Bergman shift operators for Jordan domains

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1 [ 1 ] University of Cyprus Bergman shift operators for Jordan domains Nikos Stylianopoulos University of Cyprus Num An 2012 Fifth Conference in Numerical Analysis University of Ioannina September 2012 Bergman Toeplitz Comparison

2 [ 2 ] University of Cyprus Bergman polynomials {p n } on an Jordan domain G Γ Ω Φ Ψ Δ D G 0 Γ := G Ω := C \ G f, g := f (z)g(z)da(z), f L2 (G) := f, f 1/2. G The Bergman polynomials {p n } n=0 of G are the orthonormal polynomials w.r.t. the area measure on G: p m, p n = p m (z)p n (z)da(z) = δ m,n, G with p n (z) = λ n z n +, λ n > 0, n = 0, 1, 2,.... Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

3 [ 3 ] University of Cyprus Shift Operator Let L 2 a(g) denote the Bergman space of square integrable and analytic functions in G and consider the Bergman shift operator on L 2 a(g). That is, S z : L 2 a(g) L 2 a(g) with S z f = zf. Properties of S z (i) S z defines a subnormal operator on L 2 a(g). (ii) σ(s z ) = G and σ ess (S z ) = G (Axler, Conway & McDonald, Can. J. Math.,1982). (iii) S z (f ) = P G (zf ), where P G denotes the orthogonal projection from L 2 (G) to L 2 a(g). Proof of (iii): For any f, g L 2 a(g) it holds that S z f, g = f, S z g = f, zg = zf, g = P G (zf ), g. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

4 [ 4 ] University of Cyprus Short-term recurrences for Bergman polynomials {p n } In general it holds that n+1 zp n (z) = b k,n p k (z), where b k,n := zp n, p k. k=0 Definition We say that the polynomials {p n } n=0 satisfy an m + 1-term recurrence relation, if for any n m 1, zp n (z) = b n+1,n p n+1 (z) + b n,n p n (z) b n m+1,n p n m+1 (z). Lemma (Putinar & St, CAOT, 2007) If the Bergman polynomials {p n } satisfy an m + 1-term recurrence relation, then for any p P d, it holds that S z p P d+m 1. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

5 [ 5 ] University of Cyprus Proof of Lemma Using the short-term recurrence relation we have: n m+1 Sz p, p n = p, S z p n = p, zp n = p, b k,n p k = k=0 n m+1 k=0 b k,n p, p k. Thus, S z p, p n = 0, for any d < n m + 1, i.e., for n > d + m 1. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

6 [ 6 ] University of Cyprus Only ellipses carry 3-term recurrence relations Theorem (Putinar & St, CAOT, 2007) If the Bergman polynomials {p n } satisfy a 3-term recurrence relation, then Γ = G is an ellipse. Proof. We use Havin s Lemma: L 2 (G) = L 2 a(g) W 1,2 0 (G). z = P G (z) + g = S z (z) + g = p + g, where g W 1,2 0 (G) and deg(p) 1. Hence, by integration we obtain zz = Q(z) + g(z) + f (z), Q = p, f L 2 a(g). From uniqueness in W 1,2 0 (G) we further obtain f = Q + const. Hence, z 2 = Q(z) + Q(z) + c + g(z), for z G, and from regularity of G: z 2 = Q(z) + Q(z) + c, z Γ. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

7 [ 7 ] University of Cyprus Only ellipses carry short-term recurrences for p n In fact, a great deal more can be obtained: Theorem (St, C. R. Acad. Sci. Paris, 2010) Assume that: (i) Γ = G is piecewise analytic without cusps. (ii) The Bergman polynomials {p n } n=0 satisfy an m + 1-term recurrence relation, with some m 2. Then m = 2 and Γ is an ellipse. The above theorem refines some deep results of Putinar & St (CAOT, 2007) and Khavinson & St (Springer, 2010). Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

8 [ 8 ] University of Cyprus Matrix representation for S z The Bergman operator S z has the following upper Hessenberg matrix representation with respect to the Bergman polynomials {p n } n=0 of G: b 00 b 01 b 02 b 03 b 04 b 05 b 10 b 11 b 12 b 13 b 14 b 15 0 b 21 b 22 b 23 b 24 b 25 M = 0 0 b 32 b 33 b 34 b 35, b 43 b 44 b b 54 b where b k,n = zp n, p k are the Fourier coefficients of S z p n = zp n. Note The eigenvalues of the n n principal submatrix M n of M coincide with the zeros of p n. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

9 [ 9 ] University of Cyprus Banded Hessenberg matrices for OP s are Jacobi In the Numerical Linear Algebra jargon the short-term recurrence theorem reads as follows: Theorem If the upper Hessenberg matrix M is banded, with constant bandwidth 3, then it is tridiagonal, i.e., a Jacobi matrix. This result should put an end to the long search in Numerical Linear Algebra, for practical semi-iterative methods (aka polynomial iteration methods) based on short-term recurrence relations of orthogonal polynomials. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

10 [ 10 ] University of Cyprus Example: G D This example shows why modern text books on Functional Analysis or Operators Theory do not refer to matrices: Indeed, in this case we have: n + 1 p n (z) = π zn, n = 0, 1,.... Therefore, in the matrix representation M of S z the only non-zero diagonals are the main subdiagonal, and hence for any n N, M n is a nilpotent matrix. As a result, the Caley-Hamilton theorem implies: This is in sharp contrast to: σ(m n ) = {0}. σ ess (M) = σ ess (S z ) = {w : w = 1} and σ(m) = σ(s z ) = {w : w 1}. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

11 [ 11 ] University of Cyprus The inverse conformal map Ψ Γ Ω Φ Ψ Δ D G 0 Recall that Φ(z) = γz + γ 0 + γ 1 z + γ 2 z 2 +, and let Ψ := Φ 1 : {w : w > 1} Ω, denote the inverse conformal map. Then, where Ψ(w) = bw + b 0 + b 1 w + b 2 +, w < 1, w 2 b = cap(γ) = 1/γ. Bergman Toeplitz Comparison Polynomials Shift Short-term Matrix

12 [ 12 ] University of Cyprus The Toeplitz matrix with (continuous) symbol Ψ T ψ = b 0 b 1 b 2 b 3 b 4 b 5 b 6 b b 0 b 1 b 2 b 3 b 4 b 5 0 b b 0 b 1 b 2 b 3 b b b 0 b 1 b 2 b b b 0 b 1 b b b 0 b b b Bergman Toeplitz Comparison Matrix

13 [ 13 ] University of Cyprus Spectral properties Theorem (St, Constr, Approx., to appear) If Γ is piecewise analytic without cusps, then 1 b n c 1 (Γ), n N, (1) n1+ω where ωπ (0 < ω < 2) is the smallest exterior angle of Γ. Therefore, in this case, the symbol Ψ of the Toeplitz matrix T Ψ belongs to the Wiener algebra. Thus, T Ψ defines a bounded linear operator on the Hilbert space l 2 (N) and σ ess (T Ψ ) = Γ ; (2) see e.g., Bottcher & Grudsky, Toeplitz book, Bergman Toeplitz Comparison Matrix

14 [ 14 ] University of Cyprus Faber polynomials of G The Faber polynomial of the 2nd kind G n (z), is the polynomial part of the expansion of the Laurent series expansion of Φ n (z)φ (z) at : ( ) 1 G n (z) = Φ n (z)φ (z) + O, z. z These polynomials satisfy the recurrence relation: zg n (z) = bg n+1 (z) + n+1 Recall: zp n (z) = b k,n p k (z). Note k=0 n b k G n k (z), n = 0, 1,..., k=0 The eigenvalues of the n n principal submatrix T n of T ψ coincide with the zeros of G n. Bergman Toeplitz Comparison Matrix

15 [ 15 ] University of Cyprus Ratio asymptotics for λ n Theorem (St, C. R. Acad. Sci. Paris, 2010 & Constr. Approx.) Assume that Γ is piecewise analytic without cusps.then, for n N, n + 2 λ n = cap(γ) + ξ n, where ξ n c(γ) 1 n + 1 λ n+1 n. The above relation provides the means for computing approximations to the capacity of Γ, by using only the leading coefficients of the Bergman polynomials. Bergman Toeplitz Comparison Theory A Question Examples

16 [ 16 ] University of Cyprus Some coincidence Recall σ ess (M) = Γ = σ ess (T Ψ ). Consider the main subdiagonal b n+1,n of M. Then: b n+1,n = zp n, p n+1 = λ n z n+1 +, p n+1 = λ n z n+1, p n+1 = Since cap(γ) = b, it follows from the ratio asymptotics for λ n, that: λ n λ n+1. Corollary n + 2 n + 1 b n+1,n = b + O ( ) 1, n N. n That is, the main subdiagonal of the upper Hessenberg matrix M tends to the main subdiagonal of the Toeplitz matrix T ψ. Bergman Toeplitz Comparison Theory A Question Examples

17 [ 17 ] University of Cyprus M T ψ diagonally The next series of theorems show that the connection between the two matrices M and T Ψ is much more substantial. Theorem (Saff & St., CAOT, 2012) Assume that Γ is piecewise analytic without cusps. Then, it holds as n, ( ) n n + 1 b n+1,n = b + O, (3) n and for k 0, n k + 1 n + 1 where O depends on k but not on n. ( ) 1 b n k,n = b k + O n, (4) Bergman Toeplitz Comparison Theory A Question Examples

18 [ 18 ] University of Cyprus M T ψ diagonally: Smooth curve Improvements in the order of convergence occur in cases when Γ is smooth. Theorem (Saff & St., CAOT, 2012) Assume that Γ C(p + 1, α), with p + α > 1/2. Then, it holds as n, ( ) n n + 1 b n+1,n = b + O, (5) n 2(p+α) and for k 0, n k + 1 n + 1 where O depends on k but not on n. ( ) 1 b n k,n = b k + O n p+α, (6) Bergman Toeplitz Comparison Theory A Question Examples

19 [ 19 ] University of Cyprus M T ψ diagonally: Analytic curve For the case of an analytic boundary Γ further improved asymptotic results can be obtained. Theorem (Saff & St., CAOT, 2012) Assume that the boundary Γ is analytic and let ρ < 1 be the smallest index for which Φ is conformal in the exterior of L ρ. Then, it holds as n, n + 2 n + 1 b n+1,n = b + O(ρ 2n ), (7) and for k 0, n k + 1 n + 1 b n k,n = b k + O( nρ n ), (8) where O depends on k but not on n. Bergman Toeplitz Comparison Theory A Question Examples

20 [ 20 ] University of Cyprus Is M T ψ compact? Corollary If the upper Hessenberg matrix M is banded, with constant bandwidth, then M T ψ defines a compact operator on l 2 (N). Is the following true? The adjoint operator S z : L 2 a(g) L 2 a(g) maps P into itself if and only if Γ := G is an ellipse. The above is equivalent to: Conjecture, Khavinson & Shapiro, 1992 The Dirichlet problem for G has a polynomial solution for any polynomial data on Γ if an only if Γ = G is an ellipse. Bergman Toeplitz Comparison Theory A Question Examples

21 [ 20 ] University of Cyprus Is M T ψ compact? Corollary If the upper Hessenberg matrix M is banded, with constant bandwidth, then M T ψ defines a compact operator on l 2 (N). Is the following true? The adjoint operator S z : L 2 a(g) L 2 a(g) maps P into itself if and only if Γ := G is an ellipse. The above is equivalent to: Conjecture, Khavinson & Shapiro, 1992 The Dirichlet problem for G has a polynomial solution for any polynomial data on Γ if an only if Γ = G is an ellipse. Bergman Toeplitz Comparison Theory A Question Examples

22 [ 21 ] University of Cyprus Example: G is a 3-cusped hypocycloid 1 0,5-0, ,5 1 1,5-0,5-1 Assume that ν(p) denotes the normalized counting measure of zeros of the polynomial P. Also let µ Γ denote the equilibrium measure on Γ, note that supp(µ Γ ) = Γ and recall σ ess (M) = Γ = σ ess (T Ψ ). Then: Levin, Saff & St., Constr. Approx. (2003): ν(p n ) He & Saff, JAT (1994): µ Γ, n, n N, N N. σ(t n ) [0, 1.5] [0, 1.5e i2π/3 ] [0, 1.5e i4π/3 ]. Bergman Toeplitz Comparison Theory A Question Examples

23 [ 22 ] University of Cyprus Example: G is the square 0,6 0,4 0,2 0-0,6-0,4-0,2 0 0,2 0,4 0,6-0,2-0,4-0,6 Maymeskul & Saff, JAT (2003): Recall: σ ess (M) = Γ = σ ess (T Ψ ). σ(m n ) the two diagonals. Kuijlaars & Saff, Math. Proc. Cambrigde Phil. Soc. (1995): ν(g n ) µ Γ, n, n N, N N Bergman Toeplitz Comparison Theory A Question Examples

24 [ 23 ] University of Cyprus Example: G is the canonical pentagon 1 Regular 5 gon: zeros of orthogonal polynomials up to degree 50. Recall: σ ess (M) = Γ = σ ess (T Ψ ). Levin, Saff & St., Constr. Approx. (2003): ν(p n ) µ Γ, n, n N, N N Kuijlaars & Saff, Math. Proc. Cambrigde Phil. Soc. (1995): ν(g n ) µ Γ, n, n N, N N Bergman Toeplitz Comparison Theory A Question Examples

25 [ 24 ] University of Cyprus Asymptotics Γ Ω Φ Ψ Δ D G 0 Ω := C \ G Φ(z) = γz + γ 0 + γ 1 z + γ 2 +. z2 cap(γ) = 1/γ The Bergman polynomials of G: p n (z) = λ n z n +, λ n > 0, n = 0, 1, 2,.... Bergman Toeplitz Comparison Theory A Question Examples

26 [ 25 ] University of Cyprus Strong asymptotics when Γ is analytic Ω Φ Δ G D 0 ρ 1 Lρ Γ T. Carleman, Ark. Mat. Astr. Fys. (1922) If ρ < 1 is the smallest index for which Φ is conformal in ext(l ρ ), then n + 1 γ 2(n+1) π λ 2 = 1 α n, where 0 α n c 1 (Γ) ρ 2n, n where p n (z) = n + 1 π Φn (z)φ (z){1 + A n (z)}, n N, A n (z) c 2 (Γ) n ρ n, z Ω. Bergman Toeplitz Comparison Theory A Question Examples

27 [ 26 ] University of Cyprus Strong asymptotics when Γ is smooth We say that Γ C(p, α), for some p N and 0 < α < 1, if Γ is given by z = g(s), where s is the arclength, with g (p) Lipα. Then both Φ and Ψ := Φ 1 are p times continuously differentiable in Ω \ { } and \ { } respectively, with Φ (p) and Ψ (p) Lipα. P.K. Suetin, Proc. Steklov Inst. Math. AMS (1974) Assume that Γ C(p + 1, α), with p + α > 1/2. Then, for n N, n + 1 γ 2(n+1) 1 π λ 2 = 1 α n, where 0 α n c 1 (Γ) n n, 2(p+α) where p n (z) = n + 1 π Φn (z)φ (z){1 + A n (z)}, A n (z) c 2 (Γ) log n, z Ω. np+α Bergman Toeplitz Comparison Theory A Question Examples

28 [ 27 ] University of Cyprus Strong asymptotics for Γ non-smooth Theorem (St, C. R. Acad. Sci. Paris, 2010) Assume that Γ is piecewise analytic without cusps.then, for n N, and for any z Ω, n + 1 γ 2(n+1) π λ 2 = 1 α n, where 0 α n c(γ) 1 n n, p n (z) = n + 1 π Φn (z)φ (z) {1 + A n (z)}, where A n (z) c 1 (Γ) dist(z, Γ) Φ (z) 1 n + c 2 (Γ) 1 n. Bergman Toeplitz Comparison Theory A Question Examples

29 [ 28 ] University of Cyprus Ratio asymptotics for p n (z) Corollary (St, C. R. Acad. Sci. Paris, 2010) For any z Ω, and sufficiently large n N, n p n (z) n + 1 p n 1 (z) = Φ(z){1 + B n(z)}, where B n (z) c 1 (Γ) dist(z, Γ) Φ (z) 1 n + c 2 (Γ) 1 n. The above relation provides the means for computing approximations to the conformal map Φ. This leads to an efficient algorithm for recovering the shape of G, from a finite collection of its power moments z m, z k n m,k=0. This method was actually commented as unsuitable by P. Henrici, in Computational Complex Analysis, Vol. III (1986), because of the instability of the Conventional GS. Bergman Toeplitz Comparison Theory A Question Examples