Let $$ Lu=-a_{ij}(x)\partial_{ij}u+b_i(x)\partial_i u $$ be a uniformly elliptic operator, with $A(x)=(a_{ij}(x))$ positive-definite. Here I'm only considering smooth coefficients, and the domain $\Omega\subset \mathbb R^d$ is as smooth as needed (but bounded). In one of its elementary versions, the classical Hopf boundary point lemma states that if $u\in C^2(\Omega)\cap C^1(\bar\Omega)$ is a supersolution $$ Lu\geq 0 $$ and attains a minimum point at $x_0\in \partial\Omega$ then $$ \partial_\nu u(x_0)<0, $$ where $\nu=\nu(x_0)$ is the outer normal to $\partial\Omega$ at $x_0$. In other words, the supersolution must grow linearly inside the domain close to a boundary minimum point. There are various possible extensions, in particular $\nu$ can be any outward pointing direction, and $\Omega$ can have corners.


Can one say anything about the behaviour of $u$ at any such boundary minimum point, assuming only that $u\in C^2(\Omega)\cap C(\bar\Omega)$?

The key point here being the lack of $C^1$ regularity up to the boudnary. I need this typically for a singular eigenvalue problem, where in fact I am trying to retrieve some information on a principal eigenpair $(\lambda_0,u_0(x))$ to a singular problem of the form $$ Lu_0=\frac{1}{\theta(x)}\lambda_0 u_0. $$ My weight $\theta(x)$ is a given, smooth function that is positive inside $\Omega$ but vanishes linearly, typically $\theta(x)=\operatorname {dist}_{\partial\Omega}(x)$.

Some probabilistic arguments tell me for free that $u_0\geq 0$ is nontrivial and $\lambda_0>0$, with $u_0$ being moreover $C^2$ in the interior and continuously vanishing at the boundary. I need to show that the vanishing is linear. Of course Hopf's lemma immediately pops up to mind, but I really cannot guarantee the $C^1$ regularity up to the boundary (and all my attempts in that direction have failed so far). Actually for my purposes it would suffice to get $$ c\operatorname{dist}_{\partial\Omega}(x)\leq u_0(x) \leq C\operatorname{dist}_{\partial\Omega}(x) $$ in a neighborhood of $x_0$. Has anyone ever heard of something like that? For example, sandwiching $u_0$ between two local lower/upper barriers vanishing linearly would do the trick, but I dont' really know how to do that.

Final kinky comment: to tell the truth, really, I'm working in the one-dimensional interval $\Omega=(0,1)$. But the question is so natural that I felt I had to write it in a slightly more general framework. So, if anyone has a specific 1D trick I'll be happy too!

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    $\begingroup$ @LSpice: thanks for the tex editing, I am indeed too lazy for real operatorname without macros (mathbb R is about as far as I usually go...) As for the text: yes, I really meant kinky, because I'm (poorly) trying to trick people into the question, whereas it's really a 1D ODE in the end. By all means feel free to edit if you feel it appropriate. $\endgroup$ Sep 11 at 23:04
  • $\begingroup$ I wouldn't want to edit to change meaning. Since the question has been answered anyway, I have deleted my comment. $\endgroup$
    – LSpice
    Sep 12 at 0:36
  • $\begingroup$ Leo, perhaps you'll find something useful also in this paper by Alvarado, R.; Brigham, D.; Maz’ya, V.; Mitrea, M.; Ziadé, E. "On the regularity of domains satisfying a uniform hour-glass condition and a sharp version of the Hopf-Oleinik boundary point principle" Journal of Mathematical Sciences (New York) 176, No. 3, 281-360 (2011), MR2839047, Zbl 1290.35046. $\endgroup$ Sep 12 at 18:37
  • $\begingroup$ it deals with a generalization of the interior sphere condition and offers some sharp results. $\endgroup$ Sep 12 at 18:41
  • $\begingroup$ Indeed this looks quite interesting to me, thanks @DanieleTampieri $\endgroup$ Sep 13 at 8:44

1 Answer 1


I think this is just the comment following Lemma 3.4 of Gilbarg and Trudinger (specifically equation 3.11).

I should add that lowering the regularity of the boundary seems like a harder problem (and is I think false in the case of a square). GT require an interior sphere condition. L. Rosales (see for instance Generalizing Hopf’s Boundary Point Lemma) seems to have the best results in this direction.

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    $\begingroup$ Oh, great, this is exactly what I need. Much appreciated! $\endgroup$ Sep 11 at 23:57

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