A sequence of mollifiers $(\rho_n)_{n \geq 1}$ is any sequence of functions on $\mathbb{R}^d$ such that $$ \rho_n \in C_c^{\infty} (\mathbb{R}^d), \quad \operatorname{supp} \rho_n \subset \overline{B(0, r_n)}, \quad \sup_n \|\rho_n\|_1 < \infty, $$ where $r_n \to 0^+$ and $B(x, r)$ denotes an open ball centered at $x$ with radius $r$.
In order to prove a result mentioned in this thread, I need to prove below approximation, i.e.,
Theorem 1 Let $f \in \mathcal C (\mathbb R^d)$. Then $\rho_n * f \xrightarrow{n \to \infty} f$ uniformly on compact sets.
Theorem 2 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 attempt?
Proof of Theorem 1 WLOG, we assume $\sup_n r_n \le 1$. Fix a compact subset $K$ of $\mathbb{R}^d$. Fix $\varepsilon>0$. Then $f$ is uniformly continuous on $K' :=K - \overline{B(0, 1)}$. There is $\delta>0$ such that if $x,y \in K'$ with $|x-y| < \delta$ then $|f(x)-f(y) |< \varepsilon$. Let $\alpha := \sup_n \|\rho_n\|_1 < \infty$. If $x \in K$ and $r_n < \delta$ then $$ \begin{align} |\rho_n * f (x)-f(x)| &\le \int_{\overline{B(0, r_n)}} |f(x-y)-f(x)| |\rho_n(y)| \, \mathrm d y \\ &\le \varepsilon \int_{\overline{B(0, r_n)}} |\rho_n(y)| \, \mathrm d y \\ &\le \varepsilon \alpha. \end{align} $$
The claim then follows by taking the limit $\varepsilon \downarrow 0$. QED.
Proof of Theorem 2 We have $\mathcal C_c (\mathbb R^d)$ is dense in $L^p (\mathbb R^d)$. Fix $\varepsilon>0$. There is $\hat f \in \mathcal C_c (\mathbb R^d)$ such that $\|f-\hat f\|_p < \varepsilon$. Then $\hat f * \rho_n \in \mathcal C_c (\mathbb R^d)$ for all $n$. WLOG, we assume $\sup_n r_n \le 1$. Then $$ \operatorname{supp} (\hat f * \rho_n) \subset \overline{\operatorname{supp} \hat f + \operatorname{supp} \rho_n} \subset K:=\operatorname{supp} \hat f + \overline{B(0, 1)}. $$
Clearly, $K$ is compact. By Theorem 1, $\hat f * \rho_n \xrightarrow{n \to \infty} \hat f$ uniformly. By triangle inequality, $$ \begin{align} \| \rho_n * f - f\|_p &\le \| \rho_n * (f - \hat f)\|_p + \| \rho_n * \hat f - \hat f\|_p + \|f-\hat f\|_p. \end{align} $$
Let $\alpha := \sup_n \|\rho_n\|_1 < \infty$. By Young's inequality, $$ \| \rho_n * (f - \hat f)\|_p \le \| \rho_n \|_1 \| f - \hat f\|_p \le \alpha \varepsilon. $$
The claim then follows from taking the limit $n \uparrow \infty$ and then $\varepsilon \downarrow 0$. QED.
Update As mentioned in comments, the condition $\int \rho_n =1$ for all $n$ is necessary.
