An Unstable Two-Phase Membrane Problem and Maximum Flux Exchange Flow

Abstract

Let U be a bounded open connected set in $\mathbb {R}^n$ ($n\ge 1$). We refer to the unique weak solution of the Poisson problem $-\Delta u = \chi _A$ on U with Dirichlet boundary conditions as $u_A$ for any measurable set A in U. The function $\psi :=u_U$ is the torsion function of U. Let V be the measure $V:=\psi \,\mathscr {L}^n$ on U where $\mathscr {L}^n$ stands for n-dimensional Lebesgue measure. We study the variational problem

$$\begin{aligned} I(U,p):=\sup \Big \{ J(A)-V(U)\,p^2\,\Big \} \end{aligned}$$

with $p\in (0,1)$ where $J(A):=\int _Au_A\,dx$ and the supremum is taken over measurable sets $A\subset U$ subject to the constraint $V(A)=pV(U)$. We relate the above problem to an unstable two-phase membrane problem. We characterise optimsers in the case $n=1$. The proof makes use of weighted isoperimetric and Pólya–Szegö inequalities.

1 Introduction and Motivation

Let U be a bounded open connected set in $\mathbb {R}^n$ ($n\ge 1$). We refer to the unique weak solution of the Poisson problem

$$\begin{aligned} -\Delta u = \chi _A\text { on }U,\,u\in W^{1,2}_0(U), \end{aligned}$$

(1.1)

as $u_A$ for any measurable set A in U. The function $\psi :=u_U$ is the torsion function of U. Let V be the measure $V:=\psi \,\mathscr {L}^n$ on U where $\mathscr {L}^n$ stands for the n-dimensional Lebesgue measure. For $p\in (0,1)$ consider the variational problem

$$\begin{aligned} I(U,p):=\sup \Big \{ J(A)-V(U)\,p^2\,\Big \} \end{aligned}$$

(1.2)

where $J(A):=(u_A,\chi _A)$ and the supremum is taken over measurable sets $A\subset U$ subject to the constraint $V(A)=pV(U)$. Here, $(\cdot ,\cdot )$ stands for the usual inner product in the real Hilbert space $L^2(U)$. Any maximiser E in (1.2) will be called an optimal configuration for the data (U, p). If E is an optimal configuration and $u=u_E$, then (u, E) will be called an optimal pair.

In Corollary 2.2 we show that for each $p\in (0,1)$ the problem (1.2) admits an optimal pair (u, E) for the data (U, p). In Proposition 3.3 we characterise the optimal configuration E as a super level set of $u/\psi$; that is, $E=\{u>c\psi \}$ for some $c\in (0,1)$ up to $\mathscr {L}^n$-a.e. equivalence. The derivation assumes that U is a $C^{1,1}$ domain. Under this last assumption, we show in Corollary 3.4 that u satisfies the following semi-linear elliptic partial differential equation with discontinuous nonlinearity. Put $v:=u-c\psi$ with c as above. Then v is a strong solution of the problem

$$\begin{aligned} -\Delta v=(1-c)\chi _{\Omega _+(v)}-c\chi _{\Omega _-(v)}\text { on }U \end{aligned}$$

where $\Omega _\pm (v):=\{\pm v>0\}$. The above equation is similar to Problem C (the two-phase membrane problem) in [20, 1.2.3] but with a sign change; see also the unstable membrane problem [20] 2.5. It is noted in [20, 1.1.7] that the composite membrane problem (see [6, 7]) is akin to the unstable membrane problem. Our terminology is adopted from [6, 7] and in places there is a similarity in method. The regular part of the free boundary $\Gamma (v)=\partial \Omega _\pm (v)\cap U$ is real-analytic (Theorem 3.7). In Sect. 4 we replace U with the unit ball B in $\mathbb {R}^n$ ($n\ge 2$). For $p\in (0,1)$ we show that any optimal configuration E for the data (B, p) possesses spherical cap symmetry $\mathscr {L}^n$-a.e. (see Theorem 4.1).

In the remainder of the article, we study the problem (1.2) in the one-dimensional case $n=1$ and take $B=(-1,1)$. In Theorem 9.5 we show that any optimal configuration E with data (B, p) is $\mathscr {L}^1$-a.e. equivalent to an open interval abutting a boundary point of B. A first step in obtaining this result is to transform the problem using an analog of the ground-state transformation (with the torsion function in place of the ground-state) (see Proposition 9.2). We then obtain an isoperimetric inequality on B with volume density $\psi$ and perimeter density $\psi ^{3/2}$ (Theorem 6.3) and a corresponding Hardy-Littlewood type inequality (Theorem 6.6) and a Pólya–Szegö inequality (Theorem 7.10). We also study the case of equality in the isoperimetric and Pólya–Szegö inequalities (Theorem 6.4 and Corollary 8.7 respectively). We have been guided by [2] in obtaining these results, though our setting and proofs are slightly different.

We have not obtained an analog of Theorem 9.5 in the case $n\ge 2$. At least part of our method transfers to higher dimensions. There is a counterpart of the isoperimetric inequality Theorem 6.3 (though its derivation is more involved with the usual difficulties around regularity and stability) and the Hardy–Littlewood inequality is a ready consequence. A potential stumbling block is the validity of a corresponding Pólya–Szegö inequality. We note that the sufficient conditons given in [22] are stringent.

The problem (1.2) is related to maximum flux exchange flow (a model of magma flow in a volcanic vent [15]). We take $n=2$ and consider a configuration of two immiscible fluids in a vertical duct with cross-section U in a state of steady flow. The densities of the fluids are labelled $\rho$, $\rho ^\prime$ with $\rho > \rho ^\prime$ and each fluid has unit viscosity. The pressure p has constant gradient $\partial p/\partial z = -G$ on U. Suppose the fluid with density $\rho$ occupies a region A in U. By the Navier–Stokes equations, the vertical component of velocity u satisfies

$$\begin{aligned} \begin{array}{lclcl} 0 &{} = &{} \Delta u + G - \rho g &{} \text { on } &{} A;\\ 0 &{} = &{} \Delta u + G - \rho ^\prime g &{} \text { on } &{} U\setminus A. \end{array} \end{aligned}$$

Dirichlet boundary conditions are imposed on the boundary of U and it is assumed that u and its gradient are continuous on the interface between the two regions A and $U\setminus A$.

The parameter G lies in the interval $(\rho ^\prime g,\rho g)$ which allows the possibility of bidirectional flow. On rescaling (and relabelling the velocities) we obtain the system

$$\begin{aligned} \begin{array}{lclcl} 0 &{} = &{} \Delta u - \lambda - 1 &{} \text { on } &{} A;\\ 0 &{} = &{} \Delta u - \lambda + 1 &{} \text { on } &{} U\setminus A; \end{array} \end{aligned}$$

(1.3)

where

$$\begin{aligned} \lambda : = \frac{(\rho ^\prime + \rho )g - 2G}{(\rho - \rho ^\prime )g}\in (-1,1) \end{aligned}$$

is a proxy for the pressure gradient. Two problems arise: one to maximise the flux $(\chi _{U\setminus A},u)$ over regions A which satisfy the flux balance condition $(u,1) = 0$ with constant $\lambda$; the other in which we optimize also over $\lambda$. In detail,

$$\begin{aligned} \gamma (U)&:=\sup \left\{ (\chi _{U\setminus A},u):(u,1) = 0,\,A\subset U\text { open},\,\lambda \in (-1,1)\right\} ,\end{aligned}$$

(1.4)

$$\begin{aligned} \gamma (U,\lambda )&:=\sup \left\{ (\chi _{U\setminus A},u):(u,1) = 0,\,A\subset U\text { open}\right\} , \end{aligned}$$

(1.5)

where in the latter $\lambda$ is fixed in the interval $(-1,1)$. In the case $n=1$ and $U=B$ we show that any optimal configuration E for the problem (1.5) with data $(B,\lambda )$ is $\mathscr {L}^1$-a.e. equivalent to an open interval abutting a boundary point of B in Theorem 9.8. Moreover, any optimal configuration E for the problem (1.4) is $\mathscr {L}^1$-a.e. equivalent to either $(-1,0)$ or (0, 1).

2 Existence of Optimal Configurations

Define

$$\begin{aligned} \mathscr {V}_t:=\Big \{f\in L^2(U):0\le f\le 1\,\mathscr {L}^n\text {-a.e. on }U\text { and }(f,\,\psi ) \le t\Big \} \end{aligned}$$

for $t\in (0,V(U))$ and consider the variational problem

$$\begin{aligned} \beta (U,t):=\sup \Big \{\,J(f):f\in \mathscr {V}_t\,\Big \} \end{aligned}$$

(2.1)

where $J(f):=(u_f,f)$ and $u_f$ is the unique solution of the Poisson problem (1.1) but with inhomogeneity f. The first main result runs as follows.

Theorem 2.1

Fix $t\in (0,V(U))$. Then

• (i) there exists $f\in \mathscr {V}_t$ such that $\beta (U,t) = J(f)$;

• (ii) $(\psi ,f)=t$;

• (iii) f has the form $f=\chi _E$ for some measurable set E in U.

Corollary 2.2

For each $p\in (0,1)$ the problem (1.2) admits an optimal pair (u, E) for the data (U, p).

Proof

Let $p\in (0,1)$ and put $t:=p V(U)$. Let E be as in Theorem 2.1 (iii). Then $V(E)=(\psi ,\chi _E)=t=pV(U)$. Let $A\subset U$ be a measurable set with $V(A)=pV(U)$. Then $f=\chi _A\in \mathscr {V}_t$ so $J(E)\ge J(A)$. $\square$

We prepare a few lemmas before proving Theorem 2.1.

Lemma 2.3

Let X, Y be (real) Banach spaces and suppose that $X\subset Y$ with continuous embedding. Let $(x_h)$ be a sequence in X which converges weakly in X to $x\in X$. Then $(x_h)$ converges weakly to x in Y.

Proof

Note that for any $g\in Y^\prime$, $g\vert _X\in X^\prime$. $\square$

We remark that the Dirichlet Laplacian $(D(\Delta ),\,\Delta )$ is associated with the Dirichlet form $(\mathscr {F},\,\mathscr {E})$ in $L^2(U)$ with form domain $\mathscr {F}:=W^{1,2}_0(U)$ and

Let G stand for the corresponding Green operator.

Lemma 2.4

Let $t\in (0,V(U))$. Then

• (i) the functional $J:\,\mathscr {V}_t\rightarrow \mathbb {R}$ is continuous in the topology of weak sequential convergence;

• (ii) $J:\,\mathscr {V}_t\rightarrow \mathbb {R}$ is convex.

Proof

(i) Suppose that a sequence $(f_h)$ in $\mathscr {V}_t$ converges weakly to an element $f\in \mathscr {V}_t$ in $L^2(U)$. Put $u_h:=G f_h\in L^2(U)$. For each h and $\varphi \in L^2(U)$,

$$\begin{aligned} (u_h,\,\varphi ) = (Gf_h,\,\varphi ) = (f_h,\,G\varphi ), \end{aligned}$$

by symmetry of G so that $(u_h,\,\varphi )\rightarrow (f,\,G\varphi )= (u,\,\varphi )$ as $h\rightarrow \infty$ where $u:=Gf$. We also have that

$$\begin{aligned} \mathscr {E}(u_h,\,\varphi )=(f_h,\,\varphi )\rightarrow (f,\,\varphi )=\mathscr {E}(u,\,\varphi )\text { as }h\rightarrow \infty \end{aligned}$$

for any $\varphi \in \mathscr {F}$. This means that $(u_h)$ converges weakly to u in $\mathscr {F}$.

Note that $W^{1,2}_0(U)\subset W^{1,1}_0(U)$ and $\Vert u\Vert _{W^{1,1}_0(U)}\le \sqrt{2\,|U|}\,\Vert u\Vert _{W^{1,2}_0(U)}$ for each $u\in W^{1,2}_0(U)$. By Lemma 2.3, $(u_h)$ converges weakly to u in $W^{1,1}_0(U)$.

If $n\ge 2$ we may use the Rellich–Kondrachov compactness theorem [10, 5.7] for example and [16, Theorem 21.2.9] to conclude that $(u_h)$ converges strongly to u in $L^1(U)$. Now

$$\begin{aligned} J(f) - J(f_h) = (u,f) - (u_h,f_h) = (u,f-f_h) + (f_h,u-u_h) \end{aligned}$$

and the right-hand side converges to zero as $h\rightarrow \infty$ as $(f_h)$ is bounded in $L^\infty (U)$. This shows that $J(f_h)\rightarrow J(f)$ as $h\rightarrow \infty$.

In the case $n=1$ we use the fact that $W^{1,2}_0(U)$ is compactly embedded in $C^0(\overline{U})$ (see [13, Theorem 7.22]) and hence in $L^1(U)$.

(ii) Let $f\in \mathscr {V}_t$. By Dirichlet’s principle [10, 2.2.5] for example,

$$\begin{aligned} E(f):=\inf _{v\in \mathscr {F}}\Big \{(1/2)\,\mathscr {E}(v,\,v)-(v,\,f)\Big \}=-(1/2)\,J(f). \end{aligned}$$

The functional E is concave so that J is convex. $\square$

Lemma 2.5

Let $t\in (0,\,V(U))$. A function f in the convex set $\mathscr {V}_t\subset L^2(U)$ is extremal only if $f=\chi _A$ $\mathscr {L}^n$-a.e. on U for some $A\subset U$ measurable with $(\psi ,\chi _A)\le t$.

Proof

The proof runs as in [11, Lemma 2]. A measurable function f on U is $\mathscr {L}^n$-a.e. equivalent to $\chi _A$ for some $A\subset U$ measurable if and only if $f(1-f)=0$ $\mathscr {L}^n$-a.e. on U. Suppose that $f\in \mathscr {V}_t$ is an extremal element and assume that $\vert \{f(1-f)\ne 0\}\vert >0$ for a contradiction. Then there exists $\varepsilon >0$ and a measurable set E in U with positive $\mathscr {L}^n$-measure such that $\varepsilon \le f\le 1-\varepsilon \text { on }E$. Decompose E into two disjoint sets $E_1,\,E_2$ each with positive $\mathscr {L}^n$-measure. Choose $\alpha =(\alpha _1,\,\alpha _2)\in \mathbb {R}^2\setminus \{0\}$ such that $\alpha _1(\psi ,\,\chi _{E_1})+\alpha _2(\psi ,\,\chi _{E_2})=0$ and define

$$\begin{aligned} f_{\tau }:=f+\tau \sum _{j=1}^2\alpha _j\chi _{E_j} \end{aligned}$$

for $\tau \in \mathbb {R}$. Then $f_\tau \in \mathscr {V}_t$ for $|\tau |\le \varepsilon /|\alpha _1|\vee |\alpha _2|$. We then derive the contradiction that f is not extremal as $f=(1/2)\left\{ f_\tau +f_{-\tau }\right\}$ for such $\tau$. $\square$

Proof of Theorem 2.1

Let $(f_h)$ be a maximising sequence for $\beta (U,t)$. Now $\mathscr {V}_t$ is weakly sequentially compact in $L^2(U)$. This follows by appeal to [16, Theorem 10.2.9] due to the fact that $\mathscr {V}_t$ is bounded, closed and convex in the reflexive Banach space $L^2(U)$. So we may assume that $(f_h)$ converges weakly in $L^2(U)$ to some $f\in \mathscr {V}_t$ as $h\rightarrow \infty$ after choosing a subsequence if necessary. By Lemma 2.4 (i),

$$\begin{aligned} \beta (U,t)=\lim _{h\rightarrow \infty }J(f_h)=J(f), \end{aligned}$$

giving item (i) of the Theorem. It is straightforward to see that $(\psi ,\,f)=t$ and hence (ii).

We now argue as in [8, Corollary 6.2]. By [5, Chapitre II §7 Proposition 1.1 (EVT II.58)], J attains its supremum on $\mathscr {V}_t$ at an extremal point f. We then invoke Lemma 2.5 to conclude that f has the form $f=\chi _A$ $\mathscr {L}^n$-a.e. on U for some measurable set A in U and hence (iii). $\square$

3 Some Partial Regularity Results

Proposition 3.1

Suppose that U is a $C^{1,1}$ domain. Let $E\subset U$ be a measurable set and $u=u_E$. Then $V(\{u=t\psi \})=0$ for each $t\in (0,1)$.

