Asymptotic Zero Distribution of Biorthogonal Polynomials D. S. Lubinsky School of Mathematics, Georgia Institute of Technology, Atlanta, GA 3332-6 USA A. Sidi Department of Computer Science, Technion-IIT, Haifa 32, Israel H. Stahl Fachbereich Mathematik, TFH Berlin, Germany Abstract Let ψ : [,] R be a strictly increasing continuous function. Let P n be a polynomial of degree n determined by the biorthogonality conditions P n (x)ψ (x) j dx =, j =,,...,n. We study the distribution of zeros of P n as n, related potential theory.. Introduction Results Let ψ : [,] [ψ (),ψ ()] be a strictly increasing continuous function, with inverse ψ [ ]. Then we may uniquely determine a monic polynomial P n of degree n by the biorthogonality conditions { P n (x) ψ (x) j, j =,,2,...,n, dx = I n, j = n. () Research supported by NSF grant DMS82 US-Israel BSF grant 28399 Email addresses: lubinsky@math.gatech.edu (D. S. Lubinsky), asidi@cs.technion.ac.il (A. Sidi) Preprint submitted to Approximation Theory December, 23
P n will have n simple zeros in (,), so we may write P n (x) = n (x x jn ). (2) j= The proof of this is the same as for classical orthogonal polynomials. Our goal in this paper is to investigate the zero distribution of P n as n. Accordingly, we define the zero counting measures µ n = n n δ xjn, (3) j= that place mass n at each of the zeros of P n, want to describe the weak limit(s) of µ n as n. This topic was initiated by the second author, in the course of his investigations on convergence acceleration [8], [24], numerical integration of singular integrs. He considered [2], [22], [23] ψ (x) = log x, x (,) found that the corresponding biorthogonal polynomials are P n (x) = n ( )( ) ( ) n j n j + j x j. j n + j= The latter are now often called the Sidi polynomials, one may represent them as a contour integral. Using steepest descent, the strong asymptotics of P n, their zero distribution, were established in [4]. Asymptotics for more general polynomials of this type were analyzed by Elbert [7]. Extensions, asymptotics, applications in numerical integration, convergence acceleration have been considered in [5], [6], [25], [26]. Biorthogonal polynomials of a more general form have been studied in several contexts see [5], [], []. The sorts of biorthogonal polynomials used in rom matrices [3], [6], [2] are mostly different, although there are some common ideas in the associated potential theory. Herbert Stahl s interest in this topic arose after he refereed [4]. He the first author discussed the topic at some length at a conference in honor of Paul Erdős in 995. This led to a draft paper on zero distribution in the later 99 s, with revisions in 2, 23, this paper is the partial completion of that work. For the case ψ (x) = x α, α >, we presented 2
explicit formulae in [8]. Rodrigues type representations were studied in [7]. Distribution of zeros of polynomials is closely related to potential theory [], [2], [28], accordingly we introduce some potential theoretic concepts. We let P (E) denote the set of all probability measures with compact support contained in the set E. For any positive Borel measure µ, we define its classical energy integral I (µ) = log dµ (x)dµ (t), (4) x t denote its support by supp[µ]. Where appropriate, we consider these concepts for signed measures too. For any set E in the plane, its (inner) logarithmic capacity is { } cap (E) = sup e I(µ) : µ P (E). We say that a property holds q.e. (quasi-everywhere) if it holds outside a set of capacity. We use meas to denote linear Lebesgue measure. For further orientation on potential theory, see for example [3], [9], [2]. In our setting we need a new energy integral J (µ) = K (x,t) dµ (x)dµ (t), (5) where K (x,t) = log x t + log ψ (x) ψ (t). (6) In [6], a similar energy integral was considered for ψ (t) = e t, but with an external field. The minimal energy corresponding to ψ is J (ψ) = inf {J (µ) : µ P ([,])}. (7) Under mild conditions on ψ, we shall prove that there is a unique probability measure, which we denote by ν ψ, attaining the minimum. For probability measures µ,ν, we define the classical potential the mixed potential W µ,ν (x) = U µ (x) = log log dµ (t) + x t dµ (t), (8) x t log dν (t) (9) ψ (x) ψ (t) = U µ (x) + U ν ψ[ ] ψ (x), () 3
