CHAPTER 3 Ellipticity and boundedness

So, what did we in the previous chapter? We showed that there is an ‘oscillatory integral’ interpretation of

K(x, y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, y, ξ)dξ = I(a) (3.1)

which defines, for any p, a continuous linear map

I (1 + |x − y|^{2})^{\frac{1}{2}} S_{∞}^{∞}(ℝ^{2n}; ℝ^{n}) ⟶ S′(ℝ^{2n})

the range of which is the space of pseudodifferential operators on ℝ^{n};

\begin{matrix}A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟺ A : S(ℝ^{n}) ⟶ S′(ℝ^{n}) and \\ ∃ p s.t. K_{A}(x, y) = I(a), a ∈ (1 + |x − y|^{2})^{p/2}S_{∞}^{m} (ℝ^{2n}; ℝ^{n}) . \end{matrix} (3.2)

We then showed in Proposition 2.2 that the special case, p = 0 and ∂_{y}a ≡ 0, gives an isomorphism

Ψ_{∞}^{m}(ℝ^{n}) ⟵_{q_{L}}^{⟶^{σ_{L}}} S_{∞}^{m}(ℝ^{n}; ℝ^{n}). (3.3)

The maps here, q_{L} = I, is the left quantization map and its inverse σ_{L} is the left symbol map. We also showed that Ψ_{∞}^{∞}(ℝ^{n}) forms a filtered ∗-algebra. Next shall consider invertibility, modulo Ψ_{∞}^{−∞}(ℝ^{n}). First we consider some more consequences of the reduction theorem.

3.1. Quantization and symbols

We have already noted that as well as the left quantization map leading to the isomorphism (3.3) there is a right quantization map, similarly derived from (3.1):

q_{R}(a) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(y, ξ)dξ, a ∈ S_{∞}^{m} (ℝ^{n}; ℝ^{n}) . (3.4)

In fact using the adjoint operator, ∗, on operators and writing as well ∗ for complex conjugation of symbols shows that

q_{R} = ∗ · q_{L} · ∗ (3.5)

is also an isomorphism, with inverse σ_{R}

Ψ_{∞}^{m}(ℝ^{n}) ⟵_{q_{R}}^{⟶^{σ_{R}}} S_{∞}^{m} (ℝ^{n}; ℝ^{n}) . (3.6)

These are the two “extreme” quantization procedures, see Problem 3.1 for another (more centrist). Using the proof of the reduction theorem we find:

Lemma 3.1. For any a ∈ S_{∞}^{m} (ℝ^{n}; ℝ^{n}) ,

σ_{L} (q_{R}(a)) (x, ξ) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{x}^{α}D_{ξ}^{α}a(x, ξ). (3.7)

Proof. This follows from the general formula (2.59).

One important thing to note from (3.7) is that

D_{x}^{α}D_{ξ}^{α}a(x, ξ) ∈ S_{∞}^{m−|α|} (ℝ^{n}; ℝ^{n})

so that for any pseudodifferential operator

A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ σ_{L}(A) − σ_{R}(A) ∈ S_{∞}^{m−1} (ℝ^{n}; ℝ^{n}) .

For this reason we consider the general quotient spaces

S_{∞}^{m−[1]} (ℝ^{p}; ℝ^{n}) = S_{∞}^{m} (ℝ^{p}; ℝ^{n}) /S_{∞}^{m−1} (ℝ^{p}; ℝ^{n})

and, for a ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n}), write [a] for its image, i.e. equivalence class, in the quotient space S_{∞}^{m−[1]} (ℝ^{p}; ℝ^{n}) . The “principal symbol map”

σ_{m} : Ψ_{∞}^{m}(ℝ^{n}) ⟶ S_{∞}^{m−[1]} (ℝ^{n}; ℝ^{n})
is defined by σ_{m}(A) = [σ_{L}(A)] = [σ_{R}(A)].

(3.8)

As distinct from σ_{L} or σ_{R}, σ_{m} depends on m, i.e. one needs to know that the order is at most m before it is defined.

The isomorphism (3.3) is replaced by the weaker exact sequence.

Lemma 3.2. For every m ∈ ℝ

0 ↪ Ψ_{∞}^{m−1}(ℝ^{n}) ↪ Ψ_{∞}^{m}(ℝ^{n}) ⟶^{σ_{m}} S_{∞}^{m−[1]} (ℝ^{n}; ℝ^{n}) ⟶ 0

is a short exact sequence (the ‘principal symbol sequence’ or simply the ‘symbol sequence’).

Proof. This is just the statement that the range of each map is the null space of the next i.e. that σ_{m} is surjective, which follows from (3.3), and that the null space of σ_{m} is just Ψ_{∞}^{m−1}(ℝ^{n}) and this is again (3.3) and the definition of σ_{m}.

The fundamental result proved last time is that

Ψ_{∞}^{m}(ℝ^{n}) · Ψ_{∞}^{m′}(ℝ^{n}) ⊂ Ψ_{∞}^{m+m′}(ℝ^{n}). (3.9)

In fact we showed that if A = q_{L}(a), a ∈ S_{∞}^{m} (ℝ^{n}; ℝ^{n}) and B = q_{R}(b), b ∈ S_{∞}^{m′} (ℝ^{n}; ℝ^{n}) then the composite operator has Schwartz kernel

K_{A·B}(x, y) = I (a(x, ξ)b(y, ξ))

Using the formula (2.59) again we see that

σ_{L}(A · B) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{ξ}^{α}[a(x, ξ)D_{x}^{α}b(x, ξ)]. (3.10)

Of course b = σ_{R}(B) so we really want to rewrite (3.10) in terms of σ_{L}(B).

Lemma 3.3. If A ∈ Ψ_{∞}^{m}(ℝ^{n}) and B ∈ Ψ_{∞}^{m′}(ℝ^{n}) then A ∘ B ∈ Ψ_{∞}^{m+m′}(ℝ^{n}) and

σ_{m+m′}(A ∘ B) = σ_{m}(A) · σ_{m′}(B), (3.11)

σ_{L}(A ∘ B) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{ξ}^{α}σ_{L}(A) · D_{x}^{α}σ_{L}(B). (3.12)

Proof. The simple formula (3.11) is already immediate from (3.10) since all terms with |α| ≥ 1 are of order m+m′ −|α| ≤ m+m′ −1. To get the “full” formula (3.12) we can insert into (3.10) the inverse of (3.7), namely

σ_{R}(x, ξ) ∼ ∑_{α} \frac{(−i)^{|α|}}{α!} D_{ξ}^{α}D_{x}^{α}σ_{L}(x, ξ).

