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The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

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.