiallyFrom the definition one immediately getLemma 1.2. We have(1) If mi,m2 are order functions, so are the functions mi +m2 and mim2(2) If m is an order function, so is the function m for any a E R.Sometimes it is called the "Japanese bracket" of z. It behave like z| for large z, but it has theadvantage that it is smooth and non-vanishing as z→ 0.1
LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 1. Semi-classical symbols Today we would like to extend the theory of semiclassical pseudo-differential operators with Schwartz functions as symbols to more general symbols which may depends on ~ and more importantly, which may grow as x, ξ → ∞. The key observation is the following: According to the fact (1) e i ~ (x−y)·ξ = (1 + |ξ| 2 ) −1 [1 + (~Dy) 2 ](e i ~ (x−y)·ξ ) = (1 + |x − y| 2 ) −1 [1 + (~Dξ) 2 ](e i ~ (x−y)·ξ ), we may gain factors ≈ |ξ| −m or ≈ |x − y| −m after (formally) applying integration by parts arguments many times. As a result, we may allow our symbol a(x, ξ) to grow in x or ξ, as long as the growth rate is under control, namely, in an “at most polynomially” way. ¶ Order functions. Now we introduce a conception that can be used to describe such a polynomial growth. For simplicity we will use the notation 1 hzi := (1 + |z| 2 ) 1/2 for z ∈ R d . Definition 1.1. A measurable function g : R d → R + is called an order function if there exists constants C, N so that for any z, w ∈ R d , (2) m(w) ≤ Chz − wi Nm(z). Note that if we take z = 0, we immediately get m(w) ≤ Chwi N for some constant C and N. So order functions grow at most polynomially. From the definition one immediately get Lemma 1.2. We have (1) If m1, m2 are order functions, so are the functions m1 + m2 and m1m2. (2) If m is an order function, so is the function ma for any a ∈ R. 1Sometimes it is called the “Japanese bracket” of z. It behave like |z| for large z, but it has the advantage that it is smooth and non-vanishing as z → 0. 1
2LECTURE10-10/26/2020 QUANTIZINGGENERALSYMBOLSSome simple examples of order functions:.m(z) = 1.. m(2) = (2).. m(z)=(z),where z =(z1,.,z) for somel ≤ d.. In particular, if d = 2n and a, b e R, then the functionsm(c,s) = (r)(s)bandm(r,s)=(r)°+(s)are order functions.Remark. In microlocal analysis, usually we take d = 2n and thus Rd - Rn x Rn. inwhich case the most widely used order function is (s)N, which, as we will see later,has the advantage that the corresponding symbol class (see definition below) isinvariant under coordinate changes and thus can be defined on manifolds.However,we do use other order functions in semiclassical analysis. The need for h-dependent symbols.We have seen last time that the Moyal product of two h-independent symbolswill be h-dependent in general.Here is another example showing the necessity forintroducingh-dependentsymbols:Erample.Theusual (non-semiclassical)differential operatorP=aDalal O such that for all z E IRd(3)[8°a(z)/ ≤Cam(2)(2) We say a = a(z, h) is in the symbol class S(m) if there exists ho > 0 so thatfor o < h < ho, the function a = a(,h) is in S(m) and the constant Cα in(3) is uniform for all 0 < h < ho
