Affine complete locally convex hypersurfaces

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1 Invent. math. 150, (2002) DOI: /s Affine complete locally convex hypersurfaces Neil S. Trudinger, Xu-Jia Wang Centre for Mathematics and Its Applications, The Australian National University, Canberra, ACT 0200, Australia ( Oblatum 16-VI-2001 & 27-II-2002 Published online: 29 April 2002 Springer-Verlag 2002 Abstract. An open problem in affine geometry is whether an affine complete locally uniformly convex hypersurface in Euclidean (n + 1)-space is Euclidean complete for n 2. In this paper we give the affirmative answer. As an application, it follows that an affine complete, affine maximal surface in R 3 must be an elliptic paraboloid. 1. Introduction A locally uniformly convex hypersurface M R n+1 is a locally uniformly convex immersion of a connected n-manifold N, namely a mapping T: N M R n+1, such that for any point p N, there is a neighborhood O p such that T is a diffeomorphism from O p to T(O p ) and T(O p ) is a uniformly convex C 2 graph under appropriate choice of coordinates. In affine geometry there are two different metrics defined on M, one is the Euclidean metric induced from the standard metric in R n+1, the other one is the affine metric (or Berwald-Blaschke metric [1]), given by g = K 1/(n+2) II, where K is the Gauss curvature and II is the second fundamental form on M. The hypersurface M, or more precisely the manifold N, becomes a Riemannian manifold under both metrics. Clearly if M is compact, these two metrics are equivalent. A Euclidean complete locally convex hypersurface is not necessarily affine complete [10]. Conversely, an affine complete curve (n = 1) is not necessarily Euclidean complete. An open problem is whether an affine complete locally uniformly convex hypersurface is Euclidean complete for n > 1 (see [3 5,10]). Given that this problem has existed ever since the affine metric was introduced early last century, it can Research supported by the Australian Research Council

2 46 N.S. Trudinger, X.-J. Wang be viewed as the oldest and most fundamental problem in affine differential geometry. In this paper we provide the affirmative answer to the problem. Theorem A. Let n 2.If M is an affine complete locally uniformly convex hypersurface in R n+1,thenm is Euclidean complete. Corollary 1. Let n 2.If M is an open, affine complete, locally uniformly convex hypersurface, then M can be represented as a graph of a convex function. Corollary 1 follows since a Euclidean complete, locally uniformly convex hypersurface is convex, namely it lies on the boundary of a convex body. As an application we consider the Bernstein problem for affine maximal surfaces in R 3, that is locally uniformly convex surfaces whose affine mean curvature vanishes. This problem was studied by Calabi [3 5] (see also [9,12]). In [3] he proved that an affine maximal surface is an elliptic paraboloid if it is complete in both the affine and the Euclidean metrics. In [13] we showed that Euclidean completeness alone suffices. In [4, 5], he introduced the affine Gauss map and affine Weierstrass representation, and obtained further results on the affine Bernstein problem, assuming M is affine complete as well as some other assumptions. A problem raised by Calabi (called Calabi s conjecture in [12]) is whether affine completeness alone suffices for the Bernstein problem. By Theorem A we have Corollary 2. Let M be an affine complete, affine maximal surface. Then M is an elliptic paraboloid. Corollary 2 follows immediately since by Theorem A, M is Euclidean complete and so the result in [3] (or [13]) applies. Corollary 2 also answers Problem in [9]. To prove Theorem A, we first associate to any point p M, a subset of M, Γ p, which is the maximal subset containing p where M can be represented as a graph (possibly multi-valued) over a domain in the Euclidean space R n. We then prove in Sect. 2 that if M is affine complete, there are no boundary points in any such Γ p for each p M. InSect.3,weshow that if M is not Euclidean complete, there exists a point p M such that Γ p contains a boundary point. The proof of an auxiliary lemma in Sect. 2, Lemma 2.4, which provides a criterion for a star-shaped domain to be convex in terms of boundary behaviour of convex functions, is deferred until Sect. 4. Note that in the case that M is a graph, such as the case when M is an affine sphere of hyperbolic type [2], see also [7,9,10], we can dispense with our argument in Sect. 3. Also our proof yields Theorem A and the global characterization, Corollary 1, simultaneously. Finally, we remark that an interesting alternative proof of Corollary 2, which does not depend on Theorem A, is proposed in [15].

3 Affine complete locally convex hypersurfaces Proof of Theorem A Let M be a locally uniformly convex hypersurface as above. Since M is immersed, a point p M may correspond to more than one point in N with p as their image under the mapping T. For clarity we agree with that when referring to a point p M we actually mean a point z N such that p = T(z). Similarly we say ω p M is a neighborhood of p if it is the image of a neighborhood of z in N under the mapping T. Wesayγ is a curve on M if it is the image of a curve on N,andasetE M is connected if it is the image of a connected set in N, and so on. By the immersion T there is an induced metric on N from the affine metric on M. In this paper we are concerned with affine complete hypersurfaces. Hence we may suppose that N is a complete Riemannian manifold. We say a point p R n+1 is a boundary point of M if there is a curve γ :[0, ) N such that the geodesic distance between γ(0) and γ(t) converges to infinity as t and T(γ(t)) sub-converges to p. Denote by M the set of boundary points of M.ThenM is Euclidean complete if M is empty. Suppose locally M is represented as a graph x n+1 = u(x) x = (x 1,, x n ). (2.1) Then the affine metric on the graph can also be written as ρu xi x j dx i dx j, where ρ = (detd 2 u) 1/(n+2). Therefore the affine arc-length of a curve γ on the graph is given by L = (ρu ξξ ) 1/2 ds, (2.2) l where l is the projection of γ on {x n+1 = 0}, s is the (Euclidean) arclength parameter on l, ξ = (ξ 1,,ξ n ) is the unit tangent vector on l,and u ξξ = ξ i ξ j u xi x j. Let q be an arbitrary point on M. By a rotation of coordinates we suppose the south pole of S n, i.e., the vector e n+1 = (0,, 0, 1), is the normal (on the convex side) of M at q. LetΓ = Γ q be the connected subset of M containing q such that p Γ if and only if there is a curve γ on M connecting q and p such that G(γ) is a geodesic line on the south hemisphere, where G is the Gauss mapping. The curve is unique since M is locally uniformly convex. Hence Γ can be represented as a graph (possibly multi-valued) over a domain Ω by (2.1), such that Du(l) is a line segment in Ω,wherel is the projection of γ on {x n+1 = 0}, Ω is the spherical projection on {x n+1 = 1} of G(Γ), or equivalently Ω = Du(Ω). It follows that Ω is a star-shaped domain, and Ω and Γ are maximal, namely there is no larger connected subset Γ M with the above properties. The boundary of Γ consists of two disjoint parts, Γ = 1 Γ 2 Γ. A point p 2 Γ if and only if it is an interior point of M. Therefore a point p 1 Γ(= Γ 2 Γ) is also a boundary point of M. Similarly we divide

4 48 N.S. Trudinger, X.-J. Wang Ω into two parts: y 0 2 Ω if ty 0 Ω for t (1 ε, 1 + ε) for some ε>0 small (depending on y 0 ), and 1 Ω the rest of Ω. Note that 2 Ω consists of line segments. Let u be the Legendre transformation of u, givenby u (y) = x y u(x), y Ω, (2.3) where x Ω is chosen such that Du(x) = y (x is unique by our construction of Γ). Then u is a single valued convex function and u is the Legendre transformation of u such that x = Du (y) and detd 2 u(x) detd 2 u (y) = 1. (2.4) If Ω is convex, then u is single valued. By the Legendre transformation, a curve γ on the graph of u corresponds to a curve γ of the graph of u such that a point (x, u(x)) γ corresponds to a point (y, u (y)) γ,wherey = Du(x), and vice versa. The projection of γ in Ω, l, then corresponds to the projection of γ in Ω, l, with l = Du (l ). Furthermore, a point y 0 1 Ω corresponds either to a boundary point of M in 1 Γ or to an infinite point. Indeed, letting l :[0,τ) Ω be the line segment connecting the origin to y 0 such that on the corresponding curve l = Du (l ) : [0,τ) Ω, we have either γ(t) subconverges to a boundary point of M or converges to infinity as t τ, where γ(t) = (l(t), u(l(t))). Now we derive a formula for the affine arc-length in terms of u.let l be the projection of γ as above. If ξ is a tangent vector of l at x, then ξ = D 2 u(x) ξ is a tangent vector of l at y. For any given point x 0 γ, subtracting a linear function we may suppose x 0 is the origin and Du(x 0 ) = 0. By a rotation of the coordinates we may also suppose locally u(x) = 1 2 αi x 2 i + O ( x 3), where α i > 0sinceu is locally uniformly convex. Then near the origin, u can be represented as u (y) = α i y 2 i + O ( y 3). Let ξ be the unit tangent vector of l at x 0.Thenξ = (α 1 ξ 1,,α n ξ n ).Let η = ξ / ξ. It follows u ηη (0) = u ξξ(0)/ α 2 i ξ2 i. (2.5) Hence, returning to our original coordinates, the affine arc-length of the curve γ can also be expressed as ( ) L = ρ u 1/2ds, ηη (2.6) l

5 Affine complete locally convex hypersurfaces 49 where s is the arc-length parameter of l andby(2.4),ρ =[detd 2 u ] 1/(n+2). Note that the power in ρ is positive whereas the power in ρ is negative. Now let S θ = S n 1 {x 1 > cos θ} and S θ (r) = rs θ,whereθ (0, π 2 ), r > 0. For any point y S θ (r),letl = l y be the (open) line segment joining the origin to y and C θ (r) the union of the line segments l y for all y S θ(r). C θ is a cone with vertex at the origin, radius r, aperture θ and axial direction e 1 = (1, 0,, 0). More generally, for z R n, r > 0, ξ S n, θ (0, π 2 ), we let C θ = C θ (z, r,ξ)denote the congruent cone with vertex at z,radiusr, aperture θ and axial direction ξ,ands θ = S θ (z, r,ξ)= C θ { x z =r}. Lemma 2.1. Suppose C θ = C θ (z, r,ξ) Ω. Then there exists a line segment l y = zy such that ( L y = ρ u ) 1/2ds ηη C, (2.7) where C > 0 depends on n, θ, r, and sup y Cθ/2 Du (y). Proof. Applying (2.6) to the line segment l y,wehave L y ( r ρ ds ) 1/2( r u ηη ds) 1/2 0 l y 0 ( η = y z ), y z where the second integral is less than u η (y) u η (0) C.Since r ρ ds C ( r s n 1 (ρ ) n+2 ds ) 1/(n+2), 0 we can now estimate, using the spherical polar coordinates, ( r L y dy C ρ ) 1/2 dy S θ/2 S θ/2 0 r C s n 1 detd 2 u ds ) 1/2(n+2) dy S θ/2 ( 0 0 C ( r s n 1 detd 2 u dsdy ) 1/2(n+2) S θ/2 0 = C ( C θ/2 detd 2 u ) 1/2(n+2) = C Du (C θ/2 ) 1/2(n+2). This completes the proof. We say Ω satisfies the cone condition at y 0 Ω if there is a cone C Ω with y 0 as the vertex. Since u is convex, the directional derivative u η (y 0) = η u (y 0 ) is well defined for any direction η in the cone.

6 50 N.S. Trudinger, X.-J. Wang Lemma 2.2. Suppose Ω satisfies the cone condition at y 0 with cone C θ (y 0, r,ξ) such that ξ u(y 0 ) is finite. Then Du (y) is bounded on the cone C θ/2 (y 0, 1 2 r,ξ). Proof. By the convexity of u we have u (y) u (y 0 ) + C y y 0 for any y C 3θ/4.Since ξ u (y 0 ) is bounded, we also have u (y) u (y 0 ) C y y 0 on the ray y 0 + tξ. Foranyy C θ/2, y y 0 r/2, we have dist(y, C 3θ/4 ) sin θ 2 y y 0. Therefore by the convexity of u we see that Du (y) is bounded for any y C θ/2 (y 0, 1 2 r,ξ). Corollary 2.3. Suppose there is a boundary point y 0 1 Ω such that Ω satisfies the cone condition at y 0 and ξ u (y 0 ) is finite, where ξ is the axial direction of the cone. Then M is not affine complete. Proof. Indeed, by Lemmas 2.1 and 2.2, there is a line segment l connecting y 0 to an interior point of Ω such that (2.7) holds. Let γ ={(y, u (y)) y l } be the curve on the graph of u and γ be the corresponding curve on the graph of u. Thenγ is of finite affine length and connects an interior point of Γ to a boundary point or infinite point of M. Consequently M is not affine complete. For a star-shaped domain D R n, we denote as above by 2 D the set of boundary points y such that ty D for all t (1 ε, 1 + ε) for some ε>0depending on y, and 1 D = D 2 D. Next we need a result on convex functions. The proof of the lemma is irrelevant to the argument here, and is left to Sect. 4. Lemma 2.4. Let D be a star shaped domain in R n and w be a convex function defined on D. Suppose ξ w(y) = at any boundary point y 1 D at which D satisfies the cone condition, where ξ is the axial direction of the cone. Then D is a convex domain. It follows 2 D = and y w(y) = for any y D. Applying Lemma 2.4 to D = Ω and w = u we obtain Corollary 2.5. Let M be affine complete. Let Γ, specified as above, be represented by (2.1). Then Du(x) as x Ω. Namely the Gauss mapping G(p) converges to the equator S n {x n+1 = 0} as p converges to the boundary of Γ. Indeed, since M is affine complete, by Corollary 2.3, ξ u (y) = at any boundary point y 1 Ω at which Ω satisfies the cone condition. By Lemma 2.4, Ω is convex and so u is single valued. Let {x k } Ωsuch that x k Ω. Then either y k = Du(x k ) converges to a boundary point of Ω or an infinite point (choosing a subsequence if necessary). But if y k converges to a boundary point, by Lemma 2.4 we have x k = Du (y k ) converges to infinity. Hence only the case Du(x k ) occurs. In the next section we will show that if M is not Euclidean complete, then there is a point q M such that Γ = Γ q does not satisfies Corollary 2.5, namely, Γ has a boundary point p such that G(p) S n.thisis a contradiction. Hence M is Euclidean complete and Theorem A holds.

7 Affine complete locally convex hypersurfaces A characterization of the boundary In this section we show that if a locally uniformly convex hypersurface M is not Euclidean complete, then there is a boundary point near which M is a graph. Lemma 3.1. Let M be a locally uniformly convex hypersurface in R n+1. Suppose the origin O is on M and the normal of M at O is not in the x n+1 axis. Let S be the connected subset of M {x n+1 = 0} containing O. If S M =, then the ±x n+1 -directions are not the normal of M at any point on S, and S is a locally convex, (n 1)-dimensional hypersurface. Proof. It suffices to prove the first assertion. We argue by contradiction. Let E denote the (closed) set of points on S at which the normal of M lies in the x n+1 axis. Since the x n+1 -direction is not the normal of M at the origin, it is not the normal of M at any point nearby. Let p be a point in E and let l be asmoothpathins connecting p and the origin. We may suppose l E ={p}, otherwise we may replace p by a point p l E such that E has no other point on l between the origin and p. Suppose e n+1, the negative x n+1 - direction is the normal of M at p. SinceS M =, p is an interior point of M. Hence in a neighborhood of p, M can locally be represented as a graph of a convex, nonnegative function x n+1 = g(x 1,, x n ).Thatis, {x n+1 = 0} is the tangent hyperplane of M at p and locally M stays on the side {x n+1 0}. On the other hand, since between the origin and p there is no other point of E lying on l, M intersects with {x n+1 = 0} at any point in l\{p}. Hence M can not lie on one side of {x n+1 = 0} in any neighborhood of p. We reach a contradiction. This completes the proof. Now suppose that the origin O is a point on M such that the south pole of S n is the normal of M at O.ThenL ={x n+1 = 0} is the tangent hyperplane of M at O. For any t 0, let Λ t be the connected subset of {0 x n+1 t} M containing O. ThenΛ 0 ={O}, Λ t is closed, and Λ t Λ t for any 0 t < t. Fort > 0 small, Λ t is bounded and convex since M is locally uniformly convex. Let D t denote the convex closure of Λ t,thatis,d t = {D D is a convex body containingλt }.ThenD t {0 x n+1 t}, and Λ t is convex if and only if it lies on the boundary of D t. Lemma 3.2. Suppose t > 0 is such that Λ t M = and G(Λ t ) has a positive distance from the north pole of S n,whereg: M S n is the Gauss mapping. Then Λ t is convex. Proof. We first show that the south pole e n+1 of S n cannot be the normal of M at any point in Λ t {O}. Indeed, this is true for Λ s {O} when s > 0 is small. If there is an s 0 (0, t] such that e n+1 is not the normal of M at any point in Λ s0 {0 < x n+1 < s 0 } and e n+1 is the normal of M at some point p on Λ s0 {x n+1 = s 0 }, then there is a neighborhood ω of p such that ω {p} {x n+1 > s 0 } since M is locally uniformly convex. On the other

8 52 N.S. Trudinger, X.-J. Wang hand, by our definition of Λ s0, there is a curve l connecting the origin to p such that l {0 x n+1 s 0 }; a contradiction. Since G(Λ t ) has a positive distance from the north pole of S n, Λ t is a bounded set. Let Γ s = Λ t L s (L s ={x n+1 = s}). By Lemma 3.1, Γ s, as a hypersurface in R n, is compact and locally convex. By the well known result of Hadamard, that a closed, locally convex hypersurface is convex, we see that Γ s is convex for any 0 < s t. LetΩ be the convex domain in R n ={x n+1 = t} enclosed by Γ t.thenλ t Ω, as a hypersurface in R n+1, is a compact, locally convex hypersurface, and hence is convex. We move the tangent hyperplane L ={x n+1 = 0} upward to height t.for t > 0smallwehaveΛ t M = and G(Λ t ) has a positive distance from the north pole of S n. Hence by Lemma 3.2, Λ t is a convex hypersurface. It follows that D t is the convex body enclosed by Λ t and L t. Suppose at height t > 0, G(Λ t ) has a positive distance from the north pole and Λ t contains no boundary points of M, then for ε>0small, Λ t+ε M is empty and G(Λ t+ε ) has a positive distance from the north pole. Hence by Lemma 3.2, Λ t+ε is convex. That is we can move L further upward to height t > t such that G(Λ t ) has a positive distance from the north pole and Λ t M is empty. Let τ be the largest constant such that for any t <τ, Λ t is convex, Λ t D t, G(Λ t ) has a positive distance from the north pole, and Λ t M =. We have three cases: τ = ;orτ< and there is a point p Λ {x n+1 = τ} such that the north pole is the normal of M at p, whereλ = t<τ Λ t ;orτ< and Λ M =. Case 1: τ =. Then for any t > 0, Λ t is convex by Lemma 3.2. In this case we have obviously Λ = t 0 Λ t is a Euclidean complete convex hypersurface, and M = Λ. Case 2: τ< and there is a point p Λ such that G(p) is the north pole. Let D = t<τ D t and Ω = D Λ. We may suppose there is no boundary point of M lying in Λ, the (Euclidean) closure of Λ, otherwise it is in Case 3 below. Then Ω {x n+1 = τ} is closed, and convex by Lemma 3.2. We need the following lemma. Lemma 3.3. Let z 0 N such that T(z 0 ) = O. LetA t be the connected subset of T 1 (Λ t ) containing z 0. Then T is a homeomorphism from A t to Λ t for any t (0,τ). Proof. Lemma 3.3 is obviously true for t > 0 small. Suppose the lemma is true for t, we show it is true for t + ε if t + ε<τand ε>0 is sufficiently small. Indeed, if it is not the case, then there exist sequences z k, z k A t+εk, where ε k 0, such that T(z k ) = T(z k ) = p k Λ t+εk.sinceλ t is compact, we may suppose p k p 0 Λ t. Suppose z k z and z k z, then T(z) = T(z) = p 0. By assumption, Lemma 3.3 holds at t. Hence we have z = z.howeversincet is locally a homeomorphism, we must have z k = z k when k is large. Hence Lemma 3.3 holds in an open interval (0, t ). If t <τwe show that Lemma 3.3 holds for t = t. If this is not true, then there exist two points z, z A t such that T(z) = T(z) = p 0 Λ t.bythe

9 Affine complete locally convex hypersurfaces 53 assumption that Lemma 3.3 holds for t < t, the points z, z cannot both be interior points of A t.ifz is an interior point and z is a boundary point of A t, then p 0 = T(z) is an interior point of Λ t and p 0 = T(z) is simultaneously a boundary point of Λ t. This is impossible. If both z and z are boundary points of A t, we choose sequences z k, z k A tk, k = 1, 2,,where t k t, such that z k z and z k z. Then both T(z k ) and T(z k ) converge to p 0.SinceΛ t is convex, there are curves γ k Λ t connecting T(z k ) to T(z k ) such that the (Euclidean) arc-length of γ k converges to zero. In other words, both T(z k ) and T(z k ) are in the r-neighborhood of p 0 with r 0as k.sincep 0 is an interior point of M, T is locally a homeomorphism. It follows dist(z k, z k ) 0, whence z = z. This completes the proof. We return to our discussion of Case 2. Since the north pole e n+1 of S n is a normal of M at p and M is locally uniformly convex, M can locally be represented as x n+1 = g(x 1,, x n ) + τ for a non-positive, concave function g, andg < 0 except at the point p. Hence p is an isolated point in Λ,andΩ is a single point since it is convex. Let Λ = Λ {p}. ThenΛ is a closed, convex hypersurface. Let A = t<τ A t. By Lemma 3.3, T is a homeomorphism from A to Λ.Sincep is not a boundary point of M, A N is a pre-compact set. It follows that for any z A, T(z) = p. On the other hand, arguing exactly the same as the second part of the proof of Lemma 3.3, we find that there is one and only one point z A (the closure of A ) such that T(z) = p. Hence Λ is homeomorphic to A = A {z}. But since Λ is a closed convex hypersurface, it follows that N is a closed manifold without boundary, namely M = Λ is a closed convex hypersurface without boundary. Case 3: τ< and Λ M =. In this case we have for any t <τ, Λ t is convex and G(Λ t ) has a positive distance from the north pole. We have thus proved Theorem 3.1. Let M be a locally convex hypersurface and M =.Let Λ t,t [0, ), and Λ be as above. Then there exists τ (0, ) such that Λ t is convex, Λ t M = and G(Λ t ) has a positive distance from the north pole for any 0 < t <τ, and Λ contains a boundary point of M. Proof of Theorem A. If M is not Euclidean complete, we suppose the origin O M such that e n+1 is the normal of M at O.LetΛ t = Λ t,o and D t = D t,o be defined as above. By Theorem 3.1, there exists τ (0, ) such that Λ t is convex and Λ t M = for any 0 < t <τ,and Λ contains a boundary point p of M,whereΛ = Λ O = t<τλ t,o. Let D = t<τ D t and Ω = D Λ.ThenΩ is convex and p Ω. By choosing the coordinates properly we suppose p = (a, 0,, 0,τ) such that Ω {x 1 a}. Letγ Λ beacurve,withp as an endpoint, such that G(γ) is a geodesic (meridian) line on S n.thatis,foranyq γ, the tangent hyperplane of M at q can be represented as L θ ={cos θx 1 + sin θx n+1 = c} for some constant c = c(θ). LetL θ0 be the tangent hyperplane at p (θ 0 is

10 54 N.S. Trudinger, X.-J. Wang uniquely determined by our construction). Let q be a point on γ such that θ θ 0 < π 4. Now we choose a new coordinate system such that q is the origin and L θ ={y n+1 = 0}. In the new coordinates we introduce similarly the sets Λ t,q, which is the connected component of M {0 y n+1 t} containing q. By Theorem 3.1 there is a τ = τ q > 0 such that Λ t,q M = for any 0 < t <τand Λ q contains a boundary point of M,whereΛ q = t<τλ t,q. From the above construction we have Λ q Λ O and Λ q Λ O ={p}. It follows that p Λ τ,q.letγ = Γ q be the connected subset defined in Sect. 2. Then p is a boundary point of Γ. Sinceq is chosen such that θ θ 0 < π 4,wehaveG( γ) Sn {y n+1 < 2/2},where γ γ is the arc connecting q to p. Hence p is a boundary point of Γ and G(p) S n. We reach a contradiction by Corollary 2.5. Theorem A is thus proved. Remark 1. Theorem 3.1 has been extended to nonsmooth hypersurfaces in [14], where we also found other geometric characterizations of locally convex hypersurfaces with boundary. Applications such as the existence of locally convex hypersurfaces of constant Gauss curvature are also treated there. Remark 2. Theorem A is not true for n = 1. An example is the curve x 1 = (1 + t 2 ) α/2 cos t, x 2 = (1 + t 2 ) α/2 sin t, where t (, ),andα (1, 3 ) is a constant Proof of Lemma 2.4 In this section we prove Lemma 2.4. Let ω be a convex domain in R n 1 and h a convex function in ω. Denote by µ h the Monge-Ampère measure of h, defined by µ h (E) = N h (E) for any Borel set E ω, where N h (E) is the Lebesgue measure of the set N h (E),andN h is the normal mapping of h, defined by N h (x ) ={ξ R n 1 h(y ) h(x ) + (y x ) ξ y ω} and N h (E) = x E N h (x ) for any Borel set E ω.forc 1 convex functions the normal mapping is the gradient mapping, i.e., N h (E) = Dh(E). Obviously if inf ω h < inf ω h,thenµ h (ω) > 0. By a result of Aleksandrov [11], the set {y N h (ω) x 1, x 2 ω such that y N h (x 1 ) N h(x 2 )} has measure zero. Hence if E ω is such that µ h (E) >0, then there is a point x E such that h is strictly convex at x, namely there exists ξ R n 1 such that h(y )>h(x ) + (y x ) ξ for y ω, y = x. A further result of Aleksandrov which we shall use below is the weak continuity of the Monge-Ampère measure, namely if {h k } is a sequence of convex functions

11 Affine complete locally convex hypersurfaces 55 defined on ω such that h k h locally uniformly, then µ hk µ h weakly as measures; that is ϕdµ hk ϕdµ h (4.1) ω for any continuous function ϕ with compact support in ω. We refer the reader to [6,11] for basic properties of the measure µ h. Lemma 4.1. Let h be a positive convex function in a bounded convex domain ω R n 1 such that inf h < inf h. (4.2) x ω x ω Let J ω be a Borel set such that µ h (J) >0. Then there is no convex function w defined in D ={(x, x n ) R n x ω, 1 < x n < h(x )} such that w xn (x, h(x )) = for x J. Proof. Suppose there exists such a convex function w. Thenw is locally bounded on D ={(x, h(x )) x ω} since it is convex on the tangent plane of D. Adding Cx n to w we may suppose w xn 0on {(x, h(x )) x ω}. By the convexity of w we may also suppose Dw is bounded near {(x, h(x )) x ω θ },whereω θ ={x ω dist(x, ω)>θ} and θ (0, 1) close to 0, for otherwise we can replace h by h, where h(x ) = sup{a(x ) a linear, a h in ω, and a h δ on ω}. By (4.2) we can choose δ>0 small such that µ h (J) >0. Therefore we can suppose directly that Dw is bounded near {(x, h(x )) x ω}. We may suppose J is closed, for otherwise we replace J by a closed subset J J such that µ h (J )>0. Hence for any M > 1, there exists ε>0, sufficiently small, such that w xn (x, h(x ) ε) > 2M on J. Hence in an open set E J we have w xn (x, h(x ) ε) > M.Let{h k } be a sequence of smooth convex functions such that h h k h ε and h k h as k. By the convexity of w we have w xn (x, h k (x )) > M on E. By approximation we may suppose w is smooth near {(x, h k (x )) x ω}.let w k (x ) = w(x, h k (x )).Wethenhave,fori, j = 1,, n 1, 2 w k h k h k h k h k = w xi x x i x j + w xi x n + w x j x j x n + w xn x j x n + w xn i x i x j Hence w k is convex and det 2 w k w xn n 1 det 2 h k x i x j x i x j { M n 1 det 2 h k x i x j if x E, 0ifx ω E. ω 2 h k. x i x j

12 56 N.S. Trudinger, X.-J. Wang It follows Let k we obtain by (4.1) that µ wk (ω) M n 1 µ hk (E). µ w (ω) M n 1 µ h (E) M n 1 µ h (J), where w(x ) = w(x, h(x )) is convex since it is the limit of a sequence of convex functions. By definition, µ w (ω) = N w (ω). It follows D w(ω), and so D w( ω), is unbounded, since M > 1 is arbitrary. This is in contradiction with our assumption that Dw is bounded near {(x, h(x )) x ω}. Hence Lemma 4.1 holds. For the two dimensional case Lemma 2.4 follows readily from Lemma 4.1. Indeed let e 2 = (0, 1) be a boundary point of D and locally D be represented by x 2 = g(x 1 ) for x 1 ( a, a) (see (4.3) below for a precise definition of g). If D is not convex we may suppose g(0) < 1 (g(a) + g( a)); 2 (indeed we may suppose g(0) <min{g(a), g( a)} since the conditions in Lemma 2.4 are invariant under linear transformations x Ax). Let h(x 1 ) = sup l(x 1 ), where the supremum is taken over all linear functions l g. LetJ ={g = h}. Thenµ h (J) >0. Since for any x 1 J, D satisfies the cone condition at (x 1, g(x 1 )), by assumption we have w x2 = for x 1 J. However by Lemma 4.1 there is no such convex function w. Hence Lemma 2.4 holds for n = 2. To prove Lemma 2.4 for high dimensions, we will use an induction argument and apply Lemma 4.1 in lower dimension subspaces. We will prove that Lemma 2.4 holds under weaker conditions. We say D satisfies the star-cone condition at y D if there is a cone in D with vertex at y such that y is the axial direction and the origin is contained in the cone. Obviously if D is star-shaped and D satisfies the star-cone condition at y, theny 1 D. Lemma 2.4 follows from the following Lemma 4.2 immediately. Lemma 4.2. Let D R n be a connected domain containing the origin and w a convex function defined on D. Suppose ξ w(y) = at any boundary point y D at which D satisfies the star-cone condition, where ξ = y is the axial direction of the cone. Then D is a convex domain. Proof. We may suppose D is a star-shaped domain, otherwise we consider instead the domain D ={x D Ox D}, whereox is the line segment connecting the origin O to x. If y is a boundary point of D, and D satisfies the star-cone condition at y, theny is also a boundary point of D and D also satisfies the star-cone condition at y, and vice versa. Therefore if D is proven convex, then D satisfies the star-cone condition at any boundary point and hence D D.Thatis D = D and D is convex.

13 Affine complete locally convex hypersurfaces 57 Let D be a convex domain in R n 1 such that D D {x n = 0}. Let D = D {x R n xn > 0, x D }.Let g(x ) = inf{x n (x, x n ) D}. (4.3) For given x D, if there is no t R such that (x, t) D,wedefine g(x ) =+.Theng is lower semicontinuous since the boundary D is a closed set. To prove D is convex we need to prove g is concave. 1. We will accomplish this through an induction argument. First we prove a result which will be used in each step of the induction process. Let us consider the restriction of D, g, andw in the subspace L k ={x k+1 = = x n 1 = 0}. L k is a (k + 1)-dimensional subspace of R n. Suppose there exists a bounded convex domain ω L k such that Define a convex function inf ω g < a 0 inf g. (4.4) ω ϕ(x ) = sup{a(x ) a linear, a g in ω, and a a 0 on ω}, x L k. Then ϕ = a 0 on ω and inf ω ϕ<a 0. Note that ϕ is in fact a function of k-variables since x k+1 = = x n 1 = 0foranyx L k.let J ={x ω g(x ) = ϕ(x )} χ = χ(j) ={(x, g(x )) x J} Then µ ϕ is supported on J and µ ϕ (J) >0. Let D k = L k {(x, x n ) R n x ω, 1 < x n <ϕ(x )}.SinceD R n is star shaped, so is D k (as a domain in L k ). Since ϕ is convex as a function of x 1,, x k, D k satisfies the star-cone condition on J. If there exists a subset J J with µ ϕ (J )>0 such that D satisfies the star-cone condition on χ(j ), we have by assumption that ξ w(y) = for any y χ(j ),where ξ = y is the axial direction of the cone at y. It follows by the convexity of ϕ that xn w(y) = for any y χ(j ). We reach a contradiction by applying Lemma 4.1 to w in the domain D k L k. It follows that if k < n 1, then D does not satisfy the star-cone condition for a.e. (with respect to the measure µ ϕ ) x J. Howeverifk = n 1, then D satisfies the star-cone condition automatically at any point in χ(j). Therefore Lemma 4.2 is true if (4.4) holds for k = n Let e i be the unit vector on the x i -axis. If g is not concave near x = 0, we may suppose e n ± e 1 D and g(0) <min{g(e n + e 1 ), g(e n e 1 )}, by an affine transformation of coordinates leaving the origin invariant. Let l ={ 1 < x 1 < 1} L 1 be a line segment, where L 1 is the 2-plane defined in Step 1. We identify l with the interval ( 1, 1). Let h(t) = sup{a(t) a linear function, a g on l}.

14 58 N.S. Trudinger, X.-J. Wang Then h is convex on l,andµ h (l) = µ h (J) >0, where J ={x l h(x ) = g(x )} and χ(j) ={(x, g(x )) x J} as above. When n = 2, D satisfies the star-cone condition at any point in χ(j). It follows, as in Step 1, that xn w(y) = for any y χ(j). We reach a contradiction by Lemma 4.1. Hence Lemma 2.4 is proved for n = 2. For high dimensions we have from Step 1 that D does not satisfy the star-cone condition for a.e. (with respect to the measure µ h ) x J. 3. Therefore we can suppose h is strictly convex at 0 J and D does not satisfies the star-cone condition at (0, h(0)). Then there exists a R such that h(t) at > h(0) for t = 0, where h(0) = g(0) since 0 J. By a rotation of the coordinates (x 1, x n ) (leaving x 2,, x n 1 invariant) we may suppose a = 0andh attains a strict minimum at x 0. For simplicity we suppose x 0 = 0. Then there exists ε>0suchthath(± 1 2 ) h(0) + ε since h is strictly convex at 0, and so g(± 1 2 e 1) h(0) + ε. Bythelower semicontinuity of g there exists δ>0 such that { g(x ) h(0) ε, when x e 1 g(x ) 1 2 h(0), δ or x 1 2 e 1 δ when 1 2 x 1 1, 2 ˆx δ, where ˆx = (x 2,, x n 1 ). Since D does not satisfies the star-cone condition at (0, h(0)), there is a sequence y k R n 1, y k 0, such that By choosing a subsequence we may suppose (g(y k ) g(0))/ y k. (4.5) y k = β k e 1 + α k e 2 + o(α k )e 3 + +o(α k )e n 1, where α k > 0, α k,β k 0ask. We consider the restriction of D, g and w in the subspace L 2 = {x 3 = =x n 1 = 0}. Letω = {x L 2 0 < x 2 <δ, 1 2 < x 1 < 1 }. Then for K > 1 large enough 2 we have g(x ) Kx 2 + h(0) for x ω. On the other hand, by (4.5) there exist points in the domain ω at which g < Kx 2 + h(0). Hence when restricted in L 2, the plane L ={x n = Kx 2 + h(0)} cuts a piece from D. Thatis,theset{x D L 2 x ω, x n = g(x )< Kx 2 + h(0)} is not empty. Denote Φ = L {x ω}. Φ is a rectangle in the 2-plane L ={x n = Kx 2 + h(0)} {x 3 = = x n 1 = 0}. We make a linear transformation x 1 = x 1, x i = x i for 3 i n 1, x 2 = (x 2 Kx n )/ 1 + K 2, x n = (Kx 2 + x n )/ 1 + K 2. Then L ={ x n = h(0)/ 1 + K 2 }. In the new coordinate system we define as in (4.3) a function g, thatis g( x ) = inf{ x n ( x, xn ) D}, where

15 Affine complete locally convex hypersurfaces 59 x = ( x 1,, x n 1 ) ω, and ω is the projection of Φ on { x n = 0} (both ω and ω are rectangles). From the above construction we see that g satisfies (4.4) with a 0 = h(0)/ 1 + K 2.Defineϕ, J,χ(J) as in Step 1. By (4.4) we have µ ϕ (J) >0. If n = 3, D satisfies the star-cone condition at any point in χ(j). We reach a contradiction by Lemma 4.1. Hence Lemma 2.4 is proved for n = 3. For higher dimensions we have by Step 1 that D does not satisfy the star-cone condition for a.e. (with respect to the measure µ ϕ ) x J. 4. Rewrite the function ϕ obtained in Step 3 as h. Similar to Step 3 we can suppose h is strictly convex at 0 and D does not satisfy the star-cone condition at (0, h(0)). Then arguing as in Step 3, we see that, when restricted in the subspace L 3 ={x 4 = =x n 1 = 0}, there is a plane, say, L ={x n = Kx 3 +h(0)} which cuts a piece from D.AsinStep3wecan make a linear coordinate transformation which leaves x 1, x 2, x 4,, x n 1 invariant, such that L ={ x n = h(0)/ 1 + K 2 }. Similarly we have a function g, defined on a convex cylinder ω, satisfying the condition (4.4). Therefore we can define ϕ, J,χ(J) as in Step 1. If n = 4thenD satisfies the star-cone condition on χ(j) and we reach a contradiction by Lemma 4.1 as above. Hence Lemma 2.4 holds for n = 4. For even higher dimensions we have as in Step 1 that D does not satisfy the star-cone condition for a.e. x J. 5. For a given dimension n, we continue arguing as above and at the (n 1) th step we see that D is locally represented by x n = g(x ) and g satisfies the condition (4.4) in Step 1, namely there exists a convex domain ω R n 1 such that (4.4) holds. We then define ϕ, J,χ(J) as in Step 1. In this step D satisfies the star-cone condition on any point in χ(j) and µ ϕ (J) >0. By Lemma 4.1 we reach a contradiction. Therefore g must be concave. This completes the proof. References 1. W. Blaschke, Vorlesungen über Differentialgeometrie, Berlin, E. Calabi, Complete affine hypersurfaces, Symposia Mathematica, 10 (1971), E. Calabi, Hypersurfaces with maximal affinely invariant area, Amer. J. Math., 104 (1982), E. Calabi, Convex affine maximal surfaces, Results in Math., 13 (1988), E. Calabi, Affine differential geometry and holomorphic curves, Lect. Notes Math (1990), S.Y. Cheng, S.T. Yau, On the regularity of the Monge-Ampère equation det 2 u x i x = j F(x, u), Comm. Pure Appl. Math., 30 (1977), S.Y. Cheng, S.T. Yau, Complete affine hypersurfaces I. The completeness of affine metrics, Comm. Pure Appl. Math., 39 (1986), J. van Heijenoort, On locally convex manifolds, Comm Pure Appl. Math., 5 (1952), A.M. Li, U. Simon, G. Zhao, Global affine differential geometry of hypersurfaces, Walter de Gruyter, Berlin, K. Nomizu, T. Sasaki, Affine differential geometry, Cambridge, A.V. Pogorelov, The muitidimensional Minkowski problems, New York, J. Wiley, 1978

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