Curved Cauchy kernels and complex pole cutoffs

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

A singular reciprocal can have a distributional limit even when its absolute integral diverges. The region removed around the singularity is part of the definition. We examine three such regions: a strip around a curve, a circular hole around a pole, and a sublevel set of a holomorphic function. Each limit comes with a finite test-function estimate and its complete differential source.

Write , , and Pairings are complex-linear; tests are never conjugated. The full Cauchy normalization and boundary-limit proofs are Cauchy kernels and distributional boundary limits, Corollary 1.2 and Theorem 4.1. We use the proved Taylor and point-jet results in Angular foundations, A2–A3. The holomorphic coordinate and real change-of-variables proofs are Zero hypersurfaces as curvature measures, Lemmas 2.1 and 3.0. All remaining proof locations are listed at the end.

The complete integral Taylor proof supplies the finite-order remainders used here, including signed increments, complex-valued functions, directional derivatives and uniform bounds for all required derivatives on compact neighborhoods.

A curve with only one derivative

Let be real functions on , and put . Assume

This condition excludes zeros of on each of the two open half-planes and .

Example 1.1 (an insufficient hypothesis). The weaker requirement for would allow . Its denominator is . A nonnegative compact smooth test positive near gives two divergent one-sided logarithmic integrals at , for every in a small interval about one. The absolute double integral also diverges, since it bounds below a positive multiple of the integral of across that line. Excising leaves this singularity untouched. An additional principal-value prescription there would be a different definition.

Theorem 1.1 (the curve limit and its source). Under (1.1), every compact test has the well-defined pairing

The value of is irrelevant. On smooth tests, is a distribution of order at most one. The continuous coefficient has a particular multiplier action specified in the proof. With that action define

If , this source is zero. If , let be the respective signs of on the positive and negative half-lines. Then

In particular, a real curve has its source at when .

Proof. Fix , , and let . For , substitute and set , . This test is supported inside . Separate its value at zero: The first scalar integral equals : multiply by the conjugate denominator; the odd imaginary part integrates to zero and the even real part has the indicated antiderivative. Its modulus is at most . In the other integral, the fundamental theorem gives , and . Therefore

The function is continuous away from zero by ordinary parameter integration and vanishes for . Assign any value at zero. It is a bounded measurable function, so dominated convergence proves (1.2), with

This proves continuity on the stated compact test spaces. For each positive , is bounded away from zero on the part of the compact support with . Thus is the ordinary, absolutely integrable excised double integral. The limit without excision need not be jointly absolutely integrable.

For a continuous function , define

Multiplying (1.5) by the local bound of proves that this is a distribution. This construction is specific to a coefficient depending only on ; it makes no assertion about multiplying an arbitrary distribution by a continuous function. It defines , since is continuous. For smooth it agrees with the usual multiplication and with the notation .

On a smooth test, (1.3) means exactly

Both integrals converge by (1.5). Off , is and , . Thus . Integrate (1.8) first on and , and integrate the -derivative on the whole compact support. Ordinary one-variable integration by parts is sufficient and never differentiates in . The interior terms cancel. The remaining boundary difference is , where

The positive half-strip has a positive contribution from its inner endpoint; the negative half-strip has a negative one. If , the two ordinary integrals have the same limit. Suppose . By continuity and (1.1), each half-line has a constant sign: two opposite signs on the same interval would force a zero by the intermediate value theorem. The boundary limit proved in U013, Theorem 4.1, becomes

The sign follows from ; the principal value uses symmetric excision of zero. In (1.9) the translated tests converge to in the norm on a common compact support. Uniform continuity of proves this convergence. Estimate (1.5) bounds the effect of their difference uniformly in . Consequently one may apply (1.10) to the fixed limiting test on each side. The principal values cancel and the delta terms give exactly (1.4).

Example 1.2 (crossing and tangency). The curves , , and give respective sources , , and zero. The flat crossing , with , also has source . It is smooth and all derivatives vanish at zero. Indeed each derivative on either half-line is a polynomial in times , whose limit is zero: the exponential series implies for every , and dominates any prescribed power. Induction and the fundamental theorem give the smooth extension with zero derivatives. The source is determined by the side signs.

All circularly cut poles

Theorem 2.1 (existence, order and normalization). For every positive integer , the limit

exists on every smooth compact test. It extends continuously to compact tests, and its exact distributional order is . It restricts to the function away from zero and obeys

These two covariance laws uniquely select its extension. Moreover,

and its complete source is

The cutoff gives exactly the same family, with .

Proof. At the kernel is locally integrable, since its polar absolute integral near zero is . For , choose a fixed radial smooth compact cutoff near zero and let be the Taylor polynomial of of degree . Each monomial is a linear combination of with : insert and and expand. On the circle these terms have frequency , of absolute value at most . Multiplication by leaves a nonzero integer frequency, whose integral is zero because for . Thus, for every , .

Taylor's integral remainder, proved in A2, bounds on a small disk. After multiplication by and the area factor , this is an integrable constant times . On the remaining fixed compact annulus there is no singularity. It follows that

with a fixed-support bound. The same argument works for tests, and proves the asserted upper order bound.

For precision, dilation and rotation pullbacks act by Apply the linear change of variables to the actual cut integrals. Dilation changes the radius to and introduces ; rotation leaves the radius unchanged and introduces . Passing to the limits proves (2.2), including the absence of extra point terms.

Here is a direct proof of sharp order. For , suppose the order were at most . Choose a smooth test supported in an annulus with . Such a test exists explicitly: take with nonzero nonnegative smooth annular , so . For , put . All derivatives through order are uniformly bounded; the supports lie in one fixed disk. Homogeneity gives Choose a decreasing geometric sequence whose closed support annuli are disjoint, for example with ratio smaller than . Every finite sum of the corresponding tests is smooth, supported in the same disk, and has the same uniform bound, since at any point at most one summand or its derivatives is nonzero. Its pairing is the number of summands times , which is unbounded. This contradicts order at most . Any still smaller order would imply that bound as well. Thus the exact order is . For , nonzero local integrability gives exact order zero.

We next prove uniqueness of the extension. By A3, every distribution supported at zero is a finite sum of independent delta derivatives. The invertible relations between and give an equivalent independent basis . Independence can also be tested directly on a cutoff times : , , , , so its jet isolates the indicated coefficient. The pullback definitions and test chain rule give For the rotation identity one may check and on tests, then iterate; itself is invariant under rotations.

The difference of two extensions satisfying (2.2) is supported at zero. Independence of its jets and the dilation law force for every nonzero coefficient. For there are no such indices. Otherwise , so no such jet has rotation weight . Equality of the required rotation factors for all is impossible for a nonzero coefficient; for distinct integers , some has . This proves uniqueness.

Now vanishes off zero by classical differentiation. The same test chain rules make it homogeneous of degree and of rotation weight . A point jet of that degree has , hence ; none has the required weight. Thus . Iterating this recurrence proves (2.3). The exact base identity is , proved in U013, Corollary 1.2. Distributional derivatives commute, as their test derivatives do. Applying to (2.3) proves (2.4).

Example 2.1 (the first source jets). The first two nontrivial cases are The higher pole produces a derivative of the point mass. Its rotation law is needed to select the extension, in addition to homogeneity.

A cutoff prescribed by a holomorphic function

Theorem 3.1 (the modulus cutoff). Let be connected and open and holomorphic on , not identically zero. Then

exists on every smooth compact test, defines a distribution on , and satisfies .

Proof. The isolated-zero and local-logarithm proof in Point sources and complex Gaussian kernels, Corollary 1.2, gives near each zero , with integer and . There are only finitely many zeros in any fixed compact subset of : otherwise compactness gives an accumulation point in , contradicting the isolated-zero theorem and the connected-domain identity principle.

