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An isoperimetric inequality in the plane with a log-convex density

  • I. McGillivray
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Abstract

Given a positive lower semi-continuous density f on \(\mathbb {R}^2\) the weighted volume \(V_f:=f\mathscr {L}^2\) is defined on the \(\mathscr {L}^2\)-measurable sets in \(\mathbb {R}^2\). The f-weighted perimeter of a set of finite perimeter E in \(\mathbb {R}^2\) is written \(P_f(E)\). We study minimisers for the weighted isoperimetric problem
$$\begin{aligned} I_f(v):=\inf \Big \{ P_f(E):E\text { is a set of finite perimeter in }\mathbb {R}^2\text { and }V_f(E)=v\Big \} \end{aligned}$$
for \(v>0\). Suppose f takes the form \(f:\mathbb {R}^2\rightarrow (0,+\infty );x\mapsto e^{h(|x|)}\) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Let \(v>0\) and B a centred ball in \(\mathbb {R}^2\) with \(V_f(B)=v\). We show that B is a minimiser for the above variational problem and obtain a uniqueness result.

Keywords

Isoperimetric problem Log-convex density Generalised mean curvature 

Mathematics Subject Classification

49Q20 

1 Introduction

Let f be a positive lower semi-continuous density on \(\mathbb {R}^2\). The weighted volume \(V_f:=f\mathscr {L}^2\) is defined on the \(\mathscr {L}^2\)-measurable sets in \(\mathbb {R}^2\). Let E be a set of finite perimeter in \(\mathbb {R}^2\). The weighted perimeter of E is defined by
$$\begin{aligned} P_f(E):=\int _{\mathbb {R}^2}f\,d|D\chi _E|\in [0,+\infty ]. \end{aligned}$$
(1.1)
We study minimisers for the weighted isoperimetric problem
$$\begin{aligned} I_f(v):=\inf \Big \{ P_f(E):E\text { is a set of finite perimeter in }\mathbb {R}^2\text { and }V_f(E)=v\Big \} \end{aligned}$$
(1.2)
for \(v>0\). To be more specific we suppose that f takes the form
$$\begin{aligned} f:\mathbb {R}^2\rightarrow (0,+\infty );x\mapsto e^{h(|x|)} \end{aligned}$$
(1.3)
where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Our first main result is the following. It contains the classical isoperimetric inequality (cf. [9, 12]) as a special case; namely, when h is constant on \([0,+\infty )\).

Theorem 1.1

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Let \(v>0\) and B a centred ball in \(\mathbb {R}^2\) with \(V_f(B)=v\). Then B is a minimiser for (1.2).

For \(x\ge 0\) and \(v\ge 0\) define the directional derivative of h in direction v by
$$\begin{aligned} h^\prime _+(x,v):=\lim _{t\downarrow 0}\frac{h(x+tv)-h(x)}{t}\in \mathbb {R} \end{aligned}$$
and define \(h^\prime _-(x,v)\) similarly for \(x>0\) and \(v\le 0\). We introduce the notation
$$\begin{aligned} \rho _+:=h^\prime _+(\cdot ,+1), \rho _-:=-h^\prime _+(\cdot ,-1) \text { and } \rho :=(1/2)(\rho _++\rho _-) \end{aligned}$$
on \((0,+\infty )\). The function h is locally of bounded variation and is differentiable a.e. with \(h^\prime =\rho \) a.e. on \((0,+\infty )\). Our second main result is a uniqueness theorem.

Theorem 1.2

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Suppose that \(R:=\inf \{\rho >0\}\in [0,+\infty )\) and set \(v_0:=V(B(0,R))\). Let \(v>0\) and E a minimiser for (1.2). The following hold:
  1. (i)

    if \(v\le v_0\) then E is a.e. equivalent to a ball B in \(\overline{B}(0,R)\) with \(V(B)=V(E)\);

     
  2. (ii)

    if \(v>v_0\) then E is a.e. equivalent to a centred ball B with \(V(B)=V(E)\).

     

Theorem 1.1 is a generalisation of Conjecture 3.12 in [24] (due to K. Brakke) in the sense that less regularity is required of the density f: in the latter, h is supposed to be smooth on \((0,+\infty )\) as well as convex and non-decreasing. This conjecture springs in part from the observation that the weighted perimeter of a local volume-preserving perturbation of a centred ball is non-decreasing ([24] Theorem 3.10). In addition, the conjecture holds for log-convex Gaussian densities of the form \(h:[0,+\infty )\rightarrow \mathbb {R};t\mapsto e^{ct^2}\) with \(c>0\) ([3, 24] Theorem 5.2). In subsequent work partial forms of the conjecture were proved in the literature. In [19] it is shown to hold for large v provided that h is uniformly convex in the sense that \(h^{\prime \prime }\ge 1\) on \((0,+\infty )\) (see [19] Corollary 6.8). A complemen tary result is contained in [11] Theorem 1.1 which establishes the conjecture for small v on condition that \(h^{\prime \prime }\) is locally uniformly bounded away from zero on \([0,+\infty )\). The above-mentioned conjecture is proved in large part in [7] (see Theorem 1.1) in dimension \(n\ge 2\) (see also [4]). There it is assumed that the function h is of class \(C^3\) on \((0,+\infty )\) and is convex and even (meaning that h is the restriction of an even function on \(\mathbb {R}\) to \([0,+\infty )\)). A uniqueness result is also obtained ( [7] Theorem 1.2). We obtain these results under weaker hypotheses in the 2-dimensional case and our proofs proceed along different lines.

We give a brief outline of the article. In Sect. 2 we discuss some preliminary material. In Sect. 3 we show that (1.2) admits an open minimiser E with \(C^1\) boundary M (Theorem 3.8). The argument draws upon the regularity theory for almost minimal sets (cf. [27]) and includes an adaptation of [21] Proposition 3.1. In Sect. 4 it is shown that the boundary M is of class \(C^{1,1}\) (and has weakly bounded curvature). This result is contained in [21] Corollary 3.7 (see also [8]) but we include a proof for completeness. This Section also includes the result that E may be supposed to possess spherical cap symmetry (Theorem 4.5). Section 5 contains further results on spherical cap symmetric sets useful in the sequel. The main result of Sect. 6 is Theorem 6.5 which shows that the generalised (mean) curvature is conserved along M in a weak sense. In Sect. 7 it is shown that there exist convex minimisers of (1.2). Sections 8 and 9 comprise an analytic interlude and are devoted to the study of solutions of the first-order differential equation that appears in Theorem 6.6 subject to Dirichlet boundary conditions. Section 9 for example contains a comparison theorem for solutions to a Ricatti equation (Theorem 9.15 and Corollary 9.16). These are new as far as the author is aware. Section 10 concludes the proof of our main theorems.

2 Some preliminaries

Geometric measure theory. We use \(|\cdot |\) to signify the Lebesgue measure on \(\mathbb {R}^2\) (or occasionally \(\mathscr {L}^2\)). Let E be a \(\mathscr {L}^2\)-measurable set in \(\mathbb {R}^2\). The set of points in E with density \(t\in [0,1]\) is given by
$$\begin{aligned} E^t:=\left\{ x\in \mathbb {R}^2:\,\lim _{\rho \downarrow 0}\frac{|E\cap B(x,\rho )|}{|B(x,\rho )|}=t\right\} . \end{aligned}$$
As usual \(B(x,\rho )\) denotes the open ball in \(\mathbb {R}^2\) with centre \(x\in \mathbb {R}^2\) and radius \(\rho >0\). The set \(E^1\) is the measure-theoretic interior of E while \(E^0\) is the measure-theoretic exterior of E. The essential boundary of E is the set \(\partial ^\star E:=\mathbb {R}^2{\setminus }(E^0\cup E^1)\).
Recall that an integrable function u on \(\mathbb {R}^2\) is said to have bounded variation if the distributional derivative of u is representable by a finite Radon measure Du (cf. [1] Definition 3.1 for example) with total variation |Du|; in this case, we write \(u\in \mathrm {BV}(\mathbb {R}^2)\). The set E has finite perimeter if \(\chi _E\) belongs to \(\mathrm {BV}_{\mathrm {loc}}(\mathbb {R}^2)\). The reduced boundary \(\mathscr {F}E\) of E is defined by
$$\begin{aligned} \mathscr {F}E:= & {} \Big \{x\in \mathrm {supp}|D\chi _E|:\nu ^E(x):=\lim _{\rho \downarrow 0}\frac{D\chi _E(B(x,\rho ))}{|D\chi _E|(B(x,\rho ))}\\&\quad \text { exists in }\mathbb {R}^2\text { and }|\nu ^E(x)|=1\Big \} \end{aligned}$$
(cf. [1] Definition 3.54) and is a Borel set (cf. [1] Theorem 2.22 for example). We use \(\mathscr {H}^k\) (\(k\in [0,+\infty )\)) to stand for k-dimensional Hausdorff measure. If E is a set of finite perimeter in \(\mathbb {R}^2\) then
$$\begin{aligned} \mathscr {F}E\subset E^{1/2}\subset \partial ^* E \text { and } \mathscr {H}^1(\partial ^* E{\setminus }\mathscr {F}E)=0 \end{aligned}$$
(2.1)
by [1] Theorem 3.61.
Let f be a positive locally Lipschitz density on \(\mathbb {R}^2\). Let E be a set of finite perimeter and U a bounded open set in \(\mathbb {R}^2\). The weighted perimeter of E relative to U is defined by
$$\begin{aligned} P_f(E,U):=\sup \Big \{\int _U\mathrm {div}(fX)\,dx:X\in C^\infty _c(U,\mathbb {R}^2),\Vert X\Vert _\infty \le 1\Big \}. \end{aligned}$$
By the Gauss–Green formula ( [1] Theorem 3.36 for example) and a convolution argument,
$$\begin{aligned} P_f(E,U)&=\sup \Big \{\int _{\mathbb {R}^2}f\langle \nu ^E,X\rangle \,d|D\chi _E|: X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2),\nonumber \\&\qquad \qquad \mathrm {supp}[X]\subset U,\Vert X\Vert _\infty \le 1\Big \}\nonumber \\&=\sup \Big \{\int _{\mathbb {R}^2}f\langle \nu ^E,X\rangle \,d|D\chi _E|: X\in C_c(\mathbb {R}^2,\mathbb {R}^2),\nonumber \\&\qquad \qquad \mathrm {supp}[X]\subset U,\Vert X\Vert _\infty \le 1\Big \}\nonumber \\&=\int _U f\,d|D\chi _E| \end{aligned}$$
(2.2)
where we have also used [1] Propositions 1.47 and 1.23.

Lemma 2.1

Let \(\varphi \) be a \(C^1\) diffeomeorphism of \(\mathbb {R}^2\) which coincides with the identity map on the complement of a compact set and \(E\subset \mathbb {R}^2\) with \(\chi _E\in \mathrm {BV}(\mathbb {R}^2)\). Then
  1. (i)

    \(\chi _{\varphi (E)}\in \mathrm {BV}(\mathbb {R}^2)\);

     
  2. (ii)

    \(\partial ^\star \varphi (E)=\varphi (\partial ^\star E)\);

     
  3. (iii)

    \(\mathscr {H}^1(\mathscr {F}\varphi (E)\varDelta \varphi (\mathscr {F}E))=0\).

     

Proof

Part (i) follows from [1] Theorem 3.16 as \(\varphi \) is a proper Lipschitz function. Given \(x\in E^0\) we claim that \(y:=\varphi (x)\in \varphi (E)^0\). Let M stand for the Lipschitz constant of \(\varphi \) and L stand for the Lipschitz constant of \(\varphi ^{-1}\). Note that \( B(y,r)\subset \varphi (B(x,Lr)) \) for each \(r>0\). As \(\varphi \) is a bijection and using [1] Proposition 2.49,
$$\begin{aligned} |\varphi (E)\cap B(y,r)|&\le |\varphi (E)\cap \varphi (B(x,Lr)|\\&=|\varphi (E\cap B(x,Lr))|\le M^2|E\cap B(x,Lr)|.\nonumber \end{aligned}$$
This means that
$$\begin{aligned} \frac{|\varphi (E)\cap B(y,r)|}{|B(y,r)|} \le (LM)^2 \frac{|E\cap B(x,Lr)|}{|B(x,Lr)|} \end{aligned}$$
for \(r>0\) and this proves the claim. This entails that \(\varphi (E^0)\subset [\varphi (E)]^0\). The reverse inclusion can be seen using the fact that \(\varphi \) is a bijection. In summary \(\varphi (E^0)=[\varphi (E)]^0\). The corresponding identity for \(E^1\) can be seen in a similar way. These identities entail (ii). From (2.1) and (ii) we may write \(\mathscr {F}\varphi (E)\cup N_1=\varphi (\mathscr {F}E)\cup \varphi (N_2)\) for \(\mathscr {H}^1\)-null sets \(N_1,N_2\) in \(\mathbb {R}^2\). Item (iii) follows. \(\square \)
Curves with weakly bounded curvature. Suppose the open set E in \(\mathbb {R}^2\) has \(C^1\) boundary M. Denote by \(n:M\rightarrow \mathbb {S}^1\) the inner unit normal vector field. Given \(p\in M\) we choose a tangent vector \(t(p)\in \mathbb {S}^1\) in such a way that the pair \(\{t(p),n(p)\}\) forms a positively oriented basis for \(\mathbb {R}^2\). There exists a local parametrisation \(\gamma _1:I\rightarrow M\) where \(I=(-\delta ,\delta )\) for some \(\delta >0\) of class \(C^1\) with \(\gamma _1(0)=p\). We always assume that \(\gamma _1\) is parametrised by arc-length and that \(\dot{\gamma _1}(0)=t(p)\) where the dot signifies differentiation with respect to arc-length. Let X be a vector field defined in some neighbourhood of p in M. Then
$$\begin{aligned} (D_{t}X)(p):=\frac{d}{ds}\Big |_{s=0}(X\circ \gamma _1)(s) \end{aligned}$$
(2.3)
if this limit exists and the divergence \(\mathrm {div}^M X\) of X along M at p is defined by
$$\begin{aligned} \mathrm {div}^M X:=\langle D_{t}X,t\rangle \end{aligned}$$
(2.4)
evaluated at p. Suppose that X is a vector field in \(C^1(U,\mathbb {R}^2)\) where U is an open neighbourhood of p in \(\mathbb {R}^2\). Then
$$\begin{aligned} \mathrm {div}\,X=\mathrm {div}^M X+\langle D_nX,n\rangle \end{aligned}$$
(2.5)
at p. If \(p\in M{\setminus }\{0\}\) let \(\sigma (p)\) stand for the angle measured anti-clockwise from the position vector p to the tangent vector t(p); \(\sigma (p)\) is uniquely determined up to integer multiples of \(2\pi \).
Let E be an open set in \(\mathbb {R}^2\) with \(C^{1,1}\) boundary M. Let \(x\in M\) and \(\gamma _1:I\rightarrow M\) a local parametrisation of M in a neighbourhood of x. There exists a constant \(c>0\) such that
$$\begin{aligned} |\dot{\gamma }_1(s_2)-\dot{\gamma }_1(s_1)|\le c|s_2-s_1| \end{aligned}$$
for \(s_1,s_2\in I\); a constraint on average curvature (cf. [10, 18]). That is, \(\dot{\gamma }_1\) is Lipschitz on I. So \(\dot{\gamma }_1\) is absolutely continuous and differentiable a.e. on I with
$$\begin{aligned} \dot{\gamma }_1(s_2)-\dot{\gamma }_1(s_1)=\int _{s_1}^{s_2}\ddot{\gamma }_1\,ds \end{aligned}$$
(2.6)
for any \(s_1,s_2\in I\) with \(s_1<s_2\). Moreover, \(|\ddot{\gamma }_1|\le c\) a.e. on I (cf. [1] Corollary 2.23). As \(\langle \dot{\gamma }_1,\dot{\gamma }_1\rangle =1\) on I we see that \(\langle \dot{\gamma }_1,\ddot{\gamma }_1\rangle =0\) a.e. on I. The (geodesic) curvature \(k_1\) is then defined a.e. on I via the relation
$$\begin{aligned} \ddot{\gamma }_1=k_1n_1 \end{aligned}$$
(2.7)
as in [18]. The curvature k of M is defined \(\mathscr {H}^1\)-a.e. on M by
$$\begin{aligned} k(x):=k_1(s) \end{aligned}$$
(2.8)
whenever \(x=\gamma _1(s)\) for some \(s\in I\) and \(k_1(s)\) exists. We sometimes write \(H(\cdot ,E)=k\).
Let E be an open set in \(\mathbb {R}^2\) with \(C^{1}\) boundary M. Let \(x\in M\) and \(\gamma _1:I\rightarrow M\) a local parametrisation of M in a neighbourhood of x. In case \(\gamma _1\ne 0\) let \(\theta _1\) stand for the angle measured anti-clockwise from \(e_1\) to the position vector \(\gamma _1\) and \(\sigma _1\) stand for the angle measured anti-clockwise from the position vector \(\gamma _1\) to the tangent vector \(t_1=\dot{\gamma }_1\). Put \(r_1:=|\gamma _1|\) on I. Then \(r_1,\theta _1\in C^1(I)\) and
$$\begin{aligned} \dot{r}_1&=\cos \sigma _1; \end{aligned}$$
(2.9)
$$\begin{aligned} r_1\dot{\theta }_1&=\sin \sigma _1; \end{aligned}$$
(2.10)
on I provided that \(\gamma _1\ne 0\). Now suppose that M is of class \(C^{1,1}\). Let \(\alpha _1\) stand for the angle measured anti-clockwise from the fixed vector \(e_1\) to the tangent vector \(t_1\) (uniquely determined up to integer multiples of \(2\pi \)). Then \(t_1=(\cos \alpha _1,\sin \alpha _1)\) on I so \(\alpha _1\) is absolutely continuous on I. In particular, \(\alpha _1\) is differentiable a.e. on I with \(\dot{\alpha }_1=k_1\) a.e. on I. This means that \(\alpha _1\in C^{0,1}(I)\). In virtue of the identities \(r_1\cos \sigma _1=\langle \gamma _1,t_1\rangle \) and \(r_1\sin \sigma _1=-\langle \gamma _1,n_1\rangle \) we see that \(\sigma _1\) is absolutely continuous on I and \(\sigma _1\in C^{0,1}(I)\). By choosing an appropriate branch we may assume that
$$\begin{aligned} \alpha _1&=\theta _1+\sigma _1 \end{aligned}$$
(2.11)
on I. We may choose \(\sigma \) in such a way that \(\sigma \circ \gamma _1=\sigma _1\) on I.
Flows. Recall that a diffeomorphism \(\varphi :\mathbb {R}^2\rightarrow \mathbb {R}^2\) is said to be proper if \(\varphi ^{-1}(K)\) is compact whenever \(K\subset \mathbb {R}^2\) is compact. Given \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) there exists a 1-parameter group of proper \(C^\infty \) diffeomorphisms \(\varphi :\mathbb {R}\times \mathbb {R}^2\rightarrow \mathbb {R}^2\) as in [20] Lemma 2.99 that satisfy
$$\begin{aligned} \left. \begin{array}{rl} \partial _t\varphi (t,x)&{}= X(\varphi (t,x))\text { for each }(t,x)\in \mathbb {R}\times \mathbb {R}^2;\\ \varphi (0,x) &{}= x \text { for each } x\in \mathbb {R}^2.\\ \end{array} \right. \end{aligned}$$
(2.12)
We often use \(\varphi _t\) to refer to the diffeomorphism \(\varphi (t,\cdot ):\mathbb {R}^2\rightarrow \mathbb {R}^2\).

Lemma 2.2

Let \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) and \(\varphi \) be the corresponding flow as above. Then
  1. (i)
    there exists \(R\in C^\infty (\mathbb {R}\times \mathbb {R}^2,\mathbb {R}^2)\) and \(K>0\) such that
    $$\begin{aligned} \varphi (t,x)= \left\{ \begin{array}{ll} x+tX(x)+R(t,x) &{} \text { for }x\in \mathrm {supp}[X];\\ x &{} \text { for }x\not \in \mathrm {supp}[X];\\ \end{array} \right. \end{aligned}$$
    where \(|R(t,x)|\le K t^2\) for \((t,x)\in \mathbb {R}\times \mathbb {R}^2\);
     
  2. (ii)
    there exists \(R^{(1)}\in C^\infty (\mathbb {R}\times \mathbb {R}^2,M_2(\mathbb {R}))\) and \(K_1>0\) such that
    $$\begin{aligned} d\varphi (t,x)= \left\{ \begin{array}{ll} I+tdX(x)+R^{(1)}(t,x) &{} \text { for }x\in \mathrm {supp}[X];\\ I &{} \text { for }x\not \in \mathrm {supp}[X];\\ \end{array} \right. \end{aligned}$$
    where \(|R^{(1)}(t,x)|\le K_1 t^2\) for \((t,x)\in \mathbb {R}\times \mathbb {R}^2\);
     
  3. (iii)
    there exists \(R^{(2)}\in C^\infty (\mathbb {R}\times \mathbb {R}^2,\mathbb {R})\) and \(K_2>0\) such that
    $$\begin{aligned} J_2d\varphi (t,x)= \left\{ \begin{array}{ll} 1+t\,\mathrm {div}\,X(x)+R^{(2)}(t,x) &{} \text { for }x\in \mathrm {supp}[X];\\ 1 &{} \text { for }x\not \in \mathrm {supp}[X];\\ \end{array} \right. \end{aligned}$$
    where \(|R^{(2)}(t,x)|\le K_2 t^2\) for \((t,x)\in \mathbb {R}\times \mathbb {R}^2\).
     
Let \(x\in \mathbb {R}^2,v\) a unit vector in \(\mathbb {R}^2\) and M the line though x perpendicular to v. Then
  1. (iv)
    there exists \(R^{(3)}\in C^\infty (\mathbb {R}\times \mathbb {R}^2,\mathbb {R})\) and \(K_3>0\) such that
    $$\begin{aligned} J_1d^{M}\varphi (t,x)= \left\{ \begin{array}{ll} 1+t(\mathrm {div}^{M}\,X)(x)+R^{(3)}(t,x) &{} \text { for }x\in \mathrm {supp}[X];\\ 1 &{} \text { for }x\not \in \mathrm {supp}[X];\\ \end{array} \right. \end{aligned}$$
    where \(|R^{(3)}(t,x)|\le K_3 t^2\) for \((t,x)\in \mathbb {R}\times \mathbb {R}^2\).
     

Proof

(i) First notice that \(\varphi \in C^\infty (\mathbb {R}\times \mathbb {R}^2)\) by [16] Theorem 3.3 and Exercise 3.4. The statement for \(x\not \in \mathrm {supp}[X]\) follows by uniqueness (cf. [16] Theorem 3.1); the assertion for \(x\in \mathrm {supp}[X]\) follows from Taylor’s theorem. (ii) follows likewise: note, for example, that
$$\begin{aligned}{}[\partial _{tt}d\varphi ]_{\alpha \beta }\vert _{t=0}=X^\alpha _{,\beta \delta }X^\delta +X^\alpha _{,\gamma }X^\gamma _{,\beta } \end{aligned}$$
where the subscript \(_,\) signifies partial differentiation. (iii) follows from (ii) and the definition of the 2-dimensional Jacobian (cf. [1] Definition 2.68). (iv) Using [1] Definition 2.68 together with the Cauchy–Binet formula [1] Proposition 2.69, \(J_1d^M\varphi (t,x)=|d\varphi (t,x)v|\) for \(t\in \mathbb {R}\) and the result follows from (ii). \(\square \)
Let I be an open interval in \(\mathbb {R}\) containing 0. Let \(Z:I\times \mathbb {R}^2\rightarrow \mathbb {R}^2;(t,x)\mapsto Z(t,x)\) be a continuous time-dependent vector field on \(\mathbb {R}^2\) with the properties
  1. (Z.1)

    \(Z(t,\cdot )\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\) for each \(t\in I\);

     
  2. (Z.2)

    \(\mathrm {supp}[Z(t,\cdot )]\subset K\) for each \(t\in I\) for some compact set \(K\subset \mathbb {R}^2\).

     
By [16] Theorems I.1.1, I.2.1, I.3.1, I.3.3 there exists a unique flow \(\varphi :I\times \mathbb {R}^2\rightarrow \mathbb {R}^2\) such that
  1. (F.1)

    \(\varphi :I\times \mathbb {R}^2\rightarrow \mathbb {R}^2\) is of class \(C^1\);

     
  2. (F.2)

    \(\varphi (0,x)=x\) for each \(x\in \mathbb {R}^2\);

     
  3. (F.3)

    \(\partial _t\varphi (t,x)=Z(t,\varphi (x,t))\) for each \((t,x)\in I\times \mathbb {R}^2\);

     
  4. (F.4)

    \(\varphi _t:=\varphi (t,\cdot ):\mathbb {R}^2\rightarrow \mathbb {R}^2\) is a proper diffeomorphism for each \(t\in I\).

     

Lemma 2.3

Let Z be a time-dependent vector field with the properties (Z.1)–(Z.2) and \(\varphi \) be the corresponding flow. Then
  1. (i)
    for \((t,x)\in I\times \mathbb {R}^2\),
    $$\begin{aligned} d\varphi (t,x)= \left\{ \begin{array}{ll} I+tdZ_0(x)+tR(t,x) &{} \text { for }x\in K;\\ I &{} \text { for }x\not \in K;\\ \end{array} \right. \end{aligned}$$
    where \(\sup _K|R(t,\cdot )|\rightarrow 0\) as \(t\rightarrow 0\).
     
Let \(x\in \mathbb {R}^2,v\) a unit vector in \(\mathbb {R}^2\) and M the line though x perpendicular to v. Then
  1. (ii)
    for \((t,x)\in I\times \mathbb {R}^2\),
    $$\begin{aligned} J_1d^M\varphi (t,x)= \left\{ \begin{array}{ll} 1+t(\mathrm {div}^{M}\,Z_0)(x)+tR^{(1)}(t,x) &{} \text { for }x\in K;\\ 1 &{} \text { for }x\not \in K.\\ \end{array} \right. \end{aligned}$$
    where \(\sup _K|R^{(1)}(t,\cdot )|\rightarrow 0\) as \(t\rightarrow 0\).
     

Proof

(i) We first remark that the flow \(\varphi :I\times \mathbb {R}^2\rightarrow \mathbb {R}^2\) associated to Z is continuously differentiable in tx in virtue of (Z.1) by [16] Theorem I.3.3. Put \(y(t,x):=d\varphi (t,x)\) for \((t,x)\in I\times \mathbb {R}^2\). By [16] Theorem I.3.3,
$$\begin{aligned} \dot{y}(t,x)=dZ(t,\varphi (t,x))y(t,x) \end{aligned}$$
for each \((t,x)\in I\times \mathbb {R}^2\) and \(y(0,x)=I\) for each \(x\in \mathbb {R}^2\) where I stands for the \(2\times 2\)-identity matrix. For \(x\in \ K\) and \(t\in I\),
$$\begin{aligned} d\varphi (t,x)&=I+d\varphi (t,x)-d\varphi (0,x)\nonumber \\&=I+t\dot{y}(0,x)+t\Big \{\frac{d\varphi (t,x)-d\varphi (0,x)}{t}-\dot{y}(0,x)\Big \}\nonumber \\&=I+t dZ(0,x)+t\Big \{\frac{y(t,x)-y(0,x)}{t}-\dot{y}(0,x)\Big \}\nonumber \\&=I+t dZ_0(x)+t\Big \{\frac{y(t,x)-y(0,x)}{t}-\dot{y}(0,x)\Big \}.\nonumber \end{aligned}$$
Applying the mean-value theorem component-wise and using uniform continuity of the matrix \(\dot{y}\) in its arguments we see that
$$\begin{aligned} \frac{y(t,\cdot )-y(0,\cdot )}{t}-\dot{y}(0,\cdot )\rightarrow 0 \end{aligned}$$
uniformly on K as \(t\rightarrow 0\). This leads to (i). Part (ii) follows as in Lemma 2.2. \(\square \)
Let E be a set of finite perimeter in \(\mathbb {R}^2\) with \(V_f(E)<+\infty \). The first variation of weighted volume resp. perimeter along \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) is defined by
$$\begin{aligned} \delta V_f(X)&:=\frac{d}{dt}\Big \vert _{t=0}V_f(\varphi _t(E)),\end{aligned}$$
(2.13)
$$\begin{aligned} \delta P_f^+(X)&:=\lim _{t\downarrow 0}\frac{P_f(\varphi _t(E))-P_f(E)}{t}, \end{aligned}$$
(2.14)
whenever the limit exists. By Lemma 2.1 the f-perimeter in (2.14) is well-defined.
Convex functions. Suppose that \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a convex function. For \(x\ge 0\) and \(v\ge 0\) define
$$\begin{aligned} h^\prime _+(x,v):=\lim _{t\downarrow 0}\frac{h(x+tv)-h(x)}{t}\in \mathbb {R} \end{aligned}$$
and define \(h^\prime _-(x,v)\) similarly for \(x>0\) and \(v\le 0\). For future use we introduce the notation
$$\begin{aligned} \rho _+:=h^\prime (\cdot ,+1), \rho _-:=-h^\prime (\cdot ,-1) \text { and } \rho :=(1/2)(\rho _++\rho _-) \end{aligned}$$
on \((0,+\infty )\). It holds that h is differentiable a.e. and \(h^\prime =\rho \) a.e. on \((0,+\infty )\). Define \([\rho ]:=\rho _+-\rho _-\). Then \([\rho ]\ge 0\) and vanishes a.e. on \((0,+\infty )\).

Lemma 2.4

Suppose that the function f takes the form (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a convex function. Then
  1. (i)

    the directional derivative \(f^\prime _+(x,v)\) exists in \(\mathbb {R}\) for each \(x\in \mathbb {R}^2\) and \(v\in \mathbb {R}^2\);

     
  2. (ii)
    for \(v\in \mathbb {R}^2\),
    $$\begin{aligned} f^\prime _+(x,v)= \left\{ \begin{array}{ll} f(x)h^\prime _+(|x|,\mathrm {sgn}\langle x,v\rangle )\frac{|\langle x,v\rangle |}{|x|} &{} \text { for }x\in \mathbb {R}^2{\setminus }\{0\};\\ f(0)h^\prime _+(0,+1)|v| &{} \text { for }x=0;\\ \end{array} \right. \end{aligned}$$
     
  3. (iii)
    if M is a \(C^1\) hypersurface in \(\mathbb {R}^2\) such that \(\cos \sigma \ne 0\) on M then f is differentiable \(\mathscr {H}^1\)-a.e. on M and
    $$\begin{aligned} (\nabla f)(x)=f(x)\rho (|x|)\frac{\langle x,\cdot \rangle }{|x|} \end{aligned}$$
    for \(\mathscr {H}^1\)-a.e. \(x\in M\).
     

Proof

The assertion in (i) follows from the monotonicity of chords property while (ii) is straightforward. (iii) Let \(x\in M\) and \(\gamma _1:I\rightarrow M\) be a \(C^1\)-parametrisation of M near x as above. Now \(r_1\in C^1(I)\) and \(\dot{r_1}(0)=\cos \sigma (x)\ne 0\) so we may assume that \(r_1:I\rightarrow r_1(I)\subset (0,+\infty )\) is a \(C^1\) diffeomorphism. The differentiability set D(h) of h has full Lebesgue measure in \([0,+\infty )\). It follows by [1] Proposition 2.49 that \(r_1^{-1}(D(h))\) has full measure in I. This entails that f is differentiable \(\mathscr {H}^1\)-a.e. on \(\gamma _1(I)\subset M\). \(\square \)

3 Existence and \(C^1\) regularity

We start with an existence theorem.

Theorem 3.1

Assume that f is a positive radial lower-semicontinuous non-decreasing density on \(\mathbb {R}^2\) which diverges to infinity. Then for each \(v>0\),
  1. (i)

    (1.2) admits a minimiser;

     
  2. (ii)

    any minimiser of (1.2) is essentially bounded.

     

Proof

See [22] Theorems 3.3 and 5.9. \(\square \)

But the bulk of this section will be devoted to a discussion of \(C^1\) regularity.

Proposition 3.2

Let f be a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(E\subset \mathbb {R}^2\) be a bounded set with finite perimeter. Let \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\). Then
$$\begin{aligned} \delta V_f(X) =\int _{E}\mathrm {div}(fX)\,dx =-\int _{\mathscr {F}E} f\,\langle \nu ^E, X\rangle \,d\mathscr {H}^1. \end{aligned}$$

Proof

Let \(t\in \mathbb {R}\). By the area formula ( [1] Theorem 2.71 and (2.74)),
$$\begin{aligned} V_f(\varphi _t(E)) = \int _{\varphi _t(E)}f\,dx = \int _{E}(f\circ \varphi _t)\,J_2 d(\varphi _t)_x\,dx \end{aligned}$$
(3.1)
and
$$\begin{aligned} V_f(\varphi _t(E))-V_f(E)&=\int _E(f\circ \varphi _t)J_2 d\varphi _t - f\,dx\\&=\int _E(f\circ \varphi _t)(J_2 d\varphi _t-1)\,dx+\int _Ef\circ \varphi _t-f\,dx. \end{aligned}$$
The density f is locally Lipschitz and in particular differentiable a.e. on \(\mathbb {R}^2\) (see [1] 2.3 for example). By the dominated convergence theorem and Lemma 2.2,
$$\begin{aligned} \delta V_f(X)&=\int _{E}\Big \{f\mathrm {div}(X)+\langle \nabla f,X\rangle \Big \}\,dx =\int _{E}\mathrm {div}(fX)\,dx\\&=-\int _{\mathscr {F}E} f\,\langle \nu ^E,X\rangle \,d\mathscr {H}^1 \end{aligned}$$
by the generalised Gauss–Green formula [1] Theorem 3.36. \(\square \)

Proposition 3.3

Let f be a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(E\subset \mathbb {R}^2\) be a bounded set with finite perimeter. Let \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\). Then there exist constants \(C>0\) and \(\delta >0\) such that
$$\begin{aligned} |P_f(\varphi _t(E))-P_f(E)|\le C|t| \end{aligned}$$
for \(|t|<\delta \).

Proof

Let \(t\in \mathbb {R}\). By Lemma 2.1 and [1] Theorem 3.59,
$$\begin{aligned} P_f(\varphi _t(E)) =\int _{\mathbb {R}^2}f\,d|D\chi _{\varphi _t(E)}| =\int _{\mathscr {F}\varphi _t(E)}f\,d\mathscr {H}^1 =\int _{\varphi _t(\mathscr {F}E)}f\,d\mathscr {H}^1. \end{aligned}$$
As \(\mathscr {F}E\) is countably 1-rectifiable ( [1] Theorem 3.59) we may use the generalised area formula [1] Theorem 2.91 to write
$$\begin{aligned} P_f(\varphi _t(E))=\int _{\mathscr {F}E}(f\circ \varphi _t)J_1d^{\mathscr {F}E}(\varphi _t)_x\,d\mathscr {H}^1. \end{aligned}$$
For each \(x\in \mathscr {F}E\) and any \(t\in \mathbb {R}\),
$$\begin{aligned} |(f\circ \varphi _t)(x)-f(x)|\le K|\varphi (t,x)-x|\le K\Vert X\Vert _\infty |t| \end{aligned}$$
where K is the Lipschitz constant of f on \(\mathrm {supp}[X]\). The result follows upon writing
$$\begin{aligned} P_f(\varphi _t(E))-P_f(E)&=\int _{\mathscr {F}E}(f\circ \varphi _t)(J_1d^{\mathscr {F}E}(\varphi _t)_x-1)\nonumber \\&\quad +\,[f\circ \varphi _t-f] \,d\mathscr {H}^1 \end{aligned}$$
(3.2)
and using Lemma 2.2. \(\square \)

Lemma 3.4

Let f be a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(E\subset \mathbb {R}^2\) be a bounded set with finite perimeter and \(p\in \mathscr {F}E\). For any \(r>0\) there exists \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X]\subset B(p,r)\) such that \(\delta V_f(X)=1\).

Proof

By (2.2) and [1] Theorem 3.59 and (3.57) in particular,
$$\begin{aligned} P_f(E,B(p,r))=\int _{B(p,r)\cap \mathscr {F}E}f\,d\mathscr {H}^1>0 \end{aligned}$$
for any \(r>0\). By the variational characterisation of the f-perimeter relative to B(pr) we can find \(Y\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[Y]\subset B(p,r)\) such that
$$\begin{aligned} 0<\int _{E\cap B(p,r)}\mathrm {div}(fY)\,dx=-\int _{\mathscr {F}E\cap B(p,r)}f\langle \nu ^E,Y\rangle \,d\mathscr {H}^1=:c \end{aligned}$$
where we make use of the generalised Gauss–Green formula (cf. [1] Theorem 3.36). Put \(X:=(1/c)Y\). Then \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X]\subset B(p,r)\) and \(\delta V_f(X)=1\) according to Proposition 3.2. \(\square \)

Proposition 3.5

Let f be a positive lower semi-continuous density on \(\mathbb {R}^2\). Let U be a bounded open set in \(\mathbb {R}^2\) with Lipschitz boundary. Let \(E, F_1, F_2\) be bounded sets in \(\mathbb {R}^2\) with finite perimeter. Assume that \(E\varDelta F_1\subset \subset U\) and \(E\varDelta F_2\subset \subset \mathbb {R}^2{\setminus }\overline{U}\). Define
$$\begin{aligned} F:=\Big [F_1\cap U\Big ]\cup \Big [F_2{\setminus } U\Big ]. \end{aligned}$$
Then F is a set of finite perimeter in \(\mathbb {R}^2\) and
$$\begin{aligned} P_f(E)+P_f(F)= P_f(F_1)+P_f(F_2). \end{aligned}$$

Proof

The function \(\chi _E\vert _U\in \mathrm {BV}(U)\) and \(D(\chi _E\vert _U)=(D\chi _E)\vert _U\). We write \(\chi _E^U\) for the boundary trace of \(\chi _E\vert _U\) (see [1] Theorem 3.87); then \(\chi _E^U\in L^1(\partial U,\mathscr {H}^1|\!\_\partial U)\) (cf. [1] Theorem 3.88). We use similar notation elsewhere. By [1] Corollary 3.89,
$$\begin{aligned} D\chi _E&=D\chi _E|\!\_U+(\chi _E^U-\chi _E^{\mathbb {R}^2{\setminus }\overline{U}})\nu ^U\mathscr {H}^1|\!\_{\partial U}+D\chi _E|\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ D\chi _F&=D\chi _{F_1}|\!\_U+(\chi _{F_1}^U-\chi _{F_2}^{\mathbb {R}^2{\setminus }\overline{U}})\nu ^U\mathscr {H}^1|\!\_{\partial U}+D\chi _{F_2}|\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ D\chi _{F_1}&=D\chi _{F_1}|\!\_U+(\chi _{F_1}^U-\chi _{E}^{\mathbb {R}^2{\setminus }\overline{U}})\nu ^U\mathscr {H}^1|\!\_{\partial U}+D\chi _{E}|\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ D\chi _{F_2}&=D\chi _{E}|\!\_U+(\chi _{E}^U-\chi _{F_2}^{\mathbb {R}^2{\setminus }\overline{U}})\nu ^U\mathscr {H}^1|\!\_{\partial U}+D\chi _{F_2}|\!\_(\mathbb {R}^2{\setminus }\overline{U}). \end{aligned}$$
From the definition of the total variation measure ([1] Definition 1.4),
$$\begin{aligned} |D\chi _E|&=|D\chi _E||\!\_U+|\chi _E^U-\chi _E^{\mathbb {R}^2{\setminus }\overline{U}}|\mathscr {H}^1|\!\_{\partial U}+|D\chi _E||\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ |D\chi _F|&=|D\chi _{F_1}||\!\_U+|\chi _E^U-\chi _E^{\mathbb {R}^2{\setminus }\overline{U}}|\mathscr {H}^1|\!\_{\partial U}+|D\chi _{F_2}||\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ |D\chi _{F_1}|&=|D\chi _{F_1}||\!\_U+|\chi _E^U-\chi _{E}^{\mathbb {R}^2{\setminus }\overline{U}}|\mathscr {H}^1|\!\_{\partial U}+|D\chi _{E}||\!\_(\mathbb {R}^2{\setminus }\overline{U});\nonumber \\ |D\chi _{F_2}|&=|D\chi _{E}||\!\_U+|\chi _{E}^U-\chi _E^{\mathbb {R}^2{\setminus }\overline{U}}|\mathscr {H}^1|\!\_{\partial U}+|D\chi _{F_2}||\!\_(\mathbb {R}^2{\setminus }\overline{U}); \end{aligned}$$
where we also use the fact that \(\chi _{F_1}^U=\chi _E^U\) as \(E\varDelta F_1\subset \subset U\) and similarly for \(F_2\). The result now follows. \(\square \)

Proposition 3.6

Assume that f is a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(v>0\) and suppose that the set E is a bounded minimiser of (1.2). Let U be a bounded open set in \(\mathbb {R}^2\). There exist constants \(C>0\) and \(\delta >0\) with the following property. For any \(x\in U\) and \(0<r<\delta \),
$$\begin{aligned} P_f(E)-P_f(F)\le C\big |V_f(E)-V_f(F)\big | \end{aligned}$$
(3.3)
where F is any set with finite perimeter in \(\mathbb {R}^2\) such that \(E\varDelta F\subset \subset B(x,r)\).

Proof

The proof follows that of [21] Proposition 3.1. We assume to the contrary that
$$\begin{aligned} (\forall \,C>0)(\forall \,\delta >0)(\exists \,x\in U)(\exists \,r\in (0,\delta ))(\exists \,F\subset \mathbb {R}^2) \end{aligned}$$
$$\begin{aligned} \Big [ F\varDelta E\subset \subset B(x,r) \wedge \varDelta P_f > C|\varDelta V_f| \Big ] \end{aligned}$$
(3.4)
in the language of quantifiers where we have taken some liberties with notation.
Choose \(p_1,p_2\in \mathscr {F}E\) with \(p_1\ne p_2\). Choose \(r_0>0\) such that the open balls \(B(p_1,r_0)\) and \(B(p_2,r_0)\) are disjoint. Choose vector fields \(X_j\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X_j]\subset B(p_j,r_0)\) such that
$$\begin{aligned} \delta V_f(X_j)=1 \text { and } |P_f(\varphi ^{(j)}_t(E))-P_f(E)|\le a_j|t|\text { for }|t|<\delta _j\text { and }j=1,2\nonumber \\ \end{aligned}$$
(3.5)
as in Lemma 3.4 and Proposition 3.3. Put \(a:=\max \{a_1,a_2\}\). By (3.5),
$$\begin{aligned} V_f(\varphi ^{(j)}_t(E))-V_f(E)=t + o(t)\text { as }t\rightarrow 0\text { for }j=1,2. \end{aligned}$$
So there exist \(\varepsilon >0\) and \(1>\eta >0\) such that
$$\begin{aligned}&t-\eta |t|< V_f(\varphi ^{(j)}_t(E)) - V_f(E)< t + \eta |t|;\nonumber \\&|P_f(\varphi _t^{(j)}(E)) - P_f(E) |<(a+1)|t|; \end{aligned}$$
(3.6)
for \(|t|<\varepsilon \) and \(j=1,2\). In particular,
$$\begin{aligned}&|V_f(\varphi ^{(j)}_t(E)) - V_f(E)|>(1-\eta )|t|;\nonumber \\&|P_f(\varphi _t^{(j)}(E)) - P_f(E)|<\frac{1+a}{1-\eta }|V_f(\varphi ^{(j)}_t(E)) - V_f(E)| \text { for }|t|<\varepsilon ; \end{aligned}$$
(3.7)
for \(|t|<\varepsilon \) and \(j=1,2\).
In (3.4) choose \(C=(1+a)/(1-\eta )\) and \(\delta >0\) such that
  1. (a)

    \(0<2\delta <\mathrm {dist}(B(p_1,r_0),B(p_2,r_0))\),

     
  2. (b)

    \(\sup \{V_f(B(x,\delta )):\,x\in U\}<(1-\eta )\,\varepsilon \).

     
Choose xr and \(F_1\) as in (3.4). In light of (a) we may assume that \(B(x,r)\cap B(p_1,r_0)=\emptyset \). By (b),
$$\begin{aligned} |V_f(F_1) - V_f(E)|\le V_f(B(x,r))\le V_f(B(x,\delta ))<(1-\eta )\,\varepsilon . \end{aligned}$$
(3.8)
From (3.6) and (3.8) we can find \(t\in (-\varepsilon ,\varepsilon )\) such that with \(F_2:=\varphi _t^{(1)}(E)\),
$$\begin{aligned} V_f(F_2) - V_f(E)=-\Big \{V_f(F_1) - V_f(E)\Big \} \end{aligned}$$
(3.9)
by the intermediate value theorem. From (3.4),
$$\begin{aligned} P_f(F_1) < P_f(E)-C|V_f(F_1) - V_f(E)| \end{aligned}$$
(3.10)
while from (3.7),
$$\begin{aligned} P_f(F_2) < P_f(E) + C|V_f(F_2) - V_f(E)|. \end{aligned}$$
(3.11)
Let F be the set
$$\begin{aligned} F:=\Big [F_1{\setminus } B(p_1,r_0))\Big ]\cup \Big [B(p_1,r_0)\cap F_2\Big ]. \end{aligned}$$
Note that \(E\varDelta F_2\subset \subset B(p_1,r_0)\). By Proposition 3.5, F is a bounded set of finite perimeter in \(\mathbb {R}^2\) and
$$\begin{aligned} P_f(E)+P_f(F)= P_f(F_1)+P_f(F_2). \end{aligned}$$
We then infer from (3.10), (3.11) and (3.9) that
$$\begin{aligned} P_f(F)&= P_f(F_1)+P_f(F_2)-P_f(E)\nonumber \\&<P_f(E)-C|V_f(F_1) - V_f(E)|+ P_f(E) \nonumber \\&\quad +\, C|V_f(F_2) - V_f(E)|-P_f(E)\ =P_f(E). \end{aligned}$$
On the other hand, \(V_f(F)=V_f(F_1)+V_f(F_2)-V_f(E)=V_f(E)\) by (3.9). We therefore obtain a contradiction to the f-isoperimetric property of E. \(\square \)
Let E be a set of finite perimeter in \(\mathbb {R}^2\) and U a bounded open set in \(\mathbb {R}^2\). The minimality excess is the function \(\psi \) defined by
$$\begin{aligned} \psi (E,U)&:=P(E,U)-\nu (E,U) \end{aligned}$$
(3.12)
where
$$\begin{aligned} \nu (E,U) :=\inf \{P(F,U):F\text { is a set of finite perimeter with }F\varDelta E\subset \subset U\} \end{aligned}$$
as in [27] (1.9). We recall that the boundary of E is said to be almost minimal in \(\mathbb {R}^2\) if for each bounded open set U in \(\mathbb {R}^2\) there exists \(T>0\) and a positive constant K such that for every \(x\in U\) and \(r\in (0,T)\),
$$\begin{aligned} \psi (E,B(x,r))\le K r^2. \end{aligned}$$
(3.13)
This definition corresponds to [27] Definition 1.5.

Theorem 3.7

Assume that f is a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(v>0\) and assume that E is a bounded minimiser of (1.2). Then the boundary of E is almost minimal in \(\mathbb {R}^2\).

Proof

Let U be a bounded open set in \(\mathbb {R}^2\) and \(C>0\) and \(\delta >0\) as in Proposition 3.6. The open \(\delta \)-neighbourhood of U is denoted \(I_\delta (U)\). Let \(x\in U\) and \(r\in (0,\delta )\). Put \(V:=I_{2\delta }(U)\). For the sake of brevity write \(m:=\inf _{B(x,r)}f\) and \(M:=\sup _{B(x,r)}f\). Let F be a set of finite perimeter in \(\mathbb {R}^2\) such that \(F\varDelta E\subset \subset B(x,r)\). By Proposition 3.6,
$$\begin{aligned}&P(E,B(x,r))-P(F,B(x,r))\\&\quad \le \frac{1}{m}P_f(E,B(x,r))-\frac{1}{M}P_f(F,B(x,r))\nonumber \\&\quad =\frac{1}{m} \Big ( P_f(E,B(x,r))-P_f(F,B(x,r)) \Big )+ \Big (\frac{1}{m}-\frac{1}{M} \Big ) P_f(F,B(x,r))\nonumber \\&\quad \le \frac{1}{m} \Big ( P_f(E,B(x,r))-P_f(F,B(x,r)) \Big )+ \frac{M-m}{m^2} P_f(F,B(x,r))\nonumber \\&\quad \le \frac{C}{\inf _V f}|V_f(E)-V_f(F)| + (2Lr)\frac{\sup _V f}{(\inf _V f)^2}P(F,B(x,r))\nonumber \\&\quad \le C\pi r^2\frac{\sup _V f}{\inf _V f} + (2Lr)\frac{\sup _V f}{(\inf _V f)^2}P(F,B(x,r))\nonumber \end{aligned}$$
where L stands for the Lipschitz constant of the restriction of f to V. We then derive that
$$\begin{aligned} \psi (E,B(x,r))\le C\pi r^2\frac{\sup _V f}{\inf _V f} + (2Lr)\frac{\sup _V f}{(\inf _V f)^2} \nu (E,B(x,r)). \end{aligned}$$
By [13] (5.14), \(\nu (E,B(x,r))\le \pi r\). The inequality in (3.13) now follows. \(\square \)

Theorem 3.8

Assume that f is a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(v>0\) and suppose that E is a bounded minimiser of (1.2). Then there exists a set \(\widetilde{E}\subset \mathbb {R}^2\) such that
  1. (i)

    \(\widetilde{E}\) is a bounded minimiser of (1.2);

     
  2. (ii)

    \(\widetilde{E}\) is equivalent to E;

     
  3. (iii)

    \(\widetilde{E}\) is open and \(\partial \widetilde{E}\) is a \(C^1\) hypersurface in \(\mathbb {R}^2\).

     

Proof

By [13] Proposition 3.1 there exists a Borel set F equivalent to E with the property that
$$\begin{aligned} \partial F=\{x\in \mathbb {R}^2:0<|F\cap B(x,\rho )|<\pi \rho ^2\text { for each }\rho >0\}. \end{aligned}$$
By Theorem 3.7 and [27] Theorem 1.9, \(\partial F\) is a \(C^1\) hypersurface in \(\mathbb {R}^2\) (taking note of differences in notation). The set
$$\begin{aligned} \widetilde{E}:=\{x\in \mathbb {R}^2:|F\cap B(x,\rho )|=\pi \rho ^2\text { for some }\rho >0\} \end{aligned}$$
satisfies (i)(iii). \(\square \)

4 Weakly bounded curvature and spherical cap symmetry

Theorem 4.1

Assume that f is a positive locally Lipschitz density on \(\mathbb {R}^2\). Let \(v>0\) and suppose that E is a bounded minimiser of (1.2). Then there exists a set \(\widetilde{E}\subset \mathbb {R}^2\) such that
  1. (i)

    \(\widetilde{E}\) is a bounded minimiser of (1.2);

     
  2. (ii)

    \(\widetilde{E}\) is equivalent to E;

     
  3. (iii)

    \(\widetilde{E}\) is open and \(\partial \widetilde{E}\) is a \(C^{1,1}\) hypersurface in \(\mathbb {R}^2\).

     

Proof

We may assume that E has the properties listed in Theorem 3.8. Put \(M:=\partial E\). Let \(x\in M\) and U a bounded open set containing x. Choose \(C>0\) and \(\delta >0\) as in Proposition 3.6. Let \(0<r<\delta \) and \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X]\subset B(x,r)\). Then
$$\begin{aligned} P_f(E)-P_f(\varphi _t(E))\le C|V_f(E)-V_f(\varphi _t(E))| \end{aligned}$$
for each \(t\in \mathbb {R}\). From the identity (3.2),
$$\begin{aligned}&-\int _M(f\circ \varphi _t)(J_1d^M(\varphi _t)_x-1)\,d\mathscr {H}^1\le C|V_f(E)-V_f(\varphi _t(E))| \\&\qquad +\,\int _M[f\circ \varphi _t-f]\,d\mathscr {H}^1 \nonumber \\&\quad \le \, C|V_f(E)-V_f(\varphi _t(E))| +\sqrt{2}K\Vert X\Vert _\infty \mathscr {H}^1(M\cap \mathrm {supp}[X])t \end{aligned}$$
where K stands for the Lipschitz constant of f restricted to U. On dividing by t and taking the limit \(t\rightarrow 0\) we obtain
$$\begin{aligned} -\int _M f\mathrm {div}^M X\,d\mathscr {H}^1&\le C\Big |\int _M f\langle n,X\rangle \,d\mathscr {H}^1\Big |\\&\quad +\,\sqrt{2}K\Vert X\Vert _\infty \mathscr {H}^1(M\cap \mathrm {supp}[X]) \end{aligned}$$
upon using Lemma 2.2 and Proposition 3.2. Replacing X by \(-X\) we derive that
$$\begin{aligned} \Big |\int _M f\mathrm {div}^M X\,d\mathscr {H}^1\Big |\le C_1\Vert X\Vert _\infty \mathscr {H}^1(M\cap \mathrm {supp}[X]) \end{aligned}$$
where \(C_1=C\Vert f\Vert _{L^\infty (U)}+\sqrt{2}K\). Let \(\gamma _1:I\rightarrow M\) be a local \(C^1\) parametrisation of M near x. Suppose that \(Y\in C^1_c(I,\mathbb {R}^2)\) with \(\mathrm {supp}[Y]\subset I\) and that \(\gamma _1(I)\subset M\cap B(x,r)\). Note that there exists \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X]\subset B(x,r)\) such that \(X\circ \gamma _1=Y\) on I. The above estimate entails that
$$\begin{aligned} \Big |\int _I(f\circ \gamma _1)\langle \dot{Y},t\rangle \,ds\Big |\le C_1\Big |\mathrm {supp}[Y]\Big |\Vert Y\Vert _\infty . \end{aligned}$$
This means that the function \((f\circ \gamma _1)t\) belongs to \(\mathrm {BV}(I)\) and this implies in turn that \(t\in \mathrm {BV}(I)\). For \(s_1,s_2\in I\) with \(s_1<s_2\),
$$\begin{aligned} |t(s_2)-t(s_1)|&=|Dt((s_1,s_2))|\le |Dt|((s_1,s_2))\nonumber \\&=\sup \Big \{\int _{(s_1,s_2)}\langle t,\dot{Y}\rangle \,ds:Y\in C^1_c((s_1,s_2))\text { and }\Vert Y\Vert _\infty \le 1\Big \}\nonumber \\&\le c\sup \Big \{\int _{(s_1,s_2)}(f\circ \gamma _1)\langle t,\dot{Y}\rangle \,ds:Y\in C^1_c((s_1,s_2))\text { and }\Vert Y\Vert _\infty \le 1\Big \}\nonumber \\&\le cC_1|s_2-s_1|\nonumber \end{aligned}$$
where \(1/c=\inf _{\overline{U}}f>0\). It follows that M is of class \(C^{1,1}\). \(\square \)
We turn to the topic of spherical cap symmetrisation. Denote by \(\mathbb {S}^1_\tau \) the centred circle in \(\mathbb {R}^2\) with radius \(\tau >0\). We sometimes write \(\mathbb {S}^1\) for \(\mathbb {S}^1_1\). Given \(x\in \mathbb {R}^2,v\in \mathbb {S}^1\) and \(\alpha \in (0,\pi ]\) the open cone with vertex x, axis v and opening angle \(2\alpha \) is the set
$$\begin{aligned} C(x,v,\alpha ):=\Big \{y\in \mathbb {R}^2:\langle y-x,v\rangle >|y-x|\cos \alpha \Big \}. \end{aligned}$$
Let E be an \(\mathscr {L}^2\)-measurable set in \(\mathbb {R}^2\) and \(\tau >0\). The \(\tau \)-section \(E_\tau \) of E is the set \(E_\tau :=E\cap \mathbb {S}^1_\tau \). Put
$$\begin{aligned} L(\tau )=L_E(\tau ):=\mathscr {H}^1(E_\tau )\text { for }\tau >0 \end{aligned}$$
(4.1)
and \(p(E) := \{\tau>0:L(\tau )>0\}\). The function L is \(\mathscr {L}^1\)-measurable by [1] Theorem 2.93. Given \(\tau >0\) and \(0<\alpha \le \pi \) the spherical cap \(C(\tau ,\alpha )\) is the set
$$\begin{aligned} C(\tau ,\alpha ):= \left\{ \begin{array}{ll} \mathbb {S}^1_\tau \cap C(0,e_1,\alpha ) &{} \text { if }0<\alpha <\pi ;\\ \mathbb {S}^1_\tau &{} \text { if }\alpha =\pi ;\\ \end{array} \right. \end{aligned}$$
and has \(\mathscr {H}^1\)-measure \(s(\tau ,\alpha ):=2\alpha \tau \). The spherical cap symmetral \(E^{sc}\) of the set E is defined by
$$\begin{aligned} E^{sc}:=\bigcup _{\tau \in p(E)}C(\tau ,\alpha ) \end{aligned}$$
(4.2)
where \(\alpha \in (0,\pi ]\) is determined by \(s(\tau ,\,\alpha )=L(\tau )\). Observe that \(E^{sc}\) is a \(\mathscr {L}^2\)-measurable set in \(\mathbb {R}^2\) and \(V_f(E^{sc})=V_f(E)\). Note also that if B is a centred open ball then \(B^{sc}=B{\setminus }\{0\}\). We say that E is spherical cap symmetric if \(\mathscr {H}^1((E\varDelta E^{sc})_\tau )=0\) for each \(\tau >0\). This definition is broad but suits our purposes.

The result below is stated in [22] Theorem 6.2 and a sketch proof given. A proof along the lines of [2] Theorem 1.1 can be found in [23]. First, let B be a Borel set in \((0,+\infty )\); then the annulus A(B) over B is the set \(A(B):=\{x\in \mathbb {R}^2:|x|\in B\}\).

Theorem 4.2

Let E be a set of finite perimeter in \(\mathbb {R}^2\). Then \(E^{sc}\) is a set of finite perimeter and
$$\begin{aligned} P(E^{sc},A(B))\le P(E,A(B)) \end{aligned}$$
(4.3)
for any Borel set \(B\subset (0,\infty )\) and the same inequality holds with \(E^{sc}\) replaced by any set F that is \(\mathscr {L}^2\)-equivalent to \(E^{sc}\).

Corollary 4.3

Let f be a positive lower semi-continuous radial function on \(\mathbb {R}^2\). Let E be a set of finite perimeter in \(\mathbb {R}^2\). Then \( P_f(E^{sc})\le P_f(E) \).

Proof

Assume that \(P_f(E)<+\infty \). We remark that f is Borel measurable as f is lower semi-continuous. Let \((f_h)\) be a sequence of simple Borel measurable radial functions on \(\mathbb {R}^2\) such that \(0\le f_h\le f\) and \(f_h\uparrow f\) on \(\mathbb {R}^2\) as \(h\rightarrow \infty \). By Theorem 4.2,
$$\begin{aligned} P_{f_h}(E^{sc})=\int _{\mathbb {R}^2}f_h\,d|D\chi _{E^{sc}}| \le \int _{\mathbb {R}^2}f_h\,d|D\chi _{E}|= P_{f_h}(E) \end{aligned}$$
for each h. Taking the limit \(h\rightarrow \infty \) the monotone convergence theorem gives \(P_f(E^{sc})\le P_f(E)\). \(\square \)

Lemma 4.4

Let E be an \(\mathscr {L}^2\)-measurable set in \(\mathbb {R}^2\) such that \(E{\setminus }\{0\}=E^{sc}\). Then there exists an \(\mathscr {L}^2\)-measurable set F equivalent to E such that
  1. (i)

    \(\partial F=\{x\in \mathbb {R}^2:0<|F\cap B(x,\rho )|<|B(x,\rho )|\text { for any }\rho >0\}\);

     
  2. (ii)

    F is spherical cap symmetric.

     

Proof

Put
$$\begin{aligned} E_1&:=\{x\in \mathbb {R}^2:|E\cap B(x,\rho )|=|B(x,\rho )|\text { for some }\rho>0\};\nonumber \\ E_0&:=\{x\in \mathbb {R}^2:|E\cap B(x,\rho )|=0\text { for some }\rho >0\}.\nonumber \end{aligned}$$
We claim that \(E_1\) is spherical cap symmetric. For take \(x\in E_1\) with \(\tau =|x|>0\) and \(|\theta (x)|\in (0,\pi ]\). Now \(|E\cap B(x,\rho )|=|B(x,\rho )|\) for some \(\rho >0\). Let \(y\in \mathbb {R}^2\) with \(|y|=\tau \) and \(|\theta (y)|<|\theta (x)|\). Choose a rotation \(O\in \mathrm {SO}(2)\) such that \(OB(x,\rho )=B(y,\rho )\). As \(E{\setminus }\{0\}=E^{sc},|E\cap B(y,\rho )|=|O(E\cap B(x,\rho ))|=|E\cap B(x,\rho )|=|B(x,\rho )|=|B(y,\rho )|\). The claim follows. It follows in a similar way that \(\mathbb {R}^2{\setminus } E_0\) is spherical cap symmetric. It can then be seen that the set \(F:=(E_1\cup E){\setminus } E_0\) inherits this property. As in [13] Proposition 3.1 the set F is equivalent to E and enjoys the property in (i). \(\square \)

Theorem 4.5

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a bounded minimiser of (1.2). Then there exists an \(\mathscr {L}^2\)-measurable set \(\widetilde{E}\) with the properties
  1. (i)

    \(\widetilde{E}\) is a minimiser of (1.2);

     
  2. (ii)

    \(L_{\widetilde{E}}=L\) a.e. on \((0,+\infty )\);

     
  3. (iii)

    \(\widetilde{E}\) is open, bounded and has \(C^{1,1}\) boundary;

     
  4. (iv)

    \(\widetilde{E}{\setminus }\{0\}=\widetilde{E}^{sc}\).

     

Proof

Let E be a bounded minimiser for (1.2). Then \(E_1:=E^{sc}\) is a bounded minimiser of (1.2) by Corollary 4.3 and \(L_E=L_{E_1}\) on \((0,+\infty )\). Now put \(E_2:=F\) with F as in Lemma 4.4. Then \(L_{E_2}=L\) a.e. on \((0,+\infty )\) as \(E_2\) is equivalent to \(E_1,E_2\) is a bounded minimiser of (1.2) and \(E_2\) is spherical cap symmetric. Moreover, \(\partial E_2=\{x\in \mathbb {R}^2:0<|E_2\cap B(x,\rho )|<|B(x,\rho )|\text { for any }\rho >0\}\). As in the proof of Theorem 3.8, \(\partial E_2\) is a \(C^1\) hypersurface in \(\mathbb {R}^2\). Put
$$\begin{aligned} \widetilde{E}:=\{x\in \mathbb {R}^2:|E_2\cap B(x,\rho )|=|B(x,\rho )|\text { for some }\rho >0\}. \end{aligned}$$
Then \(\widetilde{E}\) is equivalent to \(E_2\) so that (ii) holds, and is a bounded minimiser of (1.2); \(\widetilde{E}\) is open and \(\partial \widetilde{E}=\partial E_2\) is \(C^1\). In fact, \(\partial \widetilde{E}\) is of class \(C^{1,1}\) by Theorem 4.1. As \(E_2\) is spherical cap symmetric the same is true of \(\widetilde{E}\). But \(\widetilde{E}\) is open which entails that \(\widetilde{E}{\setminus }\{0\}=\widetilde{E}^{sc}\). \(\square \)

5 More on spherical cap symmetry

Let
$$\begin{aligned} H:=\{x=(x_1,x_2)\in \mathbb {R}^2:x_2>0\} \end{aligned}$$
stand for the open upper half-plane in \(\mathbb {R}^2\) and
$$\begin{aligned} S:\mathbb {R}^2\rightarrow \mathbb {R}^2;x=(x_1,x_2)\mapsto (x_1,-x_2) \end{aligned}$$
for reflection in the \(x_1\)-axis. Let \(O\in \mathrm {SO}(2)\) represent rotation anti-clockwise through \(\pi /2\).

Lemma 5.1

Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\). Let \(x\in M{\setminus }\{0\}\). Then
  1. (i)

    \(Sx\in M{\setminus }\{0\}\);

     
  2. (ii)

    \(n(Sx)=Sn(x)\);

     
  3. (iii)

    \(\cos \sigma (Sx)=-\cos \sigma (x)\).

     

Proof

(i) The closure \(\overline{E}\) of E is spherical cap symmetric. The spherical cap symmetral \(\overline{E}\) is invariant under S from the representation (4.2). (ii) is a consequence of this last observation. (iii) Note that \(t(Sx)=O^\star n(Sx)=O^\star Sn(x)\). Then
$$\begin{aligned} \cos \sigma (Sx)&=\langle Sx,t(Sx)\rangle =\langle Sx,O^\star Sn(x)\rangle =\langle x,SO^\star Sn(x)\rangle \nonumber \\&=\langle x,On(x)\rangle =-\langle x,O^\star n(x)\rangle =\cos \sigma (x)\nonumber \end{aligned}$$
as \(SO^\star S=O\) and \(O=-O^\star \). \(\square \)

We introduce the projection \(\pi :\mathbb {R}^2\rightarrow [0,+\infty );x\mapsto |x|\).

Lemma 5.2

Let E be an open set in \(\mathbb {R}^2\) with boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\).
  1. (i)

    Suppose \(0\ne x\in \mathbb {R}^2{\setminus }\overline{E}\) and \(\theta (x)\in (0,\pi ]\). Then there exists an open interval I in \((0,+\infty )\) containing \(\tau \) and \(\alpha \in (0,\theta (x))\) such that \(A(I){\setminus }\overline{S}(\alpha )\subset \mathbb {R}^2{\setminus }\overline{E}\).

     
  2. (ii)

    Suppose \(0\ne x\in E\) and \(\theta (x)\in [0,\pi )\). Then there exists an open interval I in \((0,+\infty )\) containing \(\tau \) and \(\alpha \in (\theta (x),\pi )\) such that \(A(I)\cap S(\alpha )\subset E\).

     
  3. (iii)

    For each \(0<\tau \in \pi (M),M_\tau \) is the union of two closed spherical arcs in \(\mathbb {S}^1_\tau \) symmetric about the \(x_1\)-axis.

     

Proof

(i) We can find \(\alpha \in (0,\theta (x))\) such that \(\mathbb {S}^1_\tau {\setminus } S(\alpha )\subset \mathbb {R}^2{\setminus }\overline{E}\) as can be seen from definition (4.2). This latter set is compact so \(\mathrm {dist}(\mathbb {S}^1_\tau {\setminus } S(\alpha ),\overline{E})>0\). This means that the \(\varepsilon \)-neighbourhood of \(\mathbb {S}^1_\tau {\setminus } S(\alpha )\) is contained in \(\mathbb {R}^2{\setminus }\overline{E}\) for \(\varepsilon >0\) small. The claim follows. (ii) Again from (4.2) we can find \(\alpha \in (\theta (x),\pi )\) such that \(\overline{\mathbb {S}^1_\tau \cap S(\alpha )}\subset E\) and the assertion follows as before.

(iii) Suppose \(x_1,x_2\) are distinct points in \(M_\tau \) with \(0\le \theta (x_1)<\theta (x_2)\le \pi \). Suppose y lies in the interior of the spherical arc joining \(x_1\) and \(x_2\). If \(y\in \mathbb {R}^2{\setminus }\overline{E}\) then \(x_2\in \mathbb {R}^2{\setminus }\overline{E}\) by (i) and hence \(x_2\not \in M\). If \(y\in E\) we obtain the contradiction that \(x_1\in E\) by (ii). Therefore \(y\in M\). We infer that the closed spherical arc joining \(x_1\) and \(x_2\) lies in \(M_\tau \). The claim follows noting that \(M_\tau \) is closed. \(\square \)

Lemma 5.3

Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M. Let \(x\in M\). Then
$$\begin{aligned} \liminf _{\overline{E}\ni y\rightarrow x}\Big \langle \frac{y-x}{|y-x|},n(x)\Big \rangle \ge 0. \end{aligned}$$

Proof

Assume for a contradiction that
$$\begin{aligned} \liminf _{\overline{E}\ni y\rightarrow x}\Big \langle \frac{y-x}{|y-x|},n(x)\Big \rangle \in [-1,0). \end{aligned}$$
There exists \(\eta \in (0,1)\) and a sequence \((y_h)\) in E such that \(y_h\rightarrow x\) as \(h\rightarrow \infty \) and
$$\begin{aligned} \Big \langle \frac{y_h-x}{|y_h-x|},n(x)\Big \rangle&<-\eta \end{aligned}$$
(5.1)
for each \(h\in \mathbb {N}\). Choose \(\alpha \in (0,\pi /2)\) such that \(\cos \alpha =\eta \). As M is \(C^1\) there exists \(r>0\) such that
$$\begin{aligned} B(x,r)\cap C(x,-n(x),\alpha )\cap E=\emptyset . \end{aligned}$$
By choosing h sufficiently large we can find \(y_h\in B(x,r)\) with the additional property that \(y_h\in C(x,-n(x),\alpha )\) by (5.1). We are thus led to a contradiction. \(\square \)

Lemma 5.4

Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\). For each \(0<\tau \in \pi (M)\),
  1. (i)

    \(|\cos \sigma |\) is constant on \(M_\tau \);

     
  2. (ii)

    \(\cos \sigma =0\) on \(M_\tau \cap \{x_2=0\}\);

     
  3. (iii)

    \(\langle Ox,n(x)\rangle \le 0\) for \(x\in M_\tau \cap H\)

     
  4. (iv)

    \(\cos \sigma \le 0\) on \(M_\tau \cap H\);

     
and if \(\cos \sigma \not \equiv 0\) on \(M_\tau \) then
  1. (v)

    \(\tau \in p(E)\);

     
  2. (vi)

    \(M_\tau \) consists of two disjoint singletons in \(\mathbb {S}^1_\tau \) symmetric about the \(x_1\)-axis;

     
  3. (vii)

    \(L(\tau )\in (0,2\pi \tau )\);

     
  4. (viii)

    \(M_\tau =\{(\tau \cos (L(\tau )/2\tau ),\pm \tau \sin (L(\tau )/2\tau )\}\).

     

Proof

(i) By Lemma 5.2, \(M_\tau \) is the union of two closed spherical arcs in \(\mathbb {S}^1_\tau \) symmetric about the \(x_1\)-axis. In case \(M_\tau \cap \overline{H}\) consists of a singleton the assertion follows from Lemma 5.1. Now suppose that \(M_\tau \cap \overline{H}\) consists of a spherical arc in \(\mathbb {S}^1_\tau \) with non-empty interior. It can be seen that \(\cos \sigma \) vanishes on the interior of this arc as \(0=r_1^\prime =\cos \sigma _1\) in a local parametrisation by (2.9). By continuity \(\cos \sigma =0\) on \(M_\tau \). (ii) follows from Lemma 5.1. (iii) Let \(x\in M_\tau \cap H\) so \(\theta (x)\in (0,\pi )\). Then \(S(\theta (x))\cap \mathbb {S}^1_\tau \subset \overline{E}\) as \(\overline{E}\) is spherical cap symmetric. Then
$$\begin{aligned} 0\le \lim _{S(\theta (x))\cap \mathbb {S}^1_\tau \ni y\rightarrow x}\Big \langle \frac{y-x}{|y-x|},n(x)\Big \rangle =-\langle Ox,n(x)\rangle \end{aligned}$$
by Lemma 5.3. (iv) The adjoint transformation \(O^\star \) represents rotation clockwise through \(\pi /2\). Let \(x\in M_\tau \cap H\). By (iii),
$$\begin{aligned} 0\ge \langle Ox,n(x)\rangle =\langle x,O^\star n(x)\rangle =\langle x,t(x)\rangle =\tau \cos \sigma (x) \end{aligned}$$
and this leads to the result. (v) As \(\cos \sigma \not \equiv 0\) on \(M_\tau \) we can find \(x\in M_\tau \cap H\). We claim that \(\mathbb {S}^1_\tau \cap S(\theta (x))\subset E\). For suppose that \(y\in \mathbb {S}^1_\tau \cap S(\theta (x))\) but \(y\not \in E\). We may suppose that \(0\le \theta (y)<\theta (x)<\pi \). If \(y\in \mathbb {R}^2{\setminus }\overline{E}\) then \(x\in \mathbb {R}^2{\setminus }\overline{E}\) by Lemma 5.2. On the other hand, if \(y\in M\) then the spherical arc in H joining y to x is contained in M again by Lemma 5.2. This arc also has non-empty interior in \(\mathbb {S}^1_\tau \). Now \(\cos \sigma =0\) on its interior so \(\cos (\sigma (x))=0\) by (i) contradicting the hypothesis. A similar argument deals with (vi) and this together with (v) in turn entails (vii) and (viii). \(\square \)

Lemma 5.5

Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(0\in M\). Then
  1. (i)

    \((\sin \sigma )(0+)=0\);

     
  2. (ii)

    \((\cos \sigma )(0+)=-1\).

     

Proof

(i) Let \(\gamma _1\) be a \(C^1\) parametrisation of M in a neighbourhood of 0 with \(\gamma _1(0)=0\) as above. Then \(n(0)=n_1(0)=e_1\) and hence \(t(0)=t_1(0)=-e_2\). By Taylor’s Theorem \(\gamma _1(s)=\gamma _1(0)+t_1(0)s+o(s)=-e_2s+o(s)\) for \(s\in I\). This means that \(r_1(s)=|\gamma _1(s)|=s+o(s)\) and
$$\begin{aligned} \cos \theta _1=\frac{\langle e_1,\gamma _1\rangle }{r_1}=\frac{\langle e_1,\gamma _1\rangle }{s}\frac{s}{r_1}\rightarrow 0 \end{aligned}$$
as \(s\rightarrow 0\) which entails that \((\cos \theta _1)(0-)=0\). Now \(t_1\) is continuous on I so \(t_1=-e_2+o(1)\) and \(\cos \alpha _1=\langle e_1,t_1\rangle =o(1)\). We infer that \((\cos \alpha _1)(0-)=0\). By (2.11), \(\cos \alpha _1=\cos \sigma _1\cos \theta _1-\sin \sigma _1\sin \theta _1\) on I and hence \((\sin \sigma _1)(0-)=0\). We deduce that \((\sin \sigma )(0+)=0\). Item (ii) follows from (i) and Lemma 5.4. \(\square \)
The set
$$\begin{aligned} \varOmega :=\pi \Big [(M{\setminus }\{0\})\cap \{\cos \sigma \ne 0\}\Big ] \end{aligned}$$
(5.2)
plays an important rôle in the proof of Theorem 1.1.

Lemma 5.6

Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\). Then \(\varOmega \) is an open set in \((0,+\infty )\).

Proof

Suppose \(0<\tau \in \varOmega \). Choose \(x\in M_\tau \cap \{\cos \sigma \ne 0\}\). Let \(\gamma _1:I\rightarrow M\) be a local \(C^1\) parametrisation of M in a neighbourhood of x such that \(\gamma _1(0)=x\) as before. By shrinking I if necessary we may assume that \(r_1\ne 0\) and \(\cos \sigma _1\ne 0\) on I. Then the set \(\{r_1(s):s\in I\}\subset \varOmega \) is connected and so an interval in \(\mathbb {R}\) (see for example [25] Theorems 6.A and 6.B). By (2.9), \(r_1^\prime (0)=\cos \sigma _1(0)=\cos \sigma (p)\ne 0\). This means that the set \(\{r_1(s):s\in I\}\) contains an open interval about \(\tau \). \(\square \)

6 Generalised (mean) curvature

Given a set E of finite perimeter in \(\mathbb {R}^2\) the first variation \(\delta V_f(Z)\) resp. \(\delta P_f^+(Z)\) of weighted volume and perimeter along a time-dependent vector field Z are defined as in (2.13) and (2.14).

Proposition 6.1

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Let E be a bounded open set in \(\mathbb {R}^2\) with \(C^1\) boundary M. Let Z be a time-dependent vector field. Then
$$\begin{aligned} \delta P_f^+(Z)=\int _M f^\prime _+(\cdot ,Z_0) + f\mathrm {div}^M Z_0\,d\mathscr {H}^1 \end{aligned}$$
where \(Z_0:=Z(0,\cdot )\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\).

Proof

The identity (3.2) holds for each \(t\in I\) with M in place of \(\mathscr {F}E\). The assertion follows on appealing to Lemma 2.3 and Lemma 2.4 with the help of the dominated convergence theorem. \(\square \)

Given \(X,Y\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) let \(\psi \) resp. \(\chi \) stand for the 1-parameter group of \(C^\infty \) diffeomorphisms of \(\mathbb {R}^2\) associated to the vector fields X resp. Y as in (2.12). Let I be an open interval in \(\mathbb {R}\) containing the point 0. Suppose that the function \(\sigma :I\rightarrow \mathbb {R}\) is \(C^1\). Define a flow via
$$\begin{aligned} \varphi :I\times \mathbb {R}^2\rightarrow \mathbb {R}^2; (t,x)\mapsto \chi (\sigma (t),\psi (t,x)). \end{aligned}$$

Lemma 6.2

The time-dependent vector field Z associated with the flow \(\varphi \) is given by
$$\begin{aligned} Z(t,x)=\sigma ^\prime (t)Y(\chi (\sigma (t),\psi (t,x)))+d\chi (\sigma (t),\psi (t,x))X(\psi (t,x)) \end{aligned}$$
(6.1)
for \((t,x)\in I\times \mathbb {R}^2\) and satisfies (Z.1) and (Z.2).

Proof

For \(t\in I\) and \(x\in \mathbb {R}^2\) we compute using (2.12),
$$\begin{aligned} \partial _t\varphi (t,x) =(\partial _t\chi )(\sigma (t),\psi (t,x))\sigma ^\prime (t)+d\chi (\sigma (t),\psi (t,x))\partial _t\psi (t,x) \end{aligned}$$
and this gives (6.1). Put \(K_1:=\mathrm {supp}[X],K_2:=\mathrm {supp}[Y]\) and \(K:=K_1\cup K_2\). Then (Z.2) holds with this choice of K. \(\square \)
Let E be a bounded open set in \(\mathbb {R}^2\) with \(C^1\) boundary M. Define \(\varLambda :=(M{\setminus }\{0\})\cap \{\cos \sigma =0\}\) and
$$\begin{aligned} \varLambda _1&:=\{x\in M:\mathscr {H}^1(\varLambda \cap B(x,\rho ))=\mathscr {H}^1(M\cap B(x,\rho ))\text { for some }\rho >0\}. \end{aligned}$$
(6.2)
For future reference put \(\varLambda _1^{\pm }:=\varLambda _1\cap \{x\in M:\pm \langle x,n\rangle >0\}\).