This gives the double sum (still asymptotically convergent)

σ_{L}(A ∘ B) ∼ ∑_{β} ∑_{α} \frac{(−i)^{|α|}}{α!} D_{ξ}^{α}[σ_{L}(A)D_{x}^{α} \frac{i^{_{|}β|}}{β!} D_{x}^{β}D_{ξ}^{β}σ_{L}(B)]

Setting γ = α + β this becomes

σ_{L}(A ∘ B) ∼ ∑_{γ} \frac{i^{|γ|}}{γ!} ∑_{0≤α≤γ} \frac{γ!(−1)^{|γ−α|}}{α!(γ − α)!} D_{ξ}^{α}[σ_{L}(A) × D_{ξ}^{γ−α}D_{x}^{γ}σ_{L}(B)].

Then Leibniz formula shows that this sum over α can be rewritten as

σ_{L}(A ∘ B) ∼ ∑_{γ} \frac{i^{|γ|}}{γ!} D_{ξ}^{γ}σ_{L}(A) · D_{x}^{γ}σ_{L}(B),

this is just (3.12).

The simplicity of (3.11) over (3.12) is achieved at the expense of enormous loss of information. Still many problems can be solved using (3.11) which we can think of as saying that the principal symbol maps give a homomorphism for instance from the filtered algebra Ψ_{∞}^{0}(ℝ^{n}) to the commutative algebra S_{∞}^{0−[1]} (ℝ^{n}; ℝ^{n}) .

3.2. Ellipticity

We say that an element of Ψ_{∞}^{m}(ℝ^{n}) is elliptic if it is invertible modulo an error in Ψ_{∞}^{−∞}(ℝ^{n}) with the approximate inverse of order −m i.e.

A ∈ Ψ_{∞}^{m}(ℝ^{n}) is elliptic

⟺ ∃ B ∈ Ψ_{∞}^{−m}(ℝ^{n}) s.t. A ∘ B − Id ∈ Ψ_{∞}^{−∞}(ℝ^{n}). (3.13)

Thus ellipticity, here by definition, is invertibility in Ψ_{∞}^{m}(ℝ^{n})/Ψ_{∞}^{−∞}(ℝ^{n}), with the inverse in Ψ_{∞}^{−m}(ℝ^{n})/Ψ_{∞}^{−∞}(ℝ^{n}). The point about ellipticity is that it is a phenomenon of the principal symbol.

Theorem 3.1. The following conditions on A ∈ Ψ_{∞}^{m}(ℝ^{n}) are equivalent

A is elliptic (3.14)

∃ [b] ∈ S_{∞}^{−m−[1]} (ℝ^{n}; ℝ^{n}) s.t. σ_{m}(A) · [b] ≡ 1 in S_{∞}^{0−[1]} (ℝ^{n}; ℝ^{n}) (3.15)

∃ b ∈ S_{∞}^{−m} (ℝ^{n}; ℝ^{n}) s.t. σ_{L}(A) · b − 1 ∈ S_{∞}^{−∞} (ℝ^{n}; ℝ^{n}) (3.16)

∃ ϵ > 0 s.t. | σ_{L}(A)(x, ξ) |≥ ϵ(1 + |ξ|)^{m} in |ξ| > \frac{1}{ϵ}. (3.17)

Proof. We shall show

(3.14) ⟹ (3.15) ⟹ (3.16) ⟺ (3.17) ⟹ (3.14). (3.18)

Since we know that σ_{0}(Id) = 1 applying the identity (3.11) to the definition of ellipticity in (3.13) gives

σ_{m}(A) · σ_{−m}(B) ≡ 1 in S_{∞}^{0−[1]}(ℝ^{n}, ℝ^{n}), (3.19) i.e. that (3.14) ⟹ (3.15).

Now assuming (3.19) (i.e. (3.15)), and recalling that σ_{m}(A) = [σ_{L}(A)] we find that a representation b_{1} of the class [b] must satisfy

σ_{L}(A) · b_{1} = 1 + e_{1}, e_{1} ∈ S_{∞}^{−1}(ℝ^{n}; ℝ^{n}), (3.20)

this being the meaning of the equality of residue classes. Now, since the remainder e_{1} ∈ S_{∞}^{−1}(ℝ^{n}; ℝ^{n}) the Neumann series

f ∼ ∑_{j≥1}(−1)^{j}e_{1}^{j}

is asymptotically convergent, so f ∈ S_{∞}^{0}(ℝ^{n}; ℝ^{n}) exists, and

(1 + f) · (1 + e_{1}) = 1 + e_{∞}, e_{∞} ∈ S_{∞}^{−∞}(ℝ^{n}; ℝ^{n}). (3.21)

Then multiplying (3.21) by (1 + f) gives

σ_{L}(A) · {b_{1}(1 + f)} = 1 + e_{∞} (3.22)

which proves (3.16), since b = b_{1}(1 + f) ∈ S_{∞}^{−∞}(ℝ^{n}; ℝ^{n}). Of course

sup(1 + |ξ|)^{N}|e_{∞}| < ∞ ∀ N

so

∃ C s.t. |e_{∞}(x, ξ)| < \frac{1}{2} in |ξ| > C.

From (3.22) this means

| σ_{L}(A)(x, ξ) | · | b(x, ξ) |≥ \frac{1}{2}, |ξ| > C. (3.23)

Since |b(x, ξ)| ≤ C(1 + |ξ|)^{−m} (being a symbol of order −m), (3.23) implies

inf_{|ξ|≥C} | σ_{L}(A)(x, ξ) | (1 + |ξ|)^{−m} ≤ C. (3.24)

which shows that (3.16) implies (3.17).

Conversely, suppose (3.24) holds, i.e. (3.17) holds with C = 1/ϵ. Then if ϕ ∈ C_{c}^{∞}(ℝ^{n}), has ϕ(ξ) = 1 when |ξ| ≤ 1, and ϕ(ξ) = 0 when |ξ| ≥ 2,

\frac{(1 − ϕ(ξ/C))}{σ_{L}(A)(x, ξ)} = b(x, ξ) ∈ S_{∞}^{−m}(ℝ^{n}; ℝ^{n}) (3.25)

as follows from (2.35). This gives (3.16), so (3.17) implies (3.16).

Now suppose (3.16) holds. Set B_{1} = q_{L}(b) then from (3.11) again

σ_{0}(A ∘ B_{1}) = [q_{m}(A)] · [b] ≡ 1.

That is,

A ∘ B_{1} − Id = E_{1} ∈ Ψ_{∞}^{−1}(ℝ^{n}). (3.26)

Consider the Neumann series of operators

∑_{j≥0}(−1)^{j}E_{1}^{j}.

The corresponding series of symbols is asymptotically summable so we can choose F ∈ Ψ_{∞}^{−1}(ℝ^{n}) with

σ_{L}(F) ∼ ∑_{j≥0}(−1)^{j}σ_{L}(E_{1}^{j}). (3.27)

Then

(Id +E_{1})(Id +F) = Id +E_{∞}, E_{∞} ∈ Ψ_{∞}^{−∞}(ℝ^{n}). (3.28)

Then B = B_{1}(Id +F) ∈ Ψ_{∞}^{−m}(ℝ^{n}) satisfies (3.13), i.e. A is elliptic.

In the definition of ellipticity in (3.13) we have taken B to be a ‘right parametrix’, i.e. a right inverse modulo Ψ_{∞}^{−∞}(ℝ^{n}). We can just as well take it to be a left parametrix.

Lemma 3.4. A ∈ Ψ_{∞}^{m}(ℝ^{n}) is elliptic if and only if there exists B′ ∈ Ψ_{∞}^{−m}(ℝ^{n}) such that

B′ ∘ A = Id +E′, E′ ∈ Ψ_{∞}^{−∞}(ℝ^{n}) (3.29)

and then if B satisfies (3.13), B − B′ ∈ Ψ_{∞}^{−∞}(ℝ^{n}).

Proof. Certainly (3.29) implies σ_{−m}(B′) · σ_{m}(A) ≡ 1, and the multiplication here is commutative so (3.15) holds and A is elliptic. Conversely if A is elliptic we get in place of (3.26)

B_{1} ∘ A − Id = E′_{1} ∈ Ψ_{∞}^{−1}(ℝ^{n}).

Then defining F′ as in (3.27) with E′_{1} in place of E_{1} we get (Id +F′)(Id +E′_{1}) = Id +E′_{∞} and then B′ = (Id +F′) ∘ B_{1} satisfies (3.29). Thus ‘left’ ellipticity as in (3.29) is equivalent to right ellipticity. Applying B to (3.29) gives

B′ ∘ (Id +E) = B′ ∘ (A ∘ B) = (Id +E′) ∘ B (3.30)

which shows that B − B′ ∈ Ψ_{∞}^{−∞}(ℝ^{n}).

Thus a left parametrix of an elliptic element of Ψ_{∞}^{m}(ℝ^{n}) is always a right, hence two-sided, parametrix.

3.3. The Laplacian

Suppose that g_{ij}(x) are the components of an “∞-metric” on ℝ^{n}, i.e.

\begin{matrix}g_{ij}(x) ∈ C_{∞}^{∞}(ℝ^{n}), i, j = 1, …, n \\ | ∑_{i,j=1}^{n} g_{ij}(x)ξ_{i}ξ_{j} |≥ ϵ|ξ| ∀ x ∈ ℝ^{n}, ξ ∈ ℝ^{n}. \end{matrix} (3.31)

The Laplacian of the metric is the second order differential operator

Δ_{g} = ∑_{i,j=1}^{n} \frac{1}{\sqrt{g}} D_{x_{i}}g^{ij}\sqrt{g}D_{x_{j}} (3.32)

where

g(x) = det g^{ij}(x), g^{ij}(x) = (g_{ij}(x))^{−1} .

The Laplacian is determined by the integration by parts formula

∫_{ℝ^{n}} ∑ g^{ij}(x)D_{x_{i}}ϕ · \overline{D_{x_{i}}ψ}dg = ∫ Δ_{g}ϕ · \overline{ψ}dg ∀ ϕ, ψ ∈ S(ℝ^{n}) (3.33)

where

dg = \sqrt{g}dx. (3.34)

Our assumption in (3.31) shows that Δ = Δ_{g} ∈ Diff_{∞}^{2}(ℝ^{n}) ⊂ Ψ_{∞}^{2}(ℝ^{n}) is in fact elliptic, since

σ_{2}(Δ) = ∑_{i,j=1} g^{ij}ξ_{i}ξ_{j}. (3.35)

Thus Δ has a two-sided parametrix B ∈ Ψ_{∞}^{−2}(ℝ^{n})

(3.36)  Δ ∘ B ≡ B ∘ Δ ≡ Id  mod Ψ_{∞}^{−∞}(ℝ^{n}).
Corollary 3.1. If u ∈ S′(ℝ^{n}) and Δu = 0 then u ∈ C^{∞}(ℝ^{n}).
Proof. B ∘ Δ = Id +E, E ∈ Ψ_{∞}^{−∞}(ℝ^{n}) means, given Δu = 0,

u = −Eu. (3.37)

So it suffices to prove:

Lemma 3.5. If E ∈ Ψ_{∞}^{−∞} then

E : S′(ℝ^{n}) ⟶ S′(ℝ^{n}) ∩ C^{∞}(ℝ^{n}). (3.38)

Proof. This follows from Proposition 2.4 since we can regard the kernel as a C^{∞} function of x taking values in S(ℝ_{y}^{n}).

3.4. L^{2} boundedness

So far we have thought of pseudodifferential operators, the elements of Ψ_{∞}^{m}(ℝ^{n}) for some m, as defining continuous linear operators on S(ℝ^{n}), and by duality, on S′(ℝ^{n}). Now that we have proved the composition formula we can use it to prove other ‘finite order’ regularity results. The basic one of these is L^{2} boundedness:

Proposition 3.1. If A ∈ Ψ_{∞}^{0}(ℝ^{n}) then, by continuity from S(ℝ^{n}), A defines a bounded linear operator

A : L^{2}(ℝ^{n}) ⟶ L^{2}(ℝ^{n}). (3.39)

As usual our proof is by reductio ad absurdum; we actually start there.

3.5. Residual terms

To prove boundedness (3.39) we start by assuming that A ∈ Ψ_{∞}^{−∞}(ℝ^{n}). In that case we know exactly what the Schwartz kernel looks like:

A ∈ Ψ_{∞}^{−∞}(ℝ^{n}) ⟺| sup_{x,y} |(x − y)^{α}D_{x}^{β}D_{y}^{γ}K(x, y) |< ∞ ∀ α, β, γ. (3.40)

Since S(ℝ^{n}) is dense in L^{2}(ℝ^{n}) (see Problem ) we only need to show the existence of a constant, C, such that

∫ | Aϕ(x) |^{2} dx ≤ C ∫ |ϕ|^{2} ∀ ϕ ∈ S(ℝ^{n}). (3.41)

Writing out the integral on the left

\begin{matrix}∫ | ∫ K_{A}(x, y)ϕ(y)dy |^{2} dx \\ = ∫∫∫ K_{A}(x, y)\overline{K_{A}(x, z)}ϕ(y)\overline{ϕ(z)} dydzdx \end{matrix} (3.42)

is certainly absolutely convergent and

∫ | Aϕ(x) |^{2} dx

≤ (∫∫∫ | K_{A}(x, y)K_{A}(x, z) || ϕ(y) |^{2} dydxdz)^{\frac{1}{2}}

× (∫∫∫ | K_{A}(x, y)K_{A}(x, z) || ϕ(z) |^{2} dzdxdy)^{\frac{1}{2}} .

