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Proof We first assume that X is compact. Assume (a). Fix a ∈ X. Then the map eva :C(X)→C is continuous. Therefore, eva(F) ={f(a)|f ∈ F } is relatively compact, hence bounded in C. This implies thatF is pointwise bounded.

Let ¯F denote the closure of F in C(X). Let > 0. Then by compactness of ¯F there exists a finite collection of functions fj ∈F¯, 1 ≤ j ≤ k, such that the balls B(fj;/2) in C(X) cover ¯F. By compactness of X,each fj is uniformly continuous. Hence there exists a δ >0 such that for all x, y ∈X withd(x, y)< δ and all j we have |fj(x)−fj(y)|< /3.

Let now f ∈ F¯. Then there exists a j such that kf −fjkX < /3. It follows that for all x, y ∈X with d(x, y)< δ we have

|f(x)−f(y)| ≤ |f(x)−fj(x)|+|fj(x)−fj(y)|+|fj(y)−f(y)|

≤ 2kf −fjkX +|fj(x)−fj(y)|

< .

This shows that ¯F, and hence F, is (uniformly) equicontinuous.

For the converse, assume (b). Then it is easily seen that the closure ¯F is equicontinuous and pointwise bounded as well.

SinceC(X) is metric, it suffices to show that ¯F is sequentially compact, or, equivalently, that every sequence in ¯F has a converging subsequence.

We will first show the validity of the following claim. Let (fj) be any sequence in ¯F and > 0. Then by passing to a subsequence we may arrange that for all k, l we have kfk−flkX < .

To establish the claim, let >0. Then for every a∈X there exists a δa>0 such that for for all x∈X with d(x, a)< δa and all f ∈F¯ we have |f(x)−f(a)|< /4.

By compactness ofX,there exist finitely many pointsa1, . . . , arsuch that the open balls BX(ai, δai) cover X.Fix i.Then the sequence (fj(ai)) is bounded, hence has a converging subsequence. We see that we may replace (fj) by a subsequence to arrange that (fj(ai)) converges, for every 1 ≤ i ≤ r. Thus we may pass to yet another subsequence of (fj) to arrange that for all j, k, iwe have

|fj(ai)−fk(ai)|< /4.

Letx∈X. Select i such thatd(x, ai)< δai. Then we find that for all j, k,

|fj(x)−fk(x)| ≤ |fj(x)−fj(ai)|+|fj(ai)−fk(ai)|+|fk(ai)−fk(x)|

< 3/4.

It follows that the obtained subsequence satisfies kfj −fkkX ≤3/4< , for allj, k. This establishes the claim.

Let now (fj) be a sequence in ¯F. Applying the above claim repeatedly we obtain a sequence of subsequences

(fj)(f1,j)(f2,j) · · · such that for all k, i, j we have

kfk,i−fk,jk<2−k.

The sequence (fk,k)k∈N is a subsequence of (fj) and satisfies kfk,k −fl,lkX < 2−k for all k < l. One now readily verifies that the sequence (fk,k) is Cauchy for the sup-norm on C(X) hence converges to a function f ∈ C(X). Thus ¯F is sequentially compact, and (a) follows.

The general situation can be reduced to the present one by application of a diagonal argument, see the exercise below.

Exercise 4.6.4 Let the metric space X be locally compact and σ-compact. Let (fj) be a sequence in X such that the set F :={fj |j ∈N} is equicontinuous.

(a) Show that there exists a countable sequence (Kj) of compact subsets of X such that Kj ⊂int (Kj+1) and ∪jKj =X.

(b) Use a diagonal argument to show that (fj) has a subsequence (fjν) which converges uniformly on every set Kl,for l ∈N.

(c) Show that the sequence (fjν) converges inC(X).

(d) Complete the proof of Theorem 4.6.3.

Exercise 4.6.5 Let X be a locally compact metric space. Let Cb(X) denote the set of bounded continuous functions X →C.Equipped with the supnorm k · kX this space is a Banach space.

Let F :X →[0,∞[ be a continuous function which vanishes at infinity, i.e., for every >0 there exists a compact set K ⊂X such that F(x)< for all x∈X\K.

LetF be an equicontinuous subset ofCb(X) which is dominated byF,i.e. |f(x)| ≤F(x) for all f ∈ F and x∈X.