Shrink the neighborhood so . A logarithm of the single nonzero number , together with the convergent series for , gives a holomorphic with . Define

To justify the coordinate, apply U025, Lemma 2.1, to . Its -derivative at zero is nonzero; the full contraction-and-Cauchy proof gives a holomorphic solution near zero. Let . Choose the parameter disk small enough that lies in the uniqueness disk around , and set , an open neighborhood of . We may also shrink so there. The identity and uniqueness give for . In particular ; thus and are inverse holomorphic coordinates. Differentiating gives .

For a test supported compactly within this coordinate neighborhood, the actual cut integral changes exactly to

Here the full nonlinear change-of-variables theorem is U025, Lemma 3.0. Its hypotheses hold because both maps are . If , its real derivative matrix is , so its determinant is . This proves the displayed density factor. The transformed test extends smoothly by zero outside the coordinate image: its compact support is contained in that open image, so it already vanishes on a neighborhood of its boundary. For each positive cutoff all integrals are ordinary integrable ones. Theorem 2.1 gives their limit . On a fixed compact test support, all derivatives of and needed through order are bounded. Repeated chain and product rules therefore give , proving a local finite-order bound.

We describe the finite localization explicitly. For a compact support , choose a compact neighborhood with in its interior. It contains finitely many zeros. Around each zero lying in a neighborhood of , choose disjoint small coordinate patches as above and smooth cutoffs , supported in those patches, equal to one near the zero. Such cutoffs are supplied by the scalar foundation, §13.10. The residual test has compact support away from all zeros; hence has a positive lower bound on that support. At every , the cut pairing of is the finite sum of the pairings of and . The first has an ordinary integral for small , and the other limits were just proved. The maximum of their finitely many orders gives a common finite derivative bound on tests supported in . This proves both the global limit and distributional continuity. Its value is independent of coordinates and cutoffs because every computation is a limit of the same integral (3.1).

Finally, for every compact smooth test,

Only finitely many zeros meet the compact support, so their area is zero; dominated convergence proves the last equality. Thus .

Corollary 3.2 (the full local source). On a zero patch with inverse from (3.2), let . Then

Away from the zeros the source is zero. These formulas specify the complete point-jet source, summed over the relevant zeros.

Proof. Put . The real chain rule and the Cauchy–Riemann equations give . Multiply by to obtain . Since , the exact local representation (3.3) gives By (2.4), . Its pairing supplies a second factor , leaving (3.5). The reciprocal is holomorphic off the zeros, so its classical and weak vanish there. The preceding equalities hold as distributions on each open patch; a finite smooth localization of any compact test makes them a global equality. No uncomputed boundary source remains.

Example 3.1 (a simple zero). If , choose , , so . On this patch Here is locally integrable, as (3.3) and the locally integrable show. For higher zeros, the circular cancellation occurs in the root coordinate and retains its Jacobian.

Exercises

Exercise 1 (basic). Let and . Define the curve distribution and compute its complete weak flux. Why does not force that flux to vanish?

Solution 1. Here is nonzero off zero and . Theorem 1.1 defines by its ordinary inner -integral and strip limit, with the common compact bound. The coefficient action (1.7) applies to , so The signs are , , and . Consequently The coefficient is the difference of the two limiting boundary masses in (1.9). Although the curve derivative vanishes at zero, these side masses do not agree.

Exercise 2 (basic). Let be a compact smooth cutoff equal to one near zero, and put . Evaluate .

Solution 2. Example 2.1 gives . Since and , its pairing is The sign comes from the delta derivative, with no conjugation of the test.

Exercise 3 (intermediate). Set for , , and . Prove that has zero weak flux and exact order one. Show why its kernel is not jointly locally integrable.

Solution 3. Smoothness of at zero follows from the exponential estimates in Example 1.2. Its side signs are both positive, so (1.4) gives . For fixed and , The first equality is checked by differentiating and evaluating the endpoints; the second uses . The expression is at least . Its -integral diverges on every neighborhood of zero. Tonelli therefore proves failure of joint local .