TheoremBase

Existence of Mollifier Kernels of Every Radius

Statement

Let nn be a natural number with 1≤n1\le n, let R\mathbb{R} be the real numbers, let ∥ ⋅ ∥\lVert\,\cdot\,\rVert be the Euclidean norm on Euclidean space Rn\mathbb{R}^{n}, and let dEd_{E} be the Euclidean distance, a metric on Rn\mathbb{R}^{n}; by The Open Subsets of a Metric Space Form a Topology the subsets open in (Rn,dE)(\mathbb{R}^{n},d_{E}) form a topology on Rn\mathbb{R}^{n}, to which the topological notions below refer. Let λn\lambda_{n} be Lebesgue measure on the Borel σ\sigma-algebra of Rn\mathbb{R}^{n}. For t∈Rt\in\mathbb{R} write t2t^{2} for t⋅tt\cdot t, and write t−1t^{-1} for the multiplicative inverse of t≠0t\ne 0.

Let δ∈R\delta\in\mathbb{R} with 0<δ0<\delta, let φ:R→R\varphi:\mathbb{R}\to\mathbb{R} be the function of The Exponential Bump Building Block is Smooth on the Real Line, and let ρ0:Rn→R\rho_{0}:\mathbb{R}^{n}\to\mathbb{R} be given by

ρ0(y)=φ(δ2−∥y∥2)(y∈Rn).\rho_{0}(y)=\varphi\bigl(\delta^{2}-\lVert y\rVert^{2}\bigr)\qquad(y\in\mathbb{R}^{n}).

Then the following hold.

1. (Properties of ρ0\rho_{0}) ρ0\rho_{0} is smooth on Rn\mathbb{R}^{n}; 0≤ρ0(y)0\le\rho_{0}(y) for every y∈Rny\in\mathbb{R}^{n}; ρ0(y)=0\rho_{0}(y)=0 for every yy with δ<∥y∥\delta<\lVert y\rVert; ρ0\rho_{0} is compactly supported and integrable with respect to λn\lambda_{n}; and

0<∫Rnρ0 dλn<∞.0<\int_{\mathbb{R}^{n}}\rho_{0}\,d\lambda_{n}<\infty .

2. (Normalisation) Let cc be the multiplicative inverse of ∫Rnρ0 dλn\int_{\mathbb{R}^{n}}\rho_{0}\,d\lambda_{n}, and let ρ:Rn→R\rho:\mathbb{R}^{n}\to\mathbb{R} be given by ρ(y)=c ρ0(y)\rho(y)=c\,\rho_{0}(y). Then ρ\rho is a mollifier kernel of radius δ\delta on Rn\mathbb{R}^{n}. In particular, for every natural number nn with 1≤n1\le n and every real δ>0\delta>0 a mollifier kernel of radius δ\delta on Rn\mathbb{R}^{n} exists.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…