2 LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS Some simple examples of order functions: • m(z) = 1. • m(z) = hzi. • m(z) = hz 0 i, where z 0 = (z1, · · · , zl) for some l ≤ d. • In particular, if d = 2n and a, b ∈ R, then the functions m(x, ξ) = hxi a hξi b and ˜m(x, ξ) = hxi a + hξi b are order functions. Remark. In microlocal analysis, usually we take d = 2n and thus R d z = R n x × R n ξ , in which case the most widely used order function is hξi N , which, as we will see later, has the advantage that the corresponding symbol class (see definition below) is invariant under coordinate changes and thus can be defined on manifolds. However, we do use other order functions in semiclassical analysis. ¶ The need for ~-dependent symbols. We have seen last time that the Moyal product of two ~-independent symbols will be ~-dependent in general. Here is another example showing the necessity for introducing ~-dependent symbols: Example. The usual (non-semiclassical) differential operator P = ❳ |α|≤m aαD α can be written as P = ~ −m ❳m j=0 ~ j ❳ |α|=m−j aα(~D) α , and thus is a semiclassical pseudo-differential operator with Kohn-Nirenberg symbol p(x, ξ, ~) = ~ −m ❳m j=0 ~ j ❳ |α|=m−j aα(x)ξ α . (If we don’t allow ~-dependence, we will not have these operators in our class.) ¶ Symbol classes. Now we are ready to define symbol classes associated to an order function. In what follows we will assume ~-dependence for the symbol function a. Definition 1.3. Let m be an order function on R d . (1) We say a ∈ C ∞(R d ) is in the symbol class S(m) associated to m if for any multi-index α ∈ N d , there exists constant Cα > 0 such that for all z ∈ R d , (3) |∂ α a(z)| ≤ Cαm(z). (2) We say a = a(z, ~) is in the symbol class S(m) if there exists ~0 > 0 so that for 0 < ~ < ~0, the function a = a(·, ~) is in S(m) and the constant Cα in (3) is uniform for all 0 < ~ < ~0
3LECTURE10—10/26/2020QUANTIZINGGENERALSYMBOLS(3) For any k E R and 0 ≤8 ≤1 we let S(m) be the space of functions a(z, h)whichbelongstoS(m)foreachhandsatisfies(4)[0°a(z, h)|≤ Ca-5lal-km(2),where the constants Cais again uniform in h.For example, the class S(1) contains all uniformly bounded smooth functionswith all derivatives uniformly bounded over R2n (and with respect to h e (O,hol)Wewill abbreviateSs(m)=Sg(m), S=S(1),and Ss=Ss(1),Remark. Obviously C S(m) holds for any order function m. Moreover,(1) As in the case of Schwartz functions, one could define semi-norms on S(m)to make it into a Frechet space. Under this topology, the multiplication map.: S(m1) × S(m2)→ S(m1m2),(a1,a2)→a1a2is continuous. Similar result holds for the spaces S(m)(2) Moreover, the inclusion c S(m) is dense in the topology of S((z)em) forany > 0. To see this, for each a E S(m) one just takea;(z) = x)a(zfor some x E with x(0) = 1.Note that in the case =O, we have So(m)=S(m).Wewill omit the subscriptS in S(m) if = 0. So for each k e R we haveS-*(m) = h*S(m).Obviously if ki < k2, then S-ki(m) S-k2(m). In what follows we will denotes-(m) =ns-k(m).Note that a symbol a e S-o(m) if and only if for any multi-index α e Nd and anyN e N, there exists constant Ca,N so that[0a(z, h)/ ≤ Ca,NhNm(z)In particular, a e Sk for any & and any k. It follows that for any d,(5)S-(m) =nS=*(m)
LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 3 (3) For any k ∈ R and 0 ≤ δ ≤ 1 we let S k δ (m) be the space of functions a(z, ~) which belongs to S(m) for each ~ and satisfies (4) |∂ α a(z, ~)| ≤ Cα~ −δ|α|−km(z), where the constants Cα is again uniform in ~. For example, the class S(1) contains all uniformly bounded smooth functions with all derivatives uniformly bounded over R 2n (and with respect to ~ ∈ (0, ~0]). We will abbreviate Sδ(m) = S 0 δ (m), S = S(1), and Sδ = Sδ(1), Remark. Obviously S ⊂ S(m) holds for any order function m. Moreover, (1) As in the case of Schwartz functions, one could define semi-norms on S(m) to make it into a Fr´echet space. Under this topology, the multiplication map • : S(m1) × S(m2) → S(m1m2), (a1, a2) → a1a2 is continuous. Similar result holds for the spaces S k δ (m). (2) Moreover, the inclusion S ⊂ S(m) is dense in the topology of S(hzi εm) for any ε > 0. To see this, for each a ∈ S(m) one just take aj (z) = χ( z j )a(z) for some χ ∈ S with χ(0) = 1. Note that in the case δ = 0, we have S0(m) = S(m). We will omit the subscript δ in S k δ (m) if δ = 0. So for each k ∈ R we have S −k (m) = ~ kS(m). Obviously if k1 < k2, then S −k1 (m) ⊃ S −k2 (m). In what follows we will denote S −∞(m) = ❭ k S −k (m). Note that a symbol a ∈ S −∞(m) if and only if for any multi-index α ∈ N d and any N ∈ N, there exists constant Cα,N so that |∂ α a(z, ~)| ≤ cα,N ~ Nm(z). In particular, a ∈ S k δ for any δ and any k. It follows that for any δ, (5) S −∞(m) = ❭ k S −k δ (m)
4LECTURE 10-10/26/2020 QUANTIZING GENERAL SYMBOLS2.ASYMPTOTICANALYSISIITAsymptoticseriesforgeneralsymbols.Now we extend the conception of asymptotic series that we introduced at thebeginning of Lecture 5 to asymptotic series of functions that lie in a“decreasingsequence of symbol classes associated to an order function m".Letko<ki<k2<...<kj<...→00be a sequence of increasing real numbers that tends to co.Definition 2.1. If aj E S-k (m), we say a E S-ko(m) is asymptotic to Zj=o aj, andwrite8a~aj,j=0if for every N e N,N(6)1(m)ZajEsSQj=0Note that the similar conceptions introduced in Lecture 5 is a special case withm =1 and k,=j.Again we don't require the formal series Ea, to be convergentfor any h.Remark. Obviously a ~ 0 if and only if a E S-o(m)T Borel's Lemma.The following theorem, named after E. Borel, is crucial in asymptotic analysis,which implies that every power series is the Taylor series of some smooth function:Theorem 2.2 (Borel Lemma). For any sequence aj E S-i(m), where k, is a strictlyincreasing sequence that tends to infinity, there eists a symbol a e S-ko(m) so thatj=0Moreover, a is unique up to an element in S-oo(m).Proof.UniquenessisobviousTo prove the existence, we choose a smooth cut-off function x so that.0≤x≤1on R,: X = 1 on [0, 1],. x = 0 on [2, 00).Define(7)a(r) =x(,h)a;(c),5
4 LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 2. Asymptotic analysis II ¶ Asymptotic series for general symbols. Now we extend the conception of asymptotic series that we introduced at the beginning of Lecture 5 to asymptotic series of functions that lie in a “decreasing sequence of symbol classes associated to an order function m”. Let k0 < k1 < k2 < · · · < kj < · · · → ∞ be a sequence of increasing real numbers that tends to ∞. Definition 2.1. If aj ∈ S −kj δ (m), we say a ∈ S −k0 δ (m) is asymptotic to P∞ j=0 aj , and write a ∼ ❳∞ j=0 aj , if for every N ∈ N, (6) a − ❳ N j=0 aj ∈ S −kN+1 δ (m). Note that the similar conceptions introduced in Lecture 5 is a special case with m = 1 and kj = j. Again we don’t