the ψ potential W µ (x) = W µ,µ (x) = K (x,t)dµ (t). () We note that potential theory for generalized kernels is an old topic, see for example, Chapter VI in [3]. However, there does not seem to be a comprehensive treatment covering our setting. Our most important restrictions on ψ are contained in: Definition.. Let ψ : [,] [ψ (),ψ ()] be a strictly increasing continuous function, with inverse ψ [ ]. Assume that ψ satisfies the following two conditions: (I) ( ) cap (E) = cap ψ [ ] (E) =. (2) (II) For each ε >, there exists δ > such that ( ) meas (E) δ meas ψ [ ] (E) ε. (3) Then we say that ψ preserves smallness of sets. The conditions (I), (II) are satisfied if ψ satisfies a local lower Lipschitz condition. By this we mean that we can write [,] as a countable union of intervals [a, b] such that in [a, b], there exist C, α > depending on a, b, with ψ (x) ψ (t) C t x α,x,t [a,b]. We can apply Theorem 5.3. in [9, p. 37] to ψ to deduce (2). Using classical methods, we shall prove: Theorem.2. Let ψ : [, ] [ψ (), ψ ()] be a strictly increasing continuous function that preserves smallness of sets. Define the minimal energy J = J (ψ) by (7). Then (a) J is finite there exists a unique probability measure ν ψ on [,] such that J (ν ψ ) = J. (4) (b) J q.e. in [,]. (5) In particular, this is true at each point of continuity of. 4
(c) J in supp[ν ψ ]. (6) = J q.e. in supp[ν ψ ]. (7) (d) ν ψ is absolutely continuous with respect to linear Lebesgue measure on [,]. Moreover, there are constants C C 2 depending only on ψ, such that for all compact K [,], ν ψ (K) (e) There exists ε > such that Let I n = C log capk C 2 log meas (K). (8) [,ε] [ ε,] supp[ν ψ ]. (9) P n (t) ψ (t) n dt, n. (2) Theorem.3. Let ψ : [,] [ψ (),ψ ()] be a strictly increasing continuous function that preserves smallness of sets. Let {P n } be the corresponding biorthogonal polynomials, with zero counting measures {µ n }. If supp[ν ψ ] = [,], (2) then the zero counting measures {µ n } of (P n ) satisfy µ n νψ,n (22) lim n I/n n = exp ( J ). (23) The weak convergence (22) is defined in the usual way: lim n f (t) dµ n (t) = f (t)dν ψ (t), for every continuous function f : [,] R. We can replace (2) by the more implicit, but more general, assumption that supp[ν ψ ] contains the support of every weak limit of every subsequence of (µ n ). We can at least prove it when the kernel K, hence the potential, satisfies a convexity condition: 5
Theorem.4. Let ψ : [, ] [ψ (), ψ ()] be a strictly increasing continuous function that preserves smallness of sets. In addition assume that ψ is twice continuously differentiable in (,) either (a) for x,t (,) with x t, or 2 x2k (x,t) >, (24) (b) for x,t (ψ (),ψ ()) with x t, Then Example. Let α > 2 [ ( )] x 2 K ψ [ ] (x),ψ [ ] (t) >. (25) supp[ν ψ ] = [,]. (26) ψ (x) = x α, x [,]. Then either (25) or (26) holds hence (2) holds. We show this separately for α for α <. Case I α We shall show that the hypotheses of Theorem.4 (a) are fulfilled. A straightforward calculation gives that (x,t) := (x t) 2 (ψ (x) ψ (t)) 2 2 x2k (x,t) = (x α t α ) 2 + ( αx α ) 2 (x t) 2 α(α ) x α 2 (x α t α ) (x t) 2. Writing s = tx, we see that where (x,t) = x 2α H (s), H (s) := ( s α ) 2 + α 2 ( s) 2 α (α ) ( s α )( s) 2. (27) For s >, all three terms in the right-h side of (27) are positive, so H (s) >. If s <, we see that H (s) = ( s α ) 2 + α( s) 2 {α (α ) ( s α )} ( s α ) 2 + α( s) 2 >. 6
