Convex positively one-homogeneous functional ($J(tu)=tJ(u)(t\ge 0)$)

This is a proper convex function that has degree-one positive homogeneity, with $J(0)=0$. It need not be even or nonnegative; linear functionals are examples. Its subdifferential satisfiesThe two tests $z=0$ and $z=2u$ prove equality; the subgradient inequality then gives the global upper bound. Conversely that bound and equality prove the subgradient condition. Hence a subgradient at $u$ also supports every nonnegative multiple of $u$. Requiring $J(-u)=J(u)$ gives an absolutely one-homogeneous functional.

A nonzero $u$ satisfying this relation, with $Ku\ne 0$, generalizes the eigenfunction of an absolutely one-homogeneous functional to the quadratic geometry induced by the forward map. Pairing with $u$ yields $\lambda =J(u)/\Vert Ku{\Vert }^2$. With unit data norm it is a generalized singular vector. If $\lambda >0$, $0\le \alpha \lambda \le 1$ and the exact data are $Ku$, the vector $(1-\alpha \lambda )u$ minimizes the quadratic variational regularization objective. Substitute its raywise subgradient into the subgradient optimality condition. Every minimizing branch has the same predicted data, and injectivity of $K$ gives equality of vectors. The Bregman distance along this ray is zero even though the reconstruction is shrunk. For nonnegative $J$ the branch continues as $(1-\alpha \lambda {)}_{+}u$; nonnegativity supplies $0\in \partial J(0)$ for the threshold argument.

Background on these generalized singular vectors: Benning and Burger, Ground States and Singular Vectors of Convex Variational Regularization Methods.