Measures constrained by linear PDEs [after de Philippis and Rindler]

Linear PDEs The result Proof of Theorem

Measures constrained by linear PDEs [after de Philippis and Rindler]

Pierre Pansu, Universit´e Paris-Saclay

March 1st, 2021

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

A constant coefficient linear PDE on a vector-valued functionu∶Ω⊂R^d→R^Ntakes the form

A(u) = ∑

α

Aα∂^αu=0,

whereAα∈Hom(R^N,Rⁿ)are linear maps. Here∂^αu=∂^α₁¹⋯∂^α_k^kudenote partial derivatives.

Example

Assume u∈C^∞(Ω,(R^d)^∗)is a differential1-form on open setΩ. Its exterior differential du∈C^∞(Ω,Λ²(R^d)^∗). Then du= A(u)whereAhas order1. Putting all Aαtogether into a single linear map

A∈Hom(R^d,Hom((R^d)^∗,Λ²(R^d)^∗))

=Hom((R^d)^∗⊗ (R^d)^∗,Λ²(R^d)^∗), A becomes skew-symmetrization of bilinear forms onR^d.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

A constant coefficient linear PDE on a vector-valued functionu∶Ω⊂R^d→R^Ntakes the form

A(u) = ∑

α

Aα∂^αu=0,

whereAα∈Hom(R^N,Rⁿ)are linear maps. Here∂^αu=∂^α₁¹⋯∂^α_k^kudenote partial derivatives.

Example

Assume u∈C^∞(Ω,(R^d)^∗)is a differential1-form on open setΩ. Its exterior differential du∈C^∞(Ω,Λ²(R^d)^∗). Then du= A(u)whereAhas order1. Putting all Aαtogether into a single linear map

A∈Hom(R^d,Hom((R^d)^∗,Λ²(R^d)^∗))

=Hom((R^d)^∗⊗ (R^d)^∗,Λ²(R^d)^∗), A becomes skew-symmetrization of bilinear forms onR^d.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

A constant coefficient linear PDE on a vector-valued functionu∶Ω⊂R^d→R^Ntakes the form

A(u) = ∑

α

Aα∂^αu=0,

whereAα∈Hom(R^N,Rⁿ)are linear maps. Here∂^αu=∂^α₁¹⋯∂^α_k^kudenote partial derivatives.

Example

Assume u∈C^∞(Ω,Hom(R^d,R^{`</sup>))is aR<sup>`}-valued differential1-form. Its exterior differential du∈C^∞(Ω,Hom(Λ²R^d,R^`)). Then du= A(u)whereAhas order1.

Putting all Aα together into a single linear map

A∈Hom(R^d,Hom((R^d)^∗⊗R^{`</sup>,Λ<sup>2</sup>(R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}))

=Hom((R^d)^∗⊗ (R^d)^∗⊗R^{`</sup>,Λ<sup>2</sup>(R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}), A becomes skew-symmetrization of bilinear mapsR^d→R^`.

Question. Assume a vector-valued measureµsatisfiesA(µ) =0 in distributional sense. Ellipticity should imply restrictions on the singular partµ^s ofµ. Which restrictions ?

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

A constant coefficient linear PDE on a vector-valued functionu∶Ω⊂R^d→R^Ntakes the form

A(u) = ∑

α

Aα∂^αu=0,

whereAα∈Hom(R^N,Rⁿ)are linear maps. Here∂^αu=∂^α₁¹⋯∂^α_k^kudenote partial derivatives.

Example

Assume u∈C^∞(Ω,Hom(R^d,R^{`</sup>))is aR<sup>`}-valued differential1-form. Its exterior differential du∈C^∞(Ω,Hom(Λ²R^d,R^`)). Then du= A(u)whereAhas order1.

