LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Positive smooth damping preserves the double-pole boundary

Definitions and conventions

Throughout this item, $m \in N_{0}$ . For each integer $d \geq 1$ and compact $Q \subset R^{d}$ , let $D_{Q} (R^{d}) : = {φ \in C^{\infty} (R^{d}; C) : supp φ \subseteq Q},$ with the Fr'echet topology generated by $p_{Q, N} (φ) : = \max_{∣ α ∣ \leq N} \sup_{x \in Q} ∣ \partial^{α} φ (x) ∣ (N \in N_{0}),$ where $α = (α_{1}, \dots, α_{d}) \in N_{0}^{d}$ , $∣ α ∣ = \sum_{j} α_{j}$ , and $\partial^{α} = \partial_{x_{1}}^{α_{1}} \dots \partial_{x_{d}}^{α_{d}}$ . The compact-test space $D (R^{d}) = C_{c}^{\infty} (R^{d}; C)$ has the locally convex inductive-limit topology over these fixed-support spaces (equivalently over any compact exhaustion of $R^{d}$ ). A distribution on $R^{d}$ is a continuous complex-linear functional on this topology, and its test pairing is bilinear; write $D^{'} (R^{d})$ for the space of such distributions. For $d = 0$ , use $R^{0} = {*}$ with measure one. In that case set $D (R^{0}) = C$ with its usual finite-dimensional topology.

Let $1_{m} \in D^{'} (R^{m})$ denote integration against Lebesgue measure when $m \geq 1$ , and set $⟨ 1_{0}, a ⟩ = a$ . The symbol $1$ in the Statement's tensor notation means this $1_{m}$ . Its continuity on fixed-support test spaces follows from finite Lebesgue measure of compact supports Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure; for $m = 0$ it is evaluation on $C$ . For $B \in D^{'} (R)$ define the product distribution $1_{m} \otimes B$ by $⟨ 1_{m} \otimes B, ψ ⟩ : = ⟨ B, s \mapsto \int_{R^{m}} ψ (v, s) d v ⟩,$ where the integral for $m = 0$ is evaluation at $*$ . The inner function is a compactly supported smooth test. For each fixed compact support, its seminorms are bounded by the corresponding $s$ -derivative seminorms of $ψ$ times the finite measure of its $v$ -projection Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, so this definition is a continuous distributional pairing.

Statement

Assume Countable Choice and use Euclidean Lebesgue measure. Let $m \geq 0$ . For $m = 0$ take $R^{0}$ to be a singleton with measure $1$ . For each sufficiently small $ϵ > 0$ let $d_{ϵ} : R^{m} \times R \to (0, \infty)$ be smooth and real-valued. Assume that for every compact $V \subseteq R^{m}$ and every $R > 0$ there are constants $c, C, M, ϵ_{0} > 0$ such that, on $V \times [- R, R]$ and for $0 < ϵ < ϵ_{0}$ , $c \leq d_{ϵ} (v, s) \leq C, ∣ \partial_{s} d_{ϵ} (v, s) ∣ + ∣ \partial_{s}^{2} d_{ϵ} (v, s) ∣ \leq M .$ For $φ \in C_{c}^{\infty} (R; C)$ set $\hat{φ} (k) = \int_{R} φ (s) e^{- i k s} d s$ and define $B$ by $⟨ B, φ ⟩ : = \int_{0}^{\infty} k \hat{φ} (k) d k .$ For $ψ \in C_{c}^{\infty} (R^{m} \times R; C)$ write $\hat{ψ} (v, k) = \int_{R} ψ (v, s) e^{- i k s} d s$ and define $K$ by $⟨ K, ψ ⟩ : = \int_{R^{m}} \int_{0}^{\infty} k \hat{ψ} (v, k) d k d v .$ Then $B$ and $K$ are distributions, $B$ is the lower boundary denoted $- (s - i 0)^{- 2}$ , and $K = 1 \otimes B$ in the explicitly defined pairing above. The smooth kernels $K_{ϵ} (v, s) : = - (s - i ϵ d_{ϵ} (v, s))^{- 2}$ converge to $K$ in the distributional sense: $\int K_{ϵ} (v, s) ψ (v, s) d v d s \to ⟨ K, ψ ⟩$ for every such $ψ$ . When $s$ and $ϵ$ carry units of length, $d_{ϵ}$ is dimensionless.

Facts & Assumptions

Given: Countable Choice; Lebesgue measure on $R^{m}$ and $R$ ; the stated positive damping family; and compactly supported smooth complex tests.

[A1]

Countable Choice is assumed for the stated Lebesgue-measure and product-measure constructions. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

Integration by parts applies on bounded intervals to continuously differentiable real functions; apply it separately to real and imaginary parts for complex functions. If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$

[F2]

The complex exponential is entire with derivative itself, and the chain rule gives derivatives of its compositions. The complex exponential is entire and its complex derivative is itself The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$

[F3]

For real $x, y$ , $∣ e^{x + i y} ∣ = e^{x}$ . $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$

[F4]

For real $x$ , $e^{- x} = 1 / e^{x} > 0$ , and $x / e^{a x} \to 0$ as $x \to + \infty$ for every $a > 0$ . The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ The exponential dominates every fixed nonnegative integer power at $+ \infty$

[F5]

Dominated convergence passes limits through integrals under one integrable majorant. Dominated convergence

[F6]

Fubini permits changing integration order for an $L^{1}$ function on a sigma-finite product measure. Fubini's theorem for L^1 functions on a sigma-finite product

[F7]

Under Countable Choice, Euclidean Lebesgue measure on $R^{r}$ is sigma-finite and every bounded measurable set has finite measure for $r \geq 1$ ; for $r = 0$ the parameter space is a singleton of measure $1$ . Thus every compact parameter support $V$ has finite measure. Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure

[F8]

If $G$ is differentiable on a bounded interval and $G^{'}$ is integrable, then $\int G^{'} = G (b) - G (a)$ . Smooth functions on compact intervals meet these hypotheses. The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$

Proof

1.1A1F1F7algebra

Three integrations by parts give $∣ \hat{φ} (k) ∣ \leq ∥ φ ∥_{1}$ for $0 \leq k \leq 1$ and $∣ \hat{φ} (k) ∣ \leq k^{- 3} ∥ φ^{(3)} ∥_{1}$ for $k \geq 1$ . Hence $\int_{0}^{\infty} k ∣ \hat{φ} (k) ∣ d k \leq \frac{1}{2} ∥ φ ∥_{1} + ∥ φ^{(3)} ∥_{1}$ , so $B$ is a continuous linear functional on each fixed-support test space. Applying the same bound to $ψ (v, \cdot)$ and integrating over the compact $v$ -projection bounds $∣ ⟨ K, ψ ⟩ ∣$ by a constant times the suprema of $∣ ψ ∣$ and $∣ \partial_{s}^{3} ψ ∣$ on its compact support; thus $K$ is also a distribution and its displayed pairing is absolutely convergent.

1.2F1F2F3F4algebra

If $q = η + i x$ with $η > 0$ , an antiderivative of $e^{- q k}$ is $- q^{- 1} e^{- q k}$ by [F2]. Applying integration by parts to its real and imaginary parts gives $\int_{0}^{R} e^{- q k} d k = \frac{1 - e^{- q R}}{q}, \int_{0}^{R} k e^{- q k} d k = - \frac{R e^{- q R}}{q} + \frac{1 - e^{- q R}}{q^{2}} .$ By [F3]–[F4], $∣ e^{- q R} ∣ = e^{- η R}$ and $R e^{- η R} \to 0$ ; therefore $\int_{0}^{\infty} k e^{- q k} d k = q^{- 2}$ . For $a = x_{0} + i η$ , take $q = η + i (s - x_{0}) = i (s - a)$ to obtain $- \frac{1}{(s - a)^{2}} = \int_{0}^{\infty} k e^{- k Im a} e^{- i k (s - Re a)} d k .$ In particular this represents every constant positive damping $- (s - i ϵ)^{- 2}$ .

1.3A1F8algebra

Fix a compact parameter set $V$ and a compact interval $[- R, R]$ containing the $s$ -support of a test $ψ$ . Taylor's formula with integral remainder, obtained by applying [F8] twice, gives $d_{ϵ} (v, s) = d_{0, ϵ} (v) + d_{1, ϵ} (v) s + s^{2} r_{ϵ} (v, s),$ where $d_{0, ϵ} = d_{ϵ} (v, 0)$ , $d_{1, ϵ} = \partial_{s} d_{ϵ} (v, 0)$ , and $r_{ϵ}$ is uniformly bounded on $V \times [- R, R]$ . Put $q_{ϵ} = 1 - i ϵ d_{1, ϵ}$ and $a_{ϵ} = i ϵ d_{0, ϵ} / q_{ϵ}$ . Direct multiplication gives $s - i ϵ d_{ϵ} (v, s) = q_{ϵ} (s - a_{ϵ}) - i ϵ s^{2} r_{ϵ} .$ Moreover $Im a_{ϵ} = \frac{ϵ d_{0, ϵ}}{1 + ϵ^{2} d_{1, ϵ}^{2}} ≍ ϵ, Re a_{ϵ} = \frac{- ϵ^{2} d_{0, ϵ} d_{1, ϵ}}{1 + ϵ^{2} d_{1, ϵ}^{2}} = O (ϵ^{2}),$ uniformly on $V$ .

