# Estimates on the second-order derivatives for degenerate Monge-Ampere equations

The current post comes from my previous post at stackexchange. However, I have not get any comment yet.

In a celebrated paper written by Guan, Trudinger, and Wang, authors proved the existence and uniqueness of convex $$C^{1,1}$$-solution to the Dirichlet problem for degenerate Monge-Ampere equations, and they also provided a global $$C^{1,1}$$-control to the solution. They first reduced the global $$C^2$$-estimates to the boundary estimates for the second-order derivatives, and then did the estimates on the boundary. To this end, a "simple" inequality was used for several times, which can be stated precisely as follows.

Let $$n\geq2$$ be an integer and $$\Omega\subset\mathbf R^n$$ be a bounded domain with smooth boundary. Assume $$\Omega$$ is uniformly convex if necessary. Given $$f\in C^{1,1}(\bar\Omega)$$ such that $$f>0\quad\hbox{in \Omega.}$$ Does there exist a constant $$C>0,$$ which is at most dependent on $$n,~\Omega,$$ and $$\|f\|_{C^{1,1}(\bar\Omega)},$$ such that $$$$|\nabla f(x)\cdot\tau(x)|^2\leq Cf(x)\quad\hbox{for all x\in\partial\Omega},\label{1}\tag{1}$$$$ where $$\tau(x)$$ is a unit tangent vector of $$\partial\Omega$$ at $$x.$$

Or equivalently, does the tangential derivative $$\partial_{\tau}\sqrt f$$ have a uniform upper bound?

Actually, authors used in above reference an incorrect inequality as $$|\nabla f|^2\leq Cf\quad\hbox{in \Omega.}$$ Near the boundary, it is clear that the distance function $$d(x):=\mathrm{dist}(x,\partial\Omega)$$ is a counterexample since $$|\nabla d|=1$$. Fortunately, it seems that all consequences still hold if \eqref{1} is true. If $$f(x_0)=0$$ for some boundary point $$x_0,$$ then obviously we can take above $$C=1$$ at $$x_0.$$ The remaining difficulty is to control the tangential derivative of $$\sqrt f$$ if $$f(x)>0$$ but it is very small. Probably above inequality is reasonable in the following sense. For $$x_0\in\partial\Omega.$$ Roughly, if $$f$$ behaves like some positive power of $$d_{x_0}:=d(x,x_0)$$ near $$x_0,$$ for example we say it as $$d_{x_0}^\alpha.$$ Then, $$|\nabla_\tau f|$$ might behave as $$d_{x_0}^{\alpha-1}$$ near $$x_0.$$ If \eqref{1} is not true, then $$\alpha-1<\alpha/2,$$ and thus $$\alpha<2.$$ However, this is impossible as $$f$$ is of $$C^{1,1}.$$ I hope someone could give me some comment on this topic.

This inequality comes from scaling. Assume that $$f$$ satisfies $$|D^2f| \leq 1$$ on $$\mathbb{R}^n$$ and $$f \geq 0$$. It suffices to prove that $$|\nabla f(0)|^2 \leq 2f(0).$$ Equality holds if $$f(0) = 0$$, so assume that $$f(0) > 0$$. We may assume that $$f(0) = 1$$ after taking the rescaling $$\tilde{f}(x) = \lambda^{-2}f(\lambda x)$$, with $$\lambda^2 = f(0)$$, since this rescaling preserves the Hessian of $$f$$, the sign of $$f$$, and the ratio of interest: $$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$ In the situation $$f(0) = 1$$ it is clear that $$|\nabla f(0)|^2 \leq 2$$, otherwise the Hessian bound would imply that $$f < 0$$ somewhere (follow a ray in the direction $$-\nabla f(0)$$).
The situation when $$f$$ is a nonnegative $$C^{1,\,1}$$ function on a compact manifold ($$\partial \Omega$$) is similar. One can e.g. make a nonnegative extension of $$f$$ to $$\mathbb{R}^n$$ with comparable $$C^{1,1}$$ norm and apply the previous reasoning.