(a) Let (fj) be a sequence in F. Show that for every > 0 there exists a subsequence (fjk) of (fj) such that

kfjk −fjlkX <

for all k, l.

(b) Show thatF is relatively compact in Cb(X).

Pseudo-differential operators, local theory

5.1 The space of symbols

We consider a differential operator P on Rn of the form P =p(x, Dx) = X

|α|≤d

cα(x)Dαx; (5.1.1)

here we recall that Dαx = (−i∂x)α. The coefficientscα are assumed to be smooth functions onRn. The (full) symbol ofP is the functionp∈C(Rn×Rn) given by

p(x, ξ) = X

|α|≤d

cα(x)ξα.

If f ∈Cc(Rn), then F(Dαf) = ξαFf, so that by the Fourier inversion formula we have Dαf(x) =

Z

Rn

ξαfb(ξ)eiξxdξ,

where we have writtenfb=Ff.It follows that the action of P onCc(Rn) can be described by

P f(x) = Z

Rn

p(x, ξ)f(ξ)b eiξxdξ.

Pseudo-differential operators are going to be defined by the same formula, but withpfrom a larger class of spaces of functions, the so-called symbol spaces. The idea is to make the class large enough to allow a kind of division. This in turn will allow us to construct inverses to elliptic operators modulo smoothing operators, the so-called parametrices.

We return to the full symbol p of the differential operator P of degree at most d considered above. By the polynomial nature of the symbolpin theξ-variable, there exists, for every compact subset K ⊂Rn and allα, β ∈Nn,a constant C =CK,α,β >0 such that

|∂αxξβp(x, ξ)| ≤C(1 +kξk)d−|β|, ((x, ξ)∈K×Rn).

95

Exercise 5.1.1 Prove this.

These observations motivate the following definition of the space of symbols of orderd, for d a real number.

Definition 5.1.2 LetU ⊂Rnbe an open subset and let d∈R. The space of symbols on U of order at most dis defined to be the space of functions q∈C(U×Rn) such that for each compact subset K ⊂U and all multi-indices α, β, there exists a constantC =CK,α,β such that

|∂xαξβq(x, ξ)| ≤C(1 +kξk)d−|β|, ((x, ξ)∈K×Rn).

This space is denoted bySd(U).

We note that Sd(U) can be equipped with the locally convex topology induced by the seminorms

µdK,k(q) := max

|α|,|β|≤k sup

Rn

(1 +kξk)|β|−d|∂xαξβq(x, ξ)|,

for K ⊂U compact andk ∈N.Moreover, Sm(U) is a Fr´echet space for this topology.

Exercise 5.1.3 Show that d1 ≤ d2 implies Sd1(U) ⊂ Sd2(U) with continuous inclusion map.

We agree to write

S(U) =∪d∈RSd(U), S−∞(U) = ∩d∈RSd(U).

Then S−∞(U) equals the space of smooth functions f : U ×Rn → C such that for all K ⊂U compact, N ∈N and k∈N

νK,k,N(f) := max

|α|,|β|≤k sup

Rn

(1 +kξk)N|∂xξβf(x, ξ)|<∞.

Moreover, the norms νK,k,N induce a locally convex topology onS−∞(U), which turn this space into a Fr´echet space. Here we note that a function ϕ in the usual Schwartz space S(Rn) can be viewed as the function (x, ξ)7→ϕ(ξ) in S−∞(U). The corresponding natural linear map S(Rn)→S−∞(U) is a topological linear isomorphism onto the closed subspace of functions in S−∞(U) that are constant in thex-variable. More generally, if f ∈C(U) and ϕ∈ S(Rn) then the function

f⊗ϕ: (x, ξ)7→f(x)ϕ(ξ)

belongs to S−∞(U). It can be shown thatS−∞(U) is the closure of the subspace C(U)⊗ S(Rn) generated by these elements. Accordingly, we may view S−∞(U) as a topological tensor product; this is expressed by the notation

S−∞(U) =C(U)⊗ S(b Rn).

Exercise 5.1.4 Show that the following functions are symbols on U =Rn.What can be said about their orders?

