User:Quietbritishjim/Approximate identity

From Folland's book on PDEs

Lemma

Let f∈L¹(ℝⁿ) be continuous.

Let φ∈L¹(ℝⁿ) be bounded and such that φ(x) is positive and decreasing in |x| and

$\int _{\mathbb {R} ^{n}}\varphi (\mathbf {x} )\,\mathrm {d} \mathbf {x} =1.$

Set

$\varphi _{\varepsilon }:={\frac {1}{\varepsilon ^{n}}}\varphi \left({\frac {\mathbf {x} }{\varepsilon }}\right).$

(Note that also $\textstyle \int \varphi _{\varepsilon }(\mathbf {x} )\,\mathrm {d} \mathbf {x} =1$.) Then for each x∈ℝⁿ

$\lim _{\varepsilon \to 0}{\Bigl (}\varphi _{\varepsilon }\ast f(\mathbf {x} ){\Bigr )}=f(x).$

Proof

Given η>0 we need to show that there exists α>0 such that, for all 0<ε<α, we have

$\left\vert \varphi _{\varepsilon }\ast f(\mathbf {x} )-f(\mathbf {x} )\right\vert <\eta .$

Since f is continuous there exists α₁(η)>0 such that

$\left\vert \mathbf {y} \right\vert <\alpha _{1}(\eta )\Rightarrow \left\vert f(\mathbf {x} )-f(\mathbf {x} -\mathbf {y} )\right\vert <\eta /2,$

where |B| is the volume of the unit ball in ℝⁿ; in particular

$\left\vert \int _{\vert \mathbf {y} \vert \leq \alpha _{1}}(f(\mathbf {x} -\mathbf {y} )-f(\mathbf {x} ))\varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \right\vert <{\frac {\eta }{2}}\int _{\vert \mathbf {y} \vert \leq \alpha _{1}}\varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \leq {\frac {\eta }{2}}.$

It thus remains to find an α₂(η)>0 (then set α:=min{α₁,α₂}) such that

$0<\varepsilon <\alpha _{2}\Rightarrow \left\vert \int _{\vert \mathbf {y} \vert >\alpha _{1}}(f(\mathbf {x} -\mathbf {y} )-f(\mathbf {x} ))\varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \right\vert <\eta /2.$

But this expression is bounded by the sum of

${\begin{aligned}&\int _{\left\vert \mathbf {y} \right\vert >\alpha _{1}}\left\vert f(\mathbf {x} -\mathbf {y} )\right\vert \varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \leq \int _{\mathbb {R} ^{n}}\left\vert f(\mathbf {y} )\right\vert \mathrm {d} \mathbf {y} \sup _{\left\vert \mathbf {y} \right\vert >\alpha _{1}}\varphi _{\varepsilon }(\mathbf {y} )\to 0&&\quad {\text{as }}\varepsilon \to \infty ,\\&\int _{\left\vert \mathbf {y} \right\vert >\alpha _{1}}\left\vert f(\mathbf {x} )\right\vert \varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \leq \left\vert f(\mathbf {x} )\right\vert \int _{\vert \mathbf {y} \vert >\alpha _{1}}\varphi _{\varepsilon }(\mathbf {y} )\,\mathrm {d} \mathbf {y} \to 0&&\quad {\text{as }}\varepsilon \to \infty ,\\\end{aligned}}$

so such a value for α₂(η) must exist.