Putting all Aα together into a single linear map

A∈Hom(R^d,Hom((R^d)^∗⊗R^{`</sup>,Λ<sup>2</sup>(R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}))

=Hom((R^d)^∗⊗ (R^d)^∗⊗R^{`</sup>,Λ<sup>2</sup>(R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}), A becomes skew-symmetrization of bilinear mapsR^d→R^`.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

The principal symbol. In Fourier, ifξ∈ (R^d)^∗,

F(A(u))(ξ) =A(2πiξ)(F(u)(ξ)),

whereA(ξ) = ∑αAαξ^α∈Hom(R^N,Rⁿ). It is a sum of homogeneous terms, the term of highest degreeA^k(ξ)is theprincipal symbolofAevaluated atξ.

Example

For the exterior differential, forη⊗e∈ (R^d)^∗⊗R^`, A(ξ)(η⊗e) = (ξ∧η) ⊗e.

Example

For the scalar Laplacian, A(ξ) = ∣ξ∣².

AssumeAiselliptic, i.e.A(ξ)is injective for allξ/=0. Then A(u) =0 Ô⇒ supp(F(u)) = {0}, souis a polynomial.

Conversely, assume thatAis homogeneous. If there exists nonzeroξ∈ (R^d)^∗and λ∈Rⁿsuch thatλ∈ker(A(ξ)), then for every distributionh∶R→R, the plane wave u= (h○ξ)λsolvesA(u) =0. ThusAhas nonsmooth solutions.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

The principal symbol. In Fourier, ifξ∈ (R^d)^∗,

F(A(u))(ξ) =A(2πiξ)(F(u)(ξ)),

whereA(ξ) = ∑αAαξ^α∈Hom(R^N,Rⁿ). It is a sum of homogeneous terms, the term of highest degreeA^k(ξ)is theprincipal symbolofAevaluated atξ.

Example

For the exterior differential, forη⊗e∈ (R^d)^∗⊗R^`, A(ξ)(η⊗e) = (ξ∧η) ⊗e.

Example

For the scalar Laplacian, A(ξ) = ∣ξ∣².

AssumeAiselliptic, i.e.A(ξ)is injective for allξ/=0. Then A(u) =0 Ô⇒ supp(F(u)) = {0}, souis a polynomial.

Conversely, assume thatAis homogeneous. If there exists nonzeroξ∈ (R^d)^∗and λ∈Rⁿsuch thatλ∈ker(A(ξ)), then for every distributionh∶R→R, the plane wave u= (h○ξ)λsolvesA(u) =0. ThusAhas nonsmooth solutions.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

The principal symbol. In Fourier, ifξ∈ (R^d)^∗,

F(A(u))(ξ) =A(2πiξ)(F(u)(ξ)),

whereA(ξ) = ∑αAαξ^α∈Hom(R^N,Rⁿ). It is a sum of homogeneous terms, the term of highest degreeA^k(ξ)is theprincipal symbolofAevaluated atξ.

Example

For the exterior differential, forη⊗e∈ (R^d)^∗⊗R^`, A(ξ)(η⊗e) = (ξ∧η) ⊗e.

Example

For the scalar Laplacian, A(ξ) = ∣ξ∣².

AssumeAiselliptic, i.e.A(ξ)is injective for allξ/=0. Then A(u) =0 Ô⇒ supp(F(u)) = {0}, souis a polynomial.

Conversely, assume thatAis homogeneous. If there exists nonzeroξ∈ (R^d)^∗and λ∈Rⁿsuch thatλ∈ker(A(ξ)), then for every distributionh∶R→R, the plane wave u= (h○ξ)λsolvesA(u) =0. ThusAhas nonsmooth solutions.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

The principal symbol. In Fourier, ifξ∈ (R^d)^∗,

F(A(u))(ξ) =A(2πiξ)(F(u)(ξ)),

whereA(ξ) = ∑αAαξ^α∈Hom(R^N,Rⁿ). It is a sum of homogeneous terms, the term of highest degreeA^k(ξ)is theprincipal symbolofAevaluated atξ.

Example

For the exterior differential, forη⊗e∈ (R^d)^∗⊗R^`, A(ξ)(η⊗e) = (ξ∧η) ⊗e.

Example

For the scalar Laplacian, A(ξ) = ∣ξ∣².

