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JOSÉ A. CARRILLO, MATÍAS G. DELGADINO, JEAN DOLBEAULT, RUPERT L. FRANK, AND FRANCA HOFFMANN

ABSTRACT. This paper is devoted to a new family of reverse Hardy–Littlewood–Sobolev inequalities which involve a power law kernel with positive exponent. We investigate the range of the admissible parameters and the properties of the optimal functions. A striking open question is the possibility of concentration which is analyzed and related with free energy functionals and nonlinear diffusion equations involving mean field drifts.

1. INTRODUCTION

We are concerned with the following minimization problem. For any λ > 0 and any measurable functionρ ≥ 0 on RN, let

Iλ[ρ] := Ï RN×RN|x − y| λρ(x)ρ(y)dx d y . For 0 < q < 1 we consider CN ,λ,q := inf ( Iλ[ρ] ¡R RNρ(x)dx¢α ¡R RNρ(x)qd x ¢(2−α)/q : 0 ≤ ρ ∈ L 1 ∩ Lq(RN) ,ρ 6≡ 0 ) , where α :=2 N − q (2 N + λ) N (1 − q) .

By convention, for any p > 0 we use the notation ρ ∈ Lp(RN) ifR

RN|ρ(x)|pd x is finite. Note thatα is determined by scaling and homogeneity: for given values of λ and q, the value ofα is the only one for which there is a chance that the infimum is positive. We are asking whether CN ,λ,q is equal to zero or positive and, in the latter case, whether there is a unique minimizer. As we will see, there are three regimes q < 2 N /(2 N + λ), q = 2 N /(2 N +λ) and q > 2 N /(2 N +λ), which respectively correspond to α > 0, α = 0 and α < 0. The case q = 2N/(2N + λ), in which there is an additional conformal symmetry, has already been dealt with in [16] by J. Dou and M. Zhu, and in [32] by Q.A. Ngô and V.H. Nguyen, who have explicitly computedCN ,λ,q and characterized all solutions of the corresponding Euler–Lagrange equation. Here we will mostly concentrate on the other cases. Our main result is the following.

2010 Mathematics Subject Classification. Primary: 35A23; Secondary: 26D15, 35K55, 46E35, 49J40.

Key words and phrases. Reverse Hardy–Littlewood–Sobolev inequalities; interpolation; symmetrization;

concentration; minimizer; existence of optimal functions; regularity; uniqueness; Euler–Lagrange equa-tions; free energy; nonlinear diffusion; mean field equaequa-tions; nonlinear springs; measure valued solutions.

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Theorem 1. Let N ≥ 1, λ > 0, q ∈ (0,1) and define α as above. Then the inequality Iλ[ρ] ≥ CN ,λ,q µZ RNρ(x)dxαµZ RNρ(x) q d x(2−α)/q (1) holds for any nonnegative functionρ ∈ L1∩ Lq(RN), for some positive constantCN ,λ,q, if

and only if q > N /(N + λ). In this range, if either N = 1, 2 or if N ≥ 3 and q ≥ min©1 − 2/N , 2 N /(2 N + λ)ª, there is a radial positive, nonincreasing, bounded functionρ ∈ L1 Lq(RN) which achieves the equality case.

This theorem provides a necessary and sufficient condition for the validity of the in-equality, namely q > N /(N +λ) or equivalently α < 1. Concerning the existence of an op-timizer, the theorem completely answers this question in dimensions N = 1 and N = 2. In dimensions N ≥ 3 we obtain a sufficient condition for the existence of an optimizer, namely, q ≥ min©1−2/N,2 N/(2 N +λ)ª. This is not a necessary condition and, in fact, in Proposition17we prove existence in a slightly larger, but less explicit region.

In the whole region q > N /(N + λ) we are able to prove the existence of an optimizer for the relaxed inequality

Iλ[ρ] + 2M Z RN|x| λρ(x)dx ≥ C N ,λ,q µZ RNρ(x)dx + MαµZ RNρ(x) qd x(2−α)/q (2) with the same optimal constantCN ,λ,q. Hereρ is an arbitrary nonnegative function in L1∩ Lq(RN) and M an arbitrary nonnegative real number. If M = 0, inequality (2) is re-duced to inequality (1). It is straightforward to see that (2) can be interpreted as the extension of (1) to measures with an absolutely continous partρ and an additional Dirac mass at the origin. Therefore the question about existence of an optimizer in Theorem1

is reduced to the problem of whether the optimizer for this relaxed problem in fact has a Dirac mass. Fig.1summarizes these considerations.

The optimizers have been explicitly characterized in the conformally invariant case q = q(λ) := 2 N /(2 N +λ) in [16,32] and are given, up to translations, dilations and multi-plications by constants, by

ρ(x) = ¡1 + |x|−N /q

∀ x ∈ RN. This result determines the value of the optimal constant in (1) as

CN ,λ,q(λ)= 1 πλ2 Γ³N 2 +λ2 ´ Γ³N +λ2´ Ã Γ(N) Γ¡N 2 ¢ !1+Nλ .

By a simple argument that will be exposed in Section2, we can also find the optimizers in the special caseλ = 2: if N/(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 .

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In this case we obtain that CN ,2,q =N (1 − q) πq µ (N + 2) q − N 2 q2−N (1−q)N (1−q)   Γ³ 1 1−q ´ Γ³ 1 1−qN 2 ´   2 N .

Returning to the general case (that is, q 6= 2 N /(2 N +λ) and λ 6= 2), no explicit form of the optimizers is known, but we can at least prove a uniqueness result in some cases, see also Fig.2.

Theorem 2. Assume that N /(N +λ) < q < 1 and either q ≥ 1−1/N and λ ≥ 1, or 2 ≤ λ ≤ 4.

Then the optimizer for (2) exists and is unique up to translation, dilation and multiplica-tion by a positive constant.

We refer to (1) as a reverse Hardy–Littlewood–Sobolev inequality asλ is positive. The Hardy–Littlewood–Sobolev (HLS) inequality corresponds to negative values ofλ and is named after G. Hardy and J.E. Littlewood, see [20,21], and S.L. Sobolev, see [34,35]; also see [22] for an early discussion of rearrangement methods applied to these inequalities. In 1983, E.H. Lieb in [28] proved the existence of optimal functions for negative values ofλ and established optimal constants. His proof requires an analysis of the invariances which has been systematized under the name of competing symmetries, see [9] and [29,

6] for a comprehensive introduction. Notice that rearrangement free proofs, which in some cases rely on the duality between Sobolev and HLS inequalities, have also been established more recently in various cases: see for instance [17,18,25]. Standard HLS inequalities, which correspond to negative values ofλ in Iλ[ρ], have many consequences in the theory of functional inequalities, particularly for identifying optimal constants.

Relatively few results are known in the caseλ > 0. The conformally invariant case, i.e., q = 2 N /(2 N + λ), appears in [16] and is motivated by some earlier results on the sphere (see references therein). Further results have been obtained in [32], still in the confor-mally invariant case. Another range of exponents, which has no intersection with the one considered in the present paper, was studied earlier in [36, Theorem G]. Here we fo-cus on a non-conformally invariant family of interpolation inequalities corresponding to a given L1(RN) norm. In a sense, these inequalities play for HLS inequalities a role anal-ogous to Gagliardo-Nirenberg inequalities compared to Sobolev’s conformally invariant inequality.

The study of (1) is motivated by the analysis of nonnegative solutions to the evolution equation

∂tρ = ∆ρq+ ∇ ·

¡

ρ ∇Wλ∗ ρ¢ , (3)

where the kernel is given by Wλ(x) := 1λ|x|λ. Eq. (3) is a special case of a larger family of Keller-Segel type equations, which covers the cases q = 1 (linear diffusions), q > 1 (diffusions of porous medium type) in addition to 0 < q < 1 (fast diffusions), and also the range of exponentsλ < 0. Of particular interest is the original parabolic–elliptic Keller– Segel system which corresponds in dimension N = 2 to a limit case as λ → 0, in which

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the kernel is W0(x) =21π log |x| and the diffusion exponent is q = 1. The reader is invited to refer to [24] for a global overview of this class of problems and for a detailed list of references and applications.

According to [1,33], (3) has a gradient flow structure in the Wasserstein-2 metric. The corresponding free energy functional is given by

F [ρ] := − 1 1 − q Z R q d x + 1 2λ[ρ].

As will be detailed later, optimal functions for (1) are energy minimizers forF under a mass constraint. Smooth solutionsρ(t,·) of (3) with sufficient decay properties as |x| → + ∞ conserve mass and center of mass over time while the free energy decays according to d d tF [ρ(t,·)] = − Z R ¯ ¯ ¯ q 1−q∇ρq−1− ∇Wλ∗ ρ ¯ ¯ ¯ 2 d x .

This identity allows us to identify the smooth stationary solutions as the solutions of ρs=¡C + (Wλ∗ ρs)

¢− 1 1−q

where C is a constant which has to be determined by the mass constraint. Thanks to the gradient flow structure, minimizers of the free energyF are stationary states of Eq. (3). When dealing with solutions of (3) or with minimizers of the free energy, without loss of generality we can normalize the mass to 1 in order to work in the space of probability measuresP (RN). The general case of a bounded measure with an arbitrary mass can be recovered by an appropriate change of variables. Considering the lower semicontinuous extension of the free energy to P (RN) denoted byFΓ, we obtain counterparts to Theo-rems1and2in terms ofFΓ.

Theorem 3. The free energyis bounded from below on P (RN) if and only if q > N /(N +λ). If q > N /(N +λ), then there exists a global minimizer µ∗∈ P (RN) and, modulo translations, it has the form

µ= ρ+ Mδ0

for some M∈ [0, 1). Moreover ρ∗∈ L1+∩ Lq(RN) is radially symmetric, non-increasing and supported onRN.

If M= 0, then ρis an optimizer of (1). Conversely, ifρ ∈ L1+∩ Lq(RN) is an optimizer of (1) with mass M > 0, then ρ/M is a global minimizer of FΓonP (RN).

Finally, if N /(N + λ) < q < 1 and either q ≥ 1 − 1/N and λ ≥ 1, or 2 ≤ λ ≤ 4, then the global minimizerµofonP (RN) is unique up to translation.

In the region of the parameters of Theorem1for which (1) is achieved by a radial func-tion, this optimizer is also a minimizer ofF . If the minimizer µofFΓhas a singular part, then the constantCN ,λ,qis also achieved byµin (2), up to a translation. Hence the results of Theorem3are equivalent to the results of Theorems1and2.

The use of free energies to understand the long-time asymptotics of gradient flow equations like (3) and various related models with other interaction potentials than Wλ or more general pressure variables thanρq−1has already been studied in some cases: see

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for instance [1,13,14,38]. The connection to Hardy–Littlewood–Sobolev type functional inequalities [8,3,7] is well-known for the rangeλ ∈ (−N,0]. However, the case of Wλwith λ > 0 is as far as we know entirely new.

This paper results from the merging of two earlier preprints, [15] and [11], correspond-ing to two research projects that were investigated independently.

Section2is devoted to the proof of the reverse HLS inequality (1) and also of the op-timal constant in the caseλ = 2. In Section3we study the existence of optimizers of the reverse HLS inequality via the relaxed variational problem associated with (2). The regu-larity properties of these optimizers are analysed in Section4, with the goal of providing some additional results of no-concentration. Section5is devoted to the equivalence of the reverse HLS inequalities and the existence of a lower bound ofFΓonP (RN). The rel-ative compactness of minimizing sequences of probability measures is also established as well as the uniqueness of the measure valued minimizers ofFΓ, in the same range of the parameters as in Theorem2. We conclude this paper by an appendix on a toy model for concentration which sheds some light on the threshold value q = 1 − 2/N .

2. REVERSEHLSINEQUALITY

The following proposition gives a necessary and sufficient condition for inequality (1).

Proposition 4. Letλ > 0.

(1) If 0 < q ≤ N /(N + λ), then CN ,λ,q = 0.

(2) If N /(N + λ) < q < 1, then CN ,λ,q > 0.

The result for q < N /(N + λ) was obtained in [12] using a different method. The result for q = N /(N + λ), as well as the result for 2 N /(2 N + λ) 6= q > N /(N + λ), are new.

Proof of Proposition4. Part (1). Letρ ≥ 0 be bounded with compact support and let σ ≥ 0 be a smooth function withR

RNσ(x)dx = 1. With another parameter M > 0 we consider ρε(x) = ρ(x) + M ε−Nσ(x/ε),

whereε > 0 is a small parameter. Then RRNρε(x) d x = R RNρ(x)dx + M and, by simple estimates, Z RNρε(x) q d x → Z RNρ(x) qd x as ε → 0 + (4) and Iλ[ρε] → Iλ[ρ] + 2M Z RN|x| λρ(x)dx as ε → 0 +. Thus, takingρεas a trial function,

CN ,λ,q[ρ] + 2M RR N|x|λρ(x)dx ¡R RNρ(x)dx + M¢α ¡R RNρ(x)qd x ¢(2−α)/q =: Q[ρ, M] . (5) This inequality is valid for any M and therefore we can let M → +∞. If α > 1, which is the same as q < N /(N + λ), we immediately obtain CN ,λ,q = 0 by letting M → + ∞. If α = 1,

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i.e., q = N /(N + λ), by taking the limit as M → +∞, we obtain CN ,λ,q≤ 2R RN|x|λρ(x)dx ¡R RNρ(x)qd x ¢(2−α)/q.

Let us show that by a suitable choice ofρ the right side can be made arbitrarily small. For any R > 1, we take ρR(x) := |x|−(N +λ)11≤|x|≤R(x) . Then Z RN|x| λρ Rd x = Z R q Rd x = ¯ ¯SN −1 ¯ ¯log R and, as a consequence, R RN|x|λρR(x) d x ³ R RRN /(N +λ)d x ´(N +λ)/N = ³¯ ¯SN −1 ¯ ¯log R ´−λ/N → 0 as R → ∞ .

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

In order to prove thatCN ,λ,q> 0 in the remaining cases, we need the following simple

bound, which is known as a Carlson type inequality in the literature after [10] and whose sharp form can be found in [27]. For completeness, we give a statement and a proof.

Lemma 5. Letλ > 0 and N/(N + λ) < q < 1. Then there is a constant cN ,λ,q > 0 such that

for allρ ≥ 0, µZ RNρ dx ¶1−N (1−q)λq µZ RN|x| λρ(x)dxN (1−q) λq ≥ cN ,λ,q µZ R q d x1/q . Equality is achieved if and only if

ρ(x) =³1 + |x|λ´− 1 1−q

up to translations, dilations and constant multiples, and one has

cN ,λ,q =1 λ µ (N + λ) q − N q ¶1qµ N (1 − q) (N + λ) q − NNλ 1−qq   Γ¡N 2 ¢ Γ³ 1 1−q ´ 2πN2 Γ ³ 1 1−qN λ ´ Γ¡N λ ¢   1−q q .

Proof. Let R > 0. Using Hölder’s inequality in two different ways, we obtain Z {|x|<R}ρ q d x ≤ µZ RNρ dxq |BR|1−q= C1 µZ RNρ dxq RN (1−q) and Z {|x|≥R}ρ q d x ≤ µZ RN|x| λρ(x)dxqµZ {|x|≥R}|x|1−qλq d x1−q = C2 µZ RN|x| λρ(x)dxqR−λq+N (1−q).

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The fact that C2< ∞ comes from the assumption q > N /(N + λ), which is the same as λq/(1 − q) > N. To conclude, we add these two inequalities and optimize over R.

The existence of a radial monotone non-increasing optimal function follows by stan-dard variational methods; the expression for the optimal functions is a consequence of the Euler-Lagrange equations. The expression of cN ,λ,q is then straightforward. 

