CHAPTER 2 Pseudodifferential operators

Formula (1.48) for the action of a differential operator (with coefficients in C_{∞}^{∞}(ℝ^{n})) on S(ℝ^{n}) can be written

(2.1)  P (x, D)u = (2π)^{−n} ∫ e^{i(x−y)·ξ}P (x, ξ)u(y)dydξ

= (2π)^{−n} ∫ e^{ix·ξ}P (x, ξ)û(ξ)dξ (2.2)

where û(ξ) = Fu(ξ) is the Fourier transform of u. We shall generalize this formula by generalizing P (x, ξ) from a polynomial in ξ to a symbol. In fact we shall also allow the symbol, or rather the amplitude, in the integral (2.2) to depend on the ‘incoming’ variables, y :

A(x, D)u = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, y, ξ)u(y)dydξ, u ∈ S(ℝ^{n}). (2.3)

Of course it is not clear that this integral is well-defined. To interpret (2.3) we shall first look into the properties of symbols.

2.1. Symbols

A polynomial, p, in ξ, of degree at most m, satisfies a bound

|p(ξ)| ≤ C(1 + |ξ|)^{m} ∀ ξ ∈ ℝ^{n}. (2.4)

Since successive derivatives, D_{ξ}^{α}p(ξ), are polynomials of degree m − |α|, for any multiindex α, we get the family of estimates

|D_{ξ}^{α}p(ξ)| ≤ C_{α}(1 + |ξ|)^{m−|α|} ∀ ξ ∈ ℝ^{n}, α ∈ ℕ_{0}^{n}. (2.5)

Of course if |α| > m then D_{ξ}^{α}p ≡ 0, so we can even take the constant C_{α} to be independent of α. If we consider the characteristic polynomial P (x, ξ) of a differential operator of order m with coefficients in C_{∞}^{∞}(ℝ^{n}) (i.e. all derivatives of the coefficients are bounded) (2.5) is replaced by

| D_{x}^{α}D_{ξ}^{β}P (x, ξ) |≤ C_{α,β}(1 + |ξ|)^{m−|β|} ∀ (x, ξ) ∈ ℝ^{n} × ℝ^{n}, α, β ∈ ℕ_{0}^{n}. (2.6)

There is no particular reason to have the same number of x variables as of ξ variables, so in general we define:

Definition 2.1. The space S_{∞}^{m}(ℝ^{p}; ℝ^{n}) of symbols of order m consists of those functions a ∈ C^{∞}(ℝ^{p} × ℝ^{n}) satisfying all the estimates

| D_{z}^{α}D_{ξ}^{β}a(z, ξ) |≤ C_{α,β}(1 + |ξ|)^{m−|β|} on ℝ^{p} × ℝ^{n} ∀ α ∈ ℕ_{0}^{p}, β ∈ ℕ_{0}^{n}. (2.7)

For later reference we even define S_{∞}^{m}(Ω; ℝ^{n}) when Ω ⊂ ℝ^{p} and Ω ⊂ clos(int(Ω)) as consisting of those a ∈ C^{∞}(int(Ω) × ℝ^{n}) satisfying (2.7) for (z, ξ) ∈ int(Ω) × ℝ^{n}.

The estimates (2.7) can be rewritten

(2.8)  ∥a∥_{N,m} = sup_{\begin{matrix}z∈int(Ω) \\ ξ∈ℝ^{n} \end{matrix}} max_{|α|+|β|≤N}(1 + |ξ|)^{−m+|β|}× | D_{z}^{α}D_{ξ}^{β}a(z, ξ) |< ∞.

With these norms S_{∞}^{m}(Ω; ℝ^{n}) is a Fréchet space, rather similar in structure to C_{∞}^{∞}(ℝ^{n}). Thus the topology is given by the metric

d(a, b) = ∑_{N≥0} 2^{−N} \frac{∥a − b∥_{N,m}}{1 + ∥a − b∥_{N,m}} , a, b ∈ S_{∞}^{m}(Ω; ℝ^{n}). (2.9)

The subscript ‘∞’ here is not standard notation. It refers to the assumption of uniform boundedness of the derivatives of the ‘coefficients’. More standard notation would be just S^{m}(Ω × ℝ^{n}), especially for Ω = ℝ^{p}, but I think this is too confusing.

Now to some basic properties. First notice that

(1 + |ξ|)^{m} ≤ (1 + |ξ|)^{m′} ∀ ξ ∈ ℝ^{n} ⟺ m ≤ m′.

