Uniqueness of minimizers of weighted least gradient problems arising in hybrid inverse problems

Abstract

We study the question of uniqueness of minimizers of the weighted least gradient problem

$$\begin{aligned} \min \left\{ \int _{\Omega }|Dv|_a : v\in BV_{loc}(\Omega {\setminus } S),\; v|_{\partial \Omega }= f \right\} , \end{aligned}$$

where $\int _{\Omega }|Dv|_a$ is the total variation with respect to the weight function a and S is the set of zeros of the function a. In contrast with previous results, which assume that the weight $a\in C^{1,1}(\Omega )$ and is bounded away from zero, here a is only assumed to be continuous, and is allowed to vanish and also be discontinuous in certain subsets of $\Omega$. We assume instead existence of a $C^1$ minimizer. This problem arises naturally in the hybrid inverse problem of imaging electric conductivity from interior knowledge of the magnitude of one current density vector field, where existence of a $C^1$ minimizer is known a priori.

1 Introduction

Consider the following weighted least gradient problem

$$\begin{aligned} \min \left\{ \int _{\Omega } a|Du|: u\in BV(\Omega ), u|_{\partial \Omega }=f \right\} , \end{aligned}$$

(1)

where $\Omega$ is a bounded open set in $\mathbb {R}^n$ ($n\ge 2$) with connected Lipschitz boundary $\partial \Omega$, a is a bounded non-negative function, and $f\in C(\partial \Omega )$. This problem was first studied for the case $a\equiv 1$ in [17, 19], where existence of a unique minimizer was proved under the assumption that f is continuous and $\partial \Omega$ has positive mean curvature on a dense subset of $\partial \Omega$ (see the precise hypotheses (3.1) and (3.2) in [17]). As in [6], these assumptions on $\partial \Omega$ are needed in the existence proofs but not for the uniqueness arguments. In [6] it has recently been proved that, if $a\in C^{1,1}(\Omega )$ is positive and bounded away from zero and $f\in C(\partial \Omega )$, then the weighted least gradient problem (1) has at most one minimizer in $BV(\Omega )$. The counterexample to uniqueness in [6] for $a \in C^{1,\alpha }(\Omega )$, $0<\alpha <1$, shows that the regularity $a\in C^{1,1}(\Omega )$ is sharp in general. The assumption $a>0$ in $\Omega$ is also crucial for the results in [6].

In contrast with the results in [6, 17], here we present uniqueness results that take into account the regularity of an existing minimizer. The main contribution of this paper is to show that if existence is known a priori, then the assumptions on the weight function a can be considerably weakened. The uniqueness result in this paper is based on a calibration argument and takes into account a priori existence of a $C^1$ minimizer. The following theorem is a consequence of a more general result (Theorem 1.2 below) that will be proved in Sect. 3.

Theorem 1.1

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain with connected boundary, $f\in C^1(\partial \Omega )$, and $a\in C(\overline{\Omega })$. If $a>0$ in $\overline{\Omega }$ and (7) has a minimizer $u\in C^1(\overline{\Omega })$ with $|\nabla u|>0$ in $\overline{\Omega }$, then u is the unique minimizer of (1) in $BV(\Omega )$.

Our motivation comes from a hybrid inverse problem in medical imaging, which concerns determining the conductivity of a body from knowledge of the magnitude $a=|J|$ (in $\Omega$) of one current density vector field J generated by imposing the voltage f on $\partial \Omega$, see [12]. The interior data |J| can be obtained non-invasively via a magnetic resonance technique pioneered in [7]. In [12] this problem was reduced to the weighted least gradient problem (1), by showing that the voltage potential is a minimizer. More precisely, assume $\Omega \subset \mathbb {R}^n$ is made of conductive materials with conductivity $\sigma$. If the voltage f is imposed on $\partial \Omega$, then the corresponding voltage potential u is the solution of the following conductivity equation

$$\begin{aligned} \left\{ \begin{array}{ll} \nabla \cdot \sigma \nabla u=0,&{}\quad \text{ in }\,\Omega ,\\ u=f. &{}\quad \text{ on }\,\partial \Omega \end{array} \right. \end{aligned}$$

(2)

Let $J=-\sigma \nabla u$ be the current density generated by imposing the voltage f on $\partial \Omega$. Then the voltage potential u is a minimizer of the the weighted least gradient problem

$$\begin{aligned} \min \left\{ \int _\Omega |J|\;|\nabla u|dx:\;u\in BV(\Omega ),\, u|_{\partial \Omega }=f\right\} , \end{aligned}$$

(see Proposition 1.2 in [12] and the density argument of Proposition 3 in [15]). More generally, if $\Omega$ also contains perfectly conducting and insulating inclusions ${\mathcal {U_P}}$ and ${\mathcal {U_I}}$. Then the corresponding voltage potential u is the unique solution of the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} \nabla \cdot \sigma \nabla u=0,&{}\quad \text{ in }\,\Omega {\setminus }\overline{{\mathcal {U_P}} \cup {\mathcal {U_I}}},\\ \nabla u=0, &{}\quad \text{ in } \ \ {\mathcal {U_P}},\\ u|_+=u|_-,&{}\quad \text{ on }\ \ \partial ( {\mathcal {U_P}} \cup {\mathcal {U_I}}),\\ \int _{\partial {\mathcal {U_P}}^j}\sigma \frac{\partial u}{\partial \nu }|_{+}ds=0,&{}\quad j=1,2,\ldots ,\\ \frac{\partial u}{\partial \nu }|_{+}=0,&{}\quad \text{ on }\; {\mathcal {U_I}},\\ u|_{\partial \Omega }=f, \end{array} \right. \end{aligned}$$

(3)

where ${\mathcal {U_P}} \cap {\mathcal {U_I}}=\emptyset$ and ${\mathcal {U_P}}=\bigcup _{j=1} {\mathcal {U_P}}^{j}$ is the partition of ${\mathcal {U_P}}$ into open connected components (see the appendix in [10] for more details). Moreover, if $\sigma \in C^{\alpha }(\Omega {\setminus } \overline{ {\mathcal {U_P}} \cup {\mathcal {U_I}}})$, $f \in C^{1,\alpha }(\partial \Omega )$, and the boundaries of ${\mathcal {U_P}}$, ${\mathcal {U_I}}$, and $\Omega$ are regular enough, then it follows from standard elliptic regularity results that $u\in C^{1}(\bar{\Omega } {\setminus } ( {\mathcal {U_P}} \cup {\mathcal {U_I}}))$. Under certain assumptions, it is shown in Theorem 2.1 in [10] (see also the density argument of Proposition 3 in [15]) that the solution of the Eq. (3) is a minimizer of (1), where a is the magnitude of the corresponding current density vector field. Once u is determined, the shape and locations of perfectly conducting and insulation inclusion and the conductivity outside of the inclusions can be uniquely identified. Thus existence is known a priori, and the main issue is to prove uniqueness for the variational problem. Indeed if the conductivity to be recovered is $C^{\alpha }$, then the assumption of Theorems 1.1 and 1.2 are naturally satisfied in our practical setting. See [9, 10, 13, 15] for further results on this problem with only partial data, with insulating or highly conductive inclusions, reconstruction algorithms, and stability, and also [14] for a review. From the point of view of this original application, the uniqueness result in [6] does not apply to the case of embedded insulated and perfectly conductive inclusions described in [10], where the weight a is merely continuous and may vanish in $\Omega$.