Proof of Proposition4. Part (2). By rearrangement inequalities it suffices to prove the in-equality for symmetric non-increasingρ’s. For such functions, by the simplest rearrange-ment inequality, Z RN|x − y| λρ(y)dx ≥Z RN|x| λρ(x)dx for all x ∈ RN. Thus, Iλ[ρ] ≥ Z RN|x| λρ(x)dxZ RNρ dx . (6)

In the range N +λN < q < 1 (for which α < 1), we recall that by Lemma5, for any symmetric non-increasing functionρ, we have

Iλ[ρ] ¡R RNρ(x)dx¢α ≥ µZ RNρ dx dx1−αZ RN|x| λρ(x)dx ≥ c2−α N ,λ,q µZ R qd x2−αq

because 2 − α = N (1−q)λq . As a consequence, we obtain that CN ,λ,q≥ cN ,2−αλ,q> 0 .

 Corollary 6. Letλ = 2 and N/(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= 2 c 2 q

N (1−q)

N ,2,q .

Proof. By rearrangement inequalities it is enough to prove (1) for symmetric non-increa-singρ’s, and so RRNxρ(x)dx = 0. Therefore

I2[ρ] = 2 Z RNρ(x)dx Z RN|x| 2ρ(x)dx

and the optimal function is the one of the Carlson type inequality of Lemma5.  By taking into account the fact that

cN ,2,q= 1 2 µ (N + 2) q − N qqN (1 − q) (N + 2) q − NN21−qq   Γ³ 1 1−q ´ 2πN2 Γ ³ 1 1−qN 2 ´   1−q q , we recover the expression ofCN ,2,q given in the introduction.

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Remark 7. We can now make a few observations on the reverse HLS inequality (1) and its optimal constantCN ,λ,q.

(i) The computation in the proof of Proposition4, Part (2) explains a surprising feature of (1): Iλ[ρ] controls a product of two terms. However, in the range N/(N + λ) < q < 2 N /(2 N +λ) which corresponds to α ∈ (0,1), the problem is actually reduced (with a non-optimal constant) to the interpolation ofR

RNρqd x between R

RNρ dx and RRN|x|λρ(x)dx, which has a more classical structure.

(ii) There is an alternative way to prove (1) in the range 2 N /(2 N + λ) < q < 1 using the results from [16,32]. We can indeed rely on Hölder’s inequality to get that

µZ RNρ(x) qd x1/q ≤ µZ RNρ(x) 2 N 2 N +λd xη2 N +λ2 N µZ RNρ dx1−η

withη :=2 N (1−q)λq . By applying the conformally invariant inequality Iλ[ρ] ≥ CN ,λ, 2 N 2 N +λ µZ RNρ(x) 2 N 2 N +λd x2 N +λN

shown in [16,32], we obtain that CN ,λ,q≥ CN ,λ, 2 N 2 N +λ= πλ2 Γ¡ N 2+λ2 ¢ Γ¡N +λ2¢ Ã Γ(N) Γ¡N 2 ¢ !1+Nλ .

We notice thatα = −2(1 − η)/η is negative in the range 2N/(2N + λ) < q < 1. (iii) We have

lim

q→N /(N +λ)+

CN ,λ,q= 0

because the map (λ,q) 7→ CN ,λ,qis upper semi-continuous. The proof of this last property

goes as follows. Let us rewriteQ[ρ,0] defined in (5) as

Qq,λ[ρ] :=¡R [ρ] RNρ(x)dx¢α

¡R

RNρ(x)qd x

¢(2−α)/q. (7)

In this expression of the energy quotient, we emphasize the dependence in q andλ. As before, the infimum of Qq,λ over L1+∩ Lq(RN) is CN ,λ,q. Let (q,λ) be a given point in

(0, 1) × (0,∞) and let (qn,λn)n∈Nbe a sequence converging to (q,λ). Let ε > 0 and choose

aρ which is bounded, has compact support and is such thatQq,λ[ρ] ≤ CN ,λ,q+ε. Then, by

the definition as an infimum,CN ,qn,λn ≤Qqn,λn[ρ]. On the other hand, the assumptions on ρ imply that limn→∞Qqn,λn[ρ] =Qq,λ[ρ]. We conclude that limsupn→∞CN ,qn,λn ≤ CN ,λ,q+ ε. Since ε is arbitrary, we obtain the claimed upper semi-continuity property.

3. EXISTENCE OF MINIMIZERS AND RELAXATION

We now investigate whether there are minimizers in L1+∩ Lq(RN) forCN ,λ,q if N /(N + λ) < q < 1. As mentioned before, the conformally invariant case q = 2N/(2N + λ) has

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been dealt with before and will be excluded from our considerations. We start with the simpler case 2 N /(2 N + λ) < q < 1, which corresponds to α < 0.

Proposition 8. Letλ > 0 and 2N/(2N + λ) < q < 1. Then there is a minimizer for CN ,λ,q.

Proof. Let (ρj)j ∈Nbe a minimizing sequence. By rearrangement inequalities we may

as-sume that the ρj are symmetric non-increasing. By scaling and homogeneity, we may

also assume that Z

RNρj(x) d x = Z

RNρj(x)

q

d x = 1 for all j ∈ N.

This together with the symmetric non-increasing character ofρj implies that

ρj(x) ≤ C min©|x|−N, |x|−N /q

ª

with C independent of j . By Helly’s selection theorem we may assume, after passing to a subsequence if necessary, thatρj → ρ almost everywhere. The function ρ is symmetric

non-increasing and satisfies the same upper bound asρj.

By Fatou’s lemma we have lim inf

j →∞ j] ≥ Iλ[ρ] and 1 ≥

Z

RNρ(x)dx . To complete the proof we need to show thatR

RNρ(x)qd x = 1 (which implies, in particu-lar, thatρ 6≡ 0) and then ρ will be an optimizer.

Modifying an idea from [2] we pick p ∈¡N /(N + λ), q¢ and apply (1) with the sameλ andα(p) = ¡2N − p (2N + λ)¢±¡N (1 − p)¢ to get Iλj] ≥ CN ,λ,p µZ R p jd x(2−α(p))/p .

Since the left side converges to a finite limit, namelyCN ,λ,q, we find that theρj are

uni-formly bounded in Lp(RN) and therefore we have as before ρj(x) ≤ C0|x|−N /p.

Since min©|x|−N, |x|−N /pª ∈ Lq(RN), we obtain by dominated convergence Z R q jd x → Z R qd x ,

which, in view of the normalization, implies thatRRNρ(x)qd x = 1, as claimed.  Next, we prove the existence of minimizers in the range N /(N + λ) < q < 2 N /(2 N + λ) by considering the minimization of the relaxed problem (2). The idea behind this relax-ation is to allowρ to contain a Dirac function at the origin. The motivation comes from the proof of the first part of Proposition 4. The expression ofQ[ρ,M] as defined in (5) arises precisely from a measurable functionρ together with a Dirac function of strength M at the origin. We have seen that in the regime q ≤ N /(N +λ) (that is, α ≥ 1) it is advanta-geous to increase M to infinity. This is no longer so if N /(N +λ) < q < 2 N /(2 N +λ). While it is certainly disadvantageous to move M to infinity, it has to be investigated whether the optimum M is 0 or a positive finite value.

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Let Crel N ,λ,q:= inf n Q[ρ,M] : 0 ≤ ρ ∈ L1 ∩ Lq(RN) ,ρ 6≡ 0, M ≥ 0o

whereQ[ρ,M] is defined by (5). We know that CN ,λ,qrel ≤ CN ,λ,q by restricting the

mini-mization to M = 0. On the other hand, (5) givesCN ,λ,qrel ≥ CN ,λ,q. Therefore,

Crel

N ,λ,q= CN ,λ,q,

which justifies our interpretation of CN ,relλ,q as a relaxed minimization problem. Let us start with a preliminary observation.

Lemma 9. Letλ > 0 and N/(N +λ) < q < 1. If ρ ≥ 0 is an optimal function for either CN ,relλ,q

(for an M ≥ 0) or CN ,λ,q (with M = 0), then ρ is radial (up to a translation), monotone

non-increasing and positive almost everywhere onRN.

Proof. SinceCN ,λ,q is positive, we observe thatρ is not identically 0. By rearrangement

inequalities and up to a translation, we know thatρ is radial and monotone non-increa-sing. Assume by contradiction thatρ vanishes on a set E ⊂ RN of finite, positive measure. Then Q£ρ,M + ε1E¤ = Q[ρ, M] µ 1 −2 − α q |E| R RNρ(x)qd x εq + o(εq) ¶

asε → 0+, a contradiction to the minimality for sufficiently smallε > 0. 

VaryingQ[ρ,M] with respect to ρ, we obtain the Euler–Lagrange equation on RN for any minimizer (ρ, M) forCN ,λ,qrel :

2 R RN|x − y|λρ(y) d y + M|x|λ Iλ] + 2MR RN|y|λρ(y) d y −R α Rd y + M − (2 − α) ρ(x) −1+q R R(y)qd y = 0 . (8) This equation follows from the fact that ρ is positive almost everywhere according to Lemma9.

Proposition 10. Letλ > 0 and N/(N + λ) < q < 2N/(2N + λ). Then there is a minimizer

forCN ,relλ,q.

We will later show that for N = 1 and N = 2 there is a minimizer for the original problem CN ,λ,q in the full range ofλ’s and q’s covered by Proposition10. If N ≥ 3, the same is true

under additional restrictions.

Proof of Proposition10. The beginning of the proof is similar to that of Proposition8. Let (ρj, Mj)j ∈N be a minimizing sequence. By rearrangement inequalities we may assume

that ρj is symmetric non-increasing. Moreover, by scaling and homogeneity, we may

assume that Z RNρjd x + Mj= Z R q j = 1 .

In a standard way this implies that

ρj(x) ≤ C min©|x|−N, |x|−N /q

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with C independent of j . By Helly’s selection theorem we may assume, after passing to a subsequence if necessary, that ρj → ρ almost everywhere. The function ρ is

sym-metric non-increasing and satisfies the same upper bound asρj. Passing to a further

subsequence, we can also assume that (Mj)j ∈N and

¡R

RNρjd x

¢

j ∈Nconverge and define

M := L + limj →∞Mj where L = limj →∞

R

Rjd x − R

RNρ dx, so that RRNρ dx + M = 1. In the same way as before, we show that

Z

RNρ(x)

q

d x = 1.

We now turn our attention to the L1-term. We cannot invoke Fatou’s lemma because α ∈ (0,1) and therefore this term appears in Q with a positive exponent in the denomina-tor. The problem with this term is that |x|−N is not integrable at the origin and we cannot get a better bound there. We have to argue via measures, so let dµj(x) := ρj(x) d x.

Be-cause of the upper bound onρj we have

µj ¡ RN \ BR(0)¢ = Z {|x|≥R}ρj(x) d x ≤ C Z {|x|≥R} d x |x|N /q = C 0R−N (1−q)/q.

This means that the measures are tight. After passing to a subsequence if necessary, we may assume thatµj → µ weak * in the space of measures on RN. Tightness implies that

µ(RN ) = lim j →∞ Z RNρjd x = L + Z RNρ dx .

Moreover, since the bound C |x|−N /q is integrable away from any neighborhood of the origin, we see thatµ is absolutely continuous on RN\ {0} and dµ/dx = ρ. In other words,

dµ = ρ dx + L δ.

Using weak convergence in the space of measures one can show that lim inf

j →∞ j] ≥ Iλ[ρ] + 2M

Z

RN|x|

λρ(x)dx .

Finally, by Fatou’s lemma, lim inf j →∞ Z RN|x| λρ j(x) d x ≥ Z RN|x| λ¡ ρ(x)dx + L δ¢ = Z RN|x| λρ(x)dx . Hence lim inf j →∞ Q[ρj, Mj] ≥ Q[ρ, M].

By definition of CN ,λ,qrel the right side is bounded from below by CN ,λ,qrel . On the other hand, by choice ofρj and Mj the left side is equal toCN ,λ,qrel . This proves that (ρ,M) is a

minimizer forCN ,λ,qrel . 

Next, we show that under certain assumptions a minimizer (ρ, M) for the relaxed problem must, in fact, have M= 0 and is therefore a minimizer of the original problem.

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Proposition 11. Let N ≥ 1, λ > 0 and N /(N + λ) < q < 2 N /(2 N + λ). If N ≥ 3 and λ >

2 N /(N −2), then assume in addition that q ≥ 1−2/N . If (ρ, M) is a minimizer forCN ,λ,qrel , then M= 0. In particular, there is a minimizer for CN ,λ,q.

Note that for N ≥ 3, we are implicitly assuming λ < 4N /(N −2) since otherwise the two assumptions q < 2 N /(2 N + λ) and q ≥ 1 − 2/N cannot be simultaneously satisfied. For the proof of Proposition11we need the following lemma which identifies the sub-leading term in (4).

Lemma 12. Let 0 < q < p, let f ∈ Lp∩Lq(RN) be a symmetric non-increasing function and let g ∈ Lq(RN). Then, for anyτ > 0, as ε → 0+,

Z RN ¯ ¯f (x) + ε−N /pτg(x/ε)¯¯ q d x = Z RN f q d x + εN (1−q/p)τq Z RN|g | q d x + o¡ εN (1−q/p)τq¢ .

Proof of Lemma12. We first note that

f (x) = o¡|x|−N /p¢ as x → 0 (9)

in the sense that for any c > 0 there is an r > 0 such that for all x ∈ RN with |x| ≤ r one has f (x) ≤ c |x|−N /p. To see this, we note that, since f is symmetric non-increasing,

f (x)p≤¯ 1 ¯© y ∈ RN: |y| ≤ |x| ª¯ ¯ Z |y|≤|x| f (y)pd y . The bound (9) now follows by dominated convergence.

It follows from (9) that, asε → 0+,

εN /pf (εx) → 0 for any x ∈ RN,

and therefore, in particular,τg(x) + εN /pf (εx) → τg(x) for any x ∈ RN. From the Brézis– Lieb lemma (see [5]) we know that

Z RN ¯ ¯τg(x) + εN /pf (εx)¯¯ q d x = τq Z RN|g (x)| q d x + Z RN ¡ εN /pf (εx)¢q d x + o(1).

By scaling this is equivalent to the assertion of the lemma. 

Proof of Proposition11. We argue by contradiction and assume that M> 0. Let 0 ≤ σ ∈ ¡L1∩ Lq¡

RN¢¢ ∩ L1¡ RN

, |x|λd x¢ with RRNσdx = 1. We compute the value of Q[ρ,M] for the family (ρ,M) = ¡ρ∗+ ε−Nτσ(·/ε),M− τ¢ with a parameterτ < M∗.

1) We have Iλ£ρ+ ε−Nτσ(·/ε)¤ + 2(M− τ) Z RN|x| λ¡ ρ(x) + ε−Nτσ(x/ε)¢dx = Iλ[ρ] + 2 M∗ Z RN|x| λρ(x) d x + R1

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with R1= 2 τ Ï RN×R(x) ³ |x − y|λ− |x|λ ´ ε−Nσ(y/ε)dx d y + ελτ2Iλ[σ] + 2(M− τ) τ ελ Z RN|x| λσ(x)dx .

Let us show that R1= O ¡

εβτ¢ with β := min{2,λ}. This is clear for the last two terms in the definition of R1, so it remains to consider the double integral. Ifλ ≤ 1 we use the simple inequality |x − y|λ− |x|λ≤ |y|λto conclude that

Ï RN×R(x) ³ |x − y|λ− |x|λ ´ ε−Nσ(y/ε)dx d y ≤ ελ Z RN|x| λσ(x)dxZ Rd x . Ifλ > 1 we use the fact that, with a constant C depending only on λ,

|x − y|λ− |x|λ≤ −λ|x|λ−2x · y +C³|x|(2−λ)+|y|β+ |y|λ ´

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Sinceρis radial, we obtain Ï RN×R(x) ³ |x − y|λ− |x|λ ´ ε−Nσ(y/ε)dx d y ≤ C µ εβ Z RN|x| (2−λ)+ρ(x) d x Z RN|y| βσ(y)d y + ελZ RN|x| λσ(x)dxZ R(x) d x ¶ . Using Hölder’s inequality and the fact thatρ, σ ∈ L

RN¢ ∩ L1¡ RN

, |x|λd x¢ it is easy to see that the integrals on the right side are finite, so indeed R1= O

¡ εβτ¢.