Thus we have an inclusion

S_{∞}^{m}(Ω; ℝ^{n}) ↪ S_{∞}^{m′}(Ω; ℝ^{n}) ∀ m′ ≥ m. (2.10)

Moreover this inclusion is continuous, since from (2.8), ∥a∥_{N,m′} ≤ ∥a∥_{N,m} if a ∈ S^{m}(Ω; ℝ^{n}) and m′ ≥ m. Since these spaces increase with m we think of them as a filtration of the big space

S_{∞}^{∞}(Ω; ℝ^{n}) = ⋃_{m} S_{∞}^{m}(Ω; ℝ^{n}). (2.11)

The residual space of this filtration is

S_{∞}^{−∞}(Ω; ℝ^{n}) = ⋂_{m} S_{∞}^{m}(Ω; ℝ^{n}). (2.12)

A rather useful technical result is:

Lemma 2.1. For any m′ > m the space S_{∞}^{−∞}(Ω; ℝ^{n}) is dense in the space S_{∞}^{m}(Ω; ℝ^{n}) in the topology of S_{∞}^{m′}(Ω; ℝ^{n}).

Proof. One cannot take m′ = m here, for the same reason that S(ℝ^{n}) is not dense in C_{∞}^{∞}(ℝ^{n}).

Choose ϕ ∈ C_{c}^{∞}(ℝ^{n}) with ϕ(ξ) = 1 if |ξ| < 1, ϕ(ξ) = 0 if |ξ| > 2 and consider the sequence

a_{k}(z, ξ) = ϕ(ξ/n)a(z, ξ), a ∈ S_{∞}^{m}(Ω; ℝ^{n})

I claim that a_{k} ∈ S_{∞}^{−∞}(Ω, ℝ^{n}) and a_{k} ⟶ a in S_{∞}^{m′}(Ω; ℝ^{n}) for any m′ > m. [This gives a slightly stronger result than stated, i.e. the same sequence works for all m′ > m.] Certainly for each N

|a_{k}(z, ξ)| ≤ C_{N,n}(1 + |ξ|)^{−N}

and by Leibniz formula we easily conclude that

| D_{z}^{α}D_{ξ}^{β}a_{k}(z, ξ) |≤ C_{N,α,β,n}(1 + |ξ|)^{−N} ∀ α, β, N, n.

Thus a_{k} ∈ S_{∞}^{−∞}(Ω; ℝ^{n}).

So consider the difference

(a − a_{k})(z, ξ) = (1 − ϕ)(ξ/n) a(z, ξ). (2.13)

If \frac{1}{2} > ϵ > 0 is given then

| (1 − ϕ)(\frac{ξ}{n})(1 + |ξ|)^{−ϵ} |≤ C_{ϵ}n^{−ϵ/2}, n ≥ 1. (2.14)

Indeed, in |ξ| ≥ n^{\frac{1}{2}} , (1 + |ξ|)^{−ϵ} ≤ n^{− \frac{ϵ}{2}} and (1 − ϕ) is bounded, and in |ξ| ≤ n^{\frac{1}{2}} , by Taylor’s formula

\begin{matrix}(1 − ϕ)(\frac{ξ}{n}) ≤ C|\frac{ξ}{n}| ≤ Cn^{− \frac{1}{2}} , \\ (1 + |ξ|)^{−ϵ} ≤ C. \end{matrix} (2.15)

This gives (2.14) and hence

(1 + |ξ|)^{−m′} | (a − a_{k})(z, ξ) |≤ C(1 + |ξ|)^{−ϵ}|(1 − ϕ)(\frac{ξ}{n})| ≤ Cn^{− \frac{ϵ}{2}} ⟶ 0.

This is the first symbol estimate. Next consider the ξ derivatives of (2.13). Using Leibniz’ formula

D_{ξ}^{β}(a − a_{k}) = ∑_{γ≤β} \binom{β}{γ}D_{ξ}^{β−γ}(1 − ϕ)(\frac{ξ}{n}) · D_{ξ}^{γ}a(z, ξ) (2.16)

= (1 − ϕ)(\frac{ξ}{n}) · D_{ξ}^{β}a(z, ξ) − ∑_{γ<β} \binom{β}{γ}(D^{β−γ}ϕ)(\frac{ξ}{n}) · n^{−|β−γ|}D_{ξ}^{α}a(z, ξ). (2.17)

In the first term, D_{ξ}^{β}a(z, ξ) is a symbol of order m − |β|, so

sup_{ξ}(1 + |ξ|)^{−m′+|β|}(1 − ϕ)(\frac{ξ}{n})D_{ξ}^{β}a(x, ξ) ⟶ 0

as n ⟶ ∞ if m′ > m. In all the other terms, (D^{β−γ}ϕ)(ζ) has compact support, i.e. 1 ≤ |ζ| ≤ 2. Thus for each term we get a bound

sup_{n≤|ξ|≤2n}(1 + |ξ|)^{−m′+|β|} · n^{−|β−γ|}C · (1 + |ξ|)^{m−|γ|} ≤ Cn^{−m′+m}.

