Boundary behaviour of optimal polynomial approximants

Transcription

1 Boundary behaviour of optimal polynomial approximants University of South Florida Laval University, in honour of Tom Ransford, May 2018

2 This talk is based on some recent and upcoming papers with various combinations of co-authors, including Dmitry Khavinson, Conni Liaw,Myrto Manolaki, Daniel Seco, and Brian Simanek.

3 Outline Introduction to optimal polynomial approximants

4 Outline Introduction to optimal polynomial approximants

5 Outline Introduction to optimal polynomial approximants

6 Definition of Dirichlet spaces D α For < α <, the space D α consists of all analytic functions f : D = {z C: z < 1} C whose Taylor coefficients in the expansion satisfy f (z) = f 2 α = a k z k, z D, k=0 (k + 1) α a k 2 <. k=0

7 Definition of Dirichlet spaces D α For < α <, the space D α consists of all analytic functions f : D = {z C: z < 1} C whose Taylor coefficients in the expansion satisfy f (z) = f 2 α = a k z k, z D, k=0 (k + 1) α a k 2 <. k=0 Given two functions f (z) = k=0 a kz k and g(z) = k=0 b kz k in D α, we also have the associated inner product f, g α = (k + 1) α a k b k. k=0

8 Examples α = 1 corresponds to the Bergman space A 2, consisting of functions with f (z) 2 da(z) <, da(z) = dxdy π, D

9 Examples α = 1 corresponds to the Bergman space A 2, consisting of functions with f (z) 2 da(z) <, da(z) = dxdy π, D α = 0: the Hardy space H 2, consisting of functions with sup 0<r<1 1 2π 2π 0 f (re iθ ) 2 dθ <,

10 Examples α = 1 corresponds to the Bergman space A 2, consisting of functions with f (z) 2 da(z) <, da(z) = dxdy π, D α = 0: the Hardy space H 2, consisting of functions with sup 0<r<1 1 2π 2π 0 f (re iθ ) 2 dθ <, α = 1: the (classical) Dirichlet space D of functions whose derivatives have finite area integral: f (z) 2 da(z) <. D

11 Optimal polynomial approximants: opa! Definition Let f D α. We say that a polynomial p n of degree at most n is an optimal approximant of order n to 1/f if p n minimizes pf 1 α among all polynomials p of degree at most n.

12 Optimal polynomial approximants: opa! Definition Let f D α. We say that a polynomial p n of degree at most n is an optimal approximant of order n to 1/f if p n minimizes pf 1 α among all polynomials p of degree at most n. In other words, p n is an optimal polynomial of order n to 1/f if p n f 1 α = dist Dα (1, f Pol n ), where Pol n denotes the space of polynomials of degree at most n.

13 Optimal polynomial approximants: opa! Definition Let f D α. We say that a polynomial p n of degree at most n is an optimal approximant of order n to 1/f if p n minimizes pf 1 α among all polynomials p of degree at most n. In other words, p n is an optimal polynomial of order n to 1/f if p n f 1 α = dist Dα (1, f Pol n ), where Pol n denotes the space of polynomials of degree at most n. Note: p n f is the orthogonal projection of 1 onto the subspace f Pol n.

14 Optimal polynomial approximants: opa! Definition Let f D α. We say that a polynomial p n of degree at most n is an optimal approximant of order n to 1/f if p n minimizes pf 1 α among all polynomials p of degree at most n. In other words, p n is an optimal polynomial of order n to 1/f if p n f 1 α = dist Dα (1, f Pol n ), where Pol n denotes the space of polynomials of degree at most n. Note: p n f is the orthogonal projection of 1 onto the subspace f Pol n. Therefore, optimal approximants p n always exist and are unique for any nonzero function f, and any degree n 0.

15 Initial motivation for studying opa - Cyclicity A function f D α is said to be cyclic in D α if the subspace generated by polynomials multiples of f, coincides with D α. [f ] = span{z k f : k = 0, 1, 2,...}

16 Initial motivation for studying opa - Cyclicity A function f D α is said to be cyclic in D α if the subspace generated by polynomials multiples of f, [f ] = span{z k f : k = 0, 1, 2,...} coincides with D α. Equivalently: there exists a sequence of polynomials {p n } n=1 such that p n f 1 2 α 0, as n.

17 Initial motivation for studying opa - Cyclicity A function f D α is said to be cyclic in D α if the subspace generated by polynomials multiples of f, [f ] = span{z k f : k = 0, 1, 2,...} coincides with D α. Equivalently: there exists a sequence of polynomials {p n } n=1 such that p n f 1 2 α 0, as n. The optimal polynomial approximants are the best such p n.

18 How to compute the optimal approximants? Let {ϕ k f } n k=0 be an orthonormal basis for the space f Pol n, where the degree of ϕ k is k.

19 How to compute the optimal approximants? Let {ϕ k f } n k=0 be an orthonormal basis for the space f Pol n, where the degree of ϕ k is k. Then p n f can be expressed by its Fourier coefficients in the basis ϕ k f as follows:

20 How to compute the optimal approximants? Let {ϕ k f } n k=0 be an orthonormal basis for the space f Pol n, where the degree of ϕ k is k. Then p n f can be expressed by its Fourier coefficients in the basis ϕ k f as follows: (p n f )(z) = n 1, ϕ k f α ϕ k (z)f (z). k=0

21 How to compute the optimal approximants? Let {ϕ k f } n k=0 be an orthonormal basis for the space f Pol n, where the degree of ϕ k is k. Then p n f can be expressed by its Fourier coefficients in the basis ϕ k f as follows: (p n f )(z) = p n (z) = n 1, ϕ k f α ϕ k (z)f (z). k=0 n 1, ϕ k f α ϕ k (z). k=0 Now notice that in all the D α spaces, 1, ϕ k f α = ϕ k (0)f (0).

22 Formula for the Optimal Approximants We thus have the following formula: Proposition Let α R and f D α. For integers k 0, let ϕ k be the orthonormal polynomials for the weighted space D α,f. Let p n be the optimal approximants to 1/f. Then p n (z) = f (0) n ϕ k (0)ϕ k (z). k=0

23 Connection with reproducing kernels The fact that n p n (z) = f (0) ϕ k (0)ϕ k (z) k=0 might remind you of reproducing kernels!

24 Connection with reproducing kernels The fact that n p n (z) = f (0) ϕ k (0)ϕ k (z) k=0 might remind you of reproducing kernels! Indeed, n K n (z, w) := ϕ k (w)f (w)ϕ k (z)f (z) k=0 is the reproducing kernel for the space f Pol n.

25 Connection with reproducing kernels The fact that p n (z) = f (0) n ϕ k (0)ϕ k (z) k=0 might remind you of reproducing kernels! Indeed, n K n (z, w) := ϕ k (w)f (w)ϕ k (z)f (z) k=0 is the reproducing kernel for the space f Pol n. Recall that K n is characterized by: for every g f Pol n, g(w) = g, K(, w) α, w D.

26 Connection with reproducing kernels The fact that p n (z) = f (0) n ϕ k (0)ϕ k (z) k=0 might remind you of reproducing kernels! Indeed, n K n (z, w) := ϕ k (w)f (w)ϕ k (z)f (z) k=0 is the reproducing kernel for the space f Pol n. Recall that K n is characterized by: for every g f Pol n, Therefore, K n (z, 0) = p n (z)f (z). g(w) = g, K(, w) α, w D.

27 Connection with reproducing kernels The fact that p n (z) = f (0) n ϕ k (0)ϕ k (z) k=0 might remind you of reproducing kernels! Indeed, n K n (z, w) := ϕ k (w)f (w)ϕ k (z)f (z) k=0 is the reproducing kernel for the space f Pol n. Recall that K n is characterized by: for every g f Pol n, g(w) = g, K(, w) α, w D. Therefore, K n (z, 0) = p n (z)f (z). This means that results about the behaviour of optimal approximants are actually results about the behaviour of weighted reproducing kernels!

28 Main Question of ongoing research discussed in this talk

29 Main Question of ongoing research discussed in this talk Given f D α and given e iθ T, what can we say about the limit points of p n (e iθ ) as n varies?

