Cones and derivative-growth wavefronts

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

Fourier decay can distinguish singular directions even when a distribution is singular at the same base point in every neighborhood. Here we measure that decay with a specified derivative-growth scale. The analytic case requires a sequence of compact extensions, because a fixed smooth cutoff need not preserve analytic regularity. We prove that these directional tests recover exactly the corresponding singular support, and that differentiation and multiplication by coefficients of the same regularity cannot create new singular directions. For ordinary smooth wavefront, we also identify the full frequency projection of a compact distribution and every conormal direction of a smooth density, including at zeros of its weight.

The main prerequisite is Derivative growth and finite-order localization: Lemma 1 absorbs fixed index shifts, Theorem 4 constructs cutoffs, Lemma 5 characterizes bounded compact distributions, and Theorem 6 converts a full-frequency estimate to local regularity. The Fourier foundation, F1–F5, supplies integration by parts, inversion and the exact transpose conventions. Convolution as addition of supports, B0–B2, supplies localization, smooth pairings and tensor products. Compactness, smooth bumps and Lebesgue integration have the supplied proofs linked in the preceding lesson.

Basic references are Lars Hörmander's The Analysis of Linear Partial Differential Operators I and these earlier programme readings. We use ordinary partial derivatives and the convention

What a regular direction means

Let be open, . Fix a nondecreasing positive sequence with

The class consists of smooth functions satisfying, on every compact set, a bound for all derivatives. The constants may depend on that compact set. A cone below is invariant under multiplication by every positive real number; frequency cones are subsets of .

For , a pair , , is -regular if there are an open neighborhood of , an open cone containing , and a bounded sequence such that

The constants are independent of . Boundedness is in the strong dual of ; by Lemma 5 of the preceding lesson, it means one compact support and one uniform finite-order estimate. In particular,

for fixed . The transforms refer to the zero extensions of these compact distributions.

The complement of the regular pairs is . It is closed in and conic in the second variable: a witness in (3) works throughout the product . With , write , the analytic wavefront set. With , , this is the Gevrey wavefront set of order .

The ordinary smooth wavefront set uses a single compact smooth cutoff near , requiring to decrease faster than every inverse power on a cone. This is the same Fourier definition used in the earlier smooth wavefront readings. Its constants may depend arbitrarily on the requested power. Estimate (3) prescribes that dependence through .

Keeping the growing derivative order separate

Call a sequence controlled on if its supports lie in the fixed compact , and for every fixed multi-index there are such that

These constants may depend on , but not on . The cutoffs from Theorem 4 of the preceding lesson satisfy (5), since . Taking gives uniform bounds for every fixed derivative.

Lemma 1 (the two Fourier bounds). For a sequence satisfying (5), and any fixed integer , there are constants with

In particular .

Proof. A fixed box containing bounds each derivative's norm by its supremum times the box volume. For , choose with . The first estimate follows by integrating by parts times and using (5) with . For the second, integrate by parts times, now using . The finite choice of , and the factors , are absorbed in . Replacing by costs at most .

For , the common bound controls the transform. In the second estimate choose , use , and enlarge by . This proves both estimates everywhere. The first with gives the last assertion: on the annulus , the integral is bounded by a fixed multiple of , and the unit ball has finite volume.

Lemma 2 (products of controlled sequences). Products of two sequences satisfying (5) satisfy (5) on their common support region. If and satisfies (5), then also satisfies (5).

Proof. Apply Leibniz' rule first to the fixed derivatives and then to the derivatives. There are only finitely many splittings of . For each such splitting, the sum over splittings of is bounded by

using the coordinate binomial formula. Maxima over the finitely many -splittings give the constants in (5).

For a fixed function , let be a fixed derivative order and . Its defining estimate and the fixed-shift lemma from the preceding lesson give

Here the case is included. If , monotonicity permits replacing by . Thus the derivatives of occurring on the support of obey the same estimates needed in (7). The product is compactly supported there.

The estimate on a smaller cone

If is a cone whose angular closure lies in an open cone , there is such that

Indeed, normalize . The permitted closed angular sets, including the zero endpoints, form a compact set. If the minimum of the difference were zero, the two vectors would be the same nonzero vector with a direction both inside and outside it. This is impossible. Rescaling proves (9). Also, without an angular condition,

This follows from .

Theorem 3 (local transfer). Suppose a bounded compact sequence equals on and satisfies (3) in . If is controlled on a compact , then, on each cone as above,

The sequence is bounded in . If has order at most near , its products have a uniform order-at-most- estimate.

Proof. The finite-order estimate for , the product rule, and the bounds in (5) give

on one compact neighborhood of , uniformly in . This proves the boundedness assertion by the preceding lesson's Lemma 5.