require the formal series P aj to be convergent for any ~. Remark. Obviously a ∼ 0 if and only if a ∈ S −∞(m). ¶ Borel’s Lemma. The following theorem, named after E. Borel, is crucial in asymptotic analysis, which implies that every power series is the Taylor series of some smooth function: Theorem 2.2 (Borel Lemma). For any sequence aj ∈ S −kj δ (m), where kj is a strictly increasing sequence that tends to infinity, there exists a symbol a ∈ S −k0 δ (m) so that a ∼ ❳∞ j=0 aj . Moreover, a is unique up to an element in S −∞(m). Proof. Uniqueness is obvious. To prove the existence, we choose a smooth cut-off function χ so that • 0 ≤ χ ≤ 1 on R, • χ ≡ 1 on [0, 1], • χ ≡ 0 on [2,∞). Define (7) a(x) = ❳ j χ(λj~)aj (x)
LECTURE10—10/26/2020QUANTIZING GENERAL SYMBOLS5where A, is a sequence to be determined below.We will choose A so that A→ ooas j→ o, with some additional restrictions.Note the fact A, → oo implies thatfor each h the sum (7) is a finite sum. In particular, a is smooth for each h.Additional restriction on 入,: For each multi-index α with Jal ≤ j, we havej-18a]x(,h)[0°ajl ≤Cj,ah5-8lalx(,h)m≤2k,-kj-1Ckj-kj-1mwhere we used the fact x(A,h)(,h)k-kj-1 ≤ 2ks-kj-1. We will choose >, largeenough so that , > >j-1 and so that for all [al ≤ j,2kj-kj-1Cj.a,h)0ajlj≤lalj>lal≤ZCj,a*j-0lalm+hhj-1-5jal2-3mj≤lalj>lal≤ Cahho-Jlalm,where we took Ca = Zj,h) + 10°ajl + Z[0°a;l(1 -x(A,h))j=0j=Nj=0j=N+/αl+1N+lalN-18Itkj-1-[al2-im +ZCi,ak;-8lalm+ Cj,ahk-5lal(1-x(A,h)m<j=Nj=0j=N+/αl+1N+lalN-1≤hn-8lalm +Ciahkn-3lalm+Cj,ahkj-5lal(Anh)kn-kimi=Nj=(≤ Ca,NtkN-0Jalmfor the constantN+lalN-1kN-kECjaACa.N=1+ Cjiα+j=Ni=0where inthethird lineweused thefactthatfor≤j≤N-l:. if h入w <1, then 1 - x(^,h) = 0;. if 入 ≥1, then 1-x(h)≤1≤(入)-k)
LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 5 where λj is a sequence to be determined below. We will choose λj so that λj → ∞ as j → ∞, with some additional restrictions. Note the fact λj → ∞ implies that for each ~ the sum (7) is a finite sum. In particular, a is smooth for each ~. Additional restriction on λj : For each multi-index α with |α| ≤ j, we have χ(λj~)|∂ α aj | ≤ Cj,α~ kj−δ|α|χ(λj~)m ≤ 2 kj−kj−1Cj,α ~ kj−1−δ|α| λ kj−kj−1 j m where we used the fact χ(λj~)(λj~) kj−kj−1 ≤ 2 kj−kj−1 . We will choose λj large enough so that λj > λj−1 and so that for all |α| ≤ j, 2 kj−kj−1Cj,α λ kj−kj−1 j ≤ 2 −j . It follows that for j ≥ |α|, (8) χ(λj~)|∂ α aj | ≤ ~ kj−1−δ|α| 2 −jm. It remains to prove a ∈ S −k0 δ (m) and a ∼ P aj . Fix any α, we have |∂ α a| ≤ ❳ j≤|α| |∂ α aj | + ❳ j>|α| |χ(λj~)∂ α aj | ≤ ❳ j≤|α| Cj,α~ kj−δ|α|m + ❳ j>|α| ~ kj−1−δ|α| 2 −jm ≤ Cα~ k0−δ|α|m, where we took Cα = P j≤|α| Cj,α + 1. This proves a ∈ S −k0 δ (m). To prove a ∼ P aj , we do similar calculations: For any α and any N, we have ☞ ☞ ☞ ☞ ☞ ☞ ∂ α (a − N ❳−1 j=0 aj ) ☞ ☞ ☞ ☞ ☞ ☞ ≤ ❳∞ j=N+|α|+1 |∂ α aj |χ(λj~) + N ❳ +|α| j=N |∂ α aj | + N ❳−1 j=0 |∂ α aj |(1 − χ(λj~)) ≤ ❳∞ j=N+|α|+1 ~ kj−1−δ|α| 2 −jm + N ❳ +|α| j=N Cj,α~ kj−δ|α|m + N ❳−1 j=0 Cj,α~ kj−δ|α| (1−χ(λj~))m ≤~ kN −δ|α|m + N ❳ +|α| j=N Cj,α~ kN −δ|α|m + N ❳−1 j=0 Cj,α~ kj−δ|α| (λN ~) kN −kjm ≤ Cα,N ~ kN −δ|α|m for the constant Cα,N = 1 + N ❳ +|α| j=N Cj,α + N ❳−1 j=0 Cj,αλ kN −kj N , where in the third line we used the fact that for 0 ≤ j ≤ N − 1: • if ~λN ≤ 1, then 1 − χ(λj~) = 0; • if ~λN ≥ 1, then 1 − χ(λj~) ≤ 1 ≤ (~λN ) kN −kj
