LECTURE4—09/28/2020THE FOURIER TRANSFORM1.THEFOURIERTRANSFORMON,ANDL? Some notions.We start with some abbreviations that will be used in this course..Forj=l,..,n,-0,=- D; = 10j.. For any multi-index a= (α1, ., Qn) E Nn.-lal=ai+..+an- q! = Q!...an!.-ra =291...can.- 00 - 001...0an.-Do = Dal... Dan. Schwartz functions.This is the“best"class of functions:Definition 1.1.A function EC(Rn)is called a Schwartz function (or a rapidlydecreasingfunction)if(1)sup [r*ol O, the function p(r) = e-^a is a Schwartzfunction.Erample. If is a Schwartz function, so are the functions rD,DrP, where a,βare any multi-indices. In particular, the eigenfunctions of the Harmonic oscillatorH= -+ that we get last time are all Schwartz functions.We will denote the set of all Schwartz functions by (IR), or byfor simplicity.Obviouslyisaninfinitelydimensionalvectorspace,andCo(R") C (R") C LP(R"),V1≤P≤80.1
LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM 1. The Fourier transform on S , S 0 and L 2 ¶ Some notions. We start with some abbreviations that will be used in this course. • For j = 1, · · · , n, – ∂j = ∂ ∂xj . – Dj = 1 i ∂j . • For any multi-index α = (α1, · · · , αn) ∈ N n . – |α| = α1 + · · · + αn. – α! = α1! · · · αn!. – x α = x α1 1 · · · x αn n . – ∂ α = ∂ α1 x1 · · · ∂ αn xn . – Dα = D α1 1 · · · Dαn n . ¶ Schwartz functions. This is the “best” class of functions: Definition 1.1. A function ϕ ∈ C ∞(R n ) is called a Schwartz function (or a rapidly decreasing function) if (1) sup Rn |x α ∂ βϕ| 0, the function ϕ(x) = e −λ|x| 2 is a Schwartz function. Example. If ϕ is a Schwartz function, so are the functions x αDβϕ, Dαx βϕ, where α, β are any multi-indices. In particular, the eigenfunctions of the Harmonic oscillator Hˆ = − ~ 2 2 ∆ + |x| 2 2 that we get last time are all Schwartz functions. We will denote the set of all Schwartz functions by S (R n ), or by S for simplicity. Obviously S is an infinitely dimensional vector space, and C ∞ 0 (R n ) ⊂ S (R n ) ⊂ L p (R n ), ∀1 ≤ p ≤ ∞. 1
2LECTURE4-09/28/2020 THEFOURIERTRANSFORMTo do analysis, we need to give a topology on the space . This can be done viasemi-norms. Recall that a semi-norm on a vector space V is function -|:V→Rso that for all e C and all u, e V,(1) (Absolute homogeneity) [入v|=[^//u](2)(Triangle inequality) [u+|≤[u|+[u]On one can define, for each pair of multi-indices a,β a semi-norm(2)Ipla,β = sup[raaβl.This is a separating family of semi-norms using which one can define a metric d ongviad(, 0) =2-k_ - llk(3)1+ ll -llkk=0where for each k ≥ 0.loll=max,supJr°DBplJa|+13]<KrER"is a norm on .The topology on we are going to use is the metric topologyinduced by d. It is easy to see that Pj → p in if and only ifIp-pla,B→ 0as j → oo for all a, β. With respect to this topology, is a Frechet spaceI The Fourier transform on .Let E be a Schwartz function. By definition its Fourier transform F is(4)e-ir-tp(r)dr.Fp() =() =We list several basic properties of the Fourier transform whose proofs (whichfollows from direct computations)will be omitted.Proposition 1.2. For any p, E and any multi-inder a, we have(1) F(ra) =(-1)lal D(F().(2) F(D) = F().(3) Jrn Fp($)(s)dE = Jrn (r)Fb(r)dr.(4)Forany^eR+,ifwe letpx(r)=p(Ar),then pxE and=元F0(Fpx(s) =an(5) For any a E R", if we let Tap(r) = p(r +a), then Tap E and(FTa)() = eia Fp(s)
