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Schwartz parameter pairing and integral interchange

Statement

For $φ \in S (R^{n})$ , the map

$x ⟼ T_{x} φ, (T_{x} φ) (y) = φ (x - y),$

is $C^{\infty}$ as a map from $R^{n}$ to Schwartz space, with $\partial_{x}^{γ} T_{x} φ = T_{x} (\partial^{γ} φ)$ . This clause holds in ZF.

Assume Countable Choice for the following integral clause. Let $r \geq 1$ and let $H : R^{r} \to S (R^{n})$ be continuous in every Schwartz seminorm. Suppose each $\partial_{y}^{β} H (t, y)$ is jointly measurable and, for every $α, β$ , there is $g_{α β} \in L^{1} (R^{r})$ such that

$p_{α β} (H (t)) \leq g_{α β} (t)$

for almost every $t$ . Then $G (y) = \int_{R^{r}} H (t, y) d t$ belongs to $S$ , derivatives pass under the integral, and every $u \in S^{'}$ satisfies

$⟨ u, \int H (t) d t ⟩ = \int ⟨ u, H (t) ⟩ d t .$

All integrals in this clause are Lebesgue integrals.

Facts & Assumptions

Given: A Schwartz function $φ$ ; for the second clause also Countable Choice, a family $H$ with the stated seminorm majorants, and $u \in S^{'}$ .

[F1]

Fixed translations, reflection, and derivatives preserve Schwartz space continuously (Basic operations are continuous on Schwartz space).

[F2]

The functional $u$ obeys one finite Schwartz-seminorm estimate (Finite seminorm bound characterizes tempered distributions).

[F3]

Dominated convergence and the complex integral triangle inequality hold for the stated Lebesgue integrals (Dominated convergence, The modulus of an integral is bounded by the integral of the modulus).

Proof

technique · weighted Taylor remainders and explicit finite sums

1.1

Fix a compact set of parameters $K$ . The inequality $1 + ∣ y ∣ \leq (1 + \sup_{x \in K} ∣ x ∣) (1 + ∣ x - y ∣)$ transfers every polynomial weight in $y$ to one in $x - y$ , uniformly for $x \in K$ . Apply the one-variable integral Taylor remainder along each coordinate.

F1algebra

$p_{α β} ⁣ (\frac{T_{x + h e_{j}} φ - T_{x} φ}{h} - T_{x} (\partial_{j} φ)) ⟶ 0.$

The same estimate applied to every derivative gives continuity of all iterated derivatives. [F1, algebra]

Iterating step 1.1 proves that $x \mapsto T_{x} φ$ is $C^{\infty}$ in the Schwartz topology and gives the displayed derivative formula. When $φ = 0$ every derivative is zero. No integration on parameter space and no choice principle occurred. [step 1.1]

1.2

For the integral clause, differentiate under the integral and apply the integral triangle inequality pointwise in $y$ .

$\partial_{y}^{β} G (y) = \int \partial_{y}^{β} H (t, y) d t, p_{α β} (G) \leq \int g_{α β} (t) d t .$

The derivative statement follows successively from difference quotients and dominated convergence; the seminorm estimate follows from the integral triangle inequality before taking the supremum in $y$ . Thus $G \in S$ . [F3]

1.3

Let $Q_{R} = [- R, R]^{r}$ . Subdivide it into the canonical equal mesh and form lower-corner finite sums $S_{R, m}$ for $H$ . Uniform continuity in each seminorm and [F3] make these sums converge to $G_{R} (y) = \int_{Q_{R}} H (t, y) d t$ in that seminorm. Continuity of $u$ may therefore be passed through this explicit limit.

F2F3

$u (G_{R}) = \lim_{m} u (S_{R, m}) = \lim_{m} \sum_{Q} ∣ Q ∣ u (H (t_{Q})) = \int_{Q_{R}} u (H (t)) d t .$

The last equality is the same scalar step-function approximation. [F2, F3]

2.1

The finite estimate [F2] involves only finitely many seminorms. Their $L^{1}$ majorants show both $G_{R} \to G$ in those seminorms and $\int_{Q_{R}} u (H (t)) d t \to \int_{R^{r}} u (H (t)) d t$ as $R \to \infty$ . Passing to the limit in step 1.3 proves the interchange formula. Countable Choice is used exactly through [F3]'s Lebesgue interface; the finite-sum and continuity argument adds no stronger choice.

F2F3step 1.3∎

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Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)
Radu Gelca, Functional Analysis (standard reference, not scraped)