Notes on Weyl and Minkowski Problem

Yang Guo260313651

September 6, 2011

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1 Introduction

In this note, I will mainly discuss the work of Nirenberg on Weyl andMinkowski Problem and the work of Cheng and Yau on n-dimension-alMinkowski Problem. These problems are concerned with the existence ofa convex surface of prescribing first fundamental form and Gauss curvature.The method of continuity and theories on elliptic partial differential equationare used in both works.

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2 On Weyl Problem

Weyl Problem: Given a positive definite quadratic form ds2 defined atevery point of the unit sphere such that the Gauss curvature of the form iseverywhere positive (ds2 satisfy certain conditions).Does there exist a closedconvex surface which mapped one to to one onto the sphere so that its firstfundamental form (in terms of parameters on the sphere) is ds2?

Nirenberg proved this using the method of continuity in the followingsteps.

Theorem 2.1. The given quadratic form is connected to the first fundamen-tal form of unit sphere ds2

0 by a curve of quadratic form ds2t , 0 ≤ t ≤ 1,it

equals to ds20 at t = 0 and equals to ds2 at t = 1 such that all forms on the

curve have positive Gauss curvature.

Theorem 2.2. The set of points t in [0, 1] for which the form ds2t is realizable

as a convex surface is an open set.

Theorem 2.3. The set of points t in [0, 1] for which the form ds2t is realizable

as a convex surface is a closed set.

Note that since ds20 is realizable, we infer that the set of t on which the ds2

t

is realizable as a convex surface is therefore [0, 1].And thus ds2 is realizableas a convex surface.

2.1 Preparation and Precise statement of Weyl Prob-lem

We shall now precisely state the problem by using the language of functionalanalysis.

We first introduce a parameter system on S2, notice that it is impossibleto have a single parameter system over the whole sphere. Let us considernow two overlapping region R1 and R2 containing north and south pole re-spectively which are bounded by parallels of latitude. The parameter system(u, v) for R1 is constructed by stereographic projections of R1 from south poledenoted by R′1 while the parameter system of R2 is constructed by stereo-graphic projections of R2 from north pole denoted by R′2. In the commonstrip, the two systems can be transformed to each other by a transforma-tion with a non-vanishing Jacobian. We also define the following spaces of

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functions:

mi C: f ∈mi C if ‖ f ‖im=

∑mk=0 sup |Dkf | <∞

m+αi C: f ∈m+α

i C if

‖ f ‖im+α=m∑k=0

sup |Dkf |+ supx,y∈Ri

f(x)− f(y)

|x− y|<∞

Ck : f ∈ Ck if f ∈m1 C ∩m2 C with ‖ f ‖k=‖ f ‖1k + ‖ f ‖2

k

Sk : A three vector X(u, v) ∈ Sk if each component belongs to Ck, and‖ X ‖ is defined as sum of norms in Ck for each component of X.

Mk: A quadratic form ds2 ∈ Mk if each E, F ,Gis in ki C with i = 1, 2.

And the norm defined as

‖ ds2 ‖=2∑i=1

‖ E ‖ik + ‖ F ‖ik + ‖ G ‖ik

M ′k: A positive definite quadratic form ds2 ∈ M ′

k if it is in Mk and itsGauss curvature is positive everywhere.

Mk”: ds2 ∈Mk” if it is in M ′k and is realizable as a regular surface. Now,

we formulate Weyl’s Problem as follows:Weyl Problem: Under what condition have Mk” = M ′

k? Note that if itholds for some positive integer k, then it holds for larger integers.Our main result is:

Theorem 2.4. M4” = M ′4

Theorem 2.5. Let ds2 ∈ M ′4+α ,0, α, 1. Then there exist a vector function

X(u, v) having ds2 has first fundamental form.i.e M ′4+α = M”4+α.

To prove theorem 1’ we need to show:1.M ′

4+α is connected2.M”4+α is open in M ′

4+α

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3.M”4+α is closed in M ′4+α

Now we denote the second fundamental form as L,M ,N and mean curva-ture H,unit inward normal X3 ,∆ =

√EG− F 2and other standard notations

that would be familiar to reader.

We introduce the following function.Let

ρ(u, v) =1

2(X ·X).

Then

ρuu = Γ111 ρu + Γ11

2 ρv + LX ·X3 + E

ρuv = Γ121 ρu + Γ12

2 ρv +MX ·X3 + F

ρvv = Γ221 ρu + Γ22

2 ρv +NX ·X3 +G

and

K(X ·X3)2 =LN −M2

∆2(X ·X3)

=1

∆2(ρuu − Γ11

1 ρu − Γ112 ρv − E)(ρvv − Γ22

1 ρu + Γ222 ρv −G)

− 1

∆2(ρuv − Γ12

1 ρu − Γ122 ρv − F )2

Also,

(X ·X3)2 = X2 − (X × Xu ×Xv

)∆)2

= X2 − 1

∆2[(X ·Xu)Xv − (X −Xv)Xu]

2

= 2ρ2 − 1

∆2(Eρ2

v − 2Fρuρv +Gρ2u)).

