1 Introduction
Given a metric space
$M$
, by
${\mathcal{F}}(M)$
we denote the Lipschitz-free space over
$M$
. This is a Banach space whose linear structure somehow reflects the metric structure of
$M$
. The study of Banach space theoretical properties of Lipschitz-free spaces was initiated by a paper by Godefroy and Kalton [Reference Godefroy and Kalton7], where the authors proved, using this notion, e.g. that if a separable Banach space
$Y$
is isometric (not necessarily linearly) to a subset of a Banach space
$X$
, then
$Y$
is already linearly isometric to a subspace of
$X$
. Soon after, the study of Lipschitz-free Banach spaces became an active field of study; see, e.g., [Reference Godefroy, Lancien and Zizler8], [Reference Ostrovskii21, §10] and the results mentioned below. However, the structure of these spaces is still very poorly understood to this day. For example, it is not known whether
${\mathcal{F}}(\mathbb{R}^{2})$
is isomorphic with
${\mathcal{F}}(\mathbb{R}^{3})$
.
Let us recall the construction of
${\mathcal{F}}(M)$
. Choose a distinguished “base point”
$0\in M$
and denote by
$\operatorname{Lip}_{0}(M)$
the space of all real-valued Lipschitz functions on
$M$
that map
$0\in M$
to
$0\in \mathbb{R}$
. It becomes a Banach space if we define the norm of
$f$
to be its minimal Lipschitz constant. For any
$x\in M$
, we denote by
$\unicode[STIX]{x1D6FF}(x)\in \operatorname{Lip}_{0}(M)^{\ast }$
the evaluation functional, i.e.
$\langle \unicode[STIX]{x1D6FF}(x),f\rangle =f(x)$
for
$f\in \operatorname{Lip}_{0}(M)$
. It is easy to see that
$\unicode[STIX]{x1D6FF}$
is an isometric embedding of
$M$
into
$\operatorname{Lip}_{0}(M)^{\ast }$
. The space
${\mathcal{F}}(M)$
is defined to be the closed linear span of
$\{\unicode[STIX]{x1D6FF}(x);\,x\in M\}$
with the dual space norm denoted simply by
$\Vert \cdot \Vert$
. For details and additional properties, see §2 or [Reference Cúth, Doucha and Wojtaszczyk2, §1].
There is a large number of results on the structure of Lipschitz-free spaces (some of them are recalled below), but an explicit isometric representation of
${\mathcal{F}}(M)$
is known only for very special
$M$
. It is known that
${\mathcal{F}}(\mathbb{R})$
is isometric to
$L^{1}(\mathbb{R})$
(cf. [Reference Godefroy and Kalton7, p. 128] or [Reference Godefroy, Lancien and Zizler8, p. 33]) and there is an isometric representation for certain discrete spaces. Our main result is an explicit isometric representation of
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
, where
$\unicode[STIX]{x1D6FA}$
is a non-empty convex domain in a finite-dimensional normed space. It reads as follows.
Theorem 1.1. Let
$E$
be a real normed space of dimension
$d\in \mathbb{N}$
and
$\unicode[STIX]{x1D6FA}\subset E$
be a non-empty convex open subset. Then the Lipschitz-free space
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
is canonically isometric to the quotient space
$$\begin{eqnarray}L^{1}(\unicode[STIX]{x1D6FA},E)/\{\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E);\,\operatorname{div}\boldsymbol{g}=0\text{ in the sense of distributions on }\mathbb{R}^{d}\}.\end{eqnarray}$$
Moreover, if
$\boldsymbol{o}\in \unicode[STIX]{x1D6FA}$
is the base point and
$\boldsymbol{a}\in \unicode[STIX]{x1D6FA}$
is arbitrary, then in this identification we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}(\boldsymbol{a}) & = & \displaystyle [\boldsymbol{g}]\text{ if and only if }\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)\nonumber\\ \displaystyle & & \displaystyle \text{and }\operatorname{div}\boldsymbol{g}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}\text{ in the sense of distributions on }\mathbb{R}^{d}\!,\nonumber\end{eqnarray}$$
where
$\unicode[STIX]{x1D700}_{\boldsymbol{x}}$
denotes the Dirac measure supported at
$\boldsymbol{x}$
.
Let us explain the notation and terminology used in the theorem. First, by
$[\boldsymbol{g}]$
we denote the equivalence class of
$\boldsymbol{g}$
as an element of the quotient space. Further, since
$L^{1}(\unicode[STIX]{x1D6FA},E)$
is canonically isometrically embedded into
$L^{1}(\mathbb{R}^{d},E)$
(just extend any
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
by zero outside
$\unicode[STIX]{x1D6FA}$
), any
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
can be viewed either as a regular distribution on
$\unicode[STIX]{x1D6FA}$
or as a regular distribution on
$\mathbb{R}^{d}$
. Thus,
$\operatorname{div}\boldsymbol{g}$
in the sense of distributions on
$\mathbb{R}^{d}$
is the distributional divergence of the regular distribution on
$\mathbb{R}^{d}$
induced by the described extension of
$\boldsymbol{g}$
. This is quite an important matter, since the result would become false if we considered divergence in the sense of distributions on
$\unicode[STIX]{x1D6FA}$
.
The above theorem is new for
$d\geqslant 2$
, but it covers also the known case
$d=1$
. Indeed, if
$E=\mathbb{R}$
and
$\unicode[STIX]{x1D6FA}=(a,b)$
, where
$-\infty \leqslant a<b\leqslant \infty$
, then
${\mathcal{F}}((a,b))$
is canonically identified with
$L^{1}((a,b))$
and, if
$0\in (a,b)$
, then in this identification we have
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}(x)=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D712}_{(0,x)},\quad & x\in (0,b),\\ 0,\quad & x=0,\\ -\unicode[STIX]{x1D712}_{(-x,0)},\quad & x\in (a,0).\end{array}\right.\end{eqnarray}$$
This case is covered by our theorem, since in dimension one the divergence is just the derivative and the only
$f\in L^{1}(a,b)$
, whose derivative in the sense of distributions on
$\mathbb{R}$
is zero, is the constant zero function. (We point out that we work in distributions on
$\mathbb{R}$
, not in distributions on
$(a,b)$
. If
$(a,b)$
is a bounded interval, then constant functions on
$(a,b)$
belong to
$L^{1}(a,b)$
and their derivative in the sense of distributions on
$(a,b)$
is zero unlike their derivative in the sense of distributions on
$\mathbb{R}$
.) Moreover, it is clear that the distributional derivative of the characteristic function
$\unicode[STIX]{x1D712}_{(u,v)}$
is
$\unicode[STIX]{x1D700}_{u}-\unicode[STIX]{x1D700}_{v}$
.
Our result was motivated in part by a result of Lerner [Reference Lerner15], who proved that
$\operatorname{Lip}_{0}(\mathbb{R}^{d})$
is isomorphic to the dual of the quotient described in our theorem. This is closely related to our result, since
${\mathcal{F}}(M)^{\ast }$
is isometric to
$\operatorname{Lip}_{0}(M)$
for any metric space
$M$
(see the next section). However, it was not known to us whether
${\mathcal{F}}(M)$
is the unique predual of
$\operatorname{Lip}_{0}(M)$
; therefore, to give a representation of
${\mathcal{F}}(\mathbb{R}^{d})$
, we described the mapping
$\unicode[STIX]{x1D6FF}$
. Let us note that it has been proved recently by Weaver [Reference Weaver23] that
${\mathcal{F}}(M)$
actually is the unique predual of
$\operatorname{Lip}_{0}(M)$
if
$M$
is a convex set in a Banach space; however, the description of the mapping
$\unicode[STIX]{x1D6FF}$
could be of an independent interest. We further extended the results for other norms on
$\mathbb{R}^{d}$
(which is easy) and to general convex domains (which requires some additional work).
Let us note that during the review process of this paper we found out that the problem of characterizing
${\mathcal{F}}(\mathbb{R}^{d})$
has been independently investigated in the Master thesis [Reference Flores5], where some partial results were obtained, and that similar ideas as we use in our paper were also used in [Reference Malý17, Appendix: The Sobolev space
$W^{-1,1}$
], where the author describes the Sobolev space
$W^{-1,1}$
.
Let us summarize what is known to the authors about the structure of Lipschitz-free spaces over subsets of
$\mathbb{R}^{d}$
. If
$X$
and
$Y$
are Banach spaces, we write
$X\equiv Y$
,
$X\cong Y$
and
$X{\hookrightarrow}Y$
if
$X$
is linearly isometric with
$Y$
, linearly isomorphic with
$Y$
and isomorphic to a subspace of
$Y$
, respectively. If this is not the case, we write
$X\not \equiv Y$
,
$X\ncong Y$
and
$X\not {\hookrightarrow}Y$
.
Fact 1.2.
-
(1)
${\mathcal{F}}(\mathbb{R})\equiv L^{1}(\mathbb{R})$
. -
(2) For any measure
$\unicode[STIX]{x1D707}$
, we have
${\mathcal{F}}(\mathbb{R}^{2})\not {\hookrightarrow}L^{1}(\unicode[STIX]{x1D707})$
. Moreover, there exists a convergent sequence
$K\subset \mathbb{R}^{2}$
such that
${\mathcal{F}}(K)\not {\hookrightarrow}L^{1}(\mathbb{R})$
. -
(3) If
$M\subset \mathbb{R}^{d}$
has a non-empty interior, then
${\mathcal{F}}(M)\cong {\mathcal{F}}(\mathbb{R}^{d})$
. -
(4) For every
$M\subset \mathbb{R}^{d}$
, the space
${\mathcal{F}}(M)$
is weakly sequentially complete. In particular,
$c_{0}\not {\hookrightarrow}{\mathcal{F}}(M)$
. -
(5)
${\mathcal{F}}(\mathbb{R}^{d})$
has a Schauder basis. Moreover, for every
$M\subset \mathbb{R}^{d}$
bounded and convex,
${\mathcal{F}}(M)$
has the Schauder basis as well. -
(6) For every
$M\subset \mathbb{R}^{d}$
, the space
${\mathcal{F}}(M)$
has the bounded approximation property (BAP). Moreover, if
$M$
is compact and convex or
$M=\mathbb{R}^{d}$
, then
${\mathcal{F}}(M)$
has the metric approximation property (MAP) with respect to any norm on
$\mathbb{R}^{d}$
. -
(7) If
$K\subset \mathbb{R}^{d}$
is a countable compact set, then
${\mathcal{F}}(K)$
is a dual space which has the MAP. On the other hand, if
$M\subset \mathbb{R}^{d}$
is convex and not reduced to a point, then
$L^{1}([0,1]){\hookrightarrow}{\mathcal{F}}(M)$
; in particular,
${\mathcal{F}}(M)$
is not a dual space. -
(8) For a metric space
$M$
, the space
${\mathcal{F}}(M)$
is isometric to a subspace of an
$L^{1}$
space if and only if
$M$
isometrically embeds into an
$\mathbb{R}$
-tree.
The assertion (1) is well known; see, e.g., [Reference Godefroy and Kalton7, p. 128] or [Reference Godefroy, Lancien and Zizler8, p. 33]: the proof is easy using the above description and we recall it at the beginning of the following section. The first result in (2) was shown by Naor and Schechtman [Reference Naor and Schechtman19]. As observed in [Reference Cúth, Doucha and Wojtaszczyk2, Remark 4.2], this result actually follows already from [Reference Kisljakov13] using minor modifications. The “moreover” part of (2) follows from [Reference Cúth, Doucha and Wojtaszczyk2, Theorem 1.2 and Remark 4.5]. The assertion (3) is proved in [Reference Kaufmann12, Corollary 3.5] and the assertion (4) in [Reference Cúth, Doucha and Wojtaszczyk2, Theorem 1.3]. The assertion (5) is a result of Pernecká and Hájek; see [Reference Hájek and Pernecká10, Theorem 3.1 and p. 645]. The first statement of (6) follows from [Reference Lancien and Pernecká14, Proposition 2.3], the case of
${\mathcal{F}}(\mathbb{R}^{d})$
from [Reference Godefroy and Kalton7, Proposition 5.1] and the remaining case is proved in [Reference Pernecká and Smith22, Corollary 1.2]. The first part of (7) is a special case of a result of Dalet [Reference Dalet3, Theorem 2.1]. On the other hand, if
$M$
is a convex set, then
$[0,1]$
bi-Lipschitz embeds into
$M$
; hence,
$L^{1}([0,1])\equiv {\mathcal{F}}([0,1]){\hookrightarrow}{\mathcal{F}}(M)$
. Therefore,
${\mathcal{F}}(M)$
is a separable space which fails the Radon–Nikodým property; hence, it is not a dual space. The assertion (8) is proved in [Reference Godard6, Theorem 4.2]
It seems that the main open problems are whether
${\mathcal{F}}(\ell _{1})$
is complemented in its bidual and whether
${\mathcal{F}}(M)$
has the BAP for every uniformly discrete metric space; see [Reference Godefroy, Lancien and Zizler8, Problems 16 and 18].
2 Preliminaries
In this section we introduce some notation we need to formulate and prove our results. Let us recall some basic facts concerning the Lipschitz-free spaces (for the proofs, we refer to [Reference Cúth, Doucha and Wojtaszczyk2, §1]). Let
$(M,\unicode[STIX]{x1D70C},0)$
be a pointed metric space, i.e. a metric space with a distinguished “base point” denoted by
$0$
. The space
${\mathcal{F}}(M)$
described in the introduction is uniquely characterized by the following universal property.
Let
$X$
be a Banach space and suppose that
$L:M\rightarrow X$
is a Lipschitz mapping satisfying
$L(0)=0$
. Then there exists a unique bounded linear operator
$\widehat{L}:{\mathcal{F}}(M)\rightarrow X$
extending
$L$
, i.e. the following diagram commutes:
Using this universal property of
${\mathcal{F}}(M)$
for
$X=\mathbb{R}$
, it can be rather easily shown that
${\mathcal{F}}(M)^{\ast }$
is linearly isometric to
$\operatorname{Lip}_{0}(M)$
.
The notation and terminology we use are relatively standard. For a Banach space
$X$
,
$x\in X$
and
$x^{\ast }\in X^{\ast }$
, we denote by
$\langle x^{\ast },x\rangle$
or by
$\langle x,x^{\ast }\rangle$
the application of the functional
$x^{\ast }$
on the vector
$x$
; that is,
$\langle x^{\ast },x\rangle =\langle x,x^{\ast }\rangle =x^{\ast }(x)$
.
Our basic setting is the following.
Let
$E$
be a finite-dimensional real normed space with a fixed basis
$\boldsymbol{e}_{1},\ldots ,\boldsymbol{e}_{d}$
, where
$d\geqslant 2$
. Let
$E^{\ast }$
be the dual space to
$E$
and
$\boldsymbol{e}_{1}^{\ast },\ldots \boldsymbol{e}_{d}^{\ast }$
be the dual basis. Using the coordinates with respect to these bases, we can canonically identify both
$E$
and
$E^{\ast }$
with the space
$\mathbb{R}^{d}$
equipped with the corresponding norms. On
$\mathbb{R}^{d}$
we will consider the standard Lebesgue
$d$
-dimensional measure
$\unicode[STIX]{x1D706}^{d}$
and the standard
$(d-1)$
-dimensional Hausdorff measure
${\mathcal{H}}^{d-1}$
. These measures are transferred to
$E$
and
$E^{\ast }$
via the mentioned identification.
If
$U\subset \mathbb{R}^{d}$
(or
$U\subset E$
) is a non-empty open set, by
$\mathscr{D}(U)$
we denote the space of real-valued
${\mathcal{C}}^{\infty }$
-smooth functions with a compact support in
$U$
. The symbol
$\mathscr{D}^{\,\prime }(U)$
denotes the respective space of distributions. Finally, let
$(u_{n})$
be a fixed approximate unit in
$\mathscr{D}(\mathbb{R}^{d})=\mathscr{D}(E)$
, i.e.
$u_{n}(\boldsymbol{x})=n^{d}\unicode[STIX]{x1D70C}(n\boldsymbol{x})$
, where
$\unicode[STIX]{x1D70C}\in \mathscr{D}(\mathbb{R}^{d})$
is a non-negative function with
$\int _{\mathbb{R}^{d}}\unicode[STIX]{x1D70C}=1$
.
We will also use vector-valued functions. Here
$\mathscr{D}(U,\mathbb{R}^{d})$
will denote the space of
${\mathcal{C}}^{\infty }$
-smooth vector fields with a compact support in
$U$
and with values in
$\mathbb{R}^{d}$
, i.e.
$d$
-tuples of elements of
$\mathscr{D}(U)$
. If
$\boldsymbol{g}=(g_{1},\ldots ,g_{n})$
is a
$d$
-tuple of distributions, then
$\operatorname{div}\boldsymbol{g}=\sum _{i=1}^{n}\unicode[STIX]{x2202}_{i}g_{i}$
.
By
$L^{1}(U,E)$
we denote the space of all (equivalence classes of) integrable
$E$
-valued functions defined on
$U$
. The norm on this space is defined by
$$\begin{eqnarray}\Vert \,\boldsymbol{f}\Vert =\int _{U}\Vert \,\boldsymbol{f}(\boldsymbol{x})\Vert _{E}\,d\unicode[STIX]{x1D706}^{d}(\boldsymbol{x}),\quad \boldsymbol{f}\in L^{1}(U,E).\end{eqnarray}$$
Recall that
$L^{1}(U,E)$
is canonically embedded into
$L^{1}(\mathbb{R}^{d},E)$
, where each
$f\in L^{1}(U,E)$
is extended by zero outside
$E$
.