2) For the terms in the denominator ofQ[ρ,M] we note that Z RN ¡ ρ(x) + ε−Nτσ(x/ε)¢dx + (M− τ) = Z Rd x + M∗ and, by Lemma12applied with p = 1,

Z RN ¡ ρ(x) + ε−Nτσ(x/ε)¢ q d x = Z R qd x + εN (1−q)τq Z R q d x + o¡ εN (1−q)τq¢ . Thus, µZ RN ¡ ρ(x) + ε−Nτσ(x/ε)¢ q d x ¶−2−αq = µZ R qd x ¶−2−αq à 1 −2 − α q ε N (1−q)τq R RNσqd x R RNρqd x + R2 ! with R2= o ¡ εN (1−q)τq¢.

Now we collect the estimates. Since (ρ∗, M∗) is a minimizer, we obtain that Q £ρ+ ε−Nτσ(·/ε),M− τ¤ = CN ,λ,q à 1 −2 − α q ε N (1−q)τq R RNσqd x R R qd x + R2 ! + R1 µZ Rd x + M∗ ¶−αµZ RN ¡ ρ(x) + ε−Nτσ(x/ε)¢ q d x ¶−2−αq .

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If β = min{2,λ} > N (1 − q), we can choose τ to be a fixed number in (0,M), so that R1= o ¡ εN (1−q)¢ and therefore Q £ρ+ ε−Nτσ(·/ε),M− τ¤ ≤ CN ,λ,q à 1 −2 − α q ε N (1−q)τq R RNσqd x R R qd x + o¡εN (1−q)¢ ! . Sinceα < 2, this is strictly less than CN ,λ,q forε > 0 small enough, contradicting the defi-nition ofCN ,λ,q as an infimum. Thus, M= 0.

Note that if either N = 1, 2 or if N ≥ 3 and λ ≤ 2 N /(N − 2), then the assumption q > N /(N + λ) implies that β > N (1 − q). If N ≥ 3 and λ > 2 N /(N − 2), then β = 2 ≥ N (1 − q) by assumption. Thus, it remains to deal with the case where N ≥ 3, λ > 2 N /(N − 2) and 2 = N (1 − q). In this case we have R1= O

¡ ε2τ¢ and therefore Q £ρ+ ε−Nτσ(·/ε),M− τ¤ ≤ CN ,λ,q à 1 −2 − α q ε 2τq R RNσqd x R R qd x + O¡ε2τ¢ ! .

By choosing τ small (but independent of ε) we obtain a contradiction as before. This

completes the proof of the proposition. 

Remark 13. The extra assumption q ≥ 1 − 2/N for N ≥ 3 and λ > 2 N /(N − 2) is dictated by theε2bound on R1. We claim that for anyλ ≥ 2, this bound is optimal. Namely, one has Ï RN×R(x) ³ |x − y|λ− |x|λ ´ ε−Nσ(y/ε)dx d y = ε2 λ 2 µ 1 +λ − 2 N ¶ Z RN|x| λ−2ρ(x) d x Z RN|y| 2σ(y)d y + o ¡ε

forλ ≥ 2. This follows from the fact that, for any given x 6= 0, |x − y|λ− |x|λ= −λ |x|λ−2x · y +λ 2|x| λ−2µ|y|2 + (λ − 2)(x · y) 2 |x|2 ¶ + O ³

|y|min{3,λ}+ |y|λ ´

. 4. FURTHER RESULTS OF REGULARITY

In this section we discuss the existence of a minimizer forCN ,λ,q in the regime that is

not covered by Proposition11. In particular, we will establish a connection between the regularity of minimizers for the relaxed problemCN ,λ,qrel and the presence or absence of a Dirac delta. This will allow us to establish existence of minimizers forCN ,λ,q in certain parameter regions which are not covered by Proposition11.

Proposition 14. Let N ≥ 3, λ > 2 N /(N − 2) and N /(N + λ) < q < min©1 −2/N , 2 N/(2 N + λ)ª. If (ρ, M) is a minimizer forCN ,λ,qrel such that (ρ, M∗) ∈ LN (1−q)/2(RN)×[0,+∞), then M= 0.

The condition that the minimizer (ρ, M) ofCN ,λ,qrel belongs to LN (1−q)/2(RN)×[0,+∞) has to be understood as a regularity condition onρ.

Proof. We argue by contradiction assuming that M > 0 and consider a test function ¡

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constantτ1to be determined below. As in the proof of Proposition11, one has Iλ£ ρ+ ε−Nτεσ(·/ε)¤ + 2(M− τε) Z RN|x| λ¡ ρ(x) + ε−Nσ(x/ε)¢dx = Iλ[ρ∗] + 2M∗ Z RN|x| λρ(x) d x + R1 with R1= O ¡ ε2τ

ε¢. Note here that we haveλ ≥ 2. For the terms in the denominator we

note that Z RN ¡ ρ(x) + ε−Nτεσ(x/ε)¢dx + (M− τε) = Z Rd x + M

and, by Lemma12applied with p = N (1 − q)/2 and τ = τε, i.e.,ε−Nτε= ε−N /pτ1, we have Z RN ¡ ρ(x) + ε−Nτεσ(x/ε)¢ q d x = Z R qd x + εN (1−q)τ q ε Z R q d x + o¡ εN (1−q)τq ε¢ .

Because of the choice ofτεwe have εN (1−q)τq ε = εγτq1 and ε 2τ ε= εγτ1 with γ :=N − q (N + 2) 1 − q > 0 and thus Q £ρ+ ε−Nτεσ(·/ε),M− τε¤ ≤ CN ,λ,q à 1 −2 − α q ε γτq 1 R RNσqd x R R qd x + O¡εγτ1 ¢ ! . By choosingτ1small (but independent ofε) we obtain a contradiction as before. 

Proposition14motivates the study of the regularity of the minimizer (ρ, M) ofCN ,relλ,q. We are not able to prove the regularity required in Proposition14, but we can state a di-chotomy result which is interesting by itself, and allows to deduce the existence of mini-mizers forCN ,λ,q in parameter regions not covered in Proposition11.

Proposition 15. Let N ≥ 1, λ > 0 and N /(N + λ) < q < 2 N /(2 N + λ). Let (ρ, M) be a minimizer forCN ,relλ,q. Then the following holds:

(1) IfRRd x >α2

Iλ[ρ∗] R

RN|x|λρ(x) d x, then M= 0 and ρis bounded with ρ∗(0) = Ã (2 − α)Iλ[ρ∗] R Rd x ¡R R qd x¢ ¡2RRN|x|λρ(x) d x R Rd x − αIλ[ρ] ¢ !1/(1−q) . (2) IfR Rd x =α2 R [ρ∗]

RN|x|λρ(x) d x, then M= 0 and ρis unbounded. (3) IfR

Rd x <α2

Iλ[ρ∗] R

RN|x|λρ(x) d x, thenρis unbounded and M=αIλ[ρ∗] − 2 R RN|x|λρ(x) d x R Rd x 2 (1 − α)RRN|x|λρ(x) d x > 0 . To prove Proposition15, let us begin with an elementary lemma.

Lemma 16. For constants A, B > 0 and 0 < α < 1, define

f (M ) := A + M

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Then f attains its minimum on [0, ∞) at M = 0 if αA ≤ B and at M = (αA − B)/(1 − α) > 0 ifαA > B.

Proof. We consider the function on the larger interval (−B,∞). Let us compute f0(M ) =(B + M) − α(A + M)

(B + M)α+1 =

B − αA + (1 − α)M (B + M)α+1 .

Note that the denominator of the right side vanishes exactly at M = (αA − B)/(1 − α), except possibly if this number coincides with −B.

We distinguish two cases. If A ≤ B, which is the same as (αA − B)/(1 − α) ≤ −B, then f is increasing on (−B,∞) and then f indeed attains its minimum on [0,∞) at 0. Thus it remains to deal with the other case, A > B. Then f is decreasing on¡ −B,(αA−B)/(1−α)¤ and increasing on£(αA − B)/(1 − α),∞¢. Therefore, if αA − B ≤ 0, then f is increasing on [0, ∞) and again the minimum is attained at 0. On the other hand, if αA − B > 0, then f

has a minimum at the positive number M = (αA − B)/(1 − α). 

