# A square root velocity framework for curves of bounded variation

Markus Grasmair  Department of Mathematical Sciences, Norwegian University of
Science and Technology (NTNU), 7491 Trondheim, Norway<EMAIL_ADDRESS>https://www.ntnu.edu/employees/markus.grasmair

###### Abstract.

The square root velocity transform is a powerful tool for the efficient
computation of distances between curves. Also, after factoring out
reparametrisations, it defines a distance between shapes that only depends on
their intrinsic geometry but not the concrete parametrisation. Though
originally formulated for smooth curves, the square root velocity transform
and the resulting shape distance have been thoroughly analysed for the setting
of absolutely continuous curves using a relaxed notion of reparametrisations.
In this paper, we will generalise the square root velocity distance even
further to a class of discontinuous curves. We will provide an explicit
formula for the natural extension of this distance to curves of bounded
variation and analyse the resulting quotient distance on the space of
unparametrised curves. In particular, we will discuss the existence of optimal
reparametrisations for which the minimal distance on the quotient space is
realised.

###### Key words and phrases:

Shape analysis, shape distance, square root representation, discontinuous
curves

###### 2020 Mathematics Subject Classification:

Primary: 49J45; Secondary: 58D15

##  1\. Introduction

The mathematical analysis of shapes is a topic that is required in a large
number of different applications ranging from mathematical image processing
and inverse problems over computational biology to different problems in
computer vision; see [4, 6, 8, 14, 21] to name but a few examples, see also
[19] for a larger overview. In order to perform tasks like classification and
for studying the geometry of datasets of shapes, it is necessary to define a
suitable distance on the _shape space_. Here we will focus on one-dimensional
shapes, that is, parametrised curves $c\colon I\to\mathbb{R}^{d}$, $I=[0,1]$,
where we identify two curves if they only differ by a translation or a
reparametrisation. In this setting, a particularly useful distance is the
square root velocity (SRV) distance introduced in [15, 20]. Given a
sufficiently smooth, regular curve $c$, we define its square root velocity
transform $q:=R(c)=\dot{c}/\sqrt{\lvert\dot{c}\rvert}$. Then, the distance
between two curves $c_{1}$ and $c_{2}$ with square root velocity transforms
$q_{1}$ and $q_{2}$ is defined as $d(c_{1},c_{2})=\lVert
q_{1}-q_{2}\rVert_{L^{2}}$. It is also possible to regard the space of all
smooth regular curves as a manifold with the Riemannian structure inherited
from $L^{2}(I;\mathbb{R}^{d})$ via the mapping $R$. This differential
geometric point of view of shape analysis has been studied and discussed for
instance in [2, 3, 15, 19].

With this definition, one does not yet obtain a distance on the shape space,
as $d(c_{1},c_{2})$ depends on the parametrisation of the curves $c_{1}$ and
$c_{2}$. For that, one needs to consider instead the quotient distance

(1)
$d^{S}([c_{1}],[c_{2}])=\inf_{\varphi_{1},\,\varphi_{2}}d(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2}),$

where the infimum is taken over all orientation preserving smooth
diffeomorphisms of the unit interval $I$. Various approaches for the efficient
numerical solution of (discretisations of) this optimisation problem have been
suggested, ranging from gradient based optimisation methods [12] over
dynamical programming [7, 15] and the reformulation as a
Hamilton–Jacobi–Bellman equation [22] to machine learning methods [11, 16].
There exists also an analytic algorithm for the case where the curves $c_{1}$
and $c_{2}$ are piecewise linear [13].

In addition to providing a distance between shapes, the actual solutions
$(\bar{\varphi}_{1},\bar{\varphi}_{2})$ of the optimisation problem (1) can be
used to define correspondences, or best matches, between the curves $c_{1}$
and $c_{2}$: The best match on $c_{2}$ for the point $c_{1}(t)$ is the point
$c_{2}(\bar{\varphi}_{2}(\bar{\varphi}_{1}^{-1}(t)))$. To that end, however,
it is necessary that the optimisation problem (1) actually admits a solution.
In [5], it has been shown that this is the case provided that the two curves
$c_{1}$ and $c_{2}$ are continuously differentiable and the reparametrisations
$\varphi_{1}$ and $\varphi_{2}$ are allowed to be merely absolutely continuous
and non-decreasing instead of being smooth and strictly increasing. At the
same time, an example of two Lipschitz curves was provided, where the infimum
in (1) is not attained. A closer inspection of the results of [5], however,
reveals that a convex relaxation of the problem (1) admits its minimum for
arbitrary absolutely continuous curves $c_{1}$ and $c_{2}$. Using this relaxed
distance measure, one can thus generalise shape distances to absolutely
continuous curves and still obtain best matches between arbitrary curves.
Because the reparametrisations are merely non-decreasing and not necessarily
strictly increasing, it is possible, though, that a single point on one of the
curves corresponds to a whole line segment on the other curve.

In this paper, we will generalise the analysis of [5] to discontinuous curves,
or, more specifically, to curves of bounded variation. There it is no longer
possible to define the square root velocity transform $R(c)$ in a meaningful
manner, as this would involve taking the square root of the derivative of the
curve $c$, which is a Radon measure. Instead, we work directly with the SRV
distance and provide an explicit formula for a relaxation of this distance
from absolutely continuous curves to curves of bounded variation, see Theorem
1 below. The main tool here is a generalisation of the Reshetnyak continuity
and lower semi-continuity theorems [17]. Next, we discuss how the relaxed SRV
distance can be used for defining a shape distance on curves of bounded
variation. The main challenge here is the fact that a composition of a
discontinuous curve with a non-decreasing reparametrisation is not necessarily
well-defined: If the reparametrisation $\varphi$ maps a whole interval $[a,b]$
to a single point $t$ where the curve $c$ has a jump, then it is _a–priori_
not clear how the composition $c\circ\varphi$ should be interpreted on that
interval. In Definition 2 we provide such an interpretation, which we show to
be natural in the context we are working in, see Proposition 3. Finally, we
consider the particular setting of special curves of bounded variation, where
we show that our approach gives rise to a shape distance, for which optimal
matchings exists for each pair of curves, see Theorem 6.

In Section 2 we will recall the main definitions and results concerning the
SRV transform for absolutely continuous curves, which are relevant for this
paper. The generalisation to curves and shapes of bounded variation is
presented in Section 3. Finally, the proofs of all the results are collected
in Sections 4 and 5.

##  2\. The SRV framework for absolutely continuous curves

In the following, we will provide a brief introduction into the square root
velocity framework for absolutely continuous curves following the results of
[5].

###  2.1. Square root velocity transform

Denote by $I=[0,1]$ the unit interval and by
$\operatorname{AC}(I;\mathbb{R}^{d})$ the space of absolutely continuous
curves in $\mathbb{R}^{d}$. Moreover, let
$\operatorname{AC}_{0}(I;\mathbb{R}^{d})$ the subspace of absolutely
continuous curves satisfying $c(0)=0$. The square root velocity-transform
(SRVT) of a curve $c\in\operatorname{AC}(I;\mathbb{R}^{d})$ is defined as
$R\colon\operatorname{AC}(I;\mathbb{R}^{d})\to L^{2}(I;\mathbb{R}^{d})$,

(2) $R(c)=\frac{\dot{c}}{\sqrt{\lvert\dot{c}\rvert}}.$

Here the fraction $\dot{c}/\sqrt{\lvert\dot{c}\rvert}$ is set to be zero at
points where $\dot{c}=0$. The mapping $R$ is a bijection from
$\operatorname{AC}_{0}(I;\mathbb{R}^{d})$ to $L^{2}(I;\mathbb{R}^{d})$ with
inverse

$R^{-1}(q)(x)=\int_{0}^{x}q\lvert q\rvert\,dy.$

Moreover, the SRVT is norm-preserving in the sense that

$\lVert
R(c)\rVert_{L^{2}}^{2}=\lVert\dot{c}\rVert_{L^{1}}=\operatorname{len}(c).$

Given two curves $c_{1}$, $c_{2}\in\operatorname{AC}_{0}(I;\mathbb{R}^{d})$,
we define their (squared) SRV distance as

$d(c_{1},c_{2})^{2}=\lVert R(c_{1})-R(c_{2})\rVert_{L^{2}}^{2}.$

This can be rewritten as

$d(c_{1},c_{2})^{2}=\operatorname{len}(c_{1})+\operatorname{len}(c_{2})-2S(c_{1},c_{2})$

with

$S(c_{1},c_{2})=\langle
R(c_{1}),R(c_{2})\rangle_{L^{2}}=\int_{I}\Bigl{\langle}\frac{\dot{c}_{1}}{\lvert\dot{c}_{1}\rvert},\frac{\dot{c}_{2}}{\lvert\dot{c}_{2}\rvert}\Bigr{\rangle}\sqrt{\lvert\dot{c}_{1}\rvert\lvert\dot{c}_{2}\rvert}\,dx.$

###  2.2. Shape space distance

With the definition above one obtains a distance on the space of absolutely
continuous curves. However, we are also interested in the case where one
identifies curves if they are equal up to parametrisation. Following [5], we
define

$\Gamma:=\bigl{\\{}\gamma\in\operatorname{AC}(I;I):\gamma(0)=0,\,\gamma(1)=1,\,\gamma^{\prime}>0\text{
a.e.\,}\bigr{\\}},$

the set of all absolutely continuous reparametrisations of the unit interval.
Moreover, we define

$\bar{\Gamma}:=\bigl{\\{}\gamma\in\operatorname{AC}(I;I):\gamma(0)=0,\,\gamma(1)=1,\,\gamma^{\prime}\geq
0\text{ a.e.\,}\bigr{\\}}.$

Then $\bar{\Gamma}$ is the closure of $\Gamma$ in $\operatorname{AC}(I;I)$
(with respect to the norm topology).

We say that two curves $c_{1}$,
$c_{2}\in\operatorname{AC}_{0}(I;\mathbb{R}^{d})$ are equivalent, if there
exist $\varphi_{1}$, $\varphi_{2}\in\bar{\Gamma}$ and a curve
$\hat{c}\in\operatorname{AC}_{0}(I;\mathbb{R}^{d})$ such that
$c_{1}=\hat{c}\circ\varphi_{1}$ and $c_{2}=\hat{c}\circ\varphi_{2}$. It has
been shown in [5, Prop. 12] that this defines an equivalence relation $\sim$
on $\operatorname{AC}_{0}(I;\mathbb{R}^{d})$. In the following, we will denote
the equivalence class of a curve $c$ by $[c]$. Moreover, the quotient space of
unparametrised curves is denoted by
$B(I;\mathbb{R}^{d})=\operatorname{AC}_{0}(I;\mathbb{R}^{d})/\sim$. One can
show that

$B(I;\mathbb{R}^{d})=\bigl{\\{}c\circ\bar{\Gamma}:c\in\operatorname{AC}_{0}(I;\mathbb{R}^{d}),\,\dot{c}\neq
0\text{ a.e.}\bigr{\\}}\cup\\{0\\}.$

Moreover, two curves are equivalent, if and only if they have the same
constant speed parametrisation.

On the quotient space $B(I;\mathbb{R}^{d})$ we now consider the induced
distance

$d^{\mathcal{S}}([c_{1}],[c_{2}])=\inf_{\varphi_{1},\,\varphi_{2}\in\bar{\Gamma}}d(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2}).$

Note here that, for $c\in\operatorname{AC}(I;\mathbb{R}^{d})$ and
$\varphi\in\bar{\Gamma}$, the composition $c\circ\varphi$ is again absolutely
continuous, since $\varphi$ is a non-decreasing function [10, Prop. 225C].

Since the length of a curve is invariant under reparametrisation, we can also
write

(3)
$d^{S}([c_{1}],[c_{2}])^{2}=\operatorname{len}(c_{1})+\operatorname{len}(c_{2})-\sup_{\varphi_{1},\,\varphi_{2}\in\bar{\Gamma}}S(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2})$

with

$S(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2})=\int_{I}\biggl{\langle}\frac{\dot{c}_{1}\circ\varphi_{1}}{\lvert\dot{c}_{1}\circ\varphi_{1}\rvert},\frac{\dot{c}_{2}\circ\varphi_{2}}{\lvert\dot{c}_{2}\circ\varphi_{2}\rvert}\biggr{\rangle}\sqrt{\lvert\dot{c}_{1}\circ\varphi_{1}\rvert\lvert\dot{c}_{2}\circ\varphi_{2}\rvert\varphi_{1}^{\prime}\varphi_{2}^{\prime}}\,dx.$

It has been shown in [5, Prop. 15] that the supremum in (3) is attained for
some $\varphi_{1}$, $\varphi_{2}\in\bar{\Gamma}$ provided that $c_{1}$,
$c_{2}\in C^{1}(I;\mathbb{R}^{d})$ satisfy $\dot{c}_{i}\neq 0$ almost
everywhere.

Note that the functional $S(c_{1},c_{2})$ is invariant under simultaneous
reparametrisations of the curves $c_{1}$ and $c_{2}$, that is,
$S(c_{1}\circ\varphi,c_{2}\circ\varphi)=S(c_{1},c_{2})$ for all
$\varphi\in\bar{\Gamma}$. This property is crucial for the geometric
properties of the quotient space $B(I;\mathbb{R}^{d})$, and is also often
exploited in numerical discretisations of (3).

###  2.3. Scale invariant SRV distance

It is also possible to define a scale invariant SRV distance. Here one
defines, for $c\in\operatorname{AC}_{0}(I;\mathbb{R}^{d})\setminus\\{0\\}$,

$\tilde{R}(c)=\frac{1}{\sqrt{\operatorname{len}(c)}}\frac{\dot{c}}{\sqrt{\lvert\dot{c}\rvert}}=R(c/\operatorname{len}(c)).$

This gives a mapping from
$\operatorname{AC}_{0}(I;\mathbb{R}^{d})\setminus\\{0\\}$ to the unit sphere
in $L^{2}(I;\mathbb{R}^{d})$. The corresponding scale invariant SRV distance
between two non-zero curves is therefore defined as the spherical distance
between their scale invariant SRVTs, that is,

$\tilde{d}(c_{1},c_{2})^{2}=\arccos\bigl{(}\langle
R(c_{1}/\operatorname{len}(c_{1})),R(c_{2}/\operatorname{len}(c_{2}))\rangle_{L^{2}}\bigr{)}\\\
=\arccos\bigl{(}S(c_{1}/\operatorname{len}(c_{1}),c_{2}/\operatorname{len}(c_{2}))\bigr{)}.$

Moreover, we can define a scale invariant shape distance by

$\tilde{d}^{S}([c_{1}],[c_{2}])^{2}=\inf_{\varphi_{1},\,\varphi_{2}\in\bar{\Gamma}}\tilde{d}(c_{1}\circ\varphi_{1}/\operatorname{len}(c_{1}),c_{2}\circ\varphi_{2}/\operatorname{len}(c_{2}))^{2}\\\
=\arccos\Bigl{(}\sup_{\varphi_{1},\,\varphi_{2}\in\bar{\Gamma}}S(c_{1}\circ\varphi_{1}/\operatorname{len}(c_{1}),c_{2}\circ\varphi_{2}/\operatorname{len}(c_{2}))\Bigr{)}.$

In this article, we will focus on the unscaled variant of the SRV distance.
However, all our results are based solely on the properties of the function
$S$, which is central both for the scaled and the unscaled variants. Thus all
our results hold _mutatis mutandis_ also for the scaled SRV framework.

