 Solution

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.