I'm trying to prove a result mentioned in this thread, i.e.,
Theorem Let $\rho \in L^1 (\mathbb R^d)$ such that $\int \rho = 1$. Let $\rho_n (x) := n^d \rho (nx)$ for all $x \in \mathbb R^d$. Let $f \in L^p (\mathbb R^d)$ with $p \in [1, \infty)$. Then $\rho_n * f \xrightarrow{n \to \infty} f$ in $L^p$.
Could you have a check on my below proof?
A sequence of mollifiers $(\hat \rho_n)_{n \geq 1}$ is any sequence of functions on $\mathbb{R}^d$ such that $(\int \hat \rho_n)_n$ is a constant sequence and that $$ \hat \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \hat \rho_n \subset \overline{B(0, r_n)}, $$ where $r_n \to 0^+$ and $B(x, r)$ denotes an open ball centered at $x$ with radius $r$. Then we have
Lemma Let $f \in L^p (\mathbb R^d)$ with $p \in [1, \infty)$. Then $\hat \rho_n * f \xrightarrow{n \to \infty} f \int \hat \rho_1$ in $L^p$.
Proof of the Theorem Fix $\varepsilon > 0$. We have $\mathcal C_c^\infty (\mathbb R^d)$ is dense in $L^1 (\mathbb R^d)$. There is $\hat \rho \in \mathcal C_c^\infty (\mathbb R^d)$ such that $\|\rho - \hat \rho\|_1 < \varepsilon$. Let $\hat \rho_n (x) := n^d \hat \rho (nx)$ for all $x \in \mathbb R^d$. Then $\int \hat \rho_n = \int \hat \rho$ for all $n$. Clearly, $(\hat \rho_n)_{n \geq 1}$ is a sequence of mollifiers. By Young's inequality, $$ \begin{align} \| \rho_n * f- \hat\rho_n * f\|_p &= \| (\rho_n - \hat\rho_n) * f\|_p \\ &\le \| \rho_n - \hat\rho_n\|_1 \|f\|_p \\ &= \| \rho - \hat\rho\|_1 \|f\|_p \\ &\le \varepsilon \|f\|_p. \end{align} $$
Let $a := \int \hat \rho_1$. Then $|a- 1| = |\int \rho - \int \hat \rho| \le \varepsilon$. So $$ \begin{align} \| \rho_n * f- f\|_p &\le \| \rho_n * f- \hat\rho_n * f\|_p + \| \hat \rho_n * f- af\|_p + \| af- f\|_p\\ &\le 2 \varepsilon \|f\|_p + \| \hat \rho_n * f- f\|_p. \end{align} $$
By above Lemma, $\| \hat \rho_n * f- af\|_p \xrightarrow{n \to \infty} 0$. The claim then follows by taking the limit $\varepsilon \downarrow 0$.
