An Integral Equation in Conformal Geometry

www.elsevier.com/locate/anihpc

An integral equation in conformal geometry

Fengbo Hang

, Xiaodong Wang

, Xiaodong Yan

aDepartment of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544, USA bDepartment of Mathematics, Michigan State University, East Lansing, MI 48824, USA

Received 20 January 2007; accepted 27 March 2007 Available online 31 August 2007

Abstract

Motivated by Carleman’s proof of the isoperimetric inequality in the plane, we study the problem of finding a metric with zero scalar curvature maximizing the isoperimetric ratio among all zero scalar curvature metrics in a fixed conformal class on a compact manifold with boundary. We derive a criterion for the existence and make a related conjecture.

©2007 Elsevier Masson SAS. All rights reserved.

Keywords:Isoperimetric inequalities; Poisson kernel; Yamabe type integral equations

1. Introduction

Among the many proofs of two dimensional isoperimetric inequalities, the one due to Carleman [3] is particularly interesting. Indeed by an application of Riemann mapping theorem we only need to show

D

e^2udx 1 4π

S¹

e^udθ 2

(1.1) for every harmonic functionuonD. HereDis the unit disk in the plane. Carleman deduced (1.1) by showing

D

|f|²dx 1 4π

S¹

|f|dθ 2

for every holomorphic functionf onD. Along this line, in [8] Jacobs showed that for every bounded open subsetΩ ofR²with smooth boundary, there exists a positive constantcΩ such that for every holomorphic functionf onΩ,

Ω

|f|²dxcΩ

∂Ω

|f|ds 2

.

Moreover whenΩis not simply connected, the best constantc_Ω>_4π¹ and it is achieved by some particular holomor- phic functionf. Here we formulate a higher dimensional generalization of these statements.

* Corresponding author.

E-mail addresses:fhang@math.princeton.edu (F. Hang), xwang@math.msu.edu (X. Wang), xiayan@math.msu.edu (X. Yan).

0294-1449/$ – see front matter ©2007 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2007.03.006

Assumen3,(Mⁿ, g)is a smooth compact Riemannian manifold with nonempty boundaryΣ=∂M, we write the isoperimetric ratio

I (M, g)= |M|¹ⁿ

|Σ|ⁿ⁻¹¹. (1.2)

Here|M|is the volume ofMwith respect togand|Σ|is the area ofΣ. Let[g] = {ρ²g: ρ∈C^∞(M), ρ >0}be the conformal class ofg. The set

g˜∈ [g]: the scalar curvatureR=0

is nonempty if and only if the first eigenvalue of the conformal Laplacian operatorL_g= −⁴⁽ⁿ_n−⁻₂¹⁾+Rwith respect to Dirichlet boundary condition,λ1(Lg)is strictly positive (see Section 2).

Assumeλ1(Lg) >0, we denote Θ_M,g=sup

I (M,g):˜ g˜∈ [g]withR=0

. (1.3)

Standard technique from harmonic analysis gives us ΘM,g<∞ (see Proposition 2.1). But is ΘM,g achieved? In another word, can we find a conformal metric with zero scalar curvature maximizing the isoperimetric ratio?

It follows from [7, Theorem 1.1] or Theorem 3.1 that Θ_B

1,g_Rn=I (B1, g_Rⁿ)=n⁻ⁿ⁻¹¹ω⁻

n(n−1)1

n ,

hereω_n is the volume of the unit ball inRⁿ andg_Rⁿ is the Euclidean metric onRⁿ. This just says thatΘ_B

1,g_Rn is achieved by the standard metric. In general we have the following

Theorem 1.1. Assume n3, (Mⁿ, g) is a smooth compact Riemannian manifold with nonempty boundary and λ₁(L_g) >0, then

n⁻ⁿ⁻¹¹ ω⁻

1 n(n−1)

n =Θ_B

1,g_RnΘ_M,g<∞. If in additionΘ_B

1,g_Rn< Θ_M,g, thenΘ_M,gis achieved by some conformal metrics with zero scalar curvature.

The problem illustrates very similar behavior as the Yamabe problem of finding constant scalar curvature metrics in a fixed conformal class (cf. [9]) and its boundary versions (cf. [4,5]). On the other hand, it has more nonlocal features (e.g. the Euler–Lagrange equation is a nonlinear integral equation) than the two well studied problems. In analogy with the solution of the Yamabe problem, we make the following conjecture.