The uniqueness results presented here yield global convergence in minimization schemes based on compactness (e.g. [15]) and allow for extending stability results based on the Fredholm’s alternative (e.g. in [8]) to the case of vanishing interior data.

Throughout the paper we assume that $\Omega \subset \mathbb {R}^n$ is a bounded open set with connected Lipschitz boundary $\partial \Omega$ and f is continuous on $\partial \Omega$. The following assumptions concern the most general class of weights a we allow. Let $\mathcal {I}\subset \subset \Omega$ be an open set (possibly empty) with finitely many $C^1$ connected components with finite perimeter, each of which is $C^1$-diffeomorphic with a ball. In addition, in two dimensions $\mathcal {I}$ is assumed to have at most one such component. We assume that $a=0$ in $\mathcal {I}$, $a \in C(\overline{\Omega }{\setminus } \mathcal {I})$, and that a may have at most countable many zeros in $\Omega {\setminus } \overline{\mathcal {I}}$. In other words, the set of zeros

$$\begin{aligned} S:=\{x\in \overline{\Omega }: a(x)=0 \} \end{aligned}$$

(4)

satisfy

$$\begin{aligned} \bar{S}=\overline{\mathcal {I}} \cup \Gamma , \end{aligned}$$

(5)

where $\Gamma$ is a countable set of points in $\overline{\Omega }{\setminus } \overline{\mathcal {I}}$. Note that we do not assume that a vanishes on $\partial \mathcal {I}$. Note also that a may be discontinuous at points on $\partial \mathcal {I}$.

Definition 1

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain with connected boundary. We define $BV_{loc}(\Omega {\setminus } S)$ to be the space of all functions $u \in L^1(\Omega )$ such that

$$\begin{aligned} u \in BV(\Omega {\setminus } S') \ \ \textit{for\;all\;open\;sets}\ \ S' \ \ { with} \ \ S' \supset \supset S. \end{aligned}$$

Recall the following definition from [2]: For any $u \in BV_{loc}(\Omega {\setminus } S)$ the total variation of u (with respect to the weight a) in $\Omega$ is defined as

$$\begin{aligned} \int _{\Omega } |Du|_{a}=\sup _{b \in {\mathfrak {B}}_a} \int _{\Omega } u \nabla \cdot b \ \ dx, \end{aligned}$$

(6)

where

$$\begin{aligned} {\mathfrak {B}}_a:=\left\{ b\in L^\infty (\Omega ;{\mathbb {R}}^n): \;\nabla \cdot b \in L^n(\Omega ),\; \text{ supp }(b)\subset \subset \Omega ,\; |b|\le a\; \hbox {a.e. in}\; \Omega \right\} . \end{aligned}$$

This paper considers the question of uniqueness of solutions of the weighted least gradient problem

$$\begin{aligned} \min \left\{ \int _{\Omega }|Dv|_a: v\in BV_{loc}(\Omega {\setminus } S),\; v|_{\partial \Omega }= f \right\} , \end{aligned}$$

(7)

where the boundary condition is in the sense of the trace of functions in $BV(\Omega )$. Note that, since $a\equiv 0$ in the open set ${\mathcal {I}}$, then $\int _{\Omega }|D v|_a$ is independent of $v|_{\mathcal {I}}$.

Our main uniqueness result assumes the existence of a minimizer $u \in C^1({\overline{\Omega }{\setminus } {\mathcal {I}}})$ of (7), whose set of singularities (possibly empty) satisfy

$$\begin{aligned} {\mathcal {P}}:=\{x\in \overline{\Omega } {\setminus } {\mathcal {I}}: |\nabla u|=0\} \ \ \hbox {is the union of countably many}\;C^1\hbox {-path-connected sets. } \end{aligned}$$

(8)

Now we are ready to state the more general result of this paper.

Theorem 1.2

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain with connected boundary, $f\in C^1(\partial \Omega )$. Assume that the set of zeros of a satisfy the hypothesis (5) and $a\in C(\overline{\Omega } {\setminus } {\mathcal {I}})$. If (7) has a minimizer $u\in C^1(\overline{\Omega }{\setminus } {\mathcal {I}})$ that satisfies (8), then u is the unique minimizer of (7) in $BV_{loc}(\Omega {\setminus } S)$.

Theorem 1.1 follows from the above theorem by taking $S=\emptyset$. We state other special cases in the next corollaries.

Corollary 1.3

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain with connected boundary, $f\in C^1(\partial \Omega )$, and $a\in C(\overline{\Omega })$ with a countable set S of zeros. If (7) has a minimizer $u\in C^1(\overline{\Omega })$ with a countable set of critical points, then u is the unique minimizer of (7) in $BV_{loc}(\Omega {\setminus } S)$.

Corollary 1.4

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain with connected boundary, $f\in C^1(\partial \Omega )$, and $a\in C(\overline{\Omega })$. If S satisfies (4) and the least gradient problem (7) has a minimizer $u\in C^1(\overline{\Omega })$ such that

$$\begin{aligned} \{x\in \overline{\Omega }: |\nabla u (x)|=0 \}, \end{aligned}$$

is the closure of an open set, then u is the unique minimizer of (7) in $BV_{loc}(\Omega {\setminus } S)$.

In the original application in [10], one interprets the open subsets of $\{x\in \overline{\Omega }: |\nabla u (x)|=0 \}$ in which $a>0$ as perfect conductors. To illustrate a simple case with one perfectly conducting inclusion, consider the following example from [18].

Example 1.5

Let $D=\{x\in R^2: \ \ x^2+y^2<1\}$ be the unit disk, $f(x,y)=x^2-y^2$, and ${\mathcal {P}}=(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} )\times (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. It is shown in [18] (see also [10] for a different proof) that

$$\begin{aligned} u=\left\{ \begin{array}{lll} 2x^2-1, \ \ &{}\quad { if}\;|x|\ge \frac{1}{\sqrt{2}}, \ \ |y|\le \frac{1}{\sqrt{2}},\\ 0, \ &{}\quad { if}\; (x,y)\in P,\\ 1-2y^2, \ \ &{}\quad { if} \;|x|\le \frac{1}{\sqrt{2}}, \ \ |y|\ge \frac{1}{\sqrt{2}}\\ \end{array} \right. \end{aligned}$$

is a minimizer of the least gradient problem

$$\begin{aligned} min\left\{ \int _{D}|\nabla u| dx, \ \ u\in BV(D), \ \ and\ \ u|_{\partial D}=f\right\} . \end{aligned}$$

(9)

It is easy to see that u satisfies the assumptions of Corollary 1.4 with $S={\mathcal {I}}=\emptyset$ and ${\mathcal {P}}=(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \times (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. Hence u is the only minimizer in $BV(\Omega )$.

