Fundamental solutions, continuation and approximation

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

A point-source inverse turns a differential equation into convolution. A kernel smooth away from its source gives local regularity. If that kernel is analytic there, it also supplies a common complex continuation domain, approximation across suitable complements, and solutions for arbitrary distributional forcing. We prove the approximation and existence assertions for disconnected and unbounded open sets as well.

All pairings are complex linear. Our exact earlier inputs are U021, B0–B3, Theorems 1.1, 2.1, 3.1, 3.2, 4.1 and Corollary 4.2, for localization, gluing, convolution, finite regularity and singular support; U008, Proposition 1.2 and Theorem 5.1, for finite order and uniform convergence on bounded test families; U015, Section 3, for holomorphic Taylor series and the identity principle; and functional foundations, Section 5, for the complete norm-preserving complex Hahn–Banach theorem. Point-supported distributions and uniqueness of their coefficients are proved in angular foundations, A3. The scalar, measure and algebra foundations at the end supply the elementary operations used in the proofs.

Cutoffs, smooth limits and compact duality

We first supply the additional local constructions. For an open , define using distance when . Each is compact, , and these interiors exhaust . The distance function is continuous because the triangle inequality gives ; when the complement is nonempty, the lower distance bound also keeps the closure inside .

B0 (a locally finite compact partition). Choose , compactly supported in , equal to one near ; if , take . These cutoffs come from the proved finite compact-cutoff construction. Set The finite telescoping identity gives . Every point has a neighborhood on which some ; on that neighborhood all with vanish and the sum equals one. Thus , the functions are nonnegative with compact supports in , and those supports form a locally finite family. This proves the precise partition needed below, without an unproved partition theorem.

B1 (the smooth topology). Write The seminorm on an empty compact set is defined to be zero. The countable increasing seminorms determine the same topology as all : every compact lies in some . Closure can consequently be tested by sequences, choosing errors in the -th seminorm.

If a sequence is Cauchy in every such seminorm, every derivative has a uniform limit on each compact set. These limits agree on overlaps and are continuous. On a small closed coordinate box, pass to the limit in The fundamental theorem then identifies the derivative of the first limit with the second. Induction proves that the function limit is smooth with the prescribed derivatives. This proves the completeness and smooth-series convergence used later.

B2 (compact-distribution duality). A continuous linear functional on has a bound for some compact , finite , and . Indeed continuity gives a finite intersection of seminorm balls on which ; take the union of their compact sets and the largest derivative order, then rescale. If the resulting seminorm of is zero, arbitrary scaling forces , which also proves the bound in that case.

Restriction to test functions is a distribution supported in , since tests supported outside have zero controlling seminorm. Choose a cutoff near . The bound also gives , since every derivative of vanishes on . Conversely, a distribution with compact support acts on every smooth by . U021, B0, proves independence of the cutoff. Its finite-order bound on , together with the product rule, makes this action continuous on . These constructions are inverse to each other. They also give extension by zero of a compact distribution to a larger open set: apply it to the restricted smooth test after inserting such a cutoff. Its support is unchanged, and localization proves uniqueness.

Local regularity from one point source

Let and , with constant complex coefficients. A fundamental solution is satisfying . In particular . No bound on at infinity is imposed; a compact second factor suffices for the convolutions here.

Proposition 1.1 (two inverse identities). If are compact distributions, then Proof. Differentiation of a proper convolution gives , and . Every pair has a compact factor. Equivalently, , and the triple convolution with is associative because the other two supports are compact: fixing their sum with the first coordinate in a compact output set bounds all three coordinates. These are exactly the support hypotheses in U021.

Theorem 1.2 (exact local singular support). If is smooth on , then on every open , Proof. A differential operator with smooth coefficients takes smooth functions to smooth functions, so the right side is contained in the left. For compact on the whole space, Proposition 1.1 and U021, Corollary 4.2, give For arbitrary , choose a compact cutoff equal to one near the point being considered, and extend by zero. Both and near that point; restrict the compact result there. This proves the local identity. A fundamental solution cannot be smooth at zero, since then would be smooth there whereas is not: a smooth representative of a point mass would vanish on the punctured neighborhood and by continuity at its center, contradicting a test equal to one there. Thus the stated assumption is equivalent to .

Theorem 1.3 (every derivative of a null sequence converges). Under this smooth-kernel hypothesis, if on and weakly in , then all these distributions are smooth, , and Proof. Applying the distributional derivative definition to each fixed test proves ; Theorem 1.2 gives smoothness. Subtract . For , choose equal to one near . The product rule shows that is supported in the fixed compact set , disjoint from , and tends weakly to zero. Indeed every commutator term contains a positive-order derivative of and a fixed derivative of . Proposition 1.1 gives on .

For , choose a cutoff near whose support misses . Then The formula follows from U021's local smooth convolution formula and differentiation of a smooth compact kernel test. All these tests have one compact support and bounds for every derivative uniform in , because the arguments stay in a compact set away from zero. They are a bounded test family. U008, Theorem 5.1, says that a weakly convergent sequence of distributions converges uniformly on any such family. Hence the right side tends uniformly to zero. This holds for every , proving (1.3).

The fixed null equation is essential in this conclusion; no assertion about arbitrary varying forcing terms is made.

One complex neighbourhood for every solution

A real-analytic function is locally represented by an absolutely convergent real power series, with complex coefficients allowed.

Lemma 2.1 (gluing the real series). For every real-analytic on open , there is a holomorphic extension to an open with . We may choose vertically contractible: implies for .

Proof. At each , take a positive radius such that the real ball is in , the representing series is valid there, and . Shrinking a coordinate neighborhood of absolute convergence supplies such a radius. On the complex ball the same series converges; its derivatives converge uniformly on smaller balls because is bounded for each and . It therefore defines a holomorphic function.

An intersection of two such balls is convex. If it contains , it contains the real point and an open real neighborhood of . The two functions agree on that real neighborhood. Their difference has all real derivatives zero at , so the holomorphic Taylor formula proved in U015 makes it zero on a complex neighborhood. The identity principle then makes it zero on the entire connected intersection. Thus the functions glue on the union of the balls. Its real trace is exactly , and proves vertical contractibility. No connectedness assumption on was needed.

Theorem 2.2 (one extension domain for all null solutions). Suppose and is real analytic off zero. Every open has an open complex neighborhood , independent of the solution, to which each distributional null solution on extends holomorphically. If and distributionally, the extensions converge uniformly on compact subsets of .

Proof. Use Lemma 2.1 to extend the off-zero function to on a vertically contractible with real trace . For each , choose equal to one near , and let . Define This is open: for any , the compact set has positive distance from the closed complement of , unless that complement is empty, in which case openness is immediate. It contains , and vertical contraction remains in it. If , interpret .

For a null solution set , supported in , and put This is an extended pairing in the sense of B2. More explicitly, for a compact , compactness of supplies a neighborhood of such that , after taking relatively compact. Choose one cutoff near , supported in ; use the literal test . Different choices agree by localization. If , and the pairing is zero.

Here is the holomorphy justification with the distribution order accounted for. Near any , choose a polydisk of radius in such that all its differences with are in . The Cauchy formulas and derivative bounds in U015, Section 3, give a Taylor expansion . For every fixed integer , they also give , using the larger polydisk for the derivative bounds. Derivatives in are the negatives of the corresponding real derivatives of the holomorphic kernel. The product rule gives the same bound for in the controlling norm. If is an order bound for , its pairing with the series converges absolutely for , since the bound is a product of geometric series. The paired power series thus proves holomorphy. On , Proposition 1.1 gives .

Two such extensions agree on each component of an overlap. Indeed the vertical segment from any point of the overlap to its real part remains in the overlap, so its component meets an open real subset where both functions equal . Equality of all real derivatives and the holomorphic identity principle give equality on that component. They consequently glue on This set depends only on the chosen kernel and cutoff family, and it contains .

For the convergence assertion subtract . The compact distributions tend weakly to zero with support in . On a compact complex subset of a fixed , the preceding literal kernel tests share a compact support and all their derivatives are uniformly bounded. U008's bounded-test convergence makes their pairings uniformly tend to zero. Every compact subset of is covered by finitely many neighborhoods with closures compact in individual 's; combine those finitely many estimates.

Corollary 2.3 (connected uniqueness). A null solution on connected that vanishes on a nonempty open subset is zero throughout .

Proof. It is real analytic by Theorem 2.2. The points near which it vanishes form a nonempty open set. At a limit point, every derivative vanishes by continuity, so its convergent Taylor series is zero near that point. The set is relatively closed as well; connectedness proves the assertion.

The same local kernel and finite-order series argument shows that is real analytic outside for any compact distribution , where reflection is defined by . Differentiating this definition verifies There is no conjugation in the transpose.

