An introduction to Reshetnyak's theory of subharmonic distances
François Fillastre

The aim of the present text is to provide some basics around Reshetnyak's theory of subharmonic distances, together with an overview of the main results. This text is intended to be a complement to the English translation of  [83], [84], [85], [86], [88], [87], [89], [90], [91] and [56]. They all will be published in a same volume [47].
While subharmonic distances are often confused with distances of bounded curvature, we present them as a complete autonomous theory. In turn, there is no specific prerequisite for the present text.

§ INTRODUCTION

Given a set of admissible curves in a plane domain $M$, and
a function $\lambda$ such that its square root is integrable along any admissible curve,
an intrinsic pseudo-distance can be defined over $M$. Indeed,
the integration of  the square root of $\lambda$ over an admissible curve defines the length of the curve, and the infimum of these lengths between two points is defined as the distance between the points. If $\lambda$ is $C^\infty$, one obtains a Riemannian distance, whose tensor metric is conformal to the Euclidean one. Another basic famous examples are flat metrics with conical singularities, for which $\lambda$ is a sum of $|\cdot - z_0|^\gamma$. What should be the good regularity of $\lambda$ to develop a theory of “Riemannian geometry with low regularity” on surfaces, containing the Riemannian case and the case of flat metrics? This has been answered in a series of articles by Yu. G. Reshetnyak in the $60'$s: the logarithm of $\lambda$ is a difference of subharmonic functions, and admissible curves are  rectifiable arcs (for the Euclidean distance). In this case, the pseudo-distance is a distance, called a subharmonic distance.

A subharmonic distance comes with a natural notion of curvature measure, which, in the Riemannian case, is the sectional curvature times the area measure. Actually, the $\lambda$ term of a subharmonic distance is better defined from the potential of a measure plus a harmonic mapping.
A central result in the theory relates the converge of the measures to  the (uniform) convergence of the distances. One implication of this theorem is  that, on a given domain, the set of subharmonic distances is exactly the set of the limits (for the uniform convergence) of  sequences of Riemannian distances, with uniformly bounded total variation of the curvature measure, preserving the topology.[Relaxing the curvature bound assumption would be too general, see e.g., [32]. See [27] if the condition for the limit metric to have the topology of a surface is relaxed.]

Y. Reshetnyak and A. Huber proved simultaneously that a distance preserving mapping between domains with  subharmonic distances is a conformal mapping, giving a bridge with the theory of Riemann surfaces. Also, using his convergence theorem, proved that  subharmonic distances and two-dimensional manifolds of bounded curvatures,  in the sense of A.D. Alexandrov, are (locally) the same thing, providing an analytic approaches to a classical object from the theory of metric geometry of surfaces.

Section <ref> contains a very general presentation of the two extreme cases of subharmonic metrics on a plane domain: conformal Riemannian metrics and flat cones. It also contains an introduction to  the theory of curves of bounded rotation, that form the most helpful class of curves of the theory. Section <ref> contains  basics of potential theory, that are needed in the sequel. It also contains a presentation of Reshetnyak's result about localization of $\delta$-subharmonic functions, that is one of the main technical tool used in the sequel.

Section <ref> contains basic facts about metric spaces, which are mixed with results from Section <ref> about subharmonic mappings, in order to define subharmonic metrics on plane domains.  Then, we present the heart of theory, that are convergence theorems. At the end of the section, we present a theorem about contraction onto a cone, that allows to deduce that subharmonic distances are two-dimensional manifolds of bounded curvature, in the sense of Alexandrov.

In Section <ref>, we first present the striking result mentioned above, that says that a distance preserving mapping between two plane domains, endowed with subharmonic distance, is a conformal mapping. Moreover, from a formula of Huber,
subharmonic metrics can be defined over Riemann surface. Indeed, as it was noted by M. Troyanov [110], there is a bijection between the set of Radon measure over a compact Riemann surface satisfying Gauss–Bonnet Formula, and subharmonic distances with a given total area on the surface.

In the  sections mentioned above, all the arguments should be complete, by referring to the original sources (with the exception of Theorem <ref>, see Remark <ref>). In Section <ref>, we present some related results and questions, mainly around relations between subharmonic distances and two-dimensional manifolds of bounded curvature.

The present text is not a survey about two-dimensional manifolds of bounded curvature bounded curvature, nor Alexandrov surfaces.[The term “Alexandrov surface”Alexandrov surface does not have a precise meaning. Sometimes, it means a two-dimensional manifold of bounded curvature, sometimes if means a metric surface with a curvature bound in the sense of Alexandrov, and sometimes a metric surface with a lower curvature bound. In turn, we won't use further this terminology.] Concerning surfaces with curvature bound, they are covered in many textbooks [26], [25], [5], [4]. Concerning two-dimensional manifolds of bounded curvature, we only know the original old-fashioned [11], together with [10]. The text [93] is a survey about both two-dimensional manifolds of bounded curvature and subharmonic distances, but it also contains  some  proofs, some of which we will refer to in the present text. The last pages of [93] presents the state of the art about two-dimensional manifolds of bounded curvature at the time. Let us mention another survey  related to the content of the present article: [94] —concerning the specific question of curves of bounded rotation, see [95]. And of course, the survey [110], which renewed attraction to this theory.

The present text is also not a survey about the literature around subharmonic distances. For recent results about this theory, let us only mention [97], [38]. For results using
subharmonic distances, let us mention [30], [107], [64], [68], [12], [21], [1].

§.§ Terminology

In this section, we clarify some points about terminology and choices of translation.

A first comment is that, for a metric space $(X,d)$, the function $d:X\times X\to \R$ is always called a distance in the present text, the term metric is used for $(0,2)$-tensors. We apologize that we are not following the terminology used in the articles of Reshetnyak. In those articles, what we are calling a  distance is called a metric, and what we are calling a metric is called a linear elementlinear element, as it was usual at the time.
The term distance is used by to speak about the value of $d$ between two given points.

We have kept the term polyhedron from articles, even if nowadays, the objects called polyhedra here are more often called polyhedral surfacespolyhedral surface, or more precisely in the present context, flat surfaces (with conical singularities)flat surfaces (with conical singularities).

A very confusing point is about rotation and turn of curves, which are defined in Section <ref>.
There are many different names for the turn and the rotation in the literature (and also depending on the translation). We followed the terminology in articles. It is also the one used in [93]. It is summarized in Table <ref>. In the case of a Riemannian metric, the turn defined in (<ref>) is usually called
integral curvatureintegral curvature, total curvaturetotal curvature or  integral geodesic curvatureintegral geodesic curvature, and this terminology may be used for curves in more general metric spaces.  The turn was also called swerveswerve by H. Busemann, see the footnote page 26 in [77].

General metric     Euclidean metric

turn     rotation

$\kappa_\lambda$, $\kappa_l$, $\kappa_r$     $\kappa$

поворот      вращение

povorot        vrashchenie

Turn and rotation.

For a measure $\mes$, points $z$ with $\mes(\{z\})\geq 2\pi$ will play a particular role. They can be called cusp pointscusp points, peak pointspeak point, return pointsreturn points, spike pointsspike point. However, as no special terminology is used in articles, we also won't name those points.

A statement inside a framed box has an equivalent version in an article contained in the same book as the present article, namely [47].

The author is very grateful to Alexander Lytchak for his comments, which significantly enhanced the interest and the presentation of the present text.

§ CURVES IN THE PLANE

The first section of this chapter  is a kind of lumber room. It takes the opportunity of introducing the concepts of rotation and turn of curves,  to  give some flavor about  Reshetnyak's construction. For example, we encounter the two main examples of subharmonic distances, Riemannian conformal metrics and flat cones.

The second section of this chapter is intended to be a comprehensive introduction to plane curves of bounded rotation. They are one of the main object of the theory, and, despite the fact that most of all the needed results are contained in [9], it seemed more convenient to give a detailed presentation.

§.§ Rotation, turn and Laplacian

§.§.§ Rotation of arcs in the plane

Let us recall some general definitions. Let $X$ be a Hausdorff topological space. A parameterized pathparameterized path in $X$ is a continuous mapping $\param:[a,b]\to X$, $a<b$.
We will restrict ourselves to the simpler case
of  parameterized arcsparameterized arc, where $\param$ is asked to be injective.

Sometimes we will consider that $\param(a)=\param(b)$,  in this case we will say that the arc parameterizes a  simple closed curvesimple closed curve. The restriction to arcs will facilitate the following notion of equivalence, and is not so restrictive, as any path contains an arc connecting its endpoints <cit.>.
Two parameterized arcs $\param_1:[a,b]\to X$ and $\param_2:[c,d]\to X$ are said to be equivalent, if there
exists a function $\psi:[c,d]\to [a,b]$ that is strictly monotonic and surjective (hence a homeomorphism), and
that satisfies $\param_2=\param_1\circ \psi$. This is an equivalence relation. A class of parameterized arcs will be called an arc. We will confuse this equivalence class with the subset of the plane which is the  image of a parameterization.
For an arc $\arc$, $\arc^\circ$$\arc^\circ$ is $\arc$ minus its extremities.
A choice of an element in the equivalence class is called a parameterization of the arc.
Note that our arcs are oriented.
Let $\param:[a,b]\to X$ be a parameterized arc.
As $\param([a,b])$ is compact hence closed, it is easy to see that the mapping $\param$ is indeed closed, hence a homeomorphism onto its image.

A sequence of arcs $(\arc_n)_n$  convergesconvergence (arcs) to an arc $\arc$, if there are parameterizations of these arcs on the same segment $[a,b]$,
which uniformly converge as functions.

We now look at arcs in the plane. For latter convenience, we will look at the complex plane $\C$.
Let $L$ be a broken linebroken line in the plane, that is an arc which is a union of segments. Its (Euclidean) length is the sum of the lengths of the segments composing $L$, i.e., the sum of the distances between its vertices.
A broken line $L$ is said to be inscribedinscribed in an arc $K$, if the vertices of $L$ are in the interior of $K$,  the extremities of $K$ and $L$ coincide, and the vertices of $L$ are ordered for a parameterization of $K$. The (Euclidean) lengthlength (Euclidean)
$s(\arc)$$s(\arc)$ of $\arc$ is then the supremum of the lengths of all the broken lines inscribed in $\arc$. It is well-defined in $\R\cup \{+\infty\}$. Then $\arc$ is called rectifiablerectifiable if its length is finite.
Let us recall some relevant properties of rectifiable arcs:

* the length $s$ is lower lower-semicontinuous on the set of arcs, i.e., if $\arc_n\to  \arc$, then

$$s(\arc)\leq \liminf_n s(\arc_n)~; $$

* if $\arc_n\to \arc$ and $s(\arc_n)$ are uniformly bounded from above, then $\arc$ is rectifiable;

* from any set of arcs, in a compact set with length uniformly bounded from above, one can extract a converging subsequence.

If $\param:[a,b]\to\C$ is a parameterization of an arc $\arc$, let us denote by $\arc\vert_{ t_1,t_2}$ the arc parameterized by the restriction of $\param$ to $[t_1,t_2]$, and denote $ \arc\vert_{ t}:=\arc\vert_{ a,t}$. A parameterized arc  is parameterized by arc lengthparameterized by arc length  if, for any $a\leq t_{i-1}\leq t_i\leq b$, we have $t_i-t_{i-1}=s(\arc|_{t_{i-1},t_i})$. Any rectifiable arc
has a parameterization by arc length, defined on the interval $[0,s(\arc)]$, which is $1$-Lipschitz. By an affine change of variable, one obtains from a rectifiable arc $\arc$ a parameterization on $[0,1]$, which is $s(\arc)$-Lipschitz. Such parameterization is said proportional to arc lengthproportional to arc length parameterization.

A parameterized arc $\param:[a,b]\to\C$ is regularregular if it is  $C^2$ and  $\param'(t)\not= 0$ for all $t$. The tantrixtantrix $T$ (tantrix is a contraction of tangent indicatrix) is  defined by

The curvaturecurvature (of an arc) of a regular arc
is the function

\begin{align}\label{eq:rotation coord cplxe}%%\label{eq:curvature}
k(\param)  &= \left\langle \frac{T'}{|\param'|}, \nu\right\rangle =\frac{1}{|\param'|^3}\langle \param'',\nu \rangle \nonumber \\
&= \frac{1}{|\param'|^3} \det(\param',\param'')=
\frac{\param'_1\param''_2 - \param'_2\param''_1}{|\param'|^3}= \frac{\operatorname{Re}(i\param'\bar{\param}'')}{|\param'|^3}~,
\end{align}

where we denoted  by $\nu$$\nu$ the oriented unit normal to $\param$, i.e.,

\begin{equation}\label{eq:normale}
\nu=\frac{1}{|\param'|}\param'^{\bot}
\end{equation}

where $\param'^{\bot}$

If the arc is parameterized  by  arc length, i.e., if  $\langle \param',\param'\rangle=1$, then $\langle \param'',\param'\rangle=0$ and $z''$ and $\nu$ are colinear with the same norm, and we have
$k(\param)=\pm |\param''|$, depending on the direction of $\param''$.

A regular arcregular arc is an arc whose parameterization by arc length is  a regular parameterization.
The rotationrotation (of an arc) of a regular arc $\arc$ is

\begin{equation}\label{eq:rotation}
\begin{split}
\ \kappa(K) & = \int_a^b k(\param)(t)|\param'(t)|dt~, \\
\end{split}
\end{equation}

as the definition above is independent of the choice of a regular parameterization.

If $\arc_1$ and $\arc_2$ are two regular arcs, non-overlapping but with common extremities, then using the orientation,
one can say when $\arc_2$ is on the left  of $\arc_1$, see Figure <ref> for an heuristic illustration ($\arc_1$ is the plain line and $\arc_2$ the dotted one),
and Section <ref> for more details.  Now, let us change the orientation of $\arc_2$ such that
both arc have the same starting points and the same endpoints, and let
$\alpha$ and $\beta$  be the inner angles of the open set enclosed by the curves at the extremities.
We have    [91]:

\begin{equation}\label{eq:def gen rot}\kappa(\arc_1)=\kappa(\arc_2)+\alpha+\beta~. \end{equation}

The formula above motivates the following definition.

An arc is said to be of class $\tilde{\Delta}$$\tilde{\Delta}$ (class of arcs) if it has a parameterization $z:[a,b]\to \C$ with
$z'_r(a)$ and $z'_l(b)$ that exist and are non-zero, where $z'_r$ and $z'_l$ are the right derivative and the left derivative respectively.

In short, an arc is in $\tilde{\Delta}$ if it has a semi-tangent at its extremities. Not any rectifiable arc is in the class $\tilde{\Delta}$, as shows the graph of the function $f(x)=x\sin(-\ln(x))$, which has no right derivative at $0$, but which is Lipschitz.

The inner angle at the extremities of two arcs of the class $\tilde{\Delta}$ bounding a domain can be defined. In turn, for any arc $\arc_1\in \tilde{\Delta}$, one can define its rotation  $\kappa(\arc_1)$  by the formula (<ref>), by considering any suitable regular arc $\arc_2$ on the left of
$\arc_1$ and with the same extremities, and noting  that this definition actually does not depend on the choice of $\arc_2$. It is the way the rotation for an arc of the class $\tilde{\Delta}$ is defined in [91].

§.§.§ Turn of arcs, Laplacian and curvature for Riemannian metric

We now define a  notion analogous to the rotation, when a subset of the plane is endowed with a conformal metric. We will use the standard notation $|dz|^2$ for the Euclidean metric. Indeed, if
we define the one-forms $dz=dx+idy$ and $d\bar{z}=dx-idy$, then $|dz|^2$ is a notation for
$dzd\bar{z}$, itself representing the symmetrization of the tensor product between
$dz$ and $d\bar{z}$:

$$|dz|^2=\frac{1}{2}\left( dz\tensor d\bar{z} + d\bar{z}\tensor dz\right) $$

and  an immediate computation shows that

$$|dz|^2=\frac{1}{2}\left(dx\tensor dx + dx\tensor dy +dy\tensor dx+dy\tensor dy\right)~, $$

and this last term is the usual Euclidean metric.

Let $M$ be a bounded domaindomain (i.e., a connected open set) in the plane, and let $g$ be a Riemannian metricRiemannian metric on $M$, that is, for all $z\in M$, the data of a scalar product $g(z)$ on $T_zM\simeq \R^2$, such that $z\mapsto g(z)$ is $C^3$. The Riemannian metric $g$ is said to be conformalconformal if there exists a $C^3$ function
$\lambda:M\to \R$ (necessarily positive) such that

$$g=g_\lambda:=\lambda|dz|^2~. $$

We restrict ourselves to the $C^3$ regularity for Riemannian metrics,  in order to have a $C^1$ curvature, that would be a relevant assumption. See Remark <ref>.

As an example, let $U\subset \C$ be a domain, and let $f:U\to \C$ be complex differentiable, with non-zero derivative. As

it readily follows that the pull-back over $U$ by
$f$ of the Euclidean metric is

$$|f'|^2 |dz|^2~. $$

Let $X,Y$ be smooth vector fields on $M$. If $D$ is the usual connection of the Euclidean space (the differentiation), and $\nabla$ is the Levi–Civita connection of $g_\lambda$, the Koszul formula gives

$$\nabla_XY= D_XY + \frac{1}{2\lambda} (d\lambda(X) Y + d\lambda(Y)X - \operatorname{grad}(\lambda)\langle X,Y\rangle)~, $$

where $\operatorname{grad}$ is the usual gradient.
Let $\langle\cdot,\cdot\rangle$$\langle\cdot,\cdot\rangle$ be the usual scalar product, and let
$(e_1,e_2)$ be the usual associated basis. Let $\param:[a,b]\to\C$ be a regular parameterized arc. The covariant derivative of a smooth vector field $X$ along $\param$ is given by

\begin{equation*}
\begin{split}
\ \frac{D}{dt} X& = X' + X^1\nabla_{c'}e_1 + X^2\nabla_{c'} e_2 \\
\ &=X'+\frac{1}{2\lambda}\left( X^1\binom{D\lambda(c')}{-D\lambda (c'^{\bot})} +X^2\binom{D\lambda (c'^{\bot})}{D\lambda(c')}\right) \\
\end{split}
\end{equation*}

and the geodesic curvaturegeodesic curvature$k_\lambda(\param)$ of $\param$ is, if
$\nu_\lambda=\frac{\nu}{\sqrt{\lambda}}$$\nu_\lambda$ is the unit (for $g_\lambda$) normal to $\param$ and $\nu$ is defined in  (<ref>),

\begin{equation*}\label{eq:geod curv}
\begin{split}
\ k_\lambda(\param) & = \frac{1}{\sqrt{g_\lambda(\param',\param')}}g_\lambda\left( \frac{D}{dt} \param', \nu_\lambda \right)= \frac{1}{\lambda^{1/2}}k(\param) - \frac{1}{2\lambda^{3/2}}\frac{\partial\lambda}{\partial \nu}~,
\end{split}
\end{equation*}

where $\frac{\partial }{\partial \nu}$ $\frac{\partial }{\partial \nu}$ is the derivative
in the direction of $\nu$, i.e.,

$$\frac{\partial f}{\partial \nu}  =\langle \grad f, \nu\rangle~.$$

Let $\arc$ be the arc parameterized by $\param$.
The turn $\kappa_\lambda(\arc)$turn $\kappa_\lambda$ of  $\arc$ is

\begin{equation}\label{eq:def turn r}
\begin{split}
\kappa_\lambda(\arc)& =\int_a^b k_\lambda(\param(t))\sqrt{g_\lambda(\param'(t),\param'(t))}dt = \kappa(\arc) - \frac{1}{2}\int_\arc \frac{\partial \ln \lambda}{\partial \nu}~,
\end{split}
\end{equation}

where we used the definition of the rotation $\kappa(\arc)$, see (<ref>). Note that the rotation of an arc is the turn for the Euclidean metric.

From the conformal Riemannian metric $\lambda|dz|^2$ on $M$, for a broken line $K\subset M$, we may define its length

\begin{equation}\label{def length}\tilde{s}_\lambda(K)=\int_K \sqrt{\lambda}~, \end{equation}

and then define a Riemannian distanceRiemannian distance $\rho_\lambda$ between two points, as the infimum of the lengths of the broken line contained in $M$ joining them.

For any non-negative function $\lambda$ (allowing the $+\infty$ value) that is integrable over any broken line,
(<ref>) defines a
pseudo-distance $\rho_\lambda$, i.e. a non-negative symmetric function on $M\times M$ satisfying the triangle inequality. In general, $\rho_\lambda$ is not a distance, in the sense that this function may be zero when evaluated at two distinct points.

In the case of a conformal Riemannian metric, or, more generally,
if $\lambda$ is finite, positive and continuous, then $\rho_\lambda$ is a distance. Indeed, for any $\arc$  from  a point $z_0$ to the boundary of a small Euclidean circle centered at this point, $\tilde{s}_\lambda(K)$ is greater than a positive constant times the  Euclidean length, where the positive constant is the minimum of $\lambda$ on the closed disc.
It follows that  for any $z$ belonging to the circle,  $\rho_\lambda(z,z_0)$ is positive, and then
any broken line to another point outside this disc will have larger length. Along the same line, it is easy to see that the topology induced by such a distance is the usual topology of the plane.
found lower regularity assumptions on $\lambda$ in order for $\rho_\lambda$ to be a distance, see Section <ref>.

As by definition, a  regular  parameterized arc $\arc$ is a geodesicgeodesic
if  $\frac{D}{dt}\param'=0$, it follows that a geodesic has zero turn.
Recall also that for a domain with a distance (see Remark <ref>), a shortest arcshortest arc between two points is an arc
whose length is equal to the distance between the points, supposed this latter is finite. A shortest arc may not exist, or, if it exists, may not be unique. It is classical that for a Riemannian distance, the arc length parameterization of a shortest arc is a geodesic (note that to be a shortest arc is a property of an arc, while to be a geodesic is a property of a parameterized arc). It follows that for  a Riemannian distance, a shortest arc has zero turn. One of the main result of [91] is a generalization of this fact to a larger class of metric spaces, see Theorem <ref>. More precision about shortest arcs is given in Section <ref>.

We will say that an arc is a geodesic broken linegeodesic broken line is it can be decomposed as a union of shortest arcs.

Now let  $(\tilde{e}_1,\tilde{e}_2)=(e_1/\sqrt{\lambda}, e_2/\sqrt{\lambda})$
be an orthogonal basis for $\lambda|dz|^2$.
The sectional curvature of $\lambda|dz|^2$ is given by

\begin{equation}\label{eq:def sec curv}
\begin{split}
\ \curv_\lambda & = \lambda \langle \nabla_{\tilde{e}_2}(\nabla_{\tilde{e}_1}\tilde{e}_1) -\nabla_{\tilde{e}_1}(\nabla_{\tilde{e}_2}\tilde{e}_1)+\nabla_{[\tilde{e}_1,\tilde{e}_2]}\tilde{e}_1,\tilde{e}_2\rangle~, \\
\  & =-\frac{1}{2\lambda} \Delta \ln \lambda = -\frac{1}{\lambda} \Delta \ln \sqrt{\lambda}~,
\end{split}
\end{equation}

where $\Delta$ is the usual Laplacian: $\Delta  = \frac{\partial^2 }{\partial x^2}+\frac{\partial^2}{\partial y^2} $ $\Delta$ (Euclidean Laplacian).

For a conformal Riemannian metric $\lambda|dz|^2$, the area measurearea measure $A_\lambda$$A_\lambda$ is $\lambda \mathcal{L}$, where $\mathcal{L}$$\mathcal{L}$ is the Lebesgue measure, and the curvature measurecurvature measure
$\omega_\lambda$ is $\curv_\lambda A_\lambda$, i.e.

\begin{equation}\label{def:mes riem}\mes_\lambda=\curv_\lambda \lambda \mathcal{L}=-\frac{1}{2} \Delta \ln \lambda \mathcal{L}~. \end{equation}

Note that if $h$ is a constant function, or more generally a harmonic functionharmonic function, i.e., a $C^\infty$ function such that $\Delta h=0$,
if $\bar{\lambda}=e^h\lambda$, then
$$ \omega_{\bar{\lambda}}=\omega_\lambda \mbox{ but } \curv_{\bar{\lambda}}=e^{-h}\curv_\lambda~.$$

The main point of the theory is that, up to a harmonic function, $\lambda$ is determined by the curvature measure.
This fact is based on the famous  Green formulaGreen formula, for $C^2$ functions on the plane

\begin{equation}\label{eq:green}\iint_{M} f\Delta g - g\Delta f  = \int_{\partial M} f\frac{\partial g}{\partial\nu} - g\frac{\partial f}{\partial\nu} \end{equation}

if $\partial M$ is a  piecewise $C^1$ simple closed curve, and $\nu$ is the unit outward normal to $M$.
It is worth noting that for  $z\in M$, the function  $\zeta\mapsto \ln|z-\zeta|$ is harmonic out of $\{z\}$.
This leads to the following classical result, see e.g. <cit.>.

Let $f$ be a $C^2$ function over the plane. Then for $z\in M$,

\begin{equation}\label{eq:poisson1}f(z)=\frac{1}{2\pi}\iint_M  \ln |z-\cdot| \Delta f +\frac{1}{2\pi} \int_{\partial M}  f\frac{\partial \ln|z- \cdot|}{\partial \nu}-\ln|z-\cdot| \frac{\partial f }{\partial  \nu}~.\end{equation}

Let  $\mes$ be a signed measure  with compact support in $M$, with $C^\infty$ Lebesgue density $k$.  The function $p(\mes)$ over $M$ is defined at a point $z$ by

\begin{equation}\label{eq:poisson lisse}p(z;\mes)=\frac{1}{2\pi}\iint \ln |z-\cdot| k=\frac{1}{2\pi}\iint \ln |z-\zeta| d\mes(\zeta) \end{equation}

(if nothing is specified, the integration is made over the plane). By (<ref>), on $M$ we have

\begin{equation}\label{eq:lapl curv}k=\Delta p(\mes)~. \end{equation}

The function $p(\mes)$
is the potentialpotential of $\mes$.
Comparing (<ref>) and (<ref>), it follows that if we define

$$\lambda=e^{-2p(\mes)}~, $$

then the conformal Riemannian metric $\lambda|dz|^2$
has sectional curvature $\curv=k/\lambda$, i.e., the curvature measure of $\lambda|dz|^2$ is $\mes$.

Actually, if $k$ is $C^1$, then $p(\mes)$, and in turn $\lambda$, is $C^2$ <cit.>. We hence obtain a $C^2$ Riemannian metric with a $C^1$ curvature (in general, if the metric is $C^2$, the curvature is $C^0$).
Actually, the class of $C^2$ Riemannian metrics on an open domain with a $C^1$ curvature is a relevant class for the problem of finding isothermal coordinatesIsothermal coordinates [112] —i.e., finding a $C^1$ homeomorphism such that the pull-back of the metric is conformal.
Weaker assumptions for the regularity of the Riemannian metric in order to be isometric (here an isometry is a $C^1$ mapping) to a conformal metric are given in [34], [33], [52].
See <cit.> for a proof of the existence of isothermal coordinates in the $C^\infty$ case.

However, the Riemannian metric must be more regular than only continuous. For a modern treatment, see <cit.>.  We will be concerned by conformal metrics with a different kind of regularity assumption: the logarithm of the factor $\lambda$ will be a difference of subharmonic functions. See Remark <ref> for the kind of regularity this implies.

If two $C^2$ functions $p_1$ and $p_2$ satisfy
$k=\Delta p_i$,  then
$p_1 - p_2$ is a harmonic function.
It follows that all the conformal Riemannian metrics $\lambda|dz|^2$ with curvature measure  $\mes$ are given by

\begin{equation}\label{eq:metrique conforme}\lambda =e^{-2\left(p(\mes) + h\right)}~, \end{equation}

where $h$ is a harmonic function, and $p(\mes)$ is the potential of $\mes$.

The turn  (<ref>) of an arc $\arc$, for a Riemannian conformal metric with conformal factor given by (<ref>), is given by[There is a slight difference in notation with the corresponding formula in [85], because is considering $\lambda =e^{-2p(\mes) + h}$.]

\begin{equation}\label{eq:turn riem harm}
\kappa_\lambda(\arc)=\kappa(\arc)+\int_\arc \frac{\partial  p(\mes)}{\partial \nu} +\int_K \frac{\partial h}{\partial \nu}~.
\end{equation}

We want to rewrite the expression above, in order to be able to define the turn for more general objects than a Riemannian conformal metric.  For this we need another notion, the notion of angle under which an arc is seen.

§.§.§ Angle under which an arc is seen, left and right turn of arcs

Let $\arc$ be an arc, and  $\zeta$ be a point in the plane that is not on $\arc$.
The angular function of $\arc$ angular function is a continuous function
$ \varphi(\arc,\zeta,\cdot) :[a,b]\to \R$ defined in the following way. If $\param:[a,b]\to\C$ is a parameterization of $\arc$,

$$ \varphi(\arc,\zeta,t)= \operatorname{arg} (\param(t)-\zeta)  - \operatorname{arg}(\param(a)-\zeta)~,$$

where $\operatorname{arg}(z)\in (-\pi, \pi]$arg is the principal value of the argument of $z$.

There is a priori an ambiguity in the choice of the argument.  Of course,  $\varphi(\arc,\zeta,a)=0$.
For $t$ close to $a$, $\operatorname{arg}(z(t))\in(-\pi,\pi]$ is the principal value of the argument. Then, the argument is uniquely determined by the condition that
$t\mapsto \varphi(\arc,\zeta,t)$ is continuous.

The angle under which $\arc$ is seen from the point $\zeta$angle under which an arc is seen from a point is the function $\varphi(\arc): \C\setminus K \to \R $ defined by

and it clearly does not depend on the parameterization of $\arc$.
For example, if $\arc$ is a circle, then $\varphi(\arc,\zeta)=0$ if $\zeta$ is outside the disc enclosed by the circle, and $\varphi(\arc,\zeta)=\pm 2\pi$ otherwise, depending on the orientation of the circle.

Let $\param$ be a regular parameterization of a regular arc $\arc$. If $\zeta\notin \arc$,

$$\varphi(\arc,\zeta) =-\int_a^b \frac{\partial \ln | \cdot -\zeta|}{\partial \nu}(\param(t)) |\param'(t)| dt
=-\int_\arc  \frac{\partial \ln | \cdot -\zeta|}{\partial \nu}$$

and in particular, if $0\notin \arc$,

\begin{equation}\label{eq:angle c1 zero complexe}
\varphi(\arc,0)=\int_a^b \frac{\operatorname{Re}(i
\param(t)\bar{\param}'(t))}{|\param(t)|^2}dt~.
\end{equation}

We have $\varphi(\arc,\zeta)=\int_a^b\varphi(\arc,\zeta,t)'dt$, and locally,

\begin{equation}\label{eq:der angle}\varphi(\arc,\zeta,t)'=\left(\arctan \left(\frac{\param_2(t)-\zeta_2}{\param_1(t)-\zeta_1}\right)\right)'
=-\frac{\langle \param(t)-\zeta, \param'^\bot(t)\rangle}{|\param(t)-\zeta|^2}~. \end{equation}

On the other hand,

\begin{equation}\label{eq:grad ln}\operatorname{grad} (\ln |\cdot - \zeta|)(z)=\frac{z-\zeta}{|z-\zeta|^2} \end{equation}

$$ \frac{\partial \ln |\cdot - \zeta|}{\partial \nu}(\param(t))= \frac{\langle \param(t)-\zeta , \nu\rangle}{|\param(t)-\zeta|^2}= \frac{\langle \param(t)-\zeta , \param'^\bot(t)\rangle}{|\param(t)-\zeta|^2|\param'(t)|}~.$$

Evidently, one immediately computes

$$- \frac{\partial \ln |\cdot|}{\partial \nu}(z(t))|\param'(t)|=\frac{\param_1(t)\param_2'(t)-\param_2(t)\param_1'(t)}{|\param(t)|^2}=\frac{\operatorname{Re}(i
\param(t)\bar{\param}(t)')}{|\param(t)|^2}~.$$

There is another way to interpret  Lemma <ref>. Let $h:M\to \R$ be a harmonic function on $M$. A $C^\infty$ function $h^*:M\to \R$ is said to be the (harmonic) conjugateconjugate (harmonic) of $h$ if

\begin{equation}\label{eq:CR}
\left\{
\begin{aligned}
\frac{\partial h}{\partial x}-\frac{\partial h^*}{\partial y}=0\\
\frac{\partial h}{\partial y}+\frac{\partial h^*}{\partial x}=0\\
\end{aligned}
\right.
\end{equation}

i.e., $f=h+ih^*$ is holomorphic, as (<ref>) are the Cauchy–Riemann equations. The conjugate exists on a simply connected domain if the form $- \frac{\partial h}{\partial y}dx +\frac{\partial h}{\partial x}dy$ is closed, i.e. $\Delta h=0$, that is the case as $h$ is harmonic. From (<ref>), it follows that $h^*$ is also harmonic.
Then we may compute that, for a parameterization by arc length,

$$\frac{\partial h}{\partial \normal}=\left( z_1'\frac{\partial h}{\partial y}-z'_2\frac{\partial h}{\partial x}\right)\circ z=-dh^*(z)(z')=-(h^*\circ z)'~,$$

\begin{equation}\label{eq:int conjugate}
\int_K \frac{\partial h}{\partial \normal}=h^*(z_1)-h^*(z_2)~,
%\int_K \frac{\partial h}{\partial \normal}
%& =\int_a^b z_1'(t)\frac{\partial h}{\partial %y}(z(t))-z'_2(t)\frac{\partial h}{\partial x}%%(z(t))dt \\
%& = -\int_a^b dg(z(t))(z'(t))dt =
%-\int_a^b (g\circ z)'(t) dt = g(a)-g(b)~.
\end{equation}

if $z_2$ is the endpoint of $\arc$, and $z_1$ its starting point.

The relation with Lemma <ref>
follows from the fact that on $D=\C\setminus\{\zeta +y, y<0 \}$,
the function  $z\mapsto \ln|\zeta-z|$ is the real part of $z\mapsto \operatorname{Ln}(z-\zeta)$,
with $\operatorname{Ln}$ the principal determination of the complex logarithm, i.e.,

$$\operatorname{Ln}(z-\zeta)=\ln|z-\zeta| + i \operatorname{arg}(z-\zeta)~, $$

so that on $D$, the harmonic conjugate of
$\ln|\cdot-\zeta|$ is $\operatorname{arg}(\cdot-\zeta)$, and then Lemma <ref> is essentially (<ref>).

From (<ref>)  and Lemma <ref>, we obtain the following formula,
which  is basically the content of the  first section of [85].

Let $\mes$ be a smooth Lebesgue density with compact support, and let $h$ be a harmonic function over $M$.
For $\lambda$ given by (<ref>),
the turn of a regular arc $\arc \subset M$  can be computed as follows:

\begin{equation}\label{eq:smooth turn} \kappa_\lambda(\arc)=\kappa(\arc)-\frac{1}{2\pi}\iint \varphi(\arc,\zeta) d\omega(\zeta) +\int_\arc
\frac{\partial h}{\partial \normal}~.\end{equation}

In the case of a Riemannian conformal metric, we said in Remark <ref> that $\kappa_\lambda(\arc)=0$ if $\arc$ is a shortest arc. Hence in this case:

\begin{equation}\label{eq: rot shortest riem}
\kappa(\arc)=  \frac{1}{2\pi}\iint \varphi(\arc,\zeta) d\omega(\zeta) -\int_\arc
\frac{\partial h}{\partial \normal}~.
\end{equation}

Before going on, let us mention here the following result, that is used in the proof of Theorem 9.1 in [86].

We have, for a rectifiable arc $\arc$,
if $\zeta$ is at distance $\geq \rho>0$ from $\arc$,
$$\bigvee_a^b \varphi (\arc,\zeta)\leq \frac{s(\arc)}{\rho}~.$$

Recall that the total variationtotal variation of a function $f:[a,b]\to \R$ is

\begin{equation}\label{eq:bigvee}\bigvee_a^b f = \sup \sum_i |f(t_i)-f(t_{i+1})|~, \end{equation}

where the suprememum is taken over all partitions of $[a,b]$.$\bigvee$

From the properties of the principal value of the argument, it clearly follows that if $\arc_n\to \arc$ and $\arc_n$ and $\arc$ all have the same starting point, and $\zeta\not\in K_n$,  then $\varphi (\arc_n,\zeta)\to \varphi (\arc,\zeta)$.
So  it suffices to consider the case of a broken line, the  conclusion will hold  by approximation.
Actually, due to the concatenation property of $\vee$ and $s$, it suffices to check the inequality for the segment between each vertex of the broken line. But in such a case, the conclusion follows from (<ref>), Cauchy–Schwarz inequality, and the fact that
in the smooth case, the total variation is the integral of the derivative.

