Adjoint of a pseudo-differential operator

5.5 Symbolic calculus

5.5.3 Adjoint of a pseudo-differential operator

This shows the first formula. The next formula is obtained recursively.

We can now sketch the proof of Theorem 5.5.3 in the caseρ=δ.

Sketch of the proof of Theorem 5.5.3 with ρ=δ. We assume ρ=δ∈[0,1). Writ- ing σ1 = σ1π(I +R)^−Nπ(I +R)^N and using Lemma 5.5.11, it suffices to prove (5.52) form1as negative as one wants. We proceed as in the proof of the caseρ > δ replacing Lemma 5.5.5 with Lemma 5.5.10. The details are left to the reader.

Lemma 5.5.14. Let σ∈S^−∞ and let κ: (x, y) →κ_x(y) be its associated kernel.

We set

κ⁽_x^∗)(y) := ¯κ_xy⁻¹(y⁻¹), x, y∈G.

Thenκ^(∗): (x, y)→κ⁽x^∗)(y)is smooth onG×Gand for everyα∈Nⁿ0,x→X^ακ⁽x^∗)

is continuous from GtoS(G).

The symbolσ^(∗) defined via

σ^(∗)(x, π) :=F_G(κ⁽_x^∗))(π), (x, π)∈G×G, is a smooth symbol in the sense of Definition 5.1.34 and satisfies

(Op(σ))^∗= Op(σ^(∗)).

In particular, ifσdoes not depend on x, thenσ^(∗)=σ^∗. Note that this operation is an involution since

κx(y) = ¯κ⁽_xy^∗)−1(y⁻¹).

Recall that ifσ={σ(x, π),(x, π)∈G×G} then we have defined the adjoint symbol

σ^∗={σ(x, π)^∗,(x, π)∈G×G}, (see Theorem 5.2.22). Hence we may write

σ^∗(x, π) :=σ(x, π)^∗. Proof of Lemma 5.5.14. By Corollary 3.1.30, we have

X_x^βo{κ⁽_x^∗)(y)}=X_x^βo{¯κ_xy⁻¹(y⁻¹)}= (−1)^|βo|X˜_y^βo

1=y⁻¹{¯κ_xy1(y⁻¹)}

= (−1)^|βo|

|β|≤|βo|,[β]≥[βo]

Q_βo,β(y⁻¹)X_y^β

1=y⁻¹{κ¯_xy1(y⁻¹)}

= (−1)^|βo|

|β|≤|βo|,[β]≥[βo]

Qβo,β(y⁻¹)X_x^β_1=xy−1{¯κx1(y⁻¹)},

where theQ_βo,β’s are ([β_o]−[β])-homogeneous polynomials. The regularity of κ described in Theorem 5.4.9 implies thatκ^(∗): (x, y)→κ⁽x^∗)(y) is smooth inxand y (but maybe not compactly supported inx), and it is also Schwartz iny in such a way that all the mappings G x → X_x^ακ⁽x^∗) ∈ S(G) are continuous. Clearly σ^(∗)(x, π) =π(κ⁽x^∗)) defines a smooth symbolσ^(∗).

Let φ, ψ ∈ S(G) and let x ∈ G. The regularity of κ described in Theorem 5.4.9 justifies easily the following computations:

G

(Op(σ)φ)(x)ψ(x)dx =

Gφ∗κ_x(x) ¯ψ(x)dx=

G

Gφ(z)κ_x(z⁻¹x) ¯ψ(x)dzdx

=

G

Gφ(z)¯κ⁽_x^∗₍⁾_z−1x)⁻¹((z⁻¹x)⁻¹) ¯ψ(x)dzdx

=

G

Gφ(z)κ⁽z^∗)(x⁻¹z)ψ(x)dzdx

=

Gφ(z)ψ∗κ⁽z^∗)(z)dz.

This shows that Op(σ)^∗ψ(z) =ψ∗κ⁽z^∗)(z).

In general, σ^(∗) is not the adjoint σ^∗ of the symbol σ, unless for instance it does not depend onx∈G. However, we can perform formal considerations to linkσ^(∗)withσ^∗in the following way: using the Taylor expansion forκ^∗_xinx(see equality (5.27)), we obtain

κ⁽_x^∗)(y) =κ^∗_xy−1(y)≈

α

qα(y⁻¹)X_x^ακ^∗_x(y) =

α

q˜α(y)X_x^ακ^∗_x(y).

