j. differential geometry 57 (2001) 389-407

NONPOSITIVELY CURVED SURFACES IN R^{3}

HSUNGROW CHAN & ANDREJS TREIBERGS

Abstract

We consider complete nonpositively curved surfaces with one end twice continuously differentiably immersed in Euclidean three space. If such a surface is embedded near infinity and has square integrable second fundamental form then it must lie between two parallel planes.

We generalize some results obtained by H. Chan [2], [3] about asymptotically totally geodesic, complete, nonpositively curved surfaces C^{2} immersed in ℝ^{3}. We show that if such a surface has one end which is embedded near infinity and satisfies the condition that the second fundamental form is square integrable, then it must be contained in a slab, the closed region between two parallel planes. Moreover, the surface must contact the bounding planes of the narrowest slab along locally nonconvex sets heading to infinity. This implies H. Chan’s result that a one ended surface with nonpositive curvature having only isolated parabolic points admits no isometric immersion of this type at all [2], [3].

An example of such a surface is the solution of the equation

(1 + z)x^{2} − (1 − z)y^{2} = 2z(1 − z^{2})

in three-space, which is topologically the punctured torus. This surface lies in the slab |z| ≤ 1 and for x^{2} + y^{2} > 2 is the graph of a singlevalued function z = u(x, y). The curvature is negative except at the lines z − 1 = x = 0 and z + 1 = y = 0 where it vanishes. Also both the curvature and the square length of the second fundamental form |A|^{2} decay at the order of ρ^{−4} as the distance from a point ρ → ∞.

Received March 19, 1999.

White [25] studied immersed surfaces with square integrable second fundamental form. He showed that if such a surface is nonpositively curved, then its ends are properly immersed and the Gauss map converges at the end. Thus ends contribute integer multiples of 2π to the total curvature. Other classes of nonpositively curved surfaces have also been studied. Using a method of Verner [24], Chan studied surfaces whose curvature is strictly negative, whose Gauss map is one to one and convergent near each end and whose total curvature is −2π outside every peripheral closed geodesic [2], [4].

The best known nonexistence results about immersing nonpositively curved surfaces make no extrinsic hypotheses other than regularity. Hilbert showed that a complete surface with constant negative curvature can’t be C^{4} isometrically immersed into ℝ^{3} [7]. Efimov proved that there are no C^{2} immersions of simply connected surfaces with sufficiently negative curvature. This means curvature is bounded above by a negative constant [5], [14], or is slowly varying but allowed to decay inversely with the distance squared [6]. Perelman generalized Efimov’s Theorem to halfspaces [17]. See Rozendorn’s [20] survey. Existence results are obtained by Hong [8] who constructs a global isometric immersion of a simply connected surface with sufficiently regular metric whose negative curvature decays like ρ^{−2−ϵ} and whose first and second derivatives decay correspondingly, where ρ is the distance from a point. His constructions improve previous results on isometric immersions for strips [13], [18], [21], [22], [23].

One motivation for this study comes from the general relativity question about the possibility of constructing embedding diagrams for initial metrics for the Cauchy problem for Einstein’s equations. Misner [15] constructed a complete Riemannian metric dσ^{2} on 𝕊^{1} × 𝕊^{2} − {(p_{0}, q_{0})} with vanishing scalar curvature. This manifold and its double cover represent a stationary initial time slice of a spacetime with a wormhole or with a pair of black holes. The metric is invariant under rotations about q_{0} in the second factor. If 𝕊′ = 𝕊^{1} is a great circle of 𝕊^{2} through q_{0} then the restriction to 𝕋^{2} = 𝕊^{1} × 𝕊′ is the metric

ds_{M}^{2} = dσ^{2}|_{𝕋^{2}} = a^{2}ϕ^{4} (dx^{2} + dy^{2})
ϕ(x, y) = ∑_{n=−∞}^{∞} \frac{1}{\sqrt{cosh(x + 2μn) − cos y}}

where a, μ are constants and (x, y) are coordinates in the flat torus ℝ^{2}/(2μℤ⊕2πℤ) and (p_{0}, q_{0}) = (0, 0) is the end. An embedding diagram is a C^{2} isometric immersion of the Misner surface (𝕋^{2} − {(p_{0}, q_{0})}, ds_{M}^{2}) into ℝ^{3}. The curvature of ds_{M}^{2} is strictly negative [19] and the end is asymptotic to an end of the t = 0 slice of the Schwarzschild metric (ℝ × 𝕊^{2}, dσ_{S}^{2}). In fact K = O(ρ^{−3}) so the Efimov type nonexistence theorems don’t apply. Lifting the closed geodesics around the ends to the universal cover gives directions in which the curvature fails to decay, thus the Hong type existence theorems don’t obviously apply either.

If one restricts the Schwarzschild metric to a great circle 𝕊′ ⊂ 𝕊^{2}, then the resulting complete Riemannian metric on U = ℝ×𝕊′, i.e., ds_{S}^{2} = dσ_{S}^{2}|_{U} admits an isometric embedding into Euclidean space as a parabola of revolution. In particular both K and |A|^{2} are O(ρ^{−3}) as ρ → ∞ and so the second fundamental form is in L^{2}(U, ds_{S}^{2}). Since the metric ds2 ; M̃ away from the black holes is asymptotic to the metric ds_{S}^{2} away from the single Schwarzschild hole, it is natural to postulate that if an embedding diagram for the Misner surface existed, it would be asymptotic to the embedding of the Schwarzschild surface. Price and Romano [19] tried constructing such an embedding diagram by propagating Schwarzschild data toward the black hole numerically, but the solution developed a shock suggesting that no diagram exists. Chan’s Theorem confirms nonexistence of an embedding diagram with nice end behavior.

Theorem 3 ([2], [3]). There is no C^{2} isometric immersion in Euclidean three space of the Misner wormhole surface W = (𝕋^{2} − {(p_{0}, q_{0})}, ds_{M}^{2}) under the hypothesis that it is embedded near the end and that it has square integrable mean curvature

∫_{W} H^{2} da < ∞.

Proof. The Misner surface is negatively curved with finite total curvature. Mean curvature is H = h_{11} + h_{22} in a local orthonormal frame. Hence ∫ |A|^{2} da = ∫ H^{2} − 2K da < ∞ so Theorem 1 and Theorem 2 apply. q.e.d.

Chan [4] showed nonexistence of C^{3} immersions of other physically interesting negatively curved surfaces such as the double cover of the Misner surface and the Brill-Lindquist surfaces assuming square integrable second fundamental form and one to one Gauss map near each end.

