= Solution
For a convex functional $J$ and $p\in\partial J(v)$, the <generalized Bregman distance> is
$$
\boxed{D_J^p(u,v)=J(u)-J(v)-\langle p,u-v\rangle\geq0.}
$$
The inequality is exactly the <subgradient inequality>.
Suppose $J$ is <strictly convex> and $u\ne v$, with both values finite. If the distance vanished, the supporting affine function at $v$ would also meet $J$ at $u$. At $w=(1-t)v+tu$, $0<t<1$, the subgradient inequality would give
$$
J(w)\geq J(v)+t\langle p,u-v\rangle
=(1-t)J(v)+tJ(u),
$$
contradicting strict convexity. Hence zero distance implies $u=v$.
Without strict convexity, take $J(s)=|s|$, $v=1$, and $p=1\in\partial J(1)$. For every $u\geq0$,
$$
D_J^1(u,1)=u-1-(u-1)=0,
$$
including $u=2\ne1$. Thus \b[zero Bregman distance need not identify a point]. It identifies points on the same supporting face, as explained by <zero Bregman distance and supporting faces>.
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