

A family of subcritical Caffarelli–Kohn–Nirenberg interpolation inequalities 
With the norms
‖w‖Lq,γ(Rd):=(∫Rdwqx−γdx)1/q,‖w‖Lq(Rd):=‖w‖Lq,0(Rd), let us define as the space of all measurable functions w such that is finite. Our functional framework is a space of functions such that , which is defined as the completion of the space of the smooth functions on with compact support in , with respect to the norm given by .
Now consider the family of Caffarelli–Kohn–Nirenberg interpolation inequalities given by
(1)‖w‖L2p,γ(Rd)≤Cβ,γ,p‖∇w‖L2,β(Rd)ϑ‖w‖Lp+1,γ(Rd)1−ϑ∀w∈Hβ,γp(Rd).
Here the parameters β , γ and p are subject to the restrictions
(2)d≥2,γ−2<β<d−2dγ,γ∈(−∞,d),p∈(1,p⋆]withp⋆:=d−γd−β−2 and the exponent ϑ is determined by the scaling invariance, i.e. ,ϑ=(d−γ)(p−1)p(d+β+2−2γ−p(d−β−2)). These inequalities have been introduced, among others, by L. Caffarelli, R. Kohn and L. Nirenberg in [[5]]. We observe that if , a case that has been dealt with in [[14]], and we shall focus on the subcritical case . Throughout this paper, denotes the optimal constant in ((1)). We shall say that a function is an extremal function for ((1)) if equality holds in the inequality.
Symmetry in ((1)) means that the equality case is achieved by Aubin–Talentitype functions
w⋆(x)=(1+x2+β−γ)−1/(p−1)∀x∈Rd. On the contrary, there is symmetry breaking if this is not the case, because the equality case is then achieved by a nonradial extremal function. It has been proved in [[4]] that symmetry breaking holds in ((1)) if(3)γ<0andβFS(γ)<β<d−2dγ, whereβFS(γ):=d−2−(γ−d)2−4(d−1). For completeness, we will give a short proof of this result in Section 2. Our main result shows that, under Condition ((2)), symmetry holds in the complement of the set defined by ((3)), which means that ((3)) is the sharp condition for symmetry breaking . See Fig. 1.


 Fig. 1. In dimension d =4, with p =1.2, the grey area corresponds to the cone determined by d −2+(γ −d )/p ≤β <(d −2)γ /d and γ ∈(−∞,d ) in ((2)). The light grey area is the region of symmetry, while the dark grey area is the region of symmetry breaking. The threshold is determined by the hyperbola (d −γ )^{2}−(β −d +2)^{2}−4(d −1)=0 or, equivalently β =β _{FS} (γ ). Notice that the condition p ≤p _{⋆} induces the restriction β ≥d −2+(γ −d )/p , so that the region of symmetry is bounded. The largest possible cone is achieved as p →1 and is limited from below by the condition β >γ −2. Zoom 
Assume that ((2)) holds and that
(4)β≤βFS(γ)ifγ<0. Then the extremal functions for ((1)) are radially symmetric and, up to a scaling and a multiplication by a constant, equal to .
The above result is slightly stronger than just characterizing the range of for which equality in ((1)) is achieved by radial functions. Actually our method of proof allows us to analyze the symmetry properties not only of extremal functions of ((1)), but also of all positive solutions in of the corresponding Euler–Lagrange equations, that is, up to a multiplication by a constant and a dilation, of
(5)−div(x−β∇w)=x−γ(w2p−1−wp)inRd∖{0}.
Assume that ((2)) and ((4)) hold. Then all positive solutions to ((5)) in are radially symmetric and, up to a scaling and a multiplication by a constant, equal to .
Up to a multiplication by a constant, we know that all nontrivial extremal functions for ((1)) are nonnegative solutions to ((5)). Nonnegative solutions to ((5)) are actually positive by the standard Strong Maximum principle. Theorem 1.1 is therefore a consequence of Theorem 1.2. In the particular case when , the condition ((2)) amounts to , , , and ((1)) can be written as
‖w‖L2p,γ(Rd)≤C0,γ,p‖∇w‖L2(Rd)ϑ‖w‖Lp+1,γ(Rd)1−ϑ∀w∈H0,γp(Rd). In this case, we deduce from Theorem 1.1 that symmetry always holds. This is consistent with a previous result ( and , close to 0) obtained in [[17]]. A few other cases were already known. The Caffarelli–Kohn–Nirenberg inequalities that were discussed in [[14]] correspond to the critical case , or, equivalently . Here by critical we simply mean that scales like . The limit case and , which is an endpoint for ((2)), corresponds to Hardytype inequalities: there is no extremal function, but optimality is achieved among radial functions: see [[16]]. The other endpoint is , in which case . The results of Theorem 1.1 also hold in that case with , up to existence issues: according to [[9]], either , symmetry holds and there exists a symmetric extremal function, or , and then symmetry is broken, but there is no optimal function.
Inequality ((1)) can be rewritten as an interpolation inequality with same weights on both sides using a change of variables. Here we follow the computations in [[4]] (also see [[14], [15]]). Written in spherical coordinates for a function
w˜(r,ω)=w(x),withr=xandω=xx, inequality ((1)) becomes(∫0∞∫Sd−1w˜2prd−γ−1drdω)12p≤Cβ,γ,p(∫0∞∫Sd−1∇w˜2rd−β−1drdω)ϑ2(∫0∞∫Sd−1w˜p+1rd−γ−1drdω)1−ϑp+1, where and denotes the gradient of with respect to the angular variable . Next we consider the change of variables ,(6)w˜(r,ω)=v(s,ω)∀(r,ω)∈R+×Sd−1, where α and n are two parameters such thatn=d−β−2α+2=d−γα. Our inequality can therefore be rewritten as(∫0∞∫Sd−1v2psn−1dsdω)12p≤Kα,n,p(∫0∞∫Sd−1(α2∂v∂s2+1s2∇ωv2)sn−1dsdω)ϑ2(∫0∞∫Sd−1vp+1sn−1dsdω)1−ϑp+1, withCβ,γ,p=αζKα,n,pandζ:=ϑ2+1−ϑp+1−12p=(β+2−γ)(p−1)2p(d+β+2−2γ−p(d−β−2)). Using the notationDαv=(α∂v∂s,1s∇ωv), withα=1+β−γ2andn=2d−γβ+2−γ, Inequality ((1)) is equivalent to a Gagliardo–Nirenberg type inequality corresponding to an artificial dimension n or, to be precise, to a Caffarelli–Kohn–Nirenberg inequality with weight in all terms. Notice thatp⋆=nn−2.