Conjecture 1.1.Assumen3,(Mⁿ, g)is a smooth compact Riemannian manifold with nonempty boundary and λ₁(L_g) >0. If(M, g)is not conformally diffeomorphic to(B₁, g_Rⁿ), thenΘ_M,g> Θ_B

1,g_Rn.

In Section 2 below, we will describe some basics related to the above problem and reformulate it as a maximization problem for harmonic extensions. We will also discuss some elementary estimates of the Poisson kernels and show Θ_M,g is always finite. In Section 3 we will showΘ_B

1,g_Rn is achieved by the standard metric itself and deduce some corollaries. This is a consequence of [7, Theorem 1.1]. However the approach we present here is different and of independent interest. In Section 4 we will prove the regularity of the solutions to the Euler–Lagrange equations of the maximization problem for harmonic extensions. In Section 5 we derive some asymptotic expansion formulas for the standard Poisson kernel and the Poisson kernel for the conformal Laplacian operators. These expansion formulas will be useful in the future study of Conjecture 1.1. In Section 6, we will derive the concentration compactness principle for the maximization problem and this will be used in the last section to deduce Theorem 1.1.

2. Some preparations

Assumen3,(Mⁿ, g)is a smooth compact Riemannian manifold with boundaryΣ=∂M. The conformal Lapla- cian operator is given by

L_g= −4(n−1) n−2 +R.

It satisfies the transformation law L

ρ

n−42gϕ=ρ^−nn+−22Lg(ρϕ) forρ, ϕ∈C^∞(M), ρ >0.

Let

Eg(ϕ, ψ )=

M

4(n−1)

n−2 ∇ϕ· ∇ψ+Rϕψ dμ, Eg(ϕ)=Eg(ϕ, ϕ),

heredμis the measure generated byg, then it follows from the transformation law that E

ρⁿ−⁴2g(ϕ)=E_g(ρϕ) forρ, ϕ∈C^∞(M), ρ >0, ϕ|Σ=0. (2.1) Letλ₁(L_g)be the first eigenvalue ofL_gwith respect to the Dirichlet boundary condition, then

λ₁(L_g)= inf

ϕ∈H₀¹(M)\{0}

E_g(ϕ)

Mϕ²dμg

.

Assume ρ∈C^∞(M), ρ >0. It follows from (2.1) that λ1(Lg) <0 implies λ1(L

ρ

4

n−2g) <0. On the other hand, if λ₁(L_g)0, thenλ₁(L

ρⁿ−⁴2g)(maxMρ)⁻ⁿ−⁴2λ₁(L_g). Hence the sign of the first eigenvalue of the conformal Laplacian operator does not depend on the choice of particular metric in a conformal class. This sign is useful because of the following fact:λ1(Lg) >0 if and only if we may find a scalar flat metric in the conformal class ofg. The only thing we need to verify is we can find a scalar flat conformal metric whenλ1(Lg) >0. To see this we may solve the Dirichlet problem

L_gρ=0 onM, ρ|Σ=1.

We claimρ >0 onM. To see this, we letϕbe the first eigenfunction ofL_gwithϕ >0 onM\Σ andϕ|Σ =0. Let w=^ρ_ϕ, then

−4(n−1)

n−2 w−8(n−1) n−2

∇ϕ

ϕ · ∇w+λ₁w=0 onM\Σ.

Sincew(x)→ ∞as x→Σ, it follows from strong maximum principle thatw >0 on M\Σ, henceρ >0 on M.

Note thatR

ρn−2⁴ g=ρ^−nn−+22L_gρ=0, we find the needed metric.

Assumeλ1(Lg) >0, the Green’s functionGLofLgsatisfies (L_g)_xG_L(x, y)=δ_y onM,

GL(x, y)=0 forx∈Σ.

The Poisson kernel ofL_gis given by PL(x, ξ )= −4(n−1)

n−2

∂G_L(x, y)

∂_yν

y=ξ

,

hereνis the unit outer normal direction. The solution of L_gu=0 onM,

u|Σ=f is given by

u(x)=(P_Lf )(x)=

Σ

P_L(x, ξ )f (ξ ) dS(ξ ),

heredSis the measure generated bygonΣ. Ifρ is a positive smooth function, then we have the following transfor- mation laws,

G

L,ρⁿ−⁴2g(x, y)=G_L,g(x, y) ρ(x)ρ(y) , P

L,ρⁿ−⁴2g(x, ξ )= P_L,g(x, ξ ) ρ(x)ρ(ξ )ⁿ⁻²ⁿ

,

and P

L,ρ

4 n−2g

f =ρ⁻¹P_L,g(ρf ).