To understand the main ideas of the proofs in Sects. 2 and 3, one can keep in mind the example above and assume $S={\mathcal {I}}=\emptyset$ at the first reading, to avoid some of the technicalities.

2 Preliminaries

In this section we recall and present some preliminary results that will be used in the following sections. First we recall a useful representation formula from [2]. For $u\in BV(\Omega )$

$$\begin{aligned} \int _{A}|Du|_a=\int _{A} h(x,v^u)|Du|, \end{aligned}$$

(10)

where

$$\begin{aligned} h(x,v^u)=|Du|-\hbox {ess} \sup _{b\in {\mathfrak {B_a}}}( b \cdot v^u)(x) \ \ \ \ for \ \ |Du|-a.e. \ \ x\in \Omega \end{aligned}$$

(11)

and $v^u$ denotes the Radon–Nikodym derivative

$$\begin{aligned} v^{u}(x)=\frac{d\, Du}{d\, |Du|}. \end{aligned}$$

(12)

In particular, if $u\in BV(\Omega )$, and the coefficient a is continuous in the Borel measurable subset $A \subset \Omega$, then

$$\begin{aligned} \int _{A}|Du|_a=\int _{A} a|Du|, \end{aligned}$$

(13)

as shown in [2]. The following lemma provides an extension of this formula.

Lemma 2.1

Let $\Omega \subset \mathbb {R}^n$ be a bounded open region with Lipschitz boundary. Suppose that u is a minimizer of (7) and satisfies the assumptions of Theorem 1.2. Then

$$\begin{aligned} \int _{\Omega }|D u|_a=\int _{\Omega }a |\nabla u| dx. \end{aligned}$$

Proof

Since, $a \in C(\Omega {\setminus } ({\mathcal {I}}\cup {\mathcal {P}}))$. Hence by [2, Proposition 7.1] we have that

$$\begin{aligned} h(x,v^u)= \left\{ \begin{array}{lll} a(x) \ \ &{}\hbox {in} \ \ &{}\Omega {\setminus } \overline{{\mathcal {I}}\cup {\mathcal {P}}}\\ 0 \ \ &{}\hbox {in}\ \ &{}{\mathcal {I}}. \\ \end{array} \right. \end{aligned}$$

(14)

Thus it follows from (10) that

$$\begin{aligned} \int _{\Omega } |Du|_a=\int _{\Omega {\setminus } \overline{{\mathcal {I}}\cup {\mathcal {P}}}} a|\nabla u|=\int _{\Omega } a|\nabla u|dx. \end{aligned}$$

$\square$

Let $\nu$ denote the outer unit normal vector to $\partial \Omega$ and

$$\begin{aligned} X:= \{b\in L^{\infty }(\Omega ;{\mathbb {R}}^n): \; \nabla \cdot b \in L^n(\Omega )\}. \end{aligned}$$

Then for every $b\in X$ there exists a unique function $[b\cdot \nu ] \in L^{\infty }_{{\mathcal {H}}^{n-1}}(\partial \Omega )$ such that

$$\begin{aligned} \int _{\partial \Omega } [b\cdot \nu ] u d{\mathcal {H}}^{n-1}=\int _{\Omega } u \nabla \cdot b dx +\int _{\Omega } b \cdot \nabla u dx, \ \ \forall u \in C^1(\overline{\Omega }). \end{aligned}$$

(15)

Moreover, for $u\in BV(\Omega )$ and $b\in X$, the linear functional $u\mapsto (b \cdot Du)$ gives rise to a Radon measure on $\Omega$, and

$$\begin{aligned} \int _{\partial \Omega } [b\cdot \nu ]u d{\mathcal {H}}^{n-1}=\int _{\Omega } u \nabla \cdot b dx +\int _{\Omega } (b \cdot D u), \ \ \forall u \in BV(\Omega ), \end{aligned}$$

(16)

see [1, 3] for a proof. We will need the following generalization of (16) in the proof of our uniqueness result.

Lemma 2.2

Let S be as defined in (5) and $b\in {\mathfrak {B}}_a$. If $u\in L^{\infty }(\Omega )$ and $\int _{\Omega }|Du|_a<\infty$, then

$$\begin{aligned} \int _{\partial \Omega } [{b}\cdot \nu ]u d{\mathcal {H}}^{n-1}=\int _{\Omega } u \nabla \cdot {b} dx +\int _{\Omega } ({b} \cdot D u), \end{aligned}$$

(17)

for some unique function $[b\cdot \nu ]$ in $L^{\infty }_{{\mathcal {H}}^{n-1}}(\partial \Omega )$.

Proof

By hypothesis (5) S has finite perimeter in $\Omega$. Define the set

$$\begin{aligned} {\mathcal {D}}:=\partial S \cap S. \end{aligned}$$

There exists a sequence ${\mathcal {D}}_i$ of open subsets of $\Omega$ with uniformly bounded perimeter such that ${\mathcal {D}}_{i+1}\subset {\mathcal {D}}_{i}$ and $u\in BV({\mathcal {D}}_i)$ for all $i\in \mathbb {N}$, and $\cap _{i=1}^{\infty }{\mathcal {D}}_i={\mathcal {D}}$. Now choose a sequence of cut-off functions $\varphi _i$ such that $\varphi \equiv 0$ on ${\mathcal {D}}_i$, $\varphi \equiv 1$ on the compliment of an open set ${{\mathcal {D'}}}_i \supset \supset {\mathcal {D}}_i$ with $\lim _{i\longrightarrow \infty }{\mathcal {H}}^n({\mathcal {D'}}_i)={\mathcal {H}}^n({\mathcal {D}})$, and $\int _{\Omega }|D\varphi _i|<c$ for all i. Indeed $\varphi _i$ could be chosen to be smooth approximations of the characteristic functions $\chi _{{\mathcal {D}}_i^c}$ in $BV(\mathbb {R}^n)$ and c proportional to the perimeter of ${\mathcal {D}}_i$. Then $\varphi _i u \in BV(\Omega )$ and hence by (16)

$$\begin{aligned} \int _{\partial \Omega } [{b}\cdot \nu ]\varphi _i u d{\mathcal {H}}^{n-1}= & {} \int _{\Omega } \varphi _i u \nabla \cdot {b} dx +\int _{\Omega } ({b} \cdot D (\varphi _i u))\nonumber \\= & {} \int _{\Omega } \varphi _i u \nabla \cdot {b} dx +\int _{\Omega } ({b} \cdot D \varphi _i)u+\int _{\Omega } \varphi _i ({b} \cdot D u) \end{aligned}$$

(18)

for all $b\in {\mathfrak {B}}$ and $i\in \mathbb {N}$. Note that $u\in L^{\infty }(\Omega )$ and $|b|\le a$ a.e. in $\Omega$. It follows from the continuity of a in ${\mathcal {D}} \subset S=\{x\in \overline{\Omega }: a(x)=0\}$ that $\lim _{i\rightarrow \infty } \parallel b\parallel _{L^{\infty }(D'_i{\setminus } D_i)} \rightarrow 0$. Hence

$$\begin{aligned} \left| \int _{\Omega } ({b} \cdot D \varphi _i)u \right| \le \int _{D'_i{\setminus } D_i} |({b} \cdot D \varphi _i)u| \le \int _{D'_i{\setminus } D_i}|D\varphi _i| \parallel u\parallel _{\infty } \parallel b\parallel _{L^{\infty }(D'_i{\setminus } D_i)} \longrightarrow 0, \end{aligned}$$

as $i\longrightarrow \infty$. Since ${\mathcal {I}}\subset \subset \Omega$ and $\Gamma$ [defined in (4)] is countable, by letting $i\longrightarrow \infty$ in (18) we obtain (17). $\square$

The next two results yield a calibration which will be used in the uniqueness proof.

Suppose $a\in L^2(\Omega )$ and fix $v_f \in H^1(\Omega )$ with $v_f|_{\partial \Omega }=f$. Consider the weighted least gradient problem

$$\begin{aligned} (P) \ \ \ \ \inf _{v\in H^1_0(\Omega )} \int _{\Omega } a|\nabla v+\nabla v_f|dx. \end{aligned}$$

In [9] it is shown that the dual problem to (P) is

$$\begin{aligned} (D) \ \ \ \ \max \left\{ <\nabla v_f, b>: \ \ b\in L^2(\Omega ;{\mathbb {R}}^n),\ \ |b(x)|\le a(x) \ \ a.e. \ \ \hbox {and} \ \ \nabla \cdot b\equiv 0\right\} . \end{aligned}$$

Let v(P) and v(D) be the optimal values of the primal and dual problems. It is shown in [9] that $v(P)=v(D)$ and the dual problem (D) has an optimal solution. The following proposition is an immediate consequence of Proposition 2.1 and Corollary 2.3 in [9].

Proposition 2.1

Let $a\in L^2(\Omega )$ be a non-negative function and $v_f \in H^1(\Omega )$ with $v_f|_{\partial \Omega }=f$. Then the optimal values of the primal problem (P) and dual problem (D) are equal, and the dual problem (D) has an optimal solution J with $\nabla \cdot J\equiv 0$ in $\Omega$. Moreover, if $v\in H^1_0(\Omega )$ is an optimal solution of the primal problem (P), then

$$\begin{aligned} J(x)=a(x)\frac{\nabla ( v(x)+v_f(x))}{|\nabla (v(x)+ v_f(x))|} \ \ { if} \ \ |\nabla ( v(x)+v_f(x))|\ne 0, \end{aligned}$$

for all $x\in \Omega$.

The following result is an immediate consequence of Proposition 2.1.

Corollary 2.3

Let $\Omega \subset \mathbb {R}^n$ be a bounded Lipschitz domain and assume that assumptions of Theorem 1.2 are satisfied. Then there exists an optimal solution $J\in L^{2}(\Omega ;{\mathbb {R}}^n)$ of the dual problem (D) such that $\nabla \cdot J\equiv 0$ in $\Omega$, $|J|\le a$ a.e. in $\Omega$. Moreover

$$\begin{aligned} J(x)= \left\{ \begin{array}{ll} a(x)\frac{\nabla u}{|\nabla u|}\ \ &{}\quad { if} \ \ |\nabla u|\ne 0\\ 0 \ \ &{}\quad { if}\ \ |\nabla u|=0, \\ \end{array} \right. \end{aligned}$$

(19)

where u is the solution of (7) described in the statement of Theorem 1.2.

3 Uniqueness of the minimizers

In this section we prove Theorem 1.2. To understand the main ideas of the proofs in this section one may assume $S={\mathcal {I}}=\emptyset$ at first reading.

Let u be the minimizer of the weighted least gradient problem (7) assumed in the statement of Theorem 1.2, and suppose $u^*\in BV_{loc}(\Omega {\setminus } S)$ is another minimizer. We will show that $u=u^*$ a.e. in $\Omega {\setminus } S$. We will prove Theorem 1.2 in four steps. First (Step I) we prove that $\frac{\nabla u}{|\nabla u|}=v^{u^*}$ $Du^*$-a.e. in $\Omega {\setminus } \overline{{\mathcal {I}}\cup {\mathcal {P}}}$, where $v^{u^*}$ is the Radon–Nikodym derivative in (12). In Step II we prove that almost every level set of $u^*$ is also a level set of u. In Step III, we prove that almost every level set of $u^*$ reaches $\partial \Omega$. Finally in Step IV, we show that on almost every level set of $u^*$, u and $u^*$ take the same values and therefore $u=u^*$ a.e. in $\Omega {\setminus } S$.

Step I. :

First notice that $u^*$ is bounded above and below almost everywhere. Indeed if we define

$$\begin{aligned} \bar{u}(x)= \left\{ \begin{array}{ll} u^*(x)\ \ &{}\quad \hbox {if} \ \ m_f\le u^*(x) \le M_f\\ M_f \ \ &{}\quad \hbox {if}\ \ u^*(x)>M_f, \\ m_f\ \ &{}\quad \hbox {if}\ \ u^*(x)<m_f, \end{array} \right. \end{aligned}$$

(20)

where $M_f$ and $m_f$ are the maximum and minimum values of f on $\partial \Omega$, then $\bar{u} \in BV_{loc}(\Omega {\setminus } S)$ and

$$\begin{aligned} \int _{\Omega }|D\bar{u}|_a \le \int _{\Omega }|Du^*|_a. \end{aligned}$$

(21)

Moreover the inequality is strict if $\{x\in \Omega : u^*(x)>M_f\}$ or $\{x\in \Omega : u^*(x)<m_f\}$ has positive measure. Therefore we have $\hbox {range}(u^*)= \hbox {range}(f)$.

Next we prove that

$$\begin{aligned} \frac{\nabla u}{|\nabla u|}=\frac{d Du^*}{d|Du^*|} \end{aligned}$$

$|Du^*|-a.e.$ in $\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}$.

Lemma 3.1

Suppose that the assumptions of Theorem 1.2 are satisfied and u is the minimizer of (7) assumed in the statement of Theorem 1.2. If $u^*$ is another minimizer of (7), then

$$\begin{aligned} \frac{\nabla u}{|\nabla u|}=v^{u^*} \ \ \ \ |Du^*|-a.e.\ \ \hbox {in} \ \ \Omega {\setminus } \overline{{\mathcal {I}}\cup {\mathcal {P}}}. \end{aligned}$$

Proof

It follows from the definition of $h(x, v^{u^*})$ that

$$\begin{aligned} h(x,v^{u^*})\ge J\cdot v^{u^*}, \ \ \ \ |Du^*|-a.e. \ \ \hbox {in} \ \ \Omega , \end{aligned}$$

where J is the solution of the dual problem (D) in Corollary 2.3. Hence

$$\begin{aligned} \int _{\Omega } |Du^*|_a= & {} \int _{\Omega }h(x, v^{u^*})|Du^*| \ge \int _{\Omega } J \cdot v^{u^*} |Du^*|\\= & {} \int _{\Omega } J\cdot Du^*= \int _{\partial \Omega } J\cdot \nu f d{\mathcal {H}}^{n-1}\\= & {} \int _{\Omega }\nabla u\cdot J dx= \int _{\Omega }|J| |\nabla u| \\= & {} \int _{\Omega } |Du|_a=\int _{\Omega } |Du^*|_a, \end{aligned}$$

where the third and fifth equalities follow form Lemmas 2.2 and 2.1, respectively. Therefore

$$\begin{aligned} h(x,v^{u^*})=J\cdot v^{u^*},\ \ |Du^*|-a.e. \ \ \hbox {in} \ \ \Omega . \end{aligned}$$

Since a is continuous in $\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}$, as in (14) we have

$$\begin{aligned} J\cdot v^{u^*}=h(x, v^{u^*})=a(x), \ \ |Du^*|-a.e. \ \ \hbox {in} \ \ \Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}. \end{aligned}$$

(22)

Since $|v^{u^*}|=1$ and $|J|\le a$, $|Du^*|-a.e.$ in $\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}$, we get

$$\begin{aligned} \frac{J}{|J|} =v^{u^*}, \ \ |Du^*|-a.e. \ \ \hbox {in} \ \ \Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}. \end{aligned}$$

On the other hand $|\nabla u|\ne 0$ $|Du^*|$-a.e. on $\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}$. Hence it follows from Corollary 2.3 that

$$\begin{aligned} \frac{\nabla u}{|\nabla u|}=\frac{J}{|J|} =v^{u^*}, \ \ |Du^*|-a.e. \ \ \hbox {in} \ \ \Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}. \end{aligned}$$

The proof is now complete. $\square$

Step II:

For $\lambda \in \hbox {range}(u^*)$, let

$$\begin{aligned} E_{\lambda }=\{x\in \Omega {\setminus } \overline{{\mathcal {I}}}: \ \ u^*(x) \ge \lambda \} \end{aligned}$$

and define

$$\begin{aligned} E'_\lambda :=\left\{ x\in \mathbb {R}^n: \lim _{r \rightarrow 0} \frac{{\mathcal {H}}(B(r,x)\cap E_\lambda )}{{\mathcal {H}}(B(r))}=1\right\} .\ \end{aligned}$$

(23)

By changing $u^*$ in a set of measure zero, we may assume that $E_\lambda =E'_{\lambda }$. Throughout this paper we shall always assume that $E_\lambda =E'_{\lambda }$. We also define

$$\begin{aligned} {\mathcal {Z}}=\{x\in \overline{\Omega }{\setminus } {\mathcal {I}}: u(x)\in \overline{\Gamma \cup {\mathcal {P}}}\}, \end{aligned}$$

(24)

where $\Gamma$ is defined in (4) and ${\mathcal {P}}$ is the set of critical points of u satisfying (8). Notice that if $x \notin {\mathcal {Z}}$, then $|\nabla u(x)|>0$.

Let $\Lambda$ be the set of all $\lambda \in \hbox {range}(u^*)$ such that every connected component $\Sigma$ of $\partial E_\lambda$ with $\Sigma \cap {\mathcal {Z}}=\emptyset$ is a $C^1$ hypersurface. In the next lemma we prove that

$$\begin{aligned} {\mathcal {H}}^1(\hbox {range}(u^*){\setminus } \Lambda )=0. \end{aligned}$$

(25)

Lemma 3.2

Assume that the assumptions of Theorem 1.2 are satisfied and u is the corresponding minimizer of (7). Let $u^* \in BV_{loc}(\Omega {\setminus } S)$ be another minimizer and $\Sigma$ be a connected component of $\partial E_\lambda$ for some $\lambda \in \hbox {range}(u^*)$. Then for almost every $\lambda \in \hbox {range} (u^*)$ one and only one of the following statements hold:

• (i) $\Sigma \subset {\mathcal {Z}}$.

• (ii) $\Sigma \cap {\mathcal {Z}}=\emptyset$, $\Sigma$ is a $C^1$ hypersurface, and u is constant on $\Sigma$.

Proof

By the co-area formula we have

$$\begin{aligned} 0=\int _{\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}} \varphi \left[ \frac{\nabla u}{|\nabla u|} - v^{u^*}\right] |Du^*| = \int _{0}^{\infty } \int _{\partial ^* E_{\lambda } \cap (\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}})} \varphi \left[ \frac{\nabla u}{|\nabla u|} -v^{u^*}\right] d {\mathcal {H}}^{n-1} d\lambda \end{aligned}$$

(26)

for every smooth vector field $\varphi$, where $\partial ^*E_\lambda$ is the reduced boundary of $E_\lambda$. Therefore $\frac{\nabla u}{|\nabla u|}=v^{u^*}$, ${\mathcal {H}}^{n-1}-a.e.$ in $\partial ^* E_\lambda \cap (\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}})$ for almost every $\lambda \in \hbox {range} (u^*)$. Since $|D\chi _{E_{\lambda }}|$ is the $(n-1)-$dimensional Hausdorff measure restricted to $\partial ^* E_\lambda$ (see Chapter 4 in [4]), for almost every $\lambda \in \hbox {range}(u^*)$ the generalized normal $\nu (x)$ exists for $|D\chi _{E_{\lambda }}|-a.e.$ $x\in \partial E_\lambda \cap (\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}})$ and coincides there with the continuous vector field $\frac{\nabla u}{|\nabla u|}$.

Now let $x\in \partial E_\lambda \cap \Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}}$ such that $x \notin {\mathcal {Z}}$. Since ${\mathcal {Z}}$ is closed, there exists $\epsilon >0$ such that $B_{\epsilon }(x)\cap {\mathcal {Z}}=\emptyset$. By Theorem 4.8 in [4], $\partial E_\lambda \cap B_{\epsilon }(x)$ can be represented as the graph of a Lipschitz continuous function g. Thus the derivative of g coincides almost everywhere with a continuous function and therefore g must be $C^1$. Hence we conclude that if $x \in [\partial E_\lambda \cap (\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}})] {\setminus } Z$, then $\partial E_\lambda$ is a $C^1$ hyperspace near x for almost every $\lambda \in \hbox {range}(u^*)$.

Next we show that u is constant on every $C^1$ connected open subset of $\Sigma$ of $\partial E_\lambda \cap (\Omega \backslash \overline{{\mathcal {I}} \cup {\mathcal {P}}})$. Let $\gamma :(-\epsilon ,+\epsilon ) \rightarrow \Sigma$ be an arbitrary $C^1$ curve. Then

$$\begin{aligned} \frac{d}{dt}u(\gamma (s))=|\nabla u(\gamma (s))|\nu (\gamma (s)).\gamma '(s)=0, \end{aligned}$$

because either $|\nabla u(\gamma (s))|=0$ or $\nu (\gamma (s)).\gamma '(s)=0$ on $\Sigma$. Thus u is constant along $\gamma$ and hence u is constant on $\Sigma$. Therefore it follows from continuity of u and the definition of the set ${\mathcal {Z}}$ that, for almost every $\lambda \in \hbox {range}(u^*)$, if $\Sigma \not \subset Z$ is a connected component of $\partial E_\lambda \cap (\Omega {\setminus } \overline{{\mathcal {I}} \cup {\mathcal {P}}})$, then $\Sigma$ is a $C^1$ hypersurface, u is constant on $\Sigma$, and $\Sigma \cap {\mathcal {Z}}=\emptyset$. The proof is now complete. $\square$

Step III:

We show next that every connected component of $\partial E_{\lambda }$ intersects the boundary $\partial \Omega$.

