# Lie brackets of nonsmooth vector fields and commutation of their flows

Chiara Rigoni  Chiara Rigoni, Institut für Angewandte Mathematik, Universität
Bonn, Endenicher Allee 60
D-53115, Bonn<EMAIL_ADDRESS>,  Eugene Stepanov  Eugene Stepanov,
St.Petersburg Branch of the Steklov Mathematical Institute of the Russian
Academy of Sciences, Fontanka 27, 191023 St.Petersburg, Russia and HSE
University, Moscow and ITMO University<EMAIL_ADDRESS>and  Dario
Trevisan  Dario Trevisan, Dipartimento di Matematica, Università di Pisa
Largo Bruno Pontecorvo 5
I-56127, Pisa<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

It is well-known that the flows generated by two smooth vector fields commute,
if the Lie bracket of these vector fields vanishes. This assertion is known to
extend to Lipschitz continuous vector fields, up to interpreting the vanishing
of their Lie bracket in the sense of almost everywhere equality. We show that
this cannot be extended to general a.e. differentiable vector fields admitting
a.e. unique flows. We show however that the extension holds when one field is
Lipschitz continuous and the other one is merely Sobolev regular (but
admitting a regular Lagrangian flow).

The first author gratefully acknowledges support by the European Union through
the ERC-AdG 694405 RicciBounds. The work of the second author has been also
partially financed by the Program of the Presidium of the Russian Academy of
Sciences #01 ’Fundamental Mathematics and its Applications’ under grant
PRAS-18-01 and RFBR grant #20-01-00630A. The third author was partially
supported by GNAMPA-INdAM 2019 project ”Proprietà analitiche e geometriche di
campi aleatori” and University of Pisa, Project PRA 2018-49.

##  1\. Introduction

One of the well-known basic facts of differential geometry, ultimately leading
to the Frobenius theorem on integral manifolds, is that the flows of two
smooth vector fields $V^{1}$, $V^{2}$ over a smooth finite dimensional
manifold $M$ commute, if and only if their Lie bracket $[V^{1},V^{2}]$
vanishes. In a more formal way, denoting the flow of $V^{i}$ by
$F_{t}^{i}(x):=y(t)$, where $y(\cdot)$ is the integral curve of the
differential equation

(1.1) $\dot{y}=V^{i}(y),\quad y(0)=x,$

one has

(1.2) $F_{t}^{1}(F_{s}^{2}(x))=F_{s}^{2}(F_{t}^{1}(x))\quad\mbox{for all $x\in
M$, whenever }[V^{1},V^{2}]=0,$

once $t\in\mathbb{R}$ and $s\in\mathbb{R}$ are such that the respective
expressions are defined. What happens with this statement for possibly
nonsmooth vector fields $V^{i}$, even when the underlying manifold $M$ itself
remains smooth (in the sequel we will for simplicity consider just the case of
$M=\mathbb{R}^{d}$, a finite-dimensional Euclidean space)?

Of course, to be meaningful, the question posed has to be restricted to the
cases when all the objects present in (1.2) are well-defined. This is however
nontrivial already in the case when $V^{i}$ are Lipschitz. Namely, in this
case the flows of $V^{i}$ are still defined because the Cauchy problem (1.1)
has a unique solution for every initial datum $x\in\mathbb{R}^{d}$. However,
there is a problem with the meaning of the Lie bracket $[V^{1},V^{2}]$; in
fact, the latter is defined for smooth vector fields by the formula

(1.3) $[V^{1},V^{2}](x):=DV^{2}(x)V^{1}(x)-DV^{1}(x)V^{2}(x),$

where $DV^{i}$ are Jacobi matrices of $V^{i}$, $i=1,2$, which is meaningless
when $V^{i}$ are only Lipschitz, because then their derivatives are not
necessarily defined for all $x\in\mathbb{R}^{d}$ but only for almost all with
respect to the Lebesgue measure. The natural question would be then: does
(1.2) hold for a.e. - rather than for all, $x\in\mathbb{R}^{d}$ \- once the
vanishing of the commutator $[V^{1},V^{2}]=0$ is also understood in the sense
of a.e. equality? The answer is positive as shown in the seminal paper by F.
Rampazzo and H.J. Sussmann [9].

The problem is therefore what happens when one descends in regularity of
$V^{i}$ further beyond Lipschitz continuity. In fact, there are many cases
when Cauchy problems (1.1) admit unique solutions for every or just for a.e.
intial datum. For instance, as shown in [5, corollary 5.2], for a vector field
$V\in W^{1,p}(\mathbb{R}^{d};\mathbb{R}^{d})$ with $p>d$ and
$\operatorname{div}V$ bounded, a unique flow $x\mapsto F_{t}(x)$ is defined
for a.e. $x\in\mathbb{R}^{d}$ and all $t\in\mathbb{R}$ as $F_{t}(x):=y(t)$
where $y(\cdot)$ is the unique solution to the ODE

$\dot{y}=V(y)$

satisfying $y(0)=x$. Further, even if $p\leq d$, then a solution to the latter
ODE may be not unique for a.e. initial datum, but in this case there is a
natural selection of such solutions called regular Lagrangian flow [2].
Moreover, there are many other cases besides Sobolev regularity when the
vector field $V$ admits a regular Lagrangian flow. In view of the Rampazzo-
Sussmann result for Lipschitz vector fields, it is tempting to conjecture that
also in the cases in which the fields $V^{i}\in
W^{1,p_{i}}(\mathbb{R}^{d};\mathbb{R}^{d})$ generating unique flows for a.e.
initial datum (or at least regular Lagrangian flows) $F^{i}$, it is true that

(1.4) $F_{t}^{1}(F_{s}^{2}(x))=F_{s}^{2}(F_{t}^{1}(x))\quad\mbox{for a.e.
$x\in\mathbb{R}^{d}$, whenever }[V^{1},V^{2}]=0\quad\mbox{a.e.},$

of course when the Lie bracket $[V^{1},V^{2}](x)$ is well defined for a.e.
$x\in\mathbb{R}^{d}$. We will show however by means of a counterexample that
in general this is false. Nevertheless we are able to show (Theorem 4.1 and
Theorem 4.2) that this conjecture is true, if one of the vector fields is
Lipschitz (which generalizes the result of [9]).

The main difficulty here is that when the vector fields $V^{i}$ are just
Sobolev, the respective Lagrangian flows have only very weak regularity
properties: in particular, they are just summable but not Sobolev (not even of
fractional order) as shown in [1, 4], so that their weak derivatives do not
have any pointwise (even a.e.) meaning. This is in sharp contrast with the
situation when $V^{i}$ are Lipschitz, since in the latter case it is a
textbook result that the respective flows are Lipschitz too. That is why the
technique we adopt in this paper is essentially different from that used in
[9]. In fact, in the latter paper a set-valued Lie bracket of Lipschitz vector
fields is introduced and then it is shown, substantially, that the usual
expansions (of course adapted now to the set-valued setting) of compositions
of flow maps in each point are still valid for this case with the set-valued
Lie bracket instead of the classical one. As a side remark, we mention that
another quite natural extension of both the classical Lie bracket and its set-
valued version, the measure valued Lie bracket, has been introduced recently
in [6]. In the general case of Sobolev $V^{i}$, in view of the mentioned lack
of regularity of flows, no technique based on pointwise expansions would work
and therefore we are forced to develop a completely different PDE/measure
theory-style approach.

##  2\. Notation and preliminaries

###  2.1. General notation

The finite-dimensional space $\mathbb{R}^{d}$ is assumed to be equipped with
the Euclidean norm $|\cdot|$; the notation $B_{r}(x)\subset\mathbb{R}^{d}$
stands for the usual open Euclidean ball of radius $r$ centered at $x$. In
general, the norm in the normed space $E$ will be denoted by $\|\cdot\|_{E}$.

All the measures over a metric space considered in the sequel are positive
Radon measures, not necessarily finite. The notation $\mathcal{L}^{d}$ stands
for the Lebesgue measure in $\mathbb{R}^{d}$. If $\mu$ is a measure over a
metric space $X$, then for a Borel map $T\colon X\to Y$ between metric spaces
$X$ and $Y$ we denote by $T_{\\#}\mu$ the push-forward of $\mu$, i.e. the
measure over $Y$ defined by $(T_{\\#}\mu)(B):=\mu(T^{-1}(B))$ for every Borel
$B\subset Y$.

We use the notation $A\lesssim B$ for possibly vector valued functions $A$,
$B$ defined on a subset $\Omega\subseteq\mathbb{R}^{d}$, when there exists a
constant $C\geq 0$ such that

$|A|(x)\leq C|B|(x)\quad\text{for every $x\in\Omega$}$

with some constant $C>0$ depending possibly on parameters such as dimension of
the space or integrability exponents, but not on $A$ and $B$.

###  2.2. Spaces

For a measure $\mu$ in a metric space $E$ we denote by
$L^{p}(E,\mu;\mathbb{R}^{m})$ the usual Lebesgue space of $\mu$-integrable
with exponent $p\geq 1$ functions $f\colon E\to\mathbb{R}^{m}$
($\mu$-essentially bounded when $p=+\infty$); the reference to $\mu$ will be
omitted in the case $\mu=\mathcal{L}^{d}$. The reference to $E$ may be also
omitted when no confusion is possible. Analogously,
$W^{k,p}(\mathbb{R}^{d};\mathbb{R}^{m})$ (resp.
$W^{k,p}_{\operatorname{loc}}(\mathbb{R}^{d};\mathbb{R}^{m})$) will stand for
the usual Sobolev (resp. locally Sobolev) class of functions over
$\mathbb{R}^{d}$ with values in $\mathbb{R}^{m}$. In all the cases the
reference to $\mathbb{R}^{m}$ will be omitted when $m=1$, i.e. for real valued
functions. The norm in $L^{p}(\mathbb{R}^{d})$ is denoted for brevity just
$\|\cdot\|_{p}$.

