Proposition: Correct cone-promotion boundary

For any norm $\nu$ on ${\mathbb{R}}^2$,

is a three-dimensional closed proper convex cone. If its base is strictly convex, it is strictly convex in the usual proper-face sense. The dimension requirement in the Itô–Lourenço classification is therefore satisfied after this promotion. The missing hypothesis is homogeneity, not dimension.

Strict convexity and a $C^1$ base alone do not imply that $K_{\nu }$ is linearly isomorphic to a Lorentz cone.

Proof. Convexity follows from the triangle inequality and homogeneity of $\nu$. The cone is closed, pointed, and has nonempty interior. A nontrivial higher-dimensional proper face would cut the section $t=1$ in a nontrivial boundary segment; strict convexity excludes it.

The classification of strictly convex homogeneous proper cones requires homogeneity ItoLourenco. No such property follows from the preceding facts.

For a counterexample, take $\nu (x)=({x_1}^4+{x_2}^4{)}^{1/4}$. Its unit ball is smooth and strictly convex. The associated $4$-cone is not homogeneous and is not a Lorentz cone up to linear isomorphism ItoLourenco.