Distributions as kernels of continuous 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.

An integral kernel records how an operator transfers information from an input point to an output point. A function kernel cannot describe evaluation or differentiation. A distribution on the product space can describe both. This lesson proves Laurent Schwartz's kernel theorem and explains precisely which continuity assumptions it needs. The local theorem has no growth condition at infinity. Its tempered counterpart has a global polynomial bound.

The supplied scalar calculus foundation, §§12–13, proves the compactness, derivative, integral, exponential and smooth-cutoff facts used below. The finite algebra foundation, §§10.1–10.6, supplies the real and complex finite-coordinate rules. The measure foundation, §§15.0–15.1 and 16.1–16.2, supplies Lebesgue integration, dominated convergence, Fubini and the linear change of variables used in the examples. The complete Baire proof, its selected logical base, the seminorm metric and smooth-space completeness are in Functional extension and complete test spaces, §§6, 14.1–14.2 and 19. These are supplied earlier programme proofs with their stated licences.

The locally finite partition, inductive-limit criterion, periodic Fourier reconstruction and Schwartz completeness needed here are proved below. Definitions of test spaces and distributions are recalled explicitly.

Basic references are [Dyatlov 2026], [Melrose 2016] and [Schwartz 1952]. The argument here uses localized periodic expansions, first on bounded rectangles and then on a spatial lattice. It also proves the bounded-set facts needed to pass from weak continuity to strong continuity.

What continuity means

Let and be open. For a compact set , write These increasing seminorms give its Fréchet topology. Completeness follows by taking uniform limits of every derivative and using the fundamental theorem of calculus to identify successive derivatives. The limit still vanishes outside .

Give the test space the topology generated by all seminorms whose restriction to every is continuous. This is a locally convex topology: finite intersections of seminorm balls are convex neighborhoods, and the triangle and homogeneity inequalities prove continuity of addition and scalar multiplication. Evaluation seminorms separate points. Each inclusion is continuous by construction.

Here is its linear mapping property. A linear map , where the topology of is given by seminorms , is continuous if and only if every restriction is continuous. One direction follows by composing with the inclusions. Conversely, each is then one of the defining seminorms, so the inverse image of every finite intersection of target seminorm balls is open. An exhaustion by compact sets with nested interiors gives the same test-space topology: every compact is contained in one member of the exhaustion, and the inclusion between the two fixed-support spaces preserves the derivative seminorms. This is the locally convex inductive-limit topology. A distribution is a continuous complex-linear functional on this space. We use the bilinear convention , with no conjugation of .

Thus exactly when, for every , there are and such that The order and constant may vary with .

The weak dual topology on tests each fixed . The strong dual topology uses where runs over bounded subsets of . A subset of a locally convex space is bounded when every continuous seminorm is bounded on it.

These are different definitions of topology. The continuity equivalence proved below concerns this particular class of operators; it does not assert equality of the two dual topologies.

Lemma 1.1 (bounded test families). A subset is bounded if and only if all its elements are supported in one compact set and every derivative seminorm is uniformly bounded on .

Proof. Suppose has no common compact support. For a compact exhaustion , choose and with . Enlarging the exhaustion along the construction, we can make escape every compact subset of . Choose positive with . Then is finite on every test function. On each fixed support space it is the maximum of finitely many continuous evaluations, so it is a continuous seminorm on the inductive limit. It is unbounded on , a contradiction.

For every , the seminorm is continuous on every support space and hence on . Boundedness therefore gives uniform derivative bounds. Conversely, if has a common compact support and uniform derivative bounds, every continuous seminorm restricted to that Fréchet support space is bounded by a constant times one of its increasing derivative seminorms. It is bounded on . This proves the converse.

For a linear map , weak continuity means that is continuous for every fixed . It does not initially require a uniform estimate as varies. The next lemma supplies that estimate on fixed supports.

Two continuous variables give one estimate

Lemma 2.1 (the Baire estimate). Let be Fréchet spaces with increasing seminorms . If is a separately continuous bilinear form, there are indices and a constant such that

Proof. For positive integers , put Each set is closed because is continuous. Continuity in , for each fixed , shows that these countably many sets cover . Baire's theorem gives one with nonempty interior. Choose and a neighborhood of zero such that and . Subtraction gives There are and with . For , apply the preceding bound to . For , every positive multiple of belongs to that seminorm ball; letting the multiple increase gives . This proves (2.1), with .

For a weakly continuous , apply the lemma to on . Separate continuity in follows because each is a distribution. We obtain The indices depend on both compact sets. No global order bound has appeared.

