Asymptotics of Polynomials Orthogonal with respect to a Logarithmic Weight

In this paper we compute the asymptotic behavior of the recurrence coefficients for polynomials orthogonal with respect to a logarithmic weight $w(x){\rm d}x = \log \frac{2k}{1-x}{\rm d}x$ on $(-1,1)$, $k>1$, and verify a conjecture of A. Magnus for these coefficients. We use Riemann-Hilbert/steepest-descent methods, but not in the standard way as there is no known parametrix for the Riemann-Hilbert problem in a neighborhood of the logarithmic singularity at $x=1$.


Introduction
In this paper we consider the sequence of polynomials {p n (x)} that are orthonormal with respect to a logarithmic measure w(x)dx on [−1, 1) and 0 elsewhere 1 −1 found in the neighborhood of singularities of the underlying measure for the associated Riemann-Hilbert problem. Because no such local solution is known near x = 1 for the logarithmic case, the standard approach cannot be applied directly. The approach in this paper hinges instead on a comparison with the Legendre polynomials {p n } which are orthogonal with respect to the weight w(x) = 1, for x ∈ [−1, 1], 0, otherwise, and which have recurrence coefficientsã n ,b n . The asymptotics ofã n andb n are well known. We have, from [13], a n = 0,b n = 1 2 The result of this comparison, and the main result of this paper is that Theorem 1.2. As n → ∞, the differences between the log-orthogonal coefficients and the Legendre coefficients have the following asymptotic behavior a n −ã n = 2C (n log n) 2
This result, and more, was conjectured by Alphonse Magnus in [14], up to the precise value for the constant C. We are indebted to his work, which provided the inspiration for the above result. For the weight w(x) = − log x on [0, 1], in particular, Magnus conjectured that as n → ∞ a n = 1 2 and on numerical grounds he conjectured further that C ∼ −0.044. The weight − log x on [0, 1] corresponds to the case k = 1 for which our analysis is not yet complete. Nevertheless, our prelimary calculations confirm (1.3) and (1.4) with stronger estimates O 1 n 2 (log n) 3 on the error terms, and with the explicit value −3/64 for C. As −3/64 ∼ −0.047 this is close to Magnus's conjectured value.
The proof proceeds as follows: The polynomials {p n } have a well-known expression in terms of the solution Y of a Riemann-Hilbert problem due to Fokas-Its-Kitaev, see Section 3.1. We follow a now standard series of transformations as in [7,13], that transform the original Riemann-Hilbert problem for Y into a modified problem that is close to a Riemann-Hilbert problem whose asymptotics can be inferred directly.
Here is where our analysis diverges from the standard analysis. The standard analysis compares the modified Riemann-Hilbert problem to a model Riemann-Hilbert problem (parametrix) that explicitly solves the Riemann-Hilbert problem locally around the critical points at the edge of the measure. The new difficulty of our problem is that no explicit local solution is known near the logarithmic singularity. The main new feature of our analysis is that, for the problem at hand, no local solution is needed.
The strategy here is to compare the modified Riemann-Hilbert problem above to the corresponding modified Riemann-Hilbert problem for the Legendre case. In the standard approach, a comparison is made by examining the quotient of solutions to two Riemann-Hilbert problems as the solution to a Riemann-Hilbert problem in its own right. However, for the case at hand, the log singularity renders this approach ineffective. Instead we compare the problems through the operator theory that underlies Riemann-Hilbert problems, see Section 4, or see [10] for more details.
In Theorem 4.4 we prove a general and very useful formula that allows for an effective comparison of two Riemann-Hilbert problems on the same contour. In our case, where we compare the log case with the Legendre case, this formula can be approximated by quantities solely related to the Legendre case, for which the asymptotics are well known, and this leads, eventually, to the formulas (1.1) and (1.2). It is a remarkable, and unexpected, development that the log problem can be approximated adequately by the Legendre problem.
The proof is laid out within this paper as follows: • In Section 2, we prove several preliminary properties and estimates on the auxilliary functions that will arise in our analysis.
• In Section 3, we introduce the Riemann-Hilbert problem for the orthogonal polynomials and transform this problem into one which can be controlled as n → ∞.
• In Section 4, we prove bounds, uniform in n, on the resolvent operators that are associated with the Riemann-Hilbert problem in the standard way and prove Theorem 4.4.
• In Section 5, we calculate the asymptotics of the recurrence coefficients by combining the results of the previous sections with calculations derived from the known asymptotics of Legendre quanties.
• The above calculations are given in Appendices A, B, and C.
Note that the case k = 1 corresponding to the weight log 2 1−x on the interval [−1, 1), for which w(x) > 0 on its support, but for which w(x) c > 0 fails, is not considered in this paper. Recall that questions concerning orthogonal polynomials on the line with respect to a measure of compact support correspond to questions involving associated orthogonal polynomials on the unit circle, where the vanishing of the weight log 2 1−x at the point −1 corresponds to a Fisher-Hartwig singularity for the related problem on the circle, see [8]. The methods in this paper do not extend immediately to cover this case, but will be addressed in a future paper. This paper is in the line of questioning concerning the effect that singularities and zeroes in the measure have on the asymptotic behavior of orthogonal polynomials. Here we have in mind the extensive history of the problem dating originally back to the work of Szegő, proving the strong limit theorem for Toeplitz determinants for non-vanishing smooth weights on the circle without singularities [15], and whose result has been extended to include a wide class of algebraic singularities, in particular, by the work of M.E. Fisher and R.E. Hartwig, see [8] for a history. Of additional interest is the recent work of Barry Simon et al. concerning sum laws of Szegő type where zeros of the form e −1/x 2 , for example, in the weight are investigated (see [2] for an up to date discussion). The logarithimic singularities explored in this paper are of practical interest in both physics and mathematics as described in [14,16]. 2 The φ function and the Szegő function In this section, we introduce several functions central to our analysis and detail their properties.

The map φ and the Szegő function F
The first part of our work closely follows the work from the papers [9,13,17], and uses the Riemann-Hilbert approach to orthogonal polynomials, first put forward by Fokas, Its, and Kitaev [11]. We apply similar sets of transformations to the Riemann-Hilbert problems associated with the weights w andw, The problems Q andQ are then close in a "certain" sense, allowing the comparison. The prob-lemQ can be asymptotically solved as in [13], as repeated in the appendix, also see Section B.1.
For the log-orthogonal polynomial case, we make use of the associated Szegő function F similar to the non-Legendre cases analyzed in [13], where ds := ds 2πi and similarly dt, dz, . . . mean dt 2πi , dz 2πi , . . . . In the calculations that follow we shall use the following lemma repeatedly without further comment: 1−s has an analytic extension from [−1, 1) to C\[1, ∞) where log takes on its principle value.
Proof . The proof is left to the reader.
uniformly as z → 1, 14) Proof . As the estimate in statement (2.12) of Proposition 2.3 is uniform in 0 < |θ| < π, we can compute for x > 1, x → 1, We will also need the following technical estimate on the difference F 2 w ± (1 + r) − F 2 w ± (1 +r) for r,r > 0 suitably small and close together. This estimate is needed to estimate the integral in Proposition 5.4. Proposition 2.5. Fix R > 0 and suppose r,r > 0 obey Then Proof . See Proposition A.4.

The Riemann-Hilbert approach
In this section we define the main Riemann-Hilbert problem and transform the problem into a form which is amenable to asymptotic analysis.
Remark 3.1. The notation O n is used as follows. For matrices we say A(z) = O n (B(z)) as z → z 0 if, for each pair i, j = 1, 2, and for any n 0, there exists ε and c n such that Similarly, for scalar functions A n (z) and B(z), we say that A n (z) = O(B(z)) if there exists c n such that