2 LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM To do analysis, we need to give a topology on the space S . This can be done via semi-norms. Recall that a semi-norm on a vector space V is function | · | : V → R so that for all λ ∈ C and all u, v ∈ V , (1) (Absolute homogeneity) |λv| = |λ||v|. (2) (Triangle inequality) |u + v| ≤ |u| + |v|. On S one can define, for each pair of multi-indices α, β a semi-norm (2) |ϕ|α,β = sup Rn |x α ∂ βϕ|. This is a separating family of semi-norms using which one can define a metric d on S via (3) d(ϕ, ψ) = ❳∞ k=0 2 −k kϕ − ψkk 1 + kϕ − ψkk , where for each k ≥ 0, kϕkk = max |α|+|β|≤k sup x∈Rn |x αD βϕ| is a norm on S . The topology on S we are going to use is the metric topology induced by d. It is easy to see that ϕj → ϕ in S if and only if |ϕj − ϕ|α,β → 0 as j → ∞ for all α, β. With respect to this topology, S is a Fr´echet space. ¶ The Fourier transform on S . Let ϕ ∈ S be a Schwartz function. By definition its Fourier transform Fϕ is (4) Fϕ(ξ) = ˆϕ(ξ) = ❩ Rn e −ix·ξϕ(x)dx. We list several basic properties of the Fourier transform whose proofs (which follows from direct computations) will be omitted. Proposition 1.2. For any ϕ, ψ ∈ S and any multi-index α, we have (1) F(x αϕ) = (−1)|α|Dα ξ (F(ϕ)). (2) F(Dα xϕ) = ξ αF(ϕ). (3) ❘ Rn Fϕ(ξ)ψ(ξ)dξ = ❘ Rn ϕ(x)Fψ(x)dx. (4) For any λ ∈ R+ , if we let ϕλ(x) = ϕ(λx), then ϕλ ∈ S and Fϕλ(ξ) = 1 λ n Fϕ( ξ λ ). (5) For any a ∈ R n , if we let Taϕ(x) = ϕ(x + a), then Taϕ ∈ S and (FTaϕ)(ξ) = e ia·ξFϕ(ξ)
3LECTURE4—09/28/2020THEFOURIERTRANSFORMNote that as a consequence of (1) and (2), we havePES-FPESOne can show that as a linearmap,F:→ is continuousAs another consequence, we can compute the Fourier transform of the Gaussianfunction. We haveF(e-1m/ /2) = (2元)e-13) /2.(5)To see this we just noticeF(e-lpP/2)(s)= / e-luP/2-ir- dr.Taking derivative, we getaL- a- / (ryF(e-P)() =-iti irdaaE=-i e-P 2(e-ir:5)da = -; F(e-)(S),ocjIt followsF(e-lP/2)() = Ce-1512/2,whereC = F(e-lapP/2)(0) = [ e-lapP/2dar = (2元)n/2UsingthesepropertieswecanproveTheorem 1.3.F:→J is an isomorphism with inverse(6)F-()- /e()sProof. We have[ Fo(E)(As)d = / 0(r)Fu()da= / (An)Fv(a)dc.Letting ^→0, we getb(0) / Fp(E)d =(0) / Fu(g)dr.In particular, if we take (6) =e-, then we get1(7)/Fo($)de5(0) =(2元)n Finally we replace by Ta in the above formula, then1[ eia-t Fp(s)de(a) = Tap(0) =(2元)n口This is exactly what we need
LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM 3 Note that as a consequence of (1) and (2), we have ϕ ∈ S =⇒ Fϕ ∈ S . One can show that as a linear map, F : S → S is continuous. As another consequence, we can compute the Fourier transform of the Gaussian function. We have (5) F(e −|x| 2/2 ) = (2π) n 2 e −|ξ| 2/2 . To see this we just notice F(e −|x| 2/2 )(ξ) = ❩ Rn e −|x| 2/2−ix·ξ dx. Taking ξ derivative, we get ∂ ∂ξj F(e − 1 2 |x| 2 )(ξ) = −ixj ❩ Rn e − 1 2 |x| 2−ix·ξ dx = i ❩ Rn ∂ ∂xj (e − 1 2 |x| 2 )e −ix·ξ dx = −i ❩ Rn e − 1 2 |x| 2 ∂ ∂xj (e −ix·ξ )dx = −ξjF(e − 1 2 |x| 2 )(ξ). It follows F(e −|x| 2/2 )(ξ) = Ce−|ξ| 2/2 , where C = F(e −|x| 2/2 )(0) = ❩ R e −|x| 2/2 dx = (2π) n/2 . Using these properties we can prove Theorem 1.3. F : S → S is an isomorphism with inverse (6) F −1ψ(x) = 1 (2π) n ❩ Rn e ix·ξψ(ξ)dξ. Proof. We have ❩ Fϕ(ξ)ψ(λξ)dξ = 1 λ n ❩ ϕ(x)Fψ( x λ )dx = ❩ ϕ(λx)Fψ(x)dx. Letting λ → 0, we get ψ(0) ❩ Fϕ(ξ)dξ = ϕ(0) ❩ Fψ(x)dx. In particular, if we take ψ(ξ) = e − |ξ| 2 2 , then we get (7) ϕ(0) = 1 (2π) n ❩ Fϕ(ξ)dξ. Finally we replace ϕ by Taϕ in the above formula, then ϕ(a) = Taϕ(0) = 1 (2π) n ❩ e ia·ξFϕ(ξ)dξ. This is exactly what we need.
