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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.

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.

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.

For a convex functional $J:U\to \mathbb{R}\cup \{+\infty \}$ on a real Banach space, its subdifferential at $v$ is empty unless $J(v)$ is finite; otherwiseGiven $p_v\in \partial J(v)$, the Bregman distance isIt can be infinite if $J(u)$ is infinite. If $J$ is differentiable at $v$, the choice is its derivative; otherwise the base subgradient must be specified. Given also $p_u\in \partial J(u)$, add both directional distances to obtain the symmetric Bregman distanceThe symmetric version here is the sum, not half the sum. These nonnegative quantities are not generally metrics and can vanish at distinct points.

Because $w_k\ge c>0$, a finite sum ${\sum }_k{w}_k|u_k|$ implies $u\in {\ell }^1$. Thus the printed weighted one-norm on l2 equals the extended nonnegative sum on every $u\in {\ell }^2$. Unbounded weights may make that sum infinite for some $u\in {\ell }^1$; membership in ${\ell }^1$ alone is not a finiteness assertion.

For each $N$, $J_N(u)={\sum }_{k=1}^N{w}_k|u_k|$ is continuous in the l2 sequence space, and $J={\sup }_N{J}_N$. If $u^{(n)}\to u$ in norm, then for every $N$,Taking the supremum over $N$ provesIt is also weakly lower semicontinuous by the same finite-coordinate argument, and convex and proper because zero and finite sequences have finite penalty.

At a finite-penalty base point $v$, coordinate variations give the subdifferential of a weighted one-norm on l2:These conditions are sufficient by summing the coordinate supporting inequalities. Since $w_k\ge c$, an infinitely supported $v$ would require infinitely many $|p_k|\ge c$, contradicting $p\in {\ell }^2$. Thus a finite-penalty point can have empty subdifferential; a Bregman base with an actual subgradient must be finitely supported.

For any such $p$, $⟨p,v⟩=J(v)$ and the Bregman distance simplifies toEach summand is nonnegative; the sum may be infinite. The pairing converges by Cauchy-Schwarz inequality. This definition does not silently assume a subgradient exists at every $v\in {\ell }^1$.

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.

For a convex function $f$, its subdifferential at $x$ is the set of supporting slopesThe subgradient optimality condition isIndeed, the defining inequality with $g=0$ is exactly the global minimum inequality. For the subdifferential of the L1 norm, the coordinates vary independently:

Let $r=Y-X{\hat{\beta }}_{\lambda }$. The Karush-Kuhn-Tucker conditions for Lasso give a vector $z\in \partial \Vert {\hat{\beta }}_{\lambda }{\Vert }_1$ such that $X^{T}r/n=\lambda z$. If ${\hat{\sigma }}_{\lambda }=\Vert r{\Vert }_2/\sqrt{n}>0$, the unsquared residual norm is differentiable at this fit, with gradientUsing the subdifferential sum rule, choose $\gamma (\lambda )=\lambda /{\hat{\sigma }}_{\lambda }$ to obtainConvexity now proves the tuning relation between Lasso and square-root Lasso:The positive residual assumption is needed both for this gradient and for the quotient defining the tuning parameter.

For the final score calculation, use the usual fixed-design interpretation of the normal linear model: regard $X,Z$ and the tuning parameter $\gamma$ in the regression of $Z$ as fixed, and choose that regression's minimizer using only those inputs. Its nonzero residual $R=R_{\gamma }$ then has the Square-root Lasso optimality conditionExpanding the response residual givesSince $R/\Vert R{\Vert }_2$ is a fixed unit vector and $\epsilon \sim N_n(0,{\sigma }^2{I}_n)$, a linear image of a multivariate normal vector gives $W\sim N(0,{\sigma }^2)$. Finally, Holder inequality and the residual score bound yieldThis is the residual-score decomposition from square-root Lasso. The normality statement also holds conditionally on a design independent of the noise. If $\gamma$ were chosen from the response noise, for example through the preceding $\gamma (\lambda )$ formula, the direction $R$ would require a separate independence argument; the final calculation treats $\gamma$ as fixed.

For probability measures $\mu ,\nu$ on ${\mathbb{R}}^d$ with finite second moments, the Knott–Smith optimality criterion states that a transport plan $\pi \in \Pi (\mu ,\nu )$ minimizes the quadratic cost $|x-y{|}^2$ if and only if there is a proper convex function $\Phi :{\mathbb{R}}^d\to \mathbb{R}\cup \{+\infty \}$ that is sequentially lower semicontinuous and satisfiesThe subdifferential is characterized bywith $\Phi (x)<\infty$. Thus the transport plan is concentrated on the graph of the subdifferential. No absolute continuity of measures assumption on $\mu$ is needed. Multiplying the cost by $1/2$ leaves the criterion unchanged. The original quadratic optimal mapping result is Knott and Smith, On the optimal mapping of distributions.

A $J$-minimizing solution is an exact solution $u_J^{†}$ satisfyingThe source condition in variational regularization says that some $p^{†}\in Y$ satisfieswhere $\partial J$ is the subdifferential.

Fix $\alpha >0$. The subgradient optimality condition forat $u_J^{†}$ isIf the source condition holds, choose $g=f+\alpha p^{†}$. Since $A{u}_J^{†}=f$, the displayed inclusion holds. Because the objective is a convex function, this condition is necessary and sufficient for global minimality.

Conversely, if the stated range condition holds for some $g$, optimality supplies $\xi \in \partial J(u_J^{†})$ withThus $\xi =A^{\ast }((g-f)/\alpha )$, which is precisely the source condition. This proves the equivalence.

For a convex function $f$, its subdifferential isFor $f(x)=\gamma |x|$,The defining inequality shows directly that $0\in \partial f(x)$ exactly when $x$ is a global minimizer.

Strict convexity meansfor distinct $x,y$ and $0<\lambda <1$. Two distinct minimizers would make the midpoint have a strictly smaller value, so a minimizer is unique.

A vector $g$ is a subgradient of the convex function $f$ at $x$ whenfor every $y$. The set of all such vectors is the subdifferential $\partial f(x)$. The proximal operator satisfiesMore generally, $u={\mathrm{prox}}_{tf}(x)$ exactly when $(x-u)/t\in \partial f(u)$.

A map $T$ is a firmly nonexpansive mapping whenfor all $x,y$. Put $p={\mathrm{prox}}_f(x)$ and $q={\mathrm{prox}}_f(y)$. Then $x-p\in \partial f(p)$ and $y-q\in \partial f(q)$. Monotonicity of the subdifferential giveswhich rearranges to

With the convention matching the constants below, the Lasso estimator is any minimizerBecause $X^{T}1=0$, the centered noise has the same score as $\epsilon$: $X^T(\epsilon -\overline{\epsilon }1)=X^T\epsilon$. Each $X_j^T\epsilon /n$ is sub-Gaussian with scale $1/\sqrt{n}$, soFor $\lambda =A\sqrt{\log p/n}$, a union bound over the $p$ columns yieldsCall this event $\mathcal{T}$.

The Karush-Kuhn-Tucker conditions for the Lasso say that there is $\hat{z}\in {\mathbb{R}}^p$ such thatEquivalently, $\hat{z}$ belongs to the subdifferential of the ${\ell }^1$ norm at $\hat{\beta }$.

The differential of the log-determinant is $d\log \det \Omega =\mathrm{Tr}({\Omega }^{-1}d\Omega )$. The subdifferential of the entrywise ${\ell }^1$ norm consists of symmetric matrices $Z$ withThe Karush-Kuhn-Tucker conditions for the Graphical Lasso are thereforeBecause $-\log \det \Omega$ is strictly convex on the positive-definite matrices, these conditions characterize the unique minimizer whenever it exists.

Write $a_i^T$ for row $i$ of $A$, and define affine functionsThenThe pointwise maximum of convex functions is convex, so $f$ is a convex function. Moreover,so $f$ has Lipschitz continuity. The simpler bound $L\le \Vert c{\Vert }_2+M{\max }_i\Vert a_i{\Vert }_2$ is also valid.

Let $I(x)=\{i:{\ell }_i(x)=f(x)\}$ be the active set. The subdifferential isthe convex hull of all active slopes. In particular, choosing any active index gives the subgradientAt a tie, every convex combination of the tied slopes is also valid.

For a convex function $f:{\mathbb{R}}^d\to \mathbb{R}$, its subdifferential at $x$ isIts elements are the subgradients of $f$ at $x$.

An element $u^{†}$ is a $J$-minimizing solution whenIt satisfies the source condition in variational regularization when there is a $w^{†}\in Y^{\ast }$ such thatwhere $\partial J(u^{†})$ is the subdifferential of the convex functional $J$.