The space $C([a,b];\mathbb{R}^{d})$ of continuous curves in $\mathbb{R}^{d}$
parameterized over the interval $[a,b]$ is endowed with the usual supremum
norm. For every $t\in[a,b]$ we let $e_{t}\colon
C([a,b];\mathbb{R}^{d})\to\mathbb{R}^{d}$ stand for the evaluation map
$e_{t}(\theta):=\theta(t)$.

###  2.3. Variation

Given a function

$\omega\colon[-T,T]^{2}\to\mathbb{R},\quad(s,t)\mapsto\omega_{st},$

we write

$[\omega]:=\limsup_{|\pi|\to 0}\sum_{t_{i}\in\pi}|\omega_{t_{i}t_{i+1}}|.$

This generalizes the notion of total variation (which corresponds to the case
$\omega_{st}:=f_{t}-f_{s}$).

###  2.4. Maximal functions

We recall the usual definitions of the sharp maximal function $g^{\sharp}_{r}$
for an $r>0$ and the maximal function $g^{*}$ of a locally integrable function
$g\colon\mathbb{R}^{d}\to\mathbb{R}$, namely,

$\displaystyle g^{\sharp}_{r}(x)$
$\displaystyle:=\sup_{s\in[0,r]}\frac{1}{\mathcal{L}^{d}(B_{s}(x))}\int_{B_{s}(x)}|g(y)-g(x)|\mathsf{d}y,$
$\displaystyle g^{*}(x)$ $\displaystyle:=\sup_{s\geq
0}\frac{1}{\mathcal{L}^{d}(B_{s}(x))}\int_{B_{s}(x)}|g(y)|\mathsf{d}y,$

so that in particular

(2.1) $g^{\sharp}_{r}\leq 2g^{*}.$

######  Remark 2.1 (convergence of maximal functions).

We notice that, if $g\in L^{p}(\mathbb{R}^{d})$ with $p>1$, then

(2.2) $g^{\sharp}_{r},g^{*}\in L^{p}(\mathbb{R}^{d}),$

and $g^{\sharp}_{r}(x)\searrow 0$ as $r\searrow 0$ for a.e.
$x\in\mathbb{R}^{d}$ (this can be proven by density of continuous functions).
In particular,

$\left\|g^{\sharp}_{r}\right\|_{p}\searrow 0\quad\mbox{as $r\searrow 0$}.$

###  2.5. Pointwise inequalities for Sobolev functions

Recall that for Sobolev functions $f\in
W^{1,1}_{\operatorname{loc}}(\mathbb{R}^{d})$, one has the following pointwise
inequality [3, corollary 1], valid for every $x$, $y\in\mathbb{R}^{d}\setminus
N$, where $N$ is Lebesgue-negligible:

(2.3)
$|f(y)-f(x)-Df(x)(y-x)|\lesssim|y-x|\left(|Df|^{\sharp}_{|y-x|}(y)+|Df|^{\sharp}_{|y-x|}(x)\right).$

From (2.1) we obtain

(2.4) $|f(y)-f(x)|\lesssim|y-x|\left(|Df|^{*}(y)+|Df|^{*}(x)\right).$

If $f\in W^{2,1}_{\operatorname{loc}}(\mathbb{R}^{d})$, then [3, corollary 1]
gives also

(2.5)
$|f(y)-f(x)-Df(x)(y-x)|\lesssim|y-x|^{2}\left(|D^{2}f|^{*}(y)+|D^{2}f|^{*}(x)\right).$

###  2.6. Regular Lagrangian flows

######  Definition 2.2.

We say that a Borel map $F\colon[a,b]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ is
a regular Lagrangian flow for the (possibly time-dependent) vector field
$V\colon[a,b]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ if

  * (i)

for a.e. $x\in\mathbb{R}^{d}$ the map $t\in[a,b]\mapsto F(t,x)$ is an
absolutely continuous solution of the ODE

$\dot{y}=V(t,y)$

for $t\in[a,b]$, satisfying $y(a)=x$;

  * (ii)

there is a constant $C>0$ independent of $t$ (called the compressibility
constant of $F$) such that

$F(t,\cdot)_{\\#}\mathcal{L}^{d}\leq C\mathcal{L}^{d}$

for every $t\in[a,b]$.

For flows we usually write $F_{t}(x)$ instead of $F(t,x)$. We will further use
a couple of obsersvations.

######  Remark 2.3.

When $\nu$ is a measure over $\mathbb{R}^{d}$ with bounded density with
respect to the Lebesgue measure, i.e.

$\nu=f\mathcal{L}^{d}\quad\mbox{with $f\in L^{\infty}(\mathbb{R}^{d})$},$

and $F\colon[a,b]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ is a regular
Lagrangian flow with compressibility constant $C$ for some vector field, then
one has

(2.6) $F(t,\cdot)_{\\#}\nu\leq C\|f\|_{\infty}\mathcal{L}^{d},$

because $F(t,\cdot)_{\\#}\nu=(f\circ
F(t,\cdot)^{-1})F(t,\cdot)_{\\#}\mathcal{L}^{d}$.

######  Lemma 2.4.

If $F\colon[a,b]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ is a regular Lagrangian
flow and $N\subset\mathbb{R}^{d}$ is Lebesgue negligible, then there is a
$B_{0}\subset\mathbb{R}^{d}$ with $\mathcal{L}^{d}(B_{0})=0$ such that
$F(t,x)\not\in N$ for every $x\not\in B_{0}$ and a.e. $t\in[a,b]$.

###### Proof.

For the set $G:=\\{(x,t)\in[a,b]\times\mathbb{R}^{d}\colon F(t,x)\in N\\}$ one
has $\mathcal{L}^{d+1}(G)=0$ by Fubini theorem since for each fixed
$t\in[a,b]$ one has

$\displaystyle\mathcal{L}^{d}(G\cap(\\{t\\}\times\mathbb{R}^{d}))=(F(t,\cdot)_{\\#}\mathcal{L}^{d})(N)\leq
C\mathcal{L}^{d}(N)=0.$

Therefore the set

$B_{0}:=\\{x\in\mathbb{R}^{d}\colon\mathcal{L}^{1}(G\cap([a,b]\times\\{x\\}))>0\\}\subset\mathbb{R}^{d}$

is Lebesgue negligible as claimed. ∎

##  3\. Vanishing Lie bracket does not imply commutativity of flows

In this section we show by means of the example below that when two vector
fields have a.e. vanishing Lie bracket, it is in general not true that their
flows commute a.e. even if we assume that they are a.e. uniquely defined, and
even if they are regular Lagrangian. The main idea comes from an example [10,
Nelson’s example] of non-commutativity of groups with generators commuting on
a common dense set. In our case, we consider a foliation of $\mathbb{R}^{d}$
into helix-like hypersurfaces so that the composition of the two flows yields
a rotation along the axis of the helix.

######  Example 3.1.

Let $d:=3$,

$f(x,y):=\arctan\left(\frac{y}{x}\right)$

and consider the vector fields

$\displaystyle V^{1}$
$\displaystyle:=\partial_{x}+(\partial_{x}f)\partial_{z}=\partial_{x}-\frac{y}{x^{2}+y^{2}}\partial_{z},$
$\displaystyle V^{2}$
$\displaystyle:=\partial_{y}+(\partial_{y}f)\partial_{z}=\partial_{y}+\frac{x}{x^{2}+y^{2}}\partial_{z},$

so that

(3.1)
$[V^{1},V^{2}]=\left(\partial_{xy}^{2}f-\partial_{yx}^{2}f\right)\partial_{z}=0,$

i.e. their Lie bracket vanishes (of course the function $f$, the vector fields
and their Lie bracket are defined everywhere in $\mathbb{R}^{d}$ outside of
the two-dimensional plane $\\{x=0\\}$, hence a.e. in $\mathbb{R}^{d}$). We now
verify the sequence of claims.

Claim 1. Both $V^{1}$ and $V^{2}$ are tangent to the level sets of the
function $(x,y,z)\in\mathbb{R}^{d}\mapsto z-f(x,y)\in\mathbb{R}$, i.e. to the
graphs of the functions $z=f(x,y)+C$ in $\mathbb{R}^{d}$, with $C$ any
constant, as observed by a direct calculation.

Claim 2. The flows $F^{i}$ for the vector fields $V^{i}$, $i=1,2$, are
uniquely defined over $[-T,T]$, for every $T>0$ and for every initial datum
outside of the plane $\\{x=0\\}$. In fact, this is quite immediate for
$F^{2}$, since the solution to the ODE $\dot{\theta}=V^{2}(\theta)$ is
uniquely defined for every initial datum
$\theta(0)\in\mathbb{R}^{d}\setminus\\{x=0\\}$ over all times.

