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Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

The required linear program isThere are $2n$ inequality constraints and one equality constraint, all independent of $x$.

The feasible set is the capped simplex, a nonempty compact set. A linear function attains its maximum at an extreme point; the previous part identifies these as the indicators of $k$-element subsets. At such a vector, the objective is the sum of the selected coordinates. Maximizing selects the $k$ largest coordinates, including when some are negative or tied. Equivalently, $f_k$ is the support function called the sum of the largest components.

Put $\delta =\exp (-{\epsilon }^{-C})$, with the absolute constant $C$ chosen sufficiently large below, and suppose for a contradiction that no $g_2$ with $0\le g_2\le 1$ satisfies $\Vert g_1-g_2{\Vert }_{\mathcal{F}}\le \epsilon$.

LetThe set $K$ is compact and convex, while $B$ is closed and convex, so $K+B$ is closed and convex. We use the following finite-dimensional form of the Hahn-Banach separation theorem: if a point lies outside a nonempty closed convex set, there is a linear functional whose value at the point is strictly greater than its supremum over that set. Identifying linear functionals on ${\mathcal{H}}_N$ through the inner product, there is therefore a function $\psi$ such thatThe separating functional cannot have zero dual norm, so rescale it to make $\Vert \psi {\Vert }_{\mathcal{F}}^{\ast }=1$. Because $\mathcal{F}$ is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of $\Vert \cdot {\Vert }_{\mathcal{F}}^{\ast }$ is precisely $\mathcal{F}$. Thus $\psi \in \mathcal{F}$ and $|\psi (n)|\le 1$.

The support function of $K$ is obtained by choosing $g(n)=1$ where $\psi (n)>0$ and $g(n)=0$ where $\psi (n)<0$. In terms of the positive part of a real-valued function ${\psi }_{+}=\max (0,\psi )$, the separating inequality becomesSince $0\le g_1\le \nu$, pointwise we have $g_1\psi \le \nu {\psi }_{+}$, and hence

Apply the supplied polynomial approximation of the positive part to $\psi$. If $P_{\epsilon }(t)={\sum }_{i=0}^m{a}_i{t}^i$, then its uniform approximation error and $\mathbb{E}\nu \le 1$ implyThe constant function $1$ and $\psi$ belong to the dual unit ball. By the assumed submultiplicativity, $\Vert {\psi }^i{\Vert }_{\mathcal{F}}^{\ast }\le 1$ for every $i\ge 0$. Dual seminorm therefore givesThe stated coefficient bound, with $C$ in $\delta$ chosen larger than the absolute constant in that bound, makes this last quantity at most $\epsilon /50$. Together with the polynomial-approximation error, this contradicts $⟨\nu -1,{\psi }_{+}⟩>\epsilon$. The required $g_2$ consequently exists. This proves the dense model theorem for a multiplicative test family.

The function $\varphi (x)={\max }_{v\in C}⟨x,v⟩$ is the support function ${\sigma }_C$. For a nonempty compact convex set,so its convex conjugate is the indicator function ${\iota }_C$. Applying the Moreau decomposition,Multiplication of an indicator function by a positive scalar does not change it, and its proximal operator is the Euclidean projection onto a convex set. Therefore

Take $a=1$ and $b=k$, sois the capped simplex. A linear objective over this convex polytope attains its maximum at a zero-one extreme point. Choosing the $k$ coordinates at which $x_i$ is largest givesEquivalently, an exchange of weight from a smaller component to a larger one never decreases the objective. Thus the sum of the largest components is the support function ${\sigma }_C$.

Part c now givesBy the projection onto a box-constrained hyperplane, $v=P_C(y/t)$ hasConsequently the proximal operator is evaluated by solving this one-dimensional equation for $\nu$, then substituting the resulting projection.

