Jets, supported distributions and local operators

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, September 2026. Checked once by GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Public domain (CC0).

Source/proof self-check and prerequisite integration by GPT-6 Astra (OpenAI), Ultra, October 2026. Historical authorship and component terms are retained.

A distribution concentrated on a point can measure derivatives of a test there. A distribution concentrated on a plane can measure normal derivatives along that plane, with distributions as tangential coefficients. The total order limits both kinds of measurement: spending one derivative in the normal direction leaves one fewer tangential derivative available. We prove that precise order bound, then use it to characterize operators whose kernels lie on the diagonal.

The preceding kernel lesson supplies test-space topology, cutoffs, support localization, product density and the local kernel theorem. The smoothing lesson, especially its integral Taylor proof, supplies compact distributions and all the Taylor estimates used below. The tensor lesson supplies products, smooth multiplication and parameter differentiation. The supplied scalar foundation, §§12–13, provides Euclidean compactness, uniform continuity, derivative rules, the fundamental theorem and smooth cutoffs; the measure foundation, §§15.0–15.1, provides integration and Fubini. Smooth approximation and the special normal-jet extension are proved below. We use ordinary derivatives and bilinear pairings; no powers of are hidden in the coefficients. The external sources at the end give exact comparisons and broader extension background, not substitutes for these programme proofs.

What a supported distribution can detect

The order of a distribution is at most if, on each fixed compact test-support set, its pairing is bounded by a constant times the supremum of derivatives through order . The constant may depend on the compact set. For a compactly supported distribution, one compact neighborhood of its support gives $|u(f)|\le C\max_{|\gamma|\le k}\sup_M|\partial^\gamma f| \tag{1.1}$ for all smooth , after a cutoff equal to one near the support. The controlling set is a neighborhood of the support; (1.1) is not an assertion that the supremum may be restricted to the support itself.

We first supply the approximation assertion. If , choose a nonnegative smooth integral-one supported in the unit ball and put . Such a is obtained from the scalar foundation's nonzero nonnegative bump by dividing by its positive integral. The support of lies in . Writing the convolution as , every -derivative exists by taking difference quotients under the integral: the segment formula bounds their remainders uniformly on the fixed compact integration region by an integrable constant times the increment. Iteration proves smoothness. The same reasoning applies to a bounded measurable function with compact support, which will be used for an indicator below.

In the alternative expression , derivatives through order may instead be taken on . Uniform continuity of each compactly supported derivative gives $\begin{gathered} \|\partial^\gamma f_\varepsilon-\partial^\gamma f\|_\infty\\ \le \sup_{|h|\le\varepsilon} \|\partial^\gamma f(\cdot-h)-\partial^\gamma f\|_\infty\\ \longrightarrow0,\qquad |\gamma|\le k. \end{gathered}$ For a compactly supported function on an open set, zero extension is , and sufficiently small keeps all supports in one compact subset of that open set. For a function specified only near a compact set, multiply it first by a smooth cutoff supported in that neighborhood. These statements also show that a function vanishing near a fixed compact set can be approximated with every approximant still vanishing on a smaller neighborhood.

An order-at-most- distribution acts on compactly supported functions. To define the pairing, approximate such a function by smooth ones, with one common compact support and uniform convergence of derivatives through order . Convolution after a slightly larger cutoff supplies these approximations. The order bound makes their pairings Cauchy and makes the limit independent of the approximation. For a compact distribution the same construction acts on arbitrary functions, by a cutoff near its support.

Theorem 1.1 (vanishing jets are invisible). Let be a distribution of order at most , and let . If $\begin{gathered} \partial^\gamma f(x)=0\\ (x\in\operatorname{supp}u,\ |\gamma|\le k), \end{gathered} \tag{1.2}$ then . If is compactly supported, need not have compact support.

Proof. The zero distribution is immediate. First assume has compact support in Euclidean space. A compact distribution on an open set can be put in this form by pairing a global test with its restriction multiplied by a fixed cutoff equal to one near . Multiplication by that cutoff preserves order and support. Functions used near can likewise be extended after a cutoff. For , write for the open -neighborhood of . Take a nonnegative smooth mollifier of integral one supported in the unit ball, and put $\chi_\varepsilon=\mathbf1_{F_{2\varepsilon}}*\rho_\varepsilon.$ For small , this is a compact smooth function supported in a fixed neighborhood of . It equals one on , is supported in , and satisfies $|\partial^\beta\chi_\varepsilon|\le C_\beta\varepsilon^{-|\beta|}. \tag{1.3}$ These estimates follow by differentiating the mollifier and taking its norm; the indicator is bounded by one.

