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
![{\displaystyle \int _{\mathbb {R} ^{n}}\varphi (\mathbf {x} )\,\mathrm {d} \mathbf {x} =1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13b13f781478a3fad8e5e308a7fcab705925e79b)
Set
![{\displaystyle \varphi _{\varepsilon }:={\frac {1}{\varepsilon ^{n}}}\varphi \left({\frac {\mathbf {x} }{\varepsilon }}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a18aedb257abe5cd2ecf1d50e92590bfe171c56b)
(Note that also
.) Then for each x∈ℝn
![{\displaystyle \lim _{\varepsilon \to 0}{\Bigl (}\varphi _{\varepsilon }\ast f(\mathbf {x} ){\Bigr )}=f(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7426e9dbdff327435549da6a59e08a314473bf3c)
- Proof
Given η>0 we need to show that there exists α>0 such that, for all 0<ε<α, we have
![{\displaystyle \left\vert \varphi _{\varepsilon }\ast f(\mathbf {x} )-f(\mathbf {x} )\right\vert <\eta .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/530956328e51b8d7e6ab1b0645ab2e4ef4c76f1a)
Since f is continuous there exists α1(η)>0 such that
![{\displaystyle \left\vert \mathbf {y} \right\vert <\alpha _{1}(\eta )\Rightarrow \left\vert f(\mathbf {x} )-f(\mathbf {x} -\mathbf {y} )\right\vert <\eta /2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf4ace54c11a2e380df27fd767cb4533d20ae45a)
where |B| is the volume of the unit ball in ℝn; in particular
![{\displaystyle \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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c005a5d2cda234848afaab51f696983d7431618)
It thus remains to find an α2(η)>0 (then set α:=min{α1,α2}) such that
![{\displaystyle 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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bfe176a355aae9767ae16edf60845bd79244893)
But this expression is bounded by the sum of
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46e43c1016c40f4496ce87c15d6db36d86d41a36)
so such a value for α2(η) must exist.