Proposition 3.1

Assume that the assumptions of Theorem 1.2 are satisfied and u is the corresponding minimizer of (7). Suppose $\Sigma _{\lambda }$ is a $C^1$ connected component of $\partial E_{\lambda }=\partial \{x \in \Omega {\setminus } {\mathcal {I}}: \ \ u^*(x)>\lambda \}$ and $\Sigma _{\lambda }\cap {\mathcal {Z}}=\emptyset$. Then

$$\begin{aligned} \overline{\Sigma }_{\lambda }\cap \partial \Omega \ne \emptyset . \end{aligned}$$

Proof

Assume $\overline{\Sigma }_{\lambda }\cap \partial \Omega = \emptyset$. Then one of the followings statements hold:

• (I) $\overline{\Sigma }_\lambda$ is a manifold without boundary in $\overline{\Omega } {\setminus } {\mathcal {I}}$.

• (II) $\overline{\Sigma }_\lambda \cap \partial {\mathcal {I}} \ne \emptyset$.

Case I:

Assume that $\overline{\Sigma }_\lambda$ is a manifold without boundary in $\overline{\Omega }$. Then, since $\partial \Omega$ is connected, $\partial \Omega \cup \Sigma _\lambda$ is a compact manifold with two connected components. By the Alexander duality theorem for $\partial \Omega \cup \Sigma _\lambda$ (see, e.g., Theorem 27.10 in [5]) we have that $\mathbb {R}^n {\setminus }(\partial \Omega \cup \Sigma _\lambda )$ is partitioned into three open connected components: $\mathbb {R}^n=(\mathbb {R}^n{\setminus } \overline{\Omega }) \cup O_1\cup O_2$. Since $\Sigma _\lambda \subset \Omega$ we have $O_1 \cup O_2=\Omega {\setminus } \Sigma _\lambda$ and then $\partial O_i \subset \partial \Omega \cup \Sigma _\lambda$, for $i=1,2$. We claim that at least one of the $\partial O_1$ or $\partial O_2$ is in $\Sigma _\lambda$. Assume not, i.e. for $i=1,2$, $\partial O_i \cap \partial \Omega \ne \emptyset$. Since $\partial \Omega$ is connected (by assumption) we have that $O_1 \cup O_2\cup \partial \Omega$ is connected which implies that $O_1 \cup O_2\cup (\mathbb {R}^n{\setminus } \Omega )$ is also connected. Again by applying the Alexander duality theorem for $\Sigma _\lambda \subset \mathbb {R}^n$, we have that $\mathbb {R}^n {\setminus } \Sigma _\lambda$ has exactly two open connected components, one of which is unbounded: $\mathbb {R}^n {\setminus } \Sigma _\lambda =O_{\infty }\cup O_0$. Since $O_1\cup O_2 \cup (\mathbb {R}^n \backslash \Omega )$ is connected and unbounded, we have that $O_1\cup O_2 \cup (\mathbb {R}^n \backslash \Omega ) \subset O_{\infty }$, which leaves $O_0 \subset \mathbb {R}^n {\setminus } (O_1 \cup O_2 \cup (\mathbb {R}^n {\setminus } \Omega )) \subset \Sigma _\lambda$. This is impossible since $O_0$ is open and $\Sigma _\lambda$ is a hypersurface. Therefore either $\partial O_1$ or $\partial O_2$ or both lie in $\Sigma _\lambda$. Assume $\partial O_1\subset \Sigma _\lambda$. We claim that u is constant in $O_1$. Indeed, by Lemma 3.2, $u=c$ on $\Sigma _{\lambda }$ for some c. Hence the new map $\tilde{u}$ defined by

$$\begin{aligned} \tilde{u}:=\left\{ \begin{array}{ll} u,\ \ &{}\quad x \in \Omega {\setminus } O_1,\\ c, \ \ \ \ &{}\quad x\in \overline{O_1}, \end{array} \right. \end{aligned}$$

is in $BV_{loc}(\Omega {\setminus } S)$ and decreases the energy, which contradicts the minimality of u. Therefore $u=c$ in $O_1$. This is a contradiction since we have assumed $\overline{\Sigma }_{\lambda } \cap {\mathcal {Z}}=\emptyset$.

Case II:

Suppose $\overline{\Sigma }_\lambda \cap \partial {\mathcal {I}} \ne \emptyset$. We first consider the dimensions $n\ge 3$. Let

$$\begin{aligned} \epsilon ^*:= { min}\left\{ \min _{i\ne j} { dist}({\mathcal {I}}_i, {\mathcal {I}}_j), \min _{i} { dist}({\mathcal {I}}_i, \partial \Omega )\right\} , \end{aligned}$$

where ${\mathcal {I}}_i$, $1\le i\le m$, are the open connected components of the set ${\mathcal {I}}$. For any $0<\epsilon <\epsilon ^*:$ define

$$\begin{aligned} {\mathcal {I}}^{\epsilon }=\{x\in \Omega : { dist}(x, {\mathcal {I}})<\epsilon \}. \end{aligned}$$

Then ${\mathcal {I}}^{\epsilon }$ is an open set with the same number of disjoint open connected components as ${\mathcal {I}}$. Now let $\Sigma _{\lambda }^{\epsilon }= \Sigma _\lambda {\setminus } {\mathcal {I}}^{\epsilon }$ which we know is $C^1$ on $\Omega {\setminus } {\mathcal {I}}^{\epsilon }$. Since $\partial \Sigma _{\lambda }^{\epsilon } \subset \partial {\mathcal {I}}^{\epsilon }$ and $\partial {\mathcal {I}}^\epsilon {\setminus } \partial \Sigma _{\lambda }^{\epsilon }$ is open, each connected component of $\partial \Sigma _\lambda ^{\epsilon }$ is the boundary of an open set in $\partial {\mathcal {I}}^{\epsilon }$ with connected boundary. Suppose M is a connected component of $\partial \Sigma _{\lambda }^{\epsilon }$. Then $M \subset \partial {\mathcal {I}}^{\epsilon }_i$ for some $1 \le i \le m$, ${\mathcal {I}}^{\epsilon }_i$ is $C^1$-diffeomorphic image of the unit ball for $\epsilon$ small, and M is an orientable manifold without boundary in $\partial {\mathcal {I}}^{\epsilon }$. Therefore it follows from Alexander’s duality theorem that

$$\begin{aligned} \partial {\mathcal {I}}_i^{\epsilon } {\setminus } M= V_1 \cup V_2, \end{aligned}$$

