1 Introduction
There is a well-known inequality proved by Carleman [5] which states that the inequality
(1.1)∫B1e2fdx≤14π(∫S1efdθ)2
holds for any smooth harmonic function 𝑓 in two-dimensional unit disk B12¯, and the equality holds if and only if f(x)=C or -2log|x-x0|+C for some x0∈R2∖B12¯ and C∈R. From the Riemann mapping theorem, we know that any simply connected compact two-dimensional manifold with boundary (M2,g) is isometric to (B12¯,e2wdx2). Obviously, if the Gauss curvature of (M2,g) is nonpositive, then 𝑤 is a subharmonic function. It follows from the maximum principle that inequality (1.1) also holds for subharmonic functions. Hence, (1.1) can be regarded as the isoperimetric inequality for the two-dimensional manifold (M2,g) with the nonpositive Gauss curvature. Naturally, a higher-dimensional form of (1.1) for the harmonic function 𝑓,
(1.2)∫B1f2nn-2dx≤n-nn-1ωn-1n-1(∫∂B1f2(n-1)n-2ds)nn-1,
could be used to represent the corresponding isoperimetric inequality for the manifold (Mn,g) equipped with the nonpositive scalar curvature, which is isometric to (B1n¯,w4n-2dx2) (𝑤 is subharmonic because of nonpositivity of scalar curvature). Inequality (1.2) was proved by Hang, Wang and Yan in [27] through the method of the subcritical approximation. At the same time, they provided an alternative proof in [26] by proving the inequality
(1.3)∥Pf∥L2nn-2(R+n)≤n-n-22(n-1)ωn-n-22n(n-1)∥f∥L2(n-1)n-2(∂R+n),
where (Pf)(x′,xn) ((x′,xn)∈R+n)) is the convolution of f(x′) with the Poisson kernel
(1.4)P(x′,xn)=2nωnxn(|x′|2+xn2)n2,
and we can see that (1.2) is equivalent to (1.3) by the Möbius transformation between R+n and B1.
It should be noted that the Poisson kernel P(x′,xn) appearing in inequality (1.3) is a fundamental solution of the Laplacian operator on half space R+n, that is to say, P(x′,xn) satisfies the equation
{-ΔP(x′,xn)=0,(x′,xn)∈R+n,P(x′,0)=δ0(x′),(x′,0)∈∂R+n.
From Caffarelli and Silvestre’s works in [4], we know that the outer normal derivatives of the Poisson integral (Pf)(x) along the boundary ∂R+n could be used to define (-Δ)12f in Rn-1 as long as 𝑓 is sufficiently smooth. In order to define general fractional Laplacian operator (-Δ)α2 for α∈(0,2), Caffarelli and Silvestre introduced an extension operator 𝐿 on the half space R+n by giving the operator L=Δ+1-αxn. It is not difficult to verify that the fundamental solution for the equation
{LP(x′,xn))=0,x=(x′,xn)∈R+n,P(x′,0)=δ0(x′),(x′,0)∈∂R+n,
is
(1.5)Pb(x′,xn)=C1xn1-b(|x′|2+xn2)n-b2,
where b=1-α. Then the fractional Laplacian (-Δ)α2f in Rn-1 could be defined as
(-Δ)α2f(x′)=limxn→0-xn1-α∂Pb(f)∂xn=αlimxn→0Pb(f)(x′,xn)-Pb(f)(x′,0)xnα,
where
Pb(f)(x′,xn)=∫Rn-1Pb(x′-y,xn)f(y)dy.
We may naturally ask whether there exists an analogue of inequality (1.3) for Pb(f). Chen [12] gave a positive answer and proved the result
(1.6)∥Pbf∥L2nn-2+b(R+n)≤Cn,b∥f∥L2(n-1)n-2+b(∂R+n)
for b<1. Dou, Guo and Zhu [15] discussed the above conformal invariant index inequality with the kernel
Pθ,γ(x′.xn)=xn(|x′|2+xn2)γ2.
For this kernel, Chen, Lu and Tao [10] established the general Lp→Lq′ boundedness
(1.7)∫R+n∫∂R+nf(x)xn(|x′-y|2+xn2)γ2g(y)dydx≤Cn,γ,p,q′∥f∥Lq′(R+n)∥g∥Lp(∂R+n),
where p,q′>1, and 2<γ≤n. By using the result in [10], Chen, Liu, Lu and Tao [8] extended it to the doubly weighted form. Recently, the above conformal invariant index inequalities have also been extended to a more general kernel in [23], which takes the form
(1.8)Pθ,γ(x′.xn)=xnθ(|x′|2+xn2)γ2
with θ>0, γ>0.
Inequalities (1.2), (1.6) and (1.7) could be also seen as a new type of HLS inequality [28, 33]. The classical HLS inequality is originally the convolution in Rn of f(x) with the Riesz kernel 1|x|γ. Many geometric inequalities, such as the Gross logarithmic Sobolev inequality [24] and the Moser–Onofri–Beckner inequality [1] can be implied by the HLS inequality. In the 1950s, Stein and Weiss [34] established the doubly weighted HLS inequality, namely, the Stein–Weiss inequality
(1.9)∫Rn∫Rn|x|-α|x-y|-γf(x)g(y)|y|-βdxdy≤Cn,α,β,p,q′∥f∥Lq′(Rn)∥g∥Lp(Rn),
where
p,q′>1,α+β≥0,α<nq,β<np′,0<γ<n.
Lieb [31] utilized the symmetric rearrangement argument to prove the existence of the extremals of the HLS inequalities. He also classified the extremal functions in the conformal index p=q′=2n2n-γ. Furthermore, Chen, Li and Ou [13] and Li [30] classified all the positive solutions of the Euler–Lagrange system related with the HLS inequality in the conformal case through the method of moving plane and the method of moving sphere respectively. Carlen and Loss [6] and Frank and Lieb [20, 21] also gave the best constant and an accurate expression of extremal functions of the HLS inequality in the conformal index without using a rearrangement argument. For more results about the HLS inequalities on the Heisenberg group and the upper half space, one can see [9, 14, 17, 22, 19, 25].