Ifg˜∈ [g]has zero scalar curvature, theng˜=uⁿ⁻²⁴ gfor some positive smooth functionu onM withL_gu=0. Let f =u|Σ, thenu=P_Lf and

I (M,g)˜ =

|PLf|ⁿ⁻²²

L

2n n−2(M)

|f|ⁿ⁻²²

L

2(n−1) n−2 (Σ )

.

Hence

Θ_M,g=sup

|P_Lf|ⁿ⁻²²

Lⁿ²ⁿ−2(M)

|f|ⁿ⁻²²

L

2(n−1) n−2 (Σ )

: f∈C^∞(Σ ), f >0

=sup

|P_Lf|ⁿ⁻²²

Lⁿ²ⁿ−2(M)

|f|ⁿ⁻²²

L

2(n−1) n−2 (Σ )

: f∈L^{2(nn−−21)}(Σ ), f =0

= sup

|PLf|

L

n−22n(M): f ∈L^{2(nn−−21)}(Σ ),|f|

L^2(n−1)n−2 (Σ )=1 ²

n−2

. (2.2)

The second equality above follows from the factP_Lis positive and an approximation procedure.

It follows easily from the definition ofΘ_M,g(see (1.3)) thatΘ_M,gdepends only on[g]. As a consequence we may choose the background metricgwith zero scalar curvature. Under this assumption the conformal Laplacian operator reduces to the constant multiple of the Laplacian operator. To continue we will need some estimates of the Poisson kernels.

2.1. Basic estimates for Poisson kernel and harmonic extensions

Let us fix some notations. Throughout this subsection, we always assume n2, (Mⁿ, g)is a smooth compact Riemannian manifold with boundary Σ=∂M. For convenience we fix a smooth compact Riemannian manifold without boundary,(Mⁿ, g)such that(M, g)is a smooth domain in(M, g). Denoted as the distance onMgenerated by g andd_Σ as the distance onΣ (when Σ is not connected andξ₁, ξ₂∈Σ lie in different components, we set d_Σ(ξ₁, ξ₂)equal to the maximal diameter of all the components ofΣ). We writet =t (x)=d(x, Σ ) for x ∈M.

Assumeδ0>0 is small enough such thatV = {x∈M: t (x) <2δ0}is a tubular neighborhood ofΣ and forξ, ζ∈Σ withd(ξ, ζ ) <2δ0, we havedΣ(ξ, ζ )2d(ξ, ζ ). Forx∈V, letπ(x)∈Σbe the unique nearest point onΣtox. For δ >0, we writeMδ= {x∈M:t (x)δ}. Forx∈M,δ >0, we useBδ(x)to denote the ball with center atx, radiusδ in(M, g).

The Green’s function of the Laplace operator satisfies −_xG(x, y)=δ_y onM,

G(x, y)=0 forx∈Σ.

Note thatG(x₁, x₂)=G(x₂, x₁)forx₁, x₂∈M.

• The solution of

−u=h onM, u|Σ=0

is given by u(x)=

M

G(x, y)h(y) dμ(y).

• The solution of

−u=0 onM, u|Σ=f

is given by u(x)= −

Σ

∂G(x, y)

∂yν

y=ξ

f (ξ ) dS(ξ ).

Hereνis the unit outer normal direction onΣ. In particular the Poisson kernel is given by P (x, ξ )= −∂G(x, y)

∂_yν

y=ξ

.

• If

−u=h onM, u|Σ=0,

then ^∂u_∂ν(ξ )= −

MP (x, ξ )h(x) dμ(x). In the future we will denote (T h)(ξ )=

M

P (x, ξ )h(x) dμ(x).

Hence ^∂u_∂ν = −T h.

Forf defined onΣ, we write (Pf )(x)=

Σ

P (x, ξ )f (ξ ) dS(ξ ).

Pf is the harmonic extension off.