In summary, if α >, we have for all x [,] s [, )\ {}, so the hypotheses (24) is fulfilled. Case II α < Here K (x,sx) > ψ [ ] (x) = x /α ( ) ψ [ ] (x),ψ [ ] (t) = log x /α t /α + log x t, which is exactly the case /α > treated above, so we see that the hypothesis (25) is fulfilled. Instead of placing an implict assumption on the support of ν ψ, we can place an implicit assumption on the zeros of {P n }, obtain a unique weak limit: Theorem.5. Let ψ : [, ] [ψ (), ψ ()] be a strictly increasing continuous function that preserves smallness of sets. Let K [,] be compact. Assume that every weak limit of every subsequence of the zero counting measures {µ n } has support K. Then there is a unique probability measure µ on K such that µ n µ,n, (28) a unique positive number A such that lim n I/n n = A, (29) Here µ is absolutely continuous with respect to linear Lebesgue measure, is the unique solution of the integral equation W µ (x) = Constant, q.e. x K, (3) Moreover, then W µ (x) = log, q.e. x K. A We note that in [6], a related integral equation to (3) appears. We shall also need the dual polynomials Q n such that Q n ψ are biorthogonal 7
to powers of x. Thus we define Q n to be a monic polynomial of degree n determined by the conditions Q n ψ (t)t j dt =, (3) j =,,2,...,n. Because of this biorthogonality condition, That is, Q n ψ (t)t n dt = I n = Q n ψ (t)p n (t)dt = P n (t)ψ (t) n dt = P n (t)ψ (t) n dt. Q n ψ (t)t n dt. (32) The orthogonality conditions ensure that Q n ψ has n distinct zeros {y jn } in (,), so we can write Q n ψ (t) = n (ψ (t) ψ (y jn )). (33) j= Let We shall prove ν n = n δ yjn. (34) n j= Theorem.6. Let ψ : [, ] [ψ (), ψ ()] be a strictly increasing continuous function that preserves smallness of sets, assume (2). We have as n, ν n νψ. We also prove the following extremal property for weak subsequential limits of {µ n }. Theorem.7. Let ψ : [, ] [ψ (), ψ ()] be a strictly increasing continuous function that preserves smallness of sets. Assume that S is an infinite subsequence of positive integers such that as n through S, µ n µ; (35) ν n ν; (36) I /n n A, (37) 8
where A R µ,ν P ([,]). Then ( ) A exp sup inf W µ,β β P([,]) [,] (38) A exp ( ) sup inf W α,ν α P([,]) [,]. (39) Remarks. (a) This extremal property is very close to a characterization of equilibrium measures for external fields. For example, with ν as above, let Q be the external field Q = U ν ψ[ ] ψ on [,]. Then the second inequality above says ( A exp sup α P([,]) inf [,] (Uα + Q) This is reminiscent of one characterization of the equilibrium measure for the external field Q [2, Theorem I.3., p. 43]. (b) Herbert Stahl sketched a proof that when ψ is strictly increasing piecewise linear, then (2) holds [27]. His expectation was that this a limiting argument could establish (2) very generally. (c) There are two principal issues left unresolved in this paper, that seem worthy of further study: (I) Find general hypotheses for supp[ν ψ ] = [,]. (II) Find an explicit representation of the solution µ of the integral equation (3), that is of + log x t µ (t)dt log ψ (x) ψ (t) µ (t) dt = Constant, x [,]. The usual methods (differentiating, solving a Cauchy singular integral equation) do not seem to work, even when ψ is analytic. ). 9
Next we show that if ψ is constant in an interval, then the support of the equilibrium measure should avoid that interval, as do most of the zeros of {P n } : Example. Let { [ ] 2x, x, 2 ψ (x) =, x [ 2,]. Then it is not difficult to see that the equilibrium measure ν ψ must have support [, 2 ]. Indeed if µ is a probability measure that has positive measure on [a,b] ( 2,), then as so Consequently, log =, x,t [a,b], ψ (x) ψ (t) J = inf ν ψ (x) = π J (µ) =. [ 2I (µ) + log ], 2 where the inf is now taken over all µ P ([, 2]). Then νψ is the classical equilibrium measure for [, 2], namely [, ], 2 x ( 2 x), x J = 2log 8 + log = log 32. 2 In this case, we can also almost explicitly determine P n. The biorthogonality conditions give for π of degree at most n, /2 P n (x) π (2x) dx + π () In particular, this is true for π, so /2 P n (x)dx = /2 /2 we obtain for any π of degree at most n, /2 P n (x)dx =. P n (x) dx, P n (x)(π (2x) π ())dx =.
Then for every polynomial S of degree n 2, /2 P n (x)s (x)( 2x) dx =, (4) which forces P n to have at least n distinct zeros in [, 2 ]. Then every weak limit of every subsequence of {µ n } has support in [, 2 ]. This paper is organized as follows: in Section 2, we present a principle of descent, a lower envelope theorem, the proof of Theorem.2. In Section 3, we prove Theorems.3.7. Throughout the sequel, we assume that ψ : [,] [ψ (),ψ ()] is a strictly increasing continuous function that preserves smallness of sets. We close this section with some extra notation. Define the companion polynomial to P n, namely R n (x) = n (x ψ (x jn )). (4) It has the property that R n ψ has the same zeros as P n. Hence Analogous to R n, we define so that j= P n (x) R n ψ (x) in [,]. (42) S n (t) = Observe that I n of (2) satisfies I n = n (t y jn ), (43) j= S n (t)q n ψ (t), t [,]. (44) P n (x) R n ψ (x)dx = 2. Proof of Theorem.2 Q n ψ (x)s n (x) dx >. (45) We begin by noting that for any positive measures α,β, W α,β is lower semicontinuous, since a potential of any positive measure is, while ψ ψ [ ] are continuous. We start with
Lemma 2. (The Principle of Descent). Let {α n } {β n }be finite positive Borel measures on [,] such that Assume moreover that as n, lim α n ([,]) = = lim β n ([,]). n n α n α; β n β. (a) If {x n } [,] x n x, n, then (b) If K [,] is compact then uniformly in K, lim inf n W αn,βn (x n ) W α,β (x ). W α,β λ in K, lim inf n W αn,βn (x) λ. Proof. (a) By the classical principle of descent, lim inf n Uαn (x n ) U α (x ), see for example, [2, Theorem I.6.8, p. 7]. Next, we see from the classical principle of descent continuity of ψ, ψ [ ] that lim inf Uβn ψ[ ] ψ (x n ) U β ψ[ ] ψ (x ). n Combining these two gives the result. (b) This follows easily from (a). If (b) fails, we can choose a sequence (x n ) in K with limit x K such that lim inf n W αn,βn (x n ) < λ W α,β (x ). Recall our notation W αn = W αn,αn. We now establish 2
Lemma 2.2 (Lower Envelope Theorem). Assume the hypotheses of Lemma 2.. Then for q.e. x [,], lim inf n,n S W αn (x) = W α (x). Proof. We already know from Lemma 2. (the principle of descent) that everywhere in [,], lim inf W αn (x) W α (x). n,n S Suppose the result is false. Then there exists ε >, a (Borel) set S of positive capacity such that lim inf n,n S W αn (x) W α (x) + ε in S. (46) Because Borel sets are inner regular, even more, capacitable, we may assume that S is compact. Then there exists a probability measure ω with support in S such that U ω is continuous in C. See, for example, [2, Corollary I.6., p. 74]. As ψ ψ [ ] are continuous, W ω = U ω + U ω ψ[ ] ψ is also continuous in [, ]. Then by Fubini s Theorem weak convergence lim inf W αn dω = lim inf W ω dα n n,n S n,n S = W ω dα = W α dω. Here since K (x,t) is bounded below in [,], we may continue this using (46) Fatou s Lemma as = (W α + ε) dω ε ( ) lim inf W αn dω ε n,n S lim inf W αn dω ε. n,n S So we have a contradiction. Next, we show that J is finite, establishing part of Theorem.2(a): 3
Lemma 2.3. J is finite. Proof. This is really a consequence of Cartan s Lemma for potentials. Let µ = meas denote Lebesgue measure on [, ]. Then for x [, ], U µ (x) = log x t dt 2 log s ds U µ is continuous. Now consider the unit measure µ ψ [ ]. By Cartan s Lemma [9, p. 366], if ε > { A ε = y R : U µ ψ[ ] (y) > log }, ε then µ (A ε ) 3eε. With a suitably small choice of ε, we then have by the hypothesis (3), With this choice of ε, let a closed set. Let ( ) µ ψ [ ] (A ε ) 2. B = [,] \ψ [ ] (A ε ), ν = µ B µ (B). As µ (B) 2, ν is a well defined probability measure. Moreover, x B ψ (x) / A ε, U ν ψ[ ] ψ (x) = [ ] U µ ψ[ ] ψ (x) U µ [,]\B ψ [ ] ψ (x) µ (B) [ log µ (B) ε + log ( ] 2 ψ L [,]) =: C <. Then J J (ν) I (ν) + C <. 4