Lemma 6.3

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Let E be a bounded open set in \(\mathbb {R}^2\) with \(C^{1,1}\) boundary M and suppose that \(E{\setminus }\{0\}=E^{sc}\). Then
  1. (i)

    \(\varLambda _1\) is a countable disjoint union of well-separated open circular arcs centred at 0;

     
  2. (ii)

    \(\mathscr {H}^1(\overline{\varLambda _1}{\setminus }\varLambda _1)=0\);

     
  3. (iii)

    f is differentiable \(\mathscr {H}^1\)-a.e. on \(M{\setminus }\overline{\varLambda _1}\).

     

The term well-separated in (i) means the following: if \(\varGamma \) is an open circular arc in \(\varLambda _1\) with \(\varGamma \cap (\varLambda _1{\setminus }\varGamma )=\emptyset \) then \(d(\varGamma ,\varLambda _1{\setminus }\varGamma )>0\).

Proof

(i) Let \(x\in \varLambda _1\) and \(\gamma _1:I\rightarrow M\) a \(C^{1,1}\) parametrisation of M near x. By shrinking I if necessary we may assume that \(\gamma _1(I)\subset M\cap B(x,\rho )\) with \(\rho \) as in (6.2). So \(\cos \sigma =0\) \(\mathscr {H}^1\)-a.e. on \(\gamma _1(I)\) and hence \(\cos \sigma _1=0\) a.e. on I. This means that \(\cos \sigma _1=0\) on I as \(\sigma _1\in C^{0,1}(I)\) and that \(r_1\) is constant on I by (2.9). Using (2.10) it can be seen that \(\gamma _1(I)\) is an open circular arc centred at 0. By compactness of M it follows that \(\varLambda _1\) is a countable disjoint union of open circular arcs centred on 0. The well-separated property flows from the fact that M is \(C^1\). (ii) follows as a consequence of this property. (iii) Let \(x\in M{\setminus }\overline{\varLambda _1}\) and \(\gamma _1:I\rightarrow M\) a \(C^{1,1}\) parametrisation of M near x with properties as before. We assume that x lies in the upper half-plane H. By shrinking I if necessary we may assume that \(\gamma _1(I)\subset (M{\setminus }\overline{\varLambda _1})\cap H\). Let \(s_1,s_2, s_3\in I\) with \(s_1<s_2<s_3\). Then \(y:=\gamma _1(s_2)\in M{\setminus }\overline{\varLambda _1}\). So \(\mathscr {H}^1(M\cap \{\cos \sigma \ne 0\}\cap B(y,\rho ))>0\) for each \(\rho >0\). This means that for small \(\eta >0\) the set \(\gamma _1((s_2-\eta ,s_2+\eta ))\cap \{\cos \sigma \ne 0\}\) has positive \(\mathscr {H}^1\)-measure. Consequently, \(r_1(s_3)-r_1(s_1)=\int _{s_1}^{s_3}\cos \sigma _1\,ds<0\) bearing in mind Lemma 5.4. This shows that \(r_1\) is strictly decreasing on I. So h is differentiable a.e. on \(r_1(I)\subset (0,+\infty )\) in virtue of the fact that h is convex and hence locally Lipschitz. This entails (iii). \(\square \)

Proposition 6.4

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a minimiser of (1.2). Assume that E is a bounded open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and suppose that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(M{\setminus }\overline{\varLambda _1}\ne \emptyset \). Then there exists \(\lambda \in \mathbb {R}\) such that for any \(X\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\),
$$\begin{aligned} 0\le \int _M\Big \{f^\prime _+(\cdot ,X) + f\,\mathrm {div}^M X-\lambda f\langle n, X\rangle \Big \} \,d\mathscr {H}^1. \end{aligned}$$

Proof

Let \(X\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\). Let \(x\in M\) and \(r>0\) such that \(M\cap B(x,r)\subset M{\setminus }\overline{\varLambda _1}\). Choose \(Y\in C^\infty _c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[Y]\subset B(x,r)\) as in Lemma 3.4. Let \(\psi \) resp. \(\chi \) stand for the 1-parameter group of \(C^\infty \) diffeomorphisms of \(\mathbb {R}^2\) associated to the vector fields X resp. Y as in (2.12). For each \((s,t)\in \mathbb {R}^2\) the set \(\chi _s(\psi _t(E))\) is an open set in \(\mathbb {R}^2\) with \(C^1\) boundary and \(\partial (\chi _s\circ \psi _t)(E)=(\chi _s\circ \psi _t)(M)\) by Lemma 2.1. Define
$$\begin{aligned} V(s,t)&:=V_f(\chi _t(\psi _s(E)))-V_f(E),\nonumber \\ P(s,t)&:=P_f(\chi _t(\psi _s(E))),\nonumber \end{aligned}$$
for \((s,t)\in \mathbb {R}^2\). We write \(F=(\chi _t\circ \psi _s)(E)\). Arguing as in Proposition 3.2,
$$\begin{aligned} \partial _t V(s,t)&=\lim _{h\rightarrow 0}(1/h)\{V_f(\chi _h(F))-V_f(F)\} =\int _F\mathrm {div}(fY)\,dx\nonumber \\&=\int _E(\mathrm {div}(fY)\circ \chi _t\circ \psi _s)\,J_2 d(\chi _t\circ \psi _s)_x\,dx\nonumber \end{aligned}$$
with an application of the area formula (cf. [1] Theorem 2.71). This last varies continuously in (st). The same holds for partial differentiation with respect to s. Indeed, put \(\eta :=\chi _t\circ \psi _s\). Then noting that \(J_2d(\eta \circ \psi _h)=(J_2d\eta )\circ \psi _h J_2d\psi _h\) and using the dominated convergence theorem,
$$\begin{aligned} \partial _s V(s,t)&=\lim _{h\rightarrow 0}(1/h) \Big \{ V_f(\eta (\psi _h(E)))-V_f(\eta (E)) \Big \}\nonumber \\&=\lim _{h\rightarrow 0}(1/h) \Big \{ \int _E(f\circ \eta \circ \psi _h)J_2d(\eta \circ \psi _h)_x\,dx-\int _E(f\circ \eta )J_2d\eta _x\,dx \Big \}\nonumber \\&=\lim _{h\rightarrow 0}(1/h) \Big \{ \int _E[(f\circ \eta \circ \psi _h)-(f\circ \eta )]J_2d(\eta \circ \psi _h)_x\,dx\nonumber \\&\quad +\int _E(f\circ \eta )[(J_2d\eta \circ \psi _h-J_2d\eta ]J_2d\psi _h\,dx\nonumber \\&\quad +\int _E(f\circ \eta )J_2d\eta [J_2d\psi _h-1]\,dx \Big \}\nonumber \\&=\int _E\langle \nabla (f\circ \eta ),X\rangle J_2d\eta _x\,dx +\int _E(f\circ \eta )\langle \nabla J_2d\eta ,X\rangle \,dx\nonumber \\&\quad +\int _E(f\circ \eta )J_2d\eta \,\mathrm {div}\,X\,dx\nonumber \end{aligned}$$
where the explanation for the last term can be found in the proof of Proposition 3.2. In this regard we note that \(d(d\chi _t)\) (for example) is continuous on \(I\times \mathbb {R}^2\) (cf. [1] Theorem 3.3 and Exercise 3.2) and in particular \(\nabla J_2d\chi _t\) is continuous on \(I\times \mathbb {R}^2\). The expression above also varies continuously in (st) as can be seen with the help of the dominated convergence theorem. This means that \(V(\cdot ,\cdot )\) is continuously differentiable on \(\mathbb {R}^2\). Note that
$$\begin{aligned} \partial _tV(0,0)=\int _E\mathrm {div}(fY)\,dx=1 \end{aligned}$$
by choice of Y. By the implicit function theorem there exists \(\eta >0\) and a \(C^1\) function \(\sigma :(-\eta ,\eta )\rightarrow \mathbb {R}\) such that \(\sigma (0)=0\) and \(V(s,\sigma (s))=0\) for \(s\in (-\eta ,\eta )\); moreover,
$$\begin{aligned} \sigma ^\prime (0)&=-\partial _s V(0,0) =-\int _E\Big \{\langle \nabla f,X\rangle +f\,\mathrm {div}\,X\Big \}\,dx\nonumber \\&=-\int _E\mathrm {div}(fX)\,dx =\int _M f\,\langle n, X\rangle \,d\mathscr {H}^1\nonumber \end{aligned}$$
by the Gauss–Green formula (cf. [1] Theorem 3.36).
The mapping
$$\begin{aligned} \varphi :(-\eta ,\eta )\times \mathbb {R}^2\rightarrow \mathbb {R}^2;t\mapsto \chi (\sigma (t),\psi (t,x)) \end{aligned}$$
satisfies conditions (F.1)–(F.4) above with \(I=(-\eta ,\eta )\) where the associated time-dependent vector field Z is given as in (6.1) and satisfies (Z.1) and (Z.2); moreover, \(Z_0=Z(0,\cdot )=\sigma ^\prime (0)Y+X\). Note that \(Z_0=X\) on \(M{\setminus } B(x,r)\).
The mapping \(I\rightarrow \mathbb {R};t\mapsto P_f(\varphi _t(E))\) is right-differentiable at \(t=0\) as can be seen from Proposition 6.1 and has non-negative right-derivative there. By Proposition 6.1 and Lemma 6.3,
$$\begin{aligned} 0\le \delta P_f^+(Z)&=\int _{M} f^\prime _+(\cdot , Z_0)+ f\,\mathrm {div}^M Z_0\,d\mathscr {H}^1\nonumber \\&=\int _{M{\setminus }\overline{\varLambda _1}} f^\prime _+(\cdot , Z_0) + f\,\mathrm {div}^M Z_0\,d\mathscr {H}^1 \nonumber \\&\quad +\,\int _{\overline{\varLambda _1}} f^\prime _+(\cdot , X) + f\,\mathrm {div}^M X\,d\mathscr {H}^1 \nonumber \\&=\int _{M{\setminus }\overline{\varLambda _1}}\sigma ^\prime (0)\langle \nabla f,Y\rangle +\langle \nabla f,X\rangle \nonumber \\&\quad +\,\sigma ^\prime (0)\,f\,\mathrm {div}^M Y+f\,\mathrm {div}^M X\,d\mathscr {H}^1\nonumber \\&\quad +\int _{\overline{\varLambda _1}} f^\prime _+(\cdot , X) + f\,\mathrm {div}^M X\,d\mathscr {H}^1\nonumber \\&=\int _{M}f^\prime _+(\cdot , X) + f\,\mathrm {div}^M X\,d\mathscr {H}^1\nonumber \\&\quad +\,\sigma ^\prime (0)\int _{M}f^\prime _+(\cdot , Y)+ f\,\mathrm {div}^M Y\,d\mathscr {H}^1. \end{aligned}$$
(6.3)
The identity then follows upon inserting the expression for \(\sigma ^\prime (0)\) above with \(\lambda =-\int _{M}f^\prime _+(\cdot ,Y) + f\,\mathrm {div}^M Y\,d\mathscr {H}^1\). The claim follows for \(X\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\) by a density argument. \(\square \)

Theorem 6.5

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a minimiser of (1.2). Assume that E is a bounded open set in \(\mathbb {R}^2\) with \(C^{1,1}\) boundary M and suppose that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(M{\setminus }\overline{\varLambda _1}\ne \emptyset \). Then there exists \(\lambda \in \mathbb {R}\) such that
  1. (i)

    \(k+\rho \sin \sigma +\lambda =0\) \(\mathscr {H}^1\)-a.e. on \(M{\setminus }\overline{\varLambda _1}\);

     
  2. (ii)

    \(\rho _--\lambda \le k\le \rho _+-\lambda \) on \(\varLambda _1^+\);

     
  3. (iii)

    \(-\rho _+-\lambda \le k\le -\rho _--\lambda \) on \(\varLambda _1^-\).

     

The expression \(k+\rho \sin \sigma \) is called the generalised (mean) curvature of M.

Proof

(i) Let \(x\in M\) and \(r>0\) such that \(M\cap B(x,r)\subset M{\setminus }\overline{\varLambda _1}\). Choose \(X\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\) with \(\mathrm {supp}[X]\subset B(x,r)\). We know from Lemma 6.3 that f is differentiable \(\mathscr {H}^1\)-a.e. on \(\mathrm {supp}[X]\). Let \(\lambda \) be as in Proposition 6.4. Replacing X by \(-X\) we deduce from Proposition 6.4 that
$$\begin{aligned} 0=\int _M\Big \{\langle \nabla f,X\rangle + f\,\mathrm {div}^M X-\lambda f\langle n, X\rangle \Big \} \,d\mathscr {H}^1. \end{aligned}$$
The divergence theorem on manifolds (cf. [1] Theorem 7.34) holds also for \(C^{1,1}\) manifolds. So
$$\begin{aligned} \int _M\langle \nabla f, X\rangle + f\,\mathrm {div}^MX\,d\mathscr {H}^1&=\int _M \partial _nf\,\langle n,\,X\rangle +\langle \nabla ^M f,\,X\rangle +f\,\mathrm {div}^M\,X \,d\mathscr {H}^1\nonumber \\&=\int _M \partial _n f\,\langle n,\,X\rangle +\mathrm {div}^M(fX) \,d\mathscr {H}^1\nonumber \\&=\int _M \partial _n f\,\langle n,\,X\rangle -Hf\langle n,\,X\rangle \,d\mathscr {H}^1\nonumber \\&=\int _M fu \left\{ \partial _n \log f-H \right\} \,d\mathscr {H}^1\nonumber \end{aligned}$$
where \(u=\langle n,X\rangle \). Combining this with the equality above we see that
$$\begin{aligned} \int _M uf \left\{ \partial _n\log f-H-\lambda \right\} \,d\mathscr {H}^1=0 \end{aligned}$$
for all \(X\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\). This leads to the result.
(ii) Let \(x\in M\) and \(r>0\) such that \(M\cap B(x,r)\subset \varLambda _1^+\). Let \(\phi \in C^1(\mathbb {S}^1_r)\) with support in \(\mathbb {S}^1_r\cap B(x,r)\). We can construct \(X\in C^1_c(\mathbb {R}^2,\mathbb {R}^2)\) with the property that \(X=\phi n\) on \(M\cap B(x,r)\). By Lemma 2.4,
$$\begin{aligned} f^\prime _+(\cdot ,X)=fh^\prime _+(|x|,\mathrm {sgn}\langle x,X\rangle )|\langle n,X\rangle | =fh^\prime _+(|x|,\mathrm {sgn}\,\phi \langle x,n\rangle )|\phi | \nonumber \end{aligned}$$
on \(\varLambda _1\). Let us assume that \(\phi \ge 0\). As \(\langle \cdot ,n\rangle >0\) on \(\varLambda _1^+\) we have that \(f^\prime _+(\cdot ,X)=f\phi h^\prime _+(|x|,+1)=f\phi \rho _+\) so by Proposition 6.4,
$$\begin{aligned} 0&\le \int _M\Big \{f^\prime _+(\cdot ,X) + f\,\mathrm {div}^M X-\lambda f\langle n, X\rangle \Big \} d\mathscr {H}^1\nonumber \\&=\int _M f\phi \Big \{\rho _+-k-\lambda \Big \} \,d\mathscr {H}^1.\nonumber \end{aligned}$$
We conclude that \(\rho _+-k-\lambda \ge 0\) on \(M\cap B(x,r)\). Now assume that \(\phi \le 0\). Then \(f^\prime _+(\cdot ,X)=-f\phi h^\prime _+(|x|,-1)=f\phi \rho _-\) so
$$\begin{aligned} 0&\le \int _M f\phi \Big \{\rho _--k-\lambda \Big \} \,d\mathscr {H}^1\nonumber \end{aligned}$$
and hence \(\rho _--k-\lambda \le 0\) on \(M\cap B(x,r)\). This shows (ii).
(iii) The argument is similar. Assume in the first instance that \(\phi \ge 0\). Then \(f^\prime _+(\cdot ,X)=f\phi h^\prime _+(|x|,-1)=-f\phi \rho _-\) so
$$\begin{aligned} 0\le \int _M f\phi \Big \{-\rho _--k-\lambda \Big \} \,d\mathscr {H}^1.\nonumber \end{aligned}$$
We conclude that \(-\rho _--k-\lambda \ge 0\) on \(M\cap B(x,r)\). Next suppose that \(\phi \le 0\). Then \(f^\prime _+(\cdot ,X)=-f\phi h^\prime _+(|x|,+1)=-f\phi \rho _+\) so
$$\begin{aligned} 0\le \int _M f\phi \Big \{-\rho _+-k-\lambda \Big \} \,d\mathscr {H}^1\nonumber \end{aligned}$$
and \(-\rho _+-k-\lambda \le 0\) on \(M\cap B(x,r)\). \(\square \)
Let E be an open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\) and that \(\varOmega \) is as in (5.2). Bearing in mind Lemma 5.4 we may define
$$\begin{aligned}&\theta _2:\varOmega \rightarrow (0,\pi );\tau \mapsto L(\tau )/2\tau ; \end{aligned}$$
(6.4)
$$\begin{aligned}&\gamma :\varOmega \rightarrow M;\tau \mapsto (\tau \cos \theta _2(\tau ),\tau \sin \theta _2(\tau )). \end{aligned}$$
(6.5)
The function
$$\begin{aligned} u:\varOmega \rightarrow [-1,1];\tau \mapsto \sin (\sigma (\gamma (\tau ))). \end{aligned}$$
(6.6)
plays a key role.

Theorem 6.6

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a bounded minimiser of (1.2). Assume that E is open with \(C^{1,1}\) boundary M and that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(M{\setminus }\overline{\varLambda _1}\ne \emptyset \) and let \(\lambda \) be as in Theorem 6.5. Then \(u\in C^{0,1}(\varOmega )\) and
$$\begin{aligned} u^\prime +(1/\tau +\rho )u+\lambda =0 \end{aligned}$$
a.e. on \(\varOmega \).

Proof

Let \(\tau \in \varOmega \) and x a point in the open upper half-plane such that \(x\in M_\tau \). There exists a \(C^{1,1}\) parametrisation \(\gamma _1:I\rightarrow M\) of M in a neighbourhood of x with \(\gamma _1(0)=x\) as above. Put \(u_1:=\sin \sigma _1\) on I. By shrinking the open interval I if necessary we may assume that \(r_1:I\rightarrow r_1(I)\) is a diffeomorphism and that \(r_1(I)\subset \subset \varOmega \). Note that \(\gamma =\gamma _1\circ r_1^{-1}\) and \(u=u_1\circ r_1^{-1}\) on \(r_1(I)\). It follows that \(u\in C^{0,1}(\varOmega )\). By (2.9),
$$\begin{aligned} u^\prime =\frac{\dot{u}_1}{\dot{r}_1}\circ r_1^{-1} =\dot{\sigma }_1\circ r_1^{-1} \end{aligned}$$
a.e. on \(r_1(I)\). As \(\dot{\alpha }_1=k_1\) a.e. on I and using the identity (2.10) we see that \(\dot{\sigma }_1=\dot{\alpha }_1-\dot{\theta }_1=k_1-(1/r_1)\sin \sigma _1\) a.e on I. Thus,
$$\begin{aligned} u^\prime =k-(1/\tau )\sin (\sigma \circ \gamma )=k-(1/\tau )u \end{aligned}$$
a.e. on \(r_1(I)\). By Theorem 6.5 there exists \(\lambda \in \mathbb {R}\) such that \(k+\rho \sin \sigma +\lambda =0\) \(\mathscr {H}^1\)-a.e. on M. So
$$\begin{aligned} u^\prime =-\rho (\tau )u-\lambda -(1/\tau )u =-(1/\tau +\rho (\tau ))u-\lambda \end{aligned}$$
a.e. on \(r_1(I)\). The result follows. \(\square \)

Lemma 6.7

Suppose that E is a bounded open set in \(\mathbb {R}^2\) with \(C^1\) boundary M and that \(E{\setminus }\{0\}=E^{sc}\). Then
  1. (i)

    \(\theta _2\in C^1(\varOmega )\);

     
  2. (ii)

    \(\theta _2^\prime =-\frac{1}{\tau }\frac{u}{\sqrt{1-u^2}}\) on \(\varOmega \).

     

Proof

Let \(\tau \in \varOmega \) and x a point in the open upper half-plane such that \(x\in M_\tau \). There exists a \(C^1\) parametrisation \(\gamma _1:I\rightarrow M\) of M in a neighbourhood of x with \(\gamma _1(0)=x\) as above. By shrinking the open interval I if necessary we may assume that \(r_1:I\rightarrow r_1(I)\) is a diffeomorphism and that \(r_1(I)\subset \subset \varOmega \). It then holds that
$$\begin{aligned} \theta _2=\theta _1\circ r_1^{-1}\text { and }\sigma \circ \gamma =\sigma _1\circ r_1^{-1} \end{aligned}$$
on \(r_1(I)\) by choosing an appropriate branch of \(\theta _1\). It follows that \(\theta _2\in C^1(\varOmega )\). By the chain-rule, (2.10) and (2.9),
$$\begin{aligned} \theta _2^\prime= & {} \frac{\dot{\theta }_1}{\dot{r}_1}\circ r_1^{-1} =\left( \frac{1}{r_1}\tan \sigma _1\right) \circ r_1^{-1}\\= & {} (1/\tau )\tan (\sigma \circ \gamma ) \end{aligned}$$
on \(r_1(I)\). By Lemma 5.4, \(\cos (\sigma \circ \gamma )=-\sqrt{1-u^2}\) on \(\varOmega \). This entails (ii). \(\square \)

7 Convexity

Lemma 7.1

Let E be a bounded open set in \(\mathbb {R}^2\) with \(C^{1,1}\) boundary M and assume that \(E{\setminus }\{0\}=E^{sc}\). Put \(d:=\sup \{|x|:x\in M\}>0\) and \(b:=(d,0)\). Let \(\gamma _1:I\rightarrow M\) be a \(C^{1,1}\) parametrisation of M near b with \(\gamma _1(0)=b\). Then
$$\begin{aligned} \lim _{\delta \downarrow 0}\Big \{\mathrm {ess\,sup}_{[-\delta ,\delta ]}k_1\Big \}\ge 1/d. \end{aligned}$$

Proof

For \(s\in I\),
$$\begin{aligned} \gamma _1(s)=de_1+se_2+\int _0^s\Big \{\dot{\gamma }_1(u)-\dot{\gamma }_1(0)\Big \}\,du \end{aligned}$$
and
$$\begin{aligned} \dot{\gamma }_1(u)-\dot{\gamma }_1(0)=\int _0^u k_1 n_1\,dv \end{aligned}$$
by (2.6). By the Fubini–Tonelli Theorem,
$$\begin{aligned} \gamma _1(s)=de_1+se_2+\int _0^s(s-u)k_1(u)n_1(u)\,du =de_1+se_2+R(s) \end{aligned}$$
for \(s\in I\). Assume for a contradiction that
$$\begin{aligned} \lim _{\delta \downarrow 0}\Big \{\mathrm {ess\,sup}_{[-\delta ,\delta ]}k_1\Big \}<l< 1/d \end{aligned}$$
for some \(l\in \mathbb {R}\). Then we can find \(\delta >0\) such that \(k_1<l\) a.e. on \([-\delta ,\delta ]\). So
$$\begin{aligned} \langle R(s),e_1\rangle&=\int _0^s(s-u)k_1(u)\langle n_1(u),e_1\rangle \,du >-(1/2)s^2l(1+o(1)) \end{aligned}$$
as \(s\downarrow 0\) and
$$\begin{aligned} r_1(s)^2-d^2&=2d\langle R(s),e_1\rangle +s^2+o(s^2)\nonumber \\&>-dls^2(1+o(1))+s^2+o(s^2) \end{aligned}$$
as \(s\downarrow 0\). Alternatively,
$$\begin{aligned} \frac{r_1(s)^2-d^2}{s^2}&>1-dl+o(1). \end{aligned}$$
As \(1-dl>0\) we can find \(s\in I\) with \(r_1(s)>d\), contradicting the definition of d. \(\square \)

Lemma 7.2

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a bounded minimiser of (1.2). Assume that E is open with \(C^{1,1}\) boundary M and that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(M{\setminus }\overline{\varLambda _1}\ne \emptyset \). Then \(\lambda \le -1/d-\rho _-(d)<0\) with \(\lambda \) as in Theorem 6.5.

Proof

We write M as the disjoint union \(M=(M{\setminus }\overline{\varLambda _1})\cup \overline{\varLambda _1}\). Let b be as above. Suppose that \(b\in \overline{\varLambda _1}\). Then \(b\in \varLambda _1\); in fact, \(b\in \varLambda _1^-\). By Theorem 6.5, \(\lambda \le -\rho _--k\) at b. By Lemma 7.1, \(\lambda \le -1/d-\rho _-(d)\) upon considering an appropriate sequence in M converging to b. Now suppose that b lies in the open set \(M{\setminus }\overline{\varLambda _1}\) in M. Let \(\gamma _1:I\rightarrow M\) be a \(C^{1,1}\) parametrisation of M near b with \(\gamma _1(I)\subset M{\setminus }\overline{\varLambda _1}\). By Theorem 6.5, \(k_1+\rho (r_1)\sin \sigma _1+\lambda =0\) a.e. on I. Now \(\sin \sigma _1(s)\rightarrow 1\) as \(s\rightarrow 0\). In light of Lemma 7.1, \(1/d+\rho (d-)+\lambda \le 0\) and \(\lambda \le -1/d-\rho _-(d)\). \(\square \)

Theorem 7.3

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a bounded minimiser of (1.2). Assume that E is open with \(C^{1,1}\) boundary M and that \(E{\setminus }\{0\}=E^{sc}\). Suppose that \(M{\setminus }\overline{\varLambda _1}\ne \emptyset \). Then E is convex.