As for $F^{1}$, we consider the solutions to the ODE
$\dot{\theta}=V^{1}(\theta)$ for every initial datum outside of the plane
$\\{x=0\\}$. If $\bar{x}:=\theta_{1}(0)<0$ then the solution $\theta(\cdot)$
to the respective Cauchy problem is defined uniquely for all
$t\in(-\infty,-\bar{x})$, and, analogously, if $\bar{x}:=\theta_{1}(0)>0$ then
it is defined uniquely for all $t\in(-\bar{x},+\infty)$. In the first case,
one has

$\lim_{t\to-\bar{x}-0}f(\theta_{1}(t),\theta_{2}(t))=\left\\{\begin{array}[]{rl}\pi/2,&\mbox{if
$\bar{y}<0$},\\\ 0,&\mbox{if $\bar{y}=0$},\\\ -\pi/2,&\mbox{if
$\bar{y}>0$},\end{array}\right.$

because $\theta_{2}(t)=\theta_{2}(0)$ for all $t\in(-\infty,-\bar{x})$ and

$\lim_{t\to-\bar{x}-0}\theta_{1}(t)=0^{-}.$

Thus, denoting $\bar{z}:=\theta_{3}(0)$ and recalling Claim 1, we get that

$\displaystyle\lim_{t\to-\bar{x}-0}\theta_{3}(t)$
$\displaystyle=\lim_{t\to-\bar{x}-0}f(\theta_{1}(t),\theta_{2}(t))+\left(\bar{z}-f(\bar{x},\bar{y})\right)$
$\displaystyle=\bar{z}-f(\bar{x},\bar{y})-\frac{\pi}{2}\mathrm{sign}\,\bar{y}.$

Symmetrically in the second case, $\theta_{2}(t)=\theta_{2}(0)$ for all
$t\in(-\bar{x},+\infty)$,

$\lim_{t\to-\bar{x}+0}\theta_{1}(t)=0^{+}$

and

$\displaystyle\lim_{t\to-\bar{x}+0}\theta_{3}(t)$
$\displaystyle=\bar{z}-f(\bar{x},\bar{y})-\frac{\pi}{2}\mathrm{sign}\,\bar{y}.$

Summing up, we have that absolutely continuous solutions $\theta(\cdot)$ to
the respective Cauchy problem are defined uniquely for every initial datum
outside of the plane $\\{x=0\\}$ and for a.e. $t\in\mathbb{R}$, or, more
precisely, for every $t\in\mathbb{R}$ except $t=-\bar{x}$, with

(3.2)
$\lim_{t\to-\bar{x}}\theta(t)=\left(0,\bar{y},\bar{z}-f(\bar{x},\bar{y})-\frac{\pi}{2}\mathrm{sign}\,\bar{y}\right).$

This defines uniquely the flow $F^{1}$.

Claim 3. One has $\operatorname{div}V^{i}=0$, and hence the flows $F^{i}$ are
regular Lagrangian (because $F^{i}_{t\\#}\mathcal{L}^{d}=\mathcal{L}^{d}$ for
all $t$ for which $F^{i}_{t}$ is defined, i.e. the compressibility constant is
one), for $i=1,2$.

Claim 4. Finally, the flows $F^{1}$ and $F^{2}$ do not commute despite (3.1).
In fact, consider an arbitrary $p=(\bar{x},\bar{y},\bar{z})\in\mathbb{R}^{d}$
with $x<0$ and $y<0$. Then for $\theta^{1}(t):=F^{1}_{t}(p)$ one has

(3.3) $\theta^{1}(t)=\left(\bar{x}+t,\bar{y},\theta^{1}_{3}(t)\right)$

with $\theta^{1}_{3}(-\bar{x})=\bar{z}-f(\bar{x},\bar{y})+\pi/2$ in view of
(3.2). By Claim 1 one has that $\theta^{1}(t)$ for $t>-\bar{x}$ belongs to the
graph of function $z=f(x,y)+C$, which means

$\theta^{1}_{3}(t)=f(\bar{x}+t,\bar{y})+C$

for all $t>-\bar{x}$. Letting $t\to-\bar{x}+0$, we get

$\bar{z}-f(\bar{x},\bar{y})+\pi/2=\theta^{1}_{3}(-\bar{x})=-\pi/2+C,$

which gives $C=\bar{z}-f(\bar{x},\bar{y})+\pi$. Therefore,
$F^{2}_{s}(F^{1}_{t}(p))$ belongs to the graph of function

$z=f(x,y)+\bar{z}-f(\bar{x},\bar{y})+\pi$

for every $s\in\mathbb{R}$ and $t>-\bar{x}$.

On the other hand, for $\theta^{2}(s):=F^{2}_{s}(p)$ one has

$\theta^{2}_{2}(s)=\left(\bar{x},\bar{y}+s,,\theta^{2}_{3}(s)\right),$

so that for every $s>-\bar{y}$ one has $\theta^{2}_{2}(s)>0$. Then for
$\sigma^{1}(t):=F^{1}_{t}(\theta^{2}(s))$ for $s>-\bar{y}$ one has

(3.4) $\sigma^{1}(t)=\left(\bar{x}+t,\bar{y}+s,\sigma^{1}_{3}(t)\right).$

Since $\theta^{2}(s)$ belongs to the graph of the function

$z=f(x,y)+\bar{z}-f(\bar{x},\bar{y})$

by Claim 1, then so is $\sigma^{1}(t)$ for $t<-\bar{x}$. By (3.2) one has

(3.5)
$\sigma^{1}(-\bar{x})=\left(\bar{x},\bar{y}+s,\bar{z}-f(\bar{x},\bar{y})-\frac{\pi}{2}\right).$

Hence for $t>-\bar{x}$ (and $s>-\bar{y})$ one has that
$F^{1}_{t}(F^{2}_{s}(p))=\sigma^{1}(t)$ belongs to the graph of function
$z=f(x,y)+C$, that is,

$\sigma^{1}_{3}(t)=f(\bar{x}+t,\bar{y})+C$

for all $t>-\bar{x}$, and the value of $C$ can be found by letting
$t\to-\bar{x}+0$, since then by (3.5) we have

$\bar{z}-f(\bar{x},\bar{y})-\pi/2=\sigma^{1}_{3}(-\bar{x})=\pi/2+C,$

which gives $C=\bar{z}-f(\bar{x},\bar{y})-\pi$. Therefore,
$F^{1}_{t}(F^{2}_{s}(p))$ belongs to the graph of function

$z=f(x,y)+\bar{z}-f(\bar{x},\bar{y})-\pi$

for every $s>-\bar{y}$ and $t>-\bar{x}$. In other words for all such pairs
$(s,t)$ one has

$F^{1}_{t}(F^{2}_{s}(p))\neq F^{2}_{s}(F^{1}_{t}(p))$

as claimed.

######  Remark 3.2.

One can easily check that both $V^{1}$, $V^{2}$ are not Sobolev (even locally)
on $\mathbb{R}^{3}$, since for $f(x,y):=\arctan(y/x)$ we have $f\notin
W^{2,1}_{\operatorname{loc}}(\mathbb{R}^{2})$. We may in fact prove more, that
is, any two two vector fields on $\mathbb{R}^{3}$ of the form

$V^{1}=\partial_{x}+\partial_{x}f\partial_{z},\quad
V^{2}=\partial_{y}+\partial_{y}f\partial_{z},$

with $f=f(x,y)\in W^{2,1}_{loc}(\mathbb{R}^{2})$ must have commuting flows
$F^{1}$, $F^{2}$. Indeed, they are explicitly given by

$\displaystyle F^{1}_{s}(x,y,z)$
$\displaystyle:=\left(x+s,y,z+\int_{0}^{s}\partial_{x}f(x+r,y)dr\right),$
$\displaystyle F^{2}_{t}(x,y,z)$
$\displaystyle:=\left(x,y+t,z+\int_{0}^{t}\partial_{y}f(x,y+r)dr\right).$

If we introduce the differential $1$-form on $\mathbb{R}^{3}$

$\omega:=df=\partial_{x}fdx+\partial_{y}fdy,$

then $F^{1}_{s}\circ F^{2}_{t}=F^{2}_{t}\circ F^{1}_{s}$, if and only if

$\int_{\partial R}df=0$

for “almost every” oriented rectangle $R\subset\mathbb{R}^{2}$ with parallel
sides to the coordinate axes and side lengths $s$, $t$ (namely, for every
rectangle of the form $[x,x+s]\times[y,y+t]$ for a.e.
$(x,y)\in\mathbb{R}^{2}$). Note that $\omega$ is independent on $z$, and,
moreover, if $f\in W^{2,1}_{\operatorname{loc}}(\mathbb{R}^{2})$, then the
components of $\omega$ belong to
$W^{1,1}_{\operatorname{loc}}(\mathbb{R}^{2})$, so that we can apply Stokes
theorem to deduce

$\int_{\partial R}\omega=\int_{\partial R}df=\int_{R}d^{2}f=0,$

thus the two flows commute, i.e. to be more precise

$F^{1}_{t}(F^{2}_{s}(x,y,z))=F^{2}_{s}(F^{1}_{t}(x,y,z))$

for a.e. $(x,y)\in\mathbb{R}^{2}$, all $z\in\mathbb{R}$ and
$(s,t)\in\mathbb{R}^{2}$.

##  4\. Vanishing Lie bracket and commutativity of regular Lagrangian flows
for Sobolev vector fields

We prove now that regular Lagrangian flows $F^{1}$, $F^{2}$ of Sobolev vector
fields $V^{1}$, $V^{2}$ respectively commute a.e., if the Lie bracket of the
latter vanishes a.e., i.e. (1.4) holds, once one of the two is Lipschitz
continuous.

