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

Let U be a bounded open connected set in Rn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^n$$\end{document} (n≥1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\ge 1$$\end{document}). We refer to the unique weak solution of the Poisson problem -Δu=χA\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-\Delta u = \chi _A$$\end{document} on U with Dirichlet boundary conditions as uA\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u_A$$\end{document} for any measurable set A in U. The function ψ:=uU\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\psi :=u_U$$\end{document} is the torsion function of U. Let V be the measure V:=ψLn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$V:=\psi \,\mathscr {L}^n$$\end{document} on U where Ln\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathscr {L}^n$$\end{document} stands for n-dimensional Lebesgue measure. We study the variational problem I(U,p):=sup{J(A)-V(U)p2}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} I(U,p):=\sup \Big \{ J(A)-V(U)\,p^2\,\Big \} \end{aligned}$$\end{document}with p∈(0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p\in (0,1)$$\end{document} where J(A):=∫AuAdx\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$J(A):=\int _Au_A\,dx$$\end{document} and the supremum is taken over measurable sets A⊂U\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A\subset U$$\end{document} subject to the constraint V(A)=pV(U)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$V(A)=pV(U)$$\end{document}. We relate the above problem to an unstable two-phase membrane problem. We characterise optimsers in the case n=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n=1$$\end{document}. The proof makes use of weighted isoperimetric and Pólya–Szegö inequalities.


Introduction and Motivation
Let U be a bounded open connected set in R n (n ≥ 1). We refer to the unique weak solution of the Poisson problem B I McGillivray maiemg@bristol.ac.uk 1 School of Mathematics, University of Bristol, University Walk, Bristol BS8 1TW, UK − u = χ A on U, u ∈ W 1,2 0 (U ), (1.1) as u A for any measurable set A in U . The function ψ := u U is the torsion function of U . Let V be the measure V := ψ L n on U where L n stands for the n-dimensional Lebesgue measure. For p ∈ (0, 1) consider the variational problem where J (A) := (u A , χ A ) and the supremum is taken over measurable sets A ⊂ U subject to the constraint V (A) = pV (U ). Here, (·, ·) 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 ∈ (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/ψ; that is, E = {u > cψ} for some c ∈ (0, 1) up to 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 − cψ with c as above. Then v is a strong solution of the problem where ± (v) := {±v > 0}. The above equation is similar to Problem C (the twophase 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 (v) = ∂ ± (v) ∩ U is real-analytic (Theorem 3.7). In Sect. 4 we replace U with the unit ball B in R n (n ≥ 2). For p ∈ (0, 1) we show that any optimal configuration E for the data (B, p) possesses spherical cap symmetry 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 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 ψ and perimeter density ψ 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 ≥ 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 ρ, ρ with ρ > ρ and each fluid has unit viscosity. The pressure p has constant gradient ∂ p/∂z = −G on U . Suppose the fluid with density ρ occupies a region A in U . By the Navier-Stokes equations, the vertical component of velocity u satisfies 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 \ A.
The parameter G lies in the interval (ρ g, ρg) which allows the possibility of bidirectional flow. On rescaling (and relabelling the velocities) we obtain the system is a proxy for the pressure gradient. Two problems arise: one to maximise the flux (χ U \A , u) over regions A which satisfy the flux balance condition (u, 1) = 0 with constant λ; the other in which we optimize also over λ. In detail, where in the latter λ 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, λ) is 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 L 1 -a.e. equivalent to either (−1, 0) or (0, 1).

Existence of Optimal Configurations
for t ∈ (0, V (U )) and consider the variational problem 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.
Proof Let p ∈ (0, 1) and put t := pV (U ). Let E be as in Theorem 2.1 (iii). Then We prepare a few lemmas before proving Theorem 2.1.

Lemma 2.3
Let X, Y be (real) Banach spaces and suppose that X ⊂ Y with continuous embedding. Let (x h ) be a sequence in X which converges weakly in X to x ∈ X.
Then (x h ) converges weakly to x in Y .
Proof Note that for any g ∈ Y , g| X ∈ X .
We remark that the Dirichlet Laplacian (D( ), ) is associated with the Dirichlet Let G stand for the corresponding Green operator.

Lemma 2.4 Let t ∈ (0, V (U )). Then
(i) the functional J : V t → R is continuous in the topology of weak sequential convergence; If n ≥ 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 In the case n = 1 we use the fact that W 1,2 0 (U ) is compactly embedded in C 0 (U ) (see [13,Theorem 7.22]) and hence in L 1 (U ).
(ii) Let f ∈ V t . By Dirichlet's principle [10, 2.2.5] for example, The functional E is concave so that J is convex.
Proof The proof runs as in [11,Lemma 2]. A measurable function f on U is L n -a.e. equivalent to Then f τ ∈ V t for |τ | ≤ ε/|α 1 | ∨ |α 2 |. We then derive the contradiction that f is not extremal Proof of Theorem 2.1 Let ( f h ) be a maximising sequence for β(U, t). Now 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 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 ∈ V t as h → ∞ after choosing a subsequence if necessary. By Lemma 2.4 (i), giving item (i) of the Theorem. It is straightforward to see that (ψ, 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 V t at an extremal point f . We then invoke Lemma 2.5 to conclude that f has the form f = χ A L n -a.e. on U for some measurable set A in U and hence (iii).
e. on U . This leads to a contradiction.
We require a version of the bathtub principle (see [17,Theorem 1.14]). Let (X, A , μ) be a finite measure space and ρ a positive A -measurable integrable function on X . Given 0 < v < μ(X ), consider the variational problem where the supremum is taken over measurable sets E ⊂ X with μ(E) = v. We say that measurable sets A, B in X are equivalent μ-a.e. and write A = B if and only if μ(A B) = 0. Proof The distribution function μ ρ : (0, ∞) → (0, V (U )); τ → μ({ρ > τ}) is non-increasing and right-continuous on (0, ∞); in fact, continuous thanks to (3.2). By right-continuity of μ ρ , μ ρ (s) ≤ v; by left-continuity, the reverse inequality holds, so according to the layer cake representation [17,Theorem 1.13]. It follows that E is an optimiser for (3.1). Thus, where the strict inequality follows from (3.3).
contradicting the assumption that E is an optimal configuration. The identity V (E {u ≥ cψ}) = 0 follows from Proposition 3.1.

Corollary 3.4
Suppose that U is a C 1,1 domain. Let p ∈ (0, 1) and suppose that (u, E) is an optimal pair for (1.2) with data (U, p). Put v := u − cψ where c is given by (3.4). Then v is a strong solution of the problem Proof By [13, Theorem 9.15], u ∈ W 2, p (U ) for any 1 < p < ∞ and u is a strong The result follows from the fact that − (cψ) = cχ U and Proposition 3.1.
as c ∈ (0, 1), a contradiction. Here, ω is an arbitrary element in the unit sphere is an optimal pair for the data (U, 1 − p). We then get a contradiction as above. Put

Spherical Cap Symmetry
In this section, we replace U by the open unit ball B in R n (n ≥ 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 ∈ (0, 1). Suppose that (u, E) is an optimal pair for the data (B, p). Then E possesses spherical cap symmetry L n -a.e.
We first discuss the operation of polarisation for integrable functions on B (see [4] and references therein). For ν ∈ S n−1 the closed half-space H = H ν is defined by for L 1 -a.e. 0 < τ < 1, and vice-versa. The Green kernel G(x, y) for B is given by where is the fundamental solution of Laplace's equation in R n as before, d stands for the diagonal in B × B and the decoration * refers to inversion in the unit sphere. We note the inequality which follows from the strong maximum principle.
and similarly S + but with the strict inequality replaced by the sign >. Put A − := τ H A + and A := A + ∪ A − . In this notation, As a consequence, and a similar identity holds for J ( f ) but without composition with reflection. We may then write It is clear from this representation with the help of (4.
In the case of equality, it holds that either |A + | = 0 or |S + | = 0. In the former case, Let ω ∈ S n−1 . Given 0 < τ < 1 and 0 < α ≤ π the spherical cap C ω (τ, α) is the set The function L is Borel measurable The spherical cap symmetrisation of E is the set for example) and |C ω E| = |E|. We say that the Borel set E ⊂ B possesses spherical cap symmetry L n -a.e. if C ω E = E up to L n -a.e. equivalence for some ω ∈ S n−1 .
The function m f (τ, ·) is non-increasing and right continuous. Define its right continuous inverse by For x ∈ B put τ = |x| and choose α ∈ (0, π] such that x · ω = τ cos α then define .
In particular, C ω f is Borel measurable and its L n -equivalence class 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 Equip Q with the 1 -norm x 1 := |x 1 | + |x 2 | and define a mapping ϕ : A geometric argument establishes the following lemma.

Lemma 4.3
For any x, y ∈ Q, ϕx − ϕy 1 ≤ x − y 1 with strict inequality if and only if x ∈ R and y ∈ S or x ∈ R and y ∈ S or the same with the rôles of x and y interchanged.

(iii) The inequality follows by (i) and (ii). On
Proof Note that |τ h x − τ x| ≤ 4|ν h − ν| for each x ∈ S n−1 and h. Now use the density of C(S n−1 ) in L 1 (S n−1 , H n−1 ) and the fact that each τ, τ h is an isometry on Proof For non-negative Borel measurable functions f, g on B, f = g if and only if |{ f > t} {g > t}| = 0 for any t > 0. As f = C ω f there exists t > 0 such that For later use we note that and strict inequality holds by Lemma 4.4 (iii).
To prove the claim, assume for a contradiction that Therefore for all r ∈ (0, 1) it holds that for the r -section of A and likewise for A . Let ν ∈ S n−1 ∩ H ω with corresponding reflection τ = τ H ν . Select a sequence (ν h ) in F which converges to ν in S n−1 and write τ h for the reflection associated to the closed half-space H ν h . For r ∈ (0, 1) \ N , and this latter converges to zero as h → ∞ by Lemma 4.5. We derive that This means that A r has density 1 at x in the sense that Choose y in A r similarly so that A r has density 1 at y.
Proof of Theorem 4.1 Let E be an optimal configuration for the data (U, p). Assume for a contradiction that E = C ω E L n -a.e. for any ω ∈ S n−1 . Then there exists ω ∈ S n−1 such that contradicting optimality of ω. 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.

Preliminaries on Weighted Dirichlet Forms
Let n = 1 and U = (a, b) be an open bounded interval in 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 L 1 -measurable set E in U is given by m(E) := E w dx. We introduce the further assumption Consider the coercive bilinear form in L 2 (U, m).  Given a real-valued function u on R + (or R) define the function θ t u on R + for each
Our next assumption is stronger than required but easy to state: (C) w is unimodal on U . m) : u is weakly differentiable on U and u ∈ L 2 (U, m) .

Lemma 5.3 Assume (A)-(C). Then
Proof Let u ∈ D(E ). There exists a Cauchy sequence (u h ) in (D, E ) which converges to u in L 2 (U, m).
Multiplying by a partition of unity we may assume that u = 0 near b. Denote by u the extension of u to R by zero.   m) : u is weakly differentiable onÛ and ϕ −1 u ∈ L 2 (Û ,m) .

An ( f, g)-Isoperimetric Inequality
Recall that an L 1 -measurable set E ⊂ B is said to be a Caccioppoli set if for each relatively compact open set in B, There then exists a unique real Radon measure Dχ E on B such that   1, x)) and We impose the additional assumptions For an L 1 -measurable set E in B the -rearrangement of E is defined by E := (−1, F −1 (V (E))). We then have that the following ( f, g)-isoperimetric inequality is valid.
by (A.4)-(A.5). The above inequality also holds for 0 ≤ p < q ≤ V (B) with > replaced by ≥ 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 ∈ (0, V (B)). Assume that E has the By Lemma 6.2, (6.2) and (A.5), The result for arbitrary E as in the statement follows by Theorem 6.1, the monotone convergence theorem and continuity of J .
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 , B). Then either E
Proof We may assume that E is the union of closed intervals E h ⊂ R (h = 1, . . . , N , N ∈ N ∪ {∞}) with non-empty interior and separated by open neighbourhoods in R as in Theorem 6.1. In virtue of (6.1) we may take N > 1. We can then find by Theorem 6.3. On the other hand, P g (E, B) = P g (E , B) = J ( p − + p + ) < J ( p − ) + J ( p + ) by (A.5), a contradiction.

Proposition 6.5 Assume (A.1)-(A.3). If g / f is strictly decreasing on B then (A.5) holds.
Proof Note that J is differentiable on (0, V (B)). Moreover, for p ∈ (0, V (B)); this shows that J is strictly concave on by concavity and the fact that 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
Proof Let ⊂ B be a relatively compact open set. Then  1]. This leads to the first claim. The second then follows straightforwardly.
Finally, we state a counterpart of Theorem 6.7.

A Pólya-Szegö Inequality
We first show that the rearrangement · is smoothing in the sense of [21] (see also [4]). Given r > 0 write E r for the r -neighbourhood of an L 1 -measurable set E in (B, d); by convention, ∅ r = ∅. The Minkowski content of E is the quantity

Lemma 7.1 Let E be a finite union of open intervals in (B, d). Then
(i) E is a Caccioppoli set inB; Pǧ(E,B). For p ∈ (0, V (B)) we may writě whereJ =ǧ •F −1 . Note thatJ = J due to the fact thatF • R = F on B.
In particular, the rearrangement · is smoothing in the sense that (E ) r ⊂ (E r ) for each L 1 -measurable set E inB and r > 0. (i) Let u be a real-valued L 1 -measurable function onB. Then ω u (t) ≥ ω u (t) for each t > 0.