These two factors are the same. Changing variables to y, t = x − y and x − z = s gives

∫ | Aϕ |^{2} dx ≤ sup_{y} ∫∫ | K_{A}(t + y, y) || K_{A}(t + y, t + y − s) | dtds · ∫ |ϕ|^{2}dy.

which gives (3.41), since from (3.40)

| K_{A}(t + y, y) |≤ C (1 + |t|)^{−n−1} ,
| K_{A} (t + y, t + y − s) |≤ C (1 + |s|)^{−n−1} .

3.6. Square root

To prove the general result, (3.39), we shall use the clever idea, due to Hörmander, of using the (approximate) square root of an operator. We shall say that an element [a] ∈ S_{∞}^{m−[1]}(ℝ^{n}; ℝ^{n}) is positive if there is some 0 < a ∈ S^{m}(ℝ^{n}; ℝ^{n}) in the equivalence class.

Proposition 3.2. Suppose A ∈ Ψ_{∞}^{m}(ℝ^{n}) is self-adjoint, A = A^{∗}, and elliptic with a positive principal symbol, then there exists B ∈ Ψ_{∞}^{m/2}(ℝ^{n}), B = B^{∗}, such that

A = B^{2} + G, G ∈ Ψ_{∞}^{−∞}(ℝ^{n}). (3.43)

Proof. This is a good exercise in the use of the symbol calculus. Let a ∈ S_{∞}^{m}(ℝ^{n}; ℝ^{n}), a > 0, be a positive representative of the principal symbol of A. Now

b_{0} = a^{\frac{1}{2}} ∈ S_{∞}^{m/2}(ℝ^{n}; ℝ^{n}). (3.44)

Let B_{0} ∈ Ψ_{∞}^{m/2}(ℝ^{n}) have principal symbol b_{0}. We can assume that B_{0} = B_{0}^{∗}, since if not we just replace it by \frac{1}{2}(B_{0} + B_{0}^{∗}) which has the same principal symbol.

The symbol calculus shows that B_{0}^{2} ∈ Ψ_{∞}^{m}(ℝ^{n}) and

σ_{m}(B_{0}^{2}) = (σ_{m/2}(B_{0}))^{2} = b_{0}^{2} = a_{0} mod S_{∞}^{m−1}.

Thus

A − B_{0}^{2} = E_{1} ∈ Ψ_{∞}^{m−1}(ℝ^{n}). (3.45)

Then we proceed inductively. Suppose we have chosen B_{j} ∈ Ψ_{∞}^{m/2−j}(ℝ^{n}), with B_{j}^{∗} = B_{j}, for j ≤ N such that

A − (∑_{j=0}^{N} B_{j})^{2} = E_{N+1} ∈ Ψ_{∞}^{m−N−1}(ℝ^{n}). (3.46)

Of course we have done this for N = 0. Then see the effect of adding B_{N+1} ∈ Ψ_{∞}^{m/2−N−1}(ℝ^{n}) :

\begin{matrix}A − ( ∑_{j=0}^{N+1} B_{j})^{2} = E_{N+1} − (∑_{j=0}^{N} B_{j}) B_{N+1} \\ −B_{N+1} (∑_{j=0}^{N} B_{j}) − B_{N+1}^{2}. \end{matrix} (3.47)

On the right side all terms are of order m − N − 2, except for E_{N+1} − B_{0}B_{N+1} − B_{N+1}B_{0} ∈ Ψ_{∞}^{m−N−1}(ℝ^{n}). (3.48) The principal symbol, of order m − N − 1, of this is just σ_{m−N−1}(E_{N+1}) − 2 b_{0} · σ_{\frac{m}{2} −N−1}(B_{N+1}). (3.49) Thus if we choose

σ_{m/2−N−1}(B_{N+1}) = \frac{1}{2} \frac{1}{b_{0}} · σ_{m−N−1}(E_{N+1})

and replace B_{N+1} by \frac{1}{2}(B_{N+1} + B_{N+1}^{∗}), we get the inductive hypothesis for N + 1. Thus we have arranged (3.46) for every N. Now define B = \frac{1}{2}(B′ + (B′)^{∗}) where

σ_{L}(B′) ∼ ∑_{j=0}^{∞} σ_{L}(B_{j}). (3.50)

Since all the B_{j} are self-adjoint B also satisfies (3.50) and from (3.46)

A − B^{2} = A − (∑_{j=0}^{N} B_{j} + B_{(N+1)})^{2} ∈ Ψ_{∞}^{m−N−1}(ℝ^{n})

for every N, since B_{(N+1)} = B− ∑_{j=0}^{N} B_{j} ∈ Ψ_{∞}^{m/2−N−1}(ℝ^{n}). Thus A−B^{2} ∈ Ψ_{∞}^{−∞}(ℝ^{n})

and we have proved (3.43), and so Proposition 3.2.

3.7. Proof of Proposition 3.1

Here is Hörmander’s clever argument. We want to show that ∥Aϕ∥ ≤ C∥ϕ∥ ∀ ϕ ∈ S(ℝ^{n}) (3.51) where A ∈ Ψ_{∞}^{0}(ℝ^{n}). The square of the left side can be written

∫ Aϕ · \overline{Aϕ}dx = ∫ ϕ · (A^{∗}Aϕ) dx.

So it suffices to show that ⟨ϕ, A^{∗}Aϕ⟩ ≤ C∥ϕ∥^{2}. (3.52) Now A^{∗}A ∈ Ψ_{∞}^{0}(ℝ^{n}). If C > 0 is a large constant,

C > sup_{x,ξ} | σ_{L}(A^{∗}A)(x, ξ) |

3.8. SOBOLEV BOUNDEDNESS 45

then C−A^{∗}A has a positive representative of its principal symbol. We can therefore apply Proposition 3.2 to it: C − A^{∗}A = B^{∗}B + G, G ∈ Ψ_{∞}^{−∞}(ℝ^{n}). (3.53) This gives

\begin{matrix}⟨ϕ, A^{∗}Aϕ⟩ = & C⟨ϕ, ϕ⟩ − ⟨ϕ, B^{∗}Bϕ⟩ − ⟨ϕ, Gϕ⟩ \\ = & C∥ϕ∥^{2} − ∥Bϕ∥^{2} − ⟨ϕ, Gϕ⟩. \end{matrix} (3.54)

The second term on the right is negative and, since G ∈ Ψ_{∞}^{−∞}(ℝ^{n}), we can use the residual case discussed above to conclude that

| ⟨ϕ, Gϕ⟩ |≤ C′∥ϕ∥^{2} ⟹ ∥Aϕ∥^{2} ≤ C∥ϕ∥^{2} + C′∥ϕ∥^{2},

so (3.39) holds and Proposition 3.1 is proved.

