Asymptotics of Toeplitz, Hankel, and Toeplitz+Hankel determinants with Fisher-Hartwig singularities

Theorem 1.1 (Ehrhardt [20]). Let f(eiθ) be defined in (1.2), V (z) be C∞on the unit circle, |||β||| < 1, ℜαj > −1/2, and αj ± βj ̸= −1, −2, . . . for j, k = 0, 1, . . . , m. Then as n →∞, Dn(f) = exp " nV0 + ∞ X k=1 kVkV−k # m Y j=0 b+(zj)−αj+βjb−(zj)−αj−βj (1.12)

Theorem 1.13 holds under condition (1.28) of Remark 1.15 below. The unifor- mity in α- and β-parameters will also hold provided s is taken large enough. In [17], we give an independent proof of Theorem 1.1, in the spirit of [16], [28], [32], using a connection of Dn(f) with the system of polynomials orthogonal with weight f(z) (1.2) on the unit circle. These polynomials also play a central role in the proofs presented here. It follows, in particular, from Theorem 1.1 that all Dk(f) ̸= 0, k = k0,

Theorem 1.8. Let f(eiθ) be defined in (1.2), V (z) be analytic in a neigh- borhood of the unit circle, and φk(z) = χkzk + · · · , bφk(z) = χkzk + · · · be the corresponding polynomials satisfying (1.16). Assume that |||β||| < 1, αj ± βj ̸= −1, −2, . . . , j, k = 0, 1, . . . , m. Let (1.20) δ = max j,k n2ℜ(βj−βk−1), where the indices j, k = 0 are omitted if α0 = β0 = 0. Then as n →∞, χ2 n−1 = exp ñ − Z 2π 0 V (eiθ) dθ

Lemma 1.12. There exist only the following two mutually exclusive pos- sibilities: • ∃bβ ∈Oβ such that ||| bβ||| < 1. Then such bβ is unique and it is the unique element of M = { bβ}. • ∃bβ ∈Oβ such that ||| bβ||| = 1. Then there are at least two such bβ’s and all of them are obtained from each other by a repeated application of the following rule: add 1 to a bβj with the smallest real part and subtract 1 from a bβj with the largest. Moreover, M = { bβ ∈Oβ : ||| bβ||| = 1}.

Theorem 1.13. Let f(z) be given in (1.2), ℜαj > −1/2, βj ∈C, j = 0, 1, . . . , m. Let M be nondegenerate. Then, as n →∞, (1.27) Dn(f) = X Ñ m Y j=0 znj j én R(f(z; n0, . . . , nm))(1 + o(1)), where the sum is over all FH-representations in M. Each R(f(z; n0, . . . , nm)) stands for the right-hand side of formula (1.12), without the error term, cor-

Theorem 1.1 holds in fact under condition (1.15). It then follows (see §6) that,

Theorem 1.18 (A particular case of Theorem 1.13). Let the symbol f±(z) be obtained from f(z) (1.2) by replacing one βj0 with βj0 ± 1 for some fixed 0 ≤j0 ≤m. Let ℜαj > −1 2, ℜβj ∈(−1/2, 1/2], j = 0, 1, . . . , m. Then, for sufficiently large n, (1.29) Dn(f+(z)) = z−n j0 φn(0) χn Dn(f(z)), Dn(f−(z)) = zn j0 bφn(0) χn

Theorem 1.18, we have Dn(f(BT)(z)) = (−1)n bφn(0) χn Dn(f(z)), where φn(z), χn, Dn(f(z)) correspond to f(z) given by (1.2) with m = 1, z0 = 1, z1 = eiπ, β0 = β1 = 1/2, α0 = α1 = 0. Observing that s = 2, j1 = j0 = 1 and j2 = 0 and using (1.30), we obtain Dn(f(BT)(z)) = (−1)n((−1)nR1,−+ R0,−). Since R1,−= R0,−= (2n)−1/2G(1/2)2G(3/2)2(1 + o(1)), we obtain (1.33) Dn(f(BT)(z)) = 1 + (−1)n 2

Theorem 1.20. Let w(x) be defined as in (1.36) with ℜβj ∈ Ä −1 2, 1 2 ä , j = 1, 2, . . . , r. Then as n →∞, Dn(w) = Dn(1)e[(n+α0+αr+1)V0−α0V (1)−αr+1V (−1)+ 1 2 P∞ k=1 kV 2 k ] (1.37) ×

Theorem 1.25. Let f(z) be defined in (1.2) with the condition f(eiθ) = f(e−iθ). Let θr+1 = π and ℜβj ∈ Ä −1 2, 1 2 ä , j = 1, 2, . . . , r, β0 = βr+1 = 0. Then as n →∞, DT+H n = enV0+ 1 2[(α0+αr+1+s+t)V0−(α0+s)V (1)−(αr+1+t)V (−1)+P∞ k=1 kV 2 k ]

Lemma 2.2 (Recurrence relations). Let Dn(f) ̸= 0, n ≥0. The orthogo- nal polynomials satisfy the following relations for n = 0, 1, . . . : χnzφn(z) = χn+1φn+1(z) −φn+1(0)zn+1 bφn+1(z−1), (2.3) χnz−1 bφn(z−1) = χn+1 bφn+1(z−1) −bφn+1(0)z−n−1φn+1(z), (2.4) χn+1z−1 bφn(z−1) = χn bφn+1(z−1) −bφn+1(0)z−nφn(z). (2.5) Moreover, (2.6) χ2 n+1 −χ2 n = φn+1(0)bφn+1(0).

Lemma 2.3 (Christoffel-Darboux identity). Let Dn(f) ̸= 0, n ≥0. For any z, a ̸= 0, n = 1, 2, . . . , (2.7) (1 −a−1z) n−1 X k=0 bφk(a−1)φk(z) = a−nφn(a)zn bφn(z−1) −bφn(a−1)φn(z). For any z ̸= 0, n = 1, 2, . . . , (2.8) n−1 X k=0 bφk(z−1)φk(z)=−nφn(z)bφn(z−1)+z Å

Lemma 2.4. Let the Toeplitz determinants Dn(f) with symbol f(z) be nonzero for all n ≥N0 with a fixed N0 ≥0. Let Φk(z) = φk(z)/χk, “Φk(z) = bφk(z)/χk, k = N0, N0 + 1, . . . be the system of monic polynomials orthogonal on the unit circle with the weight f(z). Fix an integer ℓ> 0. Then if Fk =

Lemma 2.5. Let f(z) have the property f(eiθ) = f(e−iθ), 0 ≤θ ≤2π and let w(x) = f(eiθ) | sin θ|, x = cos θ. Assume that Dn(f) ̸= 0, n ≥N0, N0 ≥0. Then the polynomials pn(x) = κnxn + · · · , n = N0, N0 + 1, . . . exist which are orthonormal with respect to weight w(x) on [−1, 1], i.e., Z 1 −1 pn(x)xmw(x)dx = κ−1 k δnm, m = 0, 1, . . . , n, n ≥N0, and, for n = N0, N0 + 1, . . . , there hold the following expressions in terms of

Theorem 11.5 in [39], and we obtain (2.17) pn(x) = 1 » 2π(1 −a2n−1) (z−nφ2n(z) + znφ2n(z−1)). Note that 1 −a2n−1 ̸= 0, n = N0, N0 + 1, . . . as follows from (2.6) which in our case can be rewritten in the form χ2 n = Ä 1 −φ2 n+1(0)/χ2 n+1

Theorem 2.6 (Connection between Toeplitz and Hankel determinants). Let N0 ≥0 and Dn(f) ̸= 0 for all n ≥N0. Let the weights f(z) and w(x) be related as in Lemma 2.5. Let, moreover, Dn(w(x)) = det ÇZ 1 −1 xj+kw(x)dx ån−1 j,k=0 , n = N0, N0 + 1, . . . be the Hankel determinant with symbol w(x) on [−1, 1]. Then, with Φn(z) = φn(z)/χn, we have (2.20) Dn(w(x))2 =

Lemma 2.4. □

Lemma 2.7 (Connection between Hankel and Toeplitz+Hankel deter- minants). Let fj be the Fourier coefficient fj = 1 2π R 2π 0 f(eiθ)e−ijθdθ. Let f(eiθ) = f(e−iθ). Then, for n = 1, 2, . . . , (2.24) det(fj−k + fj+k)n−1 j,k=0 = 2n2−2n+2 πn Dn(v(x)), where Dn(v(x)) is the Hankel determinant with symbol v(x)=f(eiθ(x))/

Proposition 4.1. Let αj ± βj ̸= −1, −2, . . . for all j. Then a solution to the above RHP (a)–(c) for Ψj(ζ), 0 < arg ζ < 2π, is given by the following function in the sector I: Ψj(ζ) = Ψ(I) j (ζ) (4.32) =

Lemma 1.12. We have to consider only the second class; i.e., |||β(r)||| = 1. We then have, relabeling β(r) j according to increasing real part, (6.1) ℜβ(r) 1 = · · · = ℜβ(r) p < ℜβ(r) p+1 ≤· · · ≤ℜβ(r) m′−ℓ< ℜβ(r) m′−ℓ+1 = · · · = ℜβ(r) m′ for some p, ℓ> 0. Here m′ is the number of singularities: m′ = m + 1 if z = 1