### Solving the Monge-Ampere equation (continued)

In a previous post I began the proof of the following theorem:

Let $\psi \in C^\alpha$ be a positive function in $B_1$, then there exists a unique function $u \in C^{2,\alpha}(B_1)$ such that

$det(D^2u(x))=\psi \mbox{ in } B_1$

$u(x) = 0 \mbox{ on } \partial B_1$

To prove the theorem, we looked at the function

$\psi_t = (1-t)+t\psi(x)$.

we noted that it is easy to solve the problem explicitely for $\psi=\psi_0$.  Then thanks to the inverse function theorem (the Banach space version) we saw that the set $A$ of $t$‘s for which we can solve the equation $A$ is open (recall that $A$ was defined as the set of those $t$ in $[0,1]$ for which we can solve the equation above with right hand side  equal to $\psi_t$).

Since $A$ is open and non-empty, if we show that its also closed the theorem would be proved.

Part II:  Showing $A$ is closed.

(see previous post of the series for Part I)

That $A$ is closed will be shown to be a consequence of the following a priori estimate:

Theorem (A priori estimate for the Monge-Ampere equation) Let $u$ be a smooth solution of

$det(D^2u)=\psi \mbox{ in } B_1$

$u= 0 \mbox{ on } \partial B_1$

There is a constant $C>0$ (depending only on $n$ and the $C^3$ norm of $\psi$) such that

$||u||_{C^{2,\alpha}}\leq C$

Proving the a priori estimate is usually the hardest step, so first let’s see that the a priori estimate in fact implies that $A$ is closed: let $t_k$ be a sequence in $A$ converging to a number $t \in [0,1]$, by the a priori estimate above the sequence of solutions $u^{(t_k)}=u^{(k)}$ is uniformly bounded in $C^{2,\alpha}$, thus by the Arzela-Ascoli theorem a subsequence converges to a function $u \in C^{2,\alpha}_0$, since $\psi_{t_k}$ converges to $\psi_t$ we see this $u$ is a solution of Problem $(t)$, i.e. $t \in A$. This proves $A$ is closed and therefore the theorem.

Of course, now we “just” have to prove the a priori estimate. ..

The broad outline of its proof is: first we prove a series of a priori estimates, each stronger than the previous one, and at each given step we  use the estimates already achieved, this will bound the first and second derivatives, the main tool here is the maximum principle used as in the classical technique of Bernstein. Then, the Holder regularity of the second derivatives will be consequence of deep regularity results from the theory of fully non-linear equations, among them the Evans-Krylov theorem. Now lets go through the argument in detail, and to try to make this just a bit readable I will number the steps

Step 1) The easiest estimate is the $L^\infty$ bound

(a priori estimate (1))

$||u||_{L^\infty(B_1)} \leq C$

where $C$ depends only on the $L^\infty$ norm of $\psi$. To prove it, consider the auxiliary function $\phi(x)=\frac{M}{2n}(|x^2|-1)$, this is a smooth function that vanishes on $\partial B_1$ and $D^2\phi = M^n$, picking $M$ large enough (and controlled by the supremum of $|\psi|$) we have

$det(D^2\phi) \geq \psi = det(D^2u)$

then the maximum principle implies that

$u \geq \phi \geq -\frac{M}{2n}$ in $B_1$

plus $u$ is a convex function  vanishing on $\partial B_1$, so we have $u \leq 0$ as well. This proves the a priori estimate (1).

Step 2) Now we go for a Lipschitz bound for $u$, which we will refer to as the

(a priori estimate (2))

$||\nabla u||_{L^\infty(B_1)} \leq C$

Let’s differentiate the equation $det(D^2u)=\psi$ in the direction of the unit vector $\xi$, and we obtain

$L u_\xi= a_{ij} (u_\xi)_{ij}=\psi_\xi$

Here $a_{ij}(x)$ correspond to the entries of the cofactor matrix of $D^2u(x)$, what the left hand side above amounts to is the linearization of $det(D^2u)$ at $u$, which is the elliptic operator $L$ (we did this before, in the previous post). Notice also that I can talk freely about third derivatives of $u$ because any strictly convex, $C^{2,\alpha}$ solution of the Monge-Ampere equation is actually $C^\infty$ (that this is true is another of the many consequences of  the Schauder theory for second order  linear elliptic equations with Holder continuous coefficients, and its not too hard to check), so the equation above holds classically. In particular, by the maximum principle (again) we conclude that for some $C$ depending on the supremum of $|\psi|$:

$\sup\limits_{B_1} |u_e| \leq \sup \limits_{\partial B_1}|\nabla u|+C$

but $u \equiv 0$ on $\partial B_1$ thus there $\nabla u = u_n n$, where $n$ is the outer normal to $\partial B_1$. Here we bring back our old friend, the function $\phi$ defined in step 1), as we know $u\geq \phi$ on $B_1$ and they agree on the boundary, therefore

$u_n \leq \phi_n$ on $\partial B_1$

also, since $u \leq 0$ and vanishes on $\partial B_1$ we also have the inequality $0 \leq u_n$. Therefore the $|u_n|$ is  controlled by $M$ times a dimensional constants, i.e.  the supremum of $|\psi|$ controls $|\nabla u|$. Putting the last three identities/inequalities together we get the a priori estimate (2).

Step 3) Notice that up to now we have only used the ellipticity of the equation and not the particular structure of the Monge-Ampere equation (even the differentiation step of Step 2) is a fact that holds for general elliptic equations). This  will change now that we are going to prove the

(a priori estimate (3))

$||D^2 u||_{L^\infty(B_1)} \leq C$

We start off as in step 2 by differentiating along a fixed, arbitrary direction $\xi$,  except now we do it twice and thus we get something much more complicated than before. We get around this noting that the map $M \to det(M)$ is a convex function on positive definite matrices, and get an inequality that allows us to drop the lower order terms:

$Lu_{\xi \xi} \geq \psi_{\xi\xi} \geq -C$

and this says that  $\sup \limits_{B_1} u_{\xi\xi}$ is controlled by the same supremum on $\partial B_1$ plus a constant $C$ controlled by $\sup |\psi|$.  Then we only have to estimate things on the boundary (just as for the first derivatives).

So lets pick a point $x_0 \in \partial B_1$ and estimate the entries of $D^2u(x_0)$. Since on $\partial B_1$ we have $u \equiv 0$ all the first, and second order angular derivatives of $u$ at $x_0$ are zero, and this says a lot about the Hessian of $u$ on the boundary, its just a matter of relating partial derivatives in polar and Cartesian coordinates to see then (after doing an adequate rotation of our coordinate system) that the entries of the Hessian satisfy the relations:

$u_{ij}(x_0)=u_n(x_0)\delta_{ij}$ for $i,j

here “$n$” stands both for the number $n$ and the exterior normal direction $\partial B_1$ since in our rotated coordinate system the vector $e_n$ is normal to $\partial B_1$ at the point $x_0$.Since we already have an a priori bound for $u_n$ this gives us an a priori bound for the second partial derivatives on the boundary, except for the partial derivatives of the form $u_{in}$, $0\leq i \leq n$.

Here we use the fact that our equation is rotation invariant (that is, the determinant is invariant under rotations of the coordinate system). Since $u_{ij}$ ($i,j) is diagonal, we can compute at once $det(D^2u(x_0))$

$det(D^2u(x_0))=\psi(x_0)=u_n^{n-1}u_{nn}$

I claim that there exists a $\delta_0>0$  such that $u_n(x_0)>\delta_0$ on $\partial B_1$, and $\delta_0$ is bounded away from zero by a quantity controlled by $\inf \psi$. Therefore there exists a  $C$ controlled by $\psi$, such that

$u_{nn}\leq C\psi(x_0)\leq C\sup |\psi|$

Plus, $u_{nn}$ is non-negative (since $u$ is convex). Thus we have given a priori estimates for all the pure second order derivatives of $u$ on $\partial B_1$, and by the remarks made at the beginning of this step we have proved the a priori estimate (3)

Step 4) Putting the 3 previous estimates together, we have:

a priori estimate (4)

$||u||_{C^{1,1}(B_1)}\leq C$

Next, we have to push this to a $C^{2,\alpha}$ estimate in  $B_1$. This will be the content of the next post (I am sure we can agree that this post is long enough). In the next post I  will talk about the Evans-Krylov theorem and the boundary behavior of solutions to elliptic equations. Hopefully that will wrap up this series of posts on the Dirichlet problem for Monge-Ampere,  after this I might go back to the Minkowski problem.