1. Preliminaries

Let M be a complete, oriented, nonpositively curved surface C^{2} immersed in Euclidean space ℝ^{3}. We assume that M satisfies

∫_{M} |A|^{2} da < ∞ (1)

where the length of the second fundamental form in local orthonomal frame is given by |A|^{2} = ∑_{i,j=1}^{2} h_{ij}^{2}. Hence M has finite total curvature, thus, by Huber’s Theorem [12], is conformal to a closed Riemann surface with finitely many punctures Σ − {q_{1}, … q_{r}}. It follows by a theorem of B. White [25] that M is properly immersed near each end and the Gauss map extends to a continuous function G : Σ → 𝕊^{2}. We orient our surface so that as a point approaches the end q_{1}, the Gauss map tends to vertical (0, 0, 1). If the surface is embedded near the q_{1} end, then in a sufficiently small disk neighborhood of q_{1}, the surface M is given as the graph of a function z = u(x, y) where (x, y) ∈ ℝ^{2} and |(x, y)| ≥ R some 0 < R < ∞.

Let γ be a round circle in the plane, D(γ) the closed disk bounded by γ, r(γ) its radius and s(γ) its center. Let a(γ) = (a_{1}, a_{2}) and c(γ) be the coordinates of the plane P_{γ} which we identify with the affine function P_{γ}(x, y) := a_{1}x + a_{2}y + c that best approximates the surface in D(γ). That is, a and c are uniquely determined by

(2)  m(γ) : = inf_{b_{1},b_{2},d∈ℝ}{ max_{(x,y)∈γ} |u(x, y) − b_{1}x − b_{2}y − d|}

= max_{(x,y)∈γ} |u(x, y) − P_{γ}(x, y)|.

Thus, P_{γ} is the unique best affine approximation in sup norm to u restricted to γ and m(γ) is the error. Let

Can (γ) = {(x, y, z) : (x, y) ∈ D(γ) and |z − P_{γ}(x, y)| ≤ m(γ)}.

Let P_{γ}^{±} denote the upper and lower planes of Can (γ), namely

P_{γ}^{±}(x, y) = a(γ) · (x, y) + c^{±}(γ)

where c^{±}(γ) = c(γ) ± m(γ). The following is an easy special case of Chebychev’s Theorem [1].

Lemma 1.1. A linear function P_{γ} which best approximates the continuous function u on the circle γ in supremum norm is uniquely characterized by the condition that there are (at least) four points in γ, in order around the circle β_{1}(γ), β_{2}(γ), β_{3}(γ), β_{4}(γ) so that β_{1} and β_{3} are in P_{γ}^{+} and β_{2} and β_{4} are in P_{γ}^{−}. In other words u − P_{γ} alternates between its extreme values +m(γ) and −m(γ) at these points.

The following is a simple case of de la Vallée-Poussin’s Theorem [1].

Lemma 1.2. If for some linear function P the difference u − P assumes the values λ_{1}, −λ_{2}, λ_{3}, −λ_{4} at four consecutive points around the circle γ, where all λ_{i} > 0, then the deviation of the best linear approximation to u ∈ C(γ) satisfies m(γ) ≥ min{λ_{1}, …, λ_{4}}.

The following is proved by S. Bernstein, E. Hopf [10] and R. Osserman [16] using the maximum principle.

Lemma 1.3. Let G ⊂ ℝ^{2} be a open set and u ∈ C^{2}(G) ∩ C(\overline{G}) such that u_{xx}u_{yy} − u_{xy}^{2} ≤ 0 in G. If u ≥ 0 on ∂G and G is bounded, then u ≥ 0 on G. If u ≥ 0 on ∂G, u > 0 somewhere in G and G is unbounded but can be placed in a sector of angle 0 ≤ θ < π/2, say G ⊂ {(x, y) : |y| < x tan θ}, then there are constants c > 0 and large r_{0} so that for all r ≥ r_{0} there holds sup{u(x, y) : (x, y) ∈ G, x^{2} + y^{2} = r^{2}} ≥ cr. If M ⊂ ℝ^{3} is C^{2} immersed surface with nonpositive curvature and if R ⊂ M is a bounded subset, then R is contained in the convex hull of ∂R.

E. Hopf [9] proved the following accessibility lemma to fill a gap in Bernstein’s proof of his theorem about entire nonpositively curved functions.

Lemma 1.4. Let Ω ⊂ ℝ^{n} be a bounded open set and Ω′ ⊂ Ω be a connected open subset whose boundary has at least one point in common with ∂Ω. Then the set C = ∂Ω ∩ ∂Ω′ contains points which are accumulation points of a Jordan curve J : [0, ∞) → Ω as t → ∞. If C is the union of two disjoint closed sets, then each contains accumulation points of such J.

2. Construction of arcs to infinity

If the error made by the best approximating plane for a circle increases as the radius increases, then it must grow at least linearly. First we wish show that there is a circle γ such that the planes P_{γ}^{±} are transverse to u(x, y) at least at one of the extreme points β_{i}.

Lemma 2.1. Let A = {(x, y) ∈ ℝ^{2} : r_{1} ≤ |(x, y)| ≤ r_{2}} be an annulus and let Γ(r) = {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} = r^{2}} and Γ_{1} and Γ_{2} be the inner and outer boundary circles of A, resp. Suppose that M is a compact piece of a C^{2} surface with K ≤ 0 immersed in the closed solid cylinder Σ = D(Γ_{2}) × ℝ of ℝ^{3} so that ∂M ⊂ ∂Σ. Further suppose that the surface is a graph given by z = u(x, y) in the region A × ℝ where u ∈ C^{2}(A). Let m(r) = m(Γ(r)) be the error function (2). Choose η > 0. Then whenever r_{1} < r < r_{2} such that m is differentiable at r and m′(r) > 0, then one of the points β_{i} ∈ Γ(r) is a point where u is not tangent to P_{γ}^{±}. That is, ∇ u(β_{i}) ≠ a(Γ(r)) for some i ∈ {1, 2, 3, 4}. If m(Γ_{1}) < m(Γ_{2}) then there is such r ∈ (r_{1}, r_{2}). In fact, r can be chosen so that

|a(Γ_{1}) − a(Γ(r))| < η. (3)

Proof. The idea of the proof is that if P^{±} is tangent to u at all minima and maxima then m(Γ(r)) would be constant.

First observe that m(Γ(r)) is a nondecreasing function. This follows from the maximum principle. For small h > 0 the Can (Γ(r + h)) contains the surface within Γ(r), i.e., M ∩ D(Γ(r)) × ℝ ⊂ Can (Γ(r + h)). Similarly m(γ′) ≤ m(γ′′) whenever the disks bounded by the circles satisfy B(γ′) ⊂ B(γ′′).

Let ∥u∥_{C^{2}(A)} < k/3 < ∞ be a bound on second derivatives in the compact set A. So for contradiction, suppose for all r ∈ (r_{1}, r_{2}) that ∇ u(β) = ∇ P_{Γ(r)}(β) whenever β ∈ Γ(r) is extremal, that is, such that u(β) ∈ P_{Γ(r)}^{+}(β) ∪ P_{Γ(r)}^{−}(β). We prove that left derivate of m exists and vanishes at r. By subtracting the affine function a(Γ(r))·(x, y)+c(Γ(r)) from u we may suppose that P^{+}(Γ(r)) = c = m(r) and that the gradient vanishes. Thus |u(β+δ)−u(β)| ≤ k|δ|^{2} whenever β is a critical point for u such as at one of the extrema. It follows for any 0 < h < r − r_{1} that if β′_{i} ∈ Γ(r − h) denote the four points radially inward from the β_{i} ∈ Γ(r) then u(β′_{i}) ≥ c − kh^{2} for i = 1, 3 and u(β′_{i}) ≤ −c + kh^{2} for i = 2, 4. It follows from de la Vallée Poussin’s Theorem, Lemma 1.2, and the fact that m is nondecreasing that m(r) − kh^{2} ≤ m(r − h) ≤ m(r). Together with the fact that m is Lipschitz so differentiable almost everywhere, one concludes that the derivative vanishes wherever it exists, proving the first part of the assertion.

If m(r) = m(r_{1}) for r_{1} ≤ r < r_{2} then a(r) is constant because Can (r_{1}) lies between the planes a(r)˙(x, y) + c(r) ± m(r_{1}) for all r. If we choose r_{3} = sup{r ∈ [r_{1}, r_{2}) : m(r) = m(r_{1})} then by continuity and constancy of a(r), we have a(r_{3}) = a(r_{1}). For ϵ < r_{2} − r_{3} sufficiently small, we have (3) whenever |r − r_{3}| < ε. Now just apply the first argument to the interval r_{1} = r_{3} and r_{2} = r_{3} + ε. q.e.d.

The circle just obtained may be modified so that the best approximating plane is not horizontal.

Lemma 2.2. Let A, Γ(r) and Σ as in Lemma 2.1. Suppose that M is a compact piece of a C^{2} surface, immersed in the closed solid cylinder Σ with ∂M ⊂ ∂Σ and K ≤ 0. Further suppose that the surface is a graph given by z = u(x, y) in the region A × ℝ where u ∈ C^{2}(A). Then whenever r_{1} < r_{2} and m(r_{1}) < m(r_{2}) then there is a circle γ ⊂ A(r_{1}, r_{2}) such that one of the β_{i} ∈ γ is not a point of tangency between P_{γ}^{±} and u. That is, ∇ u(β_{i}) ≠ a(γ) for some i ∈ {1, 2, 3, 4}. Moreover, γ can be chosen so that a(γ) ≠ 0.

Proof. If some circle from Lemma 2.1 satisfies a(Γ(r)) ≠ 0 then we are done. The idea is to show that by translating a circle it is possible to find one such that a(γ′) ≠ 0. Then with η = |a(γ′)| we apply Lemma 2.1 to circles concentric to γ′ to find one satisfying the transversality.

Choose r_{1} < r_{3} < r_{4} < r_{2} so that m(r_{1}) < m(r_{3}) < m(r_{4}) < m(r_{2}) and ε > 0 small. As before, let r ∈ (r_{3}, r_{4}) so that u is transverse to P^{+}(Γ(r)) at, say, β_{1} ∈ Γ(r). For δ > 0 small enough we have r_{3} < r−δ < r+δ < r_{4} and u(x, y) > u(β_{i})−ε for (x, y) ∈ B(β_{i}, δ), i = 1, 3 and u(x, y) < u(β_{i})+ε for (x, y) ∈ B(β_{i}, δ), i = 2, 4. Now choose r_{3} < r_{5} < r and let γ = Γ(r_{5}) + t_{1}β_{1} for small t_{1} > 0 be the translate of the circle toward β_{1} which remains in the annulus γ ⊂ A(r_{3}, r) ∪ B(β_{1}, δ). Thus L = B(β_{1}, δ) − B(Γ(r)) ⊂ A(r_{3}, r_{2}) is a crescent shaped region outside γ. By construction, ε_{2} = sup{u(x, y) − P_{Γ(r)}^{+}(x, y) : (x, y) ∈ L} > 0. By continuity, there is a smallest 0 < t_{2} < t_{1} so that sup{u(x, y) − P_{Γ(r)}^{+}(x, y) : (x, y) ∈ ∂B(t_{2}β_{1}, r_{5})} = ε_{2}/2. Observe that for this circle γ′ = ∂B(t_{2}β_{1}, r_{5}) we have a maximum of u(x, y)−P_{Γ(r)}^{+} attained in L but nowhere else. In particular, near β_{2}, β_{4} on γ′ we have u(x, y)−P_{Γ(r)}^{−} < ε and near β_{3}, u(x, y) − P_{Γ(r)}^{+} ≤ 0 since γ′ − B(β_{1}, δ) ⊂ B(0, r). Thus a(Γ(r)) ≠ a(γ′) since P_{Γ(r)}^{+} does not satisfy Chebychev’s criterion for optimality. Notice also that m(r_{1}) < m(γ′) < m(r_{2}) and B(0, r_{3}) ⊂ B(γ′) ⊂ B(0, r) ∪ B(β_{1}, δ) ⊂ B(0, r_{4}) so the lemma may be applied to circles parallel to γ′. q.e.d.

Our argument now follows the middle of E. Hopf’s proof of Bernstein’s Theorem about nonpositively curved entire graphs [10].

Lemma 2.3. Let M be a complete surface with nonpositive curvature and one end. Suppose that M ⊂ ℝ^{3} is a C^{2} immersion such that the surface is embedded near infinity as the graph of a function, i.e., for some r_{6} < ∞, M is the graph of a function z = u(x, y) outside the disk B(0, r_{6}). Suppose that u has sublinear growth. That is, there exists a continuous function ϵ : [r_{6}, ∞) → ℝ so that sup{|u(x, y)| : (x, y) ∈ D(r)} < rϵ(r) for all r ≥ r_{6} and ϵ(r) → 0 as r → ∞. Let P = P_{γ}^{+} be upper best approximating plane for a circle γ surrounding B(0, r_{6}) such that z ≤ P (x, y) for all (x, y, z) ∈ M ∩ D(r_{6}) × ℝ and ∇ P ≠ 0. Further suppose that U = {(x, y) ∈ ℝ^{2} : u(x, y) > P (x, y)} ≠ ∅. Then each connected component Ω ⊂ U contains a Jordan curve J : ℝ → Ω such that J(t) → ∞ as t → ±∞ and lim_{t→±∞} J(t)/|J(t)| = ±v where v is a unit vector perpendicular to ∇ P.

Proof. By enlarging r_{6} and translating a distance δ we may suppose γ = Γ(r_{6}). We can adjust ε(r) accordingly by replacing it with (r + δ)ε(r +δ)/r. And we may assume that ϵ is nonincreasing by replacing it by sup{ϵ(ρ) : ρ ≥ r} if necessary. By rotation and vertical translation, we arrange P (0, 0) = 0 and ∇ P = (0, q_{0}) for some q_{0} > 0 so that P (x, y) = q_{0}y. Let ξ(x, y) := u(x, y) − P (x, y). Observe that M is a graph where ξ > 0 because z ≤ P_{γ}^{+}(x, y) for (x, y, z) ∈ M ∩ D(r_{6}) × ℝ.

Choose r_{8} > r_{6} so large that ϵ(r) < q_{0} whenever r ≥ r_{8}. Define a region S = B(0, r_{8}) ∪ {(x, y) ∈ ℝ^{2} : |y| < rϵ(r)/q_{0}}. S is bounded by the continuous curves L^{±} where L^{+} = L_{−1}^{+} ∪ L_{0}^{+} ∪ L_{1}^{+}. L_{1}^{+} is the curve y = +rϵ(r)/q_{0} for r > r_{8} and x > 0. This is simply in polar coordinates sin(θ) = y/r = ϵ(r)/q_{0} which tends to zero as x → ∞. The other curves L_{1}^{−} is the reflection across the x-axis and L_{−1}^{±} are the reflections of L_{1}^{±} across the y-axis. L_{0}^{±} are the circular arcs r = r_{8} in the upper and lower halfplanes, resp. Now ξ ≤ u − q_{0}y < rϵ − rϵ = 0 on L_{±1}^{+} and ξ > 0 on L_{±1}^{−}. On L_{0}^{+} the same holds since it holds at L_{0}^{+} ∩ L_{1}^{+} and y is larger on the rest of L_{0}^{+} whereas |u| ≤ r_{8}ϵ(r_{8}) in B(0, r_{8}). Indeed ξ < 0 above L^{+} since the argument holds for arcs concentric to L_{0}^{+} of larger radius. Thus the nodal set ξ^{−1}(0) is contained in S.

Now to prove that Ω contains J. If Ω has a point in common with L^{−} then it contains all of L^{−}. Of course it contains no point of L^{+}. Thus assume that Ω does not meet L^{+} ∪ L^{−} so Ω ⊂ S. Let Ω_{s} be a nonempty component of the set ξ^{−1}(s, ∞) ∩ Ω for some small regular value s > 0. The closure is thus \overline{Ω_{s}} ⊂ Ω. Let D̃ denote the closed disk which is the geometric compactification of ℝ^{2}. Let tilde denote compactification in D and D(∞) the circle at infinity. The sets Ω_{s} and Ω can have common boundary points in the compactification only at ±v ∈ D(∞). Both ±v are common boundary points. If +v or x = +∞ is not a boundary point of Ω_{s} then Ω_{s} would lie in a halfplane x < x_{0}. S ∩ Ω_{s} would then lie in a sector and the function ξ − s would be zero on the boundary of Ω_{s} and positive inside. By Lemma 1.3, Ω_{s} is unbounded and ξ would have linear growth as x → −∞ contrary to hypothesis. It follows from E. Hopf’s accessibility Lemma 1.4 that both points ±v are limit points of Jordan arcs from within Ω. q.e.d.

Even if the error increases with radius so that by Lemma 2.1 there are points where the surface and the upper planes meet transversally, it may happen that for no circle are there points where the lower plane is actually transverse to the surface also. In this case the lower contact set

F^{−} = {(x, y) ∈ ℝ^{2} : u(x, y) = P_{γ}^{−}(x, y)} (4)

has a neighborhood in which points at infinity are accessible.

Lemma 2.4. Let M be a complete surface with nonpositive curvature and one end. Suppose that M ⊂ ℝ^{3} is a C^{2} immersion such that for some 0 < r_{1} < r_{2} ≤ ∞, M is the graph of a function z = u(x, y) outside the cylinder B(0, r_{1}) × ℝ. Let γ = Γ(r_{1}) and t_{0} ∈ γ ∩ F where F is the lower contact set (4). Let P = P_{γ}^{−} be lower best approximating plane such that z ≥ P (x, y) for all (x, y, z) ∈ M ∩ (D(r_{1}) × ℝ). Further suppose that u ≥ P for all (x, y) ∈ A(r_{1}, r_{2}). Then for all η > 0 the set F_{η} := {(x, y) ⊂ ℝ^{2} : u(x, y) < P (x, y) + η} contains a Jordan arc J : [r_{1}, r_{2}] → F_{η} − B(0, r_{1}) such that J(r_{1}) = t_{0}, J(r_{2}) ∈ Γ(r_{2}) if r_{2} finite and J is unbounded such that

lim_{t→∞} \frac{J(t)}{|J(t)|} = v (5)

converges to a unit vector v if r_{2} = ∞.

Proof. Write the complementary sets as the disjoint union of open connected components

ℝ^{2} − D(r_{1}) − F = ∐_{i∈I} U_{i}, ℝ^{2} − F = ∐_{k∈K} V_{k}, (6)

where I, K ⊂ ℤ are index sets. For eack i ∈ I there is a unique k(i) ∈ K such that U_{i} ⊂ V_{k(i)} although each V_{k} may contain several U_{i}’s.

We arrange indices so that i ≤ 0 whenever V_{k(i)} ∩ D(r_{1}) = ∅. Call T = ∂ (γ ∩ F) the set of boundary points in γ.

First we show that V_{k} are convex. In Euclidean space it suffices to show that the boundary is locally convex. If this were not the case for some k ∈ K, then there would be a open disk slice region (one of the pieces cut by a secant) X ⊂ ℝ^{2} whose boundary consists of a closed circular arc and an open line segment ∂X = A ⨿ S such that A ⊂ V_{k} and a point q ∈ F ∩ X. There is a plane λ such that λ = P over S and which cuts off a cap R from M containing q. Thus we have a contradiction since ∂R ⊂ λ and q ∉ λ violates the convex hull property Lemma 1.3.