4LECTURE4—09/28/2020THEFOURIERTRANSFORMAs a consequence, we get the following Parseval's identity:Corollary 1.4. For y E y,1I10ll22l/F0l/2(2元）Proof.We have1a la [Fp(s)Fp(E)dEFp(E)(F-1)(s)d8JRr= Jan (r)(FF-10)(a)dr=I/0l/2.口TheFouriertransform onDefinition 1.5. Any linear continuous map u : -→ C is called a tempered distri-bution. The space of tempered distributions is denoted by pSo ' is the dual of . The topology of the space ' is defined to be th weak-*topology,sothatui→uinifi(p) -→u(p), VeErample.Here are some examples of tempered distributions:.For any a ERn, the Dirac distribution d.:g C defined by8a() = (a)is a tempered distribution.. More generally, for any a e Cn and any multi-index a, the map u : J→ Cdefined byu(p) = D(a)is a tempered distribution.. Any bounded continuous function b on Rn defines a tempered distributionbyug(0) = (r)d(r)da.Note that if e,then the distribution defined by the above formula isnon-zerounless=o.ItfollowsthatcAs we can see from the examples, a tempered distribution need not be a function.However, given a tempered distribution, we can still define some operations on themas if they are functions:
4 LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM As a consequence, we get the following Parseval’s identity: Corollary 1.4. For ϕ ∈ S , kϕk 2 L2 = 1 (2π) n kFϕk 2 L2 . Proof. We have 1 (2π) n kFϕk 2 L2 = 1 (2π) n ❩ Rn Fϕ(ξ)Fϕ(ξ)dξ = ❩ Rn Fϕ(ξ)(F −1ϕ)(ξ)dξ = ❩ Rn ϕ(x)(FF −1ϕ)(x)dx = kϕk 2 L2 . ¶ The Fourier transform on S 0 . Definition 1.5. Any linear continuous map u : S → C is called a tempered distribution. The space of tempered distributions is denoted by S 0 . So S 0 is the dual of S . The topology of the space S 0 is defined to be th weak-* topology, so that uj → u in S 0 if uj (ϕ) → u(ϕ), ∀ϕ ∈ S . Example. Here are some examples of tempered distributions: • For any a ∈ R n , the Dirac distribution δa : S → C defined by δa(ϕ) = ϕ(a) is a tempered distribution. • More generally, for any a ∈ C n and any multi-index α, the map u : S → C defined by u(ϕ) = D αϕ(a) is a tempered distribution. • Any bounded continuous function ψ on R n defines a tempered distribution by uψ(ϕ) = ❩ Rn ϕ(x)ψ(x)dx. Note that if ψ ∈ S , then the distribution defined by the above formula is non-zero unless ψ = 0. It follows that S ⊂ S 0 . As we can see from the examples, a tempered distribution need not be a function. However, given a tempered distribution, we can still define some operations on them as if they are functions:
LECTURE4—09/28/2020THEFOURIER TRANSFORM5Definition 1.6. Let u Ei.We define(1) ruE via (ru)(g) =u(rp)(2) Du E " via (Dau)() = (-1)ialu(D%).(3) Fu Eg via (Fu)() = u(F)Remark. One can check that if is a bounded smooth function, then the distribu-tions raub, Dau, F(uy) (here we assume is a Schwartz function) coincides withUa, uDa and uF. In other words, these operations on tempered distributions,whenrestricted tousual functions,arethesameoperators that we arefamiliarwith.Sointhesecases,for simplicitywewill simplydenote ruw,Du,F(u)byr,Da and F(). The Fourier transform on L?(Rn).We can also view any L? function as a tempered distribution by using thesameformulaabove, namely,p(r)d(r)da.u(0) :=By this way we gets a natural embeddingC(R") C S(IR") C L2(IR") C '(R"),where one can check that all the inclusion maps are continuous. In particular, forany e L?, one can define its Fourier transform F, in the sense of distribution.ItturnsoutthatProposition 1.7 (Plancherel's Theorem).If E L?(Rn),then the tempered distri-bution Fb is also in L?(Rn), and we still have the Parsevel's identityFll12122元）mProof.By Cauchy-Schwartz inequality,for any E we haveF()()/ = [(F)≤ IL2 : IFl2 = (2)n/21l2 - /lL2Since is dense in L?(Rn), the same inequality holds for E L?(Rn). In otherwords, F() is a bounded linear map on L?(Rn).By Riesz representation theorem,Fab is represented by an L? function.Moreover, the same formula above also implies that as an L?-function,IF/2≤(2)n/2/2.ItfollowsthatIF2≤(2）2.But a direct computation shows FFb(a) = (2元)nb(-r) for Schwartz functions, andthus forL2functions bythe same arguments above. So lFFlz2 = (2)nlllL2.The Parsevel's identity follows.口
LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM 5 Definition 1.6. Let u ∈ S 0 . We define (1) x αu ∈ S 0 via (x αu)(ϕ) = u(x αϕ). (2) Dαu ∈ S 0 via (Dαu)(ϕ) = (−1)|α|u(Dαϕ). (3) Fu ∈ S 0 via (Fu)(ϕ) = u(Fϕ). Remark. One can check that if ψ is a bounded smooth function, then the distributions x αuψ, Dαuψ, F(uψ) (here we assume ψ is a Schwartz function) coincides with uxαψ, uDαψ and uFψ. In other words, these operations on tempered distributions, when restricted to usual functions, are the same operators that we are familiar with. So in these cases, for simplicity we will simply denote x αuψ, Dαuψ, F(uψ) by x αψ, Dαψ and F(ψ). ¶ The Fourier transform on L 2 (R n ). We can also view any L 2 function ψ as a tempered distribution by using the same formula above, namely, uψ(ϕ) := ❩ Rn ϕ(x)ψ(x)dx. By this way we gets a natural embedding C ∞ c (R n ) ⊂ S (R n ) ⊂ L 2 (R n ) ⊂ S 0 (R n ), where one can check that all the inclusion maps are continuous. In particular, for any ψ ∈ L 2 , one can define its Fourier transform Fψ, in the sense of distribution. It turns out that Proposition 1.7 (Plancherel’s Theorem). If ψ ∈ L 2 (R n ), then the tempered distribution Fψ is also in L 2 (R n ), and we still have the Parsevel’s identity kψk 2 L2 = 1 (2π) n kFψk 2 L2 . Proof. By Cauchy-Schwartz inequality, for any ϕ ∈ S we have |F(ψ)(ϕ)| = |ψ(Fϕ)| ≤ kψkL2 · kFϕkL2 = (2π) n/2 kψkL2 · kϕkL2 . Since S is dense in L 2 (R n ), the same inequality holds for ϕ ∈ L 2 (R n ). In other words, F(ψ) is a bounded linear map on L 2 (R n ). By Riesz representation theorem, F ψ is represented by an L 2 function. Moreover, the same formula above also implies that as an L 2 -function, kFψkL2 ≤ (2π) n/2 kψkL2 . It follows that kFFψkL2 ≤ (2π) n kψkL2 . But a direct computation shows FFψ(x) = (2π) nψ(−x) for Schwartz functions, and thus for L 2 functions by the same arguments above. So kFFψkL2 = (2π) nkψkL2 . The Parsevel’s identity follows.
