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Real powers of the Japanese bracket act on Schwartz space

Statement

For every integer n≥1 and real s, the functions ws(ξ)=⟨ξ⟩s=(1+∣ξ∣2)s/2,w−s(ξ)=⟨ξ⟩−s are smooth multipliers acting continuously on S(Rn). The multiplication maps are mutual inverses. By transposition they also act continuously and invertibly on S′(Rn), for both its weak and strong dual topologies.

Facts & Assumptions

Given: n≥1, s∈R, and the bracket ⟨ξ⟩≥1.

[F1]

Schwartz functions are actual smooth functions, and their topology is given by the seminorms pαβ(f)=sup⁡ξ∣ξα∂βf(ξ)∣ (Schwartz space and its seminorms).

[F2]

A smooth multiplier whose every derivative has polynomial growth acts continuously on S; its transpose acts continuously on S′ for both dual topologies (Smooth polynomially bounded multipliers on schwartz space).

Proof

technique · Chain-rule derivative bounds followed by transposition
1.1F1algebra

Put q(ξ)=1+∣ξ∣2. Induction on ∣α∣, differentiating either the polynomial factor or qs/2−j, expresses each derivative as a finite sum ∂αws(ξ)=∑jPα,j(ξ)q(ξ)s/2−j, where every Pα,j is a polynomial of degree at most ∣α∣. Since q=⟨ξ⟩2, each term is bounded by a constant times ⟨ξ⟩s+∣α∣, and hence by Cα,s⟨ξ⟩max⁡(0,s+∣α∣). Enlarging the exponent to an integer gives a polynomial-growth bound for this derivative.

2.1step 1.1algebra

The same induction with −s gives a polynomial-growth bound for every derivative of w−s.

3.1F1F2step 1.1step 2.1algebra

The bounds in steps 1.1 and 2.1 meet the hypotheses of [F2], so multiplication by either weight is continuous on Schwartz space. Pointwise wsw−s=1, so both compositions on S are the identity.

4.1F2step 3.1∎

For T∈S′, transposition defines ⟨w±sT,φ⟩=⟨T,w±sφ⟩. By [F2] these maps are continuous for the weak and strong dual topologies; their compositions evaluate T on wsw−sφ=φ, so they are inverse on S′.

Depends on

Used by

Dependency tree · two levels

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Sources