Lemma <ref> allows a generalization of the concept of turn of an arc in two  directions:

* the right-hand side of  (<ref>) can be defined from a finite signed Borelian measure with compact support (that has not necessarily smooth Lebesgue density) and a harmonic function;

* the right-hand side of  (<ref>) can be defined for a non-regular arc, as we saw that the rotation can be  defined for any arc of the class $\tilde{\Delta}$, and for the  term $\frac{\partial h}{\partial \normal}$, it suffices to consider the harmonic conjugate and use  (<ref>).

The point is that the angular function is not defined from a point on the arc, that may cause some problems
to define the term $\iint \varphi(\arc,\zeta) d\omega(\zeta)$ for a more general measure.
This is why
the notion of right and left turn enters the picture.

First, if $a$ is an extremity  of $K$,
if $K_n$ is a proper subarc of $K$ converging to $K$, then we define
To extend the definition of $\varphi(K,\cdot)$ to the interior $K^\circ$$K^\circ$ of $K$, we need the notion of left and right side of a curve.

The sequence of points is converging on the left to a point in the interior of the plain curve.

Let us consider a simple closed curve. By Jordan Theorem, it has an interior and an exterior.
The angle under which the arc is seen from an exterior point is equal to zero. We say that the arc is positively orientedpositively oriented if the angle under which the arc is seen from an interior point is equal to $2\pi$. The only other possibility being that the angle is equal to $-2\pi$, in which case we say that the arc is negatively oriented.

Now consider an arc $\arc$ whose extremities do not coincide. By joining it to another suitably oriented arc $\arc'$ in order to obtain a simple close curve $\arc\arc'$, as the arcs are oriented, we are able to say when the domain enclosed by $\arc\arc'$ is on the right or on the left on the left of $\arc$.
Using this,
for a sequence of points in $\C\setminus \arc$ converging to a point
in $\arc^\circ$, we can say if the convergence if on the right or on the leftconvergence on the left of $\arc$.
Figure <ref> for an heuristic argument, and paragraph 2 in [91] for more details.

It is then possible to extend the definition of the angle $\varphi(\arc,\cdot)$ to $K$ by continuity
on the left or on the right. This  leads to the definition of left angle under which the curve is seenleft angle $\varphi_l(\arc)$, as well as  right angle $\varphi_r(\arc)$,  which  satisfy the following equalities:

\begin{equation}\label{eq:left right angles}\begin{cases}   \varphi_l(\arc)-\varphi_r(\arc)= 0 & \mbox{over }  \C\setminus \arc^\circ\\
\varphi_l(K)-\varphi_r(\arc)= 2\pi &\mbox{over }   \arc^\circ
\end{cases}~.
\end{equation}

Existence of the left and right angles and the equalities above are proved in Lemma 2 in [91]. For example, if $\arc$ is a  directly oriented circle and $\zeta\in \arc$, then
$\varphi_l(\arc,\zeta)=2\pi $, and  $\varphi_r(\arc,\zeta)=0$.

We can now define the left turnleft turn
of an arc $\arc$ of the class $\tilde{\Delta}$, given  any signed Borelian measure $\mes$ with compact support, and any harmonic function $h$ defined on a domain containing $\arc$:

\begin{equation}\label{eq:def turn}\kappa_l(K):=\kappa(K)-\frac{1}{2\pi}\iint_M \varphi_r(K,z)d\mes(z)+(h^*(z_1)-h^*(z_2))~,\end{equation}

if $z_1$ and $z_2$ are the extremities of $K$,
and the right turnright turn
\begin{equation}\label{eq:def turnb}\kappa_r(K):=-\kappa(K)+\frac{1}{2\pi}\iint_M\varphi_l(K,z)d\mes(z)-(h^*(z_1)-h^*(z_2))~,\end{equation}

and we have by (<ref>)

\begin{equation}\label{eq:left+right}
\kappa_l(K)+\kappa_r(K)=\mes(K^\circ)~.
\end{equation}

Note that  we obtain a Gauss–Bonnet FormulaGauss–Bonnet Formula almost by definition: if $K$ is a regular arc enclosing a domain $D$ homeomorphic to a disc, then

\begin{equation}\label{eq:GB}\kappa_l(K)+\mes(D)=2\pi~. \end{equation}

A more general version of (<ref>) may be found in Theorem 2 in [91].

In the next paragraph, we will look at those concepts for the simplest measures which are not smooth Lebesgue densities, that are weighted Dirac measures. The motivation for looking at the turn will be made clear in Section <ref>.

§.§.§ Flat cones and weak Laplacian

As a basic and fundamental example, let us look at the turn for arcs in a flat cone metric.
Flat cones are the simplest and the main  example of subharmonic metrics which are not Riemannian. Flat cones are an elementary example of simple singularities on Riemannian surfaces, see [108].
Also,  one of the main tools in Reshetnyak's arguments, the canonical stretching of a distance (see Section <ref>), allows to reduce some problems to this  case.

For $\zeta\in \C$ and $\mes_0\in \R$, let
be the signed measure concentrated at $\zeta$ with value $\mes_0\in \R$.
Then  define

is the weightweight (cone), and
is the
cone anglecone angle.
To simplify the exposition, let us consider $\zeta=0$.  We will look at the plane endowed with the metric $\lambda|dz|^2$.
As $z\mapsto \ln |z|$ is harmonic on $\C\setminus\{0\}$, it follows
that $\lambda|dz|^2$ is actually a Riemannian flat metric on  $\C\setminus\{0\}$. Let us see that it defines however a distance on the whole plane.

We define the length of a broken line $\arc$ in the same way as for a Riemannian metric (see Remark <ref>). If $\param$ is an arc length  parameterization (for the Euclidean length),

\begin{equation}\label{eq:length curve cone}\tilde{s}_{\lambda}(\arc)=\int_0^{s(\arc)} |\param(t)|^\beta dt~,\end{equation}

where $s(\arc)$ is the Euclidean length of $\arc$.
For example, let $v$ be a unit vector and let consider the parameterized arc
$\param:[0,r]\to \C$, $\param(t)=tv$. Hence

\begin{equation}\label{eq:dist cone zero}\tilde{s}_\lambda(\param)=\int_0^r t^\beta dt=
\left\{
\begin{array}{c c l c l c l c}
+\infty &\mbox{ if }& \beta\leq -1  &\mbox{ i.e. } & \mes_0 \geq 2\pi  &\mbox{ i.e. } & \alpha \leq 0\\
\frac{1}{\beta+1}r^{\beta+1} &\mbox{ if }&  \beta > -1 & \mbox{ i.e. }&  \mes_0< 2\pi & \mbox{ i.e. } & \alpha > 0
\end{array}
\right.~.\end{equation}

Let us denote by
$\rho_\lambda(a,b)$ the infimum of  $\tilde{s}_\lambda$ for all the broken lines between $a$ and $b$. We will denote by $C_r(\zeta)$$C_r(\zeta)$ the circle of center $\zeta$ and (Euclidean) radius $r$.

$\rho_\lambda$ is a distance over the plane.

From Remark <ref>,
$\rho_\lambda$ is a pseudo-distance over the plane. Actually the argument in the  remark also gives that it is a distance on $\C\setminus\{0\}$. (<ref>)
says that the distance from $0$ to $C_r(0)$ is positive (or infinite) for any $r>0$.

The choice of the name cone angle for $\alpha$ may be explained as follows.  Let us define

\begin{equation}\label{eqValpha}V_\alpha:=\left\{ (r,\theta) \vert r\geq 0,\,
\theta\in \R / \alpha \mathbb{Z} \right\} / (0,\theta)\thicksim (0,\theta')~. \end{equation}

Then the pull-back of the metric $|z|^{2\beta}|dz|^2$ over $\mathbb{C}\setminus \{0\}$ onto $V_\alpha \setminus \{(0,0)\}$ via the mapping

$$\left(r,\theta\right) \mapsto
\left(\left(\beta+1\right)r\right)^{\frac{1}{\beta+1}} \left( \cos\left(\frac{\theta}{\beta+1}\right), \sin\left(\frac{\theta}{\beta+1}\right)\right)$$

$$dr^2 + r^2d\theta^2~. $$

It is easy to see that (<ref>) actually gives the $\rho_\lambda$ distance from  the origin to $C_r(0)$, see <cit.>.
If follows that
$C_r(0)$ is at $\rho_\lambda$ distance one  from the origin if $\mes_0< 2\pi$
and $r=\left(1+\beta\right)^{\frac{1}{1+\beta}}$.

\begin{equation}\label{eq:long cercle cone}\tilde{s}_\lambda(C_r(0))=\int_{C_r(0)} \sqrt{\lambda}= r\int_0^{2\pi}
r^{\beta}dt= 2\pi r^{1+\beta}=2\pi (1+\beta)=\alpha~,\end{equation}

hence the length for $\rho_\lambda$  of the  unit circle for $\rho_\lambda$ is $\alpha$.
From this it is easy to recognize the distance over some usual objects, see Table <ref>.

We are cheating a little in the sense that (<ref>) is a priori not the intrinsic length of $C_r(0)$ for the distance $\rho_\lambda$ (see Section <ref>). Indeed, the intrinsic length must be defined by  the supremum of the length for $\rho_\lambda$ of all the broken lines inscribed in the curve. But we will see in Section <ref> that both definitions agree in our case.

cone angle     curvature     weight

$\alpha$     $\mes_0$     $\beta=-\frac{\mes_0}{2\pi}=\frac{\alpha}{2\pi}-1$

$0$     $2\pi$     $-1$

$]0,2\pi[$     $]0,2\pi[$     $]-1,0[$

$2\pi$     $0$     $0$

$>2\pi$     $<0$     $>0$

It is easy to see that the distance given by the first line is isometric to the induced distance  over a Euclidean cylinder of radius $2\pi$. For the second line, it is a convex cone in the Euclidean $3$-space. Of course, the third line corresponds to the Euclidean plane. It is also possible to see that the distance given by the  last line corresponds to the induced distance on a convex  cone in the Lorentzian Minkowski $3$-space, i.e., $\R^3$ endowed with the bilinear form $dx^2+dy^2-dz^2$, see Example <ref>.

It follows that the point $0$ is at finite distance from the other points if $\mes(\{0\})<2\pi$, and at infinite distance if  $\mes(\{0\})\geq 2\pi$. This will be a general phenomenon for the subharmonic distances, except for the case $=2\pi$, that may behave in  different manners, depending on $\mes$. This is the source of some technical difficulties.

Note that

for any value of $\alpha>0$.

Let us mention the following basic result, which is used for example in the proof of Lemma 16 in [91].

For $\beta>-1$ i.e., $\omega_0<2\pi$,
the sector of the plane between two rays making an angle $\theta$ for the distance  $\rho_\lambda$ is isometric to a Euclidean sector of angle $\theta\left(1-\frac{\mes_0}{2\pi} \right)$.

Let us consider the function

$$f:z\to (1+\beta)^{-1}z^{\beta+1}~,$$

defined for a suitable branch cut of the plane.
For an arc length parameterization $\param$ of a rectifiable arc $K$, we have

$$\tilde{s}_\lambda(f(\arc))=\int_0^{s(\arc)} |\param(t)|^{-\frac{\omega_0}{2\pi}}dt~. $$

Comparing with (<ref>), it follows that $f$  is an isometry from
the plane with the metric $ \lambda|dz|^2$  to the Euclidean plane.
Without loss of generality we can consider that the sector is between vectors $1$ and $e^{i\theta}$, and
$f(e^{i\theta})=(1+\beta)^{-1}e^{i\theta(\beta+1)}$, so that the new sector has an angle $\theta(\beta+1)$.

As we already noted, on $\mathbb{C}\setminus\{0\}$, a flat cone metric is a Riemannian metric. Considering the function $f$ from the proof of Lemma <ref>, together with Remark <ref>,
$f$ is a Riemannian isometry
between the disc without the origin endowed with the
metric $|z|^\beta|dz|^2$ and the standard Euclidean metric.
So  if $\arc$ is a regular arc not passing through the origin, we have the following formula for the geodesic curvature of $\arc$, see Lemma <ref>,

\begin{equation}\label{eq:turn flat metric zero}
\kappa_\lambda(\arc)=\kappa(\arc)+\beta \varphi(\arc,0)~.
\end{equation}

An argument similar to the proof of Lemma <ref> gives  the following result, which is used in the beginning of the proof of Lemma 13 in [91].

Let $\arc$ be a regular arc not passing through the origin. Then

$$\kappa(\arc^{\beta+1})= \kappa_\lambda(\arc)~, $$

with $\lambda=|\cdot|^{2\beta}$, and $K^{\beta+1}$ is the image of the arc under $f(z)=z^{\beta+1}$.

It suffices to compute $\kappa(\arc^{\beta+1})$ with
(<ref>), and then use (<ref>) and (<ref>).

It is easily checked that the Euclidean length of the curve $(r,\theta=\frac{1}{r}-\pi)$, $0<r<\frac{1}{\pi}$ in polar coordinates, and $0$ at $r=0$, is infinite. Its length in the flat cone metric defined by $\lambda(z)=|z|^2$ is finite, see the last section of [90]. The distance given by $\lambda$ is isometric to the induced distance on the cone in $\R^3$, endowed with the metric $dx^2+dy^2-dz^2$, generated by the rotation preserving horizontal planes of the half-line of direction $(2,0,\sqrt{3})$.

One of the main result of [91] (see the end of Section <ref>)
applied to our flat cone situation, implies the well-known fact that a shortest arc for a flat cone metric with $0<\mes_0<2\pi$, i.e., with a cone angle  in $]0,2\pi[$, cannot contain the origin in its interior.

So far, we have considered signed measures with smooth Lebesgue density or weighted Dirac measure. It would then be natural to consider signed measures which are a sum of measures of those two kinds. Such measures are the main objects of the last part of  [91]. Indeed, they are dense (for the weak convergence of measures) in a set of measures $A$. The  Localization Theorem, see Section <ref>, will say that any of the signed measures considered in Reshetnyak's theory
can be transformed in a suitable manner to
a measure of the set $A$.

In Section <ref>, we saw the relation between the measure and the Laplacian in the Riemannian case. A similar analysis is possible here.
The weak Laplacian of a locally $L^1$ function $f:M\to \R$ over a domain $M$ of the plane  is the linear functional on $C_c^\infty(M)$, the space of $C^\infty$ functions with compact support included in $M$, defined by

$$\Delta f(\varphi)=\iint_M f \Delta\varphi~.$$

Here we are making an abuse of notation, as if $f\in C^2$, then the notation $\Delta f$ may stand for a continuous function or a linear functional on $C^\infty_c$. But if one recalls  the Green formula
(<ref>), we have the following relation for $f\in C^2$:

$$\Delta f(\varphi)=\iint_M \varphi \Delta f~,$$

i.e., if $f$ is $C^2$, the function $\Delta f$  is equal as a distribution to the distribution $\Delta f$.

Let us go back to the simplest non-smooth example.

For $\zeta\in \C$, let $p(\delta_\zeta)$ be the function defined by
$p(z;\delta_\zeta)=\frac{1}{2\pi}\ln|z-\zeta|$.  Then for $\varphi\in C_c^\infty$,

$$\Delta p(\delta_\zeta)(\varphi) =\varphi(\zeta)~.$$

It is a straightforward consequence of Lemma <ref>, by taking a domain containing the support of $\varphi$.

Hence, as a distribution, the weak Laplacian of $p(\delta_\zeta)$ is the Dirac measure $\delta_\zeta$. But it is classical that if the weak Laplacian is a positive functional, i.e., if for a $L^1$ function $f$ we have
$\Delta f(\varphi)\geq 0$ for any non-negative $\varphi\in C_c^\infty$, then it extends to a positive linear functional over $C_c^0$  [81], [53], [82]. In turn, by the Riesz Representation Theorem, it extends to a positive measure. At the end of the day, we may identify $\Delta p(\delta_\zeta)$ and $\delta_\zeta$ as positive measures.

In turn, we obtain an analogue of (<ref>), in term of measures, as if we consider the flat cone metric

$$\lambda(z)=|z-\zeta|^{-\frac{\mes_0}{\pi}}~, $$

then, in the sense of measures, by Lemma <ref>

\begin{equation}\label{cone measure weak lap flat}\mes_0\delta_\zeta=-\frac{1}{2}\Delta \ln \lambda~.\end{equation}

In the other sections, we will bring what we do for Riemannian metrics and flat metrics to its full generality. Before that, let us mention two natural generalizations of flat metrics: spherical  cone metrics
and hyperbolic cone metrics.

[Spherical cone-metrics]

Let us consider the conformal Riemannian metric

\begin{equation}\label{eq:def spher hyp}\lambda(z)|dz|^2= \frac{4}{\left(1+|z|^2\right)^2}|dz|^2~, \end{equation}

for $z$ belonging to  the plane.
A straightforward computation (see (<ref>)) shows that the curvature is constant equal to $1$.
The plane endowed with the metric (<ref>) is a conformal representation of a
spherical metricspherical metric.  On may check that
(<ref>) is actually given by the stereographic projection (from the south pole) of the induced metric on the unit sphere in the Euclidean $3$-space.

Over $\C$ minus the origin,
let us consider the function  $f(z)=\frac{1}{\beta+1}z^{\beta+1}$, for a suitable branch cut of the plane and $\beta> -1$. The spherical metric is pull-backed by $f$ (see Remark <ref>) to a conformal metric with conformal factor

$$\tilde{\lambda}(z)=\lambda(f(z))|f'(z)|^2=\frac{4|z|^{2\beta}}{\left(1+(\beta+1)^{-2}|z|^{2(\beta+1)}\right)^2}~. $$

The pull-back of the metric $\tilde{\lambda}|dz|^2$ onto the cone $V_\alpha$ given by (<ref>), where $\alpha=2\pi(\beta+1)$, via the function

$$(r,\theta)\mapsto \left(\left((\beta+1)\tan(r/2)\right)^{\frac{1}{\beta+1}}
\left(\cos\left(\frac{\theta}{\beta+1}\right),\sin\left(\frac{\theta}{\beta+1}\right)\right)   \right)$$

is written

$$dr^2+\sin(r)^2 d\theta^2~. $$

Let us compute

$$-\frac{1}{2}\Delta \ln \tilde{\lambda}(z)=
-\frac{1}{2}\Delta \ln |z|^{2\beta} + \Delta \ln \left(1+(\beta+1)^{-2}|z|^{2(\beta+1)}\right)~.

Considering the equation above in the sense of distribution,
by (<ref>), we know that the term
$-\frac{1}{2}\Delta \ln |z|^{2\beta}$ is the measure
$-2\pi\beta\delta_0$. Concerning the second term $\Delta \ln \left(1+(\beta+1)^{-2}|z|^{2(\beta+1))}\right)$, the function

$$z\mapsto\ln \left(1+(\beta+1)^{-2}|\cdot|^{2(\beta+1)}\right)$$
has no singularity on $U$, and a straightforward computation shows that

$$\Delta \ln \left(1+(\beta+1)^{-2}|z|^{2(\beta+1)}\right)=\tilde{\lambda}(z)~,$$

hence in the sense of distribution, $-\frac{1}{2}\Delta \ln \tilde{\lambda}(z)$ is equal to the measure

$$\omega_0\delta_0 + \tilde{\lambda}\mathcal{L} $$

where $\mathcal{L}$ is the Lebesgue measure over $U$ and

[Hyperbolic cone-metrics]

Let us consider the conformal Riemannian metric

\begin{equation}\label{eq:def met hyp}\lambda(z)|dz|^2= \frac{4}{\left(1-|z|^2\right)^2}|dz|^2~, \end{equation}

for $z$ belonging to a subdomain of the open unit disc.
A straightforward computation (see (<ref>)) shows that the curvature is constant equal to $-1$, so (<ref>) is a conformal representation of a
hyperbolic metrichyperbolic metric.
Suppose that $U$ contains the origin.
Over $U$ minus the origin,
let us consider the function  $f(z)=\frac{1}{\beta+1}z^{\beta+1}$, for a suitable branch cut of the plane and $\beta> -1$. The hyperbolic metric is pull-backed by $f$ (see Remark <ref>) to a conformal metric with conformal factor

$$\tilde{\lambda}(z)=\lambda(f(z))|f'(z)|^2=\frac{4|z|^{2\beta}}{\left(1-(\beta+1)^{-2}|z|^{2(\beta+1)}\right)^2}~. $$

The pull-back of the metric $\tilde{\lambda}|dz|^2$ onto the cone $V_\alpha$ given by (<ref>), where $\alpha=2\pi(\beta+1)$, via the function

$$(r,\theta)\mapsto \left(\left((\beta+1)\tanh(r/2)\right)^{\frac{1}{\beta+1}}
\left(\cos\left(\frac{\theta}{\beta+1}\right),\sin\left(\frac{\theta}{\beta+1}\right)\right)   \right)$$

is written

$$dr^2+\sinh(r)^2 d\theta^2~. $$

Let us compute

$$-\frac{1}{2}\Delta \ln \tilde{\lambda}(z)=
-\frac{1}{2}\Delta \ln |z|^{2\beta} + \Delta \ln \left(1-(\beta+1)^{-2}|z|^{2\beta+1}\right)~.

Computations similar to the ones of the spherical case (Remark <ref>) show that, in the sense of distribution, $-\frac{1}{2}\Delta \ln \tilde{\lambda}(z)$ is equal to the measure

$$\omega_0\delta_0 - \tilde{\lambda}\mathcal{L} $$

where $\mathcal{L}$ is the Lebesgue measure over $U$ and

§.§ Arcs of bounded rotation

The arcs of (absolute) bounded rotation are the basic objects of the whole theory. They form a proper subclass of the class of rectifiable arcs of the plane, which contains itself smooth arcs and broken lines.

The main property that distinguish them from the class of rectifiable arcs is the convergence of the length for a converging sequence of arcs having a uniform bound of the absolute rotation (Lemma <ref>). This is   an essential step in the proof of the Distances Convergence Theorem <ref>.

Roughly speaking, if rectifiable arcs have a Lipschitz parameterization, arcs of bounded rotation have a parameterization, see Section <ref>.

In all this section, we are considering the Euclidean plane. References for this section are [9], [72], [104], [95]. Basically, we review the properties stated in $\S$3 of [86].

§.§.§ Classical results

Let $\arc$ be a broken line, parameterized by affine functions.  Its absolute rotationabsolute rotation $|\kappa|(\arc)$ is the non-negative real number that is the sum of the angles (in $[0,\pi]$) between its left and right derivative at its interior vertices.

The absolute rotation $|\kappa|(\arc)\in\R\cup \{+\infty\}$ of an arc $\arc$ is the supremum of the absolute rotation of the broken lines inscribed in $\arc$. The arc is said to be of bounded rotationbounded rotation if $|\kappa|(\arc)$ is finite.

It is clear that if we modify a broken line $\arc$ by adding a new vertex, we obtain a new broken line $\arc'$ such that $|\kappa|(\arc')
\geq |\kappa|(\arc)$, see Figure <ref>. From this, it is easy to see that for a broken line, the two definitions of absolute rotation we have given above (as a sum of angles on the one hand, and as a supremum of absolute rotations of inscribed broken lines on the other hand) coincide.

$\alpha'+\gamma+\beta' =\alpha'+x+y+\beta'\geq \alpha+\beta $.

Let $(\arc_n)_n$ be a sequence of arcs converging to $\arc$. Then  $ |\kappa|(\arc) \leq \liminf |\kappa|(\arc_n)$.

Let $L$ be a broken line inscribed in $\arc$, and let $L_n$ be broken lines inscribed in $\arc_n$ such that the associated sequences of vertices converge to the vertices of $L$. Clearly, $|\kappa|(L_n)\to |\kappa|(L)$ and
$|\kappa|(L_n)\leq |\kappa|(\arc_n)$, so $|\kappa|(L) \leq \liminf |\kappa|(\arc_n) $, and as $L$ is arbitrary,  $  |\kappa|(\arc)=\sup |\kappa|(L)  \leq \liminf |\kappa|(\arc_n)$.

Let $(\arc_n)_n$ be a sequence of broken lines inscribed in $\arc$ and converging to $\arc$. Then $s(\arc_n)\to s(\arc)$ and $|\kappa|(\arc_n)\to |\kappa|(\arc)$.

We skip the proof, as the one for the length is classical, and
the one  for the absolute rotation is
analogous. This result gives for example that,
as the absolute rotation of a closed convex broken line is $2\pi$, the absolute rotation of a circle is $2\pi$.

If $A$ is an interior vertex of a broken line $\arc$, and $\arc_1, \arc_2$ the two broken lines obtained from $\arc$ by splitting at $A$, then if $\alpha$ is the angle at $A$, by definition

$$|\kappa|(K)=|\kappa|(K_1)+|\kappa|(K_2)+\alpha~. $$

It follows that, for any decomposition  $a=t_0<t_1 <\cdots < t_N=b$ of a parametrization of an arc $\arc$,
\begin{equation}\label{eq:sum rot subarc}\sum_i |\kappa|(K\vert_{ t_i,t_i+1})\leq |\kappa|(\arc)~.\end{equation}

A sharper version of (<ref>)  will be given  in Corollary <ref>.
From (<ref>) we deduce that  $t\mapsto |\kappa|(K\vert_t)$ is non-decreasing. This is used in the proof of
the following result, which is Theorem 5.8.1  in [9].

Let $\arc$ be an arc with $|\kappa|(\arc)< \pi$ and extremities $z_1$ and $z_2$. Then

$$\cos \left(|\kappa|(\arc) / 2 \right) s(\arc)\leq \vert z_1-z_2\vert ~.$$

From Lemma <ref>, it suffices to prove the formula for broken lines.
Let $\param:[0,s(\arc)]\to \C$ be the arc length parameterization of a  broken line. In particular, its left derivative has unit norm. Up to a rotation, suppose that the first segment of the broken line is directed by $(1,0)$. Let $e$ be a unit vector with principal argument  $|\kappa|(K)/2$. So for any $t\in [0,s(\arc)]$,
the angle between $\param'_l(t)$ and $e$ is less than  $|\kappa|(K)/2$, i.e., $\langle \param'_l(t),e \rangle \geq \cos \left(|\kappa|(\arc)/2 \right)$, and

$$\vert z_1 - z_2\vert \geq \langle z_1 - z_2 , e \rangle = \int_0^{s(\arc)} \langle \param_l', e \rangle \geq s(\arc) \cos \left(|\kappa|(\arc)/2 \right)~. $$

That's not true that for any $t_0\in [a,b]$ there exists a sufficiently small $\delta $ such that $|\kappa|(\arc_{t_0-\delta,t_0+\delta})$ is arbitrarily small, as the example of an inner vertex of a broken line shows. But this property is true on a right neighborhood $[t_0, t_0+\delta]$ for $t_0\in [a,b)$ (of course, a similar property holds
on a left neighborhood $[t_0-\delta, t_0]$ for $t_0\in (a,b]$). Indeed, we have for $t_n\to t_0\in [a,b)$, $t_0<t_n<b$, from (<ref>),

$$|\kappa|(\arc\vert_{t_0,b}) - |\kappa|(\arc\vert_{t_n,b})\geq |\kappa|(\arc\vert_{ t_0,t_n }) $$

and from $|\kappa|(\arc\vert_{t_0,b}) \geq |\kappa|(\arc\vert_{t_n,b})$ and Lemma <ref>,
it follows that  $|\kappa|(\arc\vert_{t_n,b}) \to |\kappa|(\arc\vert_{t_0,b})$, and hence

\begin{equation}\label{eq:kappa point=0}
|\kappa|(\arc\vert_{ t_0,t_n}) \xrightarrow[n\to \infty]{}  0~.
\end{equation}

Let  $\epsilon >0$,
and let us define $t_i\in [a,b)$ by
$t_0=a$ and
$|\kappa|(\arc\vert_{ t_{k-1},t_k})= \epsilon$, that is possible by (<ref>).
If $ |\kappa|(\arc) \leq A<\infty $, the number $N$ of such $k$ is finite.
Actually, we have

$$A\geq |\kappa|(\arc) \geq \sum_{k=1}^N |\kappa|(\arc\vert_{ t_{k-1},t_k})\geq
\sum_{k=1}^{N-1} |\kappa|(\arc\vert_{ t_{k-1},t_k})=(N-1)\epsilon~, $$

so that $N\leq \frac{A}{\epsilon}+1$.

Hence, for any arc with  $|\kappa|(\arc) \leq A$, for any $\epsilon >0$  there exists $N\leq \frac{A}{\epsilon}+1$ and a decomposition  $a=t_0<t_1 <\cdots < t_N=b$  such that for any $i$,  $|\kappa|(\arc\vert_{ t_i,t_i+1})\leq \epsilon$.  The decomposition depends on the arc, but $N$ depends only on $A$ and $\epsilon$.

In particular, any  arc of bounded rotation can be decomposed into a finite number of subarcs with absolute rotation less than $\pi$. Applying  Proposition <ref> to these subarcs, the concatenation property of the length gives the following result.

An arc of bounded rotation is rectifiable.

Let $A\geq 0$ and let $(\arc_n)_{n\geq 1}$, $\arc_0$ be arcs such that $\arc_n\to\arc_0$ and for any $n\geq 1$, $|\kappa|(\arc_n)\leq A$. Then  $s(\arc_n)\to s(\arc)$.

As we already know that $\liminf s(\arc_n)\geq s(\arc)$, it suffices to show that $\limsup s(\arc_n)\leq s(\arc)$. Let us consider parameterizations $\param_n$ proportional to  arc length and   a subsequence  $(\arc_{n_i})_{n_i}$ such that $s(\arc_{n_i})$ converges to $\limsup s(\arc_{n})$.  Let $\epsilon >0$. By Remark <ref>, we know that there exists   $N\leq \frac{A}{\epsilon}+1$ and a decomposition   $0=t_0^{n_i}<t_1^{n_i} <\cdots < t_N^{n_i}=1$ such that  $|\kappa|(\arc_{n_i}\vert_{ t^{n_i}_k,t^{n_i}_k+1})\leq \epsilon$. For a given $k$, we  extract from $(t^{n_i}_k)_{n_i}$ a  subsequence  converging to some number $t_k\in [0,1]$. Abusing notation, we will denote with the same indices $n_i$ this subsequence.
Let $L_{n_i}$ (resp. $L$) be the broken line with vertices  at $t^{n_i}_k$ (resp. $t_k$), so it is inscribed in $\arc_{n_i}$ (resp. $\arc$). By construction,
$$s(L_{n_i}) \to s(L)\leq s(\arc)~.$$
From Proposition <ref>,

$$s(\arc_{n_i}\vert_{t^{n_i}_k,t^{n_i}_{k+1}}) \leq \frac{1}{\cos\epsilon} \vert \param_{n_i}(t^{n_i}_{k+1})- \param_{n_i}(t^{n_i}_{k}) \vert$$

$$s(\arc_{n_i}) \leq \frac{1}{\cos\epsilon} s(L_{n_i})$$

and passing to the limit,

$$\limsup s(\arc_{n}) =  \lim s(\arc_{n_i}) \leq  \frac{1}{\cos\epsilon} \lim  s(L_{n_i}) \leq   \frac{1}{\cos\epsilon} s(\arc)  $$

and the result follows because $\epsilon$ was arbitrary.

For the proof of Lemma <ref>, we have copied the argument of Lemma 6.2 in [86]. Note that the statement in this reference is a bit more general, but only the version of Lemma <ref> will be useful in the sequel.  Lemma <ref> also follows from the following more general result (see Theorem 5.6.2 in [9]): there exists a positive constant $C$ such that, for two arcs $\arc_1, \arc_2$ of bounded rotation,

$$\vert s(\arc_1)-s(\arc_2)\vert \leq C \rho(\arc_1,\arc_2)(|\kappa|(\arc_1)+|\kappa|(\arc_2)+\pi)$$

where $\rho$ is the distance between $\arc_1$ and $\arc_2$.

Let $A\geq 0$ and let $(\arc_n)_{n\geq 1}$, $\arc_0$ be arcs such that $\arc_n\to\arc_0$ and for any $n$, $|\kappa|(\arc_n)\leq A$. Then there exists $N$ such that  for all $n$, $\arc_n$ can be decomposed into $N$ subarcs of absolute rotation less than $\pi$, and any sequence of subarcs converge to the corresponding subarc of $\arc_0$.

The equidecomposition of the $\arc_n$ is given by Remark <ref>. It remains to check that the sequence of subarcs converge to a subarc of $\arc$. But this follows from the concatenation property of the length and Lemma <ref>.

For any arc $\arc$ of bounded rotation,

\begin{equation}\label{aq: rot diam long}s(\arc) \leq \frac{\operatorname{diam}(K)}{2} \left( |\kappa|(\arc) +\pi\right)~.\end{equation}

For the proof of Proposition <ref>, we refer to Theorem 5.6.1 in [9].
It uses projection techniques, similar to the ones in paragraph 8 in [91].
There is one case for which (<ref>) is straightforward: when the absolute rotation is sufficiently small. Indeed,  consider the function over $[0,\pi)$ defined by $f(x)=\cos(x/2)(x+\pi)/2$. As $f'(0)>0$ and $f(0)>1$, there exists
$T\in (0,\pi)$ such that for $x\in (0,T)$,

\begin{equation*}\label{eq: cos 1}1\leq  \cos(x/2)(x+\pi)/2~.\end{equation*}

Then (<ref>)  follows from Proposition <ref>

For any sequence of arcs contained in a bounded domain, with absolute rotation uniformly bounded from above, one can extract a converging subsequence with converging length.

Let us present a related class of curves, that are introduced in [86].
Let $0<h<1$.
A rectifiable arc $\arc$ satisfies the condition $\Gamma(h)$$\Gamma(h)$ condition if, for its arc length parameterization $\param$, for any $t_1,t_2\in[a,b]$,

$$|z(t_1)-z(t_2)|\geq h|t_1-t_2|~. $$

If the arc is rectifiable, we already know that $z:[0,s(K)]\to \arc\subset \C$ is a $1$-Lipschitz homeomorphism. The condition  $\Gamma(h)$ means that $z^{-1}$ is $h$-Lipschitz. In turn, the arc length parameterization is bi-Lipschitz.
By definition,

$$\operatorname{diam}(\arc)\geq |z(t_1)-z(t_2)|\geq h|t_1-t_2| $$

for any $t_1,t_2$.

Let us consider an arc of bounded rotation, decomposed  it into a finite number of subarcs of absolute rotation $<\pi$. For such a subarc $\arc_i$, Proposition <ref> says that it satisfies the property $\Gamma(h)$ with

§.§.§ Arcs of bounded rotation and convex functions

Let us first prove a regularity property of arcs of bounded rotation.

Let $\param:[a,b]\to\C$ be the arc length parameterization of an arc of bounded rotation. Then $\param$ has a right derivative $\param'_r$  on $[a,b)$, which is a unit vector.

Of course  a similar result holds for the left derivative on $(a,b]$. In particular, an arc of bounded rotation belongs to the class $\tilde{\Delta}$ introduced in Definition <ref>.

Let $t_0\in [a,b)$, and $t_0<t_1<t_2<t_0+\delta$ for  a sufficiently small $\delta >0$.
Let us consider the triangle $t_0t_1t_2$, $\alpha$ being the angle at $t_0$ and $\beta$ being the angle at  $t_1$ of this triangle. Up to change $t_1,t_2$, we can suppose that $\alpha+\beta<\pi$, otherwise the right of the arc at $\param(t_0)$ is a segment and we are done. Let $k=\pi-\beta$; $k$ is the absolute rotation of the broken line with vertices $t_0$, $t_1$ and $t_2$. By definition of the absolute rotation,
$k\leq |\kappa|(\arc\vert_{t_0,t_0+\delta})$ and hence $\alpha<  |\kappa|(\arc\vert_{t_0,t_0+\delta})$, and  by (<ref>) we know that |$\kappa|(\arc\vert_{t_0,t_0+\delta})$ can be made arbitrary small. In turn, if $t_n\to t_0, t_n>t_0$,
the sequence of unit vectors $\frac{\param(t_n)-\param(t_0)}{|\param(t_n)-\param(t_0)|}$ is a Cauchy sequence on the unit circle, hence converge.

On  the other hand, as $\param$ is an arc length parameterization, from Proposition <ref> we have

$$1\geq \frac{\vert \param(t_n)-\param(t_0)\vert}{t_n-t_0} \geq \cos \left(|\kappa|(\arc\vert_{t_0,t_n})/2 \right)~,$$

so if $t_n\to t_0$, by (<ref>),  $\frac{\vert \param(t_n)-\param(t_0)\vert}{t_n-t_0}\to 1$.

At the end of the day, for $t_n>t_0$, $\frac{ \param(t_n)-\param(t_0)}{t_n-t_0}=\frac{  \param(t_n)-\param(t_0)}{\vert \param(t_n)-\param(t_0)|}\frac{\vert \param(t_n)-\param(t_0)\vert}{t_n-t_0}$ converges when $t_n\to t_0$, and it is also immediate that the limit has unit norm.

For a rectifiable  arc $\arc$ such that its arc length parameterization $\param$ admits a right derivative, we denote by $T_r(\arc):[a,b)\to \mathbb{S}^1$ its right tantrixtantrix,
that is  the function $t\mapsto \param_r'(t)$.

Let $\arc$ be
a rectifiable arc such that $T_r(K)$ is defined. Then
for $0<\eta<s(K)$,

$$\bigvee_0^{s(K)-\eta} T_r(\arc)\leq |\kappa|(\arc)~. $$

Here, the total variation of $T_r(\arc)$ is taken for the intrinsic distance $d_S$ over the circle: the definition is formally the same as in (<ref>), with $d_S$ instead of $|\cdot|$.$\bigvee$

The statement is obvious if the absolute rotation is not finite, so we can suppose that $\arc$ has bounded rotation.
It also suffices to prove the lemma for the subarc of $\arc$ parameterized on $[0,s(K)-\eta]$, for an arc length parameterization of $\arc$.
Note that $T_r$ is defined everywhere on the subarc. Abusing notation, we will denote this subarc by $\arc$ in the following.

Let $\epsilon>0$. Let $\param$ be the arc length parameterization of $\arc$
and $0=t_0<t_1<\cdots<t_n=s(K)$ be such that
$\bigvee_0^{s(K)} T_r(\arc) - \sum d_S(\param_r'(t_i),\param_r'(t_{i+1}))\leq \epsilon$.
Let $L$ be a broken line inscribed in $\arc$ such that $|\kappa|(\arc)-|\kappa|(L)\leq \epsilon$.
This last inequality remains true if we add vertices to $L$, so we can moreover suppose that the images of the $t_i$ by the arc length parameterization $\tilde{\param}$ of $L$ are vertices of $L$, and
also that
the angle between $\tilde{\param}_r'(t_i)$ and $\param_r'(t_i)$ is less than $\frac{\epsilon}{2n}$. This is possible if the vertex of $L$ next to the image of $t_i$ is sufficiently close to it, because $\param$ has a right derivative at $t_i$.
Also, the inequality we want to prove is clearly true for broken lines.
Summing up:

\begin{align}\bigvee_0^{s(K)} T_r(\arc)& \leq  \epsilon + \sum d_S(\param_r'(t_i),\param_r'(t_{i+1}))\\
& \leq  \epsilon+ \sum d_S(\param_r'(t_i),\tilde{\param}_r'(t_{i}))+d_S(\tilde{\param}_r'(t_i),\tilde{\param}_r'(t_{i+1}))+
d_S(\tilde{\param}_r'(t_{i+1}),\param_r'(t_{i+1})) \\
&\leq  2\epsilon + |\kappa|(L) \\ &\leq 3\epsilon + |\kappa|(\arc) \end{align}

and as $\epsilon$ was arbitrary we have $\bigvee_0^{s(K)} T_r(\arc)\leq |\kappa|(\arc)$.

A function $f:[a,b]\to \R$ is   if
there exist continuous convex functions $g_{1}, g_2:[a,b]\to \R$, such that $f=g_1-g_2$. A parameterized arc is called (or DC DC function) if its components are functions. See Remark <ref> about functions in metric geometry.

The arc length parameterization of an arc  of bounded rotation is $\delta$-convex.

By Lemma <ref>, the arc length parameterization $\param$ of an arc of bounded rotation has a right derivative of bounded variation in the unit circle on $[a,b-\eta]$ for any small $\eta>0$. For $x=(x_1,x_2),y=(y_1,y_2)$ unit vectors in the plane, if $d_S$ is the intrinsic distance on the circle, there is a positive constant $c$ such that $c(|x_1-y_1|+|x_2-y_2|)\leq |x-y| \leq d_S(x,y)$.  Hence  $(\param_1)_r'$ and $(\param_2)_r'$ are real-valued functions of bounded variation, and hence, for example,
$(\param_1)_r'=f_1-f_2$, where $f_1$ and $f_2$ are  two bounded non-decreasing functions on $[a,b-\eta]$.
It is known that the function $F_i$, $i=1,2$, defined by

$$F_i(t)=\int_a^t f_i~ $$

is convex on $(a,b)$, and by properties of the integral, $F_i$ is continuous and hence convex over $[a,b-\eta]$. Moreover,

$$(F_1-F_2)'_r(t)=\lim_{h\to 0, h>0}(f_1-f_2)(t+h)=\lim_{h\to 0, h>0}(\param_1)_r'(t+h)~.$$

As $f_i$ are non-decreasing, they are  continuous except on an at most countable set of points $E_i$, hence $(z_1)'_r$ is continuous on $(a,b)\setminus (E_1\cup E_2)$, hence
on this set $(F_1-F_2)'_r=(z_1)'_r$.
As $F_i$ and $\param_1$ are Lipschitz,
they have  derivative  almost everywhere on $(a,b)$ by Rademacher Theorem, so almost everywhere $(F_1-F_2)'=z_1'$,

$$\param_1(t)-\param_1(a)=\int_a^t \param_1' =\int_a^t (F_1-F_2)'  $$

so, up to add a constant to $F_1$, $\param_1=F_1-F_2$, as all those functions are continuous. Hence $\param_1$ is on $[a,b-\eta]$ for any $\eta$.

Now there are two ways to conclude. We can do the same argument with the left derivative instead of the right derivative to conclude that $\param_1$ is also on $[a+\eta,b]$, and use the fact that a locally function is [51].
Otherwise, as $\param$ has a left derivative at $b$, it is not hard to see that we can add a little segment to the endpoint of $\arc$, colinear to $\param_l'(b)$, and that the new curve has also bounded rotation. As our starting curve is a proper subarc, the partial result we obtained proves the statement of the proposition.

A function inherits many properties of convex functions. In particular, it follows that for any $t_0\in (a,b]$,

\begin{equation}\label{eqRV}\lim_{h\to 0, h>0} T_r(\arc)(t-h)=T_l(\arc)(t)~, \end{equation}

see e.g. [98]. We have a similar relation by exchanging the role of $T_r$ and $T_l$.

Hence, for an arc $\arc$ of bounded rotation,  we may extend $T_r(K)$, that is defined only on $[a,b)$, to a function $\tilde{T}_r(\arc) $ defined on $[a,b]$ by setting
$\tilde{T}_r(\arc)=T_r(\arc) $ on $[a,b)$ and

Let $\arc$ be a rectifiable arc such that
$\tilde{T}_r(\arc)$ is defined. Then

$$\bigvee_0^{s(\arc)} \tilde{T}_r(\arc)= |\kappa|(\arc)~. $$

We need to prove the converse inequality of Lemma <ref>.
The proof is similar, we leave it to the reader.

Note that it is almost immediate that the absolute rotation of the graph of a continuous convex function is equal to the angle between the directions of the tangents at its extremities.  Deeper, $\delta$-convex functions are essentially characterized by the fact that their right derivative has bounded variation [98], and this implies the following result.

An arc with a parameterization has bounded rotation.

Let $\arc$ be an arc of bounded rotation with arc length parameterization $\param$. For  $t\in (0,s(K))$, if $\arc$ is splited at $\param(t)$ into arc $\arc_1$ and $\arc_2$,

$$|\kappa|(\arc)=|\kappa|(\arc_1)+|\kappa|(\arc_2)+\angle(t)~, $$

where $\angle(t)$ is the angle between $T_r(\arc)$ and $T_l(\arc)$ at $t$.

From Proposition <ref>,

|\kappa|(\arc) = \bigvee_0^{s(K)} \tilde{T}_r(\arc) =\bigvee_0^{t} \tilde{T}_r(\arc)  + \bigvee_t^{s(K)} \tilde{T}_r(\arc) = \bigvee_0^{t} \tilde{T}_r(\arc)  + \bigvee_t^{s(K)} \tilde{T}_r(\arc_2)~.$$

Using the triangle inequality, if $d_S$ is the intrinsic distance of the circle, it is easy to see that

$$\bigvee_0^{t} \tilde{T}_r(\arc)
= \bigvee_0^{t} \tilde{T}_r(\arc_1)  +
d_S(T_r(\arc)(t),\lim_{h\to 0, h>0}T_r(\arc)(t-h))~.$$

The result follows from (<ref>) and that of course

§.§.§ Bounded rotation and bounded turn

In this part, we relate the absolute rotation to the rotation and the angular function, both introduced in Section <ref>.
We can first note that for a regular

$$\vert \kappa\vert (\arc) =\int_\arc |k|(\arc)~.$$

This is however more general. We have defined the rotation for the arcs of the class $\tilde{\Delta}$. Now let us introduce the rotation functionrotation function of the arc $\arc$:  $\kappa(t)=\kappa(K|_t)$. Of course this is not defined for any arc, as we need that for any $t$, $K|_t$ belongs to $\tilde{\Delta}$. But that is defined for an arc of the class $\Delta$:$\Delta$ (class of arcs) an arc in the class $\tilde{\Delta}$ having a parameterization with non-zero left and right derivative at each interior point.
By definition, for a parameterization on $[a,b]$, $\kappa(K)=\kappa(b)$.

For this class of arcs we can now give the following relation: an arc of bounded rotation is an arc of the class $\Delta$ with bounded total variation of the rotation function.

Let $\param:[a,b]\to \C$ be  a suitable parameterization of an arc of bounded rotation. Then

\begin{equation}\label{eq:kappa tilde delta}\kappa(K)=\operatorname{arg}(z_l'(b))-\operatorname{arg}(z_r'(a))~. \end{equation}

We have from (<ref>)

$$\operatorname{arg}(z_l'(b))-\operatorname{arg}(z_l'(a))=\int_a^b \operatorname{arg}(z'(t))'dt~, $$

$$\operatorname{arg}(z')'=\arctan\left(\frac{z_2'}{z_1'}\right)' =\frac{z_1'z_2''-z_1''z'_2}{|z'|^2}=k(z)|z'|~.$$

For an arc $\arc$ of bounded rotation, if $\kappa$ is its rotation function on $[a,b]$, then

$$|\kappa|(K)=\bigvee_a^b  \kappa~.$$

By Lemma <ref>,
$\bigvee_a^b  \kappa$ is the total variation of the principal argument of the left derivative of a parameterization of $\arc$. This is exactly the total variation of the left tantrix of the arc. The result is given by (the analog for the left tantrix of) Proposition <ref>.

Let $\arc$ be an arc of the class $\Delta$, so that the rotation function $\kappa$ is defined. Given  a  finite signed Borelian measure $\mes$ with compact support and a harmonic function $h$, let us define  the functions
$ \kappa_l$ and $\kappa_r$ on a segment $[a,b]$, for a suitable parameterization of the arc:

$$\kappa_l(t)=\kappa(t) -\frac{1}{2\pi}\iint \varphi_l(\arc,z,t)d\mes(z) +(h^*(z(a))-h^*(z(t)))~, $$

$$\kappa_r(t)=\kappa(t) +\frac{1}{2\pi}\iint \varphi_r(\arc,z,t)d\mes(z)-(h^*(z(a))-h^*(z(t)))~. $$

(Compare   (<ref>) and (<ref>).)
The arc
is said of bounded turnbounded turn if the functions $ \kappa_l$ and $\kappa_r$ are of bounded total variation. As the measure $\mes$ if finite, its total variation is finite, and it follows from (<ref>) that it suffices that only one of the functions  $ \kappa_l$ or $\kappa_r$ has bounded total variation for the arc to be of bounded turn.

Now, if $\arc$ is of bounded rotation, Lemma <ref> says that the function $\kappa$ is of bounded total variation. The function $h^*\circ z$ is also of bounded total variation, as  $\arc$ is rectifiable, so its parameterizations are of bounded total variation, and $h^*$ is smooth.

We have

$$\bigvee_a^b \varphi_l (K,z) \leq |\kappa|(K)+\pi~.$$

The above result was proved by Radon in [80]. Another proof can be found in Lemma 8.3 in [86].
Actually, the statement in this reference is only for $\varphi (K,z) $, but it is easily extended to  $\varphi_l (K,z)$ by lower semicontinuity of the total variation.
Putting things together gives  the following result.

Let $\arc$ be an arc of bounded rotation  in the plane.
Then, for any finite signed Borelian
measure $\mes$  with compact support and any harmonic function $h$, $\arc$ is an arc of bounded turn.

The main difference of nature between arcs of bounded turn and arcs of bounded rotation is that the latter class depend only on the Euclidean structure of the plane. To define arcs of bounded turn, one needs the differentiable structure of the plane (to define the $\Delta$ class), a signed measure $\mes$ and a harmonic function $h$.
Of course, for $\mes=0$ and $h=0$, bounded turn and bounded rotation are the same thing. Apart from this trivial case, the converse result to Theorem <ref> is much more harder to prove. It will be stated in Section <ref>.

§ SUBHARMONIC FUNCTIONS

We saw at the very end of Section <ref>, that if $\delta_\zeta$ is the Dirac measure with support $\{\zeta\}$,
if $p(\delta_\zeta)$ is defined on the plane by

$$p(z;\delta_\zeta)=\frac{1}{2\pi}\ln |z-\zeta| $$

then, for the weak Laplacian $\Delta$, in the sense of measures,

$$\Delta p(\delta_\zeta)=\delta_\zeta~. $$

A similar result is  actually true for any positive measure with compact support. We will only review some basic facts about potential theory in sections <ref> and <ref>. For more details, we refer to the numerous  textbooks existing on this topic, see e.g. [79], [111], [81], [53], [16], [65], [39], [55].

In Section <ref>, we will focus on the fact that the set of points where a subharmonic function has infinite values is a set of Hausdorff dimension zero. This is a fundamental fact, as it will imply  that the integral
of  the restriction of a difference of subharmonic functions on a rectifiable arc is well defined.
In Section <ref> we present in full details the Localization Theorem, which allows to
change locally a harmonic term in a difference of subharmonic functions into a constant.

§.§ Definition

Let $\mes$ be a finite positive (Borelian) measure with compact support over the plane.

It will be useful to recall that such measures are actually regular measuresregular measures:
for every Borel set $E$,

$$\mes(E)=\inf \{ \mes(V) : E \subset V, \, V\mbox{ open}\}$$

$$\mes(E)=\sup \{ \mes(K) : K\subset E,\, K \mbox{ compact}\}~.$$

For $z$ in the plane, let us denote

\begin{equation}\label{eq:potential}
p(z;\mes)=\frac{1}{2\pi} \iint \ln |z-\zeta  | d\mes(\zeta)~.
\end{equation}

We denote by $Q_r(\zeta)$$Q_r(\zeta)$ the open disc of center $\zeta$ and (Euclidean) radius $r$, and set  $Q_r=Q_r(0)$.

Let $r>1/2$ be such that the support of $\mes$ is contained in $Q_r$.
For any $z\in  Q_r$, $p(z;\mes)$ is well-defined and
$$p(z;\mes)\leq \frac{\mes(\C)}{2\pi}\ln(2r)~. $$

We have

$$p(z;\mes)=\frac{1}{2\pi}\iint_{|\zeta|\leq r}\ln|z-\zeta| d\mes(\zeta) $$

so if $z\in Q_r$, i.e. $|z|\leq r$, we have
$|z-\zeta|\leq 2r$.
In turn:

\begin{equation}\label{eq:potential bfb}\ln |z-\zeta| =\left(\ln 2r\right)- \left( \ln 2r-\ln |z-\zeta| \right)~, \end{equation}

and both functions in the brackets are non-negative,  and the Lebesgue integral of $\ln 2r$ is finite because $\mes$ is finite.

It follows that for any $z$,
$p(z;\mes)$ is well-defined in $[-\infty,+\infty)$, so we obtain a function
$p(\mes):\R^2\to [-\infty,+\infty)$, the potentialpotential of $\mes$.

Let $\mes$ be a finite positive measure with compact support. Then for any $r>0$

\begin{equation}\label{eq:supermean eq}
p(z_0;\mes)\leq \frac{1}{2\pi r}\int_{C_r(z_0)} p(\mes)~.
\end{equation}

Recall that $C_r(z_0)$ is the plane circle of center $z_0$ and radius $r$.

First note that for any $\zeta$,  the function $z\mapsto \ln\vert z-\zeta\vert$ satisfies (<ref>). Indeed, either it is harmonic (so (<ref>) is satisfied with equality) or it is equal to $-\infty$.  Then by Fubini

\begin{equation*}
\begin{split}
\frac{1}{2\pi}\int_0^{2\pi} p(z+re^{it};\mes)dt = &
\frac{1}{2\pi}\iint \frac{1}{2\pi}\int_0^{2\pi}  \ln\vert z+re^{it}- \zeta\vert dt d\mes(\zeta) \\
\geq & \frac{1}{2\pi}\iint  \ln\vert z- \zeta\vert  d\mes(\zeta) = p(z;\mes)~.
\end{split}
\end{equation*}

The potential $p(\mes)$ is upper semicontinuous,  in particular, it is bounded from above on any compact set.

By (<ref>),  for any $z\in M$,  $\zeta\mapsto \ln(2r)-\ln|z-\zeta|$ is non-negative, and  Fatou's lemma applies: for $z_n\to z$,

\begin{equation*}
\begin{split}
-p_\mes(z)& =\frac{1}{2\pi}\iint \liminf_{n\to \infty} \left(\ln(2r)- \ln |z_n-\zeta  | \right) d\mes(\zeta) \const \mes(\C)\ln(2r) \\
& \leq  \frac{1}{2\pi}\liminf_{n\to \infty} \iint \left(\ln(2r)- \ln |z_n-\zeta  | \right)d\mes(\zeta)  \const \mes(\C)\ln(2r) \\
\  &=  \liminf_{n\to \infty} -p_\mes(z_n)~.\end{split}
\end{equation*}

The assertion about boundedness is a classical property of upper semicontinuous functions, see e.g., Section 2.1 in [81].

The potential $p(\mes)$ is harmonic over $\C\setminus \operatorname{supp}\mes$.

A continuous function that satisfies equality in (<ref>) for sufficiently small circles is harmonic, see e.g., <cit.>. To prove the reverse inequality is along the same line
as  for the proof of Lemma <ref>, with the difference that
for $\C\setminus \operatorname{supp}\mes$, $ \vert z-\zeta\vert >0$ for $\zeta \in  \operatorname{supp}\mes$.

