Compact forcing, moments and positive error kernels

Reconstructed by GPT-6 Astra (OpenAI), Ultra reasoning effort, 4 October 2026. Public domain (CC0).

A compact source can produce a noncompact potential. Moment conditions remove its exterior terms. We first prove this mechanism for coordinate primitives and ordinary differential equations, then for the planar Laplacian. Two explicit potentials show why the source at a singularity must be computed distributionally. Finally, a compact fourth primitive becomes a positive kernel for a quadrature error.

Pairings are complex-linear. Write , , and for compactly supported distributions. A compact distribution acts on a smooth function by inserting a smooth cutoff equal to one near its support; the value is independent of the cutoff. Its finite-order bound and this extension are proved in U008, Proposition 1.2, and U021, B0. U021, B1–B3, Theorems 1.1, 2.1, 3.1 and Corollary 4.2, proves all convolution, parameter, associativity and smoothness statements used here.

The scalar calculus and measure proofs are in the metric, Sections 12–13, and integration, Sections 15–16, foundations. The Fourier foundations, F3, supplies the complete Gaussian integral. The constant-coordinate theorem is U032, Lemma 1.3. Further exact earlier proofs are identified at their use below; the new moment, matrix, branch, endpoint and error-kernel arguments are given in full.

A compact primitive is decided by moments

Lemma 1.1 (lower-coordinate uniqueness). Let have support in . If a product of positive powers of any selected coordinate derivatives annihilates , then .

Proof. Suppose first . U032, Lemma 1.3, gives a distribution independent of that coordinate. Pair with a mass-one bump supported strictly below and an arbitrary test in the other coordinates. The pairing is zero by support and equals the pairing of the reduced distribution with that arbitrary test. Thus the reduced distribution, and , vanish. If , apply this argument successively to ; their supports retain the same bound. For a product, treat all but one factor as already applied to , remove the remaining factor by pure-coordinate uniqueness, and repeat. This proves the assertion for every selected subset and every positive order.

We also record directly that Here . Its derivative pairs as . For , integration by parts gives , since the positive power has zero value at the endpoint. Repetition proves the formula, including all signs.

Lemma 1.2 (polynomial moments determine compact distributions). If and for every multiindex , then .

Proof. In dimension zero, the constant monomial suffices. Otherwise let be a compact neighborhood of the support on which the action is controlled by derivatives through order . For fixed , put . The exponential series is a limit of polynomials in the controlling norm. Indeed, on , bound and its first two derivatives by . A derivative of order at most of is a sum of at most products, each bounded by a fixed geometric power . Thus whose series converges because the ratio tends to zero. The finite-order action therefore gives .

Set . Its mass is one by the supplied Gaussian proof. Consequently . For a compact smooth , each derivative of is bounded and uniformly continuous, and Split at . The tail is at most , tending to zero as . On the inner set uniform continuity makes the difference uniformly small as . Thus convergence holds uniformly for every derivative.

Compact parameter integration through , justified by U021 B1 with a fixed cutoff on , now yields The even Gaussian accounts for the reflected argument. Its derivatives on the product of and the compact -support are bounded, so the parameter interchange has precisely the common finite-order bound required there. All tests have zero pairing.

Theorem 1.3 (compact multiindex primitives). For and , a compact solution of exists exactly when The inequality is coordinatewise. The compact solution is unique and lies in every closed coordinate box containing . For , the condition is empty and .

Proof. Transposition proves necessity: if , then , and the sign is .

For an active coordinate , form the compact distribution on the remaining coordinates A fixed cutoff near defines the action, bounds its order, and puts its support in the projection of that compact set. Every polynomial moment of this distribution vanishes by hypothesis. Lemma 1.2, including its dimension-zero case, gives .

Suppose . Convolve with Then . Support addition preserves the bounds in all other coordinates and the lower bound . On the region , the positive-part kernel is an ordinary polynomial in . Its binomial expansion, paired with the compact input, is the distribution-valued function This is zero. The formula also holds on joint tests, by the compact parameter-pairing identity in U021 B1; the transverse delta simply identifies the other coordinates. Hence the primitive is supported in .

Repeat this integration in each active coordinate. To justify that the next moments still vanish, let be the compact result of one step and . Pairing compactly in coordinate commutes with derivatives in , so The left input is compact on the remaining coordinates. Lemma 1.1 makes it zero. Thus each successive step retains every still-needed partial moment and remains in . Commuting coordinate derivatives proves that the final primitive solves the stated equation. Two compact solutions differ by a distribution satisfying the product equation; choose lower bounds for its compact support and apply Lemma 1.1. Empty forcing and degenerate boxes cause no change to this argument.

Theorem 1.4 (all scalar constant-coefficient ODEs). For a complex polynomial and , a compact solution of exists if and only if For , the compact solution is unique and supported in the smallest closed interval containing .

Proof. The compact transpose identity is , with no complex conjugation. Take cutoffs equal to one on a common neighborhood of the support; their derivatives contribute zero there. This proves necessity.

Let , . Define its companion matrix with superdiagonal ones and last row , and . Use, for example, the maximum row-sum matrix norm; summing products gives . The series and every derivative converge uniformly on bounded intervals by the corresponding scalar exponential series. Hence , , and gives . Differentiating in likewise gives .

For , the proved step derivative and product rule yield . Thus satisfies . Its first rows give ; the last gives .

If , then below . Above , compact smooth-kernel pairing and the exponential product identity give Every scalar component of is an adjoint null solution. To verify this without assuming a matrix-polynomial theorem, observe that for , and the last row gives . Multiplying by and using its commutation with proves for every row, so . Therefore The moment hypothesis gives , making compact in .

For uniqueness, a homogeneous distributional solution has companion vector satisfying . The product rule gives , so U032, Lemma 1.3, gives a constant vector . Thus . If is compact, this vector is zero on an exterior interval; invertibility forces . Applying the same support conclusion to every containing interval gives the stated smallest interval. For zero forcing uniqueness gives the zero solution.

For nonzero constant , the solution is and the adjoint null space is zero. For , the moment condition against all smooth functions says ; exactly then every compact solves the equation. This covers repeated and nonreal roots without factorization.

A rough kernel can detect smoothness

Theorem 2.1 (one input used twice). If and, for a fixed , is smooth for every , then is smooth.

Proof. Set and choose a smooth compact cutoff near zero. The function belongs to . Leibniz' rule and give Each term in is supported away from zero, where the remaining positive-part derivatives are smooth. In the term , . All convolutions below are proper because the two kernel factors are compact. The exact identities in U021 therefore give The left side is smooth by hypothesis. The last two terms are smooth because they have compact smooth kernel factors. Solving this equality for proves the assertion, including .

Theorem 2.2 (continuous repeated mixed derivatives). If and is continuous for every integer , then .

Proof. Denote that continuous function by . Its distributional support is contained in ; continuity makes it zero on the complement, so is compactly supported. Let Tensor differentiation gives . Both and have lower coordinate support bounds: for this follows from its explicit integral Their difference solves the homogeneous mixed equation, so Lemma 1.1 gives .

For , differentiating the successive one-dimensional integrals yields At every differentiation a positive power reaches at worst zero; its preceding endpoint value is zero, so no boundary term occurs. On bounded -sets the compact -support and bounded polynomial factors give one integrable majorant. The integration indicators converge off finitely many coordinate hyperplanes, which have measure zero. Dominated convergence proves continuity of every displayed derivative. The one-dimensional fundamental theorem at each step then identifies them as classical derivatives. For any , take ; all multiindices of total order at most are permitted. Hence for every , with its original compact support.

Look at the potential outside its source

The exact sources and satisfy , by U020, Theorem 1.1. U031, Proposition 1.1 and Theorem 1.2, proves both compact inverse identities and that a distribution harmonic on an open set is smooth there.

Theorem 3.1 (both complex moment families). For , a compact solution of exists exactly when It is unique. The forcing may be complex, so the two families are separate conditions.

Proof. Necessity follows by transposition, since both polynomial families are harmonic. For sufficiency put . Choose with . When , the scalar logarithmic series gives The logarithmic series and its identification with the real logarithm follow from the exponential-series argument in U020, Corollary 1.2. On a fixed slightly enlarged input neighborhood still inside , each derivative introduces only a polynomial factor in ; the geometric ratio stays below one. The series therefore converges in the finite derivative norm controlling . Pairing term by term gives zero. Smooth compact-kernel evaluation identifies this pairing with the potential outside the source, so for , and is compact.

If is a compact null solution, the compact inverse identity immediately gives . Thus uniqueness does not require an additional energy or maximum theorem. Because pairings are complex-linear, conjugation of a moment of would instead give a moment of its conjugate distribution; it cannot replace the second family here.

Theorem 3.2 (the exact Newton far field). Suppose is harmonic outside a compact set, and its smooth exterior representative tends to zero at infinity. Then is compact and uniformly in direction.

Proof. Put . On any fixed compact neighborhood of the input support, homogeneity of differentiated kernels gives, uniformly for large , Indeed and each derivative has the stated degree and bounded spherical profile. The compact finite-order bound gives . The difference is globally harmonic, hence smooth by U031. It tends to zero at infinity. The maximum principle in U023, Theorem 3.4, applied on balls to , bounds every fixed value by the boundary supremum on arbitrarily large spheres. These tend to zero, so .

Uniformly on , . The positive-order bounds above tend to zero after multiplication by . Hence Pair with to get the claimed limit and its sign. Decay is essential: has but the rescaled potential diverges.

Sources hidden by an ordinary formula

Theorem 4.1 (a quadratic spherical principal value). Let be a complex symmetric -by- matrix, and for . The limit of the pairings over exists as a distribution exactly when . For that trace-free case, satisfies The sum is the full double sum, so both equal off-diagonal entries contribute.

Proof. U028, Solution 8, proves by spherical excision and integration by parts the precise identity The same proof establishes the spherical moments : reflections kill the off-diagonal moments, permutations equalize the diagonal ones, and their sum is the sphere area . Consequently

If this trace is nonzero, a test equal to one near zero has truncated pairing with term , which cannot converge. If the trace is zero, subtract the test's value at zero within the unit ball. The segment fundamental theorem bounds the difference by ; multiplication by the degree- kernel and polar volume gives the integrable bound . Outside that ball, up to the fixed test radius , the bound is . These bounds prove existence and distributional continuity for exactly spherical truncations.

Contract the proved Hessian identity with . Both terms involving its trace vanish, leaving Apply and use the normalized point source to obtain the first assertion. Convolving that compact point-jet source with , and moving the derivatives to the kernel, recovers the displayed identity and hence .

Theorem 4.2 (the entire potential of a quadratic cut). On , there is a unique holomorphic such that It has , and where and at infinity. The positive harmonic function extends continuously by zero on the interval. Denote this nonnegative extension by . Its source and full potential are There are no endpoint atoms or other endpoint-supported terms.

Proof: root and boundary. The two quadratic roots have product one. Existence of the two roots uses only a square root of the nonzero complex number , obtained by halving a polar argument. A root has only if . Off this interval the roots are distinct and neither lies on the circle; exactly one has modulus greater than one. U025, Lemma 2.1, proves a holomorphic local root near each simple root. The unique exterior-modulus choices agree on overlaps, giving the required global function. On it is . The identity gives , since , and hence has the asserted asymptotic. Differentiation gives .

The local logarithm proof in U020, Corollary 1.2, shows that is harmonic. For any sequence approaching the cut, the equation bounds ; every limiting root over a cut point lies on the unit circle. Thus , including at both endpoints, proving continuity.

Over a compact interior part of the cut, the two simple local roots continue smoothly from each side. Their -limits are above and below. To fix the signs, at , , the exterior root is ; continuity along the connected interior interval keeps the sign, and complex conjugation, which preserves the defining exterior choice, gives the lower sign. Since , the normal derivatives are U011, Theorem 2.1 and Corollaries 2.2 and 2.4, supplies Green's formula on the two sides, including the finitely many boundary corners. Their equal zero function traces cancel the terms with derivatives of the test; the difference of the normal derivatives gives exactly the positive density in the statement.

At , implies , so , , and . The logarithm bound follows from its bounded derivative near modulus one. Excise circles of radius about the endpoints before applying Green's formula. Their two error bounds are and , both tending to zero. The cut density is integrable at the endpoints. Passing to the limit on every test proves the full distribution identity, with no additional endpoint term. Substitution gives total mass .

Proof: potential, including the cut. The proposed ordinary integral is The scalar substitution follows first away from the endpoints by the fundamental theorem, then for nonnegative parts by monotone convergence. We now prove absolute integrability also at its possible logarithmic singularities.

For , put . The algebraic factorization and allow the uniformly absolutely convergent series The two angular terms together integrate over a full circle. Each exponential of positive integer frequency has zero integral, by its scalar primitive, so their combined integral is zero. This gives .

For on the cut, choose either root of modulus one, so the same factorization holds. Its factors may have angular zeros. The necessary endpoint passage is elementary: For and small , the right side is at least . Indeed , and follows by integrating the continuous cosine. The positive logarithm is bounded by , while the negative part is bounded by . The integral of this bound on tends to zero, since it is at most , by the scalar logarithmic primitive. Away from those neighborhoods convergence as is uniform. Thus the full-circle logarithmic mean stays zero for , and every angular integral is absolutely finite. At an endpoint the two angular zeros can coincide, but the sum of the two integrable bounds still applies. The factorization yields at every cut point.

Finally, this ordinary potential represents the distributional convolution: for a compact output set the integral of in is uniformly bounded for , by translation into one fixed ball and planar polar integration near zero. Tonelli with the finite positive measure justifies pairing and Fubini. This identifies with , completing the pointwise and distributional conclusions.

A positive kernel measures quadrature error

Theorem 5.1 (two-node positive error kernel). For , the equation has a compact solution exactly when . That solution is unique, is positive on , and has integral . For every complex , so the absolute error is at most , with sharp constant.

Proof. A compact fourth derivative annihilates monomials of degree at most three. The second moment of the forcing is , so is necessary. At this value its zeroth and odd moments vanish as well.

We derive the kernel from the finite Taylor remainder, following the Peano-kernel method and proving its required steps. Four repeated applications of the scalar fundamental theorem give For a direct verification, differentiate the right side up to four times on ; it has fourth derivative and the same first four initial jets. The difference has fourth derivative zero and all four initial values zero, so integrating backward through those derivatives gives zero.