Compactness of the nonempty set ensures that the continuous distance function attains its minimum there. The integral Taylor formula from the preceding lesson, along the segment from that nearest point and with (1.2), gives, uniformly near , $\begin{gathered} d_F(x)=\operatorname{dist}(x,F),\\ |\partial^\gamma f(x)|\\ \le C d_F(x)^{k-|\gamma|}\omega(d_F(x)),\\ |\gamma|\le k, \end{gathered} \tag{1.4}$ where as . One can take a constant multiple of a common modulus of continuity of the order- derivatives on a fixed compact neighborhood. For , (1.4) is the continuity estimate against a derivative which is zero on . For lower orders it follows from the integral Taylor remainder along the segment to the nearest point.

The functions and agree near , so they have the same pairing with . This also holds for functions: approximate a function vanishing on a neighborhood of by smooth functions still vanishing on a smaller neighborhood, and use the order estimate. The product rule, (1.3) and (1.4) give $\max_{|\gamma|\le k}\|\partial^\gamma(\chi_\varepsilon f)\|_\infty \le C'\omega(3\varepsilon) \quad(0<\varepsilon\le1).$ The possible powers are at most one and do not obstruct convergence to zero. Estimate (1.1) proves .

For an arbitrary distribution and compactly supported , multiply by a cutoff equal to one near . The resulting compact distribution has order at most and support contained in , so the case just proved applies. Its pairing with equals that of . For a compact and a noncompact , first cut off near . Its vanishing jets there are preserved.

The theorem concerns jets that vanish. It does not provide a norm estimate using only the values of a jet on an arbitrary closed support. An extension problem enters when we prescribe a nonzero jet.

Extending one normal jet without losing derivatives

Write , with , and let be a multi-index in the variables. Put , and fix an integer . For a function on Euclidean space, denotes the maximum of the suprema of all derivatives through order .

Lemma 2.1 (a controlled normal-jet extension). Let . For each , there is a compactly supported , linear in , such that $\begin{gathered} \partial_t^\beta G(0,z)=\\ \begin{cases}g(z),&\beta=\alpha,\\0,&\beta\ne\alpha,\end{cases}\\ |\beta|\le k, \end{gathered} \tag{2.1}$ and $\|G\|_{C^k}\le C_{k,\alpha}\|g\|_{C^s}. \tag{2.2}$ If ranges over functions supported in one compact set, all these extensions have one common compact support. The assertion includes .

Proof. We give the construction and its full regularity estimate.

Choose a smooth compactly supported function on with $\begin{gathered} \int\rho=1,\\ \int w^\gamma\rho(w)\,dw=0\\ (1\le|\gamma|\le k). \end{gathered} \tag{2.3}$ Positivity is not required. Such a function is easy to construct with finitely many dilations. Start with any integral-one smooth bump . Choose distinct positive numbers , and solve the Vandermonde system $\sum_{j=0}^k c_j\lambda_j^b= \begin{cases}1,&b=0,\\0,&1\le b\le k.\end{cases}$ The coefficients have the explicit formula $c_j=\prod_{\substack{0\le i\le k\\i\ne j}} \frac{-\lambda_i}{\lambda_j-\lambda_i}.$ Let . Every polynomial of degree at most satisfies : the difference has degree at most and vanishes at the distinct nodes. A nonzero degree- polynomial has at most distinct zeros, since is divisible by by the finite geometric-sum identity, and induction removes one root at a time. Evaluating the interpolation identity at zero with , including , proves exactly the displayed system. The case uses an empty product equal to one.

Then has (2.3), since a moment of total degree scales by .

For , set , and for define $G_0(t,z)=\frac{t^\alpha}{\alpha!}(g*\rho_{|t|})(z). \tag{2.4}$ The scale is smooth off the plane. The moment conditions make convolution reproduce every polynomial of degree at most , independently of the scale. In particular, differentiating that convolution in kills each such polynomial whenever at least one derivative is taken.

Let , , and . Differentiating a scaled convolution kernel in through total order , and in through total order , costs at most a constant times in its norm. More explicitly, a derivative of the convolution kernel of total orders has the form $|t|^{-n-b-c}A_{\beta,\gamma} \left(t/|t|,w/|t|\right),$ where is smooth on the unit sphere in the normal variable and is supported in a fixed compact ball in its second variable. This follows by induction from the chain rule: differentiating the radial scale or either normalized argument supplies one further factor of , with smooth sphere coefficients. Those coefficients and the derivatives of the fixed bump are bounded on that compact product. The substitution in the absolute integral then gives the asserted bound. The differentiated kernels remain supported in , with fixed .

At a fixed , subtract the Taylor polynomial of through degree . Its remainder on that ball is bounded by $C|t|^s\omega_g(R|t|), \tag{2.5}$ where , uniformly in . For , this is the modulus of continuity of . For , it is a constant multiple of the modulus of continuity of the order- derivatives, by the integral Taylor formula. The differentiated kernel estimate therefore gives $\begin{gathered} D_{\nu,\gamma}(t,z)=\\ \partial_t^\nu\partial_z^\gamma(g*\rho_{|t|})(z),\\ D_{\nu,\gamma}(t,z)=\\ \begin{cases}\partial_z^\gamma g(z),&\nu=0,\ c\le s,\\0,&\text{otherwise},\end{cases}\\ {}+O\big(|t|^{s-|\nu|-c}\omega_g(R|t|)\big). \end{gathered} \tag{2.6}$ The leading term is found by applying the differentiated convolution to the fixed Taylor polynomial, using polynomial reproduction. If , that derivative of the polynomial is zero. If , its convolution is scale-independent, so that term is zero. This justifies (2.6) even when the derivatives of appearing in the remainder calculation do not exist beyond order .

