Analytic parametrices and straight characteristics

*Written by GPT-6 Astra (OpenAI), October 2026. Public domain (CC0). *

A directional inverse can carry a singularity forward or backward along a characteristic line. To use that inverse at analytic regularity, its error must itself be analytic. We construct two such inverses for constant-coefficient equations, prove their analytic wavefront bounds, and then propagate singularities at every derivative-growth scale. The spatial cutoff in the propagation argument is smooth; we keep track of the analytic directions it introduces. An imaginary displacement estimate quantifies the error for every further real perturbation vector.

Use , the Fourier inverse factor , and complex-linear distribution pairings. Let and let be a polynomial of degree . Its top homogeneous part is denoted by . Here real principal type means The lower coefficients may be complex. Euler's identity , obtained by differentiating at , shows that at every real , including noncharacteristic points.

The available proofs used below are holomorphic boundary directions, Theorem 2; conic decomposition, Theorem 4; analytic pullback, Theorem 2; convolution directions, Corollary 7; and analytic differential regularity, Theorem 3. Cones and derivative-growth wavefronts, Theorems 5–7, supplies projection, comparison, multiplication and differentiation. Convolution as addition of supports, B0–B3 and Theorem 1.1, supplies distributional pairings and derivative commutation. Holomorphic boundaries in convex cones supplies the polydisk proof turning coordinate holomorphy into power series. The supplied Fourier, integration, finite-algebra and metric/calculus readings justify the inversions, finite-order pairings, determinants, compactness and cutoffs used here.

1. A coordinate formula for contour deformation

The contour argument follows from a determinant identity and coordinate integration by parts.

Lemma 1 (graph homotopy identity). Let be in , and let be holomorphic near the image in question. Write , , and . Then Thus, with denoting the expression in braces on the right and with a compactly supported weight , It suffices that be defined near the portion on which or its derivatives are nonzero.

Proof. In differentiating the right side, the derivatives falling on sum to . This is the cofactor identity : expanding a component along the replaced column gives the corresponding determinant if that component agrees, and a determinant with two equal rows otherwise. It holds even when is singular.

The derivatives falling on the replaced column give , which is . The remaining terms cancel in pairs. The term where differentiates column has in column and in column . Its partner has these two columns exchanged, and the same mixed derivative, so has the opposite determinant. This proves (2). The holomorphic chain rule is what excludes derivatives in . Integrating the coordinate derivatives against , one coordinate at a time, proves (3); compact support removes the endpoint terms. Differentiation under this finite integral follows from continuity of the derivatives. A partition or a cutoff restricting to the stated portion gives the last assertion.

We will pass from compact weights to an unbounded graph only with explicit bounds on the resulting cutoff terms.

2. A displacement on which the polynomial does not vanish

Lemma 2 (positive imaginary displacement). Suppose has degree , with real principal part and nonzero gradient at every real nonzero point. Put There are and such that If a unit real vector and a closed set of directions satisfy there, with , then, after increasing , some gives on those directions.

Proof. The vector field is smooth and homogeneous of degree zero. Its derivatives of order are bounded by : differentiate the homogeneity identity and take a maximum on the compact sphere. The continuous nonzero gradient there has squared norm at least a positive number . Thus .

For a fixed , the finite polynomial Taylor expansion gives Here bounds the imaginary part of the degree- component on the real sphere. All remaining lower terms have size ; their changes under the bounded shift do too. The terms of order at least two in the principal Taylor expansion have the latter bound, because is fixed. These statements follow term by term from the finite monomial expansion, and include . Choose so , then choose so . This proves (5).

At , the derivative of differs from by at most . For , a second polynomial Taylor expansion therefore gives For large , every second derivative along this segment is bounded by a constant times , since is bounded and . The ordinary integral Taylor remainder, or the finite expansion, proves the last term. Enlarge so , and choose so . The two errors in (8) are at most . Combining with (5) proves (6).

Arbitrary imaginary perturbations

An additional imaginary displacement need not follow a positive gradient direction, and its length need not be small relative to the frequency. The following estimate retains its linear contribution and bounds the remaining terms uniformly.

