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Complete test Frechet space and uniform scaling order

Statement

Assume the Axiom of Choice. Let $n \geq 1$ , $U = R^{n} ∖ {0}$ , and $t \in D^{'} (U)$ have finite real scaling degree $δ$ . For every compact $K \subset U$ and every real $ω > δ$ , there are an integer $m \geq 0$ and $C < \infty$ such that

$∣ ⟨ t_{ρ}, ψ ⟩ ∣ \leq C ρ^{- ω} p_{m} (ψ) (0 < ρ \leq 1, ψ \in D_{K}),$

where $D_{K}$ and $p_{m} (ψ) = \max_{∣ α ∣ \leq m} \sup_{x \in K} ∣ \partial^{α} ψ (x) ∣$ are the fixed-support test-function space and seminorms of Fixed support test function frechet space. The estimate is uniform over the entire moving-scale family, not merely over each fixed test separately.

Facts & Assumptions

Given: $n \geq 1$ , $U = R^{n} ∖ {0}$ , $t \in D^{'} (U)$ with ${sd}_{0} (t) = δ \in R$ , a compact $K \subset U$ , and $ω > δ$ . Assume the Axiom of Choice.

[F1]

Scaling degree and local extension space: $t_{ρ} (ψ) = ρ^{- n} t (ψ (\cdot / ρ))$ , and $δ$ is the infimum of the real exponents for which $ρ^{η} t_{ρ} \to 0$ on every fixed test.

[F2]

Fixed support test function frechet space: $D_{K}$ has increasing derivative seminorms $p_{m}$ and the compatible metric $d (ψ, ξ) = \sum_{m \geq 0} 2^{- m - 1} \min (1, p_{m} (ψ - ξ))$ .

[F3]

Fixed support test function spaces are complete: this metric on $D_{K}$ is complete and induces the seminorm topology, without a choice axiom.

[F4]

The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain: DC supplies an $N$ -indexed chain for any serial relation and prescribed initial point.

[F5]

The Axiom of Choice: AC selects an element from every set-indexed family of nonempty sets.

[F6]

AC implies DC implies countable choice: in ZF, AC implies DC.

[F7]

Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior: under DC, a nonempty complete metric space cannot be a countable union of closed sets with empty interior.

[F8]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): AC $_{ω}$ selects a point from each sequence of nonempty sets.

[A1]

The Axiom of Choice is assumed.

Proof

Given: The hypotheses in the statement.

1.1givenF2F3F4F5F6F7F8A1

Let $X = D_{K}$ with the metric of [F2]. It is nonempty because it contains $0$ , and it is complete by [F3]. AC supplies DC by [F4]–[F6], so the Baire theorem [F7] applies. The proof of [F7] also chooses centers for its recursively constructed balls; that countable family of nonempty balls is covered by AC $_{ω}$ from [F6] and [F8].

1.2F1givenalgebra

For fixed $ψ \in X$ , set $L_{ρ} (ψ) = ρ^{ω} ⟨ t_{ρ}, ψ ⟩$ . The exponents $η$ for which $ρ^{η} t_{ρ} (ψ) \to 0$ for every fixed test form an upward-closed set: if the limit holds at $η$ , then for $ω > η$ multiply by $ρ^{ω - η} \leq 1$ . Since its infimum is $δ < ω$ , some such $η$ is below $ω$ , and consequently $L_{ρ} (ψ) \to 0$ as $ρ ↓ 0$ .

2.1F1F2step 1.2

On any compact scale interval $[ϵ, 1]$ with $ϵ > 0$ , the tests $ψ (\cdot / ρ)$ have support in one compact subset of $U$ and vary continuously in every derivative seminorm by the chain rule. Continuity of the distribution $t$ therefore makes $ρ \mapsto L_{ρ} (ψ)$ continuous there. Together with step 1.2 this proves $\sup_{0 < ρ \leq 1} ∣ L_{ρ} (ψ) ∣ < \infty$ for every $ψ \in X$ .

3.1step 1.1step 2.1F7

For each positive integer $N$ , let $E_{N} = {ψ \in X : \sup_{0 < ρ \leq 1} ∣ L_{ρ} (ψ) ∣ \leq N}$ . Each $E_{N}$ is closed because every $L_{ρ}$ is a continuous linear functional on $X$ . Step 2.1 gives $X = ⋃_{N \geq 1} E_{N}$ . By [F7], some $E_{N}$ has nonempty interior.

4.1F2step 3.1algebra∎

The topology of $X$ is generated by the increasing seminorms in [F2], so for some $ψ_{0} \in E_{N}$ , $m \geq 0$ , and $ε > 0$ , the set $ψ_{0} + {ζ : p_{m} (ζ) < ε}$ lies in $E_{N}$ . For every such $ζ$ , subtracting the two bounds for $ψ_{0} + ζ$ and $ψ_{0}$ gives $∣ L_{ρ} (ζ) ∣ \leq 2 N$ for all $ρ$ . If $p_{m} (ψ) > 0$ , apply this to $ζ = ε ψ / (2 p_{m} (ψ))$ ; if $p_{m} (ψ) = 0$ , then $p_{0} (ψ) = 0$ and hence $ψ = 0$ . Thus $∣ L_{ρ} (ψ) ∣ \leq (4 N / ε) p_{m} (ψ)$ for every $ψ$ and every $0 < ρ \leq 1$ . Replacing $L_{ρ}$ by its definition proves the claimed uniform order estimate.

Complete test Frechet space and uniform scaling order

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources