Orthogonal rational functions on the real half line with poles in

Orthogonal rational functions on the real halfline with poles in [−∞, 0]

Adhemar Bultheel, Pablo Gonzalez-Vera, Erik Hendriksen, Olav Njastad

August 2003

CORFU, Røros, Norway, August, 2003

1/16

Rational functions

Orthogonal wrt 〈f, g〉 =∫

f(x)g(x)dµ(x), supp(µ) ⊂ R = R ∪ ∞.

Rational function spaces, given ζk∞k=0 ⊂ R, (ζ0 = ∞):

Ln =

πn(x)Dn(x)

: πn ∈ Πn;Dn = r0r1 · · · rn

, rk =

ζk − z, ζk 6= ∞1, ζk = ∞

ζk moments PadeLn = Λ0,n ∞ at 0 at ∞Ln = Λ−n,0 0 at ∞ at 0

Ln = Λ−p,n−p 0,∞ strong 2 point

Ln ∈ R rational multipoint

CORFU, Røros, Norway, August, 2003 A. Bultheel

2/16

Unique OFS

Assume µ is real ⇒ OFS real.

Nesting: Ln−1 ⊂ Ln, Ln−1 ⊥ ϕn ∈ Ln \ Ln−1

Note: ϕn = Pn/Dn, Pn(ζn) 6= 0 (l.c. if ζn = ∞)1

Regular: Pn(ζn−1) 6= 0

Sign normalization: Pn(ζn−1)Pn−1(ζn−1) > 0.

1P (x) = µnxn + µn−1xn−1 + · · ·+ µ0 then P (∞) = µn.


3/16

Recurrence and Christoffel-DarbouxRecurrence regular OFS

ϕ−1 = 0, ϕ0 = 1, ϕn(x) = bn(x)ϕn−1(x) + an(x)ϕn−2(x), n ≥ 2

bn(x) =Unxγn−1 + Vnrn−1(x)

rn(x), an(x) =

Wnrn−2(x)rn(x)

γn−1 =

1, ζn−1 = ∞,0, ζn−1 6= ∞,

Un > 0, n ≥ 1Wn = − Un

Un−1< 0, n ≥ 2

Christoffel-Darboux regular OFS and Pn−2(ζn−1) 6= 0, fn = rnϕn

fn(x)fn−1(y)− fn(y)fn−1(x) = Un(x− y)∑n−1

k=0 ϕk(x)ϕn(y), x 6= y

fn(x)f ′n−1(x)− f ′n(x)fn−1(x) = −Un

∑n−1k=0 ϕ2

k(x), x = y

Corr.: ϕn(x) = 0 ⇒ ϕn+1(x)ϕn−1(x)ϕ′n(x) 6= 0


4/16

Quadrature

I(f) =∫

Rf(x)dµ(x) ≈ In(f) =

n∑k=1

Hkf(xk) =∫

RRn(x)dµ(x)

Rn ∈ Ln−1 interpolates in xknk=1 ⊂ supp(µ) \ ζkn

k=1.Hence exact in Ln−1.

Thm.: In(f) exact in larger space Ln · Lr ⇔〈Nn, g〉 = 0, ∀g ∈ Lr, Nn(x) =

∏ni=1(x− xi)/Dn(x).

Corr.: Max domain of validity Ln · Ln−1 if xknk=1 zeros of ϕn(x).

The weights are Hj = λj = 1/∑n−1

k=0 ϕ2k(xj) > 0.

This is Rational Gauss formula: In(f) =∑n

k=1 λkf(xk) =∫

R f(x)dµn(x)


5/16

Eigenvalue problem

zγnϕn(z) = dn−1rn−1(z)ϕn−1(z) + cnrn(z)ϕn(z) + dnrn+1(z)ϕn+1(z)

with cn = −Vn+1/Un+1 and dn = 1/Un+1.

Jacobi matrix:

J = tridiag

d0, d1, . . .c0, c1, . . .

d0, d1, . . .

,R = diag(r0(z), r1(z), . . .),L = diag(zγ0, zγ1, . . .), Φ =

ϕ0

ϕ1...