Riemann-Hilbert problem for Y
We will denote by π n (x) = x n + · · · ,π n = x n + · · · the monic, orthogonal polynomials associated with the weights w,w respectively. The polynomials {π n } are described by the following Riemann-Hilbert problem introduced by Fokas, Its, and Kitaev [11] (see also [7]) for a 2×2 matrix valued function Y (z) = Y (n) (z): (a) , for x ∈ R pointwise almost everywhere and in the following L 2 sense: Define the Cauchy operator, C [−1,1] by and let C ± [−1,1] denote its boundary values from C ± respectively. We give a brief summary of the properties of the Cauchy operator and its boundary values on general contours in Section 4.2.
Proposition 3.2 (Fokas-Its-Kitaev). Let γ n > 0 be the leading coefficient of p n , the orthonormal polynomials with respect to the weight w, p n = γ n π n . The function uniquely solves the RHP for Y (n) .
Proof . Aside from some special attention that must be paid to the L 2 sense in which the boundary values Y ± are achieved, the proof proceeds as in [7]. The L 2 condition in (3.2) plays the following key role in proving uniqueness: Note that, by the L 2 condition in (3.2), for any a, b ∈ R However, the jump condition in (3. and it follows, by standard calculations, that C det Y (z)dz = 0 for all closed contours C ⊂ C.
Hence det Y (z) is entire by Morera's theorem. Condition (3.3) and Liouville's theorem then implies that det Y (z) = 1 for all z. In particular, LetỸ be a second solution to the RHP (3.1)-(3.3) and define H(z) =Ỹ (z)Y −1 (z), and repeating the same argument, we find that H(z) is entire, and H(z) → I as z → ∞ and hence H(z) = 1 for all z and therefore Y =Ỹ .
We must also demonstrate that the function Y (n) (z) defined in (3.4) achieves its boundary values in the aforementioned L 2 sense. A standard argument as in [13] shows that (d) Y (n) has the following uniform asymptotics as z → −1, z ∈ C\[−1, 1] We must also analyze Y (z) for z near 1. Let z = 1 + re iθ where r > 0 and −π < θ < π, and recall from Lemma 2.2 that w(s) has to analytic extension in C\[1, ∞). The following estimate holds uniformly for −π < θ < π, where C 1 , C 2 are depicted in Fig. 1. The integrand is clearly bounded over C 2 is clearly analytic and hence the integral of C 2 is O(1) Making the change of variables s = re iθ t, uniformly for z ∈ C\[−1, 1], as z → 1. Therefore, (e) Y (n) has the following uniform asymptotics as Lastly, note that the function Y (n) (z) clearly is continuous up to the boundary for all x ∈ R\{−1, 1} from the properties of the Cauchy operator. Therefore, combining (3.5) and (3.6), it is clear that Y achieves its boundary values Y ± in the L 2 sense described in (3.2).
Let σ 3 refer to the 3rd Pauli matrix The recurrence coefficients a n , b n associated with the polynomials {π n } have the following expressions in terms of the solution Y (n) (z):

The transformation Y → T
The first step in the asymptotic analysis of Y , is to provide a reformulation of the RHP described by (3.1)-(3.3) in terms of a RHP normalized at infinity, and now with a bounded, but highly oscillatory, jump function.
Remark 3.4. In the standard Riemann-Hilbert theory, the RHP is converted to a problem involving a bounded singular operator on the contour. The replacement of the unbounded jump ( 1 w 0 1 ) with a bounded jump makes it possible to apply Riemann-Hilbert theory directly.
Recall the definitions of φ and F from Section 2 and define Since Y , φ, and F are all analytic in the region C\[−1, 1], T clearly satisfies the same analyticity condition (3.1) as Y . For x ∈ (−1, 1) where we have used property ( To summarize, T (z) solves the following standard RHP: (c) T (z) has the following asymptotics as z → ∞ (d) T (n) has the following asymptotics as z → −1 (e) T (n) has the following asymptotics as z → 1 It follows from the definitions of T (3.7) and the fact that det(Y ) ≡ 1 that det(T ) ≡ 1.
Remark 3.5. Properties (2.12) and (2.13) of Proposition 2.3 imply that F 2 w is bounded uniformly as we approach the interval [−1, 1] and therefore the jump matrix for T in (3.8) is indeed bounded.

The transformation T → Q
The transformations in these sections are well known to those familiar with the Riemann-Hilbert approach to orthogonal polynomials. The next transformation is a slight modification of the typical "lens" transformation. The difference is that it collects factors in the upper and lower lips near the singularity at 1. For our particular weight, these factors cancel each other to leading order, due to Proposition 2.4. We "deform" the RHP on [−1, 1] to a problem on the oriented contour Σ as depicted in Fig. 2.
The transformation in this section is based on the factorization of the jump matrix of T in (3.8) into a product of 3 matrices on (−1, 1): Since F , φ are analytic in C\[−1, 1], and, by Lemma 2.2, log 2k 1−s has a unique analytic continuation to C\[1, ∞), the matrix valued function is an analytic function on C\[−1, ∞) and, by Remark 3.5 and Proposition 2.1 is continuous up to the boundary. Now define (3.9) A straightforward calculations shows Q = Q (n) solves the following RHP: (d) Q has the following asymptotics as z → −1 (e) Q has the following asymptotics as z → 1 The endpoint conditions in (3.10) and (3.11) guarantee that the solution Q to the above RHP achieves its boundary values in the L 2 sense. The same argument as in Proposition 3.2 proves that this solution is unique. Therefore, any solution to the above RHP must be the solution we obtain from the RHP for Y through the transformations (3.7) and (3.9).

