Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 107 / 2

Assume an interior sphere condition at $y$, and assume the coefficients are bounded and the principal part defines a uniformly elliptic operator near its tangent Euclidean ball. Shrinking the Euclidean ball while keeping it tangent if necessary, arrange that $\overline{B_R(z)}\setminus \{y\}\subset \Omega$. Write $\nu =(y-z)/R$, the outward radial direction. The conclusion of the Hopf boundary point lemma in the regularity stated in the question isIf the one-sided normal derivative exists, this says ${\partial }_{\nu }u(y)>0$. For a $C^1$ boundary its outward normal agrees with the tangent Euclidean ball's radial normal. Continuity alone does not guarantee that the derivative exists, so its existence must be added when asserting a finite derivative.

On the annulus $R/2<r=|x-z|<R$, use the barrierThe operator has no zeroth-order term, and differentiation givesfor sufficiently large $k$: the quadratic-in-$k$ term is at least $k^2\lambda R^2$, while the remaining terms are bounded multiples of $k$.

On the inner sphere, compactness and strict inequality give a positive minimum of $u(y)-u(x)$. Choose $\epsilon >0$ small enough that $u-u(y)+\epsilon h\le 0$ there. The same inequality holds on the outer sphere since $h=0$. The weak maximum principle for elliptic operators yields $u-u(y)+\epsilon h\le 0$ throughout the annulus. Along the inward radius,This proves the lemma. The interior sphere condition supplies precisely the geometry needed for the barrier; no global boundary regularity is used.

Here a domain is connected. Suppose both a negative point and a zero point exist. The open set $D=\{v<0\}$ then has a boundary point $q$ inside $\Omega$. Choose $z\in D$ close enough to $q$ that the Euclidean ball of radius $R=\mathrm{dist}(z,Z)$ stays compactly inside $\Omega$; it is contained in $D$ and touches $Z$ at some $y$. Attainment of this distance follows by restricting to a compact neighbourhood of $q$.

On that Euclidean ball $v$ is twice differentiable andThe Hopf boundary point lemma gives a strictly positive outward radial derivative at $y$. But $y$ is an interior maximum of the $C^1$ function $v$ on $\Omega$, so $Dv(y)=0$. This contradiction proves the Hopf dichotomy with classical regularity away from the zero set:

For the requested weaker-regularity counterexample take $\Omega =(-1,1{)}^n$ andIt is a locally Lipschitz function and nonpositive. Its negative set is $x_1>0$, where $v=-\sin x_1$ is smooth and $\Delta v+v=0$. On $x_1\le 0$ it is zero. The normal derivative jumps from zero to minus one across $x_1=0$, so it is not $C^1$. All other hypotheses hold; imposing the equation also across that interface would be an additional condition absent from the question.