CHAPTER 2 Pseudodifferential operators on Euclidean space^{[C.Euclidean]}

Formula (1.92) for the action of a differential operator (with coefficients in C_{∞}^{∞}(ℝ^{n})) on S(ℝ^{n}) can be written

\begin{aligned}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 \end{aligned}^{[3.1]} (2.1)

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, which is to say a smooth function satisfying certain uniformity conditions at infinity. In fact we shall also allow the symbol, or rather the amplitude, in the integral (2.1) to depend in addition 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}).^{[3.2]} (2.2)

Of course it is not clear that this integral is well-defined.

To interpret (2.2) we shall first look into the definition and properties of symbols. Then we show how this integral can be interpreted as an oscillatory integral and that it thereby defines an operator on S(ℝ^{n}). We then investigate the properties of these pseudodifferential operators at some length.

2.1. Symbols^{[S.Symbols]}

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

|p(ξ)| ≤ C(1 + |ξ|)^{m} ∀ ξ ∈ ℝ^{n}.^{[3.3]} (2.3)

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}.^{[3.4]} (2.4)

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) (2.4) 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}.^{[3.5]} (2.5)

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

32 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}

[3.6]

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}.^{[3.7]} (2.6)

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 (2.6) for (z, ξ) ∈ int(Ω) × ℝ^{n}.

The estimates (2.6) can be rewritten

(2.7)  ∥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)| < ∞.^{[3.8]}

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}).^{[3.9]} (2.8)

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.

A more significant issue is: Why this class precisely? As we shall see below, there are other choices which are not only possible but even profitable to make. However, the present one has several virtues. It is large enough to cover most of the straightforward things we want to do (at least initially) and small enough to ‘work’ easily. It leads to what I shall refer to as the ‘traditional’ algebra of pseudodifferential operators.

Now to some basic properties. First notice that

(1 + |ξ|)^{m} ≤ C(1 + |ξ|)^{m′} ∀ ξ ∈ ℝ^{n} ⟺ m ≤ m′. (2.9)

Thus we have an inclusion

S_{∞}^{m}(Ω; ℝ^{n}) ↪ S_{∞}^{m′}(Ω; ℝ^{n}) ∀ m′ ≥ m.^{[3.10]} (2.10)

Moreover this inclusion is continuous, since from (2.7), ∥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}).^{[3.11]} (2.11)

Notice that the two ‘∞s’ here are quite different. The subscript refers to the fact that the ‘coefficients’ are bounded and stands for L^{∞} whereas the superscript ∞ stands really for ℝ. The residual space of this filtration is

S_{∞}^{−∞}(Ω; ℝ^{n}) = ⋂_{m} S_{∞}^{m}(Ω; ℝ^{n}).^{[3.12]} (2.12)

In fact the inclusion (2.10) is never dense if m′ > m. Instead we have the following rather technical, but nevertheless very useful, result.

[3.13]

Lemma 2.1. For any m ∈ ℝ and any a ∈ S_{∞}^{m}(Ω; ℝ^{n}) there is a sequence in S_{∞}^{−∞}(Ω; ℝ^{n}) which is bounded in S_{∞}^{m}(Ω; ℝ^{n}) and converges to a in the topology

2.1. SYMBOLS^{[S.SY MBOLS]} 33

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

The reason one cannot take m′ = m here is essentially the same reason that underlies the fact that S(ℝ^{n}) is not dense in C_{∞}^{∞}(ℝ^{n}). Namely any uniform limit obtained from a converging Schwartz sequence must vanish at infinity. In particular the constant function 1 ∈ S_{∞}^{0}(ℝ^{p}; ℝ^{n}) cannot be in the closure in this space of S_{∞}^{−∞}(ℝ^{p}; ℝ^{n}) if n > 0.

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

a_{k}(z, ξ) = ϕ(ξ/k)a(z, ξ), a ∈ S_{∞}^{m}(Ω; ℝ^{n}). (2.13)

We shall show that a_{k} ∈ S_{∞}^{−∞}(Ω, ℝ^{n}) is a bounded sequence in S_{∞}^{m}(Ω; ℝ^{n}) and that a_{k} ⟶ a in S_{∞}^{m′}(Ω; ℝ^{n}) for any m′ > m. Certainly for each N

|a_{k}(z, ξ)| ≤ C_{N,k}(1 + |ξ|)^{−N} (2.14)

since ϕ has compact support. Leibniz’ formula gives

D_{z}^{α}D_{ξ}^{β}a_{k}(z, ξ) = ∑_{β′≤β} \binom{β′}{β}k^{−|β′|}(D^{β′}ϕ)(ξ/k)D_{z}^{α}D_{ξ}^{β−β′}a(z, ξ).^{[1.2.2000.353]} (2.15)

On the support of ϕ(ξ/k), |ξ| ≤ k so, using the symbol estimates on a, it follows that a_{k} is bounded in S_{∞}^{m}(Ω; ℝ^{n}). We easily conclude that

|D_{z}^{α}D_{ξ}^{β}a_{k}(z, ξ)| ≤ C_{N,α,β,k}(1 + |ξ|)^{−N} ∀ α, β, N, k. (2.16)

Thus a_{k} ∈ S_{∞}^{−∞}(Ω; ℝ^{n}).

So consider the difference

(a − a_{k})(z, ξ) = (1 − ϕ)(ξ/k) a(z, ξ).^{[3.14]} (2.17)

Now, |(1 − ϕ)(ξ/k)| = 0 in |ξ| ≤ k so we only need estimate the difference in |ξ| ≥ k where this factor is bounded by 1. In this region 1 + |ξ| ≥ 1 + k so, since −m′ + m < 0,

(2.18) (1 + |ξ|)^{−m′} |(a − a_{k})(z, ξ)| ≤

(1 + k)^{−m′+m} sup_{z,ξ} |(1 + |ξ|)^{−m}|a(z, ξ)| ≤ (1 + k)^{−m′+m}∥a∥_{0,m} ⟶ 0.

This is convergence with respect to the first symbol norm.

Next consider the ξ derivatives of (2.17). Using Leibniz’ formula

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

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

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

sup_{ξ}(1 + |ξ|)^{−m′+|β|}(1 − ϕ)(\frac{ξ}{k})D_{ξ}^{β}a(x, ξ) ⟶ 0 (2.19)

34 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}

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

sup_{k≤|ξ|≤2k}(1 + |ξ|)^{−m′+|β|} · k^{−|β−γ|}C · (1 + |ξ|)^{m−|γ|} ≤ Ck^{−m′+m}. (2.20)

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′+|β|}|D_{z}^{α}D_{ξ}^{β}(a − a_{k})| ⟶ 0 as k ⟶ ∞. (2.21)

This means a_{k} ⟶ a in each of the symbol norms, and hence in the topology of S_{∞}^{m′}(ℝ^{p}; ℝ^{n}) as desired.

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}).^{[3.17]} (2.22)

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 also note the action of differentiation:

\begin{matrix}D_{z}^{α} : S_{∞}^{m}(Ω; ℝ^{n}) ⟶ S_{∞}^{m}(Ω; ℝ^{n}) and \\ D_{ξ}^{β} : S_{∞}^{m}(Ω; ℝ^{n}) ⟶ S_{∞}^{m−|β|}(Ω; ℝ^{n}). \end{matrix}^{[3.18]} (2.23)

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^{[3.19]} (2.24)

or equivalently^{1}

|a(z, ξ)| ≥ ϵ(1 + |ξ|)^{m} in |ξ| ≥ C_{ϵ}, ϵ > 0.^{[3.20]} (2.25)

[3.21]

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}).^{[3.22]} (2.26)

Proof. Using (2.25) choose ϕ as in the proof of Lemma 2.1 and set

b(z, ξ) = \begin{cases}\frac{1−ϕ(ξ/2C)}{a(z,ξ)} & |ξ| ≥ C \\ 0 & |ξ| ≤ C. \end{cases} ^{[3.23]} (2.27)

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

D_{z}^{α}D_{ξ}^{β}b = a^{−1−|α|−|β|} · G_{αβ}^{[1.2.2000.277]} (2.28)

1Note it is required that ϵ be chosen to be independent of z here, so this is a notion of uniform ellipticity.

2.2. PSEUDODIFFERENTIAL OPERATORS^{[S.P SEUDODIF F ERENT IAL.OP ERAT ORS]} 35

where G_{αβ} is a symbol of order (|α|+|β|)m−|β|. This may be proved by induction. Indeed, it is true when α = β = 0. Assuming (2.28) for some α and β, differentiation of (2.28) gives

D_{z_{j}}D_{z}^{α}D_{ξ}^{β}b = D_{z_{j}}a^{−1−|α|−|β|} · G_{αβ} = a^{−2−|α|−|β|}G′,

G′ = (−1 − |α| − |β|)(D_{z_{j}}a)G_{αβ} + aD_{z_{j}}G_{αβ}.