Proof

By [13, Theorem 9.15], $u\in W^{2,p}(U)$ for any $1<p<\infty$. Put $v:=u-t\psi$, $N_t:=\{v=0\}$ and assume that $|N_t|>0$. By [13, Lemma 7.7], we derive that $D^\alpha v=0$ $\mathscr {L}^n$-a.e. and hence V-a.e. on $N_t$ for any multi-index $\alpha$ with $|\alpha |=1$. Observe that $D^\alpha v$ belongs to $W^{1,p}(U)$ for $|\alpha |\le 1$. Applying the last-mentioned lemma once more, we see $D^\alpha v=0$ V-a.e. on $N_t$ for any multi-index $\alpha$ with $|\alpha |\le 2$. So $-\Delta v=0$ V-a.e. on $N_t$. But $-\Delta v=\chi _E-t\,\chi _U$ V-a.e. on U. This leads to a contradiction. $\square$

We require a version of the bathtub principle (see [17, Theorem 1.14]). Let $(X,\mathscr {A},\mu )$ be a finite measure space and $\rho$ a positive $\mathscr {A}$-measurable integrable function on X. Given $0<v<\mu (X)$, consider the variational problem

$$\begin{aligned} \sup \int _X\chi _E\,\rho \,d\mu \end{aligned}$$

(3.1)

where the supremum is taken over measurable sets $E\subset X$ with $\mu (E)=v$. We say that measurable sets A, B in X are equivalent $\mu$-a.e. and write $A=B$ if and only if $\mu (A\Delta B)=0$.

Theorem 3.2

Assume that

$$\begin{aligned} \mu (\{\rho =t\})=0\text { for all }t>0. \end{aligned}$$

(3.2)

Then for each $v\in (0,\mu (X))$ the problem (3.1) has a unique optimiser up to equivalence $\mu$-a.e. given by $E=\{\rho >s\}$ where

$$\begin{aligned} s:=\inf \{\tau >0:\,\mu (\{\rho >\tau \})\le v\}. \end{aligned}$$

Proof

The distribution function $\mu _\rho :(0,\infty )\rightarrow (0,V(U));\tau \mapsto \mu (\{\rho >\tau \})$ is non-increasing and right-continuous on $(0,\infty )$; in fact, continuous thanks to (3.2). By right-continuity of $\mu _\rho$, $\mu _\rho (s)\le v$; by left-continuity, the reverse inequality holds, so $\mu (E)=\mu _\rho (s)=v$. For a measurable set A in X with $\mu (A)=v$,

$$\begin{aligned} \int _X\chi _A\,\rho \,d\mu&=\int _0^\infty \mu (A\cap \{\rho >\tau \})\,d\tau \\&=\int _0^s\mu (A\cap \{\rho >\tau \})\,d\tau +\int _s^\infty \mu (A\cap \{\rho >\tau \})\,d\tau \\&\le sv+\int _s^\infty \mu (\{\rho >\tau \})\,d\tau =\int _X\chi _E\,\rho \,d\mu \end{aligned}$$

according to the layer cake representation [17, Theorem 1.13]. It follows that E is an optimiser for (3.1).

Suppose A is a measurable set in X with $\mu (A)=v$ which is not $\mu$-a.e. equivalent to E. Then

$$\begin{aligned} \mu (E)=\mu (E\setminus A)+\mu (E\cap A)=v=\mu (A\setminus E)+\mu (E\cap A) = \mu (A) \end{aligned}$$

so $\mu (E\setminus A)>0$ as otherwise $\mu (A\setminus E)=0$ and A is $\mu$-a.e. equivalent to E. By countable additivity,

$$\begin{aligned} 0<\mu (E\setminus A)=\lim _{\tau \downarrow s}\mu (\{\rho >\tau \}\setminus A). \end{aligned}$$

(3.3)

Thus,

$$\begin{aligned}&\int _X\chi _A\,\rho \,d\mu \\&\quad =\int _0^s\mu (A\cap \{\rho >\tau \})\,d\tau +\int _s^\infty \mu (\{\rho >\tau \})\,d\tau -\int _s^\infty \mu (\{\rho >\tau \}\setminus A)\,d\tau \\&\quad \le s\,v+\int _s^\infty \mu (\{\rho >\tau \})\,d\tau -\int _s^\infty \mu (\{\rho >\tau \}\setminus A)\,d\tau \\&\quad =\int _X\chi _E\,\rho \,d\mu -\int _s^\infty \mu (\{\rho >\tau \}\setminus A)\,d\tau <\int _X\chi _E\,\rho \,d\mu \end{aligned}$$

where the strict inequality follows from (3.3). $\square$

Let U be a $C^{1,1}$ domain and $p\in (0,1)$. Let (u, E) be an optimal pair for (1.2) with data (U, p). By [13, Corollary 9.18] we may assume that $u\in C^0(\overline{U})$.

Proposition 3.3

Suppose that U is a $C^{1,1}$ domain. Let $p\in (0,1)$ and suppose that (u, E) is an optimal pair for (1.2) with data (U, p). Then $V(E\Delta \{u>c\psi \})=V(E\Delta \{u\ge c\psi \})=0$ where $c\in (0,1)$ is uniquely determined by the condition

$$\begin{aligned} V(\{u>c\psi \})=pV(U). \end{aligned}$$

(3.4)

Proof

Put $F:=\{u>c\psi \}$ with c as in (3.4). Assume for a contradiction that $V(E\Delta F)>0$. We consider a version of Problem (3.1) on U with $\rho$ replaced by $w:=u_E/\psi$ and $\mu$ replaced by V. By Proposition 3.1, $V(\{w=t\})=0$ for each $t>0$; thus condition (3.2) holds. By uniqueness of the optimiser in Theorem 3.2 and the Cauchy–Schwarz inequality,

$$\begin{aligned} J(E) = (u_E,\chi _E)=\int _U w\chi _E\,dV <\int _U w\chi _F\,dV=\mathscr {E}(u_E,u_F)\le J(E)^{1/2}J(F)^{1/2} \end{aligned}$$

so that $J(E)<J(F)$, contradicting the assumption that E is an optimal configuration. The identity $V(E\Delta \{u\ge c\psi \})=0$ follows from Proposition 3.1. $\square$

Corollary 3.4

Suppose that U is a $C^{1,1}$ domain. Let $p\in (0,1)$ and suppose that (u, E) is an optimal pair for (1.2) with data (U, p). Put $v:=u-c\psi$ where c is given by (3.4). Then v is a strong solution of the problem

$$\begin{aligned} -\Delta v=(1-c)\chi _{\Omega _+(v)}-c\chi _{\Omega _-(v)}\text { on }U \end{aligned}$$

(3.5)

where $\Omega _\pm (v):=\{\pm v>0\}$ and $E=\Omega _+(v)$ $\mathscr {L}^n$-a.e.

Proof

By [13, Theorem 9.15], $u\in W^{2,p}(U)$ for any $1<p<\infty$ and u is a strong solution of $-\Delta u=\chi _E$. By Proposition 3.3, u is a strong solution of $-\Delta u=\chi _{\{u>c\psi \}}$. The result follows from the fact that $-\Delta (c\psi )=c\chi _U$ and Proposition 3.1.$\square$

Lemma 3.5

Let $p\in (0,1)$ and (u, E) be an optimal pair for the data (U, p). Then $(\psi -u,U\setminus E)$ is an optimal pair for the data $(U,1-p)$.

Proof

Let $A\subset U$ be a measurable set with $V(A)=pV(U)$. Then

$$\begin{aligned} J(U\setminus A)=J(A)-V(A)+V(U\setminus A) = J(A)+(1-2p)V(U) \end{aligned}$$

so that $J(U\setminus A)-V(U)(1-p)^2=J(A)-V(U)p^2$ and the result follows. $\square$

Put $\Gamma _\pm (v)=\partial \Omega _\pm (v)\cap U$.

Lemma 3.6

Suppose that U is a $C^{1,1}$ domain. Suppose that (u, E) is an optimal pair for the data $(U,\,p)$ and let v be as in Corollary 3.4. Then $\Gamma _+(v)=\Gamma _-(v)$.

Proof

Suppose that $x\in \Gamma _+(v)\setminus \Gamma _-(v)$. Then there exists $r>0$ such that $B(x,r)\subset U$, $u\ge c\,\psi$ on B(x, r) and $u(x)=c\psi (x)$. By Proposition 3.3, $V(B(x,r)\setminus E)\le V(\{u\ge c\psi \}\setminus E)=0$ and $B(x,r)\setminus E$ is a Lebesgue null set. Let $\Phi$ stand for the fundamental solution of Laplace’s equation in $\mathbb {R}^n$. By the mean-value formula (see [10, 2.5 Problem 3] for example), for any $0<\tau <r$,

$$\begin{aligned} u(x)&=|B(x,\tau )|^{-1}\,\int _{\partial B(x,\tau )}u\,d\mathcal {H}^{n-1}+\int _{B(x,\tau )}\Big \{\Phi (y-x)-\Phi (\tau \omega )\Big \}\chi _E\,dy \\&=|B(x,\tau )|^{-1}\,\int _{\partial B(x,\tau )}u\,d\mathcal {H}^{n-1}+\int _{B(x,\tau )}\Big \{\Phi (y-x)-\Phi (\tau \omega )\Big \}\,dy \\&\ge c\,|B(x,\tau )|^{-1}\,\int _{\partial B(x,\tau )}\psi \,d\mathcal {H}^{n-1}+\int _{B(x,\tau )}\Big \{\Phi (y-x)-\Phi (\tau \omega )\Big \}\,dy \\&> c\,\Big \{ |B(x,\tau )|^{-1}\,\int _{\partial B(x,\tau )}\psi \,d\mathcal {H}^{n-1}+\int _{B(x,\tau )}\Big \{\Phi (y-x)-\Phi (\tau \omega )\Big \}\,dy \Big \} \\&=c\psi (x) \end{aligned}$$

as $c\in (0,1)$, a contradiction. Here, $\omega$ is an arbitrary element in the unit sphere $\mathbb {S}^{n-1}$ in $\mathbb {R}^n$.

Now suppose that $x\in \Gamma _-(v)\setminus \Gamma _+(v)$. As before, there exists $r>0$ such that $u\le c\,\psi$ on B(x, r) and $u(x)=c\psi (x)$; alternatively, $\psi -u\ge (1-c)\,\psi$ on B(x, r) and $(\psi -u)(x)=(1-c)\psi (x)$. By Lemma 3.5, $(\psi -u,U\setminus E)$ is an optimal pair for the data $(U,1-p)$. We then get a contradiction as above. $\square$

Put $\Gamma (v):=\Gamma _+(v)=\Gamma _-(v)$ and $\Gamma ^*(v):=\Gamma (v)\cap \{|\nabla v|\ne 0\}$. The next theorem follows as in [20, Theorem 4.24].

Theorem 3.7

Suppose that U is a $C^{1,1}$ domain. Suppose that $(u,\,E)$ is an optimal pair for the data $(U,\,p)$ and that $x_0\in \Gamma ^*(v)$. Then there exists $r>0$ such that $\Gamma (v)\cap B(x,r)$ is a real-analytic hypersurface in B(x, r).

4 Spherical Cap Symmetry

In this section, we replace U by the open unit ball B in $\mathbb {R}^n$ ($n\ge 2$) centred at the origin. We prove the following symmetry result. The notion of spherical cap symmetry is defined below.

Theorem 4.1

Let $p\in (0,1)$. Suppose that (u, E) is an optimal pair for the data (B, p). Then E possesses spherical cap symmetry $\mathscr {L}^n$-a.e.

We first discuss the operation of polarisation for integrable functions on B (see [4] and references therein). For $\nu \in \mathbb {S}^{n-1}$ the closed half-space $H=H_\nu$ is defined by

$$\begin{aligned} H_\nu :=\left\{ x\in \mathbb {R}^n:\,x\cdot \nu \ge 0\right\} \end{aligned}$$

with an associated reflection $\tau _H:\,\mathbb {R}^n\rightarrow \mathbb {R}^n;\,x\mapsto x - 2\,(x\cdot \nu )\,\nu$. We refer to the collection of closed half-spaces H by $\mathscr {H}$. The polarisation $f_H$ of $f\in L^1_+(B)$ with respect to $H\in \mathscr {H}$ is defined as follows. Choose an $\mathscr {L}^n$-version $\widetilde{f}$ of f. Set

$$\begin{aligned} \widetilde{f}_H(x):= \left\{ \begin{array}{lcl} \widetilde{f}(x)\vee \widetilde{f}(\tau _H x) &{} \text { for } &{} x\in B\cap H,\\ \widetilde{f}(x)\wedge \widetilde{f}(\tau _H x) &{} \text { for } &{} x\in B\setminus H; \end{array} \right. \end{aligned}$$

$\widetilde{f}_H$ is $\mathscr {L}^n$-measurable and its $\mathscr {L}^n$-equivalence class $f_H:=[\widetilde{f}_H]$ is the polarisation of f. The definition is well-defined due to the fact that if $\widetilde{f}=\widetilde{g}$ $\mathscr {L}^n$-a.e. on B then $\widetilde{f}=\widetilde{g}$ $\mathcal {H}^{n-1}$-a.e. on $\mathbb {S}^{n-1}_\tau$ for $\mathscr {L}^1$-a.e. $0<\tau <1$, and vice-versa.

The Green kernel G(x, y) for B is given by

$$\begin{aligned} G(x,y)=\Phi (y-x) - \Phi (|x|(y-x^*))\text { for }(x,y)\in B\times B\setminus \mathtt {d}, \end{aligned}$$

where $\Phi$ is the fundamental solution of Laplace’s equation in $\mathbb {R}^n$ as before, $\mathtt {d}$ stands for the diagonal in $B\times B$ and the decoration $*$ refers to inversion in the unit sphere. We note the inequality

$$\begin{aligned} G(x,y) > G(\tau _H x,y)\text { for any }x,y\in B\cap \text {int}\,H, \end{aligned}$$

(4.1)

which follows from the strong maximum principle.

Theorem 4.2

Let $f\in L^1_+(B)$ and $H\in \mathscr {H}$. Then $J(f)\le J(f_H)$ with equality if and only if either $f=f_H$ or $f\circ \tau _H=f_H$ $\mathscr {L}^n$-a.e. on B.

Proof

Let $\widetilde{f}$ be an $\mathscr {L}^n$-version of f. Define

$$\begin{aligned} A^+:=\left\{ x\in B\cap H:\widetilde{f}(x)<\widetilde{f}(\tau _H x)\right\} \end{aligned}$$

and similarly $S^+$ but with the strict inequality replaced by the sign $>$. Put $A^-:=\tau _H A^+$ and $A:=A^+\cup A^-$. In this notation,

$$\begin{aligned} \widetilde{f}_H=\chi _A\widetilde{f}\circ \tau _H + \chi _{B\setminus A}\widetilde{f}. \end{aligned}$$

As a consequence,

$$\begin{aligned} J(f_H)&= J(\chi _A\,\widetilde{f}\circ \tau _H) + 2\,(\chi _A\widetilde{f}\circ \tau _H,G\chi _{B\setminus A}\widetilde{f}) + J(\chi _{B\setminus A}\widetilde{f}) \\&= J(\chi _A\widetilde{f}) + 2(\chi _A\widetilde{f}\circ \tau _H,G\chi _{B\setminus A}\widetilde{f}) + J(\chi _{B\setminus A}\widetilde{f}) \end{aligned}$$

and a similar identity holds for J(f) but without composition with reflection. We may then write

$$\begin{aligned}&J(f_H)-J(f)\\&=2(\chi _A[\widetilde{f}\circ \tau _H-\widetilde{f}],G\chi _{B\setminus A}\widetilde{f}) \\&=2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(y)\,dydx \\&\quad +2\int _{A^+}\int _{(B\setminus H)\setminus A^-}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(y)\,dydx \\&\quad +2\int _{A^-}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(y)\,dydx \\&\quad +2\int _{A^-}\int _{(B\setminus H)\setminus A^-}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(y)\,dydx \\&=2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(y)\,dydx \\&\quad +2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(\tau _H x,y)\widetilde{f}(\tau _H y)\,dydx \\&\quad -2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,\tau _H y)\widetilde{f}(y)\,dydx \\&\quad -2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))G(x,y)\widetilde{f}(\tau _H y)\,dydx \\&= 2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))(G(x,y)-G(\tau _H x,y)\widetilde{f}(y)\,dydx \\&\quad - 2\int _{A^+}\int _{B\cap H\setminus A^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))(G(x,y)-G(\tau _H x,y)\widetilde{f}(\tau _H y)\,dydx \\&= 2\int _{A^+}\int _{S^+}(\widetilde{f}(\tau _H x)-\widetilde{f}(x))(G(x,y)-G(\tau _H x,y))(\widetilde{f}(y)-\widetilde{f}(\tau _H y))\,dydx. \end{aligned}$$

It is clear from this representation with the help of (4.1) that $J(f)\le J(f_H)$.

In the case of equality, it holds that either $|A^+|=0$ or $|S^+|=0$. In the former case, $f=f_H$ while in the latter, $f\circ \tau _H = f_H$ $\mathscr {L}^n$-a.e. on B. $\square$

Let $\omega \in \mathbb {S}^{n-1}$. Given $0<\tau <1$ and $0<\alpha \le \pi$ the spherical cap $C_\omega (\tau ,\alpha )$ is the set

$$\begin{aligned} C_\omega (\tau ,\alpha ):=\big \{x=\tau \cos \theta \,\omega +\tau \sin \theta \,\eta :\,0\le \theta <\alpha ,\eta \in \mathbb {S}^{n-1}\cap \omega ^\perp \big \}\subset \mathbb {S}^{n-1}_\tau \end{aligned}$$

and has volume

$$\begin{aligned} s(\tau ,\alpha ):= \mathcal {H}^{n-1}(S_\omega (\tau ,\alpha ))=\omega _{n-2}\tau ^{n-1}\int _0^\alpha (\sin \theta )^{n-2}\,d\theta . \end{aligned}$$

For a Borel set E in B put

$$\begin{aligned} L(\tau ):=\mathcal {H}^{n-1}(E\cap \mathbb {S}^{n-1}_\tau )\text { for }0\le \tau <1\text { and } p(E) := \{0\le \tau <1:L(\tau )>0\}. \end{aligned}$$

The function L is Borel measurable The spherical cap symmetrisation of E is the set

$$\begin{aligned} C_\omega E:= \bigcup _{\tau \in p(E)}C_\omega (\tau ,\alpha ) \end{aligned}$$

(4.2)

where $\alpha \in (0,\pi ]$ is determined by $s(\tau ,\,\alpha )=L(\tau )$. Observe that $C_\omega E$ is a Borel set in B (use Fubini’s Theorem [1, 1.74] for example) and $|C_\omega E|=|E|$. We say that the Borel set $E\subset B$ possesses spherical cap symmetry $\mathscr {L}^n$-a.e. if $C_\omega E=E$ up to $\mathscr {L}^n$-a.e. equivalence for some $\omega \in \mathbb {S}^{n-1}$.

Let $f\in L^1_+(B)$ and choose an $\mathscr {L}^n$-version $\widetilde{f}$ of f. Put $m_{\widetilde{f}}(\tau ,t):=\mathcal {H}^{n-1}(\{\widetilde{f}>t\}\cap \mathbb {S}^{n-1}_\tau )$ for $t\in \mathbb {R}$ and $0\le \tau <1$. The function $m_{\widetilde{f}}(\tau ,\cdot )$ is non-increasing and right continuous. Define its right continuous inverse by

$$\begin{aligned} \widetilde{f}^\sharp (\tau ,s):=\inf \{t\in \mathbb {R}:m_{\widetilde{f}}(\tau ,t)\le s\}\text { for }0<s\le \mathcal {H}^{n-1}(\mathbb {S}^{n-1}_\tau ). \end{aligned}$$

For $x\in B$ put $\tau =|x|$ and choose $\alpha \in (0,\pi ]$ such that $x\cdot \omega =\tau \cos \alpha$ then define

$$\begin{aligned} C_\omega \widetilde{f}(x):=\widetilde{f}^\sharp (\tau ,s(\tau ,\alpha )). \end{aligned}$$

Note that $m_{\widetilde{f}}(\tau ,t)>s$ if and only if $\widetilde{f}^\sharp (\tau ,s)>t$. It follows that

$$\begin{aligned} \left\{ C_\omega \widetilde{f}>t\right\} = C_\omega \left\{ \widetilde{f}>t\right\} \text { for each }t\in \mathbb {R}. \end{aligned}$$

(4.3)

In particular, $C_\omega \widetilde{f}$ is Borel measurable and its $\mathscr {L}^n$-equivalence class $C_\omega f:=[C_\omega \widetilde{f}]$ is the spherical cap symmetrisation of f.

Before proving Theorem 4.1, we prepare a number of lemmas. We first discuss a useful two-point inequality. We introduce the notation

$$\begin{aligned} Q&:= \left\{ (x_1,x_2)\in \mathbb {R}^2:x_1\ge 0\text { and }x_2\ge 0\right\} , \\ R&:= \left\{ (x_1,x_2)\in Q:0\le x_2 < x_1\right\} , \\ S&:= \left\{ (x_1,x_2)\in Q:0\le x_1 < x_2\right\} . \end{aligned}$$

Equip Q with the $\ell ^1$-norm $\Vert x\Vert _1:=|x_1|+|x_2|$ and define a mapping $\varphi :Q\rightarrow Q$ via $(x_1,\,x_2)\mapsto (x_1\vee x_2,\,x_1\wedge x_2)$; $\varphi$ folds S onto R. A geometric argument establishes the following lemma.

Lemma 4.3

For any $x,\,y\in Q$, $\Vert \varphi x - \varphi y\Vert _1\le \Vert x-y\Vert _1$ with strict inequality if and only if $x\in R$ and $y\in \overline{S}$ or $x\in \overline{R}$ and $y\in S$ or the same with the rôles of x and y interchanged.

For $\omega \in \mathbb {S}^{n-1}$ introduce the collection of closed half-spaces $\mathscr {H}_\omega :=\Big \{H_\nu :\,\nu \in \mathbb {S}^{n-1}\text { and }\nu \cdot \omega \ge 0\Big \}$.

Lemma 4.4

Fix $\omega \in \mathbb {S}^{n-1}$. For any $H\in \mathscr {H}_\omega$ we have

• (i) for any $f,g\in L^1_+(B)$, $\Vert f_H - g_H\Vert _{L^1(B)}\le \Vert f - g\Vert _{L^1(B)};$

• (ii) for any $f\in L^1_+(B)$, $(C_\omega f)_H=C_\omega f$ $\mathscr {L}^n$-a.e. on B;

• (iii) for any $f\in L^1_+(B)$,

  $$\begin{aligned} \Vert f_H - C_\omega f\Vert _{L^1(B)}\le \Vert f - C_\omega f\Vert _{L^1(B)} \end{aligned}$$

  (4.4)

  with strict inequality if $|\left\{ f\circ \tau _H > f\right\} \cap H|>0$.

Proof

(i) By Lemma 4.3,

$$\begin{aligned}&\Vert f_H - g_H\Vert _{L^1(B)}\\&\,=\int _{B\cap H}|f_H-g_H|\,dx + \int _{B\setminus H}|f_H-g_H|\,dx \\&\,=\int _{B\cap H}|f\vee (f\circ \tau _H)-g\vee (g\circ \tau _H|\,dx \!+\! \int _{B\setminus H}|f\wedge (f\circ \tau _H)-g\wedge (g\circ \tau _H|\,dx \\&\,=\int _{B\cap H}\Big \{ |f\vee (f\circ \tau _H)-g\vee (g\circ \tau _H|+|(f\circ \tau _H)\wedge f-(g\circ \tau _H)\wedge g|\Big \}\,dx \\&\,=\int _{B\cap H}\Vert \varphi (f,f\circ \tau _H)-\varphi (g,g\circ \tau _H)\Vert _1\,dx \\&\,\le \int _{B\cap H}\Vert (f,f\circ \tau _H)-(g,g\circ \tau _H)\Vert _1\,dx \\&\,=\int _{B\cap H}\Big \{|f-g|+|f\circ \tau _H-g\circ \tau _H|\Big \}\,dx =\Vert f - g\Vert _{L^1(B)}. \end{aligned}$$

(ii) Let $f\in L^1_+(B)$ and $\widetilde{f}$ a $\mathscr {L}^n$-representative of f. For $x\in B\cap H$, $x\cdot \omega \ge (\tau _H x)\cdot \omega$ so $(C_\omega \widetilde{f})_H=C_\omega \widetilde{f}$ on B. Therefore,

$$\begin{aligned} C_\omega f=[C_\omega \widetilde{f}]=[(C_\omega \widetilde{f})_H]=[C_\omega \widetilde{f}]_H=(C_\omega f)_H. \end{aligned}$$

(iii) The inequality follows by (i) and (ii). On $B\cap H$ the pair $(C_\omega \widetilde{f},C_\omega \widetilde{f}\circ \tau _H)$ belongs to $\overline{R}$. By Lemma 4.3 if $(\widetilde{f},\widetilde{f}\circ \tau _H)\in S$ on a set of positive measure in $B\cap H$ then strict inequality holds in (4.4). This observation leads to the criterion in the lemma. $\square$

Lemma 4.5

Let $f\in L^1(\mathbb {S}^{n-1},\mathcal {H}^{n-1})$. Fix $\nu \in \mathbb {S}^{n-1}$ and let $(\nu _h)$ be a sequence in $\mathbb {S}^{n-1}$ that converges to $\nu$ in $\mathbb {S}^{n-1}$. Then

$$\begin{aligned} \Vert f\circ \tau _h-f\circ \tau \Vert _{L^1(\mathbb {S}^{n-1},\mathcal {H}^{n-1})}\rightarrow 0 \,\, \text { as }h\rightarrow \infty . \end{aligned}$$

Proof

Note that $|\tau _h x-\tau x|\le 4|\nu _h-\nu |$ for each $x\in \mathbb {S}^{n-1}$ and h. Now use the density of $C(\mathbb {S}^{n-1})$ in $L^1(\mathbb {S}^{n-1},\mathcal {H}^{n-1})$ and the fact that each $\tau ,\tau _h$ is an isometry on $L^1(\mathbb {S}^{n-1},\mathcal {H}^{n-1})$.$\square$

The next lemma is a spherical cap symmetrisation counterpart to [4, Lemma 6.3] and extends [23, Lemma 3.9].

Lemma 4.6

Let $f\in L^1_+(B)$ and $\omega \in \mathbb {S}^{n-1}$ and assume that $f\ne C_\omega f$. Then there exists $H\in \mathscr {H}_\omega$ such that

$$\begin{aligned} \Vert f_H - C_\omega f\Vert _{L^1(B)} < \Vert f - C_\omega f\Vert _{L^1(B)}. \end{aligned}$$

Proof

For non-negative Borel measurable functions f, g on B, $f=g$ if and only if $|\{f>t\}\Delta \{g>t\}|=0$ for any $t>0$. As $f\ne C_\omega f$ there exists $t>0$ such that

$$\begin{aligned} |\left\{ f>t\right\} \Delta \left\{ C_\omega f>t\right\} |>0. \end{aligned}$$

By (4.3) $|\{f>t\}|=|\{C_\omega f>t\}|$ and it follows that $|\{f\le t< C_\omega f\}|=|\{C_\omega f\le t< f\}|$. Put $A:=\left\{ f\le t<C_\omega f\right\}$ and $A^\prime :=\left\{ C_\omega f\le t<f\right\}$. For later use we note that

$$\begin{aligned} \mathcal {H}^{n-1}(A\cap \mathbb {S}^{n-1}_\tau )=\mathcal {H}^{n-1}(A^\prime \cap \mathbb {S}^{n-1}_\tau ) \end{aligned}$$

for $\mathscr {L}^1$-a.e. $\tau \in (0,1)$.

We claim there exists $H\in \mathscr {H}_\omega$ such that $|A\cap \tau _H A^\prime |>0$. Taking this as read, on $A\cap \tau _H A^\prime$ we have that $C_\omega f>t\ge C_\omega f\circ \tau _H$ so that $A\cap \tau _H A^\prime \subset H$. Also, $f\le t<f\circ \tau _H$ there. In short, $A\cap \tau _H A^\prime \subset \left\{ f\circ \tau _H>f\right\} \cap H$. So $|\left\{ f\circ \tau _H>f\right\} \cap H|>0$ and strict inequality holds by Lemma 4.4 (iii).

To prove the claim, assume for a contradiction that $|A\cap \tau _H A^\prime |=0$ for all $H\in \mathscr {H}_\omega$. Let F be a countable dense subset in $\mathbb {S}^{n-1}\cap H_\omega$. Then

$$\begin{aligned} \Big |\bigcup _{\nu \in F}(A\cap \tau _{H_\nu }A^\prime )\Big |=0. \end{aligned}$$

Therefore for all $r\in (0,1)$ it holds that

$$\begin{aligned} \mathcal {H}^{n-1}(A_r\cap \tau _{H_\nu }A^\prime _r)=0\quad \text { for every }\nu \in F, \end{aligned}$$

except on a $\mathscr {L}^1$-null set $N\subset (0,1)$. We write $A_r:=A\cap \mathbb {S}^{n-1}_r$ for the r-section of A and likewise for $A^\prime$. Let $\nu \in \mathbb {S}^{n-1}\cap H_\omega$ with corresponding reflection $\tau =\tau _{H_\nu }$. Select a sequence $(\nu _h)$ in F which converges to $\nu$ in $\mathbb {S}^{n-1}$ and write $\tau _h$ for the reflection associated to the closed half-space $H_{\nu _h}$. For $r\in (0,\,1)\setminus N$,

$$\begin{aligned} |\mathcal {H}^{n-1}(A_r\cap \tau A^\prime _r)-\mathcal {H}^{n-1}(A_r\cap \tau _h A^\prime _r)| \le \Vert \chi _{A^\prime } -\chi _{A^\prime }\circ \tau \circ \tau _h\Vert _{L^1(\mathbb {S}^{n-1}_r,\mathcal {H}^{n-1})}, \end{aligned}$$

and this latter converges to zero as $h\rightarrow \infty$ by Lemma 4.5. We derive that

$$\begin{aligned} \mathcal {H}^{n-1}(A_r\cap \tau _{H_\nu }A^\prime _r)=0\text { for every }\nu \in \mathbb {S}^{n-1}\cap H_\omega \end{aligned}$$

(4.5)

for all $r\in (0,\,1)\setminus N$.

To conclude the argument, choose $r\in (0,1)\setminus N$ such that $\mathcal {H}^{n-1}(A_r)=\mathcal {H}^{n-1}(A^\prime _r)>0$. Select a density point x for $A_r$ lying in $A_r$ using [1, Corollary 2.23] for example; that is,

$$\begin{aligned} \lim _{\rho \downarrow 0}\frac{1}{\mathcal {H}^{n-1}(B(x,\rho )_r)}\int _{B(x,\rho )_r}|\chi _{A_r}(z)-\chi _{A_r}(x)|\,\mathcal {H}^{n-1}(dz)=0. \end{aligned}$$

This means that $A_r$ has density 1 at x in the sense that

$$\begin{aligned} \frac{\mathcal {H}^{n-1}(A_r\cap B(x,\rho ))}{\mathcal {H}^{n-1}(\mathbb {S}^{n-1}_r\cap B(x,\rho ))} \rightarrow 1\text { as }\rho \downarrow 0. \end{aligned}$$

Choose y in $A^\prime _r$ similarly so that $A^\prime _r$ has density 1 at y. Then $C_\omega f(x)>t\ge C_\omega f(y)$. So there exists $\nu \in \mathbb {S}^{n-1}\cap H_\omega$ such that with $\tau =\tau _{H_\nu }$ we have that $\tau y = x$. But then

$$\begin{aligned} \lim _{\rho \downarrow 0}\frac{\mathcal {H}^{n-1}(A_r\cap \tau B_r\cap B(x,\rho ))}{\mathcal {H}^{n-1}(\mathbb {S}^{n-1}\cap B(x,\rho ))}=1, \end{aligned}$$

so that, in fact, $\mathcal {H}^{n-1}(A_r\cap \tau B_r)>0$, contradicting (4.5). $\square$

Proof of Theorem 4.1

Let E be an optimal configuration for the data (U, p). Assume for a contradiction that $E\ne C_\omega E$ $\mathscr {L}^n$-a.e. for any $\omega \in \mathbb {S}^{n-1}$. Then there exists $\omega \in \mathbb {S}^{n-1}$ such that

$$\begin{aligned} \delta :=\inf _{\nu \in \mathbb {S}^{n-1}}\Vert \chi _E-C_\nu \chi _E\Vert _{L^1(B)}= \Vert \chi _E-C_\omega \chi _E\Vert _{L^1(B)}>0. \end{aligned}$$

By Lemma 4.6 there exists $H\in \mathscr {H}_\omega$ such that $\Vert (\chi _E)_H-C_\omega \chi _E\Vert _{L^1(B)}<\Vert \chi _E-C_\omega \chi _E\Vert _{L^1(B)}$. It is plain from this that $\chi _E\ne (\chi _E)_H$; but also $\chi _E\circ \tau _H\ne (\chi _E)_H$, for otherwise,

$$\begin{aligned} \Vert \chi _E-C_{\tau _H\omega }\chi _E\Vert _{L^1(B)}= & {} \Vert \chi _E-C_{\omega }\chi _E\circ \tau _H\Vert _{L^1(B)} =\Vert (\chi _E)_H-C_{\omega }\chi _E\Vert _{L^1(B)}\\< & {} \Vert \chi _E-C_{\omega }\chi _E\Vert _{L^1(B)}, \end{aligned}$$

contradicting optimality of $\omega$. It follows by Theorem 4.2 that $J(E)<J(E_H)$, contradicting the fact that E is an optimal configuration for the data (U, p). The result now follows. $\square$

5 Preliminaries on Weighted Dirichlet Forms

Let $n=1$ and $U=(a,b)$ be an open bounded interval in $\mathbb {R}$. We are given a density function w with the property

• (A) w is a positive function in $C_0(U)$.

The weighted volume of an $\mathscr {L}^1$-measurable set E in U is given by $m(E):=\int _E w\,dx$. We introduce the further assumption

• (B) $w\in C^1(U)$ and $w^\prime /w\in L^2(U,m)$.

Consider the coercive bilinear form

$$\begin{aligned} \mathscr {E}(u,v):=\int _{U} \Big (uv+ u^\prime v^\prime \Big )\,dm \qquad (u,v\in \mathscr {D}:=C^\infty (\overline{U})) \end{aligned}$$

in $L^2(U,m)$.

Lemma 5.1

Assume (A)–(B). Then

• (i) $(\mathscr {D},\mathscr {E})$ is closable in $L^2(U,m)$ with closure denoted $(D(\mathscr {E}),\mathscr {E})$;

• (ii) $(D(\mathscr {E}),\mathscr {E})$ is a symmetric Dirichlet form in $L^2(U,m)$.

Proof

We refer to [19, Definitions I.2.3 and I.4.5] (for example). Note that $(\mathscr {D},\mathscr {E})$ satisfies the weak sector condition [19] (2.3) by the Cauchy–Schwarz inequality. Suppose $(u_h)$ is a sequence in $\mathscr {D}$ such that $u_h\rightarrow 0$ in $L^2(U,m)$. For $v\in \mathscr {D}$ an integration-by-parts gives

$$\begin{aligned} \int _U u_h^\prime v^\prime w\,dx&=\int _{\partial U}u_h v^\prime w\nu \,d\mathcal {H}^0 -\int _U u_h(v^\prime w)^\prime \,dx \\&=-\int _U u_h\frac{(v^\prime w)^\prime }{w}\,dm =-\int _U u_h\left\{ v^{\prime \prime }+v^\prime \frac{w^\prime }{w}\right\} \,dm\rightarrow 0 \end{aligned}$$

as $h\rightarrow \infty$ where $\nu =\pm 1$ is the one-dimensional unit exterior normal on $\partial U$. We have made use of the assumptions (A) and (B). The statement (i) follows by [19, Lemma I.3.4]. Then $(D(\mathscr {E}),\mathscr {E})$ is a symmetric closed form by definition (cf. [19, Definition I.2.3]). By [19, Proposition I.4.10 and II.2 (c)], $(D(\mathscr {E}),\mathscr {E})$ is a symmetric Dirichlet form. $\square$

Given a real-valued function u on $\mathbb {R}_+$ (or $\mathbb {R})$ define the function $\theta _t u$ on $\mathbb {R}_+$ for each $t>0$ by $(\theta _t u)(x):=u(x+t)$ for $x\in \mathbb {R}_+$ (or $\mathbb {R})$.

Lemma 5.2

Let $\lambda$ be a positive $\mathscr {L}^1$-integrable function on $\mathbb {R}_+$ such that

$$\begin{aligned} c:=\sup _{t>0}\sup _{x>0}\frac{\lambda (x)}{\lambda (x+t)}<\infty . \end{aligned}$$

Then

• (i) $\theta _t\in B(L^2(\mathbb {R}_+,\lambda \mathscr {L}^1))$ for each $t>0$;

• (ii) $\Vert u-\theta _t u\Vert _{L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)}\rightarrow 0$ as $t\downarrow 0$ for each $u\in L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)$.

Proof

(i) For any $u\in L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)$,

$$\begin{aligned} \Vert \theta _t u\Vert ^2_{L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)}= & {} \int _0^\infty |\theta _t u|^2\lambda \,dx \le c\int _0^\infty |u(x+t)|^2\lambda (x+t)\,dx\\\le & {} c\Vert u\Vert ^2_{L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)}. \end{aligned}$$

(ii) The statement holds for $u\in C([0,\infty ))\cap L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)$ by the dominated convergence theorem and this latter set is dense in $L^2(\mathbb {R}_+,\lambda \mathscr {L}^1)$. These observations as well as (i) lead to the result using a $3\varepsilon$-argument. $\square$

Our next assumption is stronger than required but easy to state:

• (C) w is unimodal on U.

Lemma 5.3

Assume (A)–(C). Then

$$\begin{aligned} D(\mathscr {E})=\Big \{u\in L^2(U,m): u\text { is weakly differentiable on }U\text { and }u^\prime \in L^2(U,m)\Big \}. \end{aligned}$$

Proof

Let $u\in D(\mathscr {E})$. There exists a Cauchy sequence $(u_h)$ in $(\mathscr {D},\mathscr {E})$ which converges to u in $L^2(U,m)$. Then $(u_h^\prime )$ is a Cauchy sequence in $L^2(U,m)$ with limit $v\in L^2(U,m)$ (say). For $\phi \in C^\infty _c(U)$,

$$\begin{aligned} \int _U u\phi ^\prime \,dx= & {} \lim _h\int _U u_h\phi ^\prime \,dx =-\lim _h\int _U u_h^\prime \phi \,dx =-\lim _h\int _U u_h^\prime (w^{-1}\phi )\,dm\\= & {} -\int _U v\phi \,dx; \end{aligned}$$

so u is weakly differentiable on U with weak derivative $u^\prime =v\in L^2(U,m)$.

Now let $u\in L^2(U,m)$ be weakly differentiable on U such that $u^\prime \in L^2(U,m)$. Multiplying by a partition of unity we may assume that $u=0$ near b. Denote by $\overline{u}$ the extension of u to $\mathbb {R}$ by zero. For $t>0$ put $v_t:=(\theta _t\overline{u})\vert _U$. Note that $v_t$ is weakly differentiable and $v_t^\prime =\theta _t(\overline{u^\prime })\vert _U$. For $t>0$, $v_t, v_t^\prime \in L^2(U,m)$. Let $(\rho _\varepsilon )_{\varepsilon >0}$ be a family of mollifiers on $\mathbb {R}$ (cf. [1, 2.1]). For $t>0$ and $\varepsilon >0$ small, $(\rho _\varepsilon \star (\theta _t\overline{u}))^\prime =\rho _\varepsilon \star (\theta _t\overline{u^\prime })$ on U. The operation $\star$ stands for convolution. For $t>0$ and $\varepsilon >0$ small put $w_{t,\varepsilon }:=\rho _\varepsilon \star (\theta _t\overline{u})\vert _U\in C^\infty (\overline{U})$. Now

$$\begin{aligned} \Vert u-w_{t,\varepsilon }\Vert _{L^2(U,m)}&\le \Vert u-v_t\Vert _{L^2(U,m)}+\Vert v_t-w_{t,\varepsilon }\Vert _{L^2(U,m)}; \\ \Vert u^\prime -w_{t,\varepsilon }^\prime \Vert _{L^2(U,m)}&\le \Vert u^\prime -v_t^\prime \Vert _{L^2(U,m)} +\Vert v_t^\prime -\rho _\varepsilon *(\theta _t\overline{u^\prime })\vert _U\Vert _{L^2(U,m)}. \end{aligned}$$

By Lemma 5.2 and (A)–(C) the expressions $\Vert u-v_t\Vert _{L^2(U,m)}$ and $\Vert u^\prime -v^\prime _t\Vert _{L^2(U,m)}$ are small for $t>0$ small. We also use the fact that the mollified functions are regular approximations in $L^2$ (cf. [1, 2.1]). This shows that $u\in D(\mathscr {E})$. $\square$

Suppose that $\hat{U}=(c,d)$ is an open bounded interval in $\mathbb {R}$ and $\Phi :\hat{U}\rightarrow U$ is a $C^1$ bijection such that $\Phi ^\prime =:\varphi >0$ on $\hat{U}$. Let $\hat{m}:=\Phi _\sharp m$ be the pull-back of m under $\Phi$; thus $\hat{m}=\hat{w}\mathscr {L}^1$ on $\hat{U}$ where $\hat{w}:=\varphi (w\circ \Phi )$. Define a coercive bilinear form

in $L^2(\hat{U},\hat{m})$ with domain

$$\begin{aligned} D(\hat{\mathscr {E}}):=\Big \{u\in L^2(\hat{U},\hat{m}): u\text { is weakly differentiable on }\hat{U}\text { and }\varphi ^{-1}u^\prime \in L^2(\hat{U},\hat{m})\Big \}. \end{aligned}$$

Lemma 5.4

Assume (A)–(C). Then

• (i) $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ is a symmetric Dirichlet form in $L^2(\hat{U},\hat{m})$;

• (ii) the mapping $D(\hat{\mathscr {E}})\rightarrow D(\mathscr {E});u\mapsto \overline{u}:=u\circ \Phi ^{-1}$ is a Hilbert space isomorphism.

Proof

(i) We show that $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ is closed in $L^2(\hat{U},\hat{m})$. Let $(u_h)$ be an $\hat{\mathscr {E}}^{1/2}$-Cauchy sequence in $D(\hat{\mathscr {E}})$. Then $(u_h)$ resp. $(\varphi ^{-1}u_h^\prime )$ are Cauchy sequences in $L^2(\hat{U},\hat{m})$ with limits $u\in L^2(\hat{U},\hat{m})$ resp. $v\in L^2(\hat{U},\hat{m})$. For $\phi \in C^\infty _c(\hat{U})$,

$$\begin{aligned} \int _{\hat{U}}u\phi ^\prime \,dx&=\lim _{h\rightarrow \infty }\int _{\hat{U}}u_h\phi ^\prime \,dx =-\lim _{h\rightarrow \infty }\int _{\hat{U}}u_h^\prime \phi \,dx \\&=-\lim _{h\rightarrow \infty }\int _{\hat{U}}\varphi ^{-1}u_h^\prime \frac{\phi }{w\circ \Phi }\,d\hat{m} =-\int _{\hat{U}}v\frac{\phi }{w\circ \Phi }\,d\hat{m} =-\int _{\hat{U}}(\varphi v)\phi \,dx \end{aligned}$$

so u is weakly differentiable and $\varphi ^{-1}u^\prime =v$; that is, $u\in D(\hat{\mathscr {E}})$. It then follows that $(u_h)$ converges to u in $\hat{\mathscr {E}}^{1/2}$-norm. So $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ is a symmetric closed form in $L^2(\hat{U},\hat{m})$.

Let $u\in D(\hat{\mathscr {E}})$. Given $\varepsilon >0$ let $\varphi _\varepsilon$ be as in [19, Example II.2.7]. Note that $u\in W^{1,2}_{\mathrm {loc}}(\hat{U})$. Then $\varphi _\varepsilon (u)$ is weakly differentiable on $\hat{U}$ and $\varphi _\varepsilon (u)^\prime =\varphi _\varepsilon ^\prime (u)u^\prime$ (see for example [10, 5.10, Exercise 16]) so $\varphi _\varepsilon (u)\in D(\hat{\mathscr {E}})$. We then derive that

$$\begin{aligned} \hat{\mathscr {E}}(\varphi _\varepsilon (u),\varphi _\varepsilon (u))\le \hat{\mathscr {E}}(u,u). \end{aligned}$$

By [19, Proposition I.4.7], $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ is a symmetric Dirichlet form in $L^2(\hat{U},\hat{m})$.

(ii) Let $u\in D(\hat{\mathscr {E}})$. Note that $(u^\prime /\varphi )\circ \Phi ^{-1}\in L^2(U,m)$ (use [1, (2.47)]) and $\overline{u}$ is weakly differentiable on U with $\overline{u}^\prime =(u^\prime /\varphi )\circ \Phi ^{-1}\in L^2(U,m)$. Thus the mapping is well-defined. For $u,v\in D(\hat{\mathscr {E}})$, $\hat{\mathscr {E}}(u,v)=\mathscr {E}(\overline{u},\overline{v})$ again using [1, (2.47)]. In particular, the mapping $u\mapsto \overline{u}$ is injective. Now let $u\in D(\mathscr {E})$ and put $\hat{u}:=u\circ \Phi$. Then $\hat{u}\in L^2(\hat{U},\hat{m})$, $\hat{u}$ is weakly differentiable on $\hat{U}$ with weak derivative $\hat{u}^\prime =\varphi (u^\prime \circ \Phi )$ and $\varphi ^{-1}\hat{u}^\prime \in L^2(\hat{U},\hat{m})$; in other words, $\hat{u}\in D(\hat{\mathscr {E}})$. This shows that the mapping in (ii) is surjective.$\square$

Now take $B=(-1,1)$. We are given density functions f, g with the properties

• (A.1) f is a positive function in C(B);

• (A.2) g is a positive unimodal function in $C_0(B)$.

The weighted volume of an $\mathscr {L}^1$-measurable set E in B is given by $V(E):=\int _E f\,dx$. Put $\rho :=f/g$. We introduce the further assumption

• (A.3) $\rho \in L^1(B,\mathscr {L}^1)$, $g\in C^1(B)$ and $g^\prime /f\in L^2(B,V)$.

Define

$$\begin{aligned} R:B\rightarrow \mathbb {R}; t\mapsto \int _0^t\rho (\tau )\,d\tau , \end{aligned}$$

and let $\check{B}$ denote the image of B under R; $\check{B}$ is an open bounded interval in $\mathbb {R}$. Then $R:B\rightarrow \check{B}$ is a $C^1$ bijection. Define $\check{g}:=g\circ R^{-1}$ on $\check{B}$. Define the measure $\check{V}:=\check{g}\mathscr {L}^1$ on $\check{B}$.

We introduce coercive bilinear forms

$$\begin{aligned} \check{\mathscr {E}}(u,v):=\int _{\check{B}} \Big (uv+ u^\prime v^\prime \Big )\,d\check{V} \qquad (u,\,v\in D(\check{\mathscr {E}})) \end{aligned}$$

in $L^2(\check{B},\check{V})$ with domain

$$\begin{aligned} D(\check{\mathscr {E}})=\Big \{u\in L^2(\check{B},\check{V}): u\text { is weakly differentiable on }\check{B}\text { and }u^\prime \in L^2(\check{B},\check{V})\Big \}; \end{aligned}$$

and

in $L^2(B,V)$ with domain

$$\begin{aligned} D(\hat{\mathscr {E}})=\Big \{u\in L^2(B,V): u\text { is weakly differentiable on }B\text { and }\rho ^{-1}u^\prime \in L^2(B,V)\Big \}. \end{aligned}$$

Note that with $\check{B}$ in place of U, $w:=\check{g}$ satisfies properties (A)–(C) above in light of the assumptions (A.1)–(A.3). We derive

Lemma 5.5

Assume (A.1)–(A.3). Then

• (i) $(D(\check{\mathscr {E}}),\check{\mathscr {E}})$ is a symmetric Dirichlet form in $L^2(\check{B},\check{V})$;

• (ii) $C^\infty (\overline{\check{B}})$ is dense in $D(\check{\mathscr {E}})$ with respect to the $\check{\mathscr {E}}^{1/2}$-norm;

• (iii) $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ is a symmetric Dirichlet form in $L^2(B,V)$;

• (iv) the mapping $D(\hat{\mathscr {E}})\rightarrow D(\check{\mathscr {E}});u\mapsto u\circ R^{-1}$ is a Hilbert space isomorphism.

6 An (f, g)-Isoperimetric Inequality

Recall that an $\mathscr {L}^1$-measurable set $E\subset B$ is said to be a Caccioppoli set if for each relatively compact open set $\Omega$ in B,

$$\begin{aligned} P(E,\Omega ):=\sup \Big \{\int _\Omega \chi _E\phi ^\prime \,dx:\,\phi \in C^\infty _c(\Omega ,\mathbb {R}),\Vert \phi \Vert _\infty \le 1\Big \}<\infty . \end{aligned}$$

There then exists a unique real Radon measure $D\chi _E$ on B such that

$$\begin{aligned} \int _B\chi _E \phi ^\prime \,dx=-\int _B\phi \,dD\chi _E \end{aligned}$$

for all $\phi \in C^\infty _c(B,\mathbb {R})$ [1, Corollary 1.55]. Denote by $|D\chi _E|$ the total variation measure of $D\chi _E$.

Theorem 6.1

Suppose that E is a Caccioppoli set in B with $|E|>0$. Then there exist $N\in \mathbb {N}\cup \{\infty \}$ and closed intervals $E_h=[a_{2h-1},a_{2h}]\subset \mathbb {R}$ $(h=1,\ldots ,N)$ with non-empty interior and separated by open neighbourhoods in $\mathbb {R}$ such that E is $\mathscr {L}^1$-a.e. equivalent to the union of the $E_h$.

The statement that the collection of intervals $(E_h)$ is separated by open intervals means that $\inf _{k\ne h}d(E_h,E_k)>0$ for each h. Here, d denotes the standard metric on $\mathbb {R}$.