Localizing on open sets

Partition lemma. Every open cover of an open subset of Euclidean space has a countable smooth partition of unity whose supports are compact subsets of members of the cover and form a locally finite family in .

Proof. Put

using infinite distance when , and set for . Each is compact, , and their interiors exhaust . Distance to a nonempty closed set is continuous because the triangle inequality gives ; this also justifies closedness in this formula.

The compact layer lies in the open set . Around each of its points choose concentric balls with inside that open set and inside one member of the given cover. Finitely many of the smaller balls cover . Take a smooth nonnegative bump equal to one on each smaller ball and supported in its corresponding larger ball; the exact radial bump construction is in the stated cutoff prerequisite.

The resulting countable family of supports is locally finite. In fact a point has a neighbourhood inside some , and every layer with has its selected supports outside ; the finitely many earlier layers contribute only finitely many balls. At least one bump is positive at each point: choose the smallest for which the point belongs to , and use its layer. The locally finite sum of all bumps is therefore positive and smooth. Dividing each bump by gives the required partition, with its original support unchanged.

For a distribution , say that it vanishes on an open set when it annihilates every test supported there. Its support is the complement of the union of such open sets. The preceding partition proves that vanishes on that union: a compactly supported test there is a finite sum of tests localized inside the individual zero neighborhoods. Thus a test whose support avoids has zero pairing. The same finite partition argument proves that distributions agreeing on an open cover agree on its union.

For later use, distributional differentiation means This is again continuous: differentiation preserves a fixed support and increases its required derivative seminorm by one. A locally integrable function defines a distribution by integration, since its pairing with a test supported in is bounded by its norm times .

Breaking a test function into products

Choose rectangles and whose closures lie in and . Choose larger rectangles , , containing those closures, and cutoffs , equal to one near , . Regard the larger rectangles as fundamental cells of tori. Let , be their exponential Fourier modes. The frequencies include the scale factors determined by the side lengths of the rectangles.

For , extend by zero to the larger cells and then periodically. If denotes its normalized Fourier coefficient, then The series converges with every derivative, with support in one fixed compact product.

Integration by parts using powers of the periodic Laplacian gives The eigenvalues of that Laplacian are comparable to , with constants depending on the cells. A derivative of order at most of a term in (3.1) costs at most . Choose ; the lattice sum of these bounds is finite. Explicitly, in dimension , the shell contains at most integer points by counting the surrounding coordinate box. For , the sum of over that shell is at most . The geometric series converges, and the finitely many points with do not affect convergence. Thus the periodic series converges absolutely with every derivative.

Here is why its value is the original periodic function. On the circle of length , the Fejér kernels are Expanding the finite square gives Fourier coefficients , where . The integral of is zero for a nonzero integer , by evaluating its primitive at the endpoints, and is one for under normalized measure. They are therefore nonnegative and have integral one for the normalized measure , by orthogonality of the exponential modes. Outside a fixed neighborhood of zero their values are , from the geometric-sum identity. Their products in several variables therefore form an approximate identity: split the convolution into a small neighborhood of zero, controlled by uniform continuity, and its complement, whose mass tends to zero. For a product kernel, the complement of a small coordinate box is contained in the union of the corresponding one-coordinate complements; positivity and the unit integral in the other coordinates give the same vanishing-mass bound. Consequently Fejér means converge uniformly to every continuous periodic function. For the absolutely convergent Fourier series above, its Fejér means also converge uniformly to its sum, because their coefficients are bounded by one and tend to one at every fixed frequency. Both limits agree. Multiplying by , which equals one on the support of , proves (3.1). Rectangles of other side lengths follow by rescaling.

Lemma 3.1 (product tests detect distributions). Finite sums of tests are dense in . A distribution on the product that vanishes on every such product is zero.

Proof. On a relatively compact product rectangle, (3.1) is precisely an approximation by finite sums of products, converging in one support space. For a general compactly supported , choose smooth partitions of unity on and subordinate to relatively compact rectangles. Only finitely many products of partition functions meet . Apply (3.1) to each resulting piece. The sum of these finite approximations converges in one compact support space. Continuity of a distribution proves the last assertion.

The local kernel theorem

Theorem 4.1 (Schwartz's kernel theorem on open sets). For arbitrary open , , the following data are equivalent:

1. A distribution .
2. A weakly continuous linear map .
3. A linear map continuous into the strong dual topology.
4. A bilinear form on separately continuous in each test variable.

The correspondence is The kernel is unique. In particular, continuity of each scalar pairing is enough; no estimate uniform over all supports is required.

The same class of operators is obtained if weak continuity is specified only using convergent test sequences. On a fixed support space, every scalar pairing is then a sequentially continuous linear functional on a metrizable Fréchet space, hence is continuous. To see the last implication, failure of continuity would give vectors tending to zero in the metric with functional values bounded away from zero. The inductive-limit property then gives continuity on all of . Conversely a continuous scalar pairing preserves every convergent test sequence.

Proof. Start with . On fixed compact supports , its distribution estimate and the product rule give for some . For fixed , this defines a distribution . For fixed , it defines a continuous functional of . Thus it supplies both weak continuity and separate continuity.

To prove strong continuity, let be bounded. Lemma 1.1 puts its elements in one , with . Formula (4.2) implies The restriction of to every support space is therefore continuous into the strong dual. The inductive-limit property proves global strong continuity. Strong continuity implies weak continuity because a singleton test family is bounded.

Conversely, start with a separately continuous bilinear form . It gives a weakly continuous map by . Use the rectangles, cutoffs and modes of Section 3. Define, for , Lemma 2.1 on the supports of bounds the second factor by With , (3.2) shows that (4.3) converges absolutely and is bounded by . It is a distribution on .

If , its Fourier coefficients factor into the coefficients of and . Their cutoff expansions converge in the two fixed support spaces. The joint estimate (2.2) then gives Lemma 3.1 implies uniqueness on that rectangle. In particular the construction is independent of the larger cells and cutoffs.

On overlaps of two product rectangles, cover the overlap by smaller product rectangles. The two distributions agree on product tests in each smaller rectangle and hence agree there by Lemma 3.1. They therefore glue. Explicitly, choose a locally finite partition subordinate to the product-rectangle cover and define the value on a global test by the finite sum of the corresponding localized values. For two such partitions, refine each term by the other partition. Only finitely many terms meet the test support, the two local distributions agree on every intersection, and both double sums therefore coincide. This proves independence of the partition. On each compact support only finitely many distribution estimates are needed, so the result is continuous. This gives . To verify (4.1) on an arbitrary product test, choose product partitions on and . The two test functions split into finite sums, each localized product has the already proved pairing, and bilinearity recombines them to . Lemma 3.1 gives global uniqueness. All four correspondences are inverse.

Polynomial control on the whole space

For , let be the Schwartz space, with increasing seminorms Its continuous dual is , the tempered distributions. The weak and strong dual topologies are defined by fixed tests and bounded test families, respectively. A family in is bounded exactly when every is uniformly bounded on it.

Completeness of the Schwartz space. The separate points, so the seminorm metric in the supplied §14.1 applies. Suppose is Cauchy in every . For every multi-index , the functions converge uniformly to a continuous . Passing to the limit in the coordinate identity shows that . Continuous partial derivatives and their repeated coordinate identities show that is smooth with . For fixed and , the functions are uniformly Cauchy, so they have a bounded uniform limit. Pointwise that limit is . Consequently and for every . The seminorm metric is complete. Thus is Fréchet, and the full Baire theorem used in Lemma 2.1 applies to it.

Lemma 5.1 (a spatial and frequency expansion). There are compactly supported smooth functions on , supported in , with on and For such functions in dimensions , every has an expansion Here , . The series converges absolutely in every Schwartz seminorm, and, for all integers ,

Proof. Choose a nonnegative bump , supported in , positive on . Its integer translates have a positive smooth periodic sum . Put , and choose supported in , equal to one on a neighborhood of . This gives the required partition.

Let be the normalized Fourier coefficient on the -periodic product cell of This function is supported away from the boundary of the cell. Integration by parts with gives the frequency factor in (5.2). On its fixed support, is comparable to , uniformly in the lattice points. The product rule gives the spatial factor and the displayed derivative bound.

For any integer , the Schwartz seminorm of an atom in (5.1) is at most Choose and in (5.2). Summation over the two lattices is then finite, proving absolute convergence in . For a fixed , Fourier reconstruction gives the localized function because on its support. Summing those functions gives by the spatial partition. The convergence just proved also justifies this equality in .

Theorem 5.2 (the tempered kernel theorem). Distributions correspond uniquely, by (4.1), to weakly continuous linear maps Every such map is also continuous into the strong dual topology. Equivalently, kernels correspond to separately continuous bilinear forms on the two Schwartz spaces.

Proof. A tempered kernel satisfies . The product rule and give This defines a weakly continuous map. Taking the supremum over a bounded family of 's proves strong continuity.

Conversely, the Baire estimate gives For the atoms in Lemma 5.1, the value of is at most Define by replacing each atom in (5.1) by this bilinear value and summing with its coefficient. Choose and in (5.2). The sum converges absolutely and is bounded by a fixed Schwartz seminorm of ; hence is tempered.