By the inductive hypothesis, G′ is a symbol of order (|α|+1+|β|)m−|β|. A similar argument applies to derivatives with respect to the ξ variables.

2.2. Pseudodifferential operators^{[S.P seudodifferential.operators]}

Now we proceed to discuss the formula (2.2) where we shall assume that, for some w, m ∈ ℝ,

\begin{matrix}a(x, y, ξ) = (1 + |x − y|^{2})^{w/2}ã(x, y, ξ) \\ ã ∈ S_{∞}^{m}(ℝ_{(x,y)}^{2n}; ℝ_{ξ}^{n}). \end{matrix}^{[3.24]} (2.29)

The extra ‘weight’ 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^{2} that if a ∈ C^{∞}(ℝ^{2n} × ℝ^{n}) then a ∈ (1 + |x − y|^{2})^{w/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|)^{w}(1 + |ξ|)^{m−|γ|} ∀ α, β, γ ∈ ℕ_{0}^{n}.^{[3.32]} (2.30)

If m < −n then, for each u ∈ S(ℝ^{n}) the integral in (2.2) is absolutely convergent, locally uniformly in x, since

\begin{matrix}|a(x, y, ξ)u(y)| ≤ C(1 + |x − y|)^{w}(1 + |ξ|)^{m}(1 + |y|)^{−N} \\ ≤ C(1 + |x|)^{w}(1 + |ξ|)^{m}(1 + |y|)^{m}, m < −n. \end{matrix} (2.31)

Here we have used the following simple consequence of the triangle inequality

(1 + |x − y|) ≤ (1 + |x|)(1 + |y|)

from which it follows that

(1 + |x − y|)^{w} ≤ \begin{cases}(1 + |x|)^{w}(1 + |y|)^{w} & if w > 0 \\ (1 + |x|)^{w}(1 + |y|)^{−w} & if w ≤ 0. \end{cases} ^{[19.2.1998.102]} (2.32)

Thus we conclude that, provided m < −n,

A : S(ℝ^{n}) ⟶ (1 + |x|^{2})^{w/2}C_{∞}^{0}(ℝ^{n}).^{[3.33]} (2.33)

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

[3.25]

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

\begin{aligned}(1 + |x − y|^{2})^{w/2}S_{∞}^{m} & (ℝ^{2n}; ℝ^{n}) ∋ a ⟼ I(a) = \\ & (2π)^{−n} ∫ e^{i(x−y)·ξ}a(x, y, ξ)dξ ∈ (1 + |x|^{2} + |y|^{2})^{w/2}C_{∞}^{0}(ℝ^{2n}) \end{aligned}^{[3.26]} (2.34)

2See Problem 2.5.

extends by continuity to

I : (1 + |x − y|^{2})^{w/2}S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) ⟶ S′(ℝ^{2n})^{[1.2.2000.302]} (2.35)

for each w, m ∈ ℝ in the topology of S_{∞}^{m′}(ℝ^{2n}; ℝ^{n}) for any m′ > m.

Proof. Since we already have the density of S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}) in S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) in the toplogy of S_{∞}^{m′}(ℝ^{2n}; ℝ^{n}) for any m′ > m, we only need to show the continuity of the map (2.34) on this residual subspace with respect to the topology of S_{∞}^{m′}(ℝ^{2n}; ℝ^{n}) for any m′, which we may as well write as m. What we shall show is that, for each w, m ∈ ℝ, there are integers N, k ∈ ℕ such that, in terms of the norms in (2.7) and (1.6)

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

a = (1 + |x − y|^{2})^{w/2}ã, ã ∈ S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}).^{[3.27]}

To see this we just use integration by parts.

Set ϕ̃(x, y) = (1 + |x − y|^{2})^{w/2}ϕ(x, y). Observe that

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

Thus we can write, for ã ∈ S_{∞}^{−∞}, with a = (1 + |x − y|^{2})^{w/2}ã and for any q ∈ ℕ

\begin{matrix}I(a)(ϕ) = & ∫∫ (2π)^{−n} ∫ e^{i(x−y)·ξ}(1 + |ξ|^{2})^{−2q} \\ (1 − ξ · D_{x})^{q}(1 + ξ · D_{y})^{q}[ã(x, y, ξ)ϕ̃(x, y)]dξdxdy \\ = & ∑_{|γ|≤2q} ∫∫ ( ∫ e^{i(x−y)·ξ}a_{γ}^{(q)}(x, y, ξ)dξ)D_{(x,y)}^{γ}ϕ̃(x, y)dxdy. \end{matrix}^{[3.28]} (2.37)

Here the a_{γ}^{(q)} arise by expanding the powers of the operator

(1 − ξ · D_{x})^{q}(1 + ξ · D_{x})^{q} = ∑_{|μ|,|ν|≤q} C_{μ,ν}ξ^{μ+ν}D_{x}^{μ}D_{y}^{ν}

and applying Leibniz’ formula. Thus a_{γ}^{(q)} arises from terms in which 2q −|γ| derivatives act on ã so it is of the form

a_{γ} = (1 + |ξ|^{2})^{−2q} ∑_{|μ|≤|γ|,|γ|≤2q} C_{μ,γ}ξ^{γ}D_{(x,y)}^{μ}ã

⟹ ∥a_{γ}∥_{N,m} ≤ C_{m,q,N}∥ã∥_{N+2q,m−2q} ∀ m, N, q.

So (for given m) if we take −2q + m < −n, e.g. q > max(\frac{n+m}{2}, 0) and use the integrability of (1 + |x| + |y|)^{−2n−1} on ℝ^{2n}, then

|I(a)(ϕ)| ≤ C∥ã∥_{2q,m}∥ϕ̃∥_{2q+2n+1} ≤ C∥ã∥_{2q,m}∥ϕ∥_{2q+w+2n+1}. (2.38)

This is the estimate (2.36), which proves the desired continuity.

In showing the existence of the Schwartz’ kernel in this proof we do not really need to integrate by parts in both x and y; either separately will do the trick. We can use this observation to show that these pseudodifferential operator act on S(ℝ^{n}).

2.3. COMPOSITION^{[S.COMP OSIT ION]} 37

[3.29]

Lemma 2.3. If a ∈ (1 + |x − y|^{2})^{w/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}).^{[3.30]} (2.39)

We shall denote by Ψ_{∞}^{m}(ℝ^{n}) the linear space of operators (2.39), corresponding to (1 + |x − y|^{2})^{−w/2}a ∈ S_{∞}^{m}(ℝ^{2n}; ℝ^{n}) for some w. I call them pseudodifferential operators ‘of traditional type’ – or type ‘1,0’, the meaning of which is explained in Problem 2.16 below.

Proof. Proceeding as in (2.37) 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)dξψ(x)dydx,

a_{γ} ∈ (1 + |x − y|^{2})^{w/2}S^{m−q}(ℝ^{2n}; ℝ^{n}) if a ∈ (1 + |x − y|^{2})^{w/2}S^{m}(ℝ^{2n}; ℝ^{n}).

The integration by parts is justified by continuity from S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}). Taking −q+ m < −n − |w|, this shows that Au is given by the convergent integral

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

A : S(ℝ^{n}) ⟶ (1 + |x|^{2})^{\frac{|w|}{2}} C_{∞}^{0}(ℝ^{n})^{[1.2.2000.282]}

which is really just (2.33) again. The important point is that the weight depends on w but not on m. Notice that

D_{x_{j}}Au(x) = (2π)^{−n} ∑_{|γ|≤2q} ∫∫ e^{i(x−y)·ξ}(ξ_{j} + D_{x_{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 (2.39) follows from (2.33) or (2.40) since we conclude that

x^{α}D_{x}^{β}Au ∈ (1 + |x|^{2})^{\frac{|w|}{2}} C_{∞}^{0}(ℝ^{n}), ∀ α, β ∈ ℕ_{0}^{n}

and this implies that Au ∈ S(ℝ^{n}).

2.3. Composition^{[S.Composition]}

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

[3.31]

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. Thus our aim is to demonstrate the fundamental

Theorem 2.2. [Composition] The space Ψ_{∞}^{∞}(ℝ^{n}) is an order-filtered ∗-algebra on S(ℝ^{n}).