AssumeAiselliptic, i.e.A(ξ)is injective for allξ/=0. Then A(u) =0 Ô⇒ supp(F(u)) = {0}, souis a polynomial.

Conversely, assume thatAis homogeneous. If there exists nonzeroξ∈ (R^d)^∗and λ∈Rⁿsuch thatλ∈ker(A(ξ)), then for every distributionh∶R→R, the plane wave u= (h○ξ)λsolvesA(u) =0. ThusAhas nonsmooth solutions.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

The principal symbol. In Fourier, ifξ∈ (R^d)^∗,

F(A(u))(ξ) =A(2πiξ)(F(u)(ξ)),

whereA(ξ) = ∑αAαξ^α∈Hom(R^N,Rⁿ). It is a sum of homogeneous terms, the term of highest degreeA^k(ξ)is theprincipal symbolofAevaluated atξ.

Example

For the exterior differential, forη⊗e∈ (R^d)^∗⊗R^`, A(ξ)(η⊗e) = (ξ∧η) ⊗e.

Example

For the scalar Laplacian, A(ξ) = ∣ξ∣².

AssumeAiselliptic, i.e.A(ξ)is injective for allξ/=0. Then A(u) =0 Ô⇒ supp(F(u)) = {0}, souis a polynomial.

Conversely, assume thatAis homogeneous. If there exists nonzeroξ∈ (R^d)^∗and λ∈Rⁿsuch thatλ∈ker(A(ξ)), then for every distributionh∶R→R, the plane wave u= (h○ξ)λsolvesA(u) =0. ThusAhas nonsmooth solutions.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

Nonelliptic behaviours are concentrated in thewave cone.

Definition (Wave cone)

The wave cone of an order k operatorAisΛA∶= ⋃

ξ/=0

kerA^k(ξ) ⊂R^N.

Example

For the exterior differential d ,ΛA= ⋃

ξ/=0

ξ⊗R^{`</sup>⊂ (R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}=Hom(R^d,R^{`</sup>)consists of all rank≤1linear mapsR<sup>d</sup>→R<sup>`}.

Indeed, the kernel ofξ∧ ⋅ ∶Λ¹→Λ²coincides with the image ofξ∧ ⋅ ∶Λ⁰→Λ¹. This reflects (at the symbol level) the (local) exactness of the de Rham complex.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

Nonelliptic behaviours are concentrated in thewave cone.

Definition (Wave cone)

The wave cone of an order k operatorAisΛA∶= ⋃

ξ/=0

kerA^k(ξ) ⊂R^N.

Example

For the exterior differential d ,ΛA= ⋃

ξ/=0

ξ⊗R^{`</sup>⊂ (R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}=Hom(R^d,R^{`</sup>)consists of all rank≤1linear mapsR<sup>d</sup>→R<sup>`}.

Indeed, the kernel ofξ∧ ⋅ ∶Λ¹→Λ²coincides with the image ofξ∧ ⋅ ∶Λ⁰→Λ¹. This reflects (at the symbol level) the (local) exactness of the de Rham complex.

Linear PDEs The result Proof of Theorem

The problem Ellipticity The wave cone

Nonelliptic behaviours are concentrated in thewave cone.

Definition (Wave cone)

The wave cone of an order k operatorAisΛA∶= ⋃

ξ/=0

kerA^k(ξ) ⊂R^N.

Example

For the exterior differential d ,ΛA= ⋃

ξ/=0

ξ⊗R^{`</sup>⊂ (R<sup>d</sup>)<sup>∗</sup>⊗R<sup>`}=Hom(R^d,R^{`</sup>)consists of all rank≤1linear mapsR<sup>d</sup>→R<sup>`}.

Indeed, the kernel ofξ∧ ⋅ ∶Λ¹→Λ²coincides with the image ofξ∧ ⋅ ∶Λ⁰→Λ¹. This reflects (at the symbol level) the (local) exactness of the de Rham complex.