Assume that α, n and p are such that
d≥2,α>0,n>dandp∈(1,p⋆]. Then the inequality (7)‖v‖L2p,d−n(Rd)≤Kα,n,p‖Dαv‖L2,d−n(Rd)ϑ‖v‖Lp+1,d−n(Rd)1−ϑ∀v∈Hd−n,d−np(Rd), holds with optimal constant as above and optimality is achieved among radial functions if and only if (8)α≤αFSwithαFS:=d−1n−1. When symmetry holds, optimal functions are equal, up to a scaling and a multiplication by a constant, to v⋆(x):=(1+x2)−1/(p−1)∀x∈Rd.
We may notice that neither nor depend on p and that the curve determines the same threshold for the symmetrybreaking region as in the critical case . In the case , this curve was found by V. Felli and M. Schneider, who proved in [[19]] the linear instability of all radial critical points if . When , symmetry holds under Condition ((8)) as was proved in [[14]]. Our goal is to extend this last result to the subcritical regime .
The change of variables is an important intermediate step, because it allows one to recast the problem as a more standard interpolation inequality in which the dimension n is, however, not necessarily an integer. Actually n plays the role of a dimension in view of the scaling properties of the inequalities and, with respect to this dimension , they are critical if and subcritical otherwise. The critical case has been studied in [[14]] using tools of entropy methods, a critical fast diffusion flow and, in particular, a reformulation in terms of a generalized Fisher information . In the subcritical range, we shall replace the entropy by a Rényi entropy power as in [[18], [21]], and make use of the corresponding fast diffusion flow. As in [[14]], the flow is used only at the heuristic level in order to produce a welladapted test function. The core of the method is based on the Bakry–Emery computation, also known as the carré du champ method , which is well adapted to optimal interpolation inequalities: see for instance [[2]] for a general exposition of the method and [[12], [13]] for its use in the presence of nonlinear flows. Also see [[6]] for earlier considerations on the Bakry–Emery method applied to nonlinear flows and related functional inequalities in unbounded domains. However, in noncompact manifolds and in the presence of weights, integrations by parts have to be justified. In the critical case, one can rely on an additional invariance to use an Emden–Fowler transformation and rewrite the problem as an autonomous equation on a cylinder, which simplifies the estimates a lot. In the subcritical regime, estimates have to be adapted, since after the Emden–Fowler transformation, the problem in the cylinder is no longer autonomous.
This paper is organized as follows. We recall the computations that characterize the linear instability of radially symmetric minimizers in Section 2. In Section 3, we expose the strategy for proving symmetry in the subcritical regime when there are no weights. Section 4 is devoted to the Bakry–Emery computation applied to Rényi entropy powers, in the presence of weights. This provides a proof of our main results, if we admit that no boundary term appears in the integrations by parts in Section 4. To prove this last result, regularity and decay estimates of positive solutions to ((5)) are established in Section 5, which indeed show that no boundary term has to be taken into account (see Proposition 5.1).
For completeness, we summarize known results on symmetry breaking for ((1)). Details can be found in [[4]]. With the notations of Corollary 1.3, let us define the functional
J[v]:=ϑlog(‖Dαv‖L2,d−n(Rd))+(1−ϑ)log(‖v‖Lp+1,d−n(Rd))+logKα,n,p−log(‖v‖L2p,d−n(Rd)) obtained by taking the difference of the logarithm of the two terms in ((7)). Let us define , whereμδ(x):=1(1+x2)δ. Since as defined in Corollary 1.3 is a critical point of , a Taylor expansion at order shows that‖Dαv⋆‖L2,d−n(Rd)2J[v⋆+εμδ/2f]=12ε2ϑQ[f]+o(ε2) with andQ[f]=∫RdDαf2xn−ddμδ−4pα2p−1∫Rdf2xn−ddμδ+1. The following Hardy–Poincaré inequality has been established in [[4]].
Let , , and . Then
(9)∫RdDαf2xn−ddμδ≥Λ∫Rdf2xn−ddμδ+1 holds for any , with , such that , with an optimal constant Λ given by Λ={2α2(2δ−n)if0<α2≤(d−1)δ2n(2δ−n)(δ−1),2α2δηifα2>(d−1)δ2n(2δ−n)(δ−1), where η is the unique positive solution to η(η+n−2)=d−1α2. Moreover, Λ is achieved by a nontrivial eigenfunction corresponding to the equality in ((9)). If , the eigenspace is generated by , with , 2,…d and the eigenfunctions are not radially symmetric, while in the other case the eigenspace is generated by the radially symmetric eigenfunction .
As a consequence, is a nonnegative quadratic form if and only if . Otherwise, takes negative values, and a careful analysis shows that symmetry breaking occurs in ((1)) if
2α2δη<4pα2p−1⟺η<1, which meansd−1α2=η(η+n−2)<n−1, and this is equivalent to .


The strategy for proving symmetry without weights 
Before going into the details of the proof, we explain the strategy for the case of the Gagliardo–Nirenberg inequalities without weights. There are several ways to compute the optimizers, and the relevant papers are [[2], [6], [7], [8], [11], [18]] (also see additional references therein). The inequality is of the form
(10)‖w‖L2p(Rd)≤C0,0,p‖∇w‖L2(Rd)ϑ‖w‖Lp+1(Rd)1−ϑwith1<p<dd−2 andϑ=d(p−1)p(d+2−p(d−2)).
It is known through the work in [[11]] that the optimizers of this inequality are, up to multiplications by a constant, scalings and translations, given by
w⋆(x)=(1+x2)−1p−1∀x∈Rd. In our perspective, the idea is to use a version of the carré du champ or Bakry–Emery method introduced in [[1]]: by differentiating a relevant quantity along the flow, we recover the inequality in a form that turns out to be sharp. The version of the carré du champ we shall use is based on the Rényi entropy powers whose concavity as a function of t has been studied by M. Costa in [[10]] in the case of linear diffusions (see [[21]] and references therein for more recent papers). In [[23]], C. Villani observed that the carré du champ method gives a proof of the logarithmic Sobolev inequality in the Blachman–Stam form, also known as the Weissler form: see [[3], [24]]. G. Savaré and G. Toscani observed in [[21]] that the concavity also holds in the nonlinear case, which has been used in [[18]] to give an alternative proof of the Gagliardo–Nirenberg inequalities, that we are now going to sketch.
