Vector measures as metric currents

Question

Currents in metric spaces were introduced by Ambrosio and Kirchheim in 2000 as a generalization of currents in euclidean spaces. Very roughly, a principle idea is to replace smooth test functions (and smooth differential forms) by Lipschitz test functions (and forms). Rademacher's theorem about a.e. differentiability of Lipschitz functions usually gives the link between both concepts.

There is a quite spectacular theorem of S. Smirnov from 1994 about the decomposition of vector measures $\mu=(\mu_1,\ldots,\mu_n)$ with compact support whose distributional divergence div$(\mu)=\sum_{j=1}^n \partial_j\mu_j$ is again a measure. Such charges decompose as $\mu=\int_C [\theta] d\nu(\theta)$ for a positive measure $\nu$ on a set $C$ of Lipschitz curves with charges $[\theta]$ associated to curves (whose components are $\theta_\#(\dot\theta_j\mathscr L_{[0,1]})$) such that also the total variation measure $\|\mu\|$ decomposes as $\|\mu\|=\int_C \| [\theta]\| d\nu(\theta)$.

This theorem was transfered to normal metric $1$-currents by Paolini and Stepanov in 2012 and 2013 and their formulations strongly suggests that their theorems contain those of Smirnov. One should thus interpret a charge as a metric $1$-current. The only way I can think of is to define for a $1$-form $fd\pi$ (which just means that $f$ and $\pi$ are real-valued Lipschitz functions on $\mathbb R^n$ and $f$ is also bounded) $$ T_\mu(fd\pi)=\sum_{j=1}^n \int f \partial_j \pi\ d\mu_j.$$ If $\pi$ is continuously differentiable everything is fine and quite simple calculations show that the measure representing div$(\mu)$ corresponds to the boundary $\partial T_\mu$.

However, for Lipschitz functions $\pi$, the derivatives of $\pi$ only exist $\mathscr L^n$-almost everywhere so that the definition of $T_\mu$ above only makes sense for charges $\mu$ which are continuous with respect to the Lebesgue measure $\mathscr L^n$. Adding this hypothesis to Smirnov's theorem is a rather severe restriction of generality.

Question. Is there another way to associate to every charge $\mu$ a current $T_\mu$ so that Smirnov's theorems directly follow from the analogous results of Paolini and Stepanov?

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Edit.

In the article Divergence measure vector fields M. Silhavy claims that for every charge $\mu$ with div$(\mu)$ being a measure and every bounded $f\in W^{1,\infty}(\mathbb R^n)$ (in particular, for every bounded Lipschitz functions) the charge $f\cdot\mu$ has the same property that $\lambda=$div$(f\cdot\mu)$ is a measure. A presentation of this result including proofs is in H. Frid's Remarks on the theory of divergence measure fields.

As in mlk's answer one can then define $$T_\mu(fd\pi)=-\int \pi d\lambda.$$ However, Silhavy's result uses quite heavy mashinery so that the reduction of Smirnov's theorems to those of Paolini and Stepanov is quite indirect (to say the least). The question whether there is a direct way from Paolini and Stepanov to Smirnov thus remains.

Answer 1

To me your definition seems to be the right one, you just need to prove that it is well defined when approximating Lipschitz with $C^1$-functions. For that you probably need the distributional divergence. Specifically, if you set $f\equiv 1$, then $$ \sum_j \int \partial_j \pi d\mu_j = \int \pi d \nu$$ for $\nu = -\operatorname{Div} \mu$.

So if $(\pi_k)_k \subset C^1 $ with $\pi_k \to \pi$ uniformly and $Lip(\pi_k) <C$, then $$T_\mu(1 d\pi_k - 1 d\pi_l ) =\sum_j \int \partial_j (\pi_k-\pi_l) d\mu_j = \int \pi_k-\pi_l d \nu$$ which is a Cauchy-sequence. From we get that $T_\mu(1 d\pi)$ is well defined as $\lim_{k\to \infty} T_\mu(1d\pi_k)$.

Now there is probably some clever way to involve $f$ again, but in the worst case, the same trick should work when replacing $f$ with characteristic functions of nice sets, which then in turn can be used to approximate general $f$.

Answer 2

There is a rather simple argument from the article of Frid mentioned in the (edited) question based on smoothing by convolution with standard mollifiers $\chi_\varepsilon(x)=\varepsilon^{-n}\chi(x/\varepsilon)$ (the Rademacher theorem was thus a red herring): For $\pi_\varepsilon=\pi\ast \chi_\varepsilon$ the product rule gives $$ \mathrm{div}(\pi_\varepsilon\mu)=\pi_\varepsilon \mathrm{div}(\mu)+\langle\nabla\pi_\varepsilon,\mu\rangle$$ and we want to show that $\nu_\varepsilon=\langle\nabla\pi_\varepsilon,\mu\rangle= \sum\limits_{j=1}^n \partial_j\pi_\varepsilon \mu_j$ converges for $\varepsilon\to 0$ in the weak$^*$-topology of signed measures (i.e., the topology of pointwise convergence on all continuous functions vanishing at infinity). Since $\pi$ is Lipschitz, the densities $\nabla\pi_\varepsilon$ are uniformly bounded so that $\nu_\varepsilon$ is a uniformly bounded family of measures and hence relatively weak$^*$-compact by Alaoglu's theorem. On the other hand, $\mathrm{div}(\pi_\varepsilon\mu)$ converges in the much coarser weak$^*$ topology of $\mathscr D'(\mathbb R^n)$ since taking partial derivatives is continuous on $\mathscr D'(\mathbb R^n)$, and $\pi_\varepsilon \mathrm{div}(\mu)$ converges to the measure $\pi\mathrm{div}(\mu)$. Therefore, $\nu_\varepsilon=\mathrm{div}(\pi_\varepsilon\mu)-\pi_\varepsilon\mathrm{div}(\mu)$ converges in $\mathscr D'(\mathbb R^n)$ so that all accumulation points in the weak$^*$ topology of measures coincide. This proves the weak$^*$-convergence of $\nu_\varepsilon$ to a measure $\nu$ which can finally be used to define $$ T_\mu(fd\pi) =\int f d\langle\nabla\pi,\mu\rangle \text{ as } \int fd\nu.$$