Lemma 2.1.For0δ < δ₀, denoteΣ_δ= {x∈M: d(x, Σ )=δ}. Ifu∈C^∞(M)is a nonnegative harmonic function, then

Σ_δ

u dSc(M, g)

Σ

u dS.

Proof. Denoteνas the unit outer normal direction. Sinceδ₀is small, for 0δ < δ₀, the mapψ_δ:Σ→Σ_δgiven by ψ_δ(ξ )=exp_ξ(−δν(ξ ))is a diffeomorphism and

Σδu dS=

Σu◦ψ_δ·J_ψδdS. Hence d

dδ

Σ_δ

u dS=

Σ_δ

∂u

∂t dS+

Σ

u◦ψ_δ·dJ_ψδ dδ dS

c(M, g)

Σ

u◦ψ_δ·J_ψδdS

=c(M, g)

Σδ

u dS.

Here we have used the equation

Σδ

∂u

∂t dS=0 which follows from the divergence theorem and the factuis harmonic.

It follows that

Σ_δu dSc(M, g)

Σu dS. 2

To avoid confusion we emphasize that the constantsc(M, g)’s are different in different formulas. This convention applies throughout the article. We will need the following classical estimate for Poisson kernels.

Lemma 2.2.The Poisson kernelP (x, ξ )satisfies 0P (x, ξ )c(M, g) t (x)

[t (x)²+dΣ(π(x), ξ )²]ⁿ² forx∈M_δ0 andξ∈Σ.

Proof. It follows from Lemma 2.1 and an approximation procedure that for 0< δδ₀,

Mδ

P (x, ξ ) dμ(x)c(M, g)δ.

SinceP (x, ξ )is nonnegative, harmonic inx andP (x, ξ )=0 forx∈Σ\{ξ}, it follows from the elliptic estimates of harmonic function that we only need to consider the caset (x)+d_Σ(π(x), ξ )is small. Lett (x)+d_Σ(π(x), ξ )=δ. If t (x)^δ₇, by mean value inequality

P (x, ξ )c(M, g) δⁿ

B_δ

7

(x)

P (y, ξ ) dμ(y)c(M, g) δⁿ⁻¹

c(M, g) t (x)

[t (x)²+d_Σ(π(x), ξ )²]ⁿ².

Assumet (x) <^δ₇, thend(π(x), ξ ) >^3δ₇. By the gradient estimate of harmonic functions we know ∇P (·, ξ )

L∞(B_2δ/7(π(x))∩M)c(M, g) δⁿ⁺¹

B3δ/7(π(x))∩M

P (y, ξ ) dμ(y)c(M, g) δⁿ ,

henceP (x, ξ )c(M, g)^{t (x)}_δn . The lemma follows. 2

As an application of Lemma 2.2 we may derive the following inequality for harmonic extensions. Recall ifXis a measure space,p >0 anduis a measurable function onX, then

|u|L^p_W(X)=sup

t >0

t|u|> t^1p.

Here||u|> t|is the measure of the set{|u|> t}.

Proposition 2.1.The harmonic extension operatorP satisfies

|Pf|

L

n n−1

W (M)c(M, g)|f|L¹(Σ )

and

|Pf|

L

np

n−1(M)c(M, g, p)|f|L^p(Σ )

for1< p∞.

Proof. We only need to prove the weak type estimate. The strong estimate follows from Marcinkiewicz interpolation theorem [11, p. 197] and the basic fact |Pf|L^∞(M)|f|L^∞(Σ ). To prove the weak type estimate we may assume f 0 and|f|L¹(Σ )=1. It follows from Lemma 2.2 that

0(Pf )(x)c(M, g) t (x)ⁿ⁻¹.

Forδ₀=δ₀(M, g) >0 small, it follows from Lemma 2.1 that

M_δ

P (x, ξ ) dμ(x)c(M, g)δ forξ∈Σand 0< δ < δ0. Hence forδ∈(0, δ0),

Mδ

(Pf )(x) dμ(x)=

Σ

dS(ξ )

f (ξ )

Mδ

P (x, ξ ) dμ(x)

c(M, g)δ.

Forλc(M, g), we have

|Pf > λ| =x∈M: t (x) < c(M, g)λ⁻ⁿ⁻¹¹ , (Pf )(x) > λ

1

λ

M

c(M,g)λ− 1 n−1

(Pf )(x) dμ(x)

c(M, g)λ⁻ⁿⁿ−1. The proposition follows. 2

For 1< p <∞, if we write c_M,g,p=sup

|Pf|

L

np

n−1(M): f ∈L^p(Σ ),|f|L^p(Σ )=1

, (2.3)

thenc_M,g,p<∞. In view of (2.2), when the background metricghas zero scalar curvature, Θ_M,g=c

2 n−2

M,g,²⁽ⁿ_n−⁻₂¹⁾ <∞. (2.4)

In the future we will also need the following compactness property.

Corollary 2.1.For1p <∞,1q <_n^np₋₁, the operatorP:L^p(Σ )→L^q(M)is compact.

Proof. First assume 1< p <∞. Iff_i∈L^p(Σ )such that|f_i|L^p(Σ )1, it follows from Lemma 2.2 that (Pf_i)(x)c(M, g)

t (x)ⁿ⁻¹ forx∈M\Σ.

Using elliptic estimates of harmonic functions we know after passing to a subsequence we may find au∈C^∞(M\Σ ) such thatPfi→uinC_loc^∞(M\Σ ). Forδ >0 small, we have

|Pf_i−Pf_j|L^q(M)|Pf_i−Pf_j|L^q(M\Mδ)+ |Pf_i−Pf_j|L^q(Mδ)

|Pf_i−Pf_j|L^q(M\Mδ)+ |Pf_i−Pf_j|

L

np

n−1(M_δ)|M_δ|^1q−nnp−1 |Pf_i−Pf_j|L^q(M\Mδ)+c(M, g, p)|M_δ|^1q−nnp−1.

Hence lim sup

i,j→∞|Pfi−Pfj|L^q(M)c(M, g, p)|Mδ|^1q−n−1np .

Lettingδ→0⁺, we seePf_i is a Cauchy sequence inL^q(M). In another word,P:L^p(Σ )→L^q(M)is compact.

Whenp=1, the argument is similar. We only need to observe that for any 1q <q <˜ _n−ⁿ₁,P:L¹(Σ )→L^q˜(M) is bounded. 2

Let h be a function on M, recall (T h)(ξ )=

MP (x, ξ )h(x) dμ(x). We have the following dual statement to Proposition 2.1.

Proposition 2.2.For1p < nandh∈L^p(M),

|T h|

L

(n−1)p

n−p (Σ )c(M, g, p)|h|L^p(M).

Proof. We may prove the inequality by a duality argument. Indeed for any nonnegative functionshonMandfonΣ, we have

0

Σ

(T h)(ξ )f (ξ ) dS(ξ )=

Σ

dS(ξ )

M

P (x, ξ )h(x)f (ξ ) dμ(x)

=

M

(Pf )(x)h(x) dμ(x)|Pf|

L

p

p−1(M)|h|L^p(M)

c(M, g, p)|h|L^p(M)|f|

L

(n−1)p n(p−1)(Σ )

,

the proposition follows. One may also prove the inequality directly. Indeed it follows from Lemma 2.2 that

|P (·, ξ )|

Ln−1ⁿ ,∞(M)c(M, g) <∞forξ∈Σ. HenceT:L^n,1(M)→L^∞(Σ )is a bounded linear map. On the other hand forh∈L¹(M),

Σ

(T h)(ξ )dS(ξ )

Σ

dS(ξ )

M

P (x, ξ )h(x)dμ(x)=

M

h(x)dμ(x).

HenceT:L¹(M)→L¹(Σ )is also bounded. The proposition follows from the Marcinkiewicz interpolation theorem.

Finally we point out for 1< p < n, we may solve −u=h onM,

u|Σ=0

and(T h)(ξ )= −^∂u_∂ν(ξ ). By theL^ptheory we know|u|W^2,p(M)c(M, g, p)|h|L^p(M). It follows from boundary trace embedding theorem [1, p. 164] that

|T h|

L

(n−1)p n−p (Σ )=

∂u

∂ν

L^(n−1)pn−p (Σ )

c(M, g, p)|u|_W^2,p_(M)c(M, g, p)|h|L^p(M). 2

2.2. Miscellaneous

Later on we will need the following Hausdorff–Young type inequality to estimate some nonmajor terms.

Lemma 2.3.LetXandY be measure spaces,1p, q₀, q₁, r∞,pr,q₀rand 1

p+ 1 q₁ = q₀

q₁r +1.