######  Theorem 4.1.

Let $p\in[1,\infty]$, and assume that $V^{1}\in W^{1,p}(\mathbb{R}^{d})$ is
such that its regular Lagrangian flow $F^{1}_{s}$ is uniquely defined for $s$
in some interval containing zero, and $V^{2}\in W^{1,\infty}(\mathbb{R}^{d})$.
Let $(F^{1}_{t})_{t\in\mathbb{R}}$, $(F^{2}_{t})_{t\in\mathbb{R}}$ be the
corresponding regular Lagrangian flows. If

$[V^{1},V^{2}](x):=(DV^{2})(x)V^{1}(x)-(DV^{1})(x)V^{2}(x)=0\quad\mbox{for
a.e.\ $x\in\mathbb{R}^{d}$},$

then the flows commute, i.e., for a.e. $x\in\mathbb{R}^{d}$, it holds

$F^{2}_{t}\left(F^{1}_{s}(x)\right)=F^{1}_{s}\left(F^{2}_{t}(x)\right),\quad\text{for
every $s$, $t\in\mathbb{R}$.}$

###### Proof.

Since $V^{2}$ is Lipschitz continuous, its regular Lagrangian flow coincides
a.e. with the classical flow. Therefore, we may assume that $x\mapsto
F^{2}_{t}(x)$ is defined everywhere and Lipschitz continuous. It is also
sufficient to prove the thesis for $s$, $t\in[-T,T]$ for an arbitrary $T>0$
such that both $F^{i}_{t}$ are defined for $t\in[-T,T]$, $i=1,2$.

Let $\nu$ stand for a finite measure on $\mathbb{R}^{d}$ with bounded and
strictly positive density with respect to $\mathcal{L}^{d}$ (the $L^{\infty}$
norm of the latter will be denoted by $\|\nu\|_{\infty}$), e.g. a standard
Gaussian. For $x\in\mathbb{R}^{d}$ and $s,t\in[-T,T]$ we set

(4.1) $X_{s,t}(x):=F^{2}_{t}(F^{1}_{s}(x)).$

Since the flow $F^{2}$ is Lipschitz continuous, we have that for a.e.
$x\in\mathbb{R}^{d}$ and for every $s,t\in[-T,T]$, the map $(s,t)\mapsto
X_{s,t}(x)$ is continuous. Moreover, for a.e. $x\in\mathbb{R}^{d}$ one has

$\displaystyle\partial_{t}X_{s,t}(x)=V^{2}(X_{s,t}(x))$ for a.e. $t\in(-T,T)$,
$\displaystyle\partial_{s}X_{s,0}(x)=V^{1}(X_{s,0}(x))$ for a.e. $s\in(-T,T)$.

Now, for $s$, $s^{\prime}$, $t\in[-T,T]$, we write

$A_{s,s^{\prime};t}(x):=X_{s^{\prime},t}(x)-X_{s,t}(x),\quad
B_{s,s^{\prime};t}(x):=A_{s,s^{\prime};t}(x)-(s^{\prime}-s)V^{1}(X_{s,t}(x)),$

frequently omitting the reference to $x$ for brevity, and for every
$t\in[-T,T]$ we consider the measure $\eta:=\hat{X}_{t\\#}\nu$ over
$C([-T,T],\mathbb{R}^{d})$, where $\hat{X}_{t}\colon\mathbb{R}^{d}\to
C([-T,T],\mathbb{R}^{d})$ is defined by the formula

$\hat{X}_{t}(x):=X_{(\cdot,t)}(x).$

Note that $\eta$ is concentrated over curves $[-T,T]\ni s\mapsto X_{s,t}(x)$
for a set of full $\nu$-measure of $x\in\mathbb{R}^{d}$. Further,

$e_{s\\#}\eta=X_{s,t\\#}\nu=F^{2}_{t\\#}(F^{1}_{s\\#}\nu)\leq
C_{2}C_{1}\mathcal{L}^{d},$

where $C_{i}$ are the compressibility constants for $F^{i}$, $i=1,2$.

We will show that for some function
$(\omega_{ss^{\prime}})_{s,s^{\prime}\in[-T,T]}$, with $[\omega]=0$ one has

(4.2)
$\left\|B_{s,s^{\prime};t}(\cdot)\right\|_{L^{1}(\nu)}\lesssim\omega_{ss^{\prime}},\quad\text{for
$s,s^{\prime}\in[-T,T]$.}$

This would give

$\left\|\theta_{s^{\prime}}-\theta_{s}-(s^{\prime}-s)V^{1}(\theta_{s})\right\|_{L^{1}(C([-T,T],\mathbb{R}^{d});\eta)}\lesssim\omega_{ss^{\prime}}$

for every $s,s^{\prime}\in[-T,T]$, which means in view of Proposition A.1
(applied with $V^{1}$ instead of $V$ and $C_{1}C_{2}$ instead of $C$) that
$\eta$-a.e. $\theta\in C([-T,T],\mathbb{R}^{d})$ is an integral curve of
$V^{1}$. In other words, for $\nu$-a.e. (hence for $\mathcal{L}^{d}$-a.e.,
because the density of $\nu$ is strictly positive) $x\in\mathbb{R}^{d}$ the
curve $\theta$ defined by $\theta(s):=X_{s,t}(x)$ satisfies
$\partial_{s}\theta=V^{1}(\theta(s))$ for every $s\in[-T,T]$. Since for such
curves one has $\theta(0)=F^{2}_{t}(x)$, then for a.e. $x\in\mathbb{R}^{d}$
one has

$X_{s,t}(x)=F^{1}_{s}(F^{2}_{t}(x)),$

which is the claim of the theorem being proven.

The rest of the proof is therefore dedicated to showing the claim (4.2). It
will be done in six steps.

_Step 1: Estimate on $A_{s,s^{\prime};0}$._ In the subsequent estimates we
suppose without loss of generality that $s^{\prime}\geq s$ (the case $s\geq
s^{\prime}$ is completely symmetric). One has

$\displaystyle|A_{s,s^{\prime};0}(x)|^{p}$
$\displaystyle=|X_{s^{\prime},0}(x)-X_{s,0}(x)|^{p}=|F^{1}_{s}(x)-F^{1}_{s^{\prime}}(x)|^{p}$
$\displaystyle\leq\left|\int_{s}^{s^{\prime}}|V^{1}(y(\tau))|\,d\tau\right|^{p}$
$\displaystyle\leq|s-s^{\prime}|^{p/p^{\prime}}\int_{s}^{s^{\prime}}|V^{1}(y(\tau))|^{p}\,d\tau\quad\mbox{by
H\"{o}lder inequality}.$

Integrating with respect to $\tilde{\eta}:=\nu\llcorner
B\otimes\delta_{F^{1}(\cdot,x)}$, where $B\subset\mathbb{R}^{d}$ is an
arbitrary fixed Borel set, gives

(4.3) $\displaystyle\int_{B}|A_{s,s^{\prime};0}(x)|^{p}\mathsf{d}\nu(x)$
$\displaystyle\leq|s-s^{\prime}|^{p/p^{\prime}}\left|\int_{C([-T,T];\mathbb{R}^{d})}\,\mathsf{d}\tilde{\eta}(y)\int_{s}^{s^{\prime}}|V^{1}(y(\tau))|^{p}\,\mathsf{d}\tau\right|$
$\displaystyle=|s-s^{\prime}|^{p/p^{\prime}}\left|\int_{s}^{s^{\prime}}\mathsf{d}\tau\int_{C([-T,T];\mathbb{R}^{d})}\,\mathsf{d}\tilde{\eta}(y)|V^{1}(y(\tau))|^{p}\right|$
$\displaystyle=|s-s^{\prime}|^{p/p^{\prime}}\left|\int_{s}^{s^{\prime}}\,\mathsf{d}\tau\int_{\mathbb{R}^{d}}\,\mathsf{d}e_{\tau\\#}\tilde{\eta}(x)|V^{1}(x)|^{p}\right|$
$\displaystyle\leq
C\|\nu\|_{\infty}|s-s^{\prime}|^{1+p/p^{\prime}}\int_{F^{1}_{\tau}(B)}|V^{1}(x)|^{p}\,\mathsf{d}x,$

because $e_{\tau\\#}\tilde{\eta}=F^{1}_{\tau\\#}(\nu\llcorner
B)=\mathbf{1}_{F^{1}_{\tau}(B)}F^{1}_{\tau\\#}\nu$ and in view of (2.6). In
particular, with $B:=\mathbb{R}^{d}$ we get

(4.4) $\displaystyle\|A_{s,s^{\prime};0}(x)\|_{L^{p}(\nu)}$
$\displaystyle\lesssim|s-s^{\prime}|^{1/p+1/p^{\prime}}\left\|V^{1}\right\|_{p}=|s-s^{\prime}|\left\|V^{1}\right\|_{p}.$

Further, since $F^{1}_{\tau}(B)=(F^{1}_{-\tau})^{-1}(B)$, we get

$\mathcal{L}^{d}(F^{1}_{\tau}(B))=(F^{1}_{-\tau\\#}\mathcal{L}^{d})(B)\leq
C\mathcal{L}^{d}(B),$

and hence from (4.3) with $p=1$ we get that the functions
$A_{s,s^{\prime};0}(\cdot)/|s-s^{\prime}|$ are equiintegrable in $L^{1}(\nu)$,
if $V^{1}\in L^{1}(\mathbb{R}^{d})$.

