# 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?

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*}
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.