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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.

Under the finite-valued assumptions, subgradients of an infimal convolution are obtained as its attained dual optimizers. Equivalently $z\in \partial (h^{\ast }+k^{\ast })(p)$. Splitting the latter subdifferential into two separate ones needs a subdifferential sum rule qualification; solving the aggregate dual objective avoids that extra assumption.

Compute the convex conjugate of $f_z$ at a general pair $(v,y)$. With $w=z+u-x$, the independent variables become $(x,w)$ and $u=w+x-z$, soTherefore the Lagrange dual function isUsing strong duality from the preceding solution givesEquivalently the conjugate of an infimal convolution is $F^{\ast }=k^{\ast }+h^{\ast }$, and the continuous convex $F$ equals its biconjugate.

For practical subgradient computation, minimize the known convex dual objectiveEvery optimizer, and only an optimizer, belongs to $\partial F(z)$ by Fenchel–Young inequality. The full characterization isThis is infimal-convolution dual subgradients; the set is nonempty because $F$ is finite convex everywhere. If both conjugates are differentiable at the optimizer, solve $\nabla k^{\ast }(y)+\nabla h^{\ast }(y)=z$. For nonsmooth conjugates, use the displayed aggregate subdifferential or a convex minimization algorithm. Replacing it by $\partial k^{\ast }(y)+\partial h^{\ast }(y)$ requires the usual subdifferential sum rule qualification; it is not automatically justified solely by knowing the two conjugates.

The conjugate of an infimal convolution is the sum of the conjugates:Here $g^{\ast }(p)=\frac{1}{2}\Vert p{\Vert }_2^2$, and $h^{\ast }(p)={\delta }_{[-1,1{]}^n}(p)$, the indicator functional of the unit infinity-norm ball. ThusThe infimal convolution is finite convex and continuous, so the Fenchel-Moreau theorem applies without a closure defect. By equality in the Fenchel–Young inequality and the subdifferential sum rule,This is precisely the variational characterization of projecting $x$ onto the cube. ConsequentlyFor the primal split, $y_i=\mathrm{clip}(x_i,-1,1)$ and $z_i=\mathrm{sign}(x_i)(|x_i|-1{)}_{+}$. These directly minimize the two scalar terms. The Huber loss is continuously differentiable, including at $x_i=\pm 1$, but its second derivative changes there. The one-dimensional sketch shows a quadratic center joined tangentially to linear tails:

Applied to a discrete gradient, the Huber gradient regularizer penalizes small slopes quadratically and large slopes linearly. Compared with pure squared-gradient smoothing it preserves large edges better; compared with pure total variation denoising it encourages small smooth variations and reduces the strong preference for piecewise-constant plateaus. It can therefore be useful for denoising signals or images containing both smooth regions and sharp transitions. It still penalizes edges and can bias their amplitude, and it does not guarantee complete elimination of staircasing in total variation denoising. The unit threshold must be scaled appropriately for data units and grid spacing.

Fix $x$ and minimize $J_x(y)=f(y)+\Vert y-x{\Vert }^2/(2\tau )$. A proper lower semicontinuous convex function has an affine minorant $f(y)\ge a\cdot y+b$, as follows by separating a point below its closed epigraph. HenceThe quadratic dominates the linear term, proving coercivity. Properness supplies at least one finite trial value, lower semicontinuity passes to limits, and finite-dimensional compactness makes a bounded minimizing sequence converge along a subsequence to a minimizer. Thus the proximal operator exists at every $x$.

The subdifferential sum rule applies because the quadratic is finite and continuous everywhere. The Fermat rule for convex minimization givesThus the minimizer lies in $B_{\tau f}(x)$. The uniqueness proved in part (a), or strict convexity of the quadratic sum, now yieldsThis proof displays the separate roles of properness, lower semicontinuity, convexity and finite dimension. In particular, compactness here is not inferred merely from strict convexity.

