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THE SPHERE COVERING INEQUALITY AND ITS DUAL

CHANGFENG GUI, FENGBO HANG, AND AMIR MORADIFAM

Abstract. We present a new proof of the sphere covering inequality in thespirit of comparison geometry, and as a byproduct we find another sphere cov-

ering inequality which can be viewed as the dual of the original one. We also

prove sphere covering inequalities on surfaces satisfying general isoperimetricinequalities, and discuss their applications to elliptic equations with exponen-

tial nonlinearities in dimension two. The approach in this paper extends,improves, and unifies several inequalities about solutions of elliptic equations

with exponential nonlinearities.

1. Introduction

Second order nonlinear elliptic equations with exponential nonlinearity of theform

(1.1) ∆u+ eu = f(x) in Ω ⊂ R2,

arise in many important problems in mathematics, mathematical physics and bi-ology. Such equations have been extensively studied in the context of Moser-Trudinger inequalities, Chern-Simons self-dual vortices, Toda systems, conformalgeometry, statistical mechanics of two-dimensional turbulence, self-gravitating cos-mic strings, theory of elliptic functions and hyperelliptic curves and free boundarymodels of cell motility, see [BC1, BC2, BCLT, BL1, BL2, BLT, BT1, BT2, Be,BeFR, CLS, CaY, CagLMP1, CagLMP2, ChanFL, ChangY1, ChangY2, ChangY3,ChaniK, DJLW, GL, Li1, Li2, LiL, LiL2, LiW1, LiW2, LiWY, Y] and the referencescited therein.

The sphere covering inequality was recently introduced in [GuM1], and has beenapplied to solve various problems about symmetry and uniqueness of solutions ofelliptic equations with exponential nonlinearity in dimension n = 2. It particular,it was applied to prove a long-standing conjecture of Chang-Yang ([ChangY1])concerning the best constant in Moser-Trudinger type inequalities [GuM1], and hasled to several symmetry and uniqueness results for mean field equations, Onsagervortices, Sinh-Gordon equation, cosmic string equation, Toda systems and rigidityof Hawking mass in general relativity [GuHMW, GuJM, GuM1, GuM2, GuM3,LLTY, SSTW, WWY].

Theorem 1.1 (The Sphere Covering Inequality [GuM1, Theorem 1.1]). Let Ω0 ⊂R2 be a simply connected domain. Assume u1 ∈ C2

(Ω0

)such that

(1.2) ∆u1 + e2u1 ≥ 0 on Ω0,

∫Ω0

e2u1dx ≤ 4π.

1

2 CHANGFENG GUI, FENGBO HANG, AND AMIR MORADIFAM

Let Ω ⊂ Ω0 be a bounded open set. Assume u2 ∈ C2(Ω)

such that

∆u2 + e2u2 ≥ ∆u1 + e2u1 in Ω,

u2 > u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(1.3)

∫Ω

e2u1dx+

∫Ω

e2u2dx ≥ 4π.

In this paper, we present an approach that completes, simplifies and improvesthe sphere covering inequality and several other inequalities about solutions of theelliptic equations with exponential nonlinearities. In particular, we will prove thefollowing generalization of the sphere covering inequality with a method differentfrom the one in [GuM1].

Theorem 1.2. Let Ω0 ⊂ R2 be a simply connected domain. Assume u1 ∈ C2(Ω0

)such that

(1.4) ∆u1 + e2u1 ≥ 0 on Ω0,

∫Ω0

e2u1dx ≤ 4π.


and 0 < λ ≤ 1 such that

∆u2 + λe2u2 ≥ ∆u1 + e2u1 in Ω,

u2 > u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(1.5)

∫Ω

e2u1dx+

∫Ω

e2u2dx ≥ 4π

λ.

We shall also prove the following inequality which can be viewed as the dual ofthe sphere covering inequality.

Theorem 1.3. Let Ω0 ⊂ R2 be a simply connected domain. Assume u1 ∈ C2(Ω0

)such that

(1.6) ∆u1 + e2u1 ≥ 0 on Ω0,

∫Ω0

e2u1dx ≤ 4π.


such that

∆u2 + e2u2 ≤ ∆u1 + e2u1 in Ω,

u2 < u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(1.7)

∫Ω

e2u1dx+

∫Ω

e2u2dx ≥ 4π.

