centroaffine Minkowski problem A remark on rotationally symmetric solutions to the ScienceDirect

ScienceDirect

J. Differential Equations 266 (2019) 4394–4431

www.elsevier.com/locate/jde

A remark on rotationally symmetric solutions to the centroaffine Minkowski problem

Jian Lu

DepartmentofAppliedMathematics,ZhejiangUniversityofTechnology,Hangzhou310023,China Received 1April2018

Availableonline 6November2018

Abstract

Inthispaperwestudythe solvabilityoftherotationallysymmetriccentroaffineMinkowskiproblem.

Bydelicateblow-upanalyses,weremoveatechnicalconditionintheexistenceresultobtainedbyLuand Wang [30].

©2018ElsevierInc.Allrightsreserved.

MSC:35J96;35J75;53A15;53A07

Keywords:Monge–Ampèreequation;CentroaffineMinkowskiproblem;Existenceofsolutions;Blow-upanalyses

1. Introduction

Given a convex body Xin the Euclidean space Rⁿ⁺¹containing the origin, the centroaffine curvatureof ∂Xat point pis by definition equal to K/dⁿ⁺², where Kis the Gauss curvature and d is the distance from the origin to the tangent hyperplane of ∂Xat p. The centroaffine curvature is invariant under unimodular linear transforms in Rⁿ⁺¹and has received much attention in ge- ometry [36,37]. The centroaffine Minkowski problem[11] is a prescribed centroaffine curvature problem, which in the smooth case is equivalent to solving the following Monge–Ampère type equation

✩ ThisworkwassupportedbyNaturalScienceFoundationofChina(11871432,11401527).

E-mailaddress:lj-tshu04@163.com.

https://doi.org/10.1016/j.jde.2018.09.034

0022-0396/©2018ElsevierInc.Allrightsreserved.

det(∇²H+H I )= f

Hⁿ⁺² onSⁿ, (1)

where f is a given positive function, His the support function of a bounded convex body Xin Rⁿ⁺¹, I is the unit matrix, ∇²H =(∇ijH )is the Hessian matrix of covariant derivatives of H with respect to an orthonormal frame on Sⁿ. When f is a constant, this equation describes affine hyperspheres of elliptic type, and all its solutions are ellipsoids centered at the origin [8].

Equation (1) is also the special case of the Lp-Minkowski problem with p= −n −1. The L_p-Minkowski problem, introduced by Lutwak [31], is an important generalization of the clas- sical Minkowski problem, and is a basic problem in the Lp-Brunn–Minkowski theory in modern convex geometry. It has attracted great attention over the last two decades, see e.g. [5,6,10,11,14, 16,18–21,26,32–34,39,40,42,44,46] and references therein.

Equation (1) naturally arises in anisotropic Gauss curvature flows and describes their self- similar solutions [4,7,12,17,41]. Besides, its parabolic form can be used for image processing [2].

Eq. (1) can be reduced to a singular Monge–Ampère equation in the half Euclidean space Rⁿ₊⁺¹, the regularity of which was strongly studied in [22,23].

Equation (1) corresponds to the critical case of the famous Blaschke–Santaló inequality in convex geometry [35]:

vol(X)inf

ξ∈X

1 n+1

Sⁿ

dS(x)

(H (x)−ξ·x)ⁿ⁺¹ ≤κ_n²₊₁, (2) where X is any convex body in Rⁿ⁺¹, vol(X)is the volume of the convex body X, H is the support function of X, and κn+1is the volume of the unit ball in Rⁿ⁺¹. Also Eq. (1) remains invariant under projective transforms on Sⁿ[11,30]. When f is a constant function, it only has constant solutions up to projective transformations. This result has been known for a long time, see e.g. [8], which implies that there is no a priori estimates on solutions for general f with- out additional assumptions. Besides, Chou and Wang [11] found an obstruction for solutions to Eq. (1), which means it may have no solution for some f. On the other hand, it may also have many solutions for some f [15]. This situation is similar, in some aspects, to the prescribed scalar curvature problem on Sⁿ, which involves critical exponents of Sobolev inequalities and the Kazdan–Warner obstruction [9,38]. So the solvability of Eq. (1) is a rather complicated problem due to these features.

For n =1, the existence of solutions to Eq. (1) was investigated in [1,3,10,12,13,24,25,40,43].