3.8. Sobolev boundedness Using the basic boundedness result, Proposition 3.1, and the calculus of pseu-

dodifferential operators we can prove more general results on the action of pseudodifferential operators on Sobolev spaces.

Recall that for any positive integer, k, H^{k}(ℝ^{n}) = {u ∈ L^{2}(ℝ^{n}); D^{α}u ∈ L^{2}(ℝ^{n}) ∀ |α| ≤ k}. (3.55) Using the Fourier transform we find u ∈ H^{k}(ℝ^{n}) ⟹ ξ^{α}û(ξ) ∈ L^{2}(ℝ^{n}) ∀ |α| ≤ k. (3.56) Now these finitely many conditions can be written just (1 + |ξ|^{2})^{k/2} û(ξ) ∈ L^{2}(ℝ^{n}). (3.57)

Notice that a(ξ) = (1 + |ξ|^{2})^{k/2} = ⟨ξ⟩^{k} ∈ S_{∞}^{k}(ℝ^{n}). Here we use the notation ⟨ξ⟩ = (1 + |ξ|^{2})^{\frac{1}{2}} (3.58) for a smooth (symbol) of the size of 1 + |ξ|, thus (3.57) just says u ∈ H^{k}(ℝ^{n}) ⟺ ⟨D⟩^{k}u ∈ L^{2}(ℝ^{n}). (3.59)

For negative integers

H^{k}(ℝ^{n}) = {u ∈ S′(ℝ^{n}); u = ∑_{|β|≤−k} D^{β}u_{β}, u_{β} ∈ L^{2}(ℝ^{n})}, −k ∈ ℕ. (3.60)

The same sort of discussion applies, showing that u ∈ H^{k}(ℝ^{n}) ⟺ ⟨D⟩^{k}u ∈ L^{2}(ℝ^{n}). (3.61) In view of this we define the Sobolev space H^{m}(ℝ^{n}), for any real order, by u ∈ H^{k}, (ℝ^{n}) ⟺ ⟨D⟩^{m}u ∈ L^{2}(ℝ^{n}). (3.62) It is a Hilbert space with

∥u||_{m}^{2} = ∥⟨D⟩^{m}u∥_{L^{2}}^{2} = ∫ (1 + |ξ|^{2})^{m}|û(ξ)|dξ. (3.63)

Clearly we have H^{m}(ℝ^{n}) ⊇ H^{m′}(ℝ^{n}) if m′ ≥ m. (3.64)

Notice that it is rather unfortunate that these spaces get smaller as m gets bigger, as opposed to the spaces Ψ_{∞}^{m}(ℝ^{n}) which get bigger with m. Anyway that’s life and we have to think of

H^{∞}(ℝ^{n}) = ⋂_{m} H^{m}(ℝ^{n}) as the residual space
H^{−∞}(ℝ^{n}) = ⋃_{m} H^{m}(ℝ^{n}) as the big space.

(3.65)

It is important to note that

S(ℝ^{n}) ⊊ H^{∞}(ℝ^{n}) ⊊ H^{−∞}(ℝ^{n}) ⊊ S′(ℝ^{n}). (3.66)

In particular we do not capture all the tempered distributions in H^{−∞}(ℝ^{n}). We therefore consider weighted versions of these Sobolev spaces:

⟨x⟩^{q}H^{m}(ℝ^{n}) = {u ∈ S′(ℝ^{n}); ⟨x⟩^{−q}u ∈ H^{m}(ℝ^{n})}. (3.67)

Theorem 3.2. For each q, m, M ∈ ℝ each A ∈ Ψ_{∞}^{M}(ℝ^{n}) defines a continuous linear map

A : ⟨x⟩^{q}H^{m}(ℝ^{n}) ⟶ ⟨x⟩^{q}H^{m−M}(ℝ^{n}). (3.68)

Proof. Let us start off with q = 0, so we want to show that

A : H^{m}(ℝ^{n}) ⟶ H^{m−M}(ℝ^{n}), A ∈ Ψ_{∞}^{M}(ℝ^{n}) (3.69)

Now from (3.62) we see that

(3.70) u ∈ H^{m}(ℝ^{n}) ⟺ ⟨D⟩^{m}u ∈ L^{2}(ℝ^{n})

⟺ ⟨D⟩^{m−M}⟨D⟩^{M}u ∈ L^{2}(ℝ^{n}) ⟺ ⟨D⟩^{M}u ∈ H^{m−M}(ℝ^{n}) ∀ m, M.

That is,

⟨D⟩^{M} : H^{m}(ℝ^{n}) ⟷ H^{m−M}(ℝ^{n}) ∀ m, M. (3.71)

To prove (3.69) it suffices to show that

B = ⟨D⟩^{−M+m} · A · ⟨D⟩^{−m} : L^{2}(ℝ^{n}) ⟶ L^{2}(ℝ^{n}) (3.72)

since then A = ⟨D⟩^{−m+M} · B · ⟨D⟩^{m} maps H^{m}(ℝ^{n}) to H^{m−M}(ℝ^{n}) :

\xymatrix{H^{m}(ℝ^{n}) \ar[d]_{⟨D⟩^{m}} \ar[r]^{A} & H^{m−M}(ℝ^{n}) \ar[d]^{⟨D⟩^{m−M}} \\ L^{2}(ℝ^{n}) \ar[r]_{B} & L^{2}(ℝ^{n}).} (3.73)

Since B ∈ Ψ_{∞}^{0}(ℝ^{n}), by the composition theorem, we already know (3.72).

Thus we have proved (3.69). To prove the general case, (3.68), we proceed in the same spirit. Thus ⟨x⟩^{q} is an isomorphism from H^{m}(ℝ^{n}) to ⟨x⟩^{q}H^{m}(ℝ^{n}), by definition. So to get (3.68) we need to show that

Q = ⟨x⟩^{q} · A · ⟨x⟩^{−q} : H^{m}(ℝ^{n}) ⟶ H^{m−M}(ℝ^{n}), (3.74)

i.e. satisfies (3.69). Consider the Schwartz kernel of Q. Writing A in left-reduced form, with symbol a,

K_{Q}(x, y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}⟨x⟩^{q}a(x, ξ)dξ · ⟨y⟩^{−q}. (3.75)

Now if we check that

⟨x⟩^{q}⟨y⟩^{−q}a(x, ξ) ∈ (1 + |x − y|^{2})^{\frac{|q|}{2}} S_{∞}^{M} (ℝ^{2n}; ℝ^{n}) (3.76)

then we know that Q ∈ Ψ_{∞}^{M}(ℝ^{n}) and we get (3.74) from (3.69). Thus we want to show that

⟨x − y⟩^{−|q|} \frac{⟨x⟩^{q}}{⟨y⟩^{q}} a(x, ξ) ∈ S_{∞}^{M}(ℝ^{2n}; ℝ^{n}) (3.77)

assuming of course that a(x, ξ) ∈ S_{∞}^{M}(ℝ^{n}; ℝ^{n}). By interchanging the variables x and y if necessary we can assume that q > 0. Consider separately the two regions

\begin{aligned}\{(x, y); |x − y| < \frac{1}{4}(|x| + |y|)\} & = Ω_{1} \\ \{(x, y); |x − y| > \frac{1}{8}(|x| + |y|)\} & = Ω_{2} \end{aligned} (3.78)

In Ω_{1}, x is “close” to y, in the sense that

|x| ≤ |x − y| + |y| ≤ \frac{1}{4}(|x| + |y|) + |y| ⟹ |x| ≤ \frac{4}{3} · \frac{5}{4}|y| ≤ 2|y|.

