Reverse Hardy-Littlewood-Sobolev inequalities

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Jean Dolbeault

http://www.ceremade.dauphine.fr/∼dolbeaul Ceremade, Université Paris-Dauphine

June 23, 2018

Analysis of Effective One-Particle Equations and their Derivation Ludwig-Maximilians Universität(22-23/6/2018)

Joint work with J. A. Carrillo, M. G. Delgadino, R. Frank, F. Hoffmann

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Outline

Reverse HLS inequality

BThe inequality and the conformally invariant case BA proof based on Carlson’s inequality

BThe caseλ= 2

BConcentration and a relaxed inequality Existence of minimizers and relaxation BExistence minimizers ifq >2N/(2N+λ) BRlaxation and measure valued minimizers

Regions of no concentration and regularity of measure valued minimizers

BNo concentration results BRegularity issues Free Energy

BFree energy: toy model, equivalence with reverse HLS inequalities

BRelaxed free energy BUniqueness

J. Dolbeault Reverse Hardy-Littlewood-Sobolev inequalities

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Free energy point of view Concentration and a relaxed inequality

Reverse

Hardy-Littlewood-Sobolev inequality

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The reverse HLS inequality

For anyλ >0 and any measurable function ρ≥0 onR^N, let I_λ[ρ] :=

Z Z

R^N×R^N

|x−y|^λρ(x)ρ(y)dx dy

N ≥1, 0< q <1, α:= 2N−q(2N+λ) N(1−q) Convention: ρ∈L^p(R^N) ifR

R^N|ρ(x)|^pdxfor anyp >0 Theorem

The inequality

Iλ[ρ] ≥CN,λ,q

Z

R^N

ρ dx αZ

R^N

ρ^qdx

(2−α)/q

(1) holds for anyρ∈L¹₊∩L^q(R^N)with CN,λ,q>0 if and only if

q > N/(N+λ)

If eitherN= 1,2or if N ≥3andq≥min

1−2/N ,2N/(2N+λ) , then there is a radial nonnegative optimizerρ∈L¹∩L^q(R^N)

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0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

N = 4, region of the parameters(λ, q) for whichCN,λ,q>0 Optimal functions exist in the light grey area

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The conformally invariant case q = 2N/(2N + λ)

I_λ[ρ] = Z Z

R^N×R^N

|x−y|^λρ(x)ρ(y)dx dy ≥CN,λ,q

Z

R^N

ρ^qdx 2/q

2N/(2N+λ) ⇐⇒ α= 0 (Dou, Zhu 2015) (Ngô, Nguyen 2017)

The optimizers are given, up to translations, dilations and multiplications by constants, by

ρ(x) = 1 +|x|²^−N/q

∀x∈R^N and the value of the optimal constant is

CN,λ,q(λ)= 1 π^λ²

Γ ^N₂ +^λ₂ Γ N+^λ₂

Γ(N) Γ ^N₂

!¹⁺_N^λ

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0 2 4 6 8 10 12

0.2 0.4 0.6 0.8 1.0

N = 4, region of the parameters(λ, q) for whichCN,λ,q>0 The plain, red curve is the conformally invariant caseα= 0

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Z Z

R^N×R^N

|x−y|^λρ(x)ρ(y)dx dy≥CN,λ,q

Z

R^N

ρ dx ^αZ

R^N

ρ^qdx

(2−α)/q

0 2 4 6 8 10 12

0.2 0.4 0.6 0.8 1.0

α <0

0< α <1 α >1

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A Carlson type inequality

Lemma

Letλ >0 andN/(N+λ)< q <1 Z

R^N

ρ dx

1−^N^(1−q)_{λ q} Z

R^N

|x|^λρ dx ^N^(1−q)_{λ q}

≥c_N,λ,q Z

R^N

ρ^qdx ¹_q

cN,λ,q=_λ¹_(N_+λ)_q−N

q

¹_q _N_(1−q)

(N+λ)q−N

^N_λ ^1−q_q

Γ(^N2)^Γ(1−q¹ )

2π^N2 Γ(1−q¹ −^N_λ)^Γ(^Nλ) ^1−q_q

Equality is achieved if and only if

ρ(x) = 1 +|x|^λ⁻1−q¹

up to translations, dilations and constant multiples (Carlson 1934) (Levine 1948)

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An elementary proof of Carlson’s inequality