##  3\. Generalisation to BV curves

We now want to generalise the SRV distance to discontinuous curves, or, more
specifically, to curves of bounded variation. Note here that we will not
generalise the SRVT itself, since the definition of $R(c)$ involves the square
root of the derivative of $c$, which is a measure if
$c\in\operatorname{BV}(I;\mathbb{R}^{d})$ is a general function of bounded
variation. However, we will see that the resulting distance can still be
defined.

###  3.1. Curves of bounded variation

In the following, we collect some results concerning functions of bounded
variation that will be needed throughout the paper. For more details, we refer
to [1, Sec. 3.2].

For every $c\in\operatorname{BV}(I;\mathbb{R}^{d})$ and $x\in I$ the one-sided
essential limits $c^{\ell}(x):=\operatorname*{ess\,lim}_{y\to x^{-}}c(y)$ and
$c^{r}(x):=\operatorname*{ess\,lim}_{y\to x^{+}}$ are well-defined. Thus we
can define the subspace

$\operatorname{BV}_{0}(I;\mathbb{R}^{d}):=\bigl{\\{}c\in\operatorname{BV}(I;\mathbb{R}^{d}):c^{r}(0)=0\bigr{\\}}.$

For $c\in\operatorname{BV}_{0}(I;\mathbb{R}^{d})$ and every $x\in I$ we have
that $c^{\ell}(x)=Dc((0,x))$ and $c^{r}(x)=Dc((0,x])$.

We say that a pointwise defined function $\tilde{c}\colon I\to\mathbb{R}^{d}$
is a _good representative_ of $c\in\operatorname{BV}(I;\mathbb{R}^{d})$, if
for all $x\in I$ we have $\tilde{c}(x)\in[c^{\ell}(x),c^{r}(x)]$. That is,
$\tilde{c}(x)=\operatorname*{ess\,lim}_{y\to x}c(x)$ whenever $c$ is
essentially continuous, and $\tilde{c}(x)$ lies on the line segment from
$c^{\ell}(x)$ to $c^{r}(x)$ whenever $c$ has a jump at $x$. We will always
identify $c$ with any of its good representatives. Finally, for good
representatives the measure theoretic total variation of $c$ coincides with
the pointwise total variation, that is,

$\lvert Dc\rvert(I)=\sup\Bigl{\\{}\sum_{k}\lvert
c(x_{k+1})-c(x_{k})\rvert:0<x_{1}<x_{2}<\ldots<x_{N}<1\Bigr{\\}}.$

In a slight abuse of notation, we denote by $\operatorname{len}(c):=\lvert
Dc\rvert(I)$ the length of a discontinuous curve, including all of its jumps.
We say that a sequence
$\\{c^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{BV}(I;\mathbb{R}^{d})$
converges _strictly_ to $c\in\operatorname{BV}(I;\mathbb{R}^{d})$, denoted

$c^{(k)}\to^{s}c,$

if $\lVert c^{(k)}-c\rVert_{L^{1}}\to 0$ and
$\operatorname{len}(c^{(k)})\to\operatorname{len}(c)$. Equivalently, we have
that $c^{(k)}\to^{s}c$, if and only if $c^{(k)}(0)\to c(0)$,
$Dc^{(k)}\rightharpoonup^{*}Dc$ in the sense of weak convergence of Radon
mesures, and $\operatorname{len}(c^{(k)})\to\operatorname{len}(c)$.

Assume that $c\in\operatorname{BV}(I;\mathbb{R}^{d})$. Then its weak
derivative $Dc\in\mathcal{M}(I;\mathbb{R}^{d})$ can be decomposed as

$Dc=\dot{c}\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}\mathcal{L}^{1}+D^{j}c+D^{c}c,$

where $\dot{c}$ is the classical derivative of $c$ (which exists almost
everywhere), $D^{j}c$ is a purely atomic measure—the jump part of $Dc$—, and
$D^{c}c$ is a non-atomic, singular measure—the Cantor part of $Dc$. We denote
for $c\in\operatorname{BV}(I;\mathbb{R}^{d})$ by

$\Sigma(c):=\bigl{\\{}x\in I:c^{\ell}(x)\neq c^{r}(x)\bigr{\\}}$

the jump set of $c$ and define

$[c](x):=c^{r}(x)-c^{\ell}(x)\qquad\text{ for }x\in\Sigma(c).$

Then the jump part $D^{j}c$ of $Dc$ can be written as

$D^{j}c=\sum_{x\in\Sigma(c)}[c](x)\,\delta_{x}$

with $\delta_{x}$ denoting the Dirac delta centered at $x$.

Furthermore, we can decompose every function
$c\in\operatorname{BV}_{0}(I;\mathbb{R}^{d})$ uniquely as

$c=c^{(a)}+c^{(j)}+c^{(c)}$

such that $c^{(a)}$ is absolutely continuous and $(c^{(a)})^{\prime}=\dot{c}$,
$Dc^{(j)}=D^{j}c$, and $Dc^{(c)}=D^{c}c$. The set of functions $c$ where the
Cantor part $c^{(c)}$ equals zero is denoted
$\operatorname{SBV}(I;\mathbb{R}^{d})$.

###  3.2. Extension of the SRV distance

We define

$\hat{S}(c_{1},c_{2}):=\sup\Bigl{\\{}\limsup_{k}S(c_{1}^{(k)},c_{2}^{(k)}):c_{i}^{(k)}\in\operatorname{AC}(I;\mathbb{R}^{d}),\
c_{i}^{(k)}\to^{s}c_{i}\Bigr{\\}}$

and

$\hat{d}(c_{1},c_{2})=\operatorname{len}(c_{1})+\operatorname{len}(c_{2})-2\hat{S}(c_{1},c_{2}).$

That is, $\hat{d}$ is the strictly lower semi-continuous hull of the function
that is equal to $d$ on $\operatorname{AC}(I;\mathbb{R}^{d})^{2}$ and equal to
$+\infty$ else.

Our first main theorem provides an explicit expression for $\hat{S}$, and
consequently also for $\hat{d}$.

######  Theorem 1.

Assume that $c_{1}$, $c_{2}\in\operatorname{BV}(I;\mathbb{R}^{d})$. Then

$\hat{S}(c_{1},c_{2})\\\ =\int_{I}\Bigl{\langle}\frac{dDc_{1}}{\lvert
dDc_{1}\rvert},\frac{dDc_{2}}{\lvert
dDc_{2}\rvert}\Bigr{\rangle}^{+}\sqrt{\frac{d\lvert Dc_{1}\rvert}{d(\lvert
Dc_{1}\rvert+\lvert Dc_{2}\rvert)}\frac{d\lvert Dc_{2}\rvert}{d(\lvert
Dc_{1}\rvert+\lvert Dc_{2}\rvert)}}\,d(\lvert Dc_{1}\rvert+\lvert
Dc_{2}\rvert).$

Here the integrand is set to zero at points where either $d\lvert
Dc_{1}\rvert/d(\lvert Dc_{1}\rvert+\lvert Dc_{2}\rvert)=0$ or $d\lvert
Dc_{2}\rvert/d(\lvert Dc_{1}\rvert+\lvert Dc_{2}\rvert)=0$. Moreover
$(\cdot)^{+}:=\max\\{\cdot,0\\}$ denotes the positive part of the argument.

In particular, if $c_{1}$, $c_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$,
then

$\hat{S}(c_{1},c_{2})=\int_{I}\biggl{\langle}\frac{\dot{c}_{1}}{\lvert\dot{c}_{1}\rvert},\frac{\dot{c}_{2}}{\lvert\dot{c}_{2}\rvert}\biggr{\rangle}^{+}\sqrt{\lvert\dot{c}_{1}\rvert\lvert\dot{c}_{2}\rvert}\,dx\\\
+\sum_{x\in\Sigma(c_{1})\cap\Sigma(c_{2})}\biggl{\langle}\frac{[c_{1}](x)}{\lvert[c_{1}](x)\rvert},\frac{[c_{2}](x)}{\lvert[c_{2}](x)\rvert}\biggr{\rangle}^{+}\sqrt{\lvert[c_{1}](x)\rvert\,\lvert[c_{2}](x)\rvert}.$

###### Proof.

See Section 4. ∎

###  3.3. Reparametrisations of BV curves

Again, we are interested in the case of distances modulo reparametrisations.
However, in a setting with discontinuous curves $c$ and reparametrisations
$\varphi$ that are merely non-decreasing, but not necessarily strictly
increasing, the expression $c\circ\varphi$ does not always make sense: If
$\varphi$ is constant on a non-trivial interval, say $\varphi(x)=y$ for all
$x\in[a,b]$ with $b>a$ and $c$ has a jump at $y$, then $c\circ\varphi$ is not
well-defined on the interval $[a,b]$. Thus we have to use a generalised
definition of reparametrisations.

######  Definition 2.

Let $c\in\operatorname{BV}(I;\mathbb{R}^{d})$ and $\varphi\in\bar{\Gamma}$. We
define

$[c,\varphi]:=\Bigl{\\{}g\in\operatorname{BV}(I;\mathbb{R}^{d}):g(x)\in\bigl{[}c^{\ell}\bigl{(}\varphi(x)\bigr{)},c^{r}\bigl{(}\varphi(x)\bigr{)}\bigr{]}\text{
for all }x\in I\\\ \text{ and
}\operatorname{len}(c)=\operatorname{len}(g)\Bigr{\\}}.$

Here
$\bigl{[}c^{\ell}\bigl{(}\varphi(x)\bigr{)},c^{r}\bigl{(}\varphi(x)\bigr{)}\bigr{]}\subset\mathbb{R}^{d}$
denotes the line segment from $c^{\ell}\bigl{(}\varphi(x)\bigr{)}$ to
$c^{r}\bigl{(}\varphi(x)\bigr{)}$. In particular, if $x\in I$ is such that $c$
is continuous at $\varphi(x)$ and $g\in[c,\varphi]$, then $g(x)=c(\varphi(x))$
and $g$ is continuous at $x$. Moreover, note that $[c,\varphi]$ consists of
the single element $c\circ\varphi$ if either $c$ is continuous or $\varphi$
injective.

With a slight abuse of notation, we thus can define the distance between the
sets $[c_{1},\varphi_{1}]$ and $[c_{2},\varphi_{2}]$ as

$\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}]):=\inf_{g_{i}\in[c_{i},\varphi_{i}]}\hat{d}(g_{1},g_{2})=\operatorname{len}(c_{1})+\operatorname{len}(c_{2})-\sup_{g_{i}\in[c_{i},\varphi_{i}]}\hat{S}(g_{1},g_{2}).$

Our next result shows that the same distance function is obtained as the lower
semi-continuous extension of the mapping
$(c_{1},c_{2},\varphi_{1},\varphi_{2})\mapsto
d(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2})$ with respect to strict
convergence of the curves $c_{i}$ and uniform convergence of the
reparametrisations $\varphi_{i}$.

######  Proposition 3.

Define the functional
$D\colon\operatorname{BV}(I;\mathbb{R}^{d})^{2}\times\bar{\Gamma}^{2}\to\mathbb{R}\cup\\{+\infty\\}$,

$D(c_{1},c_{2};\varphi_{1},\varphi_{2}):=\begin{cases}d(c_{1}\circ\varphi_{1},c_{2}\circ\varphi_{2})&\text{
if }c_{i}\in\operatorname{AC}(I;\mathbb{R}^{d})\text{ and
}\varphi_{i}\in\Gamma,\\\ +\infty&\text{ else.}\end{cases}$

Then the lower semi-continuous hull of $D$ with respect to strict convergence
on $\operatorname{BV}(I;\mathbb{R}^{d})$ and uniform convergence on
$\bar{\Gamma}$ is the functional

$\hat{D}(c_{1},c_{2};\varphi_{1},\varphi_{2})=\inf_{g_{i}\in[c_{i},\varphi_{i}]}\hat{d}(g_{1},g_{2})=\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}]).$

###### Proof.

See Section 5.1. ∎

The next result shows that the resulting distance function is invariant under
simultaneous reparametrisations:

######  Proposition 4.

Assume that $c_{1}$, $c_{2}\in\operatorname{BV}(I;\mathbb{R}^{d})$ and
$\varphi_{1}$, $\varphi_{2}\in\bar{\Gamma}$. Then we have for all
$\psi\in\bar{\Gamma}$ that

$\hat{d}([c_{1},\varphi_{1}\circ\psi],[c_{2},\varphi_{2}\circ\psi])=\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}]).$

###### Proof.

See Section 5.1. ∎

###  3.4. A shape distance on SBV curves

We now consider the particular setting where the involved curves are special
functions of bounded variation.

Assume that $c\in\operatorname{SBV}(I;\mathbb{R}^{d})\setminus\\{0\\}$. Then
we can construct an “equivalent” function
$G(c)\in\operatorname{AC}(I;\mathbb{R}^{d})$ in the following way: Define the
function $\xi\colon I\to I$,

$\xi(x)=\frac{\lvert
D^{s}c\rvert(0,x)}{2\operatorname{len}(c)}+(1-\alpha)x,\qquad\qquad\text{ with
}\alpha=\frac{\lvert D^{s}c\rvert(I)}{2\operatorname{len}(c)}.$

Since $0\leq\alpha\leq 1/2$, it follows that $\xi$ is a strictly increasing
function with $\xi(0)=0$, $\xi(1)=1$, and $\Sigma(\xi)=\Sigma(c)$. Denote now
by $\zeta\colon I\to I$ the (unique) non-decreasing left inverse of $\xi$.
Then $\zeta\colon I\to I$ is Lipschitz continuous with Lipschitz constant
$1/(1-\alpha)\leq 2$. Now define the function
$G(c)\in\operatorname{AC}(I;\mathbb{R}^{d})$ by setting $G(c)(x)=c(\zeta(x))$
if $c$ is continuous at $\zeta(x)$, and

$G(c)(x)=c^{\ell}(\zeta(x))+\frac{x-\xi^{\ell}(\zeta(x))}{\xi^{r}(\zeta(x))-\xi^{\ell}(\zeta(x))}\bigl{(}c^{r}(\zeta(x))-c_{i}^{\ell}(\zeta(x))\bigr{)}$

if $c$ and thus also $\xi$ is discontinuous at $\zeta(x)$. That is, the jumps
of $c$ are replaced by a linear interpolation between the left and right
limits of $c$ at the jump points. In particular, we have that
$G(c)\in[c,\zeta]$.

######  Definition 5.

We say that two curves $c_{1}$, $c_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$
are equivalent, denoted $c_{1}\sim c_{2}$, if the curves $G(c_{1})$ and
$G(c_{2})$ have the same constant speed parametrisation. By $[c]$ we denote
the equivalence class of the curve $c$.

Moreover, we define the _shape distance_

$\hat{d}^{S}([c_{1}],[c_{2}])=\inf_{g_{i}\in[c_{i}]}\hat{d}(g_{1},g_{2})=\operatorname{len}(c_{1})+\operatorname{len}(c_{2})-\sup_{g_{i}\in[c_{i}]}\hat{S}(g_{1},g_{2})$

on the set of equivalence classes with respect to $\sim$.

The following result shows that the distance between the shapes $[c_{1}]$ and
$[c_{2}]$ can be computed by minimising the curve distance over all
reparametrisations of $c_{1}$ and $c_{2}$ in the sense of Definition 2.

######  Theorem 6.

Assume that $c_{1}$, $c_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$ satisfy
$\dot{c}_{i}(x)\neq 0$ for a.e. $x\in I$. Then

(4)
$\hat{d}^{S}([c_{1}],[c_{2}])=\inf_{\varphi_{i}\in\bar{\Gamma}}\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])=\inf_{\psi_{i}\in\bar{\Gamma}}\hat{d}(G(c_{1})\circ\psi_{1},G(c_{2})\circ\psi_{2}).$

Moreover, the infima in (4) are attained at some $\bar{\varphi}_{1}$,
$\bar{\varphi}_{2}\in\bar{\Gamma}$ and $\bar{\psi}_{1}$,
$\bar{\psi}_{2}\in\bar{\Gamma}$.

###### Proof.

See Section 5.2. ∎

In addition, this result shows that it is possible to compute optimal
reparametrisations for arbitrary SBV curves. In particular, it is possible to
define optimal matchings between SBV curves.

##  4\. Extension of the SRV distance to BV curves

In this section, we will prove Theorem 1, which provides an explicit form for
the extension of $S$ to curves of bounded variation. In order to do so, we
will make use of the fact that $S$ only depends on the derivatives of the
involved curves. This allows us to reformulate Theorem 1 as a result
concerning the extension of integral functionals from
$L^{1}(I;\mathbb{R}^{d})$ to $\mathcal{M}(I;\mathbb{R}^{d})$, the space of
$\mathbb{R}^{d}$-valued finite Radon measures on $I$. To that end, we define
the functional $F\colon L^{1}(I;\mathbb{R}^{d})^{2}\to\mathbb{R}$,

(5) $F(u,v):=-\int_{I}\Bigl{\langle}\frac{u}{\lvert u\rvert},\frac{v}{\lvert
v\rvert}\Bigr{\rangle}\sqrt{\lvert u\rvert\lvert v\rvert}\,dx,$

and consider its extension $\hat{F}$ to $\mathcal{M}(I;\mathbb{R}^{d})$,
defined as

$\hat{F}(\mu,\nu)=\inf\Bigl{\\{}\liminf_{k}F(u^{(k)},v^{(k)}):u^{(k)}\mathcal{L}^{1}\rightharpoonup^{*}\mu,\
v^{(k)}\mathcal{L}^{1}\rightharpoonup^{*}\nu,\\\ \lVert
u^{(k)}\rVert_{L^{1}}\to\lvert\mu\rvert(I),\ \lVert
v^{(k)}\rVert_{L^{1}}\to\lvert\nu\rvert(I)\Bigr{\\}}$

for $\mu$, $\nu\in\mathcal{M}(I;\mathbb{R}^{d})$. Then

$S(c_{1},c_{2})=-F(\dot{c}_{1},\dot{c}_{2})$

for all $c_{1}$, $c_{2}\in\operatorname{AC}(I;\mathbb{R}^{d})$. Moreover, due
to the definition of strict convergence on
$\operatorname{BV}(I;\mathbb{R}^{d})$ and since we can, up to translations,
identify a curve with its derivative, we have that

$\hat{S}(c_{1},c_{2})=-\hat{F}(Dc_{1},Dc_{2})$

for all $c_{1}$, $c_{2}\in\operatorname{BV}(I;\mathbb{R}^{d})$. Thus it is
sufficient to derive an explicit formula for the functional $\hat{F}$.

To do so, we will prove a generalisation of Reshetnyak’s continuity and lower
semi-continuity theorems [17]. These theorems essentially state that a
positively homogeneous integral functional on $\mathcal{M}(I;\mathbb{R}^{d})$
is continuous with respect to strict convergence, and that it is weakly∗ lower
semi-continuous, if and only if the integrand is convex. This result is not
immediately applicable to our situation, as we are dealing with a functional
depending on two measures, and we require lower semi-continuity with respect
to separate strict convergence. Thus, in Section 4.1, we will formulate a
generalisation of Reshetnyak’s continuity theorem that provides a lower bound
for the functional $F$. Then, in Section 4.2, we will show that this lower
bound is actually sharp. Finally, we will conclude the proof of Theorem 1 in
Section 4.3.

###  4.1. Lower semi-continuity of integral functionals

Let $\Omega\subset\mathbb{R}^{n}$ be open and bounded, and let

$f\colon\bar{\Omega}\times\Sigma\times S^{d-1}\times S^{d-1}\to\mathbb{R},$

where $\Sigma$ denotes the one-dimensional unit simplex and $S^{d-1}$ the
$d-1$-dimensional unit sphere. For simplicity, we identify $\Sigma$ with the
interval $[0,1]$. Assume that $f$ is lower semi-continuous and bounded and
that for every $x\in\Omega$ the mappings

$(\xi,\zeta)\mapsto f(x,0,\xi,\zeta)$

and

$(\xi,\zeta)\mapsto f(x,1,\xi,\zeta)$

are constant.

Then we can define the functional
$F\colon\mathcal{M}(\Omega,\mathbb{R}^{d})^{2}\to\mathbb{R}$,

$F(\mu,\nu)=\int_{\Omega}f\Bigl{(}x,\frac{d\lvert\mu\rvert}{d(\lvert\mu\rvert+\lvert\nu\rvert)},\frac{d\mu}{d\lvert\mu\rvert},\frac{d\nu}{d\lvert\nu\rvert}\Bigr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert).$

Since $f(x,\tau,\cdot,\cdot)$ is assumed to be constant for
$\tau\in\\{0,1\\}$, the integrand is independent of the choice of
$d\mu/d\lvert\mu\rvert$ and $d\nu/d\lvert\nu\rvert$ outside of the supports of
$\lvert\mu\rvert$ and $\lvert\nu\rvert$, respectively, and thus the integral
is well-defined.

######  Theorem 7.

Assume that $f$ is lower semi-continuous and bounded and that the mapping

$\tau\mapsto f(x,\tau,\xi,\zeta)$

is convex for every $x\in\Omega$ and $\xi$, $\zeta\in S^{d-1}$. Then $F$ is
lower semi-continuous with respect to strict convergence in both components.
That is, assume that $\mu_{k}\rightharpoonup^{*}\mu$ and
$\nu_{k}\rightharpoonup^{*}\nu$ in $\mathcal{M}(\Omega;\mathbb{R}^{d})$ such
that $\lvert\mu_{k}\rvert(\Omega)\to\lvert\mu\rvert(\Omega)$ and
$\lvert\nu_{k}\rvert(\Omega)\to\lvert\nu\rvert(\Omega)$. Then

$F(\mu,\nu)\leq\liminf_{k}F(\mu_{k},\nu_{k}).$

###### Proof.

We follow the proof of Reshetnyak’s continuity theorem as presented in [18,
Thm. 10.3] (see also [1, Thm. 2.38, 2.39]).

For simplicity, we write $m_{k}=d\mu_{k}/d\lvert\mu_{k}\rvert$,
$n_{k}=d\nu_{k}/d\lvert\nu_{k}\rvert$,
$t_{k}=d\lvert\mu_{k}\rvert/d(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)$, and
similarly $m=d\mu/d\lvert\mu\rvert$, $n=d\nu/d\lvert\nu\rvert$,
$t=d\lvert\mu\rvert/d(\lvert\mu\rvert+\lvert\nu\rvert)$. Next we define the
measures $\sigma_{k}$ on $\Omega\times\Sigma\times S^{d-1}\times S^{d-1}$ by

$\sigma_{k}=(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)\otimes(\delta_{t_{k}(x)}\otimes\delta_{m_{k}(x)}\otimes\delta_{n_{k}(x)}),$

that is,

$\int_{\Omega\times\Sigma\times S^{d-1}\times
S^{d-1}}\varphi(x,\tau,\xi,\zeta)\,d\sigma_{k}=\int_{\Omega}\varphi\bigl{(}x,t_{k}(x),m_{k}(x),n_{k}(x)\bigr{)}\,d(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)$

for every $\varphi\in C_{0}(\Omega\times\Sigma\times S^{d-1}\times S^{d-1})$.
Since the measures $\lvert\mu_{k}\rvert$ and $\lvert\nu_{k}\rvert$ are
uniformly bounded, it follows that the sequence $\sigma_{k}$ is bounded as
well. After possibly passing to a subsequence, we may assume without loss of
generality that $\sigma_{k}\rightharpoonup^{*}\sigma$ for some
$\sigma\in\mathcal{M}(\Omega\times\Sigma\times S^{d-1}\times S^{d-1})$.

Denoting by $\pi\colon\Omega\times\Sigma\times S^{d-1}\times S^{d-1}\to\Omega$
the projection onto the first component, we obtain that
$\pi_{\\#}\sigma_{k}=\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert$. Since $\mu_{k}$
and $\nu_{k}$ converge strictly to $\mu$ and $\nu$, it follows that
$\lvert\mu_{k}\rvert$ and $\lvert\nu_{k}\rvert$ converge weakly∗ to
$\lvert\mu\rvert$ and $\lvert\nu\rvert$ (see [18, Cor. 10.2]). Thus
$\pi_{\\#}\sigma_{k}\rightharpoonup^{*}\pi_{\\#}\sigma=\lvert\mu\rvert+\lvert\nu\rvert$.

Now (see [18, Thm. 4.4]) there exists a weakly∗ measurable family
$\rho_{x}\in\mathcal{M}(\Sigma\times S^{d-1}\times S^{d-1})$ such that
$\rho_{x}(\Sigma\times S^{d-1}\times S^{d-1})=1$ and
$\sigma=(\lvert\mu\rvert+\lvert\nu\rvert)\otimes\rho_{x}$.

Let now $\psi\in C_{0}(\Omega)$ be continuous and define $\varphi\in
C_{0}(\Omega\times\Sigma\times S^{d-1}\times S^{d-1};\mathbb{R}^{d})$,

$\varphi(x,\tau,\xi,\zeta)=\psi(x)\tau\xi.$

Then

$\int_{\Omega}\psi(x)\biggl{(}\int_{\Sigma\times S^{d-1}\times S^{d-1}}\tau\xi
d\rho_{x}(\tau,\xi,\zeta)\biggr{)}d(\lvert\mu\rvert+\lvert\nu\rvert)\\\
\begin{aligned} &=\int_{\Omega\times\Sigma\times S^{d-1}\times
S^{d-1}}\varphi(x,\tau,\xi,\zeta)\,d\sigma\\\
&=\lim_{k}\int_{\Omega\times\Sigma\times S^{d-1}\times
S^{d-1}}\varphi(x,\tau,\xi,\zeta)\,d\sigma_{k}\\\
&=\lim_{k}\int_{\Omega}\varphi\bigl{(}x,t_{k}(x),m_{k}(x),n_{k}(x)\bigr{)}\,d(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)\\\
&=\lim_{k}\int_{\Omega}\psi(x)t_{k}(x)m_{k}(x)\,d(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)\\\
&=\lim_{k}\int_{\Omega}\psi(x)\,d\mu_{k}(x)\\\
&=\int_{\Omega}\psi(x)\,d\mu(x)\\\
&=\int_{\Omega}\psi(x)t(x)m(x)\,d(\lvert\mu\rvert+\lvert\nu\rvert).\end{aligned}$

Since $\psi\in C_{0}(\Omega)$ was arbitrary, we obtain that

$\int_{\Sigma\times S^{d-1}\times
S^{d-1}}\tau\xi\,d\rho_{x}(\tau,\xi,\zeta)=t(x)m(x)$

for $(\lvert\mu\rvert+\lvert\nu\rvert)$-a.e. $x$. Similarly, using a function
$\varphi(x,\tau,\xi,\zeta)=\psi(x)(1-\tau)\zeta$, one obtains that

$\int_{\Sigma\times S^{d-1}\times S^{d-1}}(1-\tau)\zeta
d\rho_{x}(\tau,\xi,\zeta)=(1-t(x))n(x)$

for $(\lvert\mu\rvert+\lvert\nu\rvert)$-a.e. $x$. Moreover, using a function
$\varphi(x,\tau,\xi,\zeta)=\psi(x)\tau$ and recalling that
$\lvert\mu_{k}\rvert\rightharpoonup^{*}\lvert\mu\rvert$, we obtain that

$\int_{\Sigma\times S^{d-1}\times S^{d-1}}\tau d\rho_{x}(\tau,\xi,\zeta)=t(x)$

for $(\lvert\mu\rvert+\lvert\nu\rvert)$-a.e. $x$.

In particular, we obtain that

$\frac{1}{2}\int_{\Sigma\times S^{d-1}\times
S^{d-1}}\tau\lvert\xi-m(x)\rvert^{2}+(1-\tau)\lvert\zeta-n(x)\rvert^{2}\,d\rho_{x}(\tau,\xi,\zeta)\\\
\begin{aligned} &=\int_{\Sigma\times S^{d-1}\times
S^{d-1}}\tau(1-\langle\xi,m(x)\rangle)+(1-\tau)(1-\langle\zeta,n(x)\rangle\,d\rho_{x}(\tau,\xi,\zeta)\\\
&=1-\langle t(x)m(x),m(x)\rangle-\langle(1-t(x))n(x),n(x)\rangle\\\
&=0.\end{aligned}$

This shows that the measure $\rho_{x}$ is concentrated on the set

$\bigl{(}(0,1)\times\\{m(x)\\}\times\\{n(x)\\}\bigr{)}\cup\bigl{(}\\{0,1\\}\times
S^{d-1}\times S^{d-1}\bigr{)}.$

As a consequence, we can write

$\rho_{x}=T_{x}\otimes\delta_{m(x)}\otimes\delta_{n(x)}+\delta_{0}\otimes
A_{x}+\delta_{1}\otimes B_{x},$

where $T_{x}\in\mathcal{M}(\Sigma)$ satisfies $T_{x}(\\{0,1\\})=0$, and
$A_{x}$, $B_{x}\in\mathcal{M}(S^{d-1}\times S^{d-1})$. Note moreover that

$T_{x}(\Sigma)+A_{x}(S^{d-1}\times S^{d-1})+B_{x}(S^{d-1}\times
S^{d-1})=\rho_{x}(\Sigma\times S^{d-1}\times S^{d-1})=1.$

Since $f$ is lower semi-continuous and bounded, and
$\sigma_{k}\rightharpoonup^{*}\sigma$, we now obtain that

$\displaystyle\liminf_{k}F(\mu_{k},\nu_{k})$
$\displaystyle=\liminf_{k}\int_{\Omega}f\bigl{(}x,t_{k}(x),m_{k}(x),n_{k}(x)\bigr{)}\,d(\lvert\mu_{k}\rvert+\lvert\nu_{k}\rvert)$
$\displaystyle=\liminf_{k}\int_{\Omega\times\Sigma\times S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\sigma_{k}$
$\displaystyle\geq\int_{\Omega\times\Sigma\times S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\sigma$
$\displaystyle=\int_{\Omega}\biggl{(}\int_{\Sigma\times S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\rho_{x}(\tau,\xi,\zeta)\biggr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)$
$\displaystyle=\int_{\Omega}\biggl{(}\int_{\Sigma}f(x,\tau,m(x),n(x))\,dT_{x}(\tau)\biggr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)$
$\displaystyle\qquad{}+\int_{\Omega}\biggl{(}\int_{S^{d-1}\times
S^{d-1}}f(x,0,\xi,\zeta)\,dA_{x}(\xi,\zeta)\biggr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)$
$\displaystyle\qquad{}+\int_{\Omega}\biggl{(}\int_{S^{d-1}\times
S^{d-1}}f(x,1,\xi,\zeta)\,dB_{x}(\xi,\zeta)\biggr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert).$

Next we use that $f(x,0,\xi,\zeta)$ and $f(x,1,\xi,\zeta)$ are constant and
obtain that

$\displaystyle\int_{S^{d-1}\times S^{d-1}}f(x,0,\xi,\zeta)\,dA_{x}(\xi,\zeta)$
$\displaystyle=f(x,0,m(x),n(x))\,A_{x}(S^{d-1}\times S^{d-1}),$
$\displaystyle\int_{S^{d-1}\times S^{d-1}}f(x,1,\xi,\zeta)\,dB_{x}(\xi,\zeta)$
$\displaystyle=f(x,1,m(x),n(x))\,B_{x}(S^{d-1}\times S^{d-1}).$

Moreover, as the mapping $\tau\mapsto f(x,\tau,\xi,\zeta)$ is convex, we can
use Jensen’s inequality and estimate

$\int_{\Sigma}f(x,\tau,m(x),n(x))\,dT_{x}(\tau)\geq
T_{x}(\Sigma)f\biggl{(}x,\frac{1}{T_{x}(\Sigma)}\int_{\Sigma}\tau\,dT_{x}(\tau),\xi,\zeta\biggr{)}.$

Thus we see that

$\int_{\Sigma\times S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\rho_{x}(\tau,\xi,\zeta)\geq A_{x}(S^{d-1}\times
S^{d-1})f(x,0,m(x),n(x))\\\ +B_{x}(S^{d-1}\times
S^{d-1})f(x,1,m(x),n(x))+T_{x}(\Sigma)f\biggl{(}x,\frac{1}{T_{x}(\Sigma)}\int_{\Sigma}\tau\,dT_{x}(\tau),\xi,\zeta\biggr{)}.$

Now recall that

$A_{x}(S^{d-1}\times S^{d-1})+B_{x}(S^{d-1}\times S^{d-1})+T_{x}(\Sigma)=1$

and

$0\cdot A_{x}(S^{d-1}\times S^{d-1})+1\cdot B_{x}(S^{d-1}\times
S^{d-1})+T_{x}(\Sigma)\frac{1}{T_{x}(\Sigma)}\int_{\Sigma}\tau\,dT_{x}(\tau)\\\
=\int_{\Sigma\times S^{d-1}\times S^{d-1}}\tau d\rho_{x}(\tau,\xi,\zeta)=m(x)$

for $(\lvert\mu\rvert+\lvert\nu\rvert)$-a.e. $x$. Thus we can use the
convexity of $f$ w.r.t. $\tau$ and further estimate

$\int_{\Sigma\times S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\rho_{x}(\tau,\xi,\zeta)\geq
f\bigl{(}x,t(x),m(x),n(x)\bigr{)}.$

Combining these estimates, we see that

$\liminf_{k}F(\mu_{k},\nu_{k})\geq\int_{\Omega}\biggl{(}\int_{\Sigma\times
S^{d-1}\times
S^{d-1}}f(x,\tau,\xi,\zeta)\,d\rho_{x}(\tau,\xi,\zeta)\biggr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)\\\
\geq\int_{\Omega}f\bigl{(}x,t(x),m(x),n(x)\bigr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)=F(\mu,\nu),$

which concludes the proof. ∎

###  4.2. Relaxation of integral functionals

Let again $\Omega\subset\mathbb{R}^{n}$ and let
$f\colon\bar{\Omega}\times\Sigma\times S^{d-1}\times S^{d-1}\to\mathbb{R}$ be
such that the mappings $(\xi,\zeta)\mapsto f(x,0,\xi,\zeta)$ and
$(\xi,\zeta)\mapsto f(x,1,\xi,\zeta)$ are constant for every $x$. Consider
moreover the functional $F\colon
L^{1}(\Omega;\mathbb{R}^{d})^{2}\to\mathbb{R}$,

$F(u,v)=\int_{\Omega}f\Bigl{(}x,\frac{\lvert u\rvert}{\lvert u\rvert+\lvert
v\rvert},\frac{u}{\lvert u\rvert},\frac{v}{\lvert v\rvert}\Bigr{)}(\lvert
u\rvert+\lvert v\rvert)\,dx.$

In the following result, we will compute the lower semi-continuous extension
of $F$ to $\mathcal{M}(I;\mathbb{R}^{d})$ with respect to strict convergence
of measures.

######  Theorem 8.

Assume that $f\colon\bar{\Omega}\times\Sigma\times S^{d-1}\times
S^{d-1}\to\mathbb{R}$ is continuous and bounded. Define

$\hat{F}(\mu,\nu):=\inf\Bigl{\\{}\liminf_{k}F(u_{k},v_{k}):u_{k}\mathcal{L}^{n}\rightharpoonup^{*}\mu,\,v_{k}\mathcal{L}^{n}\rightharpoonup^{*}\nu,\\\
\lVert u_{k}\rVert_{L^{1}}\to\lvert\mu\rvert(\Omega),\ \lVert
v_{k}\rVert_{L^{1}}\to\lvert\nu\rvert(\Omega)\Bigr{\\}}.$

Then

$\hat{F}(\mu,\nu)=F_{c}(\mu,\nu):=\int_{\Omega}f_{c}\Bigl{(}x,\frac{d\lvert\mu\rvert}{d(\lvert\mu\rvert+\lvert\nu\rvert)},\frac{d\mu}{d\lvert\mu\rvert},\frac{d\nu}{d\lvert\nu\rvert}\Bigr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)$

for every $\mu$, $\nu\in\mathcal{M}(\Omega;\mathbb{R}^{d})$, where $f_{c}$
denotes the lower semi-continuous convex hull of $f$ with respect to the
second variable.

###### Proof.

In view of Theorem 7, we see that $F_{c}$ is lower semi-continuous with
respect to strict convergence in both components, which implies that
$F_{c}\leq\hat{F}$. Thus it is enough to find for each $\mu$,
$\nu\in\mathcal{M}(\Omega;\mathbb{R}^{d})$ sequences
$u_{k}\mathcal{L}^{n}\rightharpoonup^{*}\mu$,
$v_{k}\mathcal{L}^{n}\rightharpoonup^{*}\nu$ with $\lVert
u_{k}\rVert_{L^{1}}\to\lvert\mu\rvert(\Omega)$ and $\lVert
v_{k}\rVert_{L^{1}}\to\lvert\nu\rvert(\Omega)$ and $F(u_{k},v_{k})\to
F_{c}(\mu,\nu)$.

Assume first that $\mu=u\mathcal{L}^{n}$, $\nu=v\mathcal{L}^{n}$ with $u$,
$v\in C(\bar{\Omega};\mathbb{R}^{d})$. We can write

$F_{c}(u,v)=\int_{\Omega}f_{c}\Bigl{(}x,\frac{\lvert u\rvert}{\lvert
u\rvert+\lvert v\rvert},\frac{u}{\lvert u\rvert},\frac{v}{\lvert
v\rvert}\Bigr{)}\,(\lvert u\rvert+\lvert v\rvert)\,dx\\\
=\int_{\Omega}\Bigl{(}\alpha(x)f\Bigl{(}x,a(x),\frac{u}{\lvert
u\rvert},\frac{v}{\lvert
v\rvert}\Bigr{)}+\beta(x)f\Bigl{(}x,b(x),\frac{u}{\lvert
u\rvert},\frac{v}{\lvert v\rvert}\Bigr{)}\Bigr{)}(\lvert u\rvert+\lvert
v\rvert)\,dx$

for some $0\leq\alpha(x),\beta(x)\leq 1$ with $\alpha(x)+\beta(x)=1$ and

$\alpha(x)a(x)+\beta(x)b(x)=\frac{\lvert u(x)\rvert}{\lvert u(x)\rvert+\lvert
v(x)\rvert}.$

Now consider for $k\in\mathbb{N}$ the family of cubes
$\hat{Q}_{i}^{k}=\frac{1}{2^{k}}\Pi_{r=1}^{n}[i_{r},i_{r}+1]$,
$i\in\mathbb{Z}^{d}$ and define $Q_{i}^{k}=\hat{Q}_{i}^{k}\cap\Omega$. Then we
obtain finite partitions

$\Omega=\bigcup_{i\in I_{k}}Q_{i}^{k}$

for finite index sets $I_{k}\subset\mathbb{Z}^{n}$. Choose moreover for all
$k$ and $i\in I_{k}$ some $x_{i}^{k}\in Q_{i}^{k}$ and a partition
$Q_{i}^{k}=A_{i}^{k}\dot{\cup}B_{i}^{k}$ with disjoint sets $A_{i}^{k}$,
$B_{i}^{k}$ such that $\lvert A_{i}^{k}\rvert=\alpha(x_{i}^{k})\lvert
Q_{i}^{k}\rvert$ and $\lvert B_{i}^{k}\rvert=\beta(x_{i}^{k})\lvert
Q_{i}^{k}\rvert$.

Define now functions $u_{k}$, $v_{k}$ by

$\displaystyle u_{k}(x)$ $\displaystyle=a(x_{i}^{k})\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
u(x_{i}^{k})\rvert}u(x_{i}^{k}),$ $\displaystyle x\in A_{i}^{k},$
$\displaystyle v_{k}(x)$ $\displaystyle=(1-a(x_{i}^{k}))\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
v(x_{i}^{k})\rvert}v(x_{i}^{k}),$ $\displaystyle x\in A_{i}^{k},$
$\displaystyle u_{k}(x)$ $\displaystyle=b(x_{i}^{k})\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
u(x_{i}^{k})\rvert}u(x_{i}^{k}),$ $\displaystyle x\in B_{i}^{k},$
$\displaystyle v_{k}(x)$ $\displaystyle=(1-b(x_{i}^{k}))\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
v(x_{i}^{k})\rvert}v(x_{i}^{k}),$ $\displaystyle x\in B_{i}^{k}.$

Then

$\frac{\lvert u_{k}(x)\rvert}{\lvert u_{k}(x)\rvert+\lvert
v_{k}(x)\rvert}=\begin{cases}a(x_{i}^{k}),&\text{ if }x\in A_{i}^{k},\\\
b(x_{i}^{k}),&\text{ if }x\in B_{i}^{k}.\end{cases}$

Moreover we have for all $\ell\in\mathbb{N}$ and $k\geq\ell$ that

$\displaystyle\int_{Q_{j}^{\ell}}u_{k}(x)\,dx$
$\displaystyle=\sum_{Q_{i}^{k}\subset Q_{j}^{\ell}}\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
u(x_{i}^{k})\rvert}\bigl{(}a(x_{i}^{k})\lvert
A_{i}^{k}\rvert+b(x_{i}^{k})\lvert B_{i}^{k}\rvert\bigr{)}u(x_{i}^{k})$
$\displaystyle=\sum_{Q_{i}^{k}\subset Q_{j}^{\ell}}\frac{\lvert
u(x_{i}^{k})\rvert+\lvert v(x_{i}^{k})\rvert}{\lvert
u(x_{i}^{k})\rvert}(\alpha(x_{i}^{k})a(x_{i}^{k})+\beta(x_{i}^{k})b(x_{i}^{k}))u(x_{i}^{k})\lvert
Q_{i}^{k}\rvert$ $\displaystyle=\sum_{Q_{i}^{k}\subset
Q_{j}^{\ell}}u(x_{i}^{k})\lvert Q_{i}^{k}\rvert.$

This shows that $u_{k}\mathcal{L}^{n}\rightharpoonup^{*}u\mathcal{L}^{n}$.
Similarly, we obtain that
$v_{k}\mathcal{L}^{n}\rightharpoonup^{*}v\mathcal{L}^{n}$. Also, we have that
$\lVert u_{k}\rVert_{L^{1}}\to\lVert u\rVert_{L^{1}}$ and $\lVert
v_{k}\rVert_{L^{1}}\to\lVert v\rVert_{L^{1}}$.

In addition,

$\displaystyle F(u_{k},v_{k})$ $\displaystyle=\sum_{i\in
I_{k}}\int_{\Omega_{i}^{k}}f\Bigl{(}x,\frac{\lvert u_{k}(x)\rvert}{\lvert
u_{k}(x)\rvert+\lvert v_{k}(x)\rvert},\frac{u_{k}(x)}{\lvert
u_{k}(x)\rvert},\frac{v_{k}(x)}{\lvert v_{k}(x)\rvert}\Bigr{)}(\lvert
u_{k}(x)\rvert+\lvert v_{k}(x)\rvert)\,dx$ $\displaystyle=\sum_{i\in
I_{k}}\int_{A_{i}^{k}}f\Bigl{(}x,a(x_{i}^{k}),\frac{u(x_{i}^{k})}{\lvert
u(x_{i}^{k})\rvert},\frac{v(x_{i}^{k})}{\lvert
v(x_{i}^{k})\rvert}\Bigr{)}(\lvert v(x_{i}^{k})\rvert+\lvert
u(x_{i}^{k})\rvert)\,dx$ $\displaystyle\qquad{}+\sum_{i\in
I_{k}}\int_{B_{i}^{k}}f\Bigl{(}x,b(x_{i}^{k}),\frac{u(x_{i}^{k})}{\lvert
u(x_{i}^{k})\rvert},\frac{v(x_{i}^{k})}{\lvert
v(x_{i}^{k})\rvert}\Bigr{)}(\lvert v(x_{i}^{k})\rvert+\lvert
u(x_{i}^{k})\rvert)\,dx.$

Because of the continuity of $f$ and the fact that $\lvert
A_{i}^{k}\rvert=\alpha(x_{i}^{k})\lvert Q_{i}^{k}\rvert$ and $\lvert
B_{i}^{k}\rvert=\beta(x_{i}^{k})\lvert Q_{i}^{k}\rvert$, it follows that

$\displaystyle\lim_{k}F(u_{k},v_{k})$
$\displaystyle=\int_{\Omega}\alpha(x)f\Bigl{(}x,a(x),\frac{u(x)}{\lvert
u(x)\rvert},\frac{v(x)}{\lvert v(x)\rvert}\Bigr{)}\,(\lvert u(x)\rvert+\lvert
v(x)\rvert)\,dx$
$\displaystyle\qquad{}+\int_{\Omega}\beta(x)f\Bigl{(}x,b(x),\frac{u(x)}{\lvert
u(x)\rvert},\frac{v(x)}{\lvert v(x)\rvert}\Bigr{)}\,(\lvert u(x)\rvert+\lvert
v(x)\rvert)\,dx$ $\displaystyle=F_{c}(u,v).$

This proves the assertion for $u$, $v\in C(\bar{\Omega};\mathbb{R}^{d})$.

