is bounded and convex.

We say a point $x_0\in\partial U$ is an extreme point of a bounded convex domain $U\subset\mathbb{R}^n$ if there is a hyperplane $P$ such that $\{x_0\}=P\cap\partial U$, namely the intersection $P\cap\partial U$ is the single point $x_0$. We divide our discussion into the following two cases:

-   : $\mathcal{C}$ has an exteme point $x_0$, which is an interior point of $\Omega$;

-   : All extreme points of $\mathcal{C}$ lie on $\partial\Omega$.

We will rule out the possibility of both cases, and thus $u_0$ is strictly convex. The basic observation is that a convex function with a bounded Monge-Ampère measure is differentiable at any point on its graph, not lying on a line segment joining two boundary points, [@Caf1]. In dimension two, recall the following

\[l401\] Suppose $u$ is a nonnegative convex function in a domain $\Omega\subset\mathbb{R}^2$. The origin $0\in\Omega$ is an interior point. $u$ satisfies $u>0$ on $\partial\Omega$, $u(0)=0$ and $u(x_1,0)\geq|x_1|$. Then the Monge-Ampère measure $\mu[u]$ cannot be a bounded function.

Strict convexity I
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First we rule out the possibility that $\mathcal{G}_0$ contains a line segment with one endpoint in the interior of $\Omega$.

\[l402\] $\mathcal{C}$ contains no extreme points in the interior of $\Omega$.

The 