where $V_1, V_2$ are disjoint open connected (with respect to the topology of $\partial {\mathcal {I}}^{\epsilon }$) sets. Since $\Sigma ^{\epsilon }_{\lambda }$ can be extended to a $C^1$ hypersurface $\Sigma _\lambda$ inside ${\mathcal {I}}^{\epsilon } {\setminus } {\mathcal {I}}$, we can extend $\Sigma _\lambda ^{\epsilon }$ inside ${\mathcal {I}}^{\epsilon }_i$ to obtain a $C^1$ hypersurface $H^{\epsilon }_\lambda$ such that

$$\begin{aligned} H^{\epsilon }_\lambda \cap (\Omega {\setminus } {\mathcal {I}}^{\epsilon })=\Sigma _\lambda ^{\epsilon } \cap (\Omega {\setminus } {\mathcal {I}}^{\epsilon }) \end{aligned}$$

and $\partial (H^{\epsilon }_\lambda \cap {\mathcal {I}}^{\epsilon })=M$. Repeating this argument for other connected components of $\partial \Sigma _\lambda ^{\epsilon }$ leads to a $C^1$ orientable hypersurface ${\mathcal {S}}^{\epsilon }_\lambda$ with no boundary such that $\partial \Omega \cap {\mathcal {S}}^{\epsilon }_\lambda =\emptyset$ and ${\mathcal {S}}^{\epsilon }_\lambda \cap (\Omega {\setminus } {\mathcal {I}}^{\epsilon }) =\partial \Sigma ^{\epsilon }_\lambda$. Now apply Alexander’s duality theorem to get the partition

$$\begin{aligned} \mathbb {R}^n {\setminus } {\mathcal {S}}^{\epsilon }_\lambda =O_0^{\epsilon } \cup O^{\epsilon }_{\infty }, \end{aligned}$$

where $O_0^{\epsilon }$ and $O^{\epsilon }_{\infty }$ are open subsets of $\mathbb {R}^n$ and $O^{\epsilon }_{\infty }$ is unbounded. If $\epsilon '< \epsilon$, then $\Sigma ^{\epsilon }_\lambda \subset \Sigma ^{\epsilon '}_\lambda$ and ${\mathcal {I}}^{\epsilon '} \subset {\mathcal {I}}^{\epsilon }$. Hence

$$\begin{aligned} {\mathcal {I}}^{\epsilon } {\setminus } \bar{O}_0^{\epsilon } \subset {\mathcal {I}}^{\epsilon '} {\setminus } \bar{O}_0^{\epsilon '}. \end{aligned}$$

Now let

$$\begin{aligned} O=\cup _{0<\epsilon <\epsilon ^{*}} \left( {\mathcal {I}}^{\epsilon } {\setminus } \bar{O}_0^{\epsilon }\right) . \end{aligned}$$

Then O is open and $\partial O \subset \Sigma _\lambda \cup \bar{{\mathcal {I}}}$. We claim that u is constant in O. Indeed by Lemma 3.2, $u=c$ on $\Sigma _{\lambda }$ for some constant c. Define

$$\begin{aligned} \tilde{u}:=\left\{ \begin{array}{ll} u,\ \ &{}\quad x \in \Omega \backslash O,\\ c, \ \ \ \ &{}\quad x\in O.\\ \end{array} \right. \end{aligned}$$

(27)

Then $\tilde{u} \in BV_{loc}(\Omega {\setminus } S)$ which contradicts the minimality of u. Hence u is constant in O which is a contradiction because we have assumed $\overline{\Sigma }_{\lambda } \cap {\mathcal {Z}}=\emptyset .$ Now assume $n=2$. Since $\overline{\Sigma }_\lambda \cap \partial \Omega =\emptyset$ and ${\mathcal {I}}$ has only one connected component, there exists two distinct point $a,b \in \bar{\Sigma }_\lambda \cap \partial {\mathcal {I}}$ such that

$$\begin{aligned} \partial {\mathcal {I}} {\setminus } \{a,b\}= V_1 \cup V_2. \end{aligned}$$

Note that $\Sigma _\lambda \cup V_1$ is a continuous closed curve in $\mathbb {R}^2$. By the Jordan Curve Theorem there exists a bounded open set O such that $\partial O=\Sigma _\lambda \cup V_1$. Define $\tilde{u}$ by (27), then with a similar argument we reach a contradiction. In both cases (I) and (II) the contradiction follows from the assumption that $\overline{\Sigma }_\lambda \cap \partial \Omega = \emptyset$.$\square$

Step IV:

Since $f \in C^1(\partial \Omega )$, f can be extended to a function in $C^1(\mathbb {R}^n {\setminus } \Omega ) \cap BV(\mathbb {R}^n {\setminus } \Omega )$. We will denote the extension of f to $\Omega ^c$ by f, again. We will also denote the continuous extension of $u^*$ to $\mathbb {R}^n$ with $u^*=f$ on $\Omega ^c$ by $u^*$ again. Define

$$\begin{aligned} F_\lambda =\{x\in \mathbb {R}^n{\setminus } \bar{{\mathcal {I}}}: \ \ u^*(x)\ge \lambda \} \end{aligned}$$

and let the corresponding $F'_\lambda$ be defines as (23).

Remark 3.3

Let $\Lambda \subset \hbox {range}(u^*)$ be the set defined by Lemma 3.2 and $\lambda \in \Lambda$. By Lemma 3.2 every connected component of $\partial F'_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ is a $C^1$ hypersurface. Since $F_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ differs from $F'_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ on a set of measure zero, we may assume that $F_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ is open.

The proof of the following lemma is very similar to that of Theorem 3.7 in [17]. We include the proof for the convenience of the reader.

Lemma 3.4

Let $\Omega$ be a bounded domain with connected Lipschitz boundary. If $x \in \partial ^*F_\lambda \cap \partial \Omega$, where $\partial ^* F_\lambda$ is the reduced boundary of $F_\lambda$, and

then $\lambda =f(x)$.

Proof

Assume $f(x)<\lambda$. Then

$$\begin{aligned} 0= & {} \lim _{r\rightarrow 0} \frac{1}{|B_r(x)\cap \Omega |} \left( \int _{B_r(x)\cap \Omega \cap \{u^*< \lambda \}} |u^*(y)-f(x)|dy\right. \\&\left. + \int _{B_r(x)\cap \Omega \cap \{u^*\ge \lambda \}} |u^*(y)-f(x)|dy\right) \\\ge & {} \limsup _{r\rightarrow 0}\frac{1}{|B_r(x)\cap \Omega |} \int _{B_r(x)\cap \Omega \cap \{u^*\ge \lambda \}} |u^*(y)-f(x)|dy\\\ge & {} (\lambda -f(x)) \limsup _{r\rightarrow 0} \frac{|B_r(x)\cap \Omega \cap \{u^* \ge \lambda \}|}{|B_r(x)\cap \Omega |}. \end{aligned}$$

Consequently

$$\begin{aligned} \limsup _{r\rightarrow 0} \frac{|B_r(x)\cap \Omega \cap \{u^* \ge \lambda \}|}{|B_r(x)\cap \Omega |}=0. \end{aligned}$$