Second, all \overline{U_{i}} with i ≤ 0 meet Γ(r_{2}). Note that in this case U_{i} = V_{k(i)}. This means U_{i} with i ≤ 0 are unbounded if r_{2} = ∞. If not, since u − P = 0 on ∂U_{i}, by the maximum principle u − P ≡ 0 on U_{i} which contradicts its definition.

We prove the statement in several cases. If t_{0} is an interior point of F ∩ γ then we begin J with an arc in F ∩ γ to its boundary. Thus we may assume t_{0} ∈ T .

Suppose t_{0} is a boundary point of any of the \overline{U_{i}} which meet both γ and Γ(r_{2}). Since ∂(U_{i} ∩ D(r_{2})) is locally convex at all points outside D(r_{1}), a subarc along ∂U_{i} from γ to Γ(r_{2}) is a Lipschitz path and thus furnishes J. If U_{i} is unbounded, then (5) converges by convexity. If γ − F consists of finitely many intervals, then by proceeding in turn through bounded arcs of ∂U_{i} − D(r_{1}) or arcs of interior components of F ∩ γ we can assemble a path J ⊂ F. If this path reaches a \overline{U_{i′}} which meets Γ(r_{2}) then we continue along this ∂U_{i′} to Γ(r_{2}). If all of the U_{i} are bounded then eventually the succession of arcs reaches a point where max{|q| : q ∈ ∂U_{i} for some i} is realized, and then the radial path outward from this point connects to Γ(r_{2}) via F. Thus we suppose for the rest of the argument that γ −F has infinitely many components.

Thus in the remaining cases there is s_{∞} = lim_{j→∞} s_{i_{j}}, a limit point of some pairwise distinct boundary points s_{i_{j}} ∈ T ∩ ∂U_{i_{j}} for some subsequence {i_{j}} ⊂ I. Since the set of limit points is closed in γ, we may assume that s_{∞} is a closest limit point to t_{0} of F in γ. We may suppose, by restricting to a subsequence if necessary, that U_{i_{j}} form a strictly monotone sequence (U′_{i}s are pairwise disjoint) and that s_{i_{j}} is the closest point of \overline{U_{i_{j}}} to s_{∞} along γ. If s_{∞} ≠ t_{0}, let α be a shortest arc of γ between t_{0} and s_{∞}. We may assume that all s_{i_{j}} ∈ α, i.e., s_{i_{j}} accumulates from the α side. If not, then there are only finitely many U_{i} which meet α. By following the corresponding ∂U_{i} allows us to connect t_{0} to either Γ(r_{2}) or to s_{∞} through a path in F. If s_{∞} = t_{0} take α to be an arc ending at s_{∞} containing the limiting sequence. Let ρ_{i} = sup{|q| : q ∈ ∂U_{i}}. Let ρ′ = sup{ρ_{i} : i > 0}.

Consider now the case that r_{2} < ∞. This case is handled using compactness. Choose any t_{∞} ∈ Γ(r_{2}) ∩ F. This set is nonempty for otherwise u − P > 0 on Γ(r_{2}) so F ∩ D(r_{2}) = ∅ by the convex hull property. Since D(r_{2}) is compact, there are only finitely many of the U_{i} such that U_{i} ∩ A(r_{1}, r_{2}) ≠ ∅ which are not contained in F_{η}. That is because k = ||u||_{C^{1}(A(r_{1},r_{2}))} < ∞ so for any point q_{i} ∈ U_{i} ∩A(r_{1}, r_{2})−F_{η} has u(x, y) > s − dk by the mean value theorem, where d is the distance from q_{i} to (x, y). Thus E_{i} = B(q_{i}, η/k) ∩ A(r_{1}, r_{2}) ⊂ F_{η} so each E_{i} is disjoint for different i and has uniformly positive area. Thus only finitely many such i, say i_{1}, …, i_{m} can have all E_{i} fit inside the annulus.

J is constructed as follows. Let ν be a minimum length curve among curves in A(r_{1}, r_{2}) from t_{0} to t_{∞}. ν might follow γ for a while and then be a straight line. ν meets finitely many U_{i_{j}}, say in the order U_{i_{1}} first, U_{i_{2}} second and so on. For each U_{i_{j}} in turn, replace ν ∩ U_{i_{j}} by an arc of ∂U_{i_{j}}. If it reaches Γ(r_{2}) we stop. Continue through the j and call the result J. This replacement can happen at most m times along γ and at most m times along the straight line. The perimeters of the sets U_{i_{j}} ∩ A(r_{1}, r_{2}) are bounded by 2πr_{1} + 2πr_{2} thus the length of ν is increased by at most 4mπ(r_{1} + r_{2}). We can make it a Jordan arc by omitting loops. We have actually shown how to connect t_{0} to Γ(r_{2}) or to t_{∞} whichever comes first by a path in F_{η}.

For the next case we suppose that r_{2} = ∞ and r_{3} = ρ′ < ∞. This case is handled similarly. Choose r_{3} < r_{7} < ∞. If there are no U_{i} with i ≤ 0 then choose t_{∞} = r_{7}s_{∞} ∈ Γ(r_{7}). Then define μ to be the radially outward semiinfinite ray from t_{∞} which lies in F. If U_{i′} ≠ ∅ for some i′ ≤ 0 choose in addition a t_{∞} ∈ ∂U_{i′}, which may require r_{7} to be increased, and let μ be an arc of ∂U_{i′} heading to infinity. Since D(r_{7}) is compact, there are only finitely many of the U_{i} such that U_{i} ∩ A(r_{1}, r_{7}) ≠ ∅ which are not contained in F_{η} as before.

The next part of J is constructed as follows. Let ν be a minimum length curve A(r_{1}, r_{7}) from t_{0} to t_{∞}. Then ν meets finitely many \overline{U_{i_{j}}}. For each U_{i_{j}} replace ν ∩\overline{U_{i_{j}}} by an arc of ∂U_{i_{j}} −γ. If it reaches infinity we stop. Continue through the j and call the result ν′. This replacement can happen at most 2m times. The perimeters of the sets U_{i_{j}} ∩A(r_{1}, r_{7}) are bounded by 2πr_{1} + 2πr_{7} thus the length of ν is increased by at most 4mπ(r_{1} + r_{7}) whenever a finite arc is added. Thus the desired path is obtained by concatenating J = ν′ · μ. We can make it a Jordan arc by omitting loops. We have actually shown how to connect t_{0} to infinity or to t_{∞} whichever comes first by a path in F_{η}. Again, the limit exists as before.

In the last case we have r_{2} = ρ′ = ∞. There is a sequence of positive {i′} ∈ I so that diam(U_{i′}) is unbounded as i′ → ∞. There may be a subsequence with ρ_{i_{j}} = ∞ for all j. If not, choose a subsequence so that ρ_{i_{j}} > r_{1} + j is monotonically increasing to infinity. Choose points s_{j}, t_{j} in ∂U_{i_{j}} ∩T such that s_{j} = s_{i_{j}} ∈ γ and the arc s_{j}t_{j} of ∂U_{i_{j}} intersects both bounding circles of A(r_{1}, r_{1} + j). Let w_{j} = \overline{s_{j}t_{j}}/|\overline{s_{j}t_{j}}| be a unit vector. We may also arrange that the subsequence {i_{j}} satisfy |t_{j}| ≥ |t_{j−1}|. Then choose a subsequence {ι} so w_{ι} → w_{∞} converges monotonically. s_{ι} → s_{∞} converges monotonically by construction. Let κ_{ι} denote the ray τ ↦ s_{ι}+τw_{ι} and κ_{∞} the ray τ ↦ t_{0}+τw_{∞}. Since s_{ι} ∈ \overline{U_{ι}} are disjoint we have κ_{ι}([0, ι]) ⊂ U_{ι} so κ_{ι} converge to κ_{∞} uniformly on compact subsets and from one side. We chose s_{ι} to be the closest point in γ ∩ ∂U_{ι} to s_{∞}.

Now we shall choose the path inductively which approximates κ_{∞}. Choose ℓ = ℓ(1) > 1 so that κ_{k}([0, 3+2r_{1}]) ⊂ F_{η} ∩N for all k ≥ ℓ where N is the tubular neighborhood of points of distance less than η to κ_{∞}. In fact any other U_{i} from the original collection which lies between U_{ℓ(1)} and κ_{∞} will satisfy ∂U_{i} ∩ A(r_{1}, r_{1} + 1) ⊂ F_{η}. Using the first part of the argument, there are only finitely many of the original U_{i} which contain points in A(r_{1}, r_{1}+1)−F_{η}. Let ν be a shortest path in A(r_{1}, r_{1}+1) from t_{0} to a point of the arc s_{ℓ}t_{ℓ} in ∂U_{ℓ} ∩ Γ(r_{1} + 1). Now deform ν outward along these finitely many ∂U_{i}. If any of the U_{i} are unbounded we may stop. Call the resulting path J_{1} and parameterize it on [r_{1}, r_{1} + 1] or [r_{1}, ∞) if unbounded. It may poke outside Γ(r_{1} + 1).

Assume that J_{1}, …, J_{ι} have been chosen as finite paths. Choose ℓ = ℓ(ι + 1) > ℓ(ι) such that κ_{k}([0, ι + 3 + 2r_{1}]) ⊂ F_{η} ∩ N for all k ≥ ℓ. Let σ_{ι+1} be a path in ∂U_{ℓ} ⊂ F from Γ(r_{1} +ι) to Γ(r_{1} +ι+1) between κ_{ℓ} and κ_{∞}. Such must exist since the U_{ι} are disjoint so boundary subarcs must be between κ_{ℓ(ι)} and κ_{ℓ(ι+1)} and since |t_{ℓ(ι)}| > r_{1} + ι + 1. Now let ν_{ι+1} be a path along Γ(r_{1} + ι) from J_{ι} to σ_{ι+1} which is in F_{η} ∩ N by construction. Let the concatenation J_{ι+1} = ν_{ι+1} · σ_{ι+1} parameterized on [r_{1} + ι, r_{1} + ι + 1]. Concatenating J = J_{1} · J_{2} · . . . gives the desired path in F_{η}. Omitting loops will make it a Jordan arc. Since J_{i} ⊂ N for i > 1 also (5) follows with v = w_{∞}. q.e.d.

3. Proof of the slab theorem

The oscillation of u is defined to be

ω(r) = sup_{(x,y)∈Γ(r)} u(x, y) − inf_{(x,y)∈Γ(r)} u(x, y).

The following theorem is sharp for complete nonpositively curved surfaces in ℝ^{3} in the sense that omitting any hypothesis will allow an example not contained in a slab. The catenoid does not have one end. The Enneper’s surface does not have its end embedded. And flat K = 0 surfaces, which are generalized cylinders [11] can have arbitrary ω(r) growth, but don’t have |A| ∈ L^{2}.

Theorem 1. Let M be a complete, nonpositively curved Riemannian two manifold with one end. Suppose that M is C^{2} immersed in ℝ^{3} such that the surface is embedded near infinity and has square integrable second fundamental form (1). Then M lies between two planes, i.e., suppose the surface is oriented so that (0, 0, 1) is the limiting normal vector to M at infinity, then c^{+} = sup{z : (x, y, z) ∈ M} and c^{−} = inf{z : (x, y, z) ∈ M} are finite.

Proof. White showed that under (1) the Gauss map of a nonpositively curved surface converges as a point tends toward the end and the end is properly immersed. Thus it is possible to rotate so that the limit of the Gauss map is (0, 0, 1). Since M is also embedded near the end, for some r_{1} < ∞, M is the graph of a function z = u(x, y) outside the cylinder B(0, r_{1}) × ℝ. Furthermore, by the mean value theorem, the oscillation cannot satisfy the condition that there is a k > 0 and R > r_{1} such that ω(r_{j}) ≥ kr_{j} for a sequence R ≤ r_{j} → ∞ without violating convergence of the Gauss map. Thus, u has sublinear growth. Details are in [2], [3].

Let Γ_{1} = ∂D(r_{1}) and let P_{1}^{±} be the upper and lower planes of Can (Γ_{1}), where P_{1}^{+} = a_{1}x + a_{2}y + c_{1}. If we show that m(r) is constant for r ≥ r_{1} then we are done. In fact, then a(r) = 0 and so p_{∞}^{±} = c_{1} ± m(r). To see this, first observe that m(r) being constant implies a(r) and c(r) are constant. This is simply because the only way that Can (Γ(r)) can contain M ∩ (D(r_{1}) × ℝ) is if P_{Γ(r)}^{±} = P_{1}^{±} since the minimal planes are unique. Then a(r) = 0. If not, along the ray ζ(τ) = τa(r) we have |u(ζ(τ)) − |a(r)|^{2}τ − c(r)| ≤ m(r) thus by the mean value theorem there is a sequence τ_{i} → ∞ along which d/dτ u(ζ(τ_{i})) = |a(r)|^{2} so that in particular, the normal

G(ζ(τ_{i})) · (0, 0, 1) ≤ |a(r)|^{2}(1 + |a(r)|^{4})^{−1/2} < 1.

It follows that G(ζ(τ_{i})) does not converge to (0, 0, 1), a contradiction.

The rest of the argument rules out the possibility that m(r) is increasing. Thus suppose that m(r_{2}) > m(r_{1}) for some r_{2} > r_{1}. By Lemma 2.2, for r_{1} < r_{3} < r_{4} < r_{2} such that m(r_{1}) < m(r_{3}) < m(r_{4}) < m(r_{2}) there is a circle γ ⊂ A(r_{3}, r_{4}) such that u is transverse to one of the P_{γ}^{±} at one of the β_{i} ∈ γ and a(γ) ≠ 0. By renumbering and replacing u by −u if necessary, we may suppose that ∇ u(β_{1}) ≠ ∇ P_{γ}^{+}(β_{1}). Let its radius be r_{6} = r(γ). By translation and rotation we can arrange that the center of γ is the origin, c(γ) = 0, a(γ) = (0, q_{0}) where q_{0} > 0. Thus there are r_{5} < r_{6} < r_{7} so that the new A(r_{5}, r_{7}) is contained in the old A(r_{1}, r_{2}). Thus P_{6}^{+}(x, y) = q_{0}y + c for c = m(r_{6}) > 0. Let Ω be the component of {(x, y) ∈ ℝ^{2} : (x, y, z) ∈ M and z > P_{6}^{+}(x, y)} which contains β_{1} in its boundary. By transversality there is only one such component. By definitions Ω ∩ D(r_{6}) = ∅. By Lemma 2.3, there is a path J_{1} : ℝ → Ω ∪ {β_{1}} ⊂ S which tends to infinity for both directions x → ±∞, where S is the set defined in the proof of Lemma 2.3. Also lim_{τ→±∞} J_{1}(τ)/|J_{1}(τ)| = ±v_{1} for some unit vector v_{1} such that v_{1} · ∇ P_{6} = 0, or J_{1} → ±v_{1} as τ → ±∞ in the geometric compactification D. By splicing in a path we may assume that J_{1}(0) = β_{1}. There we also proved that u > P_{6}^{+} on the lower boundary component L^{−} of S and u < P_{6}^{+} on the upper boundary component L^{+}.

Consider the sets

G_{s} = {(x, y) ∈ ℝ^{2} : (x, y, z) ∈ M and z < q_{0}y − c + s},
F^{−} = {(x, y) ∈ ℝ^{2} : (x, y, z) ∈ M and z ≤ q_{0}y − c}.

If G_{0} is nonempty, then let ρ = inf{|t| : t ∈ G_{0} − B(0, r_{6})} and t_{1} ∈ ∂G_{0} where |t_{1}| = ρ. Let η = c/2. If ρ > r_{6} then as in the proof of Lemma 2.4 we may choose a path in G_{η} − B(0, r_{6}) connecting β_{2} to t_{1} and then by Lemma 2.3, extend it to a path J_{2} : (−∞, ∞) → G_{η} such that J_{2}(0) = β_{2} and lim_{τ→±∞} J_{2}(τ)/|J_{2}(τ)| = ±v_{1} for the same v as for J_{1}. If G_{0} = ∅ then we use Lemma 2.4 instead. This time there is a path J_{2} : [0, ∞) → G_{η} such that J_{2}(0) = β_{2} and (5) converges to the unit vector v_{2}.

Similarly we may construct a J_{4} : [0, ∞) → G_{η} such that J_{4}(0) = β_{4} and J_{4}/|J_{4}| → v_{4} as τ → ∞. Let

H_{s} = {(x, y) ∈ ℝ^{2} : (x, y, z) ∈ M and z > q_{0}y + c − s},
F^{+} = {(x, y) ∈ ℝ^{2} : (x, y, z) ∈ M and z ≥ q_{0}y + c}.

Note that H_{η} ∩ G_{η} = ∅. By replacing u by −u in Lemma 2.4 we also obtain a path J_{3} : [0, ∞) → H_{η} such that lim_{τ→∞} J_{3}(τ)/|J_{3}(τ)| = v_{3} and J_{3}(0) = β_{3}.

Now observe that these four paths must be disjoint. J_{2}, J_{4} are in G_{η} which is disjoint from H_{η} which contains J_{1}, J_{3}. To see the pairs are disjoint, suppose that J_{1} and J_{3} coincide at a point. Thus there is a simple path ψ : [0, 1] → H_{η} − B(0, r_{6}) connecting β_{1} to β_{3}. Complete the circuit with the straight line segment σ from β_{3} to β_{1} and call the enclosed region R. z is larger than P_{6}^{−} on the loop ψ · σ. Thus by the convex hull property z ≥ P_{6}^{−} for all (x, y, z) ∈ M ∩ (R × ℝ). It follows that G_{0} ∩ R = ∅. However G_{η} meets R − B(0, r_{6}) since one of β_{2} or β_{4} are in this set by the Jordan Curve Theorem, say β_{2}. This is a contradiction since J_{2} can’t reach infinity since it can’t cross ψ ⊂ H_{η}. Similarly J_{1} is disjoint from J_{3}.

We conclude the argument by analyzing two possibilities. Consider the region R^{−} bounded by the closed Jordan arc L^{−} − J_{1} in D ∪ {±v_{1}}. Either one of the β_{2} or β_{4} lies in R^{−} or neither does. Say β_{2} ∈ R^{−}. Since u−P^{−} < −c/2 on J_{2} but > 0 on J_{1}∪L^{−}, we must have J_{2} an unbounded path between L^{−} and J_{1} and thus v_{2} = ±v_{1}, say v_{2} = v_{1} = (1, 0). Now consider vertical lines x = x_{0} for large x_{0}. The vertical must intersect J_{2} at some point y_{2}. Note that (x_{0}, y_{2}) ⊂ R^{−}. For large and small y the point (x_{0}, y) ∉ R^{−}. Therefore, there must be y_{1} and y_{3}, one greater, the other less than y_{2} such that (x_{0}, y_{1}) ∈ J_{1} and (x_{0}, y_{3}) ∈ L^{−}. Now u − P_{6} is positive at (x_{0}, y_{i}) for i = 1, 3 and negative at (x_{0}, y_{2}). By the intermediate and mean value theorems, there is y_{4} between y_{1} and y_{3} where ∂u/∂y(x_{0}, y_{4}) = ∂P_{6}/∂y(x_{0}, y_{4}) = q_{0}. In particular the Gauss map is uniformly bounded away from (0, 0, 1) for a sequence tending to v_{1}, i.e., |G(x_{0}, y_{4}, u(x_{0}, y_{4}))−(0, 0, 1)|^{2} ≥ 1−(1+q_{0}^{2})^{−1/2} for all x_{0}. This means that a sequence of Gauss maps doesn’t converge to (0, 0, 1) as x_{0} → ∞ and we have reached a contradiction.

The other case is if none of the β_{2}, β_{4} are in R^{−}. Consider instead the region R^{+} bounded by the closed Jordan arc J_{1} − L^{+} in D ∪ {±v_{1}}. Since J_{3} is disjoint from J_{1} ∪ L^{+}, (5) must converge to one of the ±v_{1}, say J_{3} → +v_{1} as τ → ∞. Again, one of β_{2} or β_{4}, say β_{2}, lies in the region bounded by J′_{1} − J_{3} + σ where J′_{1} is the subarc of J_{1} from β_{1} to v_{1} and σ is a secant of Γ(r_{6}) from β_{3} to β_{1}. J_{2} is a Jordan arc which connects β_{2} to v_{1} since it can’t access other points of D(∞) being bounded by J′_{1} ∪ J_{3}. Now consider vertical lines x = x_{0} for large x_{0}. The vertical must intersect J_{2} at some point y_{2}. Note that for large and small y the point (x_{0}, y) ∉ R^{+}. Therefore, there must be y_{1} and y_{3}, one greater, the other less than y_{2} such that y_{i} ∈ J_{i} for i = 1, 3. Now u − P_{6} is positive at (x_{0}, y_{i}) for i = 1, 3 and negative at (x_{0}, y_{2}). By the intermediate and mean value theorems, there is y_{4} between y_{1} and y_{3} where ∂u/∂y(x_{0}, y_{4}) = ∂P_{6}/∂y(x_{0}, y_{4}) = q_{0}. In particular the Gauss map is uniformly bounded away from (0, 0, 1) at some (x_{0}, y_{4}) ∈ S. This means that a sequence of Gauss maps doesn’t converge to (0, 0, 1) as x_{0} → ∞ and we have reached a contradiction. q.e.d.

We now formulate a theorem which describes the lower contact set established by Theorem 1. It describes the upper contact set as well by reflecting z ↦ −z.

Theorem 2. Let M be a complete, nonpositively curved Riemannian two-manifold with one end. Suppose that M is C^{2} immersed in ℝ^{3} such that the surface is embedded near infinity and has square integrable second fundamental form (1). Suppose the surface is oriented so that (0, 0, 1) is the limiting normal vector to M at infinity and so that M is the graph of a function z = u(x, y) for x^{2} + y^{2} ≥ r_{1}^{2}. Let c^{−} = inf{z : (x, y, z) ∈ M} and F^{−} = {(x, y) ∈ ℝ^{2} : (x, y, c^{−}) ∈ M} be the lower contact set. Suppose we decompose the complements into connected components as in (6). Then

(1) There is at most one V_{i} with V_{i} ⊂ ℝ^{2} − D(r_{1}) i.e., so that i ≤ 0. Such a V_{0} must be a halfplane.

(2) There is a path J : [0, ∞) → F^{−} such that (5) converges.

(3) If (x_{0}, y_{0}) ∈ F^{−} then K(x_{0}, y_{0}, c^{−}) = 0. (The Gauss curvature vanishes on F^{−}.)

Proof. Let V_{i} be one of the domains with i ≤ 0, thus M is a graph over V_{i}. Since V_{i} is convex and not the whole plane, it must lie in a convex sector of angle θ ∈ (0, π/2]. By Lemma 1.3 this is a contradiction to sublinear growth unless θ = π/2 and V_{i} is a halfplane. Thus there are at most two disjoint V_{i} in the plane for i ≤ 0, call them V_{0}, V_{−1} which must be parallel. Let S = ℝ^{2} − V_{0} − V_{−1} be the strip between them. Note that the boundary of the region of M in the cylinder S × ℝ lies in the plane z = c^{−}. Since ∇u → 0 as distance from the origin r → ∞ we have ζ(r) = sup{z : (x, y, z) ∈ M ∩ ((Γ(r) ∩ S) × ℝ)} → c^{−} as r → ∞. It follows by the convex hull property that (D(r) ∩ S) × ℝ ∩ M ⊂ (D(r) ∩ S) × [c^{−}, ζ(r)] which tends to a plane as r → ∞, which is contradiction, proving (1).

If U_{0} ≠ ∅ then let J be a ray of ∂U_{0}. Otherwise, the construction in Lemma 2.4 constructed a path for every η > 0 in F_{η} in two steps. The first part was to connect some t_{0} ∈ F^{−} ∩Γ(r_{1}) to a point of Γ(r_{1} +1)∩F by a path in F_{η} which may poke out of D(r_{1} + 1) and which may be unbounded as η → 0. If any of the replacement ∂U_{i}’s are unbounded, then a subarc of ∂U_{i} is the desired J. If ρ′ < ∞ then a radial ray lies in F. For ρ′ = ∞, the second part of the construction was to follow segments of ∂U_{i} between κ_{∞} and κ_{ℓ}. This part of the path converged to κ_{∞} uniformly as η → 0, thus in this case, by continuity, κ_{∞} ∈ F which is the desired path.

Since z ≥ c^{−} for all (x, y, z) ∈ M near Q_{0} = (x_{0}, y_{0}, c^{−}), it follows that at Q_{0} the normal vector to M is (0, 0, 1) and the second fundamental form is nonnegative definite. The only way this can happen for nonpositive curvature is if K(q_{0}) = 0. q.e.d.

We expect every point of F may be connected to ∞ by a path in F satisfying (5). However, we cannot rule out the possibility that F is not locally path connected.

We end by constructing examples of arbitrary genus, genenralizing the surface mentioned in the second paragraph. In cylindrical coordinates of ℝ^{3}, for n ≥ 2 an integer, the surface

r^{n} (z − cos nθ) = z − z^{3}

has genus n − 1 and lies in the slab |z| ≤ 1. Its upper contact set F^{+} consists of n equally spaced rays emanating from the origin. The curvature vanishes along the rays in the intersection with the planes z = 0, ±1. Solving for r as a function of z and θ, we express the curvature in terms of derivatives of r and check that the sign of the resulting expression is nonpositive. The computation is facilitated using the symbolic computational package MAPLE.

As with the proofs of Bernstein’s Theorem, it would be desirable to find another proof here which is purely analytic.

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National Ping-Tung Teachers College, Taiwan University of Utah, Salt Lake City, Utah 84112