Lemma 12.4.9.label Suppose $\p\in L^{1},f\in L^{p}$ for $1\leq p\leq\infty$ and $\integral{}{}\pdx=a$. For each $\e>0$ define $\p_{\e}(x)\define\e^{-n}\p\parens{\frac{x}{\e}}$. Then $f*\p_{\e}\to af$ in the $L^{p}$ norm.[1]
Proof. Observe by change of variables $u=x/\e$
\begin{align*}\integral{}{}\p_{\e}(x)dx=\integral{}{}\frac{1}{\e^{n}}\p \parens{\frac{x}{\e}}dx=\integral{}{}\p(u)du=a\end{align*}
then
\begin{align*}(f*\p_{\e})(x)-af(x)&=\integral{}{}f(x-y)\p_{\e}(y)dy-\integral{}{}f(x)\p_{\e}(y)dy\\&=\integral{}{}\braks{f(x-y)-f(x)}\p_{\e}(y)dy\\&=\integral{}{}\braks{f(x-y)-f(x)}\e^{-n}\p\parens{\frac{y}{\e}}dy\\&=\integral{}{}\braks{f(x-\e y)-f(x)}\p(y)dy\end{align*}
By Minkowski’s inequality
\begin{align*}\norm{f*\p_\e-f}{p}\leq \integral{}{}\norm{f_{-\e y}-f}{p}\abs{\p(y)}dy\end{align*}
where $f_{x}(y)\define f(x+y)$ and by Dominated Convergence Theorem we are done.$\square$
- see folland for $p=\infty$keyboard_return
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