The first step consists in reformulating the inequality in new variables. We set
u=w2p, which is equivalent to , and consider the flow given by(11)∂u∂t=Δum, where m is related to p byp=12m−1. The inequalities imply that(12)1−1d<m<1. For some positive constant , one easily finds that the socalled Barenblatt–Pattle functionsu⋆(t,x)=κdt−ddm−d+2w⋆2p(κt−1dm−d+2x)=(a+bx2)−11−m are selfsimilar solutions to ((11)), where and are explicit. Thus, we see that is an optimizer for ((10)) for all t and it makes sense to rewrite ((10)) in terms of the function u . Straightforward computations show that ((10)) can be brought into the form(13)(∫Rdudx)(σ+1)m−1≤CEσ−1Iwhereσ=2d(1−m)−1 for some constant C which does not depend on u , whereE:=∫Rdumdx is a generalized Ralston–Newman entropy , also known in the literature as Tsallis entropy , andI:=∫Rdu∇P2dx is the corresponding generalized Fisher information . Here we have introduced the pressure variable P=m1−mum−1. The Rényi entropy power is defined byF:=Eσ as in [[18], [21]]. With the above choice of σ , is an affine function of t if . For an arbitrary solution to ((11)), we aim at proving that it is a concave function of t and that it is affine if and only if . For further references on related issues, see [[11], [22]]. Note that one of the motivations for choosing the variable is that it has a particular simple form for the selfsimilar solutions, namelyP⋆=m1−m(a+bx2).
Differentiating along the flow ((11)) yields
E′=(1−m)I, so thatF′=σ(1−m)GwithG:=Eσ−1I. More complicated is the derivative for the Fisher information:I′=−2∫Rdum[Tr((HessP−1dΔPId)2)+(m−1+1d)(ΔP)2]dx. Here and Id are respectively the Hessian of and the identity matrix. The computation can be found in [[18]]. Next we compute the second derivative of the Rényi entropy power with respect to t :(F)″σEσ=(σ−1)E′2E2+E″E=(σ−1)(1−m)2I2E2+(1−m)I′E=:(1−m)H. With , we obtain(14)H=−2〈Tr((HessP−1dΔPId)2)〉+(1−m)(1−σ)〈(ΔP−〈ΔP〉)2〉, where we have used the notation〈A〉:=∫RdumAdx∫Rdumdx. Note that by ((12)), we have that and hence we find that , which also means that is a nonincreasing function. In fact it is strictly decreasing unless is a polynomial function of order two in x and it is easy to see that the expression ((14)) vanishes precisely when is of the form , where a , , are constants (but a and b may still depend on t ).
Thus, while the left side of ((13)) stays constant along the flow, the right side decreases. In [[18]] it was shown that the right side decreases towards the value given by the selfsimilar solutions and hence proves ((10)) in the sharp form. In our work we pursue a different tactic. The variational equation for the optimizers of ((10)) is given by
−Δw=aw2p−1−bwp. A straightforward computation shows that this can be written in the form2mum−2div(u∇P)+∇P2+c1um−1=c2 for some constants , whose precise values are explicit. This equation can also be interpreted as the variational equation for the sharp constant in ((13)). Hence, multiplying the above equation by and integrating yields∫Rd[2mum−2div(u∇P)+∇P2]Δumdx+c1∫Rdum−1Δumdx=c2∫RdΔumdx=0. We recover the fact that, in the flow picture, is, up to a positive factor, the derivative of and hence vanishes. From the observations made above, we conclude that must be a polynomial function of order two in x . In this fashion, one obtains more than just the optimizers, namely a classification of all positive solutions to the variational equation. The main technical problem with this method is the justification of the integrations by parts, which in the case at hand, without any weight, does not offer great difficulties: see, for instance, [[6]]. This strategy can also be used to treat the problem with weights, which will be explained next. Dealing with weights, however, requires some special care, as we shall see.


The Bakry–Emery computation and Rényi entropy powers in the weighted case 
Let us adapt the above strategy to the case where there are weights in all integrals entering into the inequality, that is, let us deal with inequality ((7)) instead of inequality ((10)). In order to define a new, welladapted fast diffusion flow, we introduce the diffusion operator , which is given in spherical coordinates by
Lαu=α2(u″+n−1su′)+1s2Δωu, where denotes the Laplace–Betrami operator acting on the dimensional sphere of the angular variables, and ′ denotes here the derivative with respect to s . Consider the fast diffusion equation(15)∂u∂t=Lαum in the subcritical range . The exponents m in ((15)) and p in ((7)) are related as in Section 3 byp=12m−1⟺m=p+12p and ν is defined byν:=11−m. We consider the Fisher information defined asI[P]:=∫RduDαP2dμwithP=m1−mum−1anddμ=sn−1dsdω=sn−ddx. Here is the pressure variable . Our goal is to prove that takes the form , as in Section 3. It is useful to observe that ((15)) can be rewritten as∂u∂t=Dα⁎(uDαP) and, in order to compute , we will also use the fact that solves(16)∂P∂t=(1−m)PLαP−DαP2.
 
 First step: computation of 
Let us define
K[P]:=A[P]−(1−m)(LαP)2whereA[P]:=12LαDαP2−DαP⋅DαLαP and, on the boundary of the centered ball of radius s , the boundary term(17)b(s):=∫∂Bs(∂∂s(Pmm−1DαP2)−2(1−m)Pmm−1P′LαP)dς=sn−1(∫Sd−1(∂∂s(Pmm−1DαP2)−2(1−m)Pmm−1P′LαP)dω)(s), where by we denote the standard Hausdorff measure on .
If u solves ((15)) and if
(18)lims→0+b(s)=limS→+∞b(S)=0, then, (19)ddtI[P]=−2∫RdK[P]umdμ.
For , let us consider the set , so that . Using ((15)) and ((16)), we can compute
ddt∫A(s,S)uDαP2dμ=∫A(s,S)∂u∂tDαP2dμ+2∫A(s,S)uDαP⋅Dα∂P∂tdμ=∫A(s,S)Lα(um)DαP2dμ+2∫A(s,S)uDαP⋅Dα((1−m)PLαP−DαP2)dμ=∫A(s,S)umLαDαP2dμ+2(1−m)∫A(s,S)uPDαP⋅DαLαPdμ+2(1−m)∫A(s,S)uDαP⋅DαPLαPdμ−2∫A(s,S)uDαP⋅DαDαP2dμ+α2∫∂BS((um)′DαP2−um∂∂s(DαP2))dς−α2∫∂Bs((um)′DαP2−um∂∂s(DαP2))dς=−∫A(s,S)umLαDαP2dμ+2(1−m)∫A(s,S)uPDαP⋅DαLαPdμ+2(1−m)∫A(s,S)uDαP⋅DαPLαPdμ+α2∫∂BS((um)′DαP2+um∂∂s(DαP2))dς−α2∫∂Bs((um)′DαP2+um∂∂s(DαP2))dς, where the last line is given by an integration by parts, upon exploiting the identity :∫A(s,S)uDαP⋅DαDαP2dμ=−∫A(s,S)Dα(um)⋅DαDαP2dμ=∫A(s,S)umLαDαP2dμ−α2∫∂BSum∂∂s(DαP2)dς+α2∫∂Bsum∂∂s(DαP2)dς. 1) Using the definition of , we get that(20)−∫A(s,S)umLαDαP2dμ=−2∫A(s,S)umA[P]dμ−2∫A(s,S)umDαP⋅DαLαPdμ. 2) Taking advantage again of , an integration by parts gives∫A(s,S)uDαP⋅DαPLαPdμ=−∫A(s,S)Dα(um)⋅DαPLαPdμ=∫A(s,S)um(LαP)2dμ+∫A(s,S)umDαP⋅DαLαPdμ−α2∫∂BSumP′LαPdς+α2∫∂BsumP′LαPdς and, with , we find that(21)2(1−m)∫A(s,S)uPDαP⋅DαLαPdμ+2(1−m)∫A(s,S)uDαP⋅DαPLαPdμ=2(1−m)∫A(s,S)um(LαP)2dμ+2∫A(s,S)umDαP⋅DαLαPdμ−2(1−m)α2∫∂BSumP′LαPdς+2(1−m)α2∫∂BsumP′LαPdς. Summing ((20)) and ((21)), using ((17)) and passing to the limits as , , establishes ((19)). □
 
 Second step: two remarkable identities 
Let us define
k[P]:=12Δω∇ωP2−∇ωP⋅∇ωΔωP−1n−1(ΔωP)2−(n−2)α2∇ωP2 andR[P]:=K[P]−(1n−(1−m))(LαP)2. We observe thatR[P]=12LαDαP2−DαP⋅DαLαP−1n(LαP)2 is independent of m . We recall the result of [[14]] and give its proof for completeness.
Let , such that , and consider a function . Then,
R[P]=α4(1−1n)[P″−P′s−ΔωPα2(n−1)s2]2+2α2s2∇ωP′−∇ωPs2+k[P]s4.
By definition of , we have
R[P]=α22[α2P′2+∇ωP2s2]″+α22n−1s[α2P′2+∇ωP2s2]′+12s2Δω[α2P′2+∇ωP2s2]−α2P′(α2P″+α2n−1sP′+ΔωPs2)′−1s2∇ωP⋅∇ω(α2P″+α2n−1sP′+ΔωPs2)−1n(α2P″+α2n−1sP′+ΔωPs2)2, which can be expanded asR[P]=α22[2α2P″2+2α2P′P‴+2∇ωP′2+∇ωP⋅∇ωP″s2−8∇ωP⋅∇ωP′s3+6∇ωP2s4]+α2n−1s[α2P′P″+∇ωP⋅∇ωP′s2−∇ωP2s3]+1s2[α2P′ΔωP′+α2∇ωP′2+Δω∇ωP22s2]−α2P′(α2P‴+α2n−1sP″−α2n−1s2P′−2ΔωPs3+ΔωP′s2)−1s2(α2∇ωP⋅∇ωP″+α2n−1s∇ωP⋅∇ωP′+∇ωP⋅∇ωΔωPs2)−1n[α4P″2+α4(n−1)2s2P′2+(ΔωP)2s4+2α4n−1sP′P″+2α2P″ΔωPs2+2α2n−1s3P′ΔωP]. Collecting terms proves the result. □
Now let us study the quantity which appears in the statement of Lemma 4.2. The following computations are adapted from [[12]] and [[14]]. For completeness, we give a simplified proof in the special case of the sphere considered as a Riemannian manifold with standard metric g . We denote by the Hessian of f , which is seen as a matrix, identify its trace with the Laplace–Beltrami operator on and use the notation for the sum of the squares of the coefficients of the matrix A . It is convenient to define the trace free Hessian , the tensor and its trace free counterpart respectively by
Lf:=Hf−1d−1(Δωf)g,Zf:=∇ωf⊗∇ωffandMf:=Zf−1d−1∇ωf2fg whenever . Elementary computations show that(22)‖Lf‖2=‖Hf‖2−1d−1(Δωf)2and‖Mf‖2=‖Zf‖2−1d−1∇ωf4f2=d−2d−1∇ωf4f2. The Bochner–Lichnerowicz–Weitzenböck formula on takes the simple form(23)12Δω(∇ωf2)=‖Hf‖2+∇ω(Δωf)⋅∇ωf+(d−2)∇ωf2 where the last term, i.e. , accounts for the Ricci curvature tensor contracted with .
We recall that and . Let us introduce the notations
δ:=1d−1−1n−1 andB[P]:=∫Sd−1(12Δω(∇ωP2)−∇ω(ΔωP)⋅∇ωP−1n−1(ΔωP)2)P1−νdω, so that∫Sd−1k[P]P1−νdω=B[P]−(n−2)α2∫Sd−1∇ωP2P1−νdω.
Assume that and . There exists a positive constant such that, for any positive function ,
∫Sd−1k[P]P1−νdω≥(n−2)(αFS2−α2)∫Sd−1∇ωP2P1−νdω+c(n,m,d)∫Sd−1∇ωP4P2P1−νdω.
If , we identify with and denote by and the first and second derivatives of with respect to θ . As in [[14]], a direct computation shows that
k[P]=n−2n−1Pθθ2−(n−2)α2Pθ2=(n−2)(αFS2Pθθ2−α2Pθ2). By the Poincaré inequality, we have∫S1∂∂θ(P1−ν2Pθ)2dθ≥∫S1P1−ν2Pθ2dθ. On the other hand, an integration by parts shows that∫S1PθθPθ2PP1−νdθ=13∫S1∂∂θ(Pθ2Pθ)P−νdθ=ν3∫S1Pθ4P2P1−νdθ and, as a consequence, by expanding the square, we obtain∫S1∂∂θ(P1−ν2Pθ)2dθ=∫S1Pθθ+1−ν2Pθ2P2P1−νdθ=∫S1Pθθ2P1−νdθ−(ν−1)(ν+3)12∫S1Pθ4P2P1−νdθ. The result follows with from∫S1Pθθ2P1−νdθ≥∫S1Pθ2P1−νdθ+(ν−1)(ν+3)12∫S1Pθ4P2P1−νdθ.
