Shubin Sobolev spaces

6.4 Shubin classes

6.4.3 Shubin Sobolev spaces

In this section, we study Shubin Sobolev spaces. Many of their properties, espe- cially their equivalent characterisations, are well known. Their proofs are quite easy but often omitted in the literature. Thus we have chosen to sketch their demonstrations.

The Shubin Sobolev spaces below are a special case of Sobolev spaces for measurable fields on representation spaces, see Definition 5.1.6.

Our starting point will be the following definition for the Shubin Sobolev spaces:

Definition 6.4.15. Lets∈R. TheShubin Sobolev spaceQ_s(Rⁿ) is the subspace of S(Rⁿ) which is the completion of Dom(I +Q)^s/2 for the norm

hQs:=(I +Q)^s/2h_L²(Rⁿ). They satisfy the following properties:

Theorem 6.4.16. 1. The spaceQ_s(Rⁿ)is a Hilbert space endowed with the ses- quilinear form

(g, h)_Qs=

(I +Q)^s/2g,(I +Q)^s/2h

L²(Rⁿ). We have the inclusions

S(Rⁿ)⊂ Q_s1(Rⁿ)⊂ Q_s2(Rⁿ)⊂ S(Rⁿ), s1> s2. We also have

L²(Rⁿ) =Q0(Rⁿ) and S(Rⁿ) = #

s∈R

Q_s(Rⁿ).

2. The dual of Qs(Rⁿ) may be identified with Q−s(Rⁿ) via the distributional duality formg, h=

Rⁿgh.

3. If s∈N0,Q_s(Rⁿ)coincides with

Q_s(Rⁿ) ={h∈L²(Rⁿ) :u^α∂^β_uh∈L²(Rⁿ) ∀α, β∈Nⁿ0, |α|+|β| ≤s}.

Furthermore, the norm given by h⁽_Q^int_s ⁾=

|α|+|β|≤s

u^α∂_u^βh_L²(Rⁿ),

is equivalent to · _Qs.

4. For any s ∈ R, Qs(Rⁿ) coincides with the completion (in S(Rⁿ)) of the Schwartz spaceS(Rⁿ)for the norm

h⁽_Q^b)_s =Op^W(b^s)h_L²(Rⁿ),

whereb was given in Example 6.4.13. The norm · ⁽_Q^b)_s extended toQs(Rⁿ) is equivalent to · Qs.

5. For any s∈R, the Shubin Sobolev spaceQ_s(Rⁿ) coincides with the Sobolev space associated with the following metric weight (see Definition 6.4.10)

Qs(Rⁿ) =H (1 +|u|²+|ξ|²)^s/2, dξ²+du² 1 +|u|²+|ξ|²

.

6. For any s∈R, the operators Op^W(b^−s)(I +Q)^s/2 and(I +Q)^s/2Op^W(b^−s) are bounded and invertible onL²(Rⁿ).

7. The complex interpolation between the spaces Qs0(Rⁿ)andQs1(Rⁿ)is (Qs0(Rⁿ),Qs1(Rⁿ))_θ=Qsθ(Rⁿ), sθ= (1−θ)s0+θs1, θ∈(0,1). Before giving the proof of Theorem 6.4.16, let us recall the definition of complex interpolation:

Definition 6.4.17 (Complex interpolation). LetX0 andX1be two subspaces of a vector spaceZ. We assume thatX0andX1are Banach spaces with norms denoted by| · |_j,j = 0,1.

LetZ be the space of the functionsf defined on the strip ¯S={0≤Rez≤ 1} and valued in X0+X1 such that f is continuous on ¯S and holomorphic in S={0<Rez <1}. Forf ∈Z we define the quantity (possibly infinite)

fZ := sup

y∈R{|f(iy)|0,|f(1 +iy)|1}.

The complex interpolation space of exponentθ∈(0,1) is the space (X0, X1)_θ of vectors v ∈ X0+X1 such that there exists f ∈ Z satisfying f(θ) = v and fZ <∞.

The space (X0, X1)_θ is a subspace ofZ; it is a Banach space when endowed with the norm given by

|v|θ:= inf{fZ :f ∈Z and f(θ) =v}.

6.4. Shubin classes 463 We also refer to Appendix A.6 for the notion of analytic interpolation.

Proof of Theorem 6.4.16. From Definition 6.4.15, it is easy to prove that the space Q_s(Rⁿ) is a Hilbert space, that it is included inS(Rⁿ) and thatQ0(Rⁿ) =L²(Rⁿ).

It is a routine exercise left to the reader that the dual of Q_s(Rⁿ) isQ_−s(Rⁿ) via the distributional duality (Part (2)) and that the spaces Q_s(Rⁿ) decrease with s∈R.