Z

{|x|<R}

ρ^qdx≤ Z

R^N

ρ dx q

|BR|^1−q =C₁ Z

R^N

ρ dx q

R^N^(1−q)

and Z

{|x|≥R}

ρ^qdx≤ Z

R^N

|x|^λρ dx ^q Z

{|x|≥R}

|x|⁻^1−q^{λ q} dx

!1−q

=C₂ Z

R^N

|x|^λρ dx ^q

R^−λq+N^(1−q) and optimize overR >0

... existence of a radial monotone non-increasing optimal function;

rearrangement; Euler-Lagrange equations

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Proposition

Letλ >0. If N/(N+λ)< q <1, thenCN,λ,q>0

By rearrangement inequalities: prove the reverse HLS inequality for symmetric non-increasingρ’s so that

Z

R^N

|x−y|^λρ(y)dx≥ Z

R^N

|x|^λρ dx for all x∈R^N implies

Iλ[ρ]≥ Z

R^N

|x|^λρ dx Z

R^N

ρ dx In the range _N^N_+λ < q <1

Iλ[ρ]

R

R^Nρ(x)dxα ≥ Z

R^N

ρ dx dx 1−αZ

R^N

|x|^λρ dx≥c^2−α_N,λ,q Z

R^N

ρ^qdx ^2−α_q

and conclude with Carlson’s inequality

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The case λ = 2

Corollary

Letλ= 2 andN/(N+ 2)< q <1. Then the optimizers for (1) are given by translations, dilations and constant multiples of

ρ(x) = 1 +|x|²−_1−q¹

and the optimal constant is

CN,2,q =¹₂c

2q N(1−q)

N,2,q

By rearrangement inequalities it is enough to prove (1) for symmetric non-increasingρ’s, and soR

R^Nx ρ dx= 0. Therefore I2[ρ] = 2

Z

R^N

ρ dx Z

R^N

|x|²ρ dx

and the optimal function is optimal for Carlson’s inequality

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0 2 4 6 8 10 12

0.2 0.4 0.6 0.8 1.0

N = 4, region of the parameters (λ, q)for whichCN,λ,q>0. The dashed, red curve is the threshold caseq=N/(N+λ)

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The threshold case q = N/(N + λ) and below

Proposition

If0< q≤N/(N+λ), thenCN,λ,q= 0 = lim_q→N/(N_+λ)₊CN,λ,q

Letρ,σ≥0 such thatR

R^Nσ dx= 1, smooth (+ compact support) ρε(x) :=ρ(x) +M ε^−Nσ(x/ε)

ThenR

R^Nρ_εdx=R

R^Nρ dx+M and, by simple estimates, Z

R^N

ρ^q_εdx→ Z

R^N

ρ^qdx as ε→0+

and

Iλ[ρε]→Iλ[ρ] + 2M Z

R^N

|x|^λρ dx as ε→0+

If 0< q < N/(N+λ),i.e.,α >1, takeρε as a trial function, CN,λ,q≤ Iλ[ρ] + 2MR

R^N|x|^λρ dx R

R^Nρ dx+M^α R

R^Nρ^qdx(2−α)/q =:Q[ρ, M]

and letM →+∞ J. Dolbeault Reverse Hardy-Littlewood-Sobolev inequalities

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The threshold case: Ifα= 1,i.e.,q=N/(N+λ), by taking the limit asM →+∞, we obtain

CN,λ,q≤ 2R

R^N|x|^λρ dx R

R^Nρ^qdx(2−α)/q

For anyR >1, we take

ρR(x) :=|x|^−(N^+λ)11≤|x|≤R(x)

Then Z

R^N

|x|^λρRdx= Z

R^N

ρ^q_Rdx= S^N⁻¹

logR and, as a consequence,

R

R^N|x|^λρRdx R

R^Nρ^N/(N_R ^+λ)dx^(N+λ)/N = S^N⁻¹

logR^−λ/N

→0 as R→ ∞

This proves thatCN,λ,q= 0 forq=N/(N+λ)

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A relaxed inequality

I_λ[ρ]+2M Z

R^N

|x|^λρ dx ≥CN,λ,q

Z

R^N

ρ dx+M αZ

R^N

ρ^qdx

(2−α)/q

(2) Proposition

Ifq > N/(N+λ), the relaxed inequality (2)holds with the same optimal constantCN,λ,q as (1)and admits an optimizer(ρ, M)

