Hopf lemma

Remarks

Proof idea: let $B$ have centre $y$ and radius $R$, and set $v(x) = e^{-\lambda \abs{x - y}^2} - e^{-\lambda R^2}$. For $\lambda$ large, $L v > 0$ on the annulus $A = \set{R/2 < \abs{x - y} < R}$, and $v = 0$ on the outer sphere. For small $\varepsilon > 0$, the function $u - u(x_0) + \varepsilon v$ satisfies $L(\cdot) \ge 0$ in $A$ and is at most $0$ on $\partial A$, so it is at most $0$ in $A$ by the weak maximum principle. At $x_0$ this gives $\frac{\partial u}{\partial \nu}(x_0) \ge -\varepsilon \frac{\partial v}{\partial \nu}(x_0) = 2 \varepsilon \lambda R e^{-\lambda R^2} > 0$.

The interior ball condition is needed. Let $U$ be the part of the unit disc in the first quadrant, $x_0 = 0$ and $u(x, y) = -xy$. Then $\Delta u = 0$ and $u < 0 = u(x_0)$ in $U$, but $\nabla u(0) = 0$, so every directional derivative at $0$ vanishes. At the corner $0$ there is no ball inside $U$ touching $0$.

Going deeper

For a student

The Hopf lemma is the step that turns the weak maximum principle into the strong one. If an interior maximum occurred, one would find a small ball touching the set where the maximum is attained. The lemma would then give a nonzero gradient at a maximum point, which is impossible. It also gives uniqueness up to constants for the Neumann problem, and it is a basic tool in the moving plane method for proving symmetry of solutions. The smoothness of the boundary in the form of the interior ball condition holds, for example, for every bounded open set with $C^2$ boundary.