The space
$L^{\infty }(U,E^{\ast })$
is the space of all (equivalence classes of) essentially bounded measurable
$E^{\ast }$
-valued functions defined on
$U$
. The norm is defined by
$$\begin{eqnarray}\Vert \boldsymbol{g}\Vert =\mathop{\text{ess}\,\text{sup}}_{\boldsymbol{x}\in U}\Vert \boldsymbol{g}(\boldsymbol{x})\Vert _{E^{\ast }},\quad \boldsymbol{g}\in L^{\infty }(U,E^{\ast }).\end{eqnarray}$$
The space
$L^{\infty }(U,E^{\ast })$
is canonically isometric to the dual of
$L^{1}(U,E)$
, the duality being defined by
$$\begin{eqnarray}\langle \boldsymbol{g},\boldsymbol{f}\rangle =\int _{U}\langle \boldsymbol{g}(\boldsymbol{x}),\boldsymbol{f}(\boldsymbol{x})\rangle \,d\unicode[STIX]{x1D706}^{d}(\boldsymbol{x}),\quad \boldsymbol{g}\in L^{\infty }(U,E^{\ast }),\;\boldsymbol{f}\in L^{1}(U,E).\end{eqnarray}$$
Finally, let
$\unicode[STIX]{x1D6FA}$
be a fixed non-empty convex open subset of
$E$
with a base point
$\boldsymbol{o}\in \unicode[STIX]{x1D6FA}$
. Without loss of generality, we may and shall suppose that
$\boldsymbol{o}$
is the origin.
3 Proof of the main result
The idea of the proof of our main result, Theorem 1.1, is to mimic the known and easy proof in dimension one. Let us recall it.
It is well known that the mapping
$T:\operatorname{Lip}_{0}(\mathbb{R})\ni f\mapsto f^{\prime }\in L^{\infty }(\mathbb{R})$
is an onto isometry. Consider the adjoint
$T^{\ast }$
. If
$x>0$
, then
$T^{\ast }(\unicode[STIX]{x1D712}_{(0,x)})=\unicode[STIX]{x1D6FF}(x)$
since for any
$F\in \operatorname{Lip}_{0}(\mathbb{R})$
we have
$$\begin{eqnarray}\langle T^{\ast }\unicode[STIX]{x1D712}_{(0,x)},F\rangle =\langle \unicode[STIX]{x1D712}_{(0,x)},TF\rangle =\langle \unicode[STIX]{x1D712}_{(0,x)},F^{\prime }\rangle =\int _{0}^{x}F^{\prime }=F(x)=\langle \unicode[STIX]{x1D6FF}(x),F\rangle .\end{eqnarray}$$
Similarly,
$T^{\ast }(-\unicode[STIX]{x1D712}_{(x,0)})=\unicode[STIX]{x1D6FF}(x)$
for
$x<0$
. Since the set
$\unicode[STIX]{x1D6FF}(\mathbb{R})$
is linearly dense in
${\mathcal{F}}(\mathbb{R})$
and characteristic functions of intervals are linearly dense in
$L^{1}(\mathbb{R})$
, it easily follows that
$T^{\ast }$
isometrically maps
$L^{1}(\mathbb{R})$
onto
${\mathcal{F}}(\mathbb{R})$
.
Hence, we try to mimic this approach in higher dimension and for general convex domains. We consider the mapping
$T:\operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})\ni f\mapsto \unicode[STIX]{x1D6FB}f\in L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })$
. It is rather a standard fact that
$T$
is an into isometry (cf. Proposition 3.2(i)). However, it is not onto; the range is described in Proposition 3.2(ii). It turns out that the range is weak
$^{\ast }$
-closed in
$L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })$
and in Proposition 3.3 we describe its pre-annihilator and, hence, the respective quotient is a predual. Finally, in Proposition 3.5, we show, using the auxiliary Proposition 3.4, that the adjoint map
$T^{\ast }$
maps this predual isometrically onto
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
and give a representation of the mapping
$\unicode[STIX]{x1D6FF}$
. A large part of the first two steps, i.e. Propositions 3.2(ii) and 3.3, for the case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
, is due to [Reference Lerner15].
We start by the following proposition. The parts (i) and (ii) are classical well-known facts; we provide references for the proof. The assertion (iii) is essentially standard as well; we provide the proof for the sake of completeness.
Proposition 3.1. Let
$F:\unicode[STIX]{x1D6FA}\rightarrow \mathbb{R}$
be an
$L$
-Lipschitz function. Then the following hold.
-
(i) For almost all
$\boldsymbol{x}\in \unicode[STIX]{x1D6FA}$
, there exists the Fréchet derivative
$F^{\prime }(\boldsymbol{x})\in E^{\ast }$
. Moreover,
$\Vert F^{\prime }(\boldsymbol{x})\Vert \leqslant L$
whenever
$F^{\prime }(\boldsymbol{x})$
exists. -
(ii) The mapping
$F^{\prime }:\boldsymbol{x}\mapsto F^{\prime }(\boldsymbol{x})$
is an almost everywhere defined measurable function. Moreover, it is the gradient of
$F$
in the sense of distributions on
$\unicode[STIX]{x1D6FA}$
, i.e.
$$\begin{eqnarray}\int _{\unicode[STIX]{x1D6FA}}F^{\prime }\cdot \unicode[STIX]{x1D711}\,d\unicode[STIX]{x1D706}^{d}=-\int _{\unicode[STIX]{x1D6FA}}F\cdot \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}\,d\unicode[STIX]{x1D706}^{d},\quad \unicode[STIX]{x1D711}\in \mathscr{D}(\unicode[STIX]{x1D6FA}).\end{eqnarray}$$
-
(iii) For each
$\boldsymbol{x},\boldsymbol{y}\in \unicode[STIX]{x1D6FA}$
, we have
$$\begin{eqnarray}F(\,\boldsymbol{y})-F(\boldsymbol{x})=\lim _{n\rightarrow \infty }\int _{0}^{1}\langle (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x})),\boldsymbol{y}-\boldsymbol{x}\rangle \,dt.\end{eqnarray}$$
Proof. (i)
$F$
is differentiable almost everywhere by the classical Rademacher theorem (see, e.g., [Reference Federer4, Theorem 3.1.6] or [Reference Lukeš and Malý16, Theorem 30.3]). The estimate of the norm is obvious.
(ii) The set of Fréchet-differentiability points is of full measure, and hence Lebesgue measurable, by (i) (in fact, it is a Borel set by [Reference Lukeš and Malý16, Lemma 30.2]; for a more general result, see [Reference Zajíček24, Theorem 2]). Moreover, since the Fréchet derivative is determined by partial derivatives and partial derivatives of a continuous function are clearly of the first Baire class, the mapping
$F^{\prime }$
is measurable. The integral formula follows easily using Fubini’s theorem and integration by parts for absolutely continuous functions.
(iii) Let us first show the formula in case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
. Then, for each
$n\in \mathbb{N}$
, the convolution
$F\,\ast \,u_{n}$
is a
${\mathcal{C}}^{\infty }$
function on
$\mathbb{R}^{d}$
and its gradient satisfies
$\unicode[STIX]{x1D6FB}(F\,\ast \,u_{n})=F^{\prime }\,\ast \,u_{n}$
. Hence, we have
$$\begin{eqnarray}\displaystyle (F\,\ast \,u_{n})(\,\boldsymbol{y})-(F\,\ast \,u_{n})(\boldsymbol{x}) & = & \displaystyle \int _{0}^{1}\langle (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x})),\boldsymbol{y}-\boldsymbol{x}\rangle \,dt,\nonumber\\ \displaystyle & & \displaystyle \boldsymbol{x},\boldsymbol{y}\in \mathbb{R}^{d},n\in \mathbb{N}.\nonumber\end{eqnarray}$$
Since
$F\,\ast \,u_{n}\rightarrow F$
pointwise (in fact, uniformly on compact sets), we can take the limit to obtain the formula.
Next suppose that
$\unicode[STIX]{x1D6FA}\subsetneq \mathbb{R}^{d}$
. In this case the convolutions including functions defined on
$\unicode[STIX]{x1D6FA}$
are understood, as usually, in the sense that the respective functions are extended by zero outside
$\unicode[STIX]{x1D6FA}$
.
Let
$\widetilde{F}$
be a Lipschitz extension of
$F$
defined on
$\mathbb{R}^{d}$
. Such an extension exists, for example, due to [Reference Lukeš and Malý16, Theorem 30.5]. (In fact, one can preserve the Lipschitz constant, but it is not important at this point.) Hence, using the case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
, the formula holds with
$\widetilde{F}$
in place of
$F$
, i.e.
$$\begin{eqnarray}\widetilde{F}(\,\boldsymbol{y})-\widetilde{F}(\boldsymbol{x})=\lim _{n\rightarrow \infty }\int _{0}^{1}\langle (\widetilde{F}^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x})),\boldsymbol{y}-\boldsymbol{x}\rangle \,dt,\quad \boldsymbol{x},\boldsymbol{y}\in \mathbb{R}^{d}.\end{eqnarray}$$
Given
$\boldsymbol{x},\boldsymbol{y}\in \unicode[STIX]{x1D6FA}$
, the segment
$[\boldsymbol{x},\boldsymbol{y}]$
is a compact subset of
$\unicode[STIX]{x1D6FA}$
and hence we can find
$\unicode[STIX]{x1D700}>0$
with
$\unicode[STIX]{x1D700}<\operatorname{dist}([\boldsymbol{x},\boldsymbol{y}],\mathbb{R}^{d}\setminus \unicode[STIX]{x1D6FA})$
. Further, by the properties of the approximate unit, there is some
$n_{0}\in \mathbb{N}$
such that
$\operatorname{spt}u_{n}\subset U(\boldsymbol{o},\unicode[STIX]{x1D700})$
for
$n\geqslant n_{0}$
. Then, for
$n\geqslant n_{0}$
, one has
$$\begin{eqnarray}(F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x}))=(\widetilde{F}^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x}))\text{ for }t\in [0,1].\end{eqnarray}$$
Hence, the formula follows.◻
Proposition 3.2. For any
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
, set
$T(F)=F^{\prime }$
. Then the following hold.