Thus

⟨x − y⟩^{−q} · \frac{⟨x⟩^{q}}{⟨y⟩^{q}} ≤ C in Ω_{1}. (3.79)

On the other hand in Ω_{2},

|x| + |y| < 8|x − y| ⟹ |x| < 8|x − y|

so again

⟨x − y⟩^{−q} \frac{⟨x⟩^{q}}{⟨y⟩^{q}} ≤ C. (3.80)

In fact we easily conclude that

⟨x − y⟩^{−q} \frac{⟨x⟩^{q}}{⟨y⟩^{q}} ∈ C_{∞}^{∞}(ℝ^{n}) ∀ q, (3.81)

since differentiation by x or y makes all terms “smaller”. This proves (3.77), hence (3.76) and (3.74) and therefore (3.68), i.e. the theorem is proved.

3.9. Consequences

We can capture any tempered distribution in a weighted Sobolev space; this is really Schwartz’ representation theorem which says that any u ∈ S′(ℝ^{n}) is of the form

u = ∑_{finite} x^{α}D_{x}^{β}u_{αβ}, u_{αβ} bounded and continuous.

Clearly C_{∞}^{0}(ℝ^{n}) ⊂ ⟨x⟩^{−n}L^{2}(ℝ^{n}). Thus

Lemma 3.6.

S′(ℝ^{n}) = ⋃_{M}⟨x⟩^{M}H^{−M}(ℝ^{n}).

The elliptic regularity result we found before can now be refined:

Proposition 3.3. If A ∈ Ψ_{∞}^{m}(ℝ^{n}) is elliptic then

\begin{matrix}Au ∈ ⟨x⟩^{p}H^{q}(ℝ^{n}), u ∈ ⟨x⟩^{p′}H^{q′}(ℝ^{n}) \\ ⟹ u ∈ ⟨x⟩^{p′′}H^{q+m}(ℝ^{n}), p′′ = max(p, p′). \end{matrix} (3.82)

Proof. The existence of a left parametrix for A, B ∈ Ψ_{∞}^{−m}(ℝ^{n}),

B · A = Id +G, G ∈ Ψ_{∞}^{−∞}(ℝ^{n})

means that

u = B(Au) + Gu
∈ ⟨x⟩^{p}H^{q+m}(ℝ^{n}) + ⟨x⟩^{p′}H^{∞}(ℝ^{n})
⊂ ⟨x⟩^{p′′}H^{q+m}(ℝ^{n}).

3.10. Calculus of supports

Recall that we have already defined the support of a tempered distribution in the slightly round-about way:

if u ∈ S′(ℝ^{n}), supp(u) = {x ∈ ℝ^{n}; ∃ ϕ ∈ S(ℝ^{n}), ϕ(x) ≠ 0, ϕu = 0}^{∁}. (3.83)

Now if A : S(ℝ^{n}) ⟶ S′(ℝ^{n}) is any continuous linear operator we can consider the support of the kernel:

(3.84) supp(A) = supp(K_{A}) ⊂ ℝ^{n} × ℝ^{n} = ℝ^{2n}.

We write out the space as a product here to point to the fact that any subset of the product defines (is) a relation i.e. a map on subsets:

\begin{matrix}G ⊂ ℝ^{n} × ℝ^{n}, S ⊂ ℝ^{n} ⟹ \\ G ∘ S = \{x ∈ ℝ^{n}; ∃ y ∈ S s.t. (x, y) ∈ G\}. \end{matrix} (3.85)

One can write this much more geometrically in terms of the two projection maps

\xymatrix{ & ℝ^{2n} \ar[dl]_{π_{L}} \ar[dr]^{π_{R}} \\ ℝ^{n} & & ℝ^{n}.} (3.86)

Thus π_{R}(x, y) = y, π_{L}(x, y) = x. Then (3.85) can be written in terms of the action of maps on sets as

G ∘ S = π_{L} (π_{R}^{−1}(S) ∩ G) . (3.87)

From this it follows that if S is compact and G is closed, then G ∘ S is closed, since its intersection with any compact set is the image of a compact set under a continuous map, hence compact. Now, by the calculus of supports we mean the ‘trivial’ result.

Proposition 3.4. If A : S(ℝ^{n}) ⟶ S′(ℝ^{n}) is a continuous linear map then

(3.88) supp(Aϕ) ⊂ supp(A) ∘ supp(ϕ) ∀ ϕ ∈ C_{c}^{∞}(ℝ^{n}).

Proof. Since we want to bound supp(Aϕ) we can use (3.83) directly, i.e. show that

(3.89) x ∉ supp(A) ∘ supp(ϕ) ⟹ x ∉ supp(Aϕ).

Since we know supp(A) ∘ supp(ϕ) to be closed, the assumption that x is outside this set means that there exists ψ ∈ C_{c}^{∞}(ℝ^{n}) with

ψ(x) ≠ 0 and supp(ψ) ∩ supp(A) ∘ supp(ϕ) = ∅.

From (3.85) or (3.87) this means

(3.90) supp(A) ∩ (supp(ψ) × supp(ϕ)) = ∅ in ℝ^{2n}.

But this certainly implies that

\begin{matrix}K_{A}(x, y)ψ(x)ϕ(x) = 0 \\ ⟹ ψA(ϕ) = ∫ K_{A}(x, y)ψ(x)ϕ(y)dy = 0. \end{matrix} (3.91)

Thus we have proved (3.88) and the lemma.

3.11. Singular supports

As well as the support of a tempered distribution we can consider the singular support:

(3.92) sing supp(u) = {x ∈ ℝ^{n}; ∃ ϕ ∈ S(ℝ^{n}), ϕ(x) ≠ 0, ϕu ∈ S(ℝ^{n})}^{∁}.

Again this is a closed set since x ∉ sing supp(u) ⟹ ∃ ϕ ∈ S(ℝ^{n}) with ϕu ∈ S(ℝ^{n}) and ϕ(x) ≠ 0 so ϕ(x′) ≠ 0 for |x − x′| < ϵ, some ϵ > 0 and hence x′ ∉ sing supp(u) i.e. the complement of sing supp(u) is open.