Heuristically, this is the extension of the reverse HLS inequality (1) Iλ[ρ] ≥CN,λ,q

Z

R^N

ρ dx ^αZ

R^N

ρ^qdx

^(2−α)/q

to measures of the formρ+M δ

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Free energy point of view

Existence of minimizers and relaxation

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Existence of a minimizer: first case

0 2 4 6 8 10 12

0.2 0.4 0.6 0.8 1.0

Theα <0 case: dark grey region Proposition

Ifλ >0 and _2N^2N_+λ < q <1, there is a minimizerρforCN,λ,q

The limit caseα= 0, q= _2N+λ^2N is theconformally invariantcase: see (Dou, Zhu 2015)and(Ngô, Nguyen 2017)

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A minimizing sequenceρ_j can be taken radially symmetric non-increasing by rearrangement, and such that

Z

R^N

ρj(x)dx= Z

R^N

ρj(x)^qdx= 1 for allj∈N Sinceρ_j(x)≤C min

|x|^−N,|x|^−N/q by Helly’s selection theorem we may assume thatρ_j→ρa.e., so that

lim inf

j→∞ Iλ[ρj]≥Iλ[ρ] and 1≥ Z

R^N

ρ(x)dx

by Fatou’s lemma. Pickp∈(N/(N+λ), q) and apply (1) with the sameλandα=α(p):

I_λ[ρ_j]≥CN,λ,p

Z

R^N

ρ^p_jdx

(2−α(p))/p

Hence theρj are uniformly bounded in L^p(R^N): ρj(x)≤C⁰|x|^−N/p, Z

R^N

ρ^q_jdx→ Z

R^N

ρ^qdx= 1 by dominated convergence

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Existence of a minimizer: second case

IfN/(N+λ)< q <2N/(2N+λ) we consider therelaxed inequality I_λ[ρ] + 2MR

R^N|x|^λρ dx ≥CN,λ,q

R

R^Nρ dx+M^α R

R^Nρ^qdx(2−α)/q

0 2 4 6 8 10 12

0.2 0.4 0.6 0.8 1.0

The 0< α <1 case: dark grey region Proposition

Ifq > N/(N+λ), the relaxed inequality holds with the same optimal constantCN,λ,q as (1)and admits an optimizer(ρ, M)

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Let (ρ_j, M_j) be a minimizing sequence withρ_j radially symmetric non-increasing by rearrangement, such that

Z

R^N

ρ_jdx+M_j= Z

R^N

ρ^q_j= 1

Local estimates + Helly’s selection theorem: ρj→ρalmost everywhere andMj→M :=L+ limj→∞Mj, so that R

R^Nρ dx+M = 1, andR

R^Nρ(x)^qdx= 1

We cannot invoke Fatou’s lemma becauseα∈(0,1): letdµj:=ρjdx µj R^N \BR(0)

= Z

{|x|≥R}

ρjdx≤C Z

{|x|≥R}

dx

|x|^N/q =C⁰R^−N^(1−q)/q µj are tight: up to a subsequence,µj→µweak * anddµ=ρ dx+L δ

lim inf

j→∞ Iλ[ρj]≥Iλ[ρ] + 2M Z

R^N

|x|^λρ dx , lim inf

j→∞

Z

R^N

|x|^λρjdx≥ Z

R^N

|x|^λρ dx Conclusion: lim inf_j→∞Q[ρj, Mj]≥Q[ρ, M]

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Optimizers are positive

Q[ρ, M] := I_λ[ρ] + 2MR

R^N|x|^λρ dx R

R^Nρ dx+Mα R

R^Nρ^qdx(2−α)/q

Lemma

Letλ >0 andN/(N+λ)< q <1. If ρ≥0 is an optimal function for someM >0, thenρis radial (up to a translation), monotone

non-increasing and positive a.e. onR^N

Ifρvanishes on a setE⊂R^N of finite, positive measure, then Q

ρ, M+ε1E

=Q[ρ, M]

1−2−α q

|E|

R

R^Nρ(x)^qdxε^q+o(ε^q)

asε→0+, a contradiction if (ρ, M) is a minimizer ofQ

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Euler–Lagrange equation

Euler–Lagrange equation for a minimizer (ρ∗, M∗) 2 R

R^N|x−y|^λρ∗(y)dy+M∗|x|^λ Iλ[ρ_∗] + 2M_∗R

R^N|y|^λρ_∗dy − α R

R^Nρ_∗dy+M_∗−(2−α)ρ∗(x)^−1+q R

R^Nρ_∗(y)^qdy = 0 We can reformulate the question of the optimizers of (1) as: when is it true thatM_∗= 0 ? We already know thatM_∗= 0 if

