On conformal planes of finite area

Abstract

We discuss solutions of several questions concerning the geometry of conformal planes.

1 Introduction

1.1 Applications

Recently, the Liouville equation

$$\begin{aligned} \Delta u + e ^{2u} =0 , \end{aligned}$$

(1.1)

and its (super-) solutions on $\mathbb {R}^2$ were investigated in a series of work [3, 11, 13], see also [6, 9]. Interesting facts on the geometry of the corresponding conformal planes

$$\begin{aligned} X^u= (\mathbb {R}^2, e^{2u} \cdot \delta _{Eucl}) \end{aligned}$$

were proven and the authors formulated several related questions.

Solutions of (1.1) correspond to conformal planes of constant curvature 1 and are closely related to some meromorphic functions on $\mathbb {C}$. Complex analysis can been successfully used to study the solutions and arising geometries [3, 11]. For supersolutions of (1.1), thus for conformal metrics on the plane of curvature $\ge 1$, complex analysis does not seem to be such an appropriate tool.

The theory of surfaces with integral curvature bounds in the sense of Alexandrov, see [1, 23, 28] turns out to be more helpful, especially, for questions concerning conformal planes of bounded total area and curvature. This approach implies the following solutions to four questions formulated in [13, 14].

Proposition 1.1

For a smooth $u:\mathbb {R}^2 \rightarrow \mathbb {R}$ satisfying

$$\begin{aligned} \Delta u + e ^{2u} \le 0 , \end{aligned}$$

(1.2)

let the conformal plane $X^u$ have finite area. Then the diameter $\textrm{diam}(X^u)$ of the plane $X^u$ can be any number in the interval $(0,2\pi )$.

In [13, Theorem 1.4], it was proved that (1.2) implies $\textrm{diam}(X^u) \le 2\pi$, and [13, Question 8.2] asks if the inequality $\textrm{diam}(X^u) \le \pi$ holds.

Proposition 1.2

For a smooth $u:\mathbb {R}^2 \rightarrow \mathbb {R}$ satisfying (1.2), the area of the conformal plane $X^u$ can be infinite or any positive real number.

On contrary, for solutions u of (1.1) the conformal planes $X^u$ have area $4\pi$ or infinity, [13]. It has been asked in [13, Question 8.3], whether the upper bound of $4\pi$ is valid for all conformal planes $X^u$ of finite area corresponding to solutions of (1.2). The above result has been independently observed by Alexandre Eremenko.

As a consequence, we deduce a negative answer to another question formulated in [13, Question 8.7], see Corollary 2.1 below.

1.2 From the sphere to conformal planes

The above results are easy consequences of known theorems on singular metrics on ${\mathbb {S}}^2$ with bounded integral curvature and of a simple relation between conformal planes and conformal spheres, which we are going to explain now.

By the uniformization theorem, any Riemannian metric on $\mathbb {R}^2$ is either conformally equivalent to the disc or to the plane. While it is easy to construct many (non-complete) Riemannian metrics on $\mathbb {R}^2$ with prescribed curvature properties, (for instance, with constant curvature 1), it seems difficult to verify that such a synthetically constructed metric is a conformal plane. A criterion of conformality is provided by the special case of a classical result of Cheng–Yau [10, Corollary 1]: If a complete Riemannian manifold X homeomorphic to $\mathbb {R}^2$ has at most quadratic area growth then X is a conformal plane. In particular, all complete Riemannian metrics of finite area on $\mathbb {R}^2$ are conformal planes.

An easy criterion for non-complete planes, sufficient for the Propositions stated above, is the following one.

Proposition 1.3

Let X be a Riemannian manifold homeomorphic to the plane and of finite area. Assume that the completion ${\hat{X}}$ of X is homeomorphic to ${\mathbb {S}}^2$ and that ${\hat{X}} {\setminus } X$ has just one point p. If the area of metric balls $B_r(p)$ in ${\hat{X}}$ around p grows at most quadratically,

$$\begin{aligned} \liminf _{r\rightarrow 0} \frac{\textrm{area}(B_{r}(p))}{r^2} < \infty , \end{aligned}$$

then X is conformally equivalent to the plane.