Proof of Theorem.2. (a) We can choose a sequence {α n } of probability measures on [,] such that lim J (α n) = J. n By Helly s Theorem, we can choose a subsequence converging weakly to some probability measure α on [, ], by relabelling, we may assume that the full sequence {α n } converges weakly to α. Then { α n ψ [ ]} converges weakly to α ψ [ ]. By the classical principle of descent or equivalently, lim inf n lim inf n I (α n) I (α) ( lim inf I α n ψ [ ]) ( I α ψ [ ]), n log ψ (x) ψ (t) dα n (x) dα n (t) log dα (x)dα (t). ψ (x) ψ (t) See, for example, [2, Thm. I.6.8, p. 7]. Combining these, we have J = lim inf n J (α n) J (α), so α achieves the inf, is an equilibrium distribution. If β is another such distribution, then the parallelogram law ( ) ( ) J (α + β) + J (α β) = 2 2 2 (J (α) + J (β)) = J, gives as 2 J ( ) ( ) (α β) = J J (α + β), 2 2 (α + β) is also a probability measure on [,]. Here ( ) ( ) ( ( J (α β) = I (α β) + I α ψ [ ] β ψ [ ])), 2 2 2 both terms on the right-h side are non-negative as both measures inside the energy integrals on the right have total mass. See [2, Lemma I..8, p. 29]. Hence ( ) I (α β) =, 2 5
so α = β [2, Lemma I..8, p. 29]. (b) Suppose the result is false. Then for some large enough integer n, } E := {x [,] : (x) J n, has positive capacity is compact, since is lower semi-continuous. But, dν ψ = J (ν ψ ) = J, so there exists a compact subset E 2 disjoint from E such that (x) > J 2n, x E 2, m = ν ψ (E 2 ) >. Now as E is a compact set of positive capacity, we can find a positive measure σ on E, with support in E, such that U σ is continuous in the plane [2, Cor. I.6., p. 74]. Then U σ ψ[ ] is also continuous in [ψ (),ψ ()], so W σ is continuous in [,]. We may also assume that σ (E ) = m. Define a signed measure σ on [,], by σ in E σ := ν ψ in E 2 elsewhere. Here if η (,), J (ν ψ + ησ ) = J (ν ψ ) + 2η dσ + η 2 J (σ ) { ] [ J (ν ψ ) + 2η [J n dσ + J E E 2 2n ] [ = J (ν ψ ) + 2ηm {[J n J 2n = J (ν ψ ) ηm n + η 2 J (σ ) < J (ν ψ ), ]} + η 2 J (σ ) ] } d( ν ψ ) + η 2 J (σ ) 6
for small η >. As σ has total mass, so ν ψ + ησ has total mass, we see from the identity ν ψ + ησ = ( η) ν ψ E2 + ν ψ [,]\E2 + ησ that it is non-negative. Then we have a contradiction to the minimality of J (ν ψ ). (c) Let x supp[ν ψ ] suppose that (x ) > J. By lower semi-continuity of, there exists ε > closed [a,b] containing x such that (x) > J + ε, x [a,b]. We know too that (x) J for q.e. x supp[ν ψ ]. Here as J is finite, so I (ν ψ ) must be finite (recall that K (x,t) is bounded below). Then ν ψ vanishes on sets of capacity, so this last inequality holds ν ψ a.e. (cf. [9, Theorem 3.2.3, p. 56]). Then ( b ) J = J (ν ψ ) = + (x)dν ψ (x) [,]\[a,b] a (J + ε)ν ψ ([a,b]) + J ν ψ ([,] \[a,b]) = J + εν ψ ([a,b]), a contradiction. (d) If cap (K) =, then as I (ν ψ ) <, we have also ν ψ (K) =, the inequality (8) is immediate. So assume that K supp[ν ψ ] has positive capacity, let ω be the equilibrium measure for K. We may also assume that K supp[ν ψ ], since ν ψ (K) = ν ψ (K supp[ν ψ ]). Now, there exists a positive constant C such that K (x,t) C, x,t [,]. 7
Then by (c), for x K, K (x,t) dν ψ (t) K J K (x,t)dν ψ (t) [,]\K J + C hence for x K, log x t dν ψ (t) J + C + log ( ) 2 ψ L [,] =: C. (47) K Here C is independent of K,x. Now U ω (t) = log capk for q.e. t K since ν ψ vanishes on sets of capacity zero, this also holds for ν ψ a.e. t K. Integrating (47) with respect to dω (x) using Fubini s theorem, gives U ω (t)dν ψ (t) C K hence ν ψ (K) log capk C. This gives the first inequality in (8), then well known inequalities relating cap meas give the second. In particular, that inequality implies the absolute continuity of µ with respect to linear Lebesgue measure. (e) Suppose that / supp[ν ψ ]. Let c > be the closest point in the support of ν ψ to. Then for x [, c 2 ], for all t [c,], we have from the strict monotonicity of ψ that K (x,t) < K (c,t), so for such x, (x) = < c c K (x,t)dν ψ (t) K (c,t) dν ψ (t) = (c) J. Thus in spite of the continuity of in [,c), [ < J in, c ], 2 contradicting (b). Absolute continuity of ν ψ then shows that for some ε >, we have [,ε] supp[ν ψ ]. Similarly we can show that for some ε >, [ ε,] supp[ν ψ ]. 8
3. Proof of Theorems.3.7 Recall that µ n ν n were defined respectively by (3) (34). Throughout this section, we assume that S is an infinite subsequence of positive integers such that as n through S, µ n µ; (48) ν n ν; (49) I /n n A, (5) where A R µ,ν P ([,]). In the sequel we make frequent use of identities such as P n (x) /n = exp ( U µn (x)) We begin with P n (x) R n ψ (x) /n = exp ( W µn (x)). Lemma 3. (An upper bound for W µ ). (a) With the hypotheses above, let [a,b] [,] assume that [a,b] contains two zeros of P n for infinitely many n S. Then inf W µ log [a,b] A. (b) In particular, if x is a limit of two zeros of P n as n through S, or x supp [µ], then W µ (x ) log A. Proof. (a) We may assume (by passing to a subsequence) that for all n S, P n has two zeros in [a,b]. Assume on the contrary, that for some ε >, Let x n,y n be two zeros of P n in [a,b] let inf W µ > log + ε. (5) [a,b] A R n (x) = R n (x)/[(x ψ (x n ))(x ψ (y n ))]. 9
Then we see that P n (x)rn ψ (x), x [,] \[a,b], P n (x) R n ψ (x) P n (x)r n ψ (x) ( 4 ψ L [,]) 2, x [,]. Moreover, as R n has the same asymptotic zero distribution as R n, we see from Lemma 2. (5) that lim sup P n (x)rn ψ (x) /n exp ( W µ,µ (x)) n,n S = exp ( W µ (x)) Ae ε, uniformly in [a,b]. Then by biorthogonality, positivity of P n (x) Rn ψ (x) outside [a,b], ( ) /n lim sup P n (x)rn ψ (x) dx n,n S [,]\[a,b] /n = lim sup P n (x) Rn ψ (x)dx Ae ε. n,n S [a,b] Of course Lemma 2.(b) also gives ( so A = lim sup n,n S lim sup In /n n,n S lim sup n,n S Ae ε. ) /n P n (x) Rn ψ (x) dx Ae ε, [a,b] ( ( ) /n 2/n 4 ψ L [,]) P n (x)rn ψ (x) dx This contradiction gives the result. (b) This follows from (a), lower semicontinuity of W µ. Lemma 3.2 (A Lower bound for W µ ). At each point of continuity of W µ in [,], we have In particular, this inequality holds q.e. in [,]. W µ log A. (52) 2
Proof. Assume that a [,] is a point of continuity of W µ, but for some ε >, W µ (a) log A 2ε. Then there exists an interval [a,b] containing a, such that W µ (x) log ε, x [a,b]. A By the lower envelope theorem (Lemma 2.2) lim sup n,n S = exp (P n (x)r n ψ (x)) /n ( ) lim inf W µn (x) = exp ( W µ (x)) Ae ε n,n S for q.e. x [a,b]. Let { T n = x [a,b] : (P n (x) R n ψ (x)) /n Ae ε/2}. Then for each m, n=m contains q.e. x [a,b], so has linear Lebesgue measure b a. Then for infinitely many n, T n has linear Lebesgue measure at least n 2, so so I /n n T n ( P n (x) R n ψ (x) dx T n n 2/n Ae ε/2 A = lim sup In /n Ae ε/2, n,n S ) /n a contradiction. Finally, we note that any logarithmic potential is continuous q.e. [3, p. 85], so U µ U µ ψ[ ] are continuous q.e. Our hypothesis that ψ [ ] (E) has capacity zero whenever E does ensures that U µ ψ[ ] ψ is continuous q.e. also. Hence W µ is continuous q.e. so (52) holds q.e. in [,]. Next, we establish lower upper bounds for A. 2
Lemma 3.3. (a) There exist constants C,C 2 > depending only on ψ ( not on the subsequence S above) such that C A C 2. (53) (b) In particular, (c) I (µ) <. J (µ) = log A (54) W µ = log A q.e. a.e. (µ) in supp[µ]. (55) (d) µ is absolutely continuous with respect to linear Lebesgue measure on [,]. Moreover, there are constants C C 2 depending only on ψ, not on S, such that for all compact K [,], µ (K) C log capk C 2 log meas (K). Proof. (a) Firstly as all zeros of P n R n ψ lie in [,], so I n = P n (x)r n ψ (x)dx (diamψ [,]) n. Here diam denotes the diameter of a set. So A diamψ [,]. In the other direction, we use Cartan s Lemma for polynomials [2, p. 75], [9, p. 366]. This asserts that if δ >, then ( ) δ n R n (x) 4e outside a set E of linear Lebesgue measure at most δ. Then ( ) δ n R n ψ (x), x [,] \ψ [ ] (E). 4e 22
By our hypothesis (3), we may choose δ so small that ( ) meas (E) δ meas ψ [ ] (E) 4. Next, Cartan s Lemma also shows that ( ) n P n (x), x [,] \F, 6e where Then P n (x) R n ψ (x) meas (F) 4. ( ) δ n ) 64e 2, x [,] \( ψ [ ] (E) F so I n [,]\(ψ [ ] (E) F) P n (x)r n ψ (x)dx ( ) δ n 64e 2 2. Hence (b) Since for x,t [,], log A δ 64e 2. ψ (x) ψ (t) log 2diamψ [,] >, so for x supp[µ], Lemma 3.(b) gives Then log A W µ (x) U µ (x) + log 2diamψ [,]. I (u) log A log 2diamψ [,]. (c) As µ has finite energy, it vanishes on sets of capacity zero. Then combining Lemma 3. 3.2, W µ = log A both q.e. a.e. (µ) in supp[µ]. 23
Then the first assertion (54) also follows. (d) This is almost identical to that of Theorem.2(d), following from the fact that W µ log in supp[µ]. A Proof of Theorem.5. Assume that S,µ A are as in the beginning of this section. Assume that S #, µ #, A # satisfy analogous hypotheses. We shall show that A = A # µ = µ #. Our hypothesis on the zeros shows that Then Lemma 3.3 shows that supp[µ] = supp [ µ #] = K. W µ = log A q.e. in K W µ# = log q.e. in K. A # Since I (µ) I ( µ #) are finite by Lemma 3.3, these last statements also hold µ a.e. µ # a.e. in K. Then log A = W µ dµ # = W µ# dµ = log A #. It follows that there is a unique number A that is the limit of In /n as n. Next, ( J µ µ #) ( = J (µ) + J µ #) 2 W µ dµ # As in Theorem.2(a), this then gives = log A + log A 2log A =. µ = µ #. This proof also shows that µ is the unique solution of the integral equation W µ = C q.e. in K. 24
We turn to the Proof of Theorem.3. Let µ be a weak limit of some subsequence {µ n } n S of {µ n } n=. We may also assume that (5) holds. From Lemma 3.3, µ has finite logarithmic energy, from Lemma 3.2, W µ log A q.e. in [,]. Moreover, by Theorem.2(c) our hypothesis (2), = J q.e. in [,]. Then the last relations also hold µ a.e. ν ψ a.e., so J = dµ = W µ dν ψ log A. Moreover, by Lemma 3.3(c), W µ = log A µ a.e. in supp[µ] so J (µ) = W µ dµ = log A J. Then necessarily log A = J (µ) = J µ = ν ψ. Proof of Theorem.4. Assume first that ψ is continuous in (,) that for each x, t [,] with x t, 2 x2k (x,t) >, but that the support is not all of [,]. We already know that [,ε] [ ε,] supp[ν ψ ] for some ε >. Then there exist < a < b < such that (a,b) supp[ν ψ ] =. (56) 25
We may assume that both a,b supp[ν ψ ]. (57) Then by Theorem.2(c), (a) J (b) J. But in (a,b), which lies outside the support of µ, W µ will be twice continuously differentiable, by our hypothesis, 2 ψ 2 x 2Wν (x) = x 2K (x,t) dν ψ (t) >. The convexity of forces in some (c,d) (a,b) W µ < J. This contradicts Theorem.2(b). Next, suppose that for x,t (ψ (),ψ ()) with x t, Consider We have ψ [ ] (x) = 2 [ ( )] x 2 K ψ [ ] (x),ψ [ ] (t) >. = K K ( ) ψ [ ] (x),t dν ψ (t) ( ) ψ [ ] (x),ψ [ ] (s) dν ψ ψ [ ] (s). ψ [ ] (x) J if x ψ (supp[ν ψ ]) at each point of continuity of ψ [ ], Theorem.2(b) gives ψ [ ] (x) J. We also see that for x [ψ (),ψ ()] \ψ (supp[ν ψ ]), 2 [ ] x 2 ψ [ ] 2 [ ( )] (x) = x 2 K ψ [ ] (x),ψ [ ] (s) dν ψ ψ [ ] (s) >. If < a < b < (56), (57) hold, then by Theorem.(c), ψ [ ] (ψ (a)) J ψ [ ] (ψ (b)) J 26
so in some interval (c,d) (ψ (a),ψ (b)), the convexity gives ψ [ ] < J But then < J in (ψ (c),ψ (d)), contradicting Theorem.2(b). Proof of Theorem.6. Recall from (45) that I n = S n Q n ψ S n (x) Q n ψ (x) /n = exp ( W νn (x)). Then much as in the proof of Lemma 3., 3.2, under the hypotheses (48) (5), we obtain W ν log A in supp[ν] W ν log A q.e. in [,], in particular at every point of continuity of W ν. Then the proof of Theorem.3 shows that ν = ν ψ, the result follows. We next prove an inequality for I n, assuming the hypotheses (35) (36). Below, if α,β are probability measures on [,], we set m α,β := inf [,] W α,β. Proof of Theorem.7. Let β be a probability measure on [,]. By orthogonality, for any monic polynomial Π n of degree n, we have I n = P n (x)π n ψ (x) dx. Given a probability measure on [,], we may choose a sequence of polynomials Π n such that Π n has n simple zeros in [ψ (),ψ ()], the corresponding zero counting measures converge weakly to β ψ [ ] as n. 27
(This follows easily as pure jump measures are dense in the set of probability measures.) As W µ,β m µ,β in the closed set [,], we obtain, by Lemma 2., uniformly in [,]. Then lim sup P n (x)π n ψ (x) /n exp ( m µ,β ), n,n S A = lim sup In /n exp ( m µ,β ). n,n S Taking sup s over all such β gives (38). The other relation follows similarly, because of the duality identity (32). References [] V. V. Andrievskii, H. P. Blatt, Discrepancy of Signed Measures Polynomial Approximation, Springer, New York, 22. [2] G. A. Baker, Jr., Essentials of Padé Approximants, Academic Press, 975. [3] A. Borodin, Biorthogonal Ensembles, Nuclear Physics B, 536(999), 74 732. [4] P. Borwein T. Erdelyi, Polynomials Polynomial Inequalities, Springer, New York, 995. [5] C. Brezinski, Biorthogonality its Applications to Numerical Analysis, Marcel Dekker, New York, 992. [6] T. Claeys D. Wang, Rom Matrices with Equispaced External Source, manuscript. [7] C. Elbert, Strong Asymptotics of the Generating Polynomials of the Stirling Numbers of the Second Kind, J. Approx. Theory, 9(2), 98 27. [8] W. Ford A. Sidi, An Algorithm for a Generalization of the Richardson Extrapolation Process, SIAM J. Numer. Anal., 24(987), 22 232. [9] W. K. Hayman, Subharmonic Functions, Vol. 2, Academic Press, London, 989. 28
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[23] A. Sidi, Problems 5-8, (in) Numerical Integration III, (eds. H. Brass G. Hämmerlin), Birkhäuser, Berlin, 988, pp. 32 325. [24] A. Sidi, Practical Extrapolation Methods, Cambridge University Press, Cambridge, 23. [25] A. Sidi D. S.Lubinsky, On the Zeros of Some Polynomials that Arise in Numerical Quadrature Convergence Acceleration, SIAM J. Numer. Anal., 2(983), 589 598. [26] A. Sidi D. S. Lubinsky, Biorthogonal Polynomials Numerical Integration Formulas for Infinite Intervals, Journal of Numerical Analysis, Industrial Applied Mathematics, 2(27), 29 226. [27] H. Stahl, Hwritten Notes, 23. [28] H. Stahl V. Totik, General Orthogonal Polynomials, Cambridge University Press, Cambridge, 992. 3