Linear PDEs The result Proof of Theorem

Theorem (de Philippis-Rindler 2018)

Letµbe anR^N-valued measure which solvesA(µ) =0on open setΩ. Then

dµ

d∣µ∣(x) ∈ΛAfor∣µ∣^s-almost every point x∈Ω.

Example

WhenAis the exterior differential on1-forms, Hom(R^d,R^{`</sup>)-valued measures solving dµ=0are differentials ofR<sup>`}-valued BV functions. According to the Theorem, their singular parts are supported on points where_d∣µ∣^dµ is a rank1linear mapR^d→R^`. This is Alberti’s Rank One Theorem (1993), solving a conjecture of Ambrosio-De Giorgi (1988).

Linear PDEs The result Proof of Theorem

Theorem (de Philippis-Rindler 2018)

Letµbe anR^N-valued measure which solvesA(µ) =0on open setΩ. Then

dµ

d∣µ∣(x) ∈ΛAfor∣µ∣^s-almost every point x∈Ω.

Example

WhenAis the exterior differential on1-forms, Hom(R^d,R^{`</sup>)-valued measures solving dµ=0are differentials ofR<sup>`}-valued BV functions.

According to the Theorem, their singular parts are supported on points where_d∣µ∣^dµ is a rank1linear mapR^d→R^`. This is Alberti’s Rank One Theorem (1993), solving a conjecture of Ambrosio-De Giorgi (1988).

Linear PDEs The result Proof of Theorem

Theorem (de Philippis-Rindler 2018)

Letµbe anR^N-valued measure which solvesA(µ) =0on open setΩ. Then

dµ

d∣µ∣(x) ∈ΛAfor∣µ∣^s-almost every point x∈Ω.

Example

WhenAis the exterior differential on1-forms, Hom(R^d,R^{`</sup>)-valued measures solving dµ=0are differentials ofR<sup>`}-valued BV functions. According to the Theorem, their singular parts are supported on points where_d∣µ∣^dµ is a rank1linear mapR^d→R^`. This is Alberti’s Rank One Theorem (1993), solving a conjecture of Ambrosio-De Giorgi (1988).

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

I stick to degree 1 operators. Letµbe anR^N-valued measure solvingA(µ) =0. Let∣µ∣ denote its total variation. Consider the Radon-Nikodym decomposition

µ= dµ d∣µ∣d∣µ∣.

LetE= {x∈R^d;_d∣µ∣^dµ(x) ∉ΛA}. By contradiction, assume that∣µ∣^s(E) >0.

One can pickx0∈Eand a subsequencer_k of radii which satisfy the following. Letµ_k denote the blowing up ofµatx0, at scalerk, normalized by∣µ∣(B(rk)).

1 ∣µk∣converges vaguely to a measureν.

2 λ0=_d∣µ∣^dµ(0) ∉ΛA.

3 µkconverges vaguely to the measureνλ0.

4 ∣µ∣^s(B(r_k))

∣µ∣(B(r_k)) tends to 1.

One shows thatA(νλ0) =0. Sinceλ0∉ΛA, this implies thatsupp(F(ν)) =0, hence thatνis absolutely continuous.

On the other hand, one shows that∣µk∣converges toν in total variation. This is a contradiction, since∣∣µk∣ −ν∣ ≥ ∣µk∣^s(B(rk))/∣µ∣(B(rk))which tends to 1.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

I stick to degree 1 operators. Letµbe anR^N-valued measure solvingA(µ) =0. Let∣µ∣ denote its total variation. Consider the Radon-Nikodym decomposition

µ= dµ d∣µ∣d∣µ∣.

LetE= {x∈R^d;_d∣µ∣^dµ(x) ∉ΛA}. By contradiction, assume that∣µ∣^s(E) >0.

One can pickx0∈Eand a subsequencer_k of radii which satisfy the following. Letµ_k denote the blowing up ofµatx0, at scalerk, normalized by∣µ∣(B(rk)).

1 ∣µk∣converges vaguely to a measureν.

2 λ0=_d∣µ∣^dµ(0) ∉ΛA.