It might be possible to deduce Proposition 1.3 from the theorem by Cheng–Yau mentioned above, applying a conformal change of the metric, which resembles the inversion at the point p. Instead, we observe that Proposition 1.3 is a consequence of a very general uniformization theorem in metric geometry [4, 18, 20, 22].

Remark 1.4

Some assumption in Proposition 1.3 on a neighborhood of p in ${\hat{X}}$ is needed, as the following easy example demonstrates: Consider the unit Euclidean disc with the conformal factor $f(z)=(1-|z|^2)$. The completion ${\hat{X}}$ of this conformal disc X has finite area, is homeomorphic to ${\mathbb {S}}^2$, and ${\hat{X}}\setminus X$ has just one point.

Thus, in order to construct conformal planes with prescribed properties as in Propositions 1.1, 1.2, it suffices to construct metrics on the sphere with one singularity p and prescribed geometric properties outside the singularity. We construct such a piecewise spherical metric with only 3 vertices, such that the total angle at just one of these vertices (the singularity p) is larger than $2\pi$. Note that all such metics are classified [12, 19]. Smoothing the metric at the singularites with angles smaller than $2\pi$, we obtain the desired examples. These examples have bounded integral curvature in the sense of Alexandrov, [1, 23, 28]; more classical uniformization theorems, [28], imply the conclusion of Proposition 1.3 in this case.

1.3 Completions of conformal planes

A partial converse to Proposition 1.3 is essentially contained in the proof of [13, Theorem 1.4]:

Lemma 1.5

Let the conformal plane $X=X^u$ have finite area and let ${\hat{X}}$ denote the completion of X. Then ${\hat{X}}{\setminus } X$ has at most one point.

Thus, either X is complete or ${\hat{X}} {\setminus } X$ has exactly one point p. In the latter case, the space ${\hat{X}}$ can display a rather wild behavior near p. For instance, it may not be locally compact around p, see Example 3.1 below. Even if X has curvature larger than 1 and ${\hat{X}}$ is compact, thus homeomorphic to ${\mathbb {S}}^2$, the geometry around p can be rather wild, see Example 3.2 below.

The geometry of the completion ${\hat{X}}$ at the singular point ${\hat{X}}{\setminus } X$ turns out to be much tamer if the curvature on X is assumed to be integrable.

Recall first that the Hausdorff (=canonical Riemannian) area ${\mathcal {H}}^2$ on the conformal plane $X^u$ is the multiple $e^{2u} \cdot {\mathcal {L}}^2 _{\mathbb {R}^2}$ of the Lebesgue area ${\mathcal {L}}^2$. Thus the total area of $X^u$ equals ${\mathcal {A}}(X^u)= \int _{\mathbb {R}^2} e^{2u}$.

The curvature of the conformal plane $X^u$ equals $K=e^{-2u} \cdot \Delta u$. Thus, the (integral) boundedness of the curvature of $X^u$, is the analytic assumptions $\Delta u \in L^{\infty } (\mathbb {R}^2)$ ($\Delta u \in L^{1} (\mathbb {R}^2)$). If $\Delta (u)\in L^1 (\mathbb {R}^2)$ then

$$\begin{aligned} {\mathcal {K}}(X^u):=\int _{\mathbb {R}^2} \Delta u \,d{\mathcal {L}}^2 _{\mathbb {R}^2}= \int _{X^u}K \, d{\mathcal {H}}^2 _X \end{aligned}$$

is called the total curvature of $X^u$.

Most parts of the next result are scattered through the literature:

Theorem 1.6

Let $X=X^u$ be a conformal plane of finite area ${\mathcal {A}}(X)$ and finite total curvature ${\mathcal {K}}(X)$. Then ${\mathcal {K}} (X) \ge 2\pi$. If ${\mathcal {K}}(X) >2\pi$ then $X^u$ is not complete.

If X is not complete then the completion ${\hat{X}}$ is a sphere which has bounded integral curvature in the sense of Alexandrov.

