Regularity across a distinguished variable

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

A differential equation can force smooth dependence on time while leaving point masses in space. We prove this for equations monic in the time derivative, with arbitrary differential operators in the transverse variables. The proof uses finite continuous primitives, lowers their number of time derivatives, and then applies the fundamental theorem of calculus in distribution pairings. At an endpoint, the same construction explains exactly what must extend.

All pairings are complex linear. The earlier proofs used below are U008, Proposition 1.2 and Theorem 5.1, for finite-order actions on tests and uniform convergence on bounded smooth test families; U021, B0–B3 and Theorems 1.1–2.1, for localization, gluing and finite-regularity convolution; and U031, Theorem 6.1, for the full mixed continuous-primitive construction, including its finite compact version. Completeness of the support test spaces and the complete-metric Baire theorem are proved in functional foundations, Sections 6 and 14.1–14.2. We prove the required arbitrary-family estimate here.

A curve of distributions is a joint distribution

Let be open and an open interval. Write for the transverse variable. A curve is weakly continuous if every is continuous. A bounded test family means a family with a common compact support and uniform bounds for every derivative. The strong seminorms are Strong means that successive difference quotients converge for all these seminorms, and the resulting derivatives are strongly continuous.

Lemma 1.1 (one order on a compact time interval). For compact and a compact time interval , a weakly continuous curve defined on obeys for some finite . This includes a closed interval with endpoint values.

Proof. For , set Each is closed, as an intersection of inverse images of closed disks under continuous linear functionals. For each , compactness of bounds its continuous scalar pairing, so . The supplied completeness and Baire proofs imply that some has nonempty interior. Choose in that interior and with . Since also , subtraction bounds every by in that seminorm ball. For , apply this to ; it yields . If the seminorm is zero, arbitrary scalar multiples remain in the ball, forcing every pairing to be zero. The zero test space causes no exception. No enumeration of the time family is involved.

A weakly continuous curve now defines a joint distribution by Indeed the support of a compact joint test projects into and . To prove continuity of the integrand at , split its change into the pairing with , bounded by (1.1), and the change of , which tends to zero by weak continuity. The same bound gives Linearity and this fixed-support estimate prove distributional continuity. Finite vectors are treated componentwise.

Lemma 1.2 (uniqueness and strong continuity). The map (1.2) is injective on weakly continuous curves, and each such curve is strongly continuous.

Proof. Equality on says that the continuous scalar difference integrates to zero against every compact smooth . If , multiply by a complex scalar so its real part is positive at ; it stays positive in a small interval. A nonzero nonnegative bump in that interval gives a nonzero integral, a contradiction. Thus the curves agree for every test and time.

For a sequence , weak continuity and U008, Theorem 5.1, give for every bounded . If the corresponding neighbourhood estimate for one failed, choose within of with that estimate failing; this contradicts the sequential conclusion. Hence every strong seminorm is continuous along the curve. The argument applies with one-sided time neighbourhoods as well.

We also need a joint version of the zero-derivative fact.

Lemma 1.3 (constant transverse distribution). If and , there is a unique such that .

Proof. Choose with , and define . On every compact transverse support, the joint finite-order bound for and the fixed derivatives of give a finite-order bound for . For any , put extending the compact integrand by zero. Differentiation under this finite integral makes smooth. Since , its time support lies in the convex hull of the two compact time supports, still compactly inside the interval ; its transverse support is also compact in . Therefore is a joint test and . Pairing with gives , the required tensor identity. Testing with proves uniqueness.

Removing distinguished derivatives

Lemma 2.1 (continuous transverse primitives). A continuous distribution curve is locally a finite sum of transverse derivatives of jointly continuous functions. This holds on a compact time interval, including its finite endpoints.

Proof. Fix a relatively compact transverse region and a smooth compact cutoff near its closure. Set , extended by zero to . Lemma 1.1 and the product rule give one order and one bound for all on the compact time interval. Insert one further fixed cutoff near their common support to act on noncompact kernels. For an integer , let U031, Theorem 6.1, proves by repeated scalar integration by parts. The kernel is , hence , and U008 extends the common order- action to its cut-off translates. U021, Theorem 2.1, identifies this pairing with convolution and proves .