The simplest example of potential is for a weighted mass
$\mes=\mes_0\delta_{z_0}$, for which we have already seen that $p(z;\mes)=\frac{\mes_0}{2\pi}\ln \vert z-z_0\vert$.
Now let us look at the potential of the measure $\mes_r$ which is the extension to the plane of the angular measure over $C_r(0)$ (the total measure is $2\pi$). This is the beginning of Example 2 in [86].
We have

\begin{equation*}
\begin{split}
p(z;\mes_r)&= \frac{r}{2\pi}\int_0^{2\pi}  \ln\left|z-re^{i\theta}\right|d\theta \\
&= \frac{r}{2\pi}\int_0^{2\pi}  \ln\left|\vert z\vert -re^{i(\theta-\operatorname{arg}(z))}\right|d\theta \\
&=\frac{r}{2\pi}\int_0^{2\pi}  \ln\left|\vert z\vert -re^{i\theta}\right|d\theta
\end{split}
\end{equation*}

hence $p(\mes_r)$ is radial over $Q_r$: $p(z;\mes_r)=f(t)$ with $|z|=t$. Moreover by Lemma <ref>, it is harmonic out of the circle $C_r$. For a radial function we have

$$\Delta p(z;\mes_r)= f''(t)+f'(t)/t~, $$

that is a second order Euler equation, but can be handle  easily by considering
$y=f'$. The solutions of $y'+y/r=0$ are $c/t$ for some constant $c$, hence the radial harmonic functions on the plane have the form $a\ln |z| +b$, for some constants $a,b$. As $p(\mes_r)$ is harmonic out of  the support of $\mes_r$,
it is harmonic  over $Q_r$, hence constant there (because $\ln|\cdot|$ is not harmonic at the origin), and taking $z=0$,  we have $p(z;\mes_r)=r\ln r$. As
$\lim_{\vert z\vert\to\infty} p(z,\mes_r)-r\ln \vert z\vert=0$, we obtain that out of $\bar{Q}_r$,
$p(z;\mes_r)=r\ln \vert z\vert$.

It is easy to see that $z\mapsto \ln \vert z \vert$ is in $L^p_{loc}$ for $p\geq 1$. Indeed,
$\iint_{Q_R} \vert \ln \vert^p = 2\pi \int_0^R r\vert \ln r\vert^p dr $, and the result follows
for example by the substitution $t=\ln r$. This elementary result can be generalized using the following classical result, see e.g. <cit.> or <cit.>.

Jensen inequalityCauchy inequality
Let $\mes$ be a finite positive measure with compact support, non identically equal to zero, and let $\psi:\mathbb{C}\to (-\infty,+\infty)$ be a convex function. For any $\mes$-integrable function $f:\mathbb{C}\to (-\infty,+\infty)$, we have

$$\psi \left(\frac{1}{\mes(\C)}\iint f d\mes\right)  \leq \frac{1}{\mes(\mathbb{C})} \iint \psi\circ f d\mes~. $$

A potential is in $L^p_{loc}$  for $p\geq 1$.

A potential is Borel measurable  by semicontinuity.
If $\mes=0$ then $p(\mes)=0$ and we are done. Otherwise,
let us use Lemma <ref> with $\psi(z)=\vert z\vert^p$. For a compact set $O$,

\begin{equation*}
\begin{split}\iint_O \left\vert p(\mes)\right\vert^p
&= \iint_O \left\vert \frac{1}{\mes(\C)} \iint \ln \vert z-\zeta\vert^{\frac{\mes(\C)}{2\pi}}  d\mes(\zeta) \right\vert^p dz \\
&\leq  \frac{1}{\mes(\C)} \iint_O \iint  \left\vert\ln \vert z-\zeta\vert^{\frac{\mes(\C)}{2\pi}}  \right\vert^p d\mes(\zeta)  dz \end{split}\end{equation*}

and the result follows by Fubini and because $\ln \vert \cdot\vert$ is locally in $L^p$.

It follows from Lemma <ref> that a potential  is finite almost everywhere on $M$. We will see a stronger result in Section <ref>.

Let $\mes_1, \mes_2$ be two positive finite measures with compact support, and let $h$ be a continuous function with compact support. Recall that the convolutionconvolution $\mes_1\star\mes_2$ is the measure defined by

$$\iint h(z) d (\mes_1\star\mes_2)(z)=\iint \iint h(\zeta+z)d\mes_1(\zeta)d\mes_2(z)~.$$

If $\mes_1$ has a Lebesgue density $f_1$, we can define the convolution
$f_1$ and $\mes_2$:

\begin{equation*}
\begin{split}
\iint h(z) d (f_1\star\mes_2)(z)= & \iint\iint h(\zeta+z)f_1(\zeta)d\zeta d\mes_2(z) \\
=& \iint \iint h(z)f_1(z-\zeta)dzd\mes_2(\zeta)~,
\end{split}
\end{equation*}

\begin{equation}\label{eq:conv fct mes}(f_1\star\mes_2)(z)=\iint f_1(z-\zeta)d\mes_2(\zeta) \end{equation}

and if $\mes_2$ has also a Lebesgue density $f_2$, then we recover the usual definition of convolution for functions:

$$(f_1\star f_2)(z)=\iint f_1(z-\zeta)f_2(\zeta) d\zeta~. $$

The definition (<ref>) gives in particular,

$$p(z;\mes)=(\ln \vert\cdot\vert \star \mes) (z)~.$$

Now let $\varphi$ be a smooth function with compact support. By Lemma <ref>, it is meaningful to speak about the weak Laplacian of a potential. Then a direct utilization of Fubini theorem and Lemma <ref> give

\begin{equation*}
\begin{split}
\Delta p(\mes) (\varphi) =& \iint \Delta \ln \vert \cdot -\zeta\vert (\varphi) d\mes (\zeta) \\
= & \iint \varphi (\zeta) d\mes(\zeta)~.
\end{split}
\end{equation*}

Let us state Remark <ref> as follows.

In the sense of positive measures,

\begin{equation}\label{eq:delta potentiel}\Delta p(\mes) = \mes~. \end{equation}

A function $u:M\to [-\infty,+\infty)$, defined over a domain $M$ of the plane, is a subharmonic function  subharmonic function if there   exists a
finite positive measure with compact support $\mes$ and a harmonic function $h$ over $M$ such that

$$u=p(\mes) + h~.$$

As harmonic functions are $C^\infty$
and satisfy equality in (<ref>), subharmonic functions inherit all the properties of the potential we have seen. But be careful that if a potential is defined over the entire plane, it is not necessarily the case for a subharmonic function, because of the harmonic term.

Let us recall the following classical result. It is a straightforward adaptation of a property of holomorphic functions, see e.g. [81] for details.

Let $h_1$ and $h_2$ be two harmonic functions on a domain $M$ in the plane. If $h_1=h_2$ over a non-empty open subset of $M$, then they are equal over $M$.

It follows from (<ref>) that if two subharmonic functions $p(\mes_1) + h_1$ and $p(\mes_2) + h_2$ coincide over an open set $U$, then $\mes_1\vert_U=\mes_2\vert_U$.
However there is no identity principle as Lemma <ref>, as the functions $u_1=0$ and $u_2=\max(\operatorname{Re}(z),0)$ show. As a consequence, one may change locally the measure and the harmonic function defining a subharmonic function. Section <ref> will be devoted to describe a procedure that allows to replace the harmonic term by a constant by  changing the measure over discs.

We will denote by $p(\mes,h)$$p(\mes,h)$ the subharmonic function which at $z$ takes the value

$$p(z;\mes,h)=p(z;\mes)+h(z)~. $$

Definition <ref> is the most adapted to our needs. Subharmonic functions are  more often defined as  upper semicontinuous functions satisfying  the super-mean inequality (<ref>). Actually, this alternative definition may allow the constant function equal to $-\infty$ as a subharmonic function, that is excluded by our definition. The relation between this alternative definition of subharmonic function and our's is given by
the  Riesz decomposition theorem,Riesz decomposition theorem that says  that an upper semicontinuous functions satisfying  the super-mean inequality satisfies our Definition <ref> on any relatively compact open subset of $M$, see <cit.>. (The function $-p(\mes)$ which is  used in Reshetnyak's articles is the antipotential of $\mes$, and is
a superharmonic function.)

Suppose that the boundary of $M$ is regular, and
let $u$ be a $C^2$ function over $\bar{M}$ such that
$\Delta u\geq 0$. Then $u$ is subharmonic over $M$.
Indeed, this is a straightforward consequence of Lemma <ref>: for $z\in M$

$$u(z)=p(z;\mes)+h(z) $$

$\mes=\Delta u \mathcal{L}_{\vert M}$ where  $\mathcal{L}_{\vert M}$ is the restriction of the Lebesgue measure to $M$, so that $\mes$ is a positive finite measure with compact support; and

\frac{1}{2\pi} \int_{\partial M}  u\frac{\partial \ln|z- \cdot|}{\partial \nu}-\ln|z-\cdot| \frac{\partial u }{\partial  \nu}$$

defines a harmonic function over $M$.

Recall that for a function $f$ which is locally $L^1$ over $M$, a function $v\in L^1_{loc}(M)$ is the weak derivativeweak derivative of $f$ in the direction $x$ if for any $C^1$ function $\varphi$ with compact support in $M$,

$$\iint_M v \varphi =- \iint f \frac{\partial \varphi}{\partial x}~, $$

and we define similarly the weak derivative in the direction $y$. For example, when it is defined, for $z=x+iy$,
$f(z)= \frac{x}{\vert z\vert^2}$ is the partial derivative of $\ln\vert \cdot\vert$ in the direction of $x$, and using polar coordinates it is immediate that it is locally in $L^1$, and it follows easily that $f$ is the weak derivative of $\ln\vert \cdot\vert$. Actually, $f$ is locally in $L^p$ for $1\leq p <2$. It is exactly the same
for the weak partial derivative in the direction of $y$. We then have that $\ln\vert \cdot\vert$  is locally in  the Sobolev space $W^{1,p}$ for $1\leq p <2$: it is locally in $L^p$ and its weak first partial derivatives are also locally in $L^p$ for $1\leq p <2$. See e.g. [60] for more details.

An utilization of Jensen inequality in a way similar to the proof of Lemma <ref> implies that any subharmonic function is locally in $W^{1,p}$  for $ 1\leq p <2$.

§.§ Approximation

We will use the fact that a radial subharmonic function is increasing with respect to the radius. This can be easily seen as follows:
along any line, as the weak Laplacian is non-negative, the weak second derivative is non-negative hence we have convexity with minimum at the origin by (<ref>). But we will give an argument based on the following maximum principle.

Let $u$ be an upper semicontinuous function satisfying (<ref>)
over a  domain $M$. If $u$ attains its maximal value $A$, then $u$ is constant.

In particular, if $\bar{M}$ is compact and $u$ is defined over $\bar{M}$, then
for any $z\in M$,

$$u(z)\leq \sup_{\partial M} u~.$$

$$X=\{z\in M \vert u(z)=A\}$$

and suppose that $X$ is not empty. As $A$ is the maximal value, it follows from (<ref>), by taking sufficiently small values of $r$, that $X$ is open. On the other hand,

$$M \setminus X = \{z\in M\vert u(z)<A\}$$

is open as $u$ is upper semi continuous. So $M=X$.

If $u$ is defined over the compact set $\bar{M}$, then it has an upper bound that is attained. By the preceding property, either the maximal value is attained on the boundary, or $u$ is constant, that leads to the same conclusion.

The Maximum Principle Lemma <ref> justifies the name “subharmonic”. Indeed, if $u$ is a subharmonic function over $Q_r(z)$, for any harmonic function $h$ over the disc such that $u=h$ on the boundary, then $u-h$ satisfies the condition of the Maximum Principle Lemma <ref>, so that $u\leq h$.

Let us mention the following obvious formula:
if $f$ is continuous, then for sufficiently small $r$, $\frac{1}{2\pi r}\int_{C_r(z_0)}f \leq f(z_0)+\epsilon$, from which it is easy to see that

$$\lim_{r\to 0}\frac{1}{2\pi r}\int_{C_r(z_0)} f=f(z_0)~.$$

Note also that as

$$\frac{1}{\pi r^2}\iint_{Q_r(z_0)}f
=\frac{1}{\pi r^2}\int_0^r \int_{C_R(z_0)}
f dR \leq \frac{1}{\pi r^2}\int_0^{r}2\pi R(f(z_0)+\epsilon)dR

we also have

$$\frac{1}{\pi r^2}\iint_{Q_r(z_0)} f\xrightarrow[r\to 0]{}f(z_0)~.$$

Let $\mes$ be a finite positive measure with compact support.
The function $r\mapsto \frac{1}{2\pi r}\int_{C_r(z_0)} p(\mes)$ is  non-decreasing and

$$\lim_{r\to 0}\frac{1}{2\pi r}\int_{C_r(z_0)} p(\mes)=p(z_0;\mes)~.$$

Let us define

$$\bar{v}(r)=\frac{1}{2\pi r}\int_{C_r(z_0)} p(\mes)$$

$$v(z)=\frac{1}{2\pi }\int_0^{2\pi}
p(ze^{it};\mes)dt~. $$

Similarly to the proof of Lemma <ref>, it is easy to see that as $p(\mes)$ satisfies (<ref>), then $v$ satisfies (<ref>). As $p(\mes)$ is bounded on compact sets and upper semicontinuous, it follows that $v$ is also upper semicontinuous (see the proof of Lemma <ref>). So Lemma <ref> applies, and as $v$ is radial, we obtain that for $z\in Q_r(0)$, i.e. $\vert z\vert=r'<r$,

$$\bar{v}(r')=v(z)\leq \sup_{z\in C_r(0)}v(z)=\bar{v}(r)$$

so $\bar{v}$ is non-decreasing.
It also follows that $\liminf_{r\to 0} \bar{v}(r)\geq \bar{v}(0)$, and the last result follows because $v$ is upper semicontinuous: $\limsup_{r\to 0} \bar{v}(r)\leq \bar{v}(0)$.

The following fundamental result (Proposition <ref>) is obtained from a simple convolution (or Sobolev averagingSobolev averaging as it is called in [88]), but (<ref>)
allows a strong conclusion.

Recall that a sequence of finite positive measure with compact support $(\mes_n)_n$
converges weaklyweak convergence to $\mes$ if for any continuous bounded function $f$ over the plane,

\begin{equation}\label{eq:weak conv def}\iint f d\mes_n \xrightarrow[n\to\infty]{} \iint fd\mes~. \end{equation}

The terminology of weak convergence of measures is ambiguous in the literature. However, we will be concerned only
by sequences of measures which have their support in a same compact set. In that case, the weak convergence is equivalent to
consider (<ref>) only for continuous $f$ with compact support.

Let $\mes$ be a finite positive measure with compact support. Then there exists a sequence  $\left(\mes_n\right)_n$  of finite positive measure, with compact support and smooth Lebesgue densities, such that $\mes_n$ converge weakly  to $\mes$.

Moreover,  $p(\mes_n)$ is a non-decreasing sequence pointwise converging to $p(\mes)$.

Let us take a Sobolev hat $\alpha$ defined e.g., by

c\operatorname{exp}\left(\frac{1}{|z|^2-1}\right) & \mbox{if }|z|< 1~\\
0 & \mbox{if }|z|\geq 1~
\end{cases}~,$$

where  $c$ is such that
$\iint \alpha =1.$  Let us define

$$\alpha_h(z)=\frac{1}{h^2}\alpha\left(\frac{z}{h}\right)\,~,\, f_h= \alpha_h\star \mes$$

and let $\mes_h$ be the measures with $C^\infty$ Lebesgue density $f_h$.
Let $\epsilon>0$ and let $\varphi$ be a continuous function with compact support. In particular, $\varphi$ is uniformly continuous over compact sets, and  for any $z$ and for sufficiently small $h$,
$\sup_{\zeta\in \bar{Q}_h}\vert\varphi(z+\zeta)-\varphi(z)\vert <\epsilon$. So we have

\begin{equation*}\begin{split}
\left\vert \iint \varphi d\mes_h -\iint \varphi d\mes\right\vert\leq \iint\iint_{Q_h} \vert\varphi(z+\zeta)-\varphi(z)\vert \alpha_h(\zeta) d\zeta d\mes(z) \\
\leq \epsilon \mes(\mathbb{C}) \iint_{Q_h}\alpha_h(\zeta) d\zeta =
\epsilon \mes(\mathbb{C})~,
\end{split}
\end{equation*}

i.e.,  $\mes_h$  converge weakly to $\mes$ when $h\to 0$.

$$p(\mes_h)=\ln\vert\cdot\vert \star \mes_h
=\ln\vert\cdot\vert \star f_h=\ln\vert\cdot\vert \star (\alpha_h\star \mes)=(\ln\vert\cdot\vert \star \mes)\star \alpha_h=p_\mes\star \alpha_h$$

is a $C^\infty$ function, and as $\alpha_h$ is a radial function,  substituting $t=r/h$ in the last line:

\begin{equation*}\begin{split}
p(z;\mes_h)=& \iint p(z-\zeta;\mes)\alpha_h(\zeta)d\zeta \\
=&  \int_0^{h} \left( \int_0^{2\pi}p(z-re^{i\theta};\mes) d\theta\right)  \frac{r}{h^2} \alpha\left(\frac{r}{h}\right)dr  \\
=& \int_0^1 \left( \int_0^{2\pi}p(z-hte^{i\theta};\mes) d\theta\right)  t \alpha\left(t\right)dt ~.
\end{split}
\end{equation*}

By Lemma <ref>,  the term into the parenthesis non-decreases to $2\pi p(z;\mes)$ when $h\to 0$, so by Levi's monotone convergence, $p(z;\mes_h)$ non-decreases to

It is proved in Lemma 5 in [88] that if a sequence of measures  converge weakly, then the associated sequence of potential converges in $L^p$ for $p>1$. For more results about convergence, see [65], especially Theorem 3.8.

Let us end this section with some applications of Proposition <ref>.

Let us first mention the following famous result. It  may be proved by a simple convolution, see e.g. <cit.> or <cit.>, or using the regularity of the  mean equality more explicitly <cit.>, [82].

Weyl's lemma
A locally integrable function over a domain $M$ such that its weak Laplacian is zero is equal almost everywhere to a harmonic function.

A locally integrable function over $M$ such that its weak Laplacian is a positive finite measure with compact support is called almost subharmonicalmost subharmonic.

An   almost subharmonic function is equal almost everywhere to a subharmonic function.

Let $u$ be an almost subharmonic function.  Then $\Delta (u-p(\Delta u))=0$, hence by the Weyl Lemma Theorem <ref>,
$u-p(\Delta u)$ is equal almost-everywhere to a harmonic function $h$, so an almost subharmonic function is equal almost everywhere to the subharmonic function $p(\Delta u)+h$.

A subharmonic function is an upper semicontinuous  almost subharmonic function. But those two properties do not characterize subharmonic functions,
as shows the following elementary example taken from
$v$ be the function that assigns the value $1$ to the points of a bounded closed set of measure zero in the plane, and zero otherwise. It is upper semicontinous, and its weak Laplacian is the zero measure $\mes$. So by our definition, if $v$ was subharmonic, it should be harmonic, that  is not. Let us mention the following result,
that was initially proved in [105],  see also [101].

A continuous almost subharmonic function is subharmonic.

Let $u$ be a continuous almost subharmonic function. It is equal almost everywhere to a subharmonic function
Let us consider $u_n$ and $v_n$ which are obtained from  $u$ and $v$ by convolution by the  regularizing $C^\infty$ functions from the proof of Proposition <ref>.
As $u=v$ almost everywhere, then $u_n=v_n$. As $u$ is continuous, $u_n$ converge pointwise to $u$, see e.g., [48]. By Proposition <ref>, $v_n$ converge pointwise to $v$, so $u=v$.

Let $u_1, u_2$ be two subharmonic functions equal almost everywhere over $M$. Then they are equal over $M$.

Let us first suppose that $u_1$ and $u_2$ are potentials. The approximations given by the proof of Proposition <ref> are then equal, and passing to the limit the functions are equal. In the general case, harmonic functions on $M$ are added to the potentials, they may not be integrable over the whole $M$. To overcome this, it suffices to make the convolution at the step $n$ on the set of points at distance larger than $1/n$ from the boundary of $M$.

§.§ Polar sets

Let us introduce the infinity setinfinity set of a finite positive measure over the plane, with compact support, $\mes$ :

\begin{equation}\label{eq:sing mes}
\sing =\{z\in \C : p(z;\mes)=-\infty\}~.
\end{equation}

Properties of the potential imply

* $\sing=\bigcap_{n\in \N} \{ z : p(z;\mes)< -n\}$, so
$\sing$ is a $G_\delta$ set: a countable intersection of open  sets (because $p(z;\mes)$ is upper semi-continuous by
Lemma <ref>);

* $\sing\subset \operatorname{supp}(\mes)$;

* $\sing$ is a Borel set of zero Lebesgue measure.

We will need a more precise description of $\sing$, in order to define  the restriction of the difference of two subharmonic functions to some arcs.

For a subset $A$ of a metric space, for $\alpha\geq 0$ and $\delta \in (0,+\infty]$, let us denote

$$\mathcal{H}^\alpha_\delta(A)=c_\alpha\inf \left\{ \sum_{i\in I} (\operatorname{diam}(A_i))^\alpha  : \operatorname{diam}(A_i)<\delta,\, A \subset \bigcup_{i\in I} A_i \right\}~,$$

where $c_\alpha$ is a positive constant which will be fixed below.

Note that for a given $\alpha$, the function $\delta\mapsto \mathcal{H}^\alpha_\delta(A)$ is  non-negative and non-increasing.
The $\alpha$-dimensional Hausdorff measure of $A$ is
$$\mathcal{H}^\alpha(A)=\sup_{\delta>0} \mathcal{H}^\alpha_\delta (A)~,$$
and it is actually a (Borel) measure and the Hausdorff dimension of $A$ is
$$\dim_H(A)=\inf\{\alpha>0 : \mathcal{H}^\alpha(A)=0\}~.$$

We will need the following, see e.g. [15].

For any Borel set $A\subset \R^n$, $\mathcal{H}^n(A)$ is equal to the Lebesgue measure of $A$ (for a suitable constant $c_n$).

We want to show that the infinity set has Hausdorff dimension zero. We essentially follow Section 5.9 in [16].

Let $\mes$ be a finite  positive measure over $\C$ with compact support, and let $p(\mes)$ be its potential. Then for any $\alpha > 0$ and $z\in \mathbb{C}$,

$$ p(z;\mes) \geq \frac{1}{2\pi \alpha} \inf \left\{-\mes(Q_t(z))t^{-\alpha}: t\in ]0,1]\right\}~.  $$

Let $z\in \C$. Let $\chi_1(\zeta)=1$ if $\vert z-\zeta\vert \leq 1$ and zero otherwise. Let $\chi_2(t)=1$ if $t\geq \vert z-\zeta\vert$ and zero otherwise. By Fubini,

\begin{eqnarray*}
\ 2\pi p(z;\mes)&=& \iint \ln|z-\zeta| d\mes(\zeta) \\
& \geq &  \iint_{ Q_1(z)} \ln|z-\zeta| d\mes(\zeta)   \\
\ &=& -\iint_{\vert z-\zeta\vert\leq 1}
\int_{\vert z-\zeta\vert}^1 t^{-1}dtd\mes (\zeta)\\ &=& -\iint \chi_1(\zeta) \int_0^1 t^{-1} \chi_2(t) dtd\mes(\zeta)\\
\ &=& -\int_0^1 \iint (\chi_1 \times \chi_2)(\zeta,t)d\mes(\zeta) t^{-1} dt\\
&=& -\int_0^1 \mes(\{\zeta : \vert z-\zeta\vert\leq t\}) t^{-1} dt\\
&=& -\int_0^1 \mes(Q_t(z))t^{-\alpha}t^{\alpha-1}dt~.
\end{eqnarray*}

For the following statement see
[15], [16], [43].

For any collection of discs $\{Q_{r_i}(z_i), i\in I\}$ such that $\sup_{i\in I} r_i <+\infty$, there is a countable subset $J\subset I$ such that the discs $Q_{r_j}(z_j)$ are disjoint for $j\in J$, and

$$\bigcup_{i\in I} Q_{r_i}(z_i) \subset \bigcup_{j\in J} Q_{5r_j}(z_j)~.$$

Let $\alpha >0$. For any $\varepsilon >0$, there is a countable   cover of
$\sing$ by open discs of radii $r_i$,  with $\sum r_i^\alpha <\epsilon$.

For any  $z\in \sing$,  by definition,  Lemma <ref> gives
$$\inf\left\{-\mes(Q_t(z))t^{-\alpha}, t\in]0,1] \right\}=-\infty~,$$

$$\sup\left\{\mes(Q_t(z))t^{-\alpha}, t\in]0,1] \right\}=+\infty~.$$

Hence there exists
$0<r_z\leq 1$ such that

\begin{equation}\label{eq: mu disque rz}\frac{\mes(Q_{r_z}(z))}{r_z^\alpha} > \frac{5^\alpha\mes(\C)}{\varepsilon}~.\end{equation}

As $r_z\leq 1$ for any $z$, we can apply Lemma <ref>: there exists a countable set $J$ such that
$\sing\subset \bigcup_{j\in J} Q_{5r_{z_j}}(z_j)$, and
the $Q_{r_{z_j}}(z_j)$ are disjoint. Hence by (<ref>),

$$\sum_{j\in J} (5r_{z_j})^\alpha \leq \varepsilon \frac{\sum_{j\in J} \mes(Q_{r_{z_j}}(z_j) )}{\mes(\C)}\leq  \varepsilon~,

where the last inequality holds because the family is disjoint.

The Hausdorff dimension of $\sing$ is zero, i.e. for any $\alpha >0$, the $\alpha$-dimensional Hausdorff measure of $\sing$ is zero.

Let us recall that a subset of $\R^n$ with Hausdorff dimension $<1$ is totally disconnected: its connected components are points.

Let $\alpha>0$ and $\delta>0$. For any $\delta^\alpha > \epsilon >0$, by Lemma <ref>, there exists a covering of $\sing$
by sets of diameter $r_i$ such that $\sum r_i^\alpha<\epsilon$. In particular, $r_i < \delta$. Hence $\mathcal{H}_\delta^\alpha(\sing)<\epsilon$, hence  $\mathcal{H}_\delta^\alpha(\sing)=0$ as $\epsilon$ was arbitrary. But $\delta$ is also arbitrary, hence $\mathcal{H}^\alpha(\sing)=0$.

The following result is a particular case of the Area  Formula for Lipschitz mappings, see <cit.>.
Recall that, by Rademacher Theorem, a Lipschitz mapping is differentiable almost everywhere.

Let $f:\R \to \R^2$ be an injective Lipschitz mapping,
and $E\subset \R$ a Lebesgue measurable set, then

$$\mathcal{H}^1(f(E))=\int_E | f' |~. $$

Let $\param:[0,l]\to \C$ be
the arc length parameterization of a rectifiable  arc.
Then $\param^{-1}(\sing)$ has zero Lebesgue measure in $[0,l]$.

Up to apply the Tietze Extension Theorem, we use Theorem <ref>, for $E=\param^{-1}(\sing)$, which is a Borel set.
By Proposition <ref>, $\mathcal{H}^1(\sing)=0$. Up to a set of Lebesgue zero measure, $|\param'|=1$, hence the Lebesgue measure of $\param^{-1}(\sing)$ is zero.

We can be a bit more precise in the case of curves of bounded rotation, having Remark <ref> in mind. We need the following result, whose proof is left to the reader.

Let $f$ be a $K$-Lipschitz mapping between two metric spaces $X$ and $Y$. Then for any $A\subset X$, for any $\alpha \geq 0$,
$\mathcal{H}^\alpha(f(A))\leq K^\alpha \mathcal{H}^\alpha(A)~. $

Let $\param:[0,l]\to \C$ be
the arc length parameterization of a rectifiable  arc. Suppose that
$\param^{-1}$ is Lipschitz.
Then $\param^{-1}(\sing)$ is Lebesgue measurable and has zero Hausdorff dimension.

As $\param^{-1}$ is Lipschitz, $\param^{-1}(\sing)$ is Lebesgue measurable. The result is then immediate
by Proposition <ref> and Lemma <ref>.

Let us denote by $C_\mes$$C_\mes$ the set of continuity points of $p(\mes)$. Here continuity is meant in an extended sense, allowing the $-\infty$ value. The upper semi-continuity of the potential gives the following.

With the notation above, $\sing \subset C_\mes$.

The following result is proved in [90]. See e.g.  [58] for a stronger result. Compare also with Proposition 5 in [31] (or Theorem 5.5.8  in [16]). Let $C_\mes^c$ the complement of $C_\mes$ in $M$, i.e. the set of discontinuity of $p(\mes)$.

Let $p(\mes)$ be a potential over a bounded domain $M$.
For any $\epsilon>0$,  there exists a sequence $(Q_n)_n$ of open discs with sum of diameters less than $\epsilon$ such that

$\displaystyle C_\mes^c\subset \bigcup_n Q_n$, i.e., $u$ is continuous out of the union of the discs $Q_n$.

In potential theory, a set $E$ is polarpolar set if
there exists a measure with potential  constant $=-\infty$ on $E$. Obviously $\sing$ is a polar set, and  Proposition <ref> is the classical result that a polar set has Hausdorff dimension zero.
A property holds quasi-everywherequasi-everywhere on a domain $M$ if it holds up to a polar set. In the following, the only polar set we will need to consider is $\sing$, and “quasi-everywhere” can always be read as “up to a set of zero Hausdorff dimension”. As an example,  a subharmonic function is finite quasi-everywhere (recall that the constant function equal  to $-\infty$ is not considered as a subharmonic function in the present text).

The simplest example of a subharmonic function with a non-empty polar set is, for $\mes_0\geq 0$, $p(z;\mes_0\delta_\zeta)=\ln \left| z-\zeta\right|^{\frac{\mes_0}{2\pi}}$, whose polar set is $\{\zeta\}$. This simple example can be made wilder in the following way. This is a classical example, see e.g. <cit.>, <cit.>.

On any domain $M$, there exists a subharmonic function $u$
such that $C_\mes=\sing$ is a dense uncountable set. In particular $u$ is discontinuous quasi-everywhere.

Let $(a_n)_n$ be a sequence of positive real numbers such that
$\sum^\infty a_n$ converges, and let $(\zeta_n)_n$ be a dense countable subset of $M$. Define for each $z\in M$, $u(z) = \sum a_n / |z-\zeta_n|$. The function  $u$ is a

We have $u(\zeta_n)=-\infty$, and $(\zeta_n)_n$ a dense subset of $M$, so if $u$ is continuous at a point $z$, then $u(z)$ cannot be finite, i.e., $C_\mes\subset \sing$, and by Lemma <ref>, $C_\mes=\sing$.

Let us write

$$M= \sing \bigcup \left( \bigcup_n \{z\in M | u(z)\geq n\}\right)~.$$

We have that $M$ is a union of closed sets ($\sing$ is closed, because  $\sing=C_\mes$). Moreover $\sing$ as empty interior, as well as $\{z\in M | u(z)\geq n\}$ (because $\sing$ is dense). Let us suppose that $\sing$ is countable. Then $M$ would be a union of closed sets with empty interior, and by the  Baire Category Theorem, $M$ would have empty interior, that contradicts the assumption that $M$ is a domain.

§.§ Delta-subharmonic functions

Delta-subharmonic functions are simply differences of subharmonic functions. But as a subharmonic function may be infinite at some points, a difference of subharmonic functions may not be defined everywhere.

Let $M$ be a bounded domain in the plane. A function
$v$, defined quasi-everywhere on $M$, is called  $\delta$-subharmonic function over $M$$\delta$-subharmonic function if there exists two subharmonic functions $u_1$ and $u_2$, such that

Let us write $u_i=p(\mes_i)+h_i$.
The $\delta$-subharmonic function exists only on $\C\setminus \left(\mathcal{I}_{\mes_1}\cap \mathcal{I}_{\mes_2}\right)$, i.e., it exists quasi-everywhere. Whenever it is defined, it belongs to $[-\infty,+\infty]$. The function has finite values quasi-everywhere.

We have the following  example.

Let $f:M\to \C$ be a function. Then $\ln |f|$ is

* harmonic if $f$ is a holomorphic function without zeros;

* subharmonic if $f$ is a holomorphic function;

* $\delta$-subharmonic if $f$ is meromorphic.

The first item follows from the fact that $f$ is locally the real part of a holomorphic function (considering the complex logarithm).
If $f$ is holomorphic, its zeros are isolated, and around such a point the function is the sum of an entire power plus a non-zero harmonic term.

For a real valued function $f$, let us denote
$f^+(x)=\max(0,f(x))$ and $f^-(x)=\max(0,-f(x))$,$f^+$ so that $f^+, f^-$ are non-negative and $f=f^+-f^-$.
Now, let us suppose that the boundary of $M$ is regular and that $u$ is $C^2$ over $\bar{M}$.  By Lemma <ref>, over $M$,

$$u(z)=\frac{1}{2\pi}\iint_M  \ln |z-\cdot| (\Delta u)^+ - \frac{1}{2\pi}\iint_M  \ln |z-\cdot| (\Delta u)^- + h(z)~, $$

where $h(z)=\frac{1}{2\pi} \int_{\partial M}   u\frac{\partial \ln|z- \cdot|}{\partial \nu}-\ln|z-\cdot| \frac{\partial  u }{\partial  \nu}$
is harmonic over $M$. It follows that  $u$ is $\delta$-subharmonic.

Concave functions on an open subset are characterized by the fact that their Hessian in the sense of distribution is non-positive [101], [41], [42]. Hence, they are almost superharmonic, and by continuity, they are subharmonic.
It follows  that
any function is $\delta$-subharmonic. This implies that
$C^{1,1}$ functions, i.e., functions with locally Lipschitz first derivatives, are $\delta$-subharmonic, as well as for example piecewise affine functions, see [8]. (This is also proved in [17].)  As from Remark <ref>, we know that a subharmonic function is locally in the Sobolev space $W^{1,p}$ for $p\in [1,2)$, loosely speaking
we have:

$$C^2\subset C^{1,1} \subset \mbox{ \dc} \subset \mbox{  $\delta$-subharmonic }
\subset \bigcap_{1\leq p <2} W^{1,p}_{\operatorname{loc}}~. $$

The $\delta$-convex functions (or DC functions) DC function
form a  class of functions that has some importance in metric geometry, in the wake of the Leningrad Geometric School, and  manly after [76]. See also [8], or more recently [13], [78].

We will often consider $\delta$-subharmonic functions as potentials in the following manner.
Let $\mes$ be a  signed (Borel) measuresigned measure. By definition, it takes only finite values. Recall that the total variation measure$|\mes|$ $|\mes|$ of $\mes$ is the finite positive measure defined by

$$|\mes|(E)=\sup \sum_{i=1}^\infty |\mes(E_i)|  $$

where the supremum is taken over all the partitions $ \{E_i\} $ of $E$. In general, $|\mes|(E)\geq |\mes(E)|$.
The positive and negative variations of $\mes$ are the finite positive Borel measures defined by$\mes^+$

$$\mes^+=\frac{1}{2}(|\mes|+\mes),\, \mes^-=\frac{1}{2}(|\mes|-\mes)~,$$

and we have the Jordan decomposition of the signed measure $\mes$ as a difference of two finite positive measures: $\mes=\mes^+-\mes^-$. Note also that $|\mes|=\mes^++\mes^-$.
This special decomposition has the following property: if $\mes_1$ and $\mes_2$ are positive measures, and $\mes=\mes_1-\mes_2$, then

$$\mes_1\geq \mes^+,\,\mes_2\geq \mes^-~.$$

In particular, if $\varphi=\mes_1+\mes_2$, then $\varphi \geq |\mes| $. Also, if
$\mes(E)=0$, then $\mes^+(E)=\mes^-(E)=0$. Let us also recall that $|\mes|(E)=0$ if and only if $\mes(F)=0$ for every measurable subset of $E$. The supportsupport (signed measure) of $\mes$ is defined as the support of $|\mes|$.

If a signed measure $\mes$ has compact support, then
$\mes^+$ and $\mes^-$ have compact support, and in this case we can define

$$p(\mes):=p(\mes^+) - p(\mes^-)~, $$

$p_\mes$ and similarly, if $h$ is a harmonic function,

\begin{equation}\label{eq:p mes sing}p(\mes,h):=p(\mes^+) - p(\mes^-)+h~. \end{equation}

Of course, $p(\mes,h)$ is a $\delta$-subharmonic function.

§.§ Localization of delta-subharmonic functions

We now give a detailed proof  of the
Localization Theorem. We will explain its interest at the end of the section, after its proof.
Let us  give some simple preparatory results about harmonic functions.
Let us first recall the following classical formula from the  Neumann problem for the disc.