30 Main Question of ongoing research discussed in this talk Given f D α and given e iθ T, what can we say about the limit points of p n (e iθ ) as n varies? Does p n (e iθ ) 1/f (e iθ )?

31 Main Question of ongoing research discussed in this talk Given f D α and given e iθ T, what can we say about the limit points of p n (e iθ ) as n varies? Does p n (e iθ ) 1/f (e iθ )? Can there be more than one limit point?

32 Main Question of ongoing research discussed in this talk Given f D α and given e iθ T, what can we say about the limit points of p n (e iθ ) as n varies? Does p n (e iθ ) 1/f (e iθ )? Can there be more than one limit point? Can it happen that the set {p n (e iθ ) : n = 0, 1, 2,...} is dense in C?

33 Think Taylor series for a moment, for a particular example Let f (z) = 1/(1 z).

34 Think Taylor series for a moment, for a particular example Let f (z) = 1/(1 z). The Taylor polynomials of f are P n (z) = n k=0 z k = 1 zn+1. 1 z

35 Think Taylor series for a moment, for a particular example Let f (z) = 1/(1 z). The Taylor polynomials of f are P n (z) = n k=0 z k = 1 zn+1. 1 z Their zeros are the roots of unity (except for z = 1), which are dense in the unit circle T.

36 Think Taylor series for a moment, for a particular example Let f (z) = 1/(1 z). The Taylor polynomials of f are P n (z) = n k=0 z k = 1 zn+1. 1 z Their zeros are the roots of unity (except for z = 1), which are dense in the unit circle T. It turns out that this phenomenon is quite general and was discovered by Robert Jentzsch in his Ph.D. thesis in Berlin in 1914.

37 Jentzsch Theorem Theorem (Jentzsch, 1914) Given any power series f (z) = a k z k k=0 with radius of convergence 1, every point on the unit circle is a limit point of the zeros of the partial sums of the series.

38 Historical Notes Robert Jentzsch was a talented mathematician born in Königsberg in He was a student of Georg Frobenius in Berlin. He was also a poet. He was killed in battle in There is a nice historical article Jentzsch, Mathematician and Poet by P. Duren, A.K. Herbig, and D. Khavinson in the Mathematical Intelligencer in 2008.

39 Jentzsch s Theorem for Optimal Polynomial Approximants Theorem (BKLSS, 2016) Let α R and let f D α be cyclic, such that 1/f has a singularity on the unit circle, and f (0) 0. Then every point on the unit circle is a limit point of the zeros of the optimal approximants of 1/f.

40 Jentzsch s Theorem for Optimal Polynomial Approximants Theorem (BKLSS, 2016) Let α R and let f D α be cyclic, such that 1/f has a singularity on the unit circle, and f (0) 0. Then every point on the unit circle is a limit point of the zeros of the optimal approximants of 1/f. Moral of the story: Strange things happen for optimal polynomial approximants outside the unit disk.

41 Jentzsch s Theorem for Optimal Polynomial Approximants Theorem (BKLSS, 2016) Let α R and let f D α be cyclic, such that 1/f has a singularity on the unit circle, and f (0) 0. Then every point on the unit circle is a limit point of the zeros of the optimal approximants of 1/f. Moral of the story: Strange things happen for optimal polynomial approximants outside the unit disk. Inside the unit disk, the optimal approximants converge to 1/f, if f is cyclic.

42 Jentzsch s Theorem for Optimal Polynomial Approximants Theorem (BKLSS, 2016) Let α R and let f D α be cyclic, such that 1/f has a singularity on the unit circle, and f (0) 0. Then every point on the unit circle is a limit point of the zeros of the optimal approximants of 1/f. Moral of the story: Strange things happen for optimal polynomial approximants outside the unit disk. Inside the unit disk, the optimal approximants converge to 1/f, if f is cyclic. So what happens on the circle?

43 Theorem (C.B., M. Manolaki, D. Seco, 2018) Let f be a monic polynomial of degree d with distinct zeros z 1, z 2,... z d that lie on or outside the unit disk. For each n, let p n be the optimal approximant of 1/f in D α. Write ω k := (k + 1) α and denote by d k,n the Taylor coefficients of p n f 1. Then there exists a vector A n = (A 1,n,..., A d,n ) t independent of k, such that for k = 0,..., n + d, we have d k,n = 1 ω k d A i,n zi k. i=1

44 Theorem (continued...) Moreover A n is the only solution to the linear system E Z,n A n = ( 1, 1,..., 1) t, where the matrix E Z,n is invertible and has coefficients E Z,n,l,m = n+d k=0 z m k z k l ω k.

45 Idea of the proof The proof is based on the idea that since p n f is the projection of 1 onto f Pol n, p n f 1 is orthogonal to z t f for t = 0,..., n.

46 Idea of the proof The proof is based on the idea that since p n f is the projection of 1 onto f Pol n, p n f 1 is orthogonal to z t f for t = 0,..., n. This orthogonality relation can be translated to the coefficients d k,n satisfying a recursion relation involving the coefficients of f.

47 Idea of the proof The proof is based on the idea that since p n f is the projection of 1 onto f Pol n, p n f 1 is orthogonal to z t f for t = 0,..., n. This orthogonality relation can be translated to the coefficients d k,n satisfying a recursion relation involving the coefficients of f. Then you can use an elementary linear algebra lemma describing sequences that satisfy constant term recursive relationships.

48 Lemma Let a sequence {e k } k N satisfy the recurrence relation (for k d) d e k r ˆq(r) = 0, r=0 where q is a polynomial q of degree d satisfying q(0) = 1, with simple zeros {w 1,..., w d } C. Then, there exist some constants H 1,..., H d C such that, for all k, e k is given by e k = d i=1 H i w k i.

49 Example For f (z) = 1 z, p n the n th o.p.a. of 1/f in H 2, and for z 1, we have:

50 Example For f (z) = 1 z, p n the n th o.p.a. of 1/f in H 2, and for z 1, we have: 1 p n (z) = A n (z) 1 z, where A n(z) = zn+2 (n + 2)z + n + 1 (n + 2)(1 z)

51 Example For f (z) = 1 z, p n the n th o.p.a. of 1/f in H 2, and for z 1, we have: 1 p n (z) = A n (z) 1 z, where A n(z) = zn+2 (n + 2)z + n + 1 = (n + 2)(1 z) z n+2 n z z 1 z + n + 1 n z.

52 Example For f (z) = 1 z, p n the n th o.p.a. of 1/f in H 2, and for z 1, we have: 1 p n (z) = A n (z) 1 z, where A n(z) = zn+2 (n + 2)z + n + 1 = (n + 2)(1 z) z n+2 n z z 1 z + n + 1 n For all z with z 1 we 1 z z n+2 have lim n n + 2 = 0;

53 Example For f (z) = 1 z, p n the n th o.p.a. of 1/f in H 2, and for z 1, we have: 1 p n (z) = A n (z) 1 z, where A n(z) = zn+2 (n + 2)z + n + 1 = (n + 2)(1 z) z n+2 n z z 1 z + n + 1 n For all z with z 1 we 1 z z n+2 have lim = 0; therefore n n + 2 lim A n(z) = 0 z n 1 z z = 1.

54 Bounded The previous theorem can be leveraged to prove the following. Theorem (C.B., M. Manolaki, D. Seco, 2018) Let f be a polynomial of degree d with simple zeros that all lie outside the open unit disk, and let α 1. Let p n be the n th optimal polynomial approximant of 1/f in D α. Let z 0 T, not a zero of f. Then (p n f 1)(z 0 ) 0 as n.

55 Further Remarks It turns out that this most recent result is the first step in showing that there are functions f for which the set p n (e iθ ) is dense in C! This is a preliminary result (C.B., M. Manolaki, D. Seco, 2018).

56 Further Remarks It turns out that this most recent result is the first step in showing that there are functions f for which the set p n (e iθ ) is dense in C! This is a preliminary result (C.B., M. Manolaki, D. Seco, 2018). Can we describe the functions f for which this kind of universality happens? For which the opposite happens? Are there cases in between?

57 Thank you!