The associated Legendre problem,Q
The same steps as in Section 3.3 can be repeated for the problem with the Legendre weight w(s) = 1, s ∈ [−1, 1], in which case all of the above calculations hold true provided we take F ≡ 1.

A formula for recurrence coefficients
Proposition 3.6. The following formulas hold for the recurrence coefficients in terms of the modified RHPs Q,Q, a n −ã n = Q (n) 1 Proof . Note, from Proposition 3.3, that the recurrence coefficients a n , b n are related to the RHP for Y via the formulas a n = Y (n) 1 and, again from Proposition 3.3, Y satisfies the following symmetry relation: Now for z sufficiently large (outside of the contour Σ) By Propositions 2.1 and 2.3 and so the formula for the recurrence coefficients in (3.15)-(3.16) can be expressed in terms of the RHP for Q by a n = Q and so a n = Q (n) 1 The same formulas hold forã n andb 2 n−1 , replacing Q 1 , Y 1 withQ 1 ,Ỹ 1 . Therefore, a n −ã n = Q (n) 1 as desired.

A comparison of resolvents
In this section we develop the machinery that will allow us to compare two Riemann-Hilbert problems using non-local information.

Comparing Riemann-Hilbert problem through the Cauchy transform
For a bounded, oriented, rectifiable contour Σ, consider C Σ acting on L p (Σ), 1 p < ∞, as follows where recall ds = ds 2πi . For almost all z ∈ Σ, the limits for almost all z ∈ Σ, and where H is the Hilbert transform which exists for almost all z ∈ Σ. An oriented, rectifiable contour Σ is called Carleson (or Ahlfors-David) if where U r (z) denotes a ball of radius r centered at z. The fundamental theorem, starting with the work of Calderon [3], continuing through the work of Coifman, Meyer, and McIntosh [5], and culminating in the work of Guy David [6], is the following: The operators C ± Σ are bounded from L 2 (Σ) → L 2 (Σ) (and in fact in L p (Σ) for all 1 < p < ∞) if and only if Σ is Carleson. Note, in particular, that the contour in this paper is clearly Carleson.
We will say that a pair of functions A ± ∈ L p (Σ) belongs to ∂C(L p (Σ)) iff Proposition 4.1. Let Σ be a Carleson curve, and let A ± ∈ I + ∂C(L p (Σ)) and B ± ∈ I + ∂C(L q (Σ)) for 1 < p, q < ∞ where 1 Proof . Suppose f ∈ L p (Σ) and g ∈ L q (Σ), then f (Hg), (Hf )g ∈ L r (Σ), and where we have used the fact that the Hilbert transform converges in L p (Σ) and L q (Σ). Therefore, and, since f ∈ L p (Σ), g ∈ L q (Σ), which completes the proof.
Remark 4.3. Suppose A ± ∈ I + ∂C(L p (Σ)) and suppose that A −1 ± (x) exists a.e. and lies in and A ± solves the RHP: and B ± the RHP: Suppose further that v −1 B ∈ L ∞ (Σ) and B −1 ± ∈ I + ∂C(L q (Σ)) where 1 p + 1 q = 1, then if A, B are the analytic extensions of A ± and B ± respectively, Proof . The following calculation holds almost everywhere on Σ, Therefore, almost everywhere on Σ, However, using Proposition 4.1, Thus . However, (4.1) then implies that h ≡ 0, and therefore Proposition 4.4 clearly implies for z ∈ C\Σ, If it is possible to control the L 2 (Σ) difference A − − B − then quantities related to A away from the contour Σ can be estimated by quantities that only rely on B and the jump func- The results of this section will lay the groundwork of controlling this difference, with the remainder of the work being done in the following section.
For compact Σ and f ∈ L p (Σ), p 1, we have that (C Σ f )(z) = O 1 z as z → ∞. The contour Σ defined in Section 3 is clearly compact, we will use this fact repeatedly without further comment.