3 µkconverges vaguely to the measureνλ0.

4 ∣µ∣^s(B(r_k))

∣µ∣(B(r_k)) tends to 1.

One shows thatA(νλ0) =0. Sinceλ0∉ΛA, this implies thatsupp(F(ν)) =0, hence thatνis absolutely continuous.

On the other hand, one shows that∣µk∣converges toν in total variation. This is a contradiction, since∣∣µk∣ −ν∣ ≥ ∣µk∣^s(B(rk))/∣µ∣(B(rk))which tends to 1.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

I stick to degree 1 operators. Letµbe anR^N-valued measure solvingA(µ) =0. Let∣µ∣ denote its total variation. Consider the Radon-Nikodym decomposition

µ= dµ d∣µ∣d∣µ∣.

LetE= {x∈R^d;_d∣µ∣^dµ(x) ∉ΛA}. By contradiction, assume that∣µ∣^s(E) >0.

One can pickx0∈Eand a subsequencer_k of radii which satisfy the following. Letµ_k denote the blowing up ofµatx0, at scalerk, normalized by∣µ∣(B(rk)).

1 ∣µk∣converges vaguely to a measureν.

2 λ0=_d∣µ∣^dµ(0) ∉ΛA.

3 µkconverges vaguely to the measureνλ0.

4 ∣µ∣^s(B(r_k))

∣µ∣(B(r_k)) tends to 1.

One shows thatA(νλ0) =0. Sinceλ0∉ΛA, this implies thatsupp(F(ν)) =0, hence thatνis absolutely continuous.

On the other hand, one shows that∣µk∣converges toν in total variation. This is a contradiction, since∣∣µk∣ −ν∣ ≥ ∣µk∣^s(B(rk))/∣µ∣(B(rk))which tends to 1.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

I stick to degree 1 operators. Letµbe anR^N-valued measure solvingA(µ) =0. Let∣µ∣ denote its total variation. Consider the Radon-Nikodym decomposition

µ= dµ d∣µ∣d∣µ∣.

LetE= {x∈R^d;_d∣µ∣^dµ(x) ∉ΛA}. By contradiction, assume that∣µ∣^s(E) >0.

One can pickx0∈Eand a subsequencer_k of radii which satisfy the following. Letµ_k denote the blowing up ofµatx0, at scalerk, normalized by∣µ∣(B(rk)).

1 ∣µk∣converges vaguely to a measureν.

2 λ0=_d∣µ∣^dµ(0) ∉ΛA.

3 µkconverges vaguely to the measureνλ0.

4 ∣µ∣^s(B(r_k))

∣µ∣(B(r_k)) tends to 1.

One shows thatA(νλ0) =0. Sinceλ0∉ΛA, this implies thatsupp(F(ν)) =0, hence thatνis absolutely continuous.

On the other hand, one shows that∣µk∣converges toν in total variation. This is a contradiction, since∣∣µk∣ −ν∣ ≥ ∣µk∣^s(B(rk))/∣µ∣(B(rk))which tends to 1.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

First step : absolute continuity of tangent measures In distributional sense,A(νλ0) =limA(µk) =0.

Taking Fourier transforms,

∀ξ∈ (R^d)^∗, (F(ν)(ξ))A(ξ)λ0=0. Sinceλ0∉ΛA,A(ξ)λ0/=0, henceF(ν)(ξ) =0.

Sincesupp(F(ν)) = {0},νis a (polynomial) function times Lebesgue measure.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

First step : absolute continuity of tangent measures

In distributional sense,A(νλ0) =limA(µk) =0. Taking Fourier transforms,

∀ξ∈ (R^d)^∗, (F(ν)(ξ))A(ξ)λ0=0.

Sinceλ0∉ΛA,A(ξ)λ0/=0, henceF(ν)(ξ) =0.

Sincesupp(F(ν)) = {0},νis a (polynomial) function times Lebesgue measure.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

First step : absolute continuity of tangent measures

In distributional sense,A(νλ0) =limA(µk) =0. Taking Fourier transforms,

∀ξ∈ (R^d)^∗, (F(ν)(ξ))A(ξ)λ0=0.

Sinceλ0∉ΛA,A(ξ)λ0/=0, henceF(ν)(ξ) =0.

Sincesupp(F(ν)) = {0},νis a (polynomial) function times Lebesgue measure.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

First step : absolute continuity of tangent measures

In distributional sense,A(νλ0) =limA(µk) =0. Taking Fourier transforms,

∀ξ∈ (R^d)^∗, (F(ν)(ξ))A(ξ)λ0=0.

Sinceλ0∉ΛA,A(ξ)λ0/=0, henceF(ν)(ξ) =0.

Sincesupp(F(ν)) = {0},νis a (polynomial) function times Lebesgue measure.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Second step : from vague convergence to convergence in total variation Letνk= ∣µk∣^s.

Recall that^∣µ∣_∣µ∣(B^s(B(r_(r^k))

k)) tends to 1. One can furthermore assume that

∮^B(rk)∣_d∣µ∣^dµ −λ0∣d∣µ∣^stends to 0. It follows that∣νkλ0−µk∣(B(1))tends to 0.

Pick a smooth cut-off functionχ. Then∣χνkλ0−χµk∣(R^d)tends to 0. SinceA(µ_k) =0,

A(χνkλ0) = A(χνkλ0−χµk) +A(dχ)µk.

We shall see that ellipticity allows to invertAwith estimates. By density, one can replace measures with smooth functions and total variation withL¹norm. The goal is to show thatχνk is precompact inL¹. The estimates will stem from H¨ormander- Mikhlin’s multiplier theorem.

Taking Fourier transforms,

∀ξ, F(χνk)A(ξ)λ0=A(ξ)F(Vk)(ξ) + F(Rk)(ξ),

whereV_k=χν_kλ0−χµ_ktends to 0 inL¹andR_k=A(dχ)µ_kis bounded inL¹.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Second step : from vague convergence to convergence in total variation Letνk= ∣µk∣^s.

Recall that^∣µ∣_∣µ∣(B^s(B(r_(r^k))

k)) tends to 1. One can furthermore assume that

∮^B(rk)∣_d∣µ∣^dµ −λ0∣d∣µ∣^stends to 0. It follows that∣νkλ0−µk∣(B(1))tends to 0.

Pick a smooth cut-off functionχ. Then∣χνkλ0−χµk∣(R^d)tends to 0.

SinceA(µ_k) =0,

A(χνkλ0) = A(χνkλ0−χµk) +A(dχ)µk.

We shall see that ellipticity allows to invertAwith estimates. By density, one can replace measures with smooth functions and total variation withL¹norm. The goal is to show thatχνk is precompact inL¹. The estimates will stem from H¨ormander- Mikhlin’s multiplier theorem.

Taking Fourier transforms,

∀ξ, F(χνk)A(ξ)λ0=A(ξ)F(Vk)(ξ) + F(Rk)(ξ),

whereV_k=χν_kλ0−χµ_ktends to 0 inL¹andR_k=A(dχ)µ_kis bounded inL¹.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Second step : from vague convergence to convergence in total variation Letνk= ∣µk∣^s.

Recall that^∣µ∣_∣µ∣(B^s(B(r_(r^k))

k)) tends to 1. One can furthermore assume that

∮^B(rk)∣_d∣µ∣^dµ −λ0∣d∣µ∣^stends to 0. It follows that∣νkλ0−µk∣(B(1))tends to 0.

Pick a smooth cut-off functionχ. Then∣χνkλ0−χµk∣(R^d)tends to 0.

SinceA(µ_k) =0,

A(χνkλ0) = A(χνkλ0−χµk) +A(dχ)µk.

We shall see that ellipticity allows to invertAwith estimates. By density, one can replace measures with smooth functions and total variation withL¹norm. The goal is to show thatχνk is precompact inL¹. The estimates will stem from H¨ormander- Mikhlin’s multiplier theorem.