6LECTURE10-10/26/2020QUANTIZINGGENERALSYMBOLS口This completestheproof.Remark. In many applications one take kj = j, in which case the theorem says thatfor any sequence a, E Ss(m), up to S-o(m) there exists a unique asymptotic sum-0in Ss(m).An Application:WKB solutionConsider the1-dimensional semiclassical Schrodingerequationh? d?2 dre + V(r)u = Eu.In Lecture 3 we have seen how to solve this equation for V(r) = r?/2, in which casea non-trivial solution exists if and only if E= (2n +1)h, and the solutions u areclosely related to the Hermite polynomials.In general, one can't hope to find such an exact solution.However, it is stillpossible to find an approximate solution, the so-called WKB solution2, of the formu() ~ e-i0(a)/hEt'ax(n)K≥0with ao 0, which solve the equation asymptotically:h?d?2 da2 +V(a)-E)u~oFor simplicity we will only find an approximate solution inside the region[r: V(r)<E),which isusually called the classical allowed region).Sowesupposeis afunctiondefined on aninterval Iinsidethe classical allowed region.Tofind thefunction andak's,wefirstcomputeh?de(0)a+ihs'a-a(e-;()/a) =e-0/ [i0%gio"a+22WKB is the abbreviation of three mathematicians: Wentzel, Kramers and Brillouin
6 LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS This completes the proof. Remark. In many applications one take kj = j, in which case the theorem says that for any sequence aj ∈ Sδ(m), up to S −∞(m) there exists a unique asymptotic sum a ∼ ❳∞ j=0 ~ j aj in Sδ(m). ¶ An Application: WKB solution. Consider the 1-dimensional semiclassical Schr¨odinger equation (− ~ 2 2 d 2 dx2 + V (x))u = Eu. In Lecture 3 we have seen how to solve this equation for V (x) = x 2/2, in which case a non-trivial solution exists if and only if E = (2n + 1)~, and the solutions u are closely related to the Hermite polynomials. In general, one can’t hope to find such an exact solution. However, it is still possible to find an approximate solution, the so-called WKB solution2 , of the form u(x) ∼ e −iφ(x)/~ ❳ k≥0 ~ k ak(x) with a0 6= 0, which solve the equation asymptotically: ❶ − ~ 2 2 d 2 dx2 + V (x) − E ➀ u ∼ 0. For simplicity we will only find an approximate solution inside the region {x : V (x) < E}, which is usually called the classical allowed region). So we suppose φ is a function defined on an interval I inside the classical allowed region. To find the function φ and ak’s, we first compute − ~ 2 2 d 2 dx2 (e −iφ(x)/~ a) = e −iφ/~ ➊ ~ 2 iφ00a + 1 2 (φ 0 ) 2 a + i~φ 0 a 0 − ~ 2 2 a 00➍ . 2WKB is the abbreviation of three mathematicians: Wentzel, Kramers and Brillouin.
LECTURE10-10/26/2020QUANTIZINGGENERALSYMBOLS7So after pluging in the asymptotic expansion of u into the equation, we getde-i(r)/hak(r)2 dr2 + V(a) - E) (k≥0e-i0(r)/h($)ao + V(r)ao - Eao"'ao + id'do + ((()? +V- E)ai)#L+又2+V-E)ak"ak-1+is'dk-1-0%-2+oSince ao 0, the leading term gives us (known as the eikonal equation)()?+V(a) - E= 0.(9)As a consequence, we get two different solutions of :Φ+(r) =± /VE-V(r)da.Note that the eikonal equation also simplifies the remaining equations (known asthe transport equation that determines a's:(10)2'a +id =0which determines ao (which depends on the choice of o+),and(11)20"ax-1 + id-1 --2°%-2=0for k > 2, which determines the remaining ak's (again depends on the choice of @+)Finally as a consequence of the Borel theorem, we get two asymptotic solutions:u+(a) ~ e-神+(a)/nEa(a),k≥0Remark.Forthe classical forbidden region fr :V(r)> El, one can use similarideato find an asymptotic solution of theforme-(a)/nha;(r)which solve the equation asymptotically (in a stronger sense):h2d2) u=O(h)e-0(r)/h2 dr2 +V(r)-E)The solution near the turning points V(ro) = E is more complicated. (Ref: Dimassiand Sjostrand, Spectral Asymptotics in theSemi-Classical Limit)
LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 7 So after pluging in the asymptotic expansion of u into the equation, we get (− ~ 2 2 d 2 dx2 + V (x) − E) ❸ e −iφ(x)/~ ❳ k≥0 ~ k ak(x) ➂ = e −iφ(x)/~× ✔ 1 2 (φ 0 ) 2 a0 + V (x)a0 − Ea0 + ~( i 2 φ 00a0 + iφ0 a 0 0 + (1 2 (φ 0 ) 2 + V − E)a1) + ❳ k≥2 ~ k ( i 2 φ 00ak−1 + iφ0 a 0 k−1 − 1 2 a 00 k−2 + (1 2 (φ 0 ) 2 + V − E)ak) ✸ ✺ . Since a0 6= 0, the leading term gives us (known as the eikonal equation) (9) 1 2 (φ 0 ) 2 + V (x) − E = 0. As a consequence, we get two different solutions of φ: φ±(x) = ± ❩ ➮ E − V (x)dx. Note that the eikonal equation also simplifies the remaining equations (known as the transport equation that determines ak’s: (10) i 2 φ 00a0 + iφ0 a 0 0 = 0 which determines a0 (which depends on the choice of φ±), and (11) i 2 φ 00ak−1 + iφ0 a 0 k−1 − 1 2 a 00 k−2 = 0 for k ≥ 2, which determines the remaining ak’s (again depends on the choice of φ±). Finally as a consequence of the Borel theorem, we get two asymptotic solutions: u±(x) ∼ e −iφ±(x)/~ ❳ k≥0 ~ k a ± k (x). Remark. For the classical forbidden region {x : V (x) > E}, one can use similar idea to find an asymptotic solution of the form e −φ(x)/~ ❳ j ~ j aj (x) which solve the equation asymptotically (in a stronger sense): ❶ − ~ 2 2 d 2 dx2 + V (x) − E ➀ u = O(~ ∞)e −φ(x)/~ . The solution near the turning points V (x0) = E is more complicated. (Ref: Dimassi and Sj¨ostrand, Spectral Asymptotics in the Semi-Classical Limit)
8LECTURE10—10/26/2020QUANTIZINGGENERALSYMBOLSHereisapicturefrom Wikipediashowingtheoscillationof the solution insidetheclassical allowed region and its“exponential decaying"inside the classical forbiddenregion.3.QUANTIZATIONOFGENERALSYMBOLSQuantizinggeneral symbols.Now let a E S*(R2n). We quantize a to get a semiclassical pseudo-differentialoperator as before. For example, the Kohn-Nirenberg (standard) quantization of aisCet(r-)Sa(r,E)p(y)dyde[akN 0l(a) := (2h)n Janwhile the Weyl quantization of a is[. e(-)a(*+, )o(u)dyds.[aW](a :=(2元h)n)2We have seen that if a E is a Schwartz function, then the semiclassical t-quantization Opt(a) maps continuously into, and if a pr is a tempereddistribution, then the operator Oph(a) maps continuously into '. As a result,for very general tempered distributional symbols, one can't composite semiclassicalpseudodifferential operators.Now we prove that For a symbol a St, the corresponding semiclassical pseu-dodifferential operatorsmapinto and mapintoWewill handletheL? theory later. For simplicity we take k = O and use the Weyl quantization hereTheorem 3.1. If aE Ss(m), thenaw:g-5andaw:→Moreover,both maps are continuous linear maps