The resolvent bounds
Let h, f refer to matrix valued functions, h ∈ L ∞ (Σ) and f ∈ L 2 (Σ), and define the operator C h acting on L 2 (Σ) by Clearly C h is bounded from L 2 (Σ) → L 2 (Σ). The importance of this definition lies in its role within the general theory of Riemann-Hilbert problems, see for example [4,10] and the references therein. In particular, let µ be any L 2 (Σ) solution of the equation is the unique solution to (4.2). In this case the unique solution X to the RHP normalized at infinity with jump v Σ is given by Our first step to controlling the difference Q −Q, will be showing that the resolvent in (4.3) is bounded uniformly of n for the following modified version of the Legendre Riemann-Hilbert problem: Define, as in (3.13), for the same contour Σ, and subcontours Σ 1 , Σ 2 as depicted in Fig. 2, then we have the following theorem: Proof . The following proposition follows from Proposition 2.14 from [10], see also [4].
is (uniquely) solvable for all g ∈ L 2 (Σ) with ||m ± || L 2 c||g|| L 2 with c independent of both n and g.
We want to apply Proposition 4.6 to h n =ṽ Σ . To this end, suppose a solution m ± (s) = m ± (g, n, s) exists to and note m(z) := (C Σ f )(z) for z ∈ C\Σ is the analytic continuation of m ± . It will follow from the calculation below that such a solution indeed exists.
and, in the notation of Fig. 2, set Sinceṽ(s) is uniformly bounded, it is clear that g L 2 (Σ) c g L 2 (Σ) , and it is also clear that , straightforward calculations will show that t + (s) = t − (s). Therefore t(z) is actually analytic for z ∈ C\[−1, 1]. Lastly, for z ∈ (−1, 1), first note thatg = 0, and therefore C + Σg = C − Σg , which we label C Σg . Another calculations shows: So t(z) solves a RHP: Since the restriction of C Σ to [−1, 1] is bounded as an operator from L 2 (Σ) to L 2 ([−1, 1]), and 1], and, as z → ∞, In particular, a solution to the RHP described below can be transformed in a bounded fashion to the solution m, This RHP can be solved as follows. Letĝ = (ĝ 1 ,ĝ 2 ) and u = (u 1 , u 2 ) whereĝ 1 ,ĝ 2 , u 1 , u 2 are the column vectors ofĝ and u. Then Thus, since u 1 = O n (z n−1 ) as z → ∞, any polynomial p n−1 with deg(p n−1 ) n − 1 provides a potential solution for u 1 through Plemelj's formula, and, since u 2 = O n z −n−1 , if the solution exists, it must be, again by Plemelj's formula, The condition u 2 = O n z −n−1 is therefore equivalent to a system of n equations for the n coefficients of p n−1 . Specifically, we have As det 1 −1 s k+j ds n−1 k,j=0 = 0, equation (4.9) is a system of n independent, linear equations for the n unknowns, a 0 , a 1 , . . . , a n−1 , where p n−1 (s) = a 0 + a 1 s + · · · + a n−1 s n−1 .
It follows that the solution u exists and is unique. Moreover, For all polynomials with deg(P k ) = k n − 1 So, using the fact that |φ + | = 1 and the boundedness of the Cauchy operator, we have Most importantly, we notice that c is independent of n. Therefore, by using the boundedness of the Cauchy transform and the fact that |φ ± | = 1, as well as (4.7) and (4.8), we see that, where all the constants are independent of n. Up until now, we have been assuming solutions m and t existed. However, reversing our construction, and satisfies m ± All that remains is to show m ± (z) ∈ ∂C(L 2 (Σ)). To that end, letμ(s) = m − (s) and define for z ∈ C\Σ, (4.11) Then for s ∈ Σ It follows from the same argument as in Proposition 3.2 that the function H(z) − m(z) is entire. From (4.10) and (4.11), and so m ± ∈ ∂C(L 2 (Σ)) concluding the proof of Theorem 4.5.
Theorem 4.5 has, as a corollary, the equivalent result for the modified log-orthogonal problem: So it follows from Propositions 2.1 and 2.3 that which, in particular, implies that Combined with Theorem 4.5, we see that and using (4.12), we see that for n sufficiently large, the series converges in operator norm and It follows from (4.13) that (1 − C v Σ ) −1 exists and is given by for n sufficiently large and hence So by Theorem 4.5, (1−C v Σ ) −1 is bounded in operator norm uniformly for n sufficiently large.
Remark 4.8. In fact, a standard RHP argument shows that (1 − C v Σ ) −1 exists and is bounded in L 2 → L 2 for all n 0. The point of the above theorem is that the bound is independent of n.