Apply the product rule to (2.4). When coordinatewise and , the leading term is $\frac{t^{\alpha-\beta}}{(\alpha-\beta)!}\partial_z^\gamma g(z). \tag{2.7}$ Otherwise there is no leading term. Every remainder is bounded by $C|t|^{k-b-c}\omega_g(R|t|), \tag{2.8}$ because the degree of the prefactor and add to . Thus all derivatives of through order , initially defined off the plane, extend continuously to it. Their traces are zero except for , , when the trace is .

These extensions really are the derivatives of a function. One can verify this successively in derivative order. Along a normal coordinate segment leaving the plane, use the fundamental theorem of calculus off the plane and then the continuous limiting derivative at its endpoint. Along a tangential segment in the plane, the asserted nonzero traces are derivatives of , whose required orders are at most ; all other traces are zero. This proves differentiability in each coordinate across the plane, with the prescribed continuous partial derivatives, at each order up to .

On , the same calculations with bounded by give a uniform bound for . Choose a compact smooth , equal to one near zero and supported in , and put , with the continuous extension at the plane. The product rule proves (2.2), and multiplication by preserves (2.1). If is supported in , the extension is supported in the compact product of with . This proves every assertion. If , there is no convolution: take , with the same conclusions.

The elementary extension would require too many tangential derivatives away from the plane. The scale-dependent smoothing in (2.4) supplies those missing derivatives, while the vanishing moments preserve the prescribed jet.

The exact structure on a plane

Theorem 3.1 (normal jets of a supported distribution). Let have order at most , with $\operatorname{supp}u\subset\{0\}\times\mathbb R^n.$ There are unique compactly supported distributions on , for , such that $u(f)=\sum_{|\alpha|\le k}w_\alpha\big(\partial_t^\alpha f(0,\cdot)\big) \tag{3.1}$ for every smooth . Each has order at most , and its support is contained in the tangential projection of . Equivalently, $u=\sum_{|\alpha|\le k}(-1)^{|\alpha|} (\partial_t^\alpha\delta_0)\otimes w_\alpha. \tag{3.2}$ Formula (3.1) also holds for all functions, using the continuous extensions of the pairings.

Proof. Choose near zero with compact support, and define $\begin{gathered} w_\alpha(g)=u\left(\theta(t)\frac{t^\alpha}{\alpha!}g(z)\right),\\ g\in\mathcal D(\mathbb R^n). \end{gathered} \tag{3.3}$ Initially this is a distribution of order at most . Its support lies in the stated projection, because the test in (3.3) vanishes near when does. It is independent of with the specified value near zero.

Taylor expansion in the normal variables shows that $f(t,z)-\theta(t)\sum_{|\alpha|\le k} \frac{t^\alpha}{\alpha!}\partial_t^\alpha f(0,z)$ has every total derivative through order equal to zero on the plane. Theorem 1.1 therefore makes its pairing with zero, proving (3.1) on smooth functions. Testing a proposed representation on the functions in (3.3) isolates each coefficient, proving uniqueness. The signs in (3.2) are precisely the signs of distributional differentiation of the point mass.

To obtain the sharper coefficient order, apply Lemma 2.1 to a smooth compactly supported , using . Let be its extension with only the normal jet nonzero. The difference between and the test in (3.3) has all total derivatives through order zero on the plane. Theorem 1.1 applies to that difference, so $w_\alpha(g)=u(G).$ Estimate (1.1) and (2.2) give $|w_\alpha(g)|\le C'\|g\|_{C^{k-|\alpha|}}, \tag{3.4}$ proving the claimed order. Finally, approximate an arbitrary function on a neighborhood of the compact support by smooth functions in that norm. Its normal trace of order converges in on tangential compact sets. The bounds just proved permit passage to the limit in (3.1).

When , the coefficients are scalars, and the theorem says that a distribution supported at one point is a finite sum of derivatives of a point mass, with degree bounded by its order. Translation moves the point from zero to any prescribed point. Conversely every such finite sum has that support containment and finite order. This is the point-supported structure theorem; the plane theorem gives its tangential version with the additional order information.

Localizing an arbitrary distribution supported on a plane by compact cutoffs gives the same description locally. The number of normal derivatives can increase from one region to another. Coefficient uniqueness makes those local descriptions agree on overlaps.

Corollary 3.2 (coordinate multiplication singles out the point mass). Let be open, , and suppose obeys for every . If , then for a unique scalar . If , then ; every scalar multiple of the restricted is then zero, so no uniqueness of that scalar is asserted.

Proof. Near any nonzero point some coordinate is nonzero. Multiplication there by its smooth reciprocal shows . Consequently the support is contained in . If the origin is absent, empty support gives the zero distribution. If it is present, choose equal to one near it. Multiplication by does not change . The pairing extends it to a compact distribution on supported at zero: its continuity follows from the test estimate on the fixed support of . Apply the point-supported specialization of Theorem 3.1 to this extension and restrict back to . It gives a finite sum with unique coefficients. The product rule on test jets gives

\begin{gathered} x_j\partial^\alpha\delta_0\\ =-\alpha_j\partial^{\alpha-e_j}\delta_0\quad(\alpha_j>0),\\ x_j\partial^\alpha\delta_0=0\quad(\alpha_j=0). \end{gathered} \tag{3.5}

For fixed , every resulting jet index receives exactly the coefficient . Their independence and the equation make each such coefficient zero. Every positive-degree has for at least one coordinate, so only remains. A test equal to one near zero recovers uniquely.

A finite order on an unbounded plane

Work on , with and . Say a distribution has order at most when the same integer works on every compact test-support set. The constants may depend on that set. An arbitrary distribution need not have such an integer.

Theorem 3.3 (finite-order plane distributions). Suppose has order at most and support in . There are unique distributions , for , such that $\begin{gathered} u(H)\\ =\sum_{|\alpha|\le k}w_\alpha(\partial_t^\alpha H(0,\cdot)),\\ H\in\mathcal D(\mathbb R^{d+n}). \end{gathered} \tag{3.6a}$ Their orders are at most . Their supports lie in the tangential projection of . In distribution notation, $\begin{gathered} u\\ =\sum_{|\alpha|\le k}(-1)^{|\alpha|} (\partial_t^\alpha\delta_0)\otimes w_\alpha. \end{gathered} \tag{3.6b}$ Equation (3.6a) extends to compactly supported functions. It does not give a pairing of a noncompact distribution with every noncompact function.

Proof. Fix equal to one near zero, and define $w_\alpha(g)=u\left(\theta(t)\frac{t^\alpha}{\alpha!}g(z)\right). \tag{3.6c}$ For on a fixed compact support , the function inside has support in . The product rule bounds each of its derivative seminorms by a constant times a derivative seminorm of . This is a continuous map between the fixed-support test spaces, so (3.6c) defines a distribution. Replacing by another permitted cutoff changes the test by a function vanishing near the plane; its pairing is zero. If the support of misses the projection of , the test likewise vanishes near that support. This proves the support assertion. The projection in question is closed because projection from the plane to is a homeomorphism.

Choose an arbitrary compact , a tangential cutoff near , and a compact normal cutoff near zero. The distribution $U=\eta(t)\chi(z)u$ is compactly supported, supported on the plane and of order at most . To see the last assertion, pair with and apply the order bound on its one fixed compact support; the product rule costs only derivatives through . The compact plane theorem above applies to . Its coefficient of index , calculated by (3.6c), is : the normal cutoff can be removed because it is one near the plane. The sharp coefficient estimate of that theorem gives order at most for . On every test supported in , it agrees with . As was arbitrary, this is the asserted global finite order for the coefficient, with constants still allowed to vary with .

Now fix , and choose near the tangential projection of its support. The cutoff localization above satisfies . Indeed vanishes near the intersection of the plane with the support of , and it is already zero near all other possible support points that matter. Apply the compact plane theorem to . The normal traces of are supported in the region where , so equals the summand in (3.6a). This proves that formula.

For any multi-index , the test $H_\beta(t,z)=\theta(t)t^\beta g(z)/\beta!$ has normal trace at index and zero at every other index. A monomial has a nonzero derivative at zero only when the derivative index equals its degree index, and all nonzero-order derivatives of vanish near zero. Thus these tests isolate each coefficient. They prove uniqueness even between different finite index bounds, by adding zero coefficients to the shorter family.

For ordinary derivatives and bilinear pairings, $((\partial_t^\alpha\delta_0)\otimes w_\alpha)(H) =(-1)^{|\alpha|} w_\alpha(\partial_t^\alpha H(0,\cdot)).$ The additional sign in (3.6b) therefore gives exactly (3.6a).

Finally, a compactly supported function is approximated in that norm by smooth functions with a common compact support. The left pairings converge by the order- estimate. Its normal trace of index converges in ; the coefficient estimate just proved makes every right pairing converge. The finite sum passes to the limit. For , the same proof has scalar coefficients and no tangential cutoff.

The coefficient orders determine the total order

Theorem 3.4 (exact normal-jet order). Let be a finite nonempty set of distinct transverse multi-indices, and let every , , be a nonzero distribution with finite exact order . Put $\begin{gathered} u\\ =\sum_{\alpha\in A}(-1)^{|\alpha|} (\partial_t^\alpha\delta_0)\otimes w_\alpha,\\ N=\max_{\alpha\in A}(|\alpha|+s_\alpha). \end{gathered} \tag{3.7a}$ Then and .