_Step 2: Estimate on $A_{s,s^{\prime};t}$._ Lipschitz continuity of
$F^{2}_{t}$ gives, for a.e. $x\in\mathbb{R}^{d}$,

$\partial_{t}|A_{s,s^{\prime};t}|\leq|V^{2}(X_{s,t})-V^{2}(X_{s^{\prime},t})|\leq\left\|DV^{2}\right\|_{\infty}|A_{s,s^{\prime};t}|,\quad\text{for
a.e.\ $t\in[-T,T]$.}$

Hence, by Gronwall lemma,

$\sup_{t\in[-T,T]}|A_{s,s^{\prime};t}|\leq|A_{s,s^{\prime};0}|\exp\left(\left\|DV^{2}\right\|_{\infty}T\right).$

Integrating with respect to $\nu$ gives

(4.5)
$\begin{split}\sup_{t\in[-T,T]}\left\||A_{s,s^{\prime};t}|\right\|_{L^{p}(\nu)}&\leq\left\|\sup_{t\in[-T,T]}|A_{s,s^{\prime};t}|\right\|_{L^{p}(\nu)}\\\
&\leq\left\|A_{s,s^{\prime};0}\right\|_{L^{p}(\nu)}\exp\left(\left\|DV^{2}\right\|_{\infty}T\right)\\\
&\lesssim|s^{\prime}-s|\left\|V^{1}\right\|_{p}\exp\left(\left\|DV^{2}\right\|_{\infty}T\right)\quad\mbox{in
view of\leavevmode\nobreak\ \eqref{eq_estAs0c}}\\\
&\lesssim|s^{\prime}-s|.\end{split}$

_Step 3: Integral inequality for $B_{s,s^{\prime};t}$_. We first prove that

  * (i)

for a.e. $x\in\mathbb{R}^{d}$ and for every $s\in[-T,T]$ the curve $t\mapsto
V^{1}(X_{s,t}(x))$ is absolutely continuous, and

  * (ii)

satisfies

(4.6) $\partial_{t}V^{1}(X_{s,t}(x))=DV^{1}(X_{s,t}(x))V^{2}(X_{s,t}(x)).$

In fact, for a.e. $x\in\mathbb{R}^{d}$ and for every $s\in[-T,T]$ the curve
$t\mapsto X_{s,t}(x)$ is absolutely continuous (even Lipschitz). Letting

$\displaystyle\nu^{1}_{s}:=F^{1}_{s\\#}\nu,\quad\eta^{2}_{s}:=F^{2}_{\\#}\nu^{1}_{s},$

we have that $\eta^{2}_{s}$ is a measure on $C([-T,T];\mathbb{R}^{d})$
concentrated over such curves, while

$\nu^{1}_{s}\leq C_{1}\|\nu\|_{\infty}\mathcal{L}^{d}$

by Remark 2.3, and therefore

$e_{t\\#}\eta^{2}_{s}=F^{2}_{t\\#}\nu^{1}_{s}\leq
C_{2}C_{1}\|\nu\|_{\infty}\mathcal{L}^{d}$

by the same Remark. Therefore, (i) and (ii) follow from Lemma B.3 with
$\eta^{2}_{s}$ in place of $\eta$ and $V^{1}$ in place of $V$.

Note that since $[V^{1},V^{2}](x)=0$ for a.e. $x\in\mathbb{R}^{d}$, then by
Lemma 2.4 there is a $B_{0}\subset\mathbb{R}^{d}$ with
$\mathcal{L}^{d}(B_{0})=0$ such that $[V^{1},V^{2}](X_{s,t}(x))=0$ for every
$x\not\in B_{0}$ and a.e. $s,t\in[-T,T]$.

Using (4.6) we obtain

$\begin{split}\partial_{t}B_{s,s^{\prime};t}&=\partial_{t}A_{s,s^{\prime};t}-(s^{\prime}-s)DV^{1}(X_{s,t})V^{2}(X_{s,t}),\\\
&=\partial_{t}A_{s,s^{\prime};t}-(s^{\prime}-s)DV^{2}(X_{s,t})V^{1}(X_{s,t})\quad\text{
(since $[V^{1},V^{2}](X_{s,t})=0$ a.e.)}\\\
&=DV^{2}(X_{s,t})B_{s,s^{\prime};t}+R_{s,s^{\prime};t},\end{split}$

where

$R_{s,s^{\prime};t}:=V^{2}(X_{s^{\prime},t})-V^{2}(X_{s,t})-DV^{2}(X_{s,t})(X_{s^{\prime},t}-X_{s,t}).$

At this point Gronwall lemma yields

(4.7)
$\sup_{t\in[-T,T]}|B_{s,s^{\prime};t}(x)|\leq\left(|B_{s,s^{\prime};0}(x)|+\int_{-T}^{T}|R_{s,s^{\prime};\tau}(x)|\mathsf{d}\tau\right)\exp\left(\left\|DV^{2}\right\|_{\infty}T\right),$

which implies

(4.8)
$\sup_{t\in[-T,T]}\|B_{s,s^{\prime};t}\|_{L^{1}(\nu)}\leq\left(\|B_{s,s^{\prime};0}\|_{L^{1}(\nu)}+\int_{-T}^{T}\|R_{s,s^{\prime};\tau}\|_{L^{1}(\nu)}\mathsf{d}\tau\right)\exp\left(\left\|DV^{2}\right\|_{\infty}T\right).$

_Step 4: Estimate on $B_{s,s^{\prime};0}$._ To estimate
$\|B_{s,s^{\prime};0}\|_{L^{1}(\nu)}$ we use Proposition A.1 with $V^{1}$
instead of $V$, $p_{0}:=p$, $p_{1}:=p^{\prime}$, $q:=1$, obtaining

(4.9)
$\left\|B_{s,s^{\prime};0}\right\|_{L^{1}(\nu)}\lesssim|s^{\prime}-s|^{2}.$

_Step 5: Estimate on $R_{s,s^{\prime};\tau}$._ We use (2.3) with the
components of $V^{2}$ instead of $f$, getting

$\left|V^{2}(y)-V^{2}(z)-DV^{2}(z)(y-z)\right|\lesssim|y-z|\left((DV^{2})^{\sharp}_{|y-z|}(y)+(DV^{2})^{\sharp}_{y-z}(z)\right),$

and choose $z:=X_{s,t}(x)$, $y:=X_{s^{\prime},t}(x)$, so that

$R_{s,s^{\prime};t}(x)\lesssim
A_{s,s^{\prime};t}(x)\left((DV^{2})^{\sharp}_{|A_{s,s^{\prime};t}(x)|}(X_{s^{\prime},t}(x))+(DV^{2})^{\sharp}_{|A_{s,s^{\prime};t}(x)|}(X_{s,t}(x))\right).$

Integrating the above inequality over $x\in\mathbb{R}^{d}$ with respect to
$\nu$ and dividing by $|s-s^{\prime}|$, we arrive at the estimate

(4.10)
$\begin{split}\int_{-T}^{T}&\frac{\left\|R_{s,s^{\prime};\tau}\right\|_{L^{1}(\nu)}}{|s-s^{\prime}|}\mathsf{d}\tau\lesssim\int_{-T}^{T}\mathsf{d}\tau\int_{\mathbb{R}^{d}}\frac{A_{s,s^{\prime};\tau}(x)}{|s-s^{\prime}|}D_{s,s^{\prime};\tau}(x)\mathsf{d}\nu(x),\quad\mbox{where}\\\
&D_{s,s^{\prime};\tau}(x):=\left((DV^{2})^{\sharp}_{|A_{s,s^{\prime};\tau}(x)|}(X_{s^{\prime},\tau}(x))+(DV^{2})^{\sharp}_{|A_{s,s^{\prime};\tau}(x)|}(X_{s,\tau}(x))\right).\end{split}$

We will show that one has

(4.11)
$\displaystyle\int_{-T}^{T}\mathsf{d}\tau\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{s,s^{\prime};\tau}(x)|}(X_{s^{\prime},\tau}(x))\mathsf{d}\nu(x)\to
0,$ (4.12)
$\displaystyle\int_{-T}^{T}\mathsf{d}\tau\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{s,s^{\prime};\tau}(x)|}(X_{s,\tau}(x))\mathsf{d}\nu(x)\to
0,$

which means in particular that the functions $\tilde{D}_{s,s^{\prime}}$
defined by $(\tau,x)\mapsto D_{s,s^{\prime};\tau}(x)$ converge to zero in
$L^{1}(\mathbb{R}^{d}\times(-T,T),\nu\otimes\mathcal{L}^{1})$, hence in
measure $\nu\otimes\mathcal{L}^{1}\llcorner[-T,T]$ as $\delta:=s^{\prime}-s\to
0^{+}$, uniformly in $s\in[-T,T]$. Minding that these functions are uniformly
bounded (since the maximal function $(DV^{2})^{\sharp}$ is bounded by
$\|DV^{2}\|_{\infty}$), and the functions $\tilde{A}_{s,s^{\prime}}$ defined
by $(\tau,x)\mapsto A_{s,s^{\prime};\tau}(x)/|s-s^{\prime}|$ are uniformly
bounded in $L^{1}(\mathbb{R}^{d}\times(-T,T),\nu\otimes\mathcal{L}^{1})$ and
equiintegrable by Step 1, then choosing arbitrary sequences
$s_{k},s_{k}^{\prime}\in[-T,T]$ with $\delta_{k}:=s_{k}^{\prime}-s_{k}\to
0^{+}$ from Lemma B.1 (with $f_{k}:=\tilde{A}_{s_{k},s_{k}^{\prime}}$,
$g_{k}:=\tilde{D}_{s_{k},s_{k}^{\prime}}$, $E:=\mathbb{R}^{d}\times(-T,T)$,
$\mu:=\nu\otimes\mathcal{L}^{1}$) and from (4.10) one gets

(4.13)
$\int_{-T}^{T}\left\|R_{s,s^{\prime};\tau}\right\|_{L^{1}(\nu)}\mathsf{d}\tau\lesssim|s^{\prime}-s|o(1).$

Plugging (4.9) and (4.13) into (4.8), we get

$\sup_{t\in[-T,T]}\|B_{s,s^{\prime};t}\|_{L^{1}(\nu)}\lesssim
o(|s^{\prime}-s|)$

as $s^{\prime}-s\to 0^{+}$, which gives the thesis.

Step 6. It remains to prove (4.11) and (4.12). In order to verify (4.11), let
$\nu_{s}$ be as in Step 3, i.e. $\nu_{s}:=(F^{1}_{s})_{\sharp}\nu$, so that a
change of variables gives

$\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{s,s^{\prime};\tau}(\cdot)|}(X_{s^{\prime},\tau}(\cdot))\mathsf{d}\nu=\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{0,s^{\prime}-s;\tau}(\cdot)|}(X_{s^{\prime}-s,\tau}(\cdot))\mathsf{d}\nu_{s}.$

Thus we simply observe that

$\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{0,\delta;\tau}(x)|})(X_{\delta,\tau}(x))\mathsf{d}\nu_{s}(x)\to
0$

as $\delta:=s^{\prime}-s\to 0^{+}$ uniformly in $s\in[-T,T]$ by Lemma B.2 with
$g_{\delta}(x):=(DV^{2})^{\sharp}_{|A_{0,\delta;\tau}(x)|}(X_{\delta,\tau}(x))$
(note that these functions are bounded by $\|DV^{2}\|_{\infty}$ since so are
the maximal functions $(DV^{2})^{\sharp}$ and $\lim_{\delta\to
0^{+}}g_{\delta}=0$ by Remark 2.1). Therefore by Lebesgue dominated
convergence theorem we get

$\int_{0}^{T}\mathsf{d}\tau\int_{\mathbb{R}^{d}}(DV^{2})^{\sharp}_{|A_{0,s^{\prime}-s;\tau}(x)|}(X_{s^{\prime}-s,\tau}(x))\mathsf{d}\nu_{s}(x)\to
0$

uniformly in $s\in[-T,T]$ as $s^{\prime}-s\to 0^{+}$. The proof of (4.12) is
completely analogous, and we can conclude the proof of the theorem. ∎

We give also the local version of the above Theorem 4.1.

######  Theorem 4.2.

For $p\in[1,\infty]$ assume that $V^{1}\in
W^{1,p}_{\operatorname{loc}}(\Omega;\mathbb{R}^{d})$ for some open set
$\Omega\subset\mathbb{R}^{d}$ is such that its regular Lagrangian flow
$F^{1}_{s}(x)$ is uniquely defined over $(-\bar{T}(x),\bar{T}(x))$ for some
measurable function $\bar{T}\colon\Omega\to(0,+\infty)$, and $V^{2}\in
W^{1,\infty}_{\operatorname{loc}}(\Omega;\mathbb{R}^{d})$. If

$[V^{1},V^{2}](x)=0\quad\mbox{for a.e.\ $x\in\Omega$},$

then for a.e. $x\in\Omega$ there is a $T=T(x)>0$ such that the regular
Lagrangian flows $F^{1}$, $F^{2}$ satisfy

(4.14)
$F^{2}_{t}\left(F^{1}_{s}(x)\right)=F^{1}_{s}\left(F^{2}_{t}(x)\right)\quad\text{for
every $s$, $t\in[-T,T]$.}$

###### Proof.

For every ball $B:=B_{R}(x_{0})$ such that $\bar{B}\subset\Omega$ and every
$T_{1}>0$ we find a $\rho>R$ with the property that
$B_{\rho}(x_{0})\subset\Omega$. Let $M_{B,T_{1}}\subset B$ stand for the
(possibly empty) Borel set of points $x\in B$ such that $F^{1}_{s}(x)$ is
defined for all $s\in[-T_{1},T_{1}]$ and stays in $B$. By Lemma 4.4 applied
with $V^{2}$ in place of $V$ there is a $T_{2}>0$ (depending on $B$) such that
for all $x\in B$ the classical flow induced by $V^{2}$ (which is equal to the
regular Lagrangian flow $F^{2}$) is defined for all $t\in[-T_{2},T_{2}]$ and
stays inside $B_{\rho}(x_{0})$. We set then $T:=T_{1}\wedge T_{2}$, so that
clearly $T=T(B,T_{1})$. In this way the map (4.1) is defined over
$M_{B,T_{1}}$ for all $t,s\in[-T,T]$, and takes values in $B_{\rho}(x_{0})$.
Note now that in the proof of Theorem 4.1 one only evaluates the Lie bracket
$[V_{1},V_{2}]$ along the trajectories of ODEs with right-hand sides $V_{1}$,
$V_{2}$. The latter, if started at a point of $M_{B,T_{1}}$ remain in
$B_{\rho}(x_{0})$, so that $[V_{1},V_{2}]$ vanishes along them. Thus
reiterating the proof of Theorem 4.1 with $\nu:=\mathcal{L}^{d}(M_{B,T_{1}})$
we get that for a.e. $x\in M_{B,T_{1}}$ the regular Lagrangian flows $F^{1}$,
$F^{2}$ commute, i.e. (4.14) holds. Representing now $\Omega$ as a disjoint
union of balls $\Omega:=\sqcup_{j}B_{j}$ such that $\bar{B}_{j}\subset\Omega$
for all $j\in\mathbb{N}$, one has that (4.14) holds for a.e. $x\in
M_{B_{j},1/k}$ with $T=T(B_{j},1/k)$, and since for all $j$ the sets
$\\{M_{B_{j},1/k}\\}_{k\in\mathbb{N}}$ cover almost all $B_{j}$, the claim
follows. ∎

######  Remark 4.3.

If under conditions of Theorem 4.2 one has additionally that $V^{1}\in
L^{\infty}_{\operatorname{loc}}(\Omega)$, then one can assume that $T\in
L^{\infty}_{loc}(\Omega)$. In fact, for every $x_{0}\in\Omega$, $R>0$,
$\rho>R$ as in the above proof, letting $\rho_{1}\in(R,\rho)$ apply Lemma 4.4
with $V^{1}$ instead of $V$ and $\rho_{1}$ instead of $\rho$ to get a number
$T_{1}>0$ such that $F^{1}_{s}(x)$ is defined for $s\in[-T_{1},T_{1}]$ and
$x\in B_{R}(x_{0})$ and maps almost all of $B:=B_{R}(x_{0})$ into
$B_{\rho_{1}}(x_{0})$ for every $s\in[0,T_{1}]$. Applying again Lemma 4.4 now
with $V^{2}$ in place of $V$ and $\rho_{1}$ in place of $R$ we get the
existence of a $T_{2}>0$ be such that for all $x\in B_{\rho_{1}}(x_{0})$ the
classical flow induced by $V^{2}$ (which is equal to the regular Lagrangian
flow $F^{2}$) is defined for all $t\in[-T_{2},T_{2}]$ and stays inside
$B_{\rho}(x_{0})$. We set then $T:=T_{1}\wedge T_{2}$, so that clearly
$T=T(B)$. One has then that the regular Lagrangian flows
$(F^{1}_{t})_{t\in[-T,T]}$, $(F^{2}_{t})_{t\in[-T,T]}$ defined for a.e. $x\in
B$ commute, i.e., (4.14) holds for every $s$, $t\in[-T,T]$. The claim follows
by representing $\Omega$ as a disjoint union of balls
$\Omega:=\sqcup_{j}B_{j}$ such that $\bar{B}_{j}\subset\Omega$ for all
$j\in\mathbb{N}$: in fact, every relatively compact subset of $\Omega$ is
covered by only a finite number of $B_{j}$.

The following lemma has been used in the above proof.

######  Lemma 4.4.

Let $\Omega\subset\mathbb{R}^{d}$ be an open set and the vector field $V\in
L^{\infty}_{\operatorname{loc}}(\Omega;\mathbb{R}^{d})$ be such that the
respective regular Lagrangian flow $(t,x)\mapsto F_{t}(x)$ is defined for all
$t\in(-\bar{T},\bar{T})$ whenever $x\in B$, $\bar{B}\subset\Omega$, for some
$T=T(B)>0$. Then for every $B_{\rho}(x_{0})$, and $0<R<\rho$ there is a
$T\in(0,\bar{T}(B_{R}(x_{0})))$ such that $F_{t}(x)\in B_{\rho}(x_{0})$ for
a.e. $x\in B_{R}(x_{0})$ and all $t\in[-T,T]$.