A formula for
Proposition 4.9.
Proof . The functions µ andμ are given in terms of the RHP solutions Q andQ, see [10], by and so and it is straightforward to show that So which completes the proof.
Remark 4.10. The goal of this section is to develop the machinery that allows effective comparison of the difference of the RHPs Q andQ. As mentioned in the introduction, the standard approach of comparing the RHPs for Q,Q by their quotient breaks down. Indeed, for x ∈ [1, 1 + δ], So QQ −1 is the solution to a standard RHP on Σ with jump Typically, we would like to draw conclusions about QQ −1 by demonstrating that J −I L ∞ → 0. However, while the jump functions associated with Q,Q are close in an L ∞ sense, and specifically, near the point 1, , as z → 1, this behavior is counteracted by the singular behavior ofQ. From (3.14), we have thatQ 12 ,Q 22 = O n (log(|z − 1|)) as z → 1. More is true, in fact,Q 12 ,Q 22 = c n log(|z − 1|)(1 + o(1)) as z → 1 where c n ∼ cn 1/2 , which we will not prove, but which can be inferred from the behavior ofμ described in Section B.1. Therefore, from (4.17), it is clear that Thus it is not clear how to proceed in an analysis of the quotient QQ −1 . Ultimately, this difficulty motivated our examination of the difference Q −Q.
Remark 4.11. We note that formulae (4.15) and (4.16) can be used, in principle, not only to derive our basic formula (4.14), but also to obtain asymptotic information about the orthogonal polynomials for general values of z, not just for z = ∞. Certainly for general z the analysis should go through as long as z is at a fixed distance away from the interval [−1, 1], but as z approaches the end points ±1, in particular, technical difficulties arise that we have not yet been able to address. Also, as z → ∞, one can, in principle, use (4.15) and (4.16) to obtain lower order terms for the recurrence coefficients, but again certain technical difficulties arise that we have not yet overcome.

Conclusion of the proof of Theorem 1.2
In this section we combine the previous results of the paper in order to prove the main result, Theorem 1.2.
5.1 Calculation of the integral in (4.14) The integral in (4.14) involves bothμ and µ. The asymptotics ofμ are known, see [13], or Appendix B.1 for a summary. Theorems 4.5 and 4.7 together with the known asymptotics ofμ give us control over µ in an L 2 (Σ) sense, which will be sufficient for our purposes. Namely, in Proposition B.2 we prove: Proposition 5.1. The functions µ,μ obey the following estimates: Proof . See Proposition B.2.
Proposition 5.1 together with Theorems 4.5 and 4.7 allow us to estimate the integral in (4.14) by a similar integral involving only Legendre quantities. This is done through the following proposition: Proposition 5.2. The integrals can be estimated by Proof . Recall that det(µ), det(μ) = det(Q − ), det(Q − ) = 1. Therefore the norms (5.2) and (5.4) also imply and therefore, and by (5.1) and (5.2),

26
T.O. Conway and P. Deift Somewhat similarly, and by (5.1) and (5.4), Therefore, we see that which completes the proof of Proposition 5.2.
The integrals in Proposition 5.2 can be computed asymptotically using the known asymptotics ofμ, yielding: Proof . See Proposition C.3.
Combining the results of this section, we can compute the difference Q 1 asymptotically, 1 has the following asymptotic expansions Proof . Combining Propositions 4.9, 5.2, and 5.3,

5.2
Computing the asymptotic behavior of a n , b n Proof of Theorem 1.2. Recall that by Proposition 3.6, a n −ã n = Q The calculation for a n −ã n is read directly from Proposition 5.5 a n −ã n = 3 16n 2 log 2 n To calculate b 2 n−1 −b 2 n−1 , note from (B.6) and (B.10) that as z → ∞, From (B.11), where R 1 , ∆(z) are analytic in for z ∈ C\Σ S and from which it follows that Combining Proposition 5.5 with (5.7) we see that Using Propositions 3.6 and 5.5, and substituting in (5.8) and (5.9), we see that Proof . Let z = 1 + re iθ where 0 < |θ| < π, r → 0. We will first prove that 1 −1 log w(s) where O 1 log 3 r is uniform in 0 < |θ| < π. Note that 1 −1 log w(s) The first integral is clearly bounded as z → 1. The second integral is easily seen to be controlled via the following lemma: Then as z → 1 uniformly for z ∈ C ± . Recall z = 1 + re iθ and make the change of variables t = 1 − s, then where the contours C 1 , C 2 are depicted in Fig. 3 for 0 < θ < π; there is an analogous picture for −π < θ < 0. Here t 1/2 = |t| 1/2 e iθ/2 where −π < θ < π is the analytic continuation of √ t from (0, ∞) to C\(−∞, 0]. The integral over the contour C 2 is bounded away from all singularities of the integrand, and thus, by dominated convergence, tends uniformly in 0 < |θ| < π as r → 0 to which is finite. Therefore, 1 −1 log w(s) Making the change of variables t = re iθ u, u > 0, γ = √ u, When γ r 2k 1/4 , we have that log 2k r −4 log γ. So as −2 log γ > 0, As log 2k r − 2 log γ log 2k r → ∞ as r → ∞, the integrand in the second integral is bounded above and therefore | log x| is an increasing function when x > 1, therefore since log 2k r −4 log γ and log 2k r − 2 log γ > 1 as r → 0, Integrating by parts, where, in the final step, we use the same argument as in the proof of (A.4). When r 2k 1/4 γ r 2k −1/4 , we have | log γ| 1 4 log 2k re iθ . Therefore for γ in this region, we can use the power series expansion for the logarithm where again the term O log 3 γ log 3 r is uniform in θ. Therefore The first integral in (A.6) is easily calculated The second integral, after a change of variables η = 1 γ , satisfies log w(s) Now note that for z = 1 + re iθ , Now for z = 1 + re iθ ∈ C ± , and recalling that log always refers to the principle branch, uniformly for z ∈ C ± as z → 1, which completes the proof of Proposition A.1.
Proof . Note that as z → −1, uniformly for z ∈ C\[−1, 1], 1 −1 log w(s) By the proof of Lemma A.2, the second integral is easily seen to be O(1). Therefore 1 −1 log w(s) where C refers to any clockwise contour containing [−1, 1] but not the point z. Letting C go to infinity, a residue calculation gives log w(−1)  Then n log 3 n . Therefore we would expect Proof . We define a = a(r,r) := n r r − 1 . Clearly r r = 1 + a n . In the calculations that follow, we assume without loss of generality that r >r, and hence a > 0. First, note that log F (1 + r) = (1 + r) 2 − 1 log w(s) log w(s) .
Making the change of variables t = 1 − s, where H(r,r) : The same calculations as in the proof of Proposition A.1 show that Making the changes of variables t → rt and t →rt in (A.15) within the respective integrals, we have Note that, as log k log 2k To estimate the remaining term in (A.16), but t ∈ [r 1/2 , r −1/2 ] and so | log t| 1 2 log 1 r 1 2 log 2k r . Therefore, (A.19) is ia n log 2k r r −1/2 Making the change of variables γ = t 1/2 , this becomes 2ia n log 2k r r −1/4 So, using the same calculations of the above integrals, see (A.7) and (A.8), as in the proof of Proposition A.1, which implies so a n log 2k r + 1 − w ± (1 + r) w ± (1 +r) = a n log 2k r − a n log 2k r ± iπ = ±iπa n log 2k r log 2k r ± iπ = ± iπa n log 2 2k where we have used Proposition A.1. So, in particular which concludes the proof of Proposition 2.5.

B A summary of Legendre asymptotics
In this appendix we prove a series of L 2 estimates, ultimately allowing us to replace the integral in Proposition 4.9 with an integral only involving Legendre quantities, as in Proposition 5.2.

B.1 A summary of Legendre asymptotics
In [13] the authors derive the asymptotics of the solution S(z) = S (n) (z) to the RHP (Σ S , v S ): where Σ S is depicted in Fig. 5, for s in the upper and lower lips, The asymptotics are derived for z in the regions A, B, C, D, and E, see Fig. 6. We will summarize the necessary results in this section.
where the regions Ω 1 and Ω 2 are depicted in Fig. 7.
For z ∈ C\[−1, 1], we define, as in [13], where a(z) is analytic for z ∈ C\[−1, 1] and a(z) > 0 when z > 1. Let U δ (s) refer to the closed ball of radius δ centered as s = ±1. The regions B and C are given by B = U δ (1)\Σ S and C = U δ (−1)\Σ S respectively. We define f (z) is analytic in a neighborhood of 1 and So, in particular, we can pick δ sufficiently small so that f (z) is locally conformal for z ∈ B, which clearly also implies f (−z) will be locally conformal in C. We define for −π < arg ζ < −2π/3, for ζ ∈ C\Γ, and where Γ is defined in Fig. 8. The contour Σ S is chosen within U δ (1), (resp. −1) to be the preimage of Γ under the map n 2 f (z) (resp. n 2 f (−z)). Therefore, around the point −1, Σ is the preimage of Γ under the map n 2 f (−z) as well. In other words 0 refer to the Hankel functions of the first and second kinds, see, for instance, [1]. As before, the square root in the above definition is given by its principle value.
For any fixed δ sufficiently small, the following holds uniformly, where R (n) (z) is bounded, analytic in C\Σ R , with Σ R depicted in Fig. 9, and where R 1 (z) is an n-independent, bounded, analytic function for z ∈ C\Σ R . Additionally Finally, E(z) is defined by (see Remark B.1) E(z) is analytic and bounded for z ∈ B and has determinant 1. It follows, in particular, that E −1 (z) is also analytic and bounded for z ∈ B.
Remark B.1. We use slightly different notation than in [13] in a few ways, • We refer to R(z) as R (n) (z) to emphasize its dependence on n.
As in the proof of (B.15), we have that and the same estimates as before imply For I 1 , we see that thus proving (B.17).
To move from (B.17) to (B.18), we proceed as follows Now, by (B.17) and the boundedness of the operators