We will develop an approach for Theorem 1.2 which is different from the one in[GuM1], and shall modify it to prove Theorem 1.3. Our method has the generalspirit of comparison geometry. Under the assumption of Theorem 1.2, let g =e2u1 |dx|2, here |dx|2 is the Euclidean metric. Then the Gauss curvature K ≤ 1 andthe area µ (Ω0) ≤ 4π, where µ(E) is the measure of E associated with the metric

THE SPHERE COVERING INEQUALITY AND ITS DUAL 3

g. It follows from [B] and [ChangCL, Lemma 4.2] that the following isoperimetricinequality holds on (Ω0, g) for any domain E in Ω0 with smooth boundary

(1.8) 4πµ (E)− µ2 (E) ≤ s2 (∂E) ,

where s is the 1-dimensional measure associated with g. Using g as a backgroundmetric, we can rewrite the differential inequality between u1 and u2 into a differen-tial inequality involving u = u2 − u1. Applying ideas from [B, Su] to the resultingdifferential inequality on (Ω0, g) gives us an inequality which will imply Theorem1.2. Indeed the proof of Theorem 1.2 is based on the following more general result.

Theorem 1.4. Let (M, g) be a simply connected smooth Riemann surface. AssumeK ≤ 1 and µ (M) ≤ 4π, here K is the Gauss curvature and µ is the measure of(M, g). Let Ω be a domain with compact closure and nonempty boundary, and λ bea constant. If u ∈ C2

(Ω)

such that u > 0 in Ω and

(1.9) −∆gu+ 1 ≤ λe2u, u|∂Ω = 0.

Then

(1.10) 4π

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≤ 4πµ (Ω)− µ2 (Ω) .

In particular if 0 < λ ≤ 1, then

(1.11)

∫Ω

e2udµ+ µ (Ω) ≥ 4π

λ.

It is interesting that in this comparison theorem, what is compared is not thearea itself, but the quantity 4πµ (Ω) − µ2 (Ω), which is exactly the quantity ap-peared in the isoperimetric inequality.

The proof of Theorem 1.3 also follows from the following more general result.

Theorem 1.5. Let (M, g) be a simply connected smooth Riemann surface. AssumeK ≤ 1 and µ (M) ≤ 4π, here K is the Gauss curvature and µ is the measure of(M, g). Let Ω be a domain with compact closure and nonempty boundary, and λ bea constant. If u ∈ C2

(Ω)

such that u < 0 in Ω and

(1.12) −∆gu+ 1 ≥ λe2u, u|∂Ω = 0.

Then

(1.13) 4π

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≥ 4πµ (Ω)− µ2 (Ω) .

In particular if λ = 1, then

(1.14)

∫Ω

e2udµ+ µ (Ω) ≥ 4π.

In Section 2, we present proofs of Theorems 1.2, 1.3, 1.4, and 1.5. In section3, we will prove sphere covering inequalities on surfaces satisfying general isoperi-metric inequalities and shall discuss their applications to elliptic equations withexponential nonlinearities.


2. Differential inequalities on surface with curvature at most 1

In this section we will prove Theorem 1.4 and 1.5. The main point is that theapproach in [B, Su] can be performed on simply connected surface with curvatureat most 1.

Proof of Theorem 1.4. By approximation and replacing λ with λ + ε, ε is a smallpositive number, we can assume u is a Morse function. For t > 0, let

α (t) =

∫u>t

e2udµ, β (t) =

∫u>t

dµ.

By co-area formula we get

α (t) =

∫u>t

e2u |∇u||∇u|

dµ

=

∫ ∞t

dτ

∫u=τ

e2u

|∇u|ds

=

∫ ∞t

(e2τ

∫u=τ

ds

|∇u|

)dτ

and

β (t) =

∫u>t

|∇u||∇u|

dµ =

∫ ∞t

dτ

∫u=τ

ds

|∇u|.

It follows that

α′ (t) = −e2t

∫u=t

ds

|∇u|,

and

β′ (t) = −∫u=t

ds

|∇u|.

In particular

α′ (t) = e2tβ′ (t) .

On the other hand, by the differential inequality we have∫u>t

(−∆u) dµ+ β (t) ≤ λα (t) ,

hence ∫u=t

|∇u| ds ≤ λα− β.

Multiplying both sides by −α′(t) we get

e2t

∫u=t

ds

|∇u|

∫u=t

|∇u| ds ≤ −λαα′ + e2tββ′,

which implies

e2ts (u = t)2 ≤ −λαα′ + e2tββ′.

Applying the isoperimetric inequality on (M, g) (see [B] , [ChangCL, Lemma 4.2])we get

e2t(4πβ − β2

)≤ −λαα′ + e2tββ′.

Consequently

4π(e2t)′β −

(e2t)′β2 ≤ −λ

(α2)′

+ e2t(β2)′.


In other words

4π[(e2tβ

)′ − α′] ≤ −λ (α2)′

+(e2tβ2

)′,

and hence

4π(α− e2tβ

)− λα2 + e2tβ2 is increasing.

Thus

4π (α (0)− β (0))− λα (0)2

+ β (0)2 ≤ 0.

In other words

4π

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≤ 4πµ (Ω)− µ2 (Ω) .