-
(i)
$T$
is a linear isometry of
$\operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
into
$L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })$
. -
(ii) The range of
$T$
is where the derivatives are considered in the sense of distributions on
$$\begin{eqnarray}X(\unicode[STIX]{x1D6FA})=\{\,\boldsymbol{f}=(f_{i})_{i=1}^{d}\in L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast });\,\unicode[STIX]{x2202}_{i}f_{j}=\unicode[STIX]{x2202}_{j}f_{i}\text{ for }i,j=1,\ldots ,d\},\end{eqnarray}$$
$\unicode[STIX]{x1D6FA}$
.
-
(iii) The inverse operator
$T^{-1}:X(\unicode[STIX]{x1D6FA})\rightarrow \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
is defined by
$$\begin{eqnarray}T^{-1}(\,\boldsymbol{f}\,)(\boldsymbol{x})=\lim _{n\rightarrow \infty }\int _{0}^{1}\langle (\,\boldsymbol{f}\,\ast \,u_{n})(t\boldsymbol{x}),\boldsymbol{x}\rangle \,dt,\quad \boldsymbol{f}\in X(\unicode[STIX]{x1D6FA}),~\boldsymbol{x}\in \unicode[STIX]{x1D6FA}.\end{eqnarray}$$
Proof. It follows from Proposition 3.1 that
$T$
is a linear operator from
$\operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
to
$L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })$
with
$\Vert T\Vert \leqslant 1$
. Further, the formula for the inverse mapping follows from Proposition 3.1(iii). It remains to identify the range and to show that
$T$
is an isometry.
Let us start by proving that
$T$
is an isometry. Fix
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
and set
$$\begin{eqnarray}L=\Vert F^{\prime }\Vert _{L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })}=\mathop{\text{ess}\,\text{sup}}_{x\in \unicode[STIX]{x1D6FA}}\Vert F^{\prime }(\boldsymbol{x})\Vert _{E^{\ast }}.\end{eqnarray}$$
Then
$\Vert F^{\prime }\,\ast \,u_{n}\Vert _{L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })}\leqslant L$
for each
$n\in \mathbb{N}$
. Indeed, fix
$n\in \mathbb{N}$
,
$\boldsymbol{x}\in \unicode[STIX]{x1D6FA}$
and
$\boldsymbol{h}\in E$
with
$\Vert \boldsymbol{h}\Vert \leqslant 1$
. Then
$$\begin{eqnarray}\displaystyle |\langle (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}),\boldsymbol{h}\rangle | & = & \displaystyle \biggl|\mathop{\sum }_{i=1}^{d}(\unicode[STIX]{x2202}_{i}F\,\ast \,u_{n})(\boldsymbol{x})h_{i}\biggr|\nonumber\\ \displaystyle & = & \displaystyle \biggl|\mathop{\sum }_{i=1}^{d}\int _{\mathbb{R}^{d}}\unicode[STIX]{x2202}_{i}F(\,\boldsymbol{y})u_{n}(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}\cdot h_{i}\biggr|\nonumber\\ \displaystyle & = & \displaystyle \biggl|\int _{\mathbb{R}^{d}}\mathop{\sum }_{i=1}^{d}\unicode[STIX]{x2202}_{i}F(\,\boldsymbol{y})\cdot h_{i}\cdot u_{n}(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}\biggr|\nonumber\\ \displaystyle & = & \displaystyle \biggl|\int _{\mathbb{R}^{d}}\langle F^{\prime }(\,\boldsymbol{y}),\boldsymbol{h}\rangle u_{n}(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \int _{\mathbb{R}^{d}}|\langle F^{\prime }(\,\boldsymbol{y}),\boldsymbol{h}\rangle |u_{n}(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \int _{\mathbb{R}^{d}}\Vert F^{\prime }(\,\boldsymbol{y})\Vert _{E^{\ast }}\,u_{n}(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}\leqslant L.\nonumber\end{eqnarray}$$
(During the computation, we again used the convention that the functions defined on
$\unicode[STIX]{x1D6FA}$
are extended by zero on
$\mathbb{R}^{d}\setminus \unicode[STIX]{x1D6FA}$
.) Hence, indeed,
$\Vert (F^{\prime }\ast \,u_{n})(\boldsymbol{x})\Vert \leqslant L$
for each
$n\in \mathbb{N}$
and
$\boldsymbol{x}\in \unicode[STIX]{x1D6FA}$
. Therefore, for each
$n\in \mathbb{N}$
and
$\boldsymbol{x},\boldsymbol{y}\in \unicode[STIX]{x1D6FA}$
, we have
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|\int _{0}^{1}\langle (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x})),\boldsymbol{y}-\boldsymbol{x}\rangle \,dt\biggr|\nonumber\\ \displaystyle & & \displaystyle \qquad \leqslant \int _{0}^{1}|\langle (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x})),\boldsymbol{y}-\boldsymbol{x}\rangle |\,dt\nonumber\\ \displaystyle & & \displaystyle \qquad \leqslant \int _{0}^{1}\Vert (F^{\prime }\,\ast \,u_{n})(\boldsymbol{x}+t(\,\boldsymbol{y}-\boldsymbol{x}))\Vert _{E^{\ast }}\cdot \Vert \hspace{1.0pt}\boldsymbol{y}-\boldsymbol{x}\Vert _{E}\,dt\nonumber\\ \displaystyle & & \displaystyle \qquad \leqslant L\Vert \hspace{1.0pt}\boldsymbol{y}-\boldsymbol{x}\Vert _{E}\nonumber\end{eqnarray}$$
and hence by the formula in Proposition 3.1(iii) the function
$F$
is
$L$
-Lipschitz.
It remains to prove the assertion (ii). On one hand, it is obvious that the range of
$T$
is contained in
$X(\unicode[STIX]{x1D6FA})$
, since, in the sense of distributions on
$\unicode[STIX]{x1D6FA}$
, always
$\unicode[STIX]{x2202}_{i}\unicode[STIX]{x2202}jF=\unicode[STIX]{x2202}_{j}\unicode[STIX]{x2202}_{i}F$
for
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
(in fact, for any distribution
$F\in \mathscr{D}^{\,\prime }(\unicode[STIX]{x1D6FA})$
).
Conversely, suppose that
$\boldsymbol{f}\in X(\unicode[STIX]{x1D6FA})$
. Consider
$\boldsymbol{f}$
to be extended by
$0$
on
$\mathbb{R}^{d}\setminus \unicode[STIX]{x1D6FA}$
. For
$i,j\in \{1,\ldots ,d\}$
, set
$$\begin{eqnarray}U_{i,j}=\unicode[STIX]{x2202}_{j}f_{i}\quad \text{in }\mathscr{D}^{\,\prime }(\unicode[STIX]{x1D6FA})\quad \text{and}\quad V_{i,j}=\unicode[STIX]{x2202}_{j}f_{i}\quad \text{in }\mathscr{D}^{\,\prime }(\mathbb{R}^{d}).\end{eqnarray}$$
Clearly,
$U_{i,j}=V_{i,j}|_{\mathscr{D}(\unicode[STIX]{x1D6FA})}$
and
$U_{i,j}=U_{j,i}$
for
$i,j\in \{1,\ldots ,d\}$
.