(ii) If u is uniformly continuous onB then so is u . (iii) If u is Lipschitz continuous onB then so is u and Lip(u ,B) ≤ Lip(u,B).
Proof Let t > 0. We may assume that ω u (t) > 0. Choose τ > 0 such that ω u (t) > τ. By Lemma 7.4 there exist s, s ∈ R with s > s + τ such that d({u > s},B \ {u > s }) < t. Now {u > s} = {|u| > s} and likewise for s by the counterpart of the equimeasurabilty property (6.4). By Lemma 7.3 we deduce that d({|u| > s},B \{|u| > s }) < t and again by Lemma 7.4 that ω u (t) ≥ ω |u| (t) > τ. Item (i) then follows. Part (ii) is a ready consequence. As for (iii), Let u be a Lipschitz continuous function on (B, d). By Rademacher's theorem (cf. [1, Theorem 2.14]) u is differentiable L 1 -a.e. onB and its derivative coincides with the weak derivative on a set of full measure. Put (vi)μ u is differentiable L 1 -a.e. on R with derivative given by for L 1 -a.e. t ∈ R; (vii) Ran(u) = supp(Dμ u ).
The notation above Dμ a u , Dμ s u , Dμ j u stands for the absolutely continuous resp. singular resp. jump part of the measure Dμ u (see [1, 3.2] for example).
by Fubini's theorem; soμ u ∈ BV(R) and Dμ u is the push-forward ofV under u, Dμ u = −u V (cf. [1, 1.70]). By (7.1), for any L 1 -measurable set A in R. In light of the above, we may identify Dμ a u = Dμ u (R \ N ) and Dμ s u = Dμ u N . The set of atoms of Dμ u is defined by A := {t ∈ R : Dμ u ({t}) = 0}. By [13,Lemma 7.7], we may write A as in (v). The monotone functionμ u is a good representative within its equivalence class and is differentiable L 1 -a.e. on R with derivative given by the density of Dμ u with respect to L 1 by [1,Theorem 3.28]. Item (vii) follows from (ii).

Lemma 7.7 Let u be a nonnegative Lipschitz continuous function on (B, d). Then
Proof As Z has finite L 1 -measure, A ⊂ R is a countable set. Thus and an analogous result holds for u by Lemma 7.6. The fact thatμ u =μ u entails that A = A . This leads to the result.
which is a consequence of the coarea formula (7.1). Asμ u =μ u we have that Dμ u = Dμ u . In particular, we derive thať for L 1 -a.e. t ∈ R by Lemma 7.6. Let t ∈ Ran(u) be such that Z ∩ {u = t} = ∅ and (B\Z )∩{u=t}ǧ dH 0 < ∞ and the analogous properties hold for u . We assume in addition that (7. 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.

Equality Case in the Pólya-Szegö Inequality
We now investigate the equality case in the Pólya-Szegö inequality. Suppose that u ∈ W 1,2 loc (B) is precisely represented in the sense of [18, (2.5)]. Then the set {u = t} is finite or countably infinite for L 1 -a.e. t ∈ R and the coarea formula (7.1) holds for u by [18,Theorem 1.1]. With Z as before it follows that Z ∩{u = t} = ∅ for L 1 -a.e. t ∈ R and hence N := u(Z ) ⊂ R is L 1 -negligible. Proof The proof proceeds as in Lemma 7.7.
We may then write an identity of the above form but with u , v in place of u, v.   > t},B) for L 1 -a.e. t ∈ Ran(ũ). By Corollary 6.12 the set {ũ > t} is either an open interval inB abutting a boundary point or {ũ > t} = ∅ for L 1 -a.e. t > 0. The statement follows by Lemma 8.5.
Proof Put v := u • R −1 ∈ D(Ě ) by Lemma 5.5. Also, u = v • R by Proposition 6.10. We havê by Lemma 5.5 and Corollary 7.9. SoĚ (v, v) =Ě (v , v ). By Theorem 8.6,ṽ is monotone onB and henceũ is monotone on B. Proof Let A be an L 1 -measurable set in B and put u := u A ∈ D(E ) and v := u A ∈ D(E ). By Theorem 6.6, Proposition 9.2, the Cauchy-Schwarz inequality and Corollary 7.10,