Smoothness of radial solutions to Monge-Ampere equations

We characterize when radial weak solutions to Monge-Ampere equations are smooth. This paper extends previous partial results and also covers Generalized Monge-Ampere equations and infinitely vanishing right hand side.


Introduction
It is well known that the radial homogeneous functions u = c m,n |x| 2+ 2m n provide nonsmooth solutions to the Monge-Ampère equation det D 2 u = |x| 2m with smooth right hand side when m ∈ N \ nN. This raises the question of when radial solutions u to the generalized equation are smooth, given that k is smooth and nonnegative. When u is radial, (1) reduces to a nonlinear ODE on [0, 1) that is singular at the endpoint 0. It is thus easy to prove that u is always smooth away from the origin, even where k vanishes, but smoothness at the origin is more complicated, and determined by the order of vanishing of k there. In fact, Monn [9] proves that if k = k (x) is independent of u and Du, then a radial solution u to (1) is smooth if k 1 n is smooth, and Derridj [4] has extended this criterion to the case when k (x, u, Du) = f |x| 2 2 , u, |∇u| 2 2 factors as with κ smooth and nonnegative on [0, 1), κ (0) = 0, and φ smooth and positive . Moreover, Monn also shows that u is smooth if k = k (x) vanishes to infinite order at the origin.
These results leave open the case when k has the general form k (x, u, Du) and vanishes to infinite order at the origin. The purpose of this paper is to show that radial solutions u are smooth in this remaining case as well. The following theorem encompasses all of the afore-mentioned results, and applies to generalized convex solutions u and also with f = κφ as in (2) but where φ is only assumed positive and bounded, not smooth.
Theorem 1 Suppose that u is a generalized convex radial solution (in the sense of Alexandrov) to the generalized Monge-Ampère equation (1) with Suppose moreover that there are positive constants c, C such that for (ξ, ζ) near (0, 0). Let τ ∈ Z + ∪ {∞} be the order of vanishing of f (t, 0, 0) at 0. Then u is smooth at the origin if and only if τ ∈ nZ + ∪ {∞}.
The case when k = k (x) is independent of u and Du is handled by Monn in [9] using an explicit formula for u in terms of k: where u (x) = g r 2 2 and k (x) = f r 2 2 ≥ 0 with r = |x|, x ∈ R n . In the case k vanishes to infinite order at the origin, an inequality of Hadamard is used as well. The following scale invariant version follows from Corollary 5.2 in [9]: for all 1 ≤ ℓ ≤ k − 1 and k ∈ N provided F is smooth, nondecreasing on [0, 1) and vanishes to infinite order at 0.

Proof of Theorem 1
We begin by considering Theorem 1 in the case that u is a classical C 2 solution to (1) and f satisfies (2) where f (t, 0, 0) vanishes to finite order ℓ at 0. If k is independent of u and Du, Monn uses formula (4) in [9] to show that u is smooth when f (w) 1 n is smooth. In particular this applies when ℓ ∈ nZ + . In the general case, we note that (3) implies (2), the assumption made in [4]. Indeed, using Thus the results of Derridj [4] apply to show that u is smooth for general k when ℓ ∈ nZ + .

Generalized Monge-Ampère equations
We now consider radial generalized convex solutions u to the generalized Monge-Ampère equation (1) where we assume k (·, u, q) and k (x, u, ·) are radial. We first establish that u ∈ C 2 (B n ) ∩ C ∞ (B n \ {0}). We note that results of Guan, Trudinger and Wang in [6] and [8] yield u ∈ C 1,1 (B n ) for many k in (1), but not in the generality possible in the radial case here. In order to deal with general k it would be helpful to have a formula for u in terms of k, but this is problematic. Instead we prove Theorem 1 for general k without solving for the solution explicitly, but using an inductive argument that is based on Lemma ?? when k vanishes to infinite order at the origin.
Assume that u is a generalized convex solution of (1) in the sense of Alexandrov (see [1] and [3]) and define ϕ (t) by Then ϕ is bounded since u is Lipschitz continuous. It follows that the convex radial function u is continuously differentiable at the origin, since otherwise it would have a conical singularity there and its representing measure µ u would have a Dirac component at the origin. Let g be given by formula (4) with ϕ in place of f , i.e.
and with constant C u chosen so that u and u agree on the unit sphere where We claim that u is a generalized convex solution to (1) in the sense of Alexandrov. To see this we first note that D 2 u (re 1 ) = positive semidefinite, hence u is convex. To prove that the representing measure µ e u of u is kdx it suffices to show, since both g and f are radial, that Since ∂ ∂r u (r i e i ) = g ′ r 2 In particular the convex radial function u must be continuously differentiable, since otherwise there is a jump discontinuity in the radial derivative of u at some distance r from the origin that results in a singular component in µ e u supported on the sphere of radius r. Now uniqueness of Alexandrov solutions to the Dirichlet problem (see e.g. [3]) yields u = u, and hence u ∈ C 1 (B n ). Thus ϕ ∈ C [0, 1) and from (8) we where using (7) we compute that In particular g ′ ∈ C [0, 1). We now obtain by induction that g ∈ C ∞ (0, 1), hence u ∈ C ∞ (B n \ {0}). Indeed, if g ∈ C ℓ (0, 1) then (9) implies ϕ ∈ C ℓ−1 (0, 1) and then (7) implies g ∈ C ℓ+1 (0, 1). It will be convenient to use fractional integral operators at this point. For β > 0 and f continuous define We claim that for f smooth, nonnegative and of finite type ℓ, ℓ ∈ Z + , the same is true of T β f for all β > 0. This follows immediately from the identity and the estimate When k = 1, (12) follows from differentiating and then integrating by parts, and the general case is then obtained by iteration. Now suppose that f satisfies (3) and let vanish to infinite order at 0. If κ vanishes in a neighbourhood of 0 then so does g and we have g ∈ C ∞ [0, 1) and u ∈ C ∞ (B n ). Thus we will assume t 0 κ > 0 for t > 0 in what follows. Note that (12) then implies that T n 2 κ (t) is smooth and positive on (0, 1) and vanishes to infinite order at 0. Since g ′ ∈ C [0, 1), it follows that ϕ (t) ≤ Cκ (t). Thus we have the inequality T n 2 ϕ (t) ≤ CT n 2 κ (t), and from (10) we now conclude that g ′ (t) also vanishes to infinite order at 0. Now ϕ (t) ≈ κ (t) from (3), and so also T n 2 ϕ (t) ≈ T n 2 κ (t). From we then have An application of (5) with ℓ = 1, k > n and F (t) = t 0 s n 2 −1 κ (s) ds yields t n 2 κ (t) = F ′ (t) ≤ CF (t) 1− 1 k and so the first term on the right side of (14) is bounded by a multiple of t − 1 2 F (t) 1 n − 1 k . Thus the right side of (14), and hence also g ′′ (t), vanishes to infinite order at 0. In particular g ′′ ∈ C [0, 1) and we conclude u ∈ C 2 (B n ) in this case as well.
Summarizing, we have u ∈ C ∞ (B n \ {0}), and in the case f satifies (3), we also have u ∈ C 2 (B n ). Thus from above we have that where u (x) = g |x| 2
To see this, we first note that F is smooth, nonnegative and vanishes to infinite order at 0 since the same is true of κ. Next, for any ℓ ≥ 1 and ε > 0, (5) with k large enough yields Moreover we have and an application of (17) gives We obtain similar estimates for the remaining terms in (16) and altogether this yields Using the second and third lines in (18) now shows that the first term in braces in (15) satisfies which vanishes to infinite order at 0 if 0 < ε < 1 n . Similar arguments, using (16) and the first line in (18) to estimate T n 2 +1 ϕ ′ (t), apply to the remaining terms in (15), and this completes the proof that g ′′′ vanishes to infinite order at 0 and g ′′′ ∈ C [0, 1).
We now observe that we can • continue to differentiate (15) to obtain a formula for g (ℓ) involving only appropriate powers of T n 2 ϕ (t) ≈ T n 2 κ (t) in the denominator, and derivatives of ϕ of order at most ℓ − 2 in the numerator, • and continue to differentiate (16) to obtain a formula for ϕ (ℓ−2) involving derivatives of g of order at most ℓ − 1.
It is now clear that the above arguments apply to prove that derivatives of g (t) of all orders vanish to infinite order at 0 and are continuous on [0, 1). This shows that g is smooth on [0, 1) and thus that u is smooth on B n .