Proof

The proof runs along similar lines as [22] Theorem 6.5. By Theorem 6.5, \(k+\rho \sin \sigma +\lambda =0\) \(\mathscr {H}^1\)-a.e. on \(M{\setminus }\overline{\varLambda _1}\). By Lemma 7.2,
$$\begin{aligned} 0\le k+\rho _-(d)+\lambda \le k-1/d \end{aligned}$$
and \(k\ge 1/d\) \(\mathscr {H}^1\)-a.e. on \(M{\setminus }\overline{\varLambda _1}\). On \(\varLambda _1^+,k\ge \rho _--\lambda \ge \rho _-+\rho _-(d)+1/d>0\); on the other hand, \(k<0\) on \(\varLambda _1^+\). So in fact \(\varLambda _1^+=\emptyset \). If \(b\in \varLambda _1^-\) then \(k=1/d\). On \(\varLambda _1^-\cap B(0,d),k\ge -\rho _+-\lambda \ge -\rho _++\rho _-(d)+1/d\ge 1/d\). Therefore \(k\ge 1/d>0\) \(\mathscr {H}^1\)-a.e. on M. The set E is then convex by a modification of [26] Theorem 1.8 and Proposition 1.4. It is sufficient that the function f (here \(\alpha _1\)) in the proof of the former theorem is non-decreasing. \(\square \)

8 A reverse Hermite–Hadamard inequality

Let \(0\le a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let h be a primitive of \(\rho \) on [ab] so that \(h\in C^{0,1}([a,b])\) and introduce the functions
$$\begin{aligned}&\mathtt {f}:[a,b]\rightarrow \mathbb {R};x\mapsto e^{h(x)}; \end{aligned}$$
(8.1)
$$\begin{aligned}&g:[a,b]\rightarrow \mathbb {R};x\mapsto x\mathtt {f}(x). \end{aligned}$$
(8.2)
Then
$$\begin{aligned} g^\prime =(1/x+\rho )g=\mathtt {f}+g\rho \end{aligned}$$
(8.3)
a.e. on (ab). Define
$$\begin{aligned} m=m(\rho ,a,b):=\frac{g(b)-g(a)}{\int _a^b g\,dt}. \end{aligned}$$
(8.4)
If \(\rho \) takes the constant value \(\mathbb {R}\ni \lambda \ge 0\) on [ab] we use the notation \(m(\lambda ,a,b)\) and we write \(m_0=m(0,a,b)\). A computation gives
$$\begin{aligned} m_0=m(0,a,b)=A(a,b)^{-1} \end{aligned}$$
(8.5)
where \(A(a,b):=(a+b)/2\) stands for the arithmetic mean of a and b.

Lemma 8.1

Let \(0\le a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Then \(m_0\le m\).

Proof

Note that g is convex on [ab] as can be seen from (8.3). By the Hermite-Hadamard inequality (cf. [15, 17]),
$$\begin{aligned} \frac{1}{b-a}\int _a^b g\,dt&\le \frac{g(a)+g(b)}{2}. \end{aligned}$$
(8.6)
The inequality \((b-a)(g(a)+g(b))\le (a+b)(g(b)-g(a))\) entails
$$\begin{aligned} \int _a^b g\,dt\le \frac{a+b}{2}(g(b)-g(a)) \end{aligned}$$
and the result follows on rearrangement. \(\square \)

Lemma 8.2

Let \(0\le a<b<+\infty \) and \(\lambda >0\). Then \(m(\lambda ,a,b)<\lambda +A(a,b)^{-1}\).

Proof

First suppose that \(\lambda =1\) and take \(h:[a,b]\rightarrow \mathbb {R};t\mapsto t\). In this case,
$$\begin{aligned} \int _a^b g\,dt=\int _a^b te^t\,dt=(b-1)e^b-(a-1)e^a \end{aligned}$$
and
$$\begin{aligned} m(1,a,b)=\frac{be^b-ae^a}{(b-1)e^b-(a-1)e^a}. \end{aligned}$$
The inequality in the statement is equivalent to
$$\begin{aligned} (a+b)(be^b-ae^a)<((b-1)e^b-(a-1)e^a)(2+a+b) \end{aligned}$$
which in turn is equivalent to the statement \(\tanh [(b-a)/2]<(b-a)/2\) which holds for any \(b>a\).
For \(\lambda >0\) take \(h:[a,b]\rightarrow \mathbb {R};t\mapsto \lambda t\). Substitution gives
$$\begin{aligned} \int _a^b g\,dt= & {} (1/\lambda )^2[(\lambda b-1)e^{\lambda b}-(\lambda a-1)e^{\lambda a}] \text { and } \\ g(b)-g(a)= & {} (1/\lambda )[\lambda be^{\lambda b}-\lambda ae^{\lambda a}] \end{aligned}$$
so from above
$$\begin{aligned} m(\lambda ,a,b)=\lambda m(1,\lambda a,\lambda b) <\lambda \Big \{1+A(\lambda a,\lambda b)^{-1}\Big \} =\lambda +A(a,b)^{-1}. \end{aligned}$$
\(\square \)

Theorem 8.3

Let \(0\le a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Then
  1. (i)

    \(m(\rho ,a,b)\le \rho (b-)+A(a,b)^{-1}\);

     
  2. (ii)

    equality holds if and only if \(\rho \equiv 0\) on [ab).

     

Proof

(i) Define \(h:=\int _a^\cdot \rho \,d\tau \) on [ab] so that \(h^\prime =\rho \) a.e. on (ab). Define \(h_1:[a,b]\rightarrow \mathbb {R};t\mapsto h(b)-\rho (b-)(b-t)\). Then \(h_1(b)=h(b),h_1^\prime =\rho (b-)\ge \rho =h^\prime \) a.e. on (ab) and hence \(h\ge h_1\) on [ab]. We derive
$$\begin{aligned} \int _a^b g\,dt=\int _a^b te^{h(t)}\,dt\ge \int _a^b te^{h_1(t)}\,dt=\int _a^b g_1\,dt \end{aligned}$$
and
$$\begin{aligned} g(b)-g(a)= & {} be^{h(b)}-ae^{h(a)}=be^{h_1(b)}-ae^{h(a)} \\\le & {} be^{h_1(b)}-ae^{h_1(a)}=g_1(b)-g_1(a) \end{aligned}$$
with obvious notation. This entails that \(m(\rho ,a,b)\le m(\rho (b-),a,b)\) and the result follows with the help of Lemma 8.2.

(ii) Suppose that \(\rho \not \equiv 0\) on [ab). If \(\rho \) is constant on [ab] the assertion follows from Lemma 8.2. Assume then that \(\rho \) is not constant on [ab). Then \(h\not \equiv h_1\) on [ab] in the above notation and \(\int _a^b te^{h(t)}\,dt>\int _a^b te^{h_1(t)}\,dt\) which entails strict inequality in (i). \(\square \)

With the above notation define
$$\begin{aligned} \hat{m}=\hat{m}(\rho ,a,b):=\frac{g(a)+g(b)}{\int _a^b g\,dt}. \end{aligned}$$
(8.7)
A computation gives
$$\begin{aligned} \hat{m}_0:=\hat{m}(0,a,b)=\frac{2}{b-a}. \end{aligned}$$
(8.8)

Lemma 8.4

Let \(0\le a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Then \(\hat{m}\ge \hat{m}_0\).

Proof

This follows by the Hermite-Hadamard inequality (8.6). \(\square \)

We prove a reverse Hermite-Hadamard inequality.

Theorem 8.5

Let \(0\le a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Then
  1. (i)

    \((b-a)\hat{m}(\rho ,a,b)\le 2+a\rho (a+)+b\rho (b-)\);

     
  2. (ii)

    equality holds if and only if \(\rho \equiv 0\) on [ab).

     
This last inequality can be written in the form
$$\begin{aligned} \frac{g(a)+g(b)}{2+a\rho (a+)+b\rho (b-)}&\le \frac{1}{b-a}\int _a^b g\,dt;\nonumber \end{aligned}$$
comparing with (8.6) justifies naming this a reverse Hermite-Hadamard inequality.

Proof

(i) We assume in the first instance that \(\rho \in C^1((a,b))\). We prove the above result in the form
$$\begin{aligned} \int _a^b g\,dt\ge (b-a)\frac{g(a)+g(b)}{2+a\rho (a)+b\rho (b)}. \end{aligned}$$
(8.9)
Put
$$\begin{aligned} w:=\frac{(t-a)(g(a)+g)}{2+a\rho (a)+t\rho } \end{aligned}$$
for \(t\in [a,b]\) so that
$$\begin{aligned} \int _a^b w^\prime \,dt=(b-a)\frac{g(a)+g(b)}{2+a\rho (a)+b\rho (b)}. \end{aligned}$$
Then using (8.3),
$$\begin{aligned} w^\prime&=\frac{(g(a)+g+(t-a)g^\prime )(2+a\rho (a)+t\rho )-(t-a)(g(a)+g)(\rho +t\rho ^\prime )}{(2+a\rho (a)+t\rho )^2}\nonumber \\&=\frac{(g(a)-ag^\prime +(2+t\rho )g)(2+a\rho (a)+t\rho )-(t-a)(g(a)+g)(\rho +t\rho ^\prime )}{(2+a\rho (a)+t\rho )^2}\nonumber \\&=\frac{(2+t\rho )(2+a\rho (a)+t\rho )}{(2+a\rho (a)+t\rho )^2}g\nonumber \\&\quad +\,\frac{(g(a)-ag^\prime )(2+a\rho (a)+t\rho )-(t-a)(g(a)+g)(\rho +t\rho ^\prime )}{(2+a\rho (a)+t\rho )^2}\nonumber \\&\le g -\frac{2g(a)}{(2+a\rho (a)+b\rho (b))^2}(t-a)\rho \nonumber \\&\le g \end{aligned}$$
(8.10)
on (ab) as
$$\begin{aligned} g(a)-ag^\prime =a(\mathtt {f}(a)-(1/t+\rho )g)=a(\mathtt {f}(a)-\mathtt {f}-\rho g)\le 0. \end{aligned}$$
An integration over [ab] gives the result.
Let us now assume that \(\rho \ge 0\) is a non-decreasing bounded function on [ab]. Extend \(\rho \) to \(\mathbb {R}\) via
$$\begin{aligned} \widetilde{\rho }(t):= \left\{ \begin{array}{lcl} \rho (a+) &{} \text { for } &{} t\in (-\infty ,a];\\ \rho (t) &{} \text { for } &{} t\in (a,b];\\ \rho (b-) &{} \text { for } &{} t\in (b,+\infty );\\ \end{array} \right. \end{aligned}$$
for \(t\in \mathbb {R}\). Let \((\psi _\varepsilon )_{\varepsilon >0}\) be a family of mollifiers (see e.g. [1] 2.1) and set \(\widetilde{\rho }_\varepsilon :=\widetilde{\rho }\star \psi _\varepsilon \) on \(\mathbb {R}\) for each \(\varepsilon >0\). Then \(\widetilde{\rho }_\varepsilon \in C^\infty (\mathbb {R})\) and is non-decreasing on \(\mathbb {R}\) for each \(\varepsilon >0\). Put \(\rho _\varepsilon :=\widetilde{\rho }_\varepsilon \mid _{[a,b]}\) for each \(\varepsilon >0\). Then \((\rho _\varepsilon )_{\varepsilon >0}\) converges to \(\rho \) in \(L^1((a,b))\) by [1] 2.1 for example. Note that \(h_{\varepsilon }:=\int _a^\cdot \rho _\varepsilon \,dt\rightarrow h\) pointwise on [ab] as \(\varepsilon \downarrow 0\) and that \((h_\varepsilon )\) is uniformly bounded on [ab]. Moreover, \(\rho _\varepsilon (a)\rightarrow \rho (a+)\) and \(\rho _\varepsilon (b)\rightarrow \rho (b-)\) as \(\varepsilon \downarrow 0\). By the above result,
$$\begin{aligned} (b-a)\hat{m}(\rho _{\varepsilon },a,b)\le 2+a\rho _{\varepsilon }(a)+b\rho _{\varepsilon }(b) \end{aligned}$$
for each \(\varepsilon >0\). The inequality follows on taking the limit \(\varepsilon \downarrow 0\) with the help of the dominated convergence theorem.
(ii) We now consider the equality case. We claim that
$$\begin{aligned}&(b-a)\frac{g(a)+g(b)}{2+a\rho (a+)+b\rho (b-)}\le \int _a^b g\,dt\nonumber \\&\quad -\frac{2g(a)}{(2+a\rho (a+)+b\rho (b-))^2}\int _a^b(t-a)\rho \,dt; \end{aligned}$$
(8.11)
this entails the equality condition in (ii). First suppose that \(\rho \in C^1((a,b))\). In this case the inequality in (8.10) implies (8.11) upon integration. Now suppose that \(\rho \ge 0\) is a non-decreasing bounded function on [ab]. Then (8.11) holds with \(\rho _\varepsilon \) in place of \(\rho \) for each \(\varepsilon >0\). The inequality for \(\rho \) follows by the dominated convergence theorem. \(\square \)

9 Comparison theorems for first-order differential equations

Let \(\mathscr {L}\) stand for the collection of Lebesgue measurable sets in \([0,+\infty )\). Define a measure \(\mu \) on \(([0,+\infty ),\mathscr {L})\) by \(\mu (dx):=(1/x)\,dx\). Let \(0\le a<b<+\infty \). Suppose that \(u:[a,b]\rightarrow \mathbb {R}\) is an \(\mathscr {L}^1\)-measurable function with the property that
$$\begin{aligned} \mu (\{u>t\})&<+\infty \text { for each }t>0. \end{aligned}$$
(9.1)
The distribution function \(\mu _u:(0,+\infty )\rightarrow [0,+\infty )\) of u with respect to \(\mu \) is given by
$$\begin{aligned} \mu _u(t):=\mu (\{u>t\})\text { for }t>0. \end{aligned}$$
Note that \(\mu _u\) is right-continuous and non-increasing on \((0,\infty )\) and \(\mu _u(t)\rightarrow 0\) as \(t\rightarrow \infty \).
Let u be a Lipschitz function on [ab]. Define
$$\begin{aligned} Z_1:=\{u\text { differentiable and }u^\prime =0\}, Z_2:=\{u\text { not differentiable}\}\text { and }Z:=Z_1\cup Z_2. \end{aligned}$$
By [1] Lemma 2.96, \(Z\cap \{u=t\}=\emptyset \) for \(\mathscr {L}^1\)-a.e. \(t\in \mathbb {R}\) and hence \(N:=u(Z)\subset \mathbb {R}\) is \(\mathscr {L}^1\)-negligible. We make use of the coarea formula ( [1] Theorem 2.93 and (2.74)),
$$\begin{aligned} \int _{[a,b]}\phi |u^\prime |\,dx=\int _{-\infty }^\infty \int _{\{u=t\}}\phi \,d\mathscr {H}^{0}\,dt \end{aligned}$$
(9.2)
for any \(\mathscr {L}^1\)-measurable function \(\phi :[a,b]\rightarrow [0,\infty ]\).

Lemma 9.1

Let \(0\le a<b<+\infty \) and u a Lipschitz function on [ab]. Then
  1. (i)

    \(\mu _u\in \mathrm {BV}_{\mathrm {loc}}((0,+\infty ))\);

     
  2. (ii)

    \(D\mu _u=-u_\sharp \mu \);

     
  3. (iii)

    \(D\mu _u^a=D\mu _u|\!\_((0,+\infty ){\setminus } N)\);

     
  4. (iv)

    \(D\mu _u^s=D\mu _u|\!\_N\);

     
  5. (v)

    \(A:=\Big \{t\in (0,+\infty ):\mathscr {L}^1(Z\cap \{u=t\})>0\Big \}\) is the set of atoms of \(D\mu _u\) and \(D\mu _u^j=D\mu _u|\!\_A\);

     
  6. (vi)
    \(\mu _u\) is differentiable \(\mathscr {L}^1\)-a.e. on \((0,+\infty )\) with derivative given by
    $$\begin{aligned} \mu _u^\prime (t)=-\int _{\{u=t\}{\setminus } Z}\frac{1}{|u^\prime |}\,\frac{d\mathscr {H}^{0}}{\tau } \end{aligned}$$
    for \(\mathscr {L}^1\)-a.e. \(t\in (0,+\infty )\);
     
  7. (vii)

    \(\mathrm {Ran}(u)\cap [0,+\infty )=\mathrm {supp}(D\mu _u)\).

     

The notation above \(D\mu _u^a,D\mu _u^s,D\mu _u^j\) stands for the absolutely continuous resp. singular resp. jump part of the measure \(D\mu _u\) (see [1] 3.2 for example).

Proof

For any \(\varphi \in C^\infty _c((0,+\infty ))\) with \(\mathrm {supp}[\varphi ]\subset (\tau ,+\infty )\) for some \(\tau >0\),
$$\begin{aligned} \int _0^\infty \mu _u\varphi ^\prime \,dt&=\int _{[a,b]}\varphi \circ u\,d\mu \nonumber \\&=\int _{[a,b]}\chi _{\{u>\tau \}}\varphi \circ u\,d\mu \end{aligned}$$
(9.3)
by Fubini’s theorem; so \(\mu _u\in \mathrm {BV}_{\mathrm {loc}}((0,+\infty ))\) and \(D\mu _u\) is the push-forward of \(\mu \) under \(u,D\mu _u=-u_\sharp \mu \) (cf. [1] 1.70). By (9.2),
$$\begin{aligned} D\mu _u|\!\_((0,+\infty ){\setminus } N)(A)= & {} -\mu (\{u\in A\}{\setminus } Z)\\= & {} -\int _A\int _{\{u=t\}{\setminus } Z}\frac{1}{|u^\prime |}\,\frac{d\mathcal {H}^{0}}{\tau }\,dt \end{aligned}$$
for any \(\mathscr {L}^1\)-measurable set A in \((0,+\infty )\). In light of the above, we may identify \(D\mu _u^a=D\mu _u|\!\_((0,+\infty ){\setminus } N)\) and \(D\mu _u^s=D\mu _{u}|\!\_N\). The set of atoms of \(D\mu _{u}\) is defined by \(A:=\{t\in (0,+\infty ):D\mu _u(\{t\})\ne 0\}\). For \(t>0\),
$$\begin{aligned} D\mu _u(\{t\})= & {} D\mu _u^s(\{t\})=(D\mu _u|\!\_N)((\{t\})\\= & {} -u_\sharp \mu (N\cap \{t\})=-\mu (Z\cap \{u=t\}) \end{aligned}$$
and this entails (v). The monotone function \(\mu _u\) is a good representative within its equivalence class and is differentiable \(\mathscr {L}^1\)-a.e. on \((0,+\infty )\) with derivative given by the density of \(D\mu _{u}\) with respect to \(\mathscr {L}^1\) by [1] Theorem 3.28. Item (vi) follows from (9.2) and (iii). Item (vii) follows from (ii).\(\square \)
Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \(\eta \in \{\pm 1\}^2\). We study solutions to the first-order linear ordinary differential equation
$$\begin{aligned} u^\prime +(1/x+\rho )u+\lambda =0\text { a.e. on }(a,b)\text { with }u(a)=\eta _1\text { and }u(b)=\eta _2 \end{aligned}$$
(9.4)
where \(u\in C^{0,1}([a,b])\) and \(\lambda \in \mathbb {R}\). In case \(\rho \equiv 0\) on [ab] we use the notation \(u_0\).

Lemma 9.2

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \(\eta \in \{\pm 1\}^2\). Then
  1. (i)

    there exists a solution \((u,\lambda )\) of (9.4) with \(u\in C^{0,1}([a,b])\) and \(\lambda =\lambda _\eta \in \mathbb {R}\);

     
  2. (ii)

    the pair \((u,\lambda )\) in (i) is unique;

     
  3. (iii)
    \(\lambda _\eta \) is given by
    $$\begin{aligned} -\lambda _{(1,1)}=\lambda _{(-1,-1)}=m;\,\lambda _{(1,-1)}=-\lambda _{(-1,1)}=\hat{m}; \end{aligned}$$
     
  4. (iv)

    if \(\eta =(1,1)\) or \(\eta =(-1,-1)\) then u is uniformly bounded away from zero on [ab].

     

Proof

(i) For \(\eta =(1,1)\) define \(u:[a,b]\rightarrow \mathbb {R}\) by
$$\begin{aligned} u(t):=\frac{m\int _a^t g\,ds+g(a)}{g(t)}\text { for }t\in [a,b] \end{aligned}$$
(9.5)
with m as in (8.4). Then \(u\in C^{0,1}([a,b])\) and satisfies (9.4) with \(\lambda =-m\). For \(\eta =(1,-1)\) set \(u=(-\hat{m}\int _a^\cdot g\,ds+g(a))/g\) with \(\lambda =\hat{m}\). The cases \(\eta =(-1,-1)\) and \(\eta =(-1,1)\) can be dealt with using linearity. (ii) We consider the case \(\eta =(1,1)\). Suppose that \((u_1,\lambda _1)\) resp. \((u_2,\lambda _2)\) solve (9.4). By linearity \(u:=u_1-u_2\) solves
$$\begin{aligned} u^\prime +(1/x+\rho )u+\lambda =0\text { a.e. on }(a,b) \text { with }u(a)=u(b)=0 \end{aligned}$$
where \(\lambda =\lambda _1-\lambda _2\). An integration gives that \(u=(-\lambda \int _a^\cdot g\,ds+c)/g\) for some constant \(c\in \mathbb {R}\) and the boundary conditions entail that \(\lambda =c=0\). The other cases are similar. (iii) follows as in (i). (iv) If \(\eta =(1,1)\) then \(u>0\) on [ab] from (9.5) as \(m>0\). \(\square \)

The boundary condition \(\eta _1\eta _2=-1\).

Lemma 9.3

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \((u,\lambda )\) solve (9.4) with \(\eta =(1,-1)\). Then
  1. (i)

    there exists a unique \(c\in (a,b)\) with \(u(c)=0\);

     
  2. (ii)

    \(u^\prime <0\) a.e. on [ac] and u is strictly decreasing on [ac];

     
  3. (iii)

    \(D\mu _u^s=0\).

     

Proof

(i) We first observe that \(u^\prime \le -\hat{m}<0\) a.e. on \(\{u\ge 0\}\) in view of (9.4). Suppose \(u(c_1)=u(c_2)=0\) for some \(c_1,c_2\in (a,b)\) with \(c_1<c_2\). We may assume that \(u\ge 0\) on \([c_1,c_2]\). This contradicts the above observation. Item (ii) is plain. For any \(\mathscr {L}^1\)-measurable set B in \((0,+\infty ),D\mu _u^s(B)=\mu (\{u\in B\}\cap Z)=0\) using Lemma 9.1 and (ii). \(\square \)

Lemma 9.4

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \((u,\lambda )\) solve (9.4) with \(\eta =(1,-1)\). Assume that
  1. (a)

    u is differentiable at both a and b and that (9.4) holds there;

     
  2. (b)

    \(u^\prime (a)<0\) and \(u^\prime (b)<0\);

     
  3. (c)

    \(\rho \) is differentiable at a and b.

     
Put \(v:=-u\). Then
  1. (i)

    \(\int _{\{v=1\}{\setminus } Z_v}\frac{1}{|v^\prime |}\frac{d\mathscr {H}^0}{\tau } \ge \int _{\{u=1\}{\setminus } Z_u}\frac{1}{|u^\prime |}\frac{d\mathscr {H}^0}{\tau }\);

     
  2. (ii)

    equality holds if and only if \(\rho \equiv 0\) on [ab).

     

Proof

First, \(\{u=1\}=\{a\}\) by Lemma 9.3. Further \(0<-au^\prime (a)=1+a[\hat{m}+\rho (a)]\) from (9.4). On the other hand \(\{v=1\}\supset \{b\}\) and \(0<bv^\prime (b)=-1+b[\hat{m}-\rho (b)]\). Thus
$$\begin{aligned}&\int _{\{v=1\}{\setminus } Z_v}\frac{1}{|v^\prime |}\frac{d\mathscr {H}^0}{\tau }- \int _{\{u=1\}{\setminus } Z_u}\frac{1}{|u^\prime |}\frac{d\mathscr {H}^0}{\tau }\nonumber \\&\quad \ge \frac{1}{-1+b[\hat{m}-\rho (b)]}-\frac{1}{1+a[\hat{m}+\rho (a)]}. \end{aligned}$$
By Theorem 8.5, \(0\le 2+(a-b)\hat{m}+a\rho (a)+b\rho (b)\), noting that \(\rho (a)=\rho (a+)\) in virtue of (c) and similarly at b. A rearrangement leads to the inequality. The equality assertion follows from Theorem 8.5. \(\square \)

Theorem 9.5

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Suppose that \((u,\lambda )\) solves (9.4) with \(\eta =(1,-1)\) and set \(v:=-u\). Assume that \(u>-1\) on [ab). Then
  1. (i)

    \(-\mu _v^\prime \ge -\mu _u^\prime \) for \(\mathscr {L}^1\)-a.e. \(t\in (0,1)\);

     
  2. (ii)

    if \(\rho \not \equiv 0\) on [ab) then there exists \(t_0\in (0,1)\) such that \(-\mu _v^\prime >-\mu _u^\prime \) for \(\mathscr {L}^1\)-a.e. \(t\in (t_0,1)\);

     
  3. (iii)
    for \(t\in [-1,1]\),
    $$\begin{aligned} \mu _{u_0}(t)=\log \Big \{\frac{-(b-a)t+\sqrt{(b-a)^2t^2+4ab}}{2a}\Big \} \end{aligned}$$
    and \(\mu _{v_0}=\mu _{u_0}\) on \([-1,1]\);
     
in obvious notation.