Assume next that . We follow the method of [[14]]. Applying ((23)) with and multiplying by yields, after an integration on , that can also be written as
B[P]=∫Sd−1(‖HP‖2+(d−2)∇ωP2−1n−1(ΔωP)2)P1−νdω. We recall that and set with . A straightforward computation shows that and henceB[P]=β2∫Sd−1(‖Hf+(β−1)Zf‖2+(d−2)∇ωf2−1n−1(Tr(Hf+(β−1)Zf))2)dω=β2∫Sd−1(‖Lf+(β−1)Mf‖2+(d−2)∇ωf2+δ(Tr(Hf+(β−1)Zf))2)dω. Using ((22)), we deduce from∫Sd−1Δωf∇ωf2fdω=∫Sd−1∇ωf4f2dω−2∫Sd−1Hf:Zfdω=d−1d−2∫Sd−1‖Mf‖2dω−2∫Sd−1Lf:Zfdω−2d−1∫Sd−1Δωf∇ωf2fdω that∫Sd−1Δωf∇ωf2fdω=d−1d+1[∫Sd−1d−1d−2‖Mf‖2dω−2∫Sd−1Lf:Zfdω]=d−1d+1[∫Sd−1d−1d−2‖Mf‖2dω−2∫Sd−1Lf:Mfdω] on the one hand, and from ((23)) integrated on that∫Sd−1(Δωf)2dω=d−1d−2∫Sd−1‖Lf‖2dω+(d−1)∫Sd−1∇ωf2dω on the other hand. Hence we find that∫Sd−1(Tr(Hf+(β−1)Zf))2dω=∫Sd−1((Δωf)2+2(β−1)Δωf∇ωf2f+(β−1)2∇ωf4f2)dω=d−1d−2∫Sd−1‖Lf‖2dω+(d−1)∫Sd−1∇ωf2dω+2(β−1)d−1d+1[∫Sd−1d−1d−2‖Mf‖2dω−2∫Sd−1Lf:Mfdω]+(β−1)2d−1d−2∫Sd−1‖Mf‖2dω. Altogether, we obtainB[P]=β2∫Sd−1(a‖Lf‖2+2bLf:Mf+c‖Mf‖2)dω+β2(d−2+δ(d−1))∫Sd−1∇ωf2dω, wherea=1+δd−1d−2,b=(β−1)(1−2δd−1d+1)andc=(β−1)2(1+δd−1d−2)+2(β−1)δ(d−1)2(d+1)(d−2). A tedious but elementary computation shows thatB[P]=aβ2∫Sd−1‖Lf+baMf‖2dω+(c−b2a)β2∫Sd−1‖Mf‖2dω+β2(n−2)αFS2∫Sd−1∇ωf2dω can be written in terms of asB[P]=∫Sd−1Q[P]P1−νdω+(n−2)αFS2∫Sd−1∇ωP2P1−νdω, whereQ[P]:=αFS2n−2d−2‖LP+3(ν−1)(n−d)(d+1)(n−2)(ν−3)MP‖2+(d−1)(ν−1)(n−d)[((4d−5)n+d−8)ν+9(n−d))](d−2)(d+1)2(ν−3)2(n−2)(n−1)‖MP‖2 is positive definite. This concludes the proof in the case with . □
Let us recall that
K[P]=R[P]+(1n−(1−m))(LαP)2. We can collect the two results of Lemma 4.2, Lemma 4.3 as follows.
Let , be such that , and consider a positive function . If u is related to by for some , then there exists a positive constant such that
∫RdR[P]umdμ≥α4(1−1n)∫Rd[P″−P′s−ΔωPα2(n−1)s2]2umdμ+2α2∫Rd1s2∇ωP′−∇ωPs2umdμ+(n−2)(αFS2−α2)∫Rd1s4∇ωP2umdμ+c(n,m,d)∫Rd1s4∇ωP4P2umdμ.
 
 Third step: concavity of the Rényi entropy powers and consequences 
We keep investigating the properties of the flow defined by ((15)). Let us define the entropy as
E:=∫Rdumdμ and observe thatE′=(1−m)I if u solves ((15)), after integrating by parts. The fact that boundary terms do not contribute, i.e. (24)lims→0+∫∂BsumP′dς=limS→+∞∫∂BSumP′dς=0 will be justified in Section 5: see Proposition 5.1. Note that we use ′ both for derivation w.r.t. t and w.r.t. s , at least when this does not create any ambiguity. As in Section 3, we introduce the Rényi entropy power F:=Eσ for some exponent σ to be chosen later, and find that where . With , by using Lemma 4.1, we also find that whereE2H=E2−σG′=1σ(1−m)E2−σF″=(1−m)(σ−1)(∫RduDαP2dμ)2−2∫Rdumdμ∫RdK[P]umdμ=(1−m)(σ−1)(∫RduDαP2dμ)2−2(1n−(1−m))∫Rdumdμ∫Rd(LαP)2umdμ−2∫Rdumdμ∫RdR[P]umdμ if . Using , we know that∫RduDαP2dμ=−∫RdDα(um)⋅DαPdμ=∫RdumLαPdμ and so, with the choiceσ=2n11−m−1, we may argue as in Section 3 and get thatE2H+(1−m)(σ−1)E∫RdumLαP−∫RduDαP2dμ∫Rdumdμ2dμ+2E∫RdR[P]umdμ=0 if . So, if and is of class , by Corollary 4.4, as a function of t , is concave, that is, is nonincreasing in t . Formally, converges towards a minimum, for which necessarily is a constant and , which proves that for some real constants and , according to Corollary 4.4. Since , the minimization of under the mass constraint is equivalent to the Caffarelli–Kohn–Nirenberg interpolation inequalities ((1)), since for some constant κ ,G=Eσ−1I=κ(∫Rdvp+1dμ)σ−1∫RdDαv2dμwithv=um−1/2. We emphasize that ((15)) preserves mass, that is, because, as we shall see in Proposition 5.1, no boundary terms appear when integrating by parts if v is an extremal function associated with ((7)). In particular, for mass conservation we need(25)lims→0+∫∂BsuP′dς=limS→+∞∫∂BSuP′dς=0.
The above remarks on the monotonicity of and the symmetry properties of its minimizers can in fact be extended to the analysis of the symmetry properties of all critical points of . This is actually the contents of Theorem 1.2.
Let w be a positive solution to equation ((5)). As pointed out above, by choosing
w(x)=um−1/2(rα,ω), we know that u is a critical point of under a mass constraint on , so that we can write the corresponding Euler–Lagrange equation as , for some constant C . That is, thanks to ((25)). Using as a test function amounts to apply the flow of ((15)) to with initial datum u and compute the derivative with respect to t at . This means0=∫RddG[u]⋅Lαumdμ=EσH=−(1−m)(σ−1)Eσ−1∫RdumLαP−∫RduDαP2dμ∫Rdumdμ2dμ−2Eσ−1∫RdR[P]umdμ if and ((24)) holds. Here we have used Lemma 4.1. We emphasize that this proof is purely variational and does not rely on the properties of the solutions to ((15)), although using the flow was very useful to explain our strategy. All we need is that no boundary term appears in the integrations by parts. Hence, in order to obtain a complete proof, we have to prove that ((18)), ((24)) and ((25)) hold with defined by ((17)), whenever u is a critical point of under mass constraint. This will be done in Proposition 5.1. Using Corollary 4.4, we know that , a.e. in and a.e. in , with . We conclude as in [[14]] that is an affine function of . □


Regularity and decay estimates 
In this last section, we prove the regularity and decay estimates on w (or on or u ) that are necessary to establish the absence of boundary terms in the integrations by parts of Section 4.
Under Condition ((2)), if w is a positive solution in of ((5)), then ((18)), ((24)) and ((25)) hold with as defined by ((17)), and v given by ((6)).
To prove this result, we split the proof in several steps: we will first establish a uniform bound and a decay rate for w inspired by [[17]] in Lemma 5.2, Lemma 5.3, and then follow the methodology of [[14]] in the subsequent Lemma 5.4.
Let β, γ and p satisfy the relations ((2)). Any positive solution w of ((5)) such that
(26)‖w‖L2p,γ(Rd)+‖∇w‖L2,β(Rd)+‖w‖Lp+1,γ(Rd)1−ϑ<+∞. is uniformly bounded and tends to 0 at infinity, uniformly in .
The strategy of the first part of the proof is similar to the one in [[17]], which was restricted to the case .
Let us set . For any , we multiply ((5)) by and integrate by parts (or, equivalently, plug it in the weak formulation of ((5))): we point out that the latter is indeed an admissible test function since . In that way, by letting , we obtain the identity
4(1+δ0)(2+δ0)2∫Rd∇w1+δ0/22x−βdx+∫Rdwp+1+δ0x−γdx=∫Rdw2p+δ0x−γdx. By applying ((1)) with (so that ) to the function , we deduce that‖w‖L2p+δ1,γ(Rd)2+δ0≤(2+δ0)24(1+δ0)Cβ,γ,p⋆2‖w‖L2p+δ0,γ(Rd)2p+δ0 with . Let us define the sequence by the induction relation for any , that is,δn=2p⋆−pp⋆−1(p⋆n+1−1)∀n∈N, and take . If we repeat the above estimates with replaced by and replaced by , we get‖w‖Lqn+1,γ(Rd)2+δn≤(2+δn)24(1+δn)Cβ,γ,p⋆2‖w‖Lqn,γ(Rd)qn. By iterating this estimate, we obtain the estimate‖w‖Lqn,γ(Rd)≤Cn‖w‖L2p⋆,γ(Rd)ζnwithζn:=(p⋆−1)p⋆np−1+(p⋆−p)p⋆n, where the sequence is defined by andCn+12+δn=(2+δn)24(1+δn)Cβ,γ,p⋆2Cnqn∀n∈N. The sequence converges to a finite limit . Letting we obtain the uniform bound‖w‖L∞(Rd)≤C∞‖w‖L2p⋆,γ(Rd)ζ∞≤C∞(Cβ,γ,p⋆‖∇w‖L2,β(Rd))ζ∞≤C∞(Cβ,γ,p⋆‖w‖L2p,γ(Rd)p)ζ∞, where .
In order to prove that , we can suitably adapt the above strategy. We shall do it as follows: we truncate the solution so that the truncated function is supported outside of a ball of radius and apply the iteration scheme. Up to an enlargement of the ball, that is, outside of a ball of radius for some fixed numerical constant , we get that is bounded by the energy localized in . The conclusion will hold by letting . Let us give some details.
Let be a cutoff function such that , in and in . Given , consider the sequence of radii defined by
Rn+1=(1+1n2)Rn∀n∈N. By taking logarithms, it is immediate to deduce that is monotone increasing and that there exists such thatR∞:=limn→∞Rn=aR0. Let us then define the sequence of radial cutoff functions byξn(x):=ξ2(x−RnRn+1−Rn+1)∀x∈Rd. Direct computations show that there exists some constant , which is independent of n and , such that(27)∇ξn(x)≤cn2RnχBRn+1∖BRn,∇ξn1/2(x)≤cn2RnχBRn+1∖BRn,Δξn(x)≤cn4Rn2χBRn+1∖BRn∀x∈Rd. From here on we denote by c , , etc. positive constants that are all independent of n and . We now introduce the analogue of the sequence above, which we relabel to avoid confusion. Namely, we set and , so that . If we multiply ((5)) by and integrate by parts, we obtain:∫Rd∇(ξnw1+σn)⋅∇wx−βdx+∫Rdξnwp+1+σnx−γdx=∫Rdξnw2p+σnx−γdx, whence4(1+σn)(2+σn)2∫Rdξn∇w1+σn/22x−βdx+12+σn∫Rd∇ξn⋅∇w2+σnx−βdx≤∫BRncw2p+σnx−γdx. By integrating by parts the second term in the l.h.s. and combining this estimate with∫Rd∇(ξn1/2w1+σn/2)2x−βdx≤2∫Rdξn∇w1+σn/22x−βdx+2∫Rd∇ξn1/22w2+σnx−βdx, we end up with2(1+σn)(2+σn)2∫Rd∇(ξn1/2w1+σn/2)2x−βdx−4(1+σn)(2+σn)2∫Rd∇ξn1/22w2+σnx−βdx−12+σn∫Rd(x−βΔξn−βx−β−2x⋅∇ξn)w2+σndx≤∫BRncw2p+σnx−γdx. Thanks to ((27)), we can deduce that∫Rd∇(ξn1/2w1+σn/2)2x−βdx≤∫BRn+1∖BRn(2c2+cRn2n4+βcRnn2x−1)w2+σnx−βdx+(2+σn)22(1+σn)∫BRncw2p+σnx−γdx. In particular,∫Rd∇(ξn1/2w1+σn/2)2x−βdx≤c′n4∫BRncw2+σnx−β−2dx+(2+σn)22(1+σn)‖w‖∞2p−2∫BRncw2+σnx−γdx. Since ((2)) implies that , by exploiting the explicit expression of and applying ((1)) with (and ) to the function , we can rewrite our estimate as‖w‖L2+σn+1,γ(BRn+1c)2+σn≤c″p⋆n‖w‖L2+σn,γ(BRnc)2+σn. After iterating the scheme and letting , we end up with‖w‖L∞(BR∞c)≤c‴‖w‖L2p,γ(BR0c). Since w is bounded in , in order to prove the claim, it is enough to let . □
Let β, γ and p satisfy the relations ((2)). Any positive solution w of ((5)) satisfying ((26)) is such that and there exist two positive constants, and with , such that for large enough,
C1x(γ−2−β)/(p−1)≤w(x)≤C2x(γ−2−β)/(p−1).