Separation by a compact distribution

Lemma 3.1. If is a closed linear subspace and , there is a compact distribution with and .

Proof. A basic neighborhood of missing gives one seminorm and such that for all ; combine finitely many bounds as in B2. On , set . The decomposition is unique. For , apply the separation inequality to , and for use zero, to get Thus descends to the image of this subspace in . The formula is well defined by the two triangle inequalities, and is a norm on that quotient. The norm-preserving complex Hahn–Banach theorem, proved in functional foundations Section 5, extends to the whole quotient with the same bound. Pull back to , then apply B2. This yields the required compact distribution. The normed quotient need not be complete, and its subspace need not be closed.

Hence membership in the closure of a linear space is equivalent to annihilation by every compact distribution that annihilates that space. The null space of is closed, since each derivative defining is continuous in the smooth topology. B1 makes a closure assertion here equivalent to approximation by a sequence.

Which holes prevent approximation

For open , consider the exact condition for every nonempty compact and every closed relative to . A prohibited is a compact piece both open and closed in the relative complement. This formulation includes arbitrary disconnected and unbounded sets.

Theorem 4.1 (Runge approximation). If has a fundamental solution real analytic away from zero and (4.1) holds, every null solution on is a limit of restrictions of null solutions on .

Proof. The empty-domain cases are immediate. By Lemma 3.1 it suffices to take annihilating all restrictions of null solutions on , and prove that it annihilates every null solution on . For this is immediate. Otherwise let Then , and is analytic off the nonempty compact set .

We record the component topology used here. Each component of the open set is open: a small ball centered at a point of a component is connected and lies in the same component. Its boundary is contained in , since a small ball about a boundary point outside would meet the component and then be contained in it, contradicting its being a boundary point.

If , the functions are smooth null solutions on , for every ; their possible source is at , outside . Pairing with shows that every derivative of at vanishes. Its Taylor series is zero near . Corollary 2.3, or the identical analytic open-and-closed argument, makes zero on the entire component of containing .

Now take a bounded component contained in . Its boundary lies in , so . The set is compact: a sequence in it has limits in the compact , and no limit can lie on , because is open. The remaining set is relatively closed in . Condition (4.1) gives , so .

When is bounded, every component entirely in is of this type; every other component meets and has . Thus each point outside has a neighborhood on which , since it is outside and lies in one of the zero components. Consequently . This globally closed support is bounded, being contained in bounded , and is therefore compact in . For any null solution on , extended pairings and integration by parts give To see the cutoff justification explicitly, insert a cutoff equal to one near ; its derivatives vanish there, so moving to the test leaves exactly on that support. Lemma 3.1 proves the bounded-ambient case.

We use this case to prove a needed global approximation statement before treating unbounded . Concentric balls , , satisfy (4.1). Starting at any point of the annular complement, the outward radial segment approaching the outer sphere stays in that complement. A relatively clopen piece containing its initial point must contain the entire connected segment. If the piece were compact in , its limiting outer boundary point would also belong to it, a contradiction. This works in dimension one on each of the two annular intervals. The bounded case therefore approximates a null solution on a ball by null solutions on any larger concentric ball.

Fix a closed smaller ball in an initial ball. Choose larger concentric balls with , union , and larger than the initial ball; put , . For each integer , first approximate the initial solution by on with error below on . Repeated use of the bounded case chooses a null solution on such that For each compact set and derivative order, all sufficiently late differences satisfy this summable bound there. B1 gives a smooth global limit as ; passing derivatives to the limit shows . The supremum error on is less than . Exhausting the original open ball by smaller closed balls and choosing the error to tend to zero gives global null solutions converging uniformly on every compact subset of the initial ball. This is distributional convergence, and Theorem 1.3 upgrades it to smooth convergence. This construction used only the bounded-ambient case.

Return to arbitrary , and choose with . For , the function is a null solution on . The preceding global approximation makes it a smooth limit near of global null solutions. The functional annihilates those solutions, since their restrictions are null on . Hence Every unbounded component of meets this exterior open set, and analytic uniqueness makes zero on it. In dimension one this separately covers both unbounded rays. Together with the earlier bounded-component argument, every point outside again lies in an open zero component. Thus , and (4.4) makes that globally closed support bounded and compact. Equation (4.2) and separation finish the proof.

Corollary 4.2 (necessity for a nonconstant operator). Under the same analytic-kernel hypothesis, if is nonconstant, condition (4.1) is necessary for the approximation conclusion.

