Dual space of the completion of the space of Lipschitz functions This question is a continuation of this post : Metrization of a topological vector space 
Let $C_{lip}(\mathbb R^d)$ be the space of Lipschitz functions on $\mathbb R^d$. We endow $C_{lip}(\mathbb R^d)$ with the following topology: $(f_n)_{n\ge 1} \subset C_{lip}(\mathbb R^d)$ converges to $f\in C_{lip}(\mathbb R^d)$ iff
$$\lim_{n\to\infty} \left\{\left|\int_{\mathbb R^d}(f_n-f)(x)u(x)dx\right| + \left|\int_{\mathbb R^d}\nabla(f_n-f)(x)\cdot w(x)dx\right|\right\} = 0,$$
for all $u:\mathbb R^d\to\mathbb R$ and $w:\mathbb R^d\to\mathbb R^d$ satisfying 
$$\int_{\mathbb R^d}|u(x)|(1+|x|)dx<\infty \quad\mbox{and}\quad \int_{\mathbb R^d}|w(x)|dx<\infty.$$
Consider the completion $\overline{C}_{lip}(\mathbb R^d)$ of $C_{lip}(\mathbb R^d)$ w.r.t. this topology. Could we show that any linear continuous function $T: \overline{C}_{lip}(\mathbb R^d)\to\mathbb R$ must be of the form
$$T(f)=\int_{\mathbb R^d}f(x)u(x)dx+\int_{\mathbb R^d}\nabla f(x)w(x)dx?$$
 A: The space of Lipschitz functions $\text{Lip}(\mathbb{R}^d)$ embeds via $f\mapsto (\frac{f}{1+|x|},\partial_1 f,\dots,\partial_d f)$ into the space  $ L^\infty(\mathbb{R}^d)^{d+1}$, which we isometrically identify with the dual space of $ L^1(\mathbb{R}^d)^{d+1}$. 
The image of this embedding of $\text{Lip}(\mathbb{R}^d)$ is a weakly* closed subspace $E$ of $ L^\infty(\mathbb{R}^d)^{d+1}$, in fact, presented by weak equations
$$E=\{ (g ,w_1,\dots,w_d)\in L^\infty(\mathbb{R}^d)^{d+1}: \partial_i\big((1+|x|)g \big)= w_i, \;  \partial_iw_j=\partial_jw_i,\;   \text{for all } i,j  \}$$
as the annihilator $E=F^\perp$ of the  $\|\cdot\|_1$-closed  linear span $$F:=\overline{\text{span}}\big\{(1+|x|)\phi e_0+\partial_i\phi e_i,\;\partial_i\phi e_j-\partial_j\phi e_i:\;\phi\in C^\infty_0(\mathbb{R}^d) ,\, 1\le i\le j\le d\big\}.$$
(For our needs it is not necessary to characterize better the space $F$: it sufficient to know that $E$ is weakly* closed, thus $E=F^\perp$ for $F:=E_\perp\subset L^1(\mathbb{R}^d)^{d+1}$).
Therefore $E$ is a dual space, as shown by the isometry $$E= F^\perp\sim \bigg({L^1(\mathbb{R}^d)^{d+1}\over F  }\bigg)^*$$
and induces its weak* topology $\tau^*$ on $\text{Lip}(\mathbb{R}^d)$, and the convergence structure of it. In particular, the elements $(g,w)$ of $L^1(\mathbb{R}^d)^{d+1}$ represents all $w^*$-continuous functionals on  $\text{Lip}(\mathbb{R}^d)$ via 
$$T_{(g,w)}:\text{Lip}(\mathbb{R}^d)\ni f \mapsto \int_{\mathbb{R}^d}g(x){f(x)\over 1+|x|}dx+\int_{\mathbb{R}^d}w\cdot \nabla f(x)dx.$$
So the convergence you are considering is exactly the convergence associated with this weak* topology, which is not metrizable, of course; but as a general fact, a linear functional on a dual of a separable Banach $X$ is $w^*$-continuous iff it is $w^*$-sequentially continuous (iff is in the image of the embedding $E\to E^{**}$), so at the end the answer is, yes.
A: This is my understanding of the answer given by Pietro Majer. The reasoning could be divided into three steps:


*

*The dual space of $L^1(\mathbb R^d)^{d+1}$ is identified as $L^{\infty}(\mathbb R^d)^{d+1}$, i.e. $\big(L^1(\mathbb R^d)^{d+1}\big)^*=L^{\infty}(\mathbb R^d)^{d+1}$;

*The embedding $E\subset L^{\infty}(\mathbb R^d)^{d+1}$ is weakly* closed. What does "weakly* closed" refer to? Here the definitions of $F^\perp$ and $E_\perp$ are not clear to me either;

*The elements $(g,w)$ of $L^1(\mathbb R^d)^{d+1}$ represent all $w^*$-continuous functionals on ${\rm Lip}(\mathbb R^d)$ via $T_{(g,w)}$. Why this is true? We only know $L^{\infty}(\mathbb R^d)^{d+1}$ is the dual space of $L^1(\mathbb R^d)^{d+1}$, but do not know the dual space of $L^{\infty}(\mathbb R^d)^{d+1}$.
Finally, if I only consider the space of linear continuous functionals for the space ${\rm Lip}_0(\mathbb R^d)$, could we simplify the above proof?  
