(context: in a comment to an answer of mine mentioned below, a user has asked for the proof of one of the steps)
In this answer, one of the steps mentions that
If $f\in L^p(\mathbb R^d)$, $g\in L^1(\mathbb R^d)$ with $\int_{\mathbb R^d} g=1$ and $g_n(x)=n^d\,g(nx)$, then $$\lim_{n\to\infty}\|f-f*g_n\|_p=0.$$
What would be the proof of this?
Assume for simplicity that $g\geq0$ (nothing really changes if we don't; just a few more absolute values). First, with the change of variable $x=z/n$, $$ \int_{\mathbb R^d}g_n(x)\,dx=\int_{\mathbb R^d}n^dg(nx)\,dx=\int_{\mathbb R^d} g(z)\,dz=1. $$ Then we have (applying Hölder on the last step) \begin{align} |f*g_n(x)-f(x)| &=\left|\int_{\mathbb R^d} [f(x-t) - f(x)]\, g_n(t)\, dt\right|\\[0.3cm] &\leq \int_{\mathbb R^d} |f(x-t)-f(x)| \,g_n(t)^{1/p}\,g_n(t)^{1/q}\,dt\\[0.3cm] &\leq \|g_n\|_1^{1/q}\,\left(\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,g_n(t)\,dt\right)^{1/p}. \end{align} Thus (using Tonelli and that $\|g_n\|_1=1$) \begin{align} \int_{\mathbb R^d}|f*g_n(x)-f(x)|^p\,dx &\leq\int_{\mathbb R^d}\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,g_n(t)\,dt\,dx\\[0.3cm] &=\int_{\mathbb R^d}\left(\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,dx\right)\,\,g_n(t)dt. \end{align} Recall that by $L^p$ continuity (details at the bottom), we have $$ \lim_{|t|\to0}\|f_t^{\vphantom{t}}-f\|_p=0, $$ where $f_t(x)=f(x-t)$. So, given $\varepsilon>0$, there exists $\delta>0$ such that $|t|<\delta$ implies $\|f_t-f\|_p<\varepsilon^{1/p}$. Then $$\tag1 \int_{|t|<\delta}\left(\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,dx\right)\,\,g_n(t)dt\leq \varepsilon \|g_n\|_1=\varepsilon. $$ We also have, for any $t$, \begin{align} \int_{\mathbb R^d}|f(x-t)-f(x)|^p\,dx&\leq \int_{\mathbb R^d}(|f(x-t)|+|f(x)|)^p\,dx\\[0.3cm]&\leq 2^p\int_{\mathbb R^d}(|f(x-t)|^p+|f(x)|^p)\,dx\\[0.3cm]&=2^{p+1}\|f\|_p^p. \end{align} Since (use again the change of variable $x=z/n$) $$ \int_{|t|\geq\delta} g_n(t)\,dt=\int_{|t|\geq\delta} n^d\,g(nt)\,dt=\int_{|t|\geq n\delta} g(t)\,dt\xrightarrow[n\to\infty]{}0, $$ we obtain $$\tag2\begin{aligned} \int_{|t|\geq\delta}\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,dx\,g_n(t)dt&\leq2^{p+1}\|f\|_p^p\,\int_{|t|\geq\delta} g_n \\[0.3cm]&\xrightarrow[n\to\infty]{}0. \end{aligned}$$ Combining this last equation with $(1)$, $$ \limsup_{n\to\infty}\int_{\mathbb R^d}\left(\int_{\mathbb R^d}|f(x-t)-f(x)|^p\,dx\right)\,\,g_n(t)dt\leq\varepsilon. $$ As $\varepsilon$ was arbitrary, the limit exists and is equal to zero.
$L^p$ Continuity: let $f_t(x)=f(x-t)$. Then $$\lim_{t\to0}\|f_t-f\|_p=0.$$ Indeed, assume first that $f$ is continuous with compact support $K$. Then $f$ is uniformly continuous: given $\varepsilon>0$, there exists $\delta>0$ such that $|f(x-t)-f(x)|<\varepsilon^{1/p}/|K|$ for all $x$ whenever $|t|<\delta$. Then $\|f_t-f\|_p<\varepsilon$.
Now for arbitrary $f$ there exists $g$, continuous with compact support and $\|f-g\|_p<\varepsilon$. Then, if $|t|<\delta$ (the one for $g$), $$ \|f_t-f\|_p \leq\|f_t-g_t\|_p+\|g_t-g\|_p+\|g-f\|_p<3\varepsilon. $$