Then LΦ = JRΦ. Set En = [0, . . . , 0, 1]T then taking finite sections:

(JnRn − Ln)Φn = −dnrn+1(z)ϕn+1(z)En


6/16

Take z = xk, a zero of ϕn+1(z), then

[JnRn(xk)− Ln(xk)]Φn(xk) = 0

If I∞ = diag(γ0, γ1, . . . , γn) and I0 = In − I∞ thenLn = xkI∞+I0 and Rn = I∞+diag(ζj−xk)I0. Set Zn = diag(τ0, . . . , τn)with τk = 1 if ζk = ∞, and τk = ζk otherwise. Then EVP becomes

[JnZn − I0]− λ[JnI0 + I∞]Φn(λ) = 0

All ζj = ∞: det[Jn − λIn] = 0All ζj finite: det[(JnZn − In)− λJn] = 0 or det[(Zn − J−1

n )− λIn] = 0


7/16

Special cases

case supp(µ) ζn∞n=1

general ⊂ R ⊂ RStieltjes ⊂ [0,∞) ⊂ [−∞, 0]balanced Stieltjes ⊂ [0,∞) ζn = βn ∪ αn,

−∞ ≤ βk < αl ≤ 0ζ2m = βm, ζ2m+1 = αm+1

separated balanced Stieltjes ⊂ [0,∞) βk ≤ β < α ≤ αl

monotone separated bal. Stieltjes ⊂ [0,∞) βk+1 ≤ βk, αk ≤ αk+1


8/16

Stieltjes situation

StieltjesThm.: ϕn(x) has n simple zeros in (−∞, 0).Zeros of ϕn and ϕn−1 interlace.

balanced Stieltjes: ζ2m = βm, ζ2m+1 = αm+1

ϕn(x) =

Unxγn−1+Vnrn−1(x)︷︸︸︷Qnrn−2(x) + Rnrn−1(x)

rn(x)ϕn−1(x) + Wn

rn−2(x)rn(x)

ϕn−2(x)

ζ2m−2 = βm−1 6= ∞:

Q2m < 0Q2m−1 > 0

R2m > 0R2m−1 < 0

ζ2m−2 = βm−1 = ∞:

Q2m > 0Q2m−1 > 0

R2m < 0R2m−1 < 0


9/16

Associated functions

σn(x) :=∫ ∞

0

ϕn(t)− ϕn(x)t− x

dµ(t), n = 0, 1, . . .

Stieltjes

σn(x) =Unxγn−1 + Vnrn−1(x)

rn(x)︸︷︷︸bn(x)

σn−1(x) + Wnrn−2(x)rn(x)︸︷︷︸an(x)

σn−2(x)

σ−1 = −1,σ0 = 0,

γn−1(x) =

1, ζn−1 = ∞,0, ζn−1 6= ∞,

W1 = −U1 < 0.

balanced Stieltjes : rewrite bn(x) as

bn(x) =Qnrn−2(x) + Rnrn−1(x)

rn(x)=

δn(x)rn(x)


10/16

Thus[σn(x) σn−1(x)ϕn(x) ϕn−1(x)

]=

[σn−1(x) σn−2(x)ϕn−1(x) ϕn−2(x)

] [bn(x) 1an(x) 0

],

[σ0(x) σ−1(x)ϕ0(x) ϕ−1(x)

]=

[0 −11 0

]Is a CF with convergents: σn(x)/ϕn(x).Do they converge anywhere? and if they do to what?


11/16

Monotonicityseparated balanced Stieltjes :

ζ2m−2 = βm−1 6= ∞:

δ2m(t) > 0, t ∈ (β, α)δ2m−1(t) > 0, t ∈ (β, α)

ζ2m−2 = βm−1 = ∞:

δ2m(t) > 0, t ∈ (β, α)δ2m−1(t) < 0, t ∈ (β, α)

Thm.: For t ∈ (β, α):σ2m(t)ϕ2m(t)

monotonically decreasing

σ2m+1(t)ϕ2m+1(t)

monotonically increasing

Hence (normal families) there is F0 and F∞ analytic outside [0,∞) suchthat there

limm→∞

σ2m(z)ϕ2m(z)

= F∞(z), limm→∞

σ2m+1(z)ϕ2m+1(z)

= F0(z).


12/16

Moments

Let Ωn∞n=0 be a basis for L.Consider a linear functional M defined on L · LDefine moments ck,l = M [ΩkΩl], k, l = 0, 1, . . .

A positive measure µ on [0,∞) solves the (rational) Stieltjes momentproblem on L · L if ck,l =

∫∞0

Ωk(x)Ωl(x)dµ(x), k, l = 0, 1, 2, . . . (1)

A positive measure µ on [0,∞) solves the (rational) Stieltjes momentproblem on L if ck,0 =

∫∞0

Ωk(x)dµ(x), k = 0, 1, . . . (2)

Suppose the previous µ is a solution of (1).Stieltjes function:

S(z, µ) =∫ ∞

0

dµ(t)z − t


13/16

Approximation of Stieltjes transform

Thm.: µn rational Gauss quadrature. µ solves MP in L · L. ThenS(z, µn) = σn(z)/ϕn(z), and

σn(z)ϕn(z)

− S(z, µ) =1

rn(z)ϕn(z)2

∫ ∞

0

ϕn(t)2rn(t)t− z

dµ(t)

separated balanced Stieltjes:

σ2m+1(t)ϕ2m+1(t)︸︷︷︸→F0(t)

< S(t, µ) <σ2m(t)ϕ2m(t)︸︷︷︸→F∞(t)

, t ∈ (β, α)

F0(z) ↔ µ(0), F∞(z) ↔ µ(∞) solutions of MP in L · L.F0 = F∞ ↔ µ(0) = µ(∞) then unique solution of of MP in L · L.


14/16

Canonical basis

separated balanced Stieltjes: basis for Ln:

Ω0 = 1,

Ω2m(z) =

r1(z)r3(z) · · · r2m−1(z)( zζ2− 1)( z

ζ4− 1) · · · ( z

ζ2m− 1)

Ω2m+1(z) =( zζ2− 1)( z

ζ4− 1) · · · ( z

ζ2m− 1)

r1(z)r3(z) · · · r2m+1(z)

ϕn = vnΩn+· · ·

determinate MP?:

σn(z)ϕn(z)

− σn−2(z)ϕn−2(z)

=∆n(z)

ϕn(z)ϕn−2(z).


15/16

Idea of the proof

σn(z)ϕn(z)

− σn−2(z)ϕn−2(z)

=∆n(z)

ϕn(z)ϕn−2(z).

Bound |∆n(t)| from below on (β, α)

Then since∑

[σnϕn− σn−2

ϕn−2] cvg, hence

∑|ϕn(t)ϕn−2(t)| → ∞

Thus∑|ϕ2m(t)|2 →∞ or

∑|ϕ2m+1(t)|2 →∞

Since σn(z)ϕn(z) − S(z, µ) = 1

ϕn(z)

∫∞0

ϕn(t)t−z dµ(t) for any solution µ

The Fourier coefs, of 1t−z are [σn(x)

ϕn(x) − S(x, µ)]ϕn(x)

Bessel:∑

[σn(x)ϕn(x) − S(x, µ)]2ϕn(x)2 < ∞

Since∑|ϕn(x)|2 = ∞,

∑[σn(x)ϕn(x) − S(x, µ)] = 0

Thus all solutions µ have the same Stieltjes transform on (β, α)


16/16

Determinate MP

monotone balanced separated Stieltjes:

|∆n(t)| ≥ |RnUn−1||rn(t)rn−2(t)|

≥ 1K

vnvn−2

t ∈ (β, α), n ≥ 2

Thm.: limn→∞ |vn/vn−2|1/2 = ∞ ⇒ determinate MP in L · L

Thm.: cn,n =∫∞0

Ω2n(t)dµ(t) and

∑∞n=0(cn,n)−1/2n = ∞ ⇒ determinate

MP in L · L

Strong Stieltjes moment problem:αk = 0,βk = −∞ ⇒

Ω2m(z) = zm,Ω2m+1(z) = z−(m+1),

cn =∫∞0

tndµ(t), n ∈ Z.

Thm.: Strong Stieltjes MP is determinate if∑∞m=1(c−2m)−

14m−2 = ∞ or

∑∞m=0(c2m)−

14m = ∞


Orthogonal rational functions on the real half line with poles in

Documents