Taking Fourier transforms,

∀ξ, F(χνk)A(ξ)λ0=A(ξ)F(Vk)(ξ) + F(Rk)(ξ),

whereV_k=χν_kλ0−χµ_ktends to 0 inL¹andR_k=A(dχ)µ_kis bounded inL¹.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Second step : from vague convergence to convergence in total variation Letνk= ∣µk∣^s.

Recall that^∣µ∣_∣µ∣(B^s(B(r_(r^k))

k)) tends to 1. One can furthermore assume that

∮^B(rk)∣_d∣µ∣^dµ −λ0∣d∣µ∣^stends to 0. It follows that∣νkλ0−µk∣(B(1))tends to 0.

Pick a smooth cut-off functionχ. Then∣χνkλ0−χµk∣(R^d)tends to 0.

SinceA(µ_k) =0,

A(χνkλ0) = A(χνkλ0−χµk) +A(dχ)µk.

We shall see that ellipticity allows to invertAwith estimates. By density, one can replace measures with smooth functions and total variation withL¹norm. The goal is to show thatχνk is precompact inL¹. The estimates will stem from H¨ormander- Mikhlin’s multiplier theorem.

Taking Fourier transforms,

∀ξ, F(χνk)A(ξ)λ0=A(ξ)F(Vk)(ξ) + F(Rk)(ξ),

whereV_k=χν_kλ0−χµ_ktends to 0 inL¹andR_k=A(dχ)µ_kis bounded inL¹.

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Inner multiplying withA(ξ)λ0, one gets

∣A(ξ)λ0∣²F(χνk) =A(ξ)λ0⋅A(ξ)F(Vk)(ξ) +A(ξ)λ0⋅ F(Rk)(ξ), and then

(1+ ∣A(ξ)λ0∣²)F(χνk) =A(ξ)λ0⋅A(ξ)F(Vk)(ξ) +A(ξ)λ0⋅ F(Rk)(ξ) + F(χνk). SoF(χνk) =T0(Vk) +T1((Id−∆)^−1/2Rk) +T2((Id−∆)⁻¹F(χνk))is the sum of three terms, involving operators defined by Fourier multipliers

m0(ξ) = (1+ ∣A(ξ)λ0∣²)⁻¹A(ξ)λ0⋅A(ξ), m1(ξ) = (1+ ∣A(ξ)λ0∣²)⁻¹(1+4π²∣ξ∣²)^1/2A(ξ)λ0, m2(ξ) = (1+ ∣A(ξ)λ0∣²)⁻¹(1+4π²∣ξ∣²).

(Id−∆)^−1/2is bounded and compact fromL¹to someL^q,q>1.T1andT2are bounded fromL^qtoL^q. Thereforewk∶=T1((Id−∆)^−1/2Rk) +T2((Id−∆)⁻¹F(χνk))

Linear PDEs The result Proof of Theorem

Strategy

Absolute continuity of tangent measures Convergence in total variation

Sinceλ0∉ΛA,∣A(ξ)λ0∣/∣ξ∣is bounded away from 0. This implies thatT0is bounded fromL¹to the Lorentz spaceL^1,∞(H¨ormander-Mikhlin). Henceuk∶=T0(Vk)tends to 0 in measure.

Sinceuk+wk=χνk≥0,u_k⁻is dominated by∣wk∣. By Vitali’s convergence theorem (strengthening of dominated convergence theorem where a.e. convergence is replaced with convergence in measure),u_k⁻tends to 0 inL¹.

SinceVk→0 in distributional sense, so doesuk=T0(Vk). For every smooth cut-off functionφ, 0≤φ≤1,

∫ φ∣uk∣ = ∫ φuk+2∫ φu_k⁻≤ ∫ φuk+2∫ u_k⁻ tends to 0, souk tends to 0 inL¹.

Henceχνk=uk+wkis precompact inL¹. Its vague limit isχν, so it subconverges in L¹toχν. This shows that∣µ_k∣^sconverges toνin total variation.

This contradicts the fact thatνis absolutely continuous, and completes the proof of the theorem.