LECTURE NOTES FOR 18.155, FALL 2004 6310. Sobolev embeddingThe properties of Sobolev spaces are briefly discussed above. If mis a positive integer then u ∈ Hm(Rn) ‘means’ that u has up to mderivatives in L2(Rn). The question naturally arises as to the sensein which these ‘weak’ derivatives correspond to old-fashioned ‘strong’derivatives. Of course when m is not an integer it is a little harderto imagine what these ‘fractional derivatives’ are. However the mainresult is:Theorem 10.1 (Sobolev embedding). If u ∈ Hm(Rn) where m > n/2then u ∈ C00(Rn), i.e.,(10.1) Hm(Rn) ⊂ C00(Rn) , m > n/2 .Proof. By definition, u ∈ Hm(Rn) means v ∈ S0(Rn) and hξimˆu(ξ) ∈L2(Rn). Suppose first that u ∈ S(Rn). The Fourier inversion formulashows that(2π)n|u(x)| =Zeix·ξˆu(ξ) dξ≤ZRnhξi2m|ˆu(ξ)|2dξ1/2· XRnhξi−2mdξ!1/2.Now, if m > n/2 then the second integral is finite. Since the firstintegral is the norm on Hm(Rn) we see that(10.2) supRn|u(x)| = kukL∞≤ (2π)−nkukHm, m > n/2 .This is all for u ∈ S(Rn), but S(Rn) ,→ Hm(Rn) is dense. Theestimate (10.2) shows that if uj→ u in Hm(Rn), with uj∈ S(Rn),then uj→ u0in C00(Rn). In fact u0= u in S0(Rn) since uj→ u inL2(Rn) and uj→ u0in C00(Rn) both imply thatRujϕ converges, soZRnujϕ →ZRnuϕ =ZRnu0ϕ ∀ ϕ ∈ S(Rn). Notice here the precise meaning of u = u0, u ∈ Hm(Rn) ⊂ L2(Rn),u0∈ C00(Rn). When identifying u ∈ L2(Rn) with the correspondingtempered distribution, the values on any set of measure zero ‘are lost’.Thus as functions (10.1) means that each u ∈ Hm(Rn) has a represen-tative u0∈ C00(Rn).We can extend this to higher derivatives by noting that64 RICHARD B. MELROSEProposition 10.2. If u ∈ Hm(Rn), m ∈ R, then Dαu ∈ Hm−|α|(Rn)and(10.3) Dα: Hm(Rn) → Hm−|α|(Rn)is continuous.Proof. First it is enough to show that each Djdefines a continuouslinear map(10.4) Dj: Hm(Rn) → Hm−1(Rn) ∀ jsince then (10.3) follows by composition.If m ∈ R then u ∈ Hm(Rn) means ˆu ∈ hξi−mL2(Rn). SincedDju =ξj· ˆu, and|ξj| hξi−m≤ Cmhξi−m+1∀ mwe conclude that Dju ∈ Hm−1(Rn) andkDjukHm−1≤ CmkukHm. Applying this result we seeCorollary 10.3. If k ∈ N0and m >n2+ k then(10.5) Hm(Rn) ⊂ Ck0(Rn) .Proof. If |α| ≤ k, then Dαu ∈ Hm−k(Rn) ⊂ C00(Rn). Thus the ‘weakderivatives’ Dαu are continuous. Still we have to check that this meansthat u is itself k times continuously differentiable. In fact this againfollows from the density of S(Rn) in Hm(Rn). The continuity in (10.3)implies that if uj→ u in Hm(Rn), m >n2+ k, then uj→ u0in Ck0(Rn)(using its completeness). However u = u0as before, so u ∈ Ck0(Rn). In particular we see that(10.6) H∞(Rn) =\mHm(Rn) ⊂ C∞(Rn) .These functions are not in general Schwartz test functions.Proposition 10.4. Schwartz space can be written in terms of weightedSobolev spaces(10.7) S(Rn) =\khxi−kHk(Rn) .LECTURE NOTES FOR 18.155, FALL 2004 65Proof. This follows directly from (10.5) since the left side is containedin\khxi−kCk−n0(Rn) ⊂ S(Rn). Theorem 10.5 (Schwartz representation). Any tempered distributioncan be written in the form of a finite sum(10.8) u =X|α|≤m|β|≤mxαDβxuαβ, uαβ∈ C00(Rn).or in the form(10.9) u =X|α|≤m|β|≤mDβx(xαvαβ), vαβ∈ C00(Rn).Thus every tempered distribution is a finite sum of derivatives ofcontinuous functions of poynomial growth.Proof. Essentially by definition any u ∈ S0(Rn) is continuous with re-spect to one of the norms khxikϕkCk . From the Sobolev embeddingtheorem we deduce that, with m > k + n/2,|u(ϕ)| ≤ CkhxikϕkHm∀ ϕ ∈ S(Rn).This is the same ashxi−ku(ϕ)≤ CkϕkHm∀ ϕ ∈ S(Rn).which shows that hxi−ku ∈ H−m(Rn), i.e., from Proposition 9.8,hxi−ku =X|α|≤mDαuα, uα∈ L2(Rn) .In fact, choose j > n/2 and consider vα∈ Hj(Rn) defined by ˆvα=hξi−jˆuα. As in the proof of Proposition 9.14 we conclude thatuα=X|β|≤jDβu0α,β, u0α,β∈ Hj(Rn) ⊂ C00(Rn) .Thus,17(10.10) u = hxikX|γ|≤MDγαvγ, vγ∈ C00(Rn) .To get (10.9) we ‘commute’ the factor hxikto the inside; since I havenot done such an argument carefully so far, let me do it as a lemma.17This is probably the most useful form of the representation theorem!66 RICHARD B. MELROSELemma 10.6. For any γ ∈ Nn0there are polynomials pα,γ(x) of degreesat most |γ − α| such thathxikDγv =Xα≤γDγ−αpα,γhxik−2|γ−α|v.Proof. In fact it is convenient to prove a more general result. Supposep is a polynomial of a degree at most j then there exist polynomials ofdegrees at most j + |γ − α| such that(10.11) phxikDγv =Xα≤γDγ−α(pα,γhxik−2|γ−α|v) .The lemma follows from this by taking p = 1.Furthermore, the identity (10.11) is trivial when γ = 0, and proceed-ing by induction we can suppose it is known whenever |γ| ≤ L. Taking|γ| = L + 1,Dγ= DjDγ0|γ0| = L.Writing the identity for γ0asphxikDγ0=Xα0≤γ0Dγ0−α0(pα0,γ0hxik−2|γ0−α0|v)we may differentiate with respect to xj. This givesphxikDγ= −Dj(phxik) · Dγ0v+X|α0|≤γDγ−α0(p0α0,γ0hxik−2|γ−α|+2v) .The first term on the right expands to(−(Djp) · hxikDγ0v −1ikpxjhxik−2Dγ0v) .We may apply the inductive hypothesis to each of these terms andrewrite the result in the form (10.11); it is only necessary to check theorder of the polynomials, and recall that hxi2is a polynomial of degree2. Applying Lemma 10.6 to (10.10) gives (10.9), once negative powersof hxi are absorbed into the continuous functions. Then (10.8) followsfrom (10.9) and Leibniz’s formula.