A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

For a smooth hypersurface $Y=\{f=0\}\subset {\mathbb{CP}}^n$ of degree $n+1$, let $E=\sum z_i{\partial }_{z_i}$ and $\Omega =d{z}_0\wedge \cdots \wedge d{z}_n$. The meromorphic form $({\iota }_E\Omega )/f$ descends to projective space because its scaling weights cancel and it is horizontal. Its Poincaré residue is a nowhere-zero section of $K_Y$. On the chart $z_j=1$, with remaining coordinates $t_1,\dots ,t_n$ in original-index order and $F=f{|}_{z_j=1}$, it is $(-1{)}^{j+r-1}(d{t}_1\wedge \cdots \widehat{d{t}_r}\cdots \wedge d{t}_n)/F_{t_r}$ wherever $F_{t_r}\ne 0$. Thus $K_Y\cong {\mathcal{O}}_Y$.

An ordinal $\kappa$ is a cardinal number precisely when no smaller ordinal has a bijection onto it. Ordinalhood and the graph predicate for a bijection are bounded formulas in set theory, so $\mathrm{Ord}(\kappa )\wedge \forall f\forall \alpha \in \kappa \neg \mathrm{Bij}(f,\alpha ,\kappa )$ is a Pi-one formula modulo ZF. This does not assert that every set can be assigned an ordinal cardinal without the axiom of choice.

An infinite cardinal number $\kappa$ is a regular cardinal if every function from an ordinal $\delta <\kappa$ into $\kappa$ has bounded range. Write this as $\forall f\forall \delta \in \kappa (\mathrm{Map}(f,\delta ,\kappa )\Rightarrow \exists \beta \in \kappa \forall \xi \in \delta f(\xi )<\beta )$. The graph predicates are bounded formulas in set theory, and cardinalhood is Pi-one definable, so adding infinite cardinalhood yields a Pi-one formula modulo ZF.

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 $v={\mathrm{prox}}_J(u)$ and $p=u-v$. The proximal operator optimality condition gives $p\in \partial J(v)$. By subgradient inversion under convex conjugacy, $v\in \partial J^{\ast }(p)$. Since $v=u-p$, this is exactly the optimality condition defining $p={\mathrm{prox}}_{J^{\ast }}(u)$. Uniqueness of both Hilbert proximal minimizers proves Moreau decomposition:Both terms belong to the same Hilbert space after dual identification. No orthogonality of the two terms is asserted for a general convex $J$; that stronger property pertains to special indicator/cone cases.

By the definition of the subdifferential, $p\in \partial J(u)$ meansRearrangement bounds $⟨v,p⟩-J(v)$ by $⟨u,p⟩-J(u)$, and equality is attained at $v=u$. Taking the supremum in the definition of the convex conjugate gives $J^{\ast }(p)=⟨u,p⟩-J(u)$. Conversely this equality bounds every member of that supremum and rearranges to the subgradient inequality. ThereforeApply the same argument to $J^{\ast }$ and use the Fenchel-Moreau theorem $J^{\ast \ast }=J$. With the canonical Hilbert identification of the bidual, the equality is also equivalent to $u\in \partial J^{\ast }(p)$. This proves subgradient inversion under convex conjugacy:The conditions include finiteness at the points in question; expressions involving $+\infty$ are not subgradients merely by formal subtraction.

First use the usual Hilbert space convention for the proximal operator, identifying the dual element $z$ with its representing vector. For $c\ne 0$, completing the square givesSet $v=cx-y$ and multiply the objective by $c^2$. Its variable part becomes $\Vert v-[c(q-z)-y]{\Vert }^2/2+\alpha c^{2}J(v)$. Thus the proximal operator under affine rescaling isNegative $c$ is allowed. If $c=0$ and $J(-y)<\infty$, the penalty is constant and the answer is $q-z$. If $c=0$ and $J(-y)=\infty$, $E$ is identically infinite, so there is no proper proximal problem.

The PDF actually assumes an arbitrary Banach space, so the Hilbert input shift cannot be asserted literally under its hypotheses. Define ${\mathrm{Prox}}_F(s)=\arg {\min }_v\{F(v)+\Vert v-s{\Vert }^2/2\}$ as a possibly empty or set-valued minimizer set. Absolute homogeneity of the norm still gives the valid change of variablesThis retains the linear term, rather than shifting by a dual vector. Equivalently its optimality condition uses the duality mapping $\mathcal{J}=\partial (\Vert \cdot {\Vert }^2/2)$:For a concrete failure of the Hilbert shortcut, take $({\mathbb{R}}^2,\Vert \cdot {\Vert }_1)$, $J=0$, $q=y=0$, $c=1$ and $z=(1,1)$. Every $x_1,x_2\le 0$ with $x_1+x_2=-1$ minimizes $\Vert x{\Vert }_1^2/2+x_1+x_2$ with value $-1/2$. The naive shifted answer $(-1,-1)$ has value zero and is not a minimizer. Hence the simple shifted formula requires Hilbert geometry; general Banach proximal minimizers need not be unique. For $c=0$ with finite constant penalty the Banach condition is $-z\in \mathcal{J}(x-q)$, not generally $x=q-z$.

On a real Hilbert space, for a proper lower-semicontinuous convex function $E$, the proximal operator isThe quadratic makes this objective strongly convex and coercive; the affine lower bound available for a proper closed convex function and weak lower semicontinuity give existence. The minimizer is unique. The subdifferential optimality condition isFor a positive parameter $t$, ${\mathrm{prox}}_{tE}$ uses $tE(x)$ in the same formula. On a general Banach space, the analogous norm-squared minimization is Banach-space proximal minimization; it can be set-valued and the identity-valued Hilbert derivative must be replaced by the duality mapping.