Thus, taking the group Fourier transform atπ∈G, we get σ^(∗)(x, π) =π(κ⁽_x^∗))≈

α

π(˜qα(y)X_x^ακ^∗_x(y)) =

α

Δ^αX_x^ασ(x, π)^∗.

From Theorem 5.2.22 we know that ifσ∈S_ρ,δ^m then

Δ^αX_x^ασ(x, π)^∗∈S_ρ,δ^{m−(ρ−δ)[α]}. (5.58) From these formal computations we see that the main problem is to estimate the remainder coming from the use of the Taylor expansion. This is the purpose of the following technical lemma.

Lemma 5.5.15. We fix a positive Rockland operator of homogeneous degreeν. Let m ∈ R,1 ≥ρ ≥δ ≥ 0 with ρ= 0 and δ = 1, β0 ∈ Nⁿ0, and M, M1 ∈N0. We assume that M ≥νM1 and(ρ−δ)M +ρQ > m+δ[β0] +νM1+Q. Then there exist a constantC >0, and a seminorm · _S^m_ρ,δ,a,b,0, such that for anyσ∈S^−∞

and any (x, π)∈G×G we have X_x^β0

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

π(I +R)^M1_L(Hπ)≤CσS^m_ρ,δ,a,b,0.

Proof of Lemma 5.5.15, caseβ0= 0. By Lemma 5.5.14 and the observations that follow, we have

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

=

G

⎛

⎝κ^∗_xz−1(z)−

[α]≤M

qα(z⁻¹)X_x^ακ^∗_x(z)

⎞

⎠π(z)^∗dz

=

GR^κ_x,M^∗x(z)(z⁻¹)π(z)^∗dz,

where R^κ_x,M^∗x(z) denotes the remainder of the (vector-valued) Taylor expansion of v→κ^∗_xv(z) of orderM at 0. Using (5.48), we can integrate by parts to obtain

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

π(I +R)^M1

=

[β1]+[β2]≤νM1

G

X˜_z^β₁¹=zR^X_x,M^{˜zβ2=2 zκ∗x(z2)}(z1⁻¹)π(z)^∗dz

=

[β1]+[β2]≤νM1

GR^X_x^˜₁^zβ₌²⁼²_x,M−^zXβx11_[β^κ₁^∗x_]^1(z2)(z⁻¹)π(z)^∗dz.

Taking the operator norm, we have

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

π(I +R)^M1_L(Hπ)

[β1]+[β2]≤νM1

G|R^X_x^{˜βz2=2 zXxβ11κ∗x1(z2)}

1=x,M−[β1] (z⁻¹)|dz.

For|z|<1, we will use Taylor’s theorem, see Theorem 3.1.51:

|R^X_x^˜₁^zβ₌²⁼²_x,M−^zXxβ11_[β^κ₁^∗x_]^1(z2)(z⁻¹)|

|γ|≤(M−[β1])₊+1 [γ]>(M−[β1])₊

|z|^[γ] sup

x1∈G|X_z^γX˜_z^β₂²=zX_x^β₁¹κ^∗_x1(z2)|, together with the estimate forznear the origin given in Theorem 5.4.1. The link between left and right derivatives, see (1.11), implies

sup

x1∈G|X_z^γX˜_z^β₂²=zX_x^β₁¹κ^∗_x1(z2)|= sup

x1∈G|X_z^γX_z^β₂²=zX_x^β₁¹κ_x1(z2)|.

Proceeding as in the proof of Lemma 5.5.6, we obtain that the integral

|z|<1

|R^X_x,M−^{˜zβ2=2 z}_[^X_β^{βx1κ∗x(z2)}

1] (z⁻¹)|dz

|γ|≤(M−[β1])₊+1 [γ]>(M−[β1])₊

|z|<1

|z|^[γ] sup

x1∈G|X_x^γ₁X_z^β₂²=zX_x^β₁¹κ_x1(z2)|dz

is finite whenever [γ] +Q > (m+ [β2] +δ([γ] + [β1]) +Q)/ρ with the indices as above. These conditions are implied by the hypotheses of the statement. The estimates forzlarge given in Theorem 5.4.1 show directly that the integral

|z|>1

|R^X_x^{˜zβ2=2 zXxβ11κ∗x1(z2)}

1=x,M−[β1] (z⁻¹)|dz,

is finite. Collecting the various estimates yields the statement in the case ρ= 0

andβ0= 0.

Proof of Lemma 5.5.15, general case. We proceed as above and introduce the de- rivatives with respect tox. We obtain

X_x^β0

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

=

GR^X0,M^{xβ0κ∗x·(z)}(z⁻¹)π(z)^∗dz.