We have already shown that each A ∈ Ψ_{∞}^{∞}(ℝ^{n}) defines a continuous linear map (2.39). We now want to show that

A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ A^{∗} ∈ Ψ_{∞}^{m}(ℝ^{n})^{[4.35]} (2.41)

A ∈ Ψ_{∞}^{m}(ℝ^{n}), B ∈ Ψ_{∞}^{m′}(ℝ^{n}) ⟹ A ∘ B ∈ Ψ_{∞}^{m+m′}(ℝ^{n}),^{[4.3]} (2.42)

since this is what is meant by an order-filtered (the orders add on composition) ∗-algebra (meaning (2.41) holds). In fact we will pick up some more information along the way.

2.4. Reduction^{[S.Reduction]}

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

[4.4]

Proposition 2.2. The range of (2.34) (for any w) 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})^{w/2}S_{∞}^{−∞}(ℝ^{2n}; ℝ^{n}) for some w, then

I((x_{j} − y_{j})a) = I(−D_{ξ_{j}}a) j = 1, …, n.^{[4.5]} (2.43)

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 (2.34) and integrating by parts. Since both sides of (2.43) are continuous on (1 + |x − y|^{2})^{w/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 (2.43) 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 (2.43) consider the Taylor series (with Legendre’s remainder) for a(x, y, ξ) around x = y :

(2.44) 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).^{[4.6]}

Here,

R_{N,α}(x, y, ξ) = ∫_{0}^{1}(1 − t)^{N−1}(D_{y}^{α}a)(x, (1 − t)x + ty, ξ)dt.^{[4.7]} (2.45)

Now,

(x − y)^{α}(D_{y}^{α}a)(x, y, ξ) ∈ (1 + |x − y|^{2})^{\frac{(w+|α|)}{2}} S_{∞}^{m}(ℝ^{2n}; ℝ^{n}). (2.46)

2.5. ASYMPTOTIC SUMMATION^{[S.ASY MP T OT IC.SUMMAT ION]} 39

Applying (2.43) 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})^{[4.8]} (2.47)

where the A_{j} have kernels

I( ∑_{|α|=j} \frac{i^{|α|}}{α!} (D_{y}^{α}D_{ξ}^{α}a)(x, x, ξ)).^{[4.9]} (2.48)

To proceed further we need somehow to sum this series. Of course we cannot really do this, but we can come close!

2.5. Asymptotic summation^{[S.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) (2.49)

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,

(2.50) for every N, a − ∑_{j=0}^{N−1} a_{j} ∈ S_{∞}^{m−N}(ℝ^{p}; ℝ^{n}).^{[4.10]}

We write this relation as

a ∼ ∑_{j=0}^{∞} a_{j}.^{[4.11]} (2.51)

[4.12]

Proposition 2.3. Any series a_{j} ∈ S_{∞}^{m−j}(ℝ^{p}; ℝ^{n}) is asymptotically summable, in the sense of (2.50), 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 (2.50). 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}).^{[4.13]} (2.52)

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^{3}) 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.^{[4.14]} (2.53)

3Émile Borel

We shall set

a(z, ξ) = ∑_{j=0}^{∞} ϕ(ϵ_{j}ξ)a_{j}(z, ξ).^{[4.15]} (2.54)

Since ϕ(ϵ_{j}ξ) = 0 in |ξ| < 1/ϵ_{j} → ∞ as j → ∞, only finitely many of these terms are non-zero in any ball |ξ| ≤ R. Thus a(z, ξ) is a well-defined C^{∞} function. 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)|.^{[4.16]} (2.55)

Now |ξ| ≤ \frac{1}{ϵ_{j}} on the support of ϕ(ϵ_{j}ξ)a_{j}(z, ξ) and since a_{j} is a symbol of order m − j this allows us to estimate (2.55) 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 us 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}^{[4.17]} (2.56)

where C_{N,j} is independent of ϵ_{j}.

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

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

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 (2.56) 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,m}

by diagonalization, for example setting ϵ_{j} = \frac{1}{2}ϵ_{j,j}. Thus by choosing ϵ_{j} ↓ 0 rapidly enough we ensure that the series (2.54) converges. In fact the same argument allows

2.6. RESIDUAL TERMS^{[S.RESIDUAL.T ERMS]} 41

us to ensure that for every N

(2.57)  ∑_{j≥N} ϕ(ϵ_{j}ξ)a_{j}(z, ξ) converges in S_{∞}^{m−N}(ℝ^{p}; ℝ^{n}).^{[4.18]}

This certainly gives (2.50) with a defined by (2.54).

2.6. Residual terms^{[S.Residual.terms]}

Now we can apply Proposition 2.3 to the series in (2.48), that is we can find b ∈ S_{∞}^{m}(ℝ_{x}^{n}; ℝ_{ξ}^{n}) satisfying

b(x, ξ) ∼ ∑_{α} \frac{i^{|α|}}{α!} (D_{y}^{α}a)(x, x, ξ).^{[4.19]} (2.58)

Let B = I(b) be the operator defined by this amplitude (which is independent of y). Now (2.47) says that

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

and from (2.50) applied to (2.58)

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}).^{[4.20]} (2.59)

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:

[4.21]

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} ∀ α, β, N.^{[4.22]} (2.60)

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 (2.34) with the amplitude a_{N} ∈ S_{∞}^{N}(ℝ^{2n}; ℝ^{n}). For N << −n − 1 − p the integral converges absolutely and we can integrate by parts to show that

(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_{N}(x, y, ξ)dξ

which converges absolutely, and uniformly in x, y, provided |β|+|γ|+N −|α| < −n. Thus

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

42 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}

which is another way of writing (2.60) i.e.

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

Conversely suppose that (2.60) holds. Define g(x, z) = K(x, x − z).^{[4.23]} (2.61) The estimates (2.60) become sup |D_{x}^{α}z^{γ}D_{z}^{β}g(x, z)| < ∞ ∀ α, β, γ.^{[4.24]} (2.62) 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^{[4.25]} (2.63)

the estimate (2.62) translates to

\begin{matrix}sup_{x,ξ} |D_{x}^{α}ξ^{β}D_{ξ}^{γ}b(x, ξ)| < ∞ ∀ α, β, γ \\ ⟺ b ∈ S_{∞}^{−∞}(ℝ_{x}^{n}; ℝ_{ξ}^{n}). \end{matrix}^{[4.26]} (2.64)

Now the inverse Fourier transform in (2.63), combined with (2.61) gives

K(x, y) = g(x, x − y) = (2π)^{−n} ∫ e^{i(x−y)·ξ}b(x, ξ)dξ^{[4.27]} (2.65)

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}).^{[4.127]} (2.66)

This also finishes the proof of Proposition 2.2 since in (2.58), (2.59) we have shown that A = B + R, B = I(b), R ∈ Ψ_{∞}^{−∞}(ℝ^{n})^{[4.28]} (2.67) 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).^{[4.29]} (2.68)

2.7. Proof of Composition Theorem^{[S.P roof.of.Composition.T heorem]}

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}).^{[4.30]} (2.69) Certainly A^{∗}u ∈ S′(ℝ^{n}) if u ∈ S′(ℝ^{n}) since A^{∗}u(ψ) = u(\overline{A\overline{ψ}}) and S(ℝ^{n}) ∋ ψ ⟼ \overline{Aψ̄} ∈ S(ℝ^{n})^{[4.31]} (2.70)

2.8. QUANTIZATION AND SYMBOLS^{[S.QUANT IZAT ION.AND.SY MBOLS]} 43

is clearly continuous. In terms of Schwartz kernels,

\begin{matrix}Aϕ(x) = ∫ K_{A}(x, y)ϕ(y) dy, ϕ ∈ S(ℝ^{n}) \\ A^{∗}u(x) = ∫ K_{A^{∗}}(x, y)u(y) dy, u ∈ S(ℝ^{n}). \end{matrix}^{[4.32]} (2.71)

We then 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}^{[4.33]} (2.72)

where we are using the uniqueness of Schwartz’ kernels.

This proves (2.41) 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}^{[4.34]} (2.73)

i.e. A^{∗} = I(ā(y, x, ξ)). Thus one advantage of allowing general operators (2.34) is that closure under the passage to 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)

\begin{aligned}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. \end{aligned}

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}⟨Bϕ, u⟩ = ⟨ϕ, B^{∗}u⟩ = (2π)^{−n} ∫ ∫ e^{−ix·ξ}ϕ(x)b(x, ξ)\overline{û(ξ)}dξdx \\ ⟹ B̂ϕ(ξ) = ∫ e^{−iy·ξ}b(y, ξ)ϕ(y)dy. \end{matrix}^{[21.2.1998.112]} (2.74)

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. Quantization and symbols^{[S.Quantization.and.symbols]}

So, we have now shown 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)^{[5.1]} (2.75) which defines, for any w ∈ ℝ, a continuous linear map

I : (1 + |x − y|^{2})^{\frac{w}{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 \\ ∃ w s.t. K_{A}(x, y) = I(a), a ∈ (1 + |x − y|^{2})^{\frac{w}{2}} S_{∞}^{m} (ℝ^{2n}; ℝ^{n}) . \end{matrix}^{[5.2]} (2.76)

Furthermore, we have shown in Proposition 2.2 that the special case, w = 0 and ∂_{y}a ≡ 0, gives an isomorphism

Ψ_{∞}^{m}(ℝ^{n}) ⟵_{q_{L}}^{⟶^{σ_{L}}} S_{∞}^{m}(ℝ^{n}; ℝ^{n}).^{[5.3]} (2.77)

The map here, q_{L} = I on symbols independent of y, is the left quantization map and its inverse σ_{L} is the left full symbol map. Next we consider some more consequences of this reduction theorem.

