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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.

Suppose $A{u}^{†}=f$ and $p^{†}=A^{\ast }{\mu }^{†}\in \partial J(u^{†})$. For $F_{\alpha }(u)=\Vert Au-f\Vert +\alpha J(u)$, the subgradient inequality and dual pairing bound giveThus $0<\alpha \Vert {\mu }^{†}\Vert <1$ forces every minimizer to solve $Au=f$ exactly. Comparison of penalties then shows it is a J-minimizing solution, and $D_J^{p^{†}}(u,u^{†})=0$. The conclusion also holds for all $\alpha >0$ when ${\mu }^{†}=0$. One-homogeneity is not needed for this argument.

For a proper convex function finite at $x$, global minimality is equivalent to $0\in \partial f(x)$. This follows immediately from the defining subgradient inequality. Combined with a subdifferential sum rule, it converts a convex minimization problem into an inclusion, for example the equation defining a proximal operator.

Adding the two subgradient inequalities for $p\in \partial f(x)$ and $q\in \partial f(y)$ givesThus the subdifferential is a monotone operator. This inequality proves uniqueness of its resolvent: two solutions of $z\in x+\tau \partial f(x)$ must coincide when $\tau >0$.

The forward subgradient step and backward subgradient step are the set-valued mapsThus $u\in B_{\tau f}(z)$ means $z-u\in \tau \partial f(u)$. They are respectively the explicit and implicit time discretizations of subgradient flow $\dot{x}\in -\partial f(x)$. At a differentiable point the forward update is $x-\tau \nabla f(x)$, while the backward update evaluates the gradient at the new point. For proper closed convex data the latter is the proximal operator

To prove at most one output, suppose $u,v\in B_{\tau f}(z)$. Then $(z-u)/\tau \in \partial f(u)$ and $(z-v)/\tau \in \partial f(v)$. The two subgradient inequalities imply monotonicity of a convex subdifferential, givingSince $\tau >0$, $u=v$. Convexity supplies this monotonicity; lower semicontinuity is not needed for this at-most-one argument.

Under the full printed assumptions the output actually exists at every $z$. Properness ensures a finite point and excludes $-\infty$. Proper lower-semicontinuous convex $f$ has an affine minorant, so the quadratic proximal objective is coercive. Lower semicontinuity makes its minimum attained on a compact sublevel set. Convexity and the positive quadratic curvature make it strongly convex, hence its minimizer is unique. The subgradient optimality condition, with the differentiable quadratic term, is exactly $z-u\in \tau \partial f(u)$. ThusThis also identifies precisely which assumptions provide existence, uniqueness and the update's optimality interpretation.

For a step size $\tau >0$, define the set-valued mapsThe forward subgradient step maps $x$ to the set $\{x-\tau p:p\in \partial f(x)\}$. If $f$ is differentiable this is the explicit gradient step $x-\tau \nabla f(x)$. The backward subgradient step consists of the $y$ satisfying $x-y\in \tau \partial f(y)$, an implicit step for the subgradient flow. It is the resolvent of a monotone operator associated with $\partial f$.

Suppose $y_1,y_2\in B_{\tau f}(x)$. Then $p_i=(x-y_i)/\tau \in \partial f(y_i)$. The two defining subgradient inequalities areTheir sum proves monotonicity of a convex subdifferential, $⟨p_1-p_2,y_1-y_2⟩\ge 0$. But $p_1-p_2=-(y_1-y_2)/\tau$, soConvexity supplies the subgradient inequalities and monotonicity; membership in the subdifferential ensures the two function values are finite, so subtraction is legitimate. Positivity of $\tau$ supplies the decisive sign. Properness rules out the identically infinite and negative-infinity pathologies in the overall setting, but lower semicontinuity is not needed for this at-most-one argument. Its role is in existence, proved next. The backward step cannot have two values, though uniqueness alone has not yet shown its domain is all of ${\mathbb{R}}^n$.

Use the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty ,+\infty \}$. A proper convex function never takes $-\infty$ and is finite somewhere. Its effective domain isFor a proper extended-real function, convexity meansThe endpoint cases are identities; taking $0<t<1$ avoids the ambiguous product $0\cdot (+\infty )$. Equivalently, its epigraph is a convex set. In particular, the effective domain is a convex set. Properness is an additional condition: it rules out the identically $+\infty$ function and all $-\infty$ values, rather than requiring finiteness everywhere.

