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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 $n\ge 2$?

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    $\begingroup$ Probably it's well known to the intended audience, but I didn't know: according to Godard (ams.org/journals/proc/2010-138-12/S0002-9939-2010-10421-5/…), Lipschitz-free spaces are also called Arens–Eells spaces, and are discussed in Weaver's "Lipschitz algebras" (ams.org/mathscinet-getitem?mr=1832645). $\endgroup$
    – LSpice
    Commented Mar 19, 2015 at 18:26
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    $\begingroup$ Further to @LSpice's comment: some witterings that may be of interest, if not direct relevance, are given in my MO answer here: mathoverflow.net/questions/186196/… $\endgroup$
    – Yemon Choi
    Commented Mar 19, 2015 at 18:47
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    $\begingroup$ I didn't know this. $\endgroup$
    – Nik Weaver
    Commented Mar 19, 2015 at 19:45
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    $\begingroup$ Is it known what the barycenter map is in this case? (Ie: the linear left inverse of $\delta:\mathbb{R}^d\hookrightarrow \mathcal{F}(\mathbb{R}^n)$? (Since the paper provides an explicit representation of $\delta$). $\endgroup$
    – ABIM
    Commented Jul 26, 2019 at 9:01

3 Answers 3

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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.

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  • $\begingroup$ Is there a known version when the Euclidean metric on $\mathbb{R}^d$ is snowflaked by the map $t \mapsto t^{\alpha}$ for $\alpha \in (0,1)$? $\endgroup$
    – ABIM
    Commented Oct 3, 2019 at 7:32
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    $\begingroup$ @AIM_BLB, as far as I know, there is no such result proved anywhere but please feel free to double-check by asking Marek Cúth directly. $\endgroup$
    – Tomasz Kania
    Commented Oct 3, 2019 at 7:40
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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 $X/N$.

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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 $\frac 1h ( h(x) \mathcal{L} - h(x-h) \mathcal{L}$)

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