On the other hand since f is the trace of $u^*\in BV(\mathbb {R}^n {\setminus } \Omega )$ on $\partial \Omega$, with a similar argument we conclude that

$$\begin{aligned} \limsup _{r\rightarrow 0} \frac{|B_r(x)\cap (\mathbb {R}^n {\setminus } \Omega )\cap {\{u^*}\ge \lambda \}|}{|B_r(x)\cap (\mathbb {R}^n{\setminus } \Omega )|}=0. \end{aligned}$$

Therefore

$$\begin{aligned} \lim _{r\rightarrow 0} \frac{|B_r(x) \cap \{u^* \ge \lambda \}|}{|B_r|}=0 \end{aligned}$$

and hence $x \not \in \partial ^* E_\lambda$ which is a contradiction. Similarly $f(x)>\lambda$ leads to a contradiction. Thus $f(x)=\lambda$. $\square$

Proposition 3.2

Assume that the assumptions of Theorem 1.2 hold, and let $u^*$ be a the corresponding minimizer of (7). Then for almost every $\lambda \in \Lambda$

$$\begin{aligned} u(\partial F_{\lambda }\cap (\overline{\Omega }{\setminus } {\mathcal {Z}}))=\{\lambda \}, \end{aligned}$$

where $\Lambda$ and ${\mathcal {Z}}$ are defined by (25) and (24), respectively.

Proof

In view of Remark 3.3 and Proposition 3.1, we may assume that $F_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ is open and every connected component of $\partial F_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$ is a $C^1$ hypersurface intersecting $\partial \Omega$. Now let $\Sigma$ be a connected component of $\partial F_\lambda \cap (\Omega {\setminus } {\mathcal {Z}})$. By Proposition 3.1, $\bar{\Sigma }\cap \partial \Omega \ne \emptyset$. Let $x_0\in \bar{\Sigma }\cap \partial \Omega \ne \emptyset$. Since $x_0 \not \in {\mathcal {Z}}$, $|\nabla u(x_0)|> 0$. On the other hand, note that if $x_0 \in \bar{\Sigma }\cap \partial \Omega {\setminus } \partial ^* F_\lambda$, then $x_0$ is not a regular point of the function $u^* \in BV(\mathbb {R}^n)$, i.e. $u^*$ is discontinuous at $x_0$, which is not of jump type (see §4.4 in [16] for a precise definition of regular point of BV functions). Since the set of points which are not regular points of $u^*$ has $(n-1)-$dimensional measure zero (see §4.5 [16]), for almost every $\lambda \in \Lambda$ and ${\mathcal {H}}^{n-2}$-a.e. $x_0 \in \bar{\Sigma }\cap \partial \Omega$, $x_0 \in \partial ^* F_\lambda \cap \partial \Omega$. Thus by Lemma 3.4 we conclude that $u(\Sigma )=\{\lambda \}$. $\square$

It is now straightforward to deduce uniqueness from the results established above. To make the argument rigorous it helps to work with super level sets of the solutions as in [6, 17]. Note however that we do not rely on maximum principles for minimum surfaces that are at the core of the uniqueness proofs in [6, 17].

Proof of Theorem 1.2

First we prove that $u^*=u$ a.e. in $\Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}})$. Suppose this is not true, then without loss of generality we may assume that there exists $\alpha >0$ such that

$$\begin{aligned} {\mathcal {H}}^{n}(N)>0, \end{aligned}$$

where

$$\begin{aligned} N:=\{x\in \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}}): \ \ u^*(x)\ge u(x)+\alpha \}, \end{aligned}$$

because otherwise the function f in (7) can be replaced by $-f$. Let

$$\begin{aligned} \lambda ^* =\sup \left\{ \lambda : \ \ {\mathcal {H}}^{n} (\{x\in \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}}): \ \ u(x)\ge \lambda \} \cap N )\ge \frac{{\mathcal {H}}^{n}(N)}{2}\right\} . \end{aligned}$$

Since $u\in L^1(\Omega {\setminus } \bar{{\mathcal {I}}})$, $\lambda ^* <\infty$. For $0<\beta <1$ define

$$\begin{aligned} E_1= \{x\in \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}}) : \ \ u^*(x)\ge \lambda ^*+ (1-\beta ) \alpha \}. \end{aligned}$$

By Lemma 3.2 and Proposition 3.1 there exists $0<\beta <1$ such that $\lambda ^*+(1-\beta )\alpha \in \Lambda$. Also it follows from the definition of $\lambda ^*$ that ${\mathcal {H}}^{n}(K)>0$, where

$$\begin{aligned} K:=\{x\in \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}}): \ \ \lambda ^*-\beta \alpha< u(x) < \lambda ^* \} \cap N. \end{aligned}$$

Now let $E_2=\{x\in \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}}): \ \ u(x)\ge \lambda ^* \}$. It is easy to see that $K \subset E_1 {\setminus } \bar{E}_2 \subset \Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}})$. On the other hand by Remark 3.3 we may assume that $E_1$ is open and hence $E_1 {\setminus } \bar{E}_2$ is a non-empty open set. Also

$$\begin{aligned} \partial (E_1{\setminus } \bar{E}_2) \subset \left( \partial E_1 \cap \overline{E_2^c}\right) \cup (E_1 \cap \partial E_2) \end{aligned}$$

and in particular, $\partial (E_1{\setminus } \bar{E}_2) \subset \partial E_1 \cup \partial E_2$. Notice that $\partial (E_1{\setminus } \bar{E}_2) \not \subset \partial E_2$, because otherwise $u=\lambda ^*$ in $E_1{\setminus } \bar{E}_2$ which is in contradiction with the assumption $E_1{\setminus } \bar{E}_2 \subset (\Omega {\setminus } {\mathcal {Z}})$. Let

$$\begin{aligned} x_0 \in \partial (E_1{\setminus } \bar{E}_2). \end{aligned}$$

Then $x_0 \in \partial E_1 \cap \overline{E_2^c}$. By Proposition 3.2 we have

$$\begin{aligned} u(x_0)\in u(\partial E_1)=\{ \lambda ^*+(1-\beta )\alpha \}. \end{aligned}$$

(28)

On the other hand

$$\begin{aligned} u(x_0) \in u(\overline{E_2^c}) \subset (-\infty , \lambda ^*] \end{aligned}$$

which is in contradiction with (28). Hence $u^*=u$ a.e. in $\Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}})$.

To finish the proof let $\Sigma$ be a connected component of ${\mathcal {Z}}$. Since, $\hbox {int} (\overline{u({\mathcal {Z}})})=\emptyset$, u is continuous, $u=u^*$ in $\Omega {\setminus } ({\mathcal {Z}}\cup \bar{{\mathcal {I}}})$, and $u^*$ minimizes (7), $u=u^*$ a.e. in $\Sigma$. The proof is now complete. $\square$

Remark 3.5

Note that in domension $n=2$, if the number of components of ${\mathcal {I}}$ is bigger than one, then there may exists level curves going from one component to the other, and not intersecting $\partial \Omega$. So the uniqueness argument fails. In higher dimensions this can not happen.