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The relevant nonlinear generalized singular vector relation, without imposing unit-data normalization, isThis is a generalized eigenfunction in the forward-operator metric, rather than a distributional generalized eigenfunction of a linear operator. It supplies the normal-operator source condition with $v=\lambda u^{†}$. For a convex positively one-homogeneous functional, the subgradient inequality applied at zero and at $2u^{†}$ gives $⟨p,u^{†}⟩=J(u^{†})$. If $K{u}^{†}\ne 0$, thenAssume the usual positive eigenvalue $\lambda >0$ and $0<\alpha \lambda \le 1$. By positive homogeneity,The shifted-comparator Bregman bound therefore yieldsThe vector $(1-\alpha \lambda )u^{†}$ is itself a minimizer: $p$ remains a subgradient along the nonnegative ray, and substitution satisfies the subgradient optimality condition. If $K$ is injective, this also proves $u_{\alpha }=(1-\alpha \lambda )u^{†}$. Otherwise distinct minimizers can differ in the kernel while sharing the same predicted data.

For a nonnegative convex positively one-homogeneous functional, the branch continues as $(1-\alpha \lambda {)}_{+}{u}^{†}$ above the threshold, since $p/(\alpha \lambda )\in \partial J(0)$ there. Nonnegativity is needed for that clipping argument; bare positive homogeneity allows signed linear functionals.

Zero Bregman distance does not imply exact recovery. For $K=I$, $J(s)=|s|$, $u^{†}=a>0$ and $0<\alpha <a$, the minimizer is $u_{\alpha }=a-\alpha \ne a$, but $D_J^1(a-\alpha ,a)=0$. Both points touch the same supporting line, as described by zero Bregman distance and supporting faces.

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.