Proof

(i) The set
$$\begin{aligned} Y_u:=Z_{2,u}\cup \Big (\{u^\prime +(1/x+\rho )u+\lambda \ne 0\}{\setminus } Z_{2,u}\Big )\cup \{\rho \text { not differentiable}\}\subset [a,b] \end{aligned}$$
(in obvious notation) is a null set in [ab] and likewise for \(Y_v\). By [1] Lemma 2.95 and Lemma 2.96, \(\{u=t\}\cap (Y_u\cup Z_{1,u})=\emptyset \) for a.e. \(t\in (0,1)\) and likewise for the function v. Let \(t\in (0,1)\) and assume that \(\{u=t\}\cap (Y_u\cup Z_{1,u})=\emptyset \) and \(\{v=t\}\cap (Y_v\cup Z_{1,v})=\emptyset \). Put \(c:=\max \{u\ge t\}\). Then \(c\in (a,b),\{u>t\}=[a,c)\) by Lemma 9.3 and u is differentiable at c with \(u^\prime (c)<0\). Put \(d:=\max \{v\le t\}=\max \{u\ge -t\}\). As u is continuous on [ab] it holds that \(a<c<d<b\). Moreover, \(u^\prime (d)<0\) as \(v(d)=t\) and \(d\not \in Z_v\). Put \(\widetilde{u}:=u/t\) and \(\widetilde{v}:=v/t\) on [cd]. Then
$$\begin{aligned} \widetilde{u}^\prime +(1/\tau +\rho )\widetilde{u}+\hat{m}/t&=0\text { a.e. on }(c,d)\text { and }\widetilde{u}(c)=-\widetilde{u}(d)=1;\nonumber \\ \widetilde{v}^\prime +(1/\tau +\rho )\widetilde{v}-\hat{m}/t&=0\text { a.e. on }(c,d)\text { and }-\widetilde{v}(c)=\widetilde{v}(d)=1.\nonumber \end{aligned}$$
By Lemma 9.4,
$$\begin{aligned} \int _{\{v=t\}{\setminus } Z_v}\frac{1}{|v^\prime |}\frac{d\mathscr {H}^0}{\tau }&\ge \int _{[c,d]\cap \{v=t\}{\setminus } Z_v}\frac{1}{|v^\prime |}\frac{d\mathscr {H}^0}{\tau }\\&=(1/t)\int _{[c,d]\cap \{\widetilde{v}=1\}{\setminus } Z_v}\frac{1}{|\widetilde{v}^\prime |}\frac{d\mathscr {H}^0}{\tau }\nonumber \\&\ge (1/t)\int _{[c,d]\cap \{\widetilde{u}=1\}{\setminus } Z_u}\frac{1}{|\widetilde{u}^\prime |}\frac{d\mathscr {H}^0}{\tau }\nonumber \\&=\int _{\{u=t\}{\setminus } Z_u}\frac{1}{|u^\prime |}\frac{d\mathscr {H}^0}{\tau }.\nonumber \end{aligned}$$
By Lemma 9.1,
$$\begin{aligned} -\mu _u^\prime (t)=\int _{\{u=t\}{\setminus } Z_u}\frac{1}{|u^\prime |}\frac{d\mathscr {H}^0}{\tau } \end{aligned}$$
for \(\mathscr {L}^1\)-a.e. \(t\in (0,1)\) and a similar formula holds for v. The assertion in (i) follows.
(ii) Assume that \(\rho \not \equiv 0\) on [ab). Put \(\alpha :=\inf \{\rho >0\}\in [a,b)\). Note that \(\max \{v\le t\}\rightarrow b\) as \(t\uparrow 1\) as \(v<1\) on [ab) by assumption. Choose \(t_0\in (0,1)\) such that \(\max \{v\le t_0\}>\alpha \). Then for \(t>t_0\),
$$\begin{aligned} a<\max \{u\ge t\}<\max \{u\ge -t_0\}=\max \{v\le t_0\}<\max \{v\le t\}<d; \end{aligned}$$
that is, the interval [cd] with cd as described above intersects \((\alpha ,b]\). So for \(\mathscr {L}^1\)-a.e. \(t\in (t_0,1)\),
$$\begin{aligned} \int _{\{v=t\}{\setminus } Z_v}\frac{1}{|v^\prime |}\frac{d\mathscr {H}^0}{\tau }>\int _{\{u=t\}{\setminus } Z_u}\frac{1}{|u^\prime |}\frac{d\mathscr {H}^0}{\tau }. \end{aligned}$$
by the equality condition in Lemma 9.4. The conclusion follows from the representation of \(\mu _u\) resp. \(\mu _v\) in Lemma 9.1.
(iii) A direct computation gives
$$\begin{aligned} u_0(\tau )=\frac{1}{b-a}\Big \{-\tau +\frac{ab}{\tau }\Big \} \end{aligned}$$
for \(\tau \in [a,b]\); \(u_0\) is strictly decreasing on its domain. This leads to the formula in (iii). A similar computation gives
$$\begin{aligned} \mu _{v_0}(t)=\log \Big \{\frac{2b}{(b-a)t+\sqrt{(b-a)^2t^2+4ab}}\Big \} \end{aligned}$$
for \(t\in [-1,1]\). Rationalising the denominator results in the stated equality. \(\square \)

Corollary 9.6

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Suppose that \((u,\lambda )\) solves (9.4) with \(\eta =(1,-1)\) and set \(v:=-u\). Assume that \(u>-1\) on [ab). Then
  1. (i)

    \(\mu _u(t)\le \mu _v(t)\) for each \(t\in (0,1)\);

     
  2. (ii)

    if \(\rho \not \equiv 0\) on [ab) then \(\mu _u(t)<\mu _v(t)\) for each \(t\in (0,1)\).

     

Proof

(i) By [1] Theorem 3.28 and Lemma 9.3,
$$\begin{aligned} \mu _u(t)= & {} \mu _u(t)-\mu _u(1)=-D\mu _u((t,1])\\= & {} -D\mu _u^a((t,1])-D\mu _u^s((t,1])\\= & {} -\int _{(t,1]}\mu _u^\prime \,ds \end{aligned}$$
for each \(t\in (0,1)\) as \(\mu _u(1)=0\). On the other hand,
$$\begin{aligned} \mu _v(t)= & {} \mu _v(1)+(\mu _v(t)-\mu _v(1)) =\mu _v(1)-D\mu _v((t,1])\\= & {} \mu _v(1)-\int _{(t,1]}\mu _v^\prime \,ds-D\mu _v^s((t,1]) \end{aligned}$$
for each \(t\in (0,1)\). The claim follows from Theorem 9.5 noting that \(D\mu _v^s((t,1])\le 0\) as can be seen from Lemma 9.1. Item (ii) follows from Theorem 9.5 (ii). \(\square \)

Corollary 9.7

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Suppose that \((u,\lambda )\) solves (9.4) with \(\eta =(1,-1)\). Assume that \(u>-1\) on [ab). Let \(\varphi \in C^1((-1,1))\) be an odd strictly increasing function with \(\varphi \in L^1((-1,1))\). Then
  1. (i)

    \(\int _{\{u>0\}}\varphi (u)\,d\mu <+\infty \);

     
  2. (ii)

    \(\int _a^b\varphi (u)\,d\mu \le 0\);

     
  3. (iii)

    equality holds in (ii) if and only if \(\rho \equiv 0\) on [ab).

     
In particular,
  1. (iv)

    \(\int _a^b\frac{u}{\sqrt{1-u^2}}\,d\mu \le 0\) with equality if and only if \(\rho \equiv 0\) on [ab).

     

Proof

(i) Put \(I:=\{1>u>0\}\). The function \(u:I\rightarrow (0,1)\) is \(C^{0,1}\) and \(u^\prime \le -\hat{m}\) a.e. on I by Lemma 9.3. It has \(C^{0,1}\) inverse \(v:(0,1)\rightarrow I,v^\prime =1/(u^\prime \circ v)\) and \(|v^\prime |\le 1/\hat{m}\) a.e. on (0, 1). By a change of variables,
$$\begin{aligned} \int _{\{u>0\}}\varphi (u)\,d\mu&=\int _0^1\varphi (v^\prime /v)\,dt \end{aligned}$$
from which the claim is apparent. (ii) The integral is well-defined because \(\varphi (u)^+=\varphi (u)\chi _{\{u>0\}}\in L^1((a,b),\mu )\) by (i). By Lemma 9.3 the set \(\{u=0\}\) consists of a singleton and has \(\mu \)-measure zero. So
$$\begin{aligned} \int _a^b\varphi (u)\,d\mu&=\int _{\{u>0\}}\varphi (u)\,d\mu +\int _{\{u<0\}}\varphi (u)\,d\mu \nonumber \\&=\int _{\{u>0\}}\varphi (u)\,d\mu -\int _{\{v>0\}}\varphi (v)\,d\mu \nonumber \end{aligned}$$
where \(v:=-u\) as \(\varphi \) is an odd function. We remark that in a similar way to (9.3),
$$\begin{aligned} \int _0^1\varphi ^\prime \mu _u\,dt= & {} \int _{\{u>0\}}\Big \{\varphi (u)-\varphi (0)\Big \}\\ d\mu= & {} \int _{\{u>0\}}\varphi (u)\,d\mu \end{aligned}$$
using oddness of \(\varphi \) and an analogous formula holds with v in place of u. Thus we may write
$$\begin{aligned} \int _a^b\varphi (u)\,d\mu&=\int _0^1\varphi ^\prime \mu _u\,dt-\int _0^1\varphi ^\prime \mu _v\,dt\\&=\int _0^1\varphi ^\prime \Big \{\mu _u-\mu _v\Big \}\,dt\le 0\nonumber \end{aligned}$$
by Corollary 9.6 as \(\varphi ^\prime >0\) on (0, 1). (iii) Suppose that \(\rho \not \equiv 0\) on [ab). Then strict inequality holds in the above by Corollary 9.6. If \(\rho \equiv 0\) on [ab) the equality follows from Theorem 9.5. (iv) follows from (ii) and (iii) with the particular choice \(\varphi :(-1,1)\rightarrow \mathbb {R};t\mapsto t/\sqrt{1-t^2}\). \(\square \)
The boundary condition \(\eta _1\eta _2=1\). Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. We study solutions of the auxilliary Riccati equation
$$\begin{aligned} w^\prime +\lambda w^2=(1/x+\rho )w\text { a.e. on }(a,b) \text { with }w(a)=w(b)=1; \end{aligned}$$
(9.6)
with \(w\in C^{0,1}([a,b])\) and \(\lambda \in \mathbb {R}\). If \(\rho \equiv 0\) on [ab] then we write \(w_0\) instead of w. Suppose \((u,\lambda )\) solves (9.4) with \(\eta =(1,1)\). Then \(u>0\) on [ab] by Lemma 9.2 and we may set \(w:=1/u\). Then \((w,-\lambda )\) satisfies (9.6).

Lemma 9.8

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Then
  1. (i)

    there exists a solution \((w,\lambda )\) of (9.6) with \(w\in C^{0,1}([a,b])\) and \(\lambda \in \mathbb {R}\);

     
  2. (ii)

    the pair \((w,\lambda )\) in (i) is unique;

     
  3. (iii)

    \(\lambda =m\).

     

Proof

(i) Define \(w:[a,b]\rightarrow \mathbb {R}\) by
$$\begin{aligned} w(t):=\frac{g(t)}{m\int _a^t g\,ds+g(a)}\text { for }t\in [a,b]. \end{aligned}$$
Then \(w\in C^{0,1}([a,b])\) and (wm) satisfies (9.6). (ii) We claim that \(w>0\) on [ab] for any solution \((w,\lambda )\) of (9.6). For otherwise, \(c:=\min \{w=0\}\in (a,b)\). Then \(u:=1/w\) on [ac) satisfies
$$\begin{aligned} u^\prime +\left( \frac{1}{\tau }+\rho \right) u-\lambda =0\text { a.e. on }(a,c) \text { and }u(a)=1, u(c-)=+\infty . \end{aligned}$$
Integrating, we obtain
$$\begin{aligned} gu-g(a)-\lambda \int _a^\cdot g\,dt=0\text { on }[a,c) \end{aligned}$$
and this entails the contradiction that \(u(c-)<+\infty \). We may now use the uniqueness statement in Lemma 9.2. (iii) follows from (ii) and the particular solution given in (i). \(\square \)
We introduce the mapping
$$\begin{aligned} \omega :(0,\infty )\times (0,\infty )\rightarrow \mathbb {R};(t,x)\mapsto -(2/t)\coth (x/2). \end{aligned}$$
For \(\xi >0\),
$$\begin{aligned} |\omega (t,x)-\omega (t,y)|\le \mathrm {cosech}^2[\xi /2](1/t)|x-y| \end{aligned}$$
(9.7)
for \((t,x),(t,y)\in (0,\infty )\times (\xi ,\infty )\) and \(\omega \) is locally Lipschitzian in x on \((0,\infty )\times (0,\infty )\) in the sense of [16] I.3. Let \(0<a<b<+\infty \) and set \(\lambda :=A/G>1\). Here, \(A=A(a,b)\) stands for the arithmetic mean of ab as introduced in the previous Section while \(G=G(a,b):=\sqrt{|ab|}\) stands for their geometric mean. We refer to the inital value problem
$$\begin{aligned} z^\prime =\omega (t,z)\text { on }(0,\lambda ) \text { and }z(1)=\mu ((a,b)). \end{aligned}$$
(9.8)
Define
$$\begin{aligned} z_0:(0,\lambda )\rightarrow \mathbb {R};t\mapsto 2\log \Big \{\frac{\lambda +\sqrt{\lambda ^2-t^2}}{t}\Big \}. \end{aligned}$$

Lemma 9.9

Let \(0<a<b<+\infty \). Then
  1. (i)

    \(w_0(\tau )=\frac{2A\tau }{G^2+\tau ^2}\) for \(\tau \in [a,b]\);

     
  2. (ii)

    \(\Vert w_0\Vert _\infty =\lambda \);

     
  3. (iii)

    \(\mu _{w_0}=z_0\) on \([1,\lambda )\);

     
  4. (iv)

    \(z_0\) satisfies (9.8) and this solution is unique;

     
  5. (v)

    \(\int _{\{w_0=1\}}\frac{1}{|w_0^\prime |}\,\frac{d\mathscr {H}^0}{\tau }=2\coth (\mu ((a,b))/2)\);

     
  6. (vi)

    \(\int _a^b\frac{1}{\sqrt{w_0^2-1}}\,\frac{dx}{x}=\pi \).

     

Proof

(i) follows as in the proof of Lemma 9.8 with \(g(t)=t\) while (ii) follows by calculus. (iii) follows by solving the quadratic equation \(t\tau ^2-2A\tau +G^2t=0\) for \(\tau \) with \(t\in (0,\lambda )\). Uniqueness in (iv) follows from [16] Theorem 3.1 as \(\omega \) is locally Lipschitzian with respect to x in \((0,\infty )\times (0,\infty )\). For (v) note that \(|aw_0^\prime (a)|=1-a/A\) and \(|bw_0^\prime (b)|=b/A-1\) and
$$\begin{aligned} 2\coth (\mu ((a,b))/2)=2(a+b)/(b-a). \end{aligned}$$
(vi) We may write
$$\begin{aligned} \int _a^b\frac{1}{\sqrt{w_0^2-1}}\,\frac{d\tau }{\tau }&=\int _a^b\frac{ab+\tau ^2}{\sqrt{(a+b)^2\tau ^2-(ab+\tau ^2)^2}} \frac{d\tau }{\tau }\\&=\int _a^b\frac{ab+\tau ^2}{\sqrt{(\tau ^2-a^2)(b^2-\tau ^2)}} \frac{d\tau }{\tau }.\nonumber \end{aligned}$$
The substitution \(s=\tau ^2\) followed by the Euler substitution (cf. [14] 2.251)
$$\begin{aligned} \sqrt{(s-a^2)(b^2-s)}=t(s-a^2) \end{aligned}$$
gives
$$\begin{aligned} \int _a^b\frac{1}{\sqrt{w_0^2-1}}\,\frac{d\tau }{\tau } =\int _0^\infty \frac{1}{1+t^2}+\frac{ab}{b^2+a^2t^2}\,dt=\pi . \end{aligned}$$
\(\square \)

Lemma 9.10

Let \(0<a<b<+\infty \). Then
  1. (i)

    for \(y>a\) the function \(x\mapsto \frac{by-ax}{(y-a)(b-x)}\) is strictly increasing on \((-\infty ,b]\);

     
  2. (ii)

    the function \(y\mapsto \frac{(b-a)y}{(y-a)(b-y)}\) is strictly increasing on [Gb];

     
  3. (iii)

    for \(x<b\) the function \(y\mapsto \frac{by-ax}{(y-a)(b-x)}\) is strictly decreasing on \([a,+\infty )\)

     

Proof

The proof is an exercise in calculus. \(\square \)

Lemma 9.11

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \((w,\lambda )\) solve (9.6). Assume
  1. (i)

    w is differentiable at both a and b and that (9.6) holds there;

     
  2. (ii)

    \(w^\prime (a)>0\) and \(w^\prime (b)<0\);

     
  3. (iii)

    \(w>1\) on (ab);

     
  4. (iv)

    \(\rho \) is differentiable at a and b.

     
Then
$$\begin{aligned} \int _{\{w=1\}{\setminus } Z_w}\frac{1}{|w^\prime |}\frac{d\mathscr {H}^0}{\tau }\ge 2\coth (\mu ((a,b))/2) \end{aligned}$$
with equality if and only if \(\rho \equiv 0\) on [ab).

Proof

At the end-points \(x=a,b\) the condition (i) entails that \(w^\prime +m-\rho =1/x=w_0^\prime +m_0\) so that
$$\begin{aligned} w^\prime -w_0^\prime&=m_0-m+\rho \text { at }x=a,b. \end{aligned}$$
(9.9)
We consider the four cases
  1. (a)

    \(w^\prime (a)\ge w_0^\prime (a)\text { and }w^\prime (b)\ge w_0^\prime (b)\);

     
  2. (b)

    \(w^\prime (a)\ge w_0^\prime (a)\text { and }w^\prime (b)\le w_0^\prime (b)\);

     
  3. (c)

    \(w^\prime (a)\le w_0^\prime (a)\text { and }w^\prime (b)\ge w_0^\prime (b)\);

     
  4. (d)

    \(w^\prime (a)\le w_0^\prime (a)\text { and }w^\prime (b)\le w_0^\prime (b)\);

     
in turn.
  1. (a)
    Condition (a) together with (9.9) means that \(m_0-m+\rho (a)\ge 0\); that is, \(m-\rho (a)\le m_0\). By (i) and (ii), \(bm-b\rho (b)-1=-bw^\prime (b)> 0\); or \(m-\rho (b)>1/b\). Therefore,
    $$\begin{aligned} 0<1/b<m-\rho (b)\le m-\rho (a)\le 1/A \end{aligned}$$
    by (8.5). Put \(x:=1/(m-\rho (b))\) and \(y:=1/(m-\rho (a))\). Then
    $$\begin{aligned} a<A\le y\le x<b. \end{aligned}$$
    We write
    $$\begin{aligned} aw^\prime (a)&=-(m-\rho (a))a+1=-(1/y)a+1>0;\nonumber \\ bw^\prime (b)&=-(m-\rho (b))b+1=-(1/x)b+1<0.\nonumber \end{aligned}$$
    Making use of assumption (iii),
    $$\begin{aligned} \int _{\{w=1\}{\setminus } Z_w}\frac{1}{|w^\prime |}\frac{d\mathscr {H}^0}{x}&=\frac{1}{-(1/y)a+1}-\frac{1}{-(1/x)b+1}\\&=\frac{by-ax}{(y-a)(b-x)}. \end{aligned}$$
    By Lemma 9.10 (i) then (ii),
    $$\begin{aligned} \int _{\{w=1\}}\frac{1}{|w^\prime |}\frac{d\mathscr {H}^0}{x}&\ge \frac{(b-a)y}{(y-a)(b-y)} \ge \frac{(b-a)A}{(A-a)(b-A)}\\&=2\frac{a+b}{b-a} =2\coth (\mu ((a,b))/2). \end{aligned}$$
    If equality holds then \(\rho (a)=\rho (b)\) and \(\rho \) is constant on [ab]. By Theorem 8.3 we conclude that \(\rho \equiv 0\) on [ab).
     
  2. (b)

    Condition (b) together with (9.9) entails that \(0\le m_0-m+\rho (a)\) and \(0\le -m_0+m-\rho (b)\) whence \(0\le \rho (a)-\rho (b)\) upon adding; so \(\rho \) is constant on the interval [ab] by monotonicity. Define x and y as above. Then \(x=y\) and \(y\ge A\). The result now follows in a similar way to case (a).

     
  3. (c)
    In this case,
    $$\begin{aligned} \frac{1}{aw^\prime (a)}-\frac{1}{bw^\prime (b)}\ge & {} \frac{1}{aw_0^\prime (a)} -\frac{1}{bw_0^\prime (b)}\\= & {} 2\coth (\mu ((a,b))/2) \end{aligned}$$
    by Lemma 9.9. If equality holds then \(w^\prime (b)=w_0^\prime (b)\) so that \(m_0-m+\rho (b)=0\) and \(\rho \) vanishes on [ab] by Theorem 8.3.
     
  4. (d)

    Condition (d) together with (9.9) means that \(m_0-m+\rho (b)\le 0\); that is, \(m\ge \rho (b)+m_0\). On the other hand, by Theorem 8.3, \(m\le \rho (b)+m_0\). In consequence, \(m=\rho (b)+m_0\). It then follows that \(\rho \equiv 0\) on [ab] by Theorem 8.3. Now use Lemma 9.9.

     
\(\square \)

Lemma 9.12

Let \(\phi :(0,+\infty )\rightarrow (0,+\infty )\) be a convex non-increasing function with \(\inf _{(0,+\infty )}\phi >0\). Let \(\varLambda \) be an at most countably infinite index set and \((x_h)_{h\in \varLambda }\) a sequence of points in \((0,+\infty )\) with \(\sum _{h\in \varLambda }x_h<+\infty \). Then
$$\begin{aligned} \sum _{h\in \varLambda }\phi (x_h)\ge \phi \left( \sum _{h\in \varLambda } x_h\right) \end{aligned}$$
and the left-hand side takes the value \(+\infty \) in case \(\varLambda \) is countably infinite and is otherwise finite.

Proof

Suppose \(0<x_1<x_2<+\infty \). By convexity \(\phi (x_1)+\phi (x_2)\ge 2\phi (\frac{x_1+x_2}{2})\ge \phi (x_1+x_2)\) as \(\phi \) is non-increasing. The result for finite \(\varLambda \) follows by induction. \(\square \)

Theorem 9.13

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \((w,\lambda )\) solve (9.6). Assume that \(w>1\) on (ab). Then
  1. (i)
    for \(\mathscr {L}^1\)-a.e. \(t\in (1,\Vert w\Vert _\infty )\),
    $$\begin{aligned} -\mu _w^\prime \ge (2/t)\coth ((1/2)\mu _w); \end{aligned}$$
    (9.10)
     
  2. (ii)

    if \(\rho \not \equiv 0\) on [ab) then there exists \(t_0\in (1,\Vert w\Vert _\infty )\) such that strict inequality holds in (9.10) for \(\mathscr {L}^1\)-a.e. \(t\in (1,t_0)\).

     

Proof

(i) The set
$$\begin{aligned} Y_w:= & {} Z_{2,w}\cup \Big (\{w^\prime +m w^2\ne (1/x+\rho )w\}{\setminus } Z_{2,w}\Big )\\&\cup \{\rho \text { not differentiable}\}\subset [a,b] \end{aligned}$$
is a null set in [ab]. By [1] Lemma 2.95 and Lemma 2.96, \(\{w=t\}\cap (Y_w\cap Z_{1,w})=\emptyset \) for a.e. \(t>1\). Let \(t\in (1,\Vert w\Vert _\infty )\) and assume that \(\{w=t\}\cap (Y_w\cap Z_{1,w})=\emptyset \). We write \(\{w>t\}=\bigcup _{h\in \varLambda }I_h\) where \(\varLambda \) is an at most countably infinite index set and \((I_h)_{h\in \varLambda }\) are disjoint non-empty well-separated open intervals in (ab). The term well-separated means that for each \(h\in \varLambda ,\inf _{k\in \varLambda {\setminus }\{h\}}d(I_h,I_k)>0\). This follows from the fact that \(w^\prime \ne 0\) on \(\partial I_h\) for each \(h\in \varLambda \). Put \(\widetilde{w}:=w/t\) on \(\overline{\{w>t\}}\) so
$$\begin{aligned} \widetilde{w}^\prime +(mt)\widetilde{w}^2=(1/x+\rho )\widetilde{w}\text { a.e. on }\{w>t\}\text { and }\widetilde{w}=1\text { on }\{w=t\}. \end{aligned}$$
We use the fact that the mapping \(\phi :(0,+\infty )\rightarrow (0,+\infty );t\mapsto \coth t\) satisfies the hypotheses of Lemma 9.12. By Lemmas 9.11 and 9.12,
$$\begin{aligned} (0,+\infty ]\ni \int _{\{w=t\}{\setminus } Z_w}\frac{1}{|w^\prime |}\frac{d\mathscr {H}^0}{x}&=(1/t)\int _{\{\widetilde{w}=1\}}\frac{1}{|\widetilde{w}^\prime |}\frac{d\mathscr {H}^0}{\tau }\nonumber \\&=(1/t)\sum _{h\in \varLambda }\int _{\partial I_h}\frac{1}{|\widetilde{w}^\prime |}\frac{d\mathscr {H}^0}{\tau }\nonumber \\&\ge (2/t)\sum _{h\in \varLambda }\coth ((1/2)\mu (I_h))\nonumber \\&\ge (2/t)\coth \left( (1/2)\sum _{h\in \varLambda }\mu (I_h)\right) \nonumber \\&=(2/t)\coth ((1/2)\mu (\{w>t\})))\nonumber \\&=(2/t)\coth ((1/2)\mu _w(t)).\nonumber \end{aligned}$$
The statement now follows from Lemma 9.1.

(ii) Suppose that \(\rho \not \equiv 0\) on [ab). Put \(\alpha :=\min \{\rho >0\}\in [a,b)\). Now that \(\{w>t\}\uparrow (a,b)\) as \(t\downarrow 1\) as \(w>1\) on (ab). Choose \(t_0\in (1,\Vert w\Vert _\infty )\) such that \(\{w>t_0\}\cap (\alpha ,b)\ne \emptyset \). Then for each \(t\in (1,t_0)\) there exists \(h\in \varLambda \) such that \(\rho \not \equiv 0\) on \(I_h\). The statement then follows by Lemma 9.11. \(\square \)

Lemma 9.14

Let \(\emptyset \ne S\subset \mathbb {R}\) be bounded and suppose S has the property that for each \(s\in S\) there exists \(\delta >0\) such that \([s,s+\delta )\subset S\). Then S is \(\mathscr {L}^1\)-measurable and \(|S|>0\).

Proof

For each \(s\in S\) put \(t_s:=\inf \{t>s:t\not \in S\}\). Then \(s<t_s<+\infty ,[s,t_s)\subset S\) and \(t_s\not \in S\). Define
$$\begin{aligned} \mathscr {C}:=\Big \{[s,t]:s\in S\text { and }t\in (s,t_s)\Big \}. \end{aligned}$$
Then \(\mathscr {C}\) is a Vitali cover of S (see [6] Chapter 16 for example). By Vitali’s Covering Theorem (cf. [6] Theorem 16.27) there exists an at most countably infinite subset \(\varLambda \subset \mathscr {C}\) consisting of pairwise disjoint intervals such that
$$\begin{aligned} \left| S{\setminus }\bigcup _{I\in \varLambda }I\right| =0. \end{aligned}$$
Note that \(I\subset S\) for each \(I\in \varLambda \). Consequently, \(S=\bigcup _{I\in \varLambda }I\cup N\) where N is an \(\mathscr {L}^1\)-null set and hence S is \(\mathscr {L}^1\)-measurable. The positivity assertion is clear. \(\square \)

Theorem 9.15

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Let \((w,\lambda )\) solve (9.6). Assume that \(w>1\) on (ab). Put \(T:=\min \{\Vert w_0\Vert _\infty ,\Vert w\Vert _\infty \}>1\). Then
  1. (i)

    \(\mu _w(t)\le \mu _{w_0}(t)\) for each \(t\in [1,T)\);

     
  2. (ii)

    \(\Vert w\Vert _\infty \le \Vert w_0\Vert _\infty \);

     
  3. (iii)

    if \(\rho \not \equiv 0\) on [ab) then there exists \(t_0\in (1,\Vert w\Vert _\infty )\) such that \(\mu _w(t)<\mu _{w_0}(t)\) for each \(t\in (1,t_0)\).

     

Proof

(i) We adapt the proof of [16] Theorem I.6.1. The assumption entails that \(\mu _w(1)=\mu _{w_0}(1)=\mu ((a,b))\). Suppose for a contradiction that \(\mu _w(t)>\mu _{w_0}(t)\) for some \(t\in (1,T)\).

For \(\varepsilon >0\) consider the initial value problem
$$\begin{aligned} z^\prime =\omega (t,z)+\varepsilon \text { and }z(1)=\mu ((a,b))+\varepsilon \end{aligned}$$
(9.11)
on (0, T). Choose \(\upsilon \in (0,1)\) and \(\tau \in (t,T)\). By [16] Lemma I.3.1 there exists \(\varepsilon _0>0\) such that for each \(0\le \varepsilon <\varepsilon _0\) (9.11) has a continuously differentiable solution \(z_\varepsilon \) defined on \([\upsilon ,\tau ]\) and this solution is unique by [16] Theorem I.3.1. Moreover, the sequence \((z_\varepsilon )_{0<\varepsilon <\varepsilon _0}\) converges uniformly to \(z_0\) on \([\upsilon ,\tau ]\).

Given \(0<\varepsilon<\eta <\varepsilon _0\) it holds that \(z_0\le z_\varepsilon \le z_\eta \) on \([1,\tau ]\) by [16] Theorem I.6.1. Note for example that \(z_0^\prime \le \omega (\cdot ,z_0)+\varepsilon \) on \((1,\tau )\). In fact, \((z_\varepsilon )_{0<\varepsilon <\varepsilon _0}\) decreases strictly to \(z_0\) on \((1,\tau )\). For if, say, \(z_0(s)=z_\varepsilon (s)\) for some \(s\in (1,\tau )\) then \(z_\varepsilon ^\prime (s)=\omega (s,z_\varepsilon (s))+\varepsilon >\omega (s,z_0(s))=z_0^\prime (s)\) by (9.11); while on the other hand \(z_\varepsilon ^\prime (s)\le z_0^\prime (s)\) by considering the left-derivative at s and using the fact that \(z_\varepsilon \ge z_0\) on \([1,\tau ]\). This contradicts the strict inequality.

Choose \(\varepsilon _1\in (0,\varepsilon _0)\) such that \(z_\varepsilon (t)<\mu _w(t)\) for each \(0<\varepsilon <\varepsilon _1\). Now \(\mu _w\) is right-continuous and strictly decreasing as \(\mu _w(t)-\mu _w(s)=-\mu (\{s<w\le t\})<0\) for \(1\le s<t<\Vert w\Vert _\infty \) by continuity of w. So the set \(\{z_\varepsilon <\mu _w\}\cap (1,t)\) is open and non-empty in \((0,+\infty )\) for each \(\varepsilon \in (0,\varepsilon _1)\). Thus there exists a unique \(s_\varepsilon \in [1,t)\) such that
$$\begin{aligned} \mu _w>z_\varepsilon \text { on }(s_\varepsilon ,t] \text { and }\mu _w(s_\varepsilon )=z_\varepsilon (s_\varepsilon ) \end{aligned}$$
for each \(\varepsilon \in (0,\varepsilon _1)\). As \(z_\varepsilon (1)>\mu ((a,b))\) it holds that each \(s_\varepsilon >1\). Note that \(1<s_\varepsilon <s_\eta \) whenever \(0<\varepsilon <\eta \) as \((z_\varepsilon )_{_{0<\varepsilon <\varepsilon _0}}\) decreases strictly to \(z_0\) as \(\varepsilon \downarrow 0\).
Define
$$\begin{aligned} S:=\Big \{s_\varepsilon :0<\varepsilon <\varepsilon _1\Big \}\subset (1,t). \end{aligned}$$
We claim that for each \(s\in S\) there exists \(\delta >0\) such that \([s,s+\delta )\subset S\). This entails that S is \(\mathscr {L}^1\)-measurable with positive \(\mathscr {L}^1\)-measure by Lemma 9.14.
Suppose \(s=s_\varepsilon \in S\) for some \(\varepsilon \in (0,\varepsilon _1)\) and put \(z:=z_\varepsilon (s)=\mu _w(s)\). Put \(k:=\mathrm {cosech}^2(z_0(t)/2)\). For \(0\le \zeta<\eta <\varepsilon _1\) define
$$\begin{aligned} \varOmega _{\zeta ,\eta }:=\Big \{(u,y)\in \mathbb {R}^2:u\in (0,t)\text { and }z_\zeta (u)<y<z_\eta (u)\Big \} \end{aligned}$$
and note that this is an open set in \(\mathbb {R}^2\). We remark that for each \((u,y)\in \varOmega _{\zeta ,\eta }\) there exists a unique \(\nu \in (\zeta ,\eta )\) such that \(y=z_\nu (u)\). Given \(r>0\) with \(s+r<t\) set
$$\begin{aligned} Q=Q_r:=\Big \{(u,y)\in \mathbb {R}^2:s\le u<s+r\text { and }|y-z|<\Vert z_\varepsilon -z\Vert _{C([s,s+r])}\Big \}. \end{aligned}$$
Choose \(r\in (0,t-s)\) and \(\varepsilon _2\in (\varepsilon ,\varepsilon _1)\) such that
  1. (a)

    \(Q_r\subset \varOmega _{0,\varepsilon _1}\);

     
  2. (b)

    \(\Vert z_\varepsilon -z\Vert _{C([s,s+r])}<s\varepsilon /(2k)\);

     
  3. (c)

    \(\sup _{\eta \in (\varepsilon ,\varepsilon _2)}\Vert z_\eta -z\Vert _{C([s,s+r])}\le \Vert z_\varepsilon -z\Vert _{C([s,s+r])}\);

     
  4. (d)

    \(z_\eta <\mu _w\) on \([s+r,t]\) for each \(\eta \in (\varepsilon ,\varepsilon _2)\).

     
We can find \(\delta \in (0,r)\) such that \(z_\varepsilon<\mu _w<z_{\varepsilon _2}\) on \((s,s+\delta )\) as \(z_{\varepsilon _2}(s)>z\); in other words, the graph of \(\mu _w\) restricted to \((s,s+\delta )\) is contained in \(\varOmega _{\varepsilon ,\varepsilon _2}\).
Let \(u\in (s,s+\delta )\). Then \(\mu _w(u)=z_\eta (u)\) for some \(\eta \in (\varepsilon ,\varepsilon _2)\) as above. We claim that \(u=s_\eta \) so that \(u\in S\). This implies in turn that \([s,s+\delta )\subset S\). Suppose for a contradiction that \(z_\eta \not <\mu _w\) on (ut]. Then there exists \(v\in (u,t]\) such that \(\mu _w(v)=z_\eta (v)\). In view of condition (d), \(v\in (u,s+r)\). By [1] Theorem 3.28 and Theorem 9.13,
$$\begin{aligned} \mu _w(v)-\mu _w(u)&=D\mu _w((u,v]) =D\mu _w^a((u,v])+D\mu _w^s((u,v]) \nonumber \\&\le D\mu _w^a((u,v]) =\int _u^v\mu _w^\prime \,d\tau \le \int _u^v\omega (\cdot ,\mu _w)\,d\tau .\nonumber \end{aligned}$$
On the other hand,
$$\begin{aligned} z_\eta (v)-z_\eta (u)&=\int _u^v z_\eta ^\prime \,d\tau =\int _u^v\omega (\cdot ,z_\eta )\,d\tau +\eta (v-u).\nonumber \end{aligned}$$
We derive that
$$\begin{aligned} \varepsilon (v-u)&\le \eta (v-u) \le \int _u^v\Big \{\omega (\cdot ,\mu _w)-\omega (\cdot ,z_\eta )\Big \}\,d\tau \\&\le k\int _u^v|\mu _w-z_\eta |\,d\mu \nonumber \end{aligned}$$
using the estimate (9.7). Thus
$$\begin{aligned} \varepsilon&\le k\frac{1}{v-u}\int _u^v|\mu _w-z_\eta |\,d\mu \nonumber \\&\le (k/s)\Vert \mu _w-z_\eta \Vert _{C([u,v])}\nonumber \\&\le (k/s)\Big \{\Vert \mu _w-z\Vert _{C([s,s+r])}+\Vert z_\eta -z\Vert _{C([s,s+r])}\Big \}\nonumber \\&\le (2k/s)\Vert z_\varepsilon -z\Vert _{C([s,s+r])} <\varepsilon \nonumber \end{aligned}$$
by (b) and (c) giving rise to the desired contradiction.
By Theorem 9.13, \(\mu _w^\prime \le \omega (\cdot ,\mu _w)\) for \(\mathscr {L}^1\)-a.e. \(t\in S\). Choose \(s\in S\) such that \(\mu _w\) is differentiable at s and the latter inequality holds at s. Let \(\varepsilon \in (0,\varepsilon _1)\) such that \(s=s_\varepsilon \). For any \(u\in (s,t)\),
$$\begin{aligned} \mu _w(u)-\mu _w(s)>z_\varepsilon (u)-z_\varepsilon (s). \end{aligned}$$
We deduce that \(\mu _w^\prime (s)\ge z_\varepsilon ^\prime (s)\). But then
$$\begin{aligned} \mu _w^\prime (s)\ge z_\varepsilon ^\prime (s)=\omega (s,z_\epsilon (s))+\varepsilon >\omega (s,\mu _w(s)). \end{aligned}$$
This strict inequality holds on a set of full measure in S. This contradicts Theorem 9.13.
  • (ii) Use the fact that \(\Vert w\Vert _\infty =\sup \{t>0:\mu _w(t)>0\}\).

  • (iii) Assume that \(\rho \not \equiv 0\) on [ab). Let \(t_0\in (1,\Vert w\Vert _\infty )\) be as in Lemma 9.13. Then for \(t\in (1,t_0)\),
    $$\begin{aligned} \mu _w(t)-\mu _w(1)&=D\mu _w((1,t])=D\mu _w^a((1,t])\nonumber \\&\quad +D\mu _w^s((1,t])\le D\mu _w^a((1,t])\nonumber \\&=\int _{(1,t]}\mu _w^\prime \,ds<\int _{(1,t]}\omega (s,\mu _w)\,ds\nonumber \\&\le \int _{(1,t]}\omega (s,\mu _{w_0})\,ds=\mu _{w_0}(t)-\mu _{w_0}(1)\nonumber \end{aligned}$$
    by Theorem 9.13, Lemma 9.9 and the inequality in (i).
\(\square \)

Corollary 9.16

Let \(0<a<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [ab]. Suppose that \((w,\lambda )\) solves (9.6). Assume that \(w>1\) on (ab). Let \(0\le \varphi \in C^1((1,+\infty ))\) be strictly decreasing with \(\int _a^b\varphi (w_0)\,d\mu <+\infty \). Then
  1. (i)

    \(\int _a^b\varphi (w)\,d\mu \ge \int _a^b\varphi (w_0)\,d\mu \);

     
  2. (ii)

    equality holds in (i) if and only if \(\rho \equiv 0\) on [ab).

     
In particular,
  1. (iii)

    \(\int _a^b\frac{1}{\sqrt{w^2-1}}\,d\mu \ge \pi \) with equality if and only if \(\rho \equiv 0\) on [ab).

     

Proof

(i) Let \(\varphi \ge 0\) be a decreasing function on \((1,+\infty )\) which is piecewise \(C^1\). Suppose that \(\varphi (1+)<+\infty \). By Tonelli’s Theorem,
$$\begin{aligned} \int _{[1,+\infty )}\varphi ^\prime \mu _w\,ds&=\int _{[1,+\infty )}\varphi ^\prime \Big \{\int _{(a,b)}\chi _{\{w>s\}}\,d\mu \Big \}\,ds\nonumber \\&=\int _{(a,b)}\Big \{\int _{[1,+\infty )}\varphi ^\prime \chi _{\{w>s\}}\,ds\Big \}\,d\mu \nonumber \\&=\int _{(a,b)}\Big \{\varphi (w)-\varphi (1)\Big \}\,d\mu \\&=\int _{(a,b)}\varphi (w)\,d\mu -\varphi (1)\mu ((a,b))\nonumber \end{aligned}$$
and a similar identity holds for \(\mu _{w_0}\). By Theorem 9.15, \(\int _a^b\varphi (w)\,d\mu \ge \int _a^b\varphi (w_0)\,d\mu \). Now suppose that \(0\le \varphi \in C^1((1,+\infty ))\) is strictly decreasing with \(\int _a^b\varphi (w_0)\,d\mu <+\infty \). The inequality holds for the truncated function \(\varphi \wedge n\) for each \(n\in \mathbb {N}\). An application of the monotone convergence theorem establishes the result for \(\varphi \).
(ii) Suppose that equality holds in (i). For \(c\in (1,+\infty )\) put \(\varphi _1:=\varphi \vee \varphi (c)-\varphi (c)\) and \(\varphi _2:=\varphi \wedge \varphi (c)\). By (i) we deduce
$$\begin{aligned} \int _a^b\varphi _2(w)\,d\mu =\int _a^b\varphi _2(w_0)\,d\mu ; \end{aligned}$$
and hence by the above that
$$\begin{aligned} \int _{[c,+\infty )}\varphi ^\prime \Big \{\mu _w-\mu _{w_0}\Big \}\,ds=0. \end{aligned}$$
This means that \(\mu _w=\mu _{w_0}\) on \((c,+\infty )\) and hence on \((1,+\infty )\). By Theorem 9.15 we conclude that \(\rho \equiv 0\) on [ab). (iii) flows from (i) and (ii) noting that the function \(\varphi :(1,+\infty )\rightarrow \mathbb {R};t\mapsto 1/\sqrt{t^2-1}\) satisfies the integral condition by Lemma 9.9. \(\square \)
The case \(a=0\). Let \(0<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [0, b]. We study solutions to the first-order linear ordinary differential equation
$$\begin{aligned} u^\prime +(1/x+\rho )u+\lambda =0\text { a.e. on }(0,b)\text { with }u(0)=0\text { and }u(b)=1 \end{aligned}$$
(9.12)
where \(u\in C^{0,1}([0,b])\) and \(\lambda \in \mathbb {R}\). If \(\rho \equiv 0\) on [0, b] then we write \(u_0\) instead of u.

Lemma 9.17

Let \(0<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [0, b]. Then
  1. (i)

    there exists a solution \((u,\lambda )\) of (9.12) with \(u\in C^{0,1}([0,b])\) and \(\lambda \in \mathbb {R}\);

     
  2. (ii)

    \(\lambda \) is given by \(\lambda =-g(b)/G(b)\) where \(G:=\int _0^\cdot g\,ds\);

     
  3. (iii)

    the pair \((u,\lambda )\) in (i) is unique;

     
  4. (iv)

    \(u>0\) on (0, b].

     

Proof

(i) The function \(u:[a,b]\rightarrow \mathbb {R}\) given by
$$\begin{aligned} u=\frac{g(b)}{G(b)}\frac{G}{g} \end{aligned}$$
(9.13)
on [0, b] solves (9.12) with \(\lambda \) as in (ii). (iii) Suppose that \((u_1,\lambda _1)\) resp. \((u_2,\lambda _2)\) solve (9.12). By linearity \(u:=u_1-u_2\) solves
$$\begin{aligned} u^\prime +(1/x+\rho )u+\lambda =0\text { a.e. on }(0,b) \text { with }u(0)=u(b)=0 \end{aligned}$$
where \(\lambda =\lambda _1-\lambda _2\). An integration gives that \(u=(-\lambda G+c)/g\) for some constant \(c\in \mathbb {R}\) and the boundary conditions entail that \(\lambda =c=0\). (iv) follows from the formula (9.13) and unicity. \(\square \)

Lemma 9.18

Suppose \(-\infty<a<b<+\infty \) and that \(\phi :[a,b]\rightarrow \mathbb {R}\) is convex. Suppose that there exists \(\xi \in (a,b)\) such that
$$\begin{aligned} \phi (\xi )=\frac{b-\xi }{b-a}\phi (a)+\frac{\xi -a}{b-a}\phi (b). \end{aligned}$$
Then
$$\begin{aligned} \phi (c)=\frac{b-c}{b-a}\phi (a)+\frac{c-a}{b-a}\phi (b) \end{aligned}$$
for each \(c\in [a,b]\).

Proof

Let \(c\in (\xi ,b)\). By monotonicity of chords,
$$\begin{aligned} \frac{\phi (\xi )-\phi (a)}{\xi -a}\le \frac{\phi (c)-\phi (\xi )}{c-\xi } \end{aligned}$$
so
$$\begin{aligned} \phi (c)&\ge \frac{c-a}{\xi -a}\phi (\xi )-\frac{c-\xi }{\xi -a}\phi (a)\nonumber \\&=\frac{c-a}{\xi -a}\Big \{\frac{b-\xi }{b-a}\phi (a)+\frac{\xi -a}{b-a}\phi (b)\Big \} -\frac{c-\xi }{\xi -a}\phi (a)\nonumber \\&=\frac{b-c}{b-a}\phi (a)+\frac{c-a}{b-a}\phi (b)\nonumber \end{aligned}$$
and equality follows. The case \(c\in (a,\xi )\) is similar. \(\square \)

Lemma 9.19

Let \(0<b<+\infty \) and \(\rho \ge 0\) be a non-decreasing bounded function on [0, b]. Let \((u,\lambda )\) satisfy (9.12). Then
  1. (i)

    \(u\ge u_0\) on [0, b];

     
  2. (ii)

    if \(\rho \not \equiv 0\) on [0, b) then \(u>u_0\) on (0, b).

     

Proof

(i) The mapping \(G:[0,b]\rightarrow [0,G(b)]\) is a bijection with inverse \(G^{-1}\). Define \(\eta :[0,G(b)]\rightarrow \mathbb {R}\) via \(\eta :=(tg)\circ G^{-1}\). Then
$$\begin{aligned} \eta ^\prime =\frac{(tg)^\prime }{g}\circ G^{-1} =(2+t\rho )\circ G^{-1} \end{aligned}$$
a.e. on (0, G(b)) so \(\eta ^\prime \) is non-decreasing there. This means that \(\eta \) is convex on [0, G(b)]. In particular, \(\eta (s)\le [\eta (G(b))/G(b)]s\) for each \(s\in [0,G(b)]\). For \(t\in [0,b]\) put \(s:=G(t)\) to obtain \(tg(t)\le (bg(b)/G(b))G(t)\). A rearrangement gives \(u\ge u_0\) on [0, b] noting that \(u_0:[0,b]\rightarrow \mathbb {R};t\mapsto t/b\). (ii) Assume \(\rho \not \equiv 0\) on [0, b). Suppose that \(u(c)=u_0(c)\) for some \(c\in (0,b)\). Then \(\eta (G(c))=[\eta (G(b))/G(b)]G(c)\). By Lemma 9.18, \(\eta ^\prime =0\) on (0, G(b)). This implies that \(\rho \equiv 0\) on [0, b). \(\square \)

Lemma 9.20

Let \(0<b<+\infty \). Then \(\int _0^b\frac{u_0}{\sqrt{1-u_0^2}}\,d\mu =\pi /2\).

Proof

The integral is elementary as \(u_0(t)=t/b\) for \(t\in [0,b]\). \(\square \)

10 Proof of main results

Lemma 10.1

Let \(x\in H\) and v be a unit vector in \(\mathbb {R}^2\) such that the pair \(\{x,v\}\) forms a positively oriented orthogonal basis for \(\mathbb {R}^2\). Put \(b:=(\tau ,0)\) where \(|x|=\tau \) and \(\gamma :=\theta (x)\in (0,\pi )\). Let \(\alpha \in (0,\pi /2)\) such that
$$\begin{aligned} \frac{\langle v,x-b\rangle }{|x-b|}=\cos \alpha . \end{aligned}$$
Then
  1. (i)

    \(C(x,v,\alpha )\cap H\cap \overline{C}(0,e_1,\gamma )=\emptyset \);

     
  2. (ii)

    for any \(y\in C(x,v,\alpha )\cap H{\setminus }\overline{B}(0,\tau )\) the line segment [by] intersects \(\mathbb {S}^1_\tau \) outside the closed cone \(\overline{C}(0,e_1,\gamma )\).

     
We point out that \(C(0,e_1,\gamma )\) is the open cone with vertex 0 and axis \(e_1\) which contains the point x on its boundary. We note that \(\cos \alpha \in (0,1)\) because
$$\begin{aligned} \langle v,x-b\rangle= & {} -\langle v,b\rangle =-\langle (1/\tau )Ox,b\rangle \nonumber \\= & {} -\langle Op,e_1\rangle =\langle x,O^\star e_1\rangle =\langle x,e_2\rangle >0 \end{aligned}$$
(10.1)
and if \(|x-b|=\langle v,x-b\rangle \) then \(b=x-\lambda v\) for some \(\lambda \in \mathbb {R}\) and hence \(x_1=\langle e_1,x\rangle =\tau \) and \(x_2=0\).

Proof

(i) For \(\omega \in \mathbb {S}^1\) define the open half-space
$$\begin{aligned} H_{\omega }:=\{y\in \mathbb {R}^2:\langle y,\omega \rangle >0\}. \end{aligned}$$
We claim that \(C(x,v,\alpha )\subset H_v\). For given \(y\in C(x,v,\alpha )\),
$$\begin{aligned} \langle y,v\rangle =\langle y-x,v\rangle>|y-x|\cos \alpha >0. \end{aligned}$$
On the other hand, it holds that \(\overline{C}(0,e_1,\gamma )\cap H\subset \overline{H}_{-v}\). This establishes (i).
(ii) By some trigonometry \(\gamma =2\alpha \). Suppose that \(\omega \) is a unit vector in \(C(b,-e_1,\pi /2-\alpha )\). Then \(\lambda :=\langle \omega ,e_1\rangle <\cos \alpha \) since upon rewriting the membership condition for \(C(b,-e_1,\pi /2-\alpha )\) we obtain the quadratic inequality
$$\begin{aligned} \lambda ^2-2\cos ^2\alpha \lambda +\cos \gamma >0. \end{aligned}$$
For \(\omega \) a unit vector in \(\overline{C}(0,e_1,\gamma )\) the opposite inequality \(\langle \omega ,e_1\rangle \ge \cos \alpha \) holds. This shows that
$$\begin{aligned} C(b,-e_1,\pi /2-\alpha )\cap \overline{C}(0,e_1,\gamma )\cap \mathbb {S}^1_\tau =\emptyset . \end{aligned}$$
The set \(C(x,v,\alpha )\cap H\) is contained in the open convex cone \(C(b,-e_1,\pi /2-\alpha )\). Suppose \(y\in C(x,v,\alpha )\cap H{\setminus }\overline{B}(0,\tau )\). Then the line segment [by] is contained in \(C(b,-e_1,\pi /2-\alpha )\cup \{b\}\). Now the set \(C(b,-e_1,\pi /2-\alpha )\cap \mathbb {S}^1_\tau \) disconnects \(C(b,-e_1,\pi /2-\alpha )\cup \{b\}\). This entails that \((b,y]\cap C(b,-e_1,\pi /2-\alpha )\cap \mathbb {S}^1_\tau \ne \emptyset \). The foregoing paragraph entails that \((b,y]\cap \overline{C}(0,e_1,\gamma )\cap \mathbb {S}^1_\tau =\emptyset \). This establishes the result. \(\square \)

Lemma 10.2

Let E be an open set in \(\mathbb {R}^2\) such that \(M:=\partial E\) is a \(C^{1,1}\) hypersurface in \(\mathbb {R}^2\). Assume that \(E{\setminus }\{0\}=E^{sc}\). Suppose
  1. (i)

    \(x\in (M{\setminus }\{0\})\cap H\);

     
  2. (ii)

    \(\sin (\sigma (x))=-1\).

     
Then E is not convex.

Proof

Let \(\gamma _1:I\rightarrow M\) be a \(C^{1,1}\) parametrisation of M in a neighbourhood of x with \(\gamma _1(0)=x\) as above. As \(\sin (\sigma (x))=-1,n(x)\) and hence \(n_1(0)\) point in the direction of x. Put \(v:=-t_1(0)=-t(x)\). We may write
$$\begin{aligned} \gamma _1(s)=\gamma _1(0)+st_1(0)+R_1(s)=x-sv+R_1(s) \end{aligned}$$
for \(s\in I\) where \(R_1(s)=s\int _0^1\dot{\gamma }_1(ts)-\dot{\gamma }_1(0)\,dt\) and we can find a finite positive constant K such that \(|R_1(s)|\le Ks^2\) on a symmetric open interval \(I_0\) about 0 with \(I_0\subset \subset I\). Then
$$\begin{aligned} \frac{\langle \gamma _1(s)-x,v\rangle }{|\gamma _1(s)-x|}&=\frac{\langle -sv+R_1,v\rangle }{|-sv+R_1|}\\&=\frac{1-\langle (R_1/s),v\rangle }{|v-R_1/s|}\rightarrow 1 \end{aligned}$$
as \(s\uparrow 0\). Let \(\alpha \) be as in Lemma 10.1 with x and v as just mentioned. The above estimate entails that \(\gamma _1(s)\in C(x,v,\alpha )\) for small \(s<0\). By (2.9) and Lemma 5.4 the function \(r_1\) is non-increasing on I. In particular, \(r_1(s)\ge r_1(0)=|x|=:\tau \) for \(I\ni s<0\) and \(\gamma _1(s)\not \in B(0,\tau )\).

Choose \(\delta _1>0\) such that \(\gamma _1(s)\in C(x,v,\alpha )\cap H\) for each \(s\in [-\delta _1,0)\). Put \(\beta :=\inf \{s\in [-\delta _1,0]:r_1(s)=\tau \}\). Suppose first that \(\beta \in [-\delta _1,0)\). Then E is not convex (see Lemma 5.2). Now suppose that \(\beta =0\). Let \(\gamma \) be as in Lemma 10.1. Then the open circular arc \(\mathbb {S}^1_\tau {\setminus } \overline{C}(0,e_1,\gamma )\) does not intersect \(\overline{E}\): for otherwise, M intersects \(\mathbb {S}^1_\tau {\setminus } \overline{C}(0,e_1,\gamma )\) and \(\beta <0\) bearing in mind Lemma 5.2. Choose \(s\in [-\delta _1,0)\). Then the points b and \(\gamma _1(s)\) lie in \(\overline{E}\). But by Lemma 10.1 the line segment \([b,\gamma _1(s)]\) intersects \(\mathbb {S}^1_\tau \) in \(\mathbb {S}^1_\tau {\setminus } \overline{C}(0,e_1,\gamma )\). Let \(c\in [b,\gamma _1(s)]\cap \mathbb {S}^1_\tau \). Then \(c\not \in \overline{E}\). This shows that \(\overline{E}\) is not convex. But if E is convex then \(\overline{E}\) is convex. Therefore E is not convex. \(\square \)

Theorem 10.3

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Given \(v>0\) let E be a bounded minimiser of (1.2). Assume that E is open, \(M:=\partial E\) is a \(C^{1,1}\) hypersurface in \(\mathbb {R}^2\) and \(E{\setminus }\{0\}=E^{sc}\). Put
$$\begin{aligned} R:=\inf \{\rho >0\}\in [0,+\infty ). \end{aligned}$$
(10.2)
Then \(\varOmega \cap (R,+\infty )=\emptyset \) with \(\varOmega \) as in (5.2).

Proof

Suppose that \(\varOmega \cap (R,+\infty )\ne \emptyset \). As \(\varOmega \) is open in \((0,+\infty )\) by Lemma 5.6 we may write \(\varOmega \) as a countable union of disjoint open intervals in \((0,+\infty )\). By a suitable choice of one of these intervals we may assume that \(\varOmega =(a,b)\) for some \(0\le a<b<+\infty \) and that \(\varOmega \cap (R,+\infty )\ne \emptyset \). Let us assume for the time being that \(a>0\). Note that \([a,b]\subset \pi (M)\) and \(\cos \sigma \) vanishes on \(M_a\cup M_b\).

Let \(u:\varOmega \rightarrow [-1,1]\) be as in (6.6). Then u has a continuous extension to [ab] and \(u=\pm 1\) at \(\tau =a,b\). This may be seen as follows. For \(\tau \in (a,b)\) the set \(M_\tau \cap \overline{H}\) consists of a singleton by Lemma 5.4. The limit \(x:=\lim _{\tau \downarrow a}M_\tau \cap \overline{H}\in \mathbb {S}^1_a\cap \overline{H}\) exists as M is \(C^1\). There exists a \(C^{1,1}\) parametrisation \(\gamma _1:I\rightarrow M\) with \(\gamma _1(0)=x\) as above. By (2.9) and Lemma 5.4, \(r_1\) is decreasing on I. So \(r_1>a\) on \(I\cap \{s<0\}\) for otherwise the \(C^1\) property fails at x. It follows that \(\gamma _1=\gamma \circ r_1\) and \(\sigma _1=\sigma \circ \gamma \circ r_1\) on \(I\cap \{s<0\}\). Thus \(\sin (\sigma \circ \gamma )\circ r_1=\sin \sigma _1\) on \(I\cap \{s<0\}\). Now the function \(\sin \sigma _1\) is continuous on I. So \(u\rightarrow \sin \sigma _1(0)\in \{\pm 1\}\) as \(\tau \downarrow a\). Put \(\eta _1:=u(a)\) and \(\eta _2:=u(b)\).

Let us consider the case \(\eta =(\eta _1,\eta _2)=(1,1)\). According to Theorem 6.5 the generalised (mean) curvature is constant \(\mathscr {H}^1\)-a.e. on M with value \(-\lambda \), say. Note that \(u<1\) on (ab) for otherwise \(\cos (\sigma \circ \gamma )\) vanishes at some point in (ab) bearing in mind Lemma 5.4. By Theorem 6.6 the pair \((u,\lambda )\) satisfies (9.4) with \(\eta =(1,1)\). By Lemma 9.2, \(u>0\) on [ab]. Put \(w:=1/u\). Then \((w,-\lambda )\) satisfies (9.6) and \(w>1\) on (ab). By Lemma 6.7,
$$\begin{aligned} \theta _2(b)-\theta _2(a)= & {} \int _a^b\theta _2^\prime \,d\tau =-\int _a^b\frac{u}{\sqrt{1-u^2}}\,\frac{d\tau }{\tau }\\= & {} -\int _a^b\frac{1}{\sqrt{w^2-1}}\,\frac{d\tau }{\tau }. \end{aligned}$$
By Corollary 9.16, \(|\theta _2(b)-\theta _2(a)|>\pi \). But this contradicts the definition of \(\theta _2\) in (6.4) as \(\theta _2\) takes values in \((0,\pi )\) on (ab). If \(\eta =(-1,-1)\) then \(\lambda >0\) by Lemma 9.2; this contradicts Lemma 7.2.

Now let us consider the case \(\eta =(-1,1)\). Using the same formula as above, \(\theta _2(b)-\theta _2(a)<0\) by Corollary 9.7. This means that \(\theta _2(a)\in (0,\pi ]\). As before the limit \(x:=\lim _{\tau \downarrow a}M_\tau \cap \overline{H}\in \mathbb {S}^1_a\cap \overline{H}\) exists as M is \(C^1\). Using a local parametrisation it can be seen that \(\theta _2(a)=\theta (x)\) and \(\sin (\sigma (x))=-1\). If \(\theta _2(a)\in (0,\pi )\) then E is not convex by Lemma 10.2. This contradicts Theorem 7.3. Note that we may assume that \(\theta _2(a)\in (0,\pi )\). For otherwise, \(\langle \gamma ,e_2\rangle <0\) for \(\tau >a\) near a, contradicting the definition of \(\gamma \) (6.5). If \(\eta =(1,-1)\) then \(\lambda >0\) by Lemma 9.2 and this contradicts Lemma 7.2 as before.

Suppose finally that \(a=0\). By Lemma 5.5, \(u(0)=0\) and \(u(b)=\pm 1\). Suppose \(u(b)=1\). Again employing the formula above, \(\theta _2(b)-\theta _2(0)<-\pi /2\) by Lemma 9.19, the fact that the function \(\phi :(0,1)\rightarrow \mathbb {R};t\mapsto t/\sqrt{1-t^2}\) is strictly increasing and Lemma 9.20. This means that \(\theta _2(0)>\pi /2\). This contradicts the \(C^1\) property at \(0\in M\). If \(u(b)=-1\) then then \(\lambda >0\) by Lemma 9.2 giving a contradiction. \(\square \)

Lemma 10.4

Let f be as in (1.3) where \(h:[0,+\infty )\rightarrow \mathbb {R}\) is a non-decreasing convex function. Let \(v>0\).
  1. (i)

    Let E be a bounded minimiser of (1.2). Assume that E is open, \(M:=\partial E\) is a \(C^{1,1}\) hypersurface in \(\mathbb {R}^2\) and \(E{\setminus }\{0\}=E^{sc}\). Then for any \(r>0\) with \(r\ge R,M{\setminus }\overline{B}(0,r)\) consists of a finite union of disjoint centred circles.

     
  2. (ii)

    There exists a minimiser E of (1.2) such that \(\partial E\) consists of a finite union of disjoint centred circles.

     

Proof

(i) First observe that
$$\begin{aligned} \emptyset \ne \pi (M)&=\Big [\pi (M)\cap [0,r]\Big ]\cup \Big [\pi (M)\cap (r,+\infty )\Big ]{\setminus }\varOmega \end{aligned}$$
by Lemma 10.3. We assume that the latter member is non-empty. By definition of \(\varOmega ,\cos \sigma =0\) on \(M\cap A((r,+\infty ))\). Let \(\tau \in \pi (M)\cap (r,+\infty )\). We claim that \(M_\tau =\mathbb {S}^1_\tau \). Suppose for a contradiction that \(M_\tau \ne \mathbb {S}^1_\tau \). By Lemma 5.2, \(M_\tau \) is the union of two closed spherical arcs in \(\mathbb {S}^1_\tau \). Let x be a point on the boundary of one of these spherical arcs relative to \(\mathbb {S}^1_\tau \). There exists a \(C^{1,1}\) parametrisation \(\gamma _1:I\rightarrow M\) of M in a neighbourhood of x with \(\gamma _1(0)=x\) as before. By shrinking I if necessary we may assume that \(\gamma _1(I)\subset A((r,+\infty ))\) as \(\tau >r\). By (2.9), \(\dot{r}_1=0\) on I as \(\cos \sigma _1=0\) on I because \(\cos \sigma =0\) on \(M\cap A((r,+\infty ))\); that is, \(r_1\) is constant on I. This means that \(\gamma _1(I)\subset \mathbb {S}^1_\tau \). As the function \(\sin \sigma _1\) is continuous on I it takes the value \(\pm 1\) there. By (2.10), \(r_1\dot{\theta }_1=\sin \sigma _1=\pm 1\) on I. This means that \(\theta _1\) is either strictly decreasing or strictly increasing on I. This entails that the point x is not a boundary point of \(M_\tau \) in \(\mathbb {S}^1_\tau \) and this proves the claim.

It follows from these considerations that \(M{\setminus }\overline{B}(0,r)\) consists of a finite union of disjoint centred circles. Note that \(f\ge e^{h(0)}=:c>0\) on \(\mathbb {R}^2\). As a result, \(+\infty >P_f(E)\ge cP(E)\) and in particular the relative perimeter \(P(E,\mathbb {R}^2{\setminus }\overline{B}(0,r))<+\infty \). This explains why \(M{\setminus }\overline{B}(0,r)\) comprises only finitely many circles.

(ii) Let E be a bounded minimiser of (1.2) such that E is open, \(M:=\partial E\) is a \(C^{1,1}\) hypersurface in \(\mathbb {R}^2\) and \(E{\setminus }\{0\}=E^{sc}\) as in Theorem 4.5. Assume that \(R>0\). By (i), \(M{\setminus }\overline{B}(0,R)\) consists of a finite union of disjoint centred circles. We claim that only one of the possibilities
$$\begin{aligned} M_R&=\emptyset ,\,M_R=\mathbb {S}^1_R,\,M_R=\{Re_1\}\text { or }M_R=\{-Re_1\} \end{aligned}$$
(10.3)
holds. To prove this suppose that \(M_R\ne \emptyset \) and \(M_R\ne \mathbb {S}^1_R\). Bearing in mind Lemma 5.2 we may choose \(x\in M_R\) such that x lies on the boundary of \(M_R\) relative to \(\mathbb {S}^1_R\). Assume that \(x\in H\). Let \(\gamma _1:I\rightarrow M\) be a local parametrisation of M with \(\gamma _1(0)=x\) with the usual conventions. We first notice that \(\cos (\sigma (x))=0\) for otherwise we obtain a contradiction to Theorem 10.3. As \(r_1\) is decreasing on I and x is a relative boundary point it holds that \(r_1<R\) on \(I^+:=I\cap \{s>0\}\). As \(M{\setminus }\overline{\varLambda _1}\) is open in M we may suppose that \(\gamma _1(I^+)\subset M{\setminus }\overline{\varLambda _1}\). According to Theorem 6.5 the curvature k of \(\gamma _1(I^+)\cap B(0,R)\) is a.e. constant as \(\rho \) vanishes on (0, R). Hence \(\gamma _1(I^+)\cap B(0,R)\) consists of a line or circular arc. The fact that \(\cos (\sigma (x))=0\) means that \(\gamma _1(I^+)\cap B(0,R)\) cannot be a line. So \(\gamma _1(I^+)\cap B(0,R)\) is an open arc of a circle C containing x in its closure with centre on the line-segment [0, x] and radius \(r\in (0,R)\). By considering a local parametrisation, it can be seen that \(C\cap B(0,R)\subset M\). But this contradicts the fact that \(E{\setminus }\{0\}=E^{sc}\). In summary, \(M_R\subset \{\pm Re_1\}\). Finally note that if \(M_R=\{\pm Re_1\}\) then \(M_R=\mathbb {S}^1_R\) by Lemma 5.2. This establishes (10.3).

Suppose that \(M_R=\emptyset \). As both sets M and \(\mathbb {S}^1_R\) are compact, \(d(M,\mathbb {S}^1_R)>0\). Assume first that \(\mathbb {S}^1_R\subset E\). Put \(F:=B(0,R){\setminus } E\) and suppose \(F\ne \emptyset \). Then F is a set of finite perimeter, \(F\subset \subset B(0,R)\) and \(P(F)=P(E,B(0,R))\). Let B be a centred ball with \(|B|=|F|\). By the classical isoperimetric inequality, \(P(B)\le P(F)\). Define \(E_1:=(\mathbb {R}^2{\setminus } B)\cap (B(0,R)\cup E)\). Then \(V_f(E_1)=V_f(E)\) and \(P_f(E_1)\le P_f(E)\). That is, \(E_1\) is a minimiser of (1.2) such that \(\partial E_1\) consists of a finite union of disjoint centred circles. Now suppose that \(\mathbb {S}^1_R\subset \mathbb {R}^2{\setminus }\overline{E}\). In like fashion we may redefine E via \(E_1:=B\cup (E{\setminus }\overline{B}(0,R))\) with B a centred ball in B(0, R). The remaining cases in (10.3) can be dealt with in a similar way. The upshot of this argument is that there exists a m inimiser of (1.2) whose boundary M consists of a finite union of disjoint centred circles in case \(R>0\).

Now suppose that \(R=0\). By (i), \(M{\setminus }\overline{B}(0,r)\) consists of a finite union of disjoint centred circles for any \(r\in (0,1)\). If these accumulate at 0 then M fails to be \(C^1\) at the origin. The assertion follows. \(\square \)

Lemma 10.5

Suppose that the function \(J:[0,+\infty )\rightarrow [0,+\infty )\) is continuous non-decreasing and \(J(0)=0\). Let \(N\in \mathbb {N}\cup \{+\infty \}\) and \(\{t_h:h=0,\ldots ,2N+1\}\) a sequence of points in \([0,+\infty )\) with
$$\begin{aligned} t_0>t_1>\cdots>t_{2h}>t_{2h+1}>\cdots \ge 0. \end{aligned}$$
Then
$$\begin{aligned} +\,\infty \ge \sum _{h=0}^{2N+1}J(t_h)\ge J\left( \sum _{h=0}^{2N+1}(-1)^{h}t_h\right) . \end{aligned}$$

Proof

We suppose that \(N=+\infty \). The series \(\sum _{h=0}^\infty (-1)^{h}t_h\) converges by the alternating series test. For each \(n\in \mathbb {N}\),
$$\begin{aligned} \sum _{h=0}^{2n+1}(-1)^{h}t_h\le t_0 \end{aligned}$$
and the same inequality holds for the infinite sum. As in Step 2 in [5] Theorem 2.1,
$$\begin{aligned} +\,\infty \ge \sum _{h=0}^\infty J(t_h) \ge J(t_0) \ge J\left( \sum _{h=0}^\infty (-1)^{h}t_h\right) \end{aligned}$$
as J is non-decreasing. \(\square \)

Proof of Theorem 1.1

There exists a minimiser E of (1.2) with the property that \(\partial E\) consists of a finite union of disjoint centred circles according to Lemma 10.4. As such we may write
$$\begin{aligned} E=\bigcup _{h=0}^NA((a_{2h+1},a_{2h})) \end{aligned}$$
where \(N\in \mathbb {N}\) and \(+\infty>a_0>a_1>\cdots>a_{2N}>a_{2N+1}>0\). Define
$$\begin{aligned}&\mathtt {f}:[0,+\infty )\rightarrow \mathbb {R};t\mapsto e^{h(t)};\nonumber \\&g:[0,+\infty )\rightarrow \mathbb {R};t\mapsto t\mathtt {f}(t);\nonumber \\&G:[0,+\infty )\rightarrow \mathbb {R};t\mapsto \int _0^t g\,d\tau .\nonumber \end{aligned}$$
Then \(G:[0,+\infty )\rightarrow [0,+\infty )\) is a bijection with inverse \(G^{-1}\). Define the strictly increasing function
$$\begin{aligned} J:[0,+\infty )\rightarrow \mathbb {R};t\mapsto g\circ G^{-1}. \end{aligned}$$
Put \(t_h:=G(a_h)\) for \(h=0,\ldots ,2N+1\). Then \(+\infty>t_0>t_1>\cdots>t_{2N}>t_{2N+1}>> 0\). Put \(B:=B(0,r)\) where \(r:=G^{-1}(v/2\pi )\) so that \(V_f(B)=v\). Note that
$$\begin{aligned} v= & {} V_f(E)=2\pi \sum _{h=0}^N\Big \{G(a_{2h})-G(a_{2h+1})\Big \}\\= & {} 2\pi \sum _{h=0}^{2N+1}(-1)^{h}t_h. \end{aligned}$$
By Lemma 10.5,
$$\begin{aligned} P_f(E)= & {} 2\pi \sum _{h=0}^{2N+1}g(a_h)=2\pi \sum _{h=0}^{2N+1}J(t_h) \\\ge & {} 2\pi J\left( \sum _{h=0}^{2N+1}(-1)^h t_h\right) \\= & {} 2\pi J(v/2\pi )=P_f(B). \end{aligned}$$
\(\square \)

Proof of Theorem 1.2

Let \(v>0\) and E be a minimiser for (1.2). Then E is essentially bounded by Theorem 3.1. By Theorem 4.5 there exists an \(\mathscr {L}^2\)-measurable set \(\widetilde{E}\) with the properties
  1. (a)

    \(\widetilde{E}\) is a minimiser of (1.2);

     
  2. (b)

    \(L_{\widetilde{E}}=L_E\) a.e. on \((0,+\infty )\);

     
  3. (c)

    \(\widetilde{E}\) is open, bounded and has \(C^{1,1}\) boundary;

     
  4. (d)

    \(\widetilde{E}{\setminus }\{0\}=\widetilde{E}^{sc}\).

     
(i) Suppose that \(0<v\le v_0\) so that \(R>0\). Choose \(r\in (0,R]\) such that \(V(B(0,r))=V(E)=v\). Suppose that \(\widetilde{E}{\setminus }\overline{B}(0,R)\ne \emptyset \). By Lemma 10.4 there exists \(t>R\) such that \(\mathbb {S}^1_t\subset M\). As g is strictly increasing, \(g(t)>g(r)\). So \(P_f(E)=P_f(\widetilde{E})\ge \pi g(t)>\pi g(r)=P_f(B(0,r))\). This contradicts the fact that E is a minimiser for (1.2). So \(\widetilde{E}\subset \overline{B}(0,R)\) and \(L_{\widetilde{E}}=0\) on \((R,+\infty )\). By property (b), \(|E{\setminus }\overline{B}(0,R)|=0\). By the uniqueness property in the classical isoperimetric theorem (see for example [12] Theorem 4.11) the set E is equivalent to a ball B in \(\overline{B}(0,R)\).

(ii) With \(r>0\) as before, \(V(B(0,r))=V(E)=v>v_0=V(B(0,R))\) so \(r>R\). If \(\widetilde{E}{\setminus }\overline{B}(0,r)\ne \emptyset \) we derive a contradiction in the same way as above. Consequently, \(\widetilde{E}=B:=B(0,r)\). Thus, \(L_E=L_B\) a.e. on \((0,+\infty )\); in particular, \(|E{\setminus } B|=0\). This entails that E is equivalent to B. \(\square \)

References

  1. 1.
    Ambrosio, L., Fusco, N., Pallara, D.: Functions of Bounded Variation and Free Discontinuity Problems. Oxford University Press, Oxford (2000)zbMATHGoogle Scholar
  2. 2.
    Barchiesi, M., Cagnetti, F., Fusco, N.: Stability of the Steiner symmetrization of convex sets. J. Eur. Math. Soc. 15, 1245–1278 (2013)MathSciNetCrossRefzbMATHGoogle Scholar
  3. 3.
    Borell, C.: The Ornstein–Uhlenbeck Velocity Process in Backward Time and Isoperimetry. Chalmers Tekniska Hógskola, Göteborgs Universitet, Gothenburg (1986)Google Scholar
  4. 4.
    Boyer, W., Brown, B., Chambers, G., Loving, A., Tammen, S.: Isoperimetric regions in \({\mathbb{R}}^n\) with density \(r^p\), Analysis and Geometry in Metric Spaces, ISSN (Online) pp. 2299–3274 (2016)Google Scholar
  5. 5.
    Betta, F.M., Brock, F., Mercaldo, A., Posteraro, M.R.: Weighted isoperimetric inequalities on \(\mathbb{R}^n\) and applications to rearrangements. Math. Nachr. 281(4), 466–498 (2008)MathSciNetCrossRefzbMATHGoogle Scholar
  6. 6.
    Carothers, N.L.: Real Analysis. Cambridge University Press, Cambridge (2000)CrossRefzbMATHGoogle Scholar
  7. 7.
    Chambers, G.R.: Proof of the log-convex density conjecture. To appear in J. Eur. Math. Soc (2016)Google Scholar
  8. 8.
    Cinti, E., Pratelli, A.: Regularity of isoperimetric sets in \({\mathbb{R}}^2\) with density. To appear in Math. Ann. (2017)Google Scholar
  9. 9.
    De Giorgi, E.: Sulla proprietá isoperimetrica dell’ipersfera, nella classe degli insiemi aventi frontiera orientata di misura finita. Atti Accad. Naz. Lincei Mem. Cl. Sci. Fis. Mat. Mat. Sez. I 5, 33–44 (1958)MathSciNetzbMATHGoogle Scholar
  10. 10.
    Dubins, L.E.: On curves of minimal length with a constraint on average curvature, and with prescribed initial and terminal positions and tangents. Am. J. Math. 79(3), 497–516 (1957)MathSciNetCrossRefzbMATHGoogle Scholar
  11. 11.
    Figalli, A., Maggi, F.: On the isoperimetric problem for radial log-convex densities. Calc. Var. Partial Differ. Equ. 48(3–4), 447–489 (2013)MathSciNetCrossRefzbMATHGoogle Scholar
  12. 12.
    Fusco, N.: The classical isoperimetric theorem. Rend. Acc. Sci. Fis. Mat. Napoli 4(71), 63–107 (2004)MathSciNetzbMATHGoogle Scholar
  13. 13.
    Giusti, E.: Minimal Surfaces and Functions of Bounded Variation. Birkhäuser, Boston (1984)CrossRefzbMATHGoogle Scholar
  14. 14.
    Gradshteyn, I.S., Ryzhik, I.M.: Tables of Integrals, Series and Products. Academic Press, New York (1965)zbMATHGoogle Scholar
  15. 15.
    Hadamard, J.: Étude sur les propriétés des fonctions entieres et en particulier d’une fonction considérée par Riemann. J. Math. Pures et Appl. 58, 171–215 (1893)zbMATHGoogle Scholar
  16. 16.
    Hale, J.: Ordinary Differential Equations. Wiley, Hoboken (1969)zbMATHGoogle Scholar
  17. 17.
    Hermite, C.: Sur deux limites d’une intégrale définie. Mathesis 3, 82 (1883)Google Scholar
  18. 18.
    Howard, H., Treibergs, A.: A reverse isoperimetric inequality, stability and extremal theorems for plane curves with bounded curvature. Rocky Mt. J. Math. 25(2), 635–684 (1995)MathSciNetCrossRefzbMATHGoogle Scholar
  19. 19.
    Kolesnikov, A.V., Zhdanov, R.I.: On Isoperimetric Sets of Radially Symmetric Measures, Concentration, Functional Inequalities and Isoperimetry, Contemporary Mathematics, vol. 545, pp. 123–154. American Mathematical Society, Providence (2011)CrossRefzbMATHGoogle Scholar
  20. 20.
    Lee, J.M.: Manifolds and Differential Geometry. American Mathematical Society, Providence (2009)CrossRefzbMATHGoogle Scholar
  21. 21.
    Morgan, F.: Regularity of isoperimetric hypersurfaces in Riemannian manifolds. Trans. AMS 355(12), 5041–5052 (2003)MathSciNetCrossRefzbMATHGoogle Scholar
  22. 22.
    Morgan, F., Pratelli, A.: Existence of isoperimetric regions in \(\mathbb{R}^n\) with density. Ann. Glob. Anal. Geom. 43(4), 331–365 (2013)CrossRefzbMATHGoogle Scholar
  23. 23.
    Perugini, M.: Ph.D. Thesis (in preparation), University of Sussex (2016)Google Scholar
  24. 24.
    Rosales, C., Cañete, A., Bale, V., Morgan, F.: On the isoperimetric problem in Euclidean space with density. Calc. Var. Partial Differ. Equ. 31(1), 27–46 (2008)MathSciNetCrossRefzbMATHGoogle Scholar
  25. 25.
    Simmons, G.F.: Introduction to Topology and Modern Analysis. McGraw-Hill, New York (1963)zbMATHGoogle Scholar
  26. 26.
    Spivak, M.: A Comprehensive Introduction to Differential Geometry, Publish or Perish, vol. 2 (1970)Google Scholar
  27. 27.
    Tamanini, I.: Regularity results for almost minimal oriented hypersurfaces in \(\mathbb{R}^n\), Quaderni del Dipartimento di Matematica dell’ Universita di Lecce (1984)Google Scholar

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Authors and Affiliations

  1. 1.School of MathematicsUniversity of BristolBristolUK

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