In general, one needs to impose some non-degenerate and topological degree conditions on f to obtain an existence result.

For higher n-dimension, only several special cases were studied, see [29,30] for the rotation- ally symmetric case, [27] for a generalized rotationally symmetric case, [21] for the mirror- symmetric case, and [45] for the discrete case. In these papers, sufficient conditions for the existence of solutions can be found. However, the solvability of Eq. (1) for a general f is still open.

In this paper, we are only concerned about the rotationally symmetric case of Eq. (1). That is, the given function f and solutions H are assumed to be rotationally symmetric with respect to the xn+1-axis in Rⁿ⁺¹with n ≥1. In the spherical coordinates, a rotationally symmetric function f on Sⁿcan be regarded as a function on [0, π], such that

f (θ ):=f (x₁,· · ·, x_n+1)withx_n+1=cosθ.

In particular, f (0)and f (π )are values of f at the north and south poles respectively. By the correspondence xn+1=cosθ, one can naturally extend f (θ )on [0, π]to be a 2π-periodic and even function on R. Observe that if f ∈C^m(Sⁿ)for some integer m, then f ∈C^m(R). Using the superscript denotes _dθ^d, we have f(0) =f(π ) =0 if it is differentiable. Throughout this paper, we will always use these conventions.

A typical existence result about the rotationally symmetric case of Eq. (1) was first established in [30] and then supplemented in [29]. To state this result, we introduce two quantities:

ni(f )=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

−f(^π₂), n≥2, π

0

[f(θ )−f(^π₂)]tanθdθ, n=1,

and

pi(f )= π

0

f(θ )cotθdθ.

Theorem A ([29,30]). Assume that f ∈C²(Sⁿ) (requiringC⁶forn =2), and that f is positive and rotationally symmetric. If f(^π₂) =0and ni(f ) ·pi(f ) <0, then Eq.(1)admits a rotation- ally symmetric solution.

The assumption f(^π₂) =0 in the above theorem is not essential, but used to reduce some difficulties in blow-up analyses. It was showed in [29] that this assumption can be removed when fis very close to a positive constant. The aim of this paper is to remove this technical assumption in a general case.

For n =1, 2, we follow the arguments in [29,30], carry out more delicate analyses, and then remove the condition f(^π₂) =0 completely.

Theorem 1. Assume that f∈C²(S¹)or f ∈C^2,α(S²)for some α∈(0, 1), and that f is positive and rotationally symmetric. If ni(f ) ·pi(f ) <0, then Eq.(1)admits a rotationally symmetric solution.

For n ≥3, the above method is no longer applicable. Inspired by [27], we carry out blow-up analyses for a variational method to obtain the following

Theorem 2. Assume that f ∈C²(Sⁿ)with n ≥3, and that f is positive and rotationally symmet- ric. If ni(f ) <−ⁿ_n⁺₊¹₂f(^π₂)²/f (^π₂)and pi(f ) >0, then Eq.(1)admits a rotationally symmetric solution.

We see in the case n ≥3, a little more restriction on ni(f )will be needed when the assump- tion f(^π₂) =0 is removed. However if f(^π₂) =0, Theorem2just becomes into the existence theorem [30, Theorem 1.3].

The paper is organized as follows. In section2, we provide some basic facts about Eq. (1) and convex bodies. Then we prove Theorem1and 2in section3and section4respectively.

2. Preliminaries

In this section we state some properties about Eq. (1) and a few facts in convex geometry, which will be used throughout this paper. One can consult [37] for more knowledge about convex geometry.

An obstruction for solutions to Eq. (1) was found by Chou and Wang [11].

Lemma 1 ([11]). Let Hbe a C³-solution to equation (1). Then we have

Sⁿ

∇ξf

Hⁿ⁺¹=0 (3)

for any projective vector field ξ, given by

ξ(x)=Bx−(x^TBx)x, x∈Sⁿ, where Bis an arbitrary matrix of order n +1.

In the rotationally symmetric case, (3) is reduced to π

0

f(θ )sinⁿθcosθ

Hⁿ⁺¹(θ ) dθ=0. (4)

See [30, Proposition 3.1].

We have a volume estimate for any solution to Eq. (1).

