LECTURE 7, 18.155, 29 SEPTEMBER 2011(1) Last time I defined the Sobolev spaces of integral order in termsof weak derivatives. To understand the L2based spaces,Hk(Rn) = {u ∈ L2(Rn); ∂αu ∈ L2(Rn) ∀ |α| ≤ k}we will use the Fourier transform. Just as with derivatives lasttime, this will be defined ‘weakly’ (so the definition above says– u in L2,→ S0(Rn) and its derivatives, ∂αu ∈ S0(Rn) happento be in (the image of) L2⊂ S0(Rn). Let me first remind youof the process.(2) We showed earlier (I think you did all the work) that multi-plication by functions of slow growth gives a map on S(Rn)– if f ∈ C∞(Rn) and for each k there exists N such that|∂α| ≤ CN(1 + |x|)Nfor all |α| ≤ k then×f : S(Rn) −→ S(Rn).We can see that this has a ‘weak definition’ as well. Namely, ifwe regard ψ ∈ S(Rn) as a distribution, denoted for the momentUψ(φ) =Zψφthen we get the identityUfψ(φ) =Zfψφ = Uψ(fφ).We want to extend multiplication by f to S0(Rn), consistentwith its action on S(Rn), so we want fUψ= Ufψand then wesee how to do it in general, just definefu(φ) = u(fφ) ∀ u ∈ S0(Rn), φ ∈ S(Rn).This makes sense since fu ∈ S0(Rn), because of the continuityon S of multiplication by f. Clearly this gives a linear map fromS0(Rn) to itself.All trivial, but we do the same thing for the Fourier transform(3) So we need to get a ‘weak’ formulation of the Fourier transformon S(Rn). What isˆψ as a distribution? For one last time letsdenote it Uˆψand compute away:Uˆψ(φ) =Zˆψφ =Zˆψ(ξ)φ(ξ).12 LECTURE 7, 18.155, 29 SEPTEMBER 2011Now, we can substitute the definition ofˆψ and not have toworry too much about convergence, since everything is in S :ˆψ(ξ) =Ze−ix·ξψ(x)dx =⇒Zˆψφ =Z Ze−ix·ξψ(x)φ(ξ)dxdξ.We can exchange the order of integration without much trouble(do a limiting argument if you are not sure) and conclude thatZˆψφ =Zψ(x)Ze−ix·ξφ(ξ)dξdx =Zψˆφ.This is the weak formulation of the Fourier transform we want,it says thatUˆψ(φ) = Uψ(ˆφ).Thus, we can extend the Fourier transform to all of S0(Rn) bydefining(1) ˆu(φ) = u(ˆφ) ∀ φ ∈ S(Rn).Proposition 1. The Fourier transform of tempered distribu-tions, defined by (1) is a bijection on S0(Rn).Proof. First observe that if u ∈ S0(Rn) then (1) does indeeddefine an element ˆu ∈ S0(Rn). In fact ˆu = u◦F : S(Rn) −→ C isthe composite of continuous linear maps, so is itself continuousan linear hence an element of S0(Rn). The fact that F is abijection on S(Rn) means that Fu = ˆu defines a bijection –since ˆu = 0 means u(ˆφ) = 0 for all φ and hence u(ψ) = 0 forall ψ ∈ S(Rn). So it is an injective linear map and its inverse isgiven by the inverse of the Fourier transform on S(Rn),v = ˆu =⇒ u = Gv, Gv(φ) = v(Gφ) ∀ v ∈ S0(Rn), φ ∈ S(Rn).This follows from the identityGˆu(φ) = ˆu(Gφ) = u(cGφ) = u(φ). (4) For the moment I will not discuss continuity of the Fouriertransform, but it is in fact continuous in the natural topologieson S0(Rn).We do want to observe that the relationship of derivationand Fourier transform, which we found on S(Rn) carries overto S0(Rn) :d∂αu = i|α|ξαˆu,dxβu = (−i)|β|∂βˆu.LECTURE 7, 18.155, 29 SEPTEMBER 2011 3These follow from the weak definition; for instanced∂αu(φ) = (∂αu)(ˆφ) = u((−1)|α|∂αˆφ)= (−1)|α|u((−i)αdξαu) = i|α|ˆu(ξαφ) = i|α|ξαˆu(φ)using the identities on S in the middle.