Here is the needed parameter check at this finite regularity. If , a fixed compact test can be approximated in on a slightly larger fixed compact support by smooth tests , using the proved mollification. The bound first lets , then . Thus the pairing converges for . If also , the cut-off kernels converge to in , by uniform continuity of the finitely many derivatives on a compact set. Split off that kernel difference and use the same bound. This proves joint continuity of . Where , . The estimates include endpoint parameters. When the assertion is simply scalar continuity; when is empty it is vacuous.

Multiplication by smooth coefficients must preserve the number of distinguished derivatives in such a representation. The exact identity, for a multi-index in all variables, is To prove the one-derivative case, pair both sides of with a test, expand , and cancel. Induction repeats that identity in each coordinate. At each step the choices of which derivatives hit give the binomial coefficient, their negative signs give , and commuting coordinate derivatives gives the displayed formula. This proves (2.2) for distributions directly.

Lemma 2.2 (a first-order equation gives continuity). Suppose a finite distribution vector satisfies where is a continuous -valued vector and is a matrix of transverse differential operators with smooth coefficients. Then has a unique strongly continuous curve representative. Arbitrary locally finite transverse orders are allowed.

Proof. Fix a product box whose closure is inside . A cutoff of equal to near a slightly larger box is compactly supported. The finite compact construction in U031 writes, on that larger box, with finitely many continuous vector coefficients. Let bound the occurring integers . We show that whenever , another representation has bound on a smaller transverse box and the same smaller time interval.

Lemma 2.1 gives a representation with only transverse derivatives. Acting by on (2.4) and using (2.2) gives with finitely many continuous coefficients. In fact the transverse differentiations cannot increase , and moving a smooth multiplier inside the derivatives subtracts time derivatives rather than adding any. Choose inside the time interval and set The new coefficients are continuous and the largest time derivative is at most . The scalar fundamental theorem, or its integration-by-parts version against a joint test, gives . Lemma 1.3 therefore gives Localize in a slightly smaller transverse box. U031's transverse continuous-primitive construction expresses it with finitely many continuous coefficients and no time derivatives. Combining this with proves the reduction.

There are only reductions. Predetermine nested transverse boxes around the point with compact inclusions; at each step choose cutoffs on the next larger box. All steps are therefore available, and the final representation has . Its scalar pairings with transverse tests are continuous: they are finite integrals of continuous coefficients against fixed differentiated tests. Fubini and the definition of distributional derivatives identify their joint distribution with .

To pass from boxes to all of , first compare local representatives on overlapping boxes. Tensor tests and Lemma 1.2 force equality at every common time on the overlap. At a fixed time, U021 B0 glues these compatible local distributions. For continuity near any time, decompose a fixed compact transverse test into finitely many subordinate tests, choose a common time neighbourhood for that finite cover, and use continuity of its finitely many local pairings. This gives a global weakly continuous curve; Lemma 1.2 makes it strongly continuous and proves uniqueness.

The full monic equation

Theorem 3.1 (regularity in the distinguished variable). If , , and where each is a transverse differential operator with smooth coefficients and is a continuous distribution curve, then has a unique representative in , in the strong topology. Its curve derivatives give its joint distribution derivatives. No commutativity of the 's is required.

Proof. First take the continuous representative supplied by Lemma 2.2. Its curve is continuous. Indeed a fixed transverse test is sent by the transpose to , with the same compact support and smoothly varying derivatives. Here the transpose is defined by ; there is no conjugation. Split its moving-test pairing into a test difference, controlled by Lemma 1.1, and a fixed-test difference, controlled by weak continuity.

Define , with oriented integrals. A common order on the compact interval between and nearby 's proves that is a distribution and a weakly continuous curve. Fubini and scalar integration by parts, with the same common-order bound, give . Lemma 1.3 applied to , and uniqueness of continuous representatives, give For every bounded test family, The right side tends to zero by strong continuity of . Thus is strongly with derivative .

For (3.1) apply this result to the companion vector Its first equations are and its last is (3.1). All entries in the resulting operator matrix are transverse operators, and its forcing is . Each component is strongly . Starting with the first component, the successive identities identify its first derivatives with the remaining components; differentiating the last gives the -th continuous derivative. Integration by parts gives all the corresponding joint derivatives. Uniqueness follows from Lemma 1.2.

For , the equation is , with the conclusion . Empty , , disconnected , and unbounded intervals are included; all arguments above are local.

We give a direct point-parameter proof for the examples. If are smooth scalar functions with real and , then is strongly smooth and Pairing with gives , whose scalar chain rule yields the formula. To justify it strongly, fix a compact time interval and a bounded test family. The product and chain rules bound every there by a common constant : the derivatives of are bounded and the test derivatives through order are uniformly bounded. Twice applying the scalar fundamental theorem gives whose absolute value is at most , for either sign of . Division by proves the first strong derivative. Each subsequent derivative is a finite sum of the same form, so induction proves all orders. Their first-derivative bounds also give strong continuity by the fundamental theorem. The uniform bounds hold when the point lies outside the common test support as well.

Example 3.2 (a smooth curve of moving point sources). For , The equation follows by integrating the total derivative of the compactly supported function . The just proved formula gives and proves all higher strong derivatives. These distributions remain spatial point masses at every time.

What a boundary hypothesis must control

At a finite endpoint, means strong continuity of the derivatives through order , with the derivatives defined one-sidedly there. No condition is imposed at an infinite endpoint.

Example 4.1 (extendible distributions with a divergent trace). On and , put and . Then . The forcing extends continuously by zero, and extends distributionally across zero. To check the latter without assuming a principal-value theorem, for a test on an interval containing zero choose inside that interval and write Extend the test by zero when needed. The first integrand has absolute value at most , by the scalar fundamental theorem. On a fixed test support the remaining integral is bounded by a constant times . The formula is therefore a distribution, restricts to for positive , and is denoted . Tensoring it with the constant transverse function gives the extension of , for example to .

Nevertheless, a transverse test of integral one pairs with as , so there is no distributional trace at zero. The coefficient fails to extend smoothly there. Distributional extendibility of the solution alone is insufficient.

Theorem 4.2 (regularity up to a boundary). In Theorem 3.1, suppose the coefficients of all extend smoothly to , where is an open neighbourhood of ; suppose extends as a distribution there; and suppose is continuous on with values in . Then The assertion is local at each finite endpoint. The extended distribution need not satisfy the equation outside .

Proof. Consider the left endpoint ; the right endpoint is identical with reversed one-sided intervals. Differentiate a chosen extension of to form an extension of its companion vector. Cut it off on a product box crossing . U031's finite compact primitive theorem gives (2.4) with finitely many coefficients continuous across , so their restrictions extend continuously to the closed half-box.

On the same one-sided compact time interval, Lemmas 1.1–2.1 give the forcing a representation by transverse derivatives of coefficients continuous up to . These functions can be extended continuously across by their endpoint values. We now perform the reduction of Lemma 2.2 only on the open side , keeping track of continuous endpoint values of every coefficient.

In forming , all coefficient derivatives in (2.2) extend smoothly, and therefore are bounded and continuous on the smaller closed box. Multiplying them by the continuous primitive coefficients preserves continuous endpoint values. The integrals in (2.5), taken from a fixed interior , also have continuous limits at , because their integrands are continuous on a compact half-box. Lemma 1.3 on the interior supplies , whose transverse continuous-primitive representation is independent of time. Thus each reduction lowers the normal degree and preserves continuous extension of the coefficients. Finitely many nested spatial boxes permit all reductions. The final degree-zero representation defines a weakly continuous curve up to , and Lemma 1.2 makes it strongly continuous.

These endpoint curves agree on overlaps: they agree for by interior uniqueness, and their scalar pairings have the same limits at . U021 B0 therefore glues them at the endpoint. The finite test decomposition used in Lemma 2.2 proves continuity of the resulting global curve on .

The right side of the companion system is now continuous on the closed half-interval, by the moving-test estimate in Theorem 3.1. Identity (3.2), initially valid for interior , extends to in every scalar pairing; a common order controls the integral. The same strong average estimate, with at , gives its one-sided derivative . The companion identities then give all one-sided derivatives and their strong continuity. Every equation used in this argument was on the original interior or was its continuous limit.

Traces create boundary sources under zero extension

Proposition 5.1 (all normal derivatives of a zero extension). Let be finite and . Its zero extension is the distribution For , The first term denotes the joint distribution of the restricted curve extended by zero, not a product of arbitrary distributions.