8 LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS Here is a picture from Wikipedia showing the oscillation of the solution inside the classical allowed region and its “exponential decaying” inside the classical forbidden region. 3. Quantization of general symbols ¶ Quantizing general symbols. Now let a ∈ S k δ (R 2n ). We quantize a to get a semiclassical pseudo-differential operator as before. For example, the Kohn-Nirenberg (standard) quantization of a is [a❜ KN ϕ](x) := 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ a(x, ξ)ϕ(y)dydξ while the Weyl quantization of a is [a❜ W ϕ](x) := 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ a( x + y 2 , ξ)ϕ(y)dydξ. We have seen that if a ∈ S is a Schwartz function, then the semiclassical tquantization Opt ~ (a) maps S 0 continuously into S , and if a ∈ S 0 is a tempered distribution, then the operator Opt ~ (a) maps S continuously into S 0 . As a result, for very general tempered distributional symbols, one can’t composite semiclassical pseudodifferential operators. Now we prove that For a symbol a ∈ S k δ , the corresponding semiclassical pseudodifferential operators map S into S and map S 0 into S 0 . We will handle the L 2 theory later. For simplicity we take k = 0 and use the Weyl quantization here. Theorem 3.1. If a ∈ Sδ(m), then a❜ W : S → S and a❜ W : S 0 → S 0 . Moreover, both maps are continuous linear maps
LECTURE10—10/26/2020QUANTIZING GENERAL SYMBOLS9Proof. Let N be a constant so that m(z) ≤ C(z)NStepl:awmapsintoLo.According tothefactset(c-)- = (1 + [312)-[1 + (hDu)1(et(-0)-)= (1 + [r - 912)-1[1 + (hDe)](et(r-)-5),we get, for p e y,([ e(-)(1+ P)--n[1+(D,)+n [a(,)()] dy(2元h)nJR(n/n et(r-u)-(1 + [r-g/12)-N×(2元h)nJR[1+ (D)] [(1 + IS1P)-N-[1+ (D,)N+n [a(,3)()] dy.2Since a e Ss(m), there exists constants C and M so that for any multi-indices a, βwith [a|≤2N and |B|≤2N+2n,Ba(,s)≤m(,)Ch(“(E)IDgDga(22Also since E , there exists constant C so that for all Jal ≤ 2N + 2n,C[D0(0)]≤ (1 + 4lu/P)N+nUsing the inequality(1+-)(1+4)≥1+()onecanseethatthereexistsconstantCsothatI[aw](r)I ≤C(/. (1 + [512)-ndE) (/ (1+ 4/g/2)-ndy) ≤CIn other words, awmaps into L.Step 2:raw maps into L.Since/ (D + )e(-)a(“+, )u()dyds,[e°aW 0(1) = (2元h) Jan the same integration by parts argument proves that rcaw maps into Lo.Step 3:ap oaw maps into L.Recall that if b(s,-r)= a(r,), thenaW = (Fh)e=obW (Fh)r→
LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS 9 Proof. Let N be a constant so that m(z) ≤ Chzi N . Step 1: a❜W maps S into L ∞. According to the facts e i ~ (x−y)·ξ = (1 + |ξ| 2 ) −1 [1 + (~Dy) 2 ](e i ~ (x−y)·ξ ) = (1 + |x − y| 2 ) −1 [1 + (~Dξ) 2 ](e i ~ (x−y)·ξ ), we get, for ϕ ∈ S , [a❜ W ϕ](x) = 1 (2π~) n ❩ Rn ❩ Rn (1 + |ξ| 2 ) −N−n [1 + (~Dy) 2 ] N+n (e i ~ (x−y)·ξ )a( x + y 2 , ξ)ϕ(y)dydξ = 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ (1 + |ξ| 2 ) −N−n [1 + (~Dy) 2 ] N+n ➉ a( x + y 2 , ξ)ϕ(y) ➌ dydξ = 1 (2π~) n ❩ Rn ❩ Rn e i ~ (x−y)·ξ (1 + |x − y| 2 ) −N × [1 + (~Dξ) 2 ] N ➜ (1 + |ξ| 2 ) −N−n [1 + (~Dy) 2 ] N+n ➉ a( x + y 2 , ξ)ϕ(y) ➌➟ dydξ. Since a ∈ Sδ(m), there exists constants C and M so that for any multi-indices α, β with |α| ≤ 2N and |β| ≤ 2N + 2n, |D α ξ D β y a( x + y 2 , ξ)| ≤ ~ MCmÜ ( x + y 2 , ξ) ≤ C~ Mh x + y 2 i N hξi N . Also since ϕ ∈ S , there exists constant C so that for all |α| ≤ 2N + 2n, |D αϕ(y)| ≤ C (1 + 4|y| 2 ) N+n . Using the inequality (1 + |x − y| 2 )(1 + 4|y| 2 ) ≥ 1 + ❶ |x + y| 2 ➀2 one can see that there exists constant C so that |[a❜ W ϕ](x)| ≤ C ⑩❩ Rn (1 + |ξ| 2 ) −n dξ❿ ⑩❩ Rn (1 + 4|y| 2 ) −n dy❿ ≤ C 0 . In other words, a❜W maps S into L ∞. Step 2: x αa❜W maps S into L ∞. Since [x α a❜ W ϕ](x) = 1 (2π~) n ❩ Rn ❩ Rn (~Dξ + y) α e i ~ (x−y)·ξ a( x + y 2 , ξ)u(y)dydξ, the same integration by parts argument proves that x αa❜W maps S into L ∞. Step 3: ∂ β ◦ a❜W maps S into L ∞. Recall that if b(ξ, −x) = a(x, ξ), then a❜ W = (F~) −1 ξ→x ◦ ❜b W ◦ (F~)x→ξ