The variables z play the rôle of parameters so we have in fact shown that

sup_{\begin{matrix}z∈Ω \\ ξ∈ℝ^{n} \end{matrix}}(1 + |ξ|)^{−m′} | D_{z}^{α}D_{ξ}^{β}(a − a_{k}) |⟶ 0 as n ⟶ ∞.

This means a_{k} ⟶ a in S_{∞}^{m′}(ℝ; ℝ^{n}) as required.

In fact this proof suggests a couple of other ‘obvious’ results. Namely

S_{∞}^{m}(Ω; ℝ^{n}) · S_{∞}^{m′}(Ω; ℝ^{n}) ⊂ S_{∞}^{m+m′}(Ω; ℝ^{n}). (2.18)

This can be proved directly using Leibniz’ formula:

sup_{ξ}(1 + |ξ|)^{−m−m′+|β|} | D_{z}^{α}D_{ξ}^{β} (a(z, ξ) · b(z, ξ)) |

\begin{matrix}≤ ∑_{\begin{matrix}μ≤α \\ γ≤β \end{matrix}} \binom{α}{μ}\binom{β}{γ} & sup_{ξ}(1 + |ξ|)^{−m+|γ|} | D_{z}^{μ}D_{ξ}^{γ}a(z, ξ) | \\ × & sup_{ξ}(1 + |ξ|)^{−m′+|β−γ|} | D_{z}^{α−μ}D_{ξ}^{β−γ}b(z, ξ) |< ∞. \end{matrix}

We note the obvious action of differentiation:

\begin{matrix}D_{z}^{α} : S_{∞}^{m}(Ω; ℝ^{n}) ⟶ S_{∞}^{m}(Ω; ℝ^{n}) \\ D_{ξ}^{β} : S_{∞}^{m}(Ω; ℝ^{n}) ⟶ S_{∞}^{m−|β|}(Ω; ℝ^{n}). \end{matrix} (2.19)

In fact, while we are thinking about these things we might as well show the important consequence of ellipticity. A symbol a ∈ S_{∞}^{m}(Ω; ℝ^{n}) is said to be (globally) elliptic if

|a(z, ξ)| ≥ ϵ(1 + |ξ|)^{m} − C(1 + |ξ|)^{m−1}, ϵ > 0 (2.20)

or equivalently

|a(z, ξ)| ≥ ϵ(1 + |ξ|)^{m} in |ξ| ≥ C_{ϵ}. (2.21)

Lemma 2.2. If a ∈ S_{∞}^{m}(Ω; ℝ^{n}) is elliptic there exists b ∈ S_{∞}^{−m}(Ω; ℝ^{n}) such that

a · b − 1 ∈ S_{∞}^{−∞}(Ω; ℝ^{n}). (2.22)

Proof. Using (2.21) choose ϕ as in the proof of Lemma 2.1 and set

b(z, ξ) = \begin{cases}\frac{1−ϕ(2ξ/C)}{a(z,ξ)} & |ξ| ≥ C \\ 0 & |ξ| ≤ C. \end{cases} (2.23)

Then b is C^{∞} since b = 0 in C ≤ |ξ| ≤ C + ϵ for ϵ > 0. The symbol estimates follow by noting that, in |ξ| ≥ C,

D_{z}^{α}D_{ξ}^{β}b = a^{−1−2|α|−|β|} · G_{αβ}

where G_{αβ} is a symbol of order (|α| + |β|)m − |β|. This can be proved by induction.

2.2. Pseudodifferential operators

Now we proceed to discuss the formula (2.3) where we shall assume that, for some p, m ∈ ℝ

\begin{matrix}a(x, y, ξ) = (1 + |x − y|^{2})^{p/2}ã(x, y, ξ) \\ ã ∈ S_{∞}^{m}(ℝ_{(x,y)}^{2n}; ℝ_{ξ}^{n}). \end{matrix} (2.24)

The extra factor (which allows polynomial growth in the direction of x − y) turns out, somewhat enigmatically, to both make no difference and be very useful! Notice that if a ∈ C^{∞}(ℝ^{2n} × ℝ^{n}) then a ∈ (1 + |x − y|^{2})^{p/2}S^{m}(ℝ^{2n}; ℝ^{n}) if and only if

|D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a(x, y, ξ)| ≤ C_{α,β,γ}(1 + |x − y|)^{p}(1 + |ξ|)^{m} ∀ α, β, γ ∈ ℕ_{0}^{n}. (2.25)

If m < −n then, for each u ∈ S(ℝ^{n}) the integral in (2.3) is absolutely convergent, uniformly in x, since

\begin{matrix}|a(x, y, ξ)u(y)| ≤ C(1 + |x − y|)^{p}(1 + |ξ|)^{m}(1 + |y|)^{−N} \\ ≤ C(1 + |x|)^{p}(1 + |ξ|)^{m}(1 + |y|)^{m}, m < −n. \end{matrix}

Thus we conclude that, provided m < −n,

A : S(ℝ^{n}) ⟶ (1 + |x|^{2})^{p}C_{∞}^{0}(ℝ^{n}).