Now let $\mu$, $\nu\in\mathcal{M}(\Omega;\mathbb{R}^{d})$ be arbitrary. Then
we can define the regularised measures
$\mu_{\varepsilon}=\varphi_{\varepsilon}\ast\mu$ and
$\nu_{\varepsilon}=\varphi_{\varepsilon}\ast\nu$, where
$\varphi_{\varepsilon}$ is a (scaled) standard mollifier. Then
$\mu_{\varepsilon}\rightharpoonup^{*}\mu$ and
$\nu_{\varepsilon}\rightharpoonup^{*}\nu$. In addition, we have that
$\lvert(\mu_{\varepsilon},\nu_{\varepsilon})\rvert(\Omega)\to\lvert(\mu,\nu)\rvert(\Omega)$
(see [1, Thm. 2.2]). As a consequence, the Reshetnyak continuity theorem [18,
Thm. 10.3] implies that $F_{c}(\mu_{\varepsilon},\nu_{\varepsilon})\to
F_{c}(\mu,\nu)$. Now note that $\mu_{\varepsilon}$ and $\nu_{\varepsilon}$ are
of the form $\mu_{\varepsilon}=u_{\varepsilon}\mathcal{L}^{n}$ and
$\nu_{\varepsilon}=v_{\varepsilon}\mathcal{L}^{n}$ with continuous functions
$u_{\varepsilon}$, $v_{\varepsilon}$. Thus we can use the first part of this
proof to find, for every $\varepsilon>0$, sequences $u_{\varepsilon}^{k}$,
$v_{\varepsilon}^{k}$ with
$u_{\varepsilon}^{k}\mathcal{L}^{n}\rightharpoonup^{*}u_{\varepsilon}\mathcal{L}^{n}$,
$v_{\varepsilon}^{k}\mathcal{L}^{n}\rightharpoonup^{*}v\mathcal{L}^{n}$ and
$\lVert u_{\varepsilon}^{k}\rVert_{L^{1}}\to\lVert
u_{\varepsilon}\rVert_{L^{1}}$, $\lVert
v_{\varepsilon}^{k}\rVert_{L^{1}}\to\lVert v_{\varepsilon}\rVert_{L^{1}}$,
such that $F(u_{\varepsilon}^{k},v_{\varepsilon}^{k})\to
F_{c}(u_{\varepsilon},v_{\varepsilon})$. By choosing an appropriate diagonal
sequence, the claim of the theorem is proven. ∎

###  4.3. Proof of Theorem 1

We will now apply the results of the previous sections to the particular
functional $F\colon L^{1}(I;\mathbb{R}^{d})^{2}\to\mathbb{R}$,

$F(u,v)=-\int_{I}\Bigl{\langle}\frac{u}{\lvert u\rvert},\frac{v}{\lvert
v\rvert}\Bigr{\rangle}\sqrt{\lvert u\rvert\lvert v\rvert}\,dx.$

This can be written in the form required by Theorems 7 and 8 by defining
$f\colon\Sigma\times S^{d-1}\times S^{d-1}\to\mathbb{R}$,

$f(t,\xi,\zeta)=-\langle\xi,\zeta\rangle\sqrt{t(1-t)}.$

Note that we do not have any dependence on the $x$-variable.

According to Theorem 8, we require the convex hull $f_{c}$ of the function $f$
with respect to the $t$ variable for fixed $\xi$ and $\zeta$. However, if
$\langle\xi,\zeta\rangle\geq 0$, then the mapping $t\mapsto f(t,\xi,\zeta)$ is
already convex and thus $f(t,\xi,\zeta)=f_{c}(t,\xi,\zeta)$ in this case.
Conversely, if $\langle\xi,\zeta\rangle<0$, then the convex hull of the
mapping $t\mapsto f(t,\xi,\zeta)$ is the constant function
$f_{c}(t,\xi,\zeta)$. This can be summarised to

$f_{c}(t,\xi,\zeta)=-\langle\xi,\zeta\rangle^{+}\sqrt{t(1-t)}$

for every $(t,\xi,\zeta)\in\Sigma\times S^{d-1}\times S^{d-1}$. As a
consequence, we have that

$\hat{F}(\mu,\nu)=-\int_{I}f_{c}\Bigl{(}\frac{d\lvert\mu\rvert}{d(\lvert\mu\rvert+\lvert\nu\rvert)},\frac{d\mu}{d\lvert\mu\rvert},\frac{d\nu}{d\lvert\nu\rvert}\Bigr{)}\,d(\lvert\mu\rvert+\lvert\nu\rvert)\\\
=\int_{I}\Bigl{\langle}\frac{d\mu}{\lvert d\mu\rvert},\frac{d\nu}{\lvert
d\nu\rvert}\Bigr{\rangle}^{+}\sqrt{\frac{d\lvert\mu\rvert}{d(\lvert\mu\rvert+\lvert\nu\rvert)}\frac{d\lvert\nu\rvert}{d(\lvert\mu\rvert+\lvert\nu\rvert)}}\,d(\lvert\mu\rvert+\lvert\nu\rvert).$

As discussed in the beginning of Section 4, this then implies that

$\hat{S}(c_{1},c_{2})=-\hat{F}(Dc_{1},Dc_{2})\\\
=\int_{I}\Bigl{\langle}\frac{dDc_{1}}{\lvert
dDc_{1}\rvert},\frac{dDc_{2}}{\lvert
dDc_{2}\rvert}\Bigr{\rangle}^{+}\sqrt{\frac{d\lvert Dc_{1}\rvert}{d(\lvert
Dc_{1}\rvert+\lvert Dc_{2}\rvert)}\frac{d\lvert Dc_{2}\rvert}{d(\lvert
Dc_{1}\rvert+\lvert Dc_{2}\rvert)}}\,d(\lvert Dc_{1}\rvert+\lvert
Dc_{2}\rvert),$

which concludes the proof of Theorem 1.

##  5\. The shape distance on $\operatorname{BV}(I;\mathbb{R}^{d})$ and
$\operatorname{SBV}(I;\mathbb{R}^{d})$

###  5.1. Distance between reparametrised BV functions

In this section, we will prove Propositions 3 and 4. To that end, we will need
some results concerning the pointwise convergence of strictly convergent
sequences of $\operatorname{BV}$-functions. The basis for these is the
following result on the properties of strictly convergent measures.

######  Lemma 9.

Let $\mu\in\mathcal{M}(I;\mathbb{R}^{d})$ and assume that
$\mu_{k}\rightharpoonup^{*}\mu$ such that
$\lvert\mu_{k}\rvert(I)\to\lvert\mu\rvert(I)$. If $U\Subset I$ is open with
$\mu(\partial U)=0$, then $\mu_{k}(U)\to\mu(U)$.

###### Proof.

By [18, Cor. 10.2] we obtain that
$\lvert\mu_{k}\rvert\rightharpoonup^{*}\lvert\mu\rvert$. Now we can use [9,
Cor. 1.204] to obtain the assertion. ∎

######  Lemma 10.

Let $c\in\operatorname{BV}(I;\mathbb{R}^{d})$, and assume that the sequence of
functions $c_{k}\in\operatorname{BV}(I;\mathbb{R}^{d})$ converges strictly to
$c$. Then for every $x\in I$ the set of accumulation points of the sequence
$c_{k}(x)$ is contained in the interval $[c^{\ell}(x),c^{r}(x)]$. In
particular, if $c$ is continuous at $x$, then $c_{k}(x)\to c(x)$.

###### Proof.

Let $z\in\mathbb{R}^{d}$ be any accumulation point of the sequence $c_{k}(x)$.
After possibly passing to a subsequence, we may then assume that $c_{k}(x)\to
z$. Assume now that $z\not\in[c^{\ell}(x),c^{r}(x)]$ and denote

$\varepsilon:=\lvert z-c^{\ell}(x)\rvert+\lvert z-c^{r}(x)\rvert-\lvert
c^{\ell}(x)-c^{r}(x)\rvert>0.$

Since $c_{k}\to c$ weakly∗ in $\operatorname{BV}(I;\mathbb{R}^{d})$, it
follows that $c_{k}(y)\to c(y)$ for almost every $y\in I$. We can therefore
choose $x_{0}<x$ and $x_{1}>x$ such that $c_{k}(x_{0})\to c(x_{0})$,
$c_{k}(x_{1})\to c(x_{1})$, all the functions $c$ and $c_{k}$ are continuous
at $x_{0}$ and $x_{1}$, $\lvert Dc\rvert(x_{0},x)+\lvert
Dc\rvert(x,x_{1})<\varepsilon/4$, $\lvert
c(x_{0})-c^{\ell}(x)\rvert<\varepsilon/4$, and $\lvert
c(x_{1})-c^{r}(x)\rvert<\varepsilon/4$. Then

$\lvert Dc\rvert(I)=\lvert Dc\rvert(0,x_{0})+\lvert
Dc\rvert(x_{0},x_{1})+\lvert Dc\rvert(x_{1},1).$

We have that $\lvert Dc\rvert(0,x_{0})=\lim_{k}\lvert Dc_{k}\rvert(0,x_{0})$
and $\lvert Dc_{x}\rvert(x_{1},1)$. Moreover,

$\lvert Dc\rvert(x_{0},x_{1})=\lvert Dc\rvert(x_{0},x)+\lvert
Dc\rvert(x,x_{1})+\lvert c^{\ell}(x)-c^{r}(x)\rvert\\\ <\lvert
z-c^{\ell}(x)\rvert+\lvert z-c^{r}(x)\rvert-\frac{3\varepsilon}{4}<\lvert
z-c(x_{0})\rvert+\lvert z-c(x_{1})\rvert-\frac{\varepsilon}{4}.$

However, we have that

$\liminf_{k}\lvert Dc_{k}\rvert(x_{0},x_{1})\geq\liminf_{k}\lvert
c_{k}(x_{0})-c_{k}(x)\rvert+\lvert c_{k}(x)-c_{k}(x_{1})\rvert\\\ =\lvert
z-c(x_{0})\rvert+\lvert z-c(x_{1})\rvert>\lvert
Dc\rvert(x_{0},x_{1})+\frac{\varepsilon}{4}.$

Combining the results above, we obtain that $\liminf_{k}\lvert
Dc_{k}\rvert(I)>\lvert Dc\rvert(I)+\varepsilon/4$, which contradicts the
assumption that $c_{k}$ converges strictly to $c$. ∎

######  Lemma 11.

Let $c\in\operatorname{BV}(I;\mathbb{R}^{d})$, and assume that the sequence of
functions $c_{k}\in\operatorname{BV}(I;\mathbb{R}^{d})$ converges strictly to
$c$. Assume moreover that $x\in I$ is such that $c$ is continuous at $x$. Then
we have for every sequence $x_{k}\to x$ that $c_{k}(x_{k})\to c(x)$.

###### Proof.

We can estimate

$\lvert c_{k}(x_{k})-c(x)\rvert\leq\lvert c_{k}(x_{k})-c_{k}(x)\rvert+\lvert
c_{k}(x)-c(x)\rvert.$

The function $c$ is continuous at $x$ and thus, in view of Lemma 10, the last
term tends to zero as $k\to\infty$. For the first term, we can estimate

$\lvert c_{k}(x_{k})-c_{k}(x)\rvert\leq\lvert Dc_{k}\rvert([x_{k},x]).$

Let now $\varepsilon>0$ be such that $\lvert
Dc\rvert(\\{x-\varepsilon,x+\varepsilon\\})=0$. Then we have for sufficiently
large $k$ that $[x_{k},x]\subset(x-\varepsilon,x+\varepsilon)$ and thus

$\limsup_{k}\lvert Dc_{k}\rvert([x_{k},x])\leq\limsup_{k}\lvert
Dc_{k}\rvert(x-\varepsilon,x+\varepsilon)=\lvert
Dc\rvert(x-\varepsilon,x+\varepsilon).$

Since this holds for almost every $\varepsilon>0$ we obtain that

$\limsup_{k}\lvert Dc_{k}\rvert([x_{k},x])\leq\lvert Dc\rvert(\\{x\\})=0.$

Combining all the estimates, we arrive at the claim. ∎

We now are able to prove Proposition 3.

###### Proof of Proposition 3.

Let $c_{i}\in\operatorname{BV}(I;\mathbb{R}^{d})$ and
$\varphi_{i}\in\bar{\Gamma}$ be fixed.

Assume that
$\\{c_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{AC}(I;\mathbb{R}^{d})$
converge strictly to $c_{i}$ and that
$\\{\varphi_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\Gamma$ converge uniformly to
$\varphi_{i}$. Since
$\operatorname{len}(c_{i}^{(k)}\circ\varphi_{i}^{(k)})=\operatorname{len}(c_{i}^{(k)})\to\operatorname{len}(c)$,
the functions $c_{i}^{(k)}\circ\varphi_{i}^{(k)}$ are uniformly bounded in
$\operatorname{BV}(I;\mathbb{R}^{d})$. After possibly passing to a
subsequence, we may therefore assume that
$c_{i}^{(k)}\circ\varphi_{i}^{(k)}\rightharpoonup^{*}g_{i}$ for some
$g_{i}\in\operatorname{BV}(I;\mathbb{R}^{d})$. Now let $\varepsilon>0$. Then
there exist $0<x_{1}<\ldots<x_{N}<1$ such that $c_{1}$ is continuous at
$\varphi_{1}(x_{\ell})$ for each $1\leq\ell\leq N$, and

$\operatorname{len}(c_{1})\leq\sum_{\ell=1}^{N-1}\lvert
c_{1}(\varphi_{1}(x_{\ell+1}))-c_{1}(\varphi_{1}(x_{\ell}))\rvert+\varepsilon.$

In addition, we can choose the points $x_{\ell}$ in such a way that
$c_{1}^{(k)}(\varphi_{1}^{(k)}(x_{\ell}))\to g_{1}(x_{\ell})$. Since
$\varphi_{1}^{(k)}(x_{\ell})\to\varphi_{1}(x_{\ell})$ for all $\ell$ and
$c_{1}^{(k)}\to^{s}c_{1}$, we obtain from Lemma 11 that

$\operatorname{len}(g_{1})\geq\sum_{\ell=1}^{N-1}\lvert
g_{1}(x_{\ell+1})-g_{1}(x_{\ell})\rvert=\lim_{k}\sum_{\ell=1}^{N-1}\lvert
c_{1}^{(k)}(\varphi_{1}^{(k)}(x_{\ell+1}))-c_{1}^{(k)}(\varphi_{1}^{(k)}(x_{\ell}))\rvert\\\
=\sum_{\ell=1}^{N-1}\lvert
c_{1}(\varphi_{1}(x_{\ell+1}))-c_{1}(\varphi_{1}(x_{\ell}))\rvert\geq\operatorname{len}(c_{1})-\varepsilon.$

Since $\varepsilon$ was arbitrary, this shows that
$\operatorname{len}(g_{1})\geq\operatorname{len}(c_{1})=\lim_{k}\operatorname{len}(c_{1}^{(k)}\circ\varphi_{1}^{(k)})$,
which in turn shows that, actually, $c_{1}^{(k)}\circ\varphi_{1}^{(k)}$
converges strictly to $g_{1}$. Similarly, we obtain that
$c_{2}^{(k)}\circ\varphi_{2}^{(k)}\to^{s}g_{2}$. Because $\hat{d}$ is strictly
lower semi-continuous, it follows that

$\hat{d}(g_{1},g_{2})\leq\liminf_{k\in\mathbb{N}}\hat{d}(c_{1}^{(k)}\circ\varphi_{1}^{(k)},c_{2}^{(k)}\circ\varphi_{2}^{(k)})\leq\liminf_{k\in\mathbb{N}}D(c_{1}^{(k)},c_{2}^{(k)};\varphi_{1}^{(k)},\varphi_{2}^{(k)}).$

Next note that it follows from Lemma 10 that $g_{i}\in[c_{i},\varphi_{i}]$,
and thus

$\inf_{g_{i}\in[c_{i},\varphi_{i}]}\hat{d}(g_{1},g_{2})\leq\liminf_{k\in\mathbb{N}}D(c_{1}^{(k)},c_{2}^{(k)};\varphi_{1}^{(k)},\varphi_{2}^{(k)}).$

Since the sequences $\\{c_{i}^{(k)}\\}_{k\in\mathbb{N}}$ and
$\\{\varphi_{i}^{(k)}\\}_{k\in\mathbb{N}}$ converging to $c_{i}$ and
$\varphi_{i}$, respectively, were arbitrary, and $\hat{D}$ is the lower semi-
continuous hull of $D$, it follows that

$\inf_{g_{i}\in[c_{i},\varphi_{i}]}\hat{d}(g_{1},g_{2})\leq\hat{D}(c_{1},c_{2};\varphi_{1},\varphi_{2}).$

Now assume that $g_{i}\in[c_{i},\varphi_{i}]$ and that
$\\{g_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{AC}(I;\mathbb{R}^{d})$
converge strictly to $g_{i}$. Define

$\varphi_{i}^{(k)}=\frac{1}{k}\operatorname{Id}+\Bigl{(}1-\frac{1}{k}\Bigr{)}\varphi_{i},$

and let

$c_{i}^{(k)}:=g_{i}^{(k)}\circ(\varphi_{i}^{(k)})^{-1}.$

Then $\varphi_{i}^{(k)}\in\Gamma$,
$c_{i}^{(k)}\in\operatorname{AC}(I;\mathbb{R}^{d})$, and
$\varphi_{i}^{(k)}\to\varphi$ uniformly. Moreover, by definition of
$c_{i}^{(k)}$, and since
$\operatorname{len}(g_{i})=\operatorname{len}(c_{i})$, we have that
$\operatorname{len}(c_{i}^{(k)})=\operatorname{len}(g_{i}^{(k)})\to\operatorname{len}(g_{i})=\operatorname{len}(c_{i})$.
Next, we show that $c_{1}^{(k)}(y)\to c_{1}(y)$ at every point $y$ where
$c_{1}$ is continuous.

To that end, let $y$ be such that $c_{1}^{\ell}(y)=c_{1}^{r}(y)$. Then
$c_{1}^{(k)}(y)=g_{1}^{(k)}((\varphi_{1}^{(k)})^{-1}(y))$. Since the sequence
$(\varphi_{1}^{(k)})^{-1}(y)$ is bounded, it has a convergent subsequence, say
$(\varphi_{1}^{(k^{\prime})})^{-1}(y)\to z$. Since
$\varphi_{1}^{(k^{\prime})}$ converges pointwise to $\varphi_{1}$, it follows
that $y=\varphi_{1}(z)$. As a consequence, we have that
$g_{1}(z)\in[c_{1}^{\ell}(\varphi_{1}(z)),c_{1}^{r}(\varphi_{1}(z))]=[c^{\ell}(y),c_{1}^{r}(y)]=\\{c_{1}(y)\\}$,
which implies in particular that $g_{1}$ is continuous at $z$. Since
$g_{1}^{(k^{\prime})}\to g_{1}$ and $(\varphi_{1}^{(k^{\prime})})^{-1}(y)\to
z$, it follows from Lemma 11 that

$c_{1}^{(k^{\prime})}(y)=g_{1}^{(k^{\prime})}((\varphi_{1}^{(k^{\prime})})^{-1}(y))\to
g_{1}(z)=c_{1}(y).$

Since this holds for every convergent subsequence, we now obtain that
$c_{1}^{(k)}(y)\to c_{1}(y)$. Together with the convergence
$\operatorname{len}(c_{1}^{(k)})\to\operatorname{len}(c_{1})$, this implies
that $c_{1}^{(k)}\to^{s}c_{1}$. Similarly we obtain that
$c_{2}^{(k)}\to^{s}c_{2}$.

Since $\hat{D}$ is the lower semi-continuous hull of $D$ with respect to
strict convergence in the $c_{i}$ components and uniform convergence in the
$\varphi_{i}$ components, it follows that

$\hat{D}(c_{1},c_{2};\varphi_{1},\varphi_{2})\leq\liminf_{k}D(c_{1}^{(k)},c_{2}^{(k)};\varphi_{1}^{(k)},\varphi_{2}^{(k)})\\\
=\liminf_{k}d(c_{1}^{(k)}\circ\varphi_{1}^{(k)},c_{2}^{(k)}\circ\varphi_{2}^{(k)})=\liminf_{k}d(g_{1}^{(k)},g_{2}^{(k)}).$

Since $g_{i}\in[c_{i},\varphi_{i}]$ were arbitrary, and
$\\{g_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{AC}(I;\mathbb{R}^{d})$
were arbitrary sequences converging strictly to $g_{i}$ it follows from the
definition of $\hat{d}$ that

$\hat{D}(c_{1},c_{2};\varphi_{1},\varphi_{2})\leq\inf_{g_{i}\in[c_{i},\varphi_{i}]}\inf_{\begin{subarray}{c}\\{g_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{AC}(I;\mathbb{R}^{d})\\\
g_{i}^{(k)}\to^{s}g_{i}\end{subarray}}d(g_{1}^{(k)},g_{2}^{(k)})=\inf_{g_{i}\in[c_{i},\varphi_{i}]}\hat{d}(g_{1},g_{2}),$

which concludes the proof. ∎

###### Proof of Proposition 4.

Define

$\psi^{(k)}:=\frac{1}{k}\operatorname{Id}+\Bigl{(}1-\frac{1}{k}\Bigr{)}\psi.$

Then $\lVert\psi^{(k)}-\psi\rVert_{\infty}\leq 2/k$. Moreover, $\psi^{(k)}$
satisfies $(\psi^{(k)})^{\prime}\geq 1/k$ almost everywhere, and thus
$\psi^{(k)}$ is invertible with inverse
$\vartheta^{(k)}:=(\psi^{(k)})^{-1}\in\Gamma$.

Now assume that the sequences
$\\{c_{i}^{(k)}\\}_{k\in\mathbb{N}}\subset\operatorname{AC}(I;\mathbb{R}^{d})$
converge strictly to $c_{i}$ and that
$(\varphi_{i}^{(k)})_{k\in\mathbb{N}}\subset\Gamma$ converge uniformly to
$\varphi_{i}$. Then

$\lVert\varphi_{i}^{(k)}\circ\psi^{(k)}-\varphi_{i}\circ\psi\rVert_{\infty}\\\
\leq\lVert\varphi_{i}^{(k)}\circ\psi^{(k)}-\varphi_{i}\circ\psi^{(k)}\rVert_{\infty}+\lVert\varphi_{i}\circ\psi^{(k)}-\varphi_{i}\circ\psi\rVert_{\infty}.$

Now the first term on the right hand side converges to zero because of the
uniform convergence of $\varphi_{i}^{(k)}$ to $\varphi_{i}$, and the second
term converges to zero because of the uniform continuity of $\varphi_{i}$ and
the uniform convergence of $\psi^{(k)}\to\psi$. Thus
$\varphi_{i}^{(k)}\circ\psi^{(k)}$ converges uniformly to
$\varphi_{i}\circ\psi$. Since $D$ is lower semi-continous with respect to
strict convergence in the $c_{i}$ variables and uniform convergence in the
$\varphi_{i}$ variables, and the functional $d$ is invariant under
simultaneous reparametrisations, it follows that

$\hat{D}(c_{1},c_{2};\varphi_{1}\circ\psi,\varphi_{2}\circ\psi)\leq\liminf_{k}d(c_{1}^{(k)}\circ\varphi_{1}^{(k)}\circ\psi^{(k)},c_{2}^{(k)}\circ\varphi_{2}^{(k)}\circ\psi^{k})\\\
=\liminf_{k}d(c_{1}^{(k)}\circ\varphi_{1}^{(k)},c_{2}^{(k)}\circ\varphi_{2}^{(k)}).$

Since this holds for all such sequences $\\{c_{i}^{(k)}\\}_{k\in\mathbb{N}}$
and $\\{\varphi_{i}^{(k)}\\}_{k\in\mathbb{N}}$, it follows that

$\hat{D}(c_{1},c_{2};\varphi_{1}\circ\psi,\varphi_{2}\circ\psi)\leq\hat{D}(c_{1},c_{2};\varphi_{1},\varphi_{2}).$

Conversely, assume that $\gamma_{i}^{(k)}\to\gamma_{i}:=\varphi_{i}\circ\psi$
uniformly. Then we can similarly estimate

(6)
$\lVert\gamma_{i}^{(k)}\circ\vartheta^{(k)}-\varphi_{i}\rVert_{\infty}\leq\lVert\gamma_{i}^{(k)}\circ\vartheta^{(k)}-\gamma_{i}\circ\vartheta^{(k)}\rVert_{\infty}+\lVert\gamma_{i}\circ\vartheta^{(k)}-\varphi_{i}\rVert_{\infty}.$

Again, the first term converges to zero because of the uniform convergence of
$\gamma_{i}^{(k)}$ to $\gamma_{i}$. For the second term, let $x\in I$ and let
$y\in I$ with $\psi(y)=x$. Then

$\lvert\gamma_{i}(\vartheta^{(k)}(x))-\varphi_{i}(x)\rvert=\lvert\varphi_{i}(\psi(\vartheta^{(k)}(\psi(y))))-\varphi_{i}(\psi(y))\rvert$

Since $\lVert\psi-\psi^{(k)}\lVert\leq 2/k$, there exists $\varepsilon_{k}$
with $\lvert\varepsilon_{k}\rvert\leq 2/k$ such that

$\psi(\vartheta^{(k)}(\psi(y)))=\psi^{(k)}(\vartheta^{(k)}(\psi(y)))+\varepsilon_{k}=\psi(y)+\varepsilon_{k}.$

Thus we have that

$\lvert\gamma_{i}(\vartheta^{(k)}(x))-\varphi_{i}(x)\rvert=\lvert\varphi_{i}(\psi(y)+\varepsilon_{k})-\varphi_{i}(\psi(y))\rvert.$

Because of the uniform continuity of $\varphi_{i}$, this implies that the
second term in (6) converges to zero, and therefore
$\lVert\gamma_{i}^{(k)}\circ\vartheta^{(k)}-\varphi_{i}\rVert_{\infty}\to 0$.
With the same argumentation as before, this now implies that

$\hat{D}(c_{1},c_{2},\varphi_{1},\varphi_{2})\leq\hat{D}(c_{1},c_{2},\varphi_{1}\circ\psi,\varphi_{2}\circ\psi).$

With Proposition 3, we now arrive at the assertion. ∎

###  5.2. Proof of Theorem 6

We start with showing that the infima in (4) are attained.

######  Proposition 12.

Assume that $c_{1}$, $c_{2}\in\operatorname{BV}(I;\mathbb{R}^{d})$. Then the
optimisation problem

$\inf_{(\varphi_{1},\varphi_{2})\in\bar{\Gamma}}\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])$

admits a solution $(\bar{\varphi}_{1},\bar{\varphi}_{2})$. Moreover, there
exist $\bar{g}_{i}\in[c_{i},\bar{\varphi}_{i}]$ such that

$\hat{d}([c_{1},\bar{\varphi}_{1}],[c_{2},\bar{\varphi}_{2}])=\hat{d}(\bar{g}_{1},\bar{g}_{2}).$

###### Proof.

We follow the proof of [5, Proposition 15].

Assume that $(\varphi_{1}^{(k)},\varphi_{2}^{(k)})_{k\in\mathbb{N}}$ is a
minimising sequence for $\hat{d}([c_{1},\cdot],[c_{2},\cdot])$. Then there
exist $\psi^{(k)}\in\bar{\Gamma}$ such that
$\varphi_{i}^{(k)}=\gamma_{i}^{(k)}\circ\psi^{(k)}$, where
$\gamma_{i}^{(k)}\in\bar{\Gamma}$ are such that
$(\gamma_{1}^{(k)})^{\prime}(x)+(\gamma_{2}^{(k)})^{\prime}(x)=2$ for a.e.
$x\in I$. By Proposition 4 we have that

$\hat{d}([c_{1},\varphi_{1}^{(k)}],[c_{2},\varphi_{2}^{(k)}])=\hat{d}([c_{1},\gamma_{1}^{(k)}\circ\psi^{(k)}],[c_{2},\gamma_{2}^{(k)}\circ\psi^{(k)}])=\hat{d}([c_{1},\gamma_{1}^{(k)}],[c_{2},\gamma_{2}^{(k)}]).$

Thus $(\gamma_{1}^{(k)},\gamma_{2}^{(k)})$ is a minimising sequence as well.
After replacing $\varphi_{i}^{(k)}$ by $\gamma_{i}^{(k)}$, we can thus assume
without loss of generality that all the functions $\varphi_{i}^{(k)}$ are
Lipschitz continuous with Lipschitz constant at most $2$. After possibly
passing to a sub-sequence, we may therefore assume without loss of generality
that $\varphi_{i}^{(k)}\to\bar{\varphi}_{i}$ for some
$\bar{\varphi}_{i}\in\bar{\Gamma}$. Since by construction the functional
$(\varphi_{1},\varphi_{2})\mapsto\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])$
is lower semi-continuous with respect to uniform convergence, it follows that
$(\bar{\varphi}_{1},\bar{\varphi}_{2})$ is a minimiser of
$\hat{d}([c_{1},\cdot],[c_{2},\cdot])$.

Now, by definition,

$\hat{d}([c_{1},\bar{\varphi}_{1}],[c_{1},\bar{\varphi}_{2}])=\inf_{g_{i}\in[c_{i},\bar{\varphi}_{i}]}\hat{S}(g_{1},g_{2}).$

Since $\hat{S}$ is strictly lower semi-continuous and the sets
$[c_{i},\bar{\varphi}_{i}]$ are compact with respect to strict convergence,
the existence of $\bar{g}_{1}$ and $\bar{g}_{2}$ follows. ∎

Next we will show that the functions $\bar{g}_{i}$ in Proposition 12 can be
chosen in $\operatorname{SBV}(I;\mathbb{R}^{d})$, if
$c_{i}\in\operatorname{SBV}(I;\mathbb{R}^{d})$. For that, we need two
preparatory results.

######  Lemma 13.

Assume that $c\in\operatorname{SBV}(I;\mathbb{R}^{d})$, that
$\varphi\in\bar{\Gamma}$, and that $g\in[c,\varphi]$. Then the singular part
$D^{s}g$ of the measure $Dg$ is concentrated on $\varphi^{-1}(\Sigma(c))$.

###### Proof.

Write $c=c^{(a)}+c^{(j)}$. Then $g=c^{(a)}\circ\varphi+h$ for some
$h\in[c^{(j)},\varphi]$. We have that
$c^{(a)}\circ\varphi\in\operatorname{AC}(I;\mathbb{R}^{d})$, which implies
that $D^{s}c=D^{s}h$. We may therefore assume without loss of generality that
$c^{(a)}=0$, $c=c^{(j)}$, and $g\in[c^{(j)},\varphi]$.

Let $x\in I$ and write $\varphi^{-1}(x)=[a,b]$ with $a\leq b$. Then there
exists a sequence $a_{k}\to a^{-}$ such that $c$ is continuous at each
$\varphi(a_{k})$. In particular, we have that $g(a_{k})=c(\varphi(a_{k}))$ for
each $k$. Since $g^{\ell}$ and $c^{\ell}$ are left continuous and
$\varphi(a_{k})\to\varphi(a)=x$, it follows that
$g^{\ell}(a)=\lim_{k\to\infty}g(a_{k})=\lim_{k\to\infty}c(\varphi(a_{k}))=c^{\ell}(x)$.
Similarly, we obtain that $g^{r}(b)=c^{r}(x)$. This implies in particular that

$\lvert Dg\rvert(\varphi^{-1}(x))\geq\lvert g^{r}(b)-g^{\ell}(a)\rvert=\lvert
c^{r}(x)-c^{\ell}(x)\rvert$

for every $x\in I$. Thus we have that

$\lvert Dc\rvert(I)=\lvert Dg\rvert(I)\geq\lvert
Dg\rvert(\varphi^{-1}(\Sigma(c)))=\sum_{x\in\Sigma(c)}\lvert
Dg\rvert(\varphi^{-1}(x))\\\ \geq\sum_{x\in\Sigma(c)}\lvert
c^{r}(x)-c^{\ell}(x)\rvert=\lvert D^{s}c\rvert(I)=\lvert Dc\rvert(I).$

This shows that $\lvert Dg\rvert(\varphi^{-1}(\Sigma(c)))=\lvert Dg\rvert(I)$,
which in turn implies that $\lvert
Dg\rvert(I\setminus\varphi^{-1}(\Sigma(c)))=0$. ∎

######  Lemma 14.

Assume that $U\subset\mathbb{R}^{d}$ is a Borel set,
$\nu\in\mathcal{M}_{+}(U)\setminus\\{0\\}$ is a non-trivial positive Radon
measure on $U$, and $g\colon U\to\mathbb{R}_{\geq 0}$ is a non-negative Borel
function on $U$ with $\int_{U}g(x)\,d\nu>0$. Assume moreover that
$\mu\in\mathcal{M}_{+}(U)$ solves the optimisation problem

(7)
$F(\mu):=\int_{U}g(x)\sqrt{\frac{d\mu}{d(\mu+\nu)}\frac{d\nu}{d(\mu+\nu)}}\,d(\mu+\nu)\to\max_{\begin{subarray}{c}\mu\in\mathcal{M}_{+}(U)\\\
\mu(U)=1\end{subarray}}$

Then $\mu\ll\nu$.

###### Proof.

Decompose $\mu=\mu^{a}+\mu^{s}$ with $\mu^{a}\ll\nu$ and $\mu^{s}\perp\nu$.
Then $\mu^{s}$ is concentrated on the set

$E:=\Bigl{\\{}x\in U:\lim_{\varepsilon\to
0}\frac{\mu(B_{\varepsilon}(x))}{\nu(B_{\varepsilon}(x))}=\infty\Bigr{\\}}.$

Moreover, we have for $(\mu+\nu)$-a.e. $x\in E$ that

$\frac{d\nu}{d(\mu+\nu)}(x)=\lim_{\varepsilon\to
0}\frac{\nu(B_{\varepsilon}(x))}{\mu(B_{\varepsilon}(x))+\nu(B_{\varepsilon}(x))}=0.$

As a consequence, $F(\mu)=F(\mu^{a})$. Now denote by $u\in L^{1}(U;\nu)$ the
density of $\mu^{a}$ with respect to $\nu$, that is, $\mu^{a}=u\,\nu$. Then

$\frac{d\mu}{d(\mu+\nu)}=\frac{u}{u+1}\quad\text{ and
}\quad\frac{d\nu}{d(\mu+\nu)}=\frac{1}{u+1}$

$(\nu+\mu)$-almost everywhere, and

$F(\mu)=F(\mu^{a})=\int_{U}g(x)\sqrt{\frac{u}{u+1}\frac{1}{u+1}}\,d((u+1)\nu)=\int_{U}g(x)\sqrt{u}\,d\nu.$

Now define $\hat{\mu}=\mu^{a}+\mu^{s}(U)\,\nu$. Then we obtain similarly that

$F(\hat{\mu})=\int_{U}g(x)\sqrt{u+\mu^{s}(U)}\,d\nu.$

Since $\int_{U}g(x)\,d\nu>0$ and $\mu$ was assumed to be a maximiser of $F$,
this implies that $\mu^{s}=0$, which proves the assertion. ∎

######  Remark 15.

Let $F$ be as in (7), but assume that $\int_{U}g(x)\,d\nu=0$. Let moreover
$\mu\in\mathcal{P}(U)$. Then we can again decompose $\mu=\mu^{a}+\mu^{s}$ with
$\mu^{a}\ll\nu$ and $\mu^{s}\perp\nu$. Moreover, we have is in the proof of
Lemma 14 that $F(\mu)=F(\mu^{a})$. However, because $\mu^{a}\ll\nu$ and
$\int_{U}g(x)\,d\nu=0$ we have that $F(\mu^{a})=0$. Thus the functional $F$ is
in this case the trivial functional $F(\mu)\equiv 0$.

######  Proposition 16.

Assume that $c_{i}\in\operatorname{SBV}(I;\mathbb{R}^{d})$ and that
$\varphi_{i}\in\bar{\Gamma}$ satisfy
$\varphi_{1}^{\prime}+\varphi_{2}^{\prime}=2$ almost everywhere. Then there
exist functions $\bar{g}_{i}\in[c_{i},\varphi_{i}]$ such that
$\bar{g}_{i}\in\operatorname{SBV}(I;\mathbb{R}^{d})$ and

$\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])=\hat{d}(g_{1},g_{2}).$

###### Proof.

Since the sets $[c_{i},\varphi_{i}]$ are strictly compact and
$\hat{d}([c_{1},\cdot],[c_{2},\cdot])$ is strictly lower semi-continuous,
there exist $g_{i}\in[c_{i},{\varphi}_{i}]$ such that
$\hat{d}([c_{1},\varphi_{1}],[c_{1},\varphi_{2}])=\hat{d}(g_{1},g_{2})$. We
will show that it is possible to replace $g_{1}$ and $g_{2}$ by functions
$\bar{g}_{1}$, $\bar{g}_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$ in such a
way that $\hat{d}(g_{1},g_{2})=\hat{d}(\bar{g}_{1},\bar{g}_{2})$, or,
equivalently, $\hat{S}(g_{1},g_{2})=\hat{S}(\bar{g}_{1},\bar{g}_{2})$.

According to Lemma 13, the singular part of $Dg_{i}$ is concentrated on
${\varphi}_{i}^{-1}(\Sigma(c_{i}))$. Now assume that $y\in\Sigma(c_{1})$.
Since ${\varphi}_{1}$ is continuous and non-decreasing it follows that
${\varphi}_{1}^{-1}(y)$ is either a single point or a closed interval. Denote
now by $R\subset\Sigma(c_{1})$ the set of jump points $y$ of $c_{1}$ for which
${\varphi}_{1}^{-1}(y)$ is an non-degenerate interval. Then the non-atomic
part $D^{c}g_{1}$ of $D^{s}g_{1}$ is concentrated on the set $E:=\bigcup_{y\in
R}\operatorname{int}{\varphi}_{1}^{-1}(y)$, since
${\varphi}_{1}^{-1}(\Sigma(c_{1}))\setminus E$ is at most countable.

Let now $y\in R$ and denote $[a,b]:={\varphi}_{1}^{-1}(y)$. By assumption, the
function $g_{1}$ solves the optimisation problem
$\max_{g\in[c_{1},\varphi_{1}]}\hat{S}(g,g_{2})$. Moreover, we have that

$Dg_{1}\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)=v\,\lvert
Dg_{1}\rvert\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)\qquad\text{ with }\qquad
v=\frac{c_{1}^{r}(y)-c_{1}^{\ell}(y)}{\lvert
c_{1}^{r}(y)-c_{1}^{\ell}(y)\rvert}\in\mathbb{R}^{d}.$

Now let $\mu\in\mathcal{M}_{+}(a,b)$ be a positive Radon measure satisfying
$\mu(a,b)=\lvert Dg_{1}\rvert(a,b)$. Then the function $\hat{g}_{1}$ defined
by

$\hat{g}_{1}=\begin{cases}g_{1}(x)&\text{ if }x\not\in(a,b),\\\
g_{1}^{\ell}(a)+v\mu(a,x)&\text{ if }x\in(a,b),\end{cases}$

satisfies $\hat{g}_{1}\in[c_{1},{\varphi}_{1}]$, and
$D\hat{g}_{1}\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)=v\mu$ and
$D\hat{g}_{1}\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(I\setminus(a,b))=Dg_{1}\mathbin{\hbox{\vrule
height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(I\setminus(a,b))$. Thus $\lvert
Dg_{1}\rvert\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)$ solves the optimisation problem

(8) $\int_{(a,b)}\Bigl{\langle}v,\frac{dDg_{2}}{d\lvert
Dg_{2}\rvert}\Bigr{\rangle}^{+}\sqrt{\frac{d\mu}{d(\mu+\lvert
dDg_{2}\rvert)}\frac{d\lvert Dg_{2}\rvert}{d(\mu+\lvert
Dg_{2}\rvert)}}\,d(\mu+\lvert Dg_{2}\rvert)\to\max$

where the maximum is taken over all $\mu\in\mathcal{M}_{+}(a,b)$ with
$\mu(a,b)=\lvert Dg_{1}\rvert(a,b)$.

Assume now that $\int_{(a,b)}\langle v,dDg_{2}/d\lvert
Dg_{2}\rvert\rangle^{+}d\lvert Dg_{2}\rvert>0$. Then we obtain from Lemma 14
that $\lvert Dg_{1}\rvert\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)\ll\lvert
Dg_{2}\rvert\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)$. Since ${\varphi}_{1}$ is
constant on $[a,b]$ and ${\varphi}_{1}^{\prime}+{\varphi}_{2}^{\prime}=2$
almost everywhere, it follows that ${\varphi}_{2}^{\prime}$ is strictly
increasing on $(a,b)$. Thus ${\varphi}_{2}^{-1}(\Sigma(c_{1})\cap(a,b))$ is an
at most countable union of single points. Since by Lemma 13 the measure
$D^{s}g_{2}\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)$ is concentrated on
${\varphi}_{2}^{-1}(\Sigma(c_{1}))\cap(a,b)$, it is purely atomic. Thus
$\lvert D^{s}g_{1}\rvert\mathbin{\hbox{\vrule height=6.0pt,depth=0.0pt\vrule
height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)$ is purely atomic as well and
therefore $\lvert D^{c}g_{1}\rvert\mathbin{\hbox{\vrule
height=6.0pt,depth=0.0pt\vrule height=0.5pt,depth=0.0pt,width=5.0pt}}(a,b)=0$.

Now assume that $\int_{(a,b)}\langle v,dDg_{2}/d\lvert
Dg_{2}\rvert\rangle^{+}d\lvert Dg_{2}\rvert=0$. As seen in Remark 15 we have
in this case that the integral in (8) is equal to zero for all choices of
$\mu$. We can therefore replace $g_{1}$ on the interval $(a,b)$ by the
function

$\bar{g}_{1}(x)=g_{1}^{\ell}(a)+\frac{x-a}{b-a}(g_{1}^{r}(b)-g_{1}^{\ell}(a)),$

corresponding to a choice of $\mu=s\mathcal{L}^{1}$ with $s=\lvert
Dg_{1}\rvert(a,b)/(b-a)$ and have that
$\hat{S}(\bar{g}_{1},g_{2})=\hat{S}(g_{1},g_{2})$.

Repeating this procedure first for each $y\in R$ and then for the function
$g_{2}$, we arrive at the claim. ∎

######  Proposition 17.

Let $c_{1}$, $c_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})\setminus\\{0\\}$.
Then

(9)
$\min_{(\varphi_{1},\varphi_{2})\in\bar{\Gamma}}\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])=\min_{(\psi_{1},\psi_{2})\in\bar{\Gamma}}\hat{d}\bigl{(}G(c_{1})\circ\psi_{1},G(c_{2})\circ\psi_{2}\bigr{)}.$

###### Proof.

We start by recalling the construction of the functions $G(c_{i})$ in Section
3.4: We set

$\xi_{i}(x)=\frac{\lvert
D^{s}c_{i}\rvert(0,x)}{2\operatorname{len}(c_{i})}+(1-\alpha_{i})x\qquad\qquad\text{
with }\alpha_{i}:=\frac{\lvert
D^{s}c_{i}\rvert(I)}{2\operatorname{len}(c_{i})}.$

Moreover, we denote by $\zeta_{i}\colon I\to I$ the non-decreasing left
inverse of $\xi_{i}$. Then the functions $G(c_{i})$ satisfy
$G(c_{i})(x)=c_{i}(\zeta_{i}(x))$ for $x\not\in\zeta(\Sigma(c_{i}))$ and
$G(c_{i})(\xi_{i}(x))=c_{i}(x)$ for $x\not\in\Sigma(c_{i})$. Moreover, we have
$G(c_{i})\in[c_{i},\zeta_{i}]$.

Assume now that $(\psi_{1},\psi_{2})\in\bar{\Gamma}$. Define
$\varphi_{i}=\zeta_{i}\circ\psi_{i}$. Since $G(c_{i})\in[c_{i},\zeta_{i}]$, it
follows that $G(c_{i})\circ\psi_{i}\in[c_{i},\varphi_{i}]$. Thus

$\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])\leq\hat{d}(G(c_{1})\circ\psi_{1},G(c_{2})\circ\psi_{2}).$

Since $(\psi_{1},\psi_{2})$ was arbitrary, this shows that the inequality
$\leq$ holds in (9).

Now assume that the maximum of $\hat{d}([c_{1},\cdot],[c_{2},\cdot])$ is
attained at $(\varphi_{1},\varphi_{2})\in\bar{\Gamma}^{2}$. In view of the
proof of Proposition 12, we may assume without loss of generality that
$\varphi_{1}^{\prime}+\varphi_{2}^{\prime}=2$ almost everywhere in $I$. Since
$c_{1}$, $c_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$, there exist by
Proposition 16 functions $g_{1}$,
$g_{2}\in\operatorname{SBV}(I;\mathbb{R}^{d})$ such that
$g_{i}\in[c_{i},\varphi_{i}]$ and
$\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])=\hat{d}(g_{1},g_{2})$.

We will now construct functions $h_{1}$,
$h_{2}\in\operatorname{AC}(I;\mathbb{R}^{d})$ in such a way that
$\hat{d}(g_{1},g_{2})=\hat{d}(h_{1},h_{2})$. The construction is similar as
for $G(c_{i})$, but we have to be careful to keep the function value of
$\hat{d}$ unchanged. We denote therefore

$\beta=\frac{\lvert D^{s}g_{1}\rvert(I)+\lvert
D^{s}g_{2}\rvert(I)}{2(\operatorname{len}(c_{1})+\operatorname{len}(c_{2}))}$

and define the function $\gamma\colon I\to I$,

$\gamma(x)=\frac{\lvert D^{s}g_{1}\rvert(0,x)+\lvert
D^{s}g_{2}\rvert(0,x)}{2(\operatorname{len}(c_{1})+\operatorname{len}(c_{2}))}+(1-\beta)x.$

Next we denote by $\vartheta\colon I\to I$ the non-decreasing left inverse of
$\gamma$, and define the functions
$h_{i}\in\operatorname{AC}(I;\mathbb{R}^{d})$ by
$h_{i}(x)=g_{i}(\vartheta(x))$ if $\gamma$ (and thus $g_{i}$) is continuous at
$\vartheta(x)$ and

$h_{i}(x)=g_{i}^{\ell}(\vartheta(x))+\frac{x-\gamma^{\ell}(\vartheta(x))}{\gamma^{r}(\vartheta(x))-\gamma^{\ell}(\vartheta(x))}\bigl{(}g_{i}^{r}(\vartheta(x))-g_{i}^{\ell}(\vartheta(x))\bigr{)}$

else.

Now assume that $x\in I$ is such that
$y:=\vartheta(x)\in\Sigma(g_{1})\cup\Sigma(g_{2})$. Then we have by
construction of $h_{i}$ that

$\dot{h}_{i}(x)=\frac{g_{i}^{r}(y)-g_{i}^{\ell}(y)}{\gamma^{r}(y)-\gamma^{\ell}(y)}=\frac{[g_{i}](y)}{[\gamma](y)}.$

Thus

$\displaystyle\hat{S}(h_{1},h_{2})$
$\displaystyle=\int_{I}\Bigl{\langle}\frac{\dot{h}_{1}}{\lvert\dot{h}_{1}\rvert},\frac{\dot{h}_{2}}{\lvert\dot{h}_{2}\rvert}\Bigr{\rangle}^{+}\sqrt{\lvert\dot{h}_{1}\rvert\lvert\dot{h}_{2}\rvert}\,dx$
$\displaystyle=\int_{\gamma(I\setminus(\Sigma(g_{1})\cup\Sigma(g_{2})))}\Bigl{\langle}\frac{\dot{g}_{1}\circ{\vartheta}}{\lvert\dot{g}_{1}\circ{\vartheta}\rvert},\frac{\dot{g}_{2}\circ{\vartheta}}{\lvert\dot{g}_{2}\circ{\vartheta}\rvert}\Bigr{\rangle}^{+}\sqrt{\lvert\dot{g}_{1}\circ{\vartheta}\rvert\lvert\dot{g}_{2}\circ\vartheta\rvert}\,\lvert\vartheta^{\prime}\rvert\,dx$
$\displaystyle\qquad+\sum_{y\in\Sigma(g_{1})\cup\Sigma(g_{2})}\int_{\gamma^{\ell}(y)}^{\gamma^{r}(y)}\Bigl{\langle}\frac{[g_{1}](y)}{\lvert[g_{1}](y)\rvert},\frac{[g_{2}](y)}{\lvert[g_{2}](y)\rvert}\Bigr{\rangle}^{+}\frac{\sqrt{\lvert[g_{1}](y)\rvert\,\lvert[g_{2}](y)\rvert}}{\lvert[\gamma](y)\rvert}\,dx$
$\displaystyle=\int_{I\setminus(\Sigma(g_{1})\cup\Sigma(g_{2}))}\Bigl{\langle}\frac{\dot{g}_{1}}{\lvert\dot{g}_{1}\rvert},\frac{\dot{g}_{2}}{\lvert\dot{g}_{2}\rvert}\Bigr{\rangle}^{+}\sqrt{\lvert\dot{g}_{1}\rvert\,\lvert\dot{g}_{2}\rvert}\,dx$
$\displaystyle\qquad+\sum_{y\in\Sigma(g_{1})\cup\Sigma(g_{2})}\Bigl{\langle}\frac{[g_{1}](y)}{\lvert[g_{1}](y)\rvert},\frac{[g_{2}](y)}{\lvert[g_{2}](y)\rvert}\Bigr{\rangle}^{+}\sqrt{\lvert[g_{1}](y)\rvert\,\lvert[g_{2}](y)\rvert}$
$\displaystyle=\hat{S}(g_{1},g_{2}).$

Since $\operatorname{len}(g_{i})=\operatorname{len}(h_{i})$, this implies that
also $\hat{d}(h_{1},h_{2})=\hat{d}(g_{1},g_{2})$. We will next construct
functions $\psi_{i}\in\bar{\Gamma}$ such that $h_{i}=G(c_{i})\circ\psi_{i}$.

We start by defining the function
$\hat{\psi}_{i}:=\xi_{i}\circ\varphi_{i}\circ\vartheta$. Since $\vartheta$ is
Lipschitz, $\varphi_{i}$ is absolutely continuous, and
$\xi_{i}\in\operatorname{SBV}(I;\mathbb{R})$, it follows that
$\hat{\psi}_{i}\in\operatorname{SBV}(I;\mathbb{R})$ as well. Moreover we have
that $G(c_{i})(\hat{\psi}_{i}(x))=h_{i}(x)$ for every $x\in I$ such that
$\varphi_{i}\circ\vartheta(x)\not\in\Sigma(c_{i})$. We may thus define
$\psi(x):=\hat{\psi}(x)$ for $x\in
I\setminus(\varphi_{i}\circ\vartheta)^{-1}(\Sigma(c_{i}))$.

Now let $y\in\Sigma(c_{i})$ and denote
$[a,b]:=(\varphi_{i}\circ\vartheta)^{-1}(y)$. Let moreover $x\in[a,b]$. Then
$h_{i}(x)$ lies on the line segment $[c_{i}^{\ell}(y),c_{i}^{r}(y)]$, that is,
we can write

$h_{i}(x)=\lambda(x)c_{i}^{r}(y)+(1-\lambda(x))c_{i}^{\ell}(y)=c_{i}^{\ell}(y)+\lambda(x)\,[c_{i}](y)$

for some $0\leq\lambda(x)\leq 1$. Moreover, since $h_{i}$ is absolutely
continuous, it follows that the mapping $x\mapsto\lambda(x)$ is absolutely
continuous. Define now

$\psi_{i}(x)=\xi^{\ell}(y)+\lambda(x)\,[\xi_{i}](y).$

Then $\zeta_{i}(\psi_{i}(x))=y$ and thus

$G(c_{i})(\psi_{i}(x))=c_{i}^{\ell}(y)+\frac{\psi_{i}(x)-\xi^{\ell}(y)}{[\xi_{i}](y)}[c_{i}](y)=c_{i}^{\ell}(y)+\lambda(x)\,[c_{i}](y)=h_{i}(x).$

By construction we have that $\psi_{i}\colon I\to I$ is non-decreasing and
$G(c_{i})\circ\psi_{i}=h_{i}$. Since the restriction of $\psi_{i}$ to
$(\varphi_{i}\circ\vartheta)^{-1}(y)$ is absolutely continuous for each
$y\in\Sigma(c_{i})$ and the restriction of $\psi_{i}$ to
$I\setminus(\varphi_{i}\circ\vartheta)^{-1}(\Sigma(c_{i}))$ is in
$\operatorname{SBV}(I;\mathbb{R})$, it follows that
$\psi_{i}\in\operatorname{SBV}(I;\mathbb{R})$ as well. Since in addition
$\psi_{i}$ is continuous, it follows that it is actually absolutely continuous
and therefore contained in $\bar{\Gamma}$.

This proves the assertion. ∎

###### Proof of Theorem 6.

By Proposition 12, both of the infima in (4) are attained at some
$\bar{\varphi}_{i}$, $\bar{\psi}_{i}\in\bar{\Gamma}$. Moreover, by Proposition
17 we have that

$\inf_{\varphi_{i}\in\bar{\Gamma}}\hat{d}([c_{1},\varphi_{1}],[c_{2},\varphi_{2}])=\inf_{\psi_{i}\in\bar{\Gamma}}\hat{d}(G(c_{1})\circ\psi_{1},G(c_{2})\circ\psi_{2}).$

It remains to show that this is further equal to

$\hat{d}^{S}([c_{1}],[c_{2}]):=\inf_{g_{i}\sim c_{i}}\hat{d}(g_{1},g_{2}).$

Assume therefore that $g_{i}\in[c_{i}]$. Denote by
$h_{i}\in\operatorname{AC}(I;\mathbb{R}^{d})$ the constant length
parametrisation of $G(g_{i})$. Then we can write
$G(g_{i})=h_{i}\circ\vartheta_{i}$ for some $\vartheta_{i}\in\bar{\Gamma}$.
Since $c_{i}\sim g_{i}$, it follows that $h_{i}$ is also the constant length
parametrisation of $G(c_{i})$. Now, since $\dot{c}_{i}\neq 0$ almost
everywhere, it follows that also $G(c_{i})^{\prime}\neq 0$ almost everywhere,
and thus we can write $h_{i}=G(c_{i})\circ\eta_{i}$ for some
$\eta_{i}\in\bar{\Gamma}$. Thus
$G(g_{i})=G(c_{i})\circ\eta_{i}\circ\vartheta_{i}$ and thus

$\hat{d}(g_{1},g_{2})\geq\inf_{\psi_{i}\in\bar{\Gamma}}\hat{d}(G(c_{1})\circ\psi_{1},G(c_{2})\circ\psi_{2}).$

Since this holds for every $g_{i}\in[c_{i}]$ and since
$G(c_{i})\circ\psi_{i}\in[c_{i}]$, the assertion follows. ∎

## Acknowlegdements

I would like to thank Esten Nicolai Wøien for valuable comments and helpful
discussions.

## References

  * [1] L. Ambrosio, N. Fusco, and D. Pallara.  Functions of bounded variation and free discontinuity problems.  Oxford Mathematical Monographs. Oxford University Press, New York, 2000\.
  * [2] M. Bauer, M. Bruveris, and P. W. Michor.  Overview of the geometries of shape spaces and diffeomorphism groups.  J. Math. Imaging Vision, 50(1-2):60–97, 2014.
  * [3] M. Bauer, N. Charon, E. Klassen, and A. Le Brigant.  Intrinsic Riemannian metrics on spaces of curves: theory and computation, 2020.  arXiv:2003.05590.
  * [4] M. Bauer, M. Eslitzbichler, and M. Grasmair.  Landmark-guided elastic shape analysis of human character motions.  Inverse Probl. Imaging, 11(4):601–621, 2017.
  * [5] M. Bruveris.  Optimal reparametrizations in the square root velocity framework.  SIAM J. Math. Anal., 48(6):4335–4354, 2016.
  * [6] E. Celledoni, M. Eslitzbichler, and A. Schmeding.  Shape analysis on Lie groups with applications in computer animation.  J. Geom. Mech., 8(3):273–304, 2016.
  * [7] G. Dogan, J. Bernal, and C.R. Hagwood.  A fast algorithm for elastic shape distances between closed planar curves.  In Proceedings of the IEEE Conference on Computer Vision and Pattern Recognition (CVPR), June 2015.
  * [8] J. Eckhardt, R. Hiptmair, T. Hohage, H. Schumacher, and M. Wardetzky.  Elastic energy regularization for inverse obstacle scattering problems.  Inverse Problems, 35(10):104009, 20, 2019.
  * [9] I. Fonseca and G. Leoni.  Modern methods in the calculus of variations: $L^{p}$ spaces.  Springer Monographs in Mathematics. Springer, New York, 2007.
  * [10] D. H. Fremlin.  Measure theory. Vol. 2.  Torres Fremlin, Colchester, 2003.  Broad foundations, Corrected second printing of the 2001 original.
  * [11] E. Hartman, Y. Sukurdeep, N. Charon, E. Klassen, and M. Bauer.  Supervised deep learning of elastic srv distances on the shape space of curves, 2021.  arXiv:2101.04929.
  * [12] W. Huang, K.A. Gallivan, A. Srivastava, and P.-A. Absil.  Riemannian optimization for registration of curves in elastic shape analysis.  J. Math. Imaging Vision, 54(3):320–343, 2016.
  * [13] S. Lahiri, D. Robinson, and E. Klassen.  Precise matching of PL curves in $\mathbb{R}^{N}$ in the square root velocity framework.  Geom. Imaging Comput., 2(3):133–186, 2015.
  * [14] W. Liu, A. Srivastava, and J. Zhang.  A mathematical framework for protein structure comparison.  PLoS computational biology, 7:e1001075, 02 2011.
  * [15] W. Mio, A. Srivastave, and S. Joshi.  On shape of plane elastic curves.  Int. J. Comput. Vision, 73(3):307–324, 2007.
  * [16] E. Nunez and S.H. Joshi.  Deep learning of warping functions for shape analysis.  In 2020 IEEE/CVF Conference on Computer Vision and Pattern Recognition Workshops (CVPRW), pages 3782–3790, 2020.
  * [17] Yu.G. Reshetnyak.  Weak convergence of completely additive vector functions on a set.  Sib. Math. J., 9(6):1039–1045, 1968.
  * [18] F. Rindler.  Calculus of variations.  Universitext. Springer, Cham, 2018.
  * [19] A. Srivastava and E. Klassen.  Functional and shape data analysis.  Springer Series in Statistics. Springer-Verlag, New York, 2016.
  * [20] A. Srivastava, E. Klassen, S.H. Joshi, and I.H. Jermyn.  Shape analysis of elastic curves in Euclidean spaces.  IEEE Trans. Pattern Anal. Mach. Intell., 33(7):1415–1428, 2011\.
  * [21] G. Sundaramoorthi, A. Mennucci, S. Soatto, and A. Yezzi.  A new geometric metric in the space of curves, and applications to tracking deforming objects by prediction and filtering.  SIAM J. Imaging Sci., 4(1):109–145, 2011.
  * [22] E. Wøien and M. Grasmair.  A PDE-based method for shape registration.  SIAM J. Imaging Sciences, 2022.  To appear.