For inequality (1.9), what will happen if p,q′∈(0,1)? Beckner [2] and Dou and Zhu [16] found that the exponent 𝛾 of the kernel becomes negative and the inequalities turn into reversed inequalities. In fact, they established the reverse type of the HLS inequality and derived the existence of extremal functions in the diagonal case p=q′=2n2n-γ with γ<0. Subsequently, Chen, Liu, Lu and Tao [7] established the reverse Stein–Weiss inequality
∫Rn∫Rn|x|α|x-y|γf(x)g(y)|y|βdxdy≥Cn,α,β,p,q′∥f∥Lq′(Rn)∥g∥Lp(Rn),
where p,q′∈(0,1), γ>0, 0≤α<-nq and 0≤β<-np′. Motivated by the above works, thus a natural questions arises: for the Poisson-type kernel such as (1.4), (1.5) and (1.8), what will happen to the inequalities when 0<p,q′<1? We find that the inequalities could become reversed if the exponents 𝜃 and 𝛾 appearing in the kernel (1.8) are negative. To be specific, we derive the following general results.
Theorem 1.1
For n>1, 0<p,q′<1, γ,θ>0 and either β<1-np′ or α<-nq+θ satisfying
(1.10)n-1np+1q′-α+β+γ-1-θn=2,
there exists some constant Cn,γ,θ,α,β,p,q′>0 such that, for any nonnegative function f∈Lq′(R+n), g∈Lp(∂R+n),
(1.11)∫R+n∫∂R+n|x|αf(x)P(x,γ,θ,y)g(y)|y|βdydx≥Cn,γ,θ,α,β,p,q′∥f∥Lq′(R+n)∥g∥Lp(∂R+n),
where
P(x,γ,θ,y)=|x-y|γxnθ,1p+1p′=1,1q+1q′=1.
Remark 1.2
Inequality (1.11) is called the reverse Stein–Weiss inequality with the Poisson-type kernel. When θ=0, the following reverse Stein–Weiss inequality on the upper half space was studied by Chen, Lu and Tao in [11]:
∫R+n∫∂R+n|x|αf(x)|x-y|γg(y)|y|βdydx≥Cn,γ,α,β,p,q′∥f∥Lq′(R+n)∥g∥Lp(∂R+n).
When θ=α=β=0, it becomes the reverse HLS inequality on the upper half space studied by Ngô [32].
If (f,g)∈Cc(R+n)×Cc(∂R+n), the left integral of (1.11) is finite, which implies the sharp constant Cn,γ,θ,α,β,p′,q′ is finite. A natural question is whether the sharp constant Cn,γ,θ,α,β,p′,q′ can be attained. To do this, we can consider the minimizing problem
(1.12)Cn,γ,θ,α,β,p,q′:=inf{∥Vγ,θ(g)∥Lq(R+n):g≥0, 0∥g∥Lp(∂R+n)=1},
where the double weighted operator Vγ(g)(x) is given by
Vγ,θ(g)(x)=∫∂R+n|x|αP(x,γ,θ,y)g(y)|y|βdy.
Then we can prove that the constant Cn,γ,θ,α,β,p,q′ could actually be achieved.
Theorem 1.3
For n>1, p,q′∈(0,1), γ>0, 0≤α<-n-1q+θ and 0≤β<1-np′ satisfying
n-1np+1q′-α+β+γ-1-θn=2,
there exists some nonnegative function g∈Lp(∂R+n) satisfying ∥g∥Lp(∂R+n)=1 and ∥Vγ(g)∥Lq(R+n)=Cn,α,β,p,q′.
Obviously, the extremals of reverse inequality (1.11) satisfy the following Euler–Lagrange system up to a constant:
(1.13){J(f,g)f(x)q′-1=∫∂R+n|x|αP(x,γ,θ,y)g(y)|y|βdy,x∈R+n,J(f,g)g(y)p-1=∫R+n|y|βP(x,γ,θ,y)f(x)|x|αdx,y∈∂R+n.
Let u=c1fq′-1, v=c2gp-1, 1q′-1=-p1 and -p2=1p-1, and pick two suitable constants c1 and c2. Then system (1.13) is simplified as
(1.14){u(x)=∫∂R+n|x|αP(x,γ,θ,y)v-p2(y)|y|βdy,x∈R+n,v(y)=∫R+n|y|βP(x,γ,θ,y)u-p1(x)|x|αdx,y∈∂R+n,
where 1p1-1+1p2-1n-1n=α+β+γ-θn.
We also give some asymptotic estimates of positive solutions to system (1.14) around the origin and near infinity.
Theorem 1.4
Suppose that (u,v) is a pair of positive Lebesgue measurable solutions of (1.14). Thenlim|x|→∞u(x)xnθ|x|γ+α=∫∂R+nv-p2(y)|y|βdy,lim|y|→∞v(y)|y|γ+β=∫R+nu-p1(x)|x|αxn-θdx,
lim|x|→0u(x)xnθ|x|α=∫∂R+nv-p2(y)|y|β+γdy,lim|y|→0v(y)|y|β=∫R+nu-p1(x)|x|α+γxn-θdx.
When α=β=0, (1.11) becomes the reverse HLS inequality with the Poisson-type kernel
(1.15)∫R+n∫∂R+nf(x)|x-y|γg(y)xnθdydx≥Cn,γ,θ,p,q′∥f∥Lq′(R+n)∥g∥Lp(∂R+n),
where 0<p,q′<1, γ>0, θ>0 satisfying
n-1np+1q′-γ-1-θn=2.
Accordingly, system (1.14) becomes
(1.16){u(x)=∫∂R+nP(x,γ,θ,y)v-p2(y)dy,x∈R+n,v(y)=∫R+nP(x,γ,θ,y)u-p1(x)dx,y∈∂R+n.
Then we have the following regularity result.
Theorem 1.5
Assume that (u,v) is a pair of positive solutions of integral system (1.16). Then
(u,v)∈C1(R+n)×C1(∂R+n).
From Theorem 1.3, we can deduce that if
(1.17)np1-1+n-1p2-1=γ-θ,
then system (1.16) must have positive solutions. A natural question arises, whether identity (1.17) is a necessary condition for the existence of positive solutions of system (1.16). We give a positive answer in the following theorem.