When 0 < λ ≤ 1, we have

4π

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≤ 4πµ (Ω)− λµ2 (Ω) ,

and

4π

(∫Ω

e2udµ− µ (Ω)

)≤ λ

(∫Ω

e2udµ+ µ (Ω)

)(∫Ω

e2udµ− µ (Ω)

).

Hence ∫Ω


λ.

We will derive Theorem 1.5 by flipping all the inequalities.

Proof of Theorem 1.5. Again we can assume u is a Morse function. For t < 0, let

α (t) =

∫u<t

e2udµ, β (t) =

∫u<t

dµ.

Then

α′ (t) = e2t

∫u=t

ds

|∇u|, β′ (t) =

∫u=t

ds

|∇u|, α′ (t) = e2tβ′ (t) .

On the other hand, since

∆u− 1 ≤ −λe2u,

integrating on u < t we get∫u=t

|∇u| ds− β (t) ≤ −λα (t) .

We have

e2t(4πβ − β2

)≤ e2ts2 (u = t)

≤ e2t

∫u=t

ds

|∇u|

∫u=t

|∇u| ds

≤ e2tββ′ − λαα′.Hence

e2t(β2)′ − λ (α2

)′ ≥ 4π(e2t)′β −

(e2t)′β2.


It follows that (e2tβ2

)′ − λ (α2)′ ≥ 4π

[(e2tβ

)′ − α′] ,and

4π(α− e2tβ

)+ e2tβ2 − λα2 is increasing.

In particular

4π (α (0)− β (0)) + β (0)2 − λα (0)

2 ≥ 0.

In other words

4π

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≥ 4πµ (Ω)− µ2 (Ω) .

When λ = 1, we have

4π

(µ (Ω)−

∫Ω

e2udµ

)≤

(µ (Ω) +

∫Ω

e2udµ

)(µ (Ω)−

∫Ω

e2udµ

).

Hence ∫Ω

e2udµ+ µ (Ω) ≥ 4π.

Theorem 1.2 easily follows from Theorem 1.4.

Proof of Theorem 1.2. Let g = e2u1 |dx|2, then

K = −e−2u1∆u1 ≤ 1

and µ (Ω0) ≤ 4π. Let u = u2 − u1. We have

−∆u ≤ e2u1(λe2u − 1

).

Hence

−∆gu ≤ λe2u − 1.

Note that u > 0 in Ω and u|∂Ω = 0. Thus by Theorem 1.4 we have∫Ω


λ.

In other words ∫Ω

e2u1dx+

∫Ω

e2u2dx ≥ 4π

λ.

By exactly the same argument as above, Theorem 1.3 follows from Theorem 1.5.

Example 2.1. Fix 0 < r < 1. We take the stereographic projection of the unitsphere S2 with respect the north pole to plane

x3 = −√

1− r2 = −h1,

then the standard metric on S2 is written as

g1 =4 (1 + h1)

2(|x|2 + (1 + h1)

2)2 |dx|

2= e2u1 |dx|2 .


For R > 1, we do stereographic projection of R · S2 with respect to the north poleto the plane

x3 =√R2 − r2 = h2,

then the metric on R · S2

g2 =4R2 (R− h2)

2(|x|2 + (R− h2)

2)2 |dx|

2= e2u2 |dx|2 .

Note that for |x| < r, u2 (x) > u1 (x),

∆u1 + e2u1 = 0,

∆u2 +R−2e2u2 = 0,∫Br

e2u1dx = 2π (1− h1) ,∫Br

e2u2dx = 2πR (R+ h2) .

We have ∫Br

e2u1dx+

∫Br

e2u2dx > 4πR2.

This is an example for Theorem 1.2 with λ = R−2.

Example 2.2. For 0 < r < R < 1, we take the stereographic projection of S2 withrespect to the north pole to the plane

x3 =√

1− r2 = h1,

then the metric on S2 is written as

g1 =4 (1− h1)

2(|x|2 + (1− h1)

2)2 |dx|

2= e2u1 |dx|2 .

We also do stereographic projection of R · S2 with respect to the north pole to theplane

x3 = −√R2 − r2 = −h2,

then the metric on R · S2 is written as

g2 =4R2 (R+ h2)

2(|x|2 + (R+ h2)

2)2 |dx|

2= e2u2 |dx|2 .

Note that for |x| < r, u2 (x) < u1 (x),

∆u1 + e2u1 = 0,

∆u2 + e2u2 = −(R−2 − 1

)e2u2 ,∫

Br

e2u1dx = 2π (1 + h1) ,∫Br

e2u2dx = 2πR (R− h2) > 2π (1− h1) .

Hence ∫Br

e2u1dx+

∫Br

e2u2dx > 4π.