As well as the left quantization map leading to the isomorphism (2.77) there is a right quantization map, similarly derived from (2.75):

q_{R}(a) = (2π)^{−n} ∫ e^{i(x−y)·ξ}a(y, ξ)dξ, a ∈ S_{∞}^{m} (ℝ^{n}; ℝ^{n}) .^{[5.4]} (2.78)

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

q_{R} = ∗ · q_{L} · ∗^{[5.5]} (2.79)

is also an isomorphism, with inverse σ_{R}

Ψ_{∞}^{m}(ℝ^{n}) ⟵_{q_{R}}^{⟶^{σ_{R}}} S_{∞}^{m} (ℝ^{n}; ℝ^{n}) .^{[5.6]} (2.80)

These are the two ‘extreme’ quantization procedures, see Problem 4.1 for another (more centrist) approach. Using the proof of the reduction theorem we find:

[5.7]

Lemma 2.4. 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) ∼ e^{i<D_{x},D_{ξ}>}a.^{[5.8]} (2.81)

For the moment the last asymptotic equality is just to help in remembering the formula, which is the same as given by the formal Taylor series expansion at the origin of the exponential.

Proof. This follows from the general formula (2.68).

2.9. Principal symbol^{[S.P rincipal.symbol]}

One important thing to note from (2.81) is that

D_{x}^{α}D_{ξ}^{α}a(x, ξ) ∈ S_{∞}^{m−|α|} (ℝ^{n}; ℝ^{n}) (2.82)

so that for any pseudodifferential operator

A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ σ_{L}(A) − σ_{R}(A) ∈ S_{∞}^{m−1} (ℝ^{n}; ℝ^{n}) . (2.83)

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}) (2.84)

2.9. PRINCIPAL SYMBOL^{[S.P RINCIP AL.SY MBOL]} 45

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)].

[5.9] (2.85)

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 (2.77) is replaced by a weaker (but very useful) exact sequence.

[5.10]

Lemma 2.5. 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 (2.77), and that the null space of σ_{m} is just Ψ_{∞}^{m−1}(ℝ^{n}) and this is again (2.77) and the definition of σ_{m}.

The fundamental result proved above is that

Ψ_{∞}^{m}(ℝ^{n}) · Ψ_{∞}^{m′}(ℝ^{n}) ⊂ Ψ_{∞}^{m+m′}(ℝ^{n}).^{[5.11]} (2.86)

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 (2.68) 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)].^{[5.12]} (2.87)

Of course b = σ_{R}(B) so we really want to rewrite (2.87) in terms of σ_{L}(B).

[5.13]

Lemma 2.6. 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),^{[5.14]} (2.88)

σ_{L}(A ∘ B) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{ξ}^{α}σ_{L}(A) · D_{x}^{α}σ_{L}(B).^{[5.15]} (2.89)

Proof. The simple formula (2.88) is already immediate from (2.87) since all terms with |α| ≥ 1 are of order m + m′ − |α| ≤ m + m′ − 1. To get the ‘full’ formula (2.89) we can insert into (2.87) the inverse of (2.81), namely

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

∼ e^{i<D_{y},D_{ξ}>}σ_{L}(A)(x, ξ)σ_{L}(B)(y, η)|_{y=x,η=ξ}.

This is just (2.89).

The simplicity of (2.88) over (2.89) is achieved at the expense of enormous loss of information. Still, many problems can be solved using (2.88) 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}) .

2.10. Ellipticity^{[S.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}).

[5.16] (2.90)

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.

[5.17]

Theorem 2.3. The following conditions on A ∈ Ψ_{∞}^{m}(ℝ^{n}) are equivalent

(2.91) A is elliptic ^{[5.18]}

∃ [b] ∈ S_{∞}^{−m−[1]} (ℝ^{n}; ℝ^{n}) s.t. σ_{m}(A) · [b] ≡ 1 in S_{∞}^{0−[1]} (ℝ^{n}; ℝ^{n})^{[5.19]} (2.92)

∃ b ∈ S_{∞}^{−m} (ℝ^{n}; ℝ^{n}) s.t. σ_{L}(A) · b − 1 ∈ S_{∞}^{−∞} (ℝ^{n}; ℝ^{n})^{[5.20]} (2.93)

∃ ϵ > 0 s.t. |σ_{L}(A)(x, ξ)| ≥ ϵ(1 + |ξ|)^{m} in |ξ| > \frac{1}{ϵ}.^{[5.21]} (2.94)

Proof. We shall show

(2.91) ⟹ (2.92) ⟹ (2.93) ⟺ (2.94) ⟹ (2.91).^{[5.22]} (2.95)

In fact Lemma 2.2 shows the equivalence of (2.93) and (2.94). Since we know that σ_{0}(Id) = 1 applying the identity (2.88) to the definition of ellipticity in (2.90) gives

σ_{m}(A) · σ_{−m}(B) ≡ 1 in S_{∞}^{0−[1]}(ℝ^{n}, ℝ^{n}),^{[5.44]} (2.96)

i.e. that (2.91) ⟹ (2.92).

Now assuming (2.96) (i.e. (2.92)), and recalling that σ_{m}(A) = [σ_{L}(A)] we find that a representative b_{1} of the class [b] must satisfy

σ_{L}(A) · b_{1} = 1 + e_{1}, e_{1} ∈ S_{∞}^{−1}(ℝ^{n}; ℝ^{n}),^{[5.23]} (2.97)

2.10. ELLIPTICITY^{[S.ELLIP T ICIT Y ]} 47

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

f ∼ ∑_{j≥1}(−1)^{j}e_{1}^{j} (2.98)

is asymptotically convergent, so f ∈ S_{∞}^{−1}(ℝ^{n}; ℝ^{n}) exists, and (1 + f) · (1 + e_{1}) = 1 + e_{∞}, e_{∞} ∈ S_{∞}^{−∞}(ℝ^{n}; ℝ^{n}).^{[5.24]} (2.99) Then multiplying (2.97) by (1 + f) gives σ_{L}(A) · {b_{1}(1 + f)} = 1 + e_{∞}^{[5.25]} (2.100) which proves (2.93), since b = b_{1}(1 + f) ∈ S_{∞}^{−m}(ℝ^{n}; ℝ^{n}). Of course sup(1 + |ξ|)^{N}|e_{∞}| < ∞ ∀ N (2.101) so

∃ C s.t. |e_{∞}(x, ξ)| < \frac{1}{2} in |ξ| > C. (2.102)

From (2.100) this means

|σ_{L}(A)(x, ξ)| · |b(x, ξ)| ≥ \frac{1}{2}, |ξ| > C.^{[5.26]} (2.103)

Since |b(x, ξ)| ≤ C(1 + |ξ|)^{−m} (being a symbol of order −m), (2.103) implies

inf_{|ξ|≥C} |σ_{L}(A)(x, ξ)|(1 + |ξ|)^{−m} ≥ C > 0.^{[5.27]} (2.104)

which shows that (2.93) implies (2.94).

Conversely, as already remarked, (2.94) implies (2.93).

Now suppose (2.93) holds. Set B_{1} = q_{L}(b) then from (2.88) again σ_{0}(A ∘ B_{1}) = [q_{m}(A)] · [b] ≡ 1. (2.105) That is, A ∘ B_{1} − Id = E_{1} ∈ Ψ_{∞}^{−1}(ℝ^{n}).^{[5.29]} (2.106) Consider the Neumann series of operators

∑_{j≥1}(−1)^{j}E_{1}^{j}. (2.107)

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

σ_{L}(F) ∼ ∑_{j≥1}(−1)^{j}σ_{L}(E_{1}^{j}).^{[5.45]} (2.108)

Then (Id +E_{1})(Id +F) = Id +E_{∞}, E_{∞} ∈ Ψ_{∞}^{−∞}(ℝ^{n}).^{[5.30]} (2.109) Thus B = B_{1}(Id +F) ∈ Ψ_{∞}^{−m}(ℝ^{n}) satisfies (2.90) and it follows that A is elliptic.