Theorem 1.6
For 𝛾, p1, p2>0, there exists a pair of positive solutions (u,v) satisfying (1.16) if and only if
np1-1+n-1p2-1=γ-θ.
When q′=2n2n+γ-2θ, p=2(n-1)2n+γ-2, using the method of moving spheres in integral forms (see [30]), we can classify the positive solutions for system (1.16). Then we can derive the accurate forms of extremals for inequality (1.15) in the conformal index.
Theorem 1.7
For γ>2θ, q′=2n2n+γ-2θ, p=2(n-1)2n+γ-2, there exists some constant Cn,γ,θ>0 such that, for any nonnegative functions f∈Lq′(R+n), g∈Lp(∂R+n), there holds
(1.18)∫R+n∫∂R+nf(x)P(x,γ,θ,y)g(y)dydx≥Cn,γ,θ∥f∥Lq′(R+n)∥g∥Lp(∂R+n),
where
Cn,γ,θ=2θ(nwn)-2n+γ-22(n-1)(∫Bn(∫∂Bn|ξ-ζ|γ(1-|ξ-zo|2)θdζ)-2nγ-2θdξ)-γ-2θ2n,z0=(0,-1)∈Rn-1×R.
The equality holds if and only if
g(y)=c(1|y-y0|2+d2)2n+γ-22 and f(x)=C(∫∂R+n|x-y|γxnθ(1|y-y0|2+d2)2n+γ-22dy)2n+γ-2θ2θ-γ
for some c,d,C>0 and y0∈∂R+n.
For x0=(0,-2), z0=(0,-1)∈Rn-1×R, consider the conformal transformation Tx0:ξ∈Bn→ξx0∈R+n given by
Tx0(ξ)=ξx0=4(ξ-x0)|ξ-x0|2+x0∈R+n.
Via easy computation, one can note that Tx0-1(x)=xx0=4(x-x0)|x-x0|2+x0∈Bn, which has the same expression with Tx0. Through this conformal transformation, we can easily show that inequality (1.18) is equivalent to the following sharp reversed integral inequality on the ball Bn with Cn,γ,θ=2θC~n,γ,θ.
Corollary 1.8
For γ>2θ, q′=2n2n+γ-2θ, p=2(n-1)2n+γ-2, the following sharp inequality holds for any nonnegative functions f∈Lq′(Bn), g∈Lp(∂Bn):
(1.19)∫Bn∫∂Bnf(ξ)|ξ-ζ|γg(ζ)(1-|ξ-zo|2)θdζdξ≥C~n,γ,θ∥f∥Lq′(Bn)∥g∥Lp(∂Bn),
where
C~n,γ,θ=(nwn)-2n+γ-22(n-1)(∫Bn(∫∂Bn|ξ-ζ|γ(1-|ξ-zo|2)θdζ)-2nγ-2θdξ)-γ-2θ2n,z0=(0,-1)∈Rn-1×R.
Remark 1.9
In the proof of Theorem 1.7, we classify the extremal functions of inequality (1.18). This together with the conformal transformation implies that g≡1 could be the extremal function of inequality (1.19). Then the sharp constant C~n,γ,θ and Cn,γ,θ can be easily computed.
3 The Proof of Theorem 1.3
In this section, we shall use the method in spirit of Lieb’s works in [31] to establish the existence of extremals of reverse inequality (1.11) involved with the Poisson-type. Let A⊂Rn be a Borel set of finite Lebesgue measure. The symmetric rearrangement A* of the set 𝐴 is the open ball centered at the origin whose volume is that of 𝐴. The decreasing rearrangement of 𝑓 is the function f*:A*→R defined by
f*(x)=inf{α:|{y:|f(y)|≥α}|≤vn|x|n},
where vn is the volume of the unit ball in Rn.
Through [3, 7, 11, 32], we derive the following lemma.
Lemma 3.1
The following statements hold.
If g(y) is radially symmetric, then we have that Vγ,θ(g)(x′,xn) is radially symmetric with respect to x′, where x=(x′,xn)∈R+n.
For any nonnegative function g∈Lp(∂R+n), we have ∥Vγ,θ(g)∥Lq(R+n)≥∥Vγ,θ(g*)∥|Lq(R+n).
We also need the following lemma.
Lemma 3.2
Under the same hypothesis as Theorem 1.3, if we furthermore assume that g∈Lp(∂R+n) is nonnegative, radially symmetric with g(r)≤εr-n-1p, then, for any 0<t<p, there holds
∥Vγ,θ(g)∥Lq(R+n)≥Cε1-pt∥g∥Lp(∂R+n)pt.
Proof
Define G:R→R by G(v)=ev(n-1)pg(ev) and H:R×R+→R by H(u,xn)=eunqVγ,θ(g)(eu,euxn). We can easily verify that(3.1)wn-21p∥G∥Lp(R)=∥g∥Lp(∂R+n),∥G∥L∞(R)≤ε,
(3.2)wn-21q∥H∥Lq(R+2)=∥Vγ,θ(g)∥Lq(R+n).
Through calculation, we can get
H(u,xn)=eunq∫∂R+n|e2u+e2uxn2|α2(|eu→-y|2+e2uxn2)γ2(euxn)θg(y)|y|βdy=eu(nq+α)∫∂R+n|1+xn2|α2(e2u-2eu→⋅y+|y|2+e2uxn2)γ2(euxn)θg(y)|y|βdy=eu(nq+α)∫-∞+∞∫Sn-2|1+xn2|α2(e2u-2eu→⋅ev→+e2v+e2uxn2)γ2(euxn)θg(ev)ev(n-1+β)dξdv=eu(nq+α+γ2)∫-∞+∞∫Sn-2|1+xn2|α2(eu-v(1+xn2)-21→⋅ξ+ev-u)γ2(euxn)θg(ev)ev(n-1+γ2+β)dξdv,
where eu→ denotes the vector sitting on ∂R+n with length eu, so do ev→ and 1→. It follows from
nq+n-1+α+β+γ-θ=n-1p
that 𝐻 can be written as
H(u,xn)=∫-∞+∞Ln,α,β(u-v,xn)G(v)dv,
where Ln,α,β(u,xn) is given by
Ln,α,β(u,xn)={eu(nq+α+γ2-θ)|1+xn2|α2xn-θ((eu(1+xn2)-2+e-u)γ2+(eu(1+xn2)+2+e-u)γ2),n=2,eu(nq+α+γ2-θ)∫Sn-2|1+xn2|α2xn-θ(eu(1+xn2)-21→⋅ξ+e-u)γ2dξ,n≥3.