Proof. Lemma 1.1 gives the fixed-support bound for (5.1). For a joint test set . Its scalar pairing obeys To verify this, split the difference quotient as The first term tends to . In the second, the test quotient tends in every derivative norm to ; its error is controlled uniformly by Lemma 1.1, and its fixed limiting test is handled by weak continuity of . The two terms on the resulting right side are continuous, by the same moving-test argument. Scalar integration by parts consequently gives on every joint test. Apply this to and differentiate the two existing endpoint masses. Repeating the step gives exactly the two sums in (5.2). Tensor derivatives are defined by , which also checks every sign.

This specifies one particular extension. Another extension can include distributions supported on the endpoint slices, and need not obey (5.2). For a homogeneous equation, the zero extension must include the displayed boundary forcing unless those terms cancel.

Exercises

Exercise 1 (basic: a curve with two spatial jets). Solve on Describe every joint-distribution solution as a curve. For the solution with , compute the first derivative of its zero extension.

Exercise 2 (basic: coefficients acting on a normal jet). For and , find with all signs. Show that is not a continuous distribution curve and cannot solve a homogeneous first-order equation (2.3) with smooth transverse coefficients.

Exercise 3 (intermediate: all initial jets for a pure normal equation). Given , a continuous curve , , and arbitrary , prove that the unique solution of with these initial derivatives is Include and all strong derivative assertions.

Exercise 4 (intermediate: sharp time regularity). For , and on , set . Find , including its value at zero. Prove that this gives continuous forcing in a pure normal equation, but .

Exercise 5 (intermediate: second-order boundary forcing). For on , let be its zero extension. Compute , distinguishing spatial derivatives from time derivatives in every endpoint term.

Exercise 6 (advanced: arbitrarily high apparent normal degree). For , put . Prove Check the transport equation and strong smoothness of the curve. Explain why the largest normal degree of a chosen representation is not intrinsic.

Exercise 7 (advanced: a variable spatial drift and every point jet). For , , , find all smooth real and complex , with these initial values at zero, for which solves . Check the joint equation and the boundary masses in the first derivative of its zero extension from .

Exercise 8 (advanced: why distributional extension is needed). For , consider Prove smoothness, periodicity and the heat equation. Construct compact spatial tests isolating each Fourier mode. Show that there is no distributional trace at zero and no joint distributional extension across zero; use time tests at scale for the latter.

Complete solutions

Solution 1. The curve has the required derivative. Lemma 1.3 applied to any other solution minus gives all solutions Multiplication of a fixed distribution by a smooth scalar function has the corresponding strong derivatives: for every bounded , the seminorm of the error is the scalar error times the finite number . Thus all these curves are strongly smooth. The initial value selects , giving . Formula (5.2) gives

Solution 2. Directly, so If came from a continuous curve, pair in with a test equal to one at zero. The resulting continuous function of would represent . Its restriction off zero would be zero, by the continuous-function test argument in Lemma 1.2; continuity would then make it zero also at zero. But , as seen by testing with times a cutoff equal to one near zero. This is a contradiction. Lemma 2.2 would make any solution of the proposed homogeneous equation a continuous curve, so cannot be one.

Solution 3. Denote the integral term by . Lemma 1.1 bounds all its test pairings by a single finite-order seminorm on compact time intervals, proving it is a distribution. Scalar differentiation of the oriented integral gives For the upper-end kernel is zero; for the next derivative is the fundamental-theorem derivative. These identities follow also for by writing the integral as minus the integral from to . Each displayed curve is weakly continuous, using the common bound and continuity of its scalar integrands, hence strongly continuous by Lemma 1.2. Differences of successive curves are integrals of the next curve by scalar calculus and Fubini; the strong average estimate from Theorem 3.1 proves that their derivatives are strong derivatives. All derivatives of of order less than vanish at .

Differentiating the polynomial term gives initial derivative at order , and zero at order . This proves existence with the claimed data. Any joint solution is by Theorem 3.1. The difference of two solutions with the same initial data has zero -th derivative. In each scalar pairing its -st derivative is constant by the scalar fundamental theorem and zero at . Repeating downward makes every lower derivative zero. Test pairings separate distributions, proving uniqueness.

Solution 4. Put . Direct differentiation on each half-line gives For any , the half-line derivative of order is a constant times a sign and . Inductively its claimed value zero at the origin is its true derivative value: the previous derivative's difference quotient has magnitude a constant times . Thus all these derivatives are continuous. Multiplication by fixed gives strong derivatives as in Solution 1, and scalar integration by parts shows that the joint derivatives are the same.