C Leading order asymptotics of the difference formula
In this appendix we calculate the integrals that arise in Proposition 5.2.
Additionally, we have the following matrix calculation: Proof . .

C.2 The first integral
Proof . As in the proof of Proposition B.2, we have The same argument as Proposition B.2, gives For I 1 Now, note that by the n-independence and analyticity of the functions E(z), E −1 (z), for z ∈ (1, 1 + 1/n), and clearly 1 0 Therefore, subtracting term by term in (C.7), (C.7) becomes −2πn which, by (C.1), yields
All that remains is to show that I 2 is of lower order. Following the same steps as the lead up to (C.7), Therefore, As in (B.29), for |z + 1| 1 n , and so Letting y = n 2 f (−z), in which case we have, as before, dz = 2dy n 2 1 + O 1 n , and recalling that n 2 f (−(Σ ∩ U 1/n (−1))) ⊂ Γ, where we have used the fact that Ψ 12 , Ψ 22 ∈ L 2 (C 0 ), where C 0 is the contour | arg y| = 2π/3, see (B.14). Therefore we see that where we have used (C.9). Thus we have proved Proposition C.3.

C.3 The second integral
Proposition C.4. Proof . While not sufficient to prove the above, note that 3 16n log 2 n 1 −i −i −1 − 3 16(n + 1) log 2 (n + 1) So this proposition essentially asserts that the leading order of the difference (C.10) is the difference of the leading orders of the two terms, which are computed in (C.6).
We have, as before, where I 1 , I 2 , I 3 refer to the difference of the integrals of the above over [1, 1 + δ], Σ ∩ U 1/n (−1), and Σ\(U 1/n (1) ∩ U 1/n (−1)) respectively. The same argument as in Proposition B.2 shows Following the same steps as in Proposition C.3, but this time using the more precise expansion R (n) (z) = I + R 1 (z) n + O 1 n 2 in the expression forμ, we see, Now, let y 1 = n 2 f (z) and y 2 = (n + 1) 2 f (z), note that if z ∈ [1, 1 + 1/n] then 0 < y 1 , y 2 < cn. Let f −1 1 refer to the inverse of f within a neighborhood of 1. Then for z ∈ [1, 1 + 1/n], and some constant a, Making the substitution y 1 = n 2 f (z) in the first integral of (C.11) and y 2 = (n + 1) 2 f (z) in the second, we obtain Note that (n+1) 2 f 1+ 1 n+1 = O(n) and (n+1) 2 f 1+ 1 n+1 −n 2 f 1+ 1 n = O(1). Therefore, since all of the terms in the above integrands are bounded for large y, and since Ψ 12 , Ψ 22 rapidly decay for large argument, we can replace the limits of the second integral in (C.12) with the limits of the first and pick up an error that is exponentially small in n. Next, recall that f −1 (y) − 1 = 2y + ay 2 + O y 3 and y is in the region f −1 y n 2 − 1 So, by Proposition 2.5, we see that Next, using the analyticity of E, R 1 , The function F 2 w + + F 2 w − − 2 is bounded in compact sets and Ψ 12 , Ψ 22 are both L 2 (0, ∞), therefore the L 1 norms are both bounded uniformly of n. Therefore (C.13) is − 4π n(n + 1) (C.14) Inverting the change of variables y = n 2 f (z), (C.14) becomes − 2πn n + 1 which is the right-hand side of (C.10).
To conclude the proof we need to show the contribution from the left endpoint, I 2 , is of lower order. Following the same steps as those leading up to (C.11), we have Recall that Ψ 2 (y) = O(log y) and Ψ 2 (y) = O 1 y as y → 0, and Ψ 2 (y), Ψ 2 (y) = O e −cy 1/2 as y → ∞, see (B.14). Therefore