Suppose without loss of generality that
$\operatorname{spt}u_{n}\subset U(\boldsymbol{o},1/n)$
for each
$n\in \mathbb{N}$
. For
$n\in \mathbb{N}$
, set
$\unicode[STIX]{x1D6FA}_{n}=\{\boldsymbol{x}\in \unicode[STIX]{x1D6FA};\,\operatorname{dist}(\boldsymbol{x},\mathbb{R}^{d}\setminus \unicode[STIX]{x1D6FA})>1/n\}$
. Then
$\unicode[STIX]{x1D6FA}_{n}$
is a convex open set and, moreover,
$\boldsymbol{o}\in \unicode[STIX]{x1D6FA}_{n}$
for
$n$
large enough. Consider functions
$u_{n}\,\ast \,\boldsymbol{f}$
for
$n\in \mathbb{N}$
. Then
$u_{n}\ast \,\boldsymbol{f}$
is a
${\mathcal{C}}^{\infty }$
mapping defined on
$\mathbb{R}^{d}$
. Moreover,
$$\begin{eqnarray}\unicode[STIX]{x2202}_{j}(u_{n}\,\ast \,f_{i})(\boldsymbol{x})=\unicode[STIX]{x2202}_{i}(u_{n}\,\ast \,f_{j})(\boldsymbol{x}),\quad i,j\in \{1,\ldots ,d\},~\boldsymbol{x}\in \unicode[STIX]{x1D6FA}_{n},~n\in \mathbb{N}.\end{eqnarray}$$
Indeed, fix
$n\in \mathbb{N}$
,
$\boldsymbol{x}\in \unicode[STIX]{x1D6FA}_{n}$
and
$i,j\in \{1,\ldots ,d\}$
. Then
$$\begin{eqnarray}\unicode[STIX]{x2202}_{j}(u_{n}\,\ast \,f_{i})(\boldsymbol{x})=(u_{n}\,\ast \,V_{i,j})(\boldsymbol{x})=V_{i,j}(\,\boldsymbol{y}\mapsto u_{n}(\boldsymbol{x}-\boldsymbol{y}))=U_{i,j}(\,\boldsymbol{y}\mapsto u_{n}(\boldsymbol{x}-\boldsymbol{y})),\end{eqnarray}$$
since the support of
$\boldsymbol{y}\mapsto u_{n}(\boldsymbol{x}-\boldsymbol{y})$
equals
$$\begin{eqnarray}\boldsymbol{x}-\operatorname{spt}u_{n}\subset U\biggl(\boldsymbol{x},\frac{1}{n}\biggr)\subset \unicode[STIX]{x1D6FA}.\end{eqnarray}$$
Thus, we can conclude by recalling that
$U_{i,j}=U_{j,i}$
.
It follows that there is a
${\mathcal{C}}^{\infty }$
-function
$F_{n}:\unicode[STIX]{x1D6FA}_{n}\rightarrow \mathbb{R}$
with
$\unicode[STIX]{x1D6FB}F_{n}=u_{n}\,\ast \,\boldsymbol{f}$
on
$\unicode[STIX]{x1D6FA}_{n}$
. Without loss of generality, we can suppose that
$F_{n}(\boldsymbol{o})=0$
for each
$n\in \mathbb{N}$
. Further, since
$\Vert u_{n}\,\ast \,\boldsymbol{f}\Vert _{L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })}\leqslant \Vert \,\boldsymbol{f}\Vert _{L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })}$
for each
$n\in \mathbb{N}$
, each
$F_{n}$
is
$L$
-Lipschitz, where
$L=\Vert \,\boldsymbol{f}\Vert _{L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })}$
. Since the sequence
$(F_{n})$
is uniformly Lipschitz and
$F_{n}(\boldsymbol{o})=0$
for each
$n\in \mathbb{N}$
, it is easy to check that it is locally uniformly bounded on
$\unicode[STIX]{x1D6FA}$
. Therefore, by the Arzelà–Ascoli theorem, one can find a subsequence which converges to an
$L$
-Lipschitz function
$F$
uniformly on compact subsets of
$\unicode[STIX]{x1D6FA}$
. (Given
$K\subset \unicode[STIX]{x1D6FA}$
compact, then
$K\subset \unicode[STIX]{x1D6FA}_{n}$
for large
$n$
and hence there is
$n_{0}$
such that
$(F_{n})_{n\geqslant n_{0}}$
is uniformly bounded and uniformly Lipschitz on
$K$
, so there is a subsequence uniformly convergent on
$K$
. A diagonal argument completes the construction.) Without loss of generality, suppose that
$F_{n}\rightarrow F$
. Then, for any
$\unicode[STIX]{x1D711}\in \mathscr{D}(\unicode[STIX]{x1D6FA})$
, we have
$$\begin{eqnarray}\int _{\unicode[STIX]{x1D6FA}}F\cdot \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}=\lim _{n\rightarrow \infty }\int _{\unicode[STIX]{x1D6FA}_{n}}F_{n}\cdot \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}=-\lim _{n\rightarrow \infty }\int _{\unicode[STIX]{x1D6FA}_{n}}(u_{n}\,\ast \,\boldsymbol{f})\cdot \unicode[STIX]{x1D711}=-\!\int _{\unicode[STIX]{x1D6FA}}\boldsymbol{f}\cdot \unicode[STIX]{x1D711},\end{eqnarray}$$
where we used that
$\operatorname{spt}\unicode[STIX]{x1D711}\subset \unicode[STIX]{x1D6FA}_{n}$
for
$n$
large enough. Hence,
$F^{\prime }=\boldsymbol{f}$
.◻
Remark.
The assertion (ii) of the previous proposition for the case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
is the content of [Reference Lerner15, Lemma 2]. The proof of the non-trivial inclusion is done in a different way. First, if
$\boldsymbol{f}\in X(\mathbb{R}^{d})$
, by [Reference Horváth11, Proposition 4.3.9], there is a distribution
$U\in \mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
such that
$\boldsymbol{f}$
is the distributional derivative of
$U$
.
This means that
$U$
belongs to the space
$L_{\infty }^{1}(\mathbb{R}^{d})$
defined in [Reference Maz’ya18, §1.1]. It is shown in [Reference Maz’ya18, Theorem 1.1.2] that then
$U\in W_{\text{loc}}^{1,p}(\mathbb{R}^{d})$
for any
$p>1$
. Let
$K=\sqrt{d}\int _{0}^{1}t^{-d/p}\,dt$
. Lemma 4.28 in [Reference Adams and Fournier1] gives
$$\begin{eqnarray}\displaystyle & \displaystyle \text{for all }\boldsymbol{x},\boldsymbol{y}\in \mathbb{R}^{d},p>d,u\in C^{\infty }(\mathbb{R}^{d}): & \displaystyle \nonumber\\ \displaystyle & \displaystyle |u(\boldsymbol{x})-u(\,\boldsymbol{y})|\leqslant 2K\Vert \boldsymbol{x}-\boldsymbol{y}\Vert ^{1-d/p}\Vert \unicode[STIX]{x1D6FB}u\Vert _{L^{p}(Q)}, & \displaystyle\end{eqnarray}$$
where
$Q$
is a cube with side length
$2\Vert \boldsymbol{x}-\boldsymbol{y}\Vert$
that contains
$\boldsymbol{x}$
and
$\boldsymbol{y}$
. From [Reference Adams and Fournier1, Theorem 4.12, Part I], the continuous embedding of
$W_{\text{loc}}^{1,p}(\mathbb{R}^{d})$
into
$C(\mathbb{R}^{d})$
follows. Consequently, (1) holds for all
$u\in W_{\text{loc}}^{1,p}(\mathbb{R}^{d})$
whenever
$p>d$
. Application of (1) to
$U$
together with Hölder’s inequality and the fact that
$\unicode[STIX]{x1D6FB}U\in L^{\infty }(\mathbb{R}^{d})$
gives
$$\begin{eqnarray}\displaystyle & & \displaystyle \hspace{-10.00002pt}\text{for all }\boldsymbol{x},\boldsymbol{y}\in \mathbb{R}^{d},p>d:\nonumber\\ \displaystyle |U(\boldsymbol{x})-U(\,\boldsymbol{y})| & {\leqslant} & \displaystyle 2^{1}K\Vert \boldsymbol{x}-\boldsymbol{y}\Vert ^{1-d/p}\Vert \unicode[STIX]{x1D6FB}U\Vert _{L^{p}(Q)}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2^{1+d/p}K\Vert \boldsymbol{x}-\boldsymbol{y}\Vert \Vert \,\boldsymbol{f}\Vert _{\infty },\nonumber\end{eqnarray}$$
i.e.
$U$
is a Lipschitz function.
A similar argument probably could be done for any convex domain, but it would be more technical and we have not found any explicit reference for it. Therefore, we give above a direct proof of the assertion (ii) for a general convex domain.
We further remark that
$X(\unicode[STIX]{x1D6FA})$
is a known object; the elements of
$X(\unicode[STIX]{x1D6FA})$
are called closed
$L^{\infty }(\unicode[STIX]{x1D6FA})$
currents.
Proposition 3.3. Set
$$\begin{eqnarray}Y(\unicode[STIX]{x1D6FA})=\{\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E);\,\operatorname{div}\boldsymbol{g}=0\text{ in }\mathscr{D}^{\,\prime }(\mathbb{R}^{d})\}.\end{eqnarray}$$
Then the following hold:
-
∙
$X(\unicode[STIX]{x1D6FA})=Y(\unicode[STIX]{x1D6FA})^{\bot }$
and
$Y(\unicode[STIX]{x1D6FA})=(X(\unicode[STIX]{x1D6FA}))_{\bot }$
in the standard duality
$(L^{1}(\unicode[STIX]{x1D6FA},E))^{\ast }=L^{\infty }(\unicode[STIX]{x1D6FA},E^{\ast })$
; -
∙
$Y(\unicode[STIX]{x1D6FA})\cap \mathscr{D}(\unicode[STIX]{x1D6FA},E)$
is dense in
$Y(\unicode[STIX]{x1D6FA})$
.
Proof. Let us recall that any
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
is canonically identified with an element of
$L^{1}(\mathbb{R}^{d},E)$
(
$\boldsymbol{g}$
is extended by zero outside
$\unicode[STIX]{x1D6FA}$
). Hence, if
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
, then
$\operatorname{div}\boldsymbol{g}$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
is the distributional divergence of the described extension of
$\boldsymbol{g}$
.