Proof

The proof is along the lines of [1, Proposition 3.52]. Put $u:=\chi _E\in \mathrm {BV}_{\mathrm {loc}}(B)$. Then $\mu :=Du$ is a real Radon measure on B [1, 1.40]. Define

$$\begin{aligned} w(t):= \left\{ \begin{array}{ll} -\mu ([t,0)) &{} \text { for }-1<t < 0;\\ 0 &{} \text { for }t=0;\\ \mu ([0,t)) &{} \text { for }0<t <1;\\ \end{array} \right. \end{aligned}$$

w is left-continuous on B by the Hahn decomposition and inner/outer regularity [1, 1.43]. By Fubini’s theorem, $w\in \mathrm {BV}_{\mathrm {loc}}(B)$ and $Dw=\mu$. By [1, Proposition 3.2], $u=c+w$ $\mathscr {L}^1$-a.e. on B for some $c\in \mathbb {R}$. Let A be the set of atoms of $\mu$ in B. Note that w is continuous on $B\setminus A$ and $w(t+)-w(t)=\mu (\{t\})$ at each $t\in A$. As $c+w\in \{0,1\}$ $\mathscr {L}^1$-a.e. on B, $\mu (\{t\})\in \{-1,1\}$ for each $t\in A$. Let $\Omega$ be a relatively compact open set in B. Then $\mathrm {Card}(A\cap \Omega )=|\mu |(A\cap \Omega )\le |\mu |(\Omega )<\infty$. Thus the set of atoms A accumulates at $\partial B$ (if at all). By the observations above, the function $c+w$ is constant on each connected component of $B\setminus A$ with values in the set $\{0,1\}$. Let the sets $E_h$ be the closure of the open intervals in $B\setminus A$ where $c+w$ takes the value 1.$\square$

Let g be a positive lower semicontinuous function on B. Let E be a Caccioppoli set in B. The g-perimeter of E relative to B is defined by

$$\begin{aligned} P_g(E,B):=\int _B g\,d|D\chi _E|. \end{aligned}$$

(6.1)

Lemma 6.2

Let g be a positive lower semicontinuous function on B and E a Caccioppoli set in B. Then

$$\begin{aligned} P_g(E,B)=\sum _{a\in A}g(a) \end{aligned}$$

where A stands for the set of atoms of $D\chi _E$ in B.

Proof

A direct computation gives $D\chi _E=\sum _{a\in A}D\chi _E(\{a\})\delta _a$ and $|D\chi _E|=\sum _{a\in A}\delta _a$ which gives the result. We use $\delta _a$ to stand for the Dirac measure at a.$\square$

Let f, g be densities on B satisfying conditions (A.1)–(A.3). The weighted volume of an $\mathscr {L}^1$-measurable set E in B is the measure given by $V(E):=\int _E f\,dx$. Define $F:[-1,1]\rightarrow [0,V(B)]$ by $F(x):=V((-1,x))$ and

$$\begin{aligned} J(p):=(g\circ F^{-1})(p)\text { for }p\in [0,V(B)]. \end{aligned}$$

We impose the additional assumptions

• (A.4) $J(p)=J(V(B)-p)$ for $0<p<V(B)$;

• (A.5) for all $p,q>0$ with $p+q<V(B)$, $J(p+q)<J(p)+J(q)$.

For an $\mathscr {L}^1$-measurable set E in B the $\bigstar$-rearrangement of E is defined by $E^\bigstar :=(-1,F^{-1}(V(E)))$. We then have that the following (f, g)-isoperimetric inequality is valid.

Theorem 6.3

Assume (A.1)–(A.5). Suppose that E is a Caccioppoli set in B. Then $P_g(E,B)\ge P_g(E^\bigstar ,B)$.

Proof

For $0<p<q<V(B)$,

$$\begin{aligned} J(p)+J(q)=J(p)+J(V(B)-q)> J(p+V(B)-q)=J(q-p) \end{aligned}$$

(6.2)

by (A.4)–(A.5). The above inequality also holds for $0\le p<q\le V(B)$ with $>$ replaced by $\ge$ by continuity; equality holds if and only if one or both of p, q are extremal.

We may suppose that $V(E)=p$ for some $p\in (0,V(B))$. Assume that E has the form $E=\bigcup _{h=1}^N[a_{2h-1},a_{2h}]$ with $-1\le a_1<a_2<\cdots <a_{2N-1}<a_{2N}\le 1$ for some $N\in \mathbb {N}$. Put $p_j:=F(a_j)$ so that $0\le p_1<\cdots < p_{2N}\le V(B)$ and $\sum _{h=1}^N(p_{2h}-p_{2h-1})=p$. By Lemma 6.2, (6.2) and (A.5),

$$\begin{aligned} P_g(E,B)&= \sum _{h=1}^N\Big ( J(p_{2h-1})+J(p_{2h})\Big ) \ge \sum _{h=1}^NJ(p_{2h}-p_{2h-1}) \\&\ge J\left( \sum _{h=1}^N(p_{2h}-p_{2h-1})\right) =J(p)=P_g(E^\bigstar ,B). \end{aligned}$$

The result for arbitrary E as in the statement follows by Theorem 6.1, the monotone convergence theorem and continuity of J. $\square$

We now investigate the equality case in the isoperimetric inequality.

Theorem 6.4

Assume (A.1)–(A.5). Suppose that E is a Caccioppoli set in B. Assume that $P_g(E,B) = P_g(E^\bigstar ,B)$. Then either $E=E^\bigstar$ or $B\setminus E=(B\setminus E)^\bigstar$ V-a.e.

Proof

We may assume that E is the union of closed intervals $E_h\subset \mathbb {R}$ $(h=1,\ldots ,N, N\in \mathbb {N}\cup \{\infty \})$ with non-empty interior and separated by open neighbourhoods in $\mathbb {R}$ as in Theorem 6.1. In virtue of (6.1) we may take $N>1$. We can then find $x\in B\setminus E$ such that $V(E\cap (x,1))>0$ and $V(E\cap (-1,x))>0$. Put $E_-:=E\cap (-1,x)$, $E_+:=E\cap (x,1)$, $F_-:=E_-^\bigstar$ and $F_+:=B\setminus (B\setminus E_+)^\bigstar$. Note that $V(E_-)=V(F_-)=p_-$ and $V(E_+)=V(F_+)=p_+$ for some $p_\pm \in (0,V(B))$. We have

$$\begin{aligned} J(p_-)+J(p_+)= & {} P_g(F_-,B)+P_g(F_+,B)\le P_g(E_-,B)+P_g(E_+,B)\\= & {} P_g(E,B) \end{aligned}$$

by Theorem 6.3. On the other hand, $P_g(E,B)=P_g(E^\bigstar ,B)=J(p_-+p_+)<J(p_-)+J(p_+)$ by (A.5), a contradiction.$\square$

Proposition 6.5

Assume (A.1)–(A.3). If $g^\prime /f$ is strictly decreasing on B then (A.5) holds.

Proof

Note that J is differentiable on (0, V(B)). Moreover, $(F^{-1})^\prime (p)=1/(f\circ F^{-1})(p)$ and

$$\begin{aligned} J^\prime (p)=g^\prime (F^{-1}(p))\cdot 1/(f\circ F^{-1})(p)=(g^\prime /f)(F^{-1}(p)) \end{aligned}$$

for $p\in (0,V(B))$; this shows that J is strictly concave on (0, V(B)) as $(0,V(B))\mapsto (-1,1):\,p\mapsto (F^{-1})(p)$ is increasing. For $0<p<q<V(B)$ we have that

$$\begin{aligned} \frac{J(p)}{p}\ge \frac{J(q)}{q} \end{aligned}$$

(6.3)

by concavity and the fact that $J(0)\ge 0$. Supposing additionally that $p+q<V(B)$ we have

$$\begin{aligned} \frac{J(q)-J(p)}{q-p}>\frac{J(p+q)-J(q)}{p} \end{aligned}$$

by considering the gradient of chords. Rearranging and using (6.3) gives (A.5).$\square$

Let u be a real-valued $\mathscr {L}^1$-measurable function on B. Put $\mu _u(t):=V(\{|u|>t\})$ for $t\ge 0$. The function $\mu _u:[0,\infty )\rightarrow [0,V(B)]$ is non-increasing, right-continuous and $\mu _u(t)\rightarrow 0$ as $t\rightarrow \infty$. Define its right-continuous inverse $u^\sharp :[0,V(B)]\rightarrow [0,\infty ]$ by

$$\begin{aligned} u^\sharp (s):=\inf \{t\ge 0:\mu _u(t)\le s\}\text { for }0\le s\le V(B), \end{aligned}$$

with the understanding that $\inf \emptyset =+\infty$. Define $u^\bigstar :=u^\sharp \circ F$ on B. Note that $\mu _u(t)>s$ if and only if $u^\sharp (s)>t$ (see Lemma 10.1). It follows that

$$\begin{aligned} V(\{|u|>t\})=V(\{u^\bigstar >t\})\text { for each }t\ge 0; \end{aligned}$$

(6.4)

in fact, $V(\{u^\bigstar >t\})=V(\{F<\mu _u(t)\})=V((-1,F^{-1}(\mu _u(t))))=\mu _u(t)$.

The following result is a Hardy–Littlewood type inequality and can be proved as in [14, 13.10] (see also [9, Theorem 3]).

Theorem 6.6

Assume (A.1)–(A.5). Let u, v be real-valued $\mathscr {L}^1$-measurable functions on B. Then

$$\begin{aligned} \int _{B} |uv|\,dV\le \int _{B} u^\bigstar v^\bigstar \,dV. \end{aligned}$$

The next non-expansivity result can be found in [9, Corollary 1].

Theorem 6.7

Assume (A.1)–(A.5). Let u, v be real-valued $\mathscr {L}^1$-measurable functions on B. Then

$$\begin{aligned} \int _{B} |u^\bigstar -v^\bigstar |^2\,dV\le \int _{B} |u-v|^2\,dV. \end{aligned}$$

Define a metric $\hat{d}$ on B as follows. The length of a piecewise $C^1$ parametrised curve $\gamma :[\alpha ,\beta ]\rightarrow B$ in $(B,\hat{d})$ is

$$\begin{aligned} \hat{L}[\gamma ]=\int _\alpha ^\beta \rho (\gamma (t))|\dot{\gamma }(t)|\,dt. \end{aligned}$$

For $x,y\in B$, $\hat{d}(x,y)$ stands for the infimum of lengths of piecewise $C^1$ parametrised curves in B connecting x to y; $\hat{d}(\cdot ,\cdot )$ is a metric on B.

Lemma 6.8

For $x,y\in B$, $\hat{d}(x,y)=d(R(x),R(y))$.

Proof

Let $\gamma :[\alpha ,\beta ]\rightarrow B$ be a piecewise $C^1$ parametrised curve in B connecting x to y. Then $\hat{L}[\gamma ]=L[R\circ \gamma ]$ in an obvious notation. So $\hat{d}(x,y)\le d(R(x),R(y))$. A similar argument gives the reverse inequality.$\square$

Note that for each $\mathscr {L}^1$-measurable set E in B, $\check{V}(R(E))=V(E)$.

Lemma 6.9

The mapping $R:B\rightarrow \check{B}$ sets up a one-to-one correspondence between Caccioppoli sets in B resp. $\check{B}$. Moreover, let E be a Caccioppoli set in B. Then

• (i) $|D\chi _{R(E)}|=R_\sharp |D\chi _{E}|$;

• (ii) $P_{\check{g}}(R(E),\check{B})=P_g(E,B)$.

Proof

Let $\Omega \subset B$ be a relatively compact open set. Then

$$\begin{aligned} |D\chi _{R(E)}|(R(\Omega ))&=\sup \left\{ \int _{R(\Omega )}\chi _{R(E)}\phi ^\prime \,dx:\phi \in C_c^1(\check{B},\mathbb {R})\text { and }\Vert \phi \Vert _\infty \le 1\right\} \\&=\sup \left\{ \int _{\Omega }\chi _{E}(\phi \circ R)^\prime \,dx:\phi \in C_c^1(\check{B},\mathbb {R})\text { and }\Vert \phi \Vert _\infty \le 1\right\} \\&=\sup \left\{ \int _{\Omega }\chi _{E}\phi ^\prime \,dx:\phi \in C_c^1(B,\mathbb {R})\text { and }\Vert \phi \Vert _\infty \le 1\right\} \\&=|D\chi _{E}|(\Omega ), \end{aligned}$$

from which the first assertion follows. Item (i) follows from the definition of the push-forward [1, Definition 1.7] and the coincidence criterion [1, Proposition 1.8], while (ii) follows from (6.1), (i) and the change of variables formula for integrals. $\square$

The function $\check{F}:\mathbb {R}\rightarrow [0,V(B)]$ defined by $\check{F}(x):=\check{V}(\check{B}\cap (-\infty ,x))$ is the cumulative distribution function of $\check{g}$. Let u be a real-valued $\mathscr {L}^1$-measurable function on $\check{B}$. Put $\check{\mu }_{u}(t):=\check{V}(\{|u|>t\})$ for $t\ge 0$ and denote by $\check{u}^{\sharp }:[0,\check{V}(\check{B})]\rightarrow [0,\infty ]$ its right-continuous inverse (as in the Appendix). Define $u^\star :=\check{u}^{\sharp }\circ \check{F}$ on $\check{B}$.

Proposition 6.10

Let u be a real-valued $\mathscr {L}^1$-measurable function on B and put $v:=u\circ R^{-1}$. Then $u^\bigstar =v^\star \circ R$. In particular, for any $\mathscr {L}^1$-measurable set $E\subset B$, $R(E)^\star =R(E^\bigstar )$.

Proof

We have that $\mu _u(t)=\check{\mu }_{v}(t)$ for each $t\ge 0$; hence $u^\sharp =\check{v}^{\sharp }$ on [0, V(B)]. Now , $\check{F}\circ R=F$ on $[-1,1]$. This leads to the first claim. The second then follows straightforwardly.$\square$

Corollary 6.11

Assume (A.1)–(A.5). Suppose that E is a Caccioppoli set in $\check{B}$. Then $P_{\check{g}}(E,\check{B})\ge P_{\check{g}}(E^\star ,\check{B})$.

Proof

Let E be a Caccioppoli set in B. By Lemma 6.9, Theorem 6.3 and Proposition 6.10,

$$\begin{aligned} P_{\check{g}}(R(E),\check{B})=P_g(E,B)\ge P_g(E^\bigstar ,B)=P_{\check{g}}(R(E^\bigstar ),\check{B}) =P_{\check{g}}(R(E)^\star ,\check{B}). \end{aligned}$$

$\square$

Corollary 6.12

Assume (A.1)–(A.5). Suppose that E is a Caccioppoli set in $\check{B}$. Assume that $P_{\check{g}}(E,\check{B}) = P_{\check{g}}(E^\star ,\check{B})$. Then either $E=E^\star$ or $\check{B}\setminus E=(\check{B}\setminus E)^\star$ $\check{V}$-a.e.

Proof

This follows from Lemma 6.9 and Theorem 6.4.$\square$

Finally, we state a counterpart of Theorem 6.7.

Theorem 6.13

Assume (A.1)–(A.5). Let u, v be real-valued $\mathscr {L}^1$-measurable functions on $\check{B}$. Then

$$\begin{aligned} \int _{\check{B}} |u^\star -v^\star |^2\,d\check{V}\le \int _{\check{B}} |u-v|^2\,d\check{V}. \end{aligned}$$

7 A Pólya–Szegö Inequality

We first show that the rearrangement $\cdot ^\star$ is smoothing in the sense of [21] (see also [4]). Given $r>0$ write $E_r$ for the r-neighbourhood of an $\mathscr {L}^1$-measurable set E in $(\check{B},d)$; by convention, $\emptyset _r=\emptyset$. The Minkowski content of E is the quantity

$$\begin{aligned} \check{V}^+(E):=\liminf _{r\downarrow 0}\frac{\check{V}(E_r)-\check{V}(E)}{r}\in [0,\infty ]. \end{aligned}$$

Lemma 7.1

Let E be a finite union of open intervals in $(\check{B},d)$. Then

• (i) E is a Caccioppoli set in $\check{B}$;

• (ii) $\check{V}^+(E)=P_{\check{g}}(E,\check{B})$.

Proof

(i) The set E is a finite union of disjoint open intervals, $\overline{E}$ is a finite union of closed intervals in $\mathbb {R}$ with non-empty interior and separated by open sets in the sense of Theorem 6.1 and $\overline{E}=E\cup I$ for a finite set $I\subset \mathbb {R}$. So E is $\mathscr {L}^1$-a.e. equivalent to $F:=\overline{E}\cap \check{B}$; in particular, E is a Caccioppoli set in $\check{B}$ by Theorem 6.1.

(ii) Suppose first that $F=[a_1,a_2]\subset \check{B}$. Then $E_r=F_r$ for each $r>0$ and $\check{V}(E)=\check{V}(F)$ as $\check{V}$ is non-atomic. For small $r>0$,

$$\begin{aligned} \frac{1}{r}\check{V}(E_r\setminus E)&= \frac{1}{r}\check{V}(F_r\setminus F)=\frac{1}{r}\int _{a_1-r}^{a_1}\check{g}\,dy+ \frac{1}{r}\int _{a_2}^{a_2+r}\check{g}\,dy \\&\rightarrow \int _{\partial F}\check{g}\,d\mathcal {H}^0=P_{\check{g}}(F,\check{B})=P_{\check{g}}(E,\check{B}) \end{aligned}$$

as $r\downarrow 0$. The result for general E as in the statement follows from the property that the closed intervals in $\overline{E}$ are separated by open sets.$\square$

For $p\in (0,V(B))$ we may write

$$\begin{aligned} \check{F}^{-1}(p)=\int _{V(B)/2}^p\frac{d\tau }{\check{J}(\tau )} \end{aligned}$$

where $\check{J}=\check{g}\circ \check{F}^{-1}$. Note that $\check{J}=J$ due to the fact that $\check{F}\circ R=F$ on B.