Let $F$ be an entire function of one variable satisfying $|F(\zeta )|\le C(1+|\zeta |{)}^N{e}^{R|\mathrm{Im}\zeta |}$, and let $P$ be a nonzero polynomial. If $F/P$ is entire, it satisfies a bound of the same exponential type, with a possibly different polynomial power. Indeed, outside a sufficiently large disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^{\mathrm{deg}P}$; inside the disk the extended quotient is bounded. Combining these estimates proves the claim. The same reasoning works with the exact support function of a compact interval. Zeros of $F$ must cancel every root of $P$ with its full multiplicity.

For $C=D^{\ast }\{p:|p_i{|}_2\le \lambda \}$, discrete total variation denoising is the proximal operator of the support function of $C$. The Moreau decomposition gives the displayed projection residual. It is generally not itself a metric projection onto a closed convex set: such a projection is idempotent, but TV denoising shrinks a sufficiently large two-level jump again when reapplied.

Robust optimization chooses a decision while accounting for every parameter in a prescribed uncertainty set. A worst-case maximization takes the form ${\sup }_x{\inf }_{r\in \mathcal{U}}F(x,r)$ with the decision chosen before the uncertain parameter. With affine dependence on the uncertainty, the support function and linear programming duality often replace the inner optimization by explicit constraints.

Let a distribution have support in a compact convex set $K$. Fixed neighborhood cutoffs initially give an exponential bound larger than the support function $H_K$. Cutoffs of width $\epsilon$ cost powers of ${\epsilon }^{-1}$ in derivative estimates but enlarge the exponential by only $e^{C\epsilon |\mathrm{Im}z|}$. Choosing $\epsilon =(1+|z|{)}^{-1}$ absorbs the cutoff cost into a polynomial and retains the exact exponential $e^{H_K(\mathrm{Im}z)}$.

The sum of the $k$ largest components of $x\in {\mathbb{R}}^n$ is the support function of the capped simplex: ${\sum }_{i=1}^k{x}_{[i]}=\max \{x^{T}v:0\le v\le 1,{\sum }_i{v}_i=k\}$.

For a real symmetric matrix $X$ with eigenvalues ${\lambda }_1\ge \cdots \ge {\lambda }_n$, the sum of its $k$ largest eigenvalues is the support function of the fantope:Here $⪯$ is the Loewner order. In an orthonormal eigenbasis for $X$, the objective depends only on the diagonal entries of $Y$, which lie in the capped simplex. Thus the identity reduces to the sum of the largest components of the eigenvalues.

If $x\notin \mathrm{range}{P}^T=(\ker P{)}^{\perp }$, the support function is infinite along a kernel line. Otherwise linear programming duality gives the displayed minimum, attained because the corresponding linear programs are feasible with finite values. Splitting $q=\alpha -\beta$ into nonnegative parts yields the equivalent minimum of $1^T(\alpha +\beta )$. Thus worst-case revenue over a polyhedral uncertainty set is $r_0^{T}x-{\sigma }_C(x)$, with value $-\infty$ off the row space.

For a nonempty closed convex set $C$, the finite-valued support function has subgradients exactly at the maximizing points of $C$. This follows from Fenchel–Young inequality and the conjugacy between the support function and the indicator functional of a constraint set. For compact $C$, the exposed maximizing set is nonempty.

For a quadratic data term and divergence maps $A_1,A_2$, the dual constraint $A_{1}v\in D$ can be split through ${\delta }_D(A_{1}v)={\sup }_w[⟨w,A_{1}v⟩-{\delta }_D^{\ast }(w)]$. For a product of Euclidean row balls of radius $\beta$, its support function is ${\delta }_D^{\ast }(w)=\beta {\sum }_i\Vert w_i{\Vert }_2$. The resulting saddle coupling is $K(u,w)=A_2^{\ast }u-A_1^{\ast }w$. The Chambolle–Pock algorithm then uses a row-ball projection, a quadratic proximal operator and radial soft thresholding, with no projection onto an intersection involving a divergence operator.