2N

2N+λ < q <1

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Regions of no concentration and regularity of measure

valued minimizers

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Free energy point of view More on regularity

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

λ q

q=^N_N⁻²

q=¯q(λ,N)

q=_2N^2N_+λ q= _N+λ^N

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No concentration 1

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

Proposition

LetN ≥1,λ >0 and N

N+λ < q < 2N 2N+λ

IfN ≥3 andλ >2N/(N−2), assume further thatq≥N−2 N If(ρ_∗, M_∗)is a minimizer, thenM_∗= 0

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Two ingredients of the proof

Based on the Brézis–Lieb lemma Lemma

Let0< q < p, letf ∈L^p∩L^q(R^N)be a symmetric non-increasing function and letg∈L^q(R^N). Then, for anyτ >0, asε→0+,

Z

R^N

f(x) +ε^−N/pτ g(x/ε)

q

dx= Z

R^N

f^qdx +ε^N(1−q/p)τ^q

Z

R^N

|g|^qdx+o

ε^N^(1−q/p)τ^q

Iλ

ρ∗+ε^−Nτ σ(·/ε)

+ 2 (M∗−τ) Z

R^N

|x|^λ ρ∗(x) +ε^−Nτ σ(x/ε) dx

=I_λ[ρ_∗]+2M_∗ Z

R^N

|x|^λρ_?dx+ (

2τRR

R^N×R^Nρ_∗(x) |x−y|^λ− |x|^λ^σ(^yε)

ε^N dx dy +ε^λτ²Iλ[σ] + 2 (M_∗−τ)τ ε^λR

R^N|x|^λσ dx

| {z }

=O(ε^βτ) with β:=min{2,λ}

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Regularity and concentration

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

Proposition

IfN ≥3,λ >2N/(N−2)and N

N+λ< q <min

N−2

N , 2N

2N+λ

,

and(ρ_∗, M_∗)∈L^N^(1−q)/2(R^N)×[0,+∞)is a minimizer, thenM_∗= 0

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Regularity

Proposition

LetN ≥1,λ >0 andN/(N+λ)< q <2N/(2N+λ) Let(ρ_∗, M_∗)be a minimizer

1 If R

R^Nρ_∗dx > ^α₂ R ^I^λ^[ρ^∗^]

RN|x|^λρ∗dx, thenM_∗= 0 andρ_∗, bounded and ρ_∗(0) = (2−α)Iλ[ρ_∗]R

R^Nρ_∗dx R

R^Nρ^q_∗dx 2R

R^N|x|^λρ_∗dxR

R^Nρ_∗dx−αI_λ[ρ_∗]

!_1−q¹

2 If R

R^Nρ_∗dx= ^α₂ R ^I^λ^[ρ^∗^]

RN|x|^λρ∗dx, thenM_∗= 0 andρ_∗ is unbounded

3 If R

R^Nρ_∗dx < ^α₂ R ^I^λ^[ρ^∗^]

RN|x|^λρ∗dx, thenρ_∗ is unbounded and M_∗= αIλ[ρ∗]−2R

R^N|x|^λρ∗dx R

R^Nρ∗dx 2 (1−α)R

R^N|x|^λρ_∗dx >0

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An ingredient of the proof

Lemma

For constantsA,B >0 and0< α <1, define f(M) = A+M

(B+M)^α for M ≥0

Thenf attains its minimum on[0,∞)atM = 0 ifα A≤B and at M = (α A−B)/(1−α)>0 if α A > B

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No concentration 2

For anyλ≥1 we deduce from

|x−y|^λ≤ |x|+|y|λ

≤2^λ−1 |x|^λ+|y|^λ that

I_λ[ρ]<2^λ Z

R^N

|x|^λρ dx Z

R^N

ρ(x)dx For allα≤2^−λ+1, we infer thatM_∗= 0 if

q≥ 2N 1−2^−λ 2N 1−2^−λ

+λ

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

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No concentration 3

Layer cake representation (superlevel sets are balls) Iλ[ρ]≤2AN,λ

Z

R^N

|x|^λρ dx Z

R^N

ρ(x)dx

AN,λ:= sup

0≤R,S<∞

RR

BR×BS|x−y|^λdx dy

|B_R|R

BS|x|^λdx+|B_S|R

BR|y|^λdy

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

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Proposition

Assume thatN ≥3and λ >2N/(N−2) and observe that N

N+λ<q(λ, N¯ )≤ 2N 1−2^−λ 2N 1−2^−λ

+λ< 2N 2N+λ forλ >2 large enough. If

max

¯

q(λ, N), N N+λ

< q < N−2 N

and if(ρ_∗, M_∗) is a minimizer, thenM_∗= 0 andρ_∗∈L^∞(R^N)

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More on regularity

Lemma

Assume thatρ_∗ is an unbounded minimizer ifλ <2, there is a constantc >0 such that

ρ∗(x)≥c|x|^{−λ/(1−q)} as x→0 ifλ≥2, there is a constantC >0 such that

ρ_∗(x) =C|x|^−2/(1−q) 1 +o(1)

as x→0 Corollary

q6= 2N

2N+λ, N

N+λ< q <1 and q≥ N−2

N ifN ≥3 Ifρ_∗ is a minimizer forCN,λ,q, thenρ_∗∈L^∞(R^N)

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0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

λ q

q=^N_N⁻²

q=¯q(λ,N)

q=_2N^2N_+λ q= _N+λ^N

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Free energy point of view

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A toy model

Assume thatusolves thefast diffusion with external driftV given by

∂u

∂t = ∆u^q+∇ · u∇V

To fix ideas: V(x) = 1 + ¹₂|x|²+_λ¹|x|^λ. Free energy functional F[u] :=

Z

R^N

V u dx− 1 1−q

Z

R^N

u^qdx Under the mass constraintM =R

R^Nu dx, smooth minimizers are uµ(x) = µ+V(x)−_1−q¹

The equation can be seen as a gradient flow d

dtF[u(t,·)] =− Z

R^N

u

q

1−q∇u^q−1− ∇V

2

dx

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A toy model (continued)

Ifλ= 2, the so-calledBarenblatt profile u_µ has finite mass if and only if

q > qc :=N−2 N

Forλ >2, the integrability condition isq >1−λ/N butq=qc is a threshold for the regularity: the mass ofuµ= (µ+V)^1/(1−q)is

M(µ) :=

Z

R^N

uµdx≤M?= Z

R^N 1

2|x|²+¹_λ|x|^λ⁻1−q¹ dx If one tries to minimize the free energy under the mass contraint R

R^Nu dx=M for an arbitrary M > M_?, the limit of a minimizing sequence is the measure

M−M?

δ+u₋₁

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A model for nonlinear springs: heuristics

V =ρ∗Wλ, Wλ(x) := _λ¹|x|^λ

is motivated by the study of the nonnegative solutions of the evolution equation

∂ρ

∂t = ∆ρ^q+∇ ·(ρ∇W_λ∗ρ)

Optimal functions for (1) are energy minimizers (eventually measure valued) for thefree energyfunctional

F[ρ] := 1 2 Z

R^N

ρ(W_λ∗ρ)dx− 1 1−q

Z

R^N

ρ^qdx= 1

2λI_λ[ρ]− 1 1−q

Z

R^N

ρ^qdx under amassconstraintM =R

R^Nρ dxwhile smooth solutions obey to d

dtF[ρ(t,·)] =− Z

R^N

ρ

q

1−q∇ρ^q−1− ∇Wλ∗ρ

2

dx

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Free energy or minimization of the quotient

F[ρ] =− 1 1−q

Z

R^N

ρ^qdx+ 1 2λIλ[ρ]

If 0< q≤N/(N+λ), thenCN,λ,q= 0: take test functions ρ_n ∈L¹₊∩L^q(R^N) such thatkρ_nk_L1(R^N)=I_λ[ρ_n] = 1 and R

R^Nρ^q_ndx=n∈N

n→+∞lim F[ρn] =− ∞

IfN/(N+λ)< q <1,ρ_`(x) :=`^−Nρ(x/`)/kρk_L1(R^N)

F[ρ`] =−`^(1−q)NA+`^λB has a minimum at`=`_? and

F[ρ]≥F[ρ`_?] =−κ?(Qq,λ[ρ])⁻

N(1−q) λ−N(1−q)

