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MIT 18 155 - Lecture Notes

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LECTURE 3, 18.155, 15 SEPTEMBER 2011(1) Since I will need them later, let me start with a little disussionof ‘bump’ functions. First, the function(1) f(x) =(exp(1|x|2−1) |x| < 10 |x| ≥ 1∈ S(Rn).Since this function vanishes in |x| ≥ 1 be definition, the onlyissue is its smoothness. It is smoot in |x| < 1 since |x|2issmooth and exp(1t−1) is smooth in t in t < 1. Moreover, as t ↑ 1,exp(1t−1) → 0 rapidly, since the argument of the exponential istending to −∞. Differentiating f(x) gives a mess but it is, byinduction, of the form(2) ∂αf(x) =Pα(x)|x|2− 1)2|α|exp(1|x|2− 1)where Pαis a polynomial. A product such as (t−1)−2kexp(1t−1) →0 as t ↑ 1 so all the derivatives converge to zero at |x| = 1. Itfollows from this that f ∈ S(Rn).(2) We can turn this into a true bump function by considering acomplementary function such as(3) g(x) =(exp(11−4|x|2) |x| > 1/20 |x| ≤ 1/2∈ C∞(Rn)by the same argument. Since f ≥ 0, g ≥ 0 and f > 0 in |x| < 1,g > 0 in |x| > 1/2, f + g > 0 so 1/(f + g) ∈ C∞(Rn). It followsthat(4)χ(x) =f(x)f(x) + g(x)∈ S(Rn), χ(x) = 0 in |x| ≥ 1 and χ(x) = 1 in |x| <12.(3) Now we use this to prove a ‘division property’.Lemma 1. If φ ∈ S(Rn) then there exist φj∈ S(Rn) such that(5) φ(x) = φ(0) exp(−|x|2) +nXj=1xjφj(x).12 LECTURE 3, 18.155, 15 SEPTEMBER 2011If you want a little exercise, show (you can use the proofhere) that one can actually find continuous linear maps Tj:S(Rn) −→ S(Rn) such that (5) holds with φj= Tjφ.(4) To prove this lemma we can start by using Taylor’s theorem. Infact let me do this constructively so just consider the obviousidentity which is just the Fundamental Theorem of Calculusagain:(6) φ(x) = φ(0) +Z10ddsφ(sx)ds.Expanding out the derivative, this becomes(7) φ(x) = φ(0) +Xjxj˜φj(x),˜φj(x) =Z10∂jφ(sx)ds.The problem here is that˜φj∈ C∞(Rn) from properties of theintegral, but there is no reason at all why they should decay atinfinity, so they are not in S(Rn). So, instead, we just cut themoff and write(8) φ(x) = φ(0)χ(x) +Xjxjχ(x)˜φj(x)+ φ0(x)where χ is the cut-off constructed above. It follows that φ0= 0in |x| <12and φ0∈ S(Rn) – the latter because all the otherterms are in S(Rn) and the former because (7) holds in |x| <12.So, now consider the function(9) ψ(x) =(|x|−2ψ0(x) |x| >140 |x| ≤14∈ S(Rn).It is smooth since the first definition makes the function zero in|x| <12. So, now we can replace (8) by(10) φ(x) = φ(0)χ(x) +Xjxjχ(x)˜φj(x) + xjψ.This is what we want, except that in (3) χ is replaced byexp(−|x|2). However, this is not a problem since(11) µ = χ(x) − exp(−|x|2) ∈ S(Rn), µ(0) = 0.Then applying (10) to µ gives some more terms and leads us to(5).LECTURE 3, 18.155, 15 SEPTEMBER 2011 3(5) Now, we will normalize the Fourier transform as(12)ˆφ(ξ) =ZRne−ix·ξφ(x)dx = limR→∞ZR−R. . .ZR−Re−ix·ξφ(x)dx1. . . dxnthinking of it as an ‘improper iterated Riemann integral’.(6) Before the limit, everything is smooth since we can differentiateunder Riemann integrals (when the integrand is smooth). Wefind for instance that(13)∂αξZR−R. . .ZR−Re−ix·ξφ(x)dx1. . . dxn=ZR−R. . .ZR−Re−ix·ξ(−ix)αφ(x)dx1. . . dxn,∂α=\(−ix)αφ.Now, xαφ ∈ S(Rn) so we can estimate as before and get(14) |∂αξˆφ(ξ)| ≤ Ckφ|2n+2+|α|.Similarly, integration by parts is possible with boundary termswhich are rapidly decreasing, so disappear in the limit as R →∞ and it follows that(15) ξβˆφ(ξ) =ZRne−ix·ξ(−i)|β|∂βφ(x)dx.(7) Combining these two identies, we see that ξβ∂αˆφ is the Fouriertransform of (−i)|α|+|β|∂β(xαφ(x)) so is bounded and continu-ous. It follows thatˆφ ∈ S(Rn) and that(16) kˆφkN≤ CNkφk2n+2+Nalthough the order is by no means optimal. This shows that(17) F : S(Rn) 3 φ 7−→ˆφ ∈ S(Rn)is a continuous linear map.(8) To examine its invertibility we consider the element of S0(Rn)which corresponds to composingRwith F. That is, we look at(18) S(Rn) 3 φ −→Zˆφ(ξ)dξ.We can insert (5) into this, to see that(19)Zˆφ(ξ)dξ = Cφ(0), C =Z\exp(−|x|2)dξsince the other terms are(20)XjZ\xjφj(x) =XjZi∂jbφj= 0.4 LECTURE 3, 18.155, 15 SEPTEMBER 2011(9)Lemma 2. The constant in (19) is C = (2π)n.Proof. Let us work out the Fourier transform(21)Ze−ix·ξexp(−|x|2)dx.In fact the integral is the product of the corresponding 1-dimensionalintegrands, so it suffices to work out(22) F (ξ) =ZRexp(−ixξ − x2)dx.Certainly this (improper Riemann) integral is rapidly conver-gent. For each ξ ∈ R we can complete the square and see that(23) F (ξ) = exp(−ξ2/4)G(ξ), G(ξ) =ZRexp(−(x + iξ/2)2)dx.In fact, G(ξ) is a constant. This can be seen using contourdeformation and Cauchy’s theorem, or by real variable methods.The rapid convergence and smoothness of the integrand meanthat G(ξ) is differentiable and we can compute its derivative(24)dG(ξ)dξ=ZR(−i(x+iξ)) exp(−(x+iξ/2)2)dx =ZR(i2ddxexp(−(x+iξ/2)2)dx = 0.Then the standard integral is(25) G(0) = G(ξ) =ZRexp(−x2)dx =√π.Thus we have computed the Fourier transform of the Gaussian:-(26) F(exp(−|x|2) = πn2exp(−|ξ|2/4) in S(Rn).Now the constant in (19) is the integral of this Fourier trans-form. Applying the same formula after a change of variables wesee that(27)ZRexp(−|ξ|2/4) = 2√πwhich proves the Lemma. (10) Thus we have finally shown that(28)ZRnˆφ = (2π)nφ(0).LECTURE 3, 18.155, 15 SEPTEMBER 2011 5From this and a little more manipulation we see that(29) G : S(Rn) −→ S(Rn), (Gφ)(ξ) = (2π)−nZeix·ξφ(x)is a two-sided continuous inverse to F.Indeed, if φ ∈ S(Rn) then ψ(x) = φ(x + y), y ∈ Rnis also anelement of S(Rn) and it has Fourier transform(30)ˆψ(ξ) =Ze−ix·ξφ(x + y) = eiy·ξˆφ(ξ).Applying (28) we deduce that(31)Zˆψ =Zeiy·ξˆφ(ξ)dξ = (2π)nψ(0) = (2π)nφ(y).This is the statement GF = Id . Note that the mapping propertyin (29) follows from the corresponding statement for F sinceGφ(ξ) = (2π)−nˆφ(−ξ).Similarly the identity FG = Id follows by changing signs,or you can just observe that F must be a bijection, since itsinjectivity has been established and its surjectivity follows fromthat of


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