Lemma: Differentiable norm gives a C 1 unit boundary

If a norm $\nu$ is differentiable at every nonzero point, then $\nabla \nu$ is continuous there and $\{\nu =1\}$ is a $C^1$ curve. This does not imply strict convexity of the unit ball.

Proof. The gradients are locally bounded because a norm is Lipschitz. If $x_j\to x\ne 0$ and a subsequence of $\nabla \nu (x_j)$ converges to $p$, the supporting inequalities

pass to the limit. Thus $p$ is a subgradient at $x$. Differentiability makes it unique, namely $\nabla \nu (x)$. Every convergent subsequence has this limit, proving continuity.

Positive homogeneity gives $\nabla \nu (x)\cdot x=\nu (x)$. The gradient is therefore nonzero on $\{\nu =1\}$, and the $C^1$ implicit-function theorem applies.

For the last assertion, take the Minkowski functional of a centrally symmetric stadium: a disk plus a nontrivial horizontal segment. Its boundary is $C^1$ but has horizontal flat segments. The unique-supporting-line criterion gives a differentiable Minkowski functional.