34 RICHARD B. MELROSE6. Test functionsSo far we have largely been dealing with integration. One thing wehave seen is that, by considering dual spaces, we can think of functionsas functionals. Let me briefly review this idea.Consider the unit ball in Rn,Bn= {x ∈ Rn; |x| ≤ 1} .I take the closed unit ball because I want to deal with a compact metricspace. We have dealt with several Banach spaces of functions on Bn,for exampleC(Bn) =u : Bn→ C ; u continuousL2(Bn) =u : Bn→ C; Borel measurable withZ|u|2dx < ∞.Here, as always below, dx is Lebesgue measure and functions are iden-tified if they are equal almost everywhere.SinceBnis compact we have a natural inclusion(6.1) C(Bn) ,→ L2(Bn) .This is also a topological inclusion, i.e., is a bounded linear map, since(6.2) kukL2≤ Cku||∞where C2is the volume of the unit ball.In general if we have such a set up thenLemma 6.1. If V ,→ U is a subspace with a stronger norm,kϕkU≤ CkϕkV∀ ϕ ∈ Vthen restriction gives a continuous linear map(6.3) U0→ V0, U03 L 7−→˜L = L|V∈ V0, k˜LkV0≤ CkLkU0.If V is dense in U then the map (6.3) is injective.Proof. By definition of the dual normk˜LkV0= supn˜L(v); kvkV≤ 1 , v ∈ Vo≤ supn˜L(v); kvkU≤ C , v ∈ Vo≤ sup {|L(u)| ; kukU≤ C , u ∈ U}= CkLkU0.If V ⊂ U is dense then the vanishing of L : U → C on V implies itsvanishing on U. LECTURE NOTES FOR 18.155, FALL 2004 35Going back to the particular case (6.1) we do indeed get a continuousmap between the dual spacesL2(Bn)∼=(L2(Bn))0→ (C(Bn))0= M(Bn) .Here we use the Riesz representation theorem and duality for Hilbertspaces. The map use here is supposed to be linear not antilinear, i.e.,(6.4) L2(Bn) 3 g 7−→Z·g dx ∈ (C(Bn))0.So the idea is to make the space of ‘test functions’ as small as reasonablypossible, while still retaining density in reasonable spaces.Recall that a function u : Rn→ C is differentiable at x ∈ Rnif thereexists a ∈ Cnsuch that(6.5) |u(x) − u(x) − a · (x − x)| = o(|x − x|) .The ‘little oh’ notation here means that given  > 0 there exists δ > 0s.t.|x − x| < δ ⇒ |u(x) − u(x) − a(x − x)| <  |x − x| .The coefficients of a = (a1, . . . , an) are the partial derivations of u atx,ai=∂u∂xj(x)since(6.6) ai= limt→0u(x + tei) − u(x)t,ei= (0, . . . , 1, 0, . . . , 0) being the ith basis vector. The function u issaid to be continuously differentiable on Rnif it is differentiable at eachpoint x ∈ Rnand each of the n partial derivatives are continuous,(6.7)∂u∂xj: Rn→ C .Definition 6.2. Let C10(Rn) be the subspace of C0(Rn) = C00(Rn) suchthat each element u ∈ C10(Rn) is continuously differentiable and∂u∂xj∈C0(Rn), j = 1, . . . , n.Proposition 6.3. The functionkukC1= kuk∞+nXi=1k∂u∂x1k∞is a norm on C10(Rn) with respect to which it is a Banach space.36 RICHARD B. MELROSEProof. That k kC1is a norm follows from the properties of k k∞. NamelykukC1= 0 certainly implies u = 0, kaukC1= |a| kukC1and the triangleinequality follows from the same inequality for k k∞.Similarly, the main part of the completeness of C10(Rn) follows fromthe completeness of C00(Rn). If {un} is a Cauchy sequence in C10(Rn)then unand the∂un∂xjare Cauchy in C00(Rn). It follows that there arelimits of these sequences,un→ v ,∂un∂xj→ vj∈ C00(Rn) .However we do have to check that v is continuously differentiable andthat∂v∂xj= vj.One way to do this is to use the Fundamental Theorem of Calculusin each variable. Thusun(x + tei) =Zt0∂un∂xj(x + sei) ds + un(x) .As n → ∞ all terms converge and so, by the continuity of the integral,u(x + tei) =Zt0vj(x + sei) ds + u(x) .This shows that the limit in (6.6) exists, so vi(x) is the partial deriva-tion of u with respect to xi. It remains only to show that u is indeeddifferentiable at each point and I leave this to you in Problem 17. So, almost by definition, we have an example of Lemma 6.1,C10(Rn) ,→ C00(Rn).It is in fact dense but I will not bother showing this (yet). So we knowthat(C00(Rn))0→ (C10(Rn))0and we expect it to be injective. Thus there are more functionals onC10(Rn) including things that are ‘more singular than measures’.An example is related to the Dirac deltaδ(x)(u) = u(x) , u ∈ C00(Rn) ,namelyC10(Rn) 3 u 7−→∂u∂xj(x) ∈ C .This is clearly a continuous linear functional which it is only just todenote∂∂xjδ(x).Of course, why stop at one derivative?LECTURE NOTES FOR 18.155, FALL 2004 37Definition 6.4. The space Ck0(Rn) ⊂ C10(Rn) k ≥ 1 is defined induc-tively by requiring that∂u∂xj∈ Ck−10(Rn) , j = 1, . . . , n .The norm on Ck0(Rn) is taken to be(6.8) kukCk = kukCk−1 +nXj=1k∂u∂xjkCk−1 .These are all Banach spaces, since if {un} is Cauchy in Ck0(Rn), it isCauchy and hence convergent in Ck−10(Rn), as is ∂un/∂xj, j = 1, . . . , n−1. Furthermore the limits of the ∂un/∂xjare the derivatives of the limitsby Proposition 6.3.This gives us a sequence of spaces getting ‘smoother and smoother’C00(Rn) ⊃ C10(Rn) ⊃ · · · ⊃ Ck0(Rn) ⊃ · · · ,with norms getting larger and larger. The duals can also be expectedto get larger and larger as k increases.As well as looking at functions getting smoother and smoother, weneed to think about ‘infinity’, since Rnis not compact. Observe thatan element g ∈ L1(Rn) (with respect to Lebesgue measure by default)defines a functional on C00(Rn) — and hence all the Ck0(Rn)s. However afunction such as the constant function 1 is not integrable on Rn. Sincewe certainly want to talk about this, and polynomials, we consider asecond condition of smallness at infinity. Let us set(6.9) hxi = (1 + |x|2)1/2a function which is the size of |x| for |x| large, but has the virtue ofbeing smooth10Definition 6.5. For any k, l ∈ N = {1, 2, · · · } sethxi−lCk0(Rn) =u ∈ Ck0(Rn) ; u = hxi−lv , v ∈ Ck0(Rn),with norm, kukk,l= kvkCk , v = hxilu.Notice that the definition just says that u = hxi−lv, with v ∈ Ck0(Rn).It follows immediately that hxi−lCk0(Rn) is a Banach space with thisnorm.Definition 6.6. Schwartz’ space11of test functions on RnisS(Rn) =u : Rn→ C; u ∈ hxi−lCk0(Rn) for all k and l ∈ N.10See Problem 18.11Laurent Schwartz – this one with a ‘t’.38 RICHARD B. MELROSEIt is not immediately apparent that this space is non-empty (well 0is in there but...); thatexp(− |x|2) ∈ S(Rn)is Problem 19. There are lots of other