Lemma 2a (all real perturbation vectors). Let satisfy (1), of degree , and set . There are positive constants , depending only on , such that, for every real with , There is no restriction on the direction or size of . The right side may be negative. In degree one the stronger exact formula in the proof has no error term.

Proof. First suppose . Lemma 2, with , supplies fixed and such that Here and below the norm of a complex vector is its Euclidean norm. To spell out the last estimate, has degree at most , while is the integral of for . The vector is bounded on the sphere and homogeneous of degree zero. Each monomial of these derivatives therefore has size at most a constant times , since and is fixed. This proves the last estimate, including , with one constant for all directions.

Write . Every point , , has norm at most . Every second derivative of is a polynomial of degree at most . Bounding its finitely many monomials, and using , gives For example, a common bound on the individual second derivatives gives this inequality because . No restriction was used.

Apply the fundamental theorem of calculus twice to the polynomial . Integrating gives The change in the order of integration is over a compact triangle with continuous integrand, separately in real and imaginary parts. In particular the sign preceding the integral is .

The imaginary part of the linear term is . Since is real, (D2)–(D4) give the lower bound The two errors are bounded by for a fixed positive . Also when . Thus (D1) holds with , after enlarging if needed.

If , write , where and . Then , and direct evaluation gives Choose so that . Take , , and any positive . Subtracting the nonnegative quantity weakens this exact bound and proves (D1). The denominator is nonzero because . This treats every allowed degree and completes the proof.

The estimate controls the error around the positive displaced graph. It does not say that every further displacement avoids zeros or keeps the imaginary part positive. Exercises 9–11 give exact examples and the uniform positive-direction consequence.

3. Two inverses with entire errors

Theorem 3 (signed analytic parametrices). For satisfying (1), there exist and entire functions on , restricted to the real space, such that The error is analytic on all of the real space. The assertion does not impose compact support or tempered growth on .

Proof, first increasing the order. Choose an integer so that , and set Its real principal part is . On the nonzero real characteristic set its gradient is , so is nonzero and points in the same direction. Euler's identity makes its gradient nonzero elsewhere too. In particular Lemma 2 applies, and .

The integral and its equation. Fix from Lemma 2. Extend to a smooth bounded field on the real space, equal to for large ; multiply by a smooth radial cutoff that vanishes near zero. All derivatives of are bounded. Choose a smooth real cutoff , zero on a sufficiently large ball and one outside a larger ball, whose nonzero support lies where (5) holds and . Put The quotient is defined only where needed and extended by zero near the inner ball. There is no pole on that support. The determinant is bounded, , and on each compact real -set the exponential has bounded modulus because is bounded. Since , annular integration proves absolute convergence, uniformly on compact real -sets. Thus is a continuous function and a distribution.

For a compact smooth test , integration by parts in gives The initial iterated integral is absolutely convergent by the preceding bound applied to . The transform on the right decreases faster than every power of , uniformly under the bounded imaginary shifts , . To verify this, choose a coordinate with and integrate its compact test times by parts. Its complex frequency has modulus at least ; the exponential's absolute value is bounded on the fixed test support. This gives the assertion for every fixed .

Apply Lemma 1 to , , and a cutoff , equal to one for and zero for . The replaced-column determinants are bounded because and are bounded. The right side of (3) is for every , uniformly in . It tends to zero. Dominated convergence and Fourier inversion give Consequently , where This is entire: the parameter set is compact, and on every compact complex -set the integrand and each derivative are bounded. Differentiation under the integral and the supplied polydisk theorem give holomorphy and power series.

Excluding a direction off the negative ray. Fix and outside the closed ray . There is a unit real vector with Indeed , and the sum of the two unit vectors in these directions is nonzero unless they are opposite. Its scalar product with each is positive; normalize that sum. Choose a sufficiently small spherical neighborhood of on whose closure .

Take a smooth angular function , with , supported inside , and equal to one on a smaller neighborhood of the marked direction. Such a function is supplied by smooth Euclidean cutoffs restricted to the sphere and radial extension. By (6), for some , the graphs have for large . Outside the extra displacement is zero and (5) applies. We may increase the inner radius in : changing that cutoff alters (11) only by an entire compact-parameter integral, so does not alter its analytic wavefront.

Work on . On compact subsets of this half-space, The matrices are bounded, since derivatives of have degree zero. The vector is . Hence the replaced-column expressions in Lemma 1, with , are , uniformly in and those . Apply (3) with weight . The terms containing are , which tend to zero because . All terms containing have compact parameter support and are entire functions of complex . Thus on where is entire. For example, the homotopy contribution to is the integral of over , with the same normalization. The denominators on that compact support have a positive lower bound. This proves both the limit and the analytic nature of the correction; a merely smooth correction would not suffice.

The portion of the integral with direction in is holomorphic in a complex neighborhood of every compact subset of . If , replacing by adds to the real exponent. For , the exponent is at most . Every derivative adds a polynomial factor, so all differentiated integrals converge there.

Cover the remaining compact set of spherical directions by finitely many small closed circular cones , each avoiding and having angle less than about its central ray. Explicitly these are sets , where and . The triangle inequality proves convexity; simultaneous membership of and forces . Every vector within distance of is in the positive dual, so that dual has interior. Choose a small such cone about each remaining unit direction with its interior covering that direction and still excluding ; compactness selects finitely many. Assign each remaining direction to the first covering cone containing it. This gives disjoint measurable pieces ; overlaps and their boundaries cause no omission.

For , the corresponding integral over is holomorphic at , , because For in a compact part of , on ; compactness of its unit section proves this positive lower bound. Thus holomorphic derivatives converge locally. For all small , even directions approaching its edge, (19) and bound the integral uniformly on each compact real -set. Dominated convergence gives its continuous real boundary value. These are precisely the full-cone growth assumptions, with exponent zero, of U075 Theorem 2.

The dual of is . Here is a direct verification of the only nontrivial inclusion. If , choose a nearest point ; the minimization may be restricted to a sufficiently large closed ball, so exists. Differentiating squared distance along segments in gives . Taking yields . Therefore lies in and . Add a sufficiently small positive multiple of an interior point of . The resulting vector is in its interior and still pairs negatively with . Thus , proving the claim.

The boundary theorem now confines each piece's analytic wavefront to , all of which avoid . The contribution and are analytic near , so the analytic sum rule gives Away from zero, is analytic. U087 Theorem 3 therefore confines any remaining wavefront covector to . This gives the negative-ray inclusion in (9) for .

Apply this same negative-ray construction to and negate its resulting distribution. This gives with ; the rays are positive because . Both errors are entire. Finally put Then . Differentiation cannot enlarge analytic wavefront. On the characteristic set the gradient multiplier is strictly positive, so the signed rays for are exactly those for . This proves (9).

4. A directional part that still satisfies the same equation

Let be admissible: , , is nondecreasing and . The analytic choice is . Earlier comparison gives .

Lemma 4 (constant-coefficient commutation). The conic-decomposition operators of U079 Theorem 4 satisfy for and every constant-coefficient polynomial .

Proof. The first two assertions are that theorem, for any finite closed conic cover. Its actual construction integrates the boundary values of the spherical representation over disjoint Borel pieces of the sphere. In the open tube, Differentiating in multiplies the test inside the pairing by the corresponding polynomial in . Thus the representation of is exactly . The boundary limit commutes with a fixed distributional derivative, since on each test this is the same limit paired with its fixed derivative. Integration over a sphere piece also commutes with that derivative: the common finite Schwartz-seminorm bound from the representation theorem applies to that derivative test, and the piece has finite measure. These facts prove the last equality in (22). No smoothness of the boundaries of the sphere pieces is needed.

5. Propagation at every derivative-growth scale

Theorem 5 (straight-characteristic propagation). Suppose satisfies (1), is open, , and . If then . On every line interval through in direction , with disjoint from , one has The interval may be unbounded. No regularity of in other directions is assumed.