(5) Next we consider the restriction of F from S0(Rn) to L2(Rn); infact we will discuss this by extension from S(Rn). Going backto the identity (2), or (3), we can set φ = ˆµ where µ ∈ S(Rn)soZˆψφ =Zˆψˆµ.Now, µ is the inverse Fourier transform of its Fourier transform:-µ(x) = (2π)−nZeix·ξˆµ(ξ)dξ.Taking complex conjutates we see thatµ(x) = (2π)−nZe−ix·ξˆµ(ξ)dξ =⇒ µ(x) = (2π)−nbµ = (2π)−nˆφ.Substituting this in (3) gives the Plancherel/Parseval identity(I am hopelessly confused about which is which)(2)Zˆφˆµ = (2π)nZψµ ∀ φ, µ ∈ S(Rn).Now, from this it follows thatProposition 2. The Fourier transform extends by continuity(using the density of S(Rn) in L2(Rn) to an essentially isomet-ric isomorphism F : L2(Rn) −→ L2(Rn);(3) kFukL2= (2π)−n/2kukL2.Proof. The norm identity follows from the discussion abovewhen u ∈ S(Rn). Then if S(Rn) 3 uk→ u in L2(Rn) it fol-lows, applying the identity to the norm of differences, that thesequence Fukis Cauchy, and hence convergent, in L2(Rn) and(3) holds for the limit – from which it follows that F : L2−→ L2is well-defined and satisfies this identity. It is necessarily injec-tive (since Fuk = 0 implies u = 0) and has inverse given by Gto which the same extension discussion applies. (6) So, now we have a simple ‘characterization’ of the image underthe Fourier transform of L2(Rn) ⊂ S0(Rn), namely it is L2(Rn)again. Thus if you believe that F ‘interchanges growth andregularity’ then L2has the same growth and regularity! We can4 LECTURE 7, 18.155, 29 SEPTEMBER 2011use this basic result to get a rather useful characterization ofthe Fourier transforms of the L2-based Sobolev spaces Hk(Rn)for k ∈ N.Before doing so, let’s recall some simple estimates. The usualnotation |x| for x ∈ Rnis the Euclidean norm, |x|2= x21+ · · · +x2n. We could also consider the ‘l1’ norm |x1| + · · · + |xn|. Asnorms on finite dimensional vector spaces these are equivalent,namely(4)1n(|x1| + · · · + |xn|) ≤ |x| ≤ |x1| + · · · + |xn|.I also introduced the notatiohxi = (1 + |x|2)12which has the virtue of being a smooth function. In fact this,and its powers, satisfy symbol estimates(5) ∂αxhxhm≤ Cm,αhxim−|α|.The point to note is that the three functions(6) (1 + |x|), (1 + |x1| + |x2| + · · · + |xn|), hxhare all the ‘same size’ on Rn– each is bounded by a positivemultiple of the other two, and hence above an below and thesame is true for any fixed power m (the same for all three ofcourse). You can check this from (4) and12(1 + |x|)2≤ (1 + |x|2) ≤ (1 + |x|)2.Since the quotients are bounded continuous functions for u ∈L2and m > 0 the three conditions(7) (1 + |x1| + · · · + |xn|)mu ∈ L2, (1 + |x|)mu ∈ L2, hximu ∈ L2are all equivalent.As already noted the advantage of the last one is that the‘multiplier’ is smooth and of slow growth so hximu ∈ S0(Rn) forany u ∈ S0(Rn) makes sense, an observation which will be usedbelow.(7) Now