AssumeKis defined onX×Y such that

X

K(x, y)^q0dx ¹

q0

A,

Y

K(x, y)^q1dy ¹

q1

A.

For a functionf defined onY, we let(Kf )(x)=

YK(x, y)f (y) dy, then

|Kf|L^r(X)A|f|L^p(Y ).

Proof. Without losing of generality we may assumeK0 andf 0, then

(Kf )(x)=

Y

K(x, y)

q0

r f (y)^prK(x, y)

r−q0

r f (y)^r−rpdy

Y

K(x, y)^q0f (y)^pdy ¹_r

Y

K(x, y)^q1dy ^r−_q^q0

1r

Y

f (y)^pdy ^r_pr^−p

A^r−qr0|f|_L^r−p^rp(Y ) Y

K(x, y)^q0f (y)^pdy ¹

r

.

Here we have used the Holder’s inequality and the fact¹_r +_q1r/(r¹_−q0)+_pr/(r¹_−p)=1. Hence (Kf )(x)^rA^r−q0|f|^r_L⁻p^p(Y )

Y

K(x, y)^q0f (y)^pdy.

Integrating both sides, we get the needed inequality. 2 3. Sharp inequalities on the unit ball

The aim of this section is to showΘ_B

1,g_Rn =c

2 n−2

B₁,g_Rn,²⁽ⁿ_n−⁻₂¹⁾ =I (B₁, g_Rⁿ)=n⁻ⁿ−¹1ω⁻

1 n(n−1)

n (see (2.3), (2.4)).

Theorem 3.1.Assumen3, then for everyf ∈L^{2(nn−−21)}(∂B₁ⁿ),

|Pf|

L

2n

n−2(B1)n^{−2(nn−2−1)}ω⁻

n−2 2n(n−1)

n |f|

L

2(n−1) n−2 (∂B1)

.

Here Pf is the harmonic extension of f, ωn is the volume of the unit ball in Rⁿ. Equality holds if and only if f (ξ )=c(1+λξ·ζ )⁻ⁿ⁻²² for some constantc,ζ ∈∂B1and0λ <1.

Note that this theorem is a consequence of [7, Theorem 1.1] (see the discussions before [7, Theorem 1.1]). Below we will present a different argument which has its own interest. Before discussing the approach, we describe some corollaries of the theorem. Note that in Proposition 2.1 the strong inequality is not true forp=1. Instead we have the following

Corollary 3.1.Assumen3, then forf ∈L^∞(∂B₁ⁿ), e^Pf

Ln−1ⁿ (B₁ⁿ)n⁻¹ω⁻

1

nne^f

L¹(∂B₁).

Moreover equality holds if and only iff is constant.

Proof. Ifuis a harmonic function, thene^u=e^u|∇u|². Hence e^uis subharmonic and not harmonic except whenu is a constant function. It follows from Theorem 3.1 that

e^{2(n−1)n−2 Pf}

Ln−2²ⁿ (B₁)P

e^{2(n−1)n−2 f}

Ln−2²ⁿ (B₁)

n^{−2(nn−2−1)}ω⁻

2n(nn−2−1)

n e^{2(nn−2−1)f}

L

2(n−1) n−2 (∂B₁)

.

Hence e^Pf

Lⁿ−ⁿ1(B₁)n⁻¹ω⁻

1

nne^f

L¹(∂B₁).

If equality holds, then e^{2(n−1)n−2 Pf} =P (e^{2(n−1)n−2 f})and e^{2(n−1)n−2 Pf} must be a harmonic function, hence Pf is equal to constant and so isf. 2

Corollary 3.2.Assumen3, then for ²⁽ⁿ_n−⁻₂¹⁾< p <∞,f∈L^p(∂B₁ⁿ),

|Pf|

L

np

n−1(B1)n^−1pω⁻

1 np

n |f|L^p(∂B₁). Equality holds if and only iff is constant.

Proof. Denoter=_2(n−1)/(n^p ₋₂₎ >1. Ifuis a harmonic function onB₁, then|u|^r is a subharmonic function and it is not harmonic except whenuis a constant function. Iff ∈L^p(∂B1), then by Theorem 3.1,

|Pf|^r

L

2n

n−2(B1)P

|f|^r

L

2n

n−2(B1)n^{−2(nn−2−1)}ω⁻

n−2 2n(n−1)

n |f|^r

L

2(n−1) n−2 (∂B1)

.

