= Solution
The relevant nonlinear <generalized singular vector> relation, without imposing unit-data normalization, is
$$
p=\lambda K^*Ku^\dagger\in\partial J(u^\dagger),\qquad u^\dagger\ne0.
$$
This 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 $v=\lambda u^\dagger$. For a <convex positively one-homogeneous functional>, the <subgradient inequality> applied at zero and at $2u^\dagger$ gives $\langle p,u^\dagger\rangle=J(u^\dagger)$. If $Ku^\dagger\ne0$, then
$$
\lambda=\frac{J(u^\dagger)}{\|Ku^\dagger\|^2}.
$$
Assume the usual positive eigenvalue $\lambda>0$ and $0<\alpha\lambda\leq1$. By <positive homogeneity>,
$$
D_J^p((1-\alpha\lambda)u^\dagger,u^\dagger)=0.
$$
The <shifted-comparator Bregman bound> therefore yields
$$
\boxed{D_J^p(u_\alpha,u^\dagger)=0,\qquad Ku_\alpha=(1-\alpha\lambda)Ku^\dagger.}
$$
The vector $(1-\alpha\lambda)u^\dagger$ 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 $u_\alpha=(1-\alpha\lambda)u^\dagger$. 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 $(1-\alpha\lambda)_+u^\dagger$ 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.
\b[Zero Bregman distance does not imply exact recovery.] For $K=I$, $J(s)=|s|$, $u^\dagger=a>0$ 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>.
Back to article page
