Convergence of Approximations of the Identity in $L^p(\mathbb R^d)$ (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? 
 A: 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 $g$),
$$
\|f_t-f\|_p
\leq\|f_t-g_t\|_p+\|g_t-g\|_p+\|g-f\|_p<3\varepsilon.
$$
