Orthogonal polynomials and zeros of optimal approximants Daniel - - PowerPoint PPT Presentation
Orthogonal polynomials and zeros of optimal approximants Daniel Seco (with Bnteau, Khavinson, Liaw, and Simanek) Universitat de Barcelona Optimal Point Configurations and Orthogonal Polynomials, CIEM, Castro Urdiales, 20/04/2017 Seco (UB)
Daniel Seco (with Bénéteau, Khavinson, Liaw, and Simanek)
Universitat de Barcelona
Optimal Point Configurations and Orthogonal Polynomials, CIEM, Castro Urdiales, 20/04/2017
Seco (UB) Approximants vs. OP CIEM 1 / 11
Definition
Let ω0 = 1, ωk > 0, and lim
ωk ωk+1 = 1. Then
H2
ω = {f ∈ Hol(D) : f(z) =
akzk, ||f||2
ω = ∞
|ak|2ωk < ∞}.
Seco (UB) Approximants vs. OP CIEM 2 / 11
Definition
Let ω0 = 1, ωk > 0, and lim
ωk ωk+1 = 1. Then
H2
ω = {f ∈ Hol(D) : f(z) =
akzk, ||f||2
ω = ∞
|ak|2ωk < ∞}.
Definition
f is cyclic (in H2
ω) if Pf is dense in H2 ω.
Seco (UB) Approximants vs. OP CIEM 2 / 11
Definition
Let ω0 = 1, ωk > 0, and lim
ωk ωk+1 = 1. Then
H2
ω = {f ∈ Hol(D) : f(z) =
akzk, ||f||2
ω = ∞
|ak|2ωk < ∞}.
Definition
f is cyclic (in H2
ω) if Pf is dense in H2 ω.
f cyclic ⇔ ∃{pn}n∈N ∈ P : ||pnf − 1||ω
n→∞
→ 0.
Seco (UB) Approximants vs. OP CIEM 2 / 11
Definition
Let ω0 = 1, ωk > 0, and lim
ωk ωk+1 = 1. Then
H2
ω = {f ∈ Hol(D) : f(z) =
akzk, ||f||2
ω = ∞
|ak|2ωk < ∞}.
Definition
f is cyclic (in H2
ω) if Pf is dense in H2 ω.
f cyclic ⇔ ∃{pn}n∈N ∈ P : ||pnf − 1||ω
n→∞
→ 0. f regular +Z(f) ∩ D = ∅ ⇒ cyclic ⇒ Z(f) ∩ D = ∅.
Seco (UB) Approximants vs. OP CIEM 2 / 11
Our approach: find p∗
n that minimizes pnf − 1 among Pn, that is,
the orthogonal projection of 1 onto Pnf.
Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach: find p∗
n that minimizes pnf − 1 among Pn, that is,
the orthogonal projection of 1 onto Pnf.
Theorem (BCLSS, ’15; FMS, ’14)
p∗
n(z) = n j=0 cjzj only solution to Mc = b where
c = (cj)n
j=0,
Mj,k =< zjf, zkf >ω, bk =< 1, zkf >ω .
Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach: find p∗
n that minimizes pnf − 1 among Pn, that is,
the orthogonal projection of 1 onto Pnf.
Theorem (BCLSS, ’15; FMS, ’14)
p∗
n(z) = n j=0 cjzj only solution to Mc = b where
c = (cj)n
j=0,
Mj,k =< zjf, zkf >ω, bk =< 1, zkf >ω . Useful techniques for studying cyclicity of a fixed function. Cyclic ⇔ p∗
nf − 1 → 0 ⇔ p∗ n(0) → 1/f(0).
Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach: find p∗
n that minimizes pnf − 1 among Pn, that is,
the orthogonal projection of 1 onto Pnf.
Theorem (BCLSS, ’15; FMS, ’14)
p∗
n(z) = n j=0 cjzj only solution to Mc = b where
c = (cj)n
j=0,
Mj,k =< zjf, zkf >ω, bk =< 1, zkf >ω . Useful techniques for studying cyclicity of a fixed function. Cyclic ⇔ p∗
nf − 1 → 0 ⇔ p∗ n(0) → 1/f(0).
So we want to know about these polynomials! Today: where are their zeros? Which points of the plane may be zeros of such polynomials? (for a fixed space)
Seco (UB) Approximants vs. OP CIEM 3 / 11
p∗
n(z)f(z) = Πn(1)(z) = n
1, ϕkf ϕk(z)f(z)
Seco (UB) Approximants vs. OP CIEM 4 / 11
p∗
n(z)f(z) = Πn(1)(z) = n
1, ϕkf ϕk(z)f(z)
Theorem (BKLSS ’16)
p∗
n(z) = n
ϕk(z)ϕk(0)f(0).
Seco (UB) Approximants vs. OP CIEM 4 / 11
p∗
n(z)f(z) = Πn(1)(z) = n
1, ϕkf ϕk(z)f(z)
Theorem (BKLSS ’16)
p∗
n(z) = n
ϕk(z)ϕk(0)f(0). We can also (in general) recover OPs from opt. apprs.
Seco (UB) Approximants vs. OP CIEM 4 / 11
p∗
n(z)f(z) = Πn(1)(z) = n
1, ϕkf ϕk(z)f(z)
Theorem (BKLSS ’16)
p∗
n(z) = n
ϕk(z)ϕk(0)f(0). We can also (in general) recover OPs from opt. apprs. In H2, Z(p∗
n) = {zj −1 : zj ∈ Z(ϕn)}.
In fact, these zero sets characterize cyclic functions.
Seco (UB) Approximants vs. OP CIEM 4 / 11
Easy trick: it is enough to study n = 1,
Seco (UB) Approximants vs. OP CIEM 5 / 11
Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f: If (z − z0) optimal for f ⇒ z0 =
zf2 <f,zf>.
Seco (UB) Approximants vs. OP CIEM 5 / 11
Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f: If (z − z0) optimal for f ⇒ z0 =
zf2 <f,zf>.
If z0 / ∈ D, z0 is achieved by f(z) = 1/(z − z0) in any space.
Seco (UB) Approximants vs. OP CIEM 5 / 11
Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f: If (z − z0) optimal for f ⇒ z0 =
zf2 <f,zf>.
If z0 / ∈ D, z0 is achieved by f(z) = 1/(z − z0) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [0, 1] are possible?
Seco (UB) Approximants vs. OP CIEM 5 / 11
Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f: If (z − z0) optimal for f ⇒ z0 =
zf2 <f,zf>.
If z0 / ∈ D, z0 is achieved by f(z) = 1/(z − z0) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [0, 1] are possible? If ω non decreasing, then
zf2 |<f,zf>| CSI
>
zf f nondec
≥ 1 ⇒ z0 / ∈ D.
Seco (UB) Approximants vs. OP CIEM 5 / 11
Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f: If (z − z0) optimal for f ⇒ z0 =
zf2 <f,zf>.
If z0 / ∈ D, z0 is achieved by f(z) = 1/(z − z0) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [0, 1] are possible? If ω non decreasing, then
zf2 |<f,zf>| CSI
>
zf f nondec
≥ 1 ⇒ z0 / ∈ D. If ∃k, n : ωk > 4ωk+n+1 then fk,n(z) = zkTn( 1+z
1−z ) makes z0 ∈ D.
Seco (UB) Approximants vs. OP CIEM 5 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness?
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness? If so, what is the extremal function?
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness? If so, what is the extremal function?
Theorem
Uω = Jω
2
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness? If so, what is the extremal function?
Theorem
Uω = Jω
2
If Uω > 1 then ∃!f ∗ (up to f ∗(z) → Cf ∗(zeiθ)).
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness? If so, what is the extremal function?
Theorem
Uω = Jω
2
If Uω > 1 then ∃!f ∗ (up to f ∗(z) → Cf ∗(zeiθ)). f ∗(z) = ∞
n=0 Pn (Jω) zn
Seco (UB) Approximants vs. OP CIEM 6 / 11
Problem
Let ω fixed (decreasing). What is the value of Uω := sup
zf2
: f ∈ H2
ω
Is the supremum a maximum? Uniqueness? If so, what is the extremal function?
Theorem
Uω = Jω
2
If Uω > 1 then ∃!f ∗ (up to f ∗(z) → Cf ∗(zeiθ)). f ∗(z) = ∞
n=0 Pn (Jω) zn
where {Pk} is a particular family of monic orthogonal polynomials given by a recurrence relationship and Jω is the Jacobi matrix (Jω)i,j = ωj
ωj+1 if |i − j| = 1; 0, o.w.
Seco (UB) Approximants vs. OP CIEM 6 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω
Seco (UB) Approximants vs. OP CIEM 7 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω Easy: functional is greater if coefficients ≥ 0.
Seco (UB) Approximants vs. OP CIEM 7 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω Easy: functional is greater if coefficients ≥ 0. Then, Lagrange multipliers and we obtain a0 = 1, a1 = λ ak+1 = λak −
ωk ωk+1 ak−1.
Seco (UB) Approximants vs. OP CIEM 7 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω Easy: functional is greater if coefficients ≥ 0. Then, Lagrange multipliers and we obtain a0 = 1, a1 = λ ak+1 = λak −
ωk ωk+1 ak−1.
By induction, aj is a monic polynomial of degree j on λ.
Seco (UB) Approximants vs. OP CIEM 7 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω Easy: functional is greater if coefficients ≥ 0. Then, Lagrange multipliers and we obtain a0 = 1, a1 = λ ak+1 = λak −
ωk ωk+1 ak−1.
By induction, aj is a monic polynomial of degree j on λ. Favard’s Theorem: ak = Pk(λ), monic polys with such a recursion are orthogonal for the spectral measure of the Jacobi matrix Jω.
Seco (UB) Approximants vs. OP CIEM 7 / 11
Enough to study f ∈ Pn and lim Uω,n = Uω Easy: functional is greater if coefficients ≥ 0. Then, Lagrange multipliers and we obtain a0 = 1, a1 = λ ak+1 = λak −
ωk ωk+1 ak−1.
By induction, aj is a monic polynomial of degree j on λ. Favard’s Theorem: ak = Pk(λ), monic polys with such a recursion are orthogonal for the spectral measure of the Jacobi matrix Jω. With other classical and modern results on orthogonal polynomials and Jacobi matrices + some work on regularity of some measures ⇒ our Thm.
Seco (UB) Approximants vs. OP CIEM 7 / 11
We were able to compute explicitly Jω for the following:
Definition
β > −1, ωk = β+k+1
k
−1, H2
ω =: A2 β, Bergman-type space (for β).
Seco (UB) Approximants vs. OP CIEM 8 / 11
We were able to compute explicitly Jω for the following:
Definition
β > −1, ωk = β+k+1
k
−1, H2
ω =: A2 β, Bergman-type space (for β).
f2
ω = (β + 1)
|f(z)|2(1 − |z|2)βdA(z)
Seco (UB) Approximants vs. OP CIEM 8 / 11
We were able to compute explicitly Jω for the following:
Definition
β > −1, ωk = β+k+1
k
−1, H2
ω =: A2 β, Bergman-type space (for β).
f2
ω = (β + 1)
|f(z)|2(1 − |z|2)βdA(z)
Theorem
β, ω as above. Jω =
β+3
√
β+2, f ∗(z) =
z
√
β+2
−(β+3)
Seco (UB) Approximants vs. OP CIEM 8 / 11
Back to the recursion (β = 0), we get: ak+1 = λak − k + 2 k + 1ak−1.
Seco (UB) Approximants vs. OP CIEM 9 / 11
Back to the recursion (β = 0), we get: ak+1 = λak − k + 2 k + 1ak−1. Sum + multiply:
∞
ak+1zk = λ
∞
akzk −
∞
k + 2 k + 1ak−1zk.
Seco (UB) Approximants vs. OP CIEM 9 / 11
Back to the recursion (β = 0), we get: ak+1 = λak − k + 2 k + 1ak−1. Sum + multiply:
∞
ak+1zk = λ
∞
akzk −
∞
k + 2 k + 1ak−1zk. This becomes a differential equation: f ∗(z)(3z − λ) + (f ∗)′(z)(1 − λz + z2) = 0.
Seco (UB) Approximants vs. OP CIEM 9 / 11
Back to the recursion (β = 0), we get: ak+1 = λak − k + 2 k + 1ak−1. Sum + multiply:
∞
ak+1zk = λ
∞
akzk −
∞
k + 2 k + 1ak−1zk. This becomes a differential equation: f ∗(z)(3z − λ) + (f ∗)′(z)(1 − λz + z2) = 0. For the solution to be analytic in D we need one zero of (1 − λz + z2) to match λ/3. This gives the value of λ. Then solve everything else.
Seco (UB) Approximants vs. OP CIEM 9 / 11
Which pairs of points happen for degree 2 and do those zeros minimize anything “reasonable” directly? ... a link with potentials?
Seco (UB) Approximants vs. OP CIEM 10 / 11
Which pairs of points happen for degree 2 and do those zeros minimize anything “reasonable” directly? ... a link with potentials? Algorithms for inversion of matrices may give characterizations of cyclicity, so... adequate algorithms?
Seco (UB) Approximants vs. OP CIEM 10 / 11
Which pairs of points happen for degree 2 and do those zeros minimize anything “reasonable” directly? ... a link with potentials? Algorithms for inversion of matrices may give characterizations of cyclicity, so... adequate algorithms? Many more problems! Ask me!
Seco (UB) Approximants vs. OP CIEM 10 / 11
Seco (UB) Approximants vs. OP CIEM 11 / 11