We can check that
Ln,α,β(u,xn)∼eu(nq+α+γ2-θ)|1+xn2|α2xn-θ(eu(1+xn2)+e-u)γ2
as u2+xn2→+∞. From the assumption of Theorem 1.3, we derive that, for any s<0,
(3.3)∫R(∫0∞Lq(u,xn)dxn)sqdu<+∞.
For any 0<t<p, we can choose s1<0 such that 1t+1s1=1+1q. Through the reverse Young inequality, we obtain
∫R|H(u,xn)|qdu≤(∫R|L(u,xn)|s1du)qs1(∫R|G|tdu)qt.
Taking the integral with respect to the variable xn over [0,+∞) for both sides, by the Minkowski inequality, (3.3) and (3.1), we have
∫R+2|H(u,xn)|qdudxn≤(∫R|G|tdu)qt∫0∞(∫RLs1(u,xn)du)qs1dxn≤(∫R|G|tdu)qt(∫R(∫0∞Lq(u,xn)dxn)s1qdu)qs1≲∥G∥L∞(R)q(1-pt)(∫R|G|pdu)qt≲εq(1-pt)(∫∂R+n|g|pdy)qt.
Through (3.2), we derive that
∥Vγ,θ(g)∥Lq(R+n)≥Cε1-pt∥g∥Lp(∂R+n)pt.
Then the proof of Lemma 3.2 is completed. ∎
Now we turn to prove Theorem 1.3.
Proof
According to Lemma 3.1, we can deduce that if 𝑔 is a minimizer for problem (1.12), 𝑔 must be radially decreasing. Likewise, if {gj}j is a minimizing sequence satisfying
limj→∞∥Vγ,θ(gj)∥Lq(R+n)=Cn,γ,θ,α,β,p.q′
with ∥gj∥Lp(∂R+n)=1, without loss of generality, we can assume that {gj}j is nonnegative, radially symmetric and monotonously decreasing.
For any R>0, we obtain
wn-2n-1gjp(R)Rn-1≤ωn-2∫0Rgjp(r)rn-2dr≤ωn-2∫0+∞gjp(r)rn-2dr=∫∂R+ngjp(x)dx=1,
which implies that there exists some constant 𝐶 independent of 𝑗 such that, for any R>0,
(3.4)0≤gj(R)≤CR-n-1p.
Set aj=supr>0rn-1pgj(r). According to Lemma 3.2, we can deduce that aj≥2c0 for some c0>0. Select γj>0 satisfying γjn-1pgj(γj)>c0, and let g~j(y)=γjn-1pgj(γjy). Then ∥g~j∥Lp=∥gj∥Lp=1, ∥Vγ,θg~j∥Lq=∥Vγ,θgj∥Lq. Replacing the sequence {gj(y)}j with the new sequence {g~j(y)}j, still denoted by {gj(y)}j, we obtain that gj(1)≥c0 for any 𝑗.
Lieb’s argument based on the Helly theorem allow us to choose a subsequence such that gj→g a.e. in ∂R+n. It is obvious that 𝑔 is nonnegative and radially decreasing.
We are left to prove that
∥Vγ,θ(g)∥Lq(R+n)=Cn,γ,θ,α,β,p.q′ and ∥g∥Lp(∂R+n)=1.
According to the definition of Vγ,θ(gj)(x), for all x∈R+n and j∈N*, we have
Vγ,θ(gj)(x)≥c0|x|α∫|y|≤1|x-y|γxnθ|y|βdy≜F(x),
where
F(x)∼{|x|αxn-θif|x|≤1,|x|γ+αxn-θif|x|>1.
For each x∈R+n, set k(x)=lim infj→∞(Vγ,θgj(x)). Fatou’s lemma gives that
(3.5)∫R+nFq(x)-kq(x)dx=∫R+nlim infj→∞[Fq(x)-(Vγ,θgj)q(x)]≤lim infj→∞∫R+nFq(x)-(Vγ,θgj)q(x)dx=∫R+nFq(x)dx-(Cn,γ,θ,α,β,p,q′)q.
Since 1q=n-1n⋅1p-α+β+γ-1-θn-1, β<1-np′, α<-n-1q+θ, we have ∫R+nFq(x)dx<∞. Combining this with (3.5), we derive that
(3.6)∫R+nFq(x)dx≥∫R+nkq(x)dx≥(Cn,α,β,p,q′)q,
which implies that the set {x∈R+n:0<k(x)<+∞} has positive measure. Hence, we can take a point x1∈R+n and extract a subsequence of Vγ,θgj, still denoted by Vγ,θgj, satisfying limj→∞Vγ,θgj(x1)=a1∈(0,+∞). We can employ the reverse Hölder inequality to obtain
∫∂R+n|x1|αP(x1,γ,θ,y)gj(y)|y|βdy≥Cα,γ∫|y|≤|x1|2gj(y)|y|βdy≥Cα,γ(∫|y|≤|x1|2gjτ(y)dy)1τ(∫|y|≤|x1|2|y|τ′βdy)1τ′,
where 0<τ<1 and 1τ+1τ′=1. Since 0<β<1-np′, we can choose 𝜏 satisfying τ>p and τ′β>1-n such that the integral ∫|y|≤|x1|2|y|τ′βdy is finite. Therefore,
(3.7)∫|y|≤|x1|2gjτ(y)dy≲1.
This together with (3.4) yields that, for any sufficiently large 𝑅, there holds
∫|y|≤Rgjτ(y)dy=∫|y|≤|x1|2gjτ(y)dy+∫|x1|2≤|y|≤Rgjτ(y)dy≤C(R).
