Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 326 / 4 / a

For a convex functional $J$ and $p\in \partial J(v)$, the generalized Bregman distance isThe 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 givecontradicting 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 $u\ge 0$,including $u=2\ne 1$. Thus 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.