Tempered growth and spectral cutoffs

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.

Rapid oscillation can make a smooth function with exponential amplitude tempered. The decisive comparison below is between the amplitude and the scale on which the phase changes. A separate global statement characterizes compact Fourier support by an exact Schwartz-convolution identity. Finally, escaping positive masses show that local order zero need not give a small Fourier order, even on one fixed ball.

All pairings are complex linear, without conjugation. The complete Schwartz and Fourier foundation, F1–F5, proves the test space, cutoff density, every Fourier seminorm estimate, both inversion identities, the exact Gaussian transform and all transposed differential identities. We use the equivalent increasing seminorms Their equivalence follows from and the monomial-weight comparison proved in F1. Thus a tempered functional satisfies for some . The proof is the finite-neighborhood rescaling argument in F5. It restricts to a distribution on compact tests by U008, Proposition 1.2.

A family is bounded when every is uniformly bounded on it. Weak convergence in means convergence on each test; the strong dual seminorms are for these bounded families. Our Fourier convention is F1–F5 proves on both spaces and the compatibility with ordinary integrable functions.

The scalar foundations, §§12, 13.1–13.5 and 13.7–13.10, supply completeness, compactness, FTC, exponential and trigonometric identities and cutoffs. The integration foundations, §§15.0–15.2 and 16, supply the absolute and dominated integrations used here. U021, B0–B2 and Theorems 1.1–2.1, supplies compact-test localization and convolution. The exact finite-jet theorem used in Solution 5 is proved in the angular foundations, A2–A3.

Two elementary weighted facts will be used repeatedly: For the first, square both sides and use For the second, the unit ball is contained in a finite cube. On each shell , , the integrand is at most and the shell lies in a cube of volume . The integral is bounded by a constant times . In particular a smooth function bounded by defines a tempered distribution, with a bound by .

When the phase absorbs the amplitude

Theorem 1.1 (an exact growth criterion). For even positive integers , the smooth function on the line has a tempered extension of its compact-test distribution if and only if . Such an extension is unique.

Proof: construction when . Put . On , define If , then for sufficiently large , on both tails because both powers are even. Differentiation of gives this same exponential times a finite sum of integer powers of . Every such term is bounded on : for any fixed polynomial power choose large in , applied to , leaving a bounded decaying factor. A finite remaining annulus causes no problem. Thus both are bounded. If , they are exactly and , also bounded.

Choose a smooth , zero on and one on . Such cutoffs follow by composing the supplied scalar bump with . Extend the following cutoff expressions by zero near the origin: The functions are bounded, and is smooth and compactly supported. Ordinary differentiation, including the cutoff terms, gives Define on All three integrals are absolutely convergent, and using the supplied arctangent integral . On a compact test, integration by parts and (1.3) identify this with . This is an actual continuous extension, without asserting absolute integrability of for every Schwartz test.

Proof: uniqueness by cutoffs. In any dimension choose equal to one on the unit ball and zero outside the radius-two ball; composition with again gives such a cutoff. For , every product-rule term in , , is supported where . A derivative of the cutoff of order also contributes at most , and is supported where . Since there for , the finite sum gives Thus in every seminorm, uniformly on bounded Schwartz families. Two continuous extensions agreeing on all compact tests agree on these approximants and therefore on every Schwartz test.

Proof: obstruction when . Suppose for a tempered extension. Choose a fixed nonnegative with . For large set The prefactor is constant in . These are legitimate compact tests.

For , the factorization of , or the FTC for , gives once . Both bounds tend to zero; exponential decay dominates every power by the series estimate used above. Also uniformly on the test support. Hence that ratio is at most two for large . The FTC then gives . The amplitude bound similarly gives . The elementary cosine bound on therefore yields Here the compact-test agreement makes the ordinary integral; all its real contributions have this lower bound.

For , differentiation of the rescaled bump gives The presumed tempered bound would thus bound . Its logarithm is Because , the first two terms are at least for large , while and . The logarithm tends to infinity. This contradicts every possible finite , proving necessity.

Thus and are tempered in this extension sense, whereas is not.

Convolution with an arbitrary Schwartz factor

Lemma 2.1 (global smoothing and the multiplier identity). If and , then is smooth. Every derivative has a polynomial growth bound with one common exponent. It defines a regular tempered distribution, and For fixed , convolution is continuous for both the weak and strong topologies of .

The cutoff proof in Theorem 1.1 and the seminorm definitions above are part of the test-space facts used in this lemma.

Proof: the ordinary Schwartz estimate. For , the integral and all its differentiated integrals converge absolutely. For each derivative, boundedness of derivatives of and integrability of justify differentiation by the FTC and dominated convergence. The weighted inequality above gives, for integers , The last integral is bounded by , whose finiteness was proved above. Thus convolution maps these tests into continuously in each variable. Absolute Fubini and also prove , since the absolute double integral is .

Proof: smooth parameter tests and growth. Choose . Reflection and the weighted inequality show Indeed , and -derivatives differ only by signs.

We verify differentiation in the test space. For a coordinate increment , the twice-FTC identity is Apply every -derivative through order , multiply by , and use (2.4) with the appropriate derivative of . On a compact set of 's and , the resulting remainder is at most . The once-FTC version gives continuity of every translated derivative in . Repeating for each coordinate and derivative proves smoothness with values in . Applying the continuous to these difference quotients gives The exponent is independent of . The preliminary weighted integral bound makes this smooth function a regular tempered distribution.

Proof: the integrated test and Fourier identity. Let and . Consider the test-valued integrand For every , (2.4) gives . This is integrable in , and its integral outside expanding cubes tends to zero by the extra Schwartz weights.

On a fixed -cube the map is uniformly continuous in every seminorm, by the translated-test calculation and the smooth scalar factor. A finite cubical Riemann sum differs from its integral in by at most the cube's volume times the corresponding modulus of continuity. It tends to zero as the mesh shrinks. To see the integral directly without assuming a vector integration theorem, differentiate its ordinary integral in ; bounded translated derivatives on the finite cube justify this, and the same weighted estimate bounds the supremum of each error. On the full space its pointwise integral is , which is Schwartz by (2.3). The tail bound proves convergence of the finite-cube integrals to this particular test in every seminorm.

Pass the Riemann sums and then the tails through using its bound. On the scalar side the integrand is absolutely integrable by (2.5). This proves the exact pairing For , absolute Fubini, bounded by , gives Both sides are Schwartz functions. Hence (2.6) and the transpose convention yield Multiplication by is legitimate on : the product rule bounds by a constant times . This proves (2.2) with its exact normalization.

Finally the test map is continuous by (2.3), and takes bounded sets to bounded sets. Formula (2.6) therefore gives continuity in the weak topology and, for every bounded , This is continuity in the strong topology. The same reasoning and the proved test continuity of give and , so both Fourier maps are also strongly continuous.

Compact Fourier support is exactly the fixed-point condition

Theorem 3.1 (spectral cutoffs). For , the following are equivalent:

1. Some satisfies .
2. The distributional support of is compact.

In this case has a smooth representative whose derivatives all obey polynomial bounds with one common exponent.

Proof. Suppose . Lemma 2.1 gives . Since is Schwartz, it tends to zero at infinity. Choose so that for . The reciprocal of is smooth there: for a nonzero complex-valued smooth function , is smooth by the real quotient rule. For every compact test supported in this exterior, Thus .

Conversely let be compact. A compact smooth cutoff equal to one on a neighborhood of exists by the supplied bump construction. Put ; then and . On compact tests : the product test vanishes near the support, and U021 B0's finite-partition support argument applies. Multiplication by is continuous on by the product rule. The cutoff density proved in Theorem 1.1 therefore extends the zero identity to all Schwartz tests. Lemma 2.1 now gives , and Fourier inversion gives . Its smoothing and common growth exponent follow from (2.5).

If is empty, the same finite-partition argument makes on all compact tests; density and inversion give , for which suffices.

The cutoff must be one on a neighborhood of the spectral support; matching a value at a point need not preserve derivatives of a point mass. Solution 4 checks the exact derivative correction.

A measure whose Fourier order is large on a small ball

Theorem 4.1 (escaping masses). Let be unbounded and any integer. There is a tempered distribution supported in , of local order zero, whose Fourier transform has order greater than in the unit ball.

Proof. Set . Inductively choose with and , using unboundedness at each step. The points escape every compact set, hence form a closed locally finite set: a convergent sequence among them must eventually lie in a compact ball, which contains only finitely many points. Define On a compact test only finitely many positive masses occur, so its pairing is bounded by a constant times the test supremum. On Schwartz tests the sum converges absolutely, since This is a tempered distribution of local order zero. Every chosen point is detected by a bump supported away from the other points, so its exact support is the selected set, contained in . The set itself need not be closed.

Choose real nonnegative with , and set . Ordinary differentiation and compact support give a smooth test supported in a compact subset of . Reflection in the integral, using that is real, gives The convolution identity here was proved in Lemma 2.1. Use the tests . They all have this one fixed compact support in the unit ball. Directly, , so The series is absolutely convergent for each by (4.2) on the Schwartz test . Its nonnegative terms cannot cancel.

If and the restriction of had order at most , its finite-order estimate on this fixed compact support would give The last inequality is the finite product rule, since each differentiated exponential contributes a component of . It contradicts (4.3), because and . If , (4.3) shows that the restriction is nonzero; its ordinary distributional order is nonnegative, hence greater than .

Exercises

Exercise 1 (foundation). For every real , prove that has a tempered extension. Give an explicit finite-seminorm bound using bounded functions and one derivative.

Exercise 2 (foundation). For a compact smooth cutoff on the unit ball and zero outside the radius-two ball, prove, for integers and , Deduce convergence uniformly on every bounded Schwartz family.

Exercise 3 (intermediate). On the line construct a Schwartz fixing by convolution. Explain why no Schwartz factor fixes .

Exercise 4 (intermediate). Let satisfy , , and put . Compute for . Determine the necessary and sufficient conditions on a general to fix this .

Exercise 5 (advanced). For , let Determine all tempered distributions with , in every dimension. Use the complete point-support proof in the angular foundations A2–A3 and the Gaussian proof in the Fourier foundation F3.

Exercise 6 (intermediate). In , take and . Prove a bound and show that has order greater than four in the unit ball.

Exercise 7 (intermediate). Solve on . Give a Schwartz factor fixing and determine its local distributional order and its exact Fourier order.

Exercise 8 (advanced). Prove that has a tempered extension by performing six integrations by parts on compact tests. Give an absolutely convergent formula on Schwartz tests; the bounded first-primitive argument is insufficient here.

Solutions

Solution 1. On put For any fixed real , for sufficiently large . The exponential-series estimate in Theorem 1.1 bounds both , including every polynomial factor produced by differentiation. Set Then . The formula has bound . It agrees with every original compact-test integral by integration by parts. Cutoff density gives uniqueness for positive, zero and negative .

Solution 2. In the product rule, the term with no cutoff derivative is supported where and is bounded, after multiplication by , by . A cutoff derivative of total order is supported in , has magnitude at most , and multiplies a derivative of of order at most . The extra weight gives . Summing the finitely many multiindex terms proves the bound. The right side tends uniformly to zero when ranges over a bounded Schwartz family. In particular every fixed continuous tempered functional has uniform cutoff convergence on that family.

Solution 3. Fourier foundation F5 gives . More generally inversion at gives, for a Schwartz test, an absolutely convergent integral since is Schwartz. Hence . Choose equal to one near , and take . Theorem 3.1 proves the required fixed point. Direct integration confirms On the other hand , , by the proved transpose differentiation identity. The density is nonzero at every real point; a phase-adjusted local bump detects it. Its support is the entire line, so Theorem 3.1 excludes any fixing Schwartz factor.

Solution 4. Since is Schwartz, the integrals with and converge absolutely. They give Differentiation of , justified in Fourier foundation F2, gives . Thus For arbitrary , the difference is . It vanishes for every exactly when , , by evaluating this affine polynomial at two values of . Equivalently , and direct test differentiation gives This verifies the derivative sign and the failure of value-only matching.