Measures, orthogonal polynomials,
and continued fractions
Michael Anshelevich
November 7, 2008
M EASURES
AND ORTHOGONAL POLYNOMIALS .
µ a positive measure on R.
A linear functional µ[P ] =
R
P (x) dµ(x),
µ : R[x] → R.
Positive:
µ[P (x)2] ≥ 0.
Inner product hP, Qi = µ[P Q].
Gram-Schmidt
n
1, x, x2, x3, . . .
o
⇒
⇒ monic orthogonal polynomials
{P0 = 1, P1, P2, P3, . . .}.
1
Theorem. (Favard, Stone, etc.) For some βi ∈ R, γi ≥ 0
xPn = Pn+1 + βnPn + γnPn−1.
2nd order recursion relation.
Two independent solutions {Pn, Qn}.
Initial conditions
P−1 = 0, P0 = 1,
Q0 = 0,
P1 = x − β0,
Q1 = 1,
Exercise.
Qn(x) = (I ⊗ µ)
"
Q2 = x − β1.
#
Pn(x) − Pn(y)
.
x−y
2
(
µ↔
(β0, β1, β2, . . .)
)
equivalent.
(γ1, γ2, γ3, . . .)
More explicit relation?
(
If know
(β0, β1, β2, . . .)
(γ1, γ2, γ3, . . .)
)
, how to recover µ?
Without going through {Pn}.
Cauchy transform
Z
Gµ(z)=
·
1
1
dµ(x) = µ
z−x
z−x
¸
µ[1]
µ[x]
µ[x2]
µ[x3]
=
+ 2 +
+
+ ...
3
4
z
z
z
z
3
µ[1]
µ[x]
µ[x2]
µ[x3]
Gµ(z) =
+ 2 +
+
+ . . ..
z
z
z3
z4
Theorem. Also
1
Gµ(z) =
γ1
z − β0 −
z − β1 −
γ2
γ3
z − β2 −
z − ...
Same coefficients as in the recursion.
Note: µ ≥ 0 ⇔ all γ ≥ 0, no determinants.
Proof II. Flajolet (1980): lattice paths.
4
Proof I. Look at
1
G(z) =
γ1
z − β0 −
γ2
z − β1 −
γ3
...
z − β2 −
z − βn−1 − γnH
polynomial
.
G=
polynomial
Claim.
Qn − γnQn−1H
G=
,
Pn − γnPn−1H
where {P, Q} with recursion {βi, γi}.
5
Proof. By induction.
G=
1
Q − γ1Q0H0
.
= 1
z − β0 − γ1H0
P1 − γ1P0H0
Assume
Qn − γnQn−1H
G=
Pn − γnPn−1H
and say
H=
1
.
z − βn − γn+1K
Then
G=
1
Qn − γnQn−1 z−βn−γ
n+1 K
1
Pn − γnPn−1 z−βn−γ
n+1 K
zQn − βnQn − γn+1QnK−γnQn−1
=
zPn − βnPn − γn+1PnK−γnPn−1
Qn+1 − γn+1QnK
=
.
Pn+1 − γn+1PnK
6
F INITE
CONTINUED FRACTIONS .
Let Gn = G cut off at level n.
1
Gn(z) =
γ1
z − β0 −
γ2
z − β1 −
γ3
...
z − β2 −
z − βn−1 − 0
H = 0.
Qn − γnQn−1H
Qn
Gn =
=
.
Pn − γnPn−1H
Pn
7
Qn
Gn =
.
Pn
Pn monic, n real roots
n
Y
Pn(x) =
(x − xi).
i=1
Z
Gn(z)=
1
Qn(z)
dµn(x) = Q
(z − xi)
z−x
= (partial fractions) =
X
ai
.
z − xi
Gn = Cauchy transform of
µn =
xi = roots of Pn,
X
ai =
aiδxi ,
Qn(xi)
.
0
Pn(xi)
8
1
Gn(z) =
γ1
z − β0 −
γ2
z − β1 −
γ3
...
z − β2 −
z − βn−1 − 0
Gn approximate G.
Gn = Gµn ,
µn =
n
X
i=1
aiδxi .
Do µn approximate µ?
Yes. In fact,
µn[P (x)] = µ[P (x)] for deg P ≤ 2n − 1.
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G AUSSIAN
QUADRATURE .
Want to evaluate
Z
n
X
f (x) dµ(x) ≈
aif (xi).
i=1
“Riemann sums”.
R
Want P (x) dµ(x) =
Pn
i=1 aiP (xi ) for P of low degree.
How to choose ai, xi?
Answer: take xi = roots of Pn.
Choose ai so that
Z
xk dµ(x)
=
X
aixki ,
k = 0, 1, . . . , n − 1
(n equations, n unknowns).
Our ai =
Qn(xi)
work.
0
Pn(xi)
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Proof. Lagrange interpolation: for any P with deg P < n,
P (x) =
Note
"
#
n
X
P (xi)Pn(x)
.
0
i=1 Pn (xi )(x − xi )
"
#
Pn(x)
Pn(x) − Pn(xi)
µ
=µ
= Qn(xi)
x − xi
x − xi
so
µ[P (x)] =

n
X
n
X

P (xi)
Qn(xi) 

Q
(x
)
=
δxi [P ]
n i
0
0
i=1 Pn (xi )
i=1 Pn (xi)

µ[P (x)] = 
n
X
i=1

aiδxi  [P (x)]
µ[P (x)] = µn[P (x)] for deg P < n.
In fact, same xi, ai work for k = n, n + 1, . . . , 2n − 1.
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For Pk , n ≤ k ≤ 2n − 1,
Pk (x) = A(x)Pn(x) + B(x),
deg A, B ≤ n − 1.
µ[Pk ] = h1, Pk i = 0.
To show: µn[Pk ] = 0.
µ[APn] = hA, Pni = 0
⇒ µ[B] = 0
⇒
deg A < n
µn[B] = 0.
Finally,
µn[APn] =
X
aiA(xi)Pn(xi) = 0,
so µn[Pk ] = 0.
So µn → µ, Gn → G, and therefore Gµ = G.
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If know {βi, γi}, can find µ?
Usually hard:
1
Gµ(z) =
γ1
z − β0 −
z − β1 −
γ2
γ3
z − β2 −
z − ...
an infinite expression.
Class of explicit examples.
Semicircle law:
q
1
4t − x21[−2√t,2√t] dx.
2πt
2.0
1.5
y 1.0
0.5
0.0
−3
−2
−1
0
1
2
3
x
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Semicircle law:
q
1
4t − x21[−2√t,2√t] dx.
2πt
2.0
1.5
y 1.0
0.5
0.0
−3
−2
−1
0
1
3
2
x
Marchenko-Pastur distributions:
q
1
·
2π
4t − (x − 1 − t)2
x
1[1+t−2√t,1+t+2√t](x) dx
+ max(1 − t, 0)δ0.
1.0
1.0
0.75
0.75
y 0.5
y 0.5
0.25
0.25
0.0
0.0
0
1
2
x
3
4
0
1
2
3
4
x
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Semicircular, Marchenko-Pastur orthogonal polynomials
satisfy
∞
X
Pn
(x)un
n=0
A(u)
=
.
1 − B(u)x
In general: free Meixner distributions
q
4(t + c) − (x − b)2
1
√
√
·
1
dx
[b−2
t+c,b+2
t+c]
2
2
2πt 1 + (b/t)x + (c/t )x
+0, 1, 2 atoms.
◦ AC support an interval
√
polynomial
◦ polynomial
◦ limited atoms
◦ outside of the AC support
◦ Not SC part
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P ERIODIC
CONTINUED FRACTIONS .
βi+n = βi,
γi+n = γi.
H = G in
1
G(z) =
γ1
z − β0 −
γ2
z − β1 −
γ3
...
z − β2 −
z − βn−1 − γnH
G=
Qn − γnQn−1G
.
Pn − γnPn−1G
γnPn−1G2 − (γnQn−1 + Pn)G + Qn = 0.
Quadratic equation!
D = (γnQn−1 + Pn)2 − 4γnQnPn−1.
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√
(γnQn−1 + Pn) − D
G=
2γnPn−1
Z
=
1
dµ(x).
z−x
Stieltjes inversion formula:
dµ(x) = −
q
dµ(x) > 0 if
1
lim Im G(x + iy).
π y→0
D(x) ∈ iR, i.e. D < 0.
D degree 2n, n intervals for D ≤ 0.
No SC.
atoms: roots of Pn−1, at most (n − 1).
Recall D = (γnQn−1 + Pn)2 − 4γnQnPn−1.
So if Pn−1(a) = 0, then D(a) ≥ 0.
Atoms outside of the AC support.
17
Eventually constant continued fractions (n = 1)
βi = β, γi = γ for i ≥ N .
p
⇔
polynomial
on one interval.
polynomial
If β = 0, γ = 1 (Peherstorfer?)
q
4 − x2
Bernstein-Szegő class
on [−2, 2].
polynomial
Weyl’s Theorem. If βi → 0, γi → 1, then
σess(µ) = [−2, 2].
Denisov-Rakhmanov Theorem. If
σess(µ) = AC support of µ = [−2, 2],
then βi → 0, γi → 1.
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√
polynomial
Eventually periodic ⇒ polynomial on n intervals.
√
polynomial
polynomial
on n intervals 6⇒ eventually periodic.
Weyl: if {βi, γi} asymptotically periodic, same essential
spectrum as for actually periodic.
Converse false.
Last, Simon: if {βi, γi} approaches the isospectral torus
of a periodic sequence, same essential spectrum.
Damanik, Killip, Simon: converse true.
19
Q UESTIONS .
◦ Connection between random matrices and (eventually)
periodic continued fractions (Pastur).
◦ Multivariate (non-commutative) orthogonal polynomials,
states, continued fractions. All exist. Continued fractions
matricial.
◦ Formulas for states with periodic continued fractions.
Free Meixner states known; “constant” after step 2. If not
formulas for states, description of their operator algebras.
◦ Connection between multi-matrix models and states with
multivariate (eventually) periodic continued fractions.
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