This is an example of Theorem 1.3.

Example 2.3. Let Ω ⊂ R2 be a bounded smooth domain, u be a smooth functionon Ω such that

−∆u+ 1 ≤ e2u in Ω,

u|∂Ω = 0,

u > 0 in Ω.

It follows from Theorem 1.4 that

(2.1)

∫Ω

e2udx+ |Ω| ≥ 4π.

Because of the usual isoperimetric inequality on R2, the assumption µ (M) ≤ 4π inTheorem 1.4 is not needed in our situation. Here we will give an example where∫

Ωe2udx+ |Ω| is arbitrary close to 4π.

For 0 < h < 1, denote r =√

1− h2. Take the stereographic projection of S2 withrespect to the north pole to the plane x3 = −h, then the metric on S2 is written as

g1 =4 (1 + h)

2[|x|2 + (1 + h)

2]2 |dx|2 = e2u1 |dx|2 .

We have

−∆u1 = e2u1 in Br;

u1|∂Br= 0;

1 ≤ e2u1 ≤ 4

(1 + h)2 .

Let

R =2

1 + hand

h2 =√R2 − r2.

Take the stereographic projection of R ·S2 with respect to the north pole to the planex3 = h2, then the metric on R · S2 is written as

g2 =4R2 (R− h2)

2[|x|2 + (R− h2)

2]2 |dx|2 = e2u2 |dx|2 .

We have

−∆u2 =(1 + h)

2

4e2u2 in Br;

u2|∂Br= 0;

u2 > u1 in Br.

Let u = u2 − u1, then u > 0 in Br, u|∂Br= 0. Moreover

−∆u =(1 + h)

2

4e2u2 − e2u1

=(1 + h)

2

4e2u1e2u − e2u1 .


It follows that

−∆u+ 1 ≤ −∆u+ e2u1

=(1 + h)

2

4e2u1e2u

≤ e2u.

On the other hand,∫Br

e2udx+ |Br| =∫Br

e2u2−2u1dx+ |Br| → 4π

as h ↑ 1−.The above example shows that one can not get any improvements to (1.11) by

assuming K ≤ a < 1. Indeed a = 0 in the example above.

3. Differential equation on surface satisfying general isoperimetricinequalities

In this section we present sphere covering type inequalities on surfaces satisfyinggeneral isoperimetric inequalities (see (3.1) below) and discuss their applications toelliptic equations with exponential nonlinearities. We find the following definitionparticularly useful.

Definition 3.1. Let M be a smooth surface and g be a metric on M . If for some0 < θ ≤ 1 and κ ∈ R, we have

(3.1) 4πθµ (E)− κµ2 (E) ≤ s2 (∂E)

for any compact smooth domain E ⊂ M , here s is the one dimensional measureassociated with g and µ is the two dimensional measure, then we say (M, g) satisfiesthe (θ, κ)-isoperimetric inequality.

Theorem 3.1. Let (M, g) be a smooth Riemann surface satisfying the (θ, κ)-isoperimetric inequality for some θ ∈ (0, 1] and κ ∈ R. If Ω ⊂ M is an opendomain with compact closure and u ∈ C∞

(Ω)

such that

(3.2) −∆gu+ κ = λe2u + f, u|∂Ω = 0, u > 0 in Ω.

Denote

(3.3) Θ =1

2π

∫Ω

f+dµ,

then

(3.4) 4π (θ −Θ)

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≤ 4πθµ (Ω)− κµ2 (Ω) .

In particular, if Θ = 0 and 0 < λ ≤ κ, then

(3.5)

∫Ω

e2udµ+ µ (Ω) ≥ 4πθ

λ.

Remark 3.1. It is worth pointing out that as long as the (θ, κ)-isoperimetric in-equality is valid, the smoothness of u and metric g is not essential to our argument.In particular f can be replaced by a signed measure. This is useful in some singularLiouville type equations. We will not elaborate this point further but refer the readerto [BC1, BC2, BGJM] and the references therein.


Proof. By approximation we can assume u is a Morse function. For t > 0, let

α (t) =

∫u>t

e2udµ, β (t) =

∫u>t

dµ.

As in the proof of Theorem 1.4, we have

α′ (t) = −e2t

∫u=t

ds

|∇u|,

β′ (t) = −∫u=t

ds

|∇u|,

α′ (t) = e2tβ′ (t) .

On the other hand,∫u>t

(−∆u) dµ+ κβ (t) = λα (t) +

∫u>t

fdµ,

hence ∫u=t

|∇u| ds ≤ λα− κβ + 2πΘ.

We have

e2t

∫u=t

ds

|∇u|

∫u=t

|∇u| ds ≤ −λαα′ + κe2tββ′ − 2πΘα′,

which implies

e2ts2 (u = t) ≤ −λαα′ + κe2tββ′ − 2πΘα′.