By Lemma 5.2 and elliptic bootstrapping methods we know that . Let us now consider the function , which satisfies the differential inequality
−div(x−β∇h)+(1−ε)x−γhp≥0∀x∈Rd∖{0} for any and C such that . On the other hand, by Lemma 5.2, is negligible compared to as and, as a consequence, for any small enough, there is an such that−div(x−β∇w)+(1−ε)x−γwp≤0ifx≥Rε. With , it follows that−div(x−β∇(h−w))+q(h−w)≥0ifx≥Rε. Hence, for C large enough, we know that for any such that , and we also have that . Using the Maximum Principle, we conclude that for any such that . The lower bound uses a similar comparison argument. Indeed, since−div(x−β∇w)+x−γwp≥0∀x∈Rd∖{0} and−div(x−β∇h)+x−γhp≤0∀x∈Rd∖{0}, if we choose C such that , we easily see thatw(x)≥(minx=1w(x)∧C)x(γ−2−β)/(p−1)∀x∈Rd∖B1. This concludes the proof. □
Our next goal is to obtain growth and decay estimates, respectively, on the functions and u as they appear in the proof of Theorem 1.2 in Section 4, in order to prove Proposition 5.1. We also need to estimate their derivatives near the origin and at infinity. Let us start by reminding the change of variables ((6)), which in particular, by Lemma 5.3, implies that for some positive constants and ,
C1s2/(1−p)≤v(s,ω)≤C2s2/(1−p)ass→+∞. Then we perform the Emden–Fowler transformation(28)v(s,ω)=saφ(z,ω)withz=−logs,a=2−n2, and see that φ satisfies the equation(29)−α2φ″−Δωφ+a2α2φ=e((n−2)p−n)zφ2p−1−e((n−2)p−n−2)z/2φp=:hinC:=R×Sd−1∋(z,ω). From here on we shall denote by ′ the derivative with respect either to z or to s , depending on the argument. By definition of φ and using Lemma 5.3, we obtain thatφ(z,ω)∼e(2−n2+2p−1)zasz→−∞, where we say that as (resp. ) if the ratio is bounded from above and from below by positive constants, independently of ω , and for z (resp. −z ) large enough. Concerning , we first note that Lemma 5.2 and ((28)) show that . The lower bound can be established by a comparison argument as in [[14]], after noticing that . Hence we obtain thatφ(z,ω)∼eaz=e2−n2zasz→+∞. Moreover, uniformly in ω , we have thath(z,ω)≤O(e−n+22z)asz→+∞,h(z,ω)∼e(−n+22+2pp−1)zasz→−∞, which in particular impliesh(z,ω)=o(φ(z,ω))asz→+∞andh(z,ω)∼φ(z,ω)asz→−∞. Finally, using [[20]] on local estimates, as we see that all first derivatives of φ converge to 0 at least with the same rate as φ . Next, [[20]] provides local estimates which, together with [[20]], up to choosing k large enough, prove that(30)φ′(z,ω),φ″(z,ω),∇ωφ(z,ω),∇ωφ′(z,ω),∇ωφ″(z,ω),Δωφ(z,ω)≤O(φ(z,ω)), uniformly in ω . Here we denote by the differentiation with respect to ω . As a consequence, we have, uniformly in ω , and for ,(31)∂zℓ∇ωth(z,ω)≤O(e−n+22z)asz→+∞,∂zℓ∇ωth(z,ω)≤O(e(−n+22+2pp−1)z)asz→−∞,(32)Δωh(z,ω)≤O(e−n+22z)asz→+∞,Δωh(z,ω)≤O(e(−n+22+2pp−1)z)asz→−∞.
Let β, γ and p satisfy the relations ((2)) and assume . For any positive solution w of ((5)) satisfying ((26)), the pressure function is such that , , , , and are of class and bounded as . On the other hand, as we have
(i)  , 
(ii)  , 
(iii)  , 
(iv)  , 
(v)  . 
By using the change of variables ((28)), we see that
P(s,ω)=p+1p−1e−12(n−2)(p−1)zφ1−p(z,ω),z=−logs. From ((30)) we easily deduce that uniformly in ω , , , , , and are of class and bounded as . Moreover, as , we obtain thatP′(s,ω)≤O(1s(φ′(z,ω)φ(z,ω)−a))and1s∇ωP(s,ω)≤O(1s(∇ωφ(z,ω)φ(z,ω))) are of order at most uniformly in ω . Similarly we obtain thatP″(s,ω)≤O(1s2(φ″(z,ω)φ(z,ω)−pφ′(z,ω)2φ(z,ω)2+(1−2a(1−p))φ′(z,ω)φ(z,ω)+a2(1−p)−a)),∇ωP′(s,ω)s−a(1−p)s2∇ωP(s,ω)≤O(1s2(∇ωφ′(z,ω)φ(z,ω)−pφ′(z,ω)∇ωφ(z,ω)φ(z,ω)2)),1s2ΔωP(s,ω)≤O(1s2(Δωφ(z,ω)φ(z,ω)−p∇ωφ(z,ω)2φ(z,ω)2)), are at most of order uniformly in ω . This shows that as and concludes the proof if . When or 3 and , more detailed estimates are needed. Properties (i)–(v) amount to prove that
(i)  , 
(ii)  , 
(iii)  , 
(iv)  , 
(v)  , 
as .

