Constrained Energy Problems with Applications to

Constrained Energy Problems with Applications to Orthogonal Polynomials of a Discrete Variable P. D. Dragnev, E. B. Say

Abstract

Given a positive measure  with kk > 1 we write  2 M if  is a probability measure and  ?  is a positive measure. Under some general assumptions on the constraining measure  and a weight function w we prove existence and uniqueness of a measure w that minimizes the weighted logarithmic energy over the class M . We also obtain a characterization theorem, a saturation result and a balayage representation for the measure w . As applications of our results we determine the (normalized) limiting zero distribution for ray sequences of a class of orthogonal polynomials of a discrete variable. Explicit results are given for the class of Krawtchouk polynomials.

1 Introduction In this paper we shall investigate constrained energy problems in the presence of an external eld. Before de ning the problem, we brie y recall the classical and the weighted energy problems of potential theory. In so doing, we introduce the terminology that will be needed for stating our results. The classical approach starts with the concepts of energy of measures and capacity of sets (cf. [T],[L]). Let E be a compact subset of the complex plane  The research done by this author is in partial ful llment of the Ph.D. requirements at the University of South Florida. y The research done by this author was supported, in part, by U. S. National Science Foundation under grant DMS-9501130.

1

C and ME be the collection of all probability measures with support in E .

The logarithmic energy of a measure  is de ned by ZZ 1 d(z)d(t) I () := log jz ? (1.1) tj and the quantity VE := inf fI () j  2 ME g is called the Robin's constant. If VE < 1, there exists unique measure E 2 ME such that I (E ) = VE . The extremal measure E is called the equilibrium measure of the set E . The logarithmic capacity of E is given by cap(E ) := e?V : (1.2) For an arbitrary nite Borel measure  with compact support, the logarithmic potential of  is de ned by Z (1.3) U  (z) := log jz ?1 tj d(t): We denote the support of  by S . The following properties of the equilibrium measure were established by O. Frostman [F]: If cap(E ) > 0, then U  (z)  VE for all z 2 C; (1.4)  U (z) = VE q.e. on E; where by a property satis ed q.e. (quasi-everywhere) we mean everywhere except on a subset of capacity zero. For the weighted energy problem we shall follow the scheme given in [ST]. Let E be a closed subset of C (we no longer require boundedness of E ). De nition 1.1 A weight function w : E ! [0; 1) is said to be admissible if it satis es the following conditions: a) w is upper semi-continuous; b) E := fz 2 E jw(z) > 0g has positive capacity; E

E

E

0

2

c) If E is unbounded, then jzjw(z) ! 0 as jzj ! 1, z 2 E . We de ne the external eld Q = Qw by w(z) = e?Q z . Then Q : E ! (?1; 1] is lower semi-continuous, Q(z) < 1 on a set of positive capacity and, if E is unbounded, then lim fQ(z) ? log jzjg = 1: jzj!1 ; z2E ( )

For  2 ME the weighted energy integral is de ned by

Iw () :=

ZZ

log [jz ? tjw(z)w(t)]? d(z)d(t) 1

Z

= I () + 2 Q d:

(1.5)

>From the admissibility of w we have that the rst integral is well-de ned. The second equality holds when both integrals are nite. Furthermore, the quantity Vw := inf fIw ()j 2 ME g is nite and there exists a unique measure w 2 ME such that Iw (w ) = Vw . The measure w is called the extremal measure associated with w and has the following properties: U  (z) + Q(z)  Fw q.e. on E; (1.6) U  (z) + Q(z)  Fw for all z 2 S ; R where Fw := Vw ? Q(t) dw (t). In the unweighted case (w  1) on the interval [?1; 1], Rakhmanov [R] introduced and studied the problem of minimal energy of measures that are subject to a constraint. Such measures arise in connection with the zero distribution of the Chebyshev polynomials of a discrete variable (cf. [Sz, x2.8]), ! ! x ? N x n n = 0; 1; :::; N ? 1: tn(x) = n!4 n n Here our goal is to study the weighted analogue of the constrained minimal energy problem in a more general geometric setting. In so doing, we also obtain new general results for the unweighted case. w

w

w

3

De nition 1.2 Let w : E ! [0; 1) be admissible weight function. A nite

Borel measure  is called an admissible constraint for w if (i) S = E ; (ii) (E ) > 1, where E is de ned in De nition 1.1(b); (iii)  has nite logarithmic energy over compact sets, i.e. for any compact set K , 0

0

ZZ

K K

1 d(z)d(t) < 1: log jz ? tj

Next we de ne the class of measures M := f j kk = 1 and 0    g; (1.7) where by    we mean that  ?  is a positive Borel measure. As a consequence we have that if  2 M , then S  S = E . The analogue of the Robin's constant for this problem is given by Vw := inf fIw () j  2 M g; (1.8) where Iw () is de ned in (1.5). De nition 1.3 A measure  2 M such that Iw () = Vw is called a constrained extremal measure for the weight w. Such a measure will be denoted by w . In the unweighted case (w  1) we omit the subscript w. The constrained energy problem deals with the existence and uniqueness of the extremal measure in question, its characterization and the investigation of its properties. The following electrostatical interpretation justi es our interest in this type of problem. Let E be a plane conductor and place a positive unit charge on E . If the force between two charged particles is proportional to the reciprocal of their distance, then the distribution of the charge with minimal energy will be E . If, in addition, an external eld Q = log(1=w) is present, then the solution will be w . Now what if we relax the condition that E is a conductor and assume, more generally, that E consists of matter with varying conductivity, what is known as continuous media (see 4

[LL])? Then one way to accommodate that generality in our mathematical model is to invoke a constraint  on E . Remark 1.4 Notice that if   w , then the extremal measure will not depend on the constraint and will coincide with w . In this sense, the constrained energy problem generalizes the weighted energy problem. The outline of this paper is the following. In Section 2 we state our main results, the analogue of Frostman's characterization theorem, the saturation property, and a balayage representation of the extremal measure, as well as some corollaries. The proofs are presented in Section 5. Section 3 is devoted to studying the asymptotic behavior of the zeros of Lp-extremal polynomials generated by a distribution with jumps at N points. In particular we obtain explicit results for ray sequences for the Krawtchouk polynomials. In Section 4 we nd explicitly the constrained extremal measure for several special cases.

2 Statements of main results. In this section we state our main results, deferring their proofs to Section 5. We start with the fundamental theorem, which asserts the existence and uniqueness of the constrained extremal measure and also provides us with a criterion for recognizing that measure.

Theorem 2.1 Let w = e?Q be an admissible weight on a closed set E

and  be an admissible constraint for w. Let Vw be de ned as in (1.8). Then the following properties hold:

(a) Vw is nite. (b) There exists a unique measure  = w 2 M such that

Iw () = Vw : (c) There exists a constant Fw such that

U  (z) + Q(z)  Fw holds ( ? ) a.e. 5

(2.1)

and

U  (z) + Q(z)  Fw for all z 2 S:

(2.2)

(d) If  2 M has compact support and there exists a constant c such that

U  (z) + Q(z)  c holds ( ? ) a.e.

(2.3)

U  (z) + Q(z)  c holds  a.e. ;

(2.4)

and then  = .

(e) For every measure  2 M with compact support we de ne

F := ess inf ? (U  (x) + Q(x)) = maxfK 2 R j U  (x) + Q(x)  K ( ? ) a:e:g:

(2.5)

Then F  F. Moreover, if the polynomial convex hull of S [ S has two-dimensional Lebesgue measure zero, then F = F implies  = . (In particular, the last assertion holds if E  R.)

Remark 2.2 From the proof we shall see that S  E for some  > 0,  w

where E := fz j w(z)  g: Further, since w(z) is admissible, E is a compact set and so the support of the constrained extremal measure is also compact. Moreover, w has nite logarithmic energy since Iw (w ) < 1 and Q is lower bounded on compact sets. Remark 2.3 Lemma 5.2 below shows that if U  is continuous on C, so is U  and if Q is also continuous, inequality (2.1) holds everywhere on S? . If, in addition, S? \ S 6= ;, which is the case in many of the  w

 w

 w

 w

6

applications, the constant Fw is uniquely determined. The next example shows that if S? \ S = ;, then the constant Fw need not be unique. Example p 2.4 Let E = [?1; 1] [ [2; 3], and let w = exp(?x ). Then dw = 2 1 ? t dt=, t 2 [?1; 1] (cf. [MS]). Let  := w +  , where  is the Lebesgue measure restricted to the interval [2; 3]. Then clearly w = w and S = [?1; 1], S? = [2; 3]. Now it is easy to see that sup (U  (z) + Q(z)) < inf; (U  (z) + Q(z))  w

 w

2

2

 w

 w

 w

?1;1]

 w

[2 3]

[

and any number between the sup and the inf can serve as a value for Fw . Remark 2.5 The following example shows that the assumption in part (e) of Theorem 2.1 that the polynomial convex hull of S [ S have twodimensional Lebesgue measure zero is essential to conclude the equality of the measures. Let E be the union of the two circles jzj = 1 and jzj = 1=2 and  := !jzj + !jzj = , where !jzj r denotes the normalized one-dimensional Lebesgue measure on the circle jzj = r. Let w  1. Then clearly  =  = !jzj , and for  = !jzj = , we have F = F = 0, but  6= . The next theorem gives more information on the location of the support of the extremal measure. Hereafter we shall assume that Q(z) and U  (z) are continuous on E . Let w be the unconstrained extremal measure associated with the weight w on E . Then the following theorem holds. Theorem 2.6 (Saturation Property) Let  be an admissible constraint for the admissible weight w = e?Q on E . If U  and Q are continuous on E , then S  S . Moreover, (2.6) (w )jS ?  (w )jS ? : =1

=1 2

=1

=

=1 2

w

 w

  w

  w

Consequently, w =  on any positive subset for a Hahn decomposition of the signed measure w ? . Remark 2.7 Theorem 2.6 has the following interpretation: Assume that the measures w and  have continuous densities,  and respectively. Then the -constrained extremal measure w saturates the constraint  on the set B where  is greater than or equal to , i.e. (w )jB = jB .

7

It is natural to ask whether a theorem of this type holds if we consider and w , where    . With the additional assumption that S = E we are able to prove the following. Theorem 2.8 Let    be two admissible constraints for the admissible weight w on E and let  and  be the corresponding  - and  -constrained extremal measures, respectively. Assume that U  and Q are continuous on E and, in addition, that S = E . Then  jS ?   jS ? : (2.7)

w1

2

1

2

1

w

2

1

2

1

2

2

1

2

1

 1 1

1 1

Remark 2.9 We note that the assumption S = E is weaker than the 1

assumption S = E because of Theorem 2.6, but the latter is easier to verify. It is open question whether the theorem remains true without this assumption. Next we derive some corollaries from the fundamental theorem. We rst introduce the following representation: (2.8) w = (w )jS nS ? + (w )jS ? :=  +  : w



1

  w

  w

2

Then  = jS nS ? =:  and we can easily derive the following fact about . Corollary 2.10 If U  and Q are continuous on E and  (S? ) > 0, then  = c , where c = 1 ? (S n S? ) > 0 and  is the weighted equilibrium measure on S? with external eld Q := fU  + Qg=c: (2.9) 1



  w

1

2

2

 w

 w

2

 w

0

1

We now introduce the dual problem. For 0 < t < kk we de ne Mt := f j kk = t and 0    g (2.10) and let  (t; w) be the solution of Iw ( (t; w)) = inf fIw () j  2 Mt g: Note that  (t; w) = t=t w . 8

Theorem 2.1 can be reformulated for this case and then we obtain the following corollary. Corollary 2.11 With the above notations, if U  and Q = log(1=w) are continuous on E and E is compact, then  ?  (t; w) =  (s; v); (2.11) where t + s = kk and v = exp((U  + tQ)=s). Remark 2.12 The proof of this corollary follows by simply rewriting the equations (2.1) and (2.2) for  ?  (t; w). The assumptions on U  , Q, and E guarantee that the weight v is admissible. Example 2.13 Let  be a probability measure with continuous logarithmic potential and compact support E and let wc denote generically the weight wc(x) := exp(U c= ). Then if c; c0 are positive real numbers with (1=c) + (1=c0) = 1 we obtain from Corollary 2.11 the formula 1 c0  = 1c c (2.12) w + c0 w 0 :  c0 0  0 ( Consider c w = c (1=c; wc) and w 0 = c  (1=c ; wc0 ) and use (2.11).) c Now suppose 1 < c < 2 and set c := w . It is easy to show that in this case the solution of the unconstrained problem on E with weight wc is given by (see (1.6)) c ) ; w = c + (1 ? (2.13) 2 2 E where E is the equilibrium measure on E (note that the continuity of U  implies cap(E ) > 0, because  has a nite logarithmic energy). Therefore, c  c=2. we have w  c=2 and by the saturation theorem we obtain Thus, if 0 := (c=(2 ? c)), then c = c=2 + (1 ? c=2)0 , where 0 is the 0-extremal measure on E with weight w  1. This fact, when combined with (2.12) in the case c > 2, yields the general formula 2 ? c c=jc? j; + (2.14) c = c 2 2 for every c > 1, c 6= 2, and c =  when c = 2. 2

c

c

c

c

c

c

c

2

9

The following theorem provides a practical method for determining w from knowledge of the set S := fz j U  (z) + Q(z)  Fw g: (2.15) Notice that S  S (see (1.6)). Theorem 2.14 (Balayage Representation) Let E be a compact set and  be an admissible constraint for the admissible weight w = e?Q on E . Suppose U  and Q are continuous on E and let v be the solution to the unconstrained extremal problem for probability measures on E with weight v = exp((U  + Q)=(kk ? 1)). Assume further that S  S ( in particular this is true if S = S = E ). Then  :=  ? w = (kk ? 1)v . Moreover, the -constrained extremal measure w has the representation w =  ? ^ + cw ; (2.16) where ^ and cw are the balayage measures (see [L], Chapter IV) of  and w onto S . Remark 2.15 This theorem completely settles the case when S = S . In the case when w  1, the set S is the whole complex plane and the following corollary holds. Corollary 2.16 Let  be an admissible constraint with respect to w  1 on the compact set E and U  be continuous on E . Then the measure  :=  ? w = (kk? 1)v , where v is the solution to the unconstrained weighted energy problem on S with weight v := exp(U = kk? ). The -constrained extremal measure  has the representation (2.17)  =  ? ^ + S : w

w

w

w

v

w

w

w

w

(

1)



Now we pay special attention to the support of w when E is a subset of the real line. Theorem 2.17 Let E  R be closed and  be admissible constraint for w = e?Q on E . Suppose I is an open interval and I  E . (a) If Q is convex on I , then S \ I is an interval. (b) If E  [0; 1) and xQ0 (x) increases on I , then S \ I is an interval.  w

 w

10

3 Zero distribution of minimal polynomials in the discrete Lp norm. In this section we present the main application - the weighted analogue of Rakhmanov's result [R] on the zero distribution of orthogonal polynomials in a discrete variable. In particular, we obtain weak asymptotics of "ray sequences" of Krawtchouk polynomials. De nition 3.1 A triangular scheme of points fi;N gNi 1;N is called admissible for the nite interval [a; b] if: (i) There exists a positive constant  such that mini ji ;N ? i;N j > =N for every N . (ii) The measures N := (1=N ) PNi (i;N ) (here (z) is the discrete measure with mass one at z) converge weak to a measure  with support S = [a; b] and continuous logarithmic potential on C. =0

=1

+1

=0

In what follows we shall assume that fwN g is a sequence of positive continuous weights on [a; b] that converge uniformly to a positive continuous weight w. Fix p > 0 and for each N  1 and 0  n  N , let pn;N be the monic polynomial of degree n that has minimal discrete Lp norm on the set N := fi;N gNi with respect to the weight wNn , that is =0

kwNn pn;N kp;N

= =

N X i=0

[wN (i;N )]pnjpn;N (i;N )jp

min kwNn pnkp; :

pn =xn +




N

 =p 1

(3.1)

In particular, when p = 2, we obtain the orthogonal polynomials with varying weight wNn on the discrete set N . In the case p = 1 the norm above is the sup norm and the extremal polynomials are the extremal Chebishev polynomials. The next theorem deals with the asymptotics of the polynomials pn;N for pairs (nj ; Nj ), where nj ! 1, Nj ! 1 and (Nj =nj ) ! C > 1 as j ! 1. Note that then j := (Nj =nj )N ! C =: , where the measure  is an admissible constraint on [a; b] for the weight w and has a continuous logarithmic potential. 2

j

11

Theorem 3.2 Let fi;N gNi 1;N be an admissible triangular scheme and let fwN g be a sequence of continuous positive weights on [a; b] uniformly converging to a positive weight w. Let fpn;N g be the associated monic Lp extremal polynomials and (nj ; Nj ) be a sequence such that nj ! 1, and Nj =nj ! C > 1 as j ! 1. Then (with n = nj ; N = Nj ) we have lim kwn p k =n = e?F ; (3.2) j !1 N n;N p; =0

=1

 w

1

N

and

p ! w ; as j ! 1; (3.3) where p := (1=n) Pp z (z) is the normalized zero counting measure of pn;N , the measure  = C (see Def. 3.1(ii)), and w , Fw are as in Theorem 2.1. We now investigate the asymptotic zero distribution and the n-th root of the discrete norms of the Krawtchouk polynomials (cf. [Sz, x2.82]) ! ! !? = n X x pn?sqs; N ? x N n ? s ? n= (?1) kn(x; p; N ) = n (pq) n?s s s where p; q > 0; p + q = 1. These polynomials are orthonormal with respect to N N! X i N ?i i p q (i); n;N

n;N ( )=0

n;N

1 2

2

=0

i=0

where (i) is the unit measure with mass point at i, i = 0; 1; :::; N . As is well known the zeros of kn are all simple and lie in the interval (0; N ). They are also separated by the mass points of the measure of orthogonality. We shall convert the problem to the interval [0; 1] by scaling the polynomials. Let Pn(x) = Pn(x; p; N ) := An;N kn(Nx; p; N ), where the factor != N An;N = n (pq)n= n!N ?n (3.4) is chosen so that Pn has leading coecient one. Then the polynomials Pn(x) satisfy the orthogonality relation N N! X i q N ?i P ( i )P ( i ) = A  ; n; m = 0; 1; :::; N: p (3.5) n n;N n;m i N mN i 1 2

2

2

=0

12

Also Pn is the discrete L -minimal polynomial, i.e. N N! X i q N ?i p ( i )g = A ; n = 0; 1; :::; N: (3.6) p kPnk = p min f n N n;N x


i i 2

2 N

2

n=

n+

2

=0

Here the triangular scheme of points is N = fi=N gNi . We shall nd the asymptotics of Pn when Nj =nj ! C > 1 as j ! 1. Consider the weights N  i Nw?Ni (=x)n to be piecewise linear with nodes at i=N and wN (i=N ) = [ i pq ] . It is not dicult to show that wN (x) ! w(x) uniformly on [0; 1], where w(x) = exp[?(C=2)(x log x + (1 ? x) log(1 ? x) + x log(q=p) ? log q)]: Since the triangular scheme N and the sequence of weights fwN g satisfy the assumptions of Theorem 3.2, we obtain that P ! w in the weak sense, and moreover, limj!1 kPnk=n = exp(?Fw ), where d = Cdx. In the case when p = q = 1=2, known as the binary case, the constrained energy problem that arises is analyzed in Example 4.2 of the next section. Thereby we obtain the following theorem. Theorem 3.3 Let kn(x; p; N ) be the Krawtchouk polynomials and Pn(x) := Ankn(Nx; p; N ) be the associated normalized monic polynomials. Consider the sequence N = Nj ; n = nj and Nj =nj ! C > 1 as j ! 1. Then the normalized zero counting measures of Pn and the n-th root of the discrete norms satisfy P ! w as j ! 1 (3.7) and lim kP k =n = e?F ; (3.8) j !1 n  =0

1 (2 )

n

1 N

n

1 N

 w

where

w(x) = [xx(1 ? x) ?x=pxq ?x]?C= and d = Cdx on [0; 1]. In particular, if p = q = 1=2, then P ! C2 dx + 2 ?2 C Cdx=jC ? j as j ! 1 1

n

1

2

2

13

(3.9)

and

C C? = =n = lim k P k j !1 n  2e(1 ? C ) C ? (

1 N

2) 2 (

=

1) 2

;

(3.10)

where Cdx=jC ?2j =: C (x) dx is the solution to the constrained energy problem on [0; 1] with weight w  1 and constraint measure Cdx=jC ? 2j. The density C is given by

8 C > x 2 [0; r] [ [1 ? r; 1]; > > > < jC ? 2j q 1 0 C (x) := > r (1 ? r ) > 2 C > > : jC ? 2j arctan @ q(r ? x)(1 ? r ? x) A x 2 [r; 1 ? r];

p

where r = 1=2 ? C ? 1=C . If C = 2, the limit measure in (3.9) is the Lebesgue measure dx and the constant on the right in (3.10) is 1=(2e). Remark p 3.4 Notice that p the limit measure in (3.9) has support [r; 1?r] = [1=2 ? C ? 1=C; 1=2 + C ? 1=C ] when C > 2, and has support [0; 1] for 1 < C  2. Thus, if x1(n;N ) denotes the smallest zero of kn(x; 1=2; N ) we have p (n;N ) C ? 1 ; for C > 2; x 1 1 lim sup  ? N 2 C j !1 and

n;N x = 0; for 1 < C  2: lim j !1 N This fact was also proved, using elementary methods, by McEliece et al [MRRW] and plays an important role in their analysis of the rate of a binary code. Recently it was shown (see [Le]) that the lim sup in the above estimate can be replaced with the ordinary limit. Remark 3.5 Note that we can nd the n-th root limit in (3.8) directly from the expression for An;N , namely C ? = ppq C =n lim A = ; j !1 n;N e(C ? 1) C ? = which in the case p = q = 1=2 is precisely what we obtain in (3.10). ( 1

1

)

(

2) 2 (

1) 2

14

4 Examples. In this section we provide some examples to illustrate the results and the ideas introduced in the previous sections. We begin with the example given by Rakhmanov in [R], which has application to the zero distribution of the Chebyshev orthogonal polynomials of a discrete variable.

Example 4.1 (Rakhmanov) Let E = S = [?1; 1], w  1 and d = C dt=2 on [?1; 1], where kk = C > 1. Obviously  is an admissible con-

straint and U  is continuous. >From Corollary 2.16 we have that ( ?  )=(C ? 1) is a solution to the weighted energy problem on [?1; 1] with weight exp(U  =(C ? 1)). The corresponding external eld Q is Z Q(x) = ?U  (x)=(C ? 1) = 2(CC? 1) log jx ? yj dy ? C f(1 ? x) log j1 ? xj + (1 + x) log j1 + xj ? 2g: (4.1) = 2(C ? 1) Since Q00 (x) = C=[(C ? 1)(1 ? x )] > 0, the function Q(x) is convex on (?1; 1) and therefore S? is an interval (cf. [ST, Theorem IV.1.10]). By the symmetry of Q, the support S? must be of the form [?r; r], where r maximizes the Maskhar-Sa functional (see [MS]) 1 Z r U  (x) p 1 F (r) = log cap([?r; r]) + C ? dx 1 ?r  r ? x Z C = log(r=2) + U  ? (y) dy (4.2) 2(C ? 1) ? p p = log(r=2) + C flog 2 ? log(1 + 1 ? r ) + 1 ? r g: C?1 Now solving Cr p = 0; F 0(r) = 1r ? (C ? 1)(1 + 1 ? r ) p we get that the maximum of F (r) is achieved when r = 1 ? C ? . >From Corollary 2.16 we obtain that  =  ? ^ +  ?r;r ; (4.3) 1

1

2





2

1

2

[ r;r ]

1

2

2

2

2

[

]

15

p

where ^ is the balayage of  on [?r; r] and  ?r;r = dx=( r ? x ) is the equilibrium measure for the interval [?r; r]. For the density of ^ we have (see [L, Chapter IV]) p t p? r d^ = 1 Z ?r + Z  C (4.4) d ( t ) + dx  ? 2 r jx ? tj r ? x Z pt ? r C = p 2t dt + C 2 2 r ? x r t ? x p p C 1 ? r 1 ? r + C ; x 2 [?r; r]: C p ? = p arctan (4.5) r ?x 2  r ?x  p Using (4.3) and the fact that C 1 ? r = 1, we obtain that d = fC (x)dx, where 8C > x 2 [?1; ?r] [ [r; 1]; > > < 2 fC (x) := > (4.6) p ! 1 ? r C > > x 2 [?r; r]: :  arctan pr ? x [

2

1

2

2

2

2

2

2

2

2

2

2

2

2

2

1

1

]

2

2

2

2

2

2

2

2

To compute the constant F  we consider U  (x) = U  (x) ? U  (x) + U  ? (x) = F  for every x 2 [?r; r]. We also have that 

^

U  (x) ? U ^ (x) = ?

[ r;r ]

Z ?r Z  ?1

1

+

g ?r;r (x; 1) d(t); [

r

]

where g ?r;r (x; 1) is the Green function for the domain C n [?r; r] with pole at 1, and U  ? (x) = log(2=r). So Z ?r Z  2  g ?r;r (x; 1) d(t): (4.7) + F = log r ? r ? The Green function is given by p j x + x ? r j; g ?r;r (x; +1) = log r [

]

[ r;r ]

1

1

[

]

2

[

2

]

16

p

where the branch of x ? r is positive for x > r. Thus p Z 1 j x + x ? r j dx 2  log F = log r ? p 1?r r p r ! = log 2r ? p 1 log 1 + r1 ? r + 1: 1?r 2

2

2

1

2

(4.8)

2

2

2

2

The example below illustrates the application of our results for nding explicitly the weak asymptotics of the Krawtchouk polynomials in the binary case (p = q = 1=2). The general case requires ner analysis and will be considered in a future paper. Example 4.2 Let E = [0; 1], the constraint measure d = Cdx, C > 1, and the weight function w(x) = exp[?(C=2)(x log x + (1 ? x) log(1 ? x) ? 1)]. Then Q(x) = (C=2)[x log x + (1 ? x) log(1 ? x) ? 1] = ?U = (x): (4.9) >From Example 2.13 with  = dxj ; we obtain that w = C2 dx + 2 ?2 C Cdx=j ?C j: (4.10) and Fw = 2 ?2 C F Cdx=j ?C j; where Cdx=jC ? j is the solution of the constrained problem on [0; 1] with constraint measure Cdx=jC ? 2j and weight w  1. The constant F Cdx=jC ? j is the corresponding extremal constant. To nd this constant we observe that C= dx F Cdx ; = log 2 + F ? ; ; where F a;b denotes the extremal constant of the -constrained energy probp lem on [a; b] with w  1. Now using (4.8) with r = 2 C ? 1=C we compute that Fw = 2 ?2 C flog 2 + log p C ? jC C? 2j log C +p jC ? 2j + 1g C ?1 2 C?1 2 ? C C ? 1 = 2 (1 + log 2 + log C ) + 2 log(C ? 1); (4.11) 2

[0 1]

2

2

2

2

( 2) [ 1 1]

[0 1] [

]

17

for the case when C 6= 2. In the case when C = 2 formula (4.10) reduces to w = dxj ; and the constant Fw = 0. 2 [0 1]

The next example deals with the case when the constraint is the wellknown Ullman distribution, with normalized support on [?1; 1]. Example 4.3 Let E = S = [?1; 1], w  1, d = C l Z p ul? du for t 2 [?1; 1]; dt  jtj u ? t where l > 0 and kk = C > 1. Then  is an admissible constraint with continuous logarithmic potential. Hence by Corollary 2.16 the dual extremal measure ( ?  )=(C ? 1) is the solution to the unconstrained weighted energy problem on [?1; 1] with external eld Z Z ul ? C 1 l  ? C ? 1 U (x) = C ? 1 ? log jx ? tj(  jtj pu ? t du)dt = ? C U  (x); (4.12) C?1 where w is the weighted equilibrium measure for the weight w = exp(? l jxjl ) with ?(l)?( 21 )

l := 2?(l + 1 ) 2 (cf. [MS]). Therefore, by the equilibrium conditions (1.6), ?U  =(C ? 1) coincides with [C=(C ? 1)] ljxjl +const. everywhere on [?1; 1]. So, by [ST, Theorem IV.5.1] we obtain that d( ?  ) = C l Z r p ul? du for t 2 [?r; r]; (4.13) dt  jtj u ? t where C (1 ? r ) = 1. Consequently the density of the -extremal measure is given by 8 Z > l p ul? du > C r < jtj  1; > d = <  jtj u ? t (4.14) ! Z ul? Z r ul? dt > > l > : C  jtj pu ? t du ? jtj pu ? t du jtj  r: 1

1

2

2

1

2

1

w

1

2

2

1

1

2

2

1

1

2

1

1

1

2

2

18

2

2

For the extremal potential we have 8 > = Fw ? Cr log 1 > < r x 2 [?r; r]  U (x) > (4.15) > 1 :  Fw ? Cr log r x 2 [?1; ?r] [ [r; 1]; where Fw = log 2 + 1=l (cf. [ST, Theorem IV.5.1]). In particular, when l = 2 we obtain 8 2C p > 1?x x 2 [?1; ?rC ] [ [rC ; 1] >  d = <  (4.16) p  > q dx > 2 C : x 2 [?rC ; rC ]:  1 ? x ? rC ? x where rC satis es C (1 ? rC ) = 1. The constant F  is given by 1 log C : F  = log 2 + 1l ? C ? (4.17) 2 C?1 

2

2

2

2

2

2

The next example deals with a case where the support of the constraining measure is the unit disk. Example 4.4 Let  = (C=)m , where m is 2-dimensional Lebesgue measure on the unit disk fz 2 C j jzj  1g and C > 1, i.e. d = C  d d; 0    1; ?   < : Clearly  is an admissible constraint with respect to w  1 and its logarithmic potential is continuous in C. Let  be the extremal measure. Then by Theorem 2.14 we get that  :=  ?  is a solution of the unconstrained weighted energy problem with weight exp(U  ) and norm k k = C ? 1. The external eld ?U  can be computed as follows: Z Z C  ?U (z) = ?  f ? log jz ?1ei j dg d 2

1

0

19

2

Z jzj

Z 1 = ?2C log jzj  d ? 2C  log 1 d jzj Z 1 1 = ?C jzj log jzj + C jzj log jzj ? C  d 1

0

2

1

2

jzj

(4.18) = ? C2 + C j2zj ; for jzj  1. Note that ?U  = C log jzj for jzj > 1. Now from we obtain that  ?  = (C=)m on q [ST, Example IV.6.2], jzj  r := (C ? 1)=C . So  =  ?  = jD , where Dr; is the annulus fz : r  jzj  1g. The constant F  is given by 2

2

r;1

1

D

F  = U  (r) = U j (r) Z = 2C  log  d r = C2 ? Cr log 1r ? Cr2 : 

r;1

1

2

2

2

In the next example we solve the constrained extremal problem in the presence of a weight w 6 1. p Example 4.5 Let w = e?px and d = Cdx=( 1 ? x ); C > 1 on E  = [?1; 1]. Then dw = 2 1 ? x dx= (cf. [MS]). We shall transfer ?x the problem to [0; 1]. The new weight q becomes w = e , the weighted equilibrium measure isq dw = (2=)( (1 ? x)=x)dx, and the constrained measure is d = C=( x(1 ? x))dx on E = [0; 1]. If C  2, the example is trivial; therefore we shall consider the case 1 < C < 2. By Theorem 2.14, we have that  = (C ? 1)v , where v is the solution to unconstrained problem on [0; 1] with weight v = exp((U  + Q)=(C ? 1)). The corresponding external eld Q is Q = ?(U  + Q)=(C ? 1) = ?(C log 2 + 2x)=(C ? 1): Since Q00 = 0, the support S is an interval. For Pn  1 we have kvnPnk ; = v(1)n, which implies by the peaking theorem (cf. [ST, Theorem IV.1.3) that 1 2 S . So the interval is of the form [a; 1]. Now [ST, 2

2

2

2

1

1

1

[0 1]

20

Theorem IV.1.11] gives us

s

1 Z (? 2 ) 1 ? x dx = ?1: (4.19)  a C ?1 x?a Using the substitution x = (1 + a)=2 ? ((1 ? a) cos )=2, we obtain 1 ? a Z  (1 + cos ) d = C ? 1 ; 2 2 from which we nd that a = 2 ? C . So S = [2 ? C; 1] and by (2.16) we obtain w =  ? ^ + cw ; (4.20) where the balayage measures are taken onto [2 ? C; 1]. Now we nd the balayage measures in question. Since  is the equilibrium measure on [0; 1] times the constant C , its balayage onto [a; 1] will be the equilibrium on [a; 1] times C , therefore d^ = q C (4.21) dx  (x ? a)(1 ? x) on [a; 1]: 1

0

Let w =  +  , where  = (w )j ;a . Then cw = c +  . So for x 2 [a; 1] we nd (see [L, Chapter IV]) 1

2

1

[0

]

1

2

q s s (a ? t)(1 ? t) dcw = 1 Z a 2 1 ? t 2 q dt + 1?x
dx  (x ? t) (x ? a)(1 ? x)  t  x s Z a 1 ? t sa ? t 1 2 2
 x ? t t dt +  1 ?x x = q  (x ? a)(1 ? x) s 2 x+ q a (4.22) =  x1 ? ? a  (x ? a)(1 ? x) : Here we used the fact that s s 1 Z a 1 a ? t dt = 1 ? x ? a ;  x?t t x 0

0

0

21

and

s

1 Z a a ? t dt = a :  t 2 Now substituting (4.21) and (4.22) in (4.20) (recall that a = 2 ? C ), we obtain the density of the -constrained extremal measure w as follows 8 > q C x 2 [0; a]; > >  x (1 ? x ) > d = < (4.23) s > dx > C 2 x?a > x 2 [a; 1] : > : qx(1 ? x) ?  1 ? x 0

If we transfer back the problem to the interval [?1; 1], the solution  := w is

8 C > > > p  d = <  1 ? x p > dx > x ?r C ? 2 x > : p1 ? x

x 2 [?r; r];

2

2

2

p

2

x 2 [?1; ?r] S[r; 1];

(4.24)

p

where r = a = 2 ? C . We see that if C ! 2?, then  tends to w , which is the solution of the unconstrained weighted problem. Indeed, if C  2, the density of the constraint measure lies above the density of w , and that is why it provides no restriction.

2

5 Proofs of main results.

Proof of Theorem 2.1: (a) First we note that the function log[jz ? tjw(z)w(t)]? 1

22

is bounded from below on E  E , which is an easy consequence of the de nition of an admissible weight; therefore Iw () is well-de ned for any  2 M and Vw > ?1. >From the de nition of E in Remark 2.2 we have

E = 0

1 [

n=1

E =n :

(5.1)

1

Since (E ) = C > 1 and  is regular Borel measure there exists n such that (E =n ) = C > 1. Then   = (EjE ) 2 M =n and Z Iw ( ) = I ( ) + Q d 0

1

0

0

0

1=n0

1

0

is nite, because I (jE

R

1=n0

) and Q d are nite. Therefore Vw is nite.

(b) The uniqueness follows exactly as in the unrestricted case (see [ST, Theorem I.1.3 (b)]). Thus we need only prove existence. First we verify the following Claim: For suciently small  > 0

Vw = inf fIw () j  2 Mj g: (5.2) Since E is compact the existence will then follow from standard compactness arguments. E

Proof of the Claim: First we observe that for every K > 0 there exists an  > 0 such that log [jz ? tjw(z)w(t)]? > K if (z; t) 2= E  E: (5.3) Indeed, assume to the contrary that there exist K > 0 and a sequence f(zn; tn)g1n such that (5.4) nlim !1 min (w(zn); w(tn)) = 0 1

=1

23

and log [jzn ? tn jw(zn)w(tn)]?  K for all n: (5.5) Now choose any convergent subsequence, say f(zn; tn)gn2N (zn; tn) ! (z; t) as n ! 1; n 2 N ; where z or t could be 1. If they are both nite, we get a contradiction with (5.5) because of (5.4). If one or both of z, t are in nite, then the contradiction with (5.5) follows from the admissibility of w; see De nition 1.1 (iii). Next we observe that for each K > 0 there exists an  > 0 such that log [jz ? tjw(z)]? > K if t 2 E and z 2 E n E: (5.6) Indeed, it is enough to prove that if the sequence f(zn; tn)g is such that w(zn) ! 0 as n ! 1 and w(tn)  w(zn), then log[jzn ? tnjw(zn)]? ! 1 as n ! 1: (5.7) To see this, choose a convergent subsequence f(zn; tn)gn2N such that zn ! z and tn ! t, where z or t could be 1. If both z and t are nite, then (5.7) follows because w(zn) ! 0. If t is nite and z = 1, then the result follows from the admissibility of w (see Def. 1.1 (iii)), because jzn ? tnjw(zn)  jznjw(zn) + jtnjw(zn) and the right-hand side tends to zero. If t = 1 and z is nite then jzn ? tnjw(zn)  jzn ? tnjw(tn) and the argument is the same. And nally, if both z and t are 1, then jzn ? tnjw(zn)  jznjw(zn) + jtnjw(tn) and we again use the admissibility of w. Now we choose n as in part (a), i.e. (E =n ) = C > 1. The constant K = K (n ) > 0 will be chosen later. It will actually depend on n , the set E =n , w, and . Next we choose 0 <  < 1=n so that (5.3) and (5.6) hold. In order to verify (5.2) for this choice of , it suces to show that if  2 M is a measure with (Ec) =  > 0 (here Ec := C n E ), then there is a measure  2 Mj with Iw () < Iw (). 1

0

0

1

1

0

1

0

1

0

0

0

0

0

E

24

Now let  be such a measure. We decompose  as  =  +  , where  := jE and  := jE . Note that k k =  > 0. Since ( ? )(E =n ) > C ? 1 +  > 0, the measure ( ? )  := ( ? )(EjE ) (5.8) =n is a positive unit measure supported on E =n , such that  (5.9)   CjE? 1 : Then the measure  :=  +  2 Mj . We will show that Iw () < Iw () which will complete the proof of (5.2). For this purpose we introduce the notation ZZ < ;  >w := log [jz ? tjw(z)w(t)]? d(z)d (t): Consider the dierence Iw () ? Iw () =<  + ;  +  >w ? <  +  ;  +  >w =  < ;  >w +2 Z< Z;  >w ? <  ;  >w ?2 < ;  >w   < ;  >w +2 log jz ? t1jw(z) d(t)d (z) c 



1

0

0

1=n0

1

0

1

0

1=n0

0

E

1

2

2

? K ? 2(1 ? )K;

(5.10) where the w(t) term in the integral is eliminated on combining the terms 2 < ;  >w and 2 < ;  >. Next we nd upper bounds independent of  and  for the two integrals above. Note that ZZ I := < ;  >w  2N + log 1 d (z)d (t); jz ? tj where N > 0 is an upper bound for Q(z) = ? log w(z) on E =n . Such a bound exists because Q(z) is lower semi-continuous and E =n is a compact set. Now, if d := diam(E =n ), the inequality can be continued as ZZ d d (z)d (t) ? log(d) I  2N + log jz ? tj Z Z d d(z)d(t)  2N ? log(d) + (C ?1 1) log jz ? tj 2

1

1

1

1

0

0

0

1

=: K ;

0

2

E1=n0 E1=n0

1

25

(5.11)

where K depends only on the choice of n . For the second integral we have ZZ I := log jz ? t1jw(z) d(t)d (z) ZZ  (1 ? )N + log jz ?1 tj d(t)d (z) Z Z  N + A\E E log jz ?1 tj d(t)d (z); where A := ft : dist(t; E =n ) < 1g. This is so because for t 2 E n A the logarithm is negative. Since diam(A)  d + 2, we have Z Z 2 d (t)d (z) I  N ? (1 ? ) log(d + 2) + log jdz + ? tj  A\E E Z Z  N ? log(d + 2) + C 1? 1 A A log jdz +? 2tj d(t)d(z) =: K : (5.12) We see that K like K depends only on n and the set E =n . Therefore by choosing K > max(K ; K ) we guarantee that the upper bound in (5.10) is negative which completes the proof of the claim. 1

0

2



1=n0

1

0

2



1=n0

0

2

2

1

1

0

1

0

2

By the de nition of Vw , there exists a sequence of measures fng  Mj such that Iw (n) ! Vw as n ! 1. Since supp(n )  E and E is compact, Helly's theorem asserts that there is a subsequence, which we again denote by n, that converges weak to a probability measure . We claim that  2 M . Indeed, if f is positive continuous function with compact support, then E

Z

f d( ? n)  0

for n ! 1, we obtain from the weak convergence that R f dall( n?. )Letting  0, which proves our claim. Now since w is upper semi-continuous, there exists a sequence of positive continuous functions wm(z) on E such that wm & w. The functions ?  Gm(z; t) := log max( m1 ; jz ? tj)wm(z)wm (t) 1

26

converge monotone increasingly on E  E to log[jz ? tjw(z)w(t)]? and so, from the monotone convergence theorem, we have 1

Iw () = mlim !1

ZZ

Gm(z; t) d(z)d(t)  ZZ  = mlim lim G ( z; t ) d ( z ) d ( t ) (5.13) m n n !1 n!1 ZZ  mlim inf log[jz ? tjw(z)w(t)]? dn(z)dn(t) !1 lim n!1  = nlim !1 Iw (n) = Vw : 1

Thus  is an extremal measure subject to the constraint  and the weight w, which proves the existence assertion.

(c) Assume to the contrary that there exists an x 2 S and a set K  S? , such that ( ? )(K ) > 0 and U  (x ) + Q(x ) > l > l > U  (x) + Q(x) for all x 2 K for some constants l and l . Since U  (x) + Q(x) is lower semi-continuous we obtain that there is a disk B(x ) = fx : jx ? x j < g, such that U  (x) + Q(x) > l for all x 2 K , where K := B(x ) \ S . Clearly, (K ) > 0 and K \ K = ;. Now we de ne a signed measure  to be (?) on K , ( ? ) on K and zero elsewhere. If we choose 0 <  < 1 and 0 < < 1 such that (K ) = ( ? )(K ), we guarantee that  is a signed measure with total mass zero and ( + ) 2 M for all  > 0 small enough. For the weighted energy of ( + ) we have Z  Iw ( + ) ? Iw () = 2 U  (x) + Q(x) d(x) +  I () < 2 [l ( ? )(K ) ? l (K )] +  I () (5.14) = ?2(l ? l )(K ) +  I () < 0; for suciently small  > 0, which contradicts the fact that  has minimal weighted energy in the class M . 0

2

2

0

0

1

2

1

2

2

0

1

1

1

1

0

1

1

0

2

2

1

2

2

2

2

1

2

27

1

1

1

2

2

(d) Let  2 M have compact support and satisfy the assumptions (2.3) and (2.4), and let  be any element of M . Then we have the representation

Z Iw ( ) ? Iw () = 2 [U  (x) + Q(x)] d( ? )(x) + I ( ? ) Z  2 [U  (x) + Q(x)] d( ? )(x):

(5.15)

Here we use the fact that I ( ? )  0 (see [H]). To complete the proof it suces to show that Z (U  + Q) d( ? )  0: (5.16) Indeed, if (5.16) is true, then Iw ()  Iw ( ) for all  2 M and therefore  = w . Set f (z) := U  (z) + Q(z) ? c; where c is the constant appearing in (2.3) and (2.4). Then, since ( ? )(C) = 0, inequality (5.16) is equivalent to

Z

f (z) d( ? )(z)  0:

(5.17)

To establish this inequality we write

Z

f d( ? ) = (

Z

E

+ ?

Z

E+

)f d( ? );

where E ? := fz 2 E j f (z) < 0g ; E := fz 2 E j f (z) > 0g: Since    and ( ? )(E ?) = 0, we have +

Z

E?

f d( ? ) =

Z

E?

f d( ?  +  ? ) =

Also, because (E ) = 0,

Z

E+

+

f d( ? ) =

Hence (5.17) holds.

Z

E+

f d  0: 28

Z

E?

f d( ? )  0:

(e) Assume that there exists  2 M with compact support such that F > F . By (2.2) we have that U  (x) + Q(x)  F for every x 2 S. Let  := jS ? ,  := jS ? , and  := jS ? . Set  :=  ?  ,  :=  ?  , and  :=  ?  . Then, by Lemma 5.1 below,  (S \ S? ) = (S? ) > 0. We have also that  =  ; therefore  ?  is a positive measure. The assumption F < F implies that there exists an  > 0, so that U  (x)  U   ? (x) ? ; ( ? ) a.e. on S \ S? = S : (5.18) Let A be the set of points x 2 S such that (5.18) is not satis ed. Then ( ? )(A) = 0, i.e.  jA = jA. This implies that  jA   jA. De ne 0 :=  jS nA and 00 :=  jA. Observe that  ? 00 is a positive measure. If 0 is the zero measure, then we are done because  ?  will be a positive measure with total mass zero, which leads to the contradiction  = . Thus, we assume that k0 k > 0. The inequality (5.18) can be rewritten as U 0 (x)  U  ?00  ? (x) ? ; 0 a.e. (5.19) The norms of the two measures in (5.19) are equal and the logarithmic energy of 0 is nite; therefore by the principal of domination the inequality holds everywhere on C, i.e. U  (x)  U  (x) ?  for all x. Letting x ! 1 we obtain the desired contradiction. This proves the rst part of (e). If F = F, we obtain in the same way that U  (x)  U  (x) everywhere in C. Then the function U ? (z), which is nonnegative and harmonic everywhere in Cn(S [ S), even at 1, achieves its minimum at 1. Therefore it is identically zero in the unbounded component of C n (S [ S ). The unicity theorem (see [ST, II.2.1]) then implies that  = . This completes the proof of the theorem. 2 Lemma 5.1 For any ;  2 M , the equality  (S? ) = 0 implies  = . Proof: Assume that the measures  and  are in the class M , and that  (S? ) = 0. Let A be any Borel subset of S . Then  (A) =  (A \ S? ) +  (A \ Sc ? ) =  (A \ Sc ? )  (A \ Sc ? ) = (A \ Sc ? )  (A): This shows that  ?  is a positive measure with total mass zero. Therefore,  = . 2 1

 

2

1

1

 

2

 

1

2

1+ 2

2

2

2

2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

2

1

1

( 1

1 )+( 2

2)

1

1

29

1

1

Now we introduce the following simple, but very useful lemma. Lemma 5.2 If  is positive measure whose logarithmic potential U  (z) is continuous on C and  is a positive measure such that   , then U  (z) is continuous on C. Proof: We shall use the fact that a potential of a positive measure is lower semi-continuous on C. Then, by the representation U  (z) = U  (z) ? U ? (z); we have that U  (z) is simultaneously lower and upper semi-continuous, and therefore continuous. 2 Proof of Theorem 2.6: Let w be the extremal measure subject to the constraint . By the continuity of U  and Q, and Lemma 5.2, the equilibrium conditions become: U  (z) + Q(z)  Fw for z 2 S? ; (5.20)   U (z) + Q(z)  Fw for z 2 S : (5.21) The equilibrium conditions for w are (see (1.6)): U  (z) + Q(z)  Fw for q.e. z 2 S (5.22)  U (z) + Q(z)  Fw for z 2 S : (5.23) Combining (5.21) and (5.22) we get U  (z) ? U  (z) + Fw  Fw for q.e. z 2 S : By the principle of domination (see [L, Theorem 1.27]) the last inequality is true everywhere. On the other hand, for z 2 S \ S? , we get from (5.20) and (5.23) U  (z) ? U  (z) + Fw  Fw ; i.e., equality holds on this set. From [ST, Theorem IV.4.5], we get (w )jS ? \S  (w )jS ? \S :  w

 w

 w

 w

w

w

w

 w

w

 w

w

 w

 w

w

  w

  w

w

w

Since on S? n S the measure w  0, we can extend the inequality to the whole set S? as stated in (2.6). This also proves that S  S .  w

w

 w

w

30

 w

To prove the last assertion, assume to the contrary that there is a positive set U for (w ? ) such that (U ) > w (U ). Let V := U \ S? . Then ( ? w )(V ) > 0 and by (2.6) we have w (V )  w (V ). But (V )  w (V ) (since V  U ), which yields a contradiction. 2  w

Proof of Theorem 2.8: The proof is a slight modi cation of the proof

of Theorem 2.6. By the continuity of U  and Q, and Lemma 5.2, the equilibrium conditions for  and  are U  (z) + Q(z)  F for z 2 S ? ; (5.24)  U (z) + Q(z)  F for z 2 S = E; (5.25) and U  (z) + Q(z)  F for z 2 S ? ; (5.26)  U (z) + Q(z)  F for z 2 S : (5.27) >From (5.25) and (5.26) we derive U  ? (z)  U  ? (z) + F ? F for z 2 S ? : By the principle of domination this inequality holds everywhere (note that  ?  is a positive measure and k ?  k = k ?  k). From (5.24) and (5.27) we get U  (z)  U  (z) ? F + F for z 2 S \ S ? : Thus U  (z)  U  (z) ? F + F for z 2 C (5.28)   U (z) = U (z) ? F + F for z 2 S \ S ? : (5.29) Again, by [ST, Theorem IV.4.5], we obtain ( )jS ?  ( )jS ? : 2 2

1

1

1

2

2

2

2

2

2

2

1

1

1

1

2

2

2

2

1

1

2

1

1

1

2

1

2

2

1 1

1

2

2

1

2

1

2

2

2

2

2

2

1

2

1

1

2

 1 1

31

2

1

1

1

1

Proof of Corollary 2.9: By Lemma 5.2 we have that U  is continuous  w

on E and since Q is also continuous on E the conditions (2.1) and (2.2) give us 1 U  (z) + Q (z)  1 F  on S ? c c w (5.30) 1 = c Fw on S ; 2

 w

0

2

that is, the probability measure  =c, which has a nite logarithmic energy, is the equilibrium measure on S? with external eld Q (cf. [ST, Theorem I.3.3]). 2 We now proceed with the proof of the Balayage Representation Theorem. Proof of Theorem 2.14: Let  := =k k ( recall that k k = kk ? 1 ). The representation 2

 w

Iw (w ) = k k2 fI () + 2

Z

0

(?U  ? Q)=k k dg + I ();

shows that Iv () = minfIv ( ) j  2 M=k k g: >From the equilibrium conditions for v we get

(5.31) (5.32)

U k k (z) ? U  (z) ? Q(z)  k kFv q.e. on S v

and

U k k (z) ? U  (z) ? Q(z)  k kFv on S : Furthermore, by the equilibrium conditions for w (see (1.6)) and the assumption that S  S , we can extend these inequalities to v

v

v

w

U k k (z) ? U  (z) + U  (z) ? Fw  k kFv q.e. on S

(5.33)

U k k (z) ? U  (z) + U  (z) ? Fw  k kFv on S :

(5.34)

v

w

and v

w

v

32

Applying the principle of domination to (5.33), we obtain that this inequality holds everywhere (notice that all measures involved have nite energy, and therefore q.e. implies a.e. with respect to the corresponding measure). Since (5.34) becomes equality, we deduce from [ST, Theorem IV.4.5] that (k kv + w )jS  jS ; (5.35) so that k kv  . This shows that v 2 M=k k , which implies that Iv (v ) = Iv (). The uniqueness of the measure v now gives that v =  and S = S . To verify the representation (2.16), we take the balayage of  and of w onto S . We have q.e. on S that U  = U  + const: and U c = U  + const. Thus, from the equality in (5.36) we can write v

v

v

^

w

w

U  (z) ? U  (z) + U c (z) = const: q.e. on S : (5.36) Finally, applying the principle of domination in both directions, we extend this equality to the whole complex plane, and by the unicity theorem ( see [L, Theorem 1.120]) we obtain ^ =  + cw , which yields (2.16). 2 ^

w

Proof of Theorem 2.17: The proof proceeds as in the unconstrained

case. (a) Assume to the contrary that S \ I is not an interval. Then there exist points x and x in the support S , such that (x ; x ) \ S = ;. From (2.2) we get U  (xi) + Q(xi)  Fw i = 1; 2: The weighted potential U  + Q is strictly convex on (x ; x ); therefore (U  + Q)(x + (1 ? )x ) < Fw for 0 <  < 1. The last inequality contradicts (2.1), thus proving part (a) of the theorem. (b) We proceed as in part (a), but instead of a convexity argument, we observe that x(U  (x) + Q(x))0 is increasing on (x ; x ). Therefore the derivative of the weighted potential U  + Q cannot change sign from positive to negative, which means that U  + Q will be either strictly monotone on (x ; x ) or will decrease and then increase. But both cases lead to a contradiction with (2.1). 2  w

1

 w

2

1

 w

2

 w

 w

 w

1

1

2

 w

 w

 w

1

2

33

1

2

2

The proof of Theorem 3.2 requires several lemmas. Lemma 5.3 With the assumptions of Theorem 3.2 we have lim sup kwNn pn;N kp;=n  e?F : j !1

(5.37)

 w

1

N

Proof: Set  := w . Since U  is continuous on C, so is U  by Lemma

5.2. Furthermore, by our assumption on w, the function Q = log(1=w) is also continuous on [a; b]. Now x  > 0 and choose A  S? to be a union of nitely many closed intervals such that U  (x) + Q(x) > Fw ?  for x 2 A and (Ac) < 1, where Ac := [a; b] n A. Let  := jA and  := jA . Also de ne j := j jA and j := j jA . Then j !  as j ! 1. Let mj := jfi;N 2 Acgj. Since (mj =n) ! (Ac) < 1, there exists a j , such that mj < n for every j > j . Now let pn be a monic polynomial with zeros fyk gnk that consist of all i;N 2 Ac and n ? mj zeros in A, chosen so that the weak convergence p !  as n ! 1 takes place (it is enough to discretize  on A). Then with QN (x) := log(1=wN (x)) we have for j > j (1)

(1)

(2)

c

(2)

(2)

c

(2)

0

0

=1

n

kwNn pn;N kp;=n  kwNn pnkp;=n = 1

1

N

N

 N

= pn) expf?

 X

k;N 2A

0

wNpn(k;N )jpn(k;N )jp

 = pn 1 (

)

min [U pn (k;N ) + QN (k;N )]g k;N 2A N 1=(pn) expf?[U pn (N ) + QN (N )]g; 1 (

= (5.38) for some N 2 N \ A. Let  2 S? be any limit point of N , say N !  as l ! 1. Then lim inf[U  (N ) + QN (N )]  U  () + Q()  Fw ? : (5.39) l!1 l

pn

l

l

Here we used the principal of descent (see [L, Theorem 1.3]) and the uniform convergence of QN . Now from (5.38) and (5.39) we see that lim sup kwNn pn;N kp;=n  e?F : j !1

1

N

 w+

Letting  ! 0 we obtain the assertion of the lemma. 2 Remark 5.4 We can modify the proof of this lemma, so that all the zeros fyk gnk of pn lie in N (see [D] for details). This shows that the restricted Lp-minimal polynomials (with zeros in N ) will also satisfy (5.37). Since any =1

34

weak limit of the normalized zero counting measures of these polynomials will be in the class M , we can use the argument in the proof of Theorem 3.2 to conclude that (3.2) is also for these restricted polynomials.. Lemma 5.5 With the assumptions of Theorem 3.2, let Pn(x) = xn +


be a polynomial of degree n with real simple zeros fxk;ngnk separated by the points of N . Assume that the normalized zero counting measures P associated with Pn have weak limit . Then lim inf kwNn Pnkp;=n  e?F ; (5.40) j !1 =1

n

1



N

where F is the constant de ned in Theorem 2.1(e) for the limiting weight w. Proof: First we note that by the separation assumption on the zeros of Pn, the limit measure  2 M . By Lemma 5.2 and the continuity of U  we obtain that U  is continuous. The weight w is positive and continuous, so Q = log(1=w) is also continuous. Let x0 2 S? be such that U  (x0) + Q(x0 ) = F . Let 0 <  and set 4 := (x0 ? ; x0 + ). Denote by (2) := j4 and (2) P := P j4 . Now we choose 0 <  <  such that the following are true: c 

c 

n

n

(i) if j > j , then jQN (y) ? Q(y)j <  for every jy ? x j < ; (ii) if jy ? x j < , then jQ(y) ? Q(x )j < ; and 0

0

0

0

(iii) if jy ? x j < , then jU  (y) ? U  (x )j < . We can guarantee (i) because of the uniform convergence of QN ; (ii) is satis ed from the continuity of Q; and (iii) follows from the continuity of U  obtained by the continuity of U  and Lemma 5.2. Let 4 := (x ? ; x + ), and write Y Y (x ? xk;n): (x ? xk;n); Pn; (x) := Pn; (x) := (2)

(2)

0

0

(2)

0

1

0

2

xk;n 24

xk;n 24c

Then Pn(x) = Pn; (x)Pn; (x). Since x 2 S? we have that q := ( ? )(4 ) > 0. Let ln be the number of zeros of Pn in 4 and mn be the number of points in N \4 . Then limj!1(mn ?ln)=n = q > 0. Since the intervals [k;N ?=(2N ); k;N +=(2N )] are disjoint (see De nition 3.1(i)), there exists for j suciently large, at least 1

2

0

35

one interval with center in 4 containing no zeros of Pn. Denote this center by N . Then lim inf kwNn Pnkp;=n  lim inf jwNn (N )Pn(N )j =n: (5.41) j !1 j !1 1

1

N

Using De nition 3.1(i) and the separation assumption on the zeros of Pn, it is easy to obtain a lower bound for jPn; (N )j =n. Indeed, let x0 ; x0 ; :::; x0l be the zeros of Pn in increasing order, lying in (x ? ; N ). We have that jN ? x0l j > =(2N ) and jN ? x0l? j > =N . In a similar fashion jN ? x0l?i j > i=N , i = 2; 3; :::; l ? 1. The same considerations are valid for the zeros in (N ; x + ). Now let sn be the number of zeros of Pn in 4. If sn = 0 we put Pn; (x)  1. If sn > 1, then the following estimate holds:  s  jPn; (N )j =n  4N s f([ s2n ] ? 1)!g =n: (5.42) We have that (sn=n) ! (4) =:  and (sn=N ) ! =C as n ! 1. So lim inf jPn; (N )j =n  e  = eC : (5.43) j !1 1

1

1

2

0

1

0

1

n

1

1

2

1

n

1

1

 log[(

) 2

]

Next we nd a lower bound for lim inf jwNn (N )Pn; (N )j =n: j !1 2

1

We have that P !  in the weak topology, and moreover U  converges uniformly on 4 to U  . Therefore, there exists j such that for j > j we have jU  (y) ? U  (y)j <  for every y 2 4. But then for j > max(j ; j ) using (i), (ii) and (iii) we deduce that (2)

(2)

(2)

n

Pn

(2)

1

(2)

1

(2)

Pn

0

lim inf jwNn (N )Pn; (N )j =n  e?Q x ?U j !1 2

1

( 0)

x0 )?4 :

(2) (

1

(5.44)

Finally, combining (5.41), (5.43) and (5.44), letting  ! 0 and using the fact that U  (x ) tends to U  (x ) ( this follows by the dominated convergence theorem ), we obtain lim inf kwNn Pnkp;=n  e?F ; j !1 (2)

0

0

1



N

which proves the lemma. 2 36

Proof of Theorem 3.2: To prove the statement of the theorem we

observe rst that the zeros of pN;n are separated by the points of N . Let  be a weak limit of the normalized counting zero measures n of pn;N . Then combining Lemmas 5.3 and 5.4 we get e?F  lim inf kwNn pn;N kp;=n  lim sup kwNn pn;N kp;=n  e?F : (5.45) n!1 

1

N

n!1

1

N

 w

Since  2 M , and F  Fw , by Theorem 2.1(e) we obtain  = w . The measure  was any weak limit, so we deduce (5.3). The equality in (5.45) implies (5.2). This completes the proof of the theorem. 2

Aknoledgement The authors would like to thank E. A. Rakhmanov and S. Damelin for their helpful comment.

References [D] P. Dragnev. Constrained energy problems for logarithmic potentials, Ph.D. Thesis, University of South Florida, Tampa (to appear). [F] O. Frostman.Potentiel d'equilibre et capacite des ensembles avec quelques applications a la theorie des fonctions, Thesis. Meddel. Lunds Univ. Mat. Sem., 3: 1-118, 1935. [H] E. Hille, Analytic Function Theory, v.2, Blaisdell Publ. Co., Waltham, 1962. [L] N. S. Landkof, Foundations of Modern Potential Theory , SpringerVerlag, New York, N.Y., (1972). [Le] V. Levenshtein, Krawtchouk polynomials and universal bounds for codes and design in Hamming spaces., IEEE Trans. Info. Theory 41 (1995), 1303-1321. [LL] L. D. Landau and E. M. Lifshitz, Electrodynamics of Continuous Media, 2nd ed., in Course of Theoretical Physics v.8, Pergamon Press (1984). 37

[MRRW] R.J. McEliece, E.R. Rodemich, H.C. Rumsey, and L.R. Welch, New upper bounds on the rate of a code via the Delsarte-Mac Williams inequalities, IEEE Trans. Inform. Theory I-23, (1977), 157-166. [MS] H. N. Maskhar and E. B. Sa, Weighted analogues of Capacity, Trans nite Diameter, and Chebyshev Constant, Constr. Approx. (1992) 8. [R] E. A. Rakhmanov, Equilibrium measure and zero distribution of extremal polynomials of discrete variable, Mat. Sb. 187(8), (1996). (In Russian) [ST] E. B. Sa and V. Totik. Logarithmic Potentials with External Fields (to appear). [Sz] G. Szeg}o. Orthogonal Polynomials, v. 23 of Colloquium Publications. Amer. Math. Soc., Providence, R. I., 1975. [T] M. Tsuji. Potential Theory in Modern Function Theory. Maruzen, Tokyo, 1959. P. D. Dragnev Department of Mathematics University of South Florida Tampa, Fl 33620 U.S.A. e-mail:[email protected]

38

E. B. Sa Inst. for Constructive Mathematics Department of Mathematics University of South Florida Tampa, Fl 33620 U.S.A. e-mail:[email protected]