A question on convergence in $\operatorname{Lip}_0(\mathbb R^n)$ $\DeclareMathOperator\Lip{Lip}$This question arose when I read Godefroy and Lerner - Some natural subspaces and quotient spaces of $L^1$.
Let $\Lip_0(\mathbb R^n)$ be the space of Lipschitz functions $f:\mathbb R^d\to\mathbb R$ vanishing at the origin, $f(0)=0$. It is known from the above paper that, endowed with the norm $\|f\|_{\Lip}\mathrel{:=}\|\nabla f\|_{\infty}$,  $\big(\Lip_0(\mathbb R^n), \|\cdot\|_{\Lip}\big)$ is a Banach space. My question is, if $f^n$ converges to $f$ under the above norm, could we deduce 
$$\lim_{n\to\infty} \int_{\mathbb R^d}\big(f^n(x)-f(x)\big)u(x)dx = 0 ,$$
where $u:\mathbb R^d \to\mathbb R_+$ is a measurable function s.t.
$$ \int_{\mathbb R^d}(1+\lvert x\rvert)u(x)dx <\infty.$$
This seems a trivial question, but I can not prove it rigorously for general dimensions. Is there any classical reference? 
 A: Denote $g_n:=f-f_n$, so that $\lim_{n\rightarrow\infty}\|\nabla g_n\|_{\infty}=0$ . Note that by the fundamental theorem of calculus we have
$$
g_n(x)=g_n(x)-g_n(0)=\int_0^1 \frac{d}{dt}(g_n(tx))dt=\int_0^1 x\cdot \nabla g_n(tx)\, dt, \qquad x\in\mathbb{R}^d.
$$
Using the above formula we get
\begin{align*}
\lim_{n\to\infty} \left|\int_{\mathbb R^d} g_n(x)u(x)dx\right| &= \lim_{n\to\infty} \left|\int_{\mathbb R^d} \left(\int_0^1 x\cdot \nabla g_n(tx)\, dt\right)\, u(x)\,dx\right|\\
&\le  \lim_{n\to\infty} \int_{\mathbb R^d} \left(\int_0^1 |x||\nabla g_n(tx)|\, dt\right)\, u(x)\,dx\\
&\le \lim_{n\to\infty} \|\nabla g_n\|_{\infty}\int_{\mathbb R^d}  |x|\, u(x)\,dx=0.
\end{align*}
A: Here is an alternative answer (also based on the control of the growth at infinity): simply use Lebesgue's Dominated Convergence theorem:
Note first that the convergence  $\|f_n-f\|\to 0$ in your $Lip_0(\mathbb R^d)$ space immediately implies pointwise a.e. convergence,
$$
f_n(x)u(x)\to f(x)u(x)
\qquad a.e.
$$
In order to apply the DCT we only need a dominating $L^1$ bound.
For this note that the Lipschitz norm controls the growth at infinity, hence $
|f_n(x)|\leq \|f_n\|\,|x|\leq 2 \|f\|\, |x|
\qquad \forall x$ uniformaly in $n$.
In particular given your assumptions on $u$ we get
$$
|f_n(x)\, u(x)|\leq 2 \|f\|\, |x|\, u(x)\in L^1
$$
and the result follows.