Using (3.1) we get

e2t(4πθβ − κβ2

)≤ −λαα′ + κe2tββ′ − 2πΘα′.

It follows that

4πθ[(e2tβ

)′ − α′] ≤ −λ (α2)′

+ κ(e2tβ2

)′ − 4πΘα′.

Integrating for t from 0 to ∞, we get

4πθ (α (0)− β (0)) ≤ λα (0)2 − κβ (0)

2+ 4πΘα (0)

In other words

4π (θ −Θ)

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

≤ 4πθµ2 (Ω)− κµ (Ω) .

If we flip the inequalities as in the proof of Theorem 1.5, we get

Theorem 3.2. Let (M, g) be a smooth Riemann surface satisfying the (θ, κ)-isoperimetric inequality for some θ ∈ (0, 1] and κ ∈ R. Assume Ω ⊂ M is anopen domain with compact closure and u ∈ C∞

(Ω)

such that

(3.6) −∆gu+ κ = λe2u + f, u|∂Ω = 0, u < 0 in Ω.

Denote

(3.7) Θ =1

2π

∫Ω

f−dµ,


then

(3.8) 4πθµ (Ω)− κµ2 (Ω) ≤ 4π (θ + Θ)

∫Ω

e2udµ− λ(∫

Ω

e2udµ

)2

.

In particular, if Θ = 0 and λ = κ > 0, then

(3.9)

∫Ω

e2udµ+ µ (Ω) ≥ 4πθ

λ.

Next we discuss some known and new applications of Theorem 3.1 and 3.2.

Example 3.1 ([B, Su]). Let (M, g) be a simply connected smooth Riemann surface

with curvature K ≤ 1. If E is a compact simply connected domain in M withnonempty smooth boundary, then we can find u ∈ C∞ (E) such that

−∆gu = K on E, u|∂E = 0.

Let g = e−2ug, then the curvature of g is zero. By Riemann mapping theorem andthe Taylor series argument for holomorphic functions in [Car], (E, g) satisfies the(1, 0)-isoperimetric inequality. On the other hand,

−∆gu ≤ e2u on E, u|∂E = 0.

If we let

Ω = p ∈ E : u (p) > 0 ,then

−∆gu ≤ e2u on Ω, u|∂Ω = 0, u > 0 in Ω.

Theorem 3.1 tells us

4π

∫Ω

e2udµ−(∫

Ω

e2udµ

)2

≤ 4πµ (Ω) .

Hence

4π

(∫E

e2udµ− µ (E)

)≤ 4π

(∫Ω

e2udµ− µ (Ω)

)≤

(∫Ω

e2udµ

)2

≤(∫

E

e2udµ

)2

i.e.

4πµ (E)− µ2 (E) ≤ 4πµ (E) ≤ s2 (∂E) = s2 (∂E) .

This is exactly the argument given in [B, Su].If we assume further that µ (M) ≤ 4π, then following [ChangCL, Lemma 4.2]

we know, for E to be a compact domain with boundary smooth but not necessarilysimply connected, there still holds

4πµ (E)− µ2 (E) ≤ s2 (∂E) .

In another word, (1, 1)-isoperimetric inequality is true for (M, g). As a consequenceTheorem 1.4 follows from Theorem 3.1.


Example 3.2. Let (M, g) be a simply connected smooth Riemann surface with

curvature K. Assume a ≥ 0 and

(3.10) Θ =1

2π

∫M

(K − a

)+

dµ < 1.

Then for any compact simply connected domain E in M with nonempty smoothboundary, we have

(3.11) 4π (1−Θ) µ (E)− aµ2 (E) ≤ s2 (∂E) .

In particular, if we assume further that µ (M) ≤ 4π(1−Θ)a , then (M, g) satisfies

(1−Θ, a)-isoperimetric inequality. In fact this is even true when g is singular, see[BC1, BC2, BGJM] and the references therein.

Indeed as in the previous example, we can find u ∈ C∞ (E) such that

−∆u = K on E, u|∂E = 0.

Let g = e−2ug, then (E, g) satisfies (1, 0)-isoperimetric inequality and

−∆gu = ae2u +(K − a

)e2u on E.

Note that

1

2π

∫E

[(K − a

)e2u]+

dµ =1

2π

∫E

(K − a

)+

dµ ≤ Θ < 1.

Let

Ω = p ∈ E : u (p) > 0 ,then

−∆gu = ae2u +(K − a

)e2u on Ω, u|∂Ω = 0, u > 0 in Ω.

Theorem 3.1 implies

4π (1−Θ)

∫Ω

e2udµ− a(∫

Ω

e2udµ

)2

≤ 4πµ (Ω) .

