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MIT 18 155 - Lecture Notes

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LECTURE 8, 18.155, 4 OCTOBER, 2011This week: Sobolev Embedding and Schwartz Structure Theorems.We now know that the Fourier transform ‘extends’ to an isomorphismof S0(Rn). This was done by using the formula, for a pair of functionsin S(Rn)Zˆφψ =Zφˆψto define the Fourier transform by(1) ˆu(φ) = u(ˆφ), u ∈ S0(Rn), φ ∈ S(Rn)since we know that S(Rn) 3 φ 7−→ˆφ ∈ S(Rn) is continuous (in fact itis a topological isomorphism).Then we showed that the Fourier transform restricts from S0(Rn) (onwhich it is a bijection, but we have not given any continuity statement)to a bijection on L2(Rn) which is essentially an isometric isomorphism,i.e. (2π)−n/2F is unitary on L2(Rn) – this follows from by the versionof (1)Zφψ = (2π)−nZˆφˆψ ∀ φ, ψ ∈ S(Rn).Next we showed that the L2based Sobolev spaces are readily char-acterized in terms of the Fourier transforms of their elements:u ∈ Hk(Rn) ⇐⇒ hξikˆu ∈ L2.Here k ∈ N but we can use this as a definition of Hk(Rn) for all k ∈ Rwhere I usually replace the integer-looking k by s or m. Then it followsthat these spaces decrease with increasing orderHs(Rn) ⊂ Hs0(Rn) ⇐⇒ s ≥ s0which may be unfortunate but that is the way things are.From this we easily deduce a form of ‘weak=strong’ for derivatives.Namely, u ∈ L2(Rn) is said to have a strong derivative, v = ∂ju if thereis a sequence un∈ S(Rn) (or C∞c(Rn) it is the same in the end) suchthat uk→ u in L2and ∂juk→ v in L2(Rn). Such a strong derivative(of any order) is automatically a weak derivative since we can integrateby parts before taking the limit sou(−∂jφ) = limk→∞uk(−∂jφ) = limk→∞(∂juk)(φ) = v(φ).12 LECTURE 8, 18.155, 4 OCTOBER, 2011In particular if a strong derivative exists then it is unique since weakderivatives are already unique.Proposition 1 (weak=strong). If k ∈ N then S(Rn) ⊂ Hk(Rn) isdense (in the norm on Hk) and each u ∈ Hk(Rn) has strong derivativesor all orders |α| ≤ k.Proof. Take u ∈ Hk(Rn), so ˆu ∈ L2(Rn) and ˆu(ξ) = hξi−kw, w ∈L2(Rn) by the result above. Since S(Rn) is dense in L2we can choosea sequence wl→ w in L2. Then choose ul∈ S(Rn) by ˆul= hξi−kwkwhere we use the fact that h·i−kis a multiplier on S and that the Fouriertransform is an isomorphism. The norm on u ∈ Hkis equivalent tokwkL2when u and w are related as above, so wl→ w in L2impliesul→ u in Hk(Rn) and we have the density.Since ξαˆul= ξαhξh−kwl, the same argument shows that ∂αulcon-verges in L2, so strong derivatives exist for |α| ≤ k. Of course if s ≥ k then u ∈ Hs(Rn) has strong derivatives of anyorder up to k – but what about the ‘extra’ s − k ‘derivatives’? We willget to this later.From this discussion the basic result relating ‘L2-derivatives’ to or-dinary derivatives also follows.Theorem 1 (Sobolev embedding). If k ∈ N thenHs(Rn) ⊂ Ck0(Rn) if s >n2+ k.This result is sharp. The spaces on the right are the spaces of functionswith k continuous, bounded derivatives each vanishing at infinity.Proof. The main point here is that hξi−sL2(Rn) ⊂ L1(Rn) if (and onlyif) s >n2. By this I mean the space of those v ∈ S0(Rn) such thatv = h·i−sw, w ∈ L2. Indeed, this is just the fact that h·i−p∈ L1(Rn) iffp > n. So, if we apply the Cauchy-Schwarz inequality,Z|v| =Zhξi−s|w| ≤ ki·i−2skL2kwkL2the result follows. Well, we really need to check that the integral makessense first! This however is the usual argument by continuity – takean approximating sequence wn→ w with wn∈ C0c(Rn) for which thecomputation does make sense, and hence deduce that vn= h·i−swn→ va.e. is an approximating sequence in L1, so v ∈ L1and the estimateholds.Now, we apply this to v = ˆu, where by assumption u ∈ Hs(Rn)means that w(ξ) = hξisˆu ∈ L2(Rn). Thus, we see that v ∈ L1and so,LECTURE 8, 18.155, 4 OCTOBER, 2011 3as you showed in the last problem set,u = Gˆu ∈ C00(Rn), if u ∈ Hs(Rn), s >n2.For higher derivatives, we just need to use density an iteration asusual. Take a sequence as in the preceding result, ul→ u in Hs(Rn)with ul∈ S(Rn). Then, ∂αulis a convergent sequence in Hs−|α|(Rn), soif s − |α| >n2it follows that ∂αul→ ∂αu ∈ C00(Rn) where the notationis for the weak (or strong) derivative of u. That is, we conclude thatthe weak derivatives up to order k are continuous. In fact we have beenthrough these arguments before, the fact that the sequence convergeswith respect to the Cknorm (supremum of derivatives up to order k)means the limit is in Ck0(Rn) – one just needs to quote the completenessof Ck0(Rn). Now, we ‘know’ that the Fourier transform exchanges regularity andgrowth. Let’s look at another (extreme) example of this.Definition 1. An element u ∈ S0(Rn) is said to be of compact support(or have compact support, or be compactly supported) if there existsR > 0 such thatu(φ) = 0 ∀ φ ∈ S(Rn) s.t. φ(x) = 0 in |x| < R.We will discuss the notion of the support of a distribution soon.Another way of looking at this property is to say that there exists χ ∈C∞c(Rn) such that χu = u. Indeed, take χ ∈ C∞c(Rn) to be identicallyequal to 1 in say |x| < R + 1 and to vanish identically in |x| > R + 2.Then if φ ∈ S(Rn) it follows that (1−χ)φ = 0 in |x| < R and (1−χ)φ ∈S(Rn) so u((1 − χ)φ) = 0 for all φ ∈ S(Rn) which is u(φ) = (χu)(φ)for all φ ∈ S(Rn). Conversely, if χu = u with χ ∈ C∞c(Rn) it followsχ(x) = 0 in |x| > R − 1 for some R and then if φ ∈ S(Rn), and φ = 0in |x| < R it follows that χφ ≡ 0 and hence χu(φ) = u(χφ) = 0 whichis the the definition.One of the Assignment Questions is: Discuss the Paley-Wiener-Schwartz Theorem. Here is an extremely watered-down version (thetrue version characterizes the Fourier transforms of compactly sup-ported distributions in terms of entire functions of several complexvariables).Lemma 1. If u ∈ S0(Rn) has compact support then ˆu ∈ C∞(Rn) is afunction of slow growth.Proof. The important properties of a distribution of compact supportare (duh!) it is a distribution and it has compact support. The first4 LECTURE 8, 18.155, 4 OCTOBER, 2011means that there are constants C and N such that|u(φ)| ≤ CkφkN= CX|α|+|β|≤Nsup |xα∂βφ|.The second means u = χu so u(φ) = u(χφ) and we see what happensto the right side when φ is replaced by χφ :sup |xα∂β(χφ)| ≤ CX|β|≤Nsup|x|≤T|∂βφ|where T is


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MIT 18 155 - Lecture Notes

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