Condensers with infinitely many touching Borel plates and minimum energy problems

TL;DR

This paper investigates minimum energy problems for infinite collections of touching Borel plates in a locally compact space, establishing conditions for solutions, their properties, and convergence, with implications for classical kernels.

Contribution

It provides new sufficient conditions for the existence, uniqueness, and properties of solutions to infinite-dimensional minimum energy problems with constraints.

Findings

Established existence and compactness of solutions.

Analyzed convergence of minimizing nets.

Illustrated phenomena of non-existence and non-uniqueness.

Abstract

Defining a condenser in a locally compact space as a locally finite, countable collection of Borel sets $A_i$, $i\in I$, with the sign $s_i=\pm1$ prescribed such that $A_i\cap A_j=\varnothing$ whenever $s_is_j=-1$, we consider a minimum energy problem with an external field over infinite dimensional vector measures $(\mu^i)_{i\in I}$, where $\mu^i$ is a suitably normalized positive Radon measure carried by $A_i$ and such that $\mu^i\leqslant\xi^i$ for all $i\in I_0$, $I_0\subset I$ and constraints $\xi^i$, $i\in I_0$, being given. If $I_0=\varnothing$, the problem reduces to the (unconstrained) Gauss variational problem, which is in general unsolvable even for a condenser of two closed, oppositely signed plates in $\mathbb R^3$ and the Coulomb kernel. Nevertheless, we provide sufficient conditions for the existence of solutions to the stated problem in its full generality, establish the vague compactness of the solutions, analyze their uniqueness, describe their weighted potentials, and single out their characteristic properties. The strong and the vague convergence of minimizing nets to the minimizers is also studied. The phenomena of non-existence and non-uniqueness of solutions to the problem are illustrated by examples. The results obtained are new even for the classical kernels on $\mathbb R^n$, $n\geqslant2$, and closed $A_i$, $i\in I$, which is important for applications.

Figures

Equations

$${\mathfrak{M}}^{+}(\mathbf{A}):=\prod _{i\in I}{\mathfrak{M}}^{+}(A_i).$$

$$⟨\mu ,\nu {⟩}_{\kappa }:=\kappa (\mu ,\nu ):=\int \kappa (x,y)d(\mu \otimes \nu )(x,y),\mu ,\nu \in {\mathcal{E}}_{\kappa }(X),$$

$$\kappa (\mathbf{\mu },\mathbf{\mu }):=\sum _{i,j\in I}{s}_i{s}_j\kappa ({\mu }^i,{\mu }^j).$$

$$G_{\kappa ,\mathbf{f}}(\mathbf{\mu }):=\kappa (\mathbf{\mu },\mathbf{\mu })+2⟨\mathbf{f},\mathbf{\mu }⟩,\mathbf{\mu }\in {\mathcal{E}}_{\kappa ,\mathbf{f}}^{+}(\mathbf{A},\mathbf{a},\mathbf{g}).$$

$${\mathfrak{M}}^{+}(\mathbf{A})\ni \mathbf{\mu }\mapsto R\mathbf{\mu }:=\sum _{i\in I}{s}_i{\mu }^i$$

$$\|\boldsymbol{\mu}_{1}-\boldsymbol{\mu}_{2}\|_{\mathcal{E}^{+}_{\kappa}(\mathbf{A})}:=\Bigl{[}\sum_{i,j\in I}\,s_{i}s_{j}\kappa(\mu^{i}_{1}-\mu^{i}_{2},\mu^{j}_{1}-\mu^{j}_{2})\Bigr{]}^{1/2},$$

$$\mathcal{E}^{+}_{\kappa}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g}):=\bigl{\{}\boldsymbol{\mu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A}):\ \langle g_{i},\mu^{i}\rangle\leqslant a_{i}\text{ \ for all\ }i\in I\bigr{\}}.$$

$$\sum _{i\in I}{a}_i{g}_{i,\inf }^{-1}:=C<\infty ,\text{ where }{g}_{i,\inf }:=\underset{x\in A_i}{\inf }{g}_i(x).$$

$$w_{\alpha }(\mathbf{A}):=\inf {\kappa }_{\alpha }({\mu }^1-{\mu }^2,{\mu }^1-{\mu }^2),$$

$${\mathcal{E}}_{{\kappa }_{\alpha }}^{+}(A_i,1):=\{\nu \in {\mathfrak{M}}^{+}(A_i)\cap {\mathcal{E}}_{{\kappa }_{\alpha }}({\mathbb{R}}^n):\nu (A_i)=1\}.$$

$$\sum _{k\in \mathbb{N}}\frac{c_{{\kappa }_{\alpha }}(F_k)}{q^{k(n-\alpha )}}<\infty .$$

$$A_{2}:=\bigl{\{}x\in\mathbb{R}^{3}:\ 0\leqslant x_{1}<\infty,\ x_{2}^{2}+x_{3}^{2}\leqslant\varrho^{2}(x_{1})\bigr{\}},$$

$$\varrho (x_1)$$

$$\underset{k\to \infty }{\lim }{\nu }_k(Q)=\nu (Q).$$

$$c_{\kappa}(Q):=\Bigl{[}\,\inf_{\nu\in\mathcal{E}_{\kappa}^{+}(Q):\ \nu(Q)=1}\,\kappa(\nu,\nu)\Bigr{]}^{-1}$$

$$|\kappa (\mu ,\nu )|⩽\Vert \mu {\Vert }_{\kappa }\cdot \Vert \nu {\Vert }_{\kappa },\text{ where }\mu ,\nu \in {\mathcal{E}}_{\kappa }(X),$$

$$c_{\kappa }(Q△\breve{Q})=0,\text{ where }Q△\breve{Q}:=(Q\setminus \breve{Q})\cup (\breve{Q}\setminus Q).$$

$$\underset{\mu \in \mathfrak{F}}{\sup }|{\mu }^i(\phi )|<\infty \text{ for all }i\in I.$$

$$\mathfrak{F}^{i}:=\bigl{\{}\mu^{i}\in\mathfrak{M}^{+}(X):\ \boldsymbol{\mu}=(\mu^{j})_{j\in I}\in\mathfrak{F}\bigr{\}}$$

$$A_i\cap A_j=\varnothing \text{ for all }i\in I^{+},j\in I^{-}.$$

$$s_{i}:={\rm sign}\,A_{i}:=\left\{\begin{array}[]{lll}+1&\text{if}&i\in I^{+},\\ -1&\text{if}&i\in I^{-},\\ \end{array}\right.$$

$${\breve{A}}^{+}:=\bigcup _{i\in I^{+}}{\breve{A}}_i\text{ and }{\breve{A}}^{-}:=\bigcup _{j\in I^{-}}{\breve{A}}_j.$$

$$R\mathbf{\mu }(\phi )=\sum _{i\in I}{s}_i{\mu }^i(\phi )\text{ for all }\phi \in C_0(X),$$

$$(R\mathbf{\mu }{)}^{+}:=\sum _{i\in I^{+}}{\mu }^i\text{ and }(R\mathbf{\mu }{)}^{-}:=\sum _{i\in I^{-}}{\mu }^i.$$

$$\kappa (\mathbf{\mu },\mathbf{\nu }):=\sum _{i,j\in I}{s}_i{s}_j\kappa ({\mu }^i,{\nu }^j)$$

$${\kappa }_{\mathbf{\mu }}^i(\cdot ):=\sum _{j\in I}{s}_i{s}_j\kappa (\cdot ,{\mu }^j),i\in I,$$

$$\sum_{i,j\in I}\,|\kappa(\mu^{i},\mu^{j})|\leqslant\sum_{i,j\in I}\,\|\mu^{i}\|_{\kappa}\|\mu^{j}\|_{\kappa}=\Bigl{(}\sum_{i\in I}\,\|\mu^{i}\|_{\kappa}\Bigr{)}^{2}.\vspace{-.2in}$$

$$⟨\psi ,R_{\mathbf{B}}\mathbf{\omega }⟩=\sum _{\ell \in L}{s}_{\ell }⟨\psi ,{\omega }^{\ell }⟩.$$

$$⟨\psi ,(R_{\mathbf{B}}\mathbf{\omega }{)}^{+}⟩⩾\sum _{\ell \in L^{+},\ell ⩽N}⟨\psi ,{\omega }^{\ell }⟩\text{ for all }N\in L^{+}.$$

Full text

Condensers with infinitely many touching Borel plates and minimum energy problems

Natalia Zorii

Dedicated to the memory of Professor Bogdan Bojarski

Abstract. Defining a condenser in a locally compact space as a locally finite, countable collection of Borel sets $A_{i}$, $i\in I$, with the sign $s_{i}=\pm 1$ prescribed such that $A_{i}\cap A_{j}=\varnothing$ whenever $s_{i}s_{j}=-1$, we consider a minimum energy problem with an external field over infinite dimensional vector measures $(\mu^{i})_{i\in I}$, where $\mu^{i}$ is a suitably normalized positive Radon measure carried by $A_{i}$ and such that $\mu^{i}\leqslant\xi^{i}$ for all $i\in I_{0}$, $I_{0}\subset I$ and constraints $\xi^{i}$, $i\in I_{0}$, being given. If $I_{0}=\varnothing$, the problem reduces to the (unconstrained) Gauss variational problem, which is in general unsolvable even for a condenser of two closed, oppositely signed plates in $\mathbb{R}^{3}$ and the Coulomb kernel. Nevertheless, we provide sufficient conditions for the existence of solutions to the stated problem in its full generality, establish the vague compactness of the solutions, analyze their uniqueness, describe their weighted potentials, and single out their characteristic properties. The strong and the vague convergence of minimizing nets to the minimizers is studied. The phenomena of non-existence and non-uniqueness of solutions to the problem are illustrated by examples. The results obtained are new even for the classical kernels on $\mathbb{R}^{n}$, $n\geqslant 2$, and closed $A_{i}$, $i\in I$, which is important for applications. 000 2010 Mathematics Subject Classification: Primary 31C15. 000 Key words: infinite dimensional vector Radon measures, consistent kernels, minimum energy problems, condensers with infinitely many touching Borel plates, external fields, constraints.

1. Introduction

The interest in minimum energy problems with external fields, initially inspired by Gauss [21] and further experiencing a new growth due to work of Frostman [14] and Polish and Japanese mathematicians (Leja et al. and Ohtsuka; see [29, 37] and the references cited therein), has been motivated by their direct relations with the Dirichlet and balayage problems. A new impulse to this part of potential theory (which is often referred to as Gauss variational problem or weighted minimum energy problems) came in the 1980’s when Gonchar and Rakhmanov [22, 24], Mhaskar and Saff [32] applied logarithmic potentials with external fields in the investigation of orthogonal polynomials and rational approximations to analytic functions.

In the present paper we study weighted minimum energy problems in a very general setting, over infinite dimensional vector measures on a locally compact (Hausdorff) space (l.c.s.) $X$ [3, Chapter I, Section 9, n∘ 7], associated with a generalized condenser. To be precise, a generalized condenser $\mathbf{A}$ in $X$ is a locally finite, countable collection of Borel sets $A_{i}\subset X$, $i\in I$, termed plates, with the sign $s_{i}:=\mathrm{sign}\,A_{i}=\pm 1$ prescribed such that $A_{i}\cap A_{j}=\varnothing$ whenever $s_{i}s_{j}=-1$. We emphasize that, although any two oppositely charged plates of a generalized condenser are disjoint, their closures in $X$ may have points in common. A generalized condenser $\mathbf{A}$ is said to be standard if the $A_{i}$, $i\in I$, are closed in $X$. The concept of a standard condenser with infinitely many (closed) plates has been introduced first in our earlier study [43], while that of a generalized condenser seems to be new. Unless explicitly stated otherwise, when speaking of a condenser, we shall tacitly assume it to be generalized.

We denote by $\mathfrak{M}(X)$ the linear space of all real-valued scalar Radon measures on $X$, equipped with the vague topology, i.e., the (Hausdorff) topology of pointwise convergence on the class $C_{0}(X)$ of all continuous functions on $X$ with compact support.111When speaking of a continuous function, we understand that the values are finite real numbers. For any set $Q\subset X$, let $\mathfrak{M}^{+}(Q)$ stand for the cone of all positive $\nu\in\mathfrak{M}(X)$ carried by $Q$ (for a definition, see Section 2.1 below). These and other notions of the theory of measures and integration on a l.c.s., to be used throughout the paper, can be found in [13, 4]; see also [15] for a short survey.

A vector measure $\boldsymbol{\mu}=(\mu^{i})_{i\in I}$ is said to be associated with a (generalized) condenser $\mathbf{A}=(A_{i})_{i\in I}$ if $\mu^{i}\in\mathfrak{M}^{+}(A_{i})$ for all $i\in I$. Denoting by $\mathfrak{M}^{+}(\mathbf{A})$ the class of all those $\boldsymbol{\mu}$, we thus have222If $I$ is a singleton, we preserve the normal fonts instead of the bold ones.

[TABLE]

The trace of the vague product space topology on $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ on $\mathfrak{M}^{+}(\mathbf{A})$ is likewise called the vague topology on $\mathfrak{M}^{+}(\mathbf{A})$.

For any topological space $Y$, let $\Psi(Y)$ consist of all lower semicontinuous (l.s.c.) functions $\psi:Y\to(-\infty,\infty]$, nonnegative unless $Y$ is compact.

A kernel on $X$ is defined as a symmetric function $\kappa\in\Psi(X\times X)$. In the present paper we shall be concerned with a positive definite kernel $\kappa$, which means that the energy $\kappa(\nu,\nu):=\int\kappa(x,y)\,d(\nu\otimes\nu)(x,y)$ of any (signed) $\nu\in\mathfrak{M}(X)$ is nonnegative whenever defined. (By definition, $\kappa(\nu,\nu)$ is well defined provided that $\kappa(\nu^{+},\nu^{+})+\kappa(\nu^{-},\nu^{-})$ or $\kappa(\nu^{+},\nu^{-})$ is finite, $\nu^{+}$ and $\nu^{-}$ being the positive and negative parts in the Hahn–Jordan decomposition of $\nu$, respectively.) Then the set $\mathcal{E}_{\kappa}(X)$ of all $\nu\in\mathfrak{M}(X)$ with finite $\kappa(\nu,\nu)$ is a pre-Hilbert space with the inner product

[TABLE]

and the seminorm $\|\nu\|_{\kappa}:=\sqrt{\kappa(\nu,\nu)}$. The topology on $\mathcal{E}_{\kappa}(X)$ determined by $\|\cdot\|_{\kappa}$ is termed strong. A (positive definite) kernel $\kappa$ is said to be strictly positive definite if the seminorm $\|\cdot\|_{\kappa}$ is a norm.

In accordance with an electrostatic interpretation of a condenser, assume that the interaction between the components $\mu^{i}$, $i\in I$, of $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ is characterized by the matrix $(s_{i}s_{j})_{i,j\in I}$, so that the energy of $\boldsymbol{\mu}$ is given by333An expression $\sum_{i\in I}c_{i}$ involving numerical values $c_{i}$ is meant to be well defined provided that every summand is so and the sum does not depend on the order of summation. By the Riemann series theorem, the sum is finite if and only if the series converges absolutely. Thus, $\kappa(\boldsymbol{\mu},\boldsymbol{\mu})$ is finite provided that $\kappa$ is $(\mu^{i}\otimes\mu^{j})$-integrable for all $i,j\in I$ and the series in (1.1) converges absolutely.

