Tensor order and integral kernel bounds

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.

An integral kernel can describe evaluation on a graph, differentiation of an input, or averaging over a region. Its absolute row and column integrals control all its bounds. For distributional kernels, a different issue is the number of test derivatives needed to bound a pairing. Tensor products can attain the sum of two orders, but sufficiently separated frequencies can reduce that exact order to their maximum.

We pair distributions complex linearly, so . The test topology and local finite-order estimates are proved in U008, the initial test-space construction and Proposition 1.2. The complete tensor construction on arbitrary Euclidean open sets is U021, B0–B2, especially (B5)–(B6). That proof includes density of sums of product tests, tensor uniqueness, exact product supports and three-factor associativity. These exact supplied results suffice here.

The scalar foundation, §§12, 13.1–13.5 and 13.7–13.10, proves compactness, calculus, exponential/trigonometric series and smooth cutoffs. The finite algebra foundation, §§10.1–10.6, proves the inner-product and matrix identities. The integration foundation, §§15.0–15.3 and 16, proves completed Lebesgue integration, Tonelli, Fubini, Hölder and measurable sections. All further estimates and sharpness arguments are proved below.

Evaluation and differentiation determine a kernel

For an operator , a kernel means a distribution on satisfying There is at most one such distribution: subtract two and apply the product-test density proved in U021 B2. In the following cases we construct it explicitly.

Theorem 1.1 (continuous graph evaluation). For every continuous real , let . Then This is an order-zero positive graph measure, with exact support . Its operator is , which is weakly continuous from tests to distributions.

Proof. The function is Borel and bounded. For a test supported in a compact rectangle , absolute Fubini and the fundamental theorem give The first integrand vanishes outside , so is absolutely integrable. The second pairing is bounded by . Thus it has order at most zero. It is integration against the pushforward of by the continuous map ; the same estimate gives local finiteness.

The graph is closed: if , continuity forces . Off it, the pairing vanishes on a sufficiently small neighborhood. At , choose a nonnegative smooth test supported in any prescribed neighborhood and positive on a smaller rectangle around that point. Continuity of places in that smaller rectangle for all in a nonempty interval about . Its pairing is strictly positive. The support is therefore exactly the graph and the measure is nonzero, so its exact order is zero.

On product tests the formula is , proving the operator identity and uniqueness. The output is continuous. For every fixed , This bound gives continuity for the test topology and hence weak continuity into distributions. The argument has made no assumption on derivatives, injectivity or properness of .

Theorem 1.2 (an input derivative fixes the sign). Suppose is continuous, and define Then its unique kernel is Here the derivative is distributional; no classical derivative of is required.

Proof. Continuous is locally integrable, so its weak derivative is a distribution. The diagonal term is the instance of Theorem 1.1. Thus the proposed sum exists as a distribution. On product tests, Both integrals are absolutely convergent on compact products. They give the required pairing, and product-test density proves uniqueness.

For completeness, is continuous. On a compact output interval , uniform continuity of on times a compact neighborhood of makes its integral continuous in . If , then for a fixed output test supported in , Thus the operator is weakly continuous on the test space. The two minus signs, one in the proposed kernel and one in weak differentiation, explain the plus sign in the input formula.

Example 1.3. Taking gives the exact order-zero kernel on that graph and output . Its corner at zero has no effect on the -derivative identity. Taking in Theorem 1.2 and differentiating the smooth density gives

Row and column masses control every exponent

Theorem 2.1 (Schur bounds, with every endpoint). Let and be open. Suppose the measurable complex kernel satisfies where . For every , the integral converges absolutely for almost every , defines a measurable output class, and obeys These bounds are valid on almost-everywhere equivalence classes and are sharp as universal bounds. If either or is zero, the kernel and every operator are zero almost everywhere.

Proof: measurable representatives. Open Euclidean sets with Lebesgue measure are sigma-finite. We first justify the choice of Borel representatives. The integration foundation, §15.0, proves that each Lebesgue measurable set differs from a Borel set inside a Borel null set. Approximate a nonnegative measurable function by the increasing finite simple functions of §16.1. Replace every level set in this countable family by its Borel version. The new simple functions agree with the old ones off one Borel null union; their pointwise limit there, extended by zero on that union, is a Borel representative. Split real and imaginary parts into positive and negative parts to treat finite complex functions. This applies on the product space as well as each factor.

Changing a representative on a null set preserves its section integrals outside a null set, by Tonelli and completion. For the input assertion directly, if off a null set , then Indeed is product null by a countable finite-measure exhaustion of . Tonelli gives zero integral on almost every row. The same argument applies to two versions of which agree almost everywhere. We may take finite representatives throughout.

Integrals of nonnegative Borel functions over one variable are measurable by Tonelli. Real and imaginary parts, split into positive and negative parts, consequently give a measurable complex integral wherever it is absolutely finite. Set the output to zero on the exceptional null set.

Proof: norm estimates. If , Tonelli on gives integral zero for ; exhaustion gives almost everywhere. If , interchange the two variables in this reasoning. A zero kernel gives zero on almost every row for every input, as above.

For , Tonelli gives Thus the inner integral is finite almost everywhere and yields the first bound. For , put Its integral is at most , so it is finite almost everywhere. Hölder in the ordinary measure, applied to and , gives on such rows Zero rows are interpreted directly as zero; all remaining rows have finite weight. Raising to and integrating proves This simultaneously proves absolute existence. At infinity choose a representative bounded by ; the absolute row integral is at most . Representative independence transfers the conclusion to the entire class. Linearity follows on the intersection of the full-measure sets where the finite collection of relevant integrals converges.

Proof: sharpness. For finite positive-volume measurable , take with . Its exact column and row bounds are and . Input gives output , with norm ratio They equal the asserted constants, so a uniformly smaller bound from these hypotheses is impossible. These are universal estimates; a particular complex kernel can have a smaller interior norm.

Example 2.2 (causal averaging). On , set Integrating over gives one, so both absolute marginal integrals are one. The norm is therefore at most one for every exponent. For , proves equality. For , use . On , direct integration gives . Therefore for , First let , then . The supremum defining the operator norm is at least one. At , equality also follows immediately by Tonelli for any nonzero nonnegative integrable input.

A nonzero factor reveals its own order

For an integer , a distribution has order at most if every compact test support has a constant such that The same must work for all supports; only the constants vary. The exact order is the least such nonnegative integer, when one exists. We assign zero distribution order zero.

Theorem 3.1 (upper bound and converse). For distributions on arbitrary Euclidean open sets , orders at most imply tensor order at most . If both factors are nonzero and their tensor has order at most , each factor has order at most .

Proof. Every compact subset of lies in the product of its compact projections. U021 (B6), with the two given order estimates, gives Each derivative has total order at most . This proves the upper bound on every compact support.

For the converse, choose a fixed test with , by rescaling a nonzero pairing. On any compact , apply the tensor order bound to . Each derivative of total order at most is a product of a derivative of through and a fixed derivative of . Hence The same argument with a normalized test of gives the estimate for . It includes . If one factor is zero, its tensor is zero regardless of the order of the other factor, so the converse hypothesis is necessary.

Concentrated jets attain the sum of orders

Theorem 4.1 (the additive bound is attained). For positive integers , the factors , have exact orders , and has exact order .

Proof. The definitions give the factor upper bounds; Theorem 3.1 gives the tensor upper bound. Choose equal to one near zero, and put All supports lie in one fixed compact rectangle. A derivative of total order has norm at most a fixed constant times , so these norms stay bounded. But Thus the tensor has no order bound . The one-variable tests have bounded derivatives through and pairing with . This proves its exact factor order; replacing by proves the other one.

Separated frequencies attain the maximum order

We will use the elementary fact that for each integer and real . The quotient of consecutive terms is , which tends to by the finite binomial formula. Choose strictly between these numbers and one. Beyond a fixed index the terms are bounded by a fixed multiple of ; the finite geometric-sum formula bounds that tail. This proves the fact without a further convergence criterion.

Theorem 5.1 (the maximum can be the exact order). For every integer , there are two distributions on , each of exact order , whose tensor product also has exact order .

Proof: defining the factors. Let and define the two series by pairing their even and odd modes with tests: For a test supported in an interval of length , integration by parts times has no boundary terms and yields After multiplying by , the summable bound is a constant times . Indeed implies , and converges, for example because the ratio of consecutive terms is at most for . The series therefore defines a linear functional with an order- estimate on every compact test support. It is a distribution by the supplied test topology; tails converge in that same seminorm estimate.

Proof: detecting the factor orders. Choose a nonnegative smooth supported compactly in and positive on . The supplied cutoff construction gives such a function. Define The sum is locally finite and smooth; a nearest integer ensures one argument lies in , so . Reindexing shows . Consequently is compact smooth and . Splitting its integral into unit intervals gives for every integer The final integral follows from the scalar exponential primitive and . There are only finitely many contributing translated pieces.

