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Smooth polynomially bounded multipliers on schwartz space

Statement

Let $a \in C^{\infty} (R^{n}; C)$ and suppose that for every multi-index $γ$ there are $C_{γ} \geq 0$ and an integer $m_{γ} \geq 0$ such that

$∣ \partial^{γ} a (x) ∣ \leq C_{γ} (1 + ∣ x ∣)^{m_{γ}} .$

Then $M_{a} : φ \mapsto a φ$ is a continuous complex-linear endomorphism of $S (R^{n})$ . Its transpose

$⟨ a u, φ ⟩ = ⟨ u, a φ ⟩$

is a tempered distribution and depends continuously on $u$ for both the weak and strong dual topologies. Polynomials and Schwartz functions satisfy the hypothesis; an arbitrary smooth function need not.

Facts & Assumptions

Given: A smooth function $a$ with the derivative-by-derivative polynomial bounds in the statement.

[F1]

Schwartz seminorms and topology are those of Schwartz space and its seminorms and Schwartz topology and convergence, with multi-indices interpreted by $C^{k}$ maps and multi-index derivative notation in Euclidean space.

[F2]

Tempered distributions are continuous functionals on $S$ , and their weak and strong topologies test singletons and bounded subsets (Tempered distribution, Weak and strong topologies on tempered distributions).

Proof

technique · Leibniz seminorm estimates and transposition

1.1

Fix $α, β$ and apply the multi-index Leibniz formula.

F1algebra

$x^{α} \partial^{β} (a φ) = \sum_{γ \leq β} (\binom{β}{γ}) x^{α} (\partial^{γ} a) \partial^{β - γ} φ .$

For each of the finitely many $γ$ , expansion of $(1 + ∣ x_{1} ∣ + \dots + ∣ x_{n} ∣)^{m_{γ}}$ bounds that summand by a finite linear combination of seminorms $p_{α + δ, β - γ} (φ)$ with $∣ δ ∣ \leq m_{γ}$ . Hence each output seminorm is bounded by finitely many input seminorms. [F1, algebra]

2.1

Step 1.1 proves simultaneously that $a φ \in S$ and that $M_{a}$ is continuous. A polynomial has only finitely many nonzero derivatives and each grows polynomially. If $a \in S$ , each derivative is bounded, so the hypothesis holds with exponent zero.

F1step 1.1

3.1

For $u \in S^{'}$ , the composition $u \circ M_{a}$ is continuous and linear, hence tempered. For a single test, $p_{φ} (a u) = p_{a φ} (u)$ , proving weak continuity of the transpose.

F2step 2.1

4.1

If $B$ is bounded in $S$ , the finite estimates of step 1.1 show that $M_{a} (B)$ is bounded. Thus $p_{B} (a u) = p_{M_{a} (B)} (u)$ , proving strong continuity. The derivative hypothesis is essential: for example $a (x) = e^{∣ x ∣^{2}}$ is smooth but does not map every Schwartz function to a Schwartz function. No choice axiom is used.

F2step 1.1∎

Depends on

Used by

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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)