Lipschitz-free spaces of $\mathbb R^n$

Question

We define $$ \text{Lip}_0(\mathbb R^n)=\{f:\mathbb R^n\rightarrow \mathbb R, \text{such that $f(0)=0$ and } \sup_{x\not=y}\frac{\vert f(x)-f(y)\vert}{\vert x-y\vert}<+\infty. \} $$ It is well-known that $\text{Lip}_0(\mathbb R^n)$ is a Banach space, which is the dual space of the so-called $\mathcal F(\mathbb R^n)$, a.k.a. the Lipschitz-free space of $\mathbb R^n$.

Claim: $\text{Lip}_0(\mathbb R^n)$ is the dual space of $X/N$ where $X$ is the space of $L^{1}({\mathbb R}^{n})$ vector fields and $N$ is the subspace of vector fields with null divergence. In other words, with $$ X=(L^{1}({\mathbb R}^{n}))^{n},\quad N=\{(f_{j})_{1\le j\le n}\in X, \ \sum_{1\le j\le n}\frac{\partial f_{j}}{\partial x_{j}}=0\}, $$ we have $ \text{Lip}_0(\mathbb R^n)=(X/N)^{*}. $ Note that in the easy case $n=1$, we find the familiar $\mathcal F(\mathbb R)=L^1(\mathbb R)$. The derivatives above are taken in the distribution sense.

Questions. (1) Is the statement of this claim well-known? (2) Could it be useful to describe more explicitly the properties of $\mathcal F(\mathbb R^n)$ when $n\ge 2$?

Answer 1

Yes, this is essentially contained in the paper

M. Cúth, O. F. K. Kalenda, P. Kaplický: Isometric representation of Lipschitz-free spaces over convex domains in finite-dimensional spaces, Mathematika, 63 (2) (2017), 538–552.

where it is proved that for any non-void, open subset $\Omega$ of $\mathbb R^n$, the space ${\rm Lip}_0(\Omega)$ is isometric to the dual of the quotient of $L_1(\Omega, \mathbb R^n)$ by the vector fields with zero divergence in the distributional sense.

Answer 2

By Helmholtz decomposition, a smooth vector field $f(x)$ can be decomposed into a curl-free component and a divergence-free component, $$f = -\nabla \phi + \nabla \times A .$$ Thus $X/N$ can be viewed as the completion of the set $\{\nabla \phi |~ \phi ~\text{is a compact supported smooth function}\}$ under the $L^1$ norm. And it is easy to check using integral by parts, for any $f \in Lip_0$, $f$ is a bounded linear functional on $X/N$.

Answer 3

I think there are much more elements in $\mathcal F(\mathbb R)$: for example we have $\mathcal{M}(\mathbb R)$ the space of finite measures is in $\mathcal F(\mathbb R)$, but also for every fixed $h \in L^p ([0,1], \mathcal{L})$ with $p>1$ you have that the functional

$$ f \mapsto \int_0^1 f'(x) h(x) \, dx $$

belongs to $\mathcal F(\mathbb R)$ (it can be seen as the strong limit of measures $\frac 1h ( h(x) \mathcal{L} - h(x-h) \mathcal{L}$)