Lemma 7.2

Let E be an $\mathscr {L}^1$-measurable set in $\check{B}$. Then $\check{V}(E_r)\ge \check{V}((E^\star )_r)$ for each $r>0$. In particular, the rearrangement $\cdot ^\star$ is smoothing in the sense that $(E^\star )_r\subset (E_r)^\star$ for each $\mathscr {L}^1$-measurable set E in $\check{B}$ and $r>0$.

Proof

We verify the conditions in [3, Theorem 2.1 (c)]. The measure $\check{V}$ is a separable non-atomic Borel measure on the metric space $(\check{B},d)$. The r-neighbourhood ($r>0$) of any open ball in $\check{B}$ is an open ball in $\check{B}$. Let E be a finite union of open intervals in $(\check{B},d)$. By Lemma 7.1 and Corollary 6.11,

$$\begin{aligned} \check{V}^+(E)=P_{\check{g}}(E,\check{B})\ge P_{\check{g}}(E^\star ,\check{B})=\check{J}(\check{V}(E)). \end{aligned}$$

Thus by [3, Theorem 2.1], $\check{V}(E_r)\ge \check{F}(\check{F}^{-1}(\check{V}(E))+r)= \check{V}((E^\star )_r)$ for any Borel set E in $\check{B}$ with $0<\check{V}(E)<V(B)$ and $r>0$. The result then extends to $\mathscr {L}^1$-measurable sets in $\check{B}$.$\square$

Lemma 7.3

Let A, E be $\mathscr {L}^1$-measurable sets in $\check{B}$ with $A\subset E$. Then $d(A^\star ,\check{B}\setminus E^\star )\ge d(A,\check{B}\setminus E)$.

Here, $d(A,E):=\inf \{d(x,y):x\in A,y\in E\}$ with the understanding that $\inf \emptyset =+\infty$.

Proof

We use the criterion that for $r>0$, $A_r\subset E$ if and only if $d(A,\check{B}\setminus E)\ge r$. Put $r:=d(A,\check{B}\setminus E)$; we may assume that $r>0$. By the criterion, $A_r\subset E$ and hence $(A_r)^\star \subset E^\star$. By Lemma 7.2, $\check{V}((A_r)^\star )=\check{V}(A_r)\ge \check{V}((A^\star )_r)$ meaning $(A^\star )_r\subset (A_r)^\star \subset E^\star$ which entails that $d(A^\star ,\check{B}\setminus E^\star )\ge r$ by the criterion. $\square$

The modulus of continuity of an arbitrary real-valued function u on $\check{B}$ is defined by

$$\begin{aligned} \omega _u(t):=\sup \Big \{|u(x)-u(y)|:\,x,\,y\in \check{B}\text { and }d(x,y)< t\Big \}\in [0,\infty ]\text { for }t>0. \end{aligned}$$

Observe that u is uniformly continuous on $\check{B}$ if and only if $\lim _{t\downarrow 0}\omega _u(t)=0$. We state the following criterion without proof.

Lemma 7.4

Let u be a real-valued function on $\check{B}$ and $t,\tau >0$. Then $\omega _u(t)>\tau$ if and only if there exist $s,s^\prime \in \mathbb {R}$ with $s>s^\prime +\tau$ such that $d(\{u>s\},\check{B}\setminus \{u>s^\prime \})<t$.

Proposition 7.5

Assume (A.1)–(A.5).

• (i) Let u be a real-valued $\mathscr {L}^1$-measurable function on $\check{B}$. Then $\omega _u(t)\ge \omega _{u^\star }(t)$ for each $t>0$.

• (ii) If u is uniformly continuous on $\check{B}$ then so is $u^\star$.

• (iii) If u is Lipschitz continuous on $\check{B}$ then so is $u^\star$ and $\mathrm {Lip}(u^\star ,\check{B})\le \mathrm {Lip}(u,\check{B})$.

Proof

Let $t>0$. We may assume that $\omega _{u^\star }(t)>0$. Choose $\tau >0$ such that $\omega _{u^\star }(t)>\tau$. By Lemma 7.4 there exist $s,s^\prime \in \mathbb {R}$ with $s>s^\prime +\tau$ such that $d(\{u^\star >s\},\check{B}\setminus \{u^\star >s^\prime \})<t$. Now $\{u^\star >s\}=\{|u|>s\}^\star$ and likewise for $s^\prime$ by the counterpart of the equimeasurabilty property (6.4). By Lemma 7.3 we deduce that $d(\{|u|>s\},\check{B}\setminus \{|u|>s^\prime \})<t$ and again by Lemma 7.4 that $\omega _{u}(t)\ge \omega _{|u|}(t)>\tau$. Item (i) then follows. Part (ii) is a ready consequence. As for (iii),

$$\begin{aligned} \mathrm {Lip}(u,\check{B})&:=\sup \left\{ \frac{|u(x)-u(y)|}{d(x,y)}:x,y\in \check{B}, x\ne y\right\} \\&=\sup _{t>0}(1/t)\omega _u(t)\ge \sup _{t>0}(1/t)\omega _{u^\star }(t)=\mathrm {Lip}(u^\star ,\check{B}). \end{aligned}$$

$\square$

Let u be a Lipschitz continuous function on $(\check{B},d)$. By Rademacher’s theorem (cf. [1, Theorem 2.14]) u is differentiable $\mathscr {L}^1$-a.e. on $\check{B}$ and its derivative coincides with the weak derivative on a set of full measure. Put

$$\begin{aligned} Z_1&:=\big \{x\in \check{B}:\,u\text { is differentiable at }x\text { and }u^\prime (x)=0\big \}, \\ Z_2&:=\big \{x\in \check{B}:\,u\text { is not differentiable at }x\big \}\text { and }Z:=Z_1\cup Z_2. \end{aligned}$$

By [1, Lemmas 2.95 and 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. The analogous sets corresponding to $u^\star$ will be decorated with the subscript $\star$.

We shall make use of the coarea formula [1, Theorem 2.93 and (2.74)],

$$\begin{aligned} \int _{\check{B}}\phi |u^\prime |\,dx = \int _{-\infty }^\infty \int _{\check{B}\cap \{u=t\}}\phi \,d\mathcal {H}^{0}\,dt \end{aligned}$$

(7.1)

for any $\mathscr {L}^1$-measurable function $\phi :\,\check{B}\rightarrow [0,\infty ]$.

Lemma 7.6

Let u be a nonnegative Lipschitz continuous function on $(\check{B},d)$. Then

• (i) $\check{\mu }_{u}\in \mathrm {BV}(\mathbb {R})$;

• (ii) $D\check{\mu }_{u}=-u_\sharp \check{V}$;

• (iii) ;

• (iv) ;

• (v) $A:=\Big \{t\in \mathbb {R}:\mathscr {L}^1(Z\cap \{u=t\})>0\Big \}$ is the set of atoms of $D\check{\mu }_{u}$ and ;

• (vi) $\check{\mu }_u$ is differentiable $\mathscr {L}^1$-a.e. on $\mathbb {R}$ with derivative given by

  $$\begin{aligned} \check{\mu }_{u}^\prime (t)=-\int _{(\check{B}\setminus Z)\cap \{u=t\}}\frac{\check{g}}{|u^\prime |}\,d\mathcal {H}^{0} \end{aligned}$$

  for $\mathscr {L}^1$-a.e. $t\in \mathbb {R}$;

• (vii) $\mathrm {Ran}(u)=\mathrm {supp}(D\check{\mu }_u)$.

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

Proof

For any $\varphi \in C^\infty _c(\mathbb {R})$,

$$\begin{aligned} \int _{-\infty }^\infty \check{\mu }_{u}\varphi ^\prime \,dt=\int _{\check{B}}\varphi \circ u\,d\check{V} \end{aligned}$$

by Fubini’s theorem; so $\check{\mu }_{u}\in \mathrm {BV}(\mathbb {R})$ and $D\check{\mu }_{u}$ is the push-forward of $\check{V}$ under u, $D\check{\mu }_{u}=-u_\sharp \check{V}$ (cf. [1, 1.70]). By (7.1),

for any $\mathscr {L}^1$-measurable set A in $\mathbb {R}$. In light of the above, we may identify and . The set of atoms of $D\check{\mu }_{u}$ is defined by $A:=\{t\in \mathbb {R}:D\check{\mu }_{u}(\{t\})\ne 0\}$. By [13, Lemma 7.7], we may write A as in (v). The monotone function $\check{\mu }_u$ is a good representative within its equivalence class and is differentiable $\mathscr {L}^1$-a.e. on $\mathbb {R}$ with derivative given by the density of $D\check{\mu }_{u}$ with respect to $\mathscr {L}^1$ by [1, Theorem 3.28]. Item (vii) follows from (ii).$\square$

Lemma 7.7

Let u be a nonnegative Lipschitz continuous function on $(\check{B},d)$. Then $\int _{\check{B}\cap Z}u^2\,d\check{V}=\int _{\check{B}\cap Z_\star }(u^\star )^2\,d\check{V}$.

Proof

As Z has finite $\mathscr {L}^1$-measure, $A\subset \mathbb {R}$ is a countable set. Thus

$$\begin{aligned} \int _{\check{B}\cap Z}u^2\,d\check{V}=\sum _{t\in A}t^2\check{V}(Z\cap \{u=t\}))=\int _A t^2 d(u_\sharp \check{V})=-\int _A t^2 dD\check{\mu }_u \end{aligned}$$

and an analogous result holds for $u^\star$ by Lemma 7.6. The fact that $\check{\mu }_u=\check{\mu }_{u^\star }$ entails that $A=A_\star$. This leads to the result. $\square$

Theorem 7.8

Assume (A.1)–(A.5). Let u be a Lipschitz continuous function on $(\check{B},d)$. Then $u,u^\star \in D(\check{\mathscr {E}})$ and $\check{\mathscr {E}}(u,u)\ge \check{\mathscr {E}}(u^\star ,u^\star )$.

Proof

Given a Lipschitz continuous function u on $(\check{B},d)$, $u\in W^{1,\infty }(\check{B})$ and $\Vert u^\prime \Vert _{L^\infty (\check{B})}=\mathrm {Lip}(u,\check{B})$ (see [1, Proposition 2.13]) so $u\in D(\check{\mathscr {E}})$. The same is true for $u^\star$ by Proposition 7.5. Replacing u by |u| and using the contraction property of the Dirichlet form $(D(\check{\mathscr {E}}),\check{\mathscr {E}})$ we may assume that u is nonnegative.

The proof hinges on the identity

$$\begin{aligned} \int _{\check{B}\setminus Z}\Big (u^2+|u^\prime |^2\Big )\,d\check{V} =\int _{\mathrm {Ran}(u)}\int _{(\check{B}\setminus Z)\cap \{u=t\}}\frac{u^2+|u^\prime |^2}{|u^\prime |}\check{g}\,d\mathcal {H}^0\,dt, \end{aligned}$$

(7.2)

which is a consequence of the coarea formula (7.1).

As $\check{\mu }_u=\check{\mu }_{u^\star }$ we have that $D\check{\mu }_u=D\check{\mu }_{u^\star }$. In particular, we derive that

$$\begin{aligned} \check{\mu }_{u}^\prime (t) =-\int _{(\check{B}\setminus Z)\cap \{u=t\}}\frac{\check{g}}{|u^\prime |}\,d\mathcal {H}^{0} =-\int _{(\check{B}\setminus Z_\star )\cap \{u^\star =t\}}\frac{\check{g}}{|u^{\star \prime }|}\,d\mathcal {H}^{0} =\check{\mu }_{u^\star }^\prime (t) \end{aligned}$$

(7.3)

for $\mathscr {L}^1$-a.e. $t\in \mathbb {R}$ by Lemma 7.6.

Let $t\in \mathrm {Ran}(u)$ be such that $Z\cap \{u=t\}=\emptyset$ and $\int _{({\check{B}\setminus Z})\cap \{u=t\}}\check{g}\,d\mathcal {H}^{0}<\infty$ and the analogous properties hold for $u^\star$. We assume in addition that (7.3) holds. Such t comprise a set of full measure in the range of u. Then $\partial \{u>t\}=\check{B}\cap \{u=t\}\cup \partial \check{B}\cap \partial \{u>t\}$ and $\{u>t\}$ is a Caccioppoli set in $\check{B}$ with finite $\check{g}$-perimeter; and likewise for $u^\star$. From Corollary 6.11,

$$\begin{aligned} \infty >\int _{\check{B}\cap \{u=t\}}\check{g}\,d\mathcal {H}^{0}&=\int _{\partial \{u>t\}}\check{g}\,d\mathcal {H}^{0} =P_{\check{g}}(\{u>t\},\check{B}) \nonumber \\&\ge P_{\check{g}}(\{u^\star >t\},\check{B})=\int _{\check{B}\cap \{u^\star =t\}}\check{g}\,d\mathcal {H}^{0}. \end{aligned}$$

(7.4)

Moreover, $\check{B}\cap \{u^\star =t\}$ consists of a singleton; thus,

$$\begin{aligned} \check{\mu }_{u^\star }^\prime (t)&=-\int _{(\check{B}\setminus Z_\star )\cap \{u^\star =t\}}\frac{\check{g}}{|(u^\star )^\prime |}\,d\mathcal {H}^{0}\nonumber \\&=-\int _{\check{B}\cap \{u^\star =t\}}\frac{\check{g}}{|(u^\star )^\prime |}\,d\mathcal {H}^{0} =-\int _{\check{B}\cap \{u^\star =t\}}\check{g}\,d\mathcal {H}^{0}/\int _{\check{B}\cap \{u^\star =t\}}|(u^\star )^\prime |\,d\mathcal {H}^{0}. \end{aligned}$$

(7.5)

By (7.3) and (7.5),

$$\begin{aligned} \int _{(\check{B}\setminus Z)\cap \{u=t\}}\frac{u^2+|u^\prime |^2}{|u^\prime |}\check{g}\,d\mathcal {H}^0&= \int _{(\check{B}\setminus Z)\cap \{u=t\}}\frac{u^2+|u^\prime |^2}{|u^\prime |}\check{g}\,d\mathcal {H}^0\nonumber \\&=\int _{(\check{B}\setminus Z)\cap \{u=t\}}\Big \{t^2+|u^\prime |^2\Big \}\,\frac{\check{g}\,d\mathcal {H}^0/|u^\prime |}{-\check{\mu }_{u}^\prime (t)}\Big (-\check{\mu }_{u}^\prime (t)\Big ) \nonumber \\&\ge \left\{ t^2+\left( \frac{\int _{(\check{B}\setminus Z)\cap \{u=t\}}\check{g}\,d\mathcal {H}^0}{-\check{\mu }_{u}^\prime (t)}\right) ^2 \right\} \left( -\check{\mu }_{u}^\prime (t)\right) \nonumber \\&=\left\{ t^2+\left( \frac{\int _{\check{B}\cap \{u=t\}}\check{g}\,d\mathcal {H}^0}{-\check{\mu }_{u}^\prime (t)}\right) ^2 \right\} \left( -\check{\mu }_{u}^\prime (t)\right) \nonumber \\&\ge \left\{ t^2+\left( \frac{\int _{\check{B}\cap \{u^\star =t\}}\check{g}\,d\mathcal {H}^0}{-\check{\mu }_{u^\star }^\prime (t)}\right) ^2 \right\} \left( -\check{\mu }_{u^\star }^\prime (t)\right) \nonumber \\&=\int _{\check{B}\cap \{u^\star =t\}}\frac{(u^\star )^2+|(u^\star )^\prime |^2}{|(u^\star )^\prime |}\check{g}\,d\mathcal {H}^0 \end{aligned}$$

(7.6)

where Jensen’s inequality has been used in the first inequality and (7.4) in the second. This inequality combined with (7.2) as well as Lemma 7.7 lead to the result. $\square$

Corollary 7.9

Let $u\in D(\check{\mathscr {E}})$. Then $u^\star \in D(\check{\mathscr {E}})$ and $\check{\mathscr {E}}(u,u)\ge \check{\mathscr {E}}(u^\star ,u^\star )$.