Let us start the proof by showing that
$X(\unicode[STIX]{x1D6FA})=(Y(\unicode[STIX]{x1D6FA})\cap \mathscr{D}(\unicode[STIX]{x1D6FA},E))^{\bot }$
.
-
$\supset$
: Fix
$\boldsymbol{f}\in (Y(\unicode[STIX]{x1D6FA})\cap \mathscr{D}(\unicode[STIX]{x1D6FA},E))^{\bot }$
. To prove that
$\boldsymbol{f}\in X(\unicode[STIX]{x1D6FA})$
, take arbitrary
$i,j\in \{1,\ldots ,d\}$
distinct and
$\unicode[STIX]{x1D711}\in \mathscr{D}(\unicode[STIX]{x1D6FA})$
. Set
$g_{i}=\unicode[STIX]{x2202}_{j}\unicode[STIX]{x1D711}$
,
$g_{j}=-\unicode[STIX]{x2202}_{i}\unicode[STIX]{x1D711}$
and
$g_{k}=0$
for
$k\in \{1,\ldots ,d\}\setminus \{i,j\}$
. Then
$$\begin{eqnarray}\langle \unicode[STIX]{x2202}_{i}f_{j}-\unicode[STIX]{x2202}_{j}f_{i},\unicode[STIX]{x1D711}\rangle =-\langle f_{j},\unicode[STIX]{x2202}_{i}\unicode[STIX]{x1D711}\rangle +\langle f_{i},\unicode[STIX]{x2202}_{j}\unicode[STIX]{x1D711}\rangle =\langle ~\boldsymbol{f},\boldsymbol{g}\rangle =0.\end{eqnarray}$$
-
$\supset$
: Suppose that
$\boldsymbol{f}\in X(\unicode[STIX]{x1D6FA})$
and take
$\boldsymbol{g}\in Y(\unicode[STIX]{x1D6FA})\cap \mathscr{D}(\unicode[STIX]{x1D6FA},E)$
. We will show that
$\langle ~\boldsymbol{f},\boldsymbol{g}\rangle =0$
. By Proposition 3.2, we know that there is
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
with
$F^{\prime }=\boldsymbol{f}$
. Then
$$\begin{eqnarray}\langle ~\boldsymbol{f},\boldsymbol{g}\rangle =\langle F^{\prime },\boldsymbol{g}\rangle =-\langle F,\operatorname{div}\boldsymbol{g}\rangle =0.\end{eqnarray}$$
Hence, by the Hahn–Banach theorem, we have
$(X(\unicode[STIX]{x1D6FA}))_{\bot }=\overline{Y(\unicode[STIX]{x1D6FA})\cap \mathscr{D}(\unicode[STIX]{x1D6FA},E)}$
. Therefore, to complete the proof it is enough to show that
$Y(\unicode[STIX]{x1D6FA})\subset (X(\unicode[STIX]{x1D6FA}))_{\bot }$
.
Let us prove it first in case
$\unicode[STIX]{x1D6FA}=E$
. Take
$\boldsymbol{f}\in X(\unicode[STIX]{x1D6FA})$
and
$\boldsymbol{g}\in Y(\unicode[STIX]{x1D6FA})$
.
In the first step, suppose that
$\boldsymbol{g}\in {\mathcal{C}}^{\infty }(\mathbb{R}^{d})$
. Fix some
$\unicode[STIX]{x1D713}\in \mathscr{D}(\mathbb{R}^{d})$
with
$\unicode[STIX]{x1D713}=1$
on
$U(\boldsymbol{o},1)$
,
$\operatorname{spt}\unicode[STIX]{x1D713}\subset U(\boldsymbol{o},2)$
and
$0\leqslant \unicode[STIX]{x1D713}\leqslant 1$
on
$\mathbb{R}^{d}$
. (The balls are taken with respect to the norm of
$E$
.) For
$n\in \mathbb{N}$
, set
$\boldsymbol{g}_{n}(\boldsymbol{x})=\unicode[STIX]{x1D713}(\boldsymbol{x}/n)\cdot \boldsymbol{g}(\boldsymbol{x})$
. Let
$F\in \operatorname{Lip}_{0}(\mathbb{R}^{d})$
be such that
$\boldsymbol{f}=F^{\prime }$
. Denote by
$L$
the Lipschitz constant of
$F$
(i.e.
$L=\Vert \,\boldsymbol{f}\Vert _{L^{\infty }(\mathbb{R}^{d},E^{\ast })}$
). Then
$$\begin{eqnarray}\displaystyle |\langle ~\boldsymbol{f},\boldsymbol{g}_{n}\rangle | & = & \displaystyle |\langle F^{\prime },\boldsymbol{g}_{n}\rangle |=|\langle F,\text{div}\,\boldsymbol{g}_{n}\rangle |\nonumber\\ \displaystyle & = & \displaystyle \biggl|\int _{\mathbb{R}^{d}}F(\boldsymbol{x})\cdot \biggl(\unicode[STIX]{x1D713}\biggl(\frac{\boldsymbol{x}}{n}\biggr)\text{div}\,\boldsymbol{g}(\boldsymbol{x})+\frac{1}{n}\langle \boldsymbol{g}(\boldsymbol{x}),\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}\biggl(\frac{\boldsymbol{x}}{n}\biggr)\rangle \biggr)\,d\boldsymbol{x}\biggr|\nonumber\\ \displaystyle & = & \displaystyle \biggl|\int _{\mathbb{R}^{d}}F(\boldsymbol{x})\cdot \frac{1}{n}\langle \boldsymbol{g}(\boldsymbol{x}),\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}\biggl(\frac{\boldsymbol{x}}{n}\biggr)\rangle \,d\boldsymbol{x}\biggr|\nonumber\\ \displaystyle & = & \displaystyle \biggl|\int _{U(\boldsymbol{o},2n)\setminus U(\boldsymbol{o},n)}F(\boldsymbol{x})\cdot \frac{1}{n}\langle \boldsymbol{g}(\boldsymbol{x}),\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}\biggl(\frac{\boldsymbol{x}}{n}\biggr)\rangle \,d\boldsymbol{x}\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{1}{n}\cdot \sup _{\boldsymbol{x}\in U(\boldsymbol{o},2n)}\cdot \,|F(\boldsymbol{x})|\cdot \Vert \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}\Vert _{L^{\infty }(\mathbb{R}^{d},E^{\ast })}\cdot \int _{\mathbb{R}^{d}\setminus U(\boldsymbol{o},n)}\Vert \boldsymbol{g}(\boldsymbol{x})\Vert _{E}\,dx\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2L\cdot \Vert \unicode[STIX]{x1D6FB}\unicode[STIX]{x1D713}\Vert _{L^{\infty }(\mathbb{R}^{d},E^{\ast })}\cdot \int _{\mathbb{R}^{d}\setminus U(\boldsymbol{o},n)}\Vert \boldsymbol{g}(\boldsymbol{x})\Vert _{E}\,dx\rightarrow 0\nonumber\end{eqnarray}$$
for
$n\rightarrow \infty$
. Since
$\boldsymbol{g}_{n}\rightarrow \boldsymbol{g}$
in
$L^{1}(\mathbb{R}^{d},E)$
, we conclude that
$\langle ~\boldsymbol{f},\boldsymbol{g}\rangle =0$
.
In the second step, let
$\boldsymbol{g}\in Y(\unicode[STIX]{x1D6FA})$
be arbitrary. Then, for each
$n\in \mathbb{N}$
, we have
$u_{n}\,\ast \,\boldsymbol{g}\in {\mathcal{C}}^{\infty }(\mathbb{R}^{d},E)$
,
$\operatorname{div}(u_{n}\,\ast \,\boldsymbol{g})=u_{n}\,\ast \,\operatorname{div}\boldsymbol{g}=0$
and hence
$\langle ~\boldsymbol{f},u_{n}\,\ast \,\boldsymbol{g}\rangle =0$
. Since
$u_{n}\,\ast \,\boldsymbol{g}\rightarrow \boldsymbol{g}$
in
$L^{1}(\mathbb{R}^{d},E)$
, necessarily
$\langle ~\boldsymbol{f},\boldsymbol{g}\rangle =0$
.
Finally, let
$\unicode[STIX]{x1D6FA}$
be arbitrary,
$\boldsymbol{f}\in X(\unicode[STIX]{x1D6FA})$
and
$\boldsymbol{g}\in Y(\unicode[STIX]{x1D6FA})$
. Let
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
be such that
$\boldsymbol{f}=F^{\prime }$
. Let
$\widetilde{F}\in \operatorname{Lip}_{0}(E)$
be an extension of
$F$
. Then, using the case
$\unicode[STIX]{x1D6FA}=E$
and the assumption
$\operatorname{div}\boldsymbol{g}=0$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
(and not just in
$\mathscr{D}^{\,\prime }(\unicode[STIX]{x1D6FA})$
), we have
$$\begin{eqnarray}\displaystyle \hspace{50.00008pt}\langle ~\boldsymbol{f},\boldsymbol{g}\rangle & = & \displaystyle \int _{\unicode[STIX]{x1D6FA}}\langle F^{\prime }(\boldsymbol{x}),\boldsymbol{g}(\boldsymbol{x})\rangle \,d\boldsymbol{x}=\int _{\unicode[STIX]{x1D6FA}}\langle \widetilde{F}^{\prime }(\boldsymbol{x}),\boldsymbol{g}(\boldsymbol{x})\rangle \,d\boldsymbol{x}\nonumber\\ \displaystyle & = & \displaystyle \int _{\mathbb{R}^{d}}\langle \widetilde{F}^{\prime }(\boldsymbol{x}),\boldsymbol{g}(\boldsymbol{x})\rangle \,d\boldsymbol{x}=\langle \widetilde{F}^{\prime },\boldsymbol{g}\rangle =0.\hspace{70.0001pt}\square\nonumber\end{eqnarray}$$
Remark.