Proof. The isolating tests from Theorem 3.3 show that would imply for every and every , contrary to the hypotheses.

For a fixed compact support of , the trace is supported in . The order of the coefficient gives $\begin{gathered} h_\alpha(z)=\partial_t^\alpha H(0,z),\\ |w_\alpha(h_\alpha)|\\ \le C_{K,\alpha}\max_{|\gamma|\le s_\alpha} \|\partial^\gamma h_\alpha\|_\infty\\ \le C_{K,\alpha}\max_{|\nu|\le N}\|\partial^\nu H\|_\infty. \end{gathered} \tag{3.7b}$ Summing finitely many bounds proves order at most .

If , nonzero and order at most zero already give exact order zero. Suppose and, for a contradiction, has order at most . Theorem 3.3 gives its expansion with normal indices of degree at most and coefficient orders at most . Uniqueness identifies those coefficients with the family in (3.7a), after adjoining zeros on both sides. Choose an index attaining . If , uniqueness makes its coefficient zero, a contradiction. Otherwise uniqueness and the sharp coefficient bound give , also a contradiction. Order at most is impossible, so the exact order is .

Distinct normal indices prevent cancellation of the highest total order. Tangential distributions can have the same support, or fail to have compact support, without changing this argument.

Corollary 3.5 (any finite-order tensor partner). For every nonzero of finite exact order , $\operatorname{ord} ((\partial_t^\alpha\delta_0)\otimes v) =|\alpha|+s. \tag{3.8}$

Proof. Use Theorem 3.4 with the one coefficient . Multiplication by a nonzero scalar preserves exact order. The two signs multiply to one, so the resulting distribution is exactly the tensor in (3.8). Theorem 3.4 gives the equality. No compactness, point support or temperedness of is required.

The point-supported problem below asks for a direct scaled-test proof. Formula (3.8) also covers noncompact or non-point-supported partners.

Kernels on the diagonal

Let be open. A linear map is support-preserving if $\begin{gathered} \operatorname{supp}T\phi\subset\operatorname{supp}\phi\\ (\phi\in\mathcal D(X)). \end{gathered} \tag{4.1}$

Theorem 4.1 (continuous local operators with distributional coefficients). Suppose is weakly continuous. The following conditions are equivalent:

is support-preserving.
Its kernel is supported in the diagonal .
There are unique distributions , locally finite as a family, such that $T\phi=\sum_\alpha a_\alpha\,\partial^\alpha\phi. \tag{4.2}$

Here locally finite means that every point has a neighborhood on which all but finitely many vanish. Consequently every compact set meets the supports of only finitely many nonzero terms. The order need not be bounded on all of . Every locally finite expression (4.2) defines a weakly continuous support-preserving map.

Proof. If (4.1) holds, take disjoint open neighborhoods . For , , $K(\psi\otimes\phi)=\langle T\phi,\psi\rangle=0.$ Product tests detect distributions, so vanishes on . Such rectangles cover the complement of the diagonal, proving condition 2. Conversely, if the kernel is supported in the diagonal, its support relation from Proposition 6.5 of the kernel lesson sends into itself, proving condition 1.

For condition 2 to imply condition 3, change variables by $z=x,\qquad t=y-x.$ This invertible linear map has determinant of absolute value one. Define the transformed distribution explicitly by $\widetilde K(F)=K\big((x,y)\longmapsto F(y-x,x)\big).$ A compactly supported test on the transformed domain becomes a compactly supported test on ; every derivative seminorm is bounded by a finite sum of seminorms of the same order, by the linear chain rule. The inverse change has the same property. Hence this is a distributional change of variables, and with . Vanishing on open sets is transported by these inverse test maps, so the transformed kernel is supported on . Near any fixed relatively compact output region, multiply it by a compact cutoff in and a cutoff in equal to one near zero. Choose the latter with sufficiently small support that on the chosen coordinate region. Extend that compact localization by zero into the full Euclidean product. It has some finite order , so Theorem 3.1 applies. Choose a smaller output region where the localizing cutoff is one. For an original test whose -projection is contained in that region, write . Then the formula is $K(H)=\sum_{|\alpha|\le k} w_\alpha\big(\partial_t^\alpha\widetilde H(0,\cdot)\big).$ Only the germ of the test near the diagonal contributes, so inserting a normal cutoff equal to one there does not affect this formula. On a product test , with supported in that smaller output region, it becomes $K(\psi\otimes\phi)=\sum_{|\alpha|\le k} w_\alpha(\psi\,\partial^\alpha\phi).$ Thus gives (4.2) locally. Changing local cutoffs does not affect the distribution near the diagonal over a smaller output region. Uniqueness in Theorem 3.1 makes the coefficients agree on these smaller regions, including zero coefficients beyond the locally required order. They therefore glue to unique distributions on . Their orders and the number of terms are locally bounded, giving the stated local finiteness.