For $x\in \mathrm{dom}f$, the subdifferential is the set of supporting slopesEach element is a subgradient. Define $\partial f(x)=\varnothing$ outside $\mathrm{dom}f$. The defining subgradient inequality must hold globally, not merely in a neighborhood of $x$. For a differentiable finite convex function, $\partial f(x)=\{\nabla f(x)\}$.

For a proper convex function $E$ on a real Hilbert space, its subdifferential at $y$ isIn a Banach space, interpret the pairing with an element of the dual space. For a chosen subgradient $p\in \partial E(y)$, the Bregman distance isIf also $q\in \partial E(x)$, its symmetric Bregman distance isThe subgradient inequality proves nonnegativity. These quantities depend on the chosen subgradients and generally are not metrics: they need not distinguish different points or obey the triangle inequality.

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 $\lambda >0$ and labels $Y_i\in \{-1,1\}$, with squared-norm penalty normalized as $\lambda {\alpha }^{T}K\alpha$, the kernel soft-margin support vector machine solvesHere $(u{)}_{+}=\max (u,0)$ is the hinge loss building block, and $K_i$ is column $i$ of the symmetric kernel matrix. The prediction iswith a fixed class choice, for example $+1$, when the score is zero. Other positive rescalings of the penalty change numerical coefficients in the optimality equation but not the zero-coefficient conclusion.

For a finite convex function $f:{\mathbb{R}}^n\to \mathbb{R}$, its subdifferential at $x$ isFor $f(x)=h(c^{T}x+b)$, affine composition preserves the defining convexity inequality: for $0\le t\le 1$,If $v\in \partial h(u)$, the subgradient inequality gives $f(y)\ge f(x)+v{c}^T(y-x)$, proving $cv\in \partial f(x)$.

Conversely, let $g\in \partial f(x)$ and first suppose $c\ne 0$. For every $z\in \ker c^T$, the function is constant along $x+tz$. Applying the subgradient inequality for both signs of $t$ shows $g^{T}z=0$. Thus $g$ is in the span of $c$, or $g=cv$ with $v=g^{T}c/\Vert c{\Vert }_2^2$. Substituting $y=x+(a-u)c/\Vert c{\Vert }_2^2$ gives $h(a)\ge h(u)+v(a-u)$ for every real $a$. Hence $v\in \partial h(u)$. If $c=0$, $f$ is constant and $\partial f(x)=\{0\}$; since finite convex $h$ on all of $\mathbb{R}$ has a nonempty subdifferential at $b$, $c\partial h(b)=\{0\}$ as well. This handles the case in which the suggested orthogonal projection would be undefined. ThereforeThis proves the subdifferential under scalar affine composition identity with all its degenerate cases.

Let $m_i=Y_i(\hat{\mu }+K_i^T\hat{\alpha })$ be a fitted support-vector-machine margin. The subdifferential of the hinge loss yields choicesApplying the affine-composition identity and the subdifferential sum rule to the convex objective, its optimum satisfiesThe second equation is the intercept condition. Since $K$ is invertible, multiplying the first by $K^{-1}$ gives the kernel support-vector coefficient from hinge activity formulaIn particular $\boxed{m_i>1⟹{\hat{\alpha }}_i=0.}$ Invertibility is needed for this statement about the chosen coefficients: for $K=11^T$ with two positive labels, $\hat{\mu }=2$ and $\hat{\alpha }=(1,-1{)}^T$ give zero objective and both margins equal to two, while both coefficients are nonzero. A singular kernel matrix allows coefficient changes in the null space without changing the fitted function.

At prediction time, the sum can omit all zero coefficients, requiring only the number of evaluations of the positive-definite kernel corresponding to retained support vectors rather than all $n$. This need not remove the cost of forming or solving with a dense training kernel matrix. Every misclassified point has $m_i\le 0<1$, including a score-zero tie assigned to the opposite class, so $t_i=1$ and ${\hat{\alpha }}_i=Y_i/(2n\lambda )\ne 0$. Thus many classification errors imply many active support vectors and little prediction-time sparsity benefit. Correctly classified points inside the unit margin can also have nonzero coefficients; a point exactly on the margin need not have a nonzero coefficient.