[TABLE]

Let $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ consist of all $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ with finite $\kappa(\boldsymbol{\mu},\boldsymbol{\mu})$ (see footnote 3).

To define admissible measures in the extremal problem we shall be dealing with, fix a numerical vector $\mathbf{a}=(a_{i})_{i\in I}$ with $a_{i}>0$, a vector-valued function $\mathbf{g}=(g_{i})_{i\in I}$ with continuous $g_{i}:X\to(0,\infty)$, and a vector-valued external field $\mathbf{f}=(f_{i})_{i\in I}$ with universally measurable $f_{i}:X\to[-\infty,\infty]$. Let $\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ consist of all $\boldsymbol{\mu}\in\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ such that $\langle g_{i},\mu^{i}\rangle:=\int g_{i}\,d\mu^{i}=a_{i}$ for all $i\in I$ and $\langle\mathbf{f},\boldsymbol{\mu}\rangle:=\sum_{i\in I}\,\langle f_{i},\mu^{i}\rangle$ is finite (see footnote 3); then so is the weighted energy

[TABLE]

Fix also $I_{0}\subset I$ and $\xi^{i}\in\mathfrak{M}^{+}(A_{i})$, $i\in I_{0}$, such that $\langle g_{i},\xi^{i}\rangle>a_{i}$; these $\xi^{i}$, $i\in I_{0}$, will serve as (upper) constraints acting on positive measures carried by $A_{i}$, $i\in I_{0}$. We shall be concerned with the problem of minimizing the weighted energy $G_{\kappa,\mathbf{f}}(\boldsymbol{\mu})$ over all $\boldsymbol{\mu}\in\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ with the additional property that $\mu^{i}\leqslant\xi^{i}$ for all $i\in I_{0}$.

If $I_{0}=\varnothing$, the problem reduces to the (unconstrained) Gauss variational problem, which is in general unsolvable even for a standard condenser of two closed, oppositely charged plates in $\mathbb{R}^{n}$, $n\geqslant 3$, and the Riesz kernels $\kappa_{\alpha}(x,y):=|x-y|^{\alpha-n}$, $\alpha\in(0,n)$. (Here, $|x-y|$ denotes the Euclidean distance between $x,y\in\mathbb{R}^{n}$.) See Theorem 1.6 below providing necessary and sufficient conditions for the solvability of this problem for $\alpha\in(0,2]$. The phenomenon of unsolvability is illustrated by Example 1.7.

Nevertheless, we provide sufficient conditions for the existence of solutions to the stated problem in its full generality and establish the vague compactness of the solutions (Theorems 6.1, 6.3, and 6.5), analyze their uniqueness (Section 4.2), describe their weighted potentials, and single out their characteristic properties (Theorem 8.2 and Corollary 8.3). The strong and the vague convergence of minimizing nets to the minimizers is also studied (Eq. (6.5) and Corollary 6.7). We discover the phenomenon of non-uniqueness of solutions to the problem, which is illustrated by Example 4.6.

Remark 1.1**.**

The results obtained are new even for the classical kernels on $\mathbb{R}^{n}$, $n\geqslant 2$ (in particular, for $-\log|x-y|$ on $\mathbb{R}^{2}$), and closed $A_{i}$, $i\in I$, which is important for applications. While our investigation is focused on theoretical aspects in a very general context, and possible applications are so far outside the frames of the present paper, it is noteworthy to remark that minimum energy problems in the constrained and unconstrained settings for the logarithmic kernel and finite dimensional vector measures have been considered for several decades in relation to Hermite–Padé approximants [23, 1] and random matrix ensembles [27, 2].

The results of the present paper, mentioned above, are obtained for a condenser with nearly closed plates, which differ from closed sets in a set of zero inner capacity $c_{\kappa}(\cdot)$ (Definition 2.8).444These closed sets may not form a condenser. Nevertheless, we develop an efficient approach to the study of energies and potentials of infinite dimensional vector measures for an arbitrary generalized condenser (Section 3), which we intend to use in our further work.

The approach developed is based on the observation that, since $(A_{i})_{i\in I}$ is locally finite, the $A_{i}$, $i\in I$, are Borel, and $A_{i}\cap A_{j}=\varnothing$ whenever $s_{i}s_{j}=-1$, the mapping

[TABLE]

maps $\mathfrak{M}^{+}(\mathbf{A})$ onto a certain set of signed scalar Radon measures on $X$. Furthermore, $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ becomes a semimetric space with the semimetric

[TABLE]

and $R$ maps $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ isometrically onto its (scalar) $R$-image, contained in the pre-Hilbert space $\mathcal{E}_{\kappa}(X)$ (see Section 3.5). In view of this isometry, the topology on the semimetric space $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ is likewise termed strong.

Another fact crucial to our approach is a strong completeness result for a certain subspace of $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$, where $\mathbf{A}$ is a standard condenser (see Theorem 1.2 below, established in our earlier paper [43]). Let $A^{+}$, resp. $A^{-}$, denote the union of the $A_{i}$, $i\in I$, with $s_{i}=+1$, resp. $s_{i}=-1$. Write

[TABLE]

Theorem 1.2**.**

Assume the $A_{i}$, $i\in I$, are closed, $\kappa$ is consistent,555We refer to [15, 17] for the concept of consistency (see also Section 2.2 below). and

[TABLE]

If, moreover, $\kappa|_{A^{+}\times A^{-}}$ is upper bounded, then the following assertions hold.

• $\bullet$

$\mathcal{E}^{+}_{\kappa}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g})$ is complete in the induced strong topology. In more detail, any strong Cauchy net in $\mathcal{E}^{+}_{\kappa}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g})$ converges strongly to any of its vague cluster points.

• $\bullet$

If, moreover, $\kappa$ is strictly positive definite and the $A_{i}$, $i\in I$, are mutually disjoint, then the strong topology on $\mathcal{E}^{+}_{\kappa}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g})$ is finer than the induced vague topology.

1.1. Minimum $\alpha$-Riesz energy problem for a standard condenser

We next show that the problem in question is in general unsolvable even in the case where $\mathbf{A}=(A_{1},A_{2})$ is a standard condenser in $\mathbb{R}^{n}$, $n\geqslant 3$, with $s_{1}=+1$, $s_{2}=-1$, $f_{1}\equiv f_{2}\equiv 0$, $g_{1}\equiv g_{2}\equiv 1$, $a_{1}=a_{2}=1$, $I_{0}=\varnothing$, and $\kappa(x,y):=\kappa_{\alpha}(x,y):=|x-y|^{\alpha-n}$, $\alpha\in(0,2]$. Under these requirements, the problem can equivalently be rewritten as follows:

[TABLE]

where $\mu^{i}$, $i=1,2$, ranges over the class

[TABLE]

To formulate the corresponding result and to explain in brief the reason for the phenomenon of unsolvability, we first recall the concept of $\alpha$-thinness at infinity.

Throughout Section 1.1, $F$ denotes a closed set in $\mathbb{R}^{n}$, $n\geqslant 3$, such that $F^{c}:=\mathbb{R}^{n}\setminus F\neq\varnothing$, and $F^{*}$ the inverse of $F$ relative to $\{x\in\mathbb{R}^{n}:|x-x_{0}|=1\}$, $x_{0}\in F^{c}$ being fixed. Let $\nu^{F}$ stand for the $\alpha$-Riesz swept measure of $\nu\in\mathfrak{M}^{+}(\mathbb{R}^{n})$ onto $F$, determined uniquely by [19, Theorem 3.6].

Definition 1.3**.**

$F$ is said to be $\alpha$-thin at infinity if any of the following four equivalent assertions holds:

• (i)

$F^{*}$ is $\alpha$-thin at $x_{0}$.

• (ii)

Either $F$ is compact, or $x_{0}$ is an $\alpha$-irregular boundary point of $F^{*}$.

• (iii)

If $F_{k}$ denotes $F\cap\{x\in\mathbb{R}^{n}:q^{k}\leqslant|x-x_{0}|<q^{k+1}\}$, where $q\in(1,\infty)$, then

[TABLE]

• (iv)

There exists a connected component $D$ of $F^{c}$ such that for every $\nu\in\mathfrak{M}^{+}(D)$ with $\nu(\mathbb{R}^{n})<\infty$ we have $\nu^{F}(\mathbb{R}^{n})<\nu(\mathbb{R}^{n})$.

The equivalence of (i) and (ii) is due to [5, Theorem VII.13] or [28, Theorem 5.10], that of (ii) and (iii) holds by the Wiener criterion, and that of (iii) and (iv) has been established in [19, Theorem 3.22] (see also earlier papers [39, Theorem B] and [40, Theorem 4]).

Theorem 1.4**.**

If $F$ is not $\alpha$-thin at infinity, then $c_{\kappa_{\alpha}}(F)=\infty$. This cannot be reversed, i.e., there is $F$ with $c_{\kappa_{\alpha}}(F)=\infty$ that is $\alpha$-thin at infinity.

Proof.

According to [28, Lemma 5.5], c_{\kappa_{\alpha}}(F)<\infty$$\iff$$\sum_{k\in\mathbb{N}}\,c_{\kappa_{\alpha}}(F_{k})<\infty, $F_{k}$ being defined in Definition 1.3(iii). When compared with (1.5), this yields the theorem. ∎

Remark 1.5**.**

There is a false statement in [5, Chapter IX, Section 6] and [31, p. 216] that a closed set $F\subset\mathbb{R}^{n}$, $n\geqslant 3$, is $2$-thin at infinity if and only if $c_{\kappa_{2}}(F)<\infty$. Although for any $\alpha\in(0,2]$, $c_{\kappa_{\alpha}}(F)<\infty$ implies, indeed, the $\alpha$-thinness of $F$ at infinity, the converse is incorrect (see Theorem 1.4 above, as well as [7, pp. 276–277] pertaining to $\alpha=2$). We emphasize that the $\alpha$-thinness of $F$ at infinity is indeed equivalent to the existence of the $\alpha$-Riesz equilibrium measure $\gamma_{F}$ on $F$, but treated in an extended sense where $\gamma_{F}(F)=\kappa_{\alpha}(\gamma_{F},\gamma_{F})=\infty$ is allowed [28, Theorem 5.1].666Mizuta [33] has shown that the $2$-thinness of a closed planar set at infinity does not necessarily imply the finiteness of the logarithmic capacity, giving thus an answer in the negative to a conjecture by Ninomiya related to the kernel $-\log|x-y|$ on $\mathbb{R}^{2}$ [31, p. 216].

Returning to problem (1.4), we can certainly assume that $c_{\kappa_{\alpha}}(A_{i})>0$, $i=1,2$, for if not, then $w_{\alpha}(\mathbf{A})=+\infty$, and hence the problem makes no sense. There is also no loss of generality in assuming $c_{\kappa_{\alpha}}(A_{1})<\infty$, because if $c_{\kappa_{\alpha}}(A_{i})=\infty$ for $i=1,2$, then $w_{\alpha}(\mathbf{A})=0$; and hence this infimum cannot be an actual minimum, $\kappa_{\alpha}$ being strictly positive definite [28, Theorem 1.15].

Theorem 1.6** (see [40, Theorem 5]).**

Assume, for simplicity, $A^{c}_{2}$ is connected. If, moreover, the Euclidean distance between $A_{1}$ and $A_{2}$ is ${}>0$, then problem (1.4) is (uniquely) solvable if and only if either $c_{\kappa_{\alpha}}(A_{2})<\infty$, or $A_{2}$ is not $\alpha$-thin at infinity.

It follows that, if $A_{2}$ is $\alpha$-thin at infinity, but $c_{\kappa_{\alpha}}(A_{2})=\infty$ (such $A_{2}$ exists by Theorem 1.4), then $w_{\alpha}(\mathbf{A})$ cannot be attained among admissible measures. The reason for this phenomenon is that, under the quoted assumptions, any minimizing sequence converges strongly and vaguely to a (unique) $\gamma=\gamma^{+}-\gamma^{-}$ such that $\gamma^{+}\in\mathcal{E}^{+}_{\kappa_{\alpha}}(A_{1},1)$, while $\gamma^{-}=(\gamma^{+})^{A_{2}}$ [40, Eq. (27)]. Since $A_{2}$ is $\alpha$-thin at infinity, we get $(\gamma^{+})^{A_{2}}(A_{2})<1$ by Definition 1.3(iv), and problem (1.4) therefore has no solution.

Example 1.7**.**

Let $n=3$ and $\alpha=2$. Define $A_{2}$ to be a rotation body

[TABLE]

where $\varrho$ is given by one of the following three formulae:

[TABLE]

and let $A_{1}$ be a closed ball in $\mathbb{R}^{3}\setminus A_{2}$. Then $A_{2}$ is not $2$-thin at infinity if $\varrho$ is defined by (1.6), $A_{2}$ is $2$-thin at infinity but has infinite Newtonian capacity if $\varrho$ is given by (1.7), and finally $c_{\kappa_{2}}(A_{2})<\infty$ if (1.8) holds [41, Example 5.3]. By Theorem 1.6, problem (1.4) is therefore solvable for $\mathbf{A}=(A_{1},A_{2})$ if $A_{2}$ is determined either by (1.6), or by (1.8), but problem (1.4) is unsolvable if $A_{2}$ is given by (1.7) (see Figure 1).

Remark 1.8**.**

Theorem 1.6 and Example 1.7 have been illustrated in [35, 25] by means of numerical experiments.

2. Preliminaries

2.1. Measures, vague convergence, capacity

We shall tacitly use the notation of Section 1. The vague topology on $\mathfrak{M}(X)$ in general does not possess a countable base, and hence it cannot be described in terms of convergence of sequences. We follow Moore and Smith’s theory of convergence, based on the concept of nets [34] (see also [26, Chapter 2] and [13, Chapter 0]). However, if $X$ is metrizable and countable at infinity, where the latter means that $X$ can be written as a countable union of compact sets [3, Chapter I, Section 9, n∘ 9], then $\mathfrak{M}(X)$ satisfies the first axiom of countability [18, Remark 2.4], and the use of nets may be avoided.

Lemma 2.1** (see, e.g., [15, Section 1.1]).**

For any $\psi\in\Psi(X)$ the map $\nu\mapsto\langle\psi,\nu\rangle$ is vaguely l.s.c. on $\mathfrak{M}^{+}(X)$.

Let a set $Q\subset X$ and a measure $\nu\in\mathfrak{M}^{+}(X)$ be given. If $Q$ is $\nu$-measurable, then the indicator function $1_{Q}$ of $Q$ is locally $\nu$-integrable, and hence one can consider the trace (restriction) $\nu|_{Q}=1_{Q}\cdot\nu$ of $\nu$ on $Q$ [13, Section 4.14.7]. As in [13, Section 4.7.3], $Q$ is said to be $\nu$-$\sigma$-finite if $Q$ is contained in a countable union of $\nu$-integrable open sets.777This necessarily holds if $X$ is countable at infinity or $\nu$ is bounded, i.e., with $\nu(X)<\infty$. If $Q$ is open or $\nu$-measurable and $\nu$-$\sigma$-finite, then $\nu_{*}(Q)=\nu^{*}(Q)\in[0,\infty]$, where $\nu_{*}(Q)$ and $\nu^{*}(Q)$ denote the inner and the outer $\nu$-measure of $Q$, respectively [13, Eqs. (4.7.3), (4.7.4)]; and we write $\nu(Q):=\nu_{*}(Q)=\nu^{*}(Q)$.