2.1A1F1F2F5F6F7step 1.1step 1.2algebra

The linearized kernel $- q_{ϵ}^{- 2} (s - a_{ϵ})^{- 2}$ has the representation of step 1.2. Pairing it with $ψ (v, s)$ and using Fubini gives the $v, k$ integrand $q_{ϵ} (v)^{- 2} k e^{- k Im a_{ϵ} (v)} e^{i k Re a_{ϵ} (v)} \hat{ψ} (v, k) .$ Three integrations by parts in $s$ show that $k ∣ \hat{ψ} (v, k) ∣$ is integrable over $V \times (0, \infty)$ ; sigma-finiteness follows from [F7] and [F6] justifies the order change. The integrand converges pointwise to $k \hat{ψ} (v, k)$ and is bounded by an integrable constant multiple of it. Dominated convergence [F5] therefore gives the limit $⟨ K, ψ ⟩$ .

2.2givenF7step 1.3algebra

Let $D_{ϵ} = s - i ϵ d_{ϵ} (v, s)$ and $L_{ϵ} = q_{ϵ} (s - a_{ϵ})$ . The given real-valued damping bounds $c \leq d_{ϵ} \leq C$ imply $∣ D_{ϵ} ∣^{2} = s^{2} + ϵ^{2} d_{ϵ}^{2}$ on the compact set, hence $∣ D_{ϵ} ∣ \geq \min (1, c) \sqrt{s^{2} + ϵ^{2}}$ and $∣ D_{ϵ} ∣ \leq \max (1, C) \sqrt{s^{2} + ϵ^{2}}$ . step 1.3 gives $Im a_{ϵ} \geq c ϵ$ and $∣ Re a_{ϵ} ∣ \leq C ϵ^{2}$ . If $∣ s ∣ \geq 2 C ϵ^{2}$ , then $∣ s - Re a_{ϵ} ∣ \geq ∣ s ∣ / 2$ ; otherwise $∣ s ∣ = O (ϵ^{2})$ and $∣ Im a_{ϵ} ∣ \geq c ϵ$ . Thus $∣ L_{ϵ} ∣ \geq c^{'} \sqrt{s^{2} + ϵ^{2}}$ for small $ϵ$ , while the upper bound follows from bounded $q_{ϵ}$ and $a_{ϵ} = O (ϵ)$ . The factorization in step 1.3 gives $∣ D_{ϵ} - L_{ϵ} ∣ \leq C^{'} ϵ s^{2}$ . Therefore $∣ D_{ϵ}^{- 2} - L_{ϵ}^{- 2} ∣ = \frac{∣ D_{ϵ} - L_{ϵ} ∣ ∣ D_{ϵ} + L_{ϵ} ∣}{∣ D_{ϵ} ∣^{2} ∣ L_{ϵ} ∣^{2}} \leq C^{'} \frac{ϵ s^{2}}{(s^{2} + ϵ^{2})^{3 / 2}} .$ The integral of this bound over $∣ s ∣ \leq ϵ$ is $O (ϵ)$ , and over $ϵ < ∣ s ∣ \leq R$ is $O (ϵ \log (R / ϵ))$ ; both tend to zero. Since $ψ$ is bounded and $V$ has finite measure by [F7], the pairing difference tends to zero.

3.1step 2.1step 2.2∎

step 2.2 shows that replacing the variable-damping kernel by the linearized kernel changes every compact-test pairing by a quantity tending to zero; step 2.1 gives the latter's limit as $⟨ K, ψ ⟩$ . Hence $K_{ϵ} \to K = 1 \otimes B$ in distributions.

Remarks

The cited notes develop the distribution framework and its calculus; they are not used as a substitute for the local estimate. The positive smooth damping result and its uniform error bound are proved in full above.

Positive smooth damping preserves the double-pole boundary

Definitions and conventions

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources