A Vector Equilibrium Problem for Muttalib–Borodin Biorthogonal Ensembles

A Vector Equilibrium Problem
for Muttalib–Borodin Biorthogonal Ensembles\star\star$\star$This paper is a
contribution to the Special Issue on Asymptotics and Universality in Random
Matrices, Random Growth Processes, Integrable Systems and Statistical Physics
in honor of Percy Deift and Craig Tracy. The full collection is available at
http://www.emis.de/journals/SIGMA/Deift-Tracy.html

Arno B.J. KUIJLAARS

A.B.J. Kuijlaars

Katholieke Universiteit Leuven, Department of Mathematics,
Celestijnenlaan 200B box 2400, BE-3001 Leuven, Belgium
<EMAIL_ADDRESS>http://perswww.kuleuven.be/~u0017946/

Received April 12, 2016, in final form July 03, 2016; Published online July
05, 2016

The Muttalib–Borodin biorthogonal ensemble is a joint density function for $n$
particles on the positive real line that depends on a parameter $\theta$.
There is an equilibrium problem that describes the large $n$ behavior. We show
that for rational values of $\theta$ there is an equivalent vector equilibrium
problem.

biorthogonal ensembles; vector equilibrium problem; random matrix theory;
logarithmic potential theory

31A05; 60B20

Dedicated to Percy Deift and Craig Tracy
on the occasion of their 70th birthdays

##  1 Introduction and statement of results

###  1.1 The Muttalib–Borodin ensemble

The Muttalib–Borodin biorthogonal ensemble is the following probability
density function for $n$ particles on the half line $[0,\infty)$

$\displaystyle\frac{1}{Z_{n}}\prod_{j<k}(x_{k}-x_{j})\prod_{j<k}\big{(}x_{k}^{\theta}-x_{j}^{\theta}\big{)}\prod_{j=1}^{n}e^{-nV(x_{j})},\qquad
x_{j}\geq 0,$ (1.1)

with $\theta>0$ and with an $n$-dependent weight function $w(x)=e^{-nV(x)}$
having enough decay at infinity. The model is named after Muttalib [17] who
introduced it as a simplified model for disordered conductors in the metallic
regime, and Borodin [4] who obtained profound mathematical results, in
particular for Laguerre and Jacobi weights.

The model has attracted considerable attention in recent years. Random matrix
models whose eigenvalues (or singular values) have the distribution (1.1) were
recently given in [6, 13]. The model is also related to products of random
matrices [13, 16].

In the large $n$ limit, the particles have an almost sure limiting measure
$\mu^{*}$ which is the minimizer of

$\displaystyle\frac{1}{2}\iint\log\frac{1}{|x-y|}d\mu(x)d\mu(y)+\frac{1}{2}\iint\log\frac{1}{|x^{\theta}-y^{\theta}|}d\mu(x)d\mu(y)+\int
V(x)d\mu(x)$ (1.2)

among all probability measures $\mu$ on $[0,\infty)$. This follows from large
deviation results for (1.1) and related models that were studied in [3, 5,
10]. For $\theta=1$, the functional (1.2) reduces to the usual energy in the
presence of an external field [20].

The minimizer for (1.2) was studied in detail by Claeys and Romano [7]. They
found sufficient conditions for the minimizer to be supported on an interval
$[0,a]$ for some $a>0$. Forrester and co-authors [11, 12] analyzed the
equilibrium problem for (1.2) with special potentials, and found expressions
for the minimizers as Fuss–Catalan and Raney distributions, see also [18].

It is the aim of this paper to show that for rational values of $\theta$, say
$\theta=q/r$ with $q,r\in\mathbb{N}$, there is an equivalent minimization
problem for a vector of $q+r-1$ measures. We expect that the vector
equilibrium problem will be useful for subsequent asymptotic analysis. The
special role of rational $\theta$ also appeared in the already mentioned work
[7]. This paper gives finite term recurrence relations for the biorthogonal
polynomials associated with (1.1), as well as a Christoffel–Darboux formula
for the correlation kernel are given, but only for rational $\theta$.

In order to state our results we introduce the logarithmic energy

$\displaystyle I(\mu)=\iint\log\frac{1}{|x-y|}d\mu(x)d\mu(y)$ (1.3)

of a measure $\mu$ and the mutual energy

$\displaystyle I(\mu,\nu)=\iint\log\frac{1}{|x-y|}d\mu(x)d\nu(y)$ (1.4)

of two measures $\mu$ and $\nu$. Throughout we use for $j\in\mathbb{Z}$,

$\displaystyle\Delta_{j}=\begin{cases}[0,\infty),&\text{if $j$ is even},\\\
(-\infty,0],&\text{if $j$ is odd}.\end{cases}$ (1.5)

###  1.2 Result for the case $\boldsymbol{\theta=1/r}$

We first state the result for the case $\theta=1/r$ with $r\in\mathbb{N}$.

######  Theorem 1.1.

Let $V\colon[0,\infty)\to\mathbb{R}$ be continuous and

$\displaystyle\lim_{x\to\infty}\big{(}V(x)-\log\big{(}1+x^{2}\big{)}\big{)}=+\infty.$

Then there is a unique vector $(\mu_{0}^{*},\mu_{1}^{*},\ldots,\mu_{r-1}^{*})$
of $r$ measures that minimizes the energy functional

$\displaystyle\sum_{j=0}^{r-1}I(\mu_{j})-\sum_{j=0}^{r-2}I(\mu_{j},\mu_{j+1})+\int
Vd\mu_{0}$ (1.6)

among all vectors satisfying for every $j=0,\ldots,r-1$,

  1. $i)$

$\operatorname{supp}(\mu_{j})\subset\Delta_{j}$,

  2. $ii)$

$\mu_{j}(\Delta_{j})=1-\frac{j}{r}$, and

  3. $iii)$

$I(\mu_{j})<+\infty$.

The measure $\mu_{0}^{*}$ has compact support, and it is the unique minimizer
of the functional (1.2) with $\theta=1/r$ among probability measures on
$[0,\infty)$.

The minimization problem for the energy functional (1.6) is an example of a
weakly admissible vector equilibrium problem in the sense of Hardy and
Kuijlaars, see also below. The other measures
$\mu_{1}^{*},\ldots,\mu_{r-1}^{*}$ have full unbounded support,
$\operatorname{supp}(\mu_{j}^{*})=\Delta_{j}$ for $j=1,\ldots,r-1$. In fact
$\mu_{j}^{*}=\frac{1}{2}\operatorname{Bal}\big{(}\mu_{j-1}^{*}+\mu_{j+1}^{*},\Delta_{j}\big{)}$,
where $\operatorname{Bal}$ denotes the balayage onto $\Delta_{j}$, see below
as well.

###  1.3 Result for rational $\boldsymbol{\theta}$

For general rational $\theta=q/r$ with $q,r\in\mathbb{N}$, we first make the
change of variables $x\mapsto x^{1/q}$, $y\mapsto y^{1/q}$ in the energy
functional (1.2) to obtain

$\displaystyle\frac{1}{2}\iint\\!\log\frac{1}{|x^{1/q}-y^{1/q}|}d\nu(x)d\nu(y)+\frac{1}{2}\iint\\!\log\frac{1}{|x^{1/r}-y^{1/r}|}d\nu(x)d\nu(y)+\int\\!\widehat{V}(x)d\nu(x),\\!\\!\\!\\!$
(1.7)

where $d\nu(x)=d\mu(x^{1/q})$, and $\widehat{V}(x)=V(x^{1/q})$. Note that $q$
and $r$ play a symmetric role in the energy functional (1.7).

######  Theorem 1.2.

Let $\widehat{V}\colon[0,\infty)\to\mathbb{R}$ be continuous and

$\displaystyle\lim_{x\to\infty}\big{(}\widehat{V}(x)-\log\big{(}1+x^{2}\big{)}\big{)}=+\infty.$
(1.8)

Then there is a unique vector
$(\nu_{-q+1}^{*},\ldots,\nu_{-1}^{*},\nu_{0}^{*},\nu_{1}^{*},\ldots,\nu_{r-1}^{*})$
of $q+r-1$ measures that minimizes the energy functional

$\displaystyle\sum_{j=-q+1}^{r-1}I(\nu_{j})-\sum_{j=-q+1}^{r-2}I(\nu_{j},\nu_{j+1})+\int\tilde{V}d\nu_{0}$
(1.9)

among all vectors satisfying for every $j=-q+1,\ldots,r-1$,

  1. $i)$

$\operatorname{supp}(\nu_{j})\subset\Delta_{j}$,

  2. $ii)$

$\nu_{j}(\Delta_{j})=m_{j}:=\begin{cases}1-\frac{j}{r}&\text{if }j\geq
0,\vspace{1mm}\\\ 1-\frac{|j|}{q}&\text{if }j\leq 0,\end{cases}$

  3. $iii)$

$I(\mu_{j})<+\infty$.

The measure $\nu_{0}^{*}$ has compact support and it is the unique minimizer
of the functional (1.7) among probability measures on $[0,\infty)$, and
$d\mu_{0}^{*}(x)=d\nu_{0}^{*}(x^{q})$ is the unique minimizer of (1.2) with
$\theta=q/r$.

As in Theorem 1.1 the other measures $\nu_{j}^{*}$ from Theorem 1.2 have full
unbounded support: $\operatorname{supp}(\nu_{j}^{*})=\Delta_{j}$ if $j\neq 0$.

The energy functional (1.9) with $\hat{V}=0$ and the normalizations $m_{j}$ as
in condition ii) appeared in [9, Theorem 2.3] where it describes the limiting
eigenvalue distribution of banded Toeplitz matrices. The supports $\Delta_{j}$
of the measures, however, are more general curves in that case.

Theorem 1.1 is the special case $q=1$ of Theorem 1.2 and it is enough to prove
the latter theorem. However, for sake of exposition we chose to state Theorem
1.1 separately as well.

##  2 Preliminaries

###  2.1 Vector equilibrium problems

The unique existence of a minimizing vector of measures follows from the
result of Hardy and Kuijlaars, which we recall here. The general setup of [14]
involves the following ingredients.

  1. (a)

A sequence of closed subsets $\Delta_{1},\ldots,\Delta_{d}$ of the complex
plane, each with positive logarithmic capacity.

  2. (b)

For each $i=1,\ldots,d$, a lower semi-continuous function
$V_{i}\colon\Delta_{i}\to\mathbb{R}\cup\\{+\infty\\}$ that is finite on a set
of positive capacity.

  3. (c)

A symmetric positive definite interaction matrix $C=(c_{ij})$ of size $d\times
d$.

  4. (d)

A sequence of positive numbers $m_{1},\ldots,m_{d}$, such that

$\displaystyle\liminf_{|x|\to\infty,\,x\in\Delta_{i}}\left(V_{i}(x)-\left(\sum_{j=1}^{d}c_{ij}m_{j}\right)\log\big{(}1+|x|^{2}\big{)}\right)>-\infty$
(2.1)

for every $i=1,\ldots,d$ for which $\Delta_{i}$ is unbounded.

Associated with the above data is the energy functional

$\displaystyle
J(\mu_{1},\ldots,\mu_{d})=\sum_{i,j=1}^{d}c_{ij}I(\mu_{i},\mu_{j})+\sum_{i=1}^{d}\int
V_{i}(x)d\mu_{i}(x).$ (2.2)

The problem is to minimize $J$ over
$\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$, or some subset of it.
Here we use $\mathcal{M}_{m}(\Delta)$ to denote the set of positive Borel
measures on $\Delta$ of total mass $m>0$. Such a minimization problem is
called a weakly admissible vector equilibrium problem.

The functional (2.2) is not well-defined on all of
$\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$, since there is a
problem with measures having overlapping supports, and with measures having
unbounded supports. The problem with overlapping supports was solved by
Beckermann et al. [2]. To handle the situation with unbounded supports, a
regularization of $J$ is introduced in [14, Theorem 2.6] that comes from
mapping the measures to the Riemann sphere and redefining $J$ accordingly.
This procedure involves a modification of the mutual energy (1.4) and the
corresponding logarithmic energy (1.3) to

$\displaystyle\tilde{I}(\mu,\nu)=\iint\log\left(\frac{\sqrt{1+|x|^{2}}\sqrt{1+|y|^{2}}}{|x-y|}\right)d\mu(x)d\nu(y),$

and $\tilde{I}(\mu)=\tilde{I}(\mu,\mu)$, respectively.

We may call $\tilde{I}(\mu)$ and $\tilde{I}(\mu,\nu)$ the spherical (mutual)
energy. Then

$\displaystyle\tilde{J}(\mu_{1},\ldots,\mu_{d})=\begin{cases}\displaystyle\sum\limits_{i,j=1}^{d}c_{ij}\tilde{I}(\mu_{i},\mu_{j})+\sum\limits_{i=1}^{d}\int\tilde{V}_{i}(x)d\mu_{i}(x),&\text{if
all }\tilde{I}(\mu_{i})<+\infty,\\\ +\infty,&\text{otherwise},\end{cases}$
(2.3)

where

$\displaystyle\tilde{V}_{i}(x)=V_{i}(x)-\left(\sum_{j=1}^{d}c_{ij}m_{j}\right)\log\big{(}1+|x|^{2}\big{)}$
(2.4)

is an extension of (2.2) since for vectors of measures with
$I(\mu_{i})<+\infty$ for all $i$, we have that
$\tilde{J}(\mu_{1},\ldots,\mu_{d})=J(\mu_{1},\ldots,\mu_{d})$. Note that
$\tilde{V}_{i}$ is bounded away from $-\infty$ on $\Delta_{i}$ because of the
lower-semicontinuity of $V_{i}$ and the growth condition (2.1) at infinity.

The functional (2.3) is thus defined on
$\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$ with values in
$\mathbb{R}\cup\\{+\infty\\}$ only. In addition, $\tilde{J}$ has compact
sublevel sets $\tilde{J}^{-1}((-\infty,\alpha])$ for any real $\alpha$ (which
implies that it is lower semi-continuous), and it is strictly convex on the
set where it is finite. As a consequence, there is a unique minimizer of
$\tilde{J}$ on $\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$, see
[14, Corollary 2.7].

###  2.2 Variational conditions

We use $U^{\nu}(z)=\int\log\frac{1}{|z-y|}d\nu(y)$, $z\in\mathbb{C}$, to
denote the logarithmic potential of a measure $\nu$ and

$\displaystyle\tilde{U}^{\nu}(z)=\int\log\left(\frac{\sqrt{1+|z|^{2}}\sqrt{1+|y|^{2}}}{|z-y|}\right)d\nu(y)$

to denote the spherical potential.

The variational conditions for the vector equilibrium problem were not
discussed in [14]. The following result is standard for the case $d=1$, and
its extension to $d\geq 2$ is not difficult.

######  Lemma 2.1.

Let $\vec{\mu}^{*}=(\mu_{1}^{*},\ldots,\mu_{d}^{*})$ be a vector of measures
in $\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$ with
$\tilde{I}(\mu_{i}^{*})<+\infty$ for all $i$. Suppose that for some constants
$\ell_{i}$,

$\displaystyle 2\sum_{j=1}^{d}c_{ij}\tilde{U}^{\mu_{j}^{*}}+\tilde{V}_{i}\
\begin{cases}=\ell_{i}&\text{on }\operatorname{supp}(\mu_{i}^{*}),\\\
\geq\ell_{i}&\text{on }\Delta_{i},\end{cases}$ (2.5)

for every $i=1,\ldots,d$. Then $\vec{\mu}$ is the minimizer of $\tilde{J}$ in
$\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$.

###### Proof.

The functional $\tilde{J}$ is strictly convex on the set where it is finite.
The strict convexity is due to the quadratic part
$\tilde{J}_{0}(\vec{\mu})=\sum\limits_{i,j=1}^{d}c_{ij}\tilde{I}(\mu_{i},\mu_{j})$
in the functional $\tilde{J}$ and it comes down to

$\displaystyle\tilde{J}_{0}(\vec{\mu})+\tilde{J}_{0}(\vec{\nu})\geq
2\sum_{i,j=1}^{d}c_{ij}I(\mu_{i},\nu_{j}),$ (2.6)

whenever
$\vec{\mu},\vec{\nu}\in\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$,
with strict inequality if $\vec{\mu}\neq\vec{\nu}$ and both
$\tilde{J}(\vec{\mu})<+\infty$, $\tilde{J}(\vec{\nu})<+\infty$.

Let $\vec{\mu}^{*}$ be as in the lemma. Then for any
$\vec{\mu}\in\prod\limits_{j=1}^{d}\mathcal{M}_{m_{j}}(\Delta_{j})$, we find
by integrating (2.5) with respect to $\mu_{i}$,

$\displaystyle\int\left(2\sum_{j=1}^{d}c_{ij}\tilde{U}^{\mu_{j}^{*}}+\tilde{V}_{i}\right)d\mu_{i}\geq\ell_{i}m_{i},$

and so by summing over $i$,

$\displaystyle
2\sum_{i,j=1}^{d}c_{ij}\tilde{I}(\mu_{i},\mu_{j}^{*})+\sum_{i=1}^{d}\int\tilde{V}_{i}d\mu_{i}\geq\sum_{i=1}^{d}\ell_{i}m_{i}.$
(2.7)

If we integrate (2.5) with respect to $\mu_{i}^{*}$ and sum over $i$, we find
an equality

$\displaystyle
2\sum_{i,j=1}^{d}c_{ij}\tilde{I}(\mu_{i}^{*},\mu_{j}^{*})+\sum_{i=1}^{d}\int\tilde{V}_{i}d\mu_{i}^{*}=\sum_{i=1}^{d}\ell_{i}m_{i},$

which means

$\displaystyle\sum_{i=1}^{d}\ell_{i}m_{i}=\tilde{J}(\vec{\mu}^{*})+\tilde{J}_{0}(\vec{\mu}^{*}),$
(2.8)

and in particular $\tilde{J}(\vec{\mu}^{*})<+\infty$.

Now we use (2.6) with $\vec{\nu}=\vec{\mu}^{*}$, and combine it with (2.7) and
(2.8) to find

$\displaystyle\tilde{J}(\vec{\mu})$
$\displaystyle=\tilde{J}_{0}(\vec{\mu})+\sum_{i=1}^{d}\int\tilde{V}_{i}d\mu_{i}\geq
2\int\sum_{i,j=1}^{d}c_{ij}I(\mu_{i},\mu_{j}^{*})-\tilde{J}_{0}(\vec{\mu}^{*})+\sum_{i=1}^{d}\int\tilde{V}_{i}d\mu_{i}$
$\displaystyle\geq\sum_{i=1}^{d}\ell_{i}m_{i}-\tilde{J}_{0}(\vec{\mu}^{*})=\tilde{J}(\vec{\mu}^{*}).$

Thus $\vec{\mu}^{*}$ is indeed the minimizer of $\tilde{J}$. ∎

###  2.3 Nikishin interaction and balayage

In the present paper we are dealing with the interaction matrix $C=(c_{ij})$
where

$\displaystyle c_{ij}=\begin{cases}1,&\text{if }i=j,\\\ -\frac{1}{2},&\text{if
}|i-j|=1,\\\ 0,&\text{otherwise},\end{cases}$ (2.9)

which is indeed a positive definite matrix, and sets $\Delta_{j}$ that
alternate between the positive and negative real axis as in (1.5). The
interaction matrix (2.9) is characteristic for Nikishin systems in the theory
of Hermite–Padé approximation [19]. See also [1, 15] for surveys on the
connections with Hermite-Padé approximation and random matrix theory.

In Theorem 1.2 we have $d=q+r-1$ and we use indices $i=-q+1,\ldots,r-1$,
instead of $i=1,\ldots,d$. The total masses $m_{j}$ are as in condition ii) of
Theorem 1.2. The external fields are $V_{0}=\widehat{V}$ and $V_{i}\equiv 0$
for $i\neq 0$. Then it is easy to see that the conditions (a), (b), (c), and
(d) stated in Section 2.1 are satisfied. In fact we have (2.1), since

$\displaystyle\sum_{j=-q+1}^{r-1}c_{ij}m_{j}=-\frac{1}{2}m_{i-1}+m_{i}-\frac{1}{2}m_{i+1}=\begin{cases}0,&\text{for
}i\neq 0,\\\ \frac{q^{-1}+r^{-1}}{2},&\text{for }i=0,\end{cases}$

and then (2.1) is satisfied because of (1.8).

Hence there exists a unique minimizing vector of measures
$(\nu_{-q+1}^{*},\ldots,\nu_{r-1}^{*})$ for the energy functional $\tilde{J}$
with

$\displaystyle\tilde{V}(x)=\widehat{V}(x)-\frac{q^{-1}+r^{-1}}{2}\log\big{(}1+|x|^{2}\big{)},$

see (2.4). We have to show that in addition $I(\nu_{i}^{*})<+\infty$ for all
$i$, and then we can conclude that it is also a minimizing vector for $J$.

The variational conditions (2.5) from Lemma 2.1 are in this case

$\displaystyle
2\tilde{U}^{\nu_{0}^{*}}-\tilde{U}^{\nu_{-1}^{*}}-\tilde{U}^{\nu_{1}^{*}}+\tilde{V}\
\begin{cases}=\ell_{0}&\text{on }\operatorname{supp}(\nu_{0}^{*}),\\\
\geq\ell_{0}&\text{on }[0,\infty),\end{cases}$ (2.10)

and for $j\in\\{-q+1,\ldots,r-1\\}\setminus\\{0\\}$,

$\displaystyle
2\tilde{U}^{\nu_{j}^{*}}-\tilde{U}^{\nu_{j-1}^{*}}-\tilde{U}^{\nu_{j+1}^{*}}\
\begin{cases}=\ell_{j}&\text{on }\operatorname{supp}(\nu_{j}^{*}),\\\
\geq\ell_{j}&\text{on }\Delta_{j}.\end{cases}$ (2.11)

Here we set $\nu_{-q}^{*}=\nu_{r}^{*}=0$ so that (2.10), (2.11) also hold for
$j=-q+1$ and $j=r-1$.

Note that $2\nu_{j}^{*}$ and $\nu_{j-1}^{*}+\nu_{j+1}^{*}$ have the same total
masses if $j\neq 0$. If (2.11) holds then $2\nu_{j}^{*}$ is the balayage
measure of $\nu_{j-1}^{*}+\nu_{j+1}^{*}$ onto $\Delta_{j}$. The balayage
measure has full support $\Delta_{j}$, and equality holds in (2.11) everywhere
on $\Delta_{j}$. In addition, the constant $\ell_{j}$ is zero. So for $j\neq
0$ the variational condition is

$\displaystyle
2\tilde{U}^{\nu_{j}^{*}}-\tilde{U}^{\nu_{j-1}^{*}}-\tilde{U}^{\nu_{j+1}^{*}}=0\qquad\text{on}\
\ \Delta_{j}.$ (2.12)

If $x\mapsto\log(1+|x|^{2})$ is integrable with respect to all measures, then
(2.10) reduces to

$\displaystyle 2U^{\nu_{0}^{*}}-U^{\nu_{-1}^{*}}-U^{\nu_{1}^{*}}+\widehat{V}\
\begin{cases}=\ell&\text{on }\operatorname{supp}(\nu_{0}^{*}),\\\
\geq\ell&\text{on }[0,\infty),\end{cases}$ (2.13)

for some constant $\ell$, while (2.12) reduces to

$\displaystyle
2U^{\nu_{j}^{*}}=U^{\nu_{j-1}^{*}}+U^{\nu_{j+1}^{*}}\qquad\text{on}\ \
\Delta_{j},$ (2.14)

which is the more common form for the balayage in logarithmic potential
theory, see [20] where the discussion however is restricted to measures with
compact support in $\mathbb{C}$.

Our strategy to prove Theorem 1.2 will be to establish the existence of a
vector of measures $\vec{\nu}^{*}=(\nu_{-q+1}^{*},\ldots,\nu_{r-1}^{*})$ with
$\operatorname{supp}(\nu_{j}^{*})\subset\Delta_{j}$,
$\nu_{j}^{*}(\Delta_{j})=m_{j}$, such that the conditions (2.14) and (2.13)
are satisfied. The measure $\nu_{0}^{*}$ will have compact support, and all
other measures have full support. The density of $\nu_{j}^{*}$ will behave
like

$\displaystyle\frac{d\nu_{j}^{*}(x)}{dx}=\begin{cases}O\big{(}|x|^{-1-1/q}\big{)}&\text{for
}j\geq 1,\\\ O\big{(}|x|^{-1-1/r}\big{)}&\text{for
}j\leq-1,\end{cases}\qquad\text{as}\ \ |x|\to\infty.$

##  3 Proof of Theorem 1.2

###  3.1 An auxiliary result

The following is our main auxiliary result.

######  Proposition 3.1.

Let $r\geq 2$ be an integer and let $a>0$. Then there is a unique vector
$(\mu_{1}^{*},\ldots,\mu_{r-1}^{*})$ of measures that minimizes the energy
functional

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j})-\sum_{j=1}^{r-2}I(\mu_{j},\mu_{j+1})+\int\log|x-a|d\mu_{1}(x)$
(3.1)

among all vectors of measures satisfying for every $j=1,\ldots,r-1$,

  1. $i)$

$\operatorname{supp}(\mu_{j})\subset\Delta_{j}$,

  2. $ii)$

$\mu_{j}(\Delta_{j})=1-\frac{j}{r}$, and

  3. $iii)$

$I(\mu_{j})<+\infty$.

Moreover,

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j}^{*})-\sum_{j=1}^{r-2}I(\mu_{j}^{*},\mu_{j+1}^{*})=-\frac{1}{2}\int\log|x-a|d\mu_{1}^{*}(x)$
(3.2)

and

$\displaystyle
U^{\mu_{1}^{*}}(z)=\log\left|\frac{z^{1/r}-a^{1/r}}{z-a}\right|,\qquad
z\in\mathbb{C}.$ (3.3)

###### Proof.

The minimization of (3.1) under the conditions i), ii), iii) is a weakly
admissible vector equilibrium problem for $r-1$ measures with total masses
$m_{j}=1-\frac{j}{r}$, $j=1,\ldots,r-1$. We set $\mu_{0}^{*}=\delta_{a}$, the
Dirac point mass at $a>0$ and $m_{0}=1$. Also $\mu_{r}^{*}=0$ and $m_{r}=0$.
Then by the discussion in Section 2, there is a unique minimizer
$(\mu_{1}^{*},\ldots,\mu_{r-1}^{*})$ for the extended functional. We are going
to construct the measures $\mu_{j}^{*}$ explicitly. We show that these
measures have densities with respect to Lebesgue measures that decay as
$|x|^{-1-1/r}$ as $|x|\to\infty$, and that for $j=1,\ldots,r-1$,

$\displaystyle
2U^{\mu_{j}^{*}}=U^{\mu_{j-1}^{*}}+U^{\mu_{j+1}^{*}}\qquad\text{on}\ \
\Delta_{j}.$

This implies that also also

$\displaystyle
2\tilde{U}^{\mu_{j}^{*}}=\tilde{U}^{\mu_{j-1}^{*}}+\tilde{U}^{\mu_{j+1}^{*}}\qquad\text{on}\
\ \Delta_{j}.$

Then by Lemma 2.1 it follows that $(\mu_{1}^{*},\ldots,\mu_{r-1}^{*})$ is the
minimizer for the extended functional, and since the function
$x\mapsto\log(1+|x|^{2})$ is integrable with respect to each of the measures,
it then also follows that it is the minimizer of (3.1). We will see at the end
of the proof that (3.2) and (3.3) hold as well.

We use a geometric construction based on the Riemann surface for the mapping
$w=z^{1/r}$. The Riemann surface has $r$ sheets

$\displaystyle\mathcal{R}_{j}=\begin{cases}\mathbb{C}\setminus\Delta_{1}=\mathbb{C}\setminus(-\infty,0],&\text{for
}j=1,\\\
\mathbb{C}\setminus(\Delta_{j-1}\cup\Delta_{j})=\mathbb{C}\setminus\mathbb{R},&\text{for
}j=2,\ldots,r-1,\\\ \mathbb{C}\setminus\Delta_{r-1},&\text{for
}j=r,\end{cases}$

where $\mathcal{R}_{j}$ is connected to $\mathcal{R}_{j+1}$ along $\Delta_{j}$
for $j=1,\ldots,r-1$ in the usual crosswise manner. There is one point at
infinity that connects all $r$ sheets. Note that
$\mathcal{R}_{r}=\mathbb{C}\setminus(-\infty,0]$ if $r$ is even, and
$\mathcal{R}_{r}=\mathbb{C}\setminus[0,\infty)$ if $r$ is odd.

The Riemann surface has genus zero and $z=w^{r}$ is a rational parametrization
of it. The rational function

$\displaystyle\Psi(w)=\frac{1}{rw^{r-1}(w-a^{1/r})},\qquad z=w^{r},$ (3.4)

is meromorphic on the Riemann surface with a simple pole at $z=a$, a pole of
order $r-1$ at $z=0$, and a zero of order $r$ at $z=\infty$. We use $\Psi_{j}$
to denote its restriction to $\mathcal{R}_{j}$. Explicitly, we then have

$\displaystyle\Psi_{1}(z)=\frac{1}{rz^{1-1/r}(z^{1/r}-a^{1/r})},\qquad
z\in\mathcal{R}_{1}=\mathbb{C}\setminus(-\infty,0]$ (3.5)

with the principal branch of the $r$th roots. On the other sheets we have the
same formula but with different choices of $r$th roots in $z^{1-1/r}$ and
$z^{1/r}$. We always use $a^{1/r}>0$.

The cuts $\Delta_{j}$ are oriented from left to right, and we use
$\Psi_{j,\pm}(s)$, $s\in\Delta_{j}$, to denote the limit of $\Psi_{j}(z)$ as
$z\to s$ with $\pm\operatorname{Im}z>0$. Then we define for $j=1,\ldots,r-1$,

$\displaystyle d\mu_{j}^{*}(s)=\frac{1}{2\pi
i}(\Psi_{j,+}(s)-\Psi_{j,-}(s))ds,\qquad s\in\Delta_{j}.$ (3.6)

This defines a real measure on $\Delta_{j}$ since
$\Psi_{j,-}(s)=\overline{\Psi_{j,+}(s)}$ for $s\in\Delta_{j}$, but a priori it
could be a signed measure. Suppose the density vanishes at an interior point
$s\in\Delta_{j}$. Then $\Psi_{j,\pm}(s)$ is real, which implies that $\Psi(w)$
is real for some $w\in\mathbb{C}$ with $w^{r}=s$. From the formula (3.4) for
$\Psi$ it then easily follows that $w$ is real. However, if $w$ would be real
and positive then $w^{r}$ would be on $[0,\infty)$ on the first sheet, and if
$w$ would be real and negative then $w^{r}$ would be on
$\mathbb{R}\setminus\Delta_{r-1}$ on the $r$th sheet. Since $w^{r}=s$ is on
one of the cuts, we have a contradiction and we see that the density (3.6)
does not vanish at an interior, and therefore has a constant sign.

We compute the total masses by contour integration as in [9]. We consider
$j=1$ first. Then by (3.6)

$\displaystyle\int d\mu_{j}^{*}(s)=\frac{1}{2\pi i}\int_{C}\Psi_{1}(s)ds,$

where $C$ is a contour that starts at $-\infty$ and follows the upper side of
the cut $\Delta_{1}=(-\infty,0]$, and goes back to $-\infty$ on the lower side
of the cut. We deform the contour to a big circle $|s|=R$, and we pick up a
residue condition from the pole at $s=a$. From (3.5) we calculate the residue
as

$\displaystyle\lim_{z\to a}(z-a)\Psi_{1}(z)=\frac{1}{ra^{1-1/r}}\lim_{z\to
a}\frac{z-a}{z^{1/r}-a^{1/r}}=1.$

Hence

$\displaystyle\int d\mu_{1}^{*}(s)=1-\lim_{R\to\infty}\frac{1}{2\pi
i}\int_{|s|=R}\Psi_{1}(s)ds$

with the circle $|s|=R$ oriented counterclockwise. Since for every
$j=1,\ldots,r-1$,

$\displaystyle\Psi_{j}(z)=\frac{1}{rz}+O\big{(}z^{-1-1/r}\big{)}\qquad\text{as}\
\ z\to\infty,$ (3.7)

which easily follows from (3.4), we find

$\displaystyle\int d\mu_{1}^{*}(s)=1-\frac{1}{r}.$ (3.8)

Now consider $2\leq j\leq r-1$. Then by (3.6) and the fact that
$\Psi_{j-1,\pm}=\Psi_{j,\mp}$ on $\Delta_{j-1}$,

$\displaystyle\int d\mu_{j-1}^{*}(s)-\int d\mu_{j}^{*}(s)=\frac{1}{2\pi
i}\int_{\Delta_{j-1}}\left(-\Psi_{j,-}(s)-\Psi_{j,+}(s)\right)ds$
$\displaystyle\hphantom{\int d\mu_{j-1}^{*}(s)-\int
d\mu_{j}^{*}(s)=}{}-\frac{1}{2\pi
i}\int_{\Delta_{j}}\left(\Psi_{j,+}(s)-\Psi_{j,-}(s)\right)ds$
$\displaystyle\hphantom{\int d\mu_{j-1}^{*}(s)-\int
d\mu_{j}^{*}(s)}{}=\frac{1}{2\pi
i}\int_{\mathbb{R}}\left(\Psi_{j,-}(s)-\Psi_{j,+}(s)\right)ds,$

since $\Delta_{j-1}\cup\Delta_{j}=\mathbb{R}$. Again by contour deformation
this is

$\displaystyle\int d\mu_{j-1}^{*}(s)-\int
d\mu_{j}^{*}(s)=\lim_{R\to\infty}\frac{1}{2\pi
i}\int_{|s|=R}\Psi_{j}(s)ds=\frac{1}{r},$

where we used (3.7). Together with (3.8) we conclude that

$\displaystyle\int d\mu_{j}^{*}(s)=1-\frac{j}{r},\qquad j=1,\ldots,r-1.$

Since the total masses are positive, and the densities of the measures do not
change sign, it now also follows that the measures are positive.

We introduce the Cauchy transforms of the measures

$\displaystyle F_{j}(z)=\int\frac{d\mu_{j}^{*}(s)}{z-s},\qquad
z\in\mathbb{C}\setminus\Delta_{j}.$

Then by a similar contour integration argument, where now we pick up a residue
contribution at $s=z$, while there is no contribution from infinity, we get

$\displaystyle F_{1}(z)=\frac{1}{2\pi
i}\int_{C}\frac{\Psi_{1}(s)}{z-s}ds=\frac{1}{z-a}-\Psi_{1}(z),\qquad
z\in\mathbb{C}\setminus(-\infty,0]$ (3.9)

and for $j=2,\ldots,r-1$,

$\displaystyle F_{j-1}(z)-F_{j}(z)=\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{\Psi_{j,-}(s)-\Psi_{j,+}(s)}{z-s}ds=\Psi_{j}(z),\qquad
z\in\mathbb{C}\setminus\mathbb{R},$ (3.10)

where $F_{r}(z)=0$. The identity $\Psi_{j,+}=\Psi_{j+1,-}$ on $\Delta_{j}$
then leads to

$\displaystyle F_{j,+}(x)+F_{j,-}(x)=F_{j-1}(x)+F_{j+1}(x),\qquad
x\in\Delta_{j},$ (3.11)

for $j=2,\ldots,r-1$. By (3.9) and (3.10), the identity (3.11) also holds for
$j=1$, if we agree that

$\displaystyle F_{0}(z)=\frac{1}{z-a}.$

The measures have a density that decays like $|s|^{-1-1/r}$ as $|s|\to\infty$.
This easily follows from the definitions (3.4) and (3.6). Then
$s\mapsto\log(1+s^{2})$ is integrable for these measures, and the usual
logarithmic potentials exist. By Sokhotskii–Plemelj formulas we have

$\displaystyle 2\frac{d}{dx}U^{\mu_{j}^{*}}(x)=F_{j,+}(x)+F_{j,-}(x),\qquad
x\in\Delta_{j}.$

Clearly also

$\displaystyle\frac{d}{dx}U^{\mu_{j\pm 1}^{*}}(x)=F_{j\pm 1}(x),\qquad
x\in\Delta_{j}.$

Then by integrating (3.11) we obtain

$\displaystyle
2U^{\mu_{j}^{*}}(x)=U^{\mu_{j-1}^{*}}(x)+U^{\mu_{j+1}^{*}}(x),\qquad
x\in\Delta_{j}.$ (3.12)

There is no constant of integration in (3.12) since

$\displaystyle U^{\mu_{i}^{*}}(x)=(1-\tfrac{i}{r})\log|x|+o(1)\qquad\text{as}\
\ |x|\to\infty,$

for each $i\in\\{j-1,j,j+1\\}$ and $\Delta_{j}$ is unbounded.

Thus we have reached the identity (3.12) that we aimed for, as discussed in
the beginning of the proof. It remains to verify (3.2) and (3.3).

By (3.12) we obtain for $j=1,\ldots,r-1$,

$\displaystyle I(\mu_{j}^{*})=\int
U^{\mu_{j}^{*}}d\mu_{j}^{*}=\frac{1}{2}\int\big{(}U^{\mu_{j-1}^{*}}+U^{\mu_{j+1}^{*}}\big{)}d\mu_{j}^{*}=\frac{1}{2}\big{(}I(\mu_{j-1}^{*},\mu_{j}^{*})+I(\mu_{j}^{*},\mu_{j+1}^{*})\big{)},$

which implies

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j}^{*})=\sum_{j=1}^{r-2}I(\mu_{j}^{*},\mu_{j+1}^{*})+\frac{1}{2}(I(\mu_{0}^{*},\mu_{1}^{*})+I(\mu_{r-1}^{*},\mu_{r}^{*})).$

We recall that $\mu_{0}^{*}=\delta_{a}$ and $\mu_{r}^{*}=0$ and we obtain

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j}^{*})-\sum_{j=1}^{r-2}I(\mu_{j}^{*},\mu_{j+1}^{*})=\frac{1}{2}I(\delta_{a},\mu_{1}^{*})=\frac{1}{2}\int\log\frac{1}{|x-a|}d\mu_{1}^{*}(x),$

which is the identity in (3.2).

Finally, we recall that by (3.9) and (3.5)

$\displaystyle
F_{1}(z)=\frac{1}{z-a}-\frac{1}{rz^{1-1/r}(z^{1/r}-a^{1/r})},\qquad
z\in\mathbb{C}\setminus(-\infty,0],$

which after integration leads to

$\displaystyle\int\log(z-s)d\mu_{1}^{*}(s)=\log(z-a)-\log\big{(}z^{1/r}-a^{1/r}\big{)}.$

The constant of integration vanishes since both sides behave like $(1-1/r)\log
z+o(1)$ as $z\to\infty$. Taking real parts we find (3.3). ∎

We next extend Proposition 3.1 from point masses $\delta_{a}$ with $a>0$ to
general measures with compact support on $(0,\infty)$.

######  Proposition 3.2.

Let $r\geq 2$ be an integer. Let $\mu$ be a probability measure on
$(0,\infty)$ with compact support. Then there is a unique vector
$(\mu_{1}^{*},\ldots,\mu_{r-1}^{*})$ of measures that minimizes the energy
functional

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j})-\sum_{j=1}^{r-2}I(\mu_{j},\mu_{j+1})-I(\mu,\mu_{1})$

among all vectors of measures satisfying for every $j=1,\ldots,r-1$,

  1. $i)$

$\operatorname{supp}(\mu_{j})\subset\Delta_{j}$,

  2. $ii)$

$\mu_{j}(\Delta_{j})=1-\frac{j}{r}$, and

  3. $iii)$

$I(\mu_{j})<+\infty$.

Moreover,

$\displaystyle\sum_{j=1}^{r-1}I(\mu_{j}^{*})-\sum_{j=1}^{r-2}I(\mu_{j}^{*},\mu_{j+1}^{*})-I(\mu,\mu_{1}^{*})$
$\displaystyle\qquad{}=-\frac{1}{2}I(\mu,\mu_{1}^{*})=-\frac{1}{2}\iint\log\left|\frac{x^{1/r}-y^{1/r}}{x-y}\right|d\mu(x)d\mu(y).$
(3.13)

###### Proof.

For $\mu=\delta_{a}$ this was done in Proposition 3.1.

Let $(\mu_{1}^{*}(a),\ldots,\mu_{r-1}^{*}(a))$ be the vector of measures that
we obtain from $\delta_{a}$ as in Proposition 3.1. Then for a general
probability measure $\mu$ on $(0,\infty)$ with compact support, we put

$\displaystyle\mu_{j}^{*}=\int\mu_{j}^{*}(a)d\mu(a),\qquad j=1,\ldots,r-1.$

These are well-defined positive measures satisfying i), ii) and iii) of the
proposition. The measures $\mu_{j}^{*}(a)$ have a density that decays as
$|x|^{-1-1/r}$ as $|x|\to\infty$, and the same will be true for the measures
$\mu_{j}^{*}$ since $\mu$ is compactly supported. Thus the logarithmic
potentials exist, and

$\displaystyle U^{\mu_{j}^{*}}=\int U^{\mu_{j}^{*}(a)}d\mu(a).$

The identity

$\displaystyle
2U^{\mu_{j}^{*}(a)}=U^{\mu_{j-1}^{*}(a)}+U^{\mu_{j+1}^{*}(a)}\qquad\text{on}\
\ \Delta_{j}$

holds for every $a>0$ by Proposition 3.1. Integrating this with respect to $a$
and using Fubini’s theorem, we obtain

$\displaystyle
2U^{\mu_{j}^{*}}=U^{\mu_{j-1}^{*}}+U^{\mu_{j+1}^{*}}\qquad\text{on}\ \
\Delta_{j}.$

As in the proof of Proposition 3.1 this leads to the first identity of (3.13).

For $j=1$ we get

$\displaystyle U^{\mu_{1}^{*}}(z)=\int
U^{\mu_{1}^{*}(a)}(z)d\mu(a)=\int\log\left|\frac{z^{1/r}-a^{1/r}}{z-a}\right|d\mu(a),$

see (3.3). Changing $z$ and $a$ to $x$ and $y$, and integrating over
$d\mu(x)$, we obtain the second identity of (3.13). ∎

##  4 Proofs of Theorems 1.1 and 1.2

Theorem 1.1 is the special case $q=1$ of Theorem 1.2 and so it suffices to
prove Theorem 1.2.

###### Proof.

Let $\nu_{0}$ be a probability measure on $(0,\infty)$ with compact support,
and let $(\nu_{-q+1},\ldots,\nu_{-1}$, $\nu_{0},\nu_{1},\ldots,\nu_{r-1})$ be
the minimizing vector of measures for (1.9) under the assumptions i), ii) and
iii) of Theorem 1.2, with $\nu_{0}$ fixed.

Then it is easy to see that $(\nu_{1},\ldots,\nu_{r-1})$ is the minimizer for
the vector energy problem of Proposition 3.2 with $\mu=\nu_{0}$, and hence by
(3.13),

$\displaystyle\sum_{j=1}^{r-1}I(\nu_{j})-\sum_{j=1}^{r-2}I(\nu_{j},\nu_{j+1})-I(\nu,\nu_{1})=-\frac{1}{2}\iint\log\left|\frac{x^{1/r}-y^{1/r}}{x-y}\right|d\nu_{0}(x)d\nu_{0}(y).$
(4.1)

Similarly, $(\nu_{-1},\ldots,\nu_{-q+1})$ is the minimizer for the vector
energy problem of Proposition 3.2 with $\mu=\nu_{0}$ and $q$ instead of $r$.
Thus by (3.13) again,

$\displaystyle\sum_{j=-q+1}^{-1}\\!I(\nu_{j})-\sum_{j=-q+1}^{-2}\\!I(\nu_{j},\nu_{j+1})-I(\nu_{-1},\nu_{0})=-\frac{1}{2}\iint\\!\log\left|\frac{x^{1/q}-y^{1/q}}{x-y}\right|d\nu_{0}(x)d\nu_{0}(y).\\!\\!\\!$
(4.2)

The identity also holds in case $q=1$, since then $\nu_{-1}^{*}=0$ and both
sides of (4.2) are $0$.

From (4.1) and (4.2) we find that for a fixed $\nu_{0}$, the minimum of the
energy functional (1.9) taken over all $\nu_{j}$, for $j=-q+1,-1$,
$j=1,\ldots,r-1$ satsifying items i), ii) of Theorem 1.2, is equal to

$\displaystyle
J(\nu_{-q+1},\ldots,\nu_{0},\ldots,\nu_{r-1})=I(\nu_{0})-\frac{1}{2}\iint\log\left|\frac{x^{1/q}-y^{1/q}}{x-y}\right|d\nu_{0}(x)d\nu_{0}(y)$
$\displaystyle\hphantom{J(\nu_{-q+1},\ldots,\nu_{0},\ldots,\nu_{r-1})=}{}-\frac{1}{2}\iint\log\left|\frac{x^{1/r}-y^{1/r}}{x-y}\right|d\nu_{0}(x)d\nu_{0}(y)+\int\tilde{V}(x)d\nu_{0}(x),$

which reduces to (1.7) with $\nu_{0}$ instead of $\nu$. Thus the component
$\nu_{0}^{*}$ of the minimizer for the vector energy (1.6) is also the
minimizer of (1.7) over probability measures on $[0,\infty)$.

We finally prove that $\nu_{0}^{*}$ has compact support. Let
$\rho=\nu_{-1}^{*}+\nu_{1}^{*}$. Then $\nu_{0}^{*}$ is the minimizer of

$\displaystyle\tilde{I}(\nu)-\tilde{I}(\nu,\rho)+\int\tilde{V}d\nu$ (4.3)

among probability measures $\nu$ on $[0,\infty)$.

If $x\mapsto\log(1+|x|^{2})$ would be integrable with respect to $\nu_{0}^{*}$
then it would also be the minimizer of

$\displaystyle
I(\nu)-\tilde{I}(\nu,\rho)+\int\big{(}\tilde{V}(x)+\log\big{(}1+|x|^{2}\big{)}\big{)}d\nu(x)$
$\displaystyle\qquad{}=I(\nu)+\int\left(\widehat{V}(x)-\int\log\frac{\sqrt{1+s^{2}}}{x-s}d\rho(s)\right)d\nu,$
(4.4)

which is a usual minimization problem for one measure with an external field

$\displaystyle\widehat{V}(x)+\int\log\frac{x-s}{\sqrt{1+s^{2}}}d\rho(s)$

that is continuous on $(0,\infty)$ (since $\widehat{V}$ is continuous and
$\rho$ is a measure on $(-\infty,0]$).

It is easy to see that $x-s>\sqrt{1+s^{2}}$ for $x>1$ and $s<0$. Thus
$\int\log\frac{x-s}{\sqrt{1+s^{2}}}d\rho(s)>0$ for $x>1$, and it follows from
(1.8) that

$\displaystyle\lim_{x\to\infty}\left(\widehat{V}(x)+\int\log\frac{x-s}{\sqrt{1+s^{2}}}d\rho(s)-\log\big{(}1+x^{2}\big{)}\right)=+\infty,$

which guarantees that (4.4) has a minimizer with compact support. This
minimizer also minimizes (4.3) and thus coincides with $\nu_{0}^{*}$ which
thus has compact support.

Theorem 1.2 is now fully proved. ∎

##  5 A final remark

We consider the minimization problem for (1.2) with $\theta=1/r$. From Theorem
1.1 we obtain the following result that gives conditions that guarantee that
the Cauchy transform of the minimizing measure is an algebraic function.

######  Proposition 5.1.

Let $\theta=1/r$ be rational and suppose the external field
$V\colon[0,\infty)\to\mathbb{R}$ is such that $V^{\prime}$ is a rational
function. Let $\mu^{*}$ be the probability measure that minimizes (1.2) among
all probability measures on $[0,\infty)$ and suppose that $\mu^{*}$ is
supported on a finite union of intervals Then its Cauchy transform

$\displaystyle F(z)=\int\frac{d\mu^{*}(s)}{z-s}$

is the solution of an algebraic equation of degree $r+1$. That is, there exist
rational functions $c_{j}(z)$ for $j=0,\ldots,r$, such that

$\displaystyle\sum_{j=0}^{r}c_{j}(z)F(z)^{j}+F(z)^{r+1}=0,\qquad
z\in\mathbb{C}.$

###### Proof.

We turn to the vector equilibrium problem from Theorem 1.1 and denote the
minimizing vector by $(\mu_{0}^{*},\ldots,\mu_{r-1}^{*})$ with
$\mu_{0}^{*}=\mu^{*}$. The variational conditions are

$\displaystyle 2U^{\mu^{*}}(x)-U^{\mu_{1}^{*}}(x)+V(x)\
\begin{cases}=\ell,&\text{for }x\in\operatorname{supp}(\mu^{*}),\\\
\geq\ell,&\text{for }x\in[0,\infty),\end{cases}$

and for $j=1,\ldots,r-1$,

$\displaystyle
2U^{\mu_{j}^{*}}(x)-U^{\mu_{j-1}^{*}}(x)-U^{\mu_{j+1}^{*}}(x)=0\qquad\text{for}\
\ x\in\Delta_{j},$

where $\mu_{r}^{*}=0$. We write

$\displaystyle F_{j}(z)=\int\frac{d\mu_{j}^{*}(s)}{z-s},\qquad j=1,\ldots,r,$

and we find by differentiating the variational conditions

$\displaystyle
F_{0,+}^{\prime}(x)+F_{0,-}^{\prime}(x)=F_{1}(x)+V^{\prime}(x),\qquad$
$\displaystyle x\in\operatorname{supp}(\mu^{*}),$ $\displaystyle
F_{j,+}^{\prime}(x)+F_{j,-}^{\prime}(x)=F_{j-1}(x)+F_{j+1}(x),\qquad$
$\displaystyle x\in\Delta_{j},$ (5.1)

for $j=1,\ldots,r-1$.

We construct a Riemann surface $\mathcal{R}$ with $r+1$ sheets
$\mathcal{R}_{j}$, $j=0,\ldots,r$ given by

$\displaystyle\mathcal{R}_{0}=\mathbb{C}\setminus\operatorname{supp}(\mu^{*}),$
$\displaystyle\mathcal{R}_{1}=\mathbb{C}\setminus((-\infty,0]\cup\operatorname{supp}(\mu^{*})),$
$\displaystyle\mathcal{R}_{j}=\mathbb{C}\setminus(\Delta_{j-1}\cup\Delta_{j}),\qquad\text{for}\
\ j=2,\ldots,r-1,$
$\displaystyle\mathcal{R}_{r}=\mathbb{C}\setminus\Delta_{r-1},$

where $\mathcal{R}_{0}$ is connected to $\mathcal{R}_{1}$ along
$\operatorname{supp}(\mu^{*})$ and $\mathcal{R}_{j}$ is connected to
$\mathcal{R}_{j+1}$ along $\Delta_{j}$ for $j=1,\ldots,r-1$ in the usual
crosswise manner. After adding points at infinity we obtain a compact Riemann
surface, since $\operatorname{supp}(\mu^{*})$ consists of a finite union of
intervals.

We define a function $\Psi$ on $\mathcal{R}$ by specifying on each of the
sheets

$\displaystyle\Psi(z)=\begin{cases}V^{\prime}(z)-F_{0}(z),&\text{for
}z\in\mathcal{R}_{0},\\\ F_{j-1}(z)-F_{j}(z),&\text{for
}z\in\mathcal{R}_{j},\quad j=1,\ldots,r.\end{cases}$

Then $\Psi$ is meromorphic on each of sheets (since $V^{\prime}$ is a rational
function). Moreover, the variational conditions (5.1) tell us that $\Psi$
extends to a meromorphic function on the full Riemann surface $\mathcal{R}$.
Then also $V^{\prime}-\Psi$ is a meromorphic function on $\mathcal{R}$ which
agrees with $F_{0}$ on the zero sheet. Therefore $F=F_{0}$ satisfies an
algebraic equation of degree $r+1$. ∎

Note that $\mu^{*}$ minimizes $I(\mu)+\int(V-U^{\mu_{1}^{*}})d\mu$ among all
probability measures $\mu$ on $[0,\infty)$, and the external field
$V-U^{\mu_{1}^{*}}$ is real analytic on $(0,\infty)$. If it were also real
analytic at $0$, then it would follow from results in [8] that $\mu^{*}$ is
supported on a finite union of intervals. Maybe the methods of [8] can be
adapted to the present situation, and then the assumption in Proposition 5.1
about the finite number of intervals would be unnecessary.

The Riemann surface in the proof of Proposition 5.1 has genus $0$ if and only
if $\operatorname{supp}(\mu^{*})=[0,a]$ for some $a>0$. This is the case if
$V(x)=x$, and for more general conditions see [7, Theorem 1.8]. We note that
this Riemann surface also appears in the paper of Forrester, Liu and Zinn-
Justin [12], see Fig. 1 in that paper.

### Acknowledgements

The author is supported by long term structural funding–Methusalem grant of
the Flemish Government, by KU Leuven Research Grant OT/12/073, by the Belgian
Interuniversity Attraction Pole P07/18, and by FWO Flanders projects G.0934.13
and G.0864.16.

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