###### Proof.

Take a positive

$T<\frac{\rho-R}{\|V\|_{L^{\infty}(B_{\rho}(x_{0}))}}\wedge\bar{T}(B_{R}(x_{0})).$

We have then that $F_{t}$ maps almost all of $B_{R}(x_{0})$ into
$B_{\rho}(x_{0})$ for every $t\in[-T,T]$. We show this for an arbitrary
$t\in[0,T]$, the remaining case being completely symmetric. In fact, otherwise
there is an $x\in B_{\rho}(x_{0})$ and $y\in C([0,\bar{t}];\mathbb{R}^{d})$,
$y(t):=F_{t}(x)$ such that $y(t)\in B_{\rho}(x_{0})$ for all
$t\in[0,\bar{t})$, $|y(\bar{t})-x_{0}|=\rho$ and $\bar{t}\leq T$. But

$F_{t}(x)=x+\int_{0}^{t}V(x(\tau))\mathsf{d}\tau,$

so that

$|F_{\bar{t}}(x)-x_{0}|\leq R+\bar{t}\|V\|_{L^{\infty}(B_{\rho}(x_{0}))}\leq
R+T\|V\|_{L^{\infty}(B_{\rho}(x_{0}))}<\rho,$

which is impossible because
$|F^{1}_{\bar{t}}(x)-x_{0}|=|y(\bar{t})-x_{0}|=\rho$ by assumption. ∎

##  Appendix A Concentration on integral curves

We prove here our main technical tool, namely, a criterion of when a measure
over continuous curves is concentrated over on integral curves of a given
vector field $V$. Let us remark that it is based on a discrete formulation of
the equation for integral curves.

######  Proposition A.1.

Let $p_{0}$, $p_{1}\in(1,+\infty)$ with $1/p_{0}+1/p_{1}=1/q\leq 1$. Assume
that $V\in L^{1}\left([a,b];L^{p_{0}}(\mathbb{R}^{d})\right)$, $DV\in
L^{1}\left([a,b];L^{p_{1}}(\mathbb{R}^{d})\right)$, where $a<0<b$ and $\eta$
be a finite measure on $C([a,b];\mathbb{R}^{d})$ with $e_{t\\#}\eta\leq
C\mathcal{L}^{d}$ for all $t\in[a,b]$. Then, $\eta$ is concentrated on
integral curves of $V$, if and only if there exists a variation function
$(\omega_{st})_{st\in[a,b]}$ with $[\omega]=0$ such that

(A.1)
$\left\|\theta_{t}-\theta_{s}-\left(\int_{s}^{t}V_{\tau}\mathsf{d}\tau\right)(\theta_{s})\right\|_{L^{q}(C([a,b];\mathbb{R}^{d}),\eta)}\leq\omega_{st}\quad\text{for
every $s,t\in[a,b]$.}$

In such a case one can always choose

(A.2) $\omega_{st}\lesssim
C^{1/q}\left(\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau\right)\left(\int_{s}^{t}\left\|DV_{\tau}\right\|_{p_{1}}\mathsf{d}\tau\right).$

######  Remark A.2.

If (A.1) holds, then we can always represent $\eta=F_{\\#}(e_{0\\#}\eta)$,
where $F$ denotes the regular Lagrangian flow for $V$.

###### Proof.

Assume first that $\eta$ is concentrated on integral curves of $V$. Then

(A.3)
$\left\|\theta_{t}-\theta_{s}\right\|_{p_{0}}=\left\|\int_{s}^{t}V_{\tau}(\theta_{\tau})\mathsf{d}\tau\right\|_{p_{0}}\leq
C^{1/p_{0}}\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau.$

It follows that

(A.4)
$\begin{split}&\left\|\int_{s}^{t}V_{\tau}(\theta_{r})-\left(\int_{s}^{t}V_{\tau}\mathsf{d}\tau\right)(\theta_{s})\right\|_{L^{q}(C([a,b];\mathbb{R}^{d}),\eta)}\\\
&\leq\int_{s}^{t}\left\|V_{\tau}(\theta_{\tau})-V_{\tau}(\theta_{s})\right\|_{L^{q}(C([a,b];\mathbb{R}^{d}),\eta)}\mathsf{d}\tau\\\
&\lesssim\int_{s}^{t}\left\||\theta_{\tau}-\theta_{s}|\left(|DV_{\tau}|^{*}(\theta_{\tau})+|DV_{\tau}|^{*}(\theta_{s})\right)\right\|_{L^{q}(C([a,b];\mathbb{R}^{d}),\eta)}\mathsf{d}\tau\qquad\text{by
\eqref{eq:pontiwise-sobolev-star},}\\\
&\leq\int_{s}^{t}\left\|\theta_{\tau}-\theta_{s}\right\|_{L^{p_{0}}(C([a,b];\mathbb{R}^{d}),\eta)}\left\||DV_{\tau}|^{*}(\theta_{\tau})+|DV_{\tau}|^{*}(\theta_{s})\right\|_{L^{p_{1}}(C([a,b];\mathbb{R}^{d}),\eta)}\mathsf{d}\tau\\\
&\lesssim\int_{s}^{t}C^{1/p_{1}}\left\|DV_{\tau}\right\|_{p_{1}}C^{1/p_{0}}\int_{s}^{\tau}\left\|V_{u}\right\|_{p_{0}}\mathsf{d}u\mathsf{d}\tau\quad\text{by
\eqref{eq:bound-gamma-ts} and \eqref{eq:maximal},}\\\ &\leq
C^{1/q}\int_{s}^{t}\left\|DV_{\tau}\right\|_{p_{1}}\mathsf{d}\tau\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau\end{split}$

as claimed. Conversely, assume that (A.1) holds. Then

(A.5)
$\left\|\theta_{t}-\theta_{s}\right\|_{L^{p_{0}}(\eta)}\leq\omega_{st}+\left\|\int_{s}^{t}V_{\tau}(\theta_{\tau})\mathsf{d}\tau\right\|_{L^{p_{0}}(\eta)}\leq\omega_{st}+C^{1/p_{0}}\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau.$

Applying the triangle inequality along any partition $\pi$ of $[s,t]$, we
obtain that

$\left\|\theta_{t}-\theta_{s}\right\|_{L^{p_{0}}(\eta)}\leq\sum_{t_{i}\in\pi}\omega_{t_{i}t_{i+1}}+C^{1/p_{0}}\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau.$

Therefore, (A.5) self-improves to (A.3) and we can argue then exactly as in
(A.4). It follows that

$\left\|\theta_{t}-\theta_{s}-\int_{s}^{t}V_{\tau}(\theta_{\tau})\mathsf{d}\tau\right\|_{L^{q}(C([a,b];\mathbb{R}^{d}),\eta)}\leq\omega_{st}+C^{1/q}\int_{s}^{t}\left\|DV_{\tau}\right\|_{p_{1}}\mathsf{d}\tau\int_{s}^{t}\left\|V_{\tau}\right\|_{p_{0}}\mathsf{d}\tau.$

This entails that
$f(t):=\theta_{t}-\theta_{0}-\int_{0}^{t}V_{\tau}(\theta_{\tau})\mathsf{d}\tau$
has zero total variation, i.e., it must be $\eta$-a.e. constant (and null
since $f(0)=0$). ∎

As curious consequence of Proposition A.1, we obtain the following
quantitative stability estimate for regular Lagrangian flows, which is of
interest itself though not used elsewhere in this paper.

In particular it quantifies the qualitative statement that weak convergence of
vector fields in time and strong convergence in space leads to convergence of
the respective flows, see e.g. [7, remark 2.11] or [8] for a Trotter-type
formula.

To state it we write

$\Phi^{\delta}(x):=\log(1+\delta^{-1}|x|)$

for $x\in\mathbb{R}^{d}$, $\delta>0$, and notice that, for every $x$,
$y\in\mathbb{R}^{d}$,

(A.6) $\Phi^{\delta}(y)\leq\Phi^{\delta}(x)+\frac{|y-x|}{\delta+|x|}.$

Indeed,

$\begin{split}\log(1+\delta^{-1}|y|)&\leq\log(1+\delta^{-1}(|x|+|y|))=\log\left((1+\delta^{-1}|x|)\left(1+\frac{|y-x|}{\delta+|x|}\right)\right)\\\
&\leq\log\left(1+\delta^{-1}|x|\right)+\frac{|y-x|}{\delta+|x|}.\end{split}$

######  Corollary A.3.

For $i=1,2$ let

$V^{i}\in L^{1}\left([a,b];L^{p_{0}}(\mathbb{R}^{d})\right),\quad DV^{i}\in
L^{1}\left([a,b];L^{p_{1}}(\mathbb{R}^{d})\right),$

with $p_{0}$, $p_{1}$, $a$, $b$ as in Proposition A.1, $\eta^{i}$ be finite
measures on $\Theta:=C([a,b];\mathbb{R}^{d})$ with $e_{t\\#}\eta^{i}\leq
C_{i}\mathcal{L}^{d}$ for all $t\in[a,b]$ concentrated on integral curves of
$V^{i}$. Then, for every $\delta>0$ one has

$\begin{split}\Big{\|}\sup_{i}\Phi^{\delta}(\theta^{1}_{t_{i}}-\theta^{2}_{t_{i}})&\Big{\|}_{L^{q}(\Theta\times\Theta,\eta)}\lesssim\left\|\Phi^{\delta}(\theta^{1}_{0}-\theta^{2}_{0})\right\|_{L^{q}(\Theta\times\Theta,\eta)}+\sum_{i}\left\|D\int_{t_{i}}^{t_{i+1}}V^{1}_{\tau}\mathsf{d}\tau\right\|_{p_{1}}\\\
&\quad+\frac{1}{\delta}\sum_{i}\left\|\int_{t_{i}}^{t_{i+1}}(V^{1}_{\tau}-V^{2}_{\tau})\mathsf{d}\tau\right\|_{p_{0}}+\frac{1}{\delta}\sum_{i}\left(\omega^{1}_{t_{i}t_{i+1}}+\omega^{2}_{t_{i}t_{i+1}}\right),\end{split}$

whenever $a=t_{0}<t_{1}\ldots<t_{n}\leq b$, where $\omega^{i}_{st}$ are as in
(A.2) (with $C_{i}$, $V^{i}$ instead of $C$, $V$), and $\eta$ is any coupling
between $\eta^{1}$ and $\eta^{2}$ (i.e. a Borel measure over
$\Theta\times\Theta$ with marginals $\eta^{1}$ and $\eta^{2}$).