Finally,we set

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F (u, v, · · · , ρvv) =1

∆2(ρuu − Γ11

1 ρu − Γ112

ρv − E)(ρvv − Γ221 ρu + Γ22

2 ρv −G)

− 1

∆2(ρuv − Γ12

1 ρu − Γ122 ρv − F )2

−K · [2ρ2 − 1

∆2(Eρ2

v − 2Fρuρv +Gρ2u))]

and we get

4FρuuFρvv − F 2ρuv =

4

∆2K(X ·X3) > 0

Since X,Xu and Xv are linearly independent, we may express m-th orderderivative of X with respect to u, v as a linear combination of X,Xu and Xv

involving derivative of ρ,E,F ,and G of order at most m.We cite a result by Hopf:

Lemma 2.6. If a solution ρ(u, v) of F (u, v · · · , ρvv) = 0 in domain D hascontinuous first and second derivatives satisfy Holder condition with exponentβ ,then ρ has partial derivatives up to order m+2 and m+2nd derivatives sat-isfy Holder condition with exponent β in ay closed subdomain. In particular,if F is analytic then ρ is analytic.

From this we derive the following lemma:

Lemma 2.7. For any closed convex surface X(u, v) ∈ S2+α,0 < α < 1 hasds2 ∈M ′

m+β for m ≥ 2,0 < β < 1,then X ∈ Sm+β.

Proof. Apply Hopf’s lemma to F (u, v, · · · , ρvv) = 0, since ρ has continuousfirst and second derivative which satisfy Holder condition with exponent βwhich implies ρ ∈ Cm+β since the m-th derivative of X can be expressedas linear combination of X,Xu and Xv with derivatives of E,F ,G,ρ. ThenX ∈ Sm+β as claimed.

We have a stronger form due to Nirenberg

Lemma 2.8. If X(u, v) ∈ S2 and ds2 ∈ M ′m+β then X ∈ Sm+β. Here we

don’t need Holder condition. In particular if X ∈ S2 with ds2 analytic, thenX(u, v) is analytic.

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2.2 Connectedness

We now prove the assertion that M ′4+α is connected.

Proof. By uniformization theorem for Riemann surface ,there exist a localconformal mapping of manifold in large. For ds2 defines a Riemannian man-ifold homeomorphic onto the sphere defined by the conformal mapping. Letφ(u, v) and ψ(u, v) be the parameter image of (u, v) and let ds2 denote thefirst fundamental form of the manifold in parameter system (φ, ψ). By con-formality, ds2 = gds2

0, we construct ds2t = gtds2

0 for 0 ≤ t ≤ 1. Then theGauss curvature is given as

Kt =1

gt(1− t+ tgK1)

where K1 is the Gauss curvature of ds2. We now go back to ds2t in terms

of parameter (u, v). We have ds2t have positive curvature and connect ds2

to ds20. We can also show that ds2

t ∈ M4+α. We know φ and ψ satisfy theBeltrami equation:

(E

∆φv −

F

∆φu)v − (

F

∆φv −

G

∆φu)u = 0

since E,F ,G have fourth derivative satisfy Holder condition α. ApplyHopf’s lemma ,we have φ and ψ have fifth order derivative satisfying Holderconditions. Now ds2

t as a function of φ and ψ have derivative to fourth order,so ds2

t as a function of u and v has derivative up to fourth order satisfyingHolder condition.

2.3 Openness

We shall prove the following which implies openness:

Theorem 2.9. If ds2 ∈ M ′4+α, there exist a regular closed convex surface

X(u, v) having ds2 as its line element. Then there exist a ε such that forany ds2 in M2+α satisfy ‖ ds2 − ds′2 ‖2+α< 1 is realizable as a closed convexregular surface X ′(u, v) in S2+α. And if ds′2 ∈M4+α then X ′ ∈ S4+α.

To prove the theorem, we try to describe the condition as a PDE, thesolvability of the PDE in a neighborhood indicates the openness, For suchX ′(u, v) in the theorem, we set X ′ = X + Y and ds′2 = ds2 + δds2. Now

(dX + dY ) · (dX + dY ) = ds2 + δds2

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we have,2dX · dY = δds2 − (dY )2.

Notice that this equation id nonlinear elliptic. Nirenberg solved this prob-lem by an iteration scheme and reduced the nonlinear problem to a linearproblem described as follows;

Take Y0 = 0 and define Yn recursively as :

2dX · dYn = δds2 − (dYn)2

and we cans how Yn converge to a solution Y if ‖ δds2 ‖2+α is small enoughand all all Yn lie in S2+α.

Theorem 2.10. For equation of the form 2dX · dY = dσ2 − (dZ)2 whereX ∈ S4+α, ds2 ∈ M2+α and Z(u, v) ∈ S2+α, there exist a solution Y (u, v) ∈Sα. The solution may not be unique, but we can choose a particular solutionY = φ(Z)such that:

1. ‖ φ(Z) ‖≤ K(‖ dσ2 ‖2+α + ‖ Z ‖22+α)

2. ‖ φ(Z)− φ(Z1) ‖≤ K ‖ Z + Z1 ‖2+α‖ Z − Z1 ‖2+α

where Z,Z1 ∈ S2+α, K is a positive constant depending only on X(u, v).This theorem establish openness since, from Y0 = 0, we define Yn =

φ(Yn−1) for integers n ≥ 1 which are in S2+α by the theorem. We can choosea ball ‖ Z ‖2+α< R such that ‖ φ(Z) ‖2+α< R and

‖ φ(Z1)− φ(Z2) ‖2+α< M ‖ Z1 − Z2 ‖2+α

here constant M < 1. In this way, we can apply contraction mapping prin-ciple and get a solution Y such that ‖ Y ‖2+α≤ R.(the ball is complete) TOachieve this we require:

1. ‖ δds2 ‖2+α≤ 12RK

2. 2KR < 1

If this is satisfied ,

‖ φ(Z)− φ(Z1) ‖ ≤ K ‖ Z + Z1 ‖2+α‖ Z − Z1 ‖2+α

= 2KR ‖2+α‖ Z − Z1 ‖2+α

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is a contraction mapping. and

‖ φ(Z) ‖ ≤ K(‖ dσ2 ‖ + ‖ Z ‖22+α)

< K(1

2

R

K+R2) = R

Then X ′ = X + Y ∈ S2+α has first fundamental form ds′2,and completes theproof of the theorem.

Remark : We modify the statement of the theorem a bit by requiring thatZ ∈ S3+α instead of Z ∈ S2+α since we can approximate some Z ∈ S2+α bya sequence of Zm ∈ S3+α and let Ym be the solution of the equation. Notethat Ym is cauchy in the ball and thus converge to Y for Z ∈ S2+α.

Proof. First we show the existence of solutions. Write

dσ2 − (dZ)2 = dσ2 = Edu2 + 2F dudv + Gdv2

,we know dσ2 ∈ M2+α since Z ∈ S3+α. For dσ = 0, the homogeneousequation 2dX · dY = 0 has solution of the form

Y (u, v) = A×X(u, v)

where A is a constant vector. To solve the general equation we introduce newvariables. Our goal here is to reduce the PDE con-cerning with the 3-vectorX(u, v) to a PDE concerning with only one variable.Let

p1 = X3 · Yup2 = X3 · Yvϕ(u, v) = 1

∆(Xv · Yu −Xu · Yv)

From this, we infer:

Yu = 1∆2 (EG− FF )Xu + 1

∆

(FE − EF )Xv + 1∆

(EXv − FXu)ϕ+X3p1

Yv = 1∆2 (FG− GF )Xu + 1

∆

(GE − FF )Xv + 1∆

(FXv −GXu)ϕ+X3p2

Since Xu · Yv = 12(F −∆ϕ), we have

Xuv · Yu −Xuu · Yv =1

2(Ev − Fv + ϕ∆u + ∆ϕu).

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Express everything explicitly in Christoffel symbol, we get

L

∆X3 · Yv −

M

∆X3 · Yu =

1

2(c1 − ϕu)

Here

c1 =1

∆[Fu − Ev + Γ12

1 E + (Γ122 − Γ11

1 )F − Γ112 G]

Similarly, for Xv · Yu = 12(F + ∆ϕ) , we get

M

∆X3 · Yv −

N

∆X3 · Yu =

1

2(c2 − ϕv)

Here

c2 =1

∆[Gu − Fv + Γ22

1 E + (Γ222 − Γ21

1 )F − Γ212 G]

with

ϕu = 2(M∆p1 − L

∆p2) + c1

ϕv = 2(N∆p1 − M

∆p2) + c2

Differentiate these two equation w.r.t v and u respectively and subtract, weget

1

∆(p2u − p1v) = Hϕ+

1

2T

and

T =1

∆[L1

2E + (L22 − L1

1)F − L21G]

We observe that once we solve the equation for ϕ then since LN−M2 > 0,wecan solve for p1 and p2, we can obtain a equation for ϕ along.

1

∆(N

δϕu −

M

δϕv)u −

1

∆(M

δϕu −

L

δϕv)v + 2Hϕ =

1

∆(N

δc1 −

M

δc2)u −

1

∆(M

δc1 −

L

δc2)v − T

Here δ = K∆.We define L(p1, p2) = 1

∆(Nδp1 − M

δp2)u − 1

∆(Mδp1 − L

δp2)v Then L(ϕu, ϕv) +

2Hϕ = L(c1, c2)− T

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Note that also

1

∆(N

δq1 −

M

δq2)u −

1

δ(M

δq1 −

L

δq2)v =

1

2∆(c1v − c2u)

We define L′(q1, q2) = 1∆

(Nδq1 − M

δq2)u − 1

∆(Mδq1 − L

δq2)v Then L′(p1, p2) =

12∆

(c1v−c2u) These equations are linear elliptic equations of second order.Letdω0 = ∆0dudv be the element of surface area on the sphere, consider theequation

O(Z) =1

∆0

(aZu + bZv)u +1

∆0(bZu + cZv)v + eZ = f

where a, b, c posses continuous first and second order derivative satisfy Holdercondition with exponent α.e and f also satisfy Holder condition with expo-nent α. Then Hilbert proved that for f = 0, the equation has finite numberof linearly independent solutions, say Z1, · · · , Zn, without loss of generality,we assume they’re mutually orthogonal. A necessary and sufficient conditionthat the equation is solvable for f is that f is orthogonal to Z1, · · · , Zn.

Hilbert showed that there exist a Green function of second kind G(P, P ′)having logarithmic singularity at P = P ′ such that

O(G(P, P ′)) =n∑i=1

aiZi(P )Zi(P′)

here differentiation is w.r.t P . Then

Z(P ) =

∫G(P, P ′)f(P ′)dω0(P ′)

is the solution of the equation which is orthogonal to Z1, · · · , Zn.Consider first the homogeneous equation,

L(ϕu, ϕv) + 2Hϕ = 0

It correspond to the case of infinitesimal isometric deformation Y = A ×X(u, v) then ϕ = 2A ·X3(u, v) we have also that∫

A ·X3(L(c1, c2)− T )∆dudv = 0

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which is equivalent to (f=after integration by parts)∫[c1X3 ×Xv − c2X3 ×Xu + ∆TX3]dudv = 0

Indeed, this is the case. If we consider the terms in E and its derivatives,they are∫

[E(1

∆Γ12

1 X3 ×Xv −1

∆Γ22

1 X3 ×Xu + L12X3)− 1

∆EvX3×Xv ]dudv

note that ∫EvX3×Xv ]dudv =

∫(

1

∆)vdudv.

and

(1

∆X3 ×Xv)v = −(

1

∆Γ21

1 X3 ×Xv −1

∆Γ22

1 X3 ×Xu + L12X3)

We have the integration equal zero. Similar argument are repeated for F andG and this proves the equation has a solution. It remains to show that wecan choose a particular solution Y (u, v) to satisfy the two conditions of thetheorem. Now we solve for ϕ

ϕ(P ) =

∫G(P, P ′)[L(c1(P ′), c2(P ′))− T (P ′)]dω(P ′)

we claim that such Y is chosen as follows: Use solution of ϕ given to getp1 and p2, then integrate to get Y (u, v) requiring that Y (u, v) vanish ata fixed point (u0, v0) on the sphere. Thus this Y is uniquely defined forevery dσ2 ∈ M2+α. We want an estimate of second derivative of Y does notinvolve third derivative of Z, we should consider the following, notice thatthe expression

1

∆(c1v − c2u)

does not involve third derivative of Z. This is because the only secondderivative of E,F and G occurs in the expression as

2

∆(Zuu · Zvv − Z2

uv).

Now forL′(p1, p2) = 1

2∆(c1v − c2u)

1∆

(p2u − p1v) = Hϕ+ 12T

To estimate the solution of elliptic first order systems,we consider thesystem.

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a

L′(p1, p2) =1

2∆(c1v − c2u)

1

∆(p2u − p1v) = 0

b

L′(p1, p2) = 01

∆(p2u − p1v) = Hϕ+

1

2T

For the equation L′(ψu, ψv) = 12∆

(c1v − c2u), ψ = const is a homogeneoussolution, then the equation is solvable iff∫

1

2∆(c1v − c2u)∆dudv = 0.

Integration by parts verifies this. The if Γ(P, P ′) is the Green function ofthis ,

ψ(P ′) =

∫Γ′(P, P ′)

1

2∆(c1v(P )− c2u(P ))dω(P )

Let q1 = p1 − ψu and q2 = p2 − ψv Then

L′(q1, q2) = 0

There exist ψ′ uniquely determined within an additive constant such that

M∆q1 − L

∆q2 = −ψ′u

N∆q1 − M

∆q2 = −ψ′v

and

q1 = Mδψ′u − L

δψ′v

q2 = Nδψ′u − M

δψ′v

L(ψ′u, ψ′v) = Hϕ+ 1

2T

ψ′(P ′) =∫

Γ(P, P ′)(H(P )ϕ(P ) + 12T (P ))dω(P ) + c

WLOG,we can take c = 0. Now we can express p1 and p2 in terms ofderivatives of ψ and ψ′.

p1 = ψu + q1, p2 = ψv + q2

and q1 and q2 are expressed in terms of ψ′.

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Rewrite

ϕ(P ′) = −∫

[1

∆(N

δc1 −

M

∆c2)Gu −

1

∆(M

δc1 −

L

∆c2)Gv +GT ]dω

Lemma 2.11. For Y to satisfy the two conditions, it is sufficient that thefollowing holds.

‖ Y ‖2+α≤ h[‖ dσ2 ‖1+α + ‖ 1

∆(c1v − c2u) ‖α].

where h is a constant.

Proof. Since 1∆

(c1v − c2u) involves at most second derivative of coefficient ofdσ2 ,

‖ 1

∆(c1v − c2u) ‖α≤ K(‖ dσ2 ‖4 + ‖ Z ‖2

2+α)

insert this back, then

‖ Y ‖2+α≤‖ dσ2 ‖ + ‖ Z ‖22+α .

since Y is chosen to depend linearly on dσ2.For Y ,the solution of 2dX ·dY =dσ2− (dZ)2 and Y1, the solution of 2dX · dY = dσ2− (dZ1)2 we have Y − Y1

satisfy the condition of previous lemma with dσ2 = (dZ)2 − (dZ1)2 = (dZ +dZ1) · (dZ − dZ1).That is,

‖ Y − Y1 ‖2+α≤ h[‖ (dZ)2 − (dZ21) ‖1+α + ‖ 1

∆(c1v − c2u) ‖α]

also‖ (c1v − c2u) ‖α≤ K ‖ Z + Z1 ‖2+α‖ Z − Z1 ‖2+α .

combined with above equation , the second condition holds.

Now we want to estimate ϕ,p1 and p2, that will lead to the proof of thelemma and the two conditions of previous lemma will follow.

First we find a bound for ϕ and its first derivative. Then we obtain theestimate of ψ and ψ′ and their derivative of 2nd order. We’ll also show theysatisfy Holder condition. the estimates of ϕ, ϕu, and ϕv are obtained byanalyzing integrals with singular kernels. the integrand for ϕ involves thecoefficient of dσ2 and its first derivative, G and its first derivative, whichare easily seen to satisfy |ϕ| ≤ h ‖ dσ2 ‖1 where h is a constant depend onX(u, v). For the estimate of first derivative of ϕ, we use a lemma by Hopf:

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Lemma 2.12. The integral f(P ′) =∫ ∫

H(P, P ′)ρ(P )dudv where H(P, P ′)is a kernel having first order infinity at P − P ′ and its first derivative hashas second order infinity. If g(P ′) =

∫ ∫H(P, P ′)dudv has continuous first

derivative and ρ(P ) satisfy a Holder condition, then f(P ′) has continuousfirst derivative as

∂f

∂u′= ρ(P ′)

∂g

∂u′+

∫ ∫(ρ(P )− ρ(P ′))Hu(P, P

′)dudv.

Now let Gu = H and g =∫ ∫

Gududv =∮Gdv, if P ′ lies in a closed

subdomain of Ri, g has bounded first derivative, apply Hopf’s lemma ,wehave a bound for ϕu and ϕv

|ϕu| , |ϕu| ≤ h′ ‖ dσ2 ‖1+α .

To obtain estimate for derivative ψ and ψ′,we use the following theoremin functional analysis.

Theorem 2.13. If A is a bounded ,one-to one linear operator A : B1 → B2

where B1 and B2 are Banach spaces, then the inverse of A exists and is abounded linear operator in B2.

For L′(ψu, ψv) = f solvable if f is orthogonal to ψ = const, the solution ingeneral is not unique but can be made unique if we require it to be orthogonalto the solution of homogeneous equation. Note that Banach spaces Cm+α andC ′m+α orthogonal to ψ = const. the transformation L′ maps C ′2+α to C ′α andis bounded and one to one. Then it is onto and the inverse is bounded.

‖ ψ ‖2+α ≤ h ‖ f ‖α= h ‖ 1∆c1v − c2u ‖α

‖ ψ′ ‖2+α ≤ h′ ‖ Hϕ+ 12T ‖α

Finally since,

‖ ϕ ‖1+α, ‖ p1 ‖1+α, ‖ p2 ‖1+α≤ h′′[‖ dσ2 ‖1+α + ‖ 1

∆(c1v − c2u) ‖α].

This completes the proof of 2’.

2.4 Closeness

Let X(u, v) ∈ S2 be a closed convex surface with ds2 ∈ M”4+α, and choosethe origin to be the center of the largest sphere inscribed in X. We want toshow ‖ X ‖≤ K where K depends on ‖ ds2 ‖4,max 1

Kand max 1

∆. Further,if

this is true, a previous lemma implies X ∈ S4+α.

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i From Bonnet theorem, there is a bound on the diameter of closed sur-face depends on max 1

K, so X(u, v) is bounded,

ii Since X2u = E,X2

v = G, so ‖ X ‖1< K where K1 depends on ‖ ds2 ‖4

and max 1K

.

iii The second derivative of X is also bounded if H is bounded (see below)

Notice that

H =EN − 2FM +GL

2∆2

So

HE =E2N − 2EFM + LF 2

2∆2+L

2>L

2

by positive definiteness.L ≤ 2HE.

Similarly, we can show,

N < 2HG, |M | < 2H√EG.

Once H is bounded , L,M and N is bounded

Xuu ≤∣∣∣Γ11

1

√E∣∣∣+∣∣∣Γ11

2

√G∣∣∣+ L

is bounded, the same holds for other second derivatives. Now we want to

find a bound for H, Weyl proved that

H2 ≤ maxK − frac14K[1

∆(G

∆Ku −

F

∆Kv)u −

1

∆(F

∆Ku −

E

∆Kv)u].

Let P be the point where |H| assumes its max. We use geodesic parallelcoordinate near P .i.e ds2 = du2 + g2(u, v)dv2 and g(u, v) satisfy

g(0, v) = 1

gu(0, v) = 1

gv(0, v) = 0

guv(0, v) = 0

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in a neighborhood of P . Now H is given by

2H =N

g2+ L

and

K(u, v) =−guug

We also write out Codazzi-Mainardi equation in this neighborhood of P .Mu = Lv − (gu

gM)

Mv = Nu + (gvg

)M − 2guH

From this, after simple calculation in the neighborhood near P , we have

2HK = Lvv −Nuu

and

K − 1

4K[1

∆(G

∆Ku −

F

∆Kv)u −

1

∆(F

∆Ku −

E

∆Kv)u]

can now be expressed as

1

g2Kuu +

1

g2(Kv

g2)v

Note that after simple computation and use previous relation,

1

g2Kuu +

1

g2(Kv

g2)v = 2(−L2

u − L2v −M2

u −M2v ) +

[LNuu +NLuu + 2MLuv − 2MNuv + LNvv +NLvv]

+2K2 − 2M2K

Since |H| is max at P , we have Hu = Hv = 0 and NHuu− 2MHuv +LHvv ≤0,also Lu +Nu = 0,Lv = Nv = 0. We have

0 ≥ 2(NHuu − 2MHuv + LHvv) = [NLuu + LNvv − 2MNuv

−2MLuv

+NLvv + LNuu] + (L−N)(Lvv −Nuu) + 2N2K

From this we see the bracket term is equal to

A+ (N − L)(Lvv −Nuu)− 2N2K

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where A is some nonnegative expression. Now

1

g2Kuu +

1

g2(Kv

g2)v = A+ 2K2 − 2M2K + 2HK(N − L)− 2N2K

and

K − A

4K− K

2+M2

2− H

2(N − L) +

N2

2≥ H2

gives us the claimed result.

Theorem 2.14. For ds2 ∈ M ′4 and suppose there exist a sequence ds2

i ∈M”4+αi

,0 < αi < 1, such that ‖ ds2i −ds2 ‖4 goes to zero as i goes to infinity.

Then ds2 ∈M”4. If ds2 ∈M ′m+β where m ≥ 4,β < 1 then X ∈ Sm+β

The second part follows from a previous lemma. We shall prove thetheorem with the aid of a know result concerning nonlinear elliptic equations.

Lemma 2.15. For solutions ρ(u, v) of equation F (u, v, · · · , ρvv) = 0 in somedomain P . If ρ has continuous second derivatives, then ρ satisfy a Holdercondition in any closed subdomain D exponent depend only on K,K1 andthe distance from subdomain to D. Here K is the bound for the first andsecond derivative of P ,K1 is the bound for the first derivative of F w.r.t 8arguments. m is a positive integer such that

Fρuuε2 + Fρuvεη + Fρvvη

2 ≥ m(ε2 + η2)

for all ε and η in D.

We first show our F and ρ satisfy the condition in the lemma.By previous result ‖ ρ ‖2< K. Also, we get the first derivative of F

w.r.t arguments bounded by K1. Finally, since 4FρuuFρvv − F 2ρuv = 4

∆2K(X ·X3)2 ≥ c > 0 and (X · X3) is the distance from origin to a plane tangentto X(u, v) and thus greater that the largest sphere inscribed in X whichis bounded below.Therefore we can apply the lemma, and conclude in anyclosed subdomain of R1 and R2 satisfy a Holder condition. We can choose acover of R1 and R2 by subdomains such that we can have the whole spheresatisfy a Holder condition. Therefore,combined with ‖ X ‖2< K ,we have‖ X ‖2+β< K for some β.

Now we finish the proof of closeness. Since we can choose ‖ ds2i ‖< K

for all i, and since ‖ ds2i − ds2 ‖4 goes to zero and 1

∆i, 1Ki

uniformly bounded.

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For ds2i ∈ M”4+α, Xi(u, v) ∈ S4+αi

,we have ‖ Xi ‖2+β< K. By Arzela-Ascoli theorem, we select a subsequence Xi that will converges to X with‖ X ‖2+β< K has ds2 as first fundamental form and ds2 as first fundamentalform and ds2 ∈M”4 completes the proff of closeness and main result.

3 On n-dimensional Minkowski Problem

If we’re given a closed convex hypersurface M in Rn+1, the Gauss map definesa homeomorphism between M and Sn, the unit sphere. The Gauss curvatureK of M can be identified with a function defined on Sn via the Gauss map.It follows that K satisfy: ∫

Sn

xiK−1 = 0.

Minkowski consider the converse problem : given such function K de-fined onSn , can we find a closed convex hypersurface in Rn+1 whose Gausscurvature is given by K.In fact, we are going to prove the following.

Theorem 3.1. Let K be a positive function in Ck(Sn) with k ≥ 3. Spose Kalso satisfy the above condition. Then we can find a compact strictly convexsurface in Rn+1 whose Gauss curvature is K.

3.1 Introducing support function

Given a strict convex hypersurface M , we define its support function H onSn as follows.

Let x = (x1, · · · , xn+1) ∈ Sn be any unit vector, and y = (y1, · · · , yn+1) ∈M be the inverse Gauss map of x, i.e y = N−1(x). Then define

H(x) =n+1∑i=1

xiyi.

We can extend H to be a function defined on Rn+1 0 by setting H(x) =|x|H( x

|x|) which is a homogenous function of degree 1. Note that we canrecover y ∈M from H by

yi =∂H

∂xi.

19

for i = 1, · · · , n+ 1, and since they are function of homogeneity of degreeo. They are completely determined by their value on hypersurface xi = −1for i = 1, · · · , n + 1. Restricting H to xn+1 = −1 and extending K to be afunction of homogeneity of degree 0 ,we have

(1 +n∑i=1

x2i )

n2

+1 det(∂2H

∂xi∂xj)(x1, · · · , xn,−1) = (K(x1, · · · , xn,−1))−1.

We obtain

det(Hij +Hδij) =1

K

where the derivative is in the covariant sense with respect to orthonormalframe of Sn.

To solve Minkowski Problem, we wish to solve the equation for H givenK. We use the continuity method, set 1

Kt= t

K+ (1− t) for 0 ≤ t ≤ 1 to be

a curve of postivie function satisfy (.).And consider the solution Ht of

det(Hij +Hδij) =1

Kt

Let Sα be the set of t for which (.) has a C5,α solution H. We’ll showthat Sα is both open and closed in [0, 1]. And since 0 ∈ Sα ,we have indeedSα = [0, 1]. and the equation is solvable.

3.2 Openness of Sα

Let LH be the linearized operator of the operator H → det(Hij+Hδij). Thatis, for u ∈ C(Sn),we have

LH(u) =∑i,j

c(Hij +Hδij)(uij + uδij)

where c(Hij + Hδij) denote the cofactor of matrix (Hij + Hδij). Then itfollows that ∫

Sn

uLH(v) =

∫Sn

vLH(u)

This is equivalent to

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∑j

c(Hij +Hδij)j = 0

To prove this ,we note that

det(Hpq +Hδpq)c(Hij +Hδij)l = [det(Hpq +Hδpq)]l

c(Hij+Hδij)− c(Hip +Hδip)

(Hpk +Hδpk)lc(Hkj +Hδkj)

Then we have

det(Hpq +Hδpq)∑j

c(Hij +Hδij)j =∑j,k,p

c(Hip +Hδip)c(Hkj +H

δkj)[Hjkp +Hpδjk −Hpkj −Hjδpk]

We can show that the term in the bracket is zero,thus we get the result. Wealso have for any u ∈ C5,α(Sn),

Li(u) =

∫Sn

xidet(upq + uδpq) = 0

This is because the Frechet derivative of Li at v is∫Sn

xiLu(v) =

∫Sn

vLu(xi)

since Lu(xi) = 0 ,this implies that Li is identically 0 ,so :

Range(LH) ⊥ Spanx1, · · · , xn+1

Here orthogonality is with respect to integration over Sn. We can alsohave

Lemma 3.2.

Ker(L∗H) = Ker(LH) = Spanx1, · · · , xn+1.

21

Proof. Consider Z =∑n

i=1 uiei+uen+1 and dZ =∑n

j=1(∑n

i=1(uij+uδij)ei)ωj.We let X =

∑ni=1Hiei + Hen+1 and similarly dX =

∑nj=1(

∑ni=1(Hij +

Hδij)ei)ωj.Set Ω = X ∧ Z ∧ dZ ∧ dX ∧ · · · ∧ dX, where dX appears n − 2times. Then

dΩ = dX ∧ Z ∧ dZ ∧ dX ∧ · · · ∧ dX +X ∧ dZ ∧ dZ ∧ dX ∧ · · · ∧ dX=

∑ij

c(Hij +Hδij)](e1 ∧ · · · ∧ en)⊗ ω1 ∧

· · · ∧ ωn +X ∧ dZ ∧ dZ ∧ dX ∧ · · · ∧ dX= X ∧ dZ ∧ dZ ∧ dX ∧ · · · ∧ dX

Then

0 =

∫Sn

X ∧ dZ ∧ dZ ∧ dX ∧ · · · ∧ dX = 0

Denote vkj =∑

i(uki + yδki)c(Hij + Hδij), after diagonalizing (Hij +Hδij),we can assume that vjk and vkj either have same sign or both equal tozero. Plug this back in () and (), we get

∫Sn

< X, en+1 > (∑

(

i 6= j)(viivjj − vijvji))det(Hij +Hδij)−1 = 0

since LH(u) = 0, we have∑

i vii = 0Then∑

(

i 6= j)(viivjj − vijvji) =1

2[(∑i

vii)2 −

∑i

v2ii]−

∑i 6=j

vijvji ≤ 0

Put this back in the integration,since det(Hij + Hδij) and H =< X, en+1 >are both positive we must require that vij = 0 for all i and j. Thus

uij + uδij = 0

for all i and j. Then we have dZ = 0 so Z = const and u =∑n+1

i=1 aixi.Therefore we showed LH is surjective and by Implicit function theorem

the operator defined as H → det(Hij +Hδij) is invertible near det(Hij +Hδijwhich gives the openness.

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3.3 Closeness of Sα

We first introduce two lemmas which are useful in deriving second ,thirdderivative estimate of H

Lemma 3.3. Let Omega be an n dimensional convex domain and F ∈ C2(Ω)with F > 0 Let u ∈ C4(Ω)∩C0,1(Ω) be a strictly convex solution of det(uij) =F (x) on Ω and assume u|∂Ω = a. Then there is constant α depending on n,maxΩ |Du|,maxΩ|u − a| and C1,1 norm of F on Ω such that maxi |uii| ≤α|u− a|−1.

Proof. For each i , consider w = (a − u)eu2i2 uii, so it attains maximum at

some point in Ω. Pogorelov derived the following:

− nuiia− u

− u2i

(a− u)2+ u2

ii + uiuiiF−1Fi + F−1Fii − F 2F 2

i ≤ 0

Multiply (a− u)2eu2i on both sides, we get

w2 + Aw +B ≤ 0

here A , B depend on the quantity in the lemma. Thus we get the requiredestimate of uii.

and we also have

Lemma 3.4. Let Ω be an n dimensional convex domain and F ∈ C3(Ω) withF > 0. Let u ∈ C5(Ω) ∩C1,1(Ω) be a strictly convex solution to the equationdet(uij) = F (x) and assume u|∂Ω = a. Then the third derivative of u dependson C1,1 norm of u, C2,1 norm of F , maxx∈Ω

1F (x)

and the distance from ∂Ω.

Now we start to prove closeness. Let tl be a sequence in Sα whichconverges to t0, H l be a sequence of Ck+2,α support functions with (H l

ij +H lδij) > 0 and is the solution of det(H l

ij +H lδij) = 1/Ktl , we want to showfor Kt0 there exists Ht0 satisfy det(H l

ij +H lδij) = 1/Ktl .

Lemma 3.5. Let M be a compact convex C4 hypersurface in Rn+1 and K itsGauss curvature identified as a function on Sn, then the extrinsic diameterL of M is bounded by

cn(

∫Sn

1

K)

nn−1 [infu∈Sn

∫Sn

max(0, < u, w >)K(w)−1]−1

where cn is a constant depends only on n.

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Proof. let p and q be two points on M on which the line segment joiningthem is of length L.Without loss of generality, We set the midpoint of theline segment to be our origin. And let u be the unit vector along this linesegement.For any w ∈ Sn, we have H(w) = supy∈M < y,w >≥ 1

2Lmax(0, <

u, w >). Rearrange and integrating we have:

L ≤ 2(

∫Sn

H(w)

K)(

∫Sn

max(0, < u, w >)K(w)−1)−1

since also∫Sn

H(w)K

=∫MH , and from divergence formula V ol(M) =

1n+1

∫MH where M is the volume enclosed by M . Note that

∫Sn 1/K gives

the surface area. Then the result follows from isoperimetric inequality.

Lemma 3.6. We can a positive constant r depends only on upper estimate of∫Sn 1/K and lower estimate of infu∈Sn

∫Sn max(0, < u, w >)K(w)−1 so that

w a ball of radius r can be inserted inside M .

Proof. We first find a lower bound of areas of projections of the surface onall hyperplanes.If we project the surface onto the hyperplane perpendicularto u.So the volume of M is less or equal than L times the area of projection.We get the area of projection is at least:

1

2(n+ 1)infu∈Sn

(

∫Sn

max(0, < u, w >)K(w)−1).

And since the volume of M is bounded above by Ln, we get a lowerestimate of L.Hence we can find two points p1 and p2 whose distance can beestimated from below. Then we project M onto any hyperplane containingp1 and p2.Since outer diameter is bounded and the projected area is boundedbelow, we can find a point p3 on this hyperplane such that all angles oftriangle p1p2p3 lies between ε and π − ε . Lifting p3 to p3 in M , we gettriangle p1p2p3 with the same property.

For n > 2, we project M into the hyperplane containing p1p2p3, usingthe same argument , we can find n points in M the angles between all linesegments from pi to pj with j ≤ i − 1 in the plane simplex p1, · · · , pi−1

lie between ε and π − ε.Finally, we project the hypersurface M along thedirection which is perpendicular to the hyperplane of simplex p1, · · · , pn, wecan obtain a simplex with same property p1 · · · ˜pn+1 lie in M , and we caninsert a ball inside the simplex. This proves the Lemma .

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Now, with the aid of previous lemma. We can translate M l so that theorigin lies in the interior of all M l and inf l infx∈M l |x| and supl supx∈M l |x| arepositive numbers.

Now according to

|Du| ≤ (supΩu− u(x))d(x, ∂Ω)−1.

We can therefore have a uniform gradient estimate of H l on each compactsubset of xi = −1 based on the facts inf l infx∈M l |x| and supl supx∈M l |x| arefinite.

Set Ωl(c) = x = (x1, · · · , xn,−1)|H l(x) < c, we see that all Ωl(c)lies in in a fixed compact set. And for any point , we can find c dependon supl supx∈M l |x| and

∑a2i such that it lies in Ωl(

c2) for all l. Now since

all Ωl(c) lies in in a fixed compact set,the gradient of all H l bounded, wecan apply lemma 1 and 2 to get uniform second and third derivative of H l.Thus we can choose a subsequence of H l converges to a C2,1 function H0

satisfying the equation

(1 +n∑i=1

)n2

+1 det(∂2H0

∂xi∂xj) = (Kt0(x1, · · · , xn,−1))−1.

Using Schauder estimate, one can show that H0 is Ck+2,α function andt0 ∈ Sα thus complete the proof of closeness.

Now since Sα are both open and closed in [0, 1] and 0 ∈ Sα, we haveSα = [0, 1]. This establish the existence of surface. We omitted the proof ofuniqueness.

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