Lemma 2 ([30]). There exist positive constants Cn, C˜n, depending only on n, such that for any solution Hto Eq.(1), we have

C_n

f_min≤vol(H )≤ ˜C_n f_max, where f_min=inf

Sⁿf, f_max=sup

Sⁿ

f, and vol(H ) is the volume of the convex body determined byH.

Let X be any convex body in Rⁿ⁺¹, and H be its support function. Under the action of a unimodular linear transform A^T ∈SL(n +1), Xbecomes into another convex body XA:=A^TX.

Denote the support function of XAby HA. Then H_A(x)= |Ax| ·H

Ax

|Ax| , x∈Sⁿ. (5) See e.g. [30, (2.11)].

We remark that if H is a solution to Eq. (1), then HAis a solution to the following equation det(∇²HA+HAI )= f_A

H_Aⁿ⁺², fA(x)=f Ax

|Ax| . (6)

See [11] for more details.

Related to the linear transform, there is an integral variable substitution formula.

Lemma 3 ([28]). For any integral function gon Sⁿ, and any matrix A ∈GL(n +1), we have the following variable substitution for integration:

Sⁿ

g(y)dS(y)=

Sⁿ

g Ax

|Ax|

· |detA|

|Ax|ⁿ⁺¹dS(x).

By this lemma and (5), we see for any unimodular linear transform A ∈SL(n +1), there is

Sⁿ

f Hⁿ⁺¹=

Sⁿ

f_A

H_Aⁿ⁺¹, (7)

where fAis the same as in (6).

John’s Lemma in convex geometry says that for any non-degenerate convex body Xin Rⁿ⁺¹, there is a unique ellipsoid Ewhich attains the minimum volume among all ellipsoids contain- ingX. This ellipsoid Eis called the minimum ellipsoidof X. It satisfies

1

n+1E⊂X⊂E,

where λE= {x₀+λ(x−x₀) :x∈E}with x0the center of E. We say Xis normalizedif the E is a ball.

We denote the area of Sⁿby σn, and the unit vector along xi-axis by eifor i=1, 2, · · ·, n +1.

3. Proof of Theorem1

In this section, we prove Theorem 1. To achieve this, one needs an improvement of [30, Theorem 1.2].

Theorem 3. Assume that f∈C²(S¹)or f ∈C^2,α(S²)for some α∈(0, 1), and that f is positive and rotationally symmetric. If ni(f ) ·pi(f ) =0, then there exist positive constants C,C˜ de- pending only on nand f, such that for any rotationally symmetric solution H to Eq.(1), we have

C≤H≤ ˜C.

Once we have Theorem3, we can repeat the arguments of [29] to prove that Eq. (1) admits a rotationally symmetric solution provided ni(f ) ·pi(f ) <0, yielding Theorem 1. One can consult [29] for details. So in the rest of this section, we are only concerned about Theorem3, and give its proof.

Let {H_k}be any sequence of rotationally symmetric solutions to Eq. (1). For each Hk, define a_k∈Rand Ak∈SL(n +1)as

a_k=f (^π₂)¹²/H_k(^π₂)ⁿ⁺¹, Ak=diag

a

n+11

k ,· · ·, a

n+11

k , a⁻

n+1n

k

.

(8)

Let

H_Ak(x)= |A_kx| ·H_k A_kx

|A_kx|

, x∈Sⁿ. (9)

Then HA_k is a rotationally symmetric solution to Eq. (6) with Areplaced by Ak. Note that H_Ak(^π₂)=a

1 n+1

k H_k(^π₂)=f (^π₂)²ⁿ⁺²¹ . (10) Lemma 4. There exist positive constants C, C˜ depending only on n, fmaxand fmin, such that

C≤H_Ak≤ ˜C. (11)

Proof. By the rotational symmetry of H_Ak, one can easily see that vol(HA_k)≥ κ_n

n+1H_Ak(^π₂)ⁿ[H_Ak(0)+H_Ak(π )], (12) and

maxHA_k ≤

HA_k(^π₂)²+ [HA_k(0)+HA_k(π )]². (13) Recalling HAk satisfies equation (6), by the volume estimate given in Lemma2, we have

vol(HA_k)≤C_n

maxf_Ak=C_n

f_max, (14)

which together with (12) yields

H_Ak(0)+H_Ak(π )≤ ˜C_n

f_max·H_Ak(^π₂)⁻ⁿ

= ˜C_n

f_max·f (^π₂)⁻²ⁿ⁺ⁿ²

≤ ˜C_nf

1

max2 f⁻

n 2n+2

min , where we have used (10) for the equality. Now from (13) we obtain

maxH_Ak≤f

1 2n+2

max + ˜C_nf

1

max2 f⁻

n 2n+2

min , (15)

which means the second inequality in (11) is true.

On the other hand, by virtue of [30, Lemma 2.3], there is

minH_Ak·(maxH_Ak)ⁿ·vol(HAk)≥C_nfmin.

Combining it with (14) and (15), we easily obtain the first inequality in (11). 2

To obtain uniform upper and lower bounds for {H_k}, by (9) and Lemma4, we should exclude two cases, namely ak→ +∞or ak→0⁺when k→ +∞. The second case can be still solved by the method developed in [30]. But for the first case where ak→ +∞, one needs more delicate analyses to deal with. The following are details.

First note that in the rotationally symmetric case, fAk defined in (6) can be written as

f_Ak(θ )=f (γ_ak(θ )), (16)

where

γ_ak(θ )=arccos cosθ

ia_k(θ ) , i_ak(θ )=

a_k²sin²θ+cos²θ , (17) see [30, (3.3)–(3.4)].

Lemma 5. Assume a_k → +∞ when k→ +∞. Then H_Ak converges to the constant function f (^π₂)²ⁿ¹⁺² uniformly on [0, π].

Proof. From Lemma4, we see H_Ak

is uniformly bounded. By the Blaschke selection theorem, one may assume that

H_Ak

converges uniformly to some support function H_∞on Sⁿ, which is also rotationally symmetric. It remains to prove that

H_∞≡f (^π₂)²ⁿ¹+2 onSⁿ. (18) Recall Eq. (6), namely

det(∇²H_Ak+H_AkI )= fA_k

H_Aⁿ⁺²

k

onSⁿ. (19)

Note when ak→ +∞, fA_kconverges to f (^π₂)almost everywhere on [0, π], see (16). Passing to the limit in Eq. (19), we see H_∞is a generalized solution to

det(∇²H_∞+H_∞I )=f (^π₂) H_∞ⁿ⁺² onSⁿ.

So H_∞is an elliptic affine sphere, which must be an ellipsoid [8]. By the rotational symmetry of H_∞, it should be expressed as

H_∞(x)=f (^π₂)²ⁿ¹+2|x|, x∈Sⁿ (20) for some ∈SL(n +1)of form

=diag

λⁿ+¹1,· · ·, λⁿ+¹1, λ⁻ⁿ+ⁿ1

withλ >0.

Then

H_∞(^π₂)=f (^π₂)²ⁿ⁺²¹ λⁿ⁺¹¹ .

On the other hand, recalling (10), we have

H_∞(^π₂)=f (^π₂)²ⁿ⁺²¹ .

Hence λ =1, namely is the identity matrix of order n +1. Now (20) is simplified into (18).

The proof of this lemma is completed. 2

Recall H_Aksatisfies equation (19), which in the rotationally symmetric case can be simplified into the following form:

(H_A

k+H_Ak)(H_A

kcotθ+H_Ak)ⁿ⁻¹= f_Ak H_Aⁿ⁺²

k

on[0, π], (21) see [29, (2)].

Lemma 6.

(a) There exist positive constants C, C˜ depending only on n, fmaxand fmin, such that C≤H_A

kcotθ+H_Ak≤ ˜C, (22)

C≤H_A

k+HA_k≤ ˜C. (23)

(b) If ak→ +∞when k→ +∞, then H_A

ksin¹⁴θ

converges to 0uniformly on [0, π].

Proof. (a) Recalling Lemma4, we obtain from (21) that C₁≤(H_A

k+H_Ak)(H_A

kcotθ+H_Ak)ⁿ⁻¹≤C₂ (24)

for some positive constants C1, C2depending only on n, fmaxand fmin. Note (H_A

kcosθ+H_Aksinθ )=(H_A

k+H_Ak)cosθ, the above inequality can be written as

C₁≤ 1

nsinⁿ⁻¹θcosθ · d dθ(H_A

kcosθ+H_Aksinθ )ⁿ≤C₂. (25)

When θ∈ [0, π/2], we have by (25) that d

dθC1sinⁿθ≤ d dθ(H_A

kcosθ+HA_ksinθ )ⁿ≤ d

dθC2sinⁿθ, which together with H_A

k(0) =0 implies C

1 n

1 sinθ≤H_A

kcosθ+H_Aksinθ≤C

1 n

2 sinθ, ∀θ∈ [0, π/2].