To show that, for general m, A exists as an operator, we prove that its Schwartz kernel exists.

Proposition 2.1. The map, defined for m < −n as a convergent integral,

\begin{matrix}(1 + |x − y|^{2})^{p/2}S_{∞}^{m}(ℝ^{2n}; & ℝ^{n}) ∋ a ⟼ I(a) = \\ (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, y, ξ)dξ ∈ S′(ℝ^{2n}) \end{matrix} (2.26)

extends by continuity to all m.

Proof. Since we already have the density of S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}) in S_{∞}^{m} we only need to show the continuity on this subspace. What we shall show is that, for each m ∈ ℝ there are integers N, k ∈ ℕ such that

| I(a)(ϕ) |≤ C∥a∥_{N,m}∥ϕ∥_{k} ∀ ϕ ∈ S(ℝ^{2n}). (2.27)

To see this we just use integration by parts!

Observe that

(2.28)  (1 + ξ · D_{x})e^{i(x−y)·ξ} = (1 + |ξ|^{2})e^{i(x−y)·ξ}
(2.29)  (1 − ξ · D_{y})e^{i(x−y)·ξ} = (1 + |ξ|^{2})e^{i(x−y)·ξ}.

Thus we can write (for a ∈ S_{∞}^{−∞}),

\begin{aligned}I(a)(ϕ) = ∫∫ (2π)^{−n} ∫ & e^{i(x−y)·ξ}(1 + |ξ|^{2})^{−2q} \\ & (1 − ξ · D_{x})^{q}(1 + ξ · D_{y})^{q}[a(x, y, ξ)ϕ(x, y)]dξdxdy \end{aligned} (2.30)

= ∑_{|γ|≤2q} ∫∫ ( ∫ e^{i(x−y)·ξ}a_{γ}(x, y, ξ)dξ)D_{(x,y)}^{γ}ϕ(x, y)dxdy. (2.31)

Here a_{γ} = (1 + |ξ|^{2})^{−2q} ∑\begin{matrix}μ≤γ \\ |γ|≤2q \end{matrix} C_{μ}ξ^{γ}D_{(x,y)}^{μ}a. So if we take −2q + m < −n, e.g.

q ≥ max(\frac{n+m}{2}, 0) then

| I(a)(ϕ) |≤ C∥(1 + |x − y|^{2})^{−p/2}a∥_{2q,m}∥ϕ∥_{2q+p}.

This proves the desired continuity.

In fact we do not really need to integrate by parts in both x and y, either will do the trick.

Lemma 2.3. If a ∈ (1 + |x − y|^{2})^{p/2}S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) then the operator A, with Schwartz kernel I(a), is a continuous linear map

A : S(ℝ^{n}) ⟶ S(ℝ^{n}). (2.32)

We shall denote by Ψ_{∞}^{m}(ℝ^{n}) the linear spaces of operator (2.32), corresponding to (1 + |x − y|^{2})^{p/2}a ∈ S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) for some p. They are called pseudodifferential operators.

Proof. Proceeding as in (2.31) but only integrating by parts in y we deduce that, for q large depending on m,

Au(ψ) = ∑_{γ≤2q}(2π)^{−n} ∫∫ ∫ e^{i(x−y)·ξ}a_{γ}(x, y, ξ)D_{y}^{γ}u(y)dξϕ(x)dydx.

This shows that Au is given by the convergent integral

Au(x) = ∑_{γ≤2q}(2π)^{−n} ∫∫ e^{i(x−y)·ξ}a_{γ}(x, y, ξ)D_{y}^{γ}u(y)dξdy.

Notice that

D_{x_{j}}Au(x) = (2π)^{−n} ∑_{|γ|≤2q} ∫∫ e^{i(x−y)·ξ}ξ_{j}a_{γ} · D_{y}^{γ}u(y)dydξ

and

x_{j}Au(x) = (2π)^{n} ∑_{|γ|≤2q} ∫∫ e^{i(x−y)·ξ}(−D_{ξ_{j}} + y_{j})a_{γ} · D_{y}^{γ}u(y)dydξ.

Proceeding inductively we find (2.32).

2.3. Composition

There are two extreme cases of I(a), namely where a is independent of x and of y. Below we shall prove:

Theorem 2.1 (Reduction). Each A ∈ Ψ_{∞}^{m}(ℝ^{n}) can be written uniquely as I(a′) where a′ ∈ S_{∞}^{m}(ℝ_{x}^{n}; ℝ_{ξ}^{n}).

