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I'm interested in (and a bit confused by) the following theorem of Caffarelli, proven in section $4$ of his paper Boundary regularity of maps with convex potentials II:

Assume $u$ is a convex solution to $$\det D^2u = \chi_{\Omega}$$ on $\mathbb{R}^n$, with $\Omega$ convex, $B_1(e_n) \subset \Omega \subset \{x_n > 0\} \cap B_C$ (i.e. $\partial\Omega$ is $C^{1,1}$ at $0$). Assume further that $\nabla u(\Omega)$ is bounded and convex. Then for any $\alpha < 1$, $u$ is $C^{1,\alpha}$ at $0$ in the tangential directions $\{x_n = 0\}$.

For simplicity, think we are in $\mathbb{R}^2$. The idea is that $\Omega$ being $C^{1,1}$ means that when we zoom in, $u$ looks like a solution in $B_1^+$, which has a $C^{1,1}$ estimate in the tangential directions by a Pogorelov-type computation. In the argument, Caffarelli carefully uses that $u$ is globally $C^{1,\sigma}$ for some small $\sigma,$ which follows from the convexity of $\nabla u(\Omega)$. However, I don't see why we need to use this regularity, for reasons I outline in the argument below. My main question is thus:

Is $u$ Lipschitz good enough for this argument to work, or do we need $\nabla u(\Omega)$ to be convex? (We certainly need this, and especially the $C^{1,\sigma}$ regularity it gives, for other parts of the theory, but I'm interested in this specific theorem). This would be strange to me, since then it seems we only have a condition on the boundary (and no condition on boundary data, which in the theorem above comes from prescribing $\nabla u(\Omega)$). It would be nice to know if I missed something, and if not, a convincing reason to expect this result without some sort of boundary condition on $u$!

Here is a rough outline of Caffarelli's argument, leading to my question. We can assume (by subtracting a linear function) that $u(0) = 0, \nabla u(0) = 0$. The theorem above then follows from the statement that for a centered section of $u$, $S_h = \{u < h + l_h\}$ with $l_h$ chosen so that $0$ is the center of mass of $S_h$, we have $$B_{c_{\epsilon}h^{\frac{1}{1+\alpha}}} \cap \{x_2 = 0\} \subset S_h$$ for some small $c_{\epsilon}$.

If the conclusion is false, we can find a sequence of functions $u_k$ satisfying the hypotheses and $c_k \rightarrow 0$ for which the conclusion fails at height $h_k$. The fact that $u_k$ are Lipschitz means that for $c_k$ small enough the conclusion will be true for large enough $h$, so it is easy to see that $h_k \rightarrow 0$. Now comes the key step, which says that after rescaling these sections we get a solution on $\{x_2 > 0\}$:

Roughly, dilating by $A_k = (c_kh_k^{\frac{1}{1+\alpha}})^{-1}$ in the tangential direction and by a factor $B_k$ in the vertical direction, we rescale the sections into something equivalent to the unit ball. Since quadratic scaling will preserve paraboloids, if we have $\frac{B_k}{A_k^2} \rightarrow 0$ we are in the desired situation, which gives a contradiction.

Caffarelli uses $C^{1,\sigma}$ regularity to say that the sections $S_{h}$ go out at least $h^{\frac{1}{(1+\sigma)}}$ in the vertical direction, giving $B_k \leq h_k^{-\frac{1}{1+\sigma}}$. (This is the only place where $C^{1,\sigma}$ regularity is used). We conclude that $$B_k/A_k^2 \leq c_k^2h_k^{\frac{2}{1+\alpha}-\frac{1}{1+\sigma}}.$$ (In Caffarelli's paper, in the middle of page $478$, the analagous quantity is $c_kh_k^{\frac{1}{1+\alpha}-1/4-\gamma/2}$ for some $\gamma < 1/2$, where the $\gamma$ inequality comes from $C^{1,\sigma}$ regularity.) At this point in the argument, he says "...$<c_kh_k^{S}$ for some $S(\alpha) > 0$, as long as $\alpha < 1$." However, it seems clear that even if $\sigma = 0$ (using only that $u$ is Lipschitz), if $\alpha < 1$ we get the desired result.

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