Proof

Let $u\in D(\check{\mathscr {E}})$. By Lemma 5.5 we may choose a sequence $(u_h)$ in $C^\infty (\overline{\check{B}})$ that converges to u in $(D(\check{\mathscr {E}}),\check{\mathscr {E}})$; each $u_h$ is Lipschitz continuous on $\check{B}$. By Theorem 6.13, $(u^\star _h)$ converges to $u^\star$ in $L^2(\check{B},\check{V})$. By Theorem 7.8, each $u^\star _h\in D(\check{\mathscr {E}})$ and $\check{\mathscr {E}}(u^\star _h,u^\star _h)\le \check{\mathscr {E}}(u_h,u_h)$, so the sequence $(\check{\mathscr {E}}(u^\star _h,u^\star _h))_h$ is uniformly bounded in $\mathbb {R}$. By the Banach-Alaoglu theorem (cf. [19, A2 Theorem 2.1]) we may assume that $u^\star _h\rightarrow v$ weakly as $h\rightarrow \infty$ in $(D(\check{\mathscr {E}}),\check{\mathscr {E}})$ for some $v\in D(\check{\mathscr {E}})$ by selecting a subsequence if necessary. We may identify v with $u^\star$ thanks to the $L^2(\check{B},\check{V})$ convergence and the Banach-Saks theorem (cf. [19, A2 Theorem 2.2]); hence $u^\star \in D(\check{\mathscr {E}})$. By [16, Theorem 10.1.5],

$$\begin{aligned} \check{\mathscr {E}}(u^\star ,u^\star )\le \liminf _{h\rightarrow \infty }\check{\mathscr {E}}(u_h^\star ,u_h^\star ) \le \liminf _{h\rightarrow \infty }\check{\mathscr {E}}(u_h,u_h)=\check{\mathscr {E}}(u,u). \end{aligned}$$

$\square$

Corollary 7.10

Let $u\in D(\hat{\mathscr {E}})$. Then $u^\bigstar \in D(\hat{\mathscr {E}})$ and $\hat{\mathscr {E}}(u,u)\ge \hat{\mathscr {E}}(u^\bigstar ,u^\bigstar )$.

Proof

Let $u\in D(\hat{\mathscr {E}})$. Then $v:=u\circ R^{-1}\in D(\check{\mathscr {E}})$ by Lemma 5.5. Morever, $u^\bigstar =v^\star \circ R$ by Proposition 6.10. By Lemma 5.5 and Corollary 7.9,

$$\begin{aligned} \hat{\mathscr {E}}(u,u)=\check{\mathscr {E}}(v,v)\ge \check{\mathscr {E}}(v^\star ,v^\star ) =\hat{\mathscr {E}}(v^\star \circ R,v^\star \circ R)=\hat{\mathscr {E}}(u^\bigstar ,u^\bigstar ). \end{aligned}$$

$\square$

8 Equality Case in the Pólya–Szegö Inequality

We now investigate the equality case in the Pólya–Szegö inequality.

Lemma 8.1

Let $u\in D(\check{\mathscr {E}})$ and $t\in \mathbb {R}$.

• (i) Put $v:=u\wedge t$. Then $\int _{\check{B}}|v^\prime |^2\,d\check{V}=\int _{\check{B}\cap \{u>t\}}|u^\prime |^2\,d\check{V}=\int _{\check{B}\cap \{u\ge t\}}|u^\prime |^2\,d\check{V}$.

• (ii) Put $v:=u\vee t$. Then $\int _{\check{B}}|v^\prime |^2\,d\check{V}=\int _{\check{B}\cap \{u<t\}}|u^\prime |^2\,d\check{V}=\int _{\check{B}\cap \{u\le t\}}|u^\prime |^2\,d\check{V}$.

Proof

We only prove (i). Write

$$\begin{aligned} \int _{\check{B}}|v^\prime |^2\,d\check{V} = \int _{\check{B}\cap \{u<t\}}|v^\prime |^2\,d\check{V} +\int _{\check{B}\cap \{u=t\}}|v^\prime |^2\,d\check{V} +\int _{\check{B}\cap \{u>t\}}|v^\prime |^2\,d\check{V}. \end{aligned}$$

The set $\{u<t\}$ is open in $\check{B}$ as u is continuous so $v=u$ and $v^\prime =u^\prime$ there. By [13, Lemma 7.7], $v^\prime =0$ $\mathscr {L}^1$-a.e. on $\{u\ge t\}$.$\square$

Suppose that $u\in W^{1,2}_{\mathrm {loc}}(\check{B})$ is precisely represented in the sense of [18, (2.5)]. Then the set $\{u=t\}$ is finite or countably infinite for $\mathscr {L}^1$-a.e. $t\in \mathbb {R}$ and the coarea formula (7.1) holds for u by [18, Theorem 1.1]. With Z as before it follows that $Z\cap \{u=t\}=\emptyset$ for $\mathscr {L}^1$-a.e. $t\in \mathbb {R}$ and hence $N:=u(Z)\subset \mathbb {R}$ is $\mathscr {L}^1$-negligible.

Lemma 8.2

Let u be a nonnegative function in $W^{1,2}_{\mathrm {loc}}(\check{B})$ precisely represented in the sense of [18, (2.5)]. Then statements (i)–(vii) of Lemma 7.6 hold.

Proof

This runs as in Lemma 7.6. $\square$

Lemma 8.3

Let $u\in D(\check{\mathscr {E}})$ be nonnegative. Then $\int _{\check{B}\cap Z}u^2\,d\check{V}=\int _{\check{B}\cap Z_\star }(u^\star )^2\,d\check{V}$.

Proof

The proof proceeds as in Lemma 7.7. $\square$

Lemma 8.4

Let $u\in D(\check{\mathscr {E}})$ be nonnegative. Then

• (i) for $t^\prime ,t^{\prime \prime }\in \mathbb {R}$ with $0\le t^\prime <t^{\prime \prime }$,

  $$\begin{aligned} \int _{\check{B}\cap \{t^\prime <u\le t^{\prime \prime }\}}\Big (u^2+|u^\prime |^2\Big )\,d\check{V} \ge \int _{\check{B}\cap \{t^\prime <u\le t^{\prime \prime }\}}\Big ((u^\star )^2+|(u^\star )^\prime |^2\Big )\,d\check{V}; \end{aligned}$$

• (ii) for $\mathscr {L}^1$-a.e. $t\in \mathbb {R}$,

  $$\begin{aligned} \int _{(\check{B}\setminus Z)\cap \{\tilde{u}=t\}}\frac{\tilde{u}^2+|\tilde{u}^\prime |^2}{|\tilde{u}^\prime |}\check{g}\,d\mathcal {H}^0 \ge \int _{(\check{B}\setminus Z_\star )\cap \{u^\star =t\}}\frac{(u^\star )^2+|(u^\star )^\prime |^2}{|(u^\star )^\prime |}\check{g}\,d\mathcal {H}^0 \end{aligned}$$

  where $\tilde{u}$ is the unique continuous representative of u (cf. [1, Definition 3.31 and after]).

Proof

(i) Put $v:=(u\vee t^\prime )\wedge t^{\prime \prime }\in D(\check{\mathscr {E}})$. By Lemma 8.1,

$$\begin{aligned} \int _{\check{B}}\Big (v^2+|v^\prime |^2\Big )\,d\check{V} =\int _{\check{B}\cap \{t^\prime <u\le t^{\prime \prime }\}}\Big (u^2+|u^\prime |^2\Big )\,d\check{V} +t^{\prime 2}(1-\mu _u(t^\prime )) +t^{\prime \prime 2}\mu _u(t^{\prime \prime }). \end{aligned}$$

By Lemma 10.2, $\check{v}^\sharp =(\check{u}^\sharp \vee t^\prime )\wedge t^{\prime \prime }$ on $[0,\check{V}(\check{B})]$ and hence $v^\star =(u^\star \vee t^\prime )\wedge t^{\prime \prime }$. We may then write an identity of the above form but with $u^\star ,v^\star$ in place of u, v. The statement then follows from Corollary 7.9.

(ii) Note that $\tilde{u}$ is precisely represented in the sense of [18, (2.5)]. By the coarea formula for Sobolev mappings [18, Theorem 1.1],

$$\begin{aligned} \int _{\check{B}}\phi |u^\prime |\,dx =\int _{-\infty }^\infty \int _{\check{B}\cap \{\tilde{u}=t\}}\phi \,d\mathcal {H}^0\,dt \end{aligned}$$

for any $\mathscr {L}^1$-measurable function $\phi :\check{B}\rightarrow [0,\infty ]$. In particular,

$$\begin{aligned} \int _{\check{B}\setminus Z}\phi (u)\Big (u^2+|u^\prime |^2\Big )\,d\check{V} = \int _{-\infty }^\infty \int _{(\check{B}\setminus Z)\cap \{\tilde{u}=t\}} \left\{ \frac{\tilde{u}^2+|\tilde{u}^\prime |^2}{|\tilde{u}^\prime |}\check{g}\,d\mathcal {H}^0\right\} \phi (t)\,dt \end{aligned}$$

(8.1)

for any $\mathscr {L}^1$-measurable function $\phi :\mathbb {R}\rightarrow [0,\infty ]$.

Define

$$\begin{aligned} w(t):=\int _{(\check{B}\setminus Z)\cap \{u>t\}}(u^2+|u^\prime |^2)\,d\check{V} \end{aligned}$$

for $t\in \mathbb {R}$. For $\varphi \in C^\infty _c(\mathbb {R},\mathbb {R})$,

$$\begin{aligned} \int _{-\infty }^\infty w\varphi ^\prime \,dx=\int _{\check{B}\setminus Z}\varphi (u)(u^2+|u^\prime |^2)\,d\check{V} \end{aligned}$$

so that $w\in \mathrm {BV}(\mathbb {R})$. By (8.1), $Dw=\rho \mathscr {L}^1$ where

$$\begin{aligned} \rho (t):= -\int _{(\check{B}\setminus Z)\cap \{\tilde{u}=t\}}\frac{\tilde{u}^2+|\tilde{u}^\prime |^2}{|\tilde{u}^\prime |} \check{g}\,d\mathcal {H}^0 \end{aligned}$$

for $t\in \mathbb {R}$; that is, w is absolutely continuous [1, Definition 3.31]. By [1, Theorem 3.28], w is differentiable $\mathscr {L}^1$-a.e. on $\mathbb {R}$ and $w^\prime =\rho$ $\mathscr {L}^1$-a.e. on $\mathbb {R}$. The same holds for the function $w_\star$ defined as for w but with $u^\star$ in place of u. Note that $\widetilde{u^\star }=u^\star$. The statement then follows from (i).$\square$

We state the following lemma without proof.

Lemma 8.5

Let u be a continuous real-valued function on $\check{B}$. Suppose that for $\mathscr {L}^1$-a.e. $t>0$ the set $\{u>t\}$ is either an open interval in $\check{B}$ abutting a boundary point or $\{u>t\}=\emptyset$. Then u is monotone on $\check{B}$.

Theorem 8.6

Let $u\in D(\check{\mathscr {E}})$ be nonnegative and suppose that $\check{\mathscr {E}}(u,u)=\check{\mathscr {E}}(u^\star ,u^\star )$. Then $\tilde{u}$ is monotone on $\check{B}$.

Proof

By (8.1),

$$\begin{aligned} \check{\mathscr {E}}(u,u)=\int _{\check{B}\cap Z}u^2\,d\check{V} +\int _{\mathrm {Ran}(\tilde{u})}\int _{(\check{B}\setminus Z)\cap \{\tilde{u}=t\}}\frac{\tilde{u}^2+|\tilde{u}^\prime |^2}{|\tilde{u}^\prime |}\check{g}\,d\mathcal {H}^0\,dt \end{aligned}$$

and a similar identity holds for $u^\star$. We may assume that $\mathrm {Ran}(\tilde{u})$ is a closed interval in $[0,\infty )$ with non-empty interior. By Lemmas 8.3 and 8.4,

$$\begin{aligned} \int _{(\check{B}\setminus Z)\cap \{\tilde{u}=t\}}\frac{\tilde{u}^2+|\tilde{u}^\prime |^2}{|\tilde{u}^\prime |}\check{g}\,d\mathcal {H}^0 = \int _{(\check{B}\setminus Z_\star )\cap \{u^\star =t\}}\frac{(u^\star )^2+|(u^\star )^\prime |^2}{|(u^\star )^\prime |}\check{g}\,d\mathcal {H}^0 \end{aligned}$$

for $\mathscr {L}^1$-a.e. $t\in \mathrm {Ran}(\tilde{u})$. The chain of inequalities in (7.6) is valid with $\tilde{u}$ in place of u and we may replace the sign $\ge$ with the equality sign. In particular,

$$\begin{aligned} \int _{\check{B}\cap \{\tilde{u}=t\}}\check{g}\,d\mathcal {H}^0=\int _{\check{B}\cap \{u^\star =t\}}\check{g}\,d\mathcal {H}^0 \end{aligned}$$

(8.2)

for $\mathscr {L}^1$-a.e. $t\in \mathrm {Ran}(\tilde{u})$. Suppose that $t\in \mathrm {Ran}(\tilde{u})$ such that $Z\cap \{\tilde{u}=t\}=\emptyset$ and $\int _{\check{B}\cap \{\tilde{u}=t\}}\check{g}\,d\mathcal {H}^{0}<\infty$. We remark that $\partial \{\tilde{u}>t\}=\check{B}\cap \{\tilde{u}=t\}\cup \partial \check{B}\cap \partial \{\tilde{u}>t\}$ and $\{\tilde{u}>t\}$ is a Caccioppoli set in $\check{B}$ with finite $\check{g}$-perimeter $P_{\check{g}}(\{\tilde{u}>t\},\check{B})=\int _{\check{B}\cap \{\tilde{u}=t\}}\check{g}\,d\mathcal {H}^0<\infty$. From the identity (8.2) we derive that $P_{\check{g}}(\{\tilde{u}>t\},\check{B})=P_{\check{g}}(\{u^\star >t\},\check{B})$ for $\mathscr {L}^1$-a.e. $t\in \mathrm {Ran}(\tilde{u})$. By Corollary 6.12 the set $\{\tilde{u}>t\}$ is either an open interval in $\check{B}$ abutting a boundary point or $\{\tilde{u}>t\}=\emptyset$ for $\mathscr {L}^1$-a.e. $t>0$. The statement follows by Lemma 8.5.$\square$

Corollary 8.7

Let $u\in D(\hat{\mathscr {E}})$ be nonnegative and suppose that $\hat{\mathscr {E}}(u,u)=\hat{\mathscr {E}}(u^\bigstar ,u^\bigstar )$. Then $\tilde{u}$ is monotone on B.

Proof

Put $v:=u\circ R^{-1}\in D(\check{\mathscr {E}})$ by Lemma 5.5. Also, $u^\bigstar =v^\star \circ R$ by Proposition 6.10. We have

$$\begin{aligned} \hat{\mathscr {E}}(u,u)=\check{\mathscr {E}}(v,v)\ge \check{\mathscr {E}}(v^\star ,v^\star ) =\hat{\mathscr {E}}(v^\star \circ R,v^\star \circ R)=\hat{\mathscr {E}}(u^\bigstar ,u^\bigstar ). \end{aligned}$$

by Lemma 5.5 and Corollary 7.9. So $\check{\mathscr {E}}(v,v)=\check{\mathscr {E}}(v^\star ,v^\star )$. By Theorem 8.6, $\tilde{v}$ is monotone on $\check{B}$ and hence $\tilde{u}$ is monotone on B.$\square$

9 Application to Exchange Flow

Consider the positive definite bilinear form $(\mathscr {D},\mathscr {E})$ in $L^2(B,\mathscr {L}^1)$ given by

Here, $\psi :=u_B$ is given by $\psi (x)=(1/2)(1-|x|^2)$ for $x\in B$.

Lemma 9.1

• (i) $(\mathscr {D},\mathscr {E})$ is closable in $L^2(B,\,\mathscr {L}^1)$ with closure denoted $(D(\mathscr {E}),\,\mathscr {E})$;

• (ii) $(D(\mathscr {E}),\,\mathscr {E})$ is a symmetric Dirichlet form in $L^2(B,\,\mathscr {L}^1)$;

• (iii) $D(\mathscr {E})=W^{1,2}_0(B)$.

Proof

(i) Suppose $(u_h)$ is a sequence in $\mathscr {D}$ such that $u_h\rightarrow 0$ in $L^2(B,\mathscr {L}^1)$. We write $v\in \mathscr {D}$ in the form $v=\psi w$ for some $w\in C^\infty (\overline{B})$; so $v^{\prime \prime }\in L^2(B,\mathscr {L}^1)$. An integration-by-parts gives

$$\begin{aligned} \int _B u_h^\prime v^\prime \,dx&=\int _{\partial B}u_h v^\prime \nu \,d\mathcal {H}^0-\int _B u_hv^{\prime \prime }\,dx =-\int _B u_h v^{\prime \prime }\,dx\rightarrow 0 \end{aligned}$$

as $h\rightarrow \infty$ where $\nu =\pm 1$ is the unit exterior normal on $\partial B$. The statement follows by [19, Lemma I.3.4]. Then $(D(\mathscr {E}),\mathscr {E})$ is a symmetric closed form by definition (cf. [19, Definition I.2.3]). By [19, Proposition I.4.10 and II.2 (c)], $(D(\mathscr {E}),\mathscr {E})$ is a symmetric Dirichlet form and (ii) follows. Note that $\psi =u_B\in W^{1,2}_0(B)$. Thus $C^\infty _0(B)\subset \mathscr {D}\subset W^{1,2}_0(B)$. This proves (iii).$\square$

The transient Dirichlet space $(D(\mathscr {E}),\mathscr {E})$ has reference function $\psi ^{-1}$ and

$$\begin{aligned} \int _B|u|\psi ^{-1}\,dx\le \sqrt{\mathscr {E}(u,u)} \text { for all }u\in D(\mathscr {E}). \end{aligned}$$

(9.1)

Denote by $D(\mathscr {E})_e$ the extended Dirichlet space; that is, the family of $\mathscr {L}^1$-measurable functions u on B such that $|u|<\infty$ $\mathscr {L}^1$-a.e. and there exists an $\mathscr {E}$-Cauchy sequence $(u_h)$ of functions in $D(\mathscr {E})$ such that $u_h\rightarrow u$ $\mathscr {L}^1$-a.e. on B. By [12, Lemma 1.5.5], $(D(\mathscr {E})_e,\mathscr {E})$ is a Hilbert space. The identity (9.1) extends to $D(\mathscr {E})_e$ and $D(\mathscr {E})_e\subset L^1(B,\psi ^{-1}\mathscr {L}^1)$.