Work in the real Hilbert space $L^2({\mathbb{R}}^2)$, with $g\in L^2$ and $\alpha >0$. The penalty $J$ is a proper convex function: it is finite at zero, and the bounded-variation space domain is convex. Its subdifferential at a finite-penalty $u$ consists of $q\in L^2$ satisfying $J(v)\ge J(u)+⟨q,v-u⟩$ for every $v\in L^2$.

If $q=(g-u)/\alpha \in \partial J(u)$, expand the quadratic term and use the subgradient inequality:Thus $u$ is the unique minimizer.

Conversely, let $u$ be a minimizer. Its penalty is finite since comparison with zero gives a finite objective. For any $v$ with $J(v)<\infty$, set $w_t=u+t(v-u)$, where $0<t\le 1$. Convexity gives $J(w_t)-J(u)\le t(J(v)-J(u))$. Minimality and quadratic expansion then implyLetting $t\downarrow 0$ yields $J(v)\ge J(u)+⟨(g-u)/\alpha ,v-u⟩$. The inequality is automatic if $J(v)=\infty$. HenceThis also follows from the subdifferential sum rule, since the quadratic term is everywhere continuous and differentiable, and the subgradient optimality condition. The direct proof above needs no unproved existence theorem.

On the entire plane, the printed global BV domain is not closed in the L2 geometry. The usual $BV({\mathbb{R}}^2)$ definition includes an $L^1$ condition. For $1<a\le 2$, $f(x)=(1+|x|{)}^{-a}$ belongs to $L^2$, has finite distributional variation $2\pi a{\int }_0^{\infty }r(1+r{)}^{-a-1}dr$, and fails to belong to $L^1$. The truncated $f_T=f{\chi }_{B(0,T)}$ lies in $BV\cap L^2$ and tends to $f$ in $L^2$. Its variation is the interior variation plus $2\pi T(1+T{)}^{-a}$, and tends to a finite limit. Thus $J(f_T)$ stays bounded but the literal $J(f)=\infty$, establishing failure of L2 closure of the global BV domain. One must not infer universal minimizer existence from an inapplicable closed-penalty proximal operator theorem. The standard closed extension uses the homogeneous bounded-variation space, allowing finite distributional variation without global $L^1$. The optimality equivalence just proved is valid for the literal penalty whenever a minimizer exists; the next datum has an explicit certified minimizer in its domain.

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 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.

The source condition in variational regularization at an exact solution $u^{†}$ is the existence of a source element ${\mu }^{†}\in {\mathcal{V}}^{\ast }$ such thatFor a Hilbert data space one identifies ${\mathcal{V}}^{\ast }$ with $\mathcal{V}$. If $u$ satisfies $Au=f$ and $A^{\ast }\mu \in \partial J(u)$, then every other exact solution $v$ obeys the subgradient inequalityHence $u$ is a J-minimizing solution. This is source-condition certification of a penalty-minimizing solution.

Let $u^{†}=u_J^{†}$, $p^{†}=A^{\ast }{\mu }^{†}\in \partial J(u^{†})$, and $r_{\alpha }=A{u}_{\alpha }-f$. The subgradient inequality, the dual pairing bound, and $A{u}^{†}=f$ giveMinimality of $u_{\alpha }$ in the exact penalty method givesCombining them yieldsThe coefficient is strictly positive, soNow minimality gives $J(u_{\alpha })\le J(u^{†})$, and exact feasibility plus the defining minimality of $u^{†}$ gives the reverse inequality. Thus $u_{\alpha }$ is a J-minimizing solution. Finally,This exact-penalty threshold from a source condition does not require one-homogeneity; convexity and the source subgradient suffice. If ${\mu }^{†}=0$, the same argument works for every $\alpha >0$, interpreting the upper threshold as infinite.

The stochastic gradients obeyThe nonexpansiveness of metric projection and the subgradient inequality imply, after conditioning on ${\beta }_s$,Summing telescopes. Since both ${\beta }_1$ and $\hat{\beta }$ lie in the radius-$R$ set $C$, their distance is at most $2R$. By the Jensen inequality and convexity of $f$,ChooseThe two terms are then equal, and

The proximal optimality condition givesThe subgradient inequality therefore yields, for every $u$,

The first-order subgradient inequality for a differentiable convex function givesand, after interchanging $v$ and $w$,Adding and rearranging yieldsThus the gradient $F=\nabla f$ is a monotone operator.