Planar rotations and angular spectra

Reconstructed and self-checked by GPT-6 Astra (OpenAI), Ultra reasoning effort, October 2026. Earlier edition: GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original exposition: CC0. Supplied foundations retain their stated licences.

In the plane, the Fourier transform of an even distribution of degree minus one is a quarter turn multiplied by . This statement includes continuous angular profiles, integrable profiles with jumps, measures concentrated in a pair of directions, and derivatives of angular point masses. We prove the common formula on all Schwartz tests and distinguish the different kinds of spatial distributions it describes.

We use , its bilinear transpose, and , where . The supplied Fourier foundation, F1–F5, proves every Schwartz estimate, inversion and coordinate identity. Radial powers and the logarithmic endpoint, Theorems 1.1 and 3.1 and Lemma 3.3, supplies the whole radial transform and the logarithmic gradient. Homogeneous extensions and angular moments, Lemma 1.1 and Theorems 1.2 and 3.1, supplies the angular classification and uniqueness through the origin. Its angular foundation, A1–A5, includes point jets, polar integration and circle measure.

For matrices we use the complete real symmetric diagonalization proof at the start of Section 2 in Point sources and complex Gaussian kernels, and the finite algebra foundation, Sections 10.1–10.6. The scalar, Sections 12.4–12.9, 13.1–13.5 and 13.7–13.10, and integration, Sections 15.0–15.1 and 16.1–16.2, foundations prove the calculus, cutoffs, convergence, Fubini and linear substitutions used below. The supplied measure foundation, Theorem M and M10–M11, proves positive representation and complex variation.

A linear substitution rotates a planar quadratic form

Theorem 1.1. If is real symmetric positive definite, on , and , then This is an equality of whole regular tempered distributions. For , put . Then

Proof: the whole linear transformation rule. For any invertible real , define All derivatives of a linearly substituted Schwartz test are finite linear combinations of derivatives of the original test. The bounds , follow by summing its finitely many entries. They show continuity of this test map in every weighted seminorm. The absolute Jacobian substitution in the Fourier integral gives Consequently, directly on distributions, This proof does not require the transformed distribution to have a density.

The proved real diagonalization of gives , with orthogonal and all . Set . Then , , and , by determinant multiplication. Apply the preceding rule to and U051's full radial identity. Local integrability and temperedness are preserved by the same linear change of variables. This proves (1.1).

For , direct inversion gives Both forms are positive off zero. Since , taking their positive square roots in (1.1) gives (1.2).

Continuous angular data need only a slice identity

The map is orthogonal, , and . Write . Thus . Circle measure is ; scalar translation of the integral of a periodic function proves its invariance under and under the antipodal map.

Theorem 2.1. Suppose is continuous off zero, , and for . Then has a regular tempered extension through zero and

Proof: integrability and the slice. The profile is bounded and even. Polar integration gives For , the absolute integral is at most , using the arctangent integral . The integral of on a radius- ball is . These statements prove local integrability and temperedness.

For , define . This is a one-dimensional Schwartz function, with seminorms bounded uniformly in . In detail, its -th -derivative is the integral of ; expand the finite binomial sum. With , bounds each term by an integrable multiple of , uniformly in . Dominated difference quotients prove all derivatives and their continuity. Fubini and the orthogonal coordinates now give . One-dimensional Schwartz inversion at zero therefore proves Both line integrals are absolute and uniformly bounded in direction by fixed seminorms of and , respectively.

Proof: fold the two half-lines. Apply (2.2) to . Evenness of , antipodal substitution, (2.3), and the same substitution once more give Every interchange is justified by the preceding uniform absolute bounds. Set . Because is even, . Polar integration identifies the last line with the whole regular distribution , proving (2.1).

No differentiability of was used. The result preserves an angular corner, rotating its location. Applying it twice gives , consistent with inversion and evenness.

Rough angular data and singular directions

Define the radial test average An angular distribution means a continuous complex-linear functional on the sphere test space of angular foundation A1. It is even when . Our rotation conventions are For a spatial density the latter is , since .

Theorem 2.2 (every even homogeneous distribution). For every even angular distribution , set This gives exactly all even distributions on homogeneous of degree ; in particular all such distributions are tempered. The correspondence is a linear bijection, continuous in both directions for the weak and strong dual topologies of the angular space and the corresponding subspace of . Moreover,

If is even almost everywhere and , then is the regular distribution , with Its transform has density . Angular convergence implies strong convergence of both spatial distributions and transforms. An even finite complex angular measure gives a locally finite spatial measure, with the last two bounds replacing by its total variation norm. An arbitrary angular distribution need not give a spatial measure.

Proof: the angular topology and radial estimates. On the circle the seminorms are equivalent to the radial-lift seminorms of angular A1. To see both directions explicitly, differentiation along the circle is the ambient operator . Conversely, for the degree-zero lift, On , repeated differentiation bounds every ambient derivative by finitely many angular derivatives. A smooth periodic has such a smooth lift. On use , and on add to that expression. On use ; on use . The scalar trigonometric formulas identify their unit vectors with , and the angles on overlaps differ by multiples of . Their lifts therefore agree by periodicity. The scalar foundation proves smoothness of each local expression. This also justifies the derivative formulas. Thus the stated seminorms describe precisely the existing sphere test topology.

For , repeated chain rules in yield a finite sum of with bounded smooth coefficients; for include . Induction proves this description, since a new derivative either differentiates the coefficient or adds one coordinate derivative and one factor . Put For each is integrable; for , . Dominated differentiation under the radial integral therefore proves In particular is continuous and sends bounded sets to bounded sets. Angular A1's finite-order bound, now in these equivalent seminorms, shows that is tempered.

Reflection commutes with radial averaging, proving evenness. With the dilation convention , the positive substitution gives Thus its degree is .

Proof: Fourier transformation before the angular functional. The two sides of (2.3) are smooth functions of : apply the full estimates (2.10) to , to , and to their reflected or rotated radial averages. All angular derivatives converge uniformly. Equation (2.3) is therefore an equality of smooth sphere tests. Evenness of identifies its pairing with either half-line average as half its pairing with the full-line average. Hence Since , evenness gives . Also . These two identities give both equalities for in (2.8). On even angular or spatial distributions, , so this map has inverse . It agrees with the full Fourier inverse .

Proof: bijection and both topologies. Normalize a nonnegative nonzero smooth bump supported in so that , and set It is a compact smooth test on one fixed annulus; angular A2 and the product rule bound each Schwartz seminorm by finitely many angular seminorms. Thus is continuous, preserves bounded sets, and . Therefore , proving injectivity and giving the inverse on the image.

Let now be even and homogeneous of degree . Apply U018, Theorem 1.2, to its restriction off zero. The resulting radial exponent is , so the angular test is exactly on tests supported away from zero. The inverse formula using shows that its angular functional is even. The already constructed is another homogeneous extension. U018, Theorem 3.1, gives uniqueness: degree differs from every possible point-jet degree . Its full point-jet proof therefore gives , proving surjectivity and temperedness without assuming either in advance.

Weak dual continuity follows by evaluation at the fixed test or . A strong dual seminorm takes the supremum over a bounded family of tests. The image of that family under or is bounded by the estimates just established, proving continuity in both directions for those seminorms as well.

Proof: densities, measures and limits. For , polar integration proves its asserted regular pairing and the first two bounds in (2.9), just as in (2.2). Rotation preserves and the angular norm; (2.8) gives the last bound. Apply both bounds to . On any bounded set of Schwartz tests, is uniformly bounded, proving both strong convergence statements.

For a positive finite angular measure , the formula defines a positive functional. On tests supported in a radius- ball it is bounded by . Theorem M represents it by a locally finite positive Radon measure. Equivalently it is the pushforward of : monotone convergence for continuous compact cutoffs inside an open set identifies both values on open sets, and the supplied finite-measure generating-class proof on bounded balls identifies all Borel values. For a finite complex , apply this to its four positive parts; the result does not depend on the decomposition because the integral against every test does not. Its variation on a ball is at most : apply to the radial sections of a finite Borel partition and sum; their sections are disjoint. The same inequality with the weight proves the remaining two bounds. This measure has pairing , as asserted.

The two eigenprojections. For every even , put Expanding the products and using gives idempotence, zero mixed product, sum , and . Rotation of tests is continuous, so the projections are continuous in both dual topologies. Formula (2.8) gives . If , applying again gives ; the two complex roots are . Injectivity then identifies precisely the corresponding angular eigenspace.

Exercises

Exercise 1 (foundation). Transform . Verify positivity and display the determinant before simplifying.

Exercise 2 (intermediate). Let , , and . Calculate both by the linear transformation rule and by the quarter-turn theorem. Resolve the negative determinant explicitly.

Exercise 3 (intermediate). Transform and explain the signs of the two components.

Exercise 4 (advanced). For , find the whole transform. For , approximate its profile by . Give explicit errors, measured by , for both the distributions and their transforms.

Exercise 5 (advanced). Describe every nonzero continuous even degree- Fourier eigenfunction in the plane, including both angular eigenspaces and their projections.

Exercise 6 (intermediate). Compute on all Schwartz tests and compare it with its proposed quarter-turn expression. Specify the failed hypothesis.

Exercise 7 (foundation). Invert , and verify the answer with .

Exercise 8 (advanced). Evaluate the pairing of with , once in each variable. Reduce both absolute integrals to the same elementary one-dimensional integral.

Exercise 9 (intermediate). Let be the two closed circle arcs of half-width about and . Transform the profile and give all bounds in (2.9). Construct even continuous approximations with an explicit error. Then use the atomic profile : identify its spatial measure, Fourier transform and eigenprojections. Show that narrow integrable angular caps converge strongly to this example, and explain why its spatial measure has no planar density.

Exercise 10 (advanced). With angle derivatives, let . Find as a combination of a line distribution and its transverse derivatives. Calculate its Fourier transform by two methods. Prove it is not a locally finite measure using tests of bounded supremum on a fixed compact set, and give its nonzero eigenprojections.

Solutions

Solution 1. Here , , and Theorem 1.1 gives The inverse quadratic factor contributes , cancelling the absolute Jacobian. These are whole regular tempered distributions by the theorem.

Solution 2. The profile of is , so it is even and continuous. Since , the linear rule gives For , direct multiplication of gives . Evenness and positive homogeneity imply for every real . Thus , cancelling the factor . Writing out gives Alternatively itself has all the hypotheses of Theorem 2.1, which gives immediately. Both calculations use ; orientation reversal changes no integration sign.

Solution 3. The scalar trigonometric addition formulas give and . Both profiles are antipodally even. Therefore The positive and negative signs are the two quarter-turn eigenvalues, and the theorem includes the whole locally integrable extension at zero.

Solution 4. The continuous even profile gives For real , squaring nonnegative quantities proves . The profile error has norm at most . Apply (2.9) to obtain These bounds are uniform on every bounded Schwartz family. The transformed density retains a corner along the nonzero -axis: with fixed, the two one-sided -derivatives at zero are and .

Solution 5. Set . On even functions , and . Thus have eigenvalues and . The converse follows from the eigenprojection result after Theorem 2.2. For continuous profiles the angular equalities hold pointwise: a continuous nonzero difference has, after multiplication by a constant phase, strictly positive real part on a small arc, and a nonnegative smooth bump in that arc contradicts a zero distribution. Hence the two spaces are exactly respectively. Both imply antipodal evenness after two shifts. For , the positive projection keeps , and the negative one keeps . A profile with both nonzero projections is not an eigenprofile.

Solution 6. U051, Theorem 3.1, at , gives Its whole weak gradient is the regular function , as fully proved in U051, Lemma 3.3. Explicitly, punctured integration by parts has inner boundary error bounded by ; the gradient is locally integrable, so this error tends to zero. Schwartz decay controls the outer boundary. Consequently The quarter-turn expression would be , different already at . The input is odd, so the antipodal folding in (2.4) is unavailable.

Solution 7. Formula (2.8) gives Applying again gives . This verifies the normalization and the entire distribution.

Solution 8. The frequency calculation from Solution 4 is where , using on the first quadrant and symmetry. To evaluate , put , . This is a strictly increasing smooth substitution with positive cosine; its differential cancels the square root, giving . The endpoint identity follows from the scalar trigonometric formulas.