Let us prove the complex interpolation property of Part (7). We may assume s1> s0. Forh∈ Qsθ, we consider the function

f(z) := (I +Q)^{−(zs1+(1−z)s2 0)+sθ}h, and we check easily that

f(θ) =h, f(iy)_Qs0 =f(1 +iy)_Qs1 =h_Qsθ ∀y∈R.

This shows thatQ_sθ is continuously included in (Q_s0(Rⁿ),Q_s1(Rⁿ))_θ. By duality of the complex interpolation and of the Q_s(Rⁿ), we obtain the reverse inclusion and Part (7) is proved.

Let us prove Part (4). For any s ∈ R, the operator Op^W(b^s) maps S(Rⁿ) to itself and the mapping · ⁽_Q^b)_s as defined in Part (4) is a norm onS(Rⁿ). We denote its completion inS(Rⁿ) byQ⁽s^b)(Rⁿ). From the properties of the calculus it is again a routine exercise left to the reader that the dual ofQ⁽s^b)(Rⁿ) isQ⁽_−s^b)(Rⁿ) via the distributional duality and that the spacesQ⁽s^b)(Rⁿ) decrease withs∈R.

We can prove the following property about interpolation between theQ^(b)(Rⁿ) spaces which is analogous to Part (7):

(Q⁽_s^b₀⁾(Rⁿ),Q⁽_s^b₁⁾(Rⁿ))_θ=Q⁽_s^b_θ⁾(Rⁿ), sθ= (1−θ)s0+θs1, θ∈(0,1). (6.49) Indeed we may assumes1> s0. Forh∈ Q⁽s^bθ⁾, we consider the function

f(z) =e^z(sz−sθ)Op^W

b^−sz+sθ

h where sz= (1−z)s0+zs1. Clearlyf(θ) =h. Furthermore,

f(iy)⁽_Q^b)_s

1 = |e^{iy(siy−sθ)}|Op^W(b^s1)Op^W

b^−siy+sθ

h_L²(Rⁿ)

≤ e^{−y2(s1−s0)}Op^W(b^s1)Op^W

b^−siy+sθ

Op^W(b^−sθ)_L(L²(Rⁿ))

h⁽_Q^b)

sθ, (6.50)

and

f(1 +iy)⁽_Q^b)_s

0

=|e^{(1+iy)(s1+iy−sθ)}|Op^W(b^s0)Op^W

b^−s1+iy+sθ

h_L²(Rⁿ)

≤e^{s1−sθ−y2(s1−s0)}Op^W(b^s0)Op^W

b^−s1+iy+sθ

Op^W(b^−sθ)_L(L²(Rⁿ))

h⁽_Q^b)

sθ. (6.51)

From the calculus we obtain that the two operator norms on L²(Rⁿ) in (6.50) and (6.51) are bounded by a constant of the formC(1 +|y|)^N whereC > 0 and N ∈ N0 are independent of y. This shows that Q⁽s^bθ⁾ is continuously included in (Q⁽s^b0⁾(Rⁿ),Q⁽s^b1⁾(Rⁿ))_θ. By duality of the complex interpolation and of the spaces Qs(Rⁿ), we obtain the reverse inclusion and (6.49) is proved.

Let us show that the spaces Q⁽s^b)(Rⁿ) and Q_s(Rⁿ) coincide. First let us assumes∈2N0. We have for anyh∈ Q⁽s^b)(Rⁿ):

hQs ≤ (I +Q)^s/2Op^W(b^−s)L(L²(Rⁿ))h⁽_Q^b)_s.

As Q ∈ ΨΣ²₁(Rⁿ), by Theorem 6.4.14, the operator (I +Q)^s/2Op^W(b^−s) is in ΨΣ⁰₁ and thus is bounded onL²(Rⁿ). We have obtained a continuous inclusion of Q⁽s^b)(Rⁿ) intoQs(Rⁿ). Conversely, we have for anyh∈ Qs(Rⁿ) that

h⁽_Q^b_s⁾≤ Op^W(b^s)(I +Q)^−s/2_L(L²(Rⁿ))h_Qs.

The inverse of Op^W(b^s)(I +Q)^−s/2 is (I +Q)^s/2(Op^W(b^s))⁻¹ since the opera- tors I +Qand Op^W(b^s) are invertible. Moreover, for the same reason as above, (I +Q)^s/2(Op^W(b^s))⁻¹ is bounded on L²(Rⁿ). By the inverse mapping theorem, Op^W(b^s)(I +Q)^−s/2 is bounded on L²(Rⁿ). This shows the reverse continuous inclusion. We have proved

Q⁽_s^b)(Rⁿ) =Q_s(Rⁿ)

with equivalence of norms for s ∈2N0 and this implies that this is true for any s ∈ R by the properties of duality and interpolation for Q⁽s^b)(Rⁿ) and Qs(Rⁿ).