(1) By Proposition 3.2,
$\operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
is isometric with
$X(\unicode[STIX]{x1D6FA})$
; by Proposition 3.3,
$X(\unicode[STIX]{x1D6FA})$
is isometric with
$(L^{1}(\unicode[STIX]{x1D6FA},E)/_{Y(\unicode[STIX]{x1D6FA})})^{\ast }$
. Hence,
$\operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
is isometric with
$(L^{1}(\unicode[STIX]{x1D6FA},E)/_{Y(\unicode[STIX]{x1D6FA})})^{\ast }$
. For the case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
, this is the content of [Reference Lerner15, Theorem 3].
(2) In [Reference Lerner15], the proof of the case
$\unicode[STIX]{x1D6FA}=\mathbb{R}^{d}$
of Proposition 3.3 is contained in Lemmas 4 and 5, although the equality
$X(\mathbb{R}^{d})=Y(\mathbb{R}^{d})^{\bot }$
is not explicitly mentioned there.
Proposition 3.4. Let
$\boldsymbol{a}\in \unicode[STIX]{x1D6FA}$
be fixed. Then there is
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},\mathbb{R}^{d})$
with compact support in
$\unicode[STIX]{x1D6FA}$
such that
$\operatorname{div}\boldsymbol{g}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
.
Proof. In this proof we will consider
$\mathbb{R}^{d}$
with the Euclidean norm. Set
$$\begin{eqnarray}\boldsymbol{h}(\boldsymbol{x})=\frac{\boldsymbol{x}}{d\unicode[STIX]{x1D705}_{d}\Vert \boldsymbol{x}\Vert ^{d}},\quad x\in \mathbb{R}^{d}\setminus \{\boldsymbol{o}\},\end{eqnarray}$$
where
$\unicode[STIX]{x1D705}_{d}$
is the volume of the
$d$
-dimensional ball. Then
$\boldsymbol{h}$
is the gradient of the standard fundamental solution of the Laplace equation in
$\mathbb{R}^{d}$
. In particular,
$$\begin{eqnarray}\displaystyle & \displaystyle \text{div}\,\boldsymbol{h}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}\quad \text{ in }\mathscr{D}^{\,\prime }(\mathbb{R}^{d}), & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{h}\in {\mathcal{C}}^{\infty }(\mathbb{R}^{d}\setminus \{\boldsymbol{o}\}), & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \text{div}\,\boldsymbol{h}(\boldsymbol{x})=0\quad \text{for }\boldsymbol{x}\in \mathbb{R}^{d}\setminus \{\boldsymbol{o}\}. & \displaystyle\end{eqnarray}$$
Further, for any
$r>0$
, we have
$$\begin{eqnarray}\int _{\unicode[STIX]{x2202}U(\boldsymbol{o},r)}\langle \boldsymbol{h}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})=1,\end{eqnarray}$$
where
$\unicode[STIX]{x1D708}(\boldsymbol{x})$
denotes the outer normal at the point
$\boldsymbol{x}$
and
$U(\boldsymbol{o},r)$
is the Euclidean ball centered at
$\boldsymbol{o}$
with radius
$r$
. Indeed, since
$\unicode[STIX]{x1D708}(\boldsymbol{x})=\boldsymbol{x}/\Vert \boldsymbol{x}\Vert$
, one has
$$\begin{eqnarray}\displaystyle \int _{\unicode[STIX]{x2202}U(\boldsymbol{o},r)}\langle \boldsymbol{h}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x}) & = & \displaystyle \int _{\unicode[STIX]{x2202}U(\boldsymbol{o},r)}\frac{1}{d\unicode[STIX]{x1D705}_{d}\Vert \boldsymbol{x}\Vert ^{d-1}}\,d{\mathcal{H}}^{n-1}(\boldsymbol{x})\nonumber\\ \displaystyle & = & \displaystyle \frac{{\mathcal{H}}^{n-1}(\unicode[STIX]{x2202}U(\boldsymbol{o},r))}{d\unicode[STIX]{x1D705}_{d}r^{d-1}}=\frac{{\mathcal{H}}^{n-1}(\unicode[STIX]{x2202}U(\boldsymbol{o},1))}{d\unicode[STIX]{x1D705}_{d}}=1.\nonumber\end{eqnarray}$$
The equation (5) can be extended to more general domains, as follows.
If
$U$
is a bounded convex domain containing
$\boldsymbol{o}$
, then
$$\begin{eqnarray}\int _{\unicode[STIX]{x2202}U}\langle \boldsymbol{h}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})=1.\end{eqnarray}$$
Indeed, let
$r>0$
be so small that
$\overline{U(\boldsymbol{o},r)}\subset U$
. Then, by (4) and the Gauss theorem [Reference Lukeš and Malý16, Theorem 37.22] (note that the boundary of a convex domain is Lipschitz), we have
$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \int _{U\setminus \overline{U(\boldsymbol{o},r)}}\operatorname{div}\boldsymbol{h}\nonumber\\ \displaystyle & = & \displaystyle \int _{\unicode[STIX]{x2202}U}\langle \boldsymbol{h}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})-\int _{\unicode[STIX]{x2202}U(\boldsymbol{o},r)}\langle \boldsymbol{h}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x});\nonumber\end{eqnarray}$$
We continue by setting
$$\begin{eqnarray}\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x})=\boldsymbol{h}(\boldsymbol{x})-\boldsymbol{h}(\boldsymbol{x}-\boldsymbol{a}),\quad \boldsymbol{x}\in \mathbb{R}^{d}\setminus \{\boldsymbol{a},\boldsymbol{o}\}.\end{eqnarray}$$
Then clearly
$$\begin{eqnarray}\displaystyle & \displaystyle \operatorname{div}\boldsymbol{h}_{\boldsymbol{a}}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}\quad \text{in }\mathscr{D}^{\,\prime }(\mathbb{R}^{d}), & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{h}_{\boldsymbol{a}}\in {\mathcal{C}}^{\infty }(\mathbb{R}^{d}\setminus \{\boldsymbol{o},\boldsymbol{a}\}), & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \operatorname{div}\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x})=0\quad \text{for }\boldsymbol{x}\in \mathbb{R}^{d}\setminus \{\boldsymbol{o},\boldsymbol{a}\} & \displaystyle\end{eqnarray}$$
and, moreover, by (6), we get the following result.
If
$U$
is a bounded convex domain containing
$\{\boldsymbol{o},\boldsymbol{a}\}$
, then
$$\begin{eqnarray}\int _{\unicode[STIX]{x2202}U}\langle \boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})=0.\end{eqnarray}$$
Since the segment
$[\boldsymbol{o},\boldsymbol{a}]$
is a compact subset of
$\unicode[STIX]{x1D6FA}$
, there is
$r>0$
such that
$$\begin{eqnarray}\overline{[\boldsymbol{o},\boldsymbol{a}]+U(\boldsymbol{o},r)}\subset \unicode[STIX]{x1D6FA}.\end{eqnarray}$$
Choose some
$\unicode[STIX]{x1D702}\in \mathscr{D}(\mathbb{R}^{d})$
such that
$$\begin{eqnarray}\unicode[STIX]{x1D702}=1\text{ on }[\boldsymbol{o},\boldsymbol{a}]+U\biggl(0,\frac{r}{2}\biggr)\text{ and }\operatorname{spt}\unicode[STIX]{x1D702}\subset [\boldsymbol{o},\boldsymbol{a}]+U\biggl(0,\frac{3}{4}r\biggr).\end{eqnarray}$$
Then we have:
-
∙
$\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}}\in L^{1}(\unicode[STIX]{x1D6FA},\mathbb{R}^{d})$
and has a compact support in
$\unicode[STIX]{x1D6FA}$
; -
∙
$\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}}\in {\mathcal{C}}^{\infty }(\mathbb{R}^{d}\setminus \{\boldsymbol{o},\boldsymbol{a}\})$
; -
∙
$\operatorname{div}\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x})=0$
for
$\boldsymbol{x}\in ([\boldsymbol{o},\boldsymbol{a}]+U(0,r/2))\setminus \{\boldsymbol{o},\boldsymbol{a}\}$
; -
∙
$\operatorname{div}\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x})=0$
for
$\boldsymbol{x}\in \mathbb{R}^{d}\setminus ([\boldsymbol{o},\boldsymbol{a}]+U(0,{\textstyle \frac{3}{4}}r))$
.