10LECTURE 10-10/26/2020 QUANTIZING GENERAL SYMBOLSIt follows that for any β,(hD)PaW = (Fh)2-osPW o (Fh)ar-sButPE-(F)r-E- (S)n+1gPBW o(Fh)a+P E Lo0W(Fh) E LI-(Fh)2-reW(Fh)r- E L.SoapawmapsintoL.Step4:awmapscontinuouslyto.Applying the same integration by parts arguments, one can prove that r°apaw mapsy intoL.This proves aw mapsto.The continuity can be proved by a similar argument: if all semi-norms of tends to O as j -→ oo, so do the semi-norms of aw uj.Step6:awmapstoFinally we use the fact that for any u, y e ,(awu,) = (u,aww)Since a e S(m), for any w e we have awu e . So the above formula tells usthat awu is a well-defined tempered distribution for u E '. The continuity of themap aw :i→follows from the continuity of the map aw : → 9.口 The composition law.Lasttimewe proved that if a.b E.thenWaw.bw=a*bwherea * b(r, E) = e学(De D,-De:Dn)(a(r,E)b(y, n)|(12)By using the same technique as in the proof of Theorem 3.1, one can check that thesame formula holds for symbols in Ss(m)3:namely if a E Ss(mi) and b E Ss(m2),then1awo6w=a*bwhere a * b is given by the formula (12).However,we still need to justifytheasymptotic expansion1(in)1a*b~(DeDy-DD,)[a(r, E)b(y,n)]二2k.n=3Another way to see this: since is dense in Ss(m), the result should follows from the continuityof quantization
10 LECTURE 10 — 10/26/2020 QUANTIZING GENERAL SYMBOLS It follows that for any β, (~D) β a❜ W = (F~) −1 ξ→x ◦ ξ β❜b W ◦ (F~)x→ξ. But ϕ ∈ S =⇒ (F~)x→ξϕ ∈ S =⇒ hξi n+1ξ β❜b W ◦ (F~)x→ξϕ ∈ L ∞ =⇒ ξ β❜b W (F~)x→ξϕ ∈ L 1 =⇒ (F~) −1 ξ→x ξ β❜b W (F~)x→ξϕ ∈ L ∞. So ∂ β ◦ a❜W maps S into L ∞. Step 4: a❜W maps S continuously to S . Applying the same integration by parts arguments, one can prove that x α∂ βa❜W maps S into L ∞. This proves a❜W maps S to S . The continuity can be proved by a similar argument: if all semi-norms of ϕjS tends to 0 as j → ∞, so do the semi-norms of a❜W uj . Step 6: a❜W maps S 0 to S 0 . Finally we use the fact that for any u, v ∈ S , ha❜ W u, vi = hu, a¯❜W vi. Since ¯a ∈ Sδ(m), for any v ∈ S we have a¯❜W v ∈ S . So the above formula tells us that a❜W u is a well-defined tempered distribution for u ∈ S 0 . The continuity of the map a❜W : S 0 → S 0 follows from the continuity of the map a❜W : S → S . ¶ The composition law. Last time we proved that if a, b ∈ S , then a❜ W ◦ ❜b W = a ? b ÕW , where (12) a ? b(x, ξ) = e i~ 2 (Dξ ·Dy−Dx·Dη) (a(x, ξ)b(y, η)) ☞ ☞ ☞ y=x,η=ξ . By using the same technique as in the proof of Theorem 3.1, one can check that the same formula holds for symbols in Sδ(m) 3 : namely if a ∈ Sδ(m1) and b ∈ Sδ(m2), then a❜ W ◦ ❜b W = a ? b ÕW , where a ? b is given by the formula (12). However, we still need to justify the asymptotic expansion a ? b ∼ ❳∞ k=0 1 k! (i~) k 2 k (Dξ · Dy − Dx · Dη) k [a(x, ξ)b(y, η)] ☞ ☞ ☞ ☞ ☞ y=x,η=ξ . 3Another way to see this: since S is dense in Sδ(m), the result should follows from the continuity of quantization.