Hence

|Pf|

L

np

n−1(B₁)n^−1pω⁻

1 np

n |f|L^p(∂B1).

If equality holds then|Pf|^r=P (|f|^r). In particular|Pf|^ris a harmonic function and hencePfis a constant function, so isf. 2

Remark 3.1.For 1< p <²⁽ⁿ_n−⁻₂¹⁾, 1 is still a critical point for the functional

|Pf|

L np n−1(B1)

|f|Lp (∂B1) , but calculation shows for f_ε(ξ )=1+εξ₁,

|Pf_ε|

L

n−1np(B₁)

|f_ε|L^p(∂B₁) =n^−p1ω⁻

1 np

n

1+ n−2

2n(n−1)(n+2)

2(n−1) n−2 −p

ε²+O

ε⁴ .

Hence 1 is not a local maximizer. It remains an interesting question to calculate sup

|Pf|

L

np

n−1(B₁): f∈L^p(∂B₁),|f|L^p(∂B1)=1 for thesep’s.

The new approach to Theorem 3.1 needs an interesting Kazdan–Warner type condition. To formulate the condition, we introduce the weighted isoperimetric ratio.

Assumen2,(Mⁿ, g)is a smooth compact Riemannian manifold with boundaryΣ=∂M. LetKbe a positive smooth function onΣ, then we write the weighted isoperimetric ratio

I (M, g, K)= μ(M)¹ⁿ (

ΣK dS)ⁿ−¹1

.

Heredμis the measure associated withganddSis the measure onΣ. Ifn3 and(Mⁿ, g)satisfiesλ₁(L_g) >0, for

˜

g∈ [g]with zero scalar curvature, we writeg˜=uⁿ⁻²⁴ g,u|Σ=f, then I (M,g, K)˜ = (

M(P_Lf )ⁿ²ⁿ−2dμ)¹ⁿ (

ΣKf^{2(nn−2−1)}dS)ⁿ⁻¹¹ .

The Euler–Lagrange equation of this functional reads as

M

PL(x, ξ )(PLf )(x)ⁿ⁺²ⁿ⁻²dμ(x)=const·K(ξ )f (ξ )ⁿ⁻ⁿ².

Lemma 3.1(Kazdan–Warner type condition). Assumen3,(Mⁿ, g)is a smooth compact Riemannian manifold with boundary andλ₁(L_g) >0,Kandf are positive smooth functions onΣsuch that

M

P_L(x, ξ )(P_Lf )(x)ⁿⁿ⁻²⁺²dμ(x)=K(ξ )f (ξ )ⁿ⁻²ⁿ .

LetXbe a conformal vector field onM(noteXmust be tangent toΣ), then

Σ

XK·f^{2(nn−2−1)}dS=0.

Proof. Denoteu=P_Lf. Letφ_t be the smooth 1-parameter group generated byX, then d

dt

t=0

I M, φ_t^∗

uⁿ⁻⁴²g , K

=0.

On the other hand, d

dt

t=0

I M, φ_t^∗

uⁿ−⁴2g , K

= d dt

t=0

I

M, uⁿ−⁴2g, K◦φ_−t

=I (M, uⁿ⁻²⁴ g, K) n−1

ΣXK·f^{2(n−n−21)}dS

ΣKf^{2(nn−−21)}dS .

This implies

ΣXK·f^2(n−1)n−2 dS=0. 2

Corollary 3.3.Assumen3,Kandf are positive smooth functions on∂B₁ⁿsuch that

B₁

P (x, ξ )(Pf )(x)ⁿ⁺²ⁿ⁻²dx=K(ξ )f (ξ )ⁿ⁻ⁿ²,

then

∂B₁∇K(ξ ),∇ξif (ξ )^{2(nn−−21)}dS(ξ )=0for1in.

This is because∇ξ_i is the restriction to∂B₁of a conformal vector field on(B₁, g_Rn).