On the other hand, for sufficiently large 𝑅, we also have
∫∂R+nP(x,γ,θ,y)gj(y)|y|βdy≳Rγ∫34R≤|y|≤Rgj(y)|y|βdy≳Rγgj(R)∫34R≤|y|≤R|y|βdy≳Rγ+n-1+βgj(R).
Thus, we obtain gj(R)≲R1-n-γ-β. Because of 0<α<-n-1q+θ, we have
(3.8)limR→∞limj→∞∫|y|≥Rgjp(y)dy=0.
Combining (3.7) and (3.8), we derive that
limj→∞∫∂R+ngjp(y)dy=limR→∞limj→∞∫|y|≤Rgjp(y)dy+limR→∞limj→∞∫|y|≥Rgjp(y)dy=limR→∞limj→∞∫|y|≤Rgjp(y)dy=limR→∞∫|y|≤Rgp(y)dy=∫∂R+ngp(y)dy,
which implies that ∥g∥Lp(∂R+n)=1. By Fatou’s lemma, we have
(3.9)(∫R+nkq(x)(x)dx)1q=(∫R+nlim infj→∞(Vγ,θgj)q(x)dx)1q≥(∫R+n(Vγ,θg)q(x)dx)1q.
Gathering (3.6) and (3.9), we conclude that
Cn,γ,θ,α,β,p,q′≥(∫R+nkq(x)(x)dx)1q≥(∫R+n(Vγ,θg)q(x)dx)1q≥Cn,γ,θ,α,β,p,q′,
which implies that ∥Vγ,θg∥Lq(R+n)=Cn,γ,θ,α,β,p,q′. Thus, we complete the proof of Theorem 1.3. ∎
5 The Proof of Theorem 1.5
In this section, we will give the regular estimate for positive solutions of equations (1.16). We shall prove that if (u,v) is a pair of positive solutions of system (1.16), then (u,v)∈C1(R+n)×C1(∂R+n).
We just show u(x)∈C1(R+n). For x0∈R+n, we take δ=12dist{x0,∂R+n}.
Step 1. We first prove that u(x) is continuous at x0. It is easy to obtain that, for x∈Bδ(x0), there exists a constant C(x0,γ,δ,θ) such that P(x,γ,θ,y)v-p2(y)≤(1+|y|γ)v-p2(y). From Lemma 4.1, we know that∫∂R+n(1+|y|γ)v-p2(y)dy<∞,
u(x)=∫∂R+nP(x,γ,θ,y)v-p2(y)dy≤C(x0,γ,δ,θ)∫∂R+n(1+|y|γ)v-p2(y)dy.
By the Lebesgue dominated convergence theorem, we have
limx→x0u(x)=∫∂R+nlimx→x0P(x,γ,θ,y)v-p2(y)dy=∫∂R+nP(x0,γ,θ,y)v-p2(y)dy.
Thus, we get that u(x) is continuous at x0.
Step 2. We prove that, when γ>1, u(x)∈C1(R+n). For x∈Bδ(x0), we have|∂P(x,γ,θ,y)∂xi|=|γ|x-y|γ-2(xi-yi)xnθ|≤γ|x-y|γ-1δθ≤C(x0,γ,δ,θ)(1+|y|γ-1),i=1,…,n-1,
|∂P(x,γ,θ,y)∂xn|=|γ|x-y|γ-2xnxnθ-θ|x-y|γxnθ+1|≤C(x0,γ,δ,θ)(1+|y|γ-1)+C(x0,γ,δ,θ)(1+|y|γ).
From Lemma 4.1, we have
∫∂R+n(1+|y|γ-1)v-p2(y)dy=∫|y|>1(1+|y|γ-1)v-p2(y)dy+∫|y|<1(1+|y|γ-1)v-p2(y)dy≤∫|y|>1(1+|y|γ)v-p2(y)dy+2∫|y|<1v-p2(y)dy<∞.
Likewise, By the Lebesgue dominated convergence theorem, for i=1,…,n, ∂u(x)∂xi|x=x0 exists and ∂u(x)∂xi|x=x0 is continuous at x0.
Step 3. We prove that, when 0<γ<1, for i=1,…,n-1, ∂u(x)∂xi|x=x0 exists and ∂u(x)∂xi is continuous at x0. According to the definition of derivative, for i-1,…,n-1, we have
∂u(x)∂xi|x=x0=limh→0u(x0+hei)-u(x0)h=limh→0∫∂R+n|x0+hei-y|γ-|x0-y|γ(x0)nθhv-p2(y)dy,
where ei denotes the unit vector in the direction of xi. For 0<γ<1, A,B>0, we have the inequality
(5.1)|Aγ-Bγ|≤|B|γ-1|A-B|,
which implies
||x0+hei-y|γ-|x0-y|γ(x0)nθh|⋅v-p2(y)≤|x0-y|γ-1||x0+hei-y|-|x0-y||(x0)nθ|h|v-p2(y)≤|x0-y|γ-1|hei|(x0)nθ|h|v-p2(y)=|x0-y|γ-1(x0)nθv-p2(y).
From Lemma 4.1, we know that
∫∂R+n|x0-y|γ-1(x0)nθv-p2(y)dy=∫|x0-y|<1|x0-y|γ-1(x0)nθv-p2(y)dy+∫|x0-y|>1|x0-y|γ-1(x0)nθv-p2(y)dy≤∫|x0-y|<1|x0-y|γ-1(x0)nθdy+∫|x0-y|>1v-p2(y)(x0)nθdy<∞.
By the Lebesgue dominated convergence theorem, we can get
∂u(x)∂xi|x=x0=∫∂R+nlimh→0|x0+hei-y|γ-|x0-y|γ(x0)nθhv-p2(y)dy=∫∂R+nγ|x0-y|γ-2((x0)i-yi)(x0)nθv-p2(y)dy.
So u(x) is differentiable for xi at x0, i=1,…,n-1.
Set
uiϵ(x)=∫∂R+nγ|x-y|γ-2(xi-yi)xnθv-p2(y)η(x-yϵ)dy,
where
η(t)={0if|t|≤1,0∼1if 1<|t|<2,1if|t|≥2.