Upon a conformal identification of $\mathbb {R}^2$ with ${\mathbb {S}}^2{\setminus } \{p \}$, the function u defines a $\delta$-subharmonic function on ${\mathbb {S}}^2$, in the complete and in the non-complete case.

Recall that a function is called $\delta$-subharmonic if locally around any point it can represented as a difference of two subharmonic functions.

The theory of surfaces with integral curvature bounds implies that in the non-complete case, the area growth is at most quadratic at the point $p={\hat{X}} {\setminus } X$. Moreover, limes inferior arising in Proposition 1.3 is a limit and equals $\frac{{\mathcal {K}}(X)}{2} -\pi$, see Sect. 4.1.

1.4 Uniformly bounded curvature

A final application answers the question investigated in [14] and relates this question to the theory of manifolds with both-sided curvature bounds, [5]. Slightly weaker results have been obtained in [14] by direct methods.

Proposition 1.7

Assume that the plane $X=X^u$ has finite area and that the total curvature ${\mathcal {K}}(X)$ equals $4\pi$. If the curvature K of X is uniformly bounded then the completion ${\hat{X}}$ of X is a Riemannian manifold conformally equivalent to the round sphere ${\mathbb {S}}^2$. For the conformal factor $e^{2 {\hat{u}}}$, the function ${\hat{u}}$ is of class ${\mathcal {C}}^{1,\alpha }$ on ${\mathbb {S}}^2$, for every $\alpha <1$.

Even if the curvature K is continuous on ${\hat{X}}$, the function ${\hat{u}}$ does not need to be ${\mathcal {C}}^{1,1}$. If K is $\beta$-Hoelder on ${\mathbb {S}}^2$ then ${\hat{u}}$ is ${\mathcal {C}}^{2,\beta }$.

2 From the sphere to the plane

2.1 One-point complements in spheres

In the proof of Proposition 1.3 below, we are going to freely use the vocabulary of metric geometry. We refer to [20] for the definitions and properties, in particular for the notion of weak conformality.

Proof of Proposition 1.3

By assumption, we have a geodesic metric space ${\hat{X}}$, homeomorphic to ${\mathbb {S}}^2$ and a point $p\in {\hat{X}}$ such that $X={\hat{X}}{\setminus } \{p \}$ has a smooth Riemannian metric. By assumption, the area growth at p is at most quadratic. In particular, ${\hat{X}}$ has finite 2-dimensional Hausdorff measure.

By [20, Theorem 1.3], there exists a weakly quasiconformal map $h:{\mathbb {S}}^2 \rightarrow {\hat{X}}$ from the round sphere ${\mathbb {S}}^2$.

The area growth assumption implies that h is a homeomorphism, [20, Theorem 7.4]. The map h restricts to a weakly quasiconformal map from ${\mathbb {S}}^2{\setminus } h^{-1} (p) \rightarrow X$. Since $h^{-1} (p)$ is a singleton, ${\mathbb {S}}^2{\setminus } h^{-1} (p)$ is conformally equivalent to $\mathbb {R}^2$. Therefore, we have a weakly quasiconformal map between smooth Riemannian manifolds ${\hat{h}}: \mathbb {R}^2 \rightarrow X$. If X were a conformal disc, we would obtain a weakly quasiconformal homeomorphism from $\mathbb {R}^2$ to the disc D. Such a homeomorphism cannot exist, see, for instance, [17, p. 2–4]. $\square$

Assuming that ${\hat{X}}$ has bounded integral curvature in the sense of Alexandrov, [1, 23, 28], a shorter proof of Proposition 1.3 is possible. Indeed, in this case, the uniformization theorem, [28, Section 7] states that the metric on ${\hat{X}}$ is defined as $e^v\cdot \delta _{{\mathbb {S}}^2}$, where the function v in the conformal factor is $\delta$-subharmonic on ${\mathbb {S}}^2$. This directly describes $X ={\hat{X}}{\setminus } \{p\}$ as conformally changed ${\mathbb {S}} ^2$ without a point.

2.2 Some examples of conformal planes

We are going to prove Proposition 1.1 and Proposition 1.2. Observe first, that rescaling the metric by a positive constant $\lambda \le 1$ provides again a metric in the same class (curvature at least 1, finite area). Thus, it suffices to find conformal planes of curvature $\ge 1$ and arbitrary large finite area, respectively, finite area and diameter arbitrary close to $2\pi$.

Consider a piecewise spherical metric on ${\mathbb {S}}^2$ such that the total angle is larger than $2\pi$ in at most one singularity p. In the arising metric space Y the curvature is constant 1 outside p and finitely many further points $p_1,..,p_n$. Around any point $p_i$ the metric is a spherical cone metric over a circle of length less than $2\pi$. The metric around $p_i$ can be smoothened in an arbitrary small neighborhood, such that the arising metric is smooth and has curvature $\ge 1$, [16, Lemma 2.4]. Moreover, by construction, the new smooth metric has almost the same diameter and area as the original one.

Performing this operation around every vertex $p_1,\ldots ,p_n$, we obtain a metric space $Y_{\varepsilon }$ homeomorphic to ${\mathbb {S}}^2$, such that $X:=Y_{\varepsilon } {\setminus } p$ is a smooth Riemannian manifold of curvature $\ge 1$. This manifold X is a conformal plane by Proposition 1.3; it has finite area and diameter arbitrary close (by the choice of $\varepsilon$) to the area and the diameter of Y.

Thus, in order to prove Propositions 1.1 and 1.2 it suffices to find piecewise spherical metrics Y on ${\mathbb {S}}^2$ with at most one singularity of total angle larger than $2\pi$ and arbitrary large area, respectively, diameter arbitrary close to $2\pi$.

Proof of Proposition 1.2

Consider an interval $I=[a,b]$ of large length N. Let Z be the spherical join of a point p and I. The space Z is topologically a closed disc and it has curvature one in the interior. The boundary of Z is built by two geodesics pa and pb of length $\pi /2$ and by the local geodesic I. The angle at a and b is $\frac{\pi }{2}$, the total angle at p equals N. The area of Z equals N.

Consider the doubling Y of Z along the boundary. Then Y is a piecewise spherical metric on the 2-sphere, with 3 singularities of total angles $\pi , \pi , 2N$ and with total area 2N. Due to the consideration preceding the proof, this suffices for the conclusion. $\square$

Proof of Proposition 1.1

Fix $\varepsilon <\frac{\pi }{2}$. Consider a triangle $D=pxy$ in the round sphere ${\mathbb {S}}^2$ with px of length $\varepsilon$, with $\angle pxy =\frac{\pi }{2}$ and with the length of xy equal to $\pi -\varepsilon$. Then $\angle pyx< \frac{\pi }{2} < \angle ypx$.

Consider another isometric copy $D'=pxy'$ of the triangle and glue D and $D'$ along the common side px. The arising space Z is homeomorphic to a closed disc. It has constant curvature 1 in the interior. The boundary is built by 4 geodesics py, $py'$, yx and $y'x$. The angle at x equals $\pi$, the angles at y and $y'$ are smaller than $\pi$, the angle at p is larger than $\pi$. The diameter of Z is at least the distance of y and $y'$ which is larger than $2\pi - 4\varepsilon$.

The doubling Y of Z along the boundary $\partial Z$ is homeomorphic to ${\mathbb {S}}^2$ and has diameter at least $2\pi -4\varepsilon$. Moreover, Y has piecewise constant curvature 1 and at exactly one singularity p the total angle is larger than $2\pi$. Due to the consideration preceding the proof, this suffices for the conclusion, since $\varepsilon$ can be chosen arbitrary small. $\square$

As a consequence we provide the following negative answer to [13, Question 8.7]. We refer to the discussion in [13, Section 7] for motivation and relation with the Levy–Gromov inequality.