Directly from the definition we have

(3.93) sing supp(u) ⊂ supp(u) ∀ u ∈ S(ℝ^{n}) and

(3.94) sing supp(u) = ∅ ⟺ u ∈ C^{∞}(ℝ^{n}).

3.12. Pseudolocality

We would like to have a result like (3.88) for singular support, and indeed we can get one for pseudodifferential operators. First let us work out the singular support of the kernels of pseudodifferential operators.

Proposition 3.5. If A ∈ Ψ_{∞}^{m}(ℝ^{n}) then

(3.95) sing supp(A) = sing supp(K_{A}) ⊂ {(x, y) ∈ ℝ^{2n}; x = y}.

Proof. The kernel is defined by an oscillatory integral

I(a) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, y, ξ)dξ.

If the order m is < −n we can show by integration by parts that

(x − y)^{α}I(a) = I ((−D_{ξ})^{α}a) , (3.96)

and then this must hold by continuity for all orders. If a is of order m and |α| > m + n then (−D_{ξ})^{α}a is of order less than −n, so

(x − y)^{α}I(a) ∈ C_{∞}^{0}(ℝ^{n}), |α| > m + n. (3.97)

50 3. ELLIPTICITY AND BOUNDEDNESS

In fact we can also differentiate under the integral sign: D_{x}^{β}D_{y}^{γ}(x − y)^{α}I(a) = I (D_{x}^{β}D_{y}^{γ}(−D_{ξ})^{α}a) (3.98) so generalizing (3.97) to (x − y)^{α}I(a) ∈ C_{∞}^{k}(ℝ^{n}) if |α| > m + n + k. (3.99) This implies that I(A) is C^{∞} on the complement of the diagonal, {x = y}. This proves (3.95).

An operator is said to be pseudolocal if it satisfies the condition (3.100) sing supp(Au) ⊂ sing supp(u) ∀ u ∈ C^{−∞}(ℝ^{n}).

Proposition 3.6. Pseudodifferential operators are pseudolocal.

Proof. Suppose u ∈ S′(ℝ^{n}) has compact support and \overline{x} ∉ sing supp(u). Then we can choose ϕ ∈ S(ℝ^{n}) with ϕ ≡ 1 near \overline{x} and ϕu ∈ S(ℝ^{n}) (by definition). Thus u = u_{1} + u_{2}, u_{1} = (1 − ϕ)u, u_{2} ∈ S(ℝ^{n}). (3.101) Since A : S(ℝ^{n}) ⟶ S(ℝ^{n}), Au_{2} ∈ S(ℝ^{n}) so (3.102) sing supp(Au) = sing supp(Au_{1}) and \overline{x} ∉ supp(u_{1}). Choose ψ ∈ S(ℝ^{n}) with compact support, ψ(\overline{x}) = 1 and (3.103) supp(ψ) ∩ supp(1 − ϕ) = ∅. Thus ψAu_{1} = ψA(1 − ϕ)u = Ãu (3.104) where

KÃ(x, y) = ψ(x)K_{A}(x, y)(1 − ϕ(y)).

Combining (3.103) and (3.95) shows that KÃ ∈ Ψ_{∞}^{−∞}(ℝ^{n}) so, by Lemma 3.5, Ãu ∈ C^{∞}(ℝ^{n}) and \overline{x} ∉ sing supp(Au) by (3.95). This proves the proposition.

3.13. Coordinate invariance

If Ω ⊂ ℝ^{n} is an open set, put

C_{c}^{∞}(Ω) = {u ∈ S(ℝ^{n}); supp(u) ⋐ Ω}
C_{c}^{−∞}(Ω) = {u ∈ S′(ℝ^{n}); supp(u) ⋐ Ω}

respectively the space of C^{∞} functions of compact support in Ω and of distributions of compact support in Ω. Here K ⋐ Ω indicates that K is a compact subset of Ω. Notice that if u ∈ C_{c}^{−∞}(ℝ^{n}) then u defines a continuous linear functional C^{∞}(Ω) ∋ ϕ ⟼ u(ϕ) = u(ψϕ) ∈ ℂ (3.105) where if ψ ∈ C_{c}^{∞}(Ω) is chosen to be identically one near supp(u) then (3.105) is independent of ψ. [Think about what continuity means here!]

Now suppose F : Ω ⟶ Ω′ (3.106) is a diffeomorphism between open sets of ℝ^{n}. The pull-back operation is

\begin{matrix}F^{∗} : C_{c}^{∞}(Ω′) ⟷ C_{c}^{∞}(Ω) \\ F^{∗}ϕ = ϕ ∘ F. \end{matrix}

3.13. COORDINATE INVARIANCE 51

Lemma 3.7. If F is a diffeomorphism, (3.106), between open sets of ℝ^{n} then F^{∗} : C_{c}^{−∞}(Ω′) ⟷ C_{c}^{−∞}(Ω) (3.107) by continuity.

Proof. The density of C_{c}^{∞}(Ω) in C_{c}^{−∞}(Ω), in the weak topology given by the seminorms from (3.105), can be proved in the same way as the density of S(ℝ^{n}) in S′(ℝ^{n}) (see Problem 3.14). Thus, we only need to show continuity of (3.106) in this sense. Suppose u ∈ C_{c}^{∞}(Ω) and ϕ ∈ C_{c}^{∞}(Ω′) then

\begin{aligned}(F^{∗}u)(ϕ) & = ∫ u(F (x))ϕ(x)dx \\ & = ∫ u(y)ϕ(G(y))|J_{G}(y)|dy \end{aligned} (3.108)

where J_{G}(y) = (\frac{∂G(y)}{∂y}) is the Jacobian of G, the inverse of F. Thus (3.107) can be written F^{∗}u(ϕ) = (|J_{G}|u) (G^{∗}ϕ) (3.109) and since G^{∗} : C^{∞}(Ω) ⟶ C^{∞}(Ω′) is continuous (!) we conclude that F^{∗} is continuous as desired.

Now suppose that

A : S(ℝ^{n}) ⟶ S′(ℝ^{n})

has supp(A) ⋐ Ω × Ω ⊂ ℝ^{2n}. (3.110) Then A : C_{c}^{∞}(Ω) ⟶ C_{c}^{−∞}(Ω) (3.111) by Proposition 3.4. Applying a diffeomorphism, F, as in (3.7) set A_{F} : C_{c}^{∞}(Ω′) ⟶ C_{c}^{−∞}(Ω′), A_{F} = G^{∗} ∘ A ∘ F^{∗}. (3.112)

Lemma 3.8. If A satisfies (3.110) and F is a diffeomorphism (3.7) then K_{A_{F}} (x, y) = (G × G)^{∗}K · |J_{G}(y)| on Ω′ × Ω′ (3.113) has compact support in Ω′ × Ω′.