Similarly, by (25) and H_A

k(π ) =0, we also have C

1 n

1 sinθ≤H_A

kcosθ+H_Aksinθ≤C

1 n

2 sinθ, ∀θ∈ [π/2, π]. Therefore

C

1 n

1 ≤H_A

kcotθ+H_Ak≤C

1 n

2, ∀θ∈ [0, π], which is just (22). Now recalling (24), one can obtain (23).

(b) We first note that by (11) and (23), there is

|H_A

k| ≤C₃ (26)

for some positive constant C3depending only on n, fmaxand fmin.

Now assume a_k→ +∞as k→ +∞. We claim that for any δ∈(0, π/2), H_A

k⇒0 uniformly on[δ, π−δ]. (27) In fact, by (16), fAk⇒f (^π₂)uniformly on [δ, π−δ]. By Lemma5, HAk⇒f (^π₂)²ⁿ⁺²¹ uniformly on [0, π], which implies that H_Ak⇒0 uniformly on [0, π]. Then by (21), when θ∈ [δ, π−δ], we have

H_A

k =fA_kH_A⁻ⁿ⁻²

k (H_A

kcotθ+HA_k)¹⁻ⁿ−HA_k

⇒f (^π₂)·f (^π₂)^−2nn+2+2 ·f (^π₂)^2n1−n+2 −f (^π₂)²ⁿ¹⁺²

=0.

Thus (27) is true.

We now prove

H_A

ksin¹⁴θ⇒0 uniformly on[0, π]. (28) Given any >0. By (26), there exists some δ∈(0, π/2), such that

[0,δ]∪[supπ−δ,π]|H_A

ksin¹⁴θ|< , ∀k. (29) Then by virtue of (27), there exists a k0, such that

[δ,πsup−δ]|H_A

k|< , ∀k≥k₀. (30)

Combining (29) and (30), we have

[sup0,π]|H_A

ksin¹⁴θ|< , ∀k≥k0. Thus (28) is true. 2

With a more detailed analysis, we can strengthen Lemma5for n =1, 2.

Lemma 7. Assume ak→ +∞as k→ +∞. For sufficiently large k, we have

max[0,π]H_A

k−f (^π₂)²ⁿ¹+2≤C

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎩ π

0

|fA_k−f (^π₂)|dθ, ifn=1, π

0

|fA_k−f (^π₂)|sin¹²θdθ, ifn=2,

(31)

where Cis a positive constant depending only on f (^π₂).

Proof. For simplicity, let

β:=f (^π₂)²ⁿ¹⁺² and hk(θ ):=HA_k(θ )−β.

Also we will drop the subscript k in the following proof if no confusion arises. Recall by Lemma5, hconverges uniformly to 0 on [0, π]as k→ +∞.

(a) When n =1. Now Eq. (21) is simplified as h+h+β= f_A

H_A³. (32)

Observing that

H_A⁻³=(β+h)⁻³

=β⁻³−3β⁻⁴h+6τ⁻⁵h², where τ is between βand HA(θ ), and that β=f (^π₂)^1/4, we have

f (^π₂)

H_A³ =β−3h+6β⁴τ⁻⁵h². Then (32) can be written as

h+h+3h−6β⁴τ⁻⁵h²=f_A−f (^π₂) H_A³ , namely

h+4h=f_A−f (^π₂)

H_A³ +6β⁴τ⁻⁵h². (33)

Recalling h(^π₂) =0 by (10), we can apply Lemma8to equation (33) and then obtain

max|h| ≤

f_A−f (^π₂) H_A³

L¹[0,π]

+6β⁴τ⁻⁵h²

L¹[0,π]. (34) Since HA⇒β >0 uniformly on [0, π]as k→ +∞, there exists a large integer k0, such that

max|HA−β| ≤β

2, ∀k≥k0. Then when k≥k₀we have

max|h| ≤β

2 and HA, τ∈β 2,3β

2

. (35)

Thus (34) is simplified into

max|h| ≤8β⁻³f_A−f (^π₂)

L¹[0,π]+192β⁻¹π(max|h|)². By virtue of max|h| →0 as k→ +∞, we also can assume

192β⁻¹π·max|h|<1

2 whenk≥k₀. Hence

max|h| ≤16β⁻³f_A−f (^π₂)

L¹[0,π], ∀k≥k₀, which is just (31) for n =1.