Proof of Proposition15. Step 1. We varyQ[ρ, M ] with respect to M . By the minimizing property of Mthe function

M 7→ Q[ρ, M ] = 2R RN|x|λρ(x) d x ¡R R qd x ¢(2−α)/q A + M (B + M)α with A := [ρ∗] 2R RN|x|λρ(x) d x and B := Z R(x) d x attains its minimum on [0, ∞) at M∗. From Lemma16we infer that

M= 0 if and only if α 2 Iλ[ρ] R RN|x|λρ(x) d x ≤ Z R(x) d x , and that M=αIλ[ρ∗]−2

¡R RN|x|λρ(x) d x ¢¡R RNρ(y) d y ¢ 2 (1−α)R RN|x|λρ(x) d x if α 2 Iλ[ρ∗] R RN|x|λρ(x) d x > R R(x) d x.

Step 2. We varyQ[ρ,M] with respect toρ. Letting x → 0 in the Euler–Lagrange equa-tion (8), we find that

2 R RN|y|λρ(y) d y Iλ] + 2M∗ R RN|y|λρ(y) d y − αR 1 R(y) d y + M = (2 − α) ρ∗(0) −1+q R R(y)qd y ≥ 0 , with the convention that the last inequality is an equality if and only ifρis unbounded. Consistently, we shall write thatρ(0) = +∞ in that case. We can rewrite our inequality as MαIλ[ρ∗] − 2 ¡R RN|y|λρ(y) d y ¢ ¡R Rd y ¢ 2 (1 − α)R RN|y|λρ(y) d y

with equality if and only ifρis unbounded. This completes the proof of Proposition15.

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Next, we focus on matching ranges of the parameters (N ,λ,q) with the cases (1), (2) and (3) in Proposition15. For anyλ ≥ 1 we deduce from

|x − y|λ¡|x| + |y|¢λ≤ 2λ−1¡|x|λ+ |y|λ¢ (11) that Iλ[ρ] < 2λ Z RN|x| λρ(x)dxZ RNρ(x)dx . For allα ≤ 2−λ+1, which can be translated into

q ≥ 2 N¡1 − 2 −λ¢ 2 N¡1 − 2−λ¢ + λ, that is, Z Rd x ≥ α 2 Iλ[ρ] R RN|x|λρ(x) d x ,

so that Cases (1) and (2) of Proposition15apply and we infer that M= 0. Note that this bound for q is in the range¡N /(N + λ), 2 N/(2 N + λ)¢ for all λ ≥ 1. See Fig.1.

A better range for which M= 0 can be obtained for N ≥ 3 using the fact that superlevel sets of a symmetric non-increasing function are balls. From the layer cake representation we deduce that Iλ[ρ] ≤ 2 AN ,λ Z RN|x| λρ(x)dxZ RNρ(x)dx , AN ,λ:= sup0≤R,S<∞F (R, S) , where F (R, S) := Î BR×BS|x − y| λd x d y |BR| R BS|x| λd x + |B S| R BR|y| λd y.

For anyλ ≥ 1, we have 2 AN ,λ≤ 2λby (11), and also AN ,λ≥ 1/2 because by (6) Iλ[1B1] ≥ |B1|

R

B1|y|

λd y. The bound A

N ,λ≥ 1/2 can be improved to AN ,λ> 1 for any λ > 2 as

fol-lows. We know that AN ,λ≥ F (1, 1) =N (N + λ) 2 Ï 0≤r, s≤1 rN −1sN −1 µZ π 0 ¡r2+ s2− 2 r s cos ϕ¢λ/2(sinϕ)N −2 WN d r d s with the Wallis integral WN :=

Rπ

0(sinϕ)N −2dϕ. For any λ > 2, we can apply Jensen’s inequality twice and obtain

Z π 0 ¡r2 + s2− 2 r s cos ϕ¢λ/2(sinϕ) N −2 WN ≥ µZ π 0 ¡r2 + s2− 2 r s cos ϕ¢ (sinϕ) N −2dϕ WNλ/2 =¡r2+ sλ/2 and Ï 0≤r, s≤1 rN −1sN −1¡r2 + sλ/2d r d s ≥ 1 N2 µÏ 0≤r, s≤1 rN −1sN −1¡r2 + s2¢ N2d r d sλ/2 = 1 N2 µ 2 N N + 2λ/2 .

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Hence AN ,λN + λ 2 N µ 2 N N + 2λ/2 =: BN ,λ

whereλ 7→ BN ,λis monotone increasing, so that AN ,λ≥ BN ,λ> BN ,2= 1 for any λ > 2. In

this range we can therefore define ¯ q(λ,N) := 2 N¡1 − A −1 N ,λ ¢ 2 N¡1 − A−1N ,λ¢ + λ. (12)

Based on a numerical computation, the curveλ 7→ ¯q(λ,N) is shown on Fig.1. The next result summarizes our considerations above.

Proposition 17. Assume that N ≥ 3 and λ > N −22 N . Then ¯q defined by (12) is such that ¯ q(λ,N) ≤ 2 N¡1 − 2−λ ¢ 2 N¡1 − 2−λ¢ + λ< 2 N 2 N + λ and q(λ,N) >¯ N

N + λforλ > 2 large enough. If (ρ, M) is a minimizer forCN ,relλ,qand if max© ¯q(λ,N), N

N +λª < q < N −2N , then M= 0 and

ρis bounded.

Proof. We recall that q > ¯q(λ,N) defined by (12) means that Z Rd x ≥ α 2 Iλ] R RN|x|λρ(x) d x ,

so that Case (1) of Proposition 15applies. The estimates on ¯q follow from elementary

computations. 

Next we consider the singularity ofρ at the origin in the unbounded case in more detail, in the cases which are not already covered by Propositions8,11and17.

Lemma 18. Let N ≥ 3, λ > 2 N /(N − 2) and N /(N + λ) < q < min©1 − N/2, ¯q(λ, N)ª. Let

(ρ, M) be a minimizer forCN ,relλ,q and assume that it is unbounded. Then there is a con-stant C > 0 such that

ρ(x) = C |x|−2/(1−q)¡1 + o(1)¢ as x → 0.

Proof. Sinceρ(x) → ∞ as x → 0 we can rewrite the Euler–Lagrange equation (8) as 2 R RN¡|x − y|λ− |y|λ ¢ ρ(y) d y + M|x|λ Iλ] + 2M∗ R RN|y|λρ(y) d y − (2 − α) ρ(x) −1+q R R(y)qd y = 0 . By Taylor expanding we have

Z RN ³ |x − y|λ− |y|λ ´ ρ(y) d y + M|x|λ= C1|x|2¡1 + o(1)¢ as x → 0

with C1 = 12λ(λ − 1)RRN|y|λ−2ρ(y) d y, which is finite according to (6). This gives the

claimed behavior forρat the origin. 

The proof of Lemma18relies only on (8). For this reason, we can also state the follow-ing result.

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Proposition 19. Let N ≥ 1, λ > 0 and N /(N + λ) < q < 1. If N ≥ 3 and λ > 2 N /(N −

2) we assume in addition that q ≥ min©1 − N/2, ¯q(λ, N)ª. If (ρ, M∗) ∈ L1∩ Lq ¡

RN¢ ∩

L1¡

RN, |x|λd x¢ × R+solves (8), then M

= 0 and ρis bounded.

As a consequence, under the assumptions of Proposition19, we recover that any min-imizer (ρ∗, M∗) ofCN ,relλ,q is such that M= 0 and ρ∗ is bounded. Notice that the range

¯

q(λ,N) < 1 − 2/N is covered in Proposition17but not here.

Proof. Assume by contradiction thatρis unbounded. Ifλ ≥ 2, the proof of Lemma18

applies and we know thatρ(x) ∼ |x|−2/(1−q)as x → 0. For any λ ∈ (0,1] we have that |x − y|λ≤ |x|λ+ |y|λ. Ifλ ∈ (1,2), using inequality (10) with the roles of x and y interchanged, we find thatRRN¡|x − y|λ− |y|λ

¢

ρ(y) d y ≤ C |x|λfor some C > 0. Hence, for some c > 0, ρ(x) ≥ c |x|− min{λ,2}/(1−q)

for any x ∈ RN with |x| > 0 small enough. We claim that min{λ,2}/(1 − q) ≥ N , which

contradictsRRd x < ∞. 

By recalling the results of [16] in the conformally invariant case q = 2N /(2N + λ), and the results of Propositions 4, 8, 11,19 and Lemma9, we have completed the proof of Theorem1.

5. FREEENERGY

In this section, we discuss the relation between the reverse HLS inequalities (1) and the free energy functional

F [ρ] := − 1 1 − q Z R q d x + 1 2λIλ[ρ].

We also extend the free energy functional to the set of probability measures and prove a uniqueness result in this framework.

5.1. Relaxation and extension of the free energy functional. The kernel |x − y|λis pos-itive and continuous, so there is no ambiguity with the extension of IλtoP (RN), which is simply given by

Iλ[µ] = Ï

RN×RN|x − y|

λdµ(x)dµ(y).

In this section we use the notion of weak convergence in the sense of probability theory: ifµnandµ are probability measures on RN thenµn* µ means RRNϕdµn

R

RNϕdµ for all bounded continuous functionsϕ on RN. We define the extension ofF to P (RN) by

FΓ[µ] := inf (ρn)n∈N⊂Cc∞∩P (RN)

s.t.ρn*µ

lim inf

n→∞ F [ρn] .

We also define a relaxed free energy by Frel[ρ,M] := − 1 1 − q Z RNρ(x) q d x + 1 2λIλ[ρ] + M λ Z RN|x| λρ(x)dx .

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The functionalFrelcan be characterized as the restriction ofFΓto the subset of proba-bility measures whose singular part is a multiple of aδ at the origin.

5.2. Equivalence of the optimization problems and consequences. According to Propo-sition4, we know thatCN ,λ,q = 0 if 0 < q ≤ N /(N + λ), so that one can find test functions

ρn∈ L1+∩ Lq(RN) such that kρnkL1(RN)= Iλ[ρn] = 1 and Z RNρn(x) q d x ≥ n ∈ N. As a consequence, limn→∞F [ρn] = −∞.

Next, let us consider the case N /(N + λ) < q < 1. Assume that ρ ∈ L1+∩ Lq(RN) is such that Iλ[ρ] is finite. For any ` > 0 we define ρ`(x) := `−Nρ(x/`)/kρkL1(RN)and compute

F [ρ`] = −`(1−q) NA+ `λB

whereA=1−q1 RRNρ(x)qd x/kρkq

L1(RN)andB=

1

2λIλ[ρ]/kρk2L1(RN). The function` 7→ F [ρ`] has a minimum which is achieved at` = `?where

`?:= ³

N (1−q)A λB

´λ−N (1−q)1 and, withQq,λas defined in (7), we obtain that

F [ρ] ≥ F [ρ`?] = −κ? ¡ Qq,λ[ρ]¢N (1−q) λ−N (1−q) where κ?:=λ−N (1−q) (1−q)λ (2 N ) N (1−q) λ−N (1−q). As a consequence, we have the following result.

Proposition 20. With the notations of Section5.1, for any q ∈ (0,1) and λ > 0, we have FN ,λ,q := infρ F [ρ] = infρ,MFrel[ρ,M] = infµ FΓ[µ]

where the infima are taken on L1+∩ Lq(RN),¡L1

+∩ Lq(RN)¢ × [0,∞) and P (RN) in case of, respectively, F , Frel and. Moreover FN ,λ,q > −∞ if and only if CN ,λ,q > 0, that is, if N /(N + λ) < q < 1 and, in this case,

FN ,λ,q = −κ?C−

N (1−q) λ−N (1−q)

N ,λ,q = Frel[ρ∗, M∗] = FΓ[µ∗]

for someµ= Mδ + ρ, (ρ, M∗) ∈¡L+1∩ Lq(RN)¢ × [0,1) such that RR(x) d x + M= 1. Additionally, we have that

Iλ[ρ] + 2 M∗ Z RN|x| λρ(x) d x = 2 N Z R(x) qd x .

Since (ρ, M) is also a minimizer for CN ,relλ,q, it satisfies all properties of Lemma9 and Propositions11,15and17.

Proof. This result is a simple consequence of the definitions ofFrelandFΓ. The exis-tence of the minimizer is a consequence of Propositions8and10. Ifρ ∈ L1+∩ Lq(RN) is a minimizer for FN ,λ,q, then Iλ[ρ] = 2N RRNρ(x)qd x because`?= 1, and ρ is also an opti-mizer forCN ,λ,q. Conversely, ifρ ∈ L1+∩Lq(RN) is an optimizer forCN ,λ,q, then there is an ` > 0 such that `−Nρ(·/`)/kρk

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The discussion of whether M= 0 or not in the statement of Proposition20is the same as in the discussion of the reverse Hardy–Littlewood–Sobolev inequality in Section3. Ex-cept for the question of uniqueness, this completes the proof of Theorem3.

5.3. Properties of the free energy extended to probability measures. From now on, un-less it is explicitly specified, we shall denote by ρ the absolutely continuous part of the measureµ ∈ P (RN). OnP (RN), let us define

G [µ] := 1 2λ[µ] − 1 1 − q Z RNρ(x) qd x (13) if Iλ[µ] < +∞ and extend it with the convention that G [µ] = +∞ if Iλ[µ] = +∞. Notice thatRRNρ(x)qd x is finite by Lemma5and Eq. (6) whenever Iλ[ρ] ≤ Iλ[µ] is finite. Let us start with some technical estimates. The following is a variation of [12, Lemma 2.7].

Lemma 21. Let N ≥ 1 and λ > 0, then for any a ∈ RN, r > 0 and µ ∈ P (RN) we have Iλ[µ] ≥ 21−(λ−1)+µ¡B r(a) ¢ µZ RN|y − a| λdµ(y) − 2(λ−1)+rλ ¶ .

As a consequence, if Iλ[µ] < ∞, then RRN|y − a|λdµ(y) is finite for any a ∈ RN and the infimum with respect to a is achieved.

Proof. If x ∈ Br(a) and y ∈ Br(a)c, then

|x − y|λ¡|y − a| − |x − a|¢λ¡|y − a| − r ¢λ≥ 2−(λ−1)+|y − a|λ− rλ. We can therefore bound Iλ[µ] from below by

2 Ï Br(a)×Br(a)c |x − y|λdµ(x)dµ(y) ≥ 2 µ¡Br(a) ¢ µ 2−(λ−1)+ Z Br(a)c |y − a|λdµ(y) − rλµ¡Br(a)c ¢ ¶ = 21−(λ−1)+µ¡B r(a) ¢ µZ RN|y − a| λdµ(y) −Z Br(a) |y − a|λdµ(y) − 2(λ−1)+rλµ¡B r(a)c ¢ ¶ ≥ 21−(λ−1)+µ¡B r(a) ¢ µZ RN|y − a| λdµ(y) − rλµ¡B r(a)¢ − 2(λ−1)+rλµ¡Br(a)c ¢ ¶ ≥ 21−(λ−1)+µ n¡Br(a) ¢ µZ RN|y − a| λdµ n(y) − 2(λ−1)+ ¶ . This proves the claimed inequality.

Let R > 0 be such that µ¡BR(0)¢ ≥ 1/2 and consider a ∈ BR(0)c, so that |y − a| > |a| − R

for any y ∈ BR(0). From the estimate

Z RN|y − a| λdµ(y) ≥Z BR(0) |y − a|λdµ(y) ≥1 2¡|a| − R¢ λ , we deduce that in infa∈RN

R

RN|y −a|λdµ(y), a can be restricted to a compact region of RN. Since the map a 7→RRN|y − a|λdµ(y) is lower semi-continuous, the infimum is achieved.

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Corollary 22. Letλ > 0 and N/(N + λ) < q < 1. Then there is a constant C > 0 such that G [µ] ≥ [µ]−C ≥ 1 4λa∈RinfN Z RN|x − a| λdµ(x) −C ∀µ ∈ P (RN) .