Hence

4π (1−Θ)

∫E

e2udµ− 4πµ (E)

≤ 4π (1−Θ)

∫Ω

e2udµ− 4πµ (Ω)

≤ a

(∫Ω

e2udµ

)2

≤ a

(∫E

e2udµ

)2

.

In another word,

4π (1−Θ) µ (E)− aµ2 (E) ≤ 4πµ (E) ≤ s2 (∂E) = s2 (∂E) .

Example 3.3 ([B, BC1, BC2, BGJM]). Let Ω0 ⊂ R2 be a simply connected domainand u, h ∈ C∞ (Ω0) with h (x) > 0 for any x ∈ Ω0. We write

(3.12) −∆u = he2u + f


and

(3.13) Θ =1

2π

∫Ω0

(f − 1

2∆ log h

)+

dx.

If Θ < 1 and

(3.14)

∫Ω0

he2udx ≤ 4π (1−Θ) ,

then(

Ω0, he2u |dx|2

)satisfies the (1−Θ, 1)-isoperimetric inequality. As pointed

out earlier in Remark 3.1, the regularity assumption of u and h can be weakenedand we refer the reader to [BC1, BC2, BGJM].

Proof. For convenience we denote g = he2u |dx|2, then its curvature

K = h−1e−2u

(−∆u− 1

2∆ log h

)= h−1e−2u

(he2u + f − 1

2∆ log h

)= 1 + h−1e−2u

(f − 1

2∆ log h

).

Hence1

2π

∫Ω0

(K − 1)+dµ =

1

2π

∫Ω0

(f − 1

2∆ log h

)+

dx = Θ < 1,

and

µ (Ω0) =

∫Ω0

he2udx ≤ 4π (1−Θ) .

It follows from Example 3.2 that (Ω0, g) satisfies the (1−Θ, 1)-isoperimetric in-equality

If we replace the reference metric from Euclidean metric to an arbitrary one, weend up with the following formulation.

Lemma 3.1. Let (M, g) be a simply connected Riemann surface with curvature K,and u ∈ C∞ (M). We write

(3.15) −∆gu = e2u + f

and

(3.16) Θ =1

2π

∫M

(f +K)+dµ.

If Θ < 1 and

(3.17)

∫M

e2udµ ≤ 4π (1−Θ) ,

then(M, e2ug

)satisfies the (1−Θ, 1)-isoperimetric inequality.

Proof. Let g = e2ug, then

K = e−2u (K −∆u) = 1 + e−2u (f +K) .

In particular,

1

2π

∫M

(K − 1

)+

dµ =1

2π

∫M

(f +K)+dµ = Θ < 1,


and

µ (M) =

∫M

e2udµ ≤ 4π (1−Θ) .

It follows from Example 3.2 that (M, g) satisfies the (1−Θ, 1)-isoperimetric in-equality

Note that Lemma 3.1 also follows form Example 3.3 and Riemann mappingtheorem. With Lemma 3.1 at hand, we can deduce easily a variation of spherecovering inequality.

Proposition 3.1. Let (M, g) be a simply connected Riemann surface, and u1 ∈C∞ (M). We write

(3.18) −∆gu1 = e2u1 + f

and

(3.19) Θ =1

2π

∫M

(f +K)+dµ.

Here K is the curvature of (M, g). Assume Θ < 1 and

(3.20)

∫M

e2u1dµ ≤ 4π (1−Θ) .

Let Ω ⊂ M be a domain with a compact closure and nonempty boundary. Assumeu2 ∈ C∞

(Ω)


∆gu2 + λe2u2 ≥ ∆gu1 + e2u1 in Ω,

u2 > u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(3.21)

∫Ω

e2u1dµ+

∫Ω

e2u2dµ ≥ 4π (1−Θ)

λ.

Note that Theorem 1.2 is a special case of Proposition 3.1.

Proof of Proposition 3.1. Let g = e2u1g, then it follows from Lemma 3.1 that (M, g)satisfies the (1−Θ, 1)-isoperimetric inequality. Let u = u2−u1, then on Ω we have

−∆gu ≤ e2u1(λe2u − 1

).

Hence

−∆gu ≤ λe2u − 1.

Moreover u > 0 in Ω and u|∂Ω = 0, it follows from Theorem 3.1 that∫Ω

e2udµ+ µ (Ω) ≥ 4π (1−Θ)

λ.

In another word, ∫Ω

e2u1dµ+

∫Ω

e2u2dµ ≥ 4π (1−Θ)

λ.

Using Theorem 3.2, with the same proof, we also have a dual inequality gener-alizing Theorem 1.3.


Proposition 3.2. Let (M, g) be a simply connected Riemann surface with curvatureK, and u1 ∈ C∞ (M). We write

(3.22) −∆gu1 = e2u1 + f

and

(3.23) Θ =1

2π

∫M

(f +K)+dµ.

