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If a smooth minimal surface equation for a graph solution on ${\mathbb{R}}^n\setminus \{0\}$ is homogeneous of degree one, with $n\ge 2$, then its graph extends to a plane through zero. Its gradient is homogeneous of degree zero, so $|Du{|}^2$ attains a global maximum on the unit sphere. The strong maximum principle for elliptic operators makes it constant. The differentiated equation in the gradient maximum principle for a minimal graph then forces $D^{2}u=0$, and degree-one homogeneity removes the affine constant. This argument applies to cones that are graphs of smooth functions away from their vertex; it is not a claim that every minimal surface cone is flat.

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.

The cone condition implies $u(\lambda x)=\lambda u(x)$, so $Du(\lambda x)=Du(x)$. The gradient is homogeneous of degree zero, and $v=|Du{|}^2$ attains its global maximum on the compact unit sphere. For $n\ge 2$ the punctured space is connected. Part (b) gives $\mathcal{L}v\ge 0$ for a locally uniformly elliptic operator with no zeroth-order term. The strong maximum principle for elliptic operators makes $v$ constant. HencePositive definiteness gives $D^{2}u=0$. Thus $u=a\cdot x+b$, and degree-one homogeneity forces $b=0$: $\boxed{u(x)=a\cdot x}$. This proves that a minimal graphical cone is a plane. Strictly, the graph restricted to $x\ne 0$ is the plane with its origin omitted; its closure is the full plane through the origin.

Assume $c\le 0$, bounded coefficients and uniform ellipticity near the boundary point, an interior tangent ball there, and enough regularity for the boundary normal derivative. The Hopf boundary point lemma says that a nonnegative maximum at $x_0$ with strictly smaller values inside the tangent ball satisfiesfor the outward unit normal. If $c=0$, the maximum can have either sign. Without existence of a classical derivative, one states a strict one-sided derivative bound.

The PDF also asks to deduce the strong maximum principle for elliptic operators; that request is missing from the TeX transcription. Suppose $v$ attains a nonnegative global maximum $M$ inside a connected domain but is not constant. A ball in $\{v<M\}$ can be chosen relatively compact in the domain and tangent to $\{v=M\}$ at an interior point $z$: use the distance from a nearby point to the closed maximum set. The Hopf boundary point lemma on that ball gives a nonzero outward derivative at $z$, contradicting $Dv(z)=0$ at an interior maximum. Therefore $v$ is constant. When $c=0$, the same proof works for a maximum of either sign.