Hence, if we set
$U_{1}=[\boldsymbol{o},\boldsymbol{a}]+U(0,\frac{1}{4}r)$
and
$U_{2}=[\boldsymbol{o},\boldsymbol{a}]+U(0,r)$
, we get
$$\begin{eqnarray}\displaystyle \int _{U_{2}\setminus \overline{U_{1}}}\operatorname{div}(\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}})\,d\unicode[STIX]{x1D706}^{d} & = & \displaystyle \int _{\unicode[STIX]{x2202}U_{2}}\langle \unicode[STIX]{x1D702}(\boldsymbol{x})\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})\nonumber\\ \displaystyle & & \displaystyle -\int _{\unicode[STIX]{x2202}U_{1}}\langle \unicode[STIX]{x1D702}(\boldsymbol{x})\boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})\nonumber\\ \displaystyle & = & \displaystyle 0-\int _{\unicode[STIX]{x2202}U_{1}}\langle \boldsymbol{h}_{\boldsymbol{a}}(\boldsymbol{x}),\unicode[STIX]{x1D708}(\boldsymbol{x})\rangle \,d{\mathcal{H}}^{n-1}(\boldsymbol{x})=0.\nonumber\end{eqnarray}$$
Since
$\operatorname{div}(\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}})$
restricted to
$U_{2}\setminus \overline{U_{1}}$
is a
${\mathcal{C}}^{\infty }$
-function with a compact support, by [Reference Novotný and Straškraba20, Auxiliary lemma 3.15], there is
$\boldsymbol{g}_{0}\in \mathscr{D}(U_{2}\setminus \overline{U_{1}},\mathbb{R}^{d})$
with
$\operatorname{div}\boldsymbol{g}_{0}(x)=\operatorname{div}(\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}})(x)$
for
$x\in U_{2}\setminus \overline{U_{1}}$
. Set
$\boldsymbol{g}=\unicode[STIX]{x1D702}\boldsymbol{h}_{\boldsymbol{a}}-\boldsymbol{g}_{0}$
. Then
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},\mathbb{R}^{d})$
with compact support in
$\unicode[STIX]{x1D6FA}$
. Moreover, we will show that
$\operatorname{div}\boldsymbol{g}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}$
in
$\mathscr{D}(\mathbb{R}^{d})$
. Due to (7), it is enough to show that
$\operatorname{div}(\boldsymbol{g}-\boldsymbol{h}_{\boldsymbol{a}})=0$
in
$\mathscr{D}(\mathbb{R}^{d})$
. But
$$\begin{eqnarray}\boldsymbol{g}-\boldsymbol{h}_{\boldsymbol{a}}=(\unicode[STIX]{x1D702}-1)\boldsymbol{h}_{\boldsymbol{a}}-\boldsymbol{g}_{0}\end{eqnarray}$$
is a
${\mathcal{C}}^{\infty }$
vector field on
$\mathbb{R}^{d}$
(note that
$\unicode[STIX]{x1D702}-1=0$
on
$[\boldsymbol{o},\boldsymbol{a}]+U(\boldsymbol{o},r/2)$
) and hence its distributional divergence is a
${\mathcal{C}}^{\infty }$
function and can be computed pointwise. Let us consider the following three open sets covering
$\mathbb{R}^{d}$
.
(i) On
$\overline{U_{1}}$
, we have
$\unicode[STIX]{x1D702}=1$
and
$\boldsymbol{g}_{0}=0$
; hence,
$(\unicode[STIX]{x1D702}-1)\boldsymbol{h}_{\boldsymbol{a}}-\boldsymbol{g}_{0}=0$
. Thus,
$\operatorname{div}((\unicode[STIX]{x1D702}-1)\boldsymbol{h}_{\boldsymbol{a}}-\boldsymbol{g}_{0})(\boldsymbol{x})=0$
for each
$\boldsymbol{x}\in [\boldsymbol{o},\boldsymbol{a}]+U(\boldsymbol{o},r/2)$
.
(ii) On
$\mathbb{R}^{d}\setminus U_{2}$
, we have
$\unicode[STIX]{x1D702}=0$
and
$\boldsymbol{g}_{0}=0$
; hence,
$(\unicode[STIX]{x1D702}-1)\boldsymbol{h}_{\boldsymbol{a}}-\boldsymbol{g}_{0}=-\boldsymbol{h}_{\boldsymbol{a}}$
and therefore the divergence is zero at each point of this set by (9).
(iii) For
$x\in U_{2}\setminus \overline{U_{1}}$
, we have
$$\begin{eqnarray}\operatorname{div}((\unicode[STIX]{x1D702}-1)h_{\boldsymbol{a}}-\boldsymbol{g}_{0})(\boldsymbol{x})=\operatorname{div}(\unicode[STIX]{x1D702}\boldsymbol{h}_{a})(\boldsymbol{x})-\operatorname{div}h_{\boldsymbol{a}}(\boldsymbol{x})-\operatorname{div}\boldsymbol{g}_{0}(\boldsymbol{x})=0\end{eqnarray}$$
by (9) and the choice of
$\boldsymbol{g}_{0}$
.
This completes the proof.☐
The final ingredient is the following proposition; the proof of Theorem 1.1 then immediately follows.
Proposition 3.5. Let
$T:F\mapsto F^{\prime }$
be the isometry from Proposition 3.2. Then the dual mapping
$T^{\ast }$
maps
$(L^{1}(\unicode[STIX]{x1D6FA},E)/Y(\unicode[STIX]{x1D6FA}))$
onto
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
. Moreover, for
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
, one has
$T^{\ast }([\boldsymbol{g}])=\unicode[STIX]{x1D6FF}(\boldsymbol{a})$
if and only if
$\operatorname{div}\boldsymbol{g}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
.
Proof. Fix
$\boldsymbol{a}\in \unicode[STIX]{x1D6FA}$
. Let
$\boldsymbol{g}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
be a mapping with compact support and satisfying
$\operatorname{div}\boldsymbol{g}=\unicode[STIX]{x1D700}_{\boldsymbol{o}}-\unicode[STIX]{x1D700}_{\boldsymbol{a}}$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
. (Such a mapping exists by Proposition 3.4.) For any
$F\in \operatorname{Lip}_{0}(\unicode[STIX]{x1D6FA})$
, we then have
The first three equalities are just applications of definitions. The fourth one follows from the fact that
$u_{n}\,\ast \,\boldsymbol{g}\rightarrow \boldsymbol{g}$
in the
$L^{1}$
-norm. Since
$\boldsymbol{g}$
has compact support in
$\unicode[STIX]{x1D6FA}$
,
$u_{n}\,\ast \,\boldsymbol{g}\in \mathscr{D}(\unicode[STIX]{x1D6FA},\mathbb{R}^{d})$
for large
$n$
and hence the fifth one is just rewriting the expression in the sense of distributions and the sixth one follows from the differentiation rules for distributions. Then a standard calculation follows. Let us point out that by 
we mean the function defined by a formula 
, and that we use the obvious fact that the sequence 
is also an approximate unit.
Hence, we conclude that
$T^{\ast }([\boldsymbol{g}])=\unicode[STIX]{x1D6FF}(\boldsymbol{a})$
. Now let
$\boldsymbol{g}_{1}\in L^{1}(\unicode[STIX]{x1D6FA},E)$
be arbitrary. Since
$T^{\ast }$
is an isometry,
$T^{\ast }([\boldsymbol{g}_{1}])=\unicode[STIX]{x1D6FF}(\boldsymbol{a})$
if and only if
$[\boldsymbol{g}_{1}]=[\boldsymbol{g}]$
, i.e.
$\operatorname{div}(\boldsymbol{g}_{1}-\boldsymbol{g})=0$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
or, equivalently,
$\operatorname{div}\boldsymbol{g}_{1}=\unicode[STIX]{x1D6FF}_{\boldsymbol{o}}-\unicode[STIX]{x1D6FF}_{\boldsymbol{a}}$
in
$\mathscr{D}^{\,\prime }(\mathbb{R}^{d})$
.
To conclude the proof, observe that
$(T^{\ast })^{-1}(\unicode[STIX]{x1D6FF}(\boldsymbol{a}))\in (L^{1}(\unicode[STIX]{x1D6FA},E)/Y(\unicode[STIX]{x1D6FA}))$
for any
$\boldsymbol{a}\in \unicode[STIX]{x1D6FA}$
and hence
$(T^{\ast })^{-1}$
maps
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
into
$(L^{1}(\unicode[STIX]{x1D6FA},E)/Y(\unicode[STIX]{x1D6FA}))$
. Let
$S$
be the restriction of
$(T^{\ast })^{-1}$
to
${\mathcal{F}}(\unicode[STIX]{x1D6FA})$
. Since
$S^{\ast }=T^{-1}$
is an onto isometry, necessarily
$S$
is also an onto isometry. This completes the proof.☐
Acknowledgement
We are grateful to Professor N. Lerner for sending us his preprint [Reference Lerner15].
Note added in proof
During the review process, the preprint [Reference Lerner15] has been extended and placed online, see [Reference Godefroy and Lerner9].