Assume Θ < 1 and

(3.24)

∫M

e2u1dµ ≤ 4π (1−Θ) .

Let Ω ⊂ M be a domain with compact closure and nonempty boundary. Assumeu2 ∈ C∞

(Ω)

such that

∆gu2 + e2u2 ≤ ∆gu1 + e2u1 in Ω,

u2 < u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(3.25)

∫Ω

e2u1dµ+

∫Ω

e2u2dµ ≥ 4π (1−Θ) .

Example 3.4. Let (M, g) be a simply connected Riemann surface with curvatureK, and u1, h ∈ C∞ (M) with h > 0. We write

(3.26) −∆gu1 = he2u1 + f

and

(3.27) Θ =1

2π

∫M

(f +K − 1

2∆g log h

)+

dµ.

Assume Θ < 1 and

(3.28)

∫M

he2u1dµ ≤ 4π (1−Θ) .


(Ω)


∆gu2 + λhe2u2 ≥ ∆gu1 + he2u1 in Ω,

u2 > u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(3.29)

∫Ω

he2u1dµ+

∫Ω

he2u2dµ ≥ 4π (1−Θ)

λ.

Proof. Let

v1 = u1 +1

2log h, v2 = u2 +

1

2log h,

then

(3.30) −∆gv1 = e2v1 + f − 1

2∆ log h.


Moreover

∆gv2 + λe2v2 ≥ ∆gv1 + e2v1 in Ω,

v2 > v1 in Ω,

v2|∂Ω = v1|∂Ω .

Then we can apply Proposition 3.1 to get the desired conclusion.

By a straightforward modification we can also deal with the case h changes sign.

Example 3.5. Let (M, g) be a simply connected Riemann surface with curvatureK, and u1, h,H ∈ C∞ (M) with h ≤ H and H > 0. We write

(3.31) −∆gu1 = He2u1 + f

and

(3.32) Θ =1

2π

∫M

(f +K − 1

2∆g logH

)+

dµ.

Assume Θ < 1 and

(3.33)

∫M

He2u1dµ ≤ 4π (1−Θ) .


(Ω)

such that

∆gu2 + he2u2 ≥ ∆gu1 + he2u1 in Ω,

u2 > u1 in Ω,

u2|∂Ω = u1|∂Ω .

Then

(3.34)

∫Ω

He2u1dµ+

∫Ω

He2u2dµ ≥ 4π (1−Θ) .

Proof. We have

∆gu2 ≥ ∆gu1 − h(e2u2 − e2u1

)≥ ∆gu1 −H

(e2u2 − e2u1

).

Hence

∆gu2 +He2u2 ≥ ∆gu1 +He2u1 in Ω.

Then we can apply Example 3.4.

Next we turn to solutions of semilinear equations with equal weights, see [BGJM,GuM2].

Proposition 3.3. Let Ω ⊂ R2 be a bounded open simply connected domain. As-sume u1, u2 ∈ C∞

(Ω)

such that

(3.35) ∆u1 + e2u1 ≥ 0

and

∆u1 + e2u1 ≥ ∆u2 + e2u2 in Ω,

u1 + c > u2 in Ω,

u1|∂Ω + c = u2|∂Ω .


Here c is a constant. If

(3.36)

∫Ω

e2u1dx =

∫Ω

e2u2dx = ρ,

then ρ ≥ 4π.

Proof. Note that c > 0. If ρ ≤ 4π, we will show ρ = 4π. Indeed let g = e2u1 |dx|2,then K ≤ 1 and µ (Ω) = ρ ≤ 4π. If we write u = u2 − u1 − c, then

−∆gu+ 1 ≥ e2c · e2u.

Moreover u < 0 in Ω and u|∂Ω = 0. It follows from Theorem 1.5 that

4π

∫Ω

e2udµ− e2c

(∫Ω

e2udµ

)2

≥ 4πµ (Ω)− µ2 (Ω) .

Hence

0 ≥(1− e−2c

)ρ (4π − ρ)

and we get ρ ≥ 4π.

Proposition 3.4. Let Ω ⊂ R2 be a bounded open simply connected domain. As-sume u1, u2 ∈ C∞

(Ω)

such that

∆u1 + e2u1 = ∆u2 + e2u2 ≥ 0 in Ω,

u1|∂Ω + c = u2|∂Ω .

Here c is a constant. If u1 is not identically equal to u2 and

(3.37)

∫Ω

e2u1dx =

∫Ω

e2u2dx = ρ,

then ρ ≥ 4π.

Proof. If ρ ≤ 4π, we will show ρ = 4π. Indeed let g = e2u1 |dx|2, then K ≤ 1 andµ (Ω) = ρ ≤ 4π. If we write u = u2 − u1 − c, then

(3.38) −∆gu+ 1 = e2c · e2u.