We claim that uiϵ(x) is continuous. In fact, for fixed x0,
limx→x0uiϵ(x)=limx→x0∫∂R+nγ|x-y|γ-2(xi-yi)xnθv-p2(y)η(x-yϵ)dy=limx→x0∫|x-y|>ϵγ|x-y|γ-2(xi-yi)xnθv-p2(y)η(x-yϵ)dy.
For x∈Bδ(x0), we have
||x-y|γ-2|xi-yi|xnθv-p2(y)η(x-yϵ)|≤ϵγ-1δθ(1+|y|)v-p2(y).
From Lemma 4.1, we have
∫∂R+n(1+|y|)v-p2(y)<∞.
From the Lebesgue dominated convergence theorem, we deduce that uiϵ(x) is continuous.
On the other hand, we can prove that uiϵ(x) converges uniformly to ∂u(x)∂xi on Bδ(x0) as ϵ→0. In fact, for all x∈Bδ(x0), we can write
|∂u∂xi(x)-uiϵ(x)|=∫|x-y|≤2ϵγ|x-y|γ-2(xi-yi)xnθv-p2(y)(1-η(x-yϵ))dy≤∫|x-y|≤2ϵγ|x-y|γ-2|xi-yi|xnθv-p2(y)dy≤∫|x-y|≤2ϵγ|x-y|γ-1δθv-p2(y)dy≤∫|x-y|≤2ϵγ|x-y|γ-1δθC(1+|y|γ)-p2dy≤∫|x-y|≤2ϵγ|x-y|γ-1δθCdy=C⋅γ(2ϵ)γ+n-2δθ.
So uiϵ(x) converges uniformly to ∂u(x)∂xi on Bδ(x0) as ϵ→0. From the fact that uiϵ(x) is continuous, we can deduce that ∂u(x)∂xi is continuous in Bδ(x0). Thus, ∂u(x)∂xi is continuous at x0 for i=1,…,n-1.
Step 4. We prove that, when 0<γ≤1, ∂u(x)∂xn|x=x0 exists and ∂u(x)∂xn is continuous at x0. From the definition of derivative, we get
∂u(x)∂xn=limh→0u(x+hen)-u(x)h=limh→0∫∂R+n(|x+hen-y|γ(xn+h)θ-|x-y|γxnθ)v-p2(y)dyh=limh→0∫∂R+n|x+hen-y|γ-|x-y|γ(xn+h)θhv-p2(y)dy-limh→0∫∂R+n(xn+h)θ|x-y|γ-|x-y|γxnθ(xn+h)θxnθhv-p2(y)dy=h1(x)-h2(x).
We first discuss h1(x). For fixed x0 and |h|<12(x0)n, through (5.1), we have
|
|
x
0
+
h
e
n
-
y
|
γ
-
|
x
0
-
y
|
γ
(
(
x
0
)
n
+
h
)
θ
h
|
v
-
p
2
(
y
)
≤
|
x
0
-
y
|
γ
-
1
⋅
|
|
x
0
+
h
e
n
-
y
|
-
|
x
0
-
y
|
|
|
(
x
0
)
n
+
h
|
θ
|
h
|
v
-
p
2
(
y
)
≤
2
θ
|
x
0
-
y
|
γ
-
1
(
x
0
)
n
θ
v
-
p
2
(
y
)
Using the similar arguments of step 3, we can get
h1(x0)=∫∂R+nlimh→0|x0+hen-y|γ-|x0-y|γ((x0)n+h)θhv-p2(y)dy=γ∫∂R+n|x0-y|γ-2(x0)n(x0)nθv-p2(y)dy
and that h1(x) is continuous at x0.
Now we discuss h2(x). For x∈Bδ(x0) and some t∈(0,1), we have
|(xn+h)θ|x-y|γ-|x-y|γxnθ(xn+h)θxnθh|v-p2(y)=|θ(xn+th)θ-1|x-y|γ(xn+h)θxnθh|v-p2(y)≤C(x0,γ,δ,θ)(1+|y|γ)v-p2(y).
By Lemma 4.1 and the Lebesgue dominated convergence theorem, we derive
h2(x0)=∫∂R+nlimh→0|x0+hen-y|γ-|x0-y|γ((x0)n+h)θhv-p2(y)dy=γ∫∂R+n|x-y|γ-2(x0)n(x0)nθv-p2(y)dy
and that h2(x) is continuous at x0. Thus, we can get that, when 0<γ≤1, u(x) is differentiable for xn at x0 and ∂u(x)xn is continuous at x0.
7 The Proof of Theorem 1.7
In this section, we will give the accurate forms of extremals for inequality (1.11) in the conformal index q′=2n2n+γ-2θ, p=2(n-1)2n+γ-2. Obviously, the extremals (f,g) of inequality (1.15) in the conformal index under the suitable transformation u=c1fq′-1, v=c2gp-1 satisfies the conformal integral equation
(7.1){u(x)=∫∂R+nP(x,γ,θ,y)v-2n+γ-2γ(y)dy,x∈R+n,v(y)=∫R+nP(x,γ,θ,y)u-2n+γ-2θγ-2θ(x)dx,y∈∂R+n.
Therefore, in order to give the accurate forms of extremals (f,g), it suffices to prove the following proposition.
Proposition 7.1
Assume that (u,v) is a pair of positive solutions of the conformal integral equation (7.1). Then v(y) must take the form as
v(y)=(d2+|y-y0|2c)γ2,y∈∂R+n,
for some y0∈∂R+n, c>0 and d>0.
Proof
We divide the proof into five steps. We first give some notations. Let 𝑢 and 𝑣 be positive functions defined on R+n×∂R+n. For y0∈∂R+n and λ>0, we define
uλ,y0(x)=(|x-y0|λ)γ-2θu(xλ,y0),vλ,y0(y)=(|y-y0|λ)γv(yλ,y0),
where (x,y)∈R+n×∂R+n and
xλ,y0=y0+λ2(x-y0)|x-y0|2,yλ,y0=y0+λ2(y-y0)|y-y0|2.
For R>0, defineBR(y0)={x∈Rn:|x-y0|<R},
BRn-1(y0)={y∈∂R+n:|y-y0|<R},
BR+(y0)={x=(x1,x2,…,xn)∈BR(y0):xn>0}.
For y0=0, we write BR=BR(0), BRn-1=BRn-1(0), BR+=BR+(0).
Step 1. Prove that, for any y0∈∂R+n, there exists λ0(y0)>0 such that, for any λ∈(0,λ0), there holds
uλ,y0(x)≥u(x),vλ,y0(y)≥v(y)
for x∈R+n∖Bλ+(y0), y∈∂R+n∖Bλn-1(y0).
From Theorem 1.5, we know that (u,v)∈C1(R+n)×C1(∂R+n). Then it follows that there exists s0>0 such that
∇y(|y-y0|-γ2v(y))⋅(y-y0)<0 for ally∈Bs0n-1(y0),
which implies that
vλ,y0(y)>v(y) for ally∈Bs0n-1(y0)∖Bλn-1(y0).
On the other hand, a direction application of Lemma 4.1 gives that, for small λ0∈(0,s0) and 0<λ<λ0, there holds
(7.2)vλ,y0(y)=(|y-y0|λ)γv(λ2(y-y0)|y-y0|2)≥(|y-y0|λ0)γinfBs0n-1(y0)v≥v(y) for ally∈∂Rn∖Bs0n-1(y0).
This together with (7.2) yields vλ,y0(y)≥v(y) for y∈∂R+n∖Bλn-1(y0). Similarly, we can also derive that uλ,y0(x)≥u(x) for x∈R+n∖Bλ+(y0).
Step 2. For each y0∈∂R+n, we could define
λ¯(y0)=sup{μ:uλ,y0(x)≥u(x),vλ,y0(y)≥v(y)for all 0<λ<μ,x∈R+n∖Bλ+(y0),y∈∂R+n∖Bλn-1(y0)}.
It follows from step 1 that λ¯(y0) is well defined. Now we prove that, for any y0∈∂R+n, λ¯(y0)<+∞.
In fact, it follows from the definition of λ¯(y0) that, for any 0<λ<λ¯(y0), there holds
(7.3)uλ,y0(x)≥u(x),vλ,y0(y)≥v(y) for allx∈R+n∖Bλ+(y0),y∈∂R+n∖Bλn-1(y0),
which together with (7.3) and Theorem 1.4 implies that
0<∫∂R+nv-2n+γ-2γ(y)dy=lim|x|→∞|x|-γu(x)xnθ≤lim|x|→∞|x|-γ(|x-y0|λ)γu(xλ,y0)(xλ,y0)nθ=λ-γlimx→(y0,0)u(x)xnθ=λ-γ∫∂R+nv-2n+γ-2γ(y)|y-y0|γdy.
This gives λ¯(y0)<∞.
Step 3. From the definition of λ¯, we see that there are the following four cases:
uλ¯,y0(x)=u(x),vλ¯,y0(y)=v(y),
uλ¯,y0(x)>u(x),vλ¯,y0(y)=v(y),
uλ¯,y0(x)=u(x),vλ¯,y0(y)>v(y),
uλ¯,y0(x)>u(x),vλ¯,y0(y)>v(y).
Now we prove that case (ii) and case (iii) are impossible.
Let (u,v) be a pair of positive Lebesgue measurable solutions for integral system (7.1). Through the identity
|x|x|2-y|y|2|=|x-y||x|⋅|y|,
we have
|xλ,y0-yλ,y0|=λ2|x-y||x-y0|⋅|y-y0| and |x-yλ,y0|=|(xλ,y0)λ,y0-yλ,y0|=|x-y0|⋅|xλ,y0-y||y-y0|.
Then, for x∈R+n, y∈∂R+n, straight calculations give that(7.4)uλ,y0(x)-u(x)=∫∂R+n∖Bλn-1(y0)K(y0,λ,θ,x,y)(v-2n+γ-2γ(y)-vλ,y0-2n+γ-2γ(y))dy,
(7.5)vλ,y0(y)-v(y)=∫R+n∖Bλ+(y0)K(y0,λ,θ,y,x)(u-2n+γ-2θγ-2θ(x)-uλ,y0-2n+γ-2θγ-2θ(x))dx,
whereK(y0,λ,θ,x,y)=(|x-y0|λ)γ|xλ,y0-y|γxnθ-|x-y|γxnθ,
K(y0,λ,θ,y,x)=(|y-y0|λ)γ|yλ,y0-x|γxnθ-|x-y|γxnθ.
Furthermore, we also show that K(y0,λ,θ,x,y),K(y0,λ,θ,y,x)>0 when |x-y0|,|y-y0|>λ. In fact, we may assume y0=0 and get
K(0,λ,θ,x,y)=(|x|λ)γ|xλ,0-y|γ1xnθ-|x-y|γ1xnθ=(λ2-2x⋅y+|x|2|y|2λ2)γ21xnθ-(|x|2-2x⋅y+|y|2)γ21xnθ=((|x|2-2x⋅y+|y|2+(|x|2-λ2)(|y|2λ2-1))γ2-(|x|2-2x⋅y+|y|2)γ2)1xnθ>0
if |x|,|y|>λ. Similarly, we can also derive that K(0,λ,θ,x,y)>0 for |x|,|y|>λ. Then (ii) and (iii) being impossible is a direct result of (7.4) and (7.5).
Step 4. Now we start to prove that case (iv) is also impossible. For convenience, we may assume y0=0 and use notations λ¯=λ¯(0), uλ=uλ,0 and xλ=xλ,0, yλ=yλ,0. We carry out the argument by contradiction. Suppose that case (iv) is valid. We will show that there exists an ε>0 such that
uλ(x)≥u(x),vλ(y)≥v(y) for allλ¯<λ<λ¯+εandx∈∂R+n∖Bλn-1,y∈R+n∖Bλ+.
By (7.4) and (7.5), with y0=0, λ=λ¯ and the Fatou lemma, one can calculatelim|x|→∞inf|x|-γ(uλ¯(x)-u(x))=lim|x|→∞inf|x|-γ∫∂R+n∖Bλ¯n-1K(0,λ¯,θ,x,y)(v-2n+γ-2γ-vλ¯-2n+γ-2γ)dy,
lim|y|→∞inf|y|-γ(vλ¯(y)-v(y))=lim|y|→∞inf|y|-γ∫R+n∖Bλ¯+K(0,λ¯,θ,y,x)(u-2n+γ-2θγ-2θ-uλ¯-2n+γ-2θγ-2θ)dx.
Then it follows thatlim|x|→∞inf|x|-γxnθ(uλ¯(x)-u(x))≥∫∂R+n∖Bλ¯n-1((|y|λ¯)γ-1)(v-2n+γ-2γ-vλ¯-2n+γ-2γ)dy>0,
lim|y|→∞inf|y|-γ(vλ¯(y)-v(y))≥∫R+n∖Bλ¯+((|x|λ¯)γ-1)1xnθ(u-2n+γ-2θγ-2θ-uλ-2n+γ-2θγ-2θ)dx>0.
Then, for any x∈R+n∖Bλ¯+1+ and y∈∂R+n∖Bλ¯+1n-1, there exists ε1∈(0,1) such that
uλ¯(x)-u(x)≥ε1|x|γxn-θ,vλ¯(y)-v(y)≥ε1|y|γ.
We can use continuity of 𝑢 and 𝑣 with respect to variable 𝜆 to obtainuλ(x)-u(x)≥ε1|x|γxnθ+(uλ(x)-uλ¯(x))≥ε12|x|γxnθ for allx∈R+n∖Bλ¯+1+, λ¯≤λ≤λ¯+ε2,
(7.6)vλ(y)-v(y)≥ε1|y|γ+(vλ(y)-vλ¯(y))≥ε12|y|γ for ally∈R+n∖Bλ¯+1+, λ¯≤λ≤λ¯+ε2,
for sufficiently small ε2. Thus, it suffices to verify that, for all λ¯<λ<λ¯+ϵ,uλ(x)≥u(x)forx∈Bλ¯+1+∖Bλ+,
vλ(y)≥v(y)fory∈Bλ¯+1n-1∖Bλn-1.
We only show that uλ(x)≥u(x) for all λ¯<λ<λ¯+ϵ. By step 3, we can write
(7.7)uλ(x)-u(x)=∫∂R+n∖Bλn-1K(0,λ,θ,x,y)(v-2n+γ-2γ(y)-vλ,0-2n+γ-2γ(y))dy≥∫Bλ¯+1n-1∖Bλn-1K(0,λ,θ,x,y)(v-2n+γ-2γ(y)-vλ,0-2n+γ-2γ(y))dy+∫Bλ¯+3n-1∖Bλ¯+2n-1K(0,λ,θ,x,y)(v-2n+γ-2γ(y)-vλ,0-2n+γ-2γ(y))dy.
One can employ (7.6) to derive that
(7.8)vλ(y)-v(y)≥ε2|y|γ≥ε1(λ¯+2)γ
for y∈Bλ¯+3n-1∖Bλ¯+2n-1. Obviously, there exists δ1>0 dependent of λ¯ such that
v-2n+γ-2γ(y)-vλ,0-2n+γ-2γ(y)>v-2n+γ-2γ(y)-(v+ε1(λ¯+2)γ)-2n+γ-2γ≥δ1.
For any x∈Bλ¯+1+∖Bλ+ and y∈Bλ¯+3n-1∖Bλ¯+2n-1, there exist δ~2, δ2>0 such that
(7.9)K(0,λ,θ,x,y)=xn-θ((|x|λ)γ|λ2x|x|2-y|γ-|y-x|γ)=xn-θ((|x|λ)2|λ2x|x|2-y|2)γ2-xn-θ(|y-x|2)γ2=δ~2xn-θ((|x|λ)2|λ2x|x|2-x|2-|y-x|2)≥δ2xn-θ(|x|-λ).
For any λ¯≤λ≤λ¯+ε and y∈Bλ¯+1n-1∖Bλn-1, direct calculations yield
vλ(y)-vλ¯(y)=(|y|λ)γv(λ2y|y|2)-(|y|λ¯)γv(λ¯2y|y|2)=|y|γv(λ2y|y|2)λγλ¯γ(λ¯γ-λγ)+(|y|λ¯)γ∇v(y)⋅(λ2y|y|2-λ¯2y|y|2),
which gives
(7.10)|vλ¯-2n+γ-2γ(y)-vλ-2n+γ-2γ(y)|≤C1ε,
where y∈(λ2y|y|2,λ¯2y|y|2).
For x∈Bλ¯+1+∖Bλ+, we obtain
(7.11)∫Bλ¯+1n-1∖Bλn-1K(0,λ,θ,x,y)dy=∫Bλ¯+1n-1∖Bλn-1xn-θ(|x|λ)γ|λ2x|x|2-y|γ-xn-θ|x-y|γdy≤Cxn-θ(|x|-λ)+∫Bλ¯+1n-1∖Bλn-1xn-θ|λ2x|x|2-y|γ-xn-θ|x-y|γdy≤C2xn-θ(|x|-λ).
By (7.7), (7.8), (7.9), (7.10) and (7.11), for sufficiently small ε>0, we conclude that
uλ(x)-u(x)≥xn-θ(δ1δ2∫Bλ¯+3n-1∖Bλ¯+2n-1dy-C1C2ε)(|x|-λ)≥0,
which leads to a contradiction with the definition of λ¯. This proves that case (iv) is also impossible.
Step 5. From the above analysis, we know that only (i) can happen. By a calculus lemma [29, Lemma 11.1], we know that any C1 positive solution 𝑣 must take the form
v(y)=(d2+|y-y0|2c)γ2,y∈∂R+n,
for some y0∈∂R+n, c>0 and d>0. Subsequently, the extremal function 𝑔 of inequality (1.18) must be of the form
g(y)=c(1|y-y0|2+d2)2n+γ-22
for some c>0, d>0 and y0∈∂R+n. This accomplishes the proof of Theorem 1.7. ∎