6LECTURE4-09/28/2020THEFOURIERTRANSFORM2. SEVERAL GAUSSIAN INTEGRALSFor the future purpose, we need to compute the Fourier transform of the Gauss-ian function e-rQ. For simplicity we do the computation using the standardFourier transform. One can easily rewrite these results in the semiclassical settingTheorem 2.1. Let Q be any real, symmetric, positive definite n x n matrir, then(2元)n/2(det Q)e-sTo-1s.F(e-aTQr)(S) =(8)Proof. Let's first assume n = 1, so that Q = q is a positive number. By a change ofvariable argument from (5) [or by repeating the proof if you want], one easily getsF(e-193)(5) = (2m)/2-e-$/q.gl/2Next suppose Q is a diagonal matrix. Then the left hand side of (8) is a productof n one-dimensional integrals, and the result follows from the one-dimensional case.For the general case, one only need to choose an orthogonal matrix O so thatOTQO =D= diag(>i,...,^n),where A,'s are the eigenvalues of Q. The results follows from change variables fromr to y=o-lr:F(e-+Q)(E)= /e-1TQr-in- da=/e-1Dy-iyOTEdy(2元)n/2(2元)n/2(det D)/ze-$s"OD-1oTs =(det O)e-1$"o-s.口Now suppose Q is an n n complex matrix such that Q = QT and Re(Q) ispositive definite. We observe that the integral on the left hand side of (8) is ananalytic function of the entries Qij = Qji of Q in the region ReQ > 0. The same istrue for the right hand side of (8), as long as we choose deti/2 Q to be the branchsuch that det1/2Q>0for real positive definite Q. So we immediately getTheorem 2.2. Let Q be any n × n compler matri such that Q = QT and Re(Q)is positivedefinite,then二_(2元)n/2(e)()- (-(9)Remark. Another way to describe this choice of det'/2: Since Q = QT, for anycomplex vector w E Cn we haveRe(uTQw) =wT(ReQ)w
6 LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM 2. Several Gaussian integrals For the future purpose, we need to compute the Fourier transform of the Gaussian function e − 1 2 x T Qx. For simplicity we do the computation using the standard Fourier transform. One can easily rewrite these results in the semiclassical setting. Theorem 2.1. Let Q be any real, symmetric, positive definite n × n matrix, then (8) F(e − 1 2 x T Qx)(ξ) = (2π) n/2 (det Q) 1/2 e − 1 2 ξ T Q−1 ξ . Proof. Let’s first assume n = 1, so that Q = q is a positive number. By a change of variable argument from (5) [or by repeating the proof if you want], one easily gets F(e − 1 2 qx2 )(ξ) = (2π) 1/2 q 1/2 e − 1 2 ξ 2/q . Next suppose Q is a diagonal matrix. Then the left hand side of (8) is a product of n one-dimensional integrals, and the result follows from the one-dimensional case. For the general case, one only need to choose an orthogonal matrix O so that O TQO = D = diag(λ1, · · · , λn), where λi ’s are the eigenvalues of Q. The results follows from change variables from x to y = O−1x: F(e − 1 2 x T Qx)(ξ) = ❩ Rn e − 1 2 x T Qx−ix·ξ dx = ❩ Rn e − 1 2 y T Dy−iy·OT ξ dy = (2π) n/2 (det D) 1/2 e − 1 2 ξ T OD−1OT ξ = (2π) n/2 (det Q) 1/2 e − 1 2 ξ T Q−1 ξ . Now suppose Q is an n × n complex matrix such that Q = QT and Re(Q) is positive definite. We observe that the integral on the left hand side of (8) is an analytic function of the entries Qij = Qji of Q in the region ReQ > 0. The same is true for the right hand side of (8), as long as we choose det1/2 Q to be the branch such that det1/2 Q > 0 for real positive definite Q. So we immediately get Theorem 2.2. Let Q be any n × n complex matrix such that Q = QT and Re(Q) is positive definite, then (9) F(e − 1 2 x T Qx)(ξ) = (2π) n/2 (det Q) 1/2 e − 1 2 ξ T Q−1 ξ . Remark. Another way to describe this choice of det1/2 : Since Q = QT , for any complex vector w ∈ C n we have Re( ¯w TQw) = ¯w T (ReQ)w.