Use the conventionThis proximal map is also the resolvent of a monotone operator $(I+\tau \partial f{)}^{-1}$. A proper lower semicontinuous convex function has an affine minorant, so the quadratic term makes this minimization coercive and strongly convex. A unique minimizer exists for every $x$. Its subgradient optimality condition isThe Moreau–Yosida regularisation is $f_{\tau }=f\square g^{\tau }$. The conjugate of an infimal convolution and the quadratic conjugate giveThe factor $\tau$ here is essential. Apply subgradient inversion under convex conjugacy, followed by the subdifferential sum rule with the everywhere differentiable quadratic:The last equivalence is precisely the unique proximal minimization condition. ThusThis proves both existence and uniqueness of the subgradient, rather than only identifying a possible element. The finite convex function $f_{\tau }$ is therefore differentiable, with $\nabla f_{\tau }=(I-P_{\tau })/\tau$, the gradient of a Moreau envelope.

For completeness, monotonicity of $\partial f$ applied to the two proximal conditions givesHence $P_{\tau }$ is firmly nonexpansive. Expanding the same inequality shows that $I-P_{\tau }$ is firmly nonexpansive too. In particular, $\nabla f_{\tau }$ is $1/\tau$-Lipschitz continuous. None of this requires a bounded effective domain; the result applies to the next example as well.

Write the objective as $F_a(x)=\alpha x^2/2+n^{-1}{\sum }_i{\ell }_i(x)$ with ${\ell }_i(x)=\max \{0,1-a_{i}x\}$. Each hinge loss is finite and convex, so the subdifferential sum rule applies without a qualification gap. Its subdifferential isThis formula also covers $a_i=0$, if such a sample is allowed, because the loss is then constant. For a convex function, zero in the subdifferential is necessary and sufficient for global optimality. Thus $x$ is optimal exactly when there are weights ${\lambda }_i$ such thatEquivalently, using $I(x)=\{i:a_{i}x<1\}$ and $J(x)=\{i:a_{i}x=1\}$,The notation $[0,1]a_i$ denotes the segment between $0$ and $a_i$, also when $a_i<0$. This avoids reversing interval endpoints for negative-class samples.

Because $\alpha >0$, the objective is strongly convex, and $F_a(x)\ge \alpha x^2/2$ makes it coercive. It is continuous, so a minimizer exists and is unique for every finite data vector $a$. The weights at exact margins need not be unique even though $x$ is. These are the hinge-loss optimality weights for the one-dimensional support vector machine.

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.

Assume the displayed variational minimizer exists and $K:U\to V$ is bounded and linear. The smooth data-fidelity term and the subdifferential sum rule giveUse the supplied source condition in variational regularization $p^{†}=K^{\ast }w\in \partial J(u^{†})$. Put $e=f^{\delta }-f$ and $q=K(u_{\alpha }-u^{†})$, so $\Vert e\Vert \le \delta$. By the symmetric Bregman distance identity,Dropping the last square and using $\Vert e-\alpha w{\Vert }^2\le 2\Vert e{\Vert }^2+2{\alpha }^2\Vert w{\Vert }^2$ proves the source-condition estimate for symmetric Bregman distance:The source condition also makes $u^{†}$ a $J$-minimizing exact solution: for $Ku=f$, its subgradient inequality has zero pairing with $u-u^{†}$. Existence of $u_{\alpha }$ is not implied by properness and lower semicontinuity alone; it is a hypothesis when no coercivity/compactness condition is provided.

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 a proper lower-semicontinuous convex function $f$, its proximal operator isThe squared norm is strongly convex, so the minimizer is unique. The subdifferential sum rule gives the necessary and sufficient conditionMore generally,The subgradient inversion rule for the convex conjugate says $u\in \partial f(x)$ exactly when $x\in \partial f^{\ast }(u)$. Hencewhich is precisely the proximal optimality conditionSince $x=y-tu$, this proves the generalized Moreau decomposition

The differentiable loss has gradient $X^T(X\beta -Y)/n$. The subdifferential sum rule and part (b) givewhere blockwise

A proper convex function $f$ is minimized at $x$ exactly when $0\in \partial f(x)$. For a differentiable convex term $g$ and another convex term $h$, the subdifferential sum rule gives the condition $0\in \nabla g(x)+\partial h(x)$.