This shows Part (4) and implies Parts (5) and (6).

Let us show that, for eachs∈N0, the spaceQs(Rⁿ) coincides with the space Q⁽s^int)(Rⁿ) of functionsh∈ L²(Rⁿ) such that the tempered distributions u^α∂_u^βh are in L²(Rⁿ) for every α, β ∈ Nⁿ0 such that |α|+|β| ≤ s. Endowed with the norm · ⁽_Q^int_s ⁾ defined in Part (3),Q⁽s^int)(Rⁿ) is a Banach space. We have for any h∈ Q_s(Rⁿ) =Q⁽s^b)(Rⁿ)

h⁽_Q^int_s ⁾≤

|α|+|β|≤s

u^α∂_u^βOp^W(b^−s)_L(L²(Rⁿ))h⁽_Q^b)_s.

Since the operators u^α∂_u^βOp^W(b^−s) are in ΨΣ^|α|₁ ^+|β|−s(Rⁿ) thus continuous on L²(Rⁿ) when |α|+|β| ≤ s, we see that Qs(Rⁿ) is continuously included in Q⁽s^int)(Rⁿ). For the converse, we separate the cases s even and odd. If s ∈ 2N0

then we have easily that h_Qs =

I +

j

(−∂²_uj+u²_j)_s/2

h_L²(Rⁿ)

≤ C_s

|α|+|β|≤s

u^α∂^β_uh_L²(Rⁿ)=C_sh⁽_Q^int_s ⁾.

6.4. Shubin classes 465 Now ifs∈2N0+ 1, we have, since Op^W(b⁻¹)(I +Q)^1/2is bounded and invertible (see Part (6) already proven),

h_Qs = (I +Q)^s/2h_L²(Rⁿ)≤COp^W(b⁻¹)(I +Q)^1/2(I +Q)^s/2h_L²(Rⁿ)

≤COp^W(b⁻¹)(I +

j

−∂_u²_j +u²_j)^(s+1)/2h_L²(Rⁿ)

≤C_s

|α|+|β|≤s+1

Op^W(b⁻¹)x^α∂_x^βh_L²(Rⁿ)

≤C_s

|α|+|β|≤s

u^α∂_u^βh_L²(Rⁿ)=C_sh⁽_Q^int_s ⁾,

by the property of the calculus. Therefore, for s even and odd, Q⁽s^int)(Rⁿ) is continuously included inQ_s(Rⁿ). As we have already proven the reverse inclusion, the equality holds and Part (3) is proved. This implies

#

s∈R

Q_s(Rⁿ) =S(Rⁿ)

and Part (1) is now completely proved.

These Sobolev spaces enable us to characterise the operators in the calculus.

We allow ourselves to use the shorthand notation

(adu)^α1 := (adu1)^α11. . .(adun)^α1n, and

(ad∂_u)^α2 := (ad∂_u1)^α21. . .(ad∂_un)^α2n.

Theorem 6.4.18. We assume that ρ∈(0,1]. Let A:S(Rⁿ)→ S(Rⁿ) be a linear continuous operator such that all the operators

(adu)^α1(ad∂u)^α2A, α1, α2∈Nⁿ0,

are in L(L²(Rⁿ),Q_−m+ρ(|α1|+|α2|)) in the sense that they extend to continuous operators from L²(Rⁿ) toQ_−m+ρ(|α1|+|α2|). Then A ∈ ΨΣ^m_ρ (Rⁿ). Moreover, for any∈N, there exist a constantCand an integer, both independent ofA, such that

AΨΣ^m_ρ,≤C

|α1|+|α2|≤

(adu)^α1(ad∂_u)^α2A_L(L²(Rⁿ),Q−m+ρ(|α1|+|α2|)).

Note that the converse is true, that is, givenA∈ΨΣ^m_ρ then

∀α1, α2∈Nⁿ0 (adu)^α1(ad∂_u)^α2A∈L(L²(Rⁿ),Q_−m+ρ(|α1|+|β|),).

This is just a consequence of the properties of the calculus.

The proof of Theorem 6.4.18 relies on the following characterisation of the class of symbols

Σ⁰₀(Rⁿ) :=S(1, dξ²+du²).

Theorem 6.4.19(Beals’ characterisation of Σ⁰₀(Rⁿ)). Let A :S(Rⁿ)→ S(Rⁿ) be a linear continuous operator such that all the operators

(adu)^α1(ad∂u)^α2A, α1, α2∈Nⁿ0,

are inL(L²(Rⁿ))in the sense that they extend to continuous operators onL²(Rⁿ).

Then there exits a unique functiona={a(ξ, x)} ∈Σ⁰₀(Rⁿ)such thatA= Op^W(a).