Corollary 2.1

For any $\epsilon >0$ there exist a smooth Jordan curve $\Gamma$ in $\mathbb {R}^2$ bounding a Jordan domain $\Omega$ and a smooth $u:\mathbb {R}^2 \rightarrow \mathbb {R}$ satisfying (1.2), such that $\int _{R^2} e^{2u} <\infty$ and the following holds true:

$$\begin{aligned} \left( \int _{\Gamma } e^u\right) ^2 \le \varepsilon \cdot \int _{\Omega } e^{2u} \cdot \int _{\mathbb {R}^2{\setminus } \Omega } e^{2u} . \end{aligned}$$

Proof

The construction in the proof of Proposition 1.2 provides conformal metrics $X^u$ on $\mathbb {R}^2$ with curvature $\ge 1$ and arbitrary large but finite area $A=A(u)$. Moreover, by construction, any of this metric spaces $X^u$ contains a metric ball $\Omega$ of radius $r=\frac{\pi }{10}$ in the round sphere ${\mathbb {S}}^2$. Let $l_0$ and $A_0$ denote the length of $\partial \Omega$, respectively the area of $\Omega$ (both quantities measured in $X^{u}$, hence in ${\mathbb {S}}^2$).

Then the right hand side $(\int _\Gamma e ^u )^2$ of the claimed inequality is just $l_0^2$ while the factors on the right hand side are $A_0$ and $A-A_0$ respectively. Thus, choosing u such that the area $A={\mathcal {A}} (X^u)$ satisfies

$$\begin{aligned} A\ge A_0 + \frac{l_0}{\sqrt{\varepsilon }} , \end{aligned}$$

we finish the proof. $\square$

3 Planes of finite area

The next argument is contained in the proof of [13, Theorem 1.4].

Proof of Lemma 1.5

The space X is a length space, hence so is the completion ${\hat{X}}$, [2, p. 43]. More precisely, for any $x\in {\hat{X}}{\setminus } X$ there exists a curve of finite length $\gamma _x: [0,a) \rightarrow X$, such that in ${\hat{X}}$ we have

$$\begin{aligned} \lim _{t\rightarrow a} \gamma _x (t)= x . \end{aligned}$$

Assume that we have two different points $x,y \in {\hat{X}}{\setminus } X$. Denote by $\varepsilon >0$ the distance between x and y. Consider curves $\gamma _x, \gamma _y$ of finite length in X converging to x and y, as above. By changing the starting points, we may assume that $\gamma _x$ and $\gamma _y$ have length smaller than $\frac{\varepsilon }{4}$. In order to obtain a contradiction, it suffices to find points on $\gamma _x$ and $\gamma _y$ with distance less than $\frac{\varepsilon }{4}$ from each other.

Our space X is the plane $\mathbb {R}^2$ with the Euclidean metric changed by the conformal factor $e^{2u}$. Denote by $\eta _r$ the Euclidean circle around 0 of radius r. We express the finiteness of the area in polar coordinates and obtain by the Hoelder inequality

$$\begin{aligned} \infty >{\mathcal {A}}(X) =\int _{\mathbb {R}^2} e^{2u} = \int _0 ^{\infty } \left( \int _{\eta _r}e ^{2u}\right) \, dr \ge \int _0 ^{\infty } \frac{1}{2\pi r} \left( \int _{\eta _r} e^u\right) ^2 \,dr. \end{aligned}$$

The length of $\eta _r$ in the metric space X is $\int _{\eta _r} e^u$. Therefore, we find a sequence $r_i \rightarrow \infty$ such that the length of $\eta _{r_i}$ is smaller than $\frac{\varepsilon }{4}$.

Since the curves $\gamma _x$ and $\gamma _y$ do not have limit points in X, both curves run to infinity in $\mathbb {R}^2$. Hence they both intersect $\eta _ r$, for all sufficiently large r. Thus, for sufficiently large $r_i$ as above, we find points in the intersection of $\gamma _x$ and $\gamma _y$ with $\eta _{r_i}$. The distance between these intersection points in X is less than $\frac{\varepsilon }{4}$, in contradiction to our assumption. Hence, ${\hat{X}} {\setminus } X$ contains at most one point. $\square$

We are going to explain that ${\hat{X}}$ does not need to be locally compact at the point $\{p \} = {\hat{X}} {\setminus } X$.