In the definition of ellipticity in (2.90) 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 2.7. 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})^{[5.32]} (2.110)

and then if B satisfies (2.90), B − B′ ∈ Ψ_{∞}^{−∞}(ℝ^{n}).

Proof. Certainly (2.110) implies σ_{−m}(B′)·σ_{m}(A) ≡ 1, and the multiplication here is commutative so (2.92) holds and A is elliptic. Conversely if A is elliptic we get in place of (2.106)

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

Then defining F′ as in (2.108) 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 (2.110). Thus ‘left’ ellipticity as in (2.110) is equivalent to right ellipticity. Applying B to (2.110) gives

B′ ∘ (Id +E) = B′ ∘ (A ∘ B) = (Id +E′) ∘ B^{[5.33]} (2.111)

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 and such a parametrix is unique up to an additive term in Ψ_{∞}^{−∞}(ℝ^{n}).

2.11. Elliptic regularity^{[S.Elliptic.regularity]}

One of the main reasons that the ‘residual’ terms are residual is that they are smoothing operators.

[5.42]

Lemma 2.8. If E ∈ Ψ_{∞}^{−∞}(ℝ^{n}) then

E : S′(ℝ^{n}) ⟶ S′(ℝ^{n}) ∩ C^{∞}(ℝ^{n}).^{[5.43]} (2.112)

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}).

Directly from the existence of parametrices for elliptic operators we can deduce the regularity of solutions to elliptic (pseudodifferential) equations.

[20.2.1998.103]

Proposition 2.5. If A ∈ Ψ_{∞}^{m}(ℝ^{n}) is elliptic and u ∈ S′(ℝ^{n}) satifies Au = 0 then u ∈ C^{∞}(ℝ^{n}).

Proof. Let B ∈ Ψ_{∞}^{−m}(ℝ^{n}) be a parametrix for A. Then B ∘ A = Id +E, E ∈ Ψ_{∞}^{−∞}(ℝ^{n}). Thus,

u = (BA − E)u = −Eu^{[5.41]} (2.113)

and the conclusion follows from Lemma 2.8.

2.13. L^{2} BOUNDEDNESS^{[S.L2.BOUNDEDNESS]} 49

2.12. The Laplacian^{[S.T he.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|^{2} ∀ x ∈ ℝ^{n}, ξ ∈ ℝ^{n}, ϵ > 0. \end{matrix}^{[5.34]} (2.114)

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}}^{[5.35]} (2.115)

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}} ∑_{i,j} g^{ij}(x)D_{x_{i}}ϕ · \overline{D_{x_{j}}ψ}dg = ∫ Δ_{g}ϕ · \overline{ψ}dg ∀ ϕ, ψ ∈ S(ℝ^{n})^{[5.37]} (2.116)

where

dg = \sqrt{g}dx.^{[5.38]} (2.117)

Our assumption in (2.114) 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}.^{[5.39]} (2.118)

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

(2.119)  Δ ∘ B ≡ B ∘ Δ ≡ Id  mod Ψ_{∞}^{−∞}(ℝ^{n}).^{[5.40]}

In particular we see from Proposition 2.5 that Δu = 0, u ∈ S′(ℝ^{n}) implies u ∈ C^{∞}(ℝ^{n}).

2.13. L^{2} boundedness^{[S.L2.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:

[6.1]

Proposition 2.6. [Boundedness] 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}).^{[6.2]} (2.120)

Our proof will be in two stages, the first part is by direct estimation. Namely, Schur’s lemma^{4} gives a useful criterion for an integral operator to be bounded on L^{2}.

4Schur

Lemma 2.9 (Schur). If K(x, y) is locally integrable on ℝ^{2n} and is such that

sup_{x∈ℝ^{n}} ∫_{ℝ^{n}} |K(x, y)|dy, sup_{y∈ℝ^{n}} ∫_{ℝ^{n}} |K(x, y)|dx < ∞^{[5.51]} (2.121)

then the operator K : ϕ ⟼ ∫_{ℝ^{n}} K(x, y)ϕ(y)dy is bounded on L^{2}(ℝ^{n}).

Proof. Since S(ℝ^{n}) is dense in L^{2}(ℝ^{n}) (see Problem 2.18) we only need to show the existence of a constant, C, such that

∫ |Kϕ(x)|^{2}dx ≤ C ∫ |ϕ|^{2} ∀ ϕ ∈ S(ℝ^{n}).^{[6.4]} (2.122)

Writing out the integral on the left

\begin{matrix}∫ | ∫ K(x, y)ϕ(y)dy|^{2}dx \\ = ∫∫∫ K(x, y)\overline{K(x, z)}ϕ(y)\overline{ϕ(z)} dydzdx \end{matrix}^{[6.5]} (2.123)

is certainly absolutely convergent and

∫ |Kϕ(x)|^{2}dx

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

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

These two factors are the same. Since

∫ |K(x, y)||K(x, z)| dx dz ≤ sup_{x∈ℝ^{n}} ∫ |K(x, z)|dz · sup_{y∈ℝ^{n}} ∫ |K(x, y|dx

(2.122) follows. Thus (2.121) gives (2.122).

This standard lemma immediately implies the L^{2} boundedness of the ‘residual terms.’ Thus, if K ∈ Ψ_{∞}^{−∞}(ℝ^{n}) then its kernel satisfies (2.60). This in particular implies

|K(x, y)| ≤ C (1 + |x − y|)^{−n−1}

and hence that K satisfies (2.121). Thus

(2.124) each K ∈ Ψ_{∞}^{−∞}(ℝ^{n}) is bounded on L^{2}(ℝ^{n}).^{[5.52]}

2.14. Square root^{[S.Square.root]}

To prove the general result, (2.120), 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.

2.14. SQUARE ROOT^{[S.SQUARE.ROOT]} 51

[6.6]

Proposition 2.7. Suppose A ∈ Ψ_{∞}^{m}(ℝ^{n}), m > 0, 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}).^{[6.7]} (2.125)

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 (See Problem 2.19 for an outline of the proof) b_{0} = a^{\frac{1}{2}} ∈ S_{∞}^{m/2}(ℝ^{n}; ℝ^{n}).^{[6.8]} (2.126) 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 B_{0} 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}).^{[6.9]} (2.127) 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}).^{[6.10]} (2.128)

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}) :

(2.129) 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}.^{[6.11]}

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}).^{[6.12]} (2.130) 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}).^{[6.13]} (2.131) 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 (2.128) for every N. Now define B = \frac{1}{2}(B′ + (B′)^{∗}) where

σ_{L}(B′) ∼ ∑_{j=0}^{∞} σ_{L}(B_{j}).^{[6.14]} (2.132)

52 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}

Since all the B_{j} are self-adjoint B also satisfies (2.132) and from (2.128)

A − B^{2} = A − (∑_{j=0}^{N} B_{j} + B_{(N+1)})^{2} ∈ Ψ_{∞}^{m−N−1}(ℝ^{n}) (2.133)

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 (2.125), and so Proposition 2.7.

2.15. Proof of Boundedness^{[S.P roof.of.Boundedness]}

Here is Hörmander’s argument. We want to show that ∥Aϕ∥ ≤ C∥ϕ∥ ∀ ϕ ∈ S(ℝ^{n})^{[6.15]} (2.134) where A ∈ Ψ_{∞}^{0}(ℝ^{n}). The square of the left side can be written

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

So it suffices to show that ⟨ϕ, A^{∗}Aϕ⟩ ≤ C∥ϕ∥^{2}.^{[6.16]} (2.135) Now A^{∗}A ∈ Ψ_{∞}^{0}(ℝ^{n}) with σ_{0}(A^{∗}A) = \overline{σ_{0}(A)}σ_{0}(A) ∈ ℝ. If C > 0 is a large constant,

C > sup_{x,ξ} |σ_{L}(A^{∗}A)(x, ξ)|

then C−A^{∗}A has a positive representative of its principal symbol. We can therefore apply Proposition 2.7 to it: C − A^{∗}A = B^{∗}B + G, G ∈ Ψ_{∞}^{−∞}(ℝ^{n}).^{[6.17]} (2.136) This gives

\begin{matrix}⟨ϕ, A^{∗}Aϕ⟩ = & C⟨ϕ, ϕ⟩ − ⟨ϕ, B^{∗}Bϕ⟩ − ⟨ϕ, Gϕ⟩ \\ = & C∥ϕ∥^{2} − ∥Bϕ∥^{2} − ⟨ϕ, Gϕ⟩. \end{matrix}^{[6.18]} (2.137)

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 (2.120) holds and Proposition 2.6 is proved.