And adding (I +R)^M1, we have X_x^β0

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

(I +R)^M1

=

[β1]+[β2]≤νM1

GR^X_x^˜₁^zβ₌₀²⁼² _,M−^zXβx11_[β^X₁^xβ_]^{0κ∗xx1(z2)}(z⁻¹)π(z)^∗dz.

Taking the operator norm, we have X_x^β0

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

π(I +R)^M1_L(Hπ)

[β1]+[β2]≤νM1

G|R^X_x^˜₁^zβ₌₀²⁼² _,M−^zXxβ11_[β^X₁^xβ_]^{0κ∗xx1(z2)}(z⁻¹)|dz.

For|z|<1, we use the more precise version of Taylor’s theorem than in the case β0= 0:

|R^X_x,M−^{˜βz2=2 z}_[^X_β1^xβ_]^{11Xxβ0κ∗xx1(z2)}(z⁻¹)|

|γ|≤(M−[β1])₊+1 [γ]>(M−[β1])₊

|z|^[γ] sup

|y|≤η^{(M−[β1])++1}|z|

|X_y^γX˜_z^β₂²=zX_y^β1X_x^β0κ^∗_xy(z2)|.

We proceed as in the proof of Lemma 5.5.5, that is, we use (5.51) to obtain sup

|y|≤η^{(M−[β1])++1}|z|

|X_y^γX˜_z^β₂²=zX_y^β1X_x^β0κ^∗_xy(z2)|

[β₀]≥[β0]

|β₀|≤|β0|

|z|^{[β0]−[β0]} sup

x1∈G [γ0]=[γ]+[β₀]

|X_x^γ₁⁰X˜_z^β₂²=zκ^∗_x1(z2)|.

We conclude by adapting the caseβ0= 0.

To take into account the difference operator, we will use the following obser- vation.

Lemma 5.5.16. For any α∈Nⁿ0 andσ∈S^−∞,Δ^ασ^(∗) can be written as a linear combination (independent of σ) of {Δ^ασ}^(∗) over α ∈ Nⁿ0, [α] = [α]. This is the same linear combination as when writing Δ^ασ^∗ as a linear combination of {Δ^ασ}^∗.

Proof of Lemma 5.5.16. For σ ∈ S^−∞, let κ_σ be the kernel associated with the symbolσand similarly for any other symbol.

Let us prove Part 1. We have

{q˜_ακ_σ(∗),x}(y) = ˜q_α(y)¯κ_σ,xy⁻¹(y⁻¹).

As ¯qα is a [α]-homogeneous polynomial, by Proposition 5.2.3, ˜qα is a linear com- bination of ˜q_α over multi-indicesα ∈Nⁿ0 satisfying [α] = [α]. Hence

{q˜ακ_σ^(∗)_,x}(y) =

[α]=[α]

q˜ακ_σ,xy⁻¹(y⁻¹) =

[α]=[α]

{q˜ακσ}^(∗)(y),

where

means taking a linear combination. Taking the Fourier transform, we obtain

F_G{q˜_ακ_σ(∗),x}(π) = Δ^ασ^(∗)(x, π) =

[α]=[α]

{Δ^ασ}^(∗).

We can now prove Theorem 5.5.12 in the caseρ > δ.

Proof of Theorem 5.5.12 withρ > δ. We assumeρ > δ. We fix a positive Rockland operator of homogeneous degree ν. Let us show that for any α0, β0 ∈ Nⁿ0, and M0∈N, there exists M ≥M0, a constant C >0 and a seminorm · S_ρ,δ^m,a1,b1,0, such that for anyσ∈S^−∞ we have

X_x^β0Δ^α0τM(x, π)π(I +R)^{−m−(ρ−δ)M0ν−ρ[α0]+δ[β0]}

L(Hπ)

≤Cσ_S^m_ρ,δ,a1,b1,0, (5.59) where we have denoted τM := σ^(∗)−

[α]≤MΔ^αX_x^ασ^∗. By Lemma 5.5.16, it suffices to show (5.59) only forα0= 0.

Letβ0∈N0 andM0∈N. LetM1∈N0be the smallest non-negative integer such that

−m−(ρ−δ)M0+δ[β0]

ν ≤M1.

We chooseM ≥max(M0, νM1) such that (ρ−δ)M+ρQ > m+δ[β0] +νM1+Q. This is possible asρ > δ. Then (5.59) follows from the application of Lemma 5.5.15 toM, M1and the symbolσ.