Proof. Same as (3.110).

Proposition 3.7. Suppose A ∈ Ψ_{∞}^{m}(ℝ^{n}) has kernel satisfying (3.110) and F is a diffeomorphism as in (3.7) then A_{F} , defined by (3.112), is an element of Ψ_{∞}^{m}(ℝ^{n}).

Proof. Since A ∈ Ψ_{∞}^{m}(ℝ^{n}), K_{A}(x, y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, ξ)dξ (3.114) for some a ∈ S_{∞}^{m}(ℝ^{n}; ℝ^{n}). Now choose ψ ∈ C_{c}^{∞}(Ω) such that ψ(x)ψ(y) = 1 on supp(K_{A}), possible by (3.110). Then K_{A} = I (ψ(x)ψ(y)a(x, ξ)) . (3.115)

52 3. ELLIPTICITY AND BOUNDEDNESS

In fact suppose μ_{ϵ}(x, y) ∈ C^{∞}(ℝ^{2n}) and μ ≡ 1 in |x − y| < ϵ for ϵ > 0, μ(x, y) = 0 in |x − y| > 2ϵ. Then if K_{A_{ϵ}} = I (μ_{ϵ}(x, y)ψ(x)ψ(y)a(x, ξ)) (3.116) we know that if A′_{ϵ} = A − A_{ϵ} then K_{A′_{ϵ}} = (1 − μ_{ϵ}(x, y)) K_{A} ∈ Ψ_{∞}^{−∞}(ℝ^{n}), (3.117) by (3.95). Certainly A′_{ϵ} also satisfies (3.110) and from (3.113) (A′_{ϵ})_{F} ∈ Ψ_{∞}^{−∞}(ℝ^{n}). (3.118) So we only need to consider A_{ϵ} defined by (3.116). Again using (3.113) (assuming m < −n)

K_{(A_{ϵ})_{F}} (x, y) = (2π)^{−n} ∫ e^{i(G(x)−G(y))·ξ}b(G(x), G(y), ξ) | \frac{∂G}{∂y} | dξ (3.119)

where b(x, y, ξ) = μ_{ϵ}(x − y)ψ(x)ψ(y)a(x, ξ). Applying Taylor’s formula, G(x) − G(y) = (x − y) · T (x, y) (3.120) where T (x, y) is an invertible C^{∞} matrix on K × K ∩ {|x − y| < ϵ} for ϵ < ϵ(K), where ϵ(K) > 0 depends on the compact set K ⋐ Ω′. Thus we can set

η = T^{t}(x, y) · ξ

and rewrite (3.120) as

\begin{matrix}K_{(A_{ϵ})_{F}} (x, y) = (2π)^{−n} ∫ e^{i(x−y)·η}c(x, y, η)dη \\ c(x, y, η) = b (G(x), G(y), (T^{t})^{−1}(x, y)η) | \frac{∂G}{∂y} | · | det T (x, y) |^{−1} . \end{matrix} (3.121)

So it only remains to show that c ∈ S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) and the proof is complete. We can drop all the C^{∞} factors, given by | ∂G/∂y | etc. and proceed to show that

| D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a (G(x), G(y), S(x, y)ξ) | ≤ C(1 + |ξ|)^{m−|γ|} on K × K × ℝ^{n} where K ⊂⊂ Ω′ and S is C^{∞} with | det S| ≥ ϵ. The estimates with α = β = 0 follow easily and the general case by induction: D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a (G(x), G(y), S(x, y)ξ) (3.122)

= ∑_{\begin{matrix}|μ|≤|α|+|β|+|γ| \\ |α′|≤|α|,|β|≤|β| \\ |ν|+|γ|≤|μ| \end{matrix}} M_{α,β,γ,ν}^{α′,ρ′,μ′}(x, y)ξ^{ν} (D^{α′}D^{β′}D^{μ}a) (G(x), G(y), Sξ) (3.123)

where the coefficients are C^{∞}.

3.14. Problems

Problem 3.1. Show that Weyl quantization

S_{∞}^{∞}(ℝ^{n}; ℝ^{n}) ∋ a ⟼ q_{W} (a) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(\frac{x + y}{2} , ξ)dξ

is well-defined by continuity from S_{∞}^{−∞}(ℝ^{n}; ℝ^{n}) and induces an isomorphism

S_{∞}^{m}(ℝ^{n}; ℝ^{n}) ⟵_{q_{W}}^{⟶^{σ_{W}}} Ψ_{∞}^{m}(ℝ^{n}) ∀ m ∈ ℝ.

Find an asymptotic formula relating q_{W} (A) to q_{L}(A) for any A ∈ Ψ_{∞}^{m}(ℝ^{n}).

3.14. PROBLEMS 53

Problem 3.2. Show that if A ∈ Ψ_{∞}^{m}(ℝ^{n}) then A^{∗} = A if and only if σ_{W} (A) is real-valued.

Problem 3.3. Is it true that every E ∈ Ψ_{∞}^{−∞}(ℝ^{n}) defines a map from S′(ℝ^{n}) to S(ℝ^{n})?

Problem 3.4. Show that S(ℝ^{n}) is dense in L^{2}(ℝ^{n}) by proving that if ϕ ∈ C^{∞}(ℝ^{n}) has compact support and is identically equal to 1 near the origin then

u_{n}(x) = (2π)^{−n}ϕ(\frac{x}{n}) ∫ e^{ix·ξ}ϕ(ξ/n)û(ξ)dξ ∈ S(ℝ^{n}) if u ∈ L^{2}(ℝ^{n})

and u_{n} → u in L^{2}(ℝ^{n}). Can you see any relation to pseudodifferential operators here?

Problem 3.5. Check carefully that with the definition

H^{k}(ℝ^{n}) = {u ∈ S(ℝ^{n}); u = ∑_{|α|≤−k} D^{α}u_{α}, u_{α} ∈ L^{2}(ℝ^{n})}

for −k ∈ ℕ one does have

u ∈ H^{k}(ℝ^{n}) ⟺ ⟨D⟩^{k}u ∈ L^{2}(ℝ^{n})

as claimed in the lecture.

Problem 3.6. Suppose that a(x) ∈ C_{∞}^{∞}(ℝ^{n}) and that a(x) ≥ 0. Show that the operator

A = ∑_{j=1}^{n} D_{x_{j}}^{2} + a(x)

can have no solution which is in L^{2}(ℝ^{n}).

Problem 3.7. Show that for any open set Ω ⊂ ℝ^{n}, C_{c}^{∞}(Ω) is dense in C_{c}^{−∞}(Ω) in the weak topology.

54 3. ELLIPTICITY AND BOUNDEDNESS