SECOND ASSIGNMENT WITH SOLUTIONSWAS DUE SEPTEMBER 18 IN CLASS18.155 FALL 2001RICHARD MELROSEIn the main these questions form theorems in H¨ormander’s book [1], so the proofsare available there. I suggest that you try to work them out on your own and inany case I expect written proofs, even if you need to get the idea from the book.Of course, at the very least, you will have to translate the notation.Problem 1. [H¨ormander, Theorem 3.1.4] Let I ⊂ R be an open, non-empty interval.i) Show (you may use results from class) that there exists ψ ∈ C∞c(I) withRRψ(x)ds = 1.Solution. We showed in class that there exists φ ∈ C∞c(R) which is strictlypositive on (0, 1) and vanishes on R\(0, 1). If I is a non-empty op en intervalit contains an interval [¯x − , ¯x + ] for some ¯x ∈ R and  > 0. Considerψ(x) = Cφ((x − ¯x)/) where C > 0 is to be chosen. The properties of φimply that ψ ∈ C∞c(I) and thatRRψ = 1/C, C > 0. This value of C makethe integral of ψ equal 1 as desired. ii) Show that any φ ∈ C∞c(I) may be written in the formφ =˜φ + cψ, c ∈ C,˜φ ∈ C∞c(I) withZR˜φ = 0.Solution. Set c =RIφ and observe that˜φ = φ − cψ has vanishing integral. iii) Show that if˜φ ∈ C∞c(I) andRR˜φ = 0 then there exists µ ∈ C∞c(I) such thatdµdx=˜φ in I.Solution. If we consider µ(x) =Rx−∞˜φ(s)ds then µ is infinitely differentiablesince dµ(x)/dx =˜φ(x) by the fundamental theorem of calculus. Certainlyµ(x) = 0 if x < inf I0, where I0is a compact interval outside which˜φvanishes. The assumption thatRI˜φ(x)dx =RI0˜φ(x)dx = 0 means thatµ(x) = 0 for x > sup I0so µ ∈ C∞c(I). iv) Suppose u ∈ C−∞(I) satisfiesdudx= 0, i.e.(1) u(−dφdx) = 0 ∀ φ ∈ C∞c(I),show that u = c for some constant c.12 RICHARD MELROSESolution. Using ii) and iii) we may write each element φ ∈ C∞c(I) in theform φ = cψ + dµ(x)/dx where ψ ∈ C∞c(I) is fixed and µ ∈ C∞c(I) dependson φ. Then the definition of the vanishing of the derivative in 1 shows thatu(φ) = u(cψ + dµ/dx) = cu(ψ) − u(−dµ/dx) = cu(ψ) =ZICφ(x)dxwhere we have used the value of c and set C = u(ψ). The last integral is thedefinition of the (constant) function C as a distribution on I. Thus u = Cis an equality between distributions. v) Suppose that u ∈ C−∞(I) satisfiesdudx= c, for some constant c, show thatu = cx + d for some d ∈ C.Solution. Consider the continuous function cx. This defines a distribution,which we just denote the same way. The definition of the derivative meansthat d(cx)/dx = c (as you would expect). We proved this in class, it is theintegration by parts formulad(cx)dx(φ) = cx(−dφdx) = −ZRcxdφdxdx = c(φ).Now, if we consider the difference v = u − cx it is a distribution on I whichsatisfies dv/dx = 0, so by iv) it is equal to a constant, d. That is u−cx = d,or u = cx + d. Problem 2. [H¨ormander Theorem 3.1.16]i) Use Taylor’s formula to show that there is a fixed ψ ∈ C∞c(Rn) such thatany φ ∈ C∞c(Rn) can be written in the form(2) φ = cψ +nXj=1xjψjwhere c ∈ C and the ψj∈ C∞c(Rn) depend on φ.Solution. By Taylor’s formula, or theorem, any infinitely differentiable func-tion which vanishes at 0 can be written in the form(3) φ =nXj=1xjujwhere the ujare also infinitely differeniable. Choosing ψ ∈ C∞c(Rn), asin class, with ψ(0) = 1 we can apply this to φ − cψ where c = φ(0). Theproblem is that the ujneed not have compact support. However, we canchoose another function (depending on φ), ψ0∈ C∞c(Rn) which is equal to1 on the support of φ and the support of ψ. Thus φψ0= φ and ψψ0= ψ.Then 3 gives the desired 2 where ψj− ujψ0∈ C∞c(Rn). ii) Recall that δ0is the distribution defined byδ0(φ) = φ(0) ∀ φ ∈ C∞c(Rn);explain why δ0∈ C−∞(Rn).3Solution. Certainly δ0so defined is a linear functional and it is continuoussince|δ0(φ)| ≤ supRn|φ| = kφk0. iii) Show that if u ∈ C−∞(Rn) and u(xjφ) = 0 for all φ ∈ C∞c(Rn) and j =1, . . . , n then u = cδ0for some c ∈ C.Solution. We can use 2 to evaluateu(φ) = u(cψ) +nXj=1u(xjψj) = Cφ(0)where the given identity means that the sum is zero and we write C = u(ψ)for a fixed c onstant and note that c = φ(0). This just means u = Cδ0. iv) Define the ‘Heaviside function’H(φ) =Z∞0φ(x)dx ∀ φ ∈ C∞c(R);show that H ∈ C−∞(R).Solution. The linearity of the integral shows that H is a linear functionalon C∞c(R). The basic integral estimate|H(φ)| ≤ L sup |φ| if φ(x) = 0 in |x| > Lshows that it is continuous. v) ComputeddxH ∈ C−∞(R).Solution. By definitiondHdx(φ) = −H(dφdx) = −Z∞0dφdxdx = φ(0)where we have used the fundamental theorem of calculus. Thus(4)dHdx= δ0. Problem 3. Using Problems 1 and 2, find all u ∈ C−∞(R) satisfying the differentialequationxdudx= 0 in R.Solution. By definition v =dudxis a distribution which satisfies xv = 0. Problem 2,part iii), in one dimension shows that v = cδ0. Thusdudx= cδ0. By Problem 2, partv), dH/dx = δ. Thus ˜u = u − cH satisfies d˜u/dx = 0. By Problem 1, part iv), weconclude that v = d is constant. Thus the only distributional solutions of xdudx= 0areu = cH + d, c and d constants.This means that u is constant in x < 0 and in x > 0, with possibly different values,but it m eans m ore than this as regards its ‘be haviour at 0’. 4 RICHARD MELROSEReferences[1] L. H¨ormander, The analysis of linear partial differential operators, vol. 1, Springer-Verlag,Berlin, Heidelberg, New York, Tokyo, 1983.Department of Mathematics, Massachusetts Institute of TechnologyE-mail address: