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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.

For an absolutely one-homogeneous functional $J$ that is convex and $p\in \partial J(u)$,The subgradient inequality at zero gives $⟨p,u⟩\ge J(u)$, and at $2u$ gives the reverse inequality. This is an Euler identity that needs no differentiability.

For a generalized singular vector, exact data $f=\gamma K{u}_{\lambda }$, and fixed $\alpha >0$, a compatible branch of Bregman iteration has $p^k=\min \{k\gamma /\alpha ,\lambda \}{K}^{\ast }K{u}_{\lambda }$ and the displayed $u^k$. Verify these formulas with the subgradient characterization of an absolutely one-homogeneous functional; before activation the dual coefficient remains at most $\lambda$, and after activation it is $\lambda$. The branch reaches $\gamma u_{\lambda }$ at $k_{\ast }=\lceil \alpha \lambda /\gamma \rceil +1$. Every minimizer branch has the same predicted data, since the quadratic data fidelity is strictly convex in $Ku$. Exact recovery of this particular vector for arbitrary branch choices needs uniqueness, for example injectivity of $K$. For $K(x,y)=x$ and $J(x,y)=|x|$, the second coordinate of every minimizer is arbitrary.

Let $K:U\to V$ be a bounded linear operator from a real Banach space to a real Hilbert space, and let $J$ be a proper lower-semicontinuous convex absolutely one-homogeneous functional on $U$. The normalization and source relation imply $\lambda =J(u_{\lambda })$ by the Euler identity for a convex one-homogeneous functional. For the norm penalty $J(u)=\Vert u\Vert$ on a Hilbert space, its nonzero-point subgradient is $u/\Vert u\Vert$. Hence $K^{\ast }K{u}_{\lambda }=u_{\lambda }/\Vert u_{\lambda }{\Vert }^2$: $v=u_{\lambda }/\Vert u_{\lambda }\Vert$ is a right singular vector with singular value $\sigma =1/\Vert u_{\lambda }\Vert$, and $K{u}_{\lambda }$ is the corresponding left singular vector. Thus the definition recovers the ordinary operator singular-vector relation using the one-homogeneous norm penalty. Uniqueness of the vector from its data requires extra hypotheses on $K$ and $J$.

Use the standard bounded linear operator setting for $K:U\to V$, so the adjoint operator $K^{\ast }:V\to U^{\ast }$ in the updates is defined on every datum. Put $q=K^{\ast }K{u}_{\lambda }$. The generalized singular vector relation is $\lambda q\in \partial J(u_{\lambda })$ with $\Vert K{u}_{\lambda }\Vert =1$ and $\lambda =J(u_{\lambda })\ge 0$. For an absolutely one-homogeneous functional, a subgradient $p$ at $u$ is characterized by $⟨p,u⟩=J(u)$ and $⟨p,v⟩\le J(v)$ for all $v$. Consequently $\lambda q$ is also a subgradient at every positive multiple of $u_{\lambda }$, and $bq\in \partial J(0)$ for every $0\le b\le \lambda$.

Use the initialization $u^0=p^0=0$ and start the first update at $k=0$; the PDF lists $k=1,2,\dots$ despite specifying only index-zero initial data. Suppose $p^k=b_{k}q$ with $0\le b_k\le \lambda$. DefineIf it is positive, $u^{k+1}=c_{k+1}{u}_{\lambda }$ and $p^{k+1}=\lambda q$ satisfy the optimality equation and the dual update. If it is zero, take $u^{k+1}=0$ and $p^{k+1}=(b_k+\gamma /\alpha )q\in \partial J(0)$. Thus this compatible branch hasInduction proves the finite termination of Bregman iteration on a singular vector formulasOnce the dual coefficient reaches $\lambda$, the next reconstruction is $\gamma u_{\lambda }$. Therefore a guaranteed stopping index for this branch isThe formulas also cover $\lambda =0$, where exact reconstruction occurs at the first update. The subgradient conditions are sufficient for global minimization because each objective is convex.

There is a genuine uniqueness issue in the printed claim about arbitrary iterates. Strict convexity of data fidelity in $Ku$ ensures that any two minimizers at a given step have the same image under $K$. Induction therefore makes every minimizer branch have the same dual variables and attain exact data at the stated finite index. It does not make their states equal to the specified $\gamma u_{\lambda }$ if $K$ has a null space.

For a counterexample take $U={\mathbb{R}}^2$, $V=\mathbb{R}$, $K(x,y)=x$, $J(x,y)=|x|$, $u_{\lambda }=(1,0)$, and $\lambda =1$. Every step leaves $y$ completely unconstrained. Choosing the minimizing state's second coordinate to be $k$ gives valid updates and subgradients but never reaches $(\gamma ,0)$ and does not converge as a state sequence. Thus finite exact-data recovery holds for all minimizer branches; exact recovery of the stated vector needs uniqueness or the aligned branch. If $K$ is injective, equal reconstructed data identify the state, and the literal vector conclusion follows.

Let $B=B(0,R)$, with $R>0$. Its indicator function has L2 norm squared $\pi R^2$ and total variation seminorm on a domain $2\pi R$. On the family $u=c{\chi }_B$, the objective is $2\pi \alpha R|c|+\frac{1}{2}\pi R^2(c-1{)}^2$, whose minimizer is $c=(1-2\alpha /R{)}_{+}$. To prove global optimality, a total variation calibration is needed.

Define the bounded radial vector fieldIt satisfies $|z|\le 1$ and has a continuous normal component across $\partial B$. Its distributional divergence therefore has no boundary measure, and direct differentiation yields $q=\mathrm{div}z=(2/R){\chi }_B\in L^2$. Although $z$ is not compactly supported, multiply it by a smooth radial cutoff equal to one through radius $T$ and zero beyond $2T$. The extra divergence has magnitude $O(R/T^2)$ on an annulus of area $O(T^2)$, hence L2 norm $O(R/T)$. Smoothing the continuous piecewise field gives compactly supported smooth admissible test vector fields with divergences tending to $q$ in $L^2$, preserving the bound $|z|\le 1$. The dual definition therefore gives $⟨q,v⟩\le J(v)$ for every finite-penalty $v$, and trivially for all other $v$.

Moreover $⟨q,{\chi }_B⟩=(2/R)\pi R^2=2\pi R=J({\chi }_B)$. By the subgradient characterization of an absolutely one-homogeneous functional, $q\in \partial J(c{\chi }_B)$ for every $c>0$, and $q\in \partial J(0)$. If $0<\alpha <R/2$, choose $c=1-2\alpha /R$; then $(g-c{\chi }_B)/\alpha =q$. If $\alpha \ge R/2$, choose $u=0$; then $g/\alpha =(R/(2\alpha ))q$ belongs to $\partial J(0)$, since multiplying the dual bound by a number in $[0,1]$ preserves it. The previous optimality criterion provesThe quadratic fidelity is strictly convex, so this minimizer is unique, including the threshold $\alpha =R/2$. The disk retains its radius on the positive branch and disappears on the zero branch; this is total variation denoising of a disk.

An absolutely one-homogeneous functional satisfieswith $J(0)=0$. The printed strictly positive codomain cannot hold at zero; the standard interpretation is a nonnegative, possibly extended-real functional.

For $p\in \partial J(u)$, apply the subgradient inequality at $v=0$:so $⟨p,u⟩\ge J(u)$. Apply it at $v=2u$ and use one-homogeneity:which gives the reverse inequality. ThereforeThis is the Euler identity for a convex one-homogeneous functional; differentiability is unnecessary.

For an absolutely one-homogeneous functional $J$, the subdifferential has the characterizationIf $c>0$, then $J(cu)=cJ(u)$, soThe second condition does not involve the base point. Therefore

For an absolutely one-homogeneous functional $J=\mathrm{TV}$ on a real Hilbert space, a subgradient $q$ at $u$ is characterized by $⟨q,v⟩\le J(v)$ for every $v$ and $⟨q,u⟩=J(u)$. A bounded vector field with $|z|\le 1$ and $q=\mathrm{div}z$ proves the inequality through the dual definition, provided cutoff approximation is valid in the ambient space. Equality is called a calibration. It certifies a global minimizer of total variation denoising, rather than only a minimizer within a chosen ansatz.