Finally start with a locally finite expression. On each compact output test support, only finitely many terms contribute. Pairing a term with gives , a continuous scalar functional of on every fixed input support space. The inductive-limit property gives weak continuity. Each term has support contained in , since multiplication of a distribution by a smooth function cannot create support where that function is locally zero. The locally finite sum has the same property. The kernel theorem then supplies its kernel, completing all implications.

There is a regularity refinement, provided continuity has been assumed as in this theorem.

Corollary 4.2 (smooth outputs give smooth coefficients). For a support-preserving weakly continuous , all the coefficients in (4.2) are smooth if and only if is smooth for every . In that case extends naturally to a differential operator on all distributions, by smooth multiplication and distributional differentiation.

Proof. Smooth coefficients plainly give smooth test outputs. For the converse, fix a relatively compact coordinate neighborhood where (4.2) has order at most . Choose , equal to one near . The test gives on , so is smooth there. For a multi-index , test with . On , $T(\chi x^\beta/\beta!) =\sum_{\alpha\le\beta}a_\alpha \frac{x^{\beta-\alpha}}{(\beta-\alpha)!}.$ The term is . Induction on total degree expresses it as a smooth output minus smooth polynomial multiples of previously proved smooth coefficients. Thus every locally present coefficient is smooth. These local conclusions give global smoothness. The extension to distributions is defined by the locally finite expression, with its smooth coefficients, and agrees on test functions. Writing this extension as , its bilinear transpose on tests is $P^t\psi=\sum_\alpha(-1)^{|\alpha|} \partial^\alpha(a_\alpha\psi).$ For a bounded test family , all its members have one common compact support . Only finitely many coefficients meet . The product rule and boundedness of all their derivatives on show that has support in and is bounded in every test seminorm. Thus is bounded in the test space, and $q_B(Pu)=q_{P^tB}(u)$ proves strong continuity of the extension.

The hypothesis of weak continuity was used to obtain a distribution kernel. The proof above does not assert the result without that hypothesis.

A local operator need not have a global order

On , choose nonzero smooth bumps , supported in , for integers , and positive at . Set $T\phi=\sum_{j\ge1}b_j\partial^j\phi.$ The sum is locally finite, so it defines a continuous support-preserving map. Its coefficients are smooth, but it has no finite order valid everywhere. Indeed its unique coefficient of degree is , which is nonzero for every . A competing finite-order representation would contradict coefficient uniqueness. The associated kernel lies on the diagonal and its order grows with the location along that diagonal.

A distributional coefficient can instead concentrate the output. For example, $T\phi=\phi'(2)\delta_2$ has coefficient and all other coefficients zero. Its kernel pairs as , or . It is support-preserving, but a test with has a singular output. Thus diagonal support alone does not make the coefficients smooth.

Exercises

Exercise 1 (basic: vanishing order and point masses). Let . Express in terms of the jet of at zero. Find its exact order, and show directly that it kills every function with its first three jet values zero.

Exercise 2 (intermediate: recovering tangential coefficients). On , define $\begin{gathered} u(f)=\int_0^1 f(0,z)\,dz\\ {}+\partial_t\partial_z f(0,1/2)\\ {}-2\partial_t^2 f(0,1/3). \end{gathered}$ Find the in (3.1), their orders and their supports. Determine the exact order of , and verify the coefficient-order bound.

Exercise 3 (intermediate: a moment-cancelling mollifier). Let be an even smooth integral-one bump on the line, and let . Show that has vanishing first and second moments. Explain why using a nonnegative mollifier could not give all the required moment conditions for the normal-jet extension.

Exercise 4 (advanced: normal derivatives and coefficient order). Let be a nonzero point-supported distribution of exact order on , and take a transverse multi-index . Prove that has exact order . Use suitable scaled tests, rather than inferring equality from the upper bound for a tensor product.

Exercise 5 (advanced: deciding locality from a kernel). Let be nowhere zero. Compare $\begin{gathered} T_1\phi(x)=h(x)\phi'(x),\\ T_2\phi(x)=h(x)\phi'(x+1). \end{gathered}$ Find their distributional kernels and their supports. Which operator is support-preserving? Find the bilinear transpose of each one and retain all coefficient derivatives and translation signs.

Exercise 6 (intermediate: annihilation at a translated point). Let , and suppose for all coordinates. Determine . For , compute and the other coordinate products. Explain why support at alone is weaker than simultaneous coordinate annihilation.

Exercise 7 (intermediate: noncompact coefficients and signs). On , let $\begin{gathered} u=\delta_0(t)\otimes(1+z^2)\\ {}+\delta'_0(t)\otimes(\delta'_0(z)+2\delta'_1(z))\\ {}+\delta''_0(t)\otimes e^z. \end{gathered} \tag{5.1}$ The functions are regarded as regular distributions in . Find the coefficient family in (3.6a), the full pairing on a general test, and the exact total order. Explain why noncompact coefficient supports cause no difficulty.