For of the chosen parity, put Leibniz's rule bounds all derivatives through independently of , since every resulting power of has exponent at most zero. The support is fixed. The already absolutely convergent pairing series and the integer-frequency identity give exactly All other frequencies have zero pairing. Both parities contain arbitrarily large , so neither factor has order at most .

Proof: the tensor estimate. Form the double mode series with even and odd . For a two-variable test supported in a rectangle of area , integrate by parts times in the variable with the larger frequency: The indices cannot coincide. Thus the coefficient majorant is Dropping the parity restriction while retaining distinct indices bounds this by The equality uses . The last convergence follows from for , proved by induction using , and the elementary summability just proved for the dominating . This supplies a finite order- bound for the double series.

On a product test, absolute convergence of each one-variable series factors the double sum into the two pairings. U021 B2 therefore identifies the resulting distribution with on all tests. Each factor is nonzero and has exact order ; Theorem 3.1 excludes a smaller tensor order. This proves the claim, also for .

Exercises

Exercise 1 (basic). Find the operator, exact support and half-line derivative representation of the kernel

Exercise 2 (intermediate). For , find the kernels of Explain the bounded density in one and the line measure in the other.

Exercise 3 (intermediate). On , , take Calculate its exact row and column bounds, the output on , and that input's norm ratio for every exponent. Find the endpoint operator norms.

Exercise 4 (basic). Find the exact operator norms and an attaining input for

Exercise 5 (advanced). Partition into two unit intervals and into three unit intervals. Let the six constant kernel values be Find the Schur bound and the exact operator norms. Compare the input and output norms for interval values .

Exercise 6 (intermediate). If are real measurable functions, prove that replacing by preserves every operator norm and both absolute marginal bounds.

Exercise 7 (intermediate). For compute the full tensor pairing, exact support and exact order.

Exercise 8 (advanced). Prove that is a distribution with no finite global order. What does its tensor with zero show about Theorem 3.1?

Exercise 9 (intermediate). Determine for all nonnegative integers , then find the coefficient and exact order of

Exercise 10 (advanced). For , take , and . Divide the series into the three residue classes of modulo three. Prove that each factor and their threefold tensor have exact order .

Exercise 11 (advanced). In Theorem 5.1 replace by any integer . Determine exactly when the displayed absolute double coefficient majorant converges. State both the resulting order conclusion and the limit of what divergence of this majorant proves.

Exercise 12 (advanced). Suppose from to and from to satisfy Theorem 2.1. Prove that exists absolutely almost everywhere, has column bound and row bound , and represents on every , including the endpoints and zero-bound cases.

Complete solutions

Solution 1. A product test gives . Theorem 1.1, followed by multiplication by the smooth function , gives No -derivative is taken. The kernel vanishes off the closed graph . At every graph point the positive-test argument of Theorem 1.1 has strictly positive weight , so it proves the exact support is that whole graph. The same compact-projection estimate proves order zero.

Solution 2. Splitting at , integration by parts gives for every test The two boundary terms at vanish because there. Thus its first derivative is . A second integration gives so its second derivative is . The first input derivative contributes a minus sign to the kernel, whereas two input derivatives contribute a plus. Therefore The first added density is bounded. The second kernel is supported on and gives . These identities construct the kernels directly, and product-test density proves uniqueness.

Solution 3. For , its column has , of length one. Thus . The row length, also the output , is Its essential supremum is one, so . For finite , At infinity this ratio is one. Theorem 2.1 bounds every operator norm by one. At , Tonelli and the constant column integral give for any nonzero nonnegative , so the norm is exactly one. At infinity, input attains one. The computed finite- ratio for a single input makes no additional equality assertion for the interior operator norms.

Solution 4. The absolute column and row bounds are and . Theorem 2.1 therefore gives the bound Input gives output . The norm ratio is precisely the displayed constant for each exponent, including at . Thus all bounds are attained.

Solution 5. The absolute column sums are , and the row sums . Hence , with Schur bound .

Let be the integral of an arbitrary input on its -th unit interval. Hölder at exponent two gives . Its two output values form . Conversely, the input constant at on each interval realizes these three integrals with squared norm . The integral-operator norm therefore equals the Euclidean norm of .

Here is the finite matrix argument in full. Direct multiplication gives The positive numbers solve . For either root, the vector satisfies , by direct substitution. Moreover Normalize these two nonzero vectors. They form an orthonormal basis of . Expansion in this basis proves Finite-dimensional Cauchy–Schwarz and , obtained by taking when , give For normalized , choose . Then and . Thus where the strict inequality follows from .