(b) When n =2. Now Eq. (21) is written as

(h+h+β)(hcotθ+h+β)= f_A

H_A⁴, (36)

namely

β(h+hcotθ+2h)+β²+(h+h)(hcotθ+h)= f_A

H_A⁴. (37)

Observing that

H_A⁻⁴=(β+h)⁻⁴

=β⁻⁴−4β⁻⁵h+10τ⁻⁶h², where τ is between β and HA(θ ), and that β=f (^π₂)^1/6, we have

f (^π₂)

H_A⁴ =β²−4βh+10β⁶τ⁻⁶h². (38) Then (37) can be written as

β(h+hcotθ+6h)+(h+h)(hcotθ+h)−10β⁶τ⁻⁶h²=f_A−f (^π₂) H_A⁴ , namely

h+hcotθ+6h=f_A−f (^π₂)

βH_A⁴ +Ra(θ ), (39)

where

R_a(θ )=10β⁵τ⁻⁶h²−β⁻¹(h+h)(hcotθ+h). (40) Applying Lemma10to equation (39), we have

max|h| ≤2 π

0

|f_A−f (^π₂)|

βH_A⁴ (2−log sinθ )sinθdθ +2

π

0

|R_a(θ )|(2−log sinθ )sinθdθ

≤4 π

0

|fA−f (^π₂)|

βH_A⁴ sin¹² θdθ+6 π

0

|R_a(θ )|sin³⁴θdθ.

Recalling (35), we obtain

max|h| ≤64β⁻⁵ π

0

|f_A−f (^π₂)|sin¹²θdθ+6 π

0

|R_a(θ )|sin³⁴θdθ. (41)

We see Rainvolves derivatives of h. To deal with them, we need to explore (36) more care- fully. Note that

(hcosθ+hsinθ+βsinθ )=(h+h+β)cosθ, then Eq. (36) is equivalent to

d

dθ(hcosθ+hsinθ+βsinθ )²= f_A

H_A⁴ ·2 sinθcosθ.

Therefore we have

(hcosθ+hsinθ+βsinθ )²= θ

0

f_A

H_A⁴ ·2 sintcostdt, ∀θ∈ [0, π/2]. (42) Since hcotθ+h +β >0, there is

hcosθ+hsinθ+βsinθ=

⎛

⎝ θ

0

fA

H_A⁴ ·2 sintcostdt

⎞

⎠

1/2

.

Thus we have

hcosθ+hsinθ=^θ

0

fA

H_A⁴ ·2 sintcostdt ^1/2−βsinθ

= ^θ

0 fA

H_A⁴ ·2 sintcostdt−β²sin²θ _θ

0 fA

H_A⁴ ·2 sintcostdt1/2

+βsinθ

≤ 1 βsinθ

θ

0

f_A

H_A⁴·2 sintcostdt−β²sin²θ.

(43)

Recalling (38), there is fA

H_A⁴ =fA−f (^π₂)

H_A⁴ +β²−4βh+10β⁶τ⁻⁶h², which implies that

θ

0

f_A

H_A⁴ ·2 sintcostdt= θ

0

f_A−f (^π₂)

H_A⁴ ·2 sintcostdt+β² θ

0

2 sintcostdt

+ θ

0

(−4βh+10β⁶τ⁻⁶h²)·2 sintcostdt

= θ

0

f_A−f (^π₂)

H_A⁴ ·2 sintcostdt+β²sin²θ

+ θ

0

(−4β+10β⁶τ⁻⁶h)h·2 sintcostdt.

Recalling (35), we obtain from the above equality that

θ

0

f_A

H_A⁴ ·2 sintcostdt−β²sin²θ≤32β⁻⁴ θ

0

|f_A−f (^π₂)|sintdt+324β(max|h|)sin²θ.

Then (43) is simplified into

hcosθ+hsinθ≤ 32β⁻⁵ sinθ

θ

0

|f_A−f (^π₂)|sintdt+324(max|h|)sinθ,

namely

hcotθ+hsin¹²θ≤32β⁻⁵ sin³²θ

θ

0

|fA−f (^π₂)|sintdt+324(max|h|)sin¹²θ.