Exercise 8 (advanced: a principal-value partner). Let on the line and let . Prove directly that has exact order one, and determine the exact order of .

Exercise 9 (advanced: locally finite jets without global order). For integers , choose nonzero nonnegative smooth bumps supported in . Define $\begin{gathered} U(H)\\ =\sum_{j\ge1}\int b_j(z)\partial_t^j H(0,z)\,dz. \end{gathered} \tag{5.5}$ Prove that this is a distribution supported on the plane. Determine its exact order on a region containing one whole bump but no other bump, and prove it has no finite global order.

Exercise 10 (intermediate: order and noncompact pairings). Let . Show that its exact distribution order is zero. Choose compact smooth functions with near zero, , on , and put $\begin{gathered} H_R(t,z)=\theta(t)\chi(z/R),\\ R\ge1. \end{gathered} \tag{5.6}$ Compute a lower bound for . Explain what prevents (3.6a) from assigning a finite value to the noncompact function .

Solutions

Solution 1. Distributional derivatives give $u(f)=3f(0)+2f'(0)+f''(0).$ It has order at most two and annihilates the stated vanishing jets. Choose with , and set . Its supremum and first derivative are uniformly bounded, while the pairing contains ; the lower-order terms remain bounded. Thus an order-one estimate is impossible, and the exact order is two.

Solution 2. The coefficients are $\begin{gathered} w_0=\mathbf1_{[0,1]}(z)\,dz,\\ w_1=-\delta'_{1/2},\\ w_2=-2\delta_{1/3}. \end{gathered}$ Their supports are , , , and their exact orders are zero, one and zero. The product derivatives in the displayed formula give total order at most two. Localizing near , away from , choose , with , , and supported near . First derivatives remain uniformly bounded, while the normal-second-derivative term has size proportional to ; the integral term tends to zero. Hence the exact total order is two. The bounds hold, with equality for .

Solution 3. Evenness gives zero first moments for both bumps. If the second moment of is , that of its dilation is . Thus the integral of is , and its second moment is . A nonnegative smooth integral-one function has strictly positive : if that integral were zero, it would vanish away from zero, and smoothness would force its integral to be zero. Signed kernels are therefore necessary once second-moment cancellation is required.

Solution 4. The point-supported specialization of Theorem 3.1 gives , with some coefficient of total degree nonzero. If all such coefficients vanished, the order would be at most . Choose compact smooth in with . Choose compact smooth in whose order- jet at zero makes ; a polynomial jet times a cutoff produces one. For , take $f_\varepsilon(t,z)=\varepsilon^{m-1}g(t/\varepsilon)q((z-b)/\varepsilon).$ All derivatives through total order are uniformly bounded, and the top-order pairing is a nonzero multiple of . The terms from remain bounded. Thus order at most is impossible. The tensor derivative estimate gives order at most , so the exact order is . If , the product is a nonzero multiple of the point mass at , with exact order zero.

Solution 5. Let $D_c(H)=\int h(x)H(x,x+c)\,dx.$ Then the kernels are , , because their pairings with give the derivative of on the corresponding graph. Their supports are exactly and . Containment follows from differentiation not increasing support. For equality, take a small graph neighborhood and a test , with near zero and . Its pairing is nonzero and it can be supported in that neighborhood. Such an exists because is nowhere zero and continuous.

The first operator is support-preserving by Theorem 4.1. The second is not: choose a test whose derivative is nonzero near an input point , with support in an interval of length less than . Its output is nonzero near , outside that input support. Integration by parts gives $T_1^t\psi(y)=-\partial_y(h(y)\psi(y)),$ and, after setting , $T_2^t\psi(y)=-\partial_y(h(y-1)\psi(y-1)).$ Expanding either derivative gives both the derivative of and the derivative of , each with the displayed minus sign. These are bilinear transposes; complex conjugation would belong to a separate sesquilinear adjoint convention.

Solution 6. Translate coordinates by and apply Corollary 3.2; it gives , with unique scalar since . Formula (3.5) gives , while every other coordinate product is zero. Thus has support at the single point but fails one of the annihilation equations. More general positive-degree point jets also have point support and can fail these equations; the simultaneous vanishing is what removes all such derivatives.

Solution 7. The normal coefficients are $\begin{gathered} w_0=1+z^2,\\ w_1=-(\delta'_0+2\delta'_1),\\ w_2=e^z. \end{gathered}$ Both regular functions are nonzero and locally integrable, so they have order zero. The middle coefficient has order at most one by its derivative formula and exactly one by localizing near zero: a test , with , has supremum tending to zero but nonzero derivative pairing; its support avoids for small . The full pairing is $\begin{gathered} u(H)=\int(1+z^2)H(0,z)\,dz\\ {}+H_{tz}(0,0)+2H_{tz}(0,1)\\ {}+\int e^z H_{tt}(0,z)\,dz. \end{gathered} \tag{5.2}$ There are two derivative signs in each middle tensor pairing, hence the displayed plus signs. The coefficient budgets are , , , so Theorem 3.4 gives exact order two. All integrals are finite because the trace of a compactly supported test is compactly supported. Their order constants depend on the compact tangential region; no uniform bound on over the line is required.

Solution 8. For a test supported in , , the defining symmetric pairing is $v(g)=\int_0^R\frac{g(z)-g(-z)}z\,dz. \tag{5.3}$ The mean-value estimate gives , hence . Any compact test support is contained in such an interval, proving order at most one.

Choose an even compact smooth cutoff , with , equal to one on and supported in . Choose a smooth odd function , with for , and for . Such a function is obtained by integrating a nonnegative even smooth bump supported in and normalizing its integral. Set , . These are smooth tests on one fixed support and have supremum at most one. Their principal-value pairings satisfy $\begin{gathered} v(g_\varepsilon)\\ =2\int_0^2\frac{\chi(z)q(z/\varepsilon)}z\,dz\\ \ge2\log(1/\varepsilon). \end{gathered} \tag{5.4}$ Near zero the integrand has a removable finite bound for each fixed , because and is smooth. Thus the pairings are well-defined and unbounded; no order-zero estimate holds. Consequently . Corollary 3.5 gives exact order for . Its pairing is , since the second normal derivative has positive sign. This partner has noncompact support and is not point-supported, so the older Exercise4 alone would not prove this case.

Solution 9. On any fixed compact test-support set, only finitely many intervals can contribute. Each integral has a bound by ; summing over those finitely many indices proves continuity on every fixed-support space, and hence distribution continuity. Every test supported off the plane has all the relevant normal traces zero, so the support lies on the plane. The support in fact contains every point with in the support of a bump: a tangential test near such a point with nonzero integral against that bump, multiplied by and a normal cutoff, gives a nonzero pairing.

Choose a tangential cutoff equal to one on the support of and vanishing on all the other bump supports, and a normal cutoff equal to one near zero. The compact localization of has precisely the coefficient at normal degree . It is nonzero of tangential order zero, so Theorem 3.4 gives exact total order . On any open rectangle containing that whole bump and the normal origin but no other bump, the same isolating tests are available and the exact order remains .

If had a global finite order , every fixed smooth compact localization would have order at most , by the product rule. Choosing contradicts the exact order just obtained. Thus has unbounded local orders and no finite global integer. Theorem 3.3 does not apply to it with one , although each compact localization has the plane expansion.

Solution 10. The pairing on a compact test is , whose absolute value is bounded by times the integral of on that compact tangential projection. Thus the nonzero distribution has exact order zero, also in agreement with Theorem 3.4. Because , nonnegativity and the value of on give $\begin{gathered} u(H_R)=\int e^z\chi(z/R)\,dz\\ \ge\int_0^R e^z\,dz=e^R-1. \end{gathered} \tag{5.7}$ Every is a legitimate compact test, but their supports grow with . The order-zero definition permits the controlling constant to grow with the compact support; it gives no bound for this expanding family. Substituting the constant function into the normal-trace expression would require , which diverges. The distribution's natural compact-test pairing therefore supplies no such finite value. A separate extension, if requested, would require extra hypotheses and is not part of the finite-order theorem.

References

The supplied kernel, smoothing and tensor lessons provide the earlier programme results used above, including the complete integral Taylor proof. The supplied foundational components retain their stated licences.
Lars Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, second edition (1990), 2003 reprint, ISBN 978-3-642-61497-2, §1.3, Theorems 1.3.2–1.3.3 and Corollary 1.3.4, pp. 17–19; §2.3, Theorems 2.3.3–2.3.5, pp. 46–48; §5.2, Theorems 5.2.3–5.2.4, p. 131. The exact copy supplies comparisons for regularization, vanishing jets, sharp plane-coefficient orders and diagonal kernels. This lesson supplies all its own prerequisite proofs, its simultaneous normal-variable construction, noncompact finite-order extension, exact total-order theorem and ten worked problems.
[Dyatlov 2026] Semyon Dyatlov, Lecture notes for 18.155: distributions, elliptic regularity, and applications to PDEs, MIT, 2 October 2026, §4.4, Theorem 4.19 and Lemma 4.22, pp. 53–54. Open notes. Comparison for point-supported distributions; the plane and exact coefficient-order statements are proved above.
[Melrose 2016] Richard Melrose, 18.155 Lecture 15, MIT, 1 November 2016. Open lecture. Related background for the preceding kernel theorem.
[Whitney 1934] Hassler Whitney, Analytic extensions of differentiable functions defined in closed sets, Transactions of the American Mathematical Society 36, no. 1 (1934), pp. 63–89; introduction and jet conditions, §§1–3, pp. 63–64. Full text. Broader background on extension from closed sets. No general Whitney extension theorem is assumed: Lemma 2.1 proves the particular flat normal-jet extension used here, including its norm estimate.