2.16. Sobolev boundedness^{[S.Sobolev.boundedness]}

Using the basic boundedness result, Proposition 2.6, and the calculus of pseudodifferential 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}.^{[6.19]} (2.138) Using the Fourier transform we find u ∈ H^{k}(ℝ^{n}) ⟹ ξ^{α}û(ξ) ∈ L^{2}(ℝ^{n}) ∀ |α| ≤ k.^{[6.20]} (2.139) Now these finitely many conditions can be written as just the one condition (1 + |ξ|^{2})^{k/2} û(ξ) ∈ L^{2}(ℝ^{n}).^{[6.21]} (2.140)

2.16. SOBOLEV BOUNDEDNESS^{[S.SOBOLEV.BOUNDEDNESS]} 53

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}} ^{[6.22]} (2.141) for a smooth (symbol) of the size of 1 + |ξ|, thus (2.140) just says u ∈ H^{k}(ℝ^{n}) ⟺ ⟨D⟩^{k}u ∈ L^{2}(ℝ^{n}).^{[6.23]} (2.142)

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.^{[6.24]} (2.143)

The same sort of discussion applies, showing that u ∈ H^{k}(ℝ^{n}) ⟺ ⟨D⟩^{k}u ∈ L^{2}(ℝ^{n}).^{[6.25]} (2.144) In view of this we define the Sobolev space H^{m}(ℝ^{n}), for any real order, by u ∈ H^{m}, (ℝ^{n}) ⟺ ⟨D⟩^{m}u ∈ L^{2}(ℝ^{n}).^{[6.26]} (2.145) It is a Hilbert space with ∥u||_{m}^{2} = ∥⟨D⟩^{m}u∥_{L^{2}}^{2} = ∫ (1 + |ξ|^{2})^{m}|û(ξ)|^{2}dξ.^{[6.27]} (2.146) Clearly we have H^{m}(ℝ^{n}) ⊇ H^{m′}(ℝ^{n}) if m′ ≥ m.^{[6.28]} (2.147) 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

\begin{cases}H^{∞}(ℝ^{n}) = ⋂_{m} H^{m}(ℝ^{n}) & as the residual space \\ H^{−∞}(ℝ^{n}) = ⋃_{m} H^{m}(ℝ^{n}) & as the big space. \end{cases} ^{[6.29]} (2.148)

It is important to note that S(ℝ^{n}) ⊊ H^{∞}(ℝ^{n}) ⊊ H^{−∞}(ℝ^{n}) ⊊ S′(ℝ^{n}).^{[6.30]} (2.149) 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})}.^{[6.31]} (2.150)

[6.32]

Theorem 2.4. 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}).^{[6.33]} (2.151)

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})^{[6.34]} (2.152) Now from (2.145) we see that (2.153) 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.^{[6.35]} That is,

⟨D⟩^{M} : H^{m}(ℝ^{n}) ⟷ H^{m−M}(ℝ^{n}) ∀ m, M.^{[6.36]} (2.154)

To prove (2.152) it suffices to show that

B = ⟨D⟩^{−M+m} · A · ⟨D⟩^{−m} : L^{2}(ℝ^{n}) ⟶ L^{2}(ℝ^{n})^{[6.37]} (2.155)

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}).} ^{[6.38]} (2.156)

Since B ∈ Ψ_{∞}^{0}(ℝ^{n}), by the composition theorem, we already know (2.155).

Thus we have proved (2.152). To prove the general case, (2.151), 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 (2.151) we need to show that

Q = ⟨x⟩^{−q} · A · ⟨x⟩^{q} : H^{m}(ℝ^{n}) ⟶ H^{m−M}(ℝ^{n}),^{[6.39]} (2.157)

i.e. satisfies (2.152). 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}.^{[6.40]} (2.158)

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})^{[6.41]} (2.159)

then we know that Q ∈ Ψ_{∞}^{M}(ℝ^{n}) and we get (2.157) from (2.152). Thus we want to show that

⟨x − y⟩^{−|q|} \frac{⟨y⟩^{q}}{⟨x⟩^{q}} a(x, ξ) ∈ S_{∞}^{M}(ℝ^{2n}; ℝ^{n})^{[6.42]} (2.160)

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}^{[6.43]} (2.161)

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|. (2.162)

Thus

⟨x − y⟩^{−q} · \frac{⟨x⟩^{−q}}{⟨y⟩^{−q}} ≤ C in Ω_{1}.^{[6.44]} (2.163)

On the other hand in Ω_{2},

2.18. POLYHOMOGENEITY^{[S.P OLY HOMOGENEIT Y ]} 55

|x| + |y| < 8|x − y| ⟹ |x| < 8|x − y| (2.164) so again

⟨x − y⟩^{−q} \frac{⟨x⟩^{−q}}{⟨y⟩^{−q}} ≤ C.^{[6.45]} (2.165)

In fact we easily conclude that

⟨x − y⟩^{−q} \frac{⟨y⟩^{q}}{⟨x⟩^{q}} ∈ C_{∞}^{∞}(ℝ^{n}) ∀ q,^{[6.46]} (2.166)

since differentiation by x or y makes all terms “smaller”. This proves (2.160), hence (2.159) and (2.157) and therefore (2.151), i.e. the theorem is proved.

2.17. Consequences^{[S.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

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

Clearly C_{∞}^{0}(ℝ^{n}) ⊂ ⟨x⟩^{1+n}L^{2}(ℝ^{n}). Thus as a special case of Theorem 2.4,

D_{x}^{β} : ⟨x⟩^{1+n}L^{2}(ℝ^{n}) ⟶ ⟨x⟩^{1+n}H^{−|β|}(ℝ^{n})

so

[6.47]

Lemma 2.10.

S′(ℝ^{n}) = ⋃_{M}⟨x⟩^{M}H^{−M}(ℝ^{n}). (2.168)

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

[6.48]

Proposition 2.8. 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′′}(ℝ^{n}), p′′ = max(p, p′), q′′ = max(q + m, q′). \end{matrix}^{[6.49]} (2.169)

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}).^{[6.50]} (2.170)

2.18. Polyhomogeneity^{[S.P olyhomogeneity]}

So far we have been considering operators A ∈ Ψ_{∞}^{m}(ℝ^{n}) which correspond, via (2.2), to amplitudes satisfying the symbol estimates (2.6), i.e., in S_{∞}^{m}(ℝ^{2n}; ℝ^{n}). As already remarked, there are many variants of these estimates and corresponding spaces of pseudodifferential operators. Some weakening of the estimates is discussed in the problems below, starting with Problem 2.16. Here we consider a restriction of the spaces, in that we define

S_{ph}^{m}(ℝ^{p}; ℝ) ⊂ S_{∞}^{m}(ℝ^{p}; ℝ^{n}) .^{[eq:P.1]} (2.171)

The definition of the subspace (2.171) is straightforward. First we note that if a ∈ C^{∞}(ℝ^{p}; ℝ^{n}) is homogeneous of degree m ∈ ℝ in |ξ| ≥ 1, then

a(z, tξ) = t^{m}a(z, ξ), |t|, |ξ| ≥ 1.^{[eq:P.2]} (2.172)

If it also satisfies the uniform regularity estimates

sup_{z∈ℝ^{n}, |ξ|≤2} |D_{z}^{α}D_{ξ}^{β}a(z, ξ)| < ∞ ∀ α, β ,^{[eq:P.3]} (2.173)

then in fact

a ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n}) .^{[eq:P.4]} (2.174)

Indeed, (2.173) is exactly the restriction of the symbol estimates to z ∈ ℝ^{p}, |ξ| ≤ 2. On the other hand, in |ξ| ≥ 1, a(z, ξ) is homogeneous so

|D_{z}^{α}D_{ξ}^{β}a(z, ξ)| = |ξ|^{m−|β|}|D_{z}^{α}D_{ξ}^{β}a(z, ξ̂)| , ξ̂ = \frac{ξ}{|ξ|}

from which the symbol estimates follow.