Integrating both sides over [0, π/2], we have

π

2

0

|hcotθ+h|sin¹²θdθ

≤32β⁻⁵

π

2

0

dθ sin³²θ

θ

0

|fA−f (^π₂)|sintdt+162π(max|h|)

=32β⁻⁵

π

2

0

|fA−f (^π₂)|sintdt

π

2

t

dθ

sin³²θ+162π(max|h|).

(44)

Note that

π

2

t

dθ sin³²θ ≤π

2

3 2

π

2

t

dθ θ³²

=π 2

3

2 ·2[t⁻¹² −(π/2)⁻¹²]

<4 sin⁻¹²t, then (44) is reduced into

π

2

0

|hcotθ+h|sin¹²θdθ≤128β⁻⁵

π

2

0

|f_A−f (^π₂)|sin¹²tdt+162π(max|h|). (45)

Now similar to (42), we have

(hcosθ+hsinθ+βsinθ )²= π

θ

f_A

H_A⁴ ·2 sint|cost|dt, ∀θ∈ [π/2, π]. Then following almost the same arguments used to obtain (45), one can get

π

π 2

|hcotθ+h|sin¹²θdθ≤128β⁻⁵ π

π 2

|f_A−f (^π₂)|sin¹²tdt+162π(max|h|). (46)

Adding (45) and (46) together, we have π

0

|hcotθ+h|sin¹²θdθ≤128β⁻⁵ π

0

|f_A−f (^π₂)|sin¹²tdt+324π·max|h|. (47)

Now we can estimate the integral about Rain (41). By the definition of Rain (40), there is π

0

|R_a(θ )|sin³⁴θdθ≤ π

0

10β⁵τ⁻⁶h²sin³⁴θdθ

+ π

0

β⁻¹|h+h| |hcotθ+h|sin³⁴θdθ

≤640πβ⁻¹(max|h|)²+m_k π

0

|hcotθ+h|sin¹²θdθ,

where (35) is used, and mkis defined as m_k:=β⁻¹ max

θ∈[0,π]|h(θ )+h(θ )|sin¹⁴θ.

By estimate (47), the above inequality becomes into π

0

|Ra(θ )|sin³⁴θdθ≤640πβ⁻¹(max|h|)²+324π mk·max|h|

+128β⁻⁵mk

π

0

|fA−f (^π₂)|sin¹² tdt.

(48)

Recall Lemma6(b), |h(θ ) +h(θ )| sin¹⁴θ converges uniformly to 0 on [0, π]when k→ +∞, which implies

m_k→0 ask→ +∞.

Also recall max|h| →0. We can assume when k≥k₀that 640πβ⁻¹max|h| +324π mk< 1

12. Then (48) is simplified into

π

0

|R_a(θ )|sin³⁴θdθ≤ 1

12max|h| + 1 12β⁻⁵

π

0

|f_A−f (^π₂)|sin¹²tdt. (49)

Now combining (41) and (49), we obtain

max|h| ≤129β⁻⁵ π

0

|f_A−f (^π₂)|sin¹²θdθ,

which is just (31) for n =2. 2

The following Lemmas8and 10have been used in the proof of the above Lemma7.

Lemma 8. Assume h ∈C²(R)is 2π-periodic and even. If it satisfies the following differential equation

h+4h=g, (50)

and h(^π₂) =0, then there is

maxR |h| ≤ gL¹[0,π]. Proof. One can easily solve equation (50) to obtain

h(θ )=c₁cos 2θ+c₂sin 2θ−1 2cos 2θ

θ

0

g(t)sin 2tdt+1 2sin 2θ

θ

0

g(t)cos 2tdt,

where c1and c2are constants to be determined. Then we have

h(θ )= −2c1sin 2θ+2c2cos 2θ+sin 2θ θ

0

g(t)sin 2tdt+cos 2θ θ

0

g(t)cos 2tdt.

From h(0) =0, we get c2=0. And h(^π₂) =0 implies

c₁=1 2

π

2

0

g(t)sin 2tdt.

Therefore his given by

h(θ )=1 2cos 2θ

π

2

θ

g(t )sin 2tdt+1 2sin 2θ

θ

0

g(t )cos 2tdt.