This is the main step in proving the fundamental result of this Chapter, which is that two pseudodifferential operators can be composed to give a pseudodifferential operator and that the orders are additive.

Theorem 2.2. The space Ψ_{∞}^{∞}(ℝ^{n}) is a filtered ∗-algebra on S(ℝ^{n}).

More explicitly we have to show that each A ∈ Ψ_{∞}^{∞}(ℝ^{n}) defines a continuous linear map

A : S(ℝ^{n}) ⟶ S(ℝ^{n}) (2.33)

and that

A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ A^{∗} ∈ Ψ_{∞}^{m}(ℝ^{n}) (2.34)

A ∈ Ψ_{∞}^{m}(ℝ^{n}), B ∈ Ψ_{∞}^{m′}(ℝ^{n}) ⟹ A ∘ B ∈ Ψ_{∞}^{m+m′}(ℝ^{n}). (2.35)

In fact we will pick up some more information along the way.

2.4. Reduction

We proceed to prove Theorem 2.1, which we can restate as:

Proposition 2.2. The range of (2.26) (for any p) is the same as the range of I restricted to the image of the inclusion map

S_{∞}^{m}(ℝ^{n}; ℝ^{n}) ∋ a ⟼ a(x, ξ) ∈ S_{∞}^{m}(ℝ_{(x,y)}^{2n}; ℝ^{n}).
Proof. Suppose a ∈ (1 + |x − y|^{2})^{p/2}S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}) for some p, then

I((x_{j} − y_{j})a) = I(−D_{ξ_{j}}a) j = 1, …, n. (2.36)

Indeed this is just the result of inserting the identity

D_{ξ_{j}}e^{i(x−y)·ξ} = (x_{j} − y_{j})e^{i(x−y)·ξ}

into (2.26), with N = 0, and integrating by parts. Since both sides of (2.36) are continuous on (1 + |x − y|^{2})^{p/2}S_{∞}^{∞}(ℝ^{2n}; ℝ^{n}) the identity holds in general. Notice that if a is of order m then D_{ξ_{j}}a is of order m−1, so (2.36) shows that even though the operator with amplitude (x_{j} − y_{j})a(x, y, ξ) appears to have order m, it actually has order m − 1.

To exploit (2.36) consider the Taylor series (with Legendre’s remainder) for a(x, y, ξ) around x = y :

\begin{matrix}a(x, y, ξ) = ∑_{|α|≤N−1} \frac{(−i)^{|α|}}{α!} (x − y)^{α}(D_{y}^{α}a)(x, x, ξ) \\ + ∑_{|α|=N} \frac{(−i)^{|α|}}{α!} (x − y)^{α} · R_{N,α}(x, y, ξ). \end{matrix} (2.37)

Here,

R_{N,α}(x, y, ξ) = ∫_{0}^{1}(1 − t)^{N−1}(D_{y}^{α}a)(x, (1 − t)x + ty, ξ)dt. (2.38)

Now,

(x − y)^{α}(D_{y}^{α}a)(x, y, ξ) ∈ (1 + |x − y|^{2})^{\frac{(p+|α|)}{2}} S_{∞}^{m}(ℝ^{2n}; ℝ^{n}).

Applying (2.36) repeatedly we see that if A is the operator with kernel I(a) then

A = ∑_{j=0}^{N−1} A_{j} + R_{N}, A_{j} ∈ Ψ_{∞}^{m−j}(ℝ^{n}), R_{N} ∈ Ψ_{∞}^{m−N}(ℝ^{n}) (2.39)

where the A_{j} have kernels

I( ∑_{|α|=j} \frac{i^{|α|}}{α!} (D_{y}^{α}D_{ξ}^{α}a)(x, x, ξ)). (2.40)

To proceed further we need somehow to sum this series. Of course we cannot really do this!

2.5. Asymptotic summation

Suppose a_{j} ∈ S_{∞}^{m−j}(ℝ^{p}; ℝ^{n}). The fact that the orders are decreasing means that those symbols are getting very small, for |ξ| large. The infinite series

∑_{j} a_{j}(z, ξ)

need not converge. However we shall say that it converges asymptotically, or since it is a series we say it is ‘asymptotically summable,’ if there exists a ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n}) such that

(2.41) for every N, a − ∑_{j=0}^{N−1} a_{j} ∈ S_{∞}^{m−N}(ℝ^{p}; ℝ^{n}).

We write this relation as

a ∼ ∑_{j=0}^{∞} a_{j}. (2.42)

Proposition 2.3. Any series a_{j} ∈ S_{∞}^{m−j}(ℝ^{p}; ℝ^{n}) is asymptotically summable, in the sense of (2.41), and the asymptotic sum is well defined up to an additive term in S_{∞}^{−∞}(ℝ^{p}; ℝ^{n}).

Proof. The uniqueness part is easy. Suppose a and a′ both satisfy (2.41). Taking the difference

a − a′ = (a − ∑_{j=0}^{N−1} a_{j}) − (a′ − ∑_{j=0}^{N−1} a_{j}) ∈ S_{∞}^{m−N}(ℝ^{p}; ℝ^{n}). (2.43)

Since S_{∞}^{−∞}(ℝ^{p}; ℝ^{n}) is just the intersection of the S_{∞}^{−N}(ℝ^{p}; ℝ^{n}) over N it follows that a − a′ ∈ S_{∞}^{−∞}(ℝ^{p}; ℝ^{n}), proving the uniqueness.

So to the existence of an asymptotic sum. To construct this (by Borel’s method^{1}) we cut off each term “near infinity in ξ”. Thus fix ϕ ∈ C^{∞}(ℝ^{n}) with ϕ(ξ) = 0 in |ξ| ≤ 1, ϕ(ξ) = 1 in |ξ| ≥ 2, 0 ≤ ϕ(ξ) ≤ 1. Consider a decreasing sequence

ϵ_{0} > ϵ_{1} > ⋯ > ϵ_{j} ↓ 0. (2.44)

We want to set

a(z, ξ) = ∑_{j=0}^{∞} ϕ(ϵ_{j}ξ)a_{j}(z, ξ). (2.45)

Of course we need to consider the seminorms, in S_{∞}^{m}(ℝ^{p}; ℝ^{n}) of each term. The first of these is

sup_{z} sup_{ξ}(1 + |ξ|)^{−m} | ϕ(ϵ_{j}ξ) | | a_{j}(z, ξ) | . (2.46)

Now ϕ(ϵ_{j}ξ) = 0 unless |ϵ_{j}ξ| ≤ 1, i.e. |ξ| ≤ \frac{1}{ϵ_{j}} . Since a_{j} is of order m − j this allows us to estimate (2.46) by

sup_{z} sup_{|ξ|≤ \frac{1}{ϵ_{j}}} (1 + |ξ|)^{−j} · [(1 + |ξ|)^{−m+j} | a_{j}(z, ξ) |]

≤ (1 + \frac{1}{ϵ_{j}} )^{−j} · C_{j} ≤ ϵ_{j}^{j} · C_{j}

where the C_{j}’s are fixed constants, independent of ϵ_{j}, Let’s look at the higher symbol estimates. As usual we can apply Leibniz’ formula:

sup_{z} sup_{ξ}(1 + |ξ|)^{−m+|β|} | D_{z}^{α}D_{ξ}^{β}ϕ(ϵ_{j}ξ)a_{j}(z, ξ) |

≤ ∑_{μ≤β} sup_{z} sup_{ξ}(1 + |ξ|)^{|β|−|μ|−j}ϵ_{j}^{|β|−|μ|} | (D^{β−μ}ϕ)(ϵ_{j}ξ) |

×(1 + |ξ|)^{−m+j+|μ|} | D_{z}^{α}D_{ξ}^{μ}a_{j}(z, ξ) | .

The term with μ = β we estimate as before and the others, with μ ≠ β are supported in \frac{1}{ϵ_{j}} ≤ |ξ| ≤ \frac{2}{ϵ_{j}} . Then we find that for all j

∥ϕ(ϵ_{j}ξ)a_{j}(z, ξ)∥_{N,m} ≤ C_{N,j}ϵ_{j}^{j} (2.47)

where C_{N} is independent of ϵ_{j}.

So we see that for a given N we can arrange that, for instance,

∥ϕ(ϵ_{j}ξ)a_{j}(z, ξ)∥_{N,m} ≤ C_{N} \frac{1}{j^{2}}

1Émile Borel

by choosing the ϵ_{j} to satify

C_{N,j}ϵ_{j}^{j} ≤ \frac{1}{j^{2}} ∀ j ≥ j(N).

Notice the crucial point here, we can arrange that for each N the sequence of norms in (2.47) is dominated by C_{N}j^{−2} by fixing ϵ_{j} < ϵ_{j,N} for large j. Thus we can arrange convergence of all the sums

∑_{j} ∥ϕ(ϵ_{j}ξ)a_{j}(z, ξ)∥_{N,n}

by diagonalization of the ϵ, N sequences. Thus by choosing ϵ_{j} ↓ 0 rapidly enough we ensure that the series (2.45) converges. In fact the same argument allows us to ensure that for every N

(2.48)  ∑_{j≥N} ϕ(ϵ_{j}ξ)a_{j}(z, ξ) converges in S_{∞}^{m−N}(ℝ^{p}; ℝ^{n}).

This certainly gives (2.41) with a defined by (2.45).

2.6. Residual terms

Now we can apply Proposition 2.3 to the series (2.40), i.e. we can find b ∈ S_{∞}^{m}(ℝ_{x}^{n}; ℝ_{ξ}^{n})

b(x, ξ) ∼ ∑_{α} \frac{i^{|α|}}{α!} (D_{y}^{α}a)(x, x, ξ). (2.49)

Let B = I(b) be the operator defined by the symbol (which is independent of y). Now (2.39) says that

A − B = ∑_{j=0}^{N−1} A_{j} + R_{N} − B

and from (2.41) applied to (2.49)

B = ∑_{j=0}^{N−1} A_{j} + R′_{N}, R′_{N} ∈ Ψ_{∞}^{m−N}(ℝ^{n})

Thus

A − B ∈ Ψ_{∞}^{−∞}(ℝ^{n}) = ⋂_{N} Ψ_{∞}^{N}(ℝ^{n}). (2.50)

Notice that, at this stage, we do not know that A − B has kernel I(c) with c ∈ S_{∞}^{−∞}(ℝ^{2n}, ℝ^{n}), just that it has kernel I(c_{N}) with c_{N} ∈ S_{∞}^{N}(ℝ^{2n}; ℝ^{n}) for each N.

However:

Proposition 2.4. An operator A : S(ℝ^{n}) ⟶ S′(ℝ^{n}) is an element of the space Ψ_{∞}^{−∞}(ℝ^{n}) if and only if its Schwartz kernel is C^{∞} and satisfies the estimates

| D_{x}^{α}D_{y}^{β}K(x, y) |≤ C_{N,α,β}(1 + |x − y|)^{−N} ∀ N. (2.51)

Proof. Suppose first that A ∈ Ψ_{∞}^{−∞}(ℝ^{n}), which means that A ∈ Ψ_{∞}^{N}(ℝ^{n}) for every N. The Schwartz kernel, K_{A}, of A is therefore given by (2.26) with the amplitude C_{N} ∈ S_{∞}^{N}(ℝ^{2n}; ℝ^{n}). For N << −n−1−p the integral certainly converges absolutely and we can integrate by parts to show that

\begin{matrix}(x − y)^{α}D_{x}^{β}D_{y}^{γ}K_{A}(x, y) \\ = (2π)^{−N} ∫ e^{i(x−y)·ξ}(−D_{ξ})^{α}(D_{x} + iξ)^{β}(D_{y} − iξ)^{γ}a(x, y, ξ)dζ \end{matrix}

which converges absolutely provided |β| + |γ| + N − |α| < −n. Thus

sup | (x − y)^{α}D_{x}^{β}D_{y}^{γ}K |< ∞ ∀ α, β, γ

which is another way of writing (2.51) i.e.

sup(1 + |x − y|^{2})^{N} | D_{x}^{β}D_{y}^{γ}K |< ∞ ∀ β, γ, N.

Conversely suppose that (2.51) holds. Define

g(x, z) = K(x, x − z). (2.52)

The estimates (2.51) become

sup | D_{x}^{α}z^{γ}D_{z}^{β}g(x, z) |< ∞ ∀ α, β, γ. (2.53)

That is, g is rapidly decreasing with all its derivatives in z. Taking the Fourier transform,

b(x, ξ) = ∫ e^{−iz·ξ}g(x, z)dz (2.54)

the estimate (2.53) translates to

\begin{matrix}| D_{x}^{α}ξ^{β}D_{ξ}^{γ}b(x, ξ) |< ∞ ∀ α, β, γ \\ ⟺ b ∈ S_{∞}^{−∞}(ℝ_{x}^{n}; ℝ_{ξ}^{n}). \end{matrix} (2.55)

Now the inverse Fourier transform in (2.54), combined with (2.52) gives

K(x, y) = g(x, x − y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}b(x, ξ)dξ (2.56)

i.e. K = I(b). This certainly proves the proposition and actually gives the stronger result.

A ∈ Ψ_{∞}^{−∞}(ℝ^{n}) ⟺ A = I(c), c ∈ S_{∞}^{−∞}(ℝ_{x}^{n}; ℝ_{ξ}^{n}). (2.57)

This also finishes the proof of Proposition 2.2 since in (2.49), (2.50) we have shown that

A = B + R, B = I(b), R ∈ Ψ_{∞}^{−∞}(ℝ^{n}) (2.58)

so in fact

A = I(e), e ∈ S_{∞}^{m}(ℝ_{x}^{n}; ℝ_{ξ}^{n}), e ∼ ∑_{α} \frac{i^{|α|}}{α!} (D_{y}^{α}D_{ξ}^{α}a)(x, x, ξ). (2.59)

2.7. Proof of Theorem 2.2

First consider the adjoint formula. If

A : S(ℝ^{n}) ⟶ S(ℝ^{n})

the adjoint is the operator

A^{∗} : S′(ℝ^{n}) ⟶ S′(ℝ^{n})

defined by duality:

A^{∗}u(ϕ̄) = u(\overline{Aϕ}) ∀ ϕ ∈ S(ℝ^{n}) (2.60)

Certainly A^{∗}u ∈ S′(ℝ^{n}) if u ∈ S′(ℝ^{n}) since

\begin{matrix}A^{∗}u(ψ) = u(\overline{A\overline{ψ}}) and \\ S(ℝ^{n}) ∋ ψ ⟼ \overline{Aψ̄} ∈ S(ℝ^{n}) \end{matrix} (2.61)

is clearly continuous. In terms of Schwartz kernels,

\begin{matrix}Aϕ(x) = ∫ K_{A}(x, y)ϕ(y)dy \\ A^{∗}u(x) = ∫ K_{A^{∗}}(x, y)u(y)dy \end{matrix} (2.62)

we see that

\begin{matrix}∫ K_{A^{∗}}(x, y)u(y)\overline{ϕ(x)}dydx \\ = ∫ \overline{K_{A}(x, y)ϕ(y)}dyu(x)dx \\ ⟹ K_{A^{∗}}(x, y) = \overline{K_{A}(y, x)} \end{matrix} (2.63)

where we are using the uniqueness of Schwartz kernels. This proves the first part of (2.35) since

\begin{aligned}\overline{K_{A}(y, x)} & = \overline{[\frac{1}{(2π)^{n}} ∫ e^{i(y−x)·ξ}a(y, x, ξ)dξ]} \\ & = \frac{1}{(2π)^{n}} ∫ e^{i(x−y)·ξ}ā(y, x, ξ)dξ \end{aligned} (2.64)

i.e. A^{∗} = I(ā(y, x, ξ)). Thus one advantage of allowing general operators (2.26) is that closure under adjoint is immediate.

For the composition formula we need to apply Proposition 2.2 twice. First to A ∈ Ψ_{∞}^{m}(ℝ^{n}), to write it with symbol a(x, ξ)

(2.65)  Aϕ(x) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, ξ)ϕ(y)dydξ

= (2π)^{−n} ∫ e^{ix·ξ}a(x, ξ)ϕ̂(ξ)dξ. (2.66)

Then we also apply Proposition 2.2 to B^{∗},

B^{∗}u(x) = (2π)^{−n} ∫ e^{ix·ξ}b̄(x, ξ)û(ξ)dξ

Integrating this against a test function ϕ ∈ S(ℝ^{n}) gives

\begin{matrix}⟨ϕ, B^{∗}u⟩ = (2π)^{−n} ∫ ∫ e^{−ix·ξ}ϕ(x)b(x, ξ)\overline{û(ξ)}dξdx \\ ⟹ B̂u(ξ) = ∫ e^{−iy·ξ}b(y, ξ)u(y)dy. \end{matrix}

Inserting this into the formula for Aϕ shows that

⟹ AB(u) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, ξ)b(y, ξ)u(y)dydξ.

Since a(x, ξ)b(y, ξ) ∈ S_{∞}^{m+m′}(ℝ_{(x,y)}^{2n}; ℝ_{ξ}^{n}) this shows that AB ∈ Ψ_{∞}^{m+m′}(ℝ^{n}) as claimed.

2.8. Problems

Problem 2.1. Show, in detail, that for each m ∈ ℝ

(1 + |ξ|^{2})^{− \frac{1}{2} m} ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n})

for any p. Use this to show that

S_{∞}^{m}(ℝ^{p}; ℝ^{n}) · S_{∞}^{m′}(ℝ^{p}; ℝ^{n}) = S_{∞}^{m+m′}(ℝ^{p}; ℝ^{n}).

Problem 2.2. Consider p = 0 and n = 2 in the definition of symbols and show that if a ∈ S_{∞}^{1}(ℝ^{2}) is elliptic then for r > 0 sufficiently large the integral

∫_{0}^{2π} \frac{1}{a(re^{iθ})} \frac{d}{dθ} a(re^{iθ})dθ

exists and is independent of r, where z = ξ_{1} +iξ_{2} is the complex variable in ℝ^{2} = ℂ. Conclude that there is an elliptic symbol, a on ℝ^{2}, such that there does not exist b, a symbol with

b ≠ 0 on ℝ^{2} and a(ξ) = b(ξ) for |ξ| > r

for any r.

Problem 2.3. Show that A ∈ Ψ_{∞}^{−∞}(ℝ^{n}) if and only if its Schwartz kernel is C^{∞} and satisfies all the estimates

|D_{x}^{α}D_{y}^{β}a(x, y)| < C_{α,β,N}(1 + |x − y|)^{−N}

for multiindices α, β ∈ ℕ_{0}^{n} and N ∈ ℕ_{0}.