Proof. Suppose is a prohibited decomposition. The set is open. Choose and the smooth null solution on . Assume that null solutions on approximate it smoothly on . Take equal to one on a neighborhood of , with . The commutators are smooth, supported in the fixed compact set , and converge with every derivative to a test function on that set. By U021, Theorem 2.1 and its compact-set seminorm estimates, smoothly on every compact set. Proposition 1.1 gives on . Hence the smooth function equals on and is null on . It glues with on to a smooth null solution on .

The distribution on is supported in . Extend it compactly by B2. Its equation is on the whole space: the equation holds on , and the compact extension vanishes near the complement of . U030, Corollary 4.3, gives The full point-support proof in angular foundations A3 writes , with a finite polynomial. Point-jet independence gives . This is impossible for nonconstant : the top homogeneous parts of two nonzero polynomials have nonzero product, because their largest lexicographic monomials have a unique largest product monomial with nonzero coefficient. Thus degrees add. Since , . This contradiction proves necessity.

For a nonzero constant , all null solutions are zero, and approximation holds on every pair of open sets without (4.1).

Solving on an arbitrary open set

Theorem 5.1. If has a fundamental solution real analytic away from zero, then for every open and every there is with .

Proof. There is nothing to prove for empty . Define with for the whole space, and . These open sets exhaust . Each is compact and contained in , since its points satisfy and , still strictly stronger than the inequalities for .

We verify (4.1) for . Suppose a nonempty compact relatively clopen piece exists, and choose . If , follow the outward radial ray until its first encounter with , if any. Before that encounter it is a connected path in , so relative clopenness keeps it in . A finite first encounter contradicts closedness of compact ; absence of an encounter would put an unbounded ray in , also impossible.

If , then . A nearest exists: a minimizing sequence in the nonempty closed set can be restricted to a closed bounded ball and has a convergent subsequence. For , let . This point lies in , since . Also . Thus the segment lies in , remains in by connectedness, and has limit , the same contradiction. This proves the complement condition, including cases where is empty.

Take equal to one near , extend compactly, and set . Proposition 1.1 gives on . Therefore is a null solution on , smooth by Theorem 1.2. Theorem 4.1 supplies a null solution on with If , take . Define global-on- distributions For fixed and , telescoping on gives The summands are smooth null solutions on . Since for , (5.2) gives uniform summable bounds there. The partial sums have a continuous uniform limit, hence converge as distributions: for a compact test, bound the integral of the difference by its supremum times the test's norm. Theorem 1.3 makes the limit smooth and null.

The base need not be a function, but on . Adding the smooth null limit produces a distributional solution there. These distributions agree on overlaps because all are limits of the same . More explicitly, every compact test is supported in some ; define its pairing by this local limit. Enlarging leaves it unchanged. For tests in any fixed compact set, one such works for all of them, and the local distribution estimate proves continuity. Linearity follows by choosing one for finitely many supports. This constructs a distribution on with the required equation.

The solution need not be unique. Any two solutions with the same forcing differ by a null solution, which is real analytic by Theorem 2.2.

Every distribution has continuous primitives

We now drop the hypothesis on . The one-variable identity uses . To prove it, integrate by parts against a compact test on . For the zero endpoint contributes nothing and the derivative is ; for , . Iteration gives (6.1).

Theorem 6.1 (continuous coefficients with locally finite supports). Every has continuous functions on such that and is locally finite. If has compact support, one can use finitely many coefficients, all compactly supported in .

Proof. Let first be compactly supported with order at most on the whole space. Choose an integer , and put Tensor products and (6.1) give . The function is , in particular . U008, Proposition 1.2, extends an order- distribution continuously to tests. Thus , with a fixed cutoff near , is well defined. On each compact -set its kernel tests depend continuously on in the norm, by uniform continuity of the finitely many derivatives on a larger compact set. Hence is continuous. U021, Theorem 2.1, identifies this function with the distributional convolution . Differentiating that proper convolution gives

For general , use the locally finite partition of B0. Choose equal to one near , and extend compactly to the whole space. Let be an order bound, choose , and form . Write , . Only finitely many 's meet a given compact test. Therefore (6.4) and the product rule give where Each summand is continuous and supported in . Near each point only finitely many of those supports occur, and each of their finitely many 's allows only finitely many . Thus (6.6) defines continuous functions and a locally finite family of their supports. It also justifies every interchange in (6.5). In distribution notation (6.2) means , with precisely these signs.