LECTURE4—09/28/2020THEFOURIERTRANSFORM7It follows that all eigenvalues i,.,An of Qhas positive real part.Then(det Q)1/2 = X/2...X1/2,A,/2 is thesquare root of , thathas positivereal part.whereFinally let Q be any real, symmetric, non-singular n × n matrix. Next timewe will need the Fourier transform of the function earTQr. Note that the functionearrQr is not a Schwartz function, so we are really thinking of it as a tempereddistribution and calculate its Fourier transform.Recall that the signature of the matrixQis(10)sgn(Q) = N+(Q) - N-(Q),where N+(Q)=the number of positive/negative eigenvalues of QTheorem 2.3.For any real, symmetric, non-singular n X n matrir QF(eTQr)= (2)n/2eitsgn(0)e-%TQ-15(11)I det Q]Proof. Let's first assume n = 1, so that Q = q 0 is a non-zero real number. Forany >0 we let qe= q+ ie. Applying Theorem 2.2 we getF(ez) = F(e-(e-i0)2) =(2m)/2(e - ig)1/e-1(-in)sthe square root is described after Theorem 2.2. Note that when e → 0+, we have(e- ig)/2[ Vqe-q>0 / =Vale-sgn(a),-e0so the conclusion follows.For n>1,one just proceed as before:first handlethe case of diagonal matrices,then convert the general case to the diagonal case via the diagonalization trick
LECTURE 4 — 09/28/2020 THE FOURIER TRANSFORM 7 It follows that all eigenvalues λ1, · · · , λn of Q has positive real part. Then (det Q) 1/2 = λ 1/2 1 · · · λ 1/2 n , where λ 1/2 j is the square root of λj that has positive real part. Finally let Q be any real, symmetric, non-singular n × n matrix. Next time we will need the Fourier transform of the function e i 2 x T Qx. Note that the function e i 2 x T Qx is not a Schwartz function, so we are really thinking of it as a tempered distribution and calculate its Fourier transform. Recall that the signature of the matrix Q is (10) sgn(Q) = N +(Q) − N −(Q), where N ±(Q) = the number of positive/negative eigenvalues of Q. Theorem 2.3. For any real, symmetric, non-singular n × n matrix Q, (11) F(e i 2 x T Qx) = (2π) n/2 e i π 4 sgn(Q) | det Q| 1 2 e − i 2 ξ T Q−1 ξ . Proof. Let’s first assume n = 1, so that Q = q 6= 0 is a non-zero real number. For any ε > 0 we let qε = q + iε. Applying Theorem 2.2 we get F(e i 2 qεx 2 ) = F(e − 1 2 (ε−iq)x 2 ) = (2π) 1/2 (ε − iq) 1/2 e − 1 2 (ε−iq)ξ 2 , the square root is described after Theorem 2.2. Note that when ε → 0+, we have (ε − iq) 1/2 −→ ✭ √qe− iπ 4 , q > 0 √ −qe iπ 4 , q 1, one just proceed as before: first handle the case of diagonal matrices, then convert the general case to the diagonal case via the diagonalization trick.