Proposition

Fis bounded from below if and only ifCN,λ,q>0

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Relaxed free energy F^rel[ρ, M] :=− 1

1−q Z

R^N

ρ^qdx+ 1

2λIλ[ρ] + M λ

Z

R^N

|x|^λρ dx

Corollary

Letq∈(0,1)andN/(N+λ)< q <1 inf

F^rel[ρ, M] : 0≤ρ∈L¹∩L^q(R^N), M ≥0, Z

R^N

ρ dx+M = 1

is achieved by a minimizer of (2)such that R

R^Nρ_∗dx+M_∗= 1and Iλ[ρ_∗] + 2M_∗

Z

R^N

|x|^λρ_∗dx= 2N Z

R^N

ρ^q_∗dx

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Uniqueness

Proposition

LetN/(N+λ)< q <1 and assume either that (N−1)/N < q <1 andλ≥1, or2≤λ≤4. Then the minimizer of

F^rel[ρ, M] := 1

2λIλ[ρ] +M λ

Z

R^N

|x|^λρ dx− 1 1−q

Z

R^N

ρ^qdx is unique up to translation, dilation and multiplication by a positive constant

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If (N−1)/N < q <1 andλ≥1, the lower semi-continuous extension ofFto probability measures is strictly geodesically convex in the Wasserstein-pmetric forp∈(1,2)

By strict rearrangement inequalities a minimizer (ρ, M) such that M ∈[0,1) of the relaxed free energy F^rel is (up to a translation) such thatρis radially symmetric andR

R^Nx ρ dx= 0 Let (ρ, M) and (ρ⁰, M⁰) be two minimizers and

[0,1]3t7→f(t) :=F^rel

(1−t)ρ+t ρ⁰,(1−t)M+t M⁰

f⁰⁰(t) = 1

λI_λ[ρ⁰−ρ] + 2

λ(M⁰−M) Z

R^N

|x|^λ(ρ⁰−ρ)dx +q

Z

R^N

(1−t)ρ+t ρ⁰q−2

(ρ⁰−ρ)²dx (Lopes, 2017)Iλ[h]≥0if 2≤λ≤4, for all hsuch that

R

R^N 1 +|x|^λ

|h|dx <∞ withR

R^Nh dx= 0 andR

R^Nx h dx= 0

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N = 4

0 2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

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N = 10

2 4 6 8 10 12

0.0 0.2 0.4 0.6 0.8 1.0

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References

J. Dou and M. Zhu. Reversed Hardy-Littlewood-Sobolev inequality. Int. Math. Res. Not. IMRN, 2015(19):9696-9726, 2015

Q.A. Ngô and V. Nguyen. Sharp reversed

Hardy-Littlewood-Sobolev inequality onRⁿ. Israel J. Math., 220 (1):189-223, 2017

J. A. Carrillo and M. Delgadino. Free energies and the reversed HLS inequality. ArXiv e-prints, Mar. 2018 # 1803.06232

J. Dolbeault, R. Frank, and F. Hoffmann. Reverse

Hardy-Littlewood-Sobolev inequalities. ArXiv e-prints, Mar. 2018 # 1803.06151

J. A. Carrillo, M. Delgadino, J. Dolbeault, R. Frank, and F.

Hoffmann. Reverse Hardy-Littlewood-Sobolev inequalities. In preparation.

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These slides can be found at

http://www.ceremade.dauphine.fr/∼dolbeaul/Lectures/

BLectures The papers can be found at

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Reverse Hardy-Littlewood-Sobolev inequalities

Outline

Reverse

Hardy-Littlewood-Sobolev inequality

The reverse HLS inequality

The conformally invariant case q = 2N/(2N + λ)

A Carlson type inequality

An elementary proof of Carlson’s inequality

The case λ = 2

The threshold case q = N/(N + λ) and below

A relaxed inequality

Existence of minimizers and relaxation

Existence of a minimizer: first case

Existence of a minimizer: second case

Optimizers are positive

Euler–Lagrange equation

Regions of no concentration and regularity of measure

valued minimizers

No concentration 1

Two ingredients of the proof

Regularity and concentration

Regularity

An ingredient of the proof

No concentration 2

No concentration 3

More on regularity

Free energy point of view

A toy model

A toy model (continued)

A model for nonlinear springs: heuristics

Free energy or minimization of the quotient

Uniqueness

N = 4

N = 10

References

Thank you for your attention !