Proof. Letµ ∈ P (RN) and letρ be its absolutely continuous part with respect to Lebes-gue’s measure. By Theorem1, we know that

Z RNρ(x) q d x ≤ µ Iλ[ρ] CN ,λ,qN (1−q)λ becauseR

RNρ dx ≤ µ(RN) = 1. Hence we obtain that G [µ] ≥ [µ]−C with C = min ( X 4λ− µ X CN ,λ,qN (1−q)λ : X > 0 ) . Asµ is a probability measure, the proof is completed using the inequality

inf a∈RN Z RN|x − a| λdµ(x) ≤Ï RN×RN|x − a| λdµ(x)dµ(a) = I λ[µ].  Lemma 23. Ifλ > 0 and N/(N + λ) < q < 1, then G is lower semi-continuous.

Proof. Let (µn) ⊂ P (RN) withµn* µ. We denote by ρnandρ the absolutely continuous

part of µn andµ, respectively. We have to prove that liminfn→∞G [µn] ≥ G [µ]. Either

lim infn→∞G [µn] = +∞, or it is finite and then, up to the extraction of a subsequence, we

know from Corollary22thatK := supn∈NIλ[µn] is finite. According to [33, Proposition

7.2], we also know that

lim inf

n→∞ [µn] ≥ Iλ[µ].

According to [33, Theorem 7.7] or [4, Theorem 4], for any r > 0 we have lim inf n→∞ µ − Z Br ρn(x)qd x ¶ ≥ − Z Br ρ(x)qd x .

Notice that the absolutely continuous part of the limit ofµn ¬Br coincides with the

abso-lutely continuous part ofµ ¬Br as the difference is supported on∂Br.

We choose r0> 0 to be a number such that µ(Br0) ≥ 1/2 and find n0∈ N such that for any n ≥ n0we haveµn(Br0) ≥ 1/4. By applying Lemma21, we obtain that

Z RN|x| λdµ n(x) ≤ 2(λ−1)+ ³ r0λ+ 2 Iλ[µn] ´ ≤ 2(λ−1)+¡rλ 0+ 2 K ¢

for any n ≥ n0. We apply Lemma5toρ = ρn1Brc Z Brc ρn(x)qd x ≤ cN ,λ,q−q µZ Brc ρnd xq−N (1−q)λ µZ Brc |x|λρnd xN (1−q)λ

and conclude that lim inf n→∞ µ − Z Bcr ρn(x)qd x≥ −c−qN ,λ,q¡µ¡Brc ¢¢q−N (1−q)λ ³ 2(λ−1)+¡rλ 0+ 2 K ¢´ N (1−q) λ .

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The right hand side vanishes as r → ∞, which proves the claimed lower semi-continuity.



After these preliminaries, we can now prove thatG , defined in (13), is the lower-semi-continuous envelope ofF . The precise statement goes as follows.

Proposition 24. Let 0 < q < 1 and λ > 0.Let µ ∈ P (RN) (1) If q ≤ N /(N + λ), then FΓ[µ] = −∞.

(2) If q > N /(N + λ), then FΓ[µ] = G [µ].

Proof. Assume that q ≤ N /(N + λ). Using the function ν(x) = |x|−N −λ¡ log|x|)¢−1/q, let us construct an approximation of any measure inµ ∈ P (RN) given by a sequence (ρn)n∈Nof

functions in Cc∞∩ P (RN) such that limn→∞F [ρn] = −∞.

Letη ∈ Cc(B1) be a positive mollifier with unit mass andζ ∈ Cc(B2) be a cutoff func-tion such that 1B1 ≤ ζ ≤ 1. Given any natural numbers i , j and k, we define ηi(y) := iNη(i y), ζj(y) := ζ(y/j ) and

fi , j ,k:=¡1 −k1

¢ ¡ µ ∗ ηi

¢

ζi+1kCi , j ,k(1 − ζi)ζjν

where Ci , j ,k is a positive constant that has been picked so that fi , j ,k∈ P (RN). We choose

i = n, j = enand k = k(n) such that lim

n→∞k(n) = +∞ and n→∞lim k(n)

−N qlog(n/ log n) = +∞.

By construction, ρn := fn, j (n),k(n) * µ as n → ∞ and limn→∞F [ρn] = −∞, so FΓ[µ] =

− ∞.

Assume that q > N /(N + λ) and consider a sequence of functions in Cc∞∩ P (RN) such

thatρn* µ and limn→∞F [ρn] = FΓ[µ]. If Iλ[µ] = ∞, by the lower-semicontinuity of Iλ

(see for instance [33, Proposition 7.2]), we know that limn→∞Iλ[ρn] = ∞ and deduce

from Corollary22that41λIλ[ρn] −C ≤ F [ρn] diverges, so thatFΓ[µ] = ∞ = G [µ].

Next, we assume that Iλ[µ] < ∞. According to Lemma23, we deduce from the lower semi-continuity ofG that

FΓ[µ] = lim

n→∞F [ρn] = limn→∞G [ρn] ≥ G [µ].

It remains to show the inequalityFΓ[µ] ≤ G [µ]. Let µR:= µ(BR)−1µ ¬BR. We have that

µR* µ as R → ∞ and, by monotone convergence,

lim

R→∞G [µR] = G [µ].

Letηε(x) := ε−Nη(x/ε) for a sufficiently regular, compactly supported, nonnegative func-tionη such that RRNηdx = 1. Then µR∗ηε∈ Cc∞∩P (RN) andµR∗ηε* µRasε → 0. Here we are using implicitly the metrizability of weak convergence. SinceµR∗ ηε→ ρRalmost

everywhere, Fatou’s lemma implies that lim inf ε→0 Z RN(µR∗ ηε) q d x ≥ Z R q Rd x .

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Moreover, sinceµRhas compact support, the support ofµR∗ηεis contained in a bounded

set independent ofε and therefore the interaction term is, in fact, continuous under weak convergence (see, e.g., [33, Proposition 7.2]), that is,

lim inf

ε→0 [µR∗ ηε] = Iλ[µR] .

Thus, we have shown that

lim inf

ε→0 F [µR∗ ηε] ≤ G [µR] .

Hence for any R = n ∈ N, we can find an εn> 0, small enough, such that µn∗ ηεn* µ and finally obtain that

FΓ[µ] ≤ lim

n→∞F [µn∗ ηεn] ≤ G [µ].



In Section3, using symmetric decreasing rearrangements, we proved that there is a minimizing sequence which converges to a minimizer. Here we have a stronger property.

Proposition 25. Let N /(N +λ) < q < 1. Then any minimizing sequence for FΓis relatively compact, up to translations, with respect to weak convergence. In particular, there is a minimizer for.

Proof. Let (µn)n∈N be a minimizing sequence for FΓin P (RN). After an n-dependent

translation we may assume that for any n ∈ N, Z RN|x| λdµ n(x) = inf a∈RN Z RN|x − a| λdµ n(x)

according to Lemma21. Corollary22applies Z RN|x| λdµ n(x) ≤ 4λ µ sup n F Γ[µ n] +C ¶ ,

which implies that (µn)n∈N is tight. By Prokhorov’s theorem and after passing to a

sub-sequence if necessary, (µn)n∈N converges weakly to some µ∗ ∈ P (RN). By the lower-semicontinuity property of Lemma23, we obtain that

inf µ∈P (RN)F Γ[µ] = lim n→∞F Γ[µ n] ≥ inf µ∈P (RN)F Γ[µ],

which concludes the proof. 

Remark 26. By symmetrization, Lemma9and Proposition20, we learn that, up to trans-lations, any minimizerµ of FΓis of the form µ = ρ + M δ, with M ∈ [0,1) and ρ ∈ L1+∩ Lq(RN). Moreover,ρ is radially symmetric non-increasing and strictly positive. The min-imizers ofFΓsatisfy the Euler-Lagrange conditions given by (8). This can be also shown by taking variations directly onFΓas in [11].

5.4. Uniqueness.

Theorem 27. Let N /(N + λ) < q < 1 and assume either that 1 − 1/N ≤ q < 1 and λ ≥ 1, or

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