Isometric embedding with nonnegative Gauss curvature under the graph setting

We study the regularity of the isometric embedding X: (B(O,r),g) ->(R3, gcan) of a 2-ball with nonnegatively curved C4 metric into R3. Under the assumption that X can be expressed in the graph form, we show X is C2,1 near P, which is optimal by Iaia's example.


Introduction
Weyl posted the following problem in 1916 [15]: Consider a positively curved 2-sphere (S 2 , g).Does there exist a global C 2 isometric embedding X : (S 2 , g) → (R 3 , g can ), where g can is the standard flat metric on R 3 ?Weyl himself suggested the continuity method to solve this problem and obtained a priori estimates up to the second derivatives.Lewy [9] solved the problem under the assumption the g is analytic.In 1953, Nirenberg [12] solved the Weyl problem under the mild smoothness assumption that the metric is C 4 .
P. Guan and Y.Y.Li [3] considered the question that if the Gauss curvature of the metric g is nonnegative, whether does (S 2 , g) still have a smooth isometric embedding?They proved in [3] that for any C 4 metric g on S 2 , there is a global C 1,1 isometric embedding into R 3 .Examples in Iaia [7] show that for some analytic metrics with positive Gauss curvature on S 2 except at one point, there exists only a C 2,1 but not a C 3 global isometric embedding into R 3 .
Then a natural question, posted in [3], is that if a smooth metric g on S 2 has nonnegative Gauss curvature, whether does it have a C 2,α , for some 0 < α < 1, or even a C 2,1 global isometric embedding?To study this problem, we can look at the degenerate Monge-Ampère equation where k(x, y) ≥ 0, in B r (O) for small r > 0. Guan [2] considered the case k ∈ C ∞ (B r (O)), and 1 A (x 2n + By 2m ) ≤ k(x, y) ≤ A(x 2n + By 2m ), (1.2) for some A > 0, B ≥ 0 and positive integers n ≤ m.It's shown in [2] that a C 1,1 solution u of (1.1) is smooth near the origin if (1.2) holds, and if, additionally Guan and Sawyer [4] improved this result by replacing (1.3) by a weaker condition ∆u ≥ C 0 > 0.
In this paper, we consider a C 1,1 isometric embedding X : (B(O, r), g) → (R 3 , g can ), where B(O, r) is a ball in R 2 , centred at the origin with radius r, and g is a C 4 metric with Gauss curvature k ≥ 0. We regard the image of X, as the graph of a function u.
An example is that Then g = dx 2 + dy 2 + du 2 is smooth in x, y, k(x, y) = 18(x 2 + y 2 ) > 0 except at the origin, but the embedding is only C 2,1 .First we have the following theorem when α(x, y) ≡ x, β(x, y) ≡ y, and the Gauss curvature only degenerates at a single point.
Here we only need the sign of the Gauss curvature k.By the example u = r 3 , we see that the C 2,1 smoothness of X is optimal.
Enlightened by Guan and Sawyer [4], if ∆u, or the mean curvature H, has a uniform positive lower bound near but not necessarily at the origin, we have Corollary 1.2.Assume the same assumptions as in Theorem 1.1.In addition, we assume that g ∈ C ∞ , and ∆u = u xx + u yy > C 0 > 0 for some constant C 0 , around O but not necessarily at O, then X ∈ C ∞ (B g (P, r)).
For the Monge-Ampère equation (1.1), in the case that u is radial, we have the following corollary showig u ∈ C 2,1 , which is optimal by the example u = r 3 .This result is expected to be true.We list it here as a quick corollary of lemmas in Section 2.
Corollary 1.3.Assume that a C 1,1 convex function u satisfies (1.1) in B(O, ρ), the ball of radius ρ centered at the origin, and k ≥ 0 in B(O, ρ) − O.In addition, we assume that u = Φ(r) for some function Φ, where r = x 2 + y 2 , and Here k could vanish at infinite order at r = 0. We see that Φ r is the square root of a C 4 function.
Under the general nonnegative Gauss curvature, Pogorelov's counterexample in [13] shows that a C 2,1 metric with nonnegative Gauss curvature may not have a C 2 isometric embedding.However, given a C 4 metric, under the graph setting, our result is positive.
under local coordinates x, y such that the normalization (1.5) holds.If in addition, u is (weakly) convex, then X ∈ C 2,1 (B(O, r ′ )), for any r ′ < r.
The paper is organized as follows.In Section 2, we will discuss the one dimensional model.In Section 3, we will prove Theorem 1.1.In Section 4, we prove Corollary 1.2.In Section 5, we prove Theorem 1.4.
Many thanks to Yanyan Li for introducing me this problem and the whole project.Thanks to Zheng-Chao Han for helpful discussions.

A model in dimension one
In this section, we derive C 2,1 estimates of u in an one dimension model.Assume a nonnegative function u = u(x) ∈ C 1 (2I), where I = [−1, 1], and u(0) = 0.In addition, we assume that f = u 2 x is in C 4 (2I), and f ′ (x)x ≥ 0. The goal is to show u ∈ C 2,1 (I).The condition that f ′ (x)x ≥ 0 is necessary, since there is a nonnegative function which is smooth, vanishes at infinite order at x = 0, and ( √ f ) ′′ blows up when x → 0. In fact, √ f ∈ C 1,α for any 0 < α < 1, and ( We have the following well known lemma, Proof.Assume first ||f || C 2 (2I)=1 .We only need to consider at x where f (x) > 0. If . We assume 0 < f (x) < 1 at some x ∈ I, then by the Taylor expansion, if x + t ∈ I, for some x between x and x + t.Then , and divide t on both hand sides to derive, .
, and divide t on both hand sides to derive, Notice the choice of t is valid, since |x If general, when ||f || C 2 (2I) = 1, by a scaling, we see that The following is a standard interpolation lemma, Lemma 2.2.Assume that G(y) is a C 4 function defined on [−1, 0] such that G(y) ≥ 0, and is non-decreasing.Then there exist universal constants A, B such that Proof.By the Taylor expansion, Then the lemma follows.Next is the key Theorem in this section.
Then u is C 5 in I − {f = 0}, and for every x ∈ I − {f = 0}, for some universal constant C.
) and the C 2 norms of g, h are bounded by C||f || C 4 = C for some universal constant C. Notice g, h are nonnegative.If we apply Lemma 2.1 to g, then we derive, for x ∈ I, And applying Lemma 2.1 to set h, we derive, for x ∈ I, and by (2.3), (2.5), (2.6), Case 2: Consider points x ∈ I − {f = 0}, such that f (x) ≤ x 4 .We are to prove at such x, for some universal constant C, which implies (2.2).
In sum, (2.2) is verified, and we derive (2.1) by as a scaling.Proof.We show u ∈ C 2 (I), then the theorem follows from Theorem 2.3.
First assume f = 0 only at x = 0. Taking the Taylor expansion of f (x) at 0, if f (x) = M x 2 + R(x), for some M > 0, and R(x) ∈ O(x 3 ), then for x = 0, which approaches √ M as x tends to 0. Also we check ).After a scaling, we assume f (x) ≤ x 4 .Then for x near the origin, by (2.7).So u xx (0) exists and equals 0.
Secondly, if f = 0 at some x 0 = 0, without loss of generality, assume x 0 > 0. Since f ′ (x)x ≥ 0, f is non-decreasing as x > 0. Then f ≡ 0, for 0 ≤ x ≤ x 0 .Denote x 1 = max{x ∈ 2I : f (x) = 0}.We only have to consider u xx at x 1 if x 1 ∈ I.We can use a new coordinate that translates x 1 to the origin, and apply an argument like (2.10) to show u xx (x 1 ) = 0.
For the case f > 0 in 2I, we see that u is a C 5 function on 2I.
Proof.u xx ≥ 0 implies f is non-increasing or non-decreasing on (−2, 2), depending on the sign of u x (0).Without loss of generality, we shift the origin to −2, and assume f is non-decreasing on (0, 4).

Two dimensional case with one singular point
In this section, we prove Theorem 1.1.Maybe making r a bit smaller, we assume that the C 1,1 function u in (1.6) is defined in rI ×rI.In addition, we assume that f = u 2 x > 0, u x exists and be positive, except at the origin.Here u xx exists in B(O, r) − {O}, since the Gauss curvature k > 0 except at the origin.Then by the classic theory of Monge-Ampère equations, g is C 4 implies that u is C 3,α , except at the origin, for any α ∈ (0, 1).See Section 10.3 of [14].
For (x, y) = 0, {u x = 0} is locally a curve, since at any point except the origin, u xx (x, y) > 0, then we can solve out x = A(y) as a function of y from the equation u x (x, y) = 0. Furthermore, d dy A(y) = − uxy(A(y),y) uxx(A(y),y) , which is uniformly bounded when y ∈ (δ, r) ∪ (−r, −δ) for any δ > 0. In addition, for each y, we can only have at most one x, such that u x (x, y) = 0, since u x is strictly increasing.Though the gradient of A(y) may blow up when y approaches 0, we show that {u x = 0} is a continuous curve.Lemma 3.1.Assume that u ∈ C 1,1 (rI × rI), u x (0, 0) = 0 and u x is an increasing function in x for any fixed y.Then {u x = 0} is a continuous curve near the origin.
The choice of δ depends on η, ||u|| C 1,1 .Hence, for any fixed y ∈ (−δ, δ), the zero of u x must be unique and the value of x lies in (−ǫ, ǫ), by the assumpition that u x is an increasing function in x for any fixed y.So we derive −ǫ < A(y) < ǫ as |y| < δ.

Now we prove Theorem 1.1.
Proof.By the assumption, the metric x , u 2 y , u x u y ∈ C 4 .Hence u 2 z ∈ C 4 for any z = lx + my, where l, m are fixed numbers in R. At points except the origin, k > 0. Then the classic theory of Monge-Ampère equation shows that u ∈ C 2,α at any point (x, y) away from the origin.We get u zz > 0 except at the origin.Now fix a small ǫ << r.Assume . We set s = max Now u ∈ C 3 (rI ×rI −(0, 0)), since the metric g ∈ C 4 .In rI ×δ 1 I, u has C 2,1 estimates in x, which depends only on η 1 , ǫ, ||g|| C 4 , that is, Similarly, switching x and y, we can find η 2 such that (3.1) We derived that u has uniform C 3 estimates except at the origin.Then, by a basic argument in calculus, we derive that u ∈ C 2,1 near the origin.

Proof of corrollaries
Proof of Corollary 1.2.By Theorem 1.1, u is C 2,1 in x, y.By the assumption, ∆u > C 0 > 0 around the origin but not necessarily at the origin.Without loss of generality, under a rotation of coordinates, we assume for coordinates x, ỹ, u xx (0, 0) ≥ C 0 > 0, u ỹ ỹ(0, 0) = 0 which should hold for some x, ỹ, since the Gauss curvature k = 0 at the origin.
We can rotate x, y a little bit, such that none of the x, y direction confirms with the ỹ direction, and so for some C 1 depends only on C 0 and the angle between the x, y direction and the ỹ direction.This does not change that fact that u xx u yy − u 2 xy = 0 at the origin, so we cannot apply the classic theory for Monge-Ampère equations.
Recall in the proof of Theorem 2.5, for the one dimensional model, if , where M > C 2 1 is independent of x, and R is smooth by the assumption that f = g xx − 1 is smooth.Then which has C k bounds for any integer k > 0, which only depends on ) for any k, and explicitly we have Inductively, we can prove u is C k near the origin by introducing more pairs of coordinate systems, and the corollary is verified.
In Theorem 1.1, The forms of α, β that are allowed can be slightly generalized.Proof.Under the assumption of Corrollary 4.1, in any domain which does not include the origin, we have u ∈ C 2,µ (0 < µ < 1), and if we regard u as a function of α, β.Notice we does not necessarily have u xx > 0.
The system of equations where the terms on the right hand side are C 4 in α, β by the assumption of Corrollary 4.1.Here x, y can be regarded as C 5 functions of α, β, since at the origin, we may choose x, y as normal coordinates, and after a rotation, we assume α y (0, 0) = 0, β x (0, 0) = 0. Then at the origin The Jacobian ∂(α,β) ∂(x,y) = 1 at the origin.By the implicit function theorem, we can solve out x, y as functions of α, β.
We derive that u 2 α , u 2 β , u α u β ∈ C 4 and (4.1) holds.We can apply the same method as in Section 3 to derive u ∈ C 2,1 near the origin.Then Corrollary 4.1 follows.
Proof of Corollary 1.3.Without loss of generality, we assume (1.5) holds and k = 0 at So k = Ψ(r), for some C 3 function Ψ in r, by the assumption that k is C 3 in x, y and Lemma A.2.For r < 0, we define Φ(r which further implies u is a C 2,1 function.

Nonnegative Gauss curvature case
Set δ = r 9 in this section.Assume we have the Gauss curvature k ≥ 0 on (B(P, r), g).In addition, we assume u is convex, which implies u xx , u yy , u zz , u ww ≥ 0, (5.1) (at where they exist) in B g (P, r).Here (z, w) is the new coordinate system such that z = x+y 2 , w = x−y 2 .Note that convexity of u is necessary, since we have an example where u 2 x and the metric of the graph are smooth, with nonnegative Gauss curvature, but u is not C 2 with respect to x on the y-axis.The main obstruction is that the graph of u is not convex.
For a point (x 0 , y 0 ) ∈ [−δ, δ] × [−δ, δ], if u x (x 0 , y 0 ) = 0 and u x (x, y 0 ) < 0 for every x ∈ (−δ, x 0 ), we call (x 0 , y 0 ) a left touch point of u x .Notice the left touch point is unique for any y ∈ [−δ, δ] if it exists, according to (5.1).Similarly, we can define right touch points.Proof.Without loss of generality, we only consider set of the left touch points.Define the left touch function T L (y) on [−δ, δ], where T L (y 0 ) equals x 0 if we can find an x 0 such that (x 0 , y 0 ) is the left touch point on the line y = y 0 , which is unique if any; otherwise, u x (x, y 0 ) ≥ 0 for x ∈ (−δ, δ), for which case we set T L (y 0 ) = −δ, or u x (x, y 0 ) < 0 for x ∈ (−δ, δ) for which case we set T L (y 0 ) = δ.
We check T L is a lower semi-continuous function.If T L (y) = −δ, then it's trivial since T L (x) ≥ −δ.
So T L is measurable as a lower semi-continuous function, and by the Fubini Theorem, its graph has measure zero.Lemma 5.1 shows that u xxx exists and is uniformly bounded in δI × δI minus the sets of left and right touch points, i.e. u xxx exists and is uniformly bounded, almost everywhere in rI × rI.
We are ready to prove Theorem 1.4, using mollifiers to help with applying (3.1).

Appendix A. Calculus lemmas
Lemma A.1.Assume that f satisfies conditions of Theorem 2.3.Then for g for some universal constant C.
Then g ′′ includes terms like which are all bounded by ||f || C 4 (I) .In addition, as x → 0, all these terms have limits by L'Hospital's rule, which shows g ′′ is continuous.For h, the proof is similar.Proof.It follows directly from the fact that Ψ(r) = k(r, 0).
Denote T = ∂ ∂x 1 , a tangential vector on R n .Then it's integral curves are lines (x 2 , x 3 , • • • , x n ) = const.Then we have the following lemma, Lemma A.3.Assume w is a measurable function on Ω ⊆ R n , and absolute continuous on every integral curves of T .In addition, T w exists almost everywhere, and it is integrable in Ω.Then T w is a derivative of w in the weak sense, i.e., for any smooth function v which has compact support in Ω, Proof.We need to show the one-dimensional case, then the general case is done by the Fubini's theorem.Denote T = ∂ ∂x .By the assumption, w is a continuous function, and T w exists almost everywhere.
It's integration by parts.In fact, wv is absolute continuous so we can integrate the equation using the fundamental theorem of calculus.
Consider a mollifier, for x ∈ Ω ⊆ R n , ρ(x) = c exp 1 |x| 2 − 1 , (A.1) when |x| < 1, and ρ(x) = 0 when |x| ≥ 1.Here c is selected such that R n ρ(x)dx = 1.For any w ∈ L 1 (Ω) and τ > 0, the regulation of w is defined to be x − y τ w(y)dy, where τ < dist(x, ∂Ω).Then w τ is a smooth function in a domain Ω ′ , if Ω ′ ⊆ Ω and τ < dist(∂Ω ′ , ∂Ω).If we reduce the absolute continuity condition to being continuous on integral curves of T , then the lemma is not right.Cantor function in the one dimensional case is a counterexample.

.
Apply the same argument to coordinates z = x+y 2 and w = x−y 2 , we have u zzz = u xxx + 3u xxy + 3u xyy + u yyy , u www = u xxx − 3u xxy + 3u xyy − u yyy , are uniformly bounded in two rectangular neighborhoods of the origin minus the origin, respectively, where the bounds only depend on ǫ, η 1 , η 2 , η 3 , η 4 , ||g|| C 4 .So we derive the bounds of u xxy , u xyy in a neighborhood of the origin, since u xxy = u zzz − u www − 2u yyy 6 , u xyy = u zzz + u www − 2u xxx 6 .
Then we check the two dimensional model, and derive |D k x u| ≤ B k (M, k, ||f || C k+3 ) around the origin.The estimates also hold for the two pairs of coordinate systems z = x+y 2 , w = x−y 2 , z 1 = x+2y 5 , w 1 = 2x−y 5 , if none of these four coordinates points to the ỹ or −ỹ direction.We can rotate the x, y coordinates system a little bit if one of them does.Then D 4 z u = u xxxx + 4u xxxy + 6u xxyy + 4u xyyy + u yyyy D 4 w u = u xxxx − 4u xxxy + 6u xxyy − 4u xyyy + u yyyy D 4 z 1 u = u xxxx + 8u xxxy + 24u xxyy + 32u xyyy + 16u yyyy are bounded, implying u ∈ C 4 near the origin.