In the notation of Sect. 5 we take $f=\psi$ and $g=\psi ^{3/2}$. The conditions (A.1)–(A.5) are satisfied by Proposition 6.5. Let V be the measure $V:=\psi \mathscr {L}^1$ on B. We work with the symmetric Dirichlet form

in $L^2(B,V)$ with domain

$$\begin{aligned} D(\hat{\mathscr {E}}):=\Big \{u\in L^2(B,V): u\text { is weakly differentiable on }B\text { and }\psi ^{1/2}u^\prime \in L^2(B,V)\Big \}. \end{aligned}$$

Proposition 9.2

The bijective mapping $\mathscr {D}\rightarrow C^\infty (\overline{B});\,u\mapsto \overline{u}:=u/\psi$ extends to a Hilbert space isomorphism $E:D(\mathscr {E})_e\rightarrow D(\hat{\mathscr {E}})$. In particular, for $u,v\in D(\mathscr {E})_e$ we have that

$$\begin{aligned} \mathscr {E}(u,\,v) = \hat{\mathscr {E}}(\overline{u},\,\overline{v}). \end{aligned}$$

(9.2)

Proof

Using integration-by-parts and the fact that $-\psi ^{\prime \prime }=1$ on B gives

$$\begin{aligned} \int _{B}\overline{u}^\prime \,\overline{v}^\prime \,\psi ^2\,dx&= \int _{B}(u/\psi )^\prime \,\overline{v}^\prime \,\psi ^2\,dx \\&= \int _{B}\Big (\psi \,u^\prime - u\,\psi ^\prime \Big )\,\overline{v}^\prime \,dx \\&= \int _{B}u^\prime \,\,\overline{v}^\prime \,\psi \,dx -\int _{\partial B}u\,\psi ^\prime \,\overline{v}\,\nu \,d\mathcal {H}^0 +\int _{B}(u\,\psi ^\prime )^\prime \overline{v}\,dx \\&= \int _{B} u^\prime \,\overline{v}^\prime \,\psi \,dx +\int _{B}(u\,\psi ^\prime )^\prime \overline{v}\,dx \\&= \int _{B} u^\prime \,\overline{v}^\prime \,\psi \,dx +\int _{B}(u^\prime \,\psi ^\prime )\,\overline{v} -u\,\overline{v}\,dx \\&= \int _{B}u^\prime \,v^\prime \,dx -\int _{B}\overline{u}\,\overline{v}\,\psi \,dx \end{aligned}$$

where $\nu =\pm 1$ is the unit exterior normal on $\partial B$. This establishes (9.2) on $\mathscr {D}$. The map E extends to $D(\mathscr {E})$ by density of $\mathscr {D}$ in $D(\mathscr {E})$ as does (9.2). Let $u\in D(\mathscr {E})_e$ and choose a sequence $(u_h)$ in $D(\mathscr {E})$ such that $u_h\rightarrow u$ $\mathscr {L}^1$-a.e. on B as $h\rightarrow \infty$ and $(u_h)$ is a $\mathscr {E}$-Cauchy sequence. Then $(\overline{u}_h)$ is a Cauchy sequence in $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ with limit $v\in D(\hat{\mathscr {E}})$ say. Define $Eu:=v$. This definition is well-defined and the identity (9.2) holds on $D(\mathscr {E})_e$. In particular, the mapping E is injective.

We show that E is a surjection. Let $v\in D(\hat{\mathscr {E}})$. We put $u:=\psi v$ and claim that $u\in D(\mathscr {E})_e$. Then there exists a sequence $(w_h)$ in $C^\infty (\overline{\check{B}})$ such that $v_h:=w_h\circ R\rightarrow v$ in $(D(\hat{\mathscr {E}}),\hat{\mathscr {E}})$ as $h\rightarrow \infty$. Put $u_h:=\psi v_h$ for each h. Now $u_h\in C^1(\overline{B})$ and $u_h=0$ on $\partial B$ for each h. This means that $(u_h)$ is a sequence in $D(\mathscr {E})$. By (9.2), $(u_h)$ is a $\mathscr {E}$-Cauchy sequence. As $(v_h)$ converges to v in $L^2(B,V)$ we may assume that $(v_h)$ converges to v V-a.e. on B by selecting a subsequence if necessary. Thus, $(u_h)$ converges to u $\mathscr {L}^1$-a.e. on B. This shows that $u\in D(\mathscr {E})_e$ and $Eu=v$.$\square$

Theorem 9.3

For any $\mathscr {L}^1$-measurable set A in B, $J(A)\le J(A^\bigstar )$.

Proof

Let A be an $\mathscr {L}^1$-measurable set in B and put $u:=u_A\in D(\mathscr {E})$ and $v:=u_{A^\bigstar }\in D(\mathscr {E})$. By Theorem 6.6, Proposition 9.2, the Cauchy–Schwarz inequality and Corollary 7.10,

$$\begin{aligned} J(A)&=\int _{B}\overline{u}\,\chi _A\,dV \le \int _{B}\overline{u}^\bigstar \,\chi _{A^\bigstar }\,dV =\mathscr {E}(\psi \overline{u}^\bigstar ,v) =\hat{\mathscr {E}}(\overline{u}^\bigstar ,\overline{v}) \\&\le \hat{\mathscr {E}}(\overline{u}^\bigstar ,\overline{u}^\bigstar )^{1/2}\hat{\mathscr {E}}(\overline{v},\overline{v})^{1/2} \le \hat{\mathscr {E}}(\overline{u},\overline{u})^{1/2}\hat{\mathscr {E}}(\overline{v},\overline{v})^{1/2} =J(A)^{1/2}\,J(A^\bigstar )^{1/2} \end{aligned}$$

and the result follows.$\square$

Theorem 9.4

Suppose that A is an $\mathscr {L}^1$-measurable set in B such that $J(A)= J(A^\bigstar )$. Then A is $\mathscr {L}^1$-a.e. equivalent to an open interval in B abutting a boundary point of B.

Proof

We may assume that $0<V(A)<V(B)$. From the chain of inequalities in the proof of Theorem 9.3 we derive that $\hat{\mathscr {E}}(\overline{u}^\bigstar ,\overline{u}^\bigstar )=\hat{\mathscr {E}}(\overline{u},\overline{u})$. Put $v:=\overline{u}$. From Corollary 8.7 we infer that $\widetilde{v}$ is monotone on B. Put $p:=V(A)/V(B)\in (0,1)$. By Theorem 9.3, A is an optimal configuration for the problem (1.2) for the data (B, p). By Proposition 3.3, $V(A\Delta \{\widetilde{v}>c\})=0$ for some $c\in (0,1)$. As $\widetilde{v}$ is monotone, $\{\widetilde{v}>c\}$ is an open interval in B abutting an end-point of B. This leads to the result.$\square$

We may now characterise optimal configurations for the problem (1.2).

Theorem 9.5

Let $p\in (0,1)$.

• (i) The sets $(-1,F^{-1}(pV(B)))$ and $(F^{-1}((1-p)V(B)),1)$ are optimal configurations for the problem (1.2) with data (B, p).

• (ii) If $E\subset B$ is an optimal configuration for the data (B, p) then E is $\mathscr {L}^1$-a.e. equivalent to one of the sets in (i).

Let U be a bounded open connected set in $\mathbb {R}^n$ ($n\ge 1$).

Proposition 9.6

• (i) For $\lambda \in (-1,1)$, $\gamma (U,\lambda ) = 2 I(U,p)$ where $p=(1-\lambda )/2$.

• (ii) $\gamma (U) = 2\sup _{p\in (0,1)} I(U,p)$.

Proof

(i) Let A be an open set in U. Suppose u satisfies (1.3) and the condition $(u,1)=0$. Put $f:=-(\lambda + 1)\chi _A-(\lambda - 1) \chi _{U\setminus A}$; then $u=Gf$. From the flux-balance condition and symmetry of the Green operator,

$$\begin{aligned} 0=(u,1)=(\psi ,f)=-(\lambda +1)V(A)-(\lambda -1)V(U\setminus A). \end{aligned}$$

So $\lambda = V(U)^{-1}\{V(U\setminus A)-V(U)\}$ and $V(A) = \frac{1-\lambda }{2}V(U)$. We have

$$\begin{aligned} (\chi _{U\setminus A},u)&=(G\chi _{U\setminus A},f) \\&=-(\lambda +1)(G\chi _{U\setminus A},\chi _{A})-(\lambda -1)(G\chi _{U\setminus A},\chi _{U\setminus A}) \\&=-(\lambda +1)(V(A)-J(A))-(\lambda -1)(V(U\setminus A)-V(A)+J(A)) \\&=2 J(A)-(\lambda +1)V(A)-(\lambda -1)V(U\setminus A)+(\lambda -1)V(A) \\&=2 J(A)+(\lambda -1)V(A)=2\Big \{J(A) - V(U)p^2\Big \} \end{aligned}$$

where $p=(1-\lambda )/2$. This leads to the reformulation (i). (ii) follows immediately.$\square$

Recall that for $x,y\in B$,

$$\begin{aligned} G(x,y)= \left\{ \begin{array}{lc} (1/2)(1-y)(1+x) &{} \text { for }x\le y;\\ (1/2)(1+y)(1-x) &{} \text { for }x\ge y.\\ \end{array} \right. \end{aligned}$$

Lemma 9.7

The mapping $(0,1)\rightarrow \mathbb {R};p\rightarrow I(B,p)$ has a unique global maximum at $p=1/2$.

Proof

Define $\eta :B\rightarrow \mathbb {R}$ by $\eta =J(A)-V(B)^{-1}V(A)^2$ where $A=(-1,t)$. Then $I(B,p)=\eta (F^{-1}(pV(B)))$ for $p\in (0,1)$ by Theorem 9.5. For $t\in B$,

$$\begin{aligned} \frac{d}{dt}J((-1,t))&=\frac{d}{dt}\int _{-1}^t G\chi _{(-1,t)}(x)\,dx =G\chi _{(-1,t)}(t)+\int _{-1}^t\frac{d}{dt}G\chi _{(-1,t)}(x)\,dx \\&=G\chi _{(-1,t)}(t)+\int _{-1}^t G(x,t)\,dx=2G\chi _{(-1,t)}(t), \end{aligned}$$

so that

$$\begin{aligned} \eta ^\prime (t)=2G\chi _{(-1,t)}(t)-2V(B)^{-1}V((-1,t))\psi (t). \end{aligned}$$

A further computation gives

$$\begin{aligned} G\chi _{(-1,t)}(\xi )&=(1/4)(1-t)(1+t)^2 \text { and } V(B)^{-1}V((-1,t))=(1/4)(2+3t-t^3). \end{aligned}$$

We then obtain that $\eta ^\prime (t)=-(1/4)t(1-t^2)^2$ for any $t\in B$. This proves the result.$\square$

Theorem 9.8

Let $\lambda \in (-1,1)$.

• (i) The sets $(-1,F^{-1}(pV(B)))$ and $(F^{-1}((1-p)V(B)),1)$ are optimal configurations for the problem (1.5) with data $(U,\lambda )$ where $p=(1-\lambda )/2$.

• (ii) If $E\subset B$ is an optimal configuration for the problem (1.5) with data $(U,\lambda )$ then E is $\mathscr {L}^1$-a.e. equivalent to one of the sets in (i).

• (iii) The sets $(-1,0)$ and (0, 1) are optimal configurations for the problem (1.4).

• (iv) If $E\subset B$ is an optimal configuration for the problem (1.4) then E is $\mathscr {L}^1$-a.e. equivalent to one of the sets in (iii).

Proof

Parts (i) and (ii) follow from Proposition 9.6 and Theorem 9.5. (iii) and (iv) follow from Proposition 9.6 and Lemma 9.7. $\square$

Appendix: On the Generalised Inverse Function

Let $(X,\mathscr {A},\mu )$ be a finite measure space. Let $f:X\rightarrow [0,\infty )$ be an $\mathscr {A}$-measurable function. The distribution function $\mu _f:[0,\infty )\rightarrow [0,\mu (X)]$ of f is defined by $\mu _f(t):=\mu (\{f>t\})$ for $t\ge 0$. Note that $\mu _f$ is right-continuous and non-increasing on $[0,\infty )$ and $\mu _f(t)\rightarrow 0$ as $t\rightarrow \infty$. The generalised inverse $f^\sharp :[0,\mu (X)]\rightarrow [0,\infty ]$ of $\mu _f$ is defined by $f^\sharp (s):=\inf \{t\ge 0:\mu _f(t)\le s\}$ with the understanding that $\inf \emptyset =+\infty$.

Lemma 10.1

Let $f:X\rightarrow [0,\infty )$ be an $\mathscr {A}$-measurable function. Let $t\ge 0$ and $s\in [0,\mu (X)]$. Then

• (i) $\mu _f(t)>s$ if and only if $f^\sharp (s)>t$;

• (ii) $\mu _f(t)\le s$ if and only if $f^\sharp (s)\le t$;

• (iii) $\mu _f(t-)<s$ then $f^\sharp (s)<t$ in case $t>0$.

Proof

Suppose $\mu _f(t)>s$. The set $\{\mu _f>s\}\subset [0,\infty )$ is relatively open by right-continuity of $\mu _f$. It follows that $f^\sharp (s)>t$. Conversely, if $f^\sharp (s)>t$ then $\mu _f(t)>s$ be definition of $f^\sharp$. This shows (i). (ii) follows from (i). Suppose $t>0$ and $\mu _f(t-)<s$. By definition of the left-handed limit, we may choose $0<\tau <t$ such that $\mu _f(t-)\le \mu _f(\tau )<s$. Then $f^\sharp (s)\le \tau <t$. This shows (iii). $\square$

Lemma 10.2

Let $f:X\rightarrow [0,\infty )$ be an $\mathscr {A}$-measurable function and $t>0$. Then

• (i) if $g:=f\wedge t$ then $g^\sharp =f^\sharp \wedge t$ on $[0,\mu (X))$;

• (ii) if $g:=f\vee t$ then $g^\sharp =f^\sharp \vee t$ on $[0,\mu (X))$.

Proof

(i) We have that $\mu _g=\chi _{[0,t)}\mu _f$ on $[0,\infty )$ and $g^\sharp =t\chi _{[0,\mu _f(t-))}+f^\sharp \chi _{[\mu _f(t-),\mu (X))}$. On the other hand, $w:=f^\sharp \wedge t=f^\sharp \chi _{\{f^\sharp < t\}}+t\chi _{\{f^\sharp \ge t\}}$. Let $0\le s<\mu _f(t-)$. As $\mu _f\ge \mu _f(t-)>s$ on [0, t), $\{\mu _f\le s\}\subset [t,\infty )$ and $f^\sharp (s)\ge t$; that is, $w(s)=t=g^\sharp (s)$. Now suppose that $s=\mu _f(t-)$. Note that $f^\sharp (s)\le t$. Consider the case that $f^\sharp (s)=t$. Then $w(s)=t=f^\sharp (s)=g^\sharp (s)$. If $f^\sharp (s)<t$ then $w(s)=f^\sharp (s)=g^\sharp (s)$. Finally, suppose that $s>\mu _f(t-)$. By Lemma 10.1 (iii), $w(s)=f^\sharp (s)=g^\sharp (s)$. Item (i) follows.

(ii) In this case, $\mu _g=\chi _{[0,t)}+\chi _{[t,\infty )}\mu _f$ on $[0,\infty )$ and $g^\sharp =\chi _{[0,\mu _f(t))}f^\sharp +t\chi _{[\mu _f(t),\mu (X))}$. On the other hand, $w:=f^\sharp \vee t=t\chi _{\{f^\sharp \le t\}}+f^\sharp \chi _{\{f^\sharp >t\}}$. Let $0\le s<\mu _f(t)$. By Lemma 10.1 (i), $f^\sharp (s)>t$ and $w(s)=f^\sharp (s)=g^\sharp (s)$. Now suppose that $s\ge \mu _f(t)$. By Lemma 10.1 (ii), $f^\sharp (s)\le t$ and $w(s)=t=g^\sharp (s)$.$\square$