Since , Fourier inversion on the smooth compact factor gives

For completeness, insert a fixed cutoff near the support of when writing its smooth pairing. The inverse formula for , multiplied by , converges in each of the finite smooth seminorms controlling that pairing: every differentiated integrand is bounded by a polynomial times the Schwartz function . Moving through this integral gives (13), with the factor shown in (1). The polynomial estimate (4) and Schwartz decay make the resulting integral absolutely convergent for every fixed .

For , split the integration into

On , use (3) and the uniform bound from Lemma 1:

On the complement, (9) or (10) applies, so for a fixed . Apply the second estimate (6) with , and (4). Since

the complementary integral is bounded by

The constants have been absorbed into ; is fixed. The integral is finite by the dyadic estimate in Lemma 1. Equations (15)–(16) prove (11) for . For smaller nonzero , (12) bounds the transform uniformly, while when . This completes the proof.

A single estimate over a compact set

Theorem 4 (compact localization). Let be compact and a closed cone. Suppose

For every sequence controlled on , there are with

The products are bounded compact distributions; an order bound for near is preserved uniformly.

Proof. If is empty, every cutoff is zero. If is empty, the frequency assertion is empty and boundedness follows from (12). Assume otherwise.

For a fixed , use (17) in each direction of the compact set . Shrink each witness cone slightly, retaining its direction and an angular closure inside the original cone. Finitely many of the smaller cones cover . Intersect their base neighborhoods to obtain one open on which all these finitely many witnesses equal . Choose finitely many such covering .

We need a partition that keeps the derivative estimates. Choose finitely many nonnegative smooth bumps supported compactly in the , with positive sum near , and normalize that sum after multiplying by a compact cutoff near . This gives finitely many with near . Multiple bumps assigned to the same may be added. The normalization is extended by zero outside the open set where the denominator is positive; the compact support of inside that set makes the extension smooth. This is the finite partition construction proved in B0 of the convolution lesson.

Choose smaller than all distances from to the complement of , and sufficiently small that the closed -neighborhood of lies where . With the normalized bump used in the preceding lesson, set

The total displacement is at most , so remains supported compactly in , and on a fixed neighborhood of . The derivative-distribution argument of Theorem 4 in the preceding lesson, applied to in place of its initial cutoff, gives (5) for each . The same for every is what preserves the sum exactly.

By Lemma 2, satisfies (5), has support compactly inside , and

For a fixed , apply Theorem 3 to each of its finitely many directional witnesses. Their smaller cones cover , so the largest of their constants gives one estimate for on all of that set. Sum the finitely many estimates in (20), increasing the outside constant and taking the largest exponential constant. This proves (18). Boundedness and the retained order follow directly from (12).

Recovering singular support and comparing classes

Define as the complement of the open set where is locally represented by a function. The local representatives agree on overlaps by uniqueness of continuous distributional representatives, proved in the preceding lesson. Thus this definition is unambiguous.

Theorem 5 (projection). For the projection ,

Proof. If is near , the preceding lesson's Theorem 6 supplies one bounded sequence satisfying (3) in every frequency direction. Thus no covector over is in the wavefront set.

Conversely, suppose all nonzero directions over are regular. The same finite-cover construction as in Theorem 4 gives a common base neighborhood of where every direction is regular. Choose a compact neighborhood of inside it, and construct cutoffs with support in , equal to one on a smaller fixed neighborhood. To obtain this prescribed support, apply the preceding cutoff construction within to a smaller compact neighborhood of . Theorem 4, with , gives (18) in all nonzero frequencies. The products are bounded in and agree with on that smaller neighborhood. The preceding Fourier characterization makes locally .

Theorem 6 (regularity scales). If for , then

For every admissible ,

Proof. Replacing by in (3) preserves its witnesses, with enlarged by . This proves (22). For the right inclusion in (23), use , , in the analytic witnesses.

For the left inclusion, take an -regular pair and its witnesses on . Fix once and for all , equal to one near the base point, and shrink to a cone as in (9). For any requested integer power , use the witness . Since , formula (13) holds with in place of . On , (3) at the fixed index , and , give when .

On the complement, the Schwartz estimate , together with (9)–(10) and (4), gives the same bound by the calculation in (16). Small frequencies are bounded by the finite-order estimate for the single compact distribution . Consequently on for every . The cutoff and cone were fixed throughout, so this is ordinary smooth regularity at the pair.

For Gevrey orders , the direction of (22) is : allowing faster derivative growth makes the demanded regularity weaker.

Coefficients and differential operators

Theorem 7. For distributions , a function , and a multi-index ,

Hence every finite-order differential operator , with , satisfies

Proof. Outside the union in (24), intersect the two base neighborhoods and frequency cones and add their witness sequences. The union of two compact supports is compact, the larger finite order controls the sum, and the sum of two bounds (3) has that same form. This proves the first inclusion.

For multiplication, choose controlled cutoffs equal to one near the base point and supported compactly in a witness neighborhood . Lemma 2 shows that satisfies (5). Theorem 3 makes a bounded compact sequence satisfying (3) on a smaller cone. It equals near the base point, proving the second inclusion.

For one derivative , take . This sequence is bounded in : the common compact support is unchanged, and its uniform order is at most one larger. It agrees with on . By the exact Fourier differentiation identity,

The last step is the preceding fixed-shift lemma, with both shifts equal to one. Iterate this one-derivative conclusion for a fixed . A finite sum of the proved multiplication and derivative inclusions gives (25).

The inclusions need not be equalities: on the line, , while has singular directions of every class in (2), as calculated below.

A flat hypersurface and its normal directions

On , , write and define

This is a distribution of order zero on every compact set, by the volume bound on its tangential projection. For every admissible ,

Excluding the other directions. Away from , , so Theorem 5 excludes every direction. Fix a point . Apply the preceding controlled-cutoff construction in to obtain near with one compact support. Set . The tensor pairing from the convolution lesson gives , with no extra factor of . The sequence has a uniform order-zero estimate and equals near the base point. Lemma 1, or its pure-order integration-by-parts proof, gives

Near any direction with , choose a cone on which . Since , (29) proves regularity there.

Including every normal direction. Let be any compact smooth cutoff equal to one near . Put , a nonzero smooth compact function. Its transform is nonzero at some , since Fourier inversion would otherwise give . But

for every real . As or , these frequencies enter every cone about the corresponding positive or negative normal direction. Thus no such cutoff has rapid decay on that cone. Those pairs belong to , and hence to by (23). This proves (28). In dimension one the same argument gives , because every cutoff equal to one near zero leaves .

Compact support and the frequency projection

The next three results concern ordinary smooth wavefront. We use the full cutoff-convolution estimate in Smooth pullback and the Fourier normal condition, Section 18.1, and its exact diffeomorphism rule in Section 18.3. Those proofs include their finite-order bounds and Fourier constants. The Gram surface measure and its coordinate law are proved in Regular level sets and transverse convolution, Lemma 0.2 and Theorem 1.1.

For a compactly supported distribution on , let Rapid decrease means that every weighted supremum on that cone is finite. It does not mean decrease on just one ray.

Theorem 8 (frequency projection and uniform localization). For every compactly supported , This projection is closed in . More generally, if is open and is relatively closed and conic in , then holds exactly when for every and closed cone such that .

Proof of (G2). If , the cutoff-convolution estimate makes every compact localization rapidly decreasing on a smaller cone about . Thus there is no wavefront pair with that covector.

For the converse, suppose every is regular. Put . At each , choose a cutoff equal to one near and an open regular cone containing . Finitely many smaller base neighborhoods cover . A finite smooth partition gives compact functions supported in these neighborhoods, with near . Choose one open cone about whose angular closure lies in all the finitely many regular cones. The convolution estimate makes each rapidly decreasing there. Since , their finite sum proves . If is empty, and the assertion is immediate.

For closedness, all wavefront base points lie in . Given a convergent sequence of nonzero projected covectors with nonzero limit, choose witnessing base points in . A subsequence of those points converges, and closedness of wavefront supplies a witness for the limiting covector. Compact support is what makes this projection argument valid.

Proof of (G3). Smooth multiplication and locality give . The zero extension of to all of has no additional singularities, because its support is a compact subset of . By (G2), its set misses . Cover the compact set by finitely many open cones of rapid decrease. Taking the maximum of their constants for each proves (G3) for . The transform of a compact distribution is a continuous polynomially bounded function, so the bounded-frequency part also has a finite supremum. If is empty, only that bounded part remains.

Conversely, at any pair outside , relative closedness supplies a compact cutoff equal to one near the base point and a closed cone whose interior contains the covector, with the product disjoint from . Formula (G3) makes that pair regular. This proves the characterization.

Smooth weights on every linear subspace

Theorem 9 (the full flat density formula). Let be any linear subspace, of dimension , let , and define Here is Euclidean -dimensional measure, with counting measure when . Then , viewed as a closed subset of , and No nonvanishing assumption at the selected base point is needed.

Proof. Compact test supports give finite volume and a bound for , so is a distribution of order zero locally. It vanishes off the support of . Conversely, every neighborhood of a point in that support contains a point where . In a smaller neighborhood choose a nonnegative tangential bump and use the tangential test times that bump, extended smoothly in the normal variables. Its pairing is a positive integral of , so does not vanish on the original neighborhood. This proves the support assertion.

Use orthonormal coordinates with . The orthogonal substitution has absolute determinant one. Direct Fourier substitution, or the proved diffeomorphism rule, carries covectors by the same orthogonal map. It therefore suffices to work in these coordinates.

For every compact smooth cutoff , set . The exact Fourier formula is The right side is rapidly decreasing in the tangential variable. On a cone about any covector with , shrink the cone so that . Formula (G5) then gives every full-frequency decay power. Away from the base support, vanishes. This proves the upper inclusion in (G4).