We will also need some rearrangement inequality on∂B₁which was proven in [2]. We say a functionf on∂B₁ is radially symmetric iff (ξ )is a function ofξ_n. Letf be a measurable function on∂B₁, then the symmetric rear- rangement off is a radial decreasing functionf^∗which has the same distribution asf. The following rearrangement inequality was proven in [2, Theorem 2]. Namely, ifK is a nondecreasing bounded function on[−1,1], then for all f, g∈L¹(∂B₁),

∂B1×∂B1

f (ξ )g(η)K(ξ·η) dS(ξ ) dS(η)

∂B1×∂B1

f^∗(ξ )g^∗(η)K(ξ·η) dS(ξ ) dS(η).

It follows that ifK is a bounded nonnegative nondecreasing function on[−1,1],f is nonnegative function on∂B1

and

(K∗f )(ξ )=

∂B₁

K(ξ·η)f (η) dS(η),

then for 1p <∞,|K∗f|L^p(∂B1)|K∗f^∗|L^p(∂B1). Recall the Poisson kernel on(B1, g_Rⁿ)is given by

P (x, ξ )= 1− |x|² nω_n|x−ξ|ⁿ. For 0< r <1, ξ, ζ∈∂B1,

P (rζ, ξ )= 1−r²

nωn(r²+1−2rζ·ξ )ⁿ² =Kr(ζ·ξ ).

Hence for 1p <∞andf0,

|Pf|^p_Lp(B₁)= 1 0

|K_r∗f|^p_Lp(∂B₁)rⁿ⁻¹dr 1 0

|K_r∗f^∗|^p_Lp(∂B₁)rⁿ⁻¹dr= |Pf^∗|^p_Lp(B₁).

It follows that|Pf|L^p(B₁)|Pf^∗|L^p(B₁).

Proof of Theorem 3.1. Forp >²⁽ⁿ_n−⁻₂¹⁾, we consider the variational problem sup

|Pf|

L

2n

n−2(B1): f∈L^p(∂B₁),|f|L^p(∂B₁)=1

. (3.1)

By Corollary 2.1 the operatorP:L^p(∂B₁)→Lⁿ²ⁿ−2(B₁)is compact, hence the supreme is achieved at somef_p0.

Replacingf_pbyf_p^∗we may assumef_pis radial symmetric and decreasing. After scalingf_psatisfies f_p(ξ )^p−1=

B1

P (x, ξ )(Pf_p)(x)ⁿⁿ⁺−²2 dx.

Standard bootstrap using Propositions 2.1 and 2.2 showsfp∈C^∞(∂B1)andfp>0. Rewrite the equation as

B₁

P (x, ξ )(Pfp)(x)ⁿ⁺²ⁿ⁻²dx=fp(ξ )ⁿⁿ⁻²fp(ξ )^{p−2(nn−−21)}. It follows from Corollary 3.3 that

∂B₁

∇f_p(ξ )^p−2(nn−⁻2¹⁾,∇ξ_n

f_p(ξ )²⁽ⁿⁿ−⁻2¹⁾dS(ξ )=0.

We may writeg_p(r)=f_p(0, . . . ,0,sinr,cosr)for 0rπ. Then the equality becomes π

0

g_p(r)g_p(r)^p−1sinⁿ⁻¹r dr=0.

Sinceg_p 0 andg_p>0, we getg_p =0 and hencef_p≡const. This implies

|Pf|

Lⁿ²ⁿ−2(B₁) ω

n−22n

n

(nω_n)^p1

|f|L^p(∂B₁).

Letp→ ²⁽ⁿ_n−⁻₂¹⁾, we get the needed inequality. At last we may apply [7, Theorem 1.2] to identify all the functions which achieves the equality. 2

4. Regularity of solutions to some nonlinear integral equations

Assume 1< p <∞. Iff ∈L^p(Σ )is a maximizer for the variational problem c_M,g,p=sup

|Pf|

L

np

n−1(M): f∈L^p(Σ ),|f|L^p(Σ )=1 ,

then we may assumef 0, moreover after suitable scaling it satisfies the nonlinear integral equation f (ξ )^p−1=

M

P (x, ξ )(Pf )(x)^n−1np−1dμ(x).

This section is aiming at proving all these solutions are in fact smooth.

Proposition 4.1. Assume n2, (Mⁿ, g) is a smooth compact Riemannian manifold with boundary Σ=∂M. If 1< p <∞,f ∈L^p(Σ )is nonnegative, not identically zero and it satisfies

f (ξ )^p−1=

M

P (x, ξ )(Pf )(x)^nnp−1−1dμ(x),

thenf ∈C^∞(Σ ).