###### Proof.

Write

$R_{st}^{i}:=\theta^{i}_{t}-\theta^{i}_{s}-\left(\int_{s}^{t}V_{\tau}\mathsf{d}\tau\right)(\theta_{s}^{i}),$

so that $\left\|R_{st}^{i}\right\|_{q}\leq\omega^{i}_{st}$. We use (A.6) with
$y=\theta^{1}_{t}-\theta^{2}_{t}$, $x=\theta^{1}_{s}-\theta^{2}_{s}$,
obtaining

$\begin{split}\Phi^{\delta}(\theta^{1}_{t}-\theta^{2}_{t})&\leq\Phi^{\delta}(\theta^{1}_{s}-\theta^{2}_{s})+\frac{\left|\int_{s}^{t}V_{\tau}^{1}\mathsf{d}\tau(\theta^{1}_{s})-\int_{s}^{t}V^{2}_{\tau}\mathsf{d}\tau(\theta^{2}_{s})\right|+\left|R^{1}_{st}\right|+\left|R^{2}_{st}\right|}{\delta+|\theta^{1}_{s}-\theta^{2}_{s}|}\\\
&\leq\Phi^{\delta}(\theta^{1}_{s}-\theta^{2}_{s})+\frac{\left|\int_{s}^{t}V_{\tau}^{1}\mathsf{d}\tau(\theta^{1}_{s})-\int_{s}^{t}V_{\tau}^{1}\mathsf{d}\tau(\theta^{2}_{s})\right|}{|\theta^{1}_{s}-\theta^{2}_{s}|}\\\
&\quad+\delta^{-1}\left(\left|\int_{s}^{t}(V_{\tau}^{1}-V_{\tau}^{2})\mathsf{d}\tau(\theta^{2}_{s})\right|+\left|R^{1}_{st}\right|+\left|R^{2}_{st}\right|\right).\end{split}$

To conclude, it is sufficient to choose $s:=t_{i}$, $t:=t_{i+1}$, proceed
recursively, and finally take the $L^{q}$ norms (recall that we are
considering finite measures, so that $L^{p}$ spaces are nested). ∎

##  Appendix B Auxiliary lemmata

The lemma below provides a useful version of the Lebesgue dominated
convergence theorem.

######  Lemma B.1.

If $\\{f_{k}\\}\subset L^{1}(E,\mu)$ is a bounded and equintegrable sequence
of nonnegative functions, and $g_{k}\to 0$ in measure $\mu$ with $0\leq
g_{k}\leq C$ $\mu$-a.e., then

$\int_{E}f_{k}g_{k}\mathsf{d}\mu\to 0.$

###### Proof.

For every $\varepsilon>0$ let $E_{k,\varepsilon}:=\\{x\in
E\colon|g_{k}(x)|>\varepsilon\\}$. Then

$\int_{E}f_{k}g_{k}\mathsf{d}\mu=\int_{E_{k,\varepsilon}}f_{k}g_{k}\mathsf{d}\mu+\int_{E\setminus
E_{k,\varepsilon}}f_{k}g_{k}\mathsf{d}\mu\leq
C\int_{E_{k,\varepsilon}}f_{k}\mathsf{d}\mu+\varepsilon\|f_{k}\|_{L^{1}(E,\mu)},$

the first term vanishing as $k\to\infty$ because
$\lim_{k}\mu(E_{k,\varepsilon})=0$ and in view of equiintegrability of
$f_{k}$. Since $\varepsilon>0$ is arbitrary, one gets the claim. ∎

We will use also another lemma very similar to the previous one, which we
formulate for purely notational convenience for one-parameter families of
functions rather than for sequences.

######  Lemma B.2.

Let $\left\\{\nu_{s}\right\\}_{s\in[a,b]}$ be a tight family of absolutely
continuous Borel measures $\nu_{s}=f_{s}\mathcal{L}^{d}$ such that

$\sup_{s\in[a,b]}\left\|f_{s}\right\|_{\infty}<\infty.$

Let $\left\\{g_{\delta}\right\\}_{\delta\in[0,1]}$ be a family of non-negative
Borel functions with $0\leq g_{\delta}\leq C$ for all $\delta$ and
$\lim_{\delta\to 0}g_{\delta}=0$ in measure. Then

$\lim_{\delta\to 0}\sup_{s\in[a,b]}\int g_{\delta}\mathsf{d}\nu_{s}=0.$

###### Proof.

For any $\varepsilon>0$, let $K\subseteq\mathbb{R}^{d}$ be compact with
$\sup_{s\in[a,b]}\nu_{s}(K^{c})<\varepsilon$. Then,

$\begin{split}\int_{\mathbb{R}^{d}}g_{\delta}\mathsf{d}\nu_{s}&=\int_{K}g_{\delta}f_{s}\mathsf{d}x+\int_{K^{c}}g_{\delta}f_{s}\mathsf{d}x\leq
C\int_{K}g_{\delta}\mathsf{d}x+\sup_{\delta\in[0,1]}\left\|g_{\delta}\right\|_{\infty}\nu_{s}(K^{c})\\\
&\leq C\left(\int_{K}g_{\delta}\mathsf{d}x+\varepsilon\right).\end{split}$

By Lebesgue dominated convergence theorem one has $\lim_{\delta\to
0}\int_{K}g_{\delta}\mathsf{d}x=0$, and hence

$\limsup_{\delta\to 0}\sup_{s\in[a,b]}\int g_{\delta}\mathsf{d}\nu_{s}\leq
C\varepsilon,$

and we conclude since $\varepsilon>0$ is arbitrary small. ∎

The following lemma gives a chain rule for Sobolev functions along “almost
every” integral curve of an ODE.

######  Lemma B.3.

Let $p,q\geq 1$ with $\frac{1}{p}+\frac{1}{q}\leq 1$, let $f\in
W^{1,p}(\mathbb{R}^{d})$, $V\in L^{1}([a,b];L^{q}(\mathbb{R}^{d}))$ and $\eta$
be a finite measure on $C([a,b];\mathbb{R}^{d})$ with bounded compression and
concentrated on integral curves of $V$. Then, for any a.e. representative of
$f$ one has that for $\eta$-a.e. $\theta$ the curve $t\mapsto f(\theta_{t})$
is absolutely continuous and

$\partial_{t}f(\theta_{t})=V(\theta_{t})\nabla f(\theta_{t})\quad\text{for
a.e.\leavevmode\nobreak\ $t\in[a,b]$.}$

###### Proof.

We approximate $f$ with a fast converging sequence $(f_{n})_{n\geq 1}\subseteq
C^{1}(\mathbb{R}^{d})$, i.e., such that

$\sum_{n=1}^{\infty}\left\|f-f_{n}\right\|_{p}+\left\|\nabla f-\nabla
f_{n}\right\|_{p}<\infty.$

The thesis clearly holds for every $f_{n}$ in place of $f$. We have

$\left\|f_{n}\circ e_{t}-f\circ
e_{t}\right\|_{L^{p}(\eta)}\lesssim\left\|f_{n}-f\right\|_{p}\quad\text{for
every $t\in[a,b]$,}$

so that for every $t\in[a,b]$, $\eta$-a.e.,
$\lim_{n\to\infty}f_{n}(\theta_{t})=f(\theta_{t})$. In fact the limit holds
$\eta$-a.e. and uniformly with respect to $t\in[a,b]$. To show this, we notice
that

$\begin{split}\left\|\sup_{t\in[a,b]}(f_{n+1}-f_{n})\circ
e_{t}\right\|_{L^{(}\eta)}&\leq\left\|(f_{n+1}-f_{n})\circ
e_{0}\right\|_{L^{1}(\eta)}\\\ &\quad+\left\|\int_{a}^{b}|\nabla
f_{n+1}-\nabla f_{n+1}||V_{\tau}|\circ
e_{\tau}\mathsf{d}\tau\right\|_{L^{1}(\eta)}\\\
&\lesssim\left\|f_{n+1}-f_{n}\right\|_{p}\\\ &\quad+\int_{a}^{b}\left\|\nabla
f_{n+1}-\nabla
f_{n}\right\|_{p}\left\|V_{\tau}\right\|_{q}\mathsf{d}\tau.\end{split}$

The last quantity is summable with respect to $n\geq 1$. It follows that
$\eta$-a.e. the sequence $(f_{n}(\theta_{s})_{s\in[a,b]}$ is Cauchy, hence
convergent in $C([a,b])$, towards $(f(\theta_{s}))_{s\in[a,b]}$, that is in
particular $\eta$-a.e. continuous.

To argue that it is indeed absolutely continuous, we use the bound so that,
$\eta$-a.e., for every $t\in[a,b]$,

$\lim_{n\to\infty}\int_{0}^{t}\nabla
f_{n}(\theta_{\tau})V_{\tau}(\theta_{\tau})\mathsf{d}\tau=\int_{s}^{t}\nabla
f(\theta_{\tau})V_{\tau}(\theta_{\tau})\mathsf{d}\tau.$

To conclude, we simply pass to the limit in the $\eta$-a.e. identity

$f_{n}(\theta_{t})-f_{n}(\theta_{0})=\int_{a}^{t}\nabla
f_{n}(\theta_{\tau})V_{\tau}(\theta_{\tau})\mathsf{d}\tau.$

∎

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