Let

Ω+ = x ∈ Ω : u (x) > 0 ,Ω− = x ∈ Ω : u (x) < 0 ,

then it follows from unique continuation property that |Ω\ (Ω+ ∪ Ω−)| = 0. OnΩ+, by Theorem 1.4 we have

4π

∫Ω+

e2udµ− e2c

(∫Ω+

e2udµ

)2

≤ 4πµ(Ω+)− µ2

(Ω+).

On Ω−, by Theorem 1.5 we have

4π

∫Ω−

e2udµ− e2c

(∫Ω−

e2udµ

)2

≥ 4πµ(Ω−)− µ2

(Ω−).

Using

µ(Ω+)

+ µ(Ω−)

= ρ,∫Ω+

e2udµ+

∫Ω−

e2udµ = e−2cρ,


subtracting the two inequalities we get

(4π − ρ)

(∫Ω+

e2udµ−∫

Ω−e2udµ

)≤ (4π − ρ)

(µ(Ω+)− µ

(Ω−)).

In another word

(4π − ρ)

(∫Ω+

e2udµ− µ(Ω+)

+ µ(Ω−)−∫

Ω−e2udµ

)≤ 0.

Since u is not identically equal to 0, we see∫Ω+

e2udµ− µ(Ω+)

+ µ(Ω−)−∫

Ω−e2udµ > 0.

Hence ρ ≥ 4π.

We can replace the Euclidean domain with a Riemann surface.

Example 3.6. Let (M, g) be a simply connected compact Riemann surface withnonempty boundary, and u1 ∈ C∞ (M). We write

(3.39) −∆gu1 = e2u1 + f

and

(3.40) Θ =1

2π

∫M

(f +K)+dµ.

Here K is the curvature of g. Assume u2 ∈ C∞ (M) such that

∆gu1 + e2u1 ≥ ∆gu2 + e2u2 in M ,

u1 + c > u2 in M,

u1|∂M + c = u2|∂M .


(3.41)

∫M

e2u1dµ =

∫M

e2u2dµ = ρ,

then

(3.42) ρ ≥ 4π (1−Θ) .

Proof. Without loss of generality we can assume Θ < 1 and ρ ≤ 4π (1−Θ). Let g =e2u1g, then Lemma 3.1 implies (M, g) satisfies (1−Θ, 1)-isoperimetric inequality.If we write u = u2 − u1 − c, then

−∆u+ 1 ≥ e2c · e2u on M .

Moreover u < 0 in M and u|∂M = 0. It follows from Theorem 3.2 that

4π (1−Θ)

∫M

e2udµ− e2c

(∫M

e2udµ

)2

≥ 4π (1−Θ) µ (M)− µ2 (M) .

Hence

0 ≥(1− e−2c

)ρ (4π (1−Θ)− ρ) .

Since c > 0, we get ρ ≥ 4π (1−Θ).

Using the argument in Example 3.4 we get


Example 3.7. Let (M, g) be a simply connected compact Riemann surface withnonempty boundary and curvature K, and u1, h ∈ C∞ (M) with h > 0. We write

(3.43) −∆gu1 = he2u1 + f

and

(3.44) Θ =1

2π

∫M

(f +K − 1

2∆g log h

)+

dµ.

Assume u2 ∈ C∞ (M) such that

∆gu1 + he2u1 ≥ ∆gu2 + he2u2 in M ,

u1 + c > u2 in M,

u1|∂M + c = u2|∂M .


(3.45)

∫M

he2u1dµ =

∫M

he2u2dµ = ρ,

then

(3.46) ρ ≥ 4π (1−Θ) .

In the same spirit as the proof of Proposition 3.4 but using both Theorem 3.1and 3.2 instead we have

Example 3.8. Let (M, g) be a simply connected compact Riemann surface withnonempty boundary and curvature K, and u1, u2, h ∈ C∞ (M) with h > 0. Assume

−∆u1 − he2u1 = −∆u2 − he2u2 = f in M,(3.47)

u1|∂M + c = u2|∂M .(3.48)

Here c is a constant. We denote

(3.49) Θ =1

2π

∫M

(f +K − 1

2∆g log h

)+

dµ.

If u1 is not identically equal to u2 and

(3.50)

∫M

he2u1dµ =

∫M

he2u2dµ = ρ,

then

(3.51) ρ ≥ 4π (1−Θ) .

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Department of Mathematics, The University of Texas at San Antonio, San Antonio,

TX 78249

Email address: [email protected]

Courant Institute, New York University, 251 Mercer Street, New York NY 10012


Department of Mathematics, University of California, Riverside, CA 92521


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