= Solution
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 is
$$
\boxed{\liminf_{t\downarrow0}\frac{u(y)-u(y-t\nu)}{t}>0.}
$$
If 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 barrier
$$
h(x)=e^{-kr^2}-e^{-kR^2}>0.
$$
The operator has no zeroth-order term, and differentiation gives
$$
Lh=e^{-kr^2}\left(4k^2a^{ij}(x_i-z_i)(x_j-z_j)-2k\operatorname{tr}a-2kb^i(x_i-z_i)\right)>0
$$
for 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 $\varepsilon>0$ small enough that $u-u(y)+\varepsilon h\leq0$ there. The same inequality holds on the outer sphere since $h=0$. The <weak maximum principle for elliptic operators> yields $u-u(y)+\varepsilon h\leq0$ throughout the annulus. Along the inward radius,
$$
\frac{u(y)-u(y-t\nu)}{t}\geq\varepsilon\frac{e^{-k(R-t)^2}-e^{-kR^2}}{t}
\longrightarrow2\varepsilon kRe^{-kR^2}>0.
$$
This proves the lemma. The <interior sphere condition> supplies precisely the geometry needed for the barrier; no global boundary regularity is used.
Back to article page
