Let $\Delta$ be an open triangle in $\mathbb{R}^2$ and $u\in C^0(\overline{\Delta})\cap C^\infty(\Delta)$ be the convex function satisfying $$ \det D^2u=1,\quad u|_{\partial\Delta}=0. $$ Classical results on Monge-Ampère equations imply that there exists a unique such $u$. I need informations on the boundary regularity of $u$, but could not find any literature on regularity of such Monge-Ampère equations near a boundary line segment.

Question 1. What can be said about the boundary regularity of $u$? In particular, is $\|\nabla u(x)\|$ bounded as $x\in\Delta$ approaches a point on $\partial \Delta$ which is not a vertex?


  1. The second question is equivalent to asking whether $u$ has infinite slope at a non-vertex point $p\in\partial\Delta$, or more precisely, whether the limit $$ \lim_{t\rightarrow 0^+}\frac{u((1-t)p+tq)}{t}\in [-\infty,0) $$ equals $-\infty$ for some $q\in\Delta$ (which implies the same limit for every $q\in \Delta$).

  2. It can be shown $u$ has finite slope at a vertex (namely, the above limit is finite if $p\in\partial\Delta$ is a vertex): assuming $p=0$ w.l.o.g., by Comparison Principle, $u$ is minorized by a convex function $v$ of the form $$ v(x)=c|x|^\alpha-L(x), $$ where the constant $\alpha\in (1,2)$ is arbitrary, $c>0$ is a constant to make $\det D^2v=c^2\alpha^2(\alpha-1)|x|^{2\alpha-4}$ bigger than $1$ on $\Delta$, and $L$ is a linear function on $\mathbb{R}^2$ to make $v|_{\partial\Delta}\leq 0$; whereas $v$ has value $0$ and finite slope at $x=0$.

  3. Attempting to apply Comparison Principle near non-vertex boundary points, one is lead to the following problem:

Question 2. Let $D=\{x\in\mathbb{R}^2\mid |x|<1,\, x_1>0\}$ be the half-disk. Is there a nonnegative convex function $u\in C^0(\overline{D})$ which vanishes on the boundary line segment $\{x_1=0,\, |x_2|\leq 1\}$, such that $\det D^2u\geq \lambda>0$?

I tried to construct such a $u$ of the form $u(x_1,x_2)=f(x_1)g(x_2)$ (with $f$ and $g$ nonnegative, convex and $C^2$) but was lead to the conclusion that such $u$ can never satisfy the hypotheses.


1 Answer 1


The answer to both questions is no.

For Question $2$: After subtracting a function of the form $cx_1$ we may assume that $u \geq 0$ and $u(t,0) = o(t)$. (Take $c$ to be the slope of the tangent line to $u(t,0)$ at $t = 0$; then $u - cx_1 \geq 0$ on the $x_1$ axis and vanishes on the $x_2$ axis, so by convexity is nonnegative in $D$). Then $\{u < \epsilon\}$ contains the triangle $T_{\epsilon}$ with vertices $\pm e_2$ and $4l_{\epsilon}e_1$ where $\epsilon^{-1}l_{\epsilon} \rightarrow \infty$ as $\epsilon \rightarrow 0$. It is straightforward to check that the paraboloid $P_{\epsilon} = 2\epsilon(l_{\epsilon}^{-2}(x_1 - l_{\epsilon})^2 + 4x_2^2)$ satisfies $P_{\epsilon} > \epsilon \geq u$ on $\partial T_{\epsilon}$, vanishes at $(l_{\epsilon},0)$, and $\det D^2P_{\epsilon} \rightarrow 0$ as $\epsilon \rightarrow 0$. For $\epsilon$ small this violates the comparison principle.

Question $1$ can be reduced to Question $2$ as follows: if there exists a sequence of points approaching the flat boundary where the graph of $u$ has supporting planes of uniformly bounded slope, then in the limit we get a supporting plane to the graph of $u$ with bounded slope that vanishes on one of the edges. Then subtract the plane, translate, rotate and rescale.

However, near the flat boundary I suspect there is some expansion of $u$ in powers of "singular building blocks," maybe of the form $x_1(-\log x_1)^{1/2}$ and $x_2$. (This is motivated by the observation that $w := x_1(-\log x_1)^{1/2}(x_2^2-1)$ satisfies $0 < \lambda < \det D^2w < \Lambda$ in the domain $\{|x_2| < c(-\log (x_1))^{-1/2}\}$ near the origin, which has boundary that separates from the $x_2$ axis slower than any polynomial).

  • $\begingroup$ That's really a great argument that I should have known earlier! By the way, I need to further consider the equation with right-hand side $|x|^\alpha$ instead of $1$ (still with zero boundary value, on a triangle $\Delta$ with $0\in\mathbb{R}^2$ as a vertex) and determine whether the derivative of the solution $u$ at $0$ (along a vector pointing inside $\Delta$) is finite. I can show that it is finite if $\alpha>-2$ and infinite if $\alpha<-2$, but have trouble with the $\alpha=-2$ case. Could you maybe shed some light on this? $\endgroup$
    – Xin Nie
    Nov 9, 2018 at 23:04
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    $\begingroup$ The gradient will blow up at the vertex in the case $\alpha = -2$. To see this, consider the barrier $w = -y(\log(1/y))^{1/2} + x^2(\log(1/y))^{1/2}/y$ in the quarter space $\{y > |x|\}$. A computation gives $\det D^2w = y^{-2}(1-x^2/y^2 + O(1/|\log y|))$, so $w$ is a super solution with gradient that blows up logarithmically at $0$. $\endgroup$ Nov 15, 2018 at 17:57
  • $\begingroup$ My computations show that $\Omega:=\{\det D^2w>0\}$ is the raindrop-shaped region $\left\{0<y<e^{1/2},\ x^2<\frac{y^2\left(\frac{1}{2}-\log y\right)}{\frac{3}{2}-\log y}\right\}\subset\{y>|x|\}$. Do you mean using $\tilde{w}=w+cy$ (with $c>0$ large enough to ensure $\tilde{w}\geq 0$ on $\partial\Omega$) as a barrier on $\Omega$? $\endgroup$
    – Xin Nie
    Nov 16, 2018 at 11:50
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    $\begingroup$ That is right, thanks for doing the computation carefully- in any corner of angle $< \pi$ we can take an affine rescaling of $w$, multiply by a small constant, and add a large multiple of $y$ to obtain a super-solution that is nonnegative on the boundary of a region where $u \leq 0$ and thus show blowup of the gradient at the vertex. $\endgroup$ Nov 16, 2018 at 15:08

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