A vector equilibrium problem for symmetrically located point charges on a sphere
AA vector equilibrium problem for symmetricallylocated point charges on a sphere
Juan G. Criado del Rey ∗ Arno B.J. Kuijlaars † Abstract
We study the equilibrium measure on the two dimensional sphere inthe presence of an external field generated by r + 1 equal point chargesthat are symmetrically located around the north pole. The support ofthe equilibrium measure is known as the droplet. The droplet has amotherbody which we characterize by means of a vector equilibriumproblem (VEP) for r measures in the complex plane.The model undergoes two transitions which is reflected in the sup-port of the first component of the minimizer of the VEP, namely thesupport can be a finite interval containing 0, the union of two intervals,or the full half-line. The two interval case corresponds to a droplet withtwo disjoint components, and it is analyzed by means of a genus oneRiemann surface. This paper deals with an electrostatic equilibrium problem for free chargeson the unit sphere S ⊂ R with logarithmic interaction under the influenceof a finite number of fixed point charges [7, 11, 16, 39]. Suppose there are r + 1 fixed charges at points p , . . . , p r on S , and each p j carries a charge a j >
0, leading to a charge distribution σ = r (cid:88) j =0 a j δ p j . (1.1) ∗ Department of Mathematics, Katholieke Universiteit Leuven, Belgium, Email:[email protected]. Supported by FWO Flanders project EOS 30889451. † Department of Mathematics, Katholieke Universiteit Leuven, Belgium, Email:[email protected]. Supported by long term structural funding-Methusalem grantof the Flemish Government, and by FWO Flanders projects EOS 30889451, G.0864.16 andG.0910.20. a r X i v : . [ m a t h . C A ] J u l hen there exists an equilibrium measure µ σ in the presence of the fixedcharges that is the unique probability measure on S that satisfies for someconstant (cid:96) , U µ σ + U σ = (cid:96), on D σ = supp( µ σ ) ,U µ σ + U σ ≥ (cid:96), on S , (1.2)where we use U µ ( x ) = (cid:90) log 1 (cid:107) x − y (cid:107) dµ ( y )to denote the logarithmic potential of a measure µ . The domain D σ is knownas the droplet, and it determines the measure µ σ since µ σ = ( λ ( D σ )) − λ D σ (1.3)where λ D denotes the restriction to D of the normalized Lebesgue measure λ on the sphere. It is known that λ ( D σ ) = 11 + σ ( S ) = 11 + (cid:80) rj =0 a j . (1.4)see e.g. [11, Appendix A].A motherbody (or a potential theoretic skeleton [25]) for D σ is a proba-bility measure σ ∗ supported on a one-dimensional subset of S (i.e., a curve,or a system of curves) such that for some constant (cid:96) ∗ , U σ ∗ = U µ σ + (cid:96) ∗ , on S \ D σ ,U σ ∗ ≥ U µ σ + (cid:96) ∗ , on S . (1.5)Motherbodies are connected to a variety of topics in applied complex analy-sis, such as quadrature domains and Schwarz functions [1, 12, 24, 34], partialbalayage and Hele-Shaw flows [23], orthogonal polynomials in the complexplane [3, 4, 38] and normal matrix models [5, 48].The aim of this paper is to construct such a motherbody by means of avector equilibrium problem in the special situation where the points are ina symmetric position around a distinguished point on the unit sphere, thatwithout loss of generality we can take as the north pole. More precisely, weassume that the distance to the north pole is the same for each point p j ,which means that the points are on a circle of constant latitude. On thiscircle the points are evenly distributed, like vertices of a regular r + 1-gon.We also assume a j = a, for j = 0 , . . . , r.
2n this situation we are able to compute the motherbody, which, becauseof rotational symmetry, is supported on r + 1 meridians (lines of constantlongitude) that connect the north and south poles. From the motherbodywe go on to construct the droplet D σ .With fixed points p , . . . , p r , the droplet and the support of the moth-erbody decrease as we increase a . We find three possible situations and thetransitions between them. • For small a >
0, the droplet is big and the complement S \ D σ consistsof r + 1 disjoint spherical caps, one around each of the points p j . Themotherbody is supported on the full meridians with a positive density. • For a first critical value a ,cr , the spherical caps are tangent to eachother. The density of the motherbody becomes zero at the points oftangency. • For a > a ,cr the droplet is no longer the complement of disjoint spher-ical caps. For a somewhat larger than a ,cr the droplet will have twoconnected components (provided r ≥ • For a second critical value a ,cr one of the components disappers. Ifthe points p j are in the northern hemisphere, then the componentcontaining the north pole disappears. Also the parts of the mother-body containing the north pole have disappeared at the second criticalvalue. • For larger a > a ,cr the droplet D σ is simply connected containingthe south pole (assuming again that the points p j are in the northernhemisphere). The support of the motherbody consists of r +1 segmentscontaining the south pole, one segment along each meridian. • As a → ∞ , the droplet and the support of the motherbody furthershrink to the south pole. r = 1 For r = 1 the two spherical caps are tangent at the north pole at the criticalvalue a ,cr . Then there is no second critical a -value since for each a > a ,cr the droplet is simply connected. The support of the motherbody is an3igure 1: Picture of the droplet (red region) in case r = 1 and a > a ,cr .The spherical caps centered at p and p with geodesic radii a/ (1 + 2 a )are also represented, as well as the support of the motherbody (dashed lineinside the droplet). The boundary of the droplet is mapped by stereographicprojection onto an ellipse in the complex plane [11].interval along the big circle that separates the two points p and p . SeeFigure 1 that is taken from [11] and compare also with [7, Figure 4]. Thissituation was analyzed in [11] and it was shown that the boundary of thedroplet is mapped by stereographic projection to an ellipse in the complexplane. This fact can also be deduced from earlier work by Gustafsson andTkachev in [26, Example 3].The approach of [11] is to first characterize the motherbody by meansof an equilibrium problem from logarithmic potential theory [44, 45]. Thisequilibrium problems asks for the minimizer of (cid:90) (cid:90) log 1 | x − y | dµ ( x ) dµ ( y ) + 2 (cid:90) V ( x ) dµ ( x ) (1.6)among probability measures µ on R , with V ( x ) = 1 + a x + b − ) − a x + b ) , (1.7)where ± ib , b >
1, are the images of the two points p , p under stereographicprojection onto the complex plane. The minimizer is calculated explicitly in[11, Theorem 1.6], see also [43]. The external field is only weakly admissible427, 47] and for a fixed b > a cr such that theminimizer µ V is compactly supported if and only if a > a cr . Out of theStieltjes transform of µ V a meromorphic function S is then constructed thatis shown to be the spherical Schwarz function of a certain domain Ω in thesense that its boundary is characterized by ∂ Ω : S ( z ) = ¯ z | z | . After pulling back to the sphere with inverse stereographic projection, thedomain Ω is then proved to give the droplet D σ and µ V gives the mother-body. In this paper we extend the approach of [11] to r + 1 points on the sphere.As in [11] we project onto the complex plane where we do all calculations.Instead of the equilibrium problem (1.6), (1.7) we study a vector equilibriumproblem for a vector of r measures. This will be described in section 2.1below. In this section we first describe what we aim to achieve in the complexplane.We move from the sphere to the complex plane by stereographic projec-tion, where the south pole is mapped to 0 and the north pole to ∞ . Thepoints p , . . . , p r are projected to r + 1 points with absolute value q − r +1 for some number q >
0. The projected points will be the solutions of theequation z r +1 + q = 0, namely p j (cid:55)→ q − r +1 e iθ j , θ j = πr + 1 + 2 jπr + 1 , for j = 0 , , . . . , r. (1.8)The case q < p j in the northern hemisphere, and q > θ j are chosen in such a way that the meridians separatingthe points p , . . . , p r at equal distances are mapped to the r + 1-star { z ∈ C | z r +1 ∈ [0 , ∞ ) } . The droplet D σ is mapped to a domain Ω ⊂ C ∪ {∞} , and µ σ is mapped toits pushforward on Ω which takes the form dµ Ω ( z ) = dA ( z ) tπ (1 + | z | ) (cid:12)(cid:12)(cid:12)(cid:12) Ω (1.9)5here dA ( z ) is the planar Lebesgue measure on C . and t = 11 + σ ( S ) = 11 + ( r + 1) a (1.10)The properties (1.2) translate into U µ Ω ( z ) + a log 1 | z r +1 + q − | + 1 + ( r + 1) a (cid:0) | z | (cid:1) (cid:40) = c , z ∈ Ω , ≥ c , z ∈ C . (1.11)for some constant c .The motherbody σ ∗ (that we are looking for in this paper and whoseexistence we do not a priori assume) satisfying (1.5) corresponds to a prob-ability measure µ ∗ on { z | z r +1 ∈ [0 , ∞ ) } with the property that U µ ∗ = U µ Ω + c , on C \ Ω ,U µ ∗ ≥ U µ Ω + c , on C , (1.12)for some other constant c . The aim of the paper is to construct the domainΩ and measures µ ∗ and µ Ω satisfying the conditions (1.11) and (1.12).The probability measures µ ∗ and µ Ω will be invariant under rotationsaround the origin over angle πr +1 . For our computations it will be conve-nient to remove the rotational symmetry, and change variables z (cid:55)→ z r +1 .Then µ ∗ will correspond to a probability measure µ on [0 , ∞ ), and µ Ω to aprobability measure µ U on the set U = { z r +1 | z ∈ Ω } (1.13)and µ U takes the form dµ U ( z ) = 1( r + 1) tπ dA ( z ) | z | rr +1 (cid:16) | z | r +1 (cid:17) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U . (1.14)which comes from applying the change of variables to (1.9)Our approach will be to construct µ first as the first component of theminimizer of a vector equilibrium problem (VEP) for r measures. Besides µ there will be further measures µ , . . . , µ r that play auxiliary roles. Theydo not have a direct interpretation for the problem at hand, though.In the next section we will state the VEP without trying to motivatethe form that it takes. It is actually by no means obvious that this VEP6s relevant for our problem, and it will be our main result that µ aftersymmetrization gives indeed a measure µ ∗ that can be identified as the imageof the motherbody under stereographic projection. However, for r = 1, theVEP is an equilibrium problem for one measure that, after symmetrization,can be identified with (1.7).The VEP gives rise to an algebraic structure and this will allow us tofind a domain U with a measure (1.14). Through (1.13) we find a domainΩ with rotational symmetry and the measure µ Ω as in (1.9). We prove thatit has the properties (1.11) and (1.12).The VEP depends on two parameters q > t ∈ (0 , Let r ≥ q >
0, 0 < t < E ( µ , µ , . . . , µ r ) = r (cid:88) j =1 I ( µ j ) − r − (cid:88) j =1 I ( µ j , µ j +1 )+ 1 − tt I (cid:0) µ , δ − q − (cid:1) − r + tt I (cid:0) µ r , δ ( − r q (cid:1) , (2.1)depending on r measures. Here δ − q − and δ ( − r q denote Dirac point masses.As usual we write I ( µ, ν ) = (cid:90) U µ dν = (cid:90) (cid:90) log 1 | x − y | dµ ( x ) dν ( y )for the mutual logarithmic energy of µ an ν , and I ( µ ) = I ( µ, µ ) for thelogarithmic energy of µ .Our aim is to minimize (2.1) over a vector of measures satisfying cer-tain conditions. We emphasize that a measure (without any adjective) willalways refer to a positive measure. We also encounter negative measuresor signed measures in this paper, but in such a context the adjective willalways be mentioned. 7 efinition 2.1. The vector equilibrium problem (VEP) asks to minimizethe energy functional (2.1) over vectors ( µ , . . . , µ r ) of measures subject tothe conditions(a) supp( µ j ) ⊂ ∆ j for every j , where∆ j = (cid:40) [0 , ∞ ) , if j is odd , ( −∞ , , if j is even , (2.2)(b) the total mass of µ j is µ j (∆ j ) = 1 + j − t , for j = 1 , . . . , r. (2.3)Throughout the paper we will write µ = (cid:18) − t (cid:19) δ − q − , µ r +1 = (cid:16) rt (cid:17) δ ( − r q . (2.4)Then (2.3) is also satisfied for j ∈ { , r + 1 } , but note that µ is a negativemeasure (since 0 < t < E ( µ , . . . , µ r ) = r (cid:88) j =1 I ( µ j ) − r (cid:88) j =0 I ( µ j , µ j +1 ) , (2.5)that includes µ and µ r +1 as well, but µ and µ r +1 remain fixed in the VEP.Vector equilibrium problems were first introduced by Gonchar and Rakh-manov in their study of Hermite-Pad´e approximation [21, 22], see also [42].They also appear in ensembles of random matrices that are related to mul-tiple orthogonal polynomials, see [2, 28] and references cited therein.The energy functional (2.5) involves an attraction between neighboringmeasures that is of Nikishin type, and this has appeared in a number ofsituations before. What is special is that the total masses (2.3) are in anarithmetic progression that is increasing with steps 1 /t . It is more commonthat the masses are in an arithmetic progression that decreases from 1 to 0see e.g. [17] and the examples in [2, 28].The VEP of Definition 2.1 is weakly admissible in the sense of [27] as weshow next. Lemma 2.2.
The vector equilibrium problem is weakly admissible. Thereis a unique minimizer, denoted ( µ , . . . , µ r ) . The measures µ , . . . , µ r havefull supports supp( µ j ) = ∆ j = ( − j − [0 , ∞ ) , for j = 2 , . . . , r. (2.6)8 roof. To check the conditions in Assumption 2.1 of [27], we write the energyfunctional (2.5) in the form (cid:88) ≤ i,j ≤ r c ij I ( µ i , µ j ) + r (cid:88) j =1 (cid:90) V j dµ j with c ij = i = j, − if | i − j | = 1 , , (2.7)and V j ( x ) = − U µ ( x ) = (1 − t ) log | x + q − | , if j = 1 , − U µ r +1 ( x ) = (1 + rt ) log | x − ( − r q | , if j = r, ≡ , otherwise . (2.8)The interaction matrix C = ( c ij ) is symmetric and positive definite, andeach V j is continuous on ∆ j , since − q − (cid:54)∈ ∆ and ( − r q (cid:54)∈ ∆ r .The prescribed total masses m j = µ j (∆ j ) from (2.3) come in an arith-metic progression which implies by (2.7) that r (cid:88) j =1 c ij m j = 0 , for i = 2 , . . . , r − , (2.9)and also r (cid:88) j =1 c ij m j = (cid:40) m = (1 − t ) , for i = 1 , m r +1 = (1 + rt ) , for i = r. (2.10)It follows from (2.8), (2.9) and (2.10) that, for every i = 1 , . . . , r , V i ( x ) − r (cid:88) j =1 c i,j µ j (∆ j ) log(1 + | x | ) → , as x ∈ ∆ j → ±∞ . Thus all conditions of Assumption 2.1 in [27] are satisfied, and the VEP isweakly admissible. Then there is a unique minimizer by [27, Corollary 2.7].Given the other measures, the problem for µ j (for 1 ≤ j ≤ r ) is tominimize I ( µ j ) − I ( µ j , µ j − + µ j +1 )9mong measures on ∆ j with total mass (2.3). Since µ j − + µ j +1 is a positivemeasure for j ≥
2, it follows that µ j is a balayage measure (see [45] for thenotion of balayage) µ j = 12 Bal ( µ j − + µ j +1 , ∆ j ) for j = 2 , . . . , r, (2.11)and µ j has full support for j ≥
2, see also (3.2) for the expression of thedensity of the balayage of a measure on ( −∞ ,
0] onto [0 , ∞ ). There is asimilar formula for the balayage of a measure on [0 , ∞ ) to ( −∞ ,
0] thatshows that it has indeed a full support.The balayage property (2.11) means that2 U µ j = U µ j − + U µ j +1 on ∆ j , for j = 2 , . . . , r, (2.12)and this will be important for us in what follows.The measure µ is the main player in the game. The argument in theproof of Lemma 2.2 leading to (2.11) does not work for j = 1, since µ isa negative measure. Therefore the balayage of µ + µ onto ∆ = [0 , ∞ ) isnot necessarily positive on the full half-line. However, if it is positive then(2.11) and (2.12) hold for j = 1 as well, and then also µ has a full support.It turns out that this happens for t sufficiently large (i.e., sufficiently closeto 1).Our first main result is about the structure of the support Σ = supp( µ )of µ . There are four possible cases that will be indicated with acronyms BIS = Bounded Interval Support,
UIS = Unbounded Interval Support,
TIS = Two Interval Support, and
FIS = Full Interval Support.In situations where we want to emphasize the dependence on t of thevarious notions that we introduced (and of others that are still to come), weappend a subscript t . Hence we write for example µ ,t , Σ ,t , and so on. Theorem 2.3.
Fix q > . Let ( µ , . . . , µ r ) be the minimizer of the vectorequilibrium problem depending on the parameter t ∈ (0 , . (a) There are four possible cases for Σ = supp( µ ) , depending on t ,namely there exist < x < x < ∞ such that either BIS : Σ =[0 , x ] , or UIS : Σ = [ x , ∞ ) , or TIS : Σ = [0 , x ] ∪ [ x , ∞ ) , or FIS : Σ = [0 , ∞ ) . (b) For each j = 1 , . . . , r the measure tµ j,t increases as a function of t ∈ (0 , . Suppose < q < . Then is always in the support of µ (and so UIS case does not occur for any t ∈ (0 , ). (d) The measure µ has a density that is real analytic on the interior of itssupport with a square-root vanishing at x in the BIS and
TIS cases,and at x in the UIS and
TIS cases. (e)
There exist constants c > and c ∞ > such that dµ ( x ) dx = c x − rr +1 (cid:16) O (cid:16) x r +1 (cid:17)(cid:17) as x →
0+ (2.13) in BIS , TIS and
FIS cases, and dµ ( x ) dx = c ∞ x − r +2 r +1 (cid:16) O (cid:16) x − r +1 (cid:17)(cid:17) as x → ∞ (2.14) in UIS , TIS and
FIS cases.
The proof of Theorem 2.3 is in section 3, except for the proof of part (e)which is in section 4.1.2.It follows from Theorem 2.3 that for 0 < q <
1, there are two criticalvalues 0 ≤ t ,cr ≤ t ,cr ≤ , (2.15)depending on q , such that we are in BIS case for 0 < t ≤ t ,cr , in the TIS case for t ,cr < t < t ,cr , and in FIS case for t ,cr ≤ t <
1. For r ≥
2, theinequalities in (2.15) are actually strict inequalities and each of the threepossible cases occurs for some values of t .Ultimately, t will be related by equation (1.10) to the strength a of thefixed charges on the sphere. The critical values t ,cr and t ,cr will correspondto a ,cr and a ,cr (in that order) that are used at the end of section 1.1. Remark 2.4.
There is a symmetry between q and q − that allows us torestrict attention to 0 < q < (cid:126)µ = ( µ , . . . , µ r ) be a vector of measures as in the VEP of Defini-tion 2.1. Let ν j be the image of µ j under the inversion x (cid:55)→ /x , i.e., ν j isthe measure on ∆ j with (cid:90) f dν j = (cid:90) f (cid:18) x (cid:19) dµ j ( x ) (2.16)for a function f on ∆ j . Then it is an easy calculation to show that I ( ν j , ν k ) = I ( µ j , µ k ) + (cid:90) log | x | dµ j ( x ) (cid:90) dµ k + (cid:90) log | x | dµ k ( x ) (cid:90) dµ j . (2.17)11 x x R (1) x x R (2) R (3) R in TIS case (for r = 2)Let us use E q to denote the energy functional (2.1) corresponding tothe parameter q >
0. Then using (2.17) and the total masses (2.3) of themeasures we find after straightforward calculations that E /q ( (cid:126)ν ) = E q ( (cid:126)µ ) − r ( r + 2 t − t log q. Thus whenever (cid:126)µ = ( µ , . . . , µ r ) is the minimizer of the VEP with parameter q , then (cid:126)ν = ( ν , . . . , ν r ) is the minimizer with parameter 1 /q .Due to this symmetry between q and q − , the support of µ is alwaysunbounded for q >
1, since for 0 < q <
BIS case, we then have the
UIS case for t up to the first critical value. It is then continued with the TIS case, andafter a second critical value with the
FIS case.The above also shows that for q = 1, the measure µ is invariant underthe inversion x (cid:55)→ /x . Then we do not have a BIS or UIS case, but westart with a
TIS case for t up to a critical value, followed by the FIS case.
In what follows we restrict to the case 0 < q <
1. Then 0 ∈ supp( µ ) byTheorem 2.3 (c) and we are in one of the BIS , TIS or FIS cases. We also let0 < t <
1. Our further results are based on the consideration of a Riemannsurface.
Definition 2.5.
The Riemann surface R (see Figure 2) has r + 1 sheets12 ( j ) , j = 1 , . . . , r + 1, where R (1) = C \ supp( µ ) , R ( j ) = C \ (supp( µ j − ) ∪ supp( µ j )) , for j = 2 , . . . , r + 1 , R ( r +1) = C \ supp( µ r ) (2.18)where ( µ , . . . , µ r ) is the unique minimizer for the VEP of Definition 2.1, andwe recall that supp( µ ) = Σ ⊂ [0 , ∞ ) and supp( µ j ) = ∆ j = ( − j − [0 , ∞ )for j = 2 , . . . , r . Sheet R ( j ) is connected to sheet R ( j +1) along the support of µ j in the usual crosswise manner for j = 1 , . . . , r . We also add two (in BIS case) or one (in other cases) points at infinity in order to obtain a compactRiemann surface R .A count of branch points, together with the Riemann-Hurwitz formula,see e.g. [46], shows that R has genus zero in the BIS and
FIS cases, whilethe genus is one in the
TIS case.The Stieltjes transform of the measure µ j is F j ( z ) = (cid:90) dµ j ( x ) z − x , z ∈ C \ supp( µ j ) . (2.19)This is also defined for j = 0 and j = r + 1 in which cases we have the simplerational functions F ( z ) = − tt ( z + q − ) , F r +1 ( z ) = r + tt ( z − ( − r q ) (2.20)We use the Stieltjes transforms to define a function on R . Definition 2.6.
The function Φ is defined on the Riemann surface via itsrestrictions Φ ( j ) , j = 1 , . . . , r + 1, to the various sheets, byΦ ( j ) ( z ) = tF j ( z ) − tF j − ( z ) , z ∈ R ( j ) , (2.21)for j = 1 , . . . , r + 1.Differentiating the identity (2.12) we obtain F j, + − F j, − = F j − + F j +1 on ∆ j = supp( µ j ) , for j = 2 , . . . , r, (2.22)which means in view of (2.21) that Φ ( j ) ± = Φ ( j +1) ∓ on supp( µ j ) for j =2 , . . . , r . Thus Φ is analytic across the cut connecting sheets R ( j ) and R ( j +1) j ≥
2. Φ is also analytic across the cut connecting sheets R (1) and R (2) ,as this follows from the variational condition associated with the VEP2 U µ = U µ + U µ + c on supp( µ ) , which upon differentation leads to (2.22) on supp( µ j ) for j = 1 as well. ThusΦ is meromorphic on R and it has a number of crucial properties that willbe discussed in section 4.1. U With the help of Φ we define a subset U of the complex plane that will leadto the droplet. Definition 2.7.
The set U ⊂ C ∪ {∞} is defined by U = { z ∈ C | (Im z ) · Im (cid:0) z Φ (1) ( z ) (cid:1) < } . (2.23)We write U t if we want to emphasize the dependence of U on the parameter0 < t < Theorem 2.8.
Let < q < be fixed. Then the following hold. (a) U is a closed set with the properties Σ ⊂ U and − q − . (b) For < t ≤ t ,cr (the BIS case), U is a bounded simply connected set. (c) For t ,cr < t < t ,cr (the TIS case), U consists of two disjoint com-ponents: a bounded component containing [0 , x ] and an unboundedcomponent containing [ x , ∞ ) . The complement C \ U is a boundeddoubly connected domain. (d) For t ,cr ≤ t < (the FIS case), U is unbounded and connected. Thecomplement C \ U is bounded and simply connected. (e) t (cid:55)→ U t is increasing with t . (f) z Φ (1) ( z ) is real-valued on the boundary ∂U and z Φ (1) ( z ) = | z | r +1 | z | r +1 for z ∈ ∂U. (2.24)The proof of Theorem 2.8 is in section 4. See Figure 3 for plots of U inthe three cases. 14 x − q − BIS case0 x x TIS case − q − FIS case − q − Figure 3: Domain U (shaded region) in the three cases. The blue linedenotes the support of µ and it is located within U , while − q − is locatedoutside U . Ω with spherical measure We now introduce r + 1 fold symmetry. Definition 2.9.
We define a domain Ω (see Figure 4)Ω = { z ∈ C | z r +1 ∈ U } , (2.25)and a function S ( z ) = z r Φ (1) (cid:0) z r +1 (cid:1) (2.26)which we call the spherical Schwarz function of ∂ Ω.We call S the spherical Schwarz function because of the property S ( z ) = ¯ z | z | for z ∈ ∂ Ω , (2.27)15 IS case TIS case
FIS caseFigure 4: Domain Ω (shaded region) in the three cases for r = 2. The bluelines are the support of µ ∗ .which follows from (2.24) and the definitions in Definition 2.9. It readilyfollows from (2.27) that S ( z )1 − zS ( z ) = z for z ∈ ∂ Ω, and so S ( z )1 − zS ( z ) is the usualSchwarz function of ∂ Ω, and the two notions are very much intertwined, seealso the paper [13] on vertex dynamics on the sphere.Then S is defined and meromorphic on { z ∈ C | z r +1 (cid:54)∈ supp( µ ) } withpoles at the solutions of z r +1 = − q − , with the behavior zS ( z ) → z → ∞ . Also S has an analytic continuation to a meromorphic function ona compact r + 1 sheeted Riemann surface where S ( z ) = z r Φ ( j ) ( z r +1 ) on the j th sheet. This analytic continuation has poles on the ( r + 1)st sheet givenby the solutions of z r +1 = ( − r q .We next define the two measures µ Ω and µ ∗ .16 efinition 2.10. We define a measure µ Ω on Ω by dµ Ω ( z ) = 1 πt dA ( z )(1 + | z | ) (cid:12)(cid:12)(cid:12)(cid:12) Ω , (2.28)and µ ∗ as the unique measure on the r + 1 star { z | z r +1 ∈ R + } (2.29)that is invariant under rotation z (cid:55)→ e πir +1 z and whose pushforward under z (cid:55)→ z r +1 is equal to µ (the first component of the minimizer of the VEP).This leads to the final main result of the paper. Theorem 2.11. (a) µ Ω is a probability measure on Ω , and µ ∗ is a prob-ability measure on (2.29) . (b) t (cid:55)→ Ω t , t (cid:55)→ tµ Ω ,t and t (cid:55)→ tµ ∗ t are increasing for t ∈ (0 , . (c) There is a constant c = c ,t such that U µ Ω ( z ) − − t ( r + 1) t log | z r +1 + q − | + 12 t log (cid:0) | z | (cid:1) (cid:40) = c , z ∈ Ω , ≥ c , z ∈ C . (2.30) If Ω is unbounded, then c = 0 . (d) There is a constant c = c ,t such that U µ Ω ( z ) − U µ ∗ ( z ) (cid:40) = c , z ∈ C \ Ω , ≤ c , z ∈ C . (2.31) If Ω is bounded, then c = 0 . The proof of Theorem 2.11 is in section 5.Parts (c) and (d) of Theorem 2.11 tell us that the equations (1.11) and(1.12) are satisfied provided a = 1 − t ( r + 1) t (2.32)which agrees with (1.10). Thus, as already explained, Theorem 2.11 showsthat the image of Ω under inverse stereographic projection is the droplet D σ on the unit sphere, and the pullback of µ ∗ is the motherbody σ ∗ .17 emark 2.12. The domain Ω t increases with t according to part (b) ofTheorem 2.11. It is an instance of Laplacian growth (or Hele-Shaw flow) inthe spherical metric. We refer to [25] and the references therein for more onthe interesting topic of Laplacian growth and its many connections.Figure 4 contains a plot of Ω in the various cases for the value r = 2.It is interesting to note that in the BIS case Ω coincides with the dropletin the normal matrix model with a cubic potential, see e.g. [4, 18, 48]. Theeigenvalues in this random matrix model tend to the droplet with a uniformdensity (in contrast to (2.28) which is uniform in the spherical metric),and the zeros of related orthogonal polynomials tend to the motherbody.The limiting zero counting measure is characterized by a vector equilibriumproblem in [4] that is however different from the VEP of Definition 2.1, seealso [31] for the case r ≥
3. Our VEP can be seen as a spherical analoguefrom the VEPs in [4, 31].In the normal matrix model with a cubic potential the droplet grows upto a critical time and then cusps appear on the boundary of the droplet thatcause a breakdown of the model, (see however [33, 35, 36, 37] for continu-ations beyond breakdown). In our model there is no breakdown since thetransition to the
TIS takes place before we reach the cusp situation.
Given the second component µ of the solution of the VEP of Definition2.1, µ is the probability measure µ on [0 , ∞ ) that minimizes I ( µ ) − I ( µ, σ )where σ = µ + µ is a signed measure with integral (cid:82) σ = 2. Part (a) ofTheorem 2.3 will follow from the following more general result, where it isimportant that the negative part of σ is a Dirac point mass. Note that σ used in this section is not related to σ from (1.1). Proposition 3.1.
Suppose σ = − Aδ − q − + σ + where A > and σ + is ameasure on ( −∞ , with A < (cid:82) dσ + < ∞ . Then there is a unique µ on [0 , ∞ ) that minimizes I ( µ ) − I ( µ, σ ) among all measures on [0 , ∞ ) with (cid:82) dµ = (cid:82) σ . The support Σ = supp( µ ) takes one of the forms described in Theorem 2.3 (a), namely Σ is a boundedinterval [0 , x ] containing , an unbounded interval [ x , ∞ ) not containing , the disjoint union of two intervals [0 , x ] ∪ [ x , ∞ ) , or the full half-line [0 , ∞ ) . The minimization problem in Proposition 3.1 is again weakly admissible,and there is a unique minimizer µ . If we relax the condition that µ is ameasure and also allow signed measures, then the minimizer is the balayage ν = 12 Bal( σ, [0 , ∞ )) (3.1)which is known to have the density dνdx = 12 π √ x (cid:90) −∞ (cid:112) | s | x − s dσ ( s ) , < x < ∞ . (3.2)If the density (3.2) happens to be non-negative on [0 , ∞ ), then µ = ν solvesthe minimization problem and supp µ = [0 , ∞ ).If ν is not a positive measure, then we use the idea of iterated balayage[15, 30]. This method is based on the fact that µ ≤ ν + where ν + is thepositive part of ν in its Jordan decomposition ν = ν + − ν − . In particular supp( µ ) ⊂ supp( ν + ), see [30, Lemma 3]. With this informa-tion we can restrict the minimization problem to measures supported onsupp( ν + ), and if we also allow signed measures then the minimum is at-tained by ν + − Bal( ν − , supp( ν + )) . If this happens to be a positive measure then it is equal to µ , and we canstop. Otherwise we repeat the above step, which leads to the followingiterative procedure.We put ν = ν , and iteratively for k = 1 , , . . . , we write ν k = ν + k − ν − k where ν + k and ν − k are the positive and negative parts of ν k , and we define ν k +1 = ν + k − Bal( ν − k , supp( ν + k )) . (3.3)The convergence properties of the sequence ( ν k ) k are not fully understood,but in cases where we can control the supports of the measures we will havethat ν − k → ν + k → µ as k → ∞ .Under the conditions of Proposition 3.1 we can indeed control the sup-ports, and we will show that for each k the support of ν + k takes one ofthe forms stated in the proposition, namely supp( ν + k ) is either a bounded19nterval [0 , x ,k ], an unbounded interval [ x ,k , ∞ ), a union of two intervals[0 , x ,k ] ∪ [ x ,k , ∞ ), or the full half-line [0 , ∞ ). Since the supports are de-creasing if k increases, the two sequences ( x ,k ) and ( x ,k ) are either finite(maybe even empty), or else they monotonically converge to limits x and/or x . In this way we will be able to show that supp( µ ) has one of the formsin the proposition.In the first step we show that the support of ν + = ν +1 has the requiredform. ν + For σ as in the statement of Proposition 3.1, the density (3.2) of ν takes theform v ( x )2 π √ x , with v ( x ) = (cid:90) ∞ dρ ( s ) x + s − A x + q − , < x < ∞ , (3.4)where we put A = A √ q > dρ ( s ) = √ s dσ + ( − s ). Lemma 3.2.
In the above setting the following hold. (a) v has at most two zeros in (0 , ∞ ) . (b) There exist ≤ x ≤ x ≤ ∞ such that supp( ν − ) = [ x , x ] and supp( ν + ) = (0 , x ) ∪ ( x , ∞ ) . Proof. (a) Suppose, to get a contradiction, that 0 < x < x < x < ∞ arethree zeros of v . Let s = q − and write v = f − f + f with f ( x ) = (cid:90) s dρ ( s ) x + s , f ( x ) = Ax + s , f ( x ) = (cid:90) ∞ s dρ ( s ) x + s . (3.5)Then by (3.5) and the multilinearity of the determinantdet [ f k ( x j )] j,k =0 = A (cid:90) s dρ ( s ) (cid:90) ∞ s dρ ( s ) det (cid:20) x j + s k (cid:21) j,k =0 . (3.6)There is an explicit formula for the determinant (Cauchy determinant)det (cid:20) x j + s k (cid:21) j,k =0 = (cid:81) ≤ j 20n the integral in (3.6) we have 0 < s < s < s , and since also 0 < x 0. Since A > f k ( x j )] > f k ( x j )] j,k =0 is invertible.However, since v = f − f + f by (3.4) and (3.5), and since v ( x j ) = 0for j = 0 , , 2, it follows that f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) f ( x ) − = v ( x ) v ( x ) v ( x ) = and this is a contradiction, since the matrix is invertible.(b) From part (a) we know that v has at most two zeros in (0 , ∞ ). Bycontinuity, v can also have at most two sign changes in (0 , ∞ ).If v has no sign changes then v ≥ , ∞ ), since due to the fact that σ has the density (3.4) and (cid:90) dσ = (cid:90) ∞ v ( x )2 π √ x dx > ≤ 0. Then we take x = x in the lemma.If v has one sign change, say at x ∗ > 0, and if there is no other zero of v , then v is either > , x ∗ ) and < x ∗ , ∞ ), or vice versa. In theformer case we take x = x ∗ and x = ∞ and in the latter case we take x = 0 and x = x ∗ . If there is another zero in (0 , ∞ ), then the inequalityis not strict at this one zero, but we still take x and x as above, and theconclusion of part (b) holds true if v has one sign change. [It is actually notpossible that there is another zero, but we do not need this fact.]If v has two sign changes, say at 0 < x < x < ∞ , then v is eitherpositive on (0 , x ), negative on ( x , x ), and positive again on ( x , ∞ ), orthe other way around negative on (0 , x ), positive on ( x , x ) and negative on( x , ∞ ). [Now we can be sure that the inequalities are strict since there areno more than two zeros by part (a).] The latter possibility cannot happen,which we can see by adding (cid:15)δ − (cid:15) to ν for some small (cid:15) > 0. Part (a)continues to apply and it follows that v (cid:15) ( x ) = v ( x ) + (cid:15)x + (cid:15) (3.8)has at most two sign changes on (0 , ∞ ). Since v (cid:15) (0) > (cid:15) > 0, the set where v (cid:15) < (cid:15) → v is negative on (0 , x ) ∪ ( x , ∞ ).This proves part (b) of the lemma in all cases.21 .1.3 Second step: Monotonicity of v on supp( ν + )In order to make the induction step in the iterated balayage argument thatfollows we need the following behavior of v on the parts where it is positive. Lemma 3.3. Under the same conditions as in Lemma 3.2 where we let x , x be as in part (b) of Lemma 3.2, the following hold. (a) v (cid:48) and ( xv ) (cid:48) have at most two zeros in (0 , ∞ ) . (b) If < x < x then x (cid:55)→ v ( x ) is strictly decreasing for x ∈ (0 , x ) . (c) If x < x < ∞ then x (cid:55)→ xv ( x ) is strictly increasing for x ∈ ( x , ∞ ) .Proof. (a) The proof is similar to the proof of part (a) of Lemma 3.2. Notethat from (3.4) v (cid:48) ( x ) = − (cid:90) ∞ dρ ( s )( x + s ) + A ( x + q − ) . (3.9)We let s = q − and write v (cid:48) = f (cid:48) − f (cid:48) + f (cid:48) with f , f , f as in (3.5) andthendet (cid:2) f (cid:48) j ( x k ) (cid:3) j,k =0 = − A (cid:90) s dρ ( s ) (cid:90) ∞ s dρ ( s ) det (cid:20) x k + s j ) (cid:21) j,k =0 . (3.10)A Maple calculation shows thatdet (cid:20) x k + s j ) (cid:21) j,k =0 = (cid:81) ≤ j 0, and it follows that (3.10) is negative, and inparticular non-zero, whenever 0 < x < x < x . Thus v (cid:48) cannot have morethan two positive zeros.The proof for ( xv ) (cid:48) is similar, since( xv ) (cid:48) = (cid:90) ∞ sdρ ( s )( x + s ) − A q ( x + q − ) < x < x . Then v ( x ) = 0 and v ( x ) ≥ x ∈ (0 , x )by Lemma 3.2 (b). Also v ( x ) ≤ x ∈ ( x , x ).Since v ( x ) → x → ∞ , there is a global negative minimum, say at x ∗ > x , where the derivative vanishes and changes sign. According to part(a), there is at most one other sign change of the derivative. If this werein the interval (0 , x ) then it would correspond to a local maximum of v on the interval (0 , x ). Then we modify v to v (cid:15) as in (3.8) in the proof ofLemma 3.2. Part (a) applies to v (cid:15) and it follows that v (cid:48) (cid:15) has at most two signchanges. For (cid:15) > x ∗ , say at x ∗ ( (cid:15) ),and v (cid:15) has its global minimum there. Since v (cid:48) (cid:15) (0) < (cid:15) > v (cid:48) (cid:15) in (0 , x ∗ ( (cid:15) )), and letting (cid:15) → 0+ wefind that v (cid:48) has no sign change in (0 , x ∗ ). Thus v is strictly decreasing in(0 , x ) as claimed in part (b).(c) The proof for part (c) is similar. Suppose x < x < ∞ , so that v ( x ) = 0 and v ( x ) ≥ x ∈ ( x , ∞ ) by Lemma 3.2 (b).Since xv ( x ) → x → 0, there is a global minimum of x (cid:55)→ xv ( x ), at x ∗ ∈ ( x , x ) say, where the derivative is zero and changes sign. There isat most one more zero by part (a). If there were a sign change of ( xv ) (cid:48) in( x , ∞ ), then that would give us a maximum of x (cid:55)→ xv ( x ) on ( x , ∞ ). Weagain modify v to v (cid:15) as in (3.8). Part (a), applied to v (cid:15) , tells us that ( xv (cid:15) ) (cid:48) has at most two sign changes on (0 , ∞ ). For small (cid:15) > x ∗ , say at x ∗ ( (cid:15) ) where xv (cid:15) ( x ) has its global minimum. Sincelim x →∞ ( xv (cid:15) ( x )) (cid:48) > (cid:15) > 0, this derivative then has no sign change in ( x ∗ ( (cid:15) ) , ∞ )and therefore xv (cid:15) ( x ) is strictly increasing in ( x ∗ ( (cid:15) ) , ∞ ). Letting (cid:15) → 0+ itfollows that xv ( x ) increases strictly in ( x ∗ , ∞ ), and a fortiori in ( x , ∞ ). In the final step we use the iterated balayage to complete the proof of Propo-sition 3.1. We take ν = ν where ν is the signed measure on [0 , ∞ ) withdensity (3.2). If ν ≥ ν = µ and we are in the full interval support( FIS ) case Σ = [0 , ∞ ).In the rest of the proof we assume that ν is not a positive measure.Then iteratively we construct the sequence ( ν k ) k as in (3.3). Inductively wethen have (cid:82) dν k = (cid:82) dν = (cid:82) σ and µ ≤ ν + k for every k , and in particularΣ ⊂ supp( ν + k ) . (cid:82) dν + k ) k decreases and if (cid:82) dν − k tends to 0 as k → ∞ , then ν + k → µ in the sense of weak ∗ convergence of measures on [0 , ∞ ]. In thepresent situation (with the help of Lemmas 3.2 and 3.3) we can prove thatthis is indeed the case. Lemma 3.4. For every k we have (a) supp( ν + k ) = (0 , x ,k ) ∪ ( x ,k , ∞ ) for some ≤ x ,k < x ,k ≤ ∞ , while supp( ν − k ) ⊂ [ x ,k , x ,k ] , and (b) ν k has a density v k ( x )2 π √ x where x (cid:55)→ v k ( x ) strictly decreases on (0 , x ,k ) and x (cid:55)→ xv k ( x ) strictly increases on ( x ,k , ∞ ) . Assuming that Lemma 3.4 holds, we complete the proof of Proposi-tion 3.1 as follows. The measures ( ν + k ) converges to µ , and supp( µ ) =(0 , x ) ∪ ( x , ∞ ) where x = lim k x ,k and x = lim k x ,k . This establishesProposition 3.1. Proof of Lemma 3.4. For k = 1, the statements (a) and (b) are containedin Lemmas 3.2 and 3.3.Suppose the lemma holds for a certain k ≥ 1. Let us assume that0 < x ,k < x ,k < ∞ . We use the fact that the balayage of a delta mass δ t at t ∈ ( x ,k , x ,k ) onto [0 , x ,k ] ∪ [ x ,k , ∞ ) has the density c ( t )2 π √ x x + b ( t ) | x − t | (cid:112) ( x − x ,k )( x − x ,k )with positive constants b ( t ) > c ( t ) > 0. Then Bal( ν − k , supp( ν + k )) hasthe density1 π √ x (cid:112) ( x − x ,k )( x − x ,k ) (cid:90) c ( t ) x + b ( t ) | x − t | dν − ( t ) , x ∈ (0 , x ,k ) ∪ ( x ,k , ∞ ) . In view of (3.3) and the induction hypothesis we then obtain that ν k +1 hasthe density v k +1 ( x )2 π √ x with v k +1 ( x ) = v k ( x ) − (cid:112) ( x − x ,k )( x − x ,k ) (cid:90) x ,k x ,k c ( t ) x + b ( t ) | x − t | dν − k ( t ) , (3.11)for x ∈ (0 , x ,k ) ∪ ( x ,k , ∞ ). Also by the induction hypothesis v k ( x ) isstrictly decreasing on (0 , x ,k ). The other term in the right-hand side of(3.11) (including the minus-sign) is also decreasing on (0 , x ,k ), since each24f the factors √ x ,k − x , √ x ,k − x , x + b ( t ), and t − x is positive and strictlyincreasing for x ∈ (0 , x ,k ) as 0 < x ,k < t < x ,k and b ( t ) > 0. Also notethat c ( t ) dν − k ( t ) is a positive measure on [ x ,k , x ,k ].Thus v k +1 is strictly decreasing on (0 , x ,k ). Then it is either fully neg-ative on (0 , x ,k ), in which case we take x ,k +1 = 0, or v k +1 is positive onsome interval (0 , x ,k +1 ) with 0 < x ,k +1 < x ,k and v k +1 is negative on( x ,k +1 , x ,k ).Similar arguments show that xv k +1 ( x ) is strictly increasing on ( x ,k , ∞ ).Here we need to observe that each of the factors √ x √ x − x ,k , √ x √ x ,k − x , and x + b ( t ) t − x decreases on ( x ,k , ∞ ). Thus v k +1 is either fully negative there, in which casewe put x ,k +1 = ∞ , or v k +1 is positive on some interval ( x ,k +1 , ∞ ) with x ,k < x ,k +1 < ∞ and v k +1 is negative on ( x ,k , x ,k +1 ).Parts (a) and (b) of the lemma are thus proved for k + 1 in case 0 0] with 0 < (cid:82) dσ < ∞ we define M ( σ ) = µ as the measure on [0 , ∞ ) that minimizes I ( µ ) − I ( µ, σ ) (3.12)among µ ≥ (cid:82) dµ = (cid:82) dσ .(b) Similarly, for a signed measure σ on [0 , ∞ ) with 0 < (cid:82) dσ < ∞ wedefine (cid:102) M ( σ ) = µ as the measure on (0 , −∞ ] that minimizes (3.12)among µ ≥ (cid:82) dµ = dσ .(c) Consider vectors (cid:126)ν = ( ν , . . . , ν r +1 ) of signed measures of length r + 2,such that ν j is supported on ( − j [0 , ∞ ) for j = 0 , , . . . , r + 1 and0 < (cid:82) dν j − + (cid:82) dν j +1 < ∞ for j = 1 , . . . , r . For such (cid:126)ν we define M j (cid:126)ν = ( ν , . . . , ν j − , (cid:98) ν j , ν j +1 , . . . , ν r +1 ) , j = 1 , . . . , r, (3.13)where (cid:98) ν j = (cid:40) M ( ν j − + ν j +1 ) , if j is odd , (cid:102) M ( ν j − + ν j +1 ) , if j is even , M and (cid:102) M as defined in parts (a) and (b).Some remarks are in order. Remark 3.6. (a) The measures M ( σ ) and (cid:102) M ( σ ) in parts (a) and (b) ofDefinition 3.5 are minimizers of weakly admissible equilibrium prob-lems. The minimizers uniquely exist [27].(b) If σ ≥ µ = M ( σ ) is the balayage measure µ = Bal( σ, [0 , ∞ )).In this case we have a monotonicity result0 ≤ σ ≤ (cid:101) σ = ⇒ M ( σ ) ≤ M ( (cid:101) σ ) (3.14)for measures σ and (cid:101) σ on ( −∞ , ≤ σ ≤ (cid:101) σ = ⇒ (cid:99) M ( σ ) ≤ (cid:99) M ( (cid:101) σ ) (3.15)for measures σ and (cid:101) σ on [0 , ∞ ).(d) The maps are positive homogeneous in the sense that M ( cσ ) = cM ( σ ), (cid:99) M ( cσ ) = c (cid:99) M ( σ ) and M j ( c(cid:126)ν ) = cM j ( (cid:126)ν ) if c > µ ,t , . . . , µ r,t ) is the solution of the VEP of Definition 2.1 for some q > t ∈ (0 , (cid:126)µ t = ( µ ,t , µ ,t , . . . , µ r,t , µ r +1 ,t ), then M j ( (cid:126)µ t ) = (cid:126)µ t , j = 1 , . . . , r. (3.16)That is, (cid:126)µ t is a common fixed point for the mappings M j . It is theonly common fixed point among vectors (cid:126)µ with µ and µ r +1 given by(2.4). M We are going to apply M only to positive measures and to signed measureswhose negative part is a single point mass at − q − (as in Proposition 3.1).We need the extension of the monotonicity result (3.14) to such signed mea-sure. It could be that the monotonicity result is valid more generally, butwe do not consider it here since this is all we need for our present purposes.For such signed measures σ we have the information about the supportsof M ( σ ) from Proposition 3.1, and we also rely on the iterated balayage thatwas used in the proof of Proposition 3.1.26 emma 3.7. Let σ ≤ (cid:101) σ be signed measure on ( −∞ , with < (cid:82) dσ < (cid:82) d (cid:101) σ < ∞ whose negative parts are single point masses at − q − only. Then M ( σ ) ≤ M ( (cid:101) σ ) .Proof. Under the assumptions of the lemma, the signed measures take theform σ = − Aδ − q − + σ + and (cid:101) σ = − (cid:101) Aδ − q − + (cid:101) σ + with A ≥ (cid:101) A ≥ 0, and0 ≤ σ + ≤ (cid:101) σ + . We write µ = M ( σ ) and (cid:101) µ = M ( (cid:101) σ ).We recall the iterated balayage algorithm from the proof of Proposi-tion 3.1, see in particular Lemma 3.4, and we apply it to the signed measure (cid:101) σ . That is, we start with ν = Bal( (cid:101) σ, [0 , ∞ )), and from there we constructthe sequence ( ν k ) k inductively by ν k +1 = ν + k − Bal( ν − k , supp( ν + k )) , k = 1 , , . . . . Then ( ν k ) converges to (cid:101) µ = M ( (cid:101) σ ) as was shown in the proof of Lemma 3.4.Next we define a second sequence ( ρ k ) k by ρ = Bal( σ, [0 , ∞ )), and ρ k +1 = ρ + k − Bal( ρ − k , supp( ν + k )) , k = 1 , , . . . . (3.17)Since σ ≤ (cid:101) σ we have ρ ≤ ν , and then by induction it easily follows that ρ k ≤ ν k for every k . Note that we deviate from the earlier construction bytaking in (3.17) the balayage of ρ + k onto supp( ν + k ) and not onto supp( ρ + k ).Since supp( ν + k ) ⊃ supp( ρ + k ), we however still find (by induction) that µ = M ( σ ) ≤ ρ + k for every k . Then ρ ∞ = lim k →∞ ρ k is a signed measure with ρ ∞ ≤ (cid:101) µ and µ ≤ ρ + ∞ . Thus µ ≤ (cid:101) µ as claimed in the lemma. M -convexity We need two more definitions. Note that M -convexity is not a standardterminology, but it is introduced here to help the exposition. Definition 3.8. Let (cid:126)ν be as in Definition 3.5 (c), and let M j be as in (3.13).Then we say that (cid:126)ν is M -convex if (cid:126)ν ≤ M j ( (cid:126)ν ) for every j = 1 , . . . , r. Definition 3.9. The set M q contains those vectors (cid:126)ν = ( ν , ν , . . . , ν r +1 )satisfying • ν j is a positive measure on ( − j − [0 , ∞ ) for j = 1 , . . . , r + 1, • ν = − Aδ − q − for some A < (cid:82) dν .27hen we have the following properties. Lemma 3.10. Suppose (cid:126)ν ∈ M q . (a) Then M j ( (cid:126)ν ) ∈ M q for every j = 1 , . . . , r . (b) If (cid:126)ν ≤ (cid:126)ρ ∈ M q then M j ( (cid:126)ν ) ≤ M j ( (cid:126)ρ ) for every j = 1 , . . . , r . (c) If (cid:126)ν is M -convex then so is M j (cid:126)ν for every j = 1 , . . . , r . (d) If (cid:126)ν is M -convex and c > then (cid:126)ν + ( cδ − q − , , , . . . , is M -convex. (e) If (cid:126)ν is M -convex and ρ ≥ is a measure on ( − r [0 , ∞ ) then (cid:126)ν +(0 , , . . . , , ρ ) is M -convex.Proof. (a) Obvious.(b) This follows from the monotonicity of M and (cid:102) M on positive measures,see (3.14) and (3.15), and the monotonicity of M on signed measures whosenegative part only contains a point mass at − q − , see Lemma 3.7.(c) Since (cid:126)ν is M -convex we have (cid:126)ν ≤ M k (cid:126)ν for every k . The maps M k and M j commute if | j − k | (cid:54) = 1. Thus it follows from part (b) that M j (cid:126)ν ≤ M j M k (cid:126)ν = M k M j (cid:126)ν, k (cid:54) = { j − , j + 1 } . For k ∈ { j − , j + 1 } we can verify by direct inspection that the inequalitybetween M j (cid:126)ν and M k M j (cid:126)ν also holds. The two vectors only differ at positions k = j ± 1, which for M j (cid:126)ν is equal to ν j ± , and for M j ± M j (cid:126)ν it is M ( ν j ± + (cid:98) ν j )or (cid:102) M ( ν j ± + (cid:98) ν j ) (depending on the parity of j ) with (cid:98) ν j as in (3.13). By M -convexity of (cid:126)ν we have ν j ≤ (cid:98) ν j and by the monotonicity properties of M and (cid:102) M and M -convexity once more, we have ν j − ≤ M ( ν j ± + ν j ) ≤ M ( ν j ± + (cid:98) ν j ) if j is oddwith M replaced by (cid:102) M if j is even. This proves M j (cid:126)ν ≤ M k M j (cid:126)ν also in case | j − k | = 1 and part (c) follows.(d) and (e) are straightfoward verifications. Proof. Let us take 0 < s < t < 1. Then we have to show that sµ j,s ≤ tµ j,t for every j = 0 , , . . . , r + 1. This is clear for j = 0 and j = r + 1 due to thedefinitions (2.4). 28rite (cid:126)µ s = ( µ ,s , µ ,s , . . . , µ r +1 ,s ) and similarly for (cid:126)µ t . Then by thedefinition of the operators M j , we have M j ( s(cid:126)µ s ) = s(cid:126)µ s , M j ( t(cid:126)µ t ) = t(cid:126)µ t . see also Remark 3.6 (d) and (e).Now we put (cid:126)ν = s(cid:126)µ s + (( t − s ) δ − q − , , . . . , , ( t − s ) δ ( − r q )= ( tµ ,t , sµ ,s , sµ ,s , . . . , sµ r,s , tµ r +1 ,t ) . (3.18)This is the vector s(cid:126)µ s with the 0th and r +1st components replaced by thoseof t(cid:126)µ t . Then s(cid:126)µ s ≤ (cid:126)ν and (cid:126)ν is M -convex by Lemma 3.10 (d) and (e) andthe fact that s(cid:126)µ s is M -convex.We choose an infinite sequence ( j k ) k in { , . . . , r } where we make surethat every j in { , . . . , r } appears an infinite number of times in the sequence.Then we define a sequence ( (cid:126)ν k ) k by (cid:126)ν k +1 = M j k ( (cid:126)ν k ) for k = 1 , , . . . . Inductively we find that each (cid:126)ν k is M -convex by Lemma 3.10(c) and the factthat (cid:126)ν is M -convex. Then (cid:126)ν k ≤ (cid:126)ν k +1 for every k .Also by induction it is easy to show that (cid:90) d ( (cid:126)ν k ) j ≤ t + j − k and for every j . Thus the sequence ( (cid:126)ν k ) is increasing with acomponentwise limit (cid:126)ν k → (cid:126)ν ∞ , as k → ∞ (with convergence in weak ∗ -sense).If j = j k is even, then( (cid:126)ν k +1 ) j = M (( (cid:126)ν k ) j − + ( (cid:126)ν k ) j +1 ) . while for j is odd we have to replace M by (cid:102) M .If we take the limit k → ∞ along the subsequence for which j k = j ,then it follows from this that M j ( (cid:126)ν ∞ ) = (cid:126)ν ∞ for every j . Since the 0th and r + 1st components are those of t(cid:126)µ t , we we conclude that (cid:126)ν ∞ = t(cid:126)µ t , see alsoRemark 3.6 (e).We combine the inequalities to find s(cid:126)µ s ≤ (cid:126)ν ≤ · · · ≤ (cid:126)ν k ≤ · · · ≤ (cid:126)ν ∞ = t(cid:126)µ t which indeed shows that sµ j,s ≤ tµ j,t for every j .29 .3 Proof of part (c) For θ > µ on [0 , ∞ ) we write I θ ( µ ) = (cid:90) (cid:90) log 1 | x θ − y θ | dµ ( x ) dµ ( y ) (3.19)which we may call the θ -energy of µ . For θ = 1 it reduces to the usuallogarithmic energy I ( µ ) of µ . Proposition 3.11. Let ( µ , . . . , µ r ) be the solution of the VEP of Definition2.1 with parameters q > and < t < . Then the first component µ minimizes I ( ν ) + 12 I θ ( ν ) + (cid:90) V ( x ) dν ( x ) (3.20) with θ = r and V ( x ) = − − tt log (cid:0) x + q − (cid:1) + r + tt log (cid:16) x /r + q /r (cid:17) (3.21) among all probability measures ν on [0 , ∞ ) . Energy functionals of the form (3.20) appeared before in the context ofMuttalib-Borodin ensembles [6, 41]. These are joint probability densities for n particles on the positive real line of the form1 Z n (cid:89) ≤ i For a probability measure ν on [0 , ∞ ) we define J ( ν ) = min ν ,...,ν r r (cid:88) j =2 I ( ν j ) − r (cid:88) j =1 I ( ν j , ν j +1 ) (3.23)30here ν = ν and ν r +1 = µ r +1 = r + tt δ ( − r q . The minimization is over all ν , . . . , ν r satisfying the support condition (2.2) and the total mass condition(2.3), i.e., supp( ν j ) ⊂ ∆ j and ν j (∆ j ) = 1 + j − t for j = 2 , . . . , r .This is again a weakly admissible vector equilibrium problem, similar tothe VEP from Definition 2.1, and it has a unique solution. It is simpler tosolve, since only positive measures are involved and we can be sure that theminimizers ν , . . . , ν r have full supports, with the property2 U ν j = U ν j − + U ν j +1 on ∆ j , for j = 2 , . . . , r, (3.24)see also (2.12).From (3.24) we obtain I ( ν j ) = (cid:90) U ν j dν j = 12 (cid:90) ( U ν j − − U ν j +1 ) dν j = 12 I ( ν j − , ν j ) + 12 I ( ν j , ν j +1 ) , j = 2 , . . . , r. Hence from (3.23) J ( ν ) = − I ( ν, ν ) − I ( ν r , ν r +1 )= − (cid:90) U ν dν − r + t t U ν r (( − r q ) , (3.25)where ν and ν r are from the minimizer ( ν , . . . , ν r ) associated with ν . Weare going to calculate U ν and U ν r .We first do this for a point mass ν = δ p with p > 0, and the generalcase is obtained by averaging over p . So let ( ν , . . . , ν r ) be the minimizerassociated with ν = ν = δ p . We use the Riemann surface S with r sheets S ( j ) , j = 1 , . . . , r , given by S (1) = C \ ( −∞ , , S ( j ) = C \ R , for j = 2 , . . . , r − , S ( r ) = C \ (( − r [0 , ∞ )) . (3.26)Sheet S ( j ) is connected to sheet S ( j − along the cut ∆ j = ( − j [0 , ∞ ) for j = 2 , . . . , r . We add a point at infinity to obtain a compact Riemannsurface. 31e define a function Ψ on S by its restriction Ψ ( j ) to the j th sheet asfollows.Ψ (1) ( z ) = zz − p − z (cid:90) dν ( s ) z − s , Ψ ( j ) ( z ) = z (cid:90) dν j ( s ) z − s − z (cid:90) dν j +1 ( s ) z − s , for j = 2 , . . . , r − , Ψ ( r ) ( z ) = z (cid:90) dν r ( s ) z − s − r + tt zz − ( − r q . (3.27)The conditions (3.24) imply that Ψ is meromorphic on S with poles at z = p on the first sheet and at z = ( − r q on the r th sheet, see also thediscussion after Definition 2.6 that shows why Φ is meromorphic on R . [Theconstruction of Ψ is similar to that of Φ.] Due to the total masses of themeasures we have Ψ( z ) → − t as z → ∞ .The Riemann surface (3.26) has a simple parametrization z = w r , andin the w variable the poles are at w = p /r and w = − q /r . Taking intoaccount the residues at the poles and the behavior at infinity, we find thatΨ( z ) = 1 r p /r w − p /r + r + trt q /r w + q /r − t , z = w r . (3.28)Observe also that Ψ( z ) = 0 for z = w = 0.Specifying (3.28) to the first sheet, and recalling (3.27) we find (cid:90) dν ( s ) z − s = 1 z − p − z Ψ (1) ( z )= 1 z − p − rz p /r z /r − p /r − r + trtz q /r z /r + q /r − tz (3.29)with principal branch of the fractional powers. We integrate with respect to z and find after straightforward calculation (cid:90) log( z − s ) dν ( s ) = log( z − p ) − log (cid:16) z /r − p /r (cid:17) + r + tt log (cid:16) z /r + q /r (cid:17) (3.30)There is no constant of integration since both sides behave as (1+ t − ) log z + o (1) as z → ∞ . The real part of (3.30) gives us the logarithmic potential U ν ( z ) = log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z /r − p /r z − p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − r + tt log (cid:12)(cid:12)(cid:12) z /r + q /r (cid:12)(cid:12)(cid:12) . (3.31)32n analogous calculation, based on (3.28) and the expression (3.27) ofΨ on the r th sheet, leads to the logarithmic potential of ν r , U ν r ( z ) = r + tr log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z /r + q /r z − ( − r q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − log (cid:12)(cid:12)(cid:12) z /r − p /r (cid:12)(cid:12)(cid:12) (3.32)with the branch of the r th root that is analytic on C \ ∆ r and that is realand negative for real z ∈ C \ ∆ r . However, we emphasize that p /r and q /r always denote the positive r th roots. Similarly when we write x /r with x > ν and ν r as-sociated with δ p . Associated with a general probability measure ν = ν on[0 , ∞ ), we then have measures ν and ν r whose logarithmic potentials areobtained from averaging (3.31) and (3.32) over p , that is U ν ( z ) = (cid:90) log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z /r − x /r z − x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dν ( x ) − r + tt log (cid:12)(cid:12)(cid:12) z /r + q /r (cid:12)(cid:12)(cid:12) (3.33)and U ν r ( z ) = r + tr log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z /r + q /r z − ( − r q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:90) log (cid:12)(cid:12)(cid:12) z /r − x /r (cid:12)(cid:12)(cid:12) dν ( x ) . (3.34)From (3.33) we obtain I ( ν, ν ) = I ( ν ) − I /r ( ν ) − r + tt (cid:90) log | x /r + q /r | dν ( x ) , (3.35)and from (3.34) we obtain, noting that z /r in (3.34) is negative for z ∈ R \ ∆ r , and thus in particular for z = ( − r q , U ν r (( − r q ) = r + tt log lim x → q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x /r − q /r x − q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:90) log (cid:12)(cid:12)(cid:12) − q /r − x /r (cid:12)(cid:12)(cid:12) dν ( x )= r + tt log (cid:18) r q /r − (cid:19) − (cid:90) log (cid:16) x /r + q /r (cid:17) dν ( x ) . (3.36)Using (3.35) and (3.36) in (3.25) we obtain J ( ν ) = − I ( ν ) + 12 I /r ( ν ) + r + tt (cid:90) log (cid:16) x /r + q /r (cid:17) dν ( x ) − (cid:18) r + tt (cid:19) log (cid:18) r q /r − (cid:19) . (3.37)33inally, comparing (3.23) with the energy functional (2.5) we obtain fora given ν on [0 , ∞ ) thatmin ν ,...,ν r E ( ν, ν , . . . , ν r ) = I ( ν ) − I ( ν, µ ) + J ( ν )so that in view of (3.37) and noting that µ is given by (2.4)min ν ,...,ν r E ( ν, ν , . . . , ν r ) = 12 I ( ν ) + 12 I /r ( ν ) − − tt (cid:90) log (cid:0) x + q − (cid:1) dν ( x ) + r + tt (cid:90) log (cid:16) x /r + q /r (cid:17) dν ( x ) − (cid:18) r + tt (cid:19) log (cid:18) r q /r − (cid:19) . (3.38)The left-hand side of (3.38) as a functional on probability measures ν on[0 , ∞ ) attains its minimum at ν = µ . Since the last term on the right in(3.38) is only a constant, independent of ν , the proposition follows. V is in the support In the next step we discuss a general fact about the minimizer for a Muttalib-Borodin type energy functional (3.20) with θ > 0, and where V : [0 , ∞ ) → R is continuous with lim inf x →∞ ( V ( x ) − (1 + θ ) log x ) ≥ −∞ . Under this condition there is a unique probability measure µ on [0 , ∞ ) thatminimizes (3.20).The following is well-known for the case θ = 1, but apparently has notbeen observed for general θ . Lemma 3.12. If x ≥ is such that V ( x ) = min x ≥ V ( x ) then x ∈ supp( µ ) where µ is the probability measure that minimizes (3.20) .Proof. In this proof we use the notation U µθ ( x ) = (cid:90) log 1 | x θ − s θ | dµ ( s ) . The minimizer µ satisfies, for some constant (cid:96) , U µ ( x ) + U µθ ( x ) + V ( x ) (cid:40) = (cid:96), on supp( µ ) , ≥ (cid:96) on [0 , ∞ ) . (3.39)34ow, h ( x ) = U µ ( x ) + U µθ ( x ) = (cid:90) log 1 | x − s | dµ ( s ) + (cid:90) log 1 | x θ − s θ | dµ ( s ) , extends into the complex plane where we use the principal branch of x θ ,i.e., with a branch cut along ( −∞ , h is harmonic in C \ (( −∞ , ∪ supp( µ )) and it tends to −∞ as | x | → ∞ . By the maximum principle forharmonic functions, the maximum of h is attained on supp( µ ) ∪ ( −∞ , x > s > 0, it is easy to see that | − x − s | > | x − s | and | ( − x ) θ − s θ | = | x θ e πiθ − s θ | > | x θ − s θ | . Therefore U µ ( − x ) < U µ ( x ) and U µθ ( − x ) < U µθ ( x ) for x > h ( − x ) < h ( x ) for x > 0, andtherefore the maximum of h is not attained on ( −∞ , h is attained on supp( µ ), say at x ∈ supp( µ ).If x (cid:54)∈ supp( µ ), then h ( x ) < h ( x ), and since V ( x ) ≤ V ( x ) by theassumption in the lemma, we have a strict inequality h ( x ) + V ( x ) < h ( x ) + V ( x ) . (3.40)Since x ∈ supp( µ ) the right-hand side of (3.40) is equal to (cid:96) , by the equalityin (3.39) It follows that U µ ( x ) + U µθ ( x ) + V ( x ) < (cid:96) which contradicts the inequality in (3.39). Therefore x ∈ supp( µ ). In view of Proposition 3.11 and Lemma 3.12 it is enough to show that theexternal field (3.21) attains its minimum at x = 0 in case 0 < q < 1. Thisis what we do in the next lemma, and then part (c) follows. Lemma 3.13. Let V be given by (3.21) with < q < . Then for every t ∈ (0 , it is true that V (0) = min x ≥ V ( x ) . Proof. Note that (cid:0) x /r + q /r (cid:1) r ≥ x + q for x ≥ tV ( x ) − tV (0) ≥ − (1 − t ) log (cid:0) x + q − (cid:1) + r + tr log ( x + q ) − (cid:18) − t + tr (cid:19) log q. (3.41)35ince 0 < q < 1, we have x + q ≥ q x + q for x ≥ tV ( x ) − tV (0) ≥ − (1 − t ) log (cid:0) x + q − (cid:1) + r + tr log (cid:0) q x + q (cid:1) − (cid:18) − t + tr (cid:19) log q. This means V ( x ) − V (0) ≥ (cid:0) r (cid:1) log ( qx + 1) and the lemma follows. The measure µ minimizes I ( µ ) − I ( µ, µ + µ ) among all probability mea-sures on [0 , ∞ ). Thus it is the equilibrium measure in the external field − U µ + µ which is real analytic on [0 , ∞ ). Then it follows from [14] that µ is absolutely continuous with respect to Lebesgue measure with a densitythat is real analytic in the interior of its support.From [14] it also follows that the density at the endpoint x (in BIS and TIS cases) behaves as ≈ c ( x − x ) +2 N as x → x − , for a certainnon-negative integer N . From the iterated balayage it can be seen that N = 0. Indeed, the algorithm in section 3.1.4 gives us the sequence ofsigned measures ( ν k ) k with densities that are such that √ x dν k dx is positiveand strictly decreasing on [0 , x ] by Lemma 3.4 (b). The proof of that lemma(see (3.11))) actually shows that the decrease gets stronger as k increases.Since ν k → µ as k → ∞ , it then follows that the density of µ cannot havea zero derivative at x , and thus it vanishes as a square root at x in BIS and TIS cases.Similarly, the density vanishes as a square root at x in UIS and TIS cases. Φ We start by listing a number of properties of the meromorphic function Φfrom Definition 2.6. Let q > . The function z Φ is a degree meromorphic func-tion on the Riemann surface R with the following properties. z Φ → as z → ∞ on any of the sheets. (b) It has simple poles at z = − q − on the first sheet and at z = ( − r q on the last sheet, and no other poles. (c) Suppose one of the BIS , TIS , or FIS cases, so that z ∈ supp( µ ) .Then z Φ has simple zeros at z = 0 and at a point x on the first sheetwith − q − < x < , (4.1) and no other zeros. In UIS case, there are two points on the Riemannsurface with z = 0 and z Φ has a simple zero at each of these.Proof. We already noted that Φ is meromorphic on R , see the discussionafter Definition 2.6. Thus also z Φ is meromorphic on R . In part (b) we showthat it has two simple poles, and no other poles, and therefore its degree istwo.(a) From (2.19) we have zF j ( z ) → µ j (∆ j ) as z → ∞ . In view of the totalmassses (2.3) of the measures and the definition (2.21), part (a) follows.(b) From (2.20) and (2.21) we see thatΦ (1) ( z ) = tF ( z ) + 1 − tz + q − = t (cid:90) dµ ( s ) z − s + 1 − tz + q − , (4.2)Φ ( r +1) ( z ) = − tF r ( z ) + r + tz − ( − r q , (4.3)and so z Φ has simple poles at − q − on the first sheet and at ( − r q on the r + 1-st sheet.There is no pole at z = ∞ because of part (a). There is no pole at z = 0either, since the form (2.19) of F j as a Stieltjes transform, easily impliesthat zF j ( z ) → z → 0. Thus also z Φ → z → 0. There are no othercandidates for poles, and therefore the degree is two.(c) We already remarked in part (b) that z Φ vanishes when z = 0. In UIS case there are two points on the Riemann surface with z = 0. In thatcase both of these are simple zeros, and there are no other zeros, since thedegree is two.In BIS , TIS , and FIS cases, there is only one point z = 0, and it is atmost a double zero of z Φ. Then z r +1 is the local coordinate, and z (as a37unction on the Riemann surface) has a zero of order r + 1 at z = 0. HenceΦ has a pole at z = 0 of order ≥ r − 1, and so Φ is unbounded at z = 0 (wemay assume r ≥ F is unbounded at z = 0, and it dominates the behavior of Φ (1) as z → F ( z ) < z , it then follows that Φ (1) ( z ) is negativefor negative z close to 0.From (4.2) we also see that Φ (1) ( z ) → + ∞ as z → − q − +, as the residueat the pole is positive. Thus Φ (1) changes sign on the interval ( − q − , 0) andhence there is a zero, say at x ∈ ( − q − , x is also a zero of z Φ,and we conclude that both z = x and z = 0 are simple zeros, and these arethe only zeros, as the degree of z Φ is two. Proof. Suppose we are in one of the BIS , TIS , or FIS cases, so that 0 ∈ supp( µ ). Then z = 0 is a simple zero of z Φ, by part (c) of Lemma 4.1.Since z r +1 is a local coordinate, we find that Φ has a pole of order r at z = 0. From (4.2) we then get for some non-zero constant C , F ( z ) = Cz − rr +1 (cid:16) O (cid:16) z r +1 (cid:17)(cid:17) as z → , ∞ ). By the Stieltjesinversion formula dµ dx = − π lim δ → Im F ( x + iδ ) , x > , and (2.13) follows.Finally, (2.14) follows from (2.13) and the symmetry between q and 1 /q ,see Remark 2.4. Remark 4.2. The behavior (2.13) is characteristic for the density of min-imizers of Muttalib-Borodin type energy functionals as in (3.20). This wasproved by Claeys and Romano [10, Remark 1.9] under general conditions onthe external field, which however do not cover the case (3.21). We need to know about the critical points, by which we mean the ramifica-tion points of z Φ. There is no ramification at z = 0 or z = ∞ , and so wemay alternatively characterize the critical points as those points where thederivative of z Φ ( j ) ( z ) vanishes for some j = 1 , . . . , r + 1.38 q − x y z Φ (1) in BIS case x y z Φ (1) and z Φ (2) in BIS caseFigure 5: Sketches of the graph of z Φ in BIS case. The left panel shows thegraph of z Φ (1) ( z ) on the negative real line where it has a pole at − q − , azero at x and a local maximum at y . The right panel shows the graphs of z Φ (1) ( z ) (in brown) and z Φ (2) (in blue) on the interval [ x , ∞ ). The graphof z Φ (1) ( z ) has a local minimum at y , while the graph of z Φ (2) ( z ) is strictlyincreasing. Both graphs tend to 1 at infinity. The graph of z Φ ( r +1) ( z ) is asin the right panel of Figure 7 below.Since z Φ has degree 2 the Riemann-Hurwitz formula [46] tells us thatthere are two critical points in the BIS and FIS (genus zero) cases, andfour critical points in the TIS (genus one) case. The following lemma saysthat they are all real and on the first sheet.We fix 0 < q < x , x as in Theorem 2.3 (a)depending on the various cases, and x for the zero of Φ on the first sheetas in Lemma 4.1 (c). Lemma 4.3. Let < q < . The critical points of z Φ are on the real partof the first sheet of the Riemann surface. (a) In all cases there is a critical point y with y ∈ ( x , . (b) In BIS case there is one more critical point y ∈ ( x , ∞ ) . (c) In TIS case there are three more critical points. A critical point y ∈ ( −∞ , − q − ) and two critical points y , y ∈ ( x , x ) with y < y . (d) In FIS case there is one more critical point y ∈ ( −∞ , − q − ) . − q − x y z Φ (1) in TIS case x y x y ........................ z Φ (1) and z Φ (2) in TIS caseFigure 6: Sketches of the graph of z Φ in TIS case. The left panel showsthe graph of z Φ (1) ( z ) on the negative real line where it has a pole at − q − ,zeros at x and 0, a local minimum at y , and a local maximum at y . Theright panel shows the graphs of z Φ (1) ( z ) (in brown) and z Φ (2) (in blue) onthe interval [ x , x ]. The graph of z Φ (1) ( z ) has a local minimum at y and alocal maximum at y , while the graph of z Φ (2) ( z ) is strictly increasing. Thegraph of z Φ ( r +1) ( z ) is as in the right panel of Figure 7 below. Proof. (a) By Lemma 4.1 (c) z Φ (1) ( z ) has zeros at z = x and at z = 0,and in between it is real and positive. So there is a local maximum, and thepoint y ∈ ( x , 0) where it is attained is a critical point in all cases.The proofs of parts (b)-(d) rely on an inspection of the graph of z Φ onthe real part of the Riemann surface (that is, on the part where both z andΦ are real), see Figures 5, 6, and 7 for sketches of the graphs in the variouscases. We infer the following about z Φ from the behavior at the poles andat infinity, • every value in ( −∞ , 0) is attained once in ( − q − , x ) on the first sheetand once between 0 and ( − r q on the last sheet, • every value in (1 , ∞ ) is attained once in ( −∞ , − q − ) on the first sheetand once between ( − r q and infinity on the last sheet.Since z Φ has degree two, the values in ( −∞ , 0) and (1 , ∞ ) are attainednowhere else on the Riemann surface. In particular0 < z Φ (1) ( z ) < , and0 < z Φ (2) ( z ) < , (cid:40) for z ∈ [ x , ∞ ) in BIS case , for z ∈ [ x , x ] in TIS case , (4.4)40 − q − x y z Φ (1) in FIS case q z Φ ( r +1) in all cases(when r is even)Figure 7: Sketches of the graph of z Φ in FIS case. The left panel showsthe graph of z Φ (1) ( z ) on ( −∞ , 0] where it has a pole at − q − , zeros at x and 0, a local minimum at y , and a local maximum at y . The right panelshows the graph of z Φ ( r +1) ( z ) on ( − r [0 , ∞ ) which has a pole at ( − r q .The figure is for r = 2 and it has the same features for all cases.see the right panels of Figures 5 and 6.From the fact that the density of µ vanishes as a square root at theendpoints x , x see Theorem 2.3 (d), it follows that F (cid:48) ( z ) = − (cid:90) z − x ) dµ ( x ) → −∞ (4.5)as z → x + or z → x − . Hence by (2.21) we also have (cid:16) z Φ (1) ( z ) (cid:17) (cid:48) → −∞ and (cid:16) z Φ (2) ( z ) (cid:17) (cid:48) → + ∞ (4.6)as z → x + (in BIS and TIS cases) or z → x − (in TIS case).(b) In BIS case we noted in (4.4) that z Φ (1) ( z ) takes the value x Φ (1) ( x ) ∈ (0 , 1) at x , and by (4.6) it starts to decrease if z ∈ ( x , ∞ ) increases. Sinceit tends to the value 1 at infinity, there will be a local minimum, say at y ∈ ( x , ∞ ). This is a critical point, and part (b) follows.It also follows that z Φ (2) ( z ) strictly increases for z ∈ [ x , ∞ ) since in BIS case there are no further critical points.(c) In TIS case we first observe that ddz ( zF ( z )) = − (cid:90) x ( z − x ) dµ ( x ) < , z ∈ ( x , x ) , (4.7)41 dz ( zF ( z )) = − (cid:90) x ( z − x ) dµ ( x ) > , z ∈ ( x , x ) . (4.8)The difference in sign is due to the fact that µ is supported on [0 , ∞ ), while µ is supported on ( −∞ , (cid:16) z Φ (2) ( z ) (cid:17) (cid:48) > z ∈ ( x , x ) , and therefore z Φ (2) ( z ) strictly increases on ( x , x ) and there are no criticalpoints in ( x , x ) on the second sheet. We also conclude x Φ (1) ( x ) < x Φ (1) ( x )but z Φ (1) ( z ) will not be monotonic on [ x , x ] due to (4.6). Instead it willstart to decrease at x to a local minimum, say at y , and then increasesto a local maximum, say at y , and then again decreases. This gives us thecritical points y < y in ( x , x ). We already know y ∈ ( − x , −∞ , − q − ), and this follows from the ob-servation that zF ( z ) = z (cid:90) dµ ( x ) z − x = 1 + (cid:90) xdµ ( x ) z − x with (cid:82) xdµ ( x ) z − x < z ∈ ( −∞ , 0] and (cid:90) xdµ ( x ) z − x = − Cz − r +1 (1 + O ( z − r +1 ) as z → −∞ , (4.9)with a positive constant C > 0. Also zF ( z ) = (1 − t − ) + O ( z − ) as z → ∞ ,so that by (2.21). z Φ (1) ( z ) = 1 + (cid:90) xdµ ( x ) z − x + O ( z − ) (4.10)as z → −∞ , where the second term is negative for z < O ( z − ) term as z → −∞ . Therefore z Φ (1) ( z ) decreases on an interval( −∞ , y ) for some y ∈ ( −∞ , − q − ), it reaches a local minimum at y andthen increases to + ∞ as z → − q − − . See also Figure 6.(d) In FIS case the expansions (4.9) and (4.10) remain valid, as doesthe conclusion that z Φ (1) ( z ) has a local minimum at some y ∈ ( −∞ , − q − ),and y is a critical point.After these preparations we turn to the proof of Theorem 2.8.42 .2 Proof of part (a) Proof. For x ∈ supp( µ ), we have by (2.21) and the Stieltjes inversion for-mula x Im Φ (1) ± ( x ) = xt Im ( F ) ± ( x ) = ∓ xtπ dµ ( x ) dx . (4.11)Here the subscript ± denotes the limiting value from the upper (+) or lower( − ) half plane. Then by (2.23) we find supp( µ ) ⊂ U .from the Cauchy-Riemann equations and the definition (2.23) of U , weobtain that the parts of the real line where z Φ (1) ( z ) is real and decreasingbelong to U , while those parts where z Φ (1) ( z ) is real and increasing do notbelong to U . Then in view of the behavior of z Φ (1) ( z ) on the realline thatwe see in Figures 5, 6, 7, and the fact that supp( µ ) ⊂ U , we conclude that U ∩ ( R ∪ {∞} ) = [ y , y ] in BIS case , [ −∞ , y ] ∪ [ y , y ] ∪ [ y , ∞ ] in TIS case , [ −∞ , y ] ∪ [ y , ∞ ] in FIS case . (4.12)In particular − q − (cid:54)∈ U . This proves part (a). Proof. Since the y j ’s are critical points, we have that z Φ (1) ( z ) is also real oncertain contours that emanate from each y j into the complex plane. Thesecontours are going to be the boundary ∂U of U .The labelling of the critical points in Lemma 4.3 is such that y j is a localminimum of z Φ (1) ( z ) if j is even, and a local maximum if j is odd, whenwe restrict to the real line. It means that z Φ (1) ( z ) is real and increasing on ∂U when we move away from y j with j odd, and decreasing from y j with j even.Noting that y Φ (1) ( y ) < y Φ (1) ( y ) in BIS case , (4.13)we conclude that the part of ∂U that emanates from y will end at y in BIS case. Since Σ = supp( µ ) ⊂ U , we also see that ∂U consists of asimple closed contour surrounding Σ and U is a bounded simply connecteddomain in the BIS case. 43 .4 Proof of part (c) Proof. In TIS case, we have four critical points and instead of (4.13) wehave y Φ (1) ( y ) < y Φ (1) ( y ) < y Φ (1) ( y ) < y Φ (1) ( y ) in TIS case . (4.14)Then ∂U consists of two closed contours, one containing y and y , and onecontaining y and y . Both closed contours go around Σ . It follows that U has two components, namely the bounded domain that is enclosed by theinner component of ∂U , and the unbounded domain that is outside of theouter component of ∂U . This proves part (c). Proof. In FIS case we have two critical points y < y < y Φ (1) ( y ) < y Φ (1) ( y ) , in FIS case . (4.15)Then ∂U is a closed contour containing y and y , and U is the domain thatis exterior to this contour. Proof. We know from Theorem 2.3 (b) that t (cid:55)→ tµ j,t increases with t forevery j , and tµ j,t has total mass t + j − 1. Then ρ j = ρ j,t = ∂ ( tµ j,t ) ∂t , j = 0 , . . . , r + 1 (4.16)is a probability measure on supp( µ j ) for every j . In particular ρ = δ − q − by (2.3).Thus by differentiating (2.21) for j = 1 with respect to t , ∂ Φ (1) ( z ) ∂t = (cid:90) dρ ( x ) z − x − z + q − . Since ρ and ρ are both probability measures we obtain from this ∂ ( z Φ (1) ( z )) ∂t = (cid:90) xdρ ( x ) z − x + q − z + q − . (4.17)44or Im z > xdρ ( x ) is a positive measure), while for Im z < ∂∂t Im (cid:16) z Φ (1) ( z ) (cid:17) (cid:40) < z > ,> z < , Therefore the part in the upper half plane where Im (cid:0) z Φ (1) ( z ) (cid:1) < t . Similarly, the part in the lower half plane where Im (cid:0) z Φ (1) ( z ) (cid:1) > t , which proves part (e) in view of the definition (2.23) of U . Proof. It is clear from (2.23) that z Φ (1) ( z ) is real-valued for z ∈ ∂U . Thepoint of part (f) is that ∂U is characterized by (2.24).Being a meromorphic function on a compact Riemann surface, Φ satisfiesthe algebraic equation r +1 (cid:89) j =1 (cid:16) Φ − Φ ( j ) ( z ) (cid:17) = Φ r +1 + r +1 (cid:88) k =1 ( − k e k ( z )Φ r +1 − k = 0 (4.18)where e k ( z ) is the k th elementary symmetric function in Φ (1) , . . . , Φ ( r +1) ,i.e., e k ( z ) = (cid:88) ≤ j < ··· Since µ ∗ is the symmetric pullback of the probability measure µ , itis also a probability measure. That µ Ω is a probability measure as well canbe seen from the formulas in Lemma 5.2 below, by letting z → ∞ in either(5.3) (in case Ω is bounded), or (5.4) (in case Ω is unbounded).46 .2 Proof of part (b) Proof. Since tµ ,t is increasing by Theorem 2.3 (b), also tµ ∗ t increases with t . The domains U t increase with t by Theorem 2.8. Then also Ω t increaseswith t and then tµ Ω ,t also increases with t , since by (2.28) this is just thespherical area measure π dA ( z )(1+ | z | ) restricted to Ω t . µ Ω The proofs of parts (c) and (d) are modelled after the proofs in the paper[11] that deals with the case r = 1. See in particular the proof of Proposition4.1 in [11]. As a preparation we need the following formula for the sphericalSchwarz function from (2.26), which is the analogue of [11, (5.1)]. Lemma 5.1. We have S ( z ) = (1 − t ) z r z r +1 + q − + t (cid:90) dµ ∗ ( x ) z − x (5.1) Proof. Since µ ∗ is the symmetric pullback of µ under the mapping z (cid:55)→ z r +1 , we can easily verify that their logarithmic potentials are related via U µ ∗ ( z ) = 1 r + 1 U µ ( z r +1 ) . and also, with appropriate branches of the logarithm, (cid:90) log( z − x ) dµ ∗ ( x ) = 1 r + 1 (cid:90) log (cid:0) z r +1 − x (cid:1) dµ ( x ) . Taking the z -derivative we find (cid:90) dµ ∗ ( x ) z − x = z r F ( z r +1 ) . (5.2)Then combining (2.26), (4.2), and (5.2), we find (5.1).The following lemma is the analogue of [11, Lemma 5.2] and its proof isalso very similar. Lemma 5.2. The Stieltjes transform of µ Ω satisfies (cid:90) dµ Ω ( x ) z − x = (cid:90) dµ ∗ ( x ) z − x , z ∈ C \ Ω , (5.3) (cid:90) dµ Ω ( x ) z − x = − (1 − t ) z r t ( z r +1 + q − ) + ¯ zt (1 + | z | ) , z ∈ Ω . (5.4)47 roof. Take z ∈ C \ Ω first. Then by (1.9), (2.28), the complex Green’sformula, and the property (2.27) of the spherical Schwarz function, we findif Ω is bounded, t (cid:90) dµ Ω ( s ) z − s = 1 π (cid:90) Ω dA ( s )( z − s )(1 + | s | ) = 12 πi (cid:73) ∂ Ω ¯ s ( z − s )(1 + | s | ) ds = 12 πi (cid:73) ∂ Ω S ( s ) z − s ds = − πi (cid:73) ∂ ( C \ Ω) S ( s ) z − s ds. (5.5)Since the complex Green’s formula applies to bounded domains, one hasto modify the calculation in case Ω is unbounded. Then one first makes acut-off to { z ∈ Ω | | z | ≤ R } with a large R > 0. The Green’s formula thenproduces an additional integral over | z | = R , which however tends to zeroas R → ∞ , due to the fact that S ( s ) z − s = O ( s − ) as s → ∞ . Thus (5.5) alsoholds in the unbounded case.The remaining integral in (5.5) is evaluated with the residue theoremfor C \ Ω. The spherical Schwarz function S has r + 1 simple poles at thesolutions of s r +1 = − q − , the poles are all in C \ Ω, and from (5.1) it canbe checked that S has the same residue − tr +1 at each of the poles. Togetherthey give the contribution − − tr + 1 (cid:88) s r +1 = − q − z − s = − (1 − t ) z r z r +1 + q − (5.6)to the integral (5.5). There is an additional pole in (5.5) at s = z with thecontribution S ( z ). Finally, note that there is no contribution from infinityin case C \ Ω is unbounded, since the integrand in (5.5) is O ( s − ) as s → ∞ .In total we get t (cid:90) Ω dµ Ω ( s ) z − s = − (1 − t ) z r z r +1 + q − + S ( z ) , z ∈ C \ Ω , and (5.4) follows because of (5.1).Let z ∈ Ω \ ∂ Ω. Take ε > D ( z, ε ) of radius ε around z is contained in Ω. Then by a calculation similar to (5.5), with complex48reen’s theorem and the spherical Schwarz function t (cid:90) Ω \ D ( z,ε ) dµ Ω ( s ) z − s = − πi (cid:73) ∂ ( C \ Ω) S ( s ) z − s ds − πi (cid:73) ∂D ( z,ε ) ¯ s ( z − s )(1 + | s | ) ds. (5.7)The integral over ∂ ( C \ Ω) is again evaluated using the residue theorem, butin the present situation there is no contribution from s = z , but only thecombined contribution (5.6) from the poles of S . The integral over the circle ∂D ( z, ε ) (including the prefactor − πi ) tends to ¯ z | z | as ε → ε → 0+ in (5.7) we obtain (5.4). µ Ω We will need the following result in the TIS case for the proof of Lemma5.5 below. Lemma 5.3. In FIS and TIS cases we have U µ Ω (0) + 1 − t ( r + 1) t log q = 0 . (5.8)In BIS case we have U µ Ω (0) + 1 − t ( r + 1) t log q = 12 t ( r + 1) (cid:90) ∞ x (cid:16) Φ (1) ( x ) − Φ (2) ( x ) (cid:17) dx ≤ Proof of Lemma 5.3. In the proof we use log x = log | x | + i arg x with 0 < arg x < π , and we are going to show that t (cid:90) log( x ) dµ Ω ( x ) = 1 − tr + 1 log q + tπi in FIS and TIS cases (5.9)and then (5.8) will follow by taking the real parts on both sides. The eval-uation of (5.9) follows along the lines of the proof of Lemma 5.2 but thereis a non-trivial extra step required in the TIS case.We start with the FIS case. Consider the cut-off domainΩ R,δ = { z ∈ Ω | | z | ≤ R, dist( z, [0 , ∞ )) > δ } R > δ > 0. Due to our definition of the logarithmwith the branch cut along [0 , ∞ ) we can apply the complex Green’s theoremto the integral over Ω R,δ and we find in FIS case1 π (cid:90) Ω R,δ log x dA ( x )(1 + | x | ) = 12 πi (cid:73) ∂ Ω R,δ log s s | s | ds → πi (cid:73) Ω log s s | s | ds + 12 πi (cid:73) | s | = R log s s | s | ds − (cid:90) R x x dx (5.10)as δ → (cid:90) R x x dx = 12 log (cid:0) R (cid:1) = log R + o (1) as R → ∞ . (5.11)The second integral in the right-hand side of (5.10) is evaluated withparametrization s = Re iθ , 0 < θ < π , to give12 πi (cid:73) | s | = R log( s ) s | s | ds = R log R R + R R πi = log R + πi + o (1) as R → ∞ . (5.12)In the first term we use (2.27) and then evaluate the integral by a residuecalculation over C \ Ω. The complement of Ω consists of r + 1 disjoint disksin FIS case, see Figure 4, and S ( s ) is meromorphic with one simple pole atthe solution of s r +1 + q − in each of the disks with residue − tr +1 . Therefore12 πi (cid:73) Ω log s s | s | ds = − πi (cid:73) ∂ ( C \ Ω) (log s ) S ( s ) ds = − − tr + 1 (cid:88) s : s r +1 = − q − log s = 1 − tr + 1 log q − (1 − t ) πi (5.13)Letting R → ∞ in (5.10) we find from (2.28), (5.11), (5.12) and (5.13)that t (cid:90) log( x ) dµ Ω ( x ) = lim R →∞ lim δ → π (cid:90) Ω R,δ log x dA ( x )(1 + | x | ) = 1 − tr + 1 log q + tπi 50s claimed in (5.9) in FIS case.In TIS case we have to adjust the above calculation in two ways. First,since Ω ∩ [0 , ∞ ) = [0 , y ∗ ] ∪ [ y ∗ , ∞ ), with y ∗ j = y / ( r +1) j , the integral over [0 , R ]in (5.10) is replaced by (cid:32)(cid:90) y ∗ + (cid:90) Ry ∗ (cid:33) x x dx = − (cid:90) y ∗ y ∗ x x dx + log R + o (1) as R → ∞ . Second, in the evaluation (5.13) of the integral over Ω, there is a contributionfrom the intersection [ y ∗ , y ∗ ] of C \ Ω with the positive real line, due to thediscontinuity of the logarithm. Instead of (5.13) we get12 πi (cid:73) Ω log s s | s | ds = 1 − tr + 1 log q − (1 − t ) πi − (cid:90) y ∗ y ∗ S ( x ) dx, and the result is the formula t (cid:90) log( x ) dµ Ω ( x ) = 1 − tr + 1 log q + tπi − (cid:90) y ∗ y ∗ (cid:18) S ( x ) − x x (cid:19) dx for the TIS case.To obtain (5.9) it remains to prove that the integral in the right-handside vanishes, and we do this by showing the identity (5.14) in Lemma 5.4below. The right-hand side of (5.14) is zero and to see this we recall (2.21)from which we getΦ (1) ( x ) − Φ (2) ( x ) = t (2 F ( x ) − F ( x ) − F ( x ))= − t ddx (2 U µ ( x ) − U µ ( x ) − U µ ( x )) , for x < x < x . We also recall that 2 U µ − U µ − U µ vanishes on the support of µ , andhence in particular at both x and x . Then the right-hand side is indeed 0by the fundamental theorem of calculus.Lemma 5.3 is thus proved, pending the proof of the remarkable identity(5.14). Since the proof of this identity uses new ideas that were not in [11],we decided to give it in a separate lemma. Lemma 5.4. In the TIS case we have (cid:90) y ∗ y ∗ (cid:18) S ( x ) − x x (cid:19) dx = 12( r + 1) (cid:90) x x (cid:16) Φ (1) ( x ) − Φ (2) ( x ) (cid:17) dx, (5.14) where y ∗ j = y r +1 j for j = 2 , . roof. Consider ω = Φ dz as a meromorphic differential on the Riemannsurface. Then (cid:73) a ω = (cid:90) x x (cid:16) Φ (1) ( x ) − Φ (2) ( x ) (cid:17) dx (5.15)for the cycle a that goes from x to x on the first sheet, and back from x to x on the second sheet, cf. Figure 2. The meromorphic differential hassimple poles at − q − on first sheet, at ( − r q on last sheet, and at ∞ withrespective residues 1 − t , r + t , and − − r .Since z Φ is a degree two meromorphic function we can represent theRiemann surface R by the equations η = (cid:89) j =0 ( ζ − ζ j ) , ζ = z Φ (5.16)with ζ j = y j Φ (1) ( y j ), for j = 0 , , , 3, being the four branch points of ζ with y < y < y < y by Lemma 4.3 and0 < ζ < ζ < ζ < ζ < , see also Figure 6. In the new coordinates R is a two sheeted cover of the ζ -plane, with branch cuts [ ζ , ζ ] and [ ζ , ζ ]. We label the sheets so that z = − q − corresponds to ζ = ∞ on the first sheet and z = ( − r q to ζ = ∞ on the second sheet. Then η is positive for real ζ > ζ on the first sheet.The point z = ∞ corresponds to the point ζ = 1 on the second sheet. The a -cycle goes from ζ to ζ on the first sheet and back from ζ to ζ on thesecond sheet.The meromorphic differential ω has simple poles at the two points at ζ = ∞ with residues 1 − t and r + t , and at ζ = 1 on the second sheet withresidue − − r . Then 2 ω + ( r + 1) dζζ − ± (1 − t − r ) at the two points at infinity, and residues ± ( r + 1)at the two points with ζ = 1. Thus (5.17) has an anti-symmetry with respectto the involution ( ζ, η ) (cid:55)→ ( ζ, − η ) of R . It follows that2 ω + ( r + 1) dζζ − Aζ + Bζ + Cζ − dζη for certain constants A , B , and C . From this form we conclude that (cid:73) a (cid:18) ω + ( r + 1) dζζ − (cid:19) dζ = 2 (cid:90) ζ ζ (cid:18) ω + ( r + 1) dζζ − (cid:19) dζ (cid:72) a dζζ − = 0 and therefore (cid:73) a ω = (cid:90) ζ ζ (cid:18) ω + ( r + 1) dζζ − (cid:19) = 2 (cid:90) y y Φ (1) ( z ) dz + ( r + 1) (log(1 − ζ ) − log(1 − ζ )) (5.18)since [ ζ , ζ ] on the first sheet corresponds to [ y , y ] on the first sheet in theoriginal z -variable where ω = Φ (1) ( z ) dz .Changing variable z = x r +1 and using (2.26) we have (cid:90) y y Φ (1) ( z ) dz = ( r + 1) (cid:90) y ∗ y ∗ S ( x ) dx (5.19)since y ∗ j = y r +1 j . For the last term on the right of (5.18) we recall that ζ j = y j Φ (1) ( y j ) and y j belongs to ∂U . Therefore it satisfies the equation(2.24), that is, ζ j = y r +1 j y r +1 j = ( y ∗ j ) y ∗ j ) , for j = 2 , , which we rewrite aslog (1 − ζ j ) = − log (cid:0) y ∗ j ) (cid:1) , for j = 2 , . (5.20)Hence log (1 − ζ ) − log (1 − ζ ) = log (cid:0) y ∗ ) (cid:1) − log (cid:0) y ∗ ) (cid:1) = 2 (cid:90) y ∗ y ∗ x x dx. (5.21)Combining (5.15), (5.18), (5.19), (5.21) we obtain the equality of the twointegrals in (5.14). ν t and ρ t For the proofs of parts (c) and (d), we also need to consider the dynamicalpicture where we vary t , see [11, section 6]. To emphasize the t -dependencewe attach a subscript t to the notions that vary with t .53e already observed in part (b) that tµ ∗ t and tµ Ω ,t increase with t . Thederivatives ρ t = ∂∂t ( tµ ∗ t ) , ν t = ∂∂t ( tµ Ω ,t ) (5.22)therefore exist for almost every t , as can be proved as in [8, Theorem 2], butin our case the derivatives actually exist for every t ∈ (0 , ρ t and ν t are probability measures, with supp( ρ t ) = supp( µ ∗ t ) andsupp( ν t ) = ∂ Ω t , see (2.28). Indeed ν t measures how the domain Ω t grows inthe spherical metric as t increases.Applying ∂∂t t to the identities (5.3) and (5.4) for the Stieltjes transforms,and using (5.22), we get (cid:90) dν t ( s ) z − s = (cid:90) dρ t ( s ) z − s , z ∈ C \ Ω t ,z r z r +1 + q − , z ∈ Ω t . (5.23) Lemma 5.5. There are constants C ,t and C ,t such that the following hold. (a) We have U ν t ( z ) ≤ U ρ t ( z ) + C ,t , z ∈ C , (5.24) with equality for z ∈ C \ Ω t . (b) We have U ν t ( z ) ≤ − r + 1 log | z r +1 + q − | + C ,t , z ∈ C , (5.25) with equality for z ∈ Ω t .Proof. (a) The first identity in (5.23) implies that U ν t − U ρ t is constant oneach connected component of C \ Ω t . Thus for some constant C ,t , U ν t ( z ) = U ρ t ( z ) + C ,t , z ∈ C \ Ω t , (5.26)since C \ Ω t is either connected (in BIS and TIS cases), or consists of r + 1 disjoint components (in FIS case) where due to r + 1-fold rotationalsymmetry the constant is the same on each component. If Ω t is bounded(the BIS case) then C ,t = 0, since both potentials in (5.26) behave as − log | z | + o (1) as z → ∞ .Since ν t is supported on ∂ Ω t , the function U ρ t − U ν t is superharmonicon the interior of Ω t , including at ∞ if Ω t is unbounded. By the minimum54rinciple for superharmonic functions we find the corresponding inequality(5.24) on Ω t , and part (a) follows.(b) For part (b) we argue similarly, but there is an additional twist whenΩ t is not connected (the TIS case). Using the second identity of (5.23),we apply similar reasoning to U ν t and − r +1 log | z r +1 + q − | , which is thelogarithmic potential of the discrete measure with mass r +1 at each solutionof z r +1 + q − = 0. We find that U ν t ( z ) + 1 r + 1 log | z r +1 + q − | is constant on each connected component of Ω t .In BIS and FIS cases we have that Ω t is connected and therefore forsome constant C ,t , U ν t ( z ) = − r + 1 log | z r +1 + q − | + C ,t , z ∈ Ω t , (5.27)in BIS and FIS cases. If Ω t is unbounded then we let z → ∞ in (5.27) andwe find that C ,t = 0 in FIS case.In TIS case we have that Ω t has two connected components. We findthat (5.27) holds with C ,t = 0 in the unbounded component for the samereason that C ,t = 0 in FIS case. The bounded component could potentiallyhave a different constant. However we are able to compute U ν t ( z ) at z = 0because of Lemma 5.3 which says that tU µ Ω ,t (0) = − − tr + 1 log q in the TIS case. Then taking the t -derivative and using the definition (5.22)of ν t , we obtain U ν t (0) = 1 r + 1 log q, which implies that (5.27) with C ,t = 0 holds for z = 0 and thus throughoutthe bounded component as well in the TIS case.From (5.27) and the fact that ν t is supported on ∂ Ω t , we obtain theinequality (5.25) in all cases (by the minimum principle, as in the proof ofpart (a)) and part (b) follows. Remark 5.6. The identity (5.26) and the fact that supp( ν t ) = ∂ Ω t showin fact that ν t = Bal( ρ t , ∂ Ω t ) . ν t = Bal r + 1 (cid:88) z r +1 = − q − δ z , ∂ Ω t . Thus ν t is a balayage measure onto ∂ Ω t from two sides. It is the balayageof ρ t which is supported inside Ω t , and it is also the balayage of a discretemeasure supported in the complement on Ω t . Proof. Integrating the identities (5.22) we obtain the identities tµ ∗ t = (cid:90) t ρ s ds, and tµ Ω ,t = (cid:90) t ν s ds, (5.28)which are analogous to the formulas of Buyarov and Rakhmanov [8] forvarying families of measures on the real line. We also have tµ Ω ,t = lim τ → − τ µ Ω ,τ − (cid:90) t ν s ds = dA ( z ) π (1 + | z | ) − (cid:90) t ν s ds. (5.29)We can calculate the logarithmic potential − (cid:90) C log | z − s | dA ( s ) π (1 + | s | ) = − 12 log (cid:0) | z | (cid:1) , z ∈ C . Therefore by (5.29) and (5.25), we have for every z ∈ C , tU µ Ω ,t ( z ) = − 12 log (cid:0) | z | (cid:1) − (cid:90) t U ν s ( z ) ds ≥ − 12 log (cid:0) | z | (cid:1) + (cid:90) t (cid:18) r + 1 log (cid:12)(cid:12) z r +1 + q − (cid:12)(cid:12) − C ,s (cid:19) ds = − 12 log (cid:0) | z | (cid:1) + 1 − tr + 1 log (cid:12)(cid:12) z r +1 + q − (cid:12)(cid:12) + tc ,t , (5.30)with c ,t = − t (cid:90) t C ,s ds .Equality holds in (5.25) for z ∈ Ω t which implies that equality holds in(5.30) for z ∈ (cid:92) t ≤ s< Ω s = Ω t , since Ω t ⊂ Ω s whenever t < s . The proof of part (c) is complete.56 .7 Proof of part (d) Proof. We obtain for every z ∈ C , using (5.28) and (5.24), tU µ Ω ,t ( z ) = (cid:90) t U ν s ( z ) ds ≤ (cid:90) t ( U ρ s ( z ) + C ,s ) ds = tU µ ∗ t ( z ) + tc ,t with c ,t = 1 t (cid:90) t C ,s ds. Equality holds, by Lemma 5.5 (a), for z ∈ (cid:92)