Let $h$ be a harmonic function on the closure of the disc $Q_r(z_0)$. Then

\begin{equation}\label{eq:dini}
\forall z\in Q_r(z_0),\, h(z)=h(z_0)-\frac{1}{\pi}\int_{C_r(z_0)} \ln|z-\cdot|\frac{\partial h}{\partial \nu}~.
\end{equation}

Without loss of generality, let us prove the result for $C_r(z_0)=C_1(0)=:C$.
Let $h$ be the real part of the holomorphic function $f$. As on the circle $C$ we have

$$\frac{\partial h}{\partial \nu}(z)=\left(z_1\frac{\partial h}{\partial x}(z) + z_2\frac{\partial h}{\partial y}(z)\right) $$

and as

$$z f'(z)=Df(z)(z)=z_1\frac{\partial f}{\partial x}(z)+z_2\frac{\partial f}{\partial y}(z)~, $$

we have, on the circle,

\begin{equation}\label{eq:normale harm hol}\frac{\partial h}{\partial \nu}(z)=\operatorname{Re} (zf'(z))~.\end{equation}

Now, recall the Poisson equation
that gives the solution of the Dirichlet problem for the disc: for a continuous function $H$ on $C$, define

$$F(z)=\frac{1}{2\pi }\int_{0}^{2\pi} \frac{e^{i\theta}+z}{e^{i\theta}-z} H(\theta) d\theta~.$$

Then $F$ is a holomorphic function over $Q$, and $\operatorname{Re}(F)$ is the unique harmonic function on $Q$ that extends continuously as $H$ on $C$. In particular,

\begin{equation}\label{eq:poisson}
F(z)-F(0)=\frac{z}{\pi }\int_0^{2\pi}\frac{H(\theta)}{e^{i\theta}-z}d\theta~.
\end{equation}

Now, let  $H=\frac{\partial h}{\partial \nu}$. Then by uniqueness, $F(z)=zf'(z)$ by (<ref>), and (<ref>) gives

$$f'(z)=\frac{1}{\pi } \int_0^{2\pi}\frac{\partial h}{\partial \nu}(\theta)\frac{1}{e^{i\theta}-z}d\theta~,$$

so that, if $\operatorname{Ln}$ is the principal determination of the complex logarithm,

$$f(z)=\alpha-\frac{1}{\pi } \int_0^{2\pi}\frac{\partial h}{\partial \nu}(\theta)\operatorname{Ln}( e^{i\theta}-z)d\theta~, $$

where $\alpha$ is a constant.
As $h=\operatorname{Re}f$, we obtain

$$h(z)=\operatorname{Re}\alpha - \frac{1}{\pi}\int_{C} \ln|z-\cdot|\frac{\partial h}{\partial \nu}~. $$

By (<ref>), for $z=0$,
$\int_{C} \ln|z-\cdot|\frac{\partial h}{\partial \nu}=0$, hence $\operatorname{Re}\alpha=h(0)$.

It will be suitable to write (<ref>) in the following manner.

Let $h$ be a harmonic function on a domain $M$, and let $z_0\in M$, $r>0$ such that $\bar{Q}_r(z_0)\subset M$. Then there exists a  signed measure
$\psi_r$ over the plane with

* support the circle $C_r(z_0)$,

* bounded variation, and
$|\psi_r|(\mathbb{C})\xrightarrow[r\to 0]{} 0$,

* $\psi_r(\C)=0$

such that for
$z\in Q_r(z_0)$,

\begin{equation}\label{eq:loc harm}h(z)= h(z_0) -\frac{1}{\pi}\iint  \ln|z-\zeta|d\psi_r(\zeta)~.\end{equation}

Let us write

\begin{equation*}
\begin{split}
\ \int_{C_r(z_0)}  \ln|z-\cdot|\frac{\partial h}{\partial \nu}&= r \int_0^{2\pi} \ln|z-z_0-re^{i\theta}|\frac{\partial h}{\partial \nu}(z_0+re^{i\alpha})d\theta \\
\   & = \int_0^{2\pi}\ln |z-z_0-re^{i\theta}| d\tilde{\psi}_r(\theta)
\end{split}
\end{equation*}

where  $d\tilde{\psi}_r$ is the Lebesgue–Stieltjes measures over $C_r(z_0)$ given by the (smooth) function

$$\tilde{\psi}_r(\alpha)=r\int_0^{\alpha} \frac{\partial h}{\partial \nu}(z_0+re^{i\theta})d\theta~. $$

Note that by (<ref>), we have $\tilde{\psi}_r(2\pi)=0$.
From (<ref>), we obtain

$$h(z)=h(z_0)-\frac{1}{\pi}\int_0^{2\pi} \ln |z-z_0-re^{i\theta}| d\tilde{\psi}_r(\theta)~. $$

If we denote by $\psi_r$  the measure
over the plane  with support  $C_r(z_0)$
extends $\tilde{\psi}_r$, we obtain (<ref>). We have

\begin{equation*}\label{eq:total angular measure}
\begin{split}
\ |\psi_r|(\mathbb{C})=|d\tilde{\psi}_r|([0,2\pi])= \bigvee_0^{2\pi} \tilde{\psi}_r(\alpha) & =r \int_0^{2\pi} \left|\left( \int_0^\alpha
\frac{\partial h}{\partial \nu}(z_0+re^{i\theta)}d\theta\right)' \right|d\alpha \\
\ &=r\int^{2\pi}_0 \left| \frac{\partial h}{\partial \nu} (z_0+re^{i\alpha})\right| d\alpha~,
\end{split}
\end{equation*}

and  by  regularity of $h$, the right-hand side is bounded from above by $r$ times a constant that depends only on $M$, so $|\psi_r|(\mathbb{C})\to 0$ when $r\to 0$.

The Localization of Harmonic Function Lemma <ref> will be used at several places. Indeed, given a subharmonic function $p(\mes,h)$ over $M$, this says that over an open disc whose closure is inside $M$, there are a measure $\psi$ and a constant $c$ such that

The localization of (difference of) subharmonic function, Theorem <ref> below, follows the same procedure, by applying Lemma <ref> to the harmonic part of the subharmonic function.

Let $r>0$ and $\zeta\in \C$ with
$|\zeta|>r$. Then the angular function of the circle satisfies

$$\bigvee_0^{2\pi}\varphi(C_r(0),\zeta,\cdot) = 4 \arcsin \frac{r}{|\zeta|}~.$$

Let $\param:[0,2\pi]\to \C$ be a parameterization of
$C_r(0)$ such that $\param(0)$ is the orthogonal projection of $\zeta$ onto $C_r(0)$. Then
$\varphi:=\varphi(C_r(0),\zeta,\cdot)$ is decreasing on $[0,\pi/2]$ and $[3\pi/2, 2\pi]$, and increasing on $[\pi/2,3\pi/2]$ hence

\begin{align*}
\bigvee_0^{2\pi}\varphi &=\bigvee_0^{\pi/2}\varphi+\bigvee_{\pi/2}^{\pi}\varphi
\\ &=-2\varphi(C_r(0),\zeta,\pi/2)+2\varphi(C_r(0),\zeta,3\pi/2)\end{align*}

and the result follows, see Figure <ref>.

To Lemma <ref>: $\displaystyle \alpha=-\varphi(C_r(0),\zeta,\pi/2)=\arcsin \frac{r}{|\zeta|}$.

[Localization Theorem, <cit.>]
Let $h$ be a harmonic function on a domain $M$, and let $\mes$ be a signed measure with compact support.
For any $\epsilon,\eta >0$, there exist $\delta >0$, $k\in \R$ and a signed measure
$\tilde{\omega}$ such that

* $\tilde{\omega}=\omega$ on $Q_\delta(z_0)$,

* $\tilde{\omega}=0$ outside of

* $|\tilde{\omega}|(Q_{\delta(1+\eta)}(z_0)\setminus\{z_0\})<\epsilon$

* on $Q_{\delta}(z_0)$,

Let $r>0$ be such that $Q_{r(1+\eta)}(z_0)\subset M$.
Let $\mes_r$ be the signed measure that is equal to $0$ on $ \bar{Q}_{r(1+\eta)}(z_0)$ and to
$\mes$ on the complementary.
We have

As $h$ is harmonic,  by Lemma <ref>, there exists a measure  $\psi_r^1$ concentrated on $C_{r(1+\eta /2)}(z_0)$ such that  for $z\in Q_{r(1+\eta /2)}(z_0)$,

$$h(z)= \iint  \ln|z-\zeta|d\psi_r^1(\zeta)+h(z_0)$$

and $|\psi^1_r|(\C)\rightarrow 0$ when $r\to 0$. This gives

$\bar{Q}_{r(1+\eta /2)}(z_0)$ is in the complementary of the support of $\mes_r$, hence in the complementary of the support of $\mes_r^+$ and $\mes_r^-$, and then  $p(\mes_r)$ is harmonic on $\bar{Q}_{r(1+\eta /2)}(z_0)$.
Lemma <ref> gives that there exists a measure  $\psi^2_r$ concentrated on $C_{r(1+\eta /2)}(z_0)$ such that
for $z\in Q_{r(1+\eta /2)}(z_0)$,

$$p(z;\mes_r)=\iint  \ln|z-\zeta|d\psi_r^2(\zeta)+p(z_0;\mes_r)~. $$

This gives

However, as the harmonic function $p(\mes_r)$ depends on $r$, we cannot directly conclude that the total variation of $\psi_r^2$ goes to $0$ when $r\to 0$. The measure $\psi_r^2$ is defined from the Lebesgue–Stieljes measure $\tilde{\psi}_r^2$ defined by (see the proof of  Lemma <ref>)

$$\tilde{\psi}_r^2(\alpha)=r(1+\eta /2)\int_0^\alpha \frac{\partial p(\mes_r)}{\partial \nu}(z_0+r(1+\eta /2)e^{it})dt~. $$

Using the definition of $p(\mes_r)$, an exchange of derivative and integration, Fubini and Lemma <ref>:

\begin{align*}
\tilde{\psi}_r^2(\alpha)&=\frac{r(1+\eta /2)}{2\pi}\int_0^\alpha \frac{\partial}{\partial \nu}   \iint_{|\zeta-z_0|>r(1+\eta)} \ln|\zeta-z_0-r(1+\eta /2)e^{it} | d\mes(\zeta) dt \\
&=\frac{1}{2\pi} \iint_{|\zeta-z_0|>r(1+\eta)} \int_0^\alpha r(1+\eta /2) \frac{\partial \ln|\zeta -\cdot |}{\partial \nu}(z_0+r(1+\eta /2)e^{it}) dt d\mes(\zeta) \\
& =-\frac{1}{2\pi} \iint_{|\zeta-z_0|>r(1+\eta)} \varphi(C_{r(1+\eta/2)}(z_0),\zeta,\alpha)d\mes(\zeta)~.
\end{align*}

By Lemma <ref>, we obtain

\begin{align*}|\psi_r^2|(\mathbb{C})&=|d\tilde{\psi}^2_r|([0,2\pi]) =\bigvee_0^{2\pi}\tilde{\psi}_r^2 \\
& \leq \frac{1}{2\pi}\iint_{|z_0-\zeta|>r(1+\eta)}\bigvee_0^{2\pi}\varphi(C_{r(1+\eta/2)}(z_0),\zeta,\cdot)d|\mes|(\zeta) \\
&= \frac{2}{\pi} \iint_{|z_0-\zeta|>r(1+\eta)} \arcsin \frac{r}{|z_0-\zeta|}d|\mes|(\zeta)\\
&= \frac{2}{\pi} \iint A_r(\zeta) d|\mes|(\zeta)
\end{align*}

\left\{
\begin{aligned}
\arcsin \frac{r}{|z_0-\zeta|} & \mbox{ when } & |z_0-\zeta|>r(1+\eta)~,\\
0 & \mbox{ } & \mbox{ otherwise}~,\\
\end{aligned}
\right.

which goes to $0$ when $r\to 0$ for any $\zeta$.
Hence $|\psi_r^2|(\mathbb{C})\rightarrow 0$ when $r\to 0$.

Let us set $\psi_r=\psi^1_r+\psi_r^2$. We have $|\psi_r|(\C)\rightarrow 0$ when $r\to 0$. Let us set

$$\tilde{\mes}_r=\mes|_{Q_{r(1+\eta)}(z_0)}+\psi_r~. $$

Note that, on $Q_r(z_0)$, $\psi_r=0$.
If we set $c_r=h(z_0)+p(z_0;\mes_r)$, we have

and one checks easily that $\tilde{\mes}_r$ satisfies the required properties for $r$ sufficiently small.

The Localization Theorem <ref> will be essential at at least two places:

* to prove the properties of the distance stretching, itself being fundamental in the proof of the Distances Convergence Theorem, see Section <ref>, or in the proof of the conformal nature of the distance preserving mappings, see Section <ref>;

* to prove that if the measure is not too concentrated, shortest arcs are arcs of bounded rotation with non-positive left and right turns, and arcs of bounded turn are of bounded rotation, see Section <ref>.

§ SUBHARMONIC DISTANCES ON THE PLANE

In all this section, unless specific precision, $M$ is  a bounded domain of the plane, $\mes$ is a signed Borelian  measure over the plane with compact support, $h$ is a harmonic function over $M$.

We defined the function
$p(\mes,h)$ over $M$ as the difference of two subharmonic functions, see (<ref>) and  Definition <ref>.  We denote

\begin{equation}\label{eq:def pl}\lambda(\mes,h)=e^{-2(p(\mes)+h)}~. \end{equation}

i.e., for $z\in M$,

$$\ln \lambda(z;\mes,h)=-\frac{1}{\pi}\iint \ln \vert z-\zeta\vert d\mes(\zeta) -2h(z)~. $$

Recall that such formula holds only quasi-everywhere on $M$, that will be implicit most of the time.

Once $\mes$ and $h$ are given, we denote $\lambda(\mes,h)$ by $\lambda$.  In turn, in the sense of distribution,  Proposition <ref> applied  to the positive and negative parts of $\mes$ gives that

\begin{equation}\label{eq:mesure PL}-\frac{1}{2}\Delta \ln \lambda =\mes~. \end{equation}

We may call $\lambda |dz|^2$ a subharmonic metricsubharmonic metric, and call $\mes$ the curvature measurecurvature measure of the metric. Apart from the fact that it is defined only quasi-everywhere, conversely to Riemannian metrics, it may reach values $0$ or $+\infty$ at some points. Standard examples are the ones of flat cones (see Lemma <ref>):

* if the curvature of the cone is positive, then the subharmonic metric is infinite at the vertex;

* if the curvature of the cone is negative, then the subharmonic metric is zero at the vertex.

We will define the $\lambda$-length of a rectifiable arc (for the Euclidean metric) in Section <ref>, and in Section <ref> we will define a distance from the length structure, and we present some preliminary convergence results. Section <ref>
presents Reshetnyak's method of canonical stretching of a distance, based on Localization  Theorem <ref>.  Then Reshetnyak's  convergence theorems are presented in Section <ref>.

§.§ Length of arcs

Let $\param$ be the arc length parameterization of a rectifiable arc.  By Lemma <ref>, $\lambda\circ \param$  is defined almost everywhere on $[0,s(K)]$, where $s(\arc)$ is  the Euclidean length of the arc.

In turn, if now $\param:[0,1]\to \C$ denotes the parameterization proportional to the arc length of a rectifiable arc $\arc$,
we can define the length of $\arc$ for the metric $\lambda |dz|^2$length for $\lambda |dz|^2$:

\begin{equation}\label{eq:tilde s}\tilde{s}_\lambda(\arc)= s(\arc)\int_0^1 \lambda^{1/2}(\param(t); \mes,h) dt~.\end{equation}

In this section, we prove a theorem about convergence of length (Theorem <ref>).

The following estimate is based on the  simple idea that, as $\lambda$ is the exponential of the integral of a logarithm, we may use Jensen inequality (Lemma <ref>).
The following result is a part of the fundamental Theorem 3.1 in [86]. We present it under the form of Lemma 4.2 in [107].

Let $\alpha$ and $\beta$ be such that

$$ 0\leq \beta = \frac{\alpha}{\pi}\mes^+(\C)~. $$

Then  if $M$ is included in $Q_R$,

$$\lambda^\alpha(z;\mes)=e^{2 \alpha(p(\mes^-)-p(\mes^+))(z)}\leq \frac{k_1^\alpha}{\omega^+(\mathbb{\C})}\iint \vert z - \zeta \vert^{-\beta} d\mes^+(\zeta)~, $$

where $k_1=2R^{\frac{\mes^-(\C)}{\pi}}$.

First,  $e^{2p(\mes^-)}$ is bounded from above by  $k_1$ by Lemma <ref>.
Then, we write

$$e^{-2\alpha p(\mes^+)}=\exp \left( \frac{1}{\omega^+(\C)}\iint \ln \vert z-\zeta\vert^{-\beta} d\mes^+(\zeta)\right)~.$$

The function $\zeta\to -\beta \ln \vert z-\zeta\vert$ is integrable because $\beta \geq 0$ and the following decomposition (up to increase $R$ to get $R>1/2$):

$$-\ln |z-\zeta| = \left(-\ln |z-\zeta| + \ln 2R\right) - \left(\ln 2R\right)~. $$

Both functions in the brackets are non-negative. We can apply Lemma <ref>, and the result follows.

An easy computation <cit.> shows that, for any arc length parameterized arc $\param$
of absolute rotation bounded from above by $A<\pi$, for any $0\leq \beta <1$, for any $\zeta\in M$, then

\begin{equation}\label{eq:inequalit gamma(h)}\int_0^{s(K)} \frac{dt}{|\zeta-\param(t)|^\beta} < \frac{2^{1+\beta}s(K)^{1-\beta}}{(1-\beta)\cos(A/2)^\beta}~. \end{equation}

Using this inequality together with an easy improvement of the proof of  Lemma <ref>, obtained the following fundamental estimate.

Let $\arc$ be an arc with absolute rotation bounded from above by $A < \pi$. Let $\arc(r)$ be the $r$-neighborhood of $\arc$. For each $\alpha, \beta$ such that

\begin{equation}\label{eq:hyp fund est}0\leq \beta = \frac{\alpha}{\pi}\omega^+(K(r)) <1~,  \end{equation}

we have

$$\int_K \lambda(\mes)^\alpha \leq  A_1 \frac{s(K)^{1-\beta}}{\cos(A/2)^\beta r^b}$$

where $b$ and $A_1$ depend only on $\alpha, \mes^+(K(r))$ and $\mes^\pm(\C)$.

In the case $\alpha=\frac{1}{2}$, the estimate in Theorem <ref> gives an upper bound for $\tilde{s}_\lambda$. The condition (<ref>) then becomes

\begin{equation}\label{eq: mes+Kr}\mes^+(K(r))<2\pi~.\end{equation}

In turn, if for a point $z\in M$,

\begin{equation}\label{eq:mes plu pet 2pi}\mes^+(\{z\}) <2\pi~,\end{equation}
as $\mes^+$ is regular, there exists a neighborhood $U$ of $z$ such that
$\mes^+(U)<2\pi$. Note that as the measures are finite, there is a finite number of points $z$ with $\mes^+(\{z\})\geq 2\pi$. The necessity of condition
(<ref>) is given by (<ref>). But it also has the following interpretation. Let us write