For , the coefficients factor into the corresponding one-variable spatial and frequency coefficients. Lemma 5.1 in each dimension, followed by joint continuity of , gives . Finite sums of Schwartz products are dense by (5.1), so is unique. Strong continuity and its converse to weak continuity follow as above.

If one dimension is zero, its test space is , and the result is the ordinary continuous-dual identification. If an open set is empty, all corresponding test and distribution spaces are zero. These cases need no Fourier construction.

Three ways to see an operator

Example 6.1 (a singular output with one input measurement). On , let , defined by For , the fundamental theorem gives . For , it is at most . Thus the absolute pairing is at most . At infinity, Schwartz decay makes the integral absolutely convergent. Its absolute value is bounded by a fixed Schwartz seminorm, so is tempered. Put The kernel theorem supplies a tempered kernel. On a general Schwartz test it is explicitly Differentiating under the integral and using Schwartz bounds verifies that the inner function is Schwartz and depends continuously on . The kernel has a singularity in its output variable although its input dependence is smooth.

Example 6.2 (a curved evaluation relation). Define for compactly supported smooth . Its local distribution kernel is For each compact support of , the integration occurs over a bounded interval. A supremum bound proves continuity. Pairing with gives the stated operator. The kernel is supported exactly on the parabola : it vanishes off it, while every neighborhood of a point of the parabola contains a nonnegative test whose pairing is positive. This kernel is a measure: it is the pushforward of the locally finite positive measure by the graph map, so countable additivity follows by taking inverse images of disjoint Borel sets. The parabola is closed and has planar measure zero by Fubini, since each vertical section is a singleton. This nonzero measure is concentrated there, so it has no density with respect to planar Lebesgue measure.

Example 6.3 (a local kernel with excessive growth). The formula defines a distribution on , and its operator is . It satisfies the local theorem. It does not define a tempered kernel. To see this, choose nonnegative bumps , with supported in , , and . The Schwartz seminorms of grow at most polynomially in , whereas No fixed Schwartz seminorm can bound these values. The global growth distinction cannot be removed from the tempered theorem.

Proposition 6.4 (graphs of continuous maps). Let be continuous. The operator , viewed as a locally integrable function on , has kernel Its support is exactly the graph of . Smoothness and injectivity of are unnecessary.

Proof. For a test supported in a compact product, its restriction to the graph is continuous and supported in the compact projection onto . Its integral is bounded by the measure of a compact neighborhood of that projection times the supremum of the test. Thus is a distribution of order zero. Its pairing with a product test is , giving the asserted operator. The graph is relatively closed by continuity of , and a test supported off it has zero pairing. At a graph point, choose a nonnegative test positive on a small product neighborhood of that point. Continuity of gives a nonempty open set of input points whose graph lies in that positive neighborhood. The pairing is positive, so every graph point is in the support.

Proposition 6.5 (the support relation of a kernel). For every local distribution kernel and every , No properness of the kernel projections is required for this test-input assertion.

Proof. The set on the right is closed in : from a convergent sequence of its output points, compactness of gives a convergent subsequence of their corresponding input points, and closedness of gives the limiting pair. On an output neighborhood disjoint from that closed set, every product test has support disjoint from , so its kernel pairing is zero. Hence vanishes on that neighborhood. This proves the containment.

Exercises

Exercise 1 (basic: locating an averaging kernel). For , define Find its kernel as a locally integrable function. Account for the Jacobian and determine its support, allowing to vanish on part of .

Exercise 2 (intermediate: which variable is differentiated?). If is the kernel of , find the kernels of and . Find the kernel of the bilinear transpose , characterized by . Give a one-dimensional check using .

Exercise 3 (intermediate: a lattice of measurements). Let have polynomial growth. Show that is tempered. Describe its operator on Schwartz functions. Prove that if all , its support is precisely .

Exercise 4 (advanced: continuity of a whole family). Suppose are weakly continuous and, for each pair of fixed compact supports , there are independent of such that Suppose all product-test pairings converge to those of a weakly continuous . Prove that their kernels converge to the kernel of on every test in . Identify the extra ingredient beyond density that makes this implication valid.

Solutions

Solution 1. Set , so . A kernel is Values at the two endpoints do not matter for this locally integrable function. Let be the support, on the real line, of , as a distribution. Then . Indeed change of variables is a diffeomorphism, and the resulting product distribution has support . To check this last assertion, a test supported outside that product has zero pairing by Fubini and localization. At a point with , every small -neighborhood contains a test detected by the one-variable distribution, by the support definition. Multiply it by an -test supported near with nonzero integral; the product pairing is nonzero. This also handles an identically zero on the integration interval, when the support is empty.