Using (5.58), classical considerations imply that (5.59) yields that for any M0∈N0, and any seminorm · _Sm−M0(ρ−δ),R

ρ,δ ,a,b, there exist a constantC >0 and a seminorm · _S^m_ρ,δ,a1,b1,0, such that for any σ1, σ2∈S^−∞ we have

τ_M0Sm−M0(ρ−δ),R

ρ,δ ,a,b≤Cσ_Sρ,δ^m_,a1,b1,0.

We can then conclude as in the proof of Theorem 5.5.3 in the caseρ > δ. In fact, we have obtained a much more precise result:

Corollary 5.5.17. We assume1≥ρ > δ≥0. Ifσ∈S_ρ,δ^m, then there exists a unique symbol σ^(∗) in S_ρ,δ^m such that

(Op(σ))^∗= Op(σ^(∗)). Furthermore, for anyM ∈N0,

{σ^(∗)(x, π)−

[α]≤M

X_x^αΔ^ασ^∗(x, π)} ∈S^m−_ρ,δ^(ρ−δ)M.

Moreover, the mapping

1 S_ρ,δ^m −→ S_ρ,δ^{m−(ρ−δ)M} σ −→ {σ^(∗)(x, π)−

[α]≤MX_x^αΔ^ασ^∗(x, π)} , is continuous.

Consequently, we can also write σ^(∗)∼^∞

j=0

⎛

⎝

[α]=j

X_x^αΔ^ασ^∗

⎞

⎠, (5.60)

where the asymptotic was defined in Definition 5.5.2.

As for composition, in the caseρ=δ, the asymptotic formula does not bring any improvement and, in this sense, is not interesting. The proof of this case is more delicate to prove but relies on the same kind of arguments as above. Using Lemma 5.5.9 instead of Lemma 5.5.6 in the proof of Lemma 5.5.15 produces the following result:

Lemma 5.5.18. We fix a positive Rockland operator of homogeneous degreeν. Let m∈R,1≤ρ≤δ≤0 with δ= 1,β0∈Nⁿ0, andM, M1∈N0. We assume that

M ≥νM1 and m+δ(M+c_β0) +νM1<−Q, where

c_β0 := max

[β0]≤[β0] [β]≥[β0],|β|≥|β0|

[β].

Then there exist a constantC >0, and a seminorm · _S^m,R

ρ,δ ,a,b, such that for any σ∈S^−∞ and any(x, π)∈G×G we have

X_x^β0

σ^(∗)(x, π)−

[α]≤M

Δ^αX_x^ασ^∗(x, π)

π(I +R)^M1_L(Hπ)≤Cσ_S^m,R

ρ,δ ,a,b. The details of the proof of Lemma 5.5.18 are left to the reader. The conditions in the statement just above show that we will require the ability to choose mas negative as one wants. We can do this thanks to the following remark.

Lemma 5.5.19. For any σ∈S^−∞ and any X ∈g, we have {π(X)σ}^(∗)=−σ^(∗)(x, π)π(X)− {X_xσ}^(∗)(x, π). More generally, for anyβ ∈Nⁿ0, we have

{π(X)^βσ}^(∗)=

[β1]+[β2]=[β]

{X_x^β1σ}^(∗)π(X)^β2,

where

denotes a linear combination independent ofσ1, σ2.

Proof of Lemma 5.5.19. We keep the notation of Lemma 5.5.14. The kernel of σ^(∗)π(X) is given via

X˜_yκ⁽_x^∗)(y) = ˜X_y{¯κ_xy⁻¹(y⁻¹)}=−X_x1=xy⁻¹κ¯_x1(y⁻¹)−X_y2=y⁻¹κ¯_xy⁻¹(y2), having used (5.50) and the Leibniz property for vector fields. Hence we recognise:

X˜_yκ⁽_x^∗)(y) =−(X_xκ_x)^(∗)(y)−(Xκ_x)^(∗)(y), and

σ^(∗)π(X) =−(Xxσ)^(∗)−(π(X)σ)^(∗).

This shows the first formula. The second formula is obtained recursively.

We can now show sketch the proof of Theorem 5.5.3 in the caseρ=δ. Sketch of the proof of Theorem 5.5.3 with ρ=δ. We assume ρ=δ∈[0,1). Writ- ingσ=π(I +R)^Nπ(I +R)^−Nσand using Lemma 5.5.19, it suffices to prove (5.59) for m as negative as one wants. We proceed as in the proof of the case ρ > δ replacing Lemma 5.5.15 with Lemma 5.5.18. The details are left to the reader.