Definition 2.2. For any m ∈ ℝ, the subspace of (one-step)^{5} polyhomogeneous symbols is defined as a subset (2.171) by the requirement that a ∈ S_{ph}^{m}(ℝ^{p}; ℝ^{n}) if and only if there exist elements a_{m−j}(z, ξ) ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n}) which are homogeneous of degree m − j in |ξ| ≥ 1, for j ∈ ℕ_{0}, such that

a ∼ ∑_{j} a_{m−j} .^{[eq:P.5]} (2.175)

Clearly

S_{ph}^{m}(ℝ^{p}; ℝ^{n}) · S_{ph}^{m′}(ℝ^{p}; ℝ^{n}) ⊂ S_{ph}^{m+m′}(ℝ^{p}; ℝ^{n}),^{[eq:P.6]} (2.176)

since the asymptotic expansion of the product is given by the formal product of the asymtotic expansion. In fact there is equality here, because

(1 + |ξ|^{2})^{m/2} ∈ S_{ph}^{m}(ℝ^{p}; ℝ^{n})^{[eq:P.7]} (2.177)

and multiplication by (1 + |ξ|^{2})^{m/2} is an isomorphism of the space S_{ph}^{0}(ℝ^{p}; ℝ^{n}) onto S_{ph}^{m}(ℝ^{p}; ℝ^{n}). Furthermore differentiation with respect to z_{j} or ξ_{l} preserves asymptotic homogeneity so

D_{x_{j}} : S_{ph}^{m}(ℝ^{p}; ℝ^{n}) ⟶ S_{ph}^{m}(ℝ^{p}; ℝ^{n}) D_{ξ_{l}} : S_{ph}^{m}(ℝ^{p}; ℝ^{n}) ⟶ S_{ph}^{m−1}(ℝ^{p}; ℝ^{n}) ∀j = 1, …, n.^{[eq:P.8]}

It is therefore no surprise that the polyhomogeneous operators form a subalgebra.

Proposition 2.9. The spaces Ψ_{ph}^{m}(ℝ^{n}) ⊂ Ψ_{∞}^{m}(ℝ^{n}) defined by the condition that the kernel of A ∈ Ψ_{ph}^{m}(ℝ^{n}) should be of the form I(a) for some

a ∈ (1 + |x − y|^{2})^{w/2}S_{ph}^{m}(ℝ^{2n}; ℝ^{n}) ,^{[eq:P.9]} (2.178)

form an order-filtered ∗-algebra.

5For a somewhat more general class of polyhomogeneous symbols, see problem 2.7.

Proof. Since the definition shows that

Ψ_{ph}^{m}(ℝ^{n}) ⊂ Ψ_{∞}^{m}(ℝ^{n})

we know already that

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

To see that products are polyhomogeneous it suffices to use (2.176) and (2.18) which together show that the asymptotic formulæ describing the left symbols of A ∈ Ψ_{ph}^{m}(ℝ^{n}) and B ∈ Ψ_{ph}^{m′}(ℝ^{m}), e.g.

σ_{L}(A) ∼ ∑_{α} \frac{i^{|α|}}{α!} D_{ξ}^{α}D_{y}^{α}a(x, y, ξ)|_{y=x}

imply that σ_{L}(A) ∈ S_{ph}^{m}(ℝ^{n}; ℝ^{n}), σ_{L}(B) ∈ S_{ph}^{m}(ℝ^{n}; ℝ^{n}). Then the asymptotic formula for the product shows that σ_{L}(A · B) ∈ S_{ph}^{m+m′}(ℝ^{n}; ℝ^{n}).

The proof of ∗-invariance is similarly elementary, since if A = I(a) then A^{∗} = I(b) with b(x, y, z) = \overline{a(y, x, ξ)} ∈ S_{ph}^{m}(ℝ^{2n}; ℝ^{n}).

This subalgebra is usually denoted simply Ψ^{m}(ℝ^{n}) and its elements are often said to be ‘classical’ pseudodifferential operators. As a small exercise in the use of the principal symbol map we shall show that

A ∈ Ψ_{ph}^{m}(ℝ^{n}), A elliptic ⟹ ∃ a parametrix

B ∈ Ψ_{ph}^{−m}(ℝ^{n}), A · B − Id, B · A − Id ∈ Ψ_{∞}^{−∞}(ℝ^{n}) .

[eq:P.10] (2.179)

In fact we already know that B ∈ Ψ_{∞}^{−m}(ℝ^{n}) exists with these properties, and even that it is unique modulo Ψ_{∞}^{−∞}(ℝ^{n}). To show that B ∈ Ψ_{ph}^{−m}(ℝ^{n}) we can use the principal symbol map.

For elements A ∈ Ψ_{ph}^{m}(ℝ^{n}) the principal symbol σ_{m}(A) ∈ S_{∞}^{m−[1]}(ℝ^{n}; ℝ^{n}) has a preferred class of representatives, namely the leading term in the expansion of σ_{L}(A)

σ_{m}(A) = σ(ξ)a_{m}(x, ξ) mod S_{ph}^{m−1}(ℝ^{n}; ℝ^{n})

where σ|ξ| = 1 in |ξ| ≥ 1, σ|ξ| = 0 in |ξ| ≤ 1/2. It is even natural to identify the principal symbol with a_{m}(x, ξ) as a homogeneous function. Then we can see that

(2.180) A ∈ Ψ_{∞}^{m}(ℝ^{n}), σ_{m}(A) homogeneous of degree m

⟺ Ψ_{ph}^{m}(ℝ^{n}) + Ψ_{∞}^{m−1}(ℝ^{n}) .^{[eq:P.11]}

Indeed, we just subtract from A an element A_{1} ∈ Ψ_{ph}^{m}(ℝ^{n}) with σ_{m}(A_{1}) = σ_{m}(A), then σ_{m}(A − A_{1}) = 0 so A − A_{1} ∈ Ψ_{∞}^{m−1}(ℝ^{n}).

So, returning to the proof of (2.179) note straight away that

σ_{−m}(B) = σ_{m}(A)^{−1}

has a homogeneous representative, namely a_{m}(x, ξ)^{−1}. Thus we have shown that for j = 1

B ∈ Ψ_{ph}^{−m}(ℝ^{n}) + Ψ_{∞}^{−m−j}(ℝ^{n}) .^{[eq:P.12]} (2.181)

We take (2.181) as an inductive hypthesis for general j. Writing this decomposition B = B′ + B_{j} it follows from the identity (2.179) that

A · B = A · B′ + AB_{j} = Id mod Ψ_{∞}^{−∞}(ℝ^{n})

so

A · B_{j} = Id −AB′ ∈ Ψ_{ph}^{0}(ℝ^{n}) ∩ Ψ_{∞}^{−j}(ℝ^{n}) = Ψ_{ph}^{−j}(ℝ^{n}).

Now applying B on the left, or using the principal symbol map, it follows that B_{j} ∈ Ψ_{ph}^{−m−j}(ℝ^{n}) + Ψ_{∞}^{−m−j−1}(ℝ^{n}) which gives the inductive hypothesis (2.181) for j + 1.

It is usually the case that a construction in Ψ_{∞}^{∗}(ℝ^{n}), applied to an element of Ψ_{ph}^{∗}(ℝ^{n}) will yield an element of Ψ_{ph}^{∗}(ℝ^{n}) and when this is the case it can generally be confirmed by an inductive argument like that used above to check (2.179).

As a subspace^{6}

S_{ph}^{m}(ℝ^{p}; ℝ^{n}) ⊂ S_{∞}^{m}(ℝ^{p}; ℝ^{n})

is not closed. Indeed, since it contains S_{∞}^{−∞}(ℝ^{p}; ℝ^{n}), its closure contains all of S_{∞}^{m′}(ℝ^{p}; ℝ^{n}) for m′ < m. In fact it is a dense subspace.^{7} To capture its properties we can strengthen the topology S_{ph}^{m}(ℝ^{p}; ℝ^{n}) inherits from S_{∞}^{m}(ℝ^{p}; ℝ^{n}).

Then, as well as the symbol norms ∥ · ∥_{N,m} in (2.7) we can add norms on the terms in the expansions in (2.175)

∥D_{x}^{α}D_{ξ}^{β}a_{m−j}(x, ξ)∥_{L^{∞}(G)}, G = ℝ^{p} × {1 ≤ |ξ| ≤ 2} .^{[eq:P.14]} (2.182)

Then we can further add the symbol norms ensuring (2.175), i.e.,

∥a − ∑_{j=0}^{k} a_{m−j}∥_{m−k−1,N} ∀ k, N .^{[eq:P.15]} (2.183)

Together these give a countable number of norms on S_{ph}^{m}(ℝ^{p}; ℝ^{n}). With respect to the metric topology defined as in (2.8) the spaces S_{ph}^{m}(ℝ^{p}; ℝ^{n}) are then complete.^{8}. Later when we wish to topologize Ψ_{ph}^{m}(ℝ^{n}), or rather related algebras, it is this type of topology we will use. Namely we identify

σ_{L} : Ψ_{ph}^{m}(ℝ^{n}) ⟷ S_{ph}^{m}(ℝ^{n}; ℝ^{n}) .^{[eq:P.16]} (2.184)

2.19. Linear invariance^{[S.Linear.invariance]}

It is rather straightforward to see that the algebra Ψ_{∞}^{∞}(ℝ^{n}) is invariant under affine transformations of ℝ^{n}. In particular if T_{a}x = x + a, for a ∈ ℝ^{n}, is translation by a and

T_{a}^{∗}f(x) = f(x + a), T_{a}^{∗} : S(ℝ^{n}) ⟶ S(ℝ^{n})

is the isomorphism on functions then a new operator is defined by

T_{a}^{∗}A_{a}f = AT_{a}^{∗}f and A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ A_{a} ∈ Ψ_{∞}^{m}(ℝ^{n}).

