Euler identity for Clarke subgradients of homogeneous margins

If $q$ is differentiable at $x$, then $q$ is differentiable at $cx$ and the homogeneity identity gives $Dq(cx)[v]=c^{L-1}Dq(x)[v]$ for every direction $v$, by writing $q(cx+tv)=c^{L}q(x+(t\mathrm{/}c)v)$ and taking the difference-quotient limit as $t\to 0$.

At every differentiability point $x\mathrm{\ne }0$, differentiating $q(cx)=c^{L}q(x)$ with respect to $c$ at $c=1$ gives $⟨x,\mathrm{\nabla }q(x)⟩=Lq(x)$.

Conversely, differentiability at $cx$ implies differentiability at $x$ by applying the same argument with scale $1\mathrm{/}c$. Thus scaling by $c$ bijects the differentiability points and sends each gradient to $c^{L-1}$ times that gradient. Taking all gradient limits as $x\to \theta$, and then their closed convex hull as in [F1], proves ${\mathrm{\partial }}^{\circ }q(c\theta )=c^{L-1}{\mathrm{\partial }}^{\circ }q(\theta )$.

For any gradient limit $h={\lim }_k\mathrm{\nabla }q(x_k)$ with differentiability points $x_k\to \theta$, continuity of $q$ gives $⟨\theta ,h⟩=Lq(\theta )$. This identity is preserved by convex combinations and limits, so [F1] yields it for every $h\in {\mathrm{\partial }}^{\circ }q(\theta )$.