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 \begin{equation}|\nabla f(x)\cdot\tau(x)|^2\leq Cf(x)\quad\hbox{for all $x\in\partial\Omega$},\label{1}\tag{1}\end{equation} 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.


1 Answer 1


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.


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