# Hypersurfaces with Null Higher Order Anisotropic Mean Curvature

Authors: Hua Wang; Yijun He
Journal: Geometry
 (2013)

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## Abstract

Given a positive functionF on 𝕊n which satisfies a convexity condition, for 1≤r≤n, we define for hypersurfaces in ℝn+1 the rth anisotropic mean curvature function Hr;F, a generalization of the usual rth mean curvature function. We call a hypersurface anisotropic minimal if HF=H1;F=0, and anisotropic r-minimal if Hr+1;F=0. Let W be the set of points which are omitted by the hyperplanes tangent to M. We will prove that if an oriented hypersurface M is anisotropic minimal, and the set W is open and nonempty, then x(M) is a part of a hyperplane of ℝn+1. We also prove that if an oriented hypersurface M is anisotropic r-minimal and its rth anisotropic mean curvature Hr;F is nonzero everywhere, and the set W is open and nonempty, then M has anisotropic relative nullity n−r.


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## Body

## 1. Introduction

LetF:𝕊n→ℝ+ be a smooth function which satisfies the following convexity condition:
(1)(D2F+FI)x>0,∀x∈𝕊n,
where 𝕊n is the standard unit sphere in ℝn+1, D2F denotes the intrinsic Hessian of F on 𝕊n, I denotes the identity on Tx𝕊n, and >0 means that the matrix is positive definite. We consider the map
(2)ϕ:𝕊n⟶ℝn+1,x⟶F(x)x+(grad𝕊nF)x;
its image WF=ϕ(𝕊n) is a smooth, convex hypersurface in ℝn+1 called the Wulff shape of F (see [1–9]). When F≡1, the Wulff shape WF is just 𝕊n.Now letx:M→ℝn+1 be a smooth immersion of an oriented hypersurface. Let N:M→𝕊n denote its Gauss map. The map ν=ϕ∘N:M→WF is called the anisotropic Gauss map of x.LetSF=-dν. SF is called the F-Weingarten operator, and the eigenvalues of SF are called anisotropic principal curvatures. Let σr be the elementary symmetric functions of the anisotropic principal curvatures κ1,κ2,…,κn:
(3)σr=∑i1<⋯<ir‍κi1⋯κir(1≤r≤n).
We set σ0=1. The rth anisotropic mean curvature Hr;F is defined by Hr;F=σr/Cnr, also see Reilly [10]. HF:=H1;F is called the anisotropic mean curvature. When F≡1, SF is just the Weingarten operator of hypersurfaces, and Hr;F is just the rth mean curvature Hr of hypersurfaces which has been studied by many authors (see [11–14]). Thus, the rth anisotropic mean curvature Hr;F generalizes the rth mean curvature Hr of hypersurfaces in the (n+1)-dimensional Euclidean space ℝn+1.We say thatx:M→ℝn+1 is anisotropic r-minimal if Hr+1;F=0.Forp∈M, we define v(p)=dimker(SF). We call v=minp∈Mv(p) the anisotropic relative nullity; it generalized the usual relative nullity.For a smooth immersionx:M→ℚcn+1 of a hypersurface into an (n+1)-dimensional space form with constant sectional curvature c, we denote by
(4)W=ℚcn+1-⋃p∈M‍(ℚcn)p,
where for every p∈M, (ℚcn)p is the totally geodesic hypersurface of ℚcn+1 tangent to x(M) at x(p). So, in the case of c=0, W is the set of points which are omitted by the hyperplanes tangent to x(M).We will study immersion withW nonempty. In this direction, Hasanis and Koutroufiotis (see [15]) proved the following.Theorem 1.
Letx:M→ℚc3 be a complete minimal immersion with c≥0. If W is nonempty, then x is totally geodesic.Later, in [16], Alencar and Frensel extended the result above assuming an extra condition. They proved the following.Theorem 2.
Letx:M→ℚcn+1 be an oriented, minimally immersed hypersurface. If  W is open and nonempty, then x is totally geodesic.In [17], Alencar and Batista studied hypersurfaces with null higher order mean curvature; they proved the following.Theorem 3.
LetM be a complete and orientable Riemannian manifold and let x:M→ℚcn+1 be an isometric immersion with Hr+1=0 and Hr≠0 everywhere, r≥1. If  W is open and nonempty, then the relative nullity v=n-r.We note that, Alencar in [18] provides examples of nontotally geodesic minimal hypersurfaces in ℝ2n, n≥4, with nonempty W; in [17], Alencar and Batista provides examples of 1-minimal hypersurfaces with H1≠0 everywhere in ℝ2n,  n≥5, with nonempty W but v≠n-1. These examples show that it is necessary to add an extra hypothesis.In this paper, we prove the anisotropic version of Theorems2 and 3 for an immersion x:M→ℝn+1. Explicitly, we prove the following two theorems.Theorem 4.
Letx:M→ℝn+1 be an oriented, anisotropic minimally immersed hypersurface. If  W is open and nonempty, then x(M) is a part of a hyperplane of ℝn+1.Theorem 5.
Letx:M→ℝn+1 be an oriented immersed hypersurface with Hr+1;F=0 and Hr;F≠0 everywhere, r≥1. If W is open and nonempty, then the anisotropic relative nullity v=n-r.


## 2. Preliminaries

In this paper, we use the summation convention of Einstein and the following convention of index ranges unless otherwise stated:(5)1≤i,j,…≤n;1≤α,β,…≤n+1.We defineF*:ℝn+1→ℝ to be
(6)F*(y)=sup{〈y,z〉F(z)∣z∈ℝn+1∖{0}};
then F* is a Minkowski norm on ℝn+1. In fact, as proved in [19], F*:ℝn+1∖{0}→ℝ is smooth and we have the following.Proposition 6.
(1)F*(y)>0, for all y∈ℝn+1∖{0};
(2)F*(ty)=tF*(y), for all y∈ℝn+1, t>0;
(3)F*(y+z)≤F*(y)+F*(z), for all y,z∈ℝn+1, and the equality holds if and only if y=0, or z=0 or y=kz for some k>0.
(4)WF={y∈ℝn+1∣F*(y)=1}.We define(7)g-αβ(y)=12∂2(F*)2∂yα∂yβ(y),gy(X,Y)=g-αβ(y)XαYβ,
where y∈ℝn+1∖{0} and X=(X1,X2,…,Xn+1), Y=(Y1,Y2,…,Yn+1)∈Tyℝn+1≅ℝn+1.From the Euler’s theorem for homogeneous functions, we have(8)∂g-αβ∂yγ(z)zβ=12∂3(F*)2∂yα∂yβ∂yγ(z)zβ=0,
where z=(z1,z2,…,zn+1)∈ℝn+1∖{0}. Thus,
(9)∂gz(X,z)∂yγ=g-αβ(z)∂Xα∂yγzβ+g-αγ(z)Xα∂zβ∂yγ,
where z=(z1,z2,…,zn+1)∈Tℝn+1 is nonzero everywhere and X=(X1,X2,…,Xn+1)∈Tℝn+1.AsF* is a Minkowski norm on ℝn+1, the following lemma holds (see [20, 21]).Lemma 7.
For anyy∈ℝn+1∖{0} and u∈ℝn+1 one has
(10)gy(y,z)≤F*(y)F*(z),
and the equality holds if and only if there exists t≥0 such that z=ty.Letx:M→ℝn+1 be an oriented hypersurface in the Euclidean space ℝn+1. Let ν:M→WF denote its anisotropic Gauss map. Then for any p∈M, ν(p) is perpendicular to x*(TpM) with respect to the inner product gν(p) and F*(ν(p))=1. Thus, we call ν(p) an anisotropic unit normal vector of TpM.


## 3. A Connection on Hypersurfaces of Minkowski Space

Letx:M→ℝn+1 be an oriented hypersurface in the Euclidean space ℝn+1 and denote ν:M→WF its anisotropic Gauss map.Let∇¯ be the standard connection on the (n+1)-dimensional Euclidean space ℝn+1. For vector fields X,Y on M, we decompose ∇¯XY as the tangent part ∇XY and the anisotropic normal part II (X,Y)ν with respect to the inner product gν. That is,
(11)∇¯XY=∇XY+II(X,Y)ν,
where gν(∇XY,ν)=0.We also have the Weingarten formula:(12)∇¯Xν=-SFX,gν(SFX,Y)=II(X,Y),
where we have used (9).It is easy to verify that∇ is a torsion free connection on M and II is a symmetric second order covariant tensor field on M. We call II the anisotropic second fundamental form.Let{ei}i=1n be a local frame of M and {ωi}i=1n its dual frame. Let gij=gν(ei,ej), ∇ei=ωij⊗ej, II(ei,ej)=hij, hij=gjkhki, where (gij) is the inverse matrix of (gij). Then we have
(13)dx=ωiei,(14)dei=ωijej+hijωjν,(15)dν=-hijωiej.Differentiating (13) and using (14), we get
(16)dωi=ωj∧ωji,hij=hji.Differentiating (14) and using (14)-(15), we get
(17)hijk=hikj,dωij-ωik∧ωkj=-12Rikljωk∧ωl,
where
(18)dhij-hikωjk-hkjωik=hijkωk,
and Riklj=-Rilkj=hikhlj-hilhkj.Differentiating (15) and using (14), we get
(19)hikj=hkij,
where
(20)dhij+hikωkj-hkjωik=hikjωk.Note(hij) is the matrix of the F-Weingarten operator SF=-dν, its eigenvalues are called the anisotropic principal curvatures, and we denote them by κ1,…,κn.We haven invariants, the elementary symmetric function σr of the anisotropic principal curvatures:
(21)σr=∑i1<⋯ir‍κi1⋯κin(1≤r≤n).
For convenience, we set σ0=1. The rth anisotropic mean curvature Hr;F is defined by
(22)Hr;F=σrCnr,Cnr=n!r!(n-r)!.Using the characteristic polynomial ofSF, σr is defined by
(23)det(tI-SF)=∑r=0n‍(-1)rσrtn-r.So, we have(24)σr=1r!∑i1,…,ir;j1,…,jr‍δi1⋯irj1⋯jrhj1i1⋯hjrir,
where δi1⋯irj1⋯jr is the usual generalized Kronecker symbol; that is, δi1⋯irj1⋯jr equals +1 (resp., −1) if i1⋯ir are distinct and (j1⋯jr) is an even (resp., odd) permutation of (i1⋯ir) and in other cases it equals zero.Definition 8.
Letf:M→ℝ be a smooth function. One defines the gradient (with respect to the induced metric gν on M) gradf of the function f by
(25)gν(gradf,X)=X(f),
where X is any smooth vector field on M.
Definefi by df=fiωi; then
(26)gradf=gijfjei.We define(27)dV=|e1,…,en,ν|ω1∧⋯∧ωn,
where |e1,…,en,ν| is the determinant of the matrix (e1,…,en,ν). Then dV is a volume element on M.Definition 9.
LetX be a smooth vector field on M. One defines the divergence (with respect to the volume element dV) divX by d{i(X)dV}=(divX)dV, where
(28)(i(X)dV)(Y1,…,Yn-1)≡dV(X,Y1,…,Yn-1),∀Y1,…,Yn-1∈𝒳(M).Lemma 10.
LetX=Xiei; then divX=Xii, where
(29)dXi+Xjωji=Xjiωj.Proof.
By (14), (15), we get
(30)d|e1,…,en,ν|=ωii|e1,…,en,ν|.
From the definition of i(X), we have
(31)i(X)dV=∑i‍(-1)i+1Xi|e1,…,en,ν|ω1∧⋯∧ω^i∧⋯∧ωn.
So,
(32)d{i(X)dV}=∑i‍(-1)i+1(dXi)∧|e1,…,en,ν|ω1∧⋯∧ω^i∧⋯∧ωn+∑i‍(-1)i+1Xi(d|e1,…,en,ν|)∧ω1∧⋯∧ω^i∧⋯∧ωn+∑j<i‍(-1)i+jXi|e1,…,en,ν|dωj∧ω1∧⋯∧ω^j∧⋯∧ω^i∧⋯∧ωn+∑j>i‍(-1)i+j+1Xi|e1,…,en,ν|dωj∧ω1∧⋯∧ω^i∧⋯∧ω^j∧⋯∧ωn=XiidV.


## 4.Lr;F Operator for Hypersurfaces

We introduce the Newton transformation defined by(33)Pr=σrI-σr-1SF+⋯+(-1)rSFr,r=0,…,n;
then
(34)P0=I,Pn=0,Pr=σrI-Pr-1SF.Lemma 11.
The matrix ofPr is given by:
(35)(Pr)ij=1r!δi1⋯irij1⋯jrjhj1i1⋯hjrir.Proof.
We prove Lemma11 inductively. For r=0, it is easy to check that (35) is true.
We can check directly(36)δi1⋯iqj1⋯jq=|δi1j1δi1j2⋯δi1jq-1δi1jqδi2j1δi2j2⋯δi2jq-1δi2jq⋮⋮⋱⋮⋮δiq-1j1δiq-1j2⋯δiq-1jq-1δiq-1jqδiqj1δiqj2⋯δiqjq-1δiqjq|.
Assume that (35) is true for r=k, we only need to show that it is also true for r=k+1. For r=k+1, using (24) and (36), we have
(37)RHSof(35)=1(k+1)!∑i1,…,ik+1;j1,…,jk+1‍δi1⋯ik+1ij1⋯jk+1jhi1j1⋯hik+1jk+1=1(k+1)!∑‍|δi1j1δi1j2⋯δi1jk+1δi1jδi2j1δi2j2⋯δi2jk+1δi2j⋮⋮⋱⋮⋮δik+1j1δik+1j2⋯δik+1jk+1δik+1jδij1δij2⋯δijk+1δij|hi1j1⋯hik+1jk+1=1(k+1)!∑‍(δijδi1⋯ik+1j1⋯jk+1-δijk+1δi1⋯ikik+1j1⋯jkj+⋯)hi1j1⋯hik+1jk+1=σk+1δij-1(k+1)!∑‍δijk+1δi1⋯ikik+1j1⋯jkjhi1j1⋯hik+1jk+1+⋯=σk+1δij-∑‍(Pk)iik+1hik+1j=(Pk+1)ij.Lemma 12.
For eachr, one has(a)
(Pr)ijj=0;(b)
Trace(PrSF)=(r+1)σr+1;(c)
Trace(Pr)=(n-r)σr;(d)
Trace(PrSF2)=σ1σr+1-(r+2)σr+2.Proof.
(a) Noting(j,jr) is skew symmetric in δi1⋯irij1⋯jrj and (j,jr) is symmetric in hj1i1⋯hjrjir (from (19), we have
(38)∑j(Pr)ijj=1(r-1)!∑i1,…,ir;j1,…,jr;j‍δi1⋯irij1⋯jrjhj1i1⋯hjrjir=0.
(b) Using (35) and (24), we have
(39)Trace(PrSF)=∑ij‍(Pr)ijhji=1r!∑i1,…,ir;j1,…,jr;i,j‍δi1⋯irij1⋯jrjhj1i1⋯hjrirhji=(r+1)σr+1.
(c) Using (b) and the definition of Pr, we have
(40)Trace(Pr)=tr(σrI)-tr(Pr-1SF)=nσr-rσr=(n-r)σr.
(d) Using (b) and the definition of Pr+1, we have
(41)Trace(PrSF2)=Trace(σr+1SF)-Trace(Pr+1SF)=σ1σr+1-(r+2)σr+2.Remark 13.
WhenF=1, Lemma 12 was a well-known result (e.g., see Barbosa and Colares [22], or Reilly [23]).Lemma 14.
One has(42)(σr)k=∑i,j‍(Pr-1)ijhjki.Proof.
From the definition ofσr, we have the following calculation:
(43)(σr)k=1r!∑i1,…,ir;j1,…,jr‍δi1⋯irj1⋯jr(hi1j1⋯hirjr)k=1(r-1)!∑i1,…,ir;j1,…,jr‍δi1⋯irj1⋯jrhi1j1⋯hirkjr=∑ir,jr‍(Pr-1)irjrhirkjr=∑i,j‍(Pr-1)ijhikj.We define an operatorLr;F:C∞(M)→C∞(M) by
(44)Lr;F(f)=div(Pr∇f).In the sequel, we will need the following lemma. Item (a) is essentially the content of Lemma1.1 and Equation (1.3) in [24], while item (b) is quoted as Proposition  1.5 in [25].Lemma 15.
Letx:M→ℝn+1 be an oriented hypersurface, and 0≤r≤n-1, p∈M.(a)
Ifσr+1(p)=0, then Pr is semidefinite at p;(b)
ifσr+1(p)=0 and σr+2(p)≠0, then Pr is definite at p.Another important result is as follows (see [26]).Lemma 16.
Letx:M→ℝn+1 be an oriented hypersurface, and p∈M.(a)
For1≤r≤n, one has Hr;F2≥Hr-1;FHr+1;F. Moreover, if equality happens for r=1 or for some 1<r<n, with Hr+1;F≠0 in this case, then p is an anisotropic umbilical point (i.e. κ1(p)=κ2(p)=⋯=κn(p));(b)
if, for some1≤r<n, one has Hr;F=Hr+1;F=0, then Hj;F=0 for all r≤j≤n. In particular, at most r-1 of the anisotropic principal curvatures are different from zero.The result below is standard, so we omit the proof.Lemma 17.
Letx:M→ℝn+1 be an oriented hypersurface. The operator Lr;F associated to the immersion x is elliptic if and only if Pr is positive definite.Definition 18.
Letf:M→ℝ be a smooth function. The Laplacian Δf is defined by Δf:=L0;Ff=div(gradf).It is easy to see thatΔ is an elliptic differential operator.Definition 19.
Letx:M→ℝn+1 be an immersed hypersurface and ν its anisotropic unit normal vector field. The function u:=gν(x,ν) is called the support function of the immersion x.Next, we computeLr;Fu for the support function u=gν(x,ν).Differentiating the decomposition(45)x=gijgν(x,ei)ej+uν,
we obtain
(46)dx={d(gijgν(x,ej))}ei+gijgν(x,ej)dei+(du)ν+udν.
So, from (13), (14), and (15) we have
(47)ωiei={d(gijgν(x,ej))+gkjgν(x,ej)ωki-uhjiωj}ei+(du+gjkgν(x,ej)hikωi)ν.
Thus, we get
(48)du=-gjkgν(x,ej)hikωi,d(gijgν(x,ej))+gkjgν(x,ej)ωki-uhjiωj=ωi.
Denote ui, (gijgν(x,ej))k, uji by
(49)gradu=uiei,(gijgν(x,ej))kωk=d(gijgν(x,ej))+(gkjgν(x,ej))ωki,ujiωj=dui+ujωji,
respectively. Then we have (using (19) the following
(50)ui=-gilhklgjkgν(x,ej)=-hkigklgν(x,el),(gikgν(x,ek))j=δji+hjigν(x,ν),uji=-hkjigklgν(x,el)-hki(gklgν(x,el))j=-hjkigklgν(x,el)-hji-hkihjku.
By using Lemmas 12 and 14, we get
(51)Lr;Fu=(Pr)ijuji=-(Pr)ijhjkigklgν(x,el)-(Pr)ijhji-(Pr)ijhkihjku=-(σr+1)kgklgν(x,el)-(Pr)ijhji-(Pr)ijhkihjku=-gν(∇σr+1,x)-(r+1)σr+1-(σ1σr+1-(r+2)σr+2)u.
Thus, we proved the following lemma.Lemma 20.
For0≤r≤n-1, one has the following
(52)Lr;Fu=-gν(∇σr+1,x)-(r+1)σr+1-(σ1σr+1-(r+2)σr+2)u.Remark 21.
Recallσ1=nHF and |II|2=σ12-2σ2; let r=0 in (52); we get
(53)Δu=-n(HF+gν(gradHF,x))-|II|2u.


## 5. Proof of Theorems4 and 5

We fix a pointo∈W as the origin of ℝn+1. Without loss of generality, we assume, for each p∈M, ν(p) is the anisotropic unit normal vector of x(M) at x(p) such that 〈x(p),ν(p)〉ν(p)>0 (otherwise we consider the function -u instead). This gives an orientation to M; indeed, the component of the position vector x perpendicular (with respect to the inner product gν) to M defines a never zero, anisotropic normal, vector field on M, such that the support function u=〈x(p),ν(p)〉ν(p) is positive on M.


### 5.1. Proof of Theorem4

Sincex is anisotropic minimal, from (53) we get
(54)Δu=-|II|2u≤0,onM.
Let u*=infMu. We claim that u* is attained at some point x0∈M. Consider a sequence {xk}⊂M such that u(xk)→u* as k→+∞. To each xk we associate yk=u(xk)ν(xk); then yk∈TxkM. Since ∥yk∥ℝn+1=u(xk)∥ν(xk)∥ℝn+1 is bounded, there exists a subsequence, which again we call {yk}, such that yk→y0 for some y0∈ℝn+1. Since ⋃p∈MTpM is closed and {yk}⊂p∈MTpM we deduce that y0∈Tx0M for some x0∈M. Thus, by the continuity of F* and Lemma 7,
(55)u*=limk→+∞u(xk)=limk→+∞F*(yk)=F*(y0)≥gν(x0)(y0,ν(x0))=u(x0),
so u*=u(x0) as needed. Now, from the usual maximum principle u is constant, u=u*=u(x0)>0. From (54) we then have II≡0 and x is totally geodesic.


### 5.2. Proof of Theorem5

SinceHr+1;F=0, from Lemma 20 we get
(56)Lr;Fu=(r+2)σr+2u.
Using Lemma 15(a) we have that Pr is semidefinite. Since Hr;F does not vanish, we have that Hr;F is positive or negative, because c(r)Hr;F=Trace(Pr), where c(r)=(n-r)Cnr. Now we use Lemma 16 and obtain the following:
(57)0=Hr+1;F2≥Hr;FHr+2;F.
Using the information above, we claim that Hr+2;F≡0.Case  (i)
(
H
r
;
F
>
0
). In this case, Pr is positive definite, and Lr;F is elliptic by Lemma 17. Using (57) we conclude that Hr+2;F≤0. Whereas from (56) we have
(58)Lr;Fu≤0.
Following exactly the proof as in Theorem 4, we conclude that u is constant, u=u*=u(x0)>0. From (56) we then have Hr+2;F≡0.Case  (ii)
(
H
r
;
F
<
0
)
. In this case, Pr is negative definite, and -Lr;F is elliptic by Lemma 17. Using (57) we conclude that Hr+2;F≥0. Whereas from (56) we have
(59)-Lr;Fu≤0.Now, following exactly the proof as in Theorem4, we conclude that u is constant, u=u*=u(x0)>0. From (56) we then have Hr+2;F≡0.Thus we conclude thatHr+2;F≡0. Now, we use Lemma 16(b) to conclude that Hj;F=0 for j≥r+1 and so that v≥n-r. Since Hr;F does not change sign we have that v=n-r.


## 5.1. Proof of Theorem4

Sincex is anisotropic minimal, from (53) we get
(54)Δu=-|II|2u≤0,onM.
Let u*=infMu. We claim that u* is attained at some point x0∈M. Consider a sequence {xk}⊂M such that u(xk)→u* as k→+∞. To each xk we associate yk=u(xk)ν(xk); then yk∈TxkM. Since ∥yk∥ℝn+1=u(xk)∥ν(xk)∥ℝn+1 is bounded, there exists a subsequence, which again we call {yk}, such that yk→y0 for some y0∈ℝn+1. Since ⋃p∈MTpM is closed and {yk}⊂p∈MTpM we deduce that y0∈Tx0M for some x0∈M. Thus, by the continuity of F* and Lemma 7,
(55)u*=limk→+∞u(xk)=limk→+∞F*(yk)=F*(y0)≥gν(x0)(y0,ν(x0))=u(x0),
so u*=u(x0) as needed. Now, from the usual maximum principle u is constant, u=u*=u(x0)>0. From (54) we then have II≡0 and x is totally geodesic.


## 5.2. Proof of Theorem5

SinceHr+1;F=0, from Lemma 20 we get
(56)Lr;Fu=(r+2)σr+2u.
Using Lemma 15(a) we have that Pr is semidefinite. Since Hr;F does not vanish, we have that Hr;F is positive or negative, because c(r)Hr;F=Trace(Pr), where c(r)=(n-r)Cnr. Now we use Lemma 16 and obtain the following:
(57)0=Hr+1;F2≥Hr;FHr+2;F.
Using the information above, we claim that Hr+2;F≡0.Case  (i)
(
H
r
;
F
>
0
). In this case, Pr is positive definite, and Lr;F is elliptic by Lemma 17. Using (57) we conclude that Hr+2;F≤0. Whereas from (56) we have
(58)Lr;Fu≤0.
Following exactly the proof as in Theorem 4, we conclude that u is constant, u=u*=u(x0)>0. From (56) we then have Hr+2;F≡0.Case  (ii)
(
H
r
;
F
<
0
)
. In this case, Pr is negative definite, and -Lr;F is elliptic by Lemma 17. Using (57) we conclude that Hr+2;F≥0. Whereas from (56) we have
(59)-Lr;Fu≤0.Now, following exactly the proof as in Theorem4, we conclude that u is constant, u=u*=u(x0)>0. From (56) we then have Hr+2;F≡0.Thus we conclude thatHr+2;F≡0. Now, we use Lemma 16(b) to conclude that Hj;F=0 for j≥r+1 and so that v≥n-r. Since Hr;F does not change sign we have that v=n-r.



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*Source: 718272-2013-06-24.xml*