Step 1: Proof of (ii) and (iv) 
If w is a positive solution to ((5)), then φ is a positive solution to ((29)). With , applying the operator to the equation ((29)) we obtain:
−α2(∇ω∂zℓφ)″−∇ωΔω∂zℓφ+a2α2∇ω∂zℓφ=∇ω∂zℓh(z,ω)inC. Defineχℓ(z):=12∫Sd−1∇ω∂zℓφ2dω, which by ((30)) converges to 0 as . Assume first that is a positive function. After multiplying the above equation by , integrating over , integrating by parts and usingχℓ′=∫Sd−1∇ω∂zℓφ∇ω∂zℓφ′dω andχℓ″=∫Sd−1∇ω∂zℓφ∇ω∂zℓφ″dω+∫Sd−1∇ω∂zℓφ′2dω, we see that satisfies−χℓ″+∫Sd−1∇ω∂zℓφ′2dω+1α2(∫Sd−1Δω∂zℓφ2dω−λ1∫Sd−1∇ω∂zℓφ2dω)+2(a2+λ1α2)χℓ=hℓα2, with . Then, using , by the Poincaré inequality we deduce∫Sd−1Δω∂zℓφ2dω≥λ1∫Sd−1∇ω∂zℓφ2dω as e.g. in [[12]], where . A Cauchy–Schwarz inequality implies that−χℓ″+χℓ′22χℓ+2(a2+λ1α2)χℓ≤hℓα2. The function satisfies−ζℓ″+(a2+λ1α2)ζℓ≤hℓ2α2ζℓ. By the Cauchy–Schwarz inequality and ((31)) we infer that for , and for . By a simple comparison argument based on the Maximum Principle, and using the convergence of to 0 at ±∞, we infer thatζℓ(z)≤−e−νz2να2∫−∞zeνthℓ(t)ζℓ(z)dt−eνz2να2∫z∞e−νthℓ(t)ζℓ(z)dt if . This is enough to deduce that as after observing that the condition−ν=−a2+λ1/α2≤a−1 is equivalent to the inequality . Hence we have shown that if is a positive function, then for ,(33)χℓ(z)≤O(e2(a−1)z)asz→+∞. In the case where is equal to 0 at some points of , it is enough to do the above comparison argument on maximal positivity intervals of to deduce the same asymptotic estimate. Finally we observe that as , which ends the proof of (ii) considering the above estimate for when . Moreover, the same estimate for together with (ii) and ((30)) proves (iv).
By applying the operator to ((29)), we obtain
−α2(Δωφ)″−Δω2φ+a2α2Δωφ=ΔωhinC. We proceed as in Step 1. With similar notations, by definingχ3(z):=12∫Sd−1Δωφ2dω, after multiplying the equation by and using the fact that−∫Sd−1ΔωφΔω2φdω=∫Sd−1∇ωΔωφ2dω≥λ1∫Sd−1Δωφ2dω, we obtain−χ3″+χ3′22χℓ+2(a2+λ1α2)χ3≤h3α2 with . Again using the same arguments as above, together with ((32)), we deduce thatχ3(z)≤O(e2(a−1)z)asz→+∞. This ends the proof of (v), using (ii), ((30)) and noticing again that as .

Step 3: Proof of (i) and (iii) 
Let us consider a positive solution φ to ((29)) and define on the function
φ0(z):=1Sd−1∫Sd−1φ(z,ω)dω. By integrating ((29)) on , we know that solves−φ0″+a2φ0=1α2Sd−1∫Sd−1h(z,ω)dω=:h0(z)α2∀z∈R, withh0(z)≤O(e−n+22z)asz→+∞,h0(z)∼e(−n+22+2pp−1)zasz→−∞. From the integral representationφ0(z)=−eaz2aα2∫−∞ze−ath0(t)dt−e−az2aα2∫z∞eath0(t)dt, we deduce that as , andφ0′(z)−aφ0(z)φ(z,ω)∼e−2az∫z∞eath0(t)dt=O(e−2z).
If we define the function , we may observe that it is bounded for z positive and, moreover,
φ′(z,ω)φ(z,ω)−a=O(e−2z)+ψ′(z,ω)e−azφ(z,ω)asz→+∞. We recall that is bounded away from 0 by a positive constant as . Hence we know that(34)φ′(z,ω)φ(z,ω)−a≤O(ψ′(z,ω))+O(e−2z). By the Poincaré inequality and estimate ((33)) with , we have∫Sd−1ψ2dω=e−2az∫Sd−1φ−φ02dω≤e−2azλ1∫Sd−1∇ωφ2dω≤O(e−2z). Moreover, by the estimate ((33)) with , we also obtaine−2az∫Sd−1φ′−φ0′2dω≤e−2azλ1∫Sd−1∇ωφ′2dω≤O(e−2z). Hence, since , the above estimates imply that∫Sd−1ψ2dω+∫Sd−1ψ′2dω≤O(e−2z), which together with ((34)) ends the proof of (i).
To prove (iii), we first check that
φ″φ−pφ′2φ2+(1−2a(1−p))φ′φ+a2(1−p)−a=O(ψ′+ψ′2+ψ″)+O(e−2z), and so it remains to prove that is of order . Sinceψ″=a2ψ−2ae−az(φ′−φ0′)+e−az(φ″−φ0″), using the above estimates, we have only to estimate the term with the second derivatives. This can be done as above by the Poincaré inequality,e−2az∫Sd−1φ″−φ0″2dω≤e−2azλ1∫Sd−1∇ωφ″2dω≤O(e−2z), based on the estimate ((33)) with . This ends the proof of (iii). □
It is straightforward to verify that the boundedness of , , , , , as and the integral estimates (i)–(v) as from Lemma 5.4 are enough in order to establish ((18)), ((24)) and ((25)). □
This research has been partially supported by the projects STAB (ANR12BS010019) (J.D.) and Kibord (ANR13BS010004) (J.D.) of the French National Research Agency (ANR), and by the NSF grant DMS1301555 (M.L.). J.D. thanks the University of Pavia for support. M.L. thanks the Humboldt Foundation for support. M.M. has been partially funded by the National Research Project “Calculus of Variations” (PRIN 201011, Italy).