Moreover, for any∈N, there exist a constantC and an integer, both indepen- dent of A, such that

aΣ⁰₀,≤C

|α1|+|α2|≤

(adu)^α1(ad∂u)^α2A_L(L²(Rⁿ)).

The converse is true, that is, givena∈Σ⁰₀(Rⁿ)thenA= Op^W(a)satisfies

∀α1, α2∈Nⁿ0 (adu)^α1(ad∂_u)^α2A∈L(L²(Rⁿ)).

We admit Beals’ theorem stated in Theorem 6.4.19, see the original article [Bea77a] for the proof.

For the sake of completeness we prove Theorem 6.4.18. This proof can also be found in [Hel84a, Th´eor`eme 1.21.1].

Sketch of the proof of Theorem 6.4.18. Let A be as in the statement andb as in Example 6.4.13. We write

Bs:= Op^W(b^s) and

Aα1,α2:= (adu)^α1(ad∂u)^α2A, α1, α2∈Nⁿ0.

We set s:= m−ρ(|α1|+|α2|). Then B_s⁻¹Aα1,α2 ∈ L(L²(Rⁿ)). Moreover, we have

ad∂_u1

B_s⁻¹A_α1,α2

= ad∂_u1

B_s⁻¹

A_α1,α2+B_s⁻¹ad∂_u1(A_α1,α2) ; the first operator of the right-hand side is in L(L²(Rⁿ),Q1(Rⁿ)) whereas the second is in L(L²(Rⁿ),Qρ(Rⁿ)). Proceeding recursively, we obtain that the op- eratorB_m−ρ⁻¹ _{(|α1|+|α2|)}Aα1,α2 satisfies the hypothesis of Beals’ Theorem (Theorem 6.4.19). Therefore, there existsc_α1,α2 ∈Σ⁰₀(Rⁿ) such that

B⁻_m−ρ¹ _(|α

1|+|α2|)A_α1,α2= Op^W(c_α1,α2)

6.4. Shubin classes 467 or, equivalently,

Aα1,α2 = Op^W(aα1,α2) with aα1,α2 =b_m−ρ(|α1|+|α2|) cα1,α2. We haveA= Op^W(a0,0) and

Op^W(a_α1,α2) = A_α1,α2= (adu)^α1(ad∂_u)^α2A

= (adu)^α1(ad∂_u)^α2Op^W(a0,0)

= Op^W

i^|α1|∂_ξ^α1∂_u^α2a0,0

, by Lemma 6.2.3, thus

aα1,α2 =i^|α1|∂_ξ^α1∂_u^α2a0,0.

Consequentlya∈Σ^m_ρ.

Looking back at the proof, we see that it can be slightly improved in the following way:

Corollary 6.4.20. We assume that ρ∈(0,1]. LetA:S(Rⁿ)→ S(Rⁿ)be a linear continuous operator.

The operator A is inΨΣ^m_ρ(Rⁿ)if and only if there exists γo ∈R such that for each α1, α2∈Nⁿ0 we have

(adu)^α1(ad∂u)^α2A∈L(Qγo(Rⁿ),Q_−m+ρ(|α1|+|α2|)+γo).

In this case this property is true for every γ∈R, that is, for eachγ∈Rand α1, α2∈Nⁿ0, we have

(adu)^α1(ad∂_u)^α2A∈L(Q_γ(Rⁿ),Q_−m+ρ(|α1|+|α2|)+γ).

Moreover, for any∈N, there exist a constantC and an integer, both indepen- dent of A, such that

AΨΣ^m_ρ,≤C

|α1|+|α2|≤

(adu)^α1(ad∂_u)^α2A_L(Qγ(Rⁿ),Q−m+ρ(|α1|+|α2|)+γ). Sketch of the proof of Corollary 6.4.20. We keep the notation of the proof of The- orem 6.4.18. Let A be as in the statement and lets:=m−ρ(|α1|+|α2|). Then B_s⁻+¹γoAα1,α2Bγo∈L(L²(Rⁿ)). Moreover, we have

ad∂u1

B⁻_s+¹γoAα1,α2Bγo

= ad∂u1

B⁻_s+¹γo

Aα1,α2Bγo

+B⁻_s+¹γoad∂u1(Aα1,α2)Bγo

+B⁻_s+¹γoA_α1,α2B_γo B_γ⁻_o¹(ad∂_u1B_γo) ; the first operator of the right-hand side is in L(L²(Rⁿ),Q1(Rⁿ)), the second is inL(L²(Rⁿ),Qρ(Rⁿ)) and the third is inL(L²(Rⁿ)). Proceeding recursively, we obtain that B_s⁻+¹γoAα1,α2Bγo satisfies the hypothesis of Theorem 6.4.19. We then

conclude as in the proof of Theorem 6.4.18.