(a) p(x, ξ) =kxk2(1 +kξk2)s,for s∈R. (b) p(x, ξ) = (1 +kxk2+kξk2)s, for s∈R. Exercise 5.1.5 LetU ⊂Rn be an open subset.

(a) Show that for each α∈ Nn the operator ∂xα gives a continuous linear map Sd(U) → Sd(U).

(b) Show that for each multi-indexα as above the operator ∂ξα restricts to a continuous linear map Sd(U)→Sd−|α|(U), for every d∈R.

(c) Show that the product map (p, q)7→pqrestricts to a continuous bilinear mapSd(U)×

Se(U)→Sd+e(U), for all d, e∈N. Discuss what happens if d=−∞ ore=−∞.

Exercise 5.1.6 Let U ⊂ Rn be an open subset and let P be an elliptic differential operator of order d on U. This means that its principal symbolσd(P) does not vanish on U ×(Rn\ {0}). Let p the full symbol of P. The purpose of this exercise is to show that there exists aq ∈S−d(U) such that pq−1∈S−∞(U). We first address the local question.

Let V ⊂U be an open subset with compact closure in U. We write pV for the restriction of p toV ×Rn.

(a) Show that there exists a constant R =RV >0 such that p(x, ξ)6= 0 for x∈ V and ξ∈Rn\B(0;R).

(b) Show that there exists a smooth function χV ∈ Cc(Rn) such that the function q:V ×Rn →Cdefined by

q(x, ξ) := (1−χ(ξ))p(x, ξ)−1 if p(x, ξ)6= 0 and by zero otherwise, is smooth.

(c) Withχ and q as above, show thatq ∈S−d(V).

(d) Show that pVq−1∈S−∞(V).

(e) Show that there exists a symbolq ∈S−d(U) such that pq−1∈S−∞(U).

The following invariance result will allow us to extend the definition of the symbol space to an arbitrary smooth manifold. Let ϕ:U →V be diffeomorphism between open subsets of Rn. We define the map ϕ :C(U ×Rn∗)→C(V ×Rn∗) by

ϕf(y, η) = f(ϕ−1(y), ηdϕ(ϕ−1(y))).

IdentifyingRn with its dual Rn∗ by using the standard inner product, we may view Sd(U) as a subspace of C(U×Rn∗).

Lemma 5.1.7 For every d∈R, the map ϕ restricts to a topological linear isomorphism Sd(U)→Sd(V).

Proof Put ψ = ϕ−1. Then, for f ∈ Sd(M), the function ϕf is given by ϕf(y, η) = f(ψ(y), ηdϕ(ψ(y))). The continuity of ϕ follows from checking that ∂yαηβf) satis-fies the required estimates by a straightfoward but tedious application of the chain rule combined with the Leibniz rule. Similarly,ψ is seen to be continuous linear.

We define the symbol spaces on a manifold as follows.

Definition 5.1.8 LetM be a smooth manifold and let d∈R.A symbol of orderd is de-fined to be smooth functionσ :TM →Rnsuch that for eachx0 ∈M there exists a coordi-nate patchUκ containingx0 such that the natural mapκ :C(TUκ)→C(κ(Uκ)×Rn∗) maps σ|TU to an element κσ of Sd(κ(Uκ)). The space of these symbols is denoted by Sd(M).

Remark 5.1.9 Letϕ:M →N be a diffeomorphism of smooth manfolds. Then it follows by application of Lemma 5.1.7 that the natural map ϕ : C(TM) → C(TN), given by

ϕf(ηϕ(x)) =ηϕ(x)Txϕ, (x∈M, ηϕ(x)∈Tϕ(x) N) restricts to a linear isomorphism

ϕ :Sd(M)−→' Sd(N).

In this lecture we will concentrate on the local theory of symbols and the associated pseudo-differential operators. The extension to manifolds will be rather straightforward, by using invariance and partitions of unity. In particular, one needs to localize on the x-variable in the symbol space. Accordingly, given A⊂U compact andd∈Rb we define

SAd(U) = {p∈Sd(U)|pr1(suppp)⊂A},

where pr1 : U ×Rn → Rn is the natural projection map. The union of these spaces, for A ⊂ U compact, is denoted by Scd(U). Here we note that Scd(U) ⊂ Scd(Rn) naturally, by extension by zero outside U×Rn.