Hence when θ∈ [0, π],

|h(θ )| ≤1 2

π

2

θ

g(t )sin 2tdt +1

2 θ

0

g(t )cos 2tdt

≤1 2 π

0

|g(t )|dt+1 2 π

0

|g(t )|dt

= π

0

|g(t )|dt,

which leads to the conclusion of this lemma. 2 Lemma 9. The homogeneous differential equation

h+hcotθ+6h=0 in(0, π ) has the following two fundamental solutions:

h1(θ )=1−3 cos²θ, h₂(θ )= −3

4cosθ+1

8(1−3 cos²θ )log1−cosθ 1+cosθ. These two solutions have the following properties:

(a) h₁(^π₂) =1, h₁(^π₂) =0and h2(^π₂) =0, h₂(^π₂) =1.

(b) Abel’s identity: h1h₂−h₁h₂=cscθ, ∀ θ∈(0, π ).

(c) h₁(θ ) =6 sinθcosθ.

(d) |h₂(θ )| ≤2 −log sinθ, ∀ θ∈(0, π ).

(e) |h₂(θ ) sinθ| ≤5/2, ∀ θ∈(0, π ).

(f) As θ→0⁺or θ→π⁻, there is

h₂(θ )=−1/2+o(1) sinθ .

Proof. Direct computations show that h1and h2are solutions to the differential equation in the lemma. And one can easily check (a), (b) and (c).

We note that 1 2

log1−cosθ 1+cosθ

≤ −log sinθ+log 2, ∀θ∈(0, π ).

Since both sides are symmetric with respect to θ=π/2, we only need to verify it for θ∈(0, π/2], which is a direct corollary of the following equality:

1 2

log1−cosθ 1+cosθ

=1 2

log 1−cos²θ (1+cosθ )²

=

log sinθ 1+cosθ

.

Now by the expression of h2, there is

|h₂(θ )| ≤3 4+1

2

log1−cosθ 1+cosθ

≤3

4−log sinθ+log 2

≤2−log sinθ, which is just (d).

Computing h₂, we have h₂(θ )=3

4sinθ+3

4sinθcosθlog1−cosθ 1+cosθ +1

4(1−3 cos²θ )cscθ.

Then

|h₂(θ )| ≤3 4+3

4sinθ

log1−cosθ 1+cosθ

+1 2cscθ

≤3 4+3

2sinθ·(−log sinθ+log 2)+1 2cscθ

≤2+1 2cscθ, which implies (e).

By the expression of h₂, we see as θ→0⁺or θ→π⁻that h₂(θ )sinθ→ −1

2, yielding (f). 2

Lemma 10. Assume h ∈C²(R)is 2π-periodic and even. If it satisfies the following differential equation

h+hcotθ+6h=g, (51)

and h(^π₂) =0, then there is

maxR |h| ≤2 π

0

|g(θ )|(2−log sinθ )sinθdθ. (52)

Proof. Recalling Lemma9, h1and h2are two fundamental solutions to the homogeneous dif- ferential equation:

h+hcotθ+6h=0 in(0, π ).

By method of variation of parameters and Lemma9(b), we solve (51) in (0, π )and obtain

h(θ )=c₁h₁+c₂h₂−h₁ θ

π/2

h₂(t)g(t)sintdt+h₂ θ

π/2

h₁(t)g(t)sintdt, (53)

where c1and c2are constants to be determined. Note the assumption h(^π₂) =0, and by (53) h(^π₂)=c1h1(^π₂)+c2h2(^π₂)=c1,

there is c1=0. Then

h(θ )=c₂h₂−h₁ θ

π/2

h₂(t)g(t)sintdt+h₂ θ

π/2

h₁(t)g(t)sintdt. (54)

To determine c2, we need to compute h(0).

By Lemma9(d), |h₂|is an integrable function in (0, π/2]. Then 0

π/2

h2(t)g(t)sintdt

is a finite number. For small θ >0, one can rewrite (54) as 1

h₂(θ )

h(θ )+h₁(θ ) θ

π/2

h2(t)g(t)sintdt

=c2+ θ

π/2

h1(t)g(t)sintdt. (55)

Letting θ→0⁺, and recalling h(0) =0, h₁(0) =0 and Lemma9(f), we obtain

0=c₂+ 0

π/2

h₁(t)g(t)sintdt,