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A compact convex set $K\subseteq {\mathbb{R}}^n$ is a compact set that is also a convex set. If $K$ is nonempty, its support function $H_K(\eta )={\max }_{x\in K}x\cdot \eta$ is finite in every direction. For such a nonempty $K$, every exterior point $x_0$ admits a strictly separating hyperplane: minimize $|x_0-y|$ over $y\in K$, and put $\omega =x_0-y$. Convexity and differentiation along the segment from $y$ to any $z\in K$ give $\omega \cdot (z-y)\le 0$, while $\omega \cdot (x_0-y)=|\omega {|}^2>0$. This is the separation used to recover spatial support from Fourier growth.

For a centrally symmetric convex body in John position, there are contact points $u_j\in \partial K\cap S^{d-1}$ and positive weights $c_j$ such that ${\sum }_j{c}_j{u}_j\otimes u_j=I$. Taking the trace gives ${\sum }_j{c}_j=d$. A supporting hyperplane at $u_j$ shows $|⟨u_j,x⟩|\le \Vert x{\Vert }_K$.

Here is the maximum-volume argument for the decomposition. If $I$ were outside the convex cone generated by the contact matrices, a separating hyperplane would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the trace remains positive. The support function of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have trace $1$ and form a compact set; finite-dimensional convexity gives a finite decomposition.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

The hard-margin constraints are feasible exactly when the classes are separated by a separating hyperplane. Failure after the corrections in part a therefore means the classes overlap.

Introduce slack variables of a support vector machine and solve the soft-margin problemsubject toLarge $C$ strongly penalizes violations and approaches the hard-margin solution when separation is possible. Small $C$ tolerates more violations in exchange for a wider, more strongly regularized margin.

A separating hyperplane for signed data satisfies $Y_i({\gamma }_0+X_i^T\beta )>0$ for every $i$; its geometric set is $\{x:{\gamma }_0+x^T\beta =0\}$. The plot marks three support vectors. At the shown fit they lie on the two support-vector-machine margin boundaries, so their signed functional margins satisfyThe solid line is the decision hyperplane $X^{\ast T}\hat{\gamma }=0$, while the dashed parallel lines are $X^{\ast T}\hat{\gamma }=1$ and $X^{\ast T}\hat{\gamma }=-1$.

If $\lambda =0$, every parameter vector with all margins at least one has zero hinge loss. Scaling or changing a separating vector can therefore give another minimizer, so the objective need not select the displayed maximum-margin direction or the same three lines. Positive quadratic regularization selects a finite, minimum-norm compromise.