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{align}

&\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{align}$$

so such a value for α2(η) must exist.