Lemma 2.2**.**

If $Q$ is $\nu$-measurable and $\nu$-$\sigma$-finite, then for any nonnegative l.s.c. function $\psi$ on $X$ we have $\langle\psi,\nu|_{Q}\rangle=\langle\psi|_{Q},\nu\rangle$.

Proof.

Applying first [13, Proposition 4.14.1(b)] and [13, Eq. (4.14.8)] to $\psi|_{Q}$ and $\nu$, and then applying [13, Proposition 4.14.1(a)] to $\psi$ and $\nu|_{Q}$, we arrive at our claim.∎

Theorem 2.3**.**

Let $X$ be metrizable and countable at infinity. If a sequence $\{\nu_{k}\}_{k\in\mathbb{N}}\subset\mathfrak{M}^{+}(X)$ converges to $\nu$ vaguely, then for any relatively compact Borel set $Q\subset X$ with $\nu(\partial_{X}Q)=0$ we have $\nu_{k}|_{Q}\to\nu|_{Q}$ vaguely as $k\to\infty$.888If $X$ is an open subset of $\mathbb{R}^{n}$, $n\geqslant 2$, then Theorem 2.3 is, in fact, [28, Theorem 0.5*′*].

Proof.

The Portmanteau theorem in the form stated in [30, Theorem 2.1] shows that under the hypotheses of Theorem 2.3,

[TABLE]

Applying now to $X$, $Q$, $\nu_{k}$ and $\nu$ the same arguments as in [28, Proof of Theorem 0.5*′*], the only difference being in using the preceding display in place of [28, Theorem 0.5], we establish the theorem.∎

Let $\mathfrak{M}^{+}(Q)$ consist of all $\nu\in\mathfrak{M}^{+}(X)$ carried by $Q$, which means that $Q^{c}:=X\setminus Q$ is locally $\nu$-negligible, or equivalently that $Q$ is $\nu$-measurable and $\nu=\nu|_{Q}$. If $Q^{c}$ is open or $\nu$-$\sigma$-finite, then the concept of local $\nu$-negligibility for $Q^{c}$ coincides with that of $\nu$-negligibility; and hence $\nu\in\mathfrak{M}^{+}(Q)$ if and only if $\nu^{*}(Q^{c})=0$. Therefore, $\nu$ is carried by a closed $Q$ if and only if it is supported by $Q$; that is, $S(\nu)\subset Q$, where $S(\nu)$ is the support of $\nu$.

In all that follows, $\kappa$ is a positive definite kernel on $X$ (Section 1). For any $Q\subset X$, write $\mathcal{E}^{+}_{\kappa}(Q):=\mathcal{E}_{\kappa}(X)\cap\mathfrak{M}^{+}(Q)$. The (inner) capacity of $Q$ is given by the formula

[TABLE]

(see, e.g., [15, 37]). Then $0\leqslant c_{\kappa}(Q)\leqslant\infty$. (As usual, the infimum over the empty set is taken to be $+\infty$. We also set 1\bigl{/}(+\infty)=0 and 1\bigl{/}0=+\infty.)

A proposition $\mathcal{P}(x)$ involving a variable point $x\in X$ is said to hold $c_{\kappa}$-nearly everywhere ($c_{\kappa}$-n.e.) on $Q$ if $c_{\kappa}(N)=0$, where $N$ consists of all $x\in Q$ for which $\mathcal{P}(x)$ fails. We write briefly ‘n.e.’ in place of ‘$c_{\kappa}$-n.e.’ if this does not cause any misunderstanding, and we omit ‘on $Q$’ if $Q=X$.

Lemma 2.4** (see [15, Lemma 2.3.1]).**

$c_{\kappa}(Q)=0$ $\iff$ $\mathcal{E}^{+}_{\kappa}(Q)=\{0\}$.

2.2. Consistent and perfect kernels

In addition to the strong topology on $\mathcal{E}_{\kappa}(X)$, determined by the seminorm $\|\cdot\|_{\kappa}$ (see Section 1), it is often useful to consider the so-called weak topology on $\mathcal{E}_{\kappa}(X)$, defined by means of the seminorms $\nu\mapsto|\kappa(\nu,\mu)|$, where $\mu\in\mathcal{E}_{\kappa}(X)$ [15]. By the Cauchy–Schwarz (Bunyakovski) inequality

[TABLE]

the strong topology on $\mathcal{E}_{\kappa}(X)$ is finer than the weak topology.

Following Fuglede [15, 17], we call a (positive definite) kernel $\kappa$ consistent if it satisfies either of the following two equivalent properties:

• (C1)

Every strong Cauchy net in $\mathcal{E}^{+}_{\kappa}(X)$ converges strongly to any of its vague cluster points (whenever these exist).

• (C2)

Every strongly bounded and vaguely convergent net in $\mathcal{E}^{+}_{\kappa}(X)$ converges weakly to its vague limit.

A kernel $\kappa$ is called perfect if it is consistent and strictly positive definite [15, Theorem 3.3], or equivalently if the following two conditions are fulfilled (see [15, p. 166]):

• (P1)

$\mathcal{E}^{+}_{\kappa}(X)$ is complete in the induced strong topology.

• (P2)

The strong topology on $\mathcal{E}^{+}_{\kappa}(X)$ is finer than the induced vague topology on $\mathcal{E}^{+}_{\kappa}(X)$.

Example 2.5**.**

On $X=\mathbb{R}^{n}$, $n\geqslant 3$, the $\alpha$-Riesz kernel $\kappa_{\alpha}$, $\alpha\in(0,n)$, is strictly positive definite and consistent, and hence altogether perfect [8]; thus so is the Newtonian kernel $\kappa_{2}(x,y)=|x-y|^{2-n}$ [6]. Recently it has been shown that, if $X=D$ where $D$ is an arbitrary open set in $\mathbb{R}^{n}$, $n\geqslant 3$, and $G^{\alpha}_{D}$, $\alpha\in(0,2]$, is the $\alpha$-Green kernel on $D$ [28, Chapter IV, Section 5], then $\kappa=G^{\alpha}_{D}$ is likewise perfect [19, Theorems 4.9, 4.11]. Furthermore, the $2$-Green kernel on a planar $2$-Greenian set is strictly positive definite by [9, Chapter XIII, Section 7] and it is consistent by [12], and hence altogether perfect. The logarithmic kernel $-\log\,|x-y|$ on a closed disc in $\mathbb{R}^{2}$ of radius ${}<1$ is strictly positive definite, as shown in [28, Theorem 1.16]. It is therefore perfect (see [36]), because it satisfies Frostman’s maximum principle by [28, Theorem 1.6], and hence is regular by [37, Eq. (1.3)]. For analogous results concerning the logarithmic kernel on closed balls of arbitrary finite dimension, see [16].

Remark 2.6**.**

In contrast to (P1), for a perfect kernel $\kappa$ the whole pre-Hilbert space $\mathcal{E}_{\kappa}(X)$ is in general strongly incomplete, and this is the case even for the $\alpha$-Riesz kernel of order $\alpha\in(1,n)$ on $\mathbb{R}^{n}$, $n\geqslant 3$ [6].

Remark 2.7**.**

The concept of consistent kernel is an efficient tool in minimum energy problems over classes of positive scalar Radon measures with finite energy. Indeed, if $Q$ is closed, $c_{\kappa}(Q)\in(0,\infty)$, and $\kappa$ is consistent, then the minimum energy problem in (2.1) has a solution $\lambda$ [15, Theorem 4.1]; we shall call this $\lambda$ an (inner) $\kappa$-capacitary measure on $Q$. (This $\lambda$ is unique if $\kappa$ is strictly positive definite.) Later the concept of consistency has been shown to be efficient also in minimum energy problems over classes of vector measures of finite or infinite dimensions associated with a standard condenser [41]–[44]. The approach developed in [41]–[44] substantially used the assumption of the boundedness of the kernel on the Cartesian product of the oppositely charged plates of a condenser, which made it possible to extend Cartan’s proof [6] of the strong completeness of the cone $\mathcal{E}_{\kappa_{2}}^{+}(\mathbb{R}^{n})$ of all positive measures on $\mathbb{R}^{n}$ with finite Newtonian energy to an arbitrary consistent kernel $\kappa$ on a l.c.s. $X$ and suitable classes of (signed) measures $\mu\in\mathcal{E}_{\kappa}(X)$ (compare with Theorem 1.2 as well as Remark 2.6 above).

2.3. Nearly closed sets

The following concept seems to be new (a private communication with Bent Fuglede).

Definition 2.8**.**

A set $Q\subset X$ is said to be nearly closed, resp. nearly compact, if there exists a closed, resp. compact, set $\breve{Q}\subset X$ such that

[TABLE]

Lemma 2.9**.**

For any nearly closed set $Q$, $\mathcal{E}^{+}_{\kappa}(Q)=\mathcal{E}^{+}_{\kappa}(\breve{Q})$.

Proof.

Note that $Q=[\breve{Q}\cup(Q\setminus\breve{Q})]\setminus(\breve{Q}\setminus Q)$. Since any set in $X$ with $c_{\kappa}(\cdot)=0$ cannot carry any nonzero measure from $\mathcal{E}_{\kappa}^{+}(X)$ (cf. Lemma 2.4), $\mathcal{E}^{+}_{\kappa}(Q)\subset\mathcal{E}^{+}_{\kappa}(\breve{Q})$. Having reversed $Q$ and $\breve{Q}$, we obtain the converse inclusion.∎

Lemma 2.10**.**

If a set $Q\subset X$ is nearly closed, then the truncated cone $\{\nu\in\mathcal{E}^{+}_{\kappa}(Q):\|\nu\|_{\kappa}\leqslant 1\}$ is closed in the induced vague topology.

Proof.

As seen from Lemma 2.9, it is enough to establish the lemma for $\breve{Q}$ in place of $Q$. Since $\mathfrak{M}^{+}(\breve{Q})$ is vaguely closed, $\breve{Q}$ being closed in $X$, and since the energy $\kappa(\nu,\nu)$ is vaguely l.s.c. on $\mathfrak{M}^{+}(X)$ [15, Lemma 2.2.1(e)], the lemma follows. ∎

3. Vector measures. Their energies and potentials

3.1. Vector

measures

Fix a countable set $I$ of indices $i\in\mathbb{N}$, and consider the Cartesian product $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$, equipped with the vague product space topology. Elements $\boldsymbol{\mu}=(\mu^{i})_{i\in I}$ of $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$, where $\mu^{i}\in\mathfrak{M}^{+}(X)$ for all $i\in I$, are termed positive $({\mathrm{Card}\,I})$-dimensional vector measures on $X$.

Definition 3.1**.**

A set $\mathfrak{F}\subset\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ is said to be vaguely bounded if for every $\varphi\in C_{0}(X)$,

[TABLE]

Lemma 3.2**.**

A vaguely bounded set $\mathfrak{F}\subset\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ is vaguely relatively compact.

Proof.

It is clear from the above definition that for every $i\in I$, the set

[TABLE]

is vaguely bounded, and hence vaguely relatively compact in $\mathfrak{M}^{+}(X)$ [4, Chapter III, Section 2, Proposition 9]. Since $\mathfrak{F}\subset\prod_{i\in I}\,\mathfrak{F}^{i}$, the lemma follows from Tychonoff’s theorem on the product of compact spaces [3, Chapter I, Section 9, Theorem 3].∎

Since $\mathfrak{M}^{+}(X)$ is Hausdorff in the vague topology, so is $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ [3, Chapter I, Section 8, Proposition 7], and hence a vague limit of any net $(\boldsymbol{\mu}_{s})_{s\in S}\subset\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ is unique if it exists. (Throughout the paper, $S$ denotes an upper directed set of indices $s$.)

3.2. Generalized and standard condensers

Assume that $I=I^{+}\cup I^{-}$, where $I^{+}\cap I^{-}=\varnothing$ and $I^{-}$ is allowed to be empty, and there corresponds to every $i\in I$ a nonempty Borel set $A_{i}\subset X$.

Definition 3.3**.**

A collection $\mathbf{A}=(A_{i})_{i\in I}$ is termed a generalized $(I^{+},I^{-})$-condenser (or simply a generalized condenser) in $X$ if every compact subset of $X$ intersects with at most finitely many $A_{i}$, and moreover

[TABLE]

Writing

[TABLE]

we call $A_{i}$, $i\in I^{+}$, and $A_{j}$, $j\in I^{-}$, positive and negative plates of the generalized condenser $\mathbf{A}$. Note that any two equally signed plates may intersect each other or even coincide. Also note that, although any two oppositely signed plates are disjoint by (3.1), their closures in $X$ may intersect each other (actually, even in a set of nonzero capacity).999This remains valid even in the case where the $A_{i}$, $i\in I$, are nearly closed. Furthermore, it follows from above definition that the sets $A^{+}:=\bigcup_{i\in I^{+}}\,A_{i}$ and $A^{-}:=\bigcup_{j\in I^{-}}\,A_{j}$ are Borel and disjoint, which will be used substantially in all that follows.

Lemma 3.4**.**

If the $A_{i}$, $i\in I$, are nearly closed, then so are $A^{+}$ and $A^{-}$.

Proof.

With $\breve{A}_{i}$, $i\in I$, determined by Definition 2.8 for $Q=A_{i}$, write

[TABLE]

Then $\breve{A}^{\pm}$ is closed, for the collection $(\breve{A}_{i})_{i\in I^{\pm}}$ of (closed) sets $\breve{A}_{i}$ is locally finite. Since $c_{\kappa}(A_{i}\bigtriangleup\breve{A}_{i})=0$ for all $i\in I$, the countable subadditivity of inner capacity on Borel sets [15, Lemma 2.3.5] yields $c_{\kappa}(A^{\pm}\bigtriangleup\breve{A}^{\pm})=0$.∎

Definition 3.5**.**

A generalized condenser is standard if its plates are closed.

Unless explicitly stated otherwise, in all that follows $\mathbf{A}=(A_{i})_{i\in I}$ is a generalized condenser in $X$. Let $\mathfrak{M}^{+}(\mathbf{A})$ consist of all $\boldsymbol{\mu}=(\mu^{i})_{i\in I}\in\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ with $\mu^{i}\in\mathfrak{M}^{+}(A_{i})$ for all $i\in I$. In other words, $\mathfrak{M}^{+}(\mathbf{A})$ stands for the Cartesian product $\prod_{i\in I}\,\mathfrak{M}^{+}(A_{i})$, equipped with the vague topology induced from $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$. Elements of $\mathfrak{M}^{+}(\mathbf{A})$ are said to be (vector) measures associated with $\mathbf{A}$.

Lemma 3.6**.**

If a condenser $\mathbf{A}$ is standard, then $\mathfrak{M}^{+}(\mathbf{A})$ is vaguely closed in $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$.

Proof.

Noting that the $\mathfrak{M}^{+}(A_{i})$, $i\in I$, are vaguely closed in $\mathfrak{M}^{+}(X)$ ($A_{i}$ being closed in $X$), we obtain the lemma from [3, Chapter I, Section 4, Corollary to Proposition 7].∎

3.3. Mapping $R:\mathfrak{M}^{+}(\mathbf{A})\to\mathfrak{M}(X)$

Since each compact subset of $X$ has points in common with at most finitely many $A_{i}$, for every $\varphi\in C_{0}(X)$ only a finite number of $\mu^{i}(\varphi)$, $\boldsymbol{\mu}=(\mu^{i})_{i\in I}\in\mathfrak{M}^{+}(\mathbf{A})$ being given, are nonzero. This implies that there corresponds to every positive vector measure $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ a unique (signed) scalar Radon measure $R\boldsymbol{\mu}=R(\boldsymbol{\mu})\in\mathfrak{M}(X)$ such that

[TABLE]

$s_{i}$ being determined by (3.2). Since the positive scalar measures $\sum_{i\in I^{+}}\mu^{i}$ and $\sum_{i\in I^{-}}\mu^{i}$ are carried by the nonintersecting Borel sets $A^{+}$ and $A^{-}$, respectively, these two measures are, in fact, the positive and negative parts in the Hahn–Jordan decomposition of $R\boldsymbol{\mu}$; i.e., $R\boldsymbol{\mu}=(R\boldsymbol{\mu})^{+}-(R\boldsymbol{\mu})^{-}$, where

[TABLE]

When the dependence of the mapping $R$ on the (generalized) condenser $\mathbf{A}$ needs to be indicated explicitly, we shall write $R_{\mathbf{A}}$ in place of $R$.

The mapping $\mathfrak{M}^{+}(\mathbf{A})\to\mathfrak{M}(X)$ thus defined is in general non-injective; i.e., there exist $\boldsymbol{\mu}_{1},\boldsymbol{\mu}_{2}\in\mathfrak{M}^{+}(\mathbf{A})$ such that $\boldsymbol{\mu}_{1}\neq\boldsymbol{\mu}_{2}$, but $R\boldsymbol{\mu}_{1}=R\boldsymbol{\mu}_{2}$. We say that $\boldsymbol{\mu}_{1},\boldsymbol{\mu}_{2}\in\mathfrak{M}^{+}(\mathbf{A})$ are $R$-equivalent if $R\boldsymbol{\mu}_{1}=R\boldsymbol{\mu}_{2}$. The relation of $R$-equivalence on $\mathfrak{M}^{+}(\mathbf{A})$ implies that of identity (and hence these two relations on $\mathfrak{M}^{+}(\mathbf{A})$ are equivalent) if and only if $A_{i}\cap A_{j}=\varnothing$ for all $i\neq j$ (compare with Lemma 3.15 below).

Lemma 3.7**.**

If a net $(\boldsymbol{\mu}_{s})_{s\in S}\subset\mathfrak{M}^{+}(\mathbf{A})$ converges vaguely to $\boldsymbol{\mu}_{0}\in\mathfrak{M}^{+}(\mathbf{A})$, then $R\boldsymbol{\mu}_{s}\to R\boldsymbol{\mu}_{0}$ vaguely in $\mathfrak{M}(X)$ as $s$ increases along $S$.

Proof.

This follows directly from the observation that the (compact) support of any $\varphi\in C_{0}(X)$ can intersect only finitely many $A_{i}$.∎

Remark 3.8**.**

Lemma 3.7 cannot in general be reversed. However, if the $A_{i}$, $i\in I$, are closed and mutually disjoint, then for any $(\boldsymbol{\mu}_{s})_{s\in S}$ and $\boldsymbol{\mu}_{0}$ in $\mathfrak{M}^{+}(\mathbf{A})$, the vague convergence of $(R\boldsymbol{\mu}_{s})_{s\in S}$ to $R\boldsymbol{\mu}_{0}$ implies the vague convergence of $(\boldsymbol{\mu}_{s})_{s\in S}$ to $\boldsymbol{\mu}_{0}$. This is seen by the Tietze–Urysohn extension theorem [13, Theorem 0.2.13].

3.4. Energies and potentials of vector measures and those of their scalar

$R$-images

For a (positive definite) kernel $\kappa$ and vector measures $\boldsymbol{\mu},\boldsymbol{\nu}\in\mathfrak{M}^{+}(\mathbf{A})$, define the mutual energy101010With regard to (3.4) and (3.5), see footnote 3.

[TABLE]

and the vector potential $\kappa_{\boldsymbol{\mu}}(\cdot)$ as a vector-valued function on $X$ with the components

[TABLE]

where $\kappa(\cdot,\nu):=\int\kappa(\cdot,y)\,d\nu(y)$ denotes the potential of $\nu\in\mathfrak{M}(X)$. For $\boldsymbol{\mu}=\boldsymbol{\nu}$, $\kappa(\boldsymbol{\mu},\boldsymbol{\nu})$ becomes the energy $\kappa(\boldsymbol{\mu},\boldsymbol{\mu})$ of $\boldsymbol{\mu}$, cf. (1.1).

Let $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ consist of all $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ with finite $\kappa(\boldsymbol{\mu},\boldsymbol{\mu})$, which means that $\kappa$ is $(\mu^{i}\otimes\mu^{j})$-integrable for all $i,j\in I$ and the series $\sum_{i,j\in I}\,|\kappa(\mu^{i},\mu^{j})|$ is convergent (the latter can be omitted if $X$ is compact, for then $I$ is finite).

Lemma 3.9**.**

For $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ to have finite energy, it is sufficient that $\mu^{i}\in\mathcal{E}^{+}_{\kappa}(A_{i})$ for all $i\in I$ and, moreover, $\sum_{i\in I}\,\|\mu^{i}\|_{\kappa}<\infty$.

Proof.

In fact, applying the Cauchy–Schwarz inequality in $\mathcal{E}_{\kappa}(X)$, we get

[TABLE]

∎

The following lemma is crucial for the establishment of relations between energies and potentials of vector measures $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A})$ and those of their (signed scalar) $R$-images $R\boldsymbol{\mu}\in\mathfrak{M}(X)$.

Lemma 3.10**.**

Given a generalized $(L^{+},L^{-})$-condenser $\mathbf{B}=(B_{\ell})_{\ell\in L}$ in a l.c.s. $Y$, consider $\boldsymbol{\omega}=(\omega^{\ell})_{\ell\in L}\in\mathfrak{M}^{+}(\mathbf{B})$ and $\psi\in\Psi(Y)$. For $\psi$ to be $|R_{\,\mathbf{B}}\boldsymbol{\omega}|$-integrable, it is necessary and sufficient that $\sum_{\ell\in L}\,|\langle\psi,\omega^{\ell}\rangle|<\infty$; and in the affirmative case,

[TABLE]

Proof.

We can certainly assume $L$ to be infinite, for otherwise the lemma is obvious. Then $Y$ is noncompact, and hence $\psi$ is nonnegative. Therefore

[TABLE]

On the other hand, since $\mathbf{B}$ is locally finite, the sum of $\omega^{\ell}$ over all $\ell\in L^{+}$ that do not exceed $N$ approaches $(R_{\,\mathbf{B}}\boldsymbol{\omega})^{+}$ vaguely as $N$ increases along $L^{+}$. Hence, by Lemma 2.1,111111The symbol $\lim_{s\in S}$ denotes a limit as $s$ increases along an upper directed set $S$.

[TABLE]

Combining these two displays and then letting $N$ along $L^{+}$, we get

[TABLE]

Since the same holds for $(R_{\,\mathbf{B}}\boldsymbol{\omega})^{-}$ and $L^{-}$, the lemma follows by subtraction.∎

Corollary 3.11**.**

Fix $\boldsymbol{\mu},\boldsymbol{\nu}\in\mathfrak{M}^{+}(\mathbf{A})$ and $x\in X$. Then

[TABLE]

each of the identities being understood in the sense that either of its sides is finite whenever so is the other and then they coincide. By (3.4) and (3.6) with $\boldsymbol{\mu}=\boldsymbol{\nu}$,

[TABLE]

Proof.

Relation (3.7) follows directly from Lemma 3.10 with $Y=X$, $\mathbf{B}=\mathbf{A}$, and $\psi(\cdot)=\kappa(x,\cdot)$. We next apply Lemma 3.10 to the (generalized) condenser $\mathbf{A}\times\mathbf{A}:=(A_{i}\times A_{j})_{(i,j)\in I\times I}$ in $X\times X$ with $s_{(i,j)}:=s_{i}s_{j}$, the function $\psi:=\kappa\in\Psi(X\times X)$, and the vector measure $\boldsymbol{\mu}\otimes\boldsymbol{\nu}\in\mathfrak{M}^{+}(\mathbf{A}\times\mathbf{A})$, where $\boldsymbol{\mu}\otimes\boldsymbol{\nu}:=(\mu^{i}\otimes\nu^{j})_{(i,j)\in I\times I}$. Noting that

[TABLE]

we arrive at (3.6).∎

Corollary 3.12**.**

Given $\boldsymbol{\mu},\boldsymbol{\nu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$, we have

[TABLE]

Furthermore, for every $i\in I$, $\kappa_{\boldsymbol{\mu}}^{i}(x)$ is finite n.e. and can be written in the form

[TABLE]

The series in (3.9) as well as in (3.10) converges absolutely, the latter being valid n.e.

Proof.

It is seen from (3.8) that $R\boldsymbol{\mu},R\boldsymbol{\nu}\in\mathcal{E}_{\kappa}(X)$; hence, $\kappa(R\boldsymbol{\mu},R\boldsymbol{\nu})$ is finite (see, e.g., [15, Lemma 3.1.1]), which yields (3.6) with the absolutely convergent series on the right-hand side. Compared with (3.4), this implies (3.9). Being the potential of a (scalar) measure of finite energy relative to the positive definite kernel, $\kappa(\cdot,R\boldsymbol{\mu})$ is finite n.e. [15, p. 164]. Hence, the series on the right-hand side in (3.7) converges absolutely n.e., which together with (3.5) establishes (3.10).∎

Remark 3.13**.**

Since the kernel is positive definite, (3.9) with $\boldsymbol{\nu}=\boldsymbol{\mu}$ yields the positivity of the energy $\kappa(\boldsymbol{\mu},\boldsymbol{\mu})$, which a priori was not obvious:

[TABLE]

Remark 3.14**.**

It is clear from the above that $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ is a convex cone. Indeed, since $\mathfrak{M}^{+}(\mathbf{A})$ is so, it is enough to observe that $R(\beta_{1}\boldsymbol{\mu}_{1}+\beta_{2}\boldsymbol{\mu}_{2})\in\mathcal{E}_{\kappa}(X)$ for any $\beta_{1},\beta_{2}\in(0,\infty)$ and $\boldsymbol{\mu}_{1},\boldsymbol{\mu}_{2}\in\mathcal{E}_{\kappa}^{+}(\mathbf{A})$. As $R\boldsymbol{\mu}_{1},R\boldsymbol{\mu}_{2}\in\mathcal{E}_{\kappa}(X)$ by (3.8), while $R(\beta_{1}\boldsymbol{\mu}_{1}+\beta_{2}\boldsymbol{\mu}_{2})=\beta_{1}R\boldsymbol{\mu}_{1}+\beta_{2}R\boldsymbol{\mu}_{2}$, the convexity of $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ follows from the linearity of $\mathcal{E}_{\kappa}(X)$.

3.5. Semimetric space of vector measures with finite

energy

We next show that the cone $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ can be thought of as a semimetric space, isometric to its (scalar) $R$-image.

Lemma 3.15**.**

The relation of $R$-equivalence on $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ is equivalent to that of identity if and only if the $A_{i}$, $i\in I$, are mutually essentially disjoint, i.e.,

[TABLE]

Proof.

The sufficiency part is obvious by Lemma 2.4. Assume now on the contrary that there are $A_{k}$ and $A_{\ell}$, $k\neq\ell$, with $c_{\kappa}(A_{k}\cap A_{\ell})>0$; then necessarily $s_{k}s_{\ell}=+1$. It follows from Lemma 2.4 that there exists a nonzero $\tau\in\mathcal{E}_{\kappa}^{+}(A_{k}\cap A_{\ell})$. Choose $\boldsymbol{\mu}=(\mu^{i})_{i\in I}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ such that $\mu^{k}|_{A_{k}\cap A_{\ell}}-\tau\geqslant 0$, and define $\boldsymbol{\mu}_{m}=(\mu_{m}^{i})_{i\in I}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$, $m=1,2$, where $\mu_{1}^{k}:=\mu^{k}-\tau$ and $\mu_{1}^{i}:=\mu^{i}$ for all $i\neq k$, while $\mu_{2}^{\ell}:=\mu^{\ell}+\tau$ and $\mu_{2}^{i}:=\mu^{i}$ for all $i\neq\ell$. Then $R\boldsymbol{\mu}_{1}=R\boldsymbol{\mu}_{2}$, and hence $\boldsymbol{\mu}_{1}$ and $\boldsymbol{\mu}_{2}$ are $R$-equivalent, but $\boldsymbol{\mu}_{1}\neq\boldsymbol{\mu}_{2}$.∎

Theorem 3.16**.**

The cone $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ is a semimetric space with the semimetric $\|\boldsymbol{\mu}_{1}-\boldsymbol{\mu}_{2}\|_{\mathcal{E}^{+}_{\kappa}(\mathbf{A})}$ defined by (1.2), and this space is isometric to its $R$-image. Assume now $\kappa$ is strictly positive definite. Then $\|\boldsymbol{\mu}_{1}-\boldsymbol{\mu}_{2}\|_{\mathcal{E}^{+}_{\kappa}(\mathbf{A})}$ becomes a metric if and only if (3.12) holds.

Proof.

Fix any $\boldsymbol{\mu}_{1},\boldsymbol{\mu}_{2}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$. Applying (3.9) to $\kappa(R\boldsymbol{\mu}_{k},R\boldsymbol{\mu}_{t})$, $k,t=1,2$, and then combining the equalities obtained, we get

[TABLE]

where the series converges absolutely. Hence, the sum on the right-hand side in (1.2) is ${}\geqslant 0$. When compared with (1.2), the last display yields

[TABLE]

Since $\|\cdot\|_{\kappa}$ is a seminorm on $\mathcal{E}_{\kappa}(X)$, the former part of the theorem follows.

Assume now $\kappa$ is strictly positive definite. By (3.13), $\|\boldsymbol{\mu}_{1}-\boldsymbol{\mu}_{2}\|_{\mathcal{E}_{\kappa}^{+}(\mathbf{A})}=0$ if and only if $\boldsymbol{\mu}_{1}$ and $\boldsymbol{\mu}_{2}$ are $R$-equivalent, while by Lemma 3.15, the relation of $R$-equivalence on $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ is equivalent to that of identity if and only if (3.12) holds.∎

In view of the isometry between $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ and its $R$-image, contained in the-Hilbert space $\mathcal{E}_{\kappa}(X)$, the topology on the semimetric space $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ is likewise termed strong. As usual, $\boldsymbol{\mu},\boldsymbol{\nu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ are said to be equivalent in the semimetric space $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ if $\|\boldsymbol{\mu}-\boldsymbol{\nu}\|_{\mathcal{E}_{\kappa}^{+}(\mathbf{A})}=0$.

Corollary 3.17**.**

$\boldsymbol{\mu},\boldsymbol{\nu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ are equivalent in $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ if and only if

[TABLE]

Proof.

In consequence of (3.13), $\boldsymbol{\mu}$ and $\boldsymbol{\nu}$ are equivalent in $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ if and only if $R\boldsymbol{\mu}$ and $R\boldsymbol{\nu}$ are equivalent in $\mathcal{E}_{\kappa}(X)$, which in turn holds if and only if $\kappa(\cdot,R\boldsymbol{\mu})=\kappa(\cdot,R\boldsymbol{\nu})$ n.e. [15, Lemma 3.2.1(a)]. Combining this with (3.10) establishes the corollary.∎

Being nonlinear, $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ is not normed. Nevertheless, for any of its elements $\boldsymbol{\mu}$ it is convenient to write $\|\boldsymbol{\mu}\|_{\mathcal{E}_{\kappa}^{+}(\mathbf{A})}:=\|\boldsymbol{\mu}-\boldsymbol{0}\|_{\mathcal{E}_{\kappa}^{+}(\mathbf{A})}$. Then

[TABLE]

4. Minimum energy problems for a generalized condenser

4.1. Formulation of the problems

For a (positive definite) kernel $\kappa$ on $X$ and a (generalized) condenser $\mathbf{A}=(A_{i})_{i\in I}$, we shall consider minimum energy problems with external fields over certain subclasses of $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$.

Fix a vector-valued external field $\mathbf{f}=(f_{i})_{i\in I}$, where each $f_{i}:X\to[-\infty,\infty]$ is $\mu$-measurable for every $\mu\in\mathfrak{M}^{+}(X)$. The $\mathbf{f}$-weighted vector potential and the $\mathbf{f}$-weighted energy of $\boldsymbol{\mu}\in\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ are defined by

[TABLE]

respectively. Let $\mathcal{E}_{\kappa,\mathbf{f}}^{+}(\mathbf{A})$ consist of all $\boldsymbol{\mu}\in\mathcal{E}_{\kappa}^{+}(\mathbf{A})$ with finite $\langle\mathbf{f},\boldsymbol{\mu}\rangle$, which means that every $f_{i}$, $i\in I$, is $\mu^{i}$-integrable and the series $\sum_{i\in I}\,\langle f_{i},\mu^{i}\rangle$ converges absolutely.

Fix a numerical vector $\mathbf{a}=(a_{i})_{i\in I}$ with $a_{i}>0$, $i\in I$, a vector-valued function $\mathbf{g}=(g_{i})_{i\in I}$, where all the $g_{i}:X\to(0,\infty)$ are (finitely) continuous, and write

[TABLE]

If $\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g}):=\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A})\cap\mathfrak{M}^{+}(\mathbf{A},\mathbf{a},\mathbf{g})$ is nonempty, or equivalently if

[TABLE]

then the following (unconstrained) $\mathbf{f}$-weighted minimum energy problem, also known in the literature as the Gauss variational problem (see, e.g., [21, 37, 38, 43, 44, 20]), makes sense.

Problem 4.1**.**

Does there exist $\boldsymbol{\lambda}_{\mathbf{A}}\in\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ with

[TABLE]

Let $\mathfrak{C}(A_{i})$, $i\in I$, consist of all $\xi^{i}\in\mathfrak{M}^{+}(A_{i})$ with $\langle g_{i},\xi^{i}\rangle>a_{i}$; those $\xi^{i}$ are said to be (upper) constraints for elements of $\mathfrak{M}^{+}(A_{i},a_{i},g_{i})$. Given $\xi^{i}\in\mathfrak{C}(A_{i})$, write

[TABLE]

where $\mu^{i}\leqslant\xi^{i}$ means that $\xi^{i}-\mu^{i}\geqslant 0$.

Fix $I_{0}\subset I$, which might be empty. We generalize Problem 4.1 by assuming that for every $i\in I_{0}$, the $i$-components $\mu^{i}$ of the (new) admissible measures $\boldsymbol{\mu}$ are now additionally required not to exceed a fixed constraint $\xi^{i}\in\mathfrak{C}(A_{i})$; that is, $\mu^{i}\in\mathfrak{M}^{\xi^{i}}(A_{i},a_{i},g_{i})$ for all $i\in I_{0}$. To be precise, write $\boldsymbol{\sigma}:=(\sigma^{i})_{i\in I}$, where

[TABLE]

and define

[TABLE]

Here the formal notation $\mathfrak{M}^{\infty}(A_{i},a_{i},g_{i})$ means that no active upper constraint is imposed on $\mu^{i}\in\mathfrak{M}^{+}(A_{i},a_{i},g_{i})$, i.e.,

[TABLE]

If $\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is nonempty (which will always be tacitly required), or equivalently if 121212See Lemma 5.5 below providing sufficient conditions for (4.3) to hold. Also note that the (nonempty) class $\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is convex (cf. Remark 3.14).

[TABLE]

then the following generalization of Problem 4.1 makes sense.

Problem 4.2**.**

Does there exist $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ with

[TABLE]

Observe that under the (permanent) assumption (4.3), Problems 4.1 also makes sense. In fact, $\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})\subset\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, and hence

[TABLE]

Problem 4.2 reduces to Problem 4.1 if $I_{0}=\varnothing$, while in the case $I_{0}=I$, Problem 4.2 is known as the constrained Gauss variational problem (see, e.g., [10, 42, 18, 20, 11]). However, the Gauss variational problem in either constrained or unconstrained setting has not been studied yet under the present requirements, where $\mathbf{A}$ is a collection of infinitely many touching Borel plates (cf. Remark 4.3 below). Finally, in the case where $I_{0}$ is a nonempty proper subset of $I$, Problem 4.2 seems to be newly introduced (even for a standard condenser), though such problem with mixed upper boundary conditions looks quite natural and also promising in relation to its possible applications (cf. Remark 1.1).

Remark 4.3**.**

The most general study of Problem 4.1 for a standard condenser of infinitely many (closed) plates seems to have been provided in [43, 44]. It includes, e.g., a complete description of the set of all $\mathbf{a}=(a_{i})_{i\in I}$ for which minimizers $\boldsymbol{\lambda}_{\mathbf{A}}$ exist as well as an analysis of their uniqueness, vague compactness, and strong and vague continuity of $\boldsymbol{\lambda}_{\mathbf{A}}$ when $\mathbf{A}$ varies. The weighted potentials of minimizers are described, and their characteristic properties are singled out.

4.2. Uniqueness of solutions

We next show that the set of the solutions to Problem 4.2 is contained in a certain equivalence class in $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$.

Lemma 4.4**.**

Any two solutions $\boldsymbol{\lambda}$ and $\widehat{\boldsymbol{\lambda}}$ to Problem 4.2 (whenever these exist) are equivalent in $\mathcal{E}_{\kappa}^{+}(\mathbf{A})$, i.e., $\|\boldsymbol{\lambda}-\widehat{\boldsymbol{\lambda}}\|_{\mathcal{E}_{\kappa}^{+}(\mathbf{A})}=0$.

Proof.

This can be shown in a way similar to that in [43, Proof of Lemma 5.1], based on the convexity of $\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, isometry between $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ and its (scalar) $R$-image, and the pre-Hilbert structure on $\mathcal{E}_{\kappa}(X)$. Indeed, we get from (4.3), (4.2), and (3.14)

[TABLE]

On the other hand, applying the parallelogram identity in $\mathcal{E}_{\kappa}(X)$ to $R\boldsymbol{\lambda}$ and $R\widehat{\boldsymbol{\lambda}}$ and then adding and subtracting $4\langle\mathbf{f},\boldsymbol{\lambda}+\widehat{\boldsymbol{\lambda}}\rangle$, we obtain

[TABLE]

When combined with the preceding relation, this yields

[TABLE]

which in view of (3.13) establishes the lemma. ∎

Corollary 4.5**.**

If $\kappa$ is strictly positive definite and the $A_{i}$, $i\in I$, are mutually essentially disjoint, then a solution to Problem 4.2 is unique.

Proof.

This follows from Lemma 4.4 when combined with Theorem 3.16. ∎

The following example shows that Corollary 4.5 fails in general if the assumption of mutual essential disjointness of the $A_{i}$, $i\in I$, is omitted from its hypotheses.

Example 4.6**.**

Let $X=\mathbb{R}^{n}$, $n\geqslant 3$, $\kappa=\kappa_{2}$, $I=I^{+}=\{1,2\}$, $I_{0}=\{1\}$, $a_{1}=a_{2}=1$, $g_{1}\equiv g_{2}\equiv 1$, $f_{1}\equiv f_{2}\equiv 0$, and let $A_{1}=A_{2}=K_{0}$, $K_{0}$ being an $(n-1)$-dimensional unit sphere. Let $\lambda$ denote the $\kappa_{2}$-capacitary measure on $K_{0}$, which exists (cf. Remark 2.7 and Example 2.5). By symmetry reasons, $\lambda$ coincides up to a normalizing factor with the $(n-1)$-dimensional surface measure $m_{n-1}$ on $K_{0}$. Define $\xi^{1}:=3\lambda$, and consider Problem 4.2 with these data. It is obvious that $\boldsymbol{\lambda}=(\lambda,\lambda)$ is one of its solutions. Choose now compact disjoint sets $K_{k}\subset K_{0}$, $k=1,2$, so that $m_{n-1}(K_{1})=m_{n-1}(K_{2})>0$, and define $\nu=\lambda|_{K_{1}}-\lambda|_{K_{2}}$. Then $\widehat{\boldsymbol{\lambda}}=(\lambda-\nu,\lambda+\nu)$ is an admissible measure for Problem 4.2 such that $R\widehat{\boldsymbol{\lambda}}=R\boldsymbol{\lambda}$, and hence $\kappa(\widehat{\boldsymbol{\lambda}},\widehat{\boldsymbol{\lambda}})=\kappa(\boldsymbol{\lambda},\boldsymbol{\lambda})$. Thus $\widehat{\boldsymbol{\lambda}}$ along with $\boldsymbol{\lambda}$ solves Problem 4.2, though $\widehat{\boldsymbol{\lambda}}\neq\boldsymbol{\lambda}$.

5. Permanent assumptions. Supplementary results

In all that follows we require that either $X$ is countable at infinity, or

[TABLE]

Lemma 5.1**.**

Let $\mu^{i}\in\mathcal{E}^{+}_{\kappa}(A_{i})$ be such that $\langle g_{i},\mu^{i}\rangle=c<\infty$. Then a proposition $\mathcal{P}(x)$ holds $\mu^{i}$-almost everywhere ($\mu^{i}$-a.e.), provided that it holds n.e. on $A_{i}$.

Proof.

The set $N$ of all $x\in A_{i}$ for which $\mathcal{P}(x)$ fails has inner capacity zero, and hence it is locally $\mu^{i}$-negligible [15, Lemma 2.3.1(iii)]. Furthermore, $N$ is $\mu^{i}$-$\sigma$-finite. This is obvious if $X$ is countable at infinity, while otherwise (5.1) holds, and therefore

[TABLE]

Being locally $\mu^{i}$-negligible and $\mu^{i}$-$\sigma$-finite, $N$ is $\mu^{i}$-negligible as claimed.∎

When speaking of an external field $\mathbf{f}=(f_{i})_{i\in I}$, we shall henceforth tacitly assume that Case I or Case II holds, where:

• I.

For every $i\in I$, $f_{i}\in\Psi(X)$;

• II.

For every $i\in I$, $f_{i}=s_{i}\kappa(\cdot,\zeta)$, where a (signed) $\zeta\in\mathcal{E}_{\kappa}(X)$ is given.

Lemma 5.2**.**

If Case II takes place, then $\mathcal{E}_{\kappa}^{+}(\mathbf{A})=\mathcal{E}_{\kappa,\mathbf{f}}^{+}(\mathbf{A})$; and moreover

[TABLE]

Proof.

Applying Lemma 3.10 to $\boldsymbol{\mu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ and each of $\kappa(\cdot,\zeta^{+}),\kappa(\cdot,\zeta^{-})\in\Psi(X)$, we get by subtraction

[TABLE]

Substituting this together with (3.14) into (4.2), we arrive at (5.3).∎

Lemma 5.3**.**

In either Case I or Case II,131313As seen from (4.3), (4.4), and (5.4), $G_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and $G^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ are both finite.

[TABLE]

Proof.

Since in Case II relation (5.4) follows directly from (5.3), it is left to consider Case I. Assume $X$ is compact, for if not, then $f_{i}\geqslant 0$ for all $i\in I$, and (5.4) holds by (3.11). But then $I$ has to be finite, while every $f_{i}$, being l.s.c., is bounded from below on the (compact) space $X$ by $-c_{i}$, where $0<c_{i}<\infty$. In addition, (5.1) and, hence, (5.2) with $c=a_{i}$ both hold for every $i\in I$ and every $\boldsymbol{\mu}\in\mathfrak{M}^{+}(\mathbf{A},\mathbf{a},\mathbf{g})$, $g_{i}$ being a strictly positive continuous function on $X$. Combining all these together gives

[TABLE]

which in view of the finiteness of $I$ again leads to (5.4). ∎

Lemma 5.4**.**

In view of the (permanent) requirement (4.3), for all $i\in I$

[TABLE]

Proof.

Let, on the contrary, $c_{\kappa}(A^{\circ}_{j})=0$ for some $j\in I$. Then, by Lemma 5.1, for every $\boldsymbol{\mu}\in\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ (which exists by (4.4)) we have $|f_{j}|=\infty$ $\mu^{j}$-a.e. But this is impossible because $\mu^{j}\neq 0$ while $f_{j}$ is $\mu^{j}$-integrable. ∎

For any $M\in(0,\infty)$ and $i\in I$, write $A_{i}^{M}:=\{x\in A^{\circ}_{i}:\ |f_{i}(x)|\leqslant M\}$.

Lemma 5.5**.**

Assume there exist $M,M_{1}\in(0,\infty)$ that are independent of $i\in I$ and possessing the properties

[TABLE]

If, moreover, (1.3) is fulfilled, then (4.3) holds.

Proof.

Fix $\varepsilon\in(0,\infty)$, and for every $i\in I\setminus I_{0}$ choose $\tau_{i}\in\mathcal{E}^{+}_{\kappa}(A_{i}^{M_{1}})$ of compact support so that $\tau_{i}(A_{i}^{M_{1}})=1$ and $\|\tau_{i}\|^{2}_{\kappa}\leqslant c_{\kappa}(A_{i}^{M_{1}})^{-1}+\varepsilon$. (Such $\tau_{i}$ exists since $c_{\kappa}(A_{i}^{M_{1}})$ would be the same if the measures $\nu$ in its definition were required to be of compact support $S(\nu)\subset A_{i}^{M_{1}}$, cf. [15, p. 153].) In view of (5.8), we thus get

[TABLE]

Write

[TABLE]

where

[TABLE]

Note that $0<\langle g_{i},\nu_{i}\rangle<\infty$ for all $i\in I$. In fact, for $i\in I\setminus I_{0}$ this holds because

[TABLE]

while for $i\in I_{0}$ it is valid by (5.7). Also observing that, again by (5.7), $\widetilde{\nu}_{i}\leqslant\xi^{i}$ for all $i\in I_{0}$, we thus get $\widetilde{\nu}_{i}\in\mathfrak{M}^{\sigma^{i}}(A_{i},a_{i},g_{i})$ for all $i\in I$. Furthermore,

[TABLE]

where the second inequality follows from (5.7) and the third from (5.6) and (1.3). Therefore, by Lemma 3.9, $\boldsymbol{\widetilde{\nu}}:=(\widetilde{\nu}_{i})_{i\in I}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa}(\mathbf{A},\mathbf{a},\mathbf{g})$. Finally,

[TABLE]

the last inequality being obtained from (1.3). Altogether, $\boldsymbol{\widetilde{\nu}}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$.∎

If $I$ is finite, Lemma 5.5 takes the following much simpler form.

Corollary 5.6**.**

Let $I$ be finite, and let $c_{\kappa}(A^{\circ}_{i})>0$ for all $i\in I\setminus I_{0}$, $A^{\circ}_{i}$ being defined in (5.5). Then (4.3) holds provided that for every $i\in I_{0}$,

[TABLE]

We drop a proof of Corollary 5.6, since it runs in a way similar to that for Lemma 5.5. Combining this with Lemma 5.4 yields the following assertion.

Corollary 5.7**.**

If $I$ is finite and $I_{0}=\varnothing$, then (4.3) and (5.5) are equivalent.

Definition 5.8**.**

A net $(\boldsymbol{\mu}_{s})_{s\in S}\subset\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is minimizing in Problem 4.2 if

[TABLE]

Let $\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ consist of all those $(\boldsymbol{\mu}_{s})_{s\in S}$; it is nonempty because of (4.3).

Lemma 5.9**.**

For any $(\boldsymbol{\mu}_{s})_{s\in S}$ and $(\boldsymbol{\nu}_{t})_{t\in T}$ in $\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$,

[TABLE]

$S\times T$ being the upper directed product 141414See, e.g., [26, Chapter 2, Section 3]. of the upper directed sets $S$ and $T$.

Proof.

In the same manner as in the proof of Lemma 4.4 we get

[TABLE]

which yields (5.10) when combined with (3.13), (4.3), (5.4), and (5.9).∎

Corollary 5.10**.**

Every $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is strong Cauchy in $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$.

6. Sufficient conditions for the solvability of

Problem 4.2

Throughout Section 6 we require the permanent assumptions stated in Sections 4.1 and 5. Furthermore, the $A_{i}$, $i\in I$, are assumed to be nearly closed. According to Lemma 3.4, then so are both $A^{+}$ and $A^{-}$. Let $\breve{A}^{+}$ and $\breve{A}^{-}$ be the (closed) sets defined by (3.3). We denote by $(\boldsymbol{\mu}_{s})^{\prime}_{s\in S}$ the cluster set of any $(\boldsymbol{\mu}_{s})_{s\in S}\subset\mathfrak{M}^{+}(\mathbf{A})$ in the vague product space topology on $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$, and $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ the class of the solutions to Problem 4.2.

Theorem 6.1**.**

Let the kernel $\kappa$ be consistent, and let the assumptions

[TABLE]

and (1.3) be both fulfilled. Also assume that

[TABLE]

while for every $i\in I\setminus I_{0}$, the following two conditions are required:

• $\bullet$

Either $A_{i}$ is nearly compact, or $c_{\kappa}(A_{i})<\infty$.151515A compact set $K\subset X$ may be of infinite capacity; $c_{\kappa}(K)$ is necessarily finite provided that $\kappa$ is strictly positive definite [15]. On the other hand, even for the Newtonian kernel, closed sets of finite capacity may be noncompact (see, e.g., Example 1.7 above).

• $\bullet$

Either $g_{i}$ is upper bounded, or there are $r_{i}\in(1,\infty)$ and $\nu_{i}\in\mathcal{E}_{\kappa}(X)$ such that

[TABLE]

Then in either Case I or Case II, $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is nonempty, vaguely compact, and given by

[TABLE]

Furthermore, for every $(\boldsymbol{\nu}_{t})_{t\in T}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and every $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}\in\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$,

[TABLE]

Definition 6.2**.**

Denoting by $\infty_{X}$ the Alexandroff point of $X$ [3, Chapter I, Section 9, n∘ 8], we say that a kernel $\kappa$ possesses the property $(\infty_{X})$ if $\kappa(\cdot,y)\to 0$ as $y\to\infty_{X}$ uniformly over compact sets in $X$.

The Riesz kernel $\kappa_{\alpha}$, $\alpha\in(0,n)$, on $\mathbb{R}^{n}$, $n\geqslant 3$, possesses the property $(\infty_{X})$. So does the $2$-Green kernel $G^{2}_{D}$ on an open bounded set $D\subset\mathbb{R}^{n}$, $n\geqslant 2$, provided that $D$ is regular in the sense of the solvability of the classical Dirichlet problem.

Theorem 6.3**.**

Assume a l.c.s. $X$ is metrizable and countable at infinity,161616Theorem 6.3 remains valid for an arbitrary l.c.s. $X$ if we assume instead that only finitely many $\breve{A}_{i}$, $i\in I^{-}$, resp. $\breve{A}_{i}$, $i\in I^{+}$, can intersect one another (see Remark 7.3). while a kernel $\kappa(x,y)$ is continuous for $x\neq y$ and possesses the property $(\infty_{X})$. Let $I^{+}$, resp. $I^{-}$, be finite, the $A_{i}$, $i\in I$, be nearly compact, and let

[TABLE]

If, moreover, Case I takes place and (1.3) holds, then for any $I_{0}$ and $\boldsymbol{\sigma}$ the class $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is nonempty, vaguely compact, and given by (6.4).

Remark 6.4**.**

If compared with Theorem 6.1, in Theorem 6.3 the kernel $\kappa$ is not required to be consistent. However, if under the hypotheses of Theorem 6.3, the consistency of $\kappa$ takes place, then Theorem 6.3 becomes valid in both Cases I and II; and moreover, then (6.5) also holds.

Recall that a kernel $\kappa$ is said to satisfy the continuity principle (or to be regular) if for any $\mu\in\mathfrak{M}^{+}(X)$ with compact support $S(\mu)$, $\kappa(\cdot,\mu)$ is continuous on $X$ whenever its restriction to $S(\mu)$ is continuous. The Riesz kernel $\kappa_{\alpha}$, $\alpha\in(0,n)$, on $\mathbb{R}^{n}$, $n\geqslant 3$, is regular [28, Theorem 1.7]. So is the $\alpha$-Green kernel $G_{D}^{\alpha}$, $\alpha\in(0,2]$, on an open set $D\subset\mathbb{R}^{n}$, $n\geqslant 3$ [19, Corollary 4.8], as well as the logarithmic kernel on $\mathbb{R}^{2}$, the latter being seen by combining [28, Theorem 1.6] and [37, Eq. (1.3)].

Theorem 6.5**.**

Assume $I$ is finite and the $A_{i}$, $i\in I$, are nearly compact. Let the kernel $\kappa$ be regular, and let the $\kappa(\cdot,\xi^{i})|_{\breve{A}_{i}}$, $i\in I_{0}$, as well as the $\kappa|_{\breve{A}_{i}\times\breve{A}_{i}}$, $i\in I\setminus I_{0}$, be continuous. Then in either Case I or Case II and for any $\mathbf{a}$ and $\mathbf{g}$, the conclusion of Theorem 6.1 remains valid.171717Theorem 6.5 is applicable to the classical kernels only provided that $I_{0}=I$.

Remark 6.6**.**

In contrast to Theorem 6.3, in Theorem 6.5 the sets $\breve{A}^{+}$ and $\breve{A}^{-}$ may have points in common. But then necessarily $c_{\kappa}(\breve{A}^{+}\cap\breve{A}^{-})=0$, and hence $\breve{A}^{+}\cap\breve{A}^{-}$ cannot carry any nonzero $\nu\in\mathcal{E}_{\kappa}(X)$ (see Lemma 2.4).

Corollary 6.7**.**

Under the hypotheses of any of Theorems 6.1, 6.3, or 6.5, if moreover $\kappa$ is strictly positive definite, while the $A_{i}$, $i\in I$, are mutually essentially disjoint, then $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ reduces to a unique element $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}$, and every $(\boldsymbol{\nu}_{t})_{t\in T}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ converges to this $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}$ vaguely.

7. Proofs of Theorems 6.1, 6.3, and 6.5 and Corollary 6.7

7.1. Auxiliary results

Throughout Section 7.1, the $A_{i}$, $i\in I$, are assumed to be nearly closed. Write

[TABLE]

Lemma 7.1**.**

If (1.3) and (6.1) both hold, then the vague cluster set $(\boldsymbol{\mu}_{s})^{\prime}_{s\in S}$ of any $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is nonempty, and moreover

[TABLE]

Proof.

Fix a net $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. It is strong Cauchy in the semimetric space $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ by Corollary 5.10, and hence strongly bounded, i.e.,

[TABLE]

the equality being valid by (3.14). Furthermore, it follows from (1.3) that (5.1) and, hence, (5.2) (with $a_{i}$ and $\mu_{s}^{i}$ in place of $c$ and $\mu^{i}$) both hold. Thus,

[TABLE]

By Lemma 2.9 with $Q=A_{i}$, the $\mu_{s}^{i}$, $s\in S$, are supported by $\breve{A}_{i}$, $A_{i}$ being nearly closed. Hence, $R\boldsymbol{\mu}_{s}^{\pm}$ is supported by $\breve{A}^{\pm}$, cf. (3.3), and therefore

[TABLE]

where the latter inequality holds by (6.1). Combining this with (7.3) establishes the inequality

[TABLE]

which together with (7.2) yields

[TABLE]

We next observe that for every $i\in I$,

[TABLE]

In view of (7.4), this will follow once we have established the inequality

[TABLE]

Assume $X$ is compact, for if not, then $\kappa\geqslant 0$ and the left-hand side in (7.6) is ${}\geqslant 0$. But then the l.s.c. function $\kappa$ on $X\times X$ is ${}\geqslant-c$, where $c\in(0,\infty)$, while $I$ is finite; and hence (7.6) follows from (7.3) in a way similar to that in the proof of Lemma 5.3.

As seen from (7.3), the net $(\boldsymbol{\mu}_{s})_{s\in S}$ is vaguely bounded, and hence, by Lemma 3.2, it is relatively compact in the vague topology on $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$. Thus, there is a subnet $(\boldsymbol{\mu}_{t})_{t\in T}$ of the net $(\boldsymbol{\mu}_{s})_{s\in S}$ such that for every $i\in I$,

[TABLE]

where $\mu^{i}\in\mathfrak{M}^{+}(X)$. It follows from (7.5) and (7.7) by Lemmas 2.9 and 2.10 that $\mu^{i}\in\mathcal{E}^{+}_{\kappa}(\breve{A}_{i})=\mathcal{E}^{+}_{\kappa}(A_{i})$, and hence $\boldsymbol{\mu}:=(\mu^{i})_{i\in I}\in\mathfrak{M}^{+}(\mathbf{A})$.

Moreover, $R\boldsymbol{\mu}^{\pm}$ is the vague limit of $R\boldsymbol{\mu}_{t}^{\pm}$ as $t$ increases along $T$, which is obtained from (7.7) according to Lemma 3.7. Applying [15, Lemma 2.2.1(e)], we therefore see from (7.4) that the energy of $R\boldsymbol{\mu}^{\pm}$ is finite. Since $\kappa$ is positive definite, so is $\kappa(R\boldsymbol{\mu}^{+},R\boldsymbol{\mu}^{-})$ (see, e.g., [15, Lemma 3.1.1]), and altogether $R\boldsymbol{\mu}\in\mathcal{E}_{\kappa}(X)$. In view of (3.8), we thus have

[TABLE]

Applying Lemma 2.1 to the continuous function $g_{i}>0$, we also obtain from (7.7)

[TABLE]

Noting that $\xi^{i}-\mu_{t}^{i}\geqslant 0$ for all $i\in I_{0}$ and $t\in T$ as well as that the vague limit of a net of positive (scalar) measures likewise is positive, we finally see from (7.7) that $\mu^{i}\leqslant\sigma^{i}$ for all $i\in I$. This together with the two preceding displays shows that, actually, $\boldsymbol{\mu}\in\mathcal{E}_{\kappa}^{\boldsymbol{\sigma}}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g})$, which establishes (7.1). ∎

Lemma 7.2**.**

Let (1.3) and (6.1) both hold, and let $\kappa$ be consistent. For every $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and every $\boldsymbol{\mu}\in(\boldsymbol{\mu}_{s})^{\prime}_{s\in S}$,

[TABLE]

[TABLE]

Proof.

We use tacitly the notation and assertions from the proof of the preceding lemma. Being consistent, the kernel $\kappa$ possesses the property (C2) (see Section 2.2). The strongly bounded net $(R\boldsymbol{\mu}_{t}^{\pm})_{t\in T}\subset\mathcal{E}^{+}_{\kappa}(X)$ therefore converges weakly to its vague limit $R\boldsymbol{\mu}^{\pm}\in\mathcal{E}^{+}_{\kappa}(X)$, which by the definition of weak convergence implies that $R\boldsymbol{\mu}_{t}\to R\boldsymbol{\mu}$ weakly as $t$ increases along $T$. By (3.13), this gives

[TABLE]

and hence, by the Cauchy–Schwarz inequality in $\mathcal{E}_{\kappa}(X)$,

[TABLE]

which establishes the relation

[TABLE]

because $\|\boldsymbol{\mu}_{t}-\boldsymbol{\mu}_{t^{\prime}}\|_{\mathcal{E}^{+}_{\kappa}(\mathbf{A})}$ becomes arbitrarily small when $t,t^{\prime}\in T$ are sufficiently large. Since a strong Cauchy net converges strongly to any of its strong cluster points, we obtain (7.9) from the last display.

As for (7.10), we first note that the equality and the third inequality here are valid by (5.9) and the permanent assumption (4.3), respectively. If Case II takes place, then the first inequality is obvious by (5.3), while the second inequality holds (with equality prevailing) again by (5.3), applied respectively to $\boldsymbol{\mu}_{s}$, $s\in S$, and $\boldsymbol{\mu}$, and the subsequent use of (7.9). Assume now Case I holds. Applying Lemma 2.1 to $f_{i}\in\Psi(X)$, we see from (7.7) after summation over $i\in I$ that181818Note that, while proving (7.11) in Case I, we have not used the assumption of consistency of the kernel.

[TABLE]

The former inequality here is obvious if $X$ is noncompact, while otherwise it can be obtained from (1.3) and (7.8) in the same manner as in the proof of Lemma 5.3. Combining (7.11) and (7.9) completes the proof of (7.10). ∎

7.2. Proof of Theorem 6.1

Fix $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and $\boldsymbol{\mu}\in(\boldsymbol{\mu}_{s})_{s\in S}^{\prime}$ (such $\boldsymbol{\mu}$ exists by Lemma 7.1). For these $(\boldsymbol{\mu}_{s})_{s\in S}$ and $\boldsymbol{\mu}$, there hold Lemmas 7.1 and 7.2 as well as the assertions in their proofs.

We assert that $\boldsymbol{\mu}$ solves Problem 4.2. We first show that, actually,

[TABLE]

As seen from (7.1) and (7.10), it is enough to prove that for any given $i\in I$,

[TABLE]

It follows from Lemma 2.9 with $Q=A_{i}$ that the $\mu^{i}_{s}$, $s\in S$, and $\mu^{i}$ are carried by $A_{i}\cap\breve{A}_{i}$. There is therefore no loss of generality in replacing each $\xi^{i}$, $i\in I_{0}$, by the extension of $\xi^{i}|_{A_{i}\cap\breve{A}_{i}}$ by [math] to all of $X$, denoted again by $\xi^{i}$.191919As $A_{i}\cap\breve{A}_{i}$ is $\xi^{i}$-measurable, $\xi^{i}|_{A_{i}\cap\breve{A}_{i}}$ exists. Given $Q\subset X$, the extension of $\nu\in\mathfrak{M}^{+}(Q)$ by [math] to all of $X$ is $\tilde{\nu}\in\mathfrak{M}^{+}(X)$ determined uniquely by the relation $\tilde{\nu}(\varphi):=\langle\varphi|_{Q},\nu\rangle$ for all $\varphi\in C_{0}(X)$. Observe that for this (new) $\xi^{i}$, assumption (6.2) remains valid.

Consider the exhaustion of the (closed) set $\breve{A}_{i}$ by the upper directed family $(K)$ of all compact subsets of $\breve{A}_{i}$.202020A family $\mathfrak{Q}$ of sets $Q\subset X$ is said to be upper directed if for any $Q_{1},Q_{2}\in\mathfrak{Q}$ there exists $Q_{3}\in\mathfrak{Q}$ such that $Q_{1}\cup Q_{2}\subset Q_{3}$. Let $(\boldsymbol{\mu}_{t})_{t\in T}$ be a subnet of $(\boldsymbol{\mu}_{s})_{s\in S}$ converging vaguely to $\boldsymbol{\mu}$ (see the proof of Lemma 7.1). Since the indicator function $1_{K}$ of $K$ is upper semicontinuous, Lemma 2.1 with $\psi=-g_{i}1_{K}=-g_{i}|_{K}$ and [15, Lemma 1.2.2] yield

[TABLE]

$T\times(K)$ being the upper directed product of the upper directed sets $T$ and $(K)$ [26, Chapter 2, Section 3]. The first inequality here holds by (7.8), while the second and third equalities follow from Lemma 2.2, the $\mu_{t}^{i}$, $t\in T$, and $\mu^{i}$ being bounded. Hence, (7.13) will be established once we have proven

[TABLE]

the former equality here being valid again according to Lemma 2.2.

Assume first that $i\in I_{0}$. Since by (6.2) and [15, Lemma 1.2.2]

[TABLE]

we have

[TABLE]

When combined with

[TABLE]

this implies the latter equality in (7.14) and, hence, (7.13).

Let now $i\in I\setminus I_{0}$. Assuming first that $\breve{A}_{i}$ is compact, we obtain (7.13) from (7.7) in view of the continuity of $g_{i}$. In fact, there is $\varphi_{i}\in C_{0}(X)$ such that $\varphi_{i}|_{\breve{A}_{i}}=g_{i}|_{\breve{A}_{i}}$. Since $\breve{A}_{i}^{c}$ is $\nu$-negligible for any $\nu\in\mathfrak{M}^{+}(\breve{A}_{i})$, we thus get

[TABLE]

Assume next $\breve{A}_{i}$ is noncompact. Then, by the stated hypotheses,

[TABLE]

Since the kernel $\kappa$ is consistent, for every $Q\subset\breve{A}_{i}$ there exists an interior equilibrium measure $\gamma_{Q}$ [15, Theorem 4.1]. Recall that, if $\Gamma_{Q}$ denotes the convex cone of all $\nu\in\mathcal{E}_{\kappa}(X)$ with $\kappa(x,\nu)\geqslant 1$ n.e. on $Q$, then $\gamma_{Q}\in\Gamma_{Q}$, i.e.,

[TABLE]

and moreover

[TABLE]

Observe that there is no loss of generality in assuming $g_{i}$ to satisfy (6.3) with some $r_{i}\in(1,\infty)$ and $\nu_{i}\in\mathcal{E}_{\kappa}(X)$. Indeed, otherwise $g_{i}$ must be bounded from above by $M\in(0,\infty)$, which combined with (7.17) for $Q=\breve{A}_{i}$ results in (6.3) with $\nu_{i}:=M^{r_{i}}\gamma_{\breve{A}_{i}}$, $r_{i}\in(1,\infty)$ being arbitrary.

Consider interior equilibrium measures $\gamma_{\breve{A}_{i}\setminus K}$ and $\gamma_{\breve{A}_{i}\setminus K^{\prime}}$, where $K,K^{\prime}\in(K)$. Because of (7.17) and (7.18), we obtain from [15, Lemma 4.1.1]

[TABLE]

As seen from (7.16) and (7.18), the net $(\|\gamma_{\breve{A}_{i}\setminus K}\|_{\kappa})_{K\in(K)}$ is bounded and decreasing, and hence it is Cauchy in $\mathbb{R}$. The preceding inequality thus shows that the net $(\gamma_{\breve{A}_{i}\setminus K})_{K\in(K)}$ is strong Cauchy in $\mathcal{E}^{+}_{\kappa}(X)$. Since, clearly, this net converges vaguely to zero, the property (C1) implies that zero is also one of its strong limits. Hence,

[TABLE]

Write $q_{i}:=r_{i}(r_{i}-1)^{-1}$, where $r_{i}\in(1,\infty)$ is the number involved in condition (6.3). Combining (6.3) with (7.17) shows that the inequality

[TABLE]

holds n.e. on $\breve{A}_{i}$, and hence $\mu_{t}^{i}$-a.e. on $X$ (see Lemma 5.1). Having integrated this relation with respect to $\mu_{t}^{i}$, we then apply the Hölder and, subsequently, the Cauchy–Schwarz inequalities to the integrals on the right. This gives

[TABLE]

Taking limits here as $(t,K)$ increases along $T\times(K)$ and using (7.5) and (7.19), we again obtain the latter equality in (7.14), and hence (7.13).

Having thus proven (7.12), we get $G_{\kappa,\mathbf{f}}(\boldsymbol{\mu})\geqslant G^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. Since the converse inequality holds by (7.10), $\boldsymbol{\mu}$ is a solution to Problem 4.2, i.e., $\boldsymbol{\mu}\in\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. As $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and $\boldsymbol{\mu}\in(\boldsymbol{\mu}_{s})^{\prime}_{s\in S}$ have been chosen arbitrarily, we obtain

[TABLE]

The converse inclusion is obvious because the trivial net $(\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}})$, where $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}$ is any element of $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, is minimizing and converges vaguely to $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}$. Thus, (6.4) indeed holds.

Any net in $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is obviously minimizing, and hence, according to (6.4), any of its vague cluster points again belongs to $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. This establishes the vague compactness of $\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. Choosing finally any $(\boldsymbol{\nu}_{t})_{t\in T}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}\in\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, we arrive at (6.5) by combining (7.9) with Lemmas 4.4 and 5.9. The proof is complete.

7.3. Proof of Theorem 6.3

Under the stated hypotheses, there is no loss of generality in assuming $I^{+}$ to be finite. Then $\breve{A}^{+}$ is compact, for $\breve{A}_{i}$, $i\in I$, are so. Fix $\varepsilon>0$. As seen from the property $(\infty_{X})$, there exists a compact set $K\subset X$ such that

[TABLE]

where $C\in(0,\infty)$ is given by (1.3). Since $\breve{A}^{-}\cap K$ is compact, it follows from (6.6) and the (finite) continuity of $\kappa(x,y)$ for $x\neq y$ that $\kappa|_{\breve{A}^{+}\times(\breve{A}^{-}\cap K)}$ is upper bounded. This together with (7.20) yields (6.1). Since (1.3) holds by assumption, we are thus able to use Lemma 7.1 as well as the assertions established in the course of its proof.

According to Lemma 2.9 with $Q=A_{i}$, for every $\boldsymbol{\nu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ we have $\nu^{i}\in\mathcal{E}^{+}_{\kappa}(A_{i}\cap\breve{A}_{i})$ for all $i\in I$. There is therefore no loss of generality in replacing each $\xi^{i}$, $i\in I_{0}$, by the extension of $\xi^{i}|_{A_{i}\cap\breve{A}_{i}}$ by [math] to all of $X$ (cf. footnote 19), denoted again by $\xi^{i}$. After having done this, we next replace (again with the notation preserved) the $A_{i}$, $i\in I$, by the (compact) sets $\breve{A}_{i}$, which again involves no loss of generality.

By (7.1), any vague cluster point $\boldsymbol{\mu}$ of any $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ belongs to $\mathcal{E}_{\kappa}^{\boldsymbol{\sigma}}(\mathbf{A},\leqslant\!\mathbf{a},\mathbf{g})$. Choose a subsequence $\{\boldsymbol{\mu}_{k}\}_{k\in\mathbb{N}}$ of $(\boldsymbol{\mu}_{s})_{s\in S}$ that converges vaguely to $\boldsymbol{\mu}$. Since $g_{i}$ is continuous and $A_{i}$ is compact, equality prevails in (7.8) (cf. (7.15)), and hence

[TABLE]

Thus, $|R\boldsymbol{\mu}|(X)\leqslant C$, cf. (7.3), and the above $K$ can be chosen so that

[TABLE]

We next use the fact that the map $(\nu_{1},\nu_{2})\mapsto\nu_{1}\otimes\nu_{2}$ from $\mathfrak{M}^{+}(X)\times\mathfrak{M}^{+}(X)$ into $\mathfrak{M}^{+}(X\times X)$ is vaguely continuous [4, Chapter 3, Section 5, Exercise 5]. Applying Lemma 2.1 to $\kappa\in\Psi(X\times X)$, we therefore obtain

[TABLE]

Furthermore,

[TABLE]

As seen from (7.20) and (7.3), each of the last two summands on the right-hand side in (7.24) is ${}<\varepsilon/3$. Since $\kappa|_{A^{+}\times(A^{-}\cap K)}$ is continuous on the (compact) space $A^{+}\times(A^{-}\cap K)$ and $R\boldsymbol{\mu}_{k}^{+}\otimes(R\boldsymbol{\mu}_{k}^{-}|_{K})\to R\boldsymbol{\mu}^{+}\otimes(R\boldsymbol{\mu}^{-}|_{K})$ vaguely, the latter being clear from Theorem 2.3 in view of (7.22), there exists $k_{0}\in\mathbb{N}$ such that for all $k\geqslant k_{0}$, the first summand in (7.24) is ${}<\varepsilon/3$. Altogether,

[TABLE]

for $\varepsilon$ has been chosen arbitrarily. Combining this with (7.23) and then substituting (3.9) and (3.11) into the inequality obtained yields

[TABLE]

Since Case I takes place, in view of footnote 18 we obtain (7.11), which together with the last display establishes the relation

[TABLE]

The equality and the third inequality here are valid by (5.9) and the permanent assumption (4.3), respectively. In view of (7.21), we thus actually have $\boldsymbol{\mu}\in\mathcal{E}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$, and therefore $G_{\kappa,\mathbf{f}}(\boldsymbol{\mu})\geqslant G^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. This together with the preceding display shows that $\boldsymbol{\mu}$ is in fact a solution to Problem 4.2.

It has thus been proven that any vague cluster point (which exists) of any minimizing net (sequence) belongs to $\mathfrak{S}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$. In the same way as it has been done at the end of the proof of Theorem 6.1, this implies (6.4) as well as the vague compactness of $\mathfrak{S}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$. The proof is complete.

Remark 7.3**.**

Assume the conditions of footnote 16 hold. Then the corresponding version of Theorem 6.3 can be proven as above (of course, with a subnet $(\boldsymbol{\mu}_{t})_{t\in T}$ in place of a subsequence), the only difference being in the fact that Theorem 2.3 may fail. Instead, choose a compact set $K$ so that $\breve{A}^{+}\cap\partial_{X}K=\varnothing$. Since this $K$ has points in common with only finitely many $\breve{A}_{i}$, $i\in I^{+}$, $(R\boldsymbol{\mu}^{+}_{t}|_{K})_{t\in T}$ again converges vaguely to $R\boldsymbol{\mu}^{+}|_{K}$. Having reversed the roles of ‘$+$’ and ‘$-$’, we arrive at our claim.

7.4. Proof of Theorem 6.5

In the same manner as this has been done in the proof of Theorem 6.3, we see that there is no loss of generality in replacing each $\xi^{i}$, $i\in I_{0}$, by the extension of $\xi^{i}|_{A_{i}\cap\breve{A}_{i}}$ by [math] to all of $X$ (denoted again by $\xi^{i}$).

We begin by showing that under the stated hypotheses the potential $\kappa(\cdot,\nu^{i})$ of any $\nu^{i}\in\mathfrak{M}^{+}(\breve{A}_{i})$, $i\in I$, such that $\nu^{i}\leqslant\sigma^{i}$ is continuous on $X$. Let first $i\in I_{0}$. Being relatively continuous on $\breve{A}_{i}\supset S(\xi^{i})$ by assumption, $\kappa(\cdot,\xi^{i})$ is continuous on $X$ by the regularity of the kernel. Since $\kappa(\cdot,\nu^{i})$ is l.s.c. and since $\kappa(\cdot,\nu^{i})=\kappa(\cdot,\xi^{i})-\kappa(\cdot,\xi^{i}-\nu^{i})$ with $\kappa(\cdot,\xi^{i})$ continuous and $\kappa(\cdot,\xi^{i}-\nu^{i})$ l.s.c., $\kappa(\cdot,\nu^{i})$ is also upper semicontinuous, hence continuous. Let now $i\in I\setminus I_{0}$. Since $-\kappa|_{\breve{A}_{i}\times\breve{A}_{i}}$ is continuous by assumption, $-\kappa(x,y)\geqslant-c$ for all $(x,y)\in\breve{A}_{i}\times\breve{A}_{i}$, where $c\in(0,\infty)$. Integrating this inequality with respect to the (bounded) $\nu^{i}\in\mathfrak{M}^{+}(\breve{A}_{i})$, we observe that $\kappa(\cdot,\nu^{i})$ is relatively upper semicontinuous on $\breve{A}_{i}$. Being also l.s.c. on $X$, it is relatively continuous on $\breve{A}_{i}\supset S(\nu^{i})$, and hence on all of $X$, again by the regularity of the kernel $\kappa$.

Choose any $(\boldsymbol{\mu}_{s})_{s\in S}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, which exists by the (permanent) assumption (4.3). By Lemma 2.9 with $Q=A_{i}$, $i\in I$, we have $(\mu_{s}^{i})_{s\in S}\subset\mathcal{E}_{\kappa}^{+}(\breve{A}_{i})$. Since $g_{i}$ is continuous and strictly positive, while $\breve{A}_{i}$ is compact,

[TABLE]

Therefore, $(\boldsymbol{\mu}_{s})_{s\in S}$ is bounded and, hence, relatively compact in the vague topology on $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ (Lemma 3.2). Fix any of its vague cluster points $\boldsymbol{\mu}=(\mu^{i})_{i\in I}\in\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$, and choose a subnet $(\boldsymbol{\mu}_{t})_{t\in T}$ of $(\boldsymbol{\mu}_{s})_{s\in S}$ converging vaguely to $\boldsymbol{\mu}$. Since $\mathfrak{M}^{\sigma^{i}}(\breve{A}_{i},a_{i},g_{i})$ is vaguely closed (cf. (7.15)),

[TABLE]

As shown in the second paragraph of the present proof, $\kappa(\cdot,\mu^{i})$ is continuous on $X$, and hence bounded on the (compact) $\breve{A}_{i}$. Combined with $\mu^{i}(\breve{A}_{i})<\infty$, this gives $\mu^{i}\in\mathcal{E}^{+}_{\kappa}(\breve{A}_{i})$. By Lemma 2.9 for $Q=A_{i}$ and the preceding display, we thus get $\boldsymbol{\mu}\in\mathcal{E}_{\kappa}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$ ($I$ being finite).

Furthermore, since every $\kappa(\cdot,\mu_{t}^{i})$, $t\in T$, is likewise continuous on $X$,

[TABLE]

Summing up these equalities, multiplied by $s_{i}s_{j}$, over all $i,j\in I$ shows that $R\boldsymbol{\mu}_{t}\to R\boldsymbol{\mu}$ strongly in $\mathcal{E}_{\kappa}(X)$; and hence, by (3.13),

[TABLE]

Since a strong Cauchy net converges strongly to any of its strong cluster points, we see that $(\boldsymbol{\mu}_{s})_{s\in S}$ converges to $\boldsymbol{\mu}$ strongly in $\mathcal{E}^{+}_{\kappa}(\mathbf{A})$, which is (7.9).

Applying now to $(\boldsymbol{\mu}_{s})_{s\in S}$ and $\boldsymbol{\mu}$ the same arguments as in the last paragraph of the proof of Lemma 7.2, we arrive at (7.10). Hence, $\boldsymbol{\mu}\in\mathcal{E}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$. The rest of the proof repeats word by word the last two paragraphs in the proof of Theorem 6.1.

7.5. Proof of Corollary 6.7

Let the assumptions of any of Theorems 6.1, 6.3, or 6.5 be fulfilled. As seen from these theorems, the class $\mathfrak{S}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$ of the solutions to Problem 4.2 is then nonempty and given by (6.4).

Assume moreover that the kernel $\kappa$ is strictly positive definite, while the $A_{i}$, $i\in I$, are mutually essentially disjoint. By Corollary 4.5, a solution to Problem 4.2 is then unique, which implies in view of (6.4) that the vague cluster sets of the minimizing nets are identical to one another, and all these reduce to a unique $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}\in\mathfrak{S}_{\kappa,\mathbf{f}}^{\boldsymbol{\sigma}}(\mathbf{A},\mathbf{a},\mathbf{g})$. Since the vague topology on $\mathfrak{M}^{+}(X)^{\mathrm{Card}\,I}$ is Hausdorff, $\boldsymbol{\lambda}^{\boldsymbol{\sigma}}_{\mathbf{A}}$ must be the vague limit of every $(\boldsymbol{\nu}_{t})_{t\in T}\in\mathbb{M}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ [3, Chapter I, Section 9, n∘ 1], as was to be proven.

8. The weighted potentials of the solutions to Problem 4.2

For any $\boldsymbol{\nu}\in\mathcal{E}^{+}_{\kappa}(\mathbf{A})$ we denote by $W^{\boldsymbol{\nu},i}_{\kappa,\mathbf{f}}$, $i\in I$, the $i$-component of the $\mathbf{f}$-weighted vector potential $W^{\boldsymbol{\nu}}_{\kappa,\mathbf{f}}$ (cf. Eq. (4.1)).

Lemma 8.1**.**

$\boldsymbol{\lambda}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ solves Problem 4.2 if and only if

[TABLE]

Proof.

For any $\boldsymbol{\mu},\boldsymbol{\nu}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and $h\in(0,1]$, we obtain by straightforward verification

[TABLE]

If $\boldsymbol{\mu}=\boldsymbol{\lambda}$ solves Problem 4.2, then the left-hand (hence, the right-hand) side of this display is $\geqslant 0$, for the class $\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ is convex, which leads to (8.1) by letting $h\to 0$. Conversely, if (8.1) holds, then the preceding formula with $\boldsymbol{\mu}=\boldsymbol{\lambda}$ and $h=1$ yields $G_{\kappa,\mathbf{f}}(\boldsymbol{\nu})\geqslant G_{\kappa,\mathbf{f}}(\boldsymbol{\lambda})$ for all $\boldsymbol{\nu}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, and hence $\boldsymbol{\lambda}\in\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. ∎

We next provide a description of the weighted potentials of the solutions to Problem 4.2 and single out their characteristic properties. The permanent assumptions stated in Sections 4.1 and 5 above are required.

Theorem 8.2**.**

Let the $A_{i}$, $i\in I$, be nearly closed, and let (1.3), (6.1), and

[TABLE]

hold. Assume also that for every $i\in I_{0}$,

[TABLE]

[TABLE]

$A^{\circ}_{i}$ being defined in (5.5). If, moreover, the $f_{i}|_{\breve{A}_{i}}$, $i\in I$, are lower bounded,212121If Case I holds, then the $f_{i}$, $i\in I$, are necessarily lower bounded on $X$. then for any $\boldsymbol{\lambda}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ the following two assertions are equivalent:

• (i)

$\boldsymbol{\lambda}\in\mathfrak{S}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$.

• (ii)

There exists $(w_{\boldsymbol{\lambda}}^{i})_{i\in I}\in\mathbb{R}^{{\rm Card}\,I}$ such that for all $i\in I_{0}$,

[TABLE]

while for all $i\in I\setminus I_{0}$,

[TABLE]

Proof.

As seen from (8.3), the $\xi^{i}$, $i\in I_{0}$, are $c_{\kappa}$-absolutely continuous, which will be permanently used in the proof.222222As in [28, p. 134], we call $\mu\in\mathfrak{M}(X)$ $c_{\kappa}$-absolutely continuous if $\mu(K)=0$ for every compact $K\subset X$ with $c_{\kappa}(K)=0$. Then $|\mu|_{*}(Q)=0$ for any $Q\subset X$ with $c_{\kappa}(Q)=0$. Every $\mu\in\mathcal{E}_{\kappa}(X)$ is $c_{\kappa}$-absolutely continuous; but not conversely [28, pp. 134–135]. There is therefore no loss of generality in replacing each $\xi^{i}$, $i\in I_{0}$, by the extension of $\xi^{i}|_{A_{i}\cap\breve{A}_{i}}$ by [math] to all of $X$ (denoted again by $\xi^{i}$). After having done this, we next replace (again with the notation preserved) the $A_{i}$, $i\in I$, by the (closed) sets $\breve{A}_{i}$, which also involves no loss of generality. Note that then (8.3) and (8.4) remain valid.

For every $\boldsymbol{\nu}=(\nu^{\ell})_{\ell\in I}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ and every $i\in I$, write $\boldsymbol{\nu}_{i}:=(\nu_{i}^{\ell})_{\ell\in I}$, where $\nu_{i}^{\ell}:=\nu^{\ell}$ for all $\ell\neq i$ and $\nu_{i}^{i}=0$; then $\boldsymbol{\nu}_{i}\in\mathcal{E}^{+}_{\kappa,\mathbf{f}}(\mathbf{A})$. According to (3.5) and (3.10), $\kappa^{i}_{\boldsymbol{\nu}_{i}}$ is given by

[TABLE]

and it is well defined and finite n.e. (Corollary 3.12).

Furthermore, under the stated assumptions, $\kappa^{i}_{\boldsymbol{\nu}_{i}}$ is lower bounded on $A_{i}$. In fact, in the same manner as in the proof of Lemma 7.1 we see from (1.3) that $|R\boldsymbol{\nu}_{i}|(X)\leqslant C$, where $C\in(0,\infty)$. This together with (6.1) implies that $\kappa(\cdot,R\boldsymbol{\nu}_{i}^{-})$, resp. $\kappa(\cdot,R\boldsymbol{\nu}_{i}^{+})$, is upper bounded on $A^{+}$, resp. on $A^{-}$, which in view of the preceding display yields our claim.

Fix $\boldsymbol{\lambda}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$. Having written $\tilde{f}_{i}:=f_{i}+\kappa^{i}_{\boldsymbol{\lambda}_{i}}$, define the function

[TABLE]

Comparing this with (3.5) and (4.1), we get

[TABLE]

Note that $W_{\kappa,\tilde{f}_{i}}^{\lambda^{i}}$ is finite n.e. on $A^{\circ}_{i}$ and lower bounded on $A_{i}$, because this is the case for each of the summands $\kappa(\cdot,\lambda^{i})$, $f_{i}$, and $\kappa^{i}_{\boldsymbol{\lambda}_{i}}$.

To establish the equivalence of (i) and (ii), suppose first that (i) holds, i.e., $\boldsymbol{\lambda}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ solves Problem 4.2. Fix $i\in I$. By (3.4) and (4.2), we get for any $\boldsymbol{\nu}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ with the additional property that $\boldsymbol{\nu}_{i}=\boldsymbol{\lambda}_{i}$ (in particular, for $\boldsymbol{\nu}=\boldsymbol{\lambda}$)

[TABLE]

Combined with $G_{\kappa,\mathbf{f}}(\boldsymbol{\nu})\geqslant G_{\kappa,\mathbf{f}}(\boldsymbol{\lambda})$, this yields $G_{\kappa,\tilde{f}_{i}}(\nu^{i})\geqslant G_{\kappa,\tilde{f}_{i}}(\lambda^{i})$, and hence $\lambda^{i}$ minimizes $G_{\kappa,\tilde{f}_{i}}(\nu)$, where $\nu$ ranges over the class $\mathcal{E}^{\sigma^{i}}_{\kappa,\tilde{f}_{i}}(A_{i},a_{i},g_{i})$. This enables us to show that there exists $w_{\lambda^{i}}\in\mathbb{R}$ such that

[TABLE]

whenever $i\in I_{0}$, while otherwise (for $i\in I\setminus I_{0}$)

[TABLE]

To this end, write for any $w\in\mathbb{R}$

[TABLE]

and assume first that $i\in I_{0}$. Then (8.11) holds with

[TABLE]

In turn, (8.11) with $w_{\lambda^{i}}=L_{i}$ yields $L_{i}<\infty$, because

[TABLE]

holds n.e. on $A^{\circ}_{i}$, hence $(\xi^{i}-\lambda^{i})$-a.e. on $A^{\circ}_{i}$, for $\xi^{i}$ and $\lambda^{i}$ are both $c_{\kappa}$-absolutely continuous, and finally $(\xi^{i}-\lambda^{i})$-a.e. on $A_{i}$ by (8.4). Also, $L_{i}>-\infty$ since, in consequence of (8.2), $\widetilde{W}_{\kappa,\tilde{f}_{i}}^{\lambda^{i}}$ along with $W_{\kappa,\tilde{f}_{i}}^{\lambda^{i}}$ is lower bounded on $A_{i}$.

We next proceed by establishing (8.12) with $w_{\lambda^{i}}=L_{i}$. Assume, on the contrary, that this fails to hold. Since $\widetilde{W}_{\kappa,\tilde{f}_{i}}^{\lambda^{i}}$ is $\lambda^{i}$-measurable, one can choose $w_{i}\in(L_{i},\infty)$ so that $\lambda^{i}(A_{i}^{+}(w_{i}))>0$. At the same time, as $w_{i}>L_{i}$, it follows from the definition of $L_{i}$ that $(\xi^{i}-\lambda^{i})(A_{i}^{-}(w_{i}))>0$. Therefore, there exist compact sets $K_{1}\subset A_{i}^{+}(w_{i})$ and $K_{2}\subset A_{i}^{-}(w_{i})$ such that

[TABLE]

Write $\tau^{i}:=(\xi^{i}-\lambda^{i})|_{K_{2}}$; then $\kappa(\tau^{i},\tau^{i})<\infty$ by (8.3). Since $\langle W_{\kappa,\tilde{f}_{i}}^{\lambda^{i}},\tau^{i}\rangle\leqslant\langle w_{i}g_{i},\tau^{i}\rangle<\infty$, in view of (8.9) we get $\langle\tilde{f}_{i},\tau^{i}\rangle<\infty$. Define

[TABLE]

Observing from (8.15) that $c_{i}\in(0,1)$, we obtain by straightforward verification $\langle g_{i},\theta^{i}\rangle=a_{i}$ and also $\theta^{i}\leqslant\xi^{i}$. Hence, $\theta^{i}\in\mathcal{E}^{\xi^{i}}_{\kappa,\tilde{f}_{i}}(A_{i},a_{i},g_{i})$. But

[TABLE]

which is impossible in view of the scalar version of Lemma 8.1 (with $I=\{i\}$). The contradiction obtained establishes (8.12).

Let now $i\in I\setminus I_{0}$. Since $\lambda^{i}$ minimizes $G_{\kappa,\tilde{f}_{i}}(\nu)$ among $\nu\in\mathcal{E}^{+}_{\kappa,\tilde{f}_{i}}(A_{i},a_{i},g_{i})$, it follows from [43, Theorem 7.1] that (8.13) and (8.14) hold with

[TABLE]

Substituting (8.11), (8.12), (8.13), and (8.14) into (8.10) establishes (8.5), (8.6), (8.7), and (8.8) with $w_{\boldsymbol{\lambda}}^{i}:=w_{\lambda^{i}}$, $i\in I$. Hence, (i)$\Rightarrow$(ii).

To complete the proof, suppose finally that (ii) holds. By (8.10), for every $i\in I_{0}$, resp. $i\in I\setminus I_{0}$, (8.11) and (8.12), resp. (8.13) and (8.14), are then fulfilled with $w_{\lambda^{i}}:=w_{\boldsymbol{\lambda}}^{i}$ and $\tilde{f}_{i}:=f_{i}+\kappa^{i}_{\boldsymbol{\lambda}_{i}}$. We observe from (8.11) and (8.12) that $\lambda^{i}(A_{i}^{+}(w_{\lambda^{i}}))=0$ and $(\xi^{i}-\lambda^{i})(A_{i}^{-}(w_{\lambda^{i}}))=0$ for all $i\in I_{0}$. Having fixed $\boldsymbol{\nu}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$, we therefore get for all $i\in I_{0}$,

[TABLE]

Furthermore, it follows from (8.13) and (8.14) that

[TABLE]

$\nu^{i}$ being $c_{\kappa}$-absolutely continuous. Hence, for all $i\in I\setminus I_{0}$,

[TABLE]

Summing up the inequalities in (8.16) and (8.17) over all $i\in I$, we see from Lemma 8.1 in view of the arbitrary choice of $\boldsymbol{\nu}\in\mathcal{E}^{\boldsymbol{\sigma}}_{\kappa,\mathbf{f}}(\mathbf{A},\mathbf{a},\mathbf{g})$ that $\boldsymbol{\lambda}$ is indeed a solution to Problem 4.2.∎

Corollary 8.3**.**

Under the hypotheses of Theorem 8.2, assume moreover that $\kappa$ is continuous on $\breve{A}^{+}\times\breve{A}^{-}$ and Case I holds. Then (8.6) and (8.8) in (ii) are equivalent to the following apparently stronger relations:

[TABLE]

Proof.

This will follow once we have proven that for any $i\in I$, $W^{\boldsymbol{\lambda},i}_{\kappa,\mathbf{f}}|_{\breve{A}_{i}}$ is l.s.c., which in turn holds if $\kappa(\cdot,R\boldsymbol{\lambda}^{-})|_{\breve{A}^{+}}$ and $\kappa(\cdot,R\boldsymbol{\lambda}^{+})|_{\breve{A}^{-}}$ are shown to be continuous. To establish the latter, write

[TABLE]

Under the stated assumptions, $\kappa^{*}$ is nonnegative and continuous, and hence

[TABLE]

is l.s.c. In the same manner as in the proof of Lemma 7.1, we observe from (1.3) that $|R\boldsymbol{\lambda}|(X)\leqslant C<\infty$. Integrating (8.18) with respect to $R\boldsymbol{\lambda}^{-}$, we therefore see that $\kappa^{*}(x,R\boldsymbol{\lambda}^{-})$, $x\in\breve{A}^{+}$, coincides up to a finite summand with the restriction of $-\kappa(x,R\boldsymbol{\lambda}^{-})$ to $\breve{A}^{+}$. It follows that $\kappa(x,R\boldsymbol{\lambda}^{-})|_{\breve{A}^{+}}$ must be upper semicontinuous. Being also l.s.c., $\kappa(x,R\boldsymbol{\lambda}^{-})|_{\breve{A}^{+}}$ is actually continuous as desired. The same holds with the indices + and - reversed. ∎

Acknowledgement. I thank Bent Fuglede for many fruitful discussions on the topic of the paper.