In fact the left-reduced symbols satisfy

σ_{L}(A_{a})(x, ξ) = σ_{L}(A)(x + a, ξ), A_{a} = T_{−a}^{∗}AT_{a}^{∗}.

6Polyhomogeneous symbols may seem to be quite sophisticated objects but they are really smooth functions on manifolds with boundary; see Problems 2.7–2.8.

7See Problem 2.9.

8See Problem 2.10.

2.21. PROBLEMS^{[S.CHAP T ER.2.P ROBLEMS]} 59

Similarly if T ∈ GL(n) is an invertible linear transformation of ℝ^{n} then

(2.185) A_{T} f = T^{∗}A(T^{∗})^{−1}f, A ∈ Ψ_{∞}^{m}(ℝ^{n}) ⟹ A_{T} ∈ Ψ_{∞}^{m}(ℝ^{n})

and σ_{L}(A_{T})(x, ξ) = σ_{L}(A)(T x, (T^{t})^{−1}ξ)| det(T)|^{[21.2.1998.104]}

where T^{t} is the transpose of T (so T x · ξ = x · T^{t}ξ) and det(T) the determinant.

This invariance means that we can define the spaces Ψ_{∞}^{m}(V) and Ψ_{ph}^{m}(V) for any vector space V (or even affine space) as operators on S(V). We are much more interested in full coordinate invariance which is discussed in Chapter 4.

2.20. Further reading^{[S.F urther.reading]}
2.21. Problems^{[S.Chapter.2.P roblems]}

[P 3.1]

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

(1 + |ξ|^{2})^{\frac{1}{2} m} ∈ S_{∞}^{m}(ℝ^{p}; ℝ^{n}) (2.186)

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}).

[P 3.2]

Problem 2.2. Consider w = 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}{2π} \frac{1}{a(re^{iθ})} \frac{d}{dθ} a(re^{iθ})dθ = \frac{1}{2π} ∫_{0}^{2π} \frac{d}{dθ} log a(re^{iθ})dθ,

exists and is an integer 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

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

for any r.

[22.2.1998.139]

Problem 2.3. Show that a symbol a ∈ S_{∞}^{m}(ℝ_{z}^{p}; ℝ_{ξ}^{n}) which satisfies an estimate

|a(z, ξ)| ≤ C(1 + |ξ|)^{m′}, m′ < m^{[22.2.1998.140]} (2.188)

is necessarily in the space S_{∞}^{m′+ϵ}(ℝ_{z}^{p}; ℝ_{ξ}^{n}) for all ϵ > 0.

[22.2.1998.142]

Problem 2.4. Show that if ϕ ∈ C_{c}^{∞}(ℝ_{z}^{p} × ℝ^{n}) and ψ ∈ C_{c}^{∞}(ℝ^{n}) with ψ(ξ) = 1 in |ξ| < 1 then

c_{ϕ}(z, ξ) = ϕ(z, \frac{ξ}{|ξ|})(1 − ψ)(ξ) ∈ S^{0}(ℝ_{z}^{p}; ℝ_{ξ}^{n}).^{[22.2.1998.143]} (2.189)

If a ∈ S^{m}(ℝ_{z}^{p}; ℝ_{ξ}^{n}) define the cone support of a in terms of its complement

(2.190) cone supp(a)^{∁} = {(z̄, ξ̄) ∈ ℝ_{z}^{p} × (ℝ_{ξ}^{n} \ {0}); ∃

ϕ ∈ C_{c}^{∞}(ℝ_{z}^{p}; ℝ^{n}), ϕ(z̄, ξ̄) ≠ 0, such that c_{ϕ}a ∈ S_{∞}^{−∞}(ℝ^{p}; ℝ^{n})}.^{[22.2.1998.144]}

60 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}

[1.2.2000.276]

Problem 2.5. Prove that (2.30) is a characterization of functions a ∈ (1 + |x − y|^{2})^{w/2}S^{m}(ℝ^{2n}; ℝ^{n}). [Hint: Use Liebniz’ formula to show instead that the equivalent estimates

|D_{x}^{α}D_{y}^{β}D_{ξ}^{γ}a(x, y, ξ)| ≤ C_{α,β,γ}(1 + |x − y|^{2})^{w/2}(1 + |ξ|)^{m−|γ|} ∀ α, β, γ ∈ ℕ_{0}^{n}

characterize this space.]

[P 3.3]

Problem 2.6. 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} (2.191)

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

[prob:MM]

Problem 2.7. General polyhomogeneous symbols and operators.

[prob:NN]

Problem 2.8. Polyhomogeneous symbols as smooth functions.

[prob:DD]

Problem 2.9. Density of polyhomogeneous symbols in L^{∞} symbols of the same order.

[prob:CC]

Problem 2.10. Completeness of the spaces of polyhomogeneous symbols.

[prob:F F]

Problem 2.11. Fourier transform??

Problem 2.12. Show that the kernel of any element of Ψ_{∞}^{∞}(ℝ^{n}) is C^{∞} away from the diagonal. Hint: Prove that (x − y)^{α}K(x, y) becomes increasingly smooth as |α| increases.

[21.2.1998.117]

Problem 2.13. Show that for any m ≥ 0 the unit ball in H^{m}(ℝ^{n}) ⊂ L^{2}(ℝ^{n}) is not precompact, i.e. there is a sequence f_{j} ∈ H^{m}(ℝ^{n}) which has ∥f_{j}∥_{m} ≤ 1 and has no subsequence convergent in L^{2}(ℝ^{n}).

[21.2.1998.118]

Problem 2.14. Show that for any R > 0 there exists N > 0 such that the Hilbert subspace of H^{N}(ℝ^{n})

{u ∈ H^{N}(ℝ^{n}); u(x) = 0 in |x| > R}^{[21.2.1998.119]} (2.192)

is compactly included in L^{2}(ℝ^{n}), i.e. the intersection of the unit ball in H^{N}(ℝ^{n}) with the subspace (2.192) is precompact in L^{2}(ℝ^{n}). Hint: This is true for any N > 0, taking N >> 0 will allow you to use the Sobolev embedding theorem and Arzela-Ascoli.

[21.2.1998.120]

Problem 2.15. Using Problem 2.14 (or otherwise) show that for any ϵ > 0

(1 + |x|)^{ϵ}H^{ϵ}(ℝ^{n}) ↪ L^{2}(ℝ^{n})

is a compact inclusion, i.e. any infinite sequence f_{n} such that (1+|x|^{2})^{−ϵ} is bounded in H^{ϵ}(ℝ^{n}) has a subsequence convergent in L^{2}(ℝ^{n}). Hint: Choose ϕ ∈ C_{c}^{∞}(ℝ^{n}) with ϕ(x) = 1 in |x| < 1 and, for each k, consider the sequence ϕ(x/k)f_{j}. Show that the Fourier transform converts this into a sequence which is bounded in (1 + |ξ|^{2})^{− \frac{1}{2} ϵ}H^{N}(ℝ_{ξ}^{n}) for any N. Deduce that it has a convergent subsequence in L^{2}(ℝ^{n}). By diagonalization (and using the rest of the assumption) show that f_{j} itself has a convergent subsequence.

[1.2.2000.279]

Problem 2.16. About ρ and δ.

Problem 2.17. Prove the formula (2.185) for the left-reduced symbol of the operator A_{T} obtained from the pseudodifferential operator A by linear change of variables. How does the right-reduced symbol transform?

[1.2.2000.280]

Problem 2.18. Density of S(ℝ^{n}) in L^{2}(ℝ^{n}).

[1.2.2000.281]

Problem 2.19. Square-root of a positive elliptic symbol is a symbol.

[21.2.1998.108]

Problem 2.20. Write out a proof to Proposition 3.2. Hint (just to do it elegantly, it is straightforward enough): Write A in right-reduced form as in (2.74) and apply it to û; this gives a formula for Âu.

[21.2.1998.110]

Problem 2.21. Show that any continuous linear operator

S′(ℝ^{n}) ⟶ S(ℝ^{n})

has Schwartz kernel in S(ℝ^{2n}.

62 2. PSEUDODIFFERENTIAL OPERATORS ON EUCLIDEAN SPACE^{[C.EUCLIDEAN]}