Problem #7.10, MA598R, Summer 2009, Weigel
Let $ \varphi \geq 0 $ with $ \int_{\mathbb{R}^n} \varphi(y)dy =1 $. Denote $ \varphi_{\epsilon}(x) = \epsilon^{-n}\varphi\left(\frac{x}{\epsilon}\right) $. Prove the following statements:
a) $ \int_{\mathbb{R}^n} \varphi_{\epsilon}(x) dx =1 $.
Proof. Using the change of variables $ y = \frac{x}{\epsilon} $ we find: $ \begin{align}\int_{\mathbb{R}^n} \varphi_{\epsilon}(x) dx &= \int_{\mathbb{R}^n} \epsilon^{-n}\varphi\left(\frac{x}{\epsilon}\right)dx\\ &= \int_{\mathbb{R}^n} \epsilon^{-n}\varphi\left(y\right)\epsilon^ndy\\ &=\int_{\mathbb{R}^n} \varphi(y)dy =1 \end{align} $
b) For any $ \delta > 0 $, $ \lim_{\epsilon \rightarrow 0} \int_{|x|> \delta} \varphi_{\epsilon}(x)dx=0 $.
Proof. Fix $ \delta $. Using the same change of variable, $ \int_{|x|> \delta} \varphi_{\epsilon}(x)dx = \int_{|y|> \frac{\delta}{\epsilon}} \varphi(y)dy $. Now we notice that $ \varphi \in L(\mathbb{R}^n) $ and that $ \frac{\delta}{\epsilon} \rightarrow \infty $ as $ \epsilon \rightarrow 0 $. Thus $ \lim_{\epsilon \rightarrow 0} \int_{|x|> \delta} \varphi_{\epsilon}(x)dx=\lim_{\epsilon \rightarrow 0}\int_{|y|> \frac{\delta}{\epsilon}} \varphi(y)dy=0 $.
c) If $ f \in L^p(\mathbb{R}^n) $, $ 1 \leq p < \infty $, then $ \lim_{\epsilon \rightarrow 0}\|f * \varphi_{\epsilon} - f\|_p = 0 $.
Proof. From part (a) we have that $ \int_{\mathbb{R}} \varphi_{\epsilon}(x) dx =1 $. Thus applying Holder's and then Fubini, $ \begin{align}\|f * \varphi_{\epsilon} - f\|_p &= \|f * \varphi_{\epsilon}(y) - f(y)\int_{\mathbb{R}^n} \varphi_{\epsilon}(x) dx\|_p\\ &= \|\int_{\mathbb{R}^n}f(y-x)\varphi_{\epsilon}(x) - f(y)\varphi_{\epsilon}(x)dx\|_p\\ &=\|\int_{\mathbb{R}^n}\left(f(y-x) - f(y)\right)\varphi_{\epsilon}(x)dx\|_p\\ &=\|\int_{\mathbb{R}^n}\left(f(y-x) - f(y)\right)\left(\varphi_{\epsilon}(x)\right)^{\frac{1}{p}+\frac{1}{p^{\prime}}}dx\|_p\\ &\leq \left[\int_{\mathbb{R}^n}\left[\int_{\mathbb{R}^n}\left|f(y-x) - f(y)\right|^p\left(\varphi_{\epsilon}(x)\right)dx\right]\left[\int_{\mathbb{R}^n}\left(\varphi_{\epsilon}(x)\right)dx\right]^{\frac{p}{p^{\prime}}}\right]^{\frac{1}{p}}dy\\ &=\|\varphi\|_1^{\frac{1}{p^{\prime}}}\left(\int_{\mathbb{R}^n}\left[\int_{\mathbb{R}^n}\left|f(y-x) - f(y)\right|^p\left(\varphi_{\epsilon}(x)\right)dx\right]dy\right)^{\frac{1}{p}}\\ &\leq \|\varphi\|_1^{\frac{1}{p^{\prime}}}\left(\int_{\mathbb{R}^n}\varphi_{\epsilon}(x)\left[\int_{\mathbb{R}^n}\left|f(y-x) - f(y)\right|^pdy\right]dx\right)^{\frac{1}{p}}\\ &= \|\varphi\|_1^{\frac{1}{p^{\prime}}}\left(\int_{|x|\leq\delta}\varphi_{\epsilon}(x)\|f(y-x) - f(y)\|_p^pdx + \int_{|x|>\delta}\varphi_{\epsilon}(x)\|f(y-x) - f(y)\|_p^pdx\right)^{\frac{1}{p}} \end{align} $
Now, for $ \gamma >0 $, we can choose $ \delta $ so that $ \|f(y-x) - f(y)\|_p^p < \gamma $ when $ |x| \leq \delta $.
$ \int_{|x|\leq\delta}\varphi_{\epsilon}(x)\|f(y-x) - f(y)\|_p^pdx \leq \gamma\|\varphi\|_1 $
Also, $ \|f(y-x) - f(y)\|_p^p \leq (2\|f\|_p)^p $ so that $ \int_{|x|>\delta}\varphi_{\epsilon}(x)\|f(y-x) - f(y)\|_p^pdx \leq (2\|f\|_p)^p\int_{|x|>\delta}\varphi_{\epsilon}(x)dx $. And from (b) we found that $ \int_{|x|>\delta}\varphi_{\epsilon}(x)dx \rightarrow 0 $ as $ \epsilon \rightarrow 0 $.
Thus we let $ \gamma $ and $ \epsilon $ go to zero to get the desired result.
d) If $ f \in C^{\infty}_0(\mathbb{R}^n) $, then $ \lim_{\epsilon \rightarrow 0} f * \varphi_{\epsilon}(x)=f(x) $.
Proof. Similar to part (c), $ \begin{align} |f * \varphi_{\epsilon}(x) - f(x)| &= \left|\int_{\mathbb{R}^n}f(y-x)\varphi_{\epsilon}(x) - f(y)\varphi_{\epsilon}(x)dx\right|\\ &\leq \int_{\mathbb{R}^n}|f(y-x) - f(y)||\varphi_{\epsilon}(x)|dx\\ \end{align} $
For $ f $ continuous at $ y $ given $ \gamma, \delta > 0 $ we have that $ |f(y-x) - f(y)|< \gamma $ when $ |x| \leq \delta $. When $ |x| > \delta $ we have that $ |f(y-x) - f(y)| \leq 2\|f\|_{\infty} $. Thus, $ \begin{align} \int_{\mathbb{R}^n}|f(y-x) - f(y)||\varphi_{\epsilon}(x)|dx &= \int_{|x|\leq \delta}|f(y-x) - f(y)||\varphi_{\epsilon}(x)|dx + \int_{|x| > \delta}|f(y-x) - f(y)||\varphi_{\epsilon}(x)|dx\\ &\leq \gamma\int_{|x|\leq \delta}|\varphi_{\epsilon}(x)|dx + 2\|f\|_{\infty}\int_{|x| > \delta}|\varphi_{\epsilon}(x)|dx\\ &\leq \gamma\|\varphi\|_1 + 2\|f\|_{\infty}\int_{|x| > \delta}|\varphi_{\epsilon}(x)|dx \end{align} $
Again we use the result from part (b) and thus we let $ \gamma $ and $ \epsilon $ go to zero to get the desired result.
