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MIT 18 155 - Lecture Notes

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LECTURE 9, 18.155, 6 OCTOBER, 2011Last time I got to the middle of the proof that the Fourier transformof a distribution of compact support is a smooth function.Lemma 1. If u ∈ S0(Rn) has compact support then ˆu ∈ C∞(Rn) is afunction of slow growth and more precisely there exists N such that forall α(1) |∂αξˆu| ≤ CαhξiN.Continued. For a distribution of compact support there exists χ ∈C∞c(Rn) such that χu = u. This allows us to define the functionF (ξ) = u(χ(x)e−ix·ξ)since χ exp(ix · ξ) ∈ C∞c(Rn) ⊂ S(Rn). This should be the Fouriertransform but we still need to check that this is so. Recall that thefact that difference quotient converges uniformly with all x derivativeson compact sets in x (namely the support of χ)(e−ix·(ξ+hej− e−ix·ξ)/h → −ixje−ix·ξis what shows that is differentiable and then we can use induction tosee that F satisfies the estimates in (1). So we just need to show thatˆu = F, i.e.u(ˆφ) = F (φ) ∀ φ ∈ S(Rn).Since both sides are tempered distributions it is enough to check thisfor φ of compact support.Think about computing u(χˆφ) in terms of φ. The integral forˆφ is aRiemann integral (or you can do it using Lebesgue integration too ofcourse) and so is the limit of a sumˆφ(x) =Zexp(−iξ · x)φ(ξ) = limj→∞2−njXkexp(−iξk,j· x)φ(ξk,j)Here ξk,jis an ordering of the finite number of points in 2−jZnin a bigball containing the support of φ and the constant is the volume of thecorresponding cube(s) (I hope).As a function of x in a compact set this approximation to the Rie-mann integral actually converges uniformly with all its derivatives –just differentiate to see this. Soχ(x)2−njXkexp(−iξk,j· x)φ(ξk,j) → χ(x)ˆφ(x) in S(Rn).12 LECTURE 9, 18.155, 6 OCTOBER, 2011By the continuity of u this impliesu(χˆφ) = limj→∞u(χ(x)2−njXkexp(−iξk,j· x)φ(ξk,j)→ χ(x)ˆφ(x) in S(Rn)=⇒ limj→∞XkF (ξkj)φ =ZF (ξ)φ(ξ)again by the properties of Riemann integrals.So indeed ˆu = F. I said we would check that the extension of the Fourier transform(etc) to S0(Rn) was forced by continuity here is the first version. Wecan say that a sequence uj∈ S0(Rn) converges weakly to u ∈ S0(Rn)if uj(φ) → u(φ) in C for all φ ∈ S(Rn). The limit is certainly uniqueand we can use the Fourier transform on S(Rn) to prove that S(Rn) ⊂S0(Rn) is weakly dense.Lemma 2. If u ∈ S0(Rn) then there exists a sequence uj∈ S(Rn) suchthat φj→ u weakly in S0(Rn).Proof. We know that if χ ∈ C∞c(Rn) has χ(x) = 1 in |x| < 1 then χkφ →φ in S(Rn), where χk(x) = χ(xk). Now, if u ∈ S0(Rn) then u(χkφ) →u(φ) for all φ ∈ S(Rn), which means that χku(φ) → u(φ) which is thestatement χku → u weakly. Here, χku has compact support, since,for instance taking χ so that χ(x) = 0 in |x| > 2, χ3kχk= χksoχ3k(χku) = χku.Thus we have shown that the subset of distributions of compactsupport, which I will denote C−∞c(Rn) ⊂ S0(Rn), is a weakly densesubspace.So, now (by a standard diagonalization argument) we just need toshow that if u ∈ C−∞c(Rn) then we can find a sequence un→ u weaklyin S0(Rn) with un∈ S(Rn). Fourier transform. By the preceedingresult, ˆu ∈ C∞(Rn) so χkˆu ∈ C∞c(Rn) ⊂ S(Rn) where we know thatχkˆu(φ) → ˆu(φ) ∀ φ ∈ S(Rn)since for instance χkˆu → ˆu in L1or C00. Anyway, this implies that uk,defined by buk= χkˆu is a sequence in S(Rn) which converges weakly tou. Another consequence of this is something you already know, namelythe Riesz Representation for L2. Here we can say it in the formLemma 3. If u ∈ S0(Rn) and|u(φ)| ≤ CkφkL2∀ φ ∈ S(Rn)LECTURE 9, 18.155, 6 OCTOBER, 2011 3then u ∈ L2(Rn).Proof. If u ∈ S0(Rn) satisfies the estimate (3) then so does χu if χ ∈C∞c(Rn) since|χu(φ)| = |u(χφ)| ≤ CkχφkL2≤ C0kφkL2.Then F = cχu also satisfies the same estimate since F is bounded withrespect to the L2norm:-|cχu(φ)| = |u(χˆφ)| ≤ CkχˆφkL2≤ C0kˆφkL2≤ C00kφkL2.However, from the result above, cχu = F is a smooth function and wecan easily check that this is in L2. Namely χ(ξk)F must be Cauchy inL2. So, using the inverse Fourier transform, which we know maps L2into itself, we conclude that χu ∈ L2. Then again we can take a limitof χ(xk)u to see that u ∈ L2(Rn). The abstract proof of the Riesz Representation Theorem in a Hilbertspace is rather neater, but this concrete one is more a guide to othersuch constructions – for instance the Schwartz kernel theorem.So, next to the structure theorem for tempered distributions. Thiscan be stated in various ways, let’s try a concise one.Theorem 1 (Schwartz Structure Theorem). Any u ∈ S(Rn) is of thethe form(2) u = hDi2N(hxi2NvN), vN∈ C00(Rn) ∩ L2(Rn)for some N ∈ N.The notation on the right here ishDh2= 1 + |D|2= 1 +XjD2j= 1 −Xj∂2jis a differential operator of order 2 so in (2) there is a differentialoperator of order 2N. A somewhat weaker (but essentially equivalent)statement is that any tempered distribution is a finite sum of productsxα∂βvα,βwith the vα,βbounded and continuous (or L2) and this followsjust by expanding out the right side of (2).Proof. The main step is a judicious application of the Sobolev embed-ding theorem and then the Riesz representation theorem.All we know initially is that u ∈ S0(Rn) and that means that forsome M,|u(φ)| ≤ CX|α|+|β|≤Msup |xαDβφ| ∀ φ ∈ S(Rn).4 LECTURE 9, 18.155, 6 OCTOBER, 2011We start by dividing u by the functon hxi−2N– which we know hasslow growth. This means of course that if v = hxi−2Nu satisfies thecorresponding estimate|v(φ)| ≤ CX|α|+|β|≤Msup |xαDβ(hxi−Mφ)| ∀ φ ∈ S(Rn).Applying Leibniz formula to commute all the derivatives through, weget a big sum on the right but of the form|v(φ)| ≤ CX|β|≤Msup |gβ(x)Dβφ)| ≤ kφ|CM∀ φ ∈ S(Rn)where if 2N > M then all the gβare bounded smooth functions thepolynomials being swamped by the decay in hxi−2N. Now, we just havea CNnorm on the right, no polynomials in it at all.Here is where Sobolev comes in, since that says|v(φ)| ≤ Ckφ|CM≤ C0kφkHM0, M0> M + n/2.Writing out the Sobolev norm explicitly we conclude that|ˆv(φ)| = |v(ˆφ)| ≤ C0kˆφ|HM0= C00khξiM0φkL2.Now consider w = hξi−2Nˆv ∈ S0(Rn) where 2N > M0, it satisfieskw(φ)k = |ˆv(hξi−2Nφ)| ≤ kφkL2so, by the Riesz representation theorem above, w ∈ L2(Rn). If we nowincrease N by


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