Example 3.1

Consider the round sphere ${\mathbb {S}}^2$ with north pole p. Take a sequence $U_j$ of small metric balls centered on a fixed meridian starting at p. We choose the metric balls pairwise disjoint, not containing p, but converging to p. Change the metric conformally on ${\mathbb {S}}^2{\setminus } \{ p\}$ in the following way. The conformal factor is constantly one outside the union of all $U_j$. The subset $U_j$ has after the conformal change diameter approximately 1 and area approximately $\frac{1}{j^2}$, thus $U_j$ becomes a long and very thin finger sticking out of the sphere. The new metric on ${\mathbb {S}}^2 {\setminus } p$ is conformally equivalent to $\mathbb {R}^2$, it has finite area and diameter. Moreover, it has infinitely many points with pairwise distances in the interval [2, 3]. Hence, the completion ${\hat{X}}$ cannot be locally compact by the theorem of Hopf–Rinow.

The next example shows that even if ${\hat{X}}$ is compact and X has curvature at least 1, the curvature does not need to be integrable and the area growth at the singularity $p={\hat{X}}{\setminus } X$ can be superquadratic.

Example 3.2

Consider the metric on $\mathbb {R}^2$ with conformal factor $e^{-\frac{2}{|z|}} \cdot |z|^{-4}$ as in [24, Section 5.1], [7, Section 4.1]. The area growth of this metric space Y at $p=0$ is superquadratic, [24, p. 19]. Euclidean balls around 0 are metric balls around $p=0$ in Y and they are convex. Y is smooth outside of 0 and direct computations reveal that the metric has positive curvature outside of p; moreover, the curvature converges to $\infty$ at p. Consider now a small closed ball B around 0 in Y such that the curvature is larger than 1 outside of $0=p$ and such that $\partial B$ has length $2\pi s < 2\pi$. Glue to B along $\partial B$ a round hemisphere of radius s. By the gluing theorem (for instance, [21]), the arising sphere has curvature $> 1$ outside the singularity 0. Smoothing the metric along $\partial B$, (see for instance, [16]), we obtain a smooth metric ${\hat{X}}$ on ${\mathbb {S}}^2$, which has curvature $\ge 1$ everywhere outside a single point p and that around p the metric is isometric to Y. By construction (and the uniformization theorem), the metric on ${\hat{X}} {\setminus } \{ p\}$ is conformally equivalent to $\mathbb {R}^2$.

It seems possible but technically more involved to construct an example of a conformal plane $X=X^u$ of curvature $\ge 1$ and finite area, such that the diameter of X is $2\pi$ (thus strengthening Proposition 1.1). In such an example the completion ${\hat{X}}$ has to be non-compact.

4 Planes of finite area and curvature

4.1 Integral bound

If the conformal plane $X=X^u$ has finite total curvature we can control the geometry at infinity much better:

Proof of Theorem 1.6

First assume that $X=X^u$ is complete. Then the curvature estimate ${\mathcal {K}}(X) \le 2\pi$ is a classical theorem of Cohn-Vossen, [8, Satz 6], valid also for complete planes of infinite area. Given that the area is finite, the equality ${\mathcal {K}}(X)=2\pi$ is proven in [25, Corollary].

Finally, due to [15, Korollar] (or, alternatively, [15, Satz 3]) if X is complete then the function u extends to a $\delta$-subharmonic function on ${\mathbb {S}}^2$, once $\mathbb {R}^2$ is identified with ${\mathbb {S}}^2$ without a point by a conformal transformation.

From now on we assume that X is not complete. We consider the completion ${\hat{X}}$ and let p be the unique point in ${\hat{X}} {\setminus } X$, Lemma 1.5.

First, we claim that ${\hat{X}}$ is compact. Otherwise, we find some $\varepsilon >0$ and infinitely many points $x_i \in X$ with pairwise distance larger than $2\varepsilon$. Removing at most one point, we can assume that the distance of any $x_i$ to p is larger than $\varepsilon$. Then the closed balls ${\bar{B}}_{\varepsilon } (x_i)$ are pairwise disjoint and compact. Moreover, removing finitely many $x_i$ and using the finiteness of total curvature, we may assume that the total positive curvature of any ${\bar{B}}_{\varepsilon } (x_i)$ is at most $\pi$. Then, for any i, we can estimate the area of the ball as

$$\begin{aligned} {\mathcal {A}}(B_{\varepsilon } (x_i) \ge \frac{\pi }{2} \cdot \varepsilon ^2, \end{aligned}$$

due to [26, Proposition 3.2], [23, Theorem 9.1]. Thus, the finiteness of ${\mathcal {A}}(X)$ contradicts the disjointness of the balls ${\bar{B}}_{\varepsilon } (x_i)$.

Therefore, ${\hat{X}}$ is compact. Due to the uniqueness of the one-point-compactification, ${\hat{X}}$ is homeomorphic to ${\mathbb {S}}^2$.

In order to prove that ${\hat{X}}$ is a surface with bounded integral curvature we present the metric on ${\hat{X}}$ as a limit of metrics with a uniform integral bound on curvature, as in [23, Section 8.4].

We claim that there exists a sequence of simple closed curves $\Gamma _j$ in X, such that for the Jordan domains $p\in O_j$ in ${\hat{X}}$ of $\Gamma _j$ the following holds true: The closure ${\bar{O}}_j =O_j \cup \Gamma _j$ is convex in ${\hat{X}}$; the diameter of ${\bar{O}}_j$ and the length of $\Gamma _j$ are at most $\frac{1}{j}$.

Note, that any such $\Gamma _j$ would be of bounded turn and the variation from the side of $X{\setminus } O_j$ (thus the mean curvature) would be non-positive, by convexity of $O_j$, cf. [23, Theorem 8.1.3]. Moreover, by the Gauss–Bonnet formula and the bound on the total curvature of X, the total curvature of $\Gamma _j$ would be uniformly bounded.

Once such $\Gamma _j$ are found, we would cut out $O_j$ and replace it by the round hemisphere ${\hat{O}}_j$ with boundary of length $\ell (\Gamma _j)$. The arising space ${\hat{X}}_j$ is a sphere with uniformly bounded integral curvature, [23, Theorems 8.3.1, 8.3.2]. Moreover, identifying ${\hat{O}}_j$ with $O_j$ by any homeomorphism fixing $\Gamma _j$, we obtain a convergence of ${\hat{X}}_j$ to ${\hat{X}}$ in the sense of [23, Section 8.4]. Thus, ${\hat{X}}$ would be of integrally bounded curvature, [23, Theorem 8.4.5].

It remains to find the required curves $\Gamma _j$. In order to find them, we fix j and set $\delta =\frac{1}{10 j}$. Consider the open ball $U=B_{\delta } (p)$. We find an index i, such that for all $k\ge i$, the curves $\eta _k:=\eta _{r_k}$ constructed in the proof of Lemma 1.5 have length $\ell (\eta _{r_k} ) <\delta$. By construction, the Jordan domains $p\in U_k$ of $\eta _k$ are nested and their intersection consists of the point p only. Choosing i large enough, we may assume in addition, that $U_k\subset U$, for all $k\ge i$.

We fix this $\eta _i$. By compactness and local contractibility, there is some $\varepsilon >0$ such that no closed curve in ${\bar{U}}_i$ of length at most $\varepsilon$ can intersect $\eta _i$ and be homotopic to $\eta _i$ within the punctured disc ${\bar{U}}_j {\setminus } \{p \}$. We now find some $k>i$ such that $\ell (\eta _k ) <\varepsilon$.

In the compact annulus A bounded by $\eta _k$ and $\eta _i$ in X we find a shortest non-contractible curve $\gamma$. This $\gamma$ is automatically simple closed. By the choice of k, this curve $\gamma$ has length at most $\epsilon$, and by the choice of $\epsilon$, the curve $\gamma$ does not intersect $\eta _i$. The Jordan domain $p\in V$ of this curve is contained in U, thus has diameter at most $2\delta$. If V were not convex, then two points on $\gamma$ could be connected within A by a shorter curve. But this would contradict the minimal property of $\gamma$. This finishes the construction of $\gamma =\gamma _j$ and, therefore, of the statement that ${\hat{X}}$ has bounded integral curvature.

The final statement that the metric of ${\hat{X}}$ is conformal to the round metric on the sphere including p is a direct consequence of the uniformization for such surfaces, [23, Section 7], [28]. $\square$

Remark 4.1

We have used some geometric arguments in the non-complete case in the proof above. Possibly, a more analytic proof of the statement using the full strength of [15, Satz 3] can be found.

Some additional comments on the structure of ${\hat{X}}$ near $p={\hat{X}}{\setminus } X$, in case that X is not complete in Theorem 1.6:

Consider the curvature measure $\hat{{\mathcal {K}}}$ on the sphere ${\hat{X}}$ with bounded integral curvature, [1, Chapter 5], [28]. This is a signed measure satisfying $\hat{{\mathcal {K}}} ({\hat{X}})=4\pi$ by the Gauss–Bonnet theorem [28, p. 20]. On the regular part X, the signed measure $\hat{{\mathcal {K}}}$ equals $K\cdot {\mathcal {H}}^2$, where K is the Gaussian curvature. Thus, $\hat{{\mathcal {K}}} (\{p \})= 4\pi - {\mathcal {K}} (X)$. On the other hand, $\hat{{\mathcal {K}}} (\{p\})$ equals $2\pi -\theta$, where $\theta$ is the total angle at the point p [23, Lemma 8.1.1]. Moreover, again by [23, Lemma 8.1.1] and the coarea formula (or using [23, Theorem 9.10])

$$\begin{aligned} \theta = \lim _{r\rightarrow 0} \frac{{\mathcal {H}}^1 (\partial B_r(p))}{r} = 2 \lim _{r\rightarrow 0} \frac{{\mathcal {H}}^2 (B_r(x))}{r^2}. \end{aligned}$$

4.2 Smoothness at infinity

We are going to provide

Proof of Proposition 1.7

We can apply Theorem 1.6. Identifying $\mathbb {R}^2$ conformally with the complement of a point p in ${\mathbb {S}}^2$, we obtain that the completion ${\hat{X}}$ is a sphere with curvature bounded in the integral sense of Alexandrov. The curvature measure $\hat{{\mathcal {K}}}$ of ${\hat{X}}$ coincides on X with the multiple $\hat{{\mathcal {K}}} = K\cdot {\mathcal {H}}^2$ of the canonical area measure ${\mathcal {H}}^2$. By the Gauss-Bonnet theorem, $\mathcal {{\hat{K}}}({\hat{X}})= 4\pi$,. Thus, by assumption, $\hat{{\mathcal {K}}} (\{ p\})=0$. Therefore, the equality $\hat{{\mathcal {K}}} = K\cdot {\mathcal {H}}^2$ is valid on all of ${\hat{X}}$.

Therefore, on all of ${\hat{X}}$, the metric is defined by a conformal change $e^{2{\hat{u}}} \cdot \delta _{{\mathbb {S}}^2}$ of the round metric on ${\mathbb {S}}^2$, such that the spherical Laplacian of ${\hat{u}}$ is a bounded function $K + 1$. Elliptic regularity implies that ${\hat{u}}$ is of class ${\mathcal {C}}^{1,\alpha }$ for any $\alpha <1$. Moreover, if the curvature $K:X=\mathbb {R}^2 \rightarrow \mathbb {R}$ extends as a $\beta$-Hoelder continuous function to ${\mathbb {S}}^2$ then ${\hat{u}}$ is ${\mathcal {C}}^{2,\beta }$-Hoelder.

An example of a conformal metric $e^{2 v} \cdot \delta _{\mathbb {R}^2}$ on a disc, which is smooth outside the origin p, not ${\mathcal {C}}^{1,1}$ at the origin and, such that the Lapalcian $\Delta v$ is continuous, is presented in [27, p. 693]. This metric (restricted to a subdisc) can clearly be extended to a metric on the sphere, which has continuous curvature but is not ${\mathcal {C}}^{1,1}$ in conformal coordinates. This example finishes the proof. $\square$