The $L^{2}$-difference of a Lipschitz function and its mollified function. Question
Let $f$ have Lipschitz constant at most $2$, and $\varphi$ be a standard mollifier. That is: $\varphi$ is smooth and has support contained in $B(0,1)$ and satisfies,
$$
\begin{cases}
0 \le \varphi \\
\int_{\mathbb{R}^{n}} \varphi = 1.
\end{cases}
$$
Let $\varphi_{\ell}$ be such that $\varphi_{\ell}(x) = \frac{1}{\ell^{n}} \varphi \left( \frac{x}{\ell} \right)$.
Suppose $\ell < 1$ and $f$ is defined on all of $B(0,2)$.
Show,
$$
\int_{B(0,1)} | f - f \star \varphi_{\ell}|^{2} \le C \ell^{2} \int_{B(0,2)} |Df|^{2}.
$$
Setting
This claim is made in some lecture notes I found online, but the notes do not elaborate on the claim at all. 
My attempt
In case I define the convolution backwards, I additionally assume that $\varphi(-x) = \varphi(x)$ without causing any problem. Then,
\begin{align*}
\int_{B(0,1)} |f - f \star \varphi_{\ell}|^{2} &= \int_{y \in B(0,1)} \left| \int_{x \in B_{\ell}(y)} (f(y) - f(x)) \varphi_{\ell}(y-x) dx \right|^{2} dy \\
& \le \ell^{2} \int_{y \in B(0,1)} \left| \int_{x \in B_{\ell}(y)} \frac{f(y) - f(x)}{|x-y|} \varphi_{\ell}(y-x) dx \right|^{2} dy \\
& \le \left( \textrm{Lip}(f) \right)^{2} \ell^{2} \int_{y \in B(0,1)} dy \\
& = C \ell^{2}.
\end{align*}
Obviously, if we additionally know $\int_{B(0,2)} |Df|^{2}$ is small then my result is not as strong as the claim.
Any thoughts or suggestions on where I made too rough of an estimate are appreciated.
 A: This is far from a complete answer but I hope you can find it helpful.
I considered the following version of the problem:
$$\int_{B_1(0)}|f-f\star \varphi_{\ell}|\lesssim  C\ell^2 \int_{B_2(0)}|Df|^2,\qquad \text{as }\ell\to 0 \qquad (*)$$
This seems the most significant part of the problem, although for some reason proving the inequality for all $\ell$ seems to pose different problems, I wonder if it is even true.
I think that your roughest estimate is the second inequality, where instead of pushing out the Lipschitz constant, an improvement is given by using 
$$\int_{y \in B_1(0)} \left| \int_{x \in B_{\ell}(y)} \frac{f(y) - f(x)}{|x-y|} \varphi_{\ell}(y-x) dx \right|^{2} dy\leq \int_{y\in B_1(0)}\left(\sup_{x\in B_{\ell}(y)}\frac{|f(y)-f(x)|}{|x-y|}\right)^2dy,\quad (1)  $$
Taking the limit for $\ell \to 0$, since the integrand in the RHS is bounded by the squared uniform Lipschitz constant $2^2=4$, it is dominated by an integrable function (i.e. the constant $4$) so that we may push the limit inside the integral. 
Moreover, assuming $f$ is differentiable, by the mean value theorem + Cauchy-Schwarz:
$$\sup_{x\in B_{\ell}(y)}\frac{|f(y)-f(x)|}{|x-y|}=\sup_{x\in B_{\ell}(y)}\left|Df(y+t(x-y))\cdot\frac{x-y}{|x-y|}\right|\leq\sup_{x\in B_{\ell}(y)}|Df(y+t(x-y))|,\quad (2)$$
Assuming additionally that $Df(y)$ is continuous for a.e. $y\in B_1(0)$, as $\ell \to 0$ in ($2$) we get $y+t(x-y)\to y$ and therefore $Df(y+t(x-y))\to Df(y)$, so that 
$$\lim_{\ell \to 0}\sup_{x\in B_{\ell}(y)}|Df(y+t(x-y))|=|Df(y)|,\qquad a.e.y\in B_1(0) ,\qquad (3)$$
(I wonder if this estimate can be obtained without these regularity assumptions on $f$? We still have room for a constant $C$ independent from $f$ and $\ell$).
And finally, putting toghether ($2$) and ($3$) and substituting in ($1$), we get
$$\int_{y\in B_1(0)}\lim_{\ell \to 0^+}\left(\sup_{x\in B_{\ell}(y)}\frac{|f(y)-f(x)|}{|y-x|}\right)^2dy \leq\int_{y\in B_1(0)}|Df(y)|^2dy $$
and ($*$) is proved with $C=1$. Moreover $|Df|^2$ need only be integrated in $B_1(0)$ instead of $B_2(0)$ (which is stronger) which is due to the fact that we took the limit for $\ell \to 0$.
