Alphabeta MathHow statement and proof provenance work
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- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
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A normalized kernel loses constant reproduction when truncated at a boundary
Statement refuted
The claim that every normalized symmetric kernel preserves constants after truncation to a domain, even at a boundary point, is false.
Facts & Assumptions
Given: In one dimension define W(s)=(35/32)(1−s²)³ for |s|≤1 and W(s)=0 otherwise, set W_ε(s)=ε^(−1)W(s/ε), take Ω=[0,∞), and define T_εf(x)=∫_0^∞f(y)W_ε(x−y)dy.
Scaled symmetric compactly supported kernel family: an admissible kernel is nonnegative, even, normalized and compactly supported.
Change of variables for an injective map on a compact Jordan set: if T is injective C¹ with invertible derivative on an open set containing compact Jordan K, then ∫_{T(K)}f(y)dy=∫_Kf(T(s))|det DT(s)|ds for Riemann-integrable f.
A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: a continuous real function on a compact interval is Riemann integrable.
Counterexample
The displayed W is nonnegative, even, supported on [−1,1], and C² at ±1 because its value and first two inside derivatives vanish there. [F3] makes it integrable on [−1,1], and its integral is (35/32)·2(1−1+3/5−1/7)=1, so it satisfies [F1].
At x=0, evenness and support reduce T_ε1(0) to ∫_0^εW_ε(y)dy. Apply [F2] on K=[0,1] with T(s)=εs; since W_ε(εs)=ε^(−1)W(s), this equals ∫0^1W(s)ds. By [F1], W is even and normalized, so the half-interval integral is (1/2)∫{−1}^1W(s)ds=1/2.
The truncated value 1/2 differs from the constant 1, so the claimed boundary reproduction fails.
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Sources
- Volker Springel, Smoothed Particle Hydrodynamics in Astrophysics, Annual Review of Astronomy and Astrophysics 48 (2010), author full text (standard reference, not scraped)
- Benjamin Seibold, MIT OpenCourseWare 18.336, Numerical Methods for Partial Differential Equations, Lecture 25: Particle Methods (Spring 2009) (standard reference, not scraped)