Free Jacobi process associated with one projection: local inverse of the flow

0.17 MB·English

by  Nizar Demni

FREE JACOBI PROCESS ASSOCIATED WITH ONE PROJECTION: LOCAL
INVERSE OF THE FLOW
N.DEMNI
5
1
0 Abstract. Wepursuethestudystartedin[8]ofthedynamicsofthespectraldistributionofthefreeJacobi
2 process associated with one orthogonal projection. More precisely, we use Lagrange inversion formula in
order to compute the Taylor coefficients of the local inverse around z = 0 of the flow determined in [8].
r
p Whentherankoftheprojectionequals1/2, theobtainedsequence reducestothemomentsequenceofthe
A free unitary Brownian motion. For general ranks in (0,1), we derive a contour integral representation for
the first derivative of the Taylor series which is a major step toward the analytic extension of the flow in

8 theopenunitdisc.

]
R 1. Reminder and motivation
P
ThefreeJacobiprocess(J ) wasintroducedin[6]asthelarge-sizelimitoftheHermitianmatrixJacobi
. t t≥0
h process ([9]). It is built as the radial part of the compression of the free unitary Brownian motion (Y )
t t≥0
t ([1]) by two orthogonalprojections P,Q :
a
{ }
m
J :=PY QY∗P.
t t t
[
In this definition, both families of operators P,Q and (Yt)t≥0 are -free (in Voiculescu’s sense) in a von
2 Neumann algebra A endowed with a finite{trace}τ and a unit 1. ∗When J is considered as a positive
v t
operator valued in the compressed algebra (PAP,τ/τ(P)), its spectral distribution µ 1 is a probability
3 t
1 distributionsupportedin[0,1]andthepositiverealnumberτ(P)µt 1 encodesthegeneralpositionproperty
{ }
0 for P,Y QY∗ ([4],[13]). Ontheotherhand,thecoupleofpapers[7]and[8]aimtodeterminetheLebesgue
0 deco{mpotsitiotn}of µ when both projections coincide P = Q. In particular, when τ(P) = 1/2, a complete
t
0
description was given in [7] (see Corollary 3.3 there and [13] for another proof): at any time t 0, µ
. ≥ t
2 coincides with the spectral distribution of
0 Y +Y∗ +21
5 2t 2t
4
1
: considered as a positive operator in (A,τ) (the spectral distribution of Y , say η , was described in [2],
v 2t 2t
Proposition 10). For an arbitrary rank τ(P) (0,1), only the discrete part in the Lebesgue decomposition
i
X ofµ wasdeterminedin[8](seeTheorem1.1).∈Astoitsabsolutely-continuouspartwithrespecttoLebesgue
t
r measure in [0,1], it was related to that of the spectral distribution, say ν , of the unitary operator
a t
U :=RY RY∗, R:=2P 1.
t t t −
Actually,thedensityoftheformerdistributionisrelatedtothedensityofthelatterthroughtheCaratheodory
extension of the Riemann map of the cut plane C [1, [ ([8], Theorem 1.1).
\ ∞
Thekeyingredientleadingtothispartialdescriptionisaflowψ sofardefinedandexploitedinaninterval
t
of the form ( 1,z ) for some z (0,1),t > 0. When τ(P) = 1/2, ψ is a one-to-one map from a Jordan
t t t
domain onto−the open unit disc D∈ and its compositional inverse coincides, up to a Cayley transform, with
the Herglotz transform of η . For arbitrary ranks, ψ is locally invertible and further information on ν
2t t t
(which in turn provide information on µ ) necessitates the investigation of the analytic extension of of the
t
localinverseofψ inD. Moreover,it wasshownin[8]that ν convergesweakly to a probabilitymeasure ν
t t ∞
whose support disconnects as soon as τ(P)=1/2 and it would be interesting to know whether this striking
6
disconnectedness happens or not at a finite time.
1Inthisintroductorypart,weommitthedependence ofournotations on{P,Q}.
1
In this paper, we use Lagrange inversion formula and derive the Taylor coefficients of the inverse of ψ ,
t
up to the elementary invertible transformations:
1+z
z D s= , s κ2+(1 κ2)s2,
∈ 7→ 1 z 7→ −
−
p
where κ:=τ(R)=2τ(P) 1. These coefficients are displayed in corollary 3.2 of proposition 3.1 below and
−
are given by sign-alternating nested (finite) sums involving Laguerre polynomials and others in the variable
κ2. In particular, we recover when κ = 0 the moment sequence of η while more generally, we can single
2t
out from the obtained Taylor series a deformation of the Herglotz transform of η which is still bounded
2t
analytic in D but no longer extends continuously to the unit circle T unless κ = 0. As to the analytic
extension of the whole Taylor series, it does not seem accessible (at least for the author) directly from the
sums alluded to above due to oscillations. For that reason, we derive a contour integral representation over
a circle centered at κ for the first derivative of the Taylor series, which is so far valid in a neighborhood
of z = 0. To this end, we use the analytic continuation of the generating series for the Jacobi polynomial
outside the interval [ 1,1] as well as a special generating series for Laguerre polynomials. The latter series
−
generalizes the Herglotz transform of η and is expressed through it. Compared to the high dissymmetry
2t
arising when κ=0, this integralrepresentationis a major step towardthe extensionof the flow in the open
6
unit disc which remains a challenging problem.
The paper is organized as follows. For sake of completeness, we briefly recall in the next section the
relation of the spectral dynamics of (J ) to those of (U ) and in particular to the flow (ψ ) . In
t t≥0 t t≥0 t t≥0
orderto make the paper self-contained,the thirdsectionincludes the variousspecialfunctions we use in the
remainder of the paper, as well as the computations leading to the Taylor coefficients. The fourth section
is devoted to the derivation of the aforementionedintegral representationand we close the paper by further
developments in relation to the extension of the derived integral in D.
2. From the free Jacobi process to the flow
Though the study of the spectral dynamics of
J =PY PY⋆P
t t t
was direct when τ(P) = 1/2, their study for arbitrary ranks τ(P) (0,1) is rather based on the following
∈
binomial-type expansion:
n
1 2n κ 1 2n
τ[(J )n]= + + τ((U )k).
t 22n+1 n 2 22n n k t
(cid:18) (cid:19) k=1(cid:18) − (cid:19)
X
Thisexpansionhasthemerittoorientourstudytothespectraldynamicsof(U )whichturnouttobeeasier
t
than those of J due to the unitarity of U . In this respect, let
t t
w+z
H (z):= ν (dw)=1+2 τ(Un)zn
κ,t T w z κ,t t
Z − n≥1
X
be the Herglotz transform of the spectral distribution ν of U . Then the key result proved in [8] states
κ,t t
that there exists a flow (t,z) ψ (z) defined in an open set of R [ 1,1] and such that
κ,t +
7→ × −
(2.1) [H (ψ (z))]2 [H (z)]2 =[H (ψ (z))]2 [H (z)]2.
κ,∞ κ,t κ,∞ κ,t κ,t κ,0
− −
Here
1+z
H (z)=H (z)=
κ,0 0
1 z
−
is the Herglotz transform of ν = δ and H is that of the weak limit ν of ν (see section 2 in [8]
κ,0 1 κ,∞ κ,∞ κ,t
for more details on ν ). Equation (2.1) was so far used to determine the discrete spectrum of J and ψ
κ,∞ t κ,t
was expressed as follows: define2
1 √1 z z
α:z − − = .
7→ 1+√1 z (1+√1 z)2
− −
2Theprincipalbranchofthesquarerootistaken.
2
Thismapextendsanalyticallyto z C [1, [andis one-to-onefromthiscutplaneontoDwhoseinverseis
∈ \ ∞
4z
α−1(z)= .
(1+z)2
Recall also from [2] (p.266-269)that the map
z 1
ξ :z − etz
2t
7→ z+1
is invertible in some Jordan domain onto D and that its compositional inverse is the Herglotz transform of
η :
2t
w+z
K (z):= η (dw)=1+2 τ(Yn)zn.
2t T w z 2t t
Z − n≥1
X
If
1+z
s:= , z D, a(s):= κ2+(1 κ2)s2,
1 z ∈ −
−
then ([8], p.283) p
a2
ψ (z)=α α−1[ξ (a(y))] .
κ,t a2 κ2 2t
(cid:18) − (cid:19)
This is a locally invertible map near z =0 so that (2.1) is equivalent to
[H (z)]2 [H (z)]2 =[H (ψ−1(z))]2+[H (ψ−1(z))]2
κ,t − κ,∞ κ,∞ κ,t κ,0 κ,t
near z = 0. Since z s,s a(s) and α are invertible transformations the inverting ψ around z = 0
κ,t
7→ 7→
amounts to the inversion of the map
a2
a α−1[ξ (a)]
7→ a2 κ2 2t
−
near a=1. This is the main task we achieve in the next section.
3. Local inverse of the flow: Lagrange inversion formula
3.1. Special functions. As claimed in the introduction, we list below the special functions occurring in
our subsequent computations (see [11], [14], [15] for further details). We start with the Gamma function
∞
Γ(x)= e−uux−1du, x>0,
Z0
and the Pochhammer symbol
(a) =(a+k 1)...(a+1)a, a R, k N,
k
− ∈ ∈
with the convention (0) =δ . The latter may be expressed as
k k0
Γ(a+k)
(a) =
k
Γ(a)
when a>0, while
( n) n
(3.1) − k =( 1)k
k! − k
(cid:18) (cid:19)
if k n and vanishes otherwise. Next comes the generalized hypergeometric function defined by the series
≤
p (a ) zm
F ((a ,1 i p),(b ,1 j q);z)= i=1 i m
p q i ≤ ≤ j ≤ ≤ q (b ) m!
m≥0 Qj=1 j m
X
whereanemptyproductequalsoneandtheparameters(a ,1 i p)Qarerealswhile(b ,1 j q) R N.
i j
Withregardto(3.1),thisseriesterminateswhenatleasta =≤ n≤ Nforsome1 i p≤,the≤refor∈ered\u−ces
i
− ∈− ≤ ≤
in this case to a polynomial of degree n. In particular, the Charlier polynomials are defined by
1
C (x,a):= F n, x; , a R 0 ,x R.
n 2 0
− − −a ∈ \{ } ∈
(cid:18) (cid:19)
3
When x Z is an integer, a generating function of these polynomials is given by
∈
(au)n
(3.2) C (x,a) =eau(1 u)x, u <1.
n
n! − | |
n≥0
X
Moreover,the n-th Laguerre polynomial with index α R defined by
∈
n
1 ( n)
(3.3) L(α)(z):= − j(α+j+1) zj,
n n! j! n−j
j=0
X
is related to the n-th Charlier polynomial via:
( a)n
(3.4) − C (x,a)=L(x−n)(a).
n! n n
When p=2,q=1, the Jacobi polynomial Pa,b of parameters a,b> 1 is represented as
n −
(a+1) 1 x
(3.5) Pa,b(x):= n F n,n+a+b+1,a+1, − .
n n! 2 1 − 2
(cid:18) (cid:19)
3.2. Inversion. Let t > 0,κ ( 1,1) be fixed. The aim of this section is to derive the Taylor coefficients
∈ −
of the inverse of the map φ defined by
κ,t
z2
φ (z)= α−1(ξ (z))
κ,t z2 κ2 2t
−
inaneighborhoodofz =1. Ofcourse,itisreadilycheckedthat∂ φ (1)=0sothatφ islocallyinvertible
z t,κ t,κ
6
there. According to Lagrange inversion formula (see [14], p.354), the Taylor coefficients of its inverse are
given by
1 z 1 n
a (κ,t):= ∂n−1 − , n 1.
n n! z φ (z) |z=1 ≥
(cid:20) κ,t (cid:21)
The issue of our computations is recorded in the proposition below:
Proposition 3.1. There exists a set of polynomials (P(m)) depending on an integer parameter m 0
n n≥1
≥
such that
n k−1
2 2n
a (κ,t)= e−kt L(m+1) (2kt)2mP(m)(κ2) .
n 22nn n k k−m−1 n
( )
k=1(cid:18) − (cid:19) m=0
X X
Proof. Set ǫ:=κ2, then we need to expand
ǫ n (1+ξ (z))2n
(3.6) (z 1)n 1 t
− − z2 4nξn(z)
t
(cid:16) (cid:17)
around z =1. To proceed, we start with
n
ǫ n n 1
1 = ( ǫ)k
− z2 k − (1+(z 1))2k
(cid:16) (cid:17) Xk=0(cid:18) (cid:19) −
n
n (2k)
= ( ǫ)k m(1 z)m
k − m! −
k=0(cid:18) (cid:19) m≥0
X X
(3.7) := (z 1)mP(m)(ǫ)
− n
m≥0
X
where we set
( 1)m n n
P(m)(ǫ):= − ( ǫ)k(2k) .
n m! k − m
k=0(cid:18) (cid:19)
X
4
Next, we expand
(1+ξ (z))2n n 2n
(z 1)n t =(z 1)n ξk
− ξn(z) − n+k t
t k=−n(cid:18) (cid:19)
X
2n n 2n (z 1)n+k
= (z 1)n+ − ektz +(z 1)n−k(1+z)ke−ktz
n − n+k (1+z)k −
(cid:18) (cid:19) k=1(cid:18) (cid:19)(cid:26) (cid:27)
X
2n 2n (z 1)n+k
= (z 1)n+ − ektz
n − n+k (1+z)k
(cid:18) (cid:19) k∈[−nX..n]\{0}(cid:18) (cid:19)
and use (3.2) together with (3.4) to derive
(z 1)n+k ekt (kt(z 1))m
− ektz = (z 1)n+k C ( k, 2kt) −
(1+z)k 2k − m − − m!
m≥0
X
(z 1)m+n+k
=ekt L(−k−m)( 2kt) − .
m − 2m+k
m≥0
X
As a result
(1+ξ (z))2n 2n 2n (z 1)m+n+k
(z 1)n t = (z 1)n+ ekt L(−k−m)( 2kt) − .
− ξn(z) n − n+k m − 2m+k
t (cid:18) (cid:19) k∈[−nX..n]\{0}(cid:18) (cid:19) mX≥0
From (3.1) and (3.3), it is clear that L(−m)(0)=0 for all m 1 while L(0)(z)=1. Hence
m ≥ 0
(z 1)n (1+ξ (z))2n n 2n (z 1)m+n+k
− =(z 1)n t = ekt L(−k−m)( 2kt) −
[α−1(ξ (z))]n − 4nξn(z) n+k m − 22n+m+k
t t k=−n(cid:18) (cid:19) m≥0
X X
2n 2n (z 1)m+k
= e(k−n)t L(n−k−m)(2(n k)t) −
k m − 2n+m+k
k=0(cid:18) (cid:19) m≥0
X X
2n 2n (z 1)m
= e(k−n)t L(n−m)(2(n k)t) −
k m−k − 2n+m
k=0(cid:18) (cid:19) m≥k
X X
m∧2n 2n (z 1)m
= e(k−n)tL(n−m)(2(n k)t) − .
m≥0( k=0 (cid:18)k (cid:19) m−k − ) 2n+m
X X
Keeping in mind (3.6) and (3.7), we end up with
n−1 2n∧m
1 1 2n
a (κ,t)= P(n−1−m)(ǫ) e(k−n)tL(n−m)(2(n k)t)
n nm=02n+m n ( k=0 (cid:18)k (cid:19) m−k − )
X X
n−1 m
1 1 2n
= P(n−1−m)(ǫ) e(k−n)tL(n−m)(2(n k)t)
nm=02n+m n (k=0(cid:18)k (cid:19) m−k − )
X X
n−1 n−1
2 2n
= e(k−n)t L(n−m)(2(n k)t)2n−1−mP(n−1−m)(ǫ)
22nn k=0(cid:18)k (cid:19) (m=k m−k − n )
X X
n n−1
2 2n
= e−kt L(n−m) (2kt)2n−1−mP(n−1−m)(ǫ)
22nn n k m−n+k n
k=1(cid:18) − (cid:19) (m=n−k )
X X
n k−1
2 2n
= e−kt L(m+1) (2kt)2mP(m)(ǫ) .
22nn n k k−m−1 n
( )
k=1(cid:18) − (cid:19) m=0
X X
(cid:3)
5
Corollary 3.2. Set Φ := α φ . Then Φ is invertible in a neighborhood of z = 1 and its inverse is
κ,t κ,t κ,t
◦
given near the origin by
( 1)n n 2n n+k
Φ−1(z)=1+ − ( 1)k b (κ,t) zn,
κ,t n≥1 n (k=1 − n+k(cid:18)n−k(cid:19) k )
X X
where b (κ,t):=k22ka (κ,t),k 1.
k k
≥
Proof. Since α is invertible in C [1, [ with α(0)=0, then
\ ∞
[α−1(z)]n
Φ−1(z)=1+ b (κ,t)
κ,t n n22n
n≥1
X
near z =0. Differentiating term-wise with respect to z and using the identity
1 zα−1(z)
∂ α−1(z)= − ,
z
1+z z
we get
1 z [α−1(z)]n
∂ Φ−1(z)= − b (κ,t) .
z κ,t z(1+z) n 22n
n≥1
X
Now recall the following fact (see [14], p.357): if (c ) ,(b ) are two sequences of real numbers related
n n≥0 n n≥0
by
n
2n
(3.8) b = c ,
n k
n k
k=0(cid:18) − (cid:19)
X
then
zn 1+α(z)
b = c [α(z)]n
n4n 1 α(z) n
n≥0 − n≥0
X X
whenever both series converge. Equivalently, the relation (3.8) is invertible with inverse given by (see [16],
p.68, Table 2.5, (2)),
n n
n+k n+k 1 2n n+k
(3.9) c =b , c = ( 1)k+n + − b = ( 1)k+n b , n 1,
0 0 n k k
− n k n k 1 − n+k n k ≥
k=0 (cid:20)(cid:18) − (cid:19) (cid:18) − − (cid:19)(cid:21) k=0 (cid:18) − (cid:19)
X X
and
[α−1(z)]n 1+z
b = c zn.
n 4n 1 z n
n≥0 − n≥0
X X
The corollary then follows from the substitutions
b =0, b =b (κ,t), n 1.
0 n n
≥
(cid:3)
Remark 3.3. If κ=0 then P(m)(0)=δ so that
n m0
n
2n
b (0,t)=2 e−ktL(1) (2kt).
n n k k−1
k=1(cid:18) − (cid:19)
X
In this case, the inverse of Φ reduces to
0,t
[α−1(z)]n 1
Φ−1(z)=1+ b (0,t) =1+2 e−ntL(1) (2nt)zn
0,t n n22n n n−1
n≥1 n≥1
X X
which is nothing else but the Herglotz transform K of η ([2], p.269).
2t 2t
6
Remark 3.4. For general ǫ [0,1),P(0) =(1 ǫ)n so that the term corresponding to m=0 in a (κ,t) is
n n
∈ −
n
2n
2(1 ǫ)n e−ktL(1) (2kt)=(1 ǫ)nb (0,t).
− n k k−1 − n
k=1(cid:18) − (cid:19)
X
Multiplying by [α−1(z)]n and summing over n 1, the preceding remark shows that
≥
[(1 ǫ)α−1(z)]n
1+2 − b (0,t)=K α[(1 ǫ)α−1(z)] .
n22n n 2t −
n≥1
X (cid:0) (cid:1)
Since K is analytic in D and since α maps C [1, [ onto D, then the map
2t
\ ∞
V :z K (α[(1 ǫ)α−1(z)])
κ,2t 2t
7→ −
extends analytically to D with values in the right half-plane (z) > 0 . It follows that there exists a
probability distribution η on T such that {ℜ }
κ,2t
w+z
V = η (dw),
κ,2t κ,2t
T w z
Z −
and η =η . However, unless ǫ=0, V is no longer continuous in the closed unit disc since the map
0,2t 2t κ,2t
z α[(1 ǫ)α−1(z)]
7→ −
is not so in C due to the presence of the square root function in the definition of α. Nonetheless, it is still a
bounded holomorphic function in D since K is continuous in D ([2], Lemma 12).
2t
4. An Integral representation
For any n 1, set
≥
n
2n n+k
(4.1) S (κ,t):= ( 1)k+n b (κ,t),
n k
− n+k n k
k=1 (cid:18) − (cid:19)
X
where we recall
k j−1
2k
b (κ,t)=2 e−jt L(m+1) (2jt)2mP(m)(ǫ) , k 1.
k k j ( j−m−1 k ) ≥
j=1(cid:18) − (cid:19) m=1
X X
For small z , consider the Taylor series
| |
M (z):= S (κ,t)zn =z∂ Φ−1(z).
κ,t n z κ,t
n≥2
X
A major part of this section is devoted to the derivation of the following integral representation:
Proposition 4.1. There exists a circle γ centered at w =κ and a neighborhood of z =0 where
κ
κ [(K (y))2 1] dw
2t
M (z)=(1 z) − .
κ,t − 2iπ [t(K (y))2+(2 t)][wK (y) κ]wR(z,w)
Zγκ 2t − 2t −
Here
R(z,w):= (1 z)2+4w2z,
−
and
p
4z(1 w2)
y =y(z,w):= − .
(1+z+R(z,w))2
Proof. It consists of four steps corresponding to summation over the indices k,n,j,m respectively. The
{ }
interchange of the summation orders is readily justified by the (uniform) estimates given below for the
generating series occurring in the proof. In the first step, we make use of the following lemma:
Lemma 4.2. For any m 0,k 1,
≥ ≥
κ wm−1(1 w2)k
(4.2) ( 1)mP(m)(ǫ)= − dw
− k 2iπ (w κ)m+1
Zγκ −
where γ is a small circle centered at w =κ.
κ
7
Proof. If m 1 then (0) =0 whence
m
≥
k
1 k
( 1)mP(m)(ǫ)= ( ǫ)l(2l)
− k m! l − m
l=1(cid:18) (cid:19)
X
k
1 k (2l+m 1)!
= ( 1)lκ2l −
m! l − (2l 1)!
l=1(cid:18) (cid:19) −
X
k
κ k
= ∂m ( 1)l(w)2l+m−1
m! w ( l − )
Xl=1(cid:18) (cid:19) |w=κ
κ
= ∂m[wm−1(1 w2)k] .
m! w − |w=κ
The lemma follows from the Cauchy integral formula and from P(0)(ǫ)=(1 ǫ)k. (cid:3)
k −
Now consider the sum over k:
n
2n n+k 2k
S(n,j):= ( 1)k+n (1 w2)k, 1 j n.
− n+k n k k j − ≤ ≤
k=j (cid:18) − (cid:19)(cid:18) − (cid:19)
X
Writing
2n n+k n+k n+k 1
= + −
n+k n k n k n k 1
(cid:18) − (cid:19) (cid:18) − (cid:19) (cid:18) − − (cid:19)
then S(n,j)=f(n,j) f(n 1,j) where
− −
n
n+k 2k
f(n,j):=( 1)n (w2 1)k
− n k k j −
k=j(cid:18) − (cid:19)(cid:18) − (cid:19)
X
with the convention that an empty sum is zero. Since
n+k 2k (n+k)!
= ,
n k k j (n k)!(k+j)!(k j)!
(cid:18) − (cid:19)(cid:18) − (cid:19) − −
then the index change k n k yields
→ −
( 1)n−j(1 w2)j n−j (n j)! (n+k+j)!(w2 1)k
f (n,j)= − − − −
1
(n j)! (n j k)! (k+2j)! k!
− k=0 − −
X
( 1)n−j(1 w2)j(n+j)!n−j (n+j+1) (1 w2)k
k
= − − (j n) −
k
(n j)!(2j)! − (2j+1) k!
k
− k=0
X
( 1)n−j(1 w2)j(2j+1) n−j (n+j+1) (1 w2)k
n−j k
= − − (j n) −
k
(n j)! − (2j+1) k!
k
− k=0
X
=( 1)n−j(1 w2)jP2j,0(2w2 1)
− − n−j −
where the last equality follows from (3.5). Besides, the symmetry relation Pa,b(z)=( 1)nPb,a( z) entails
n − n −
S(n,j)=(1 w2)j ( 1)n−jP2j,0(2w2 1) ( 1)n−j−1P2j,0 (2w2 1)
− − n−j − − − n−1−j −
n o
=(1 w2)j P0,2j(1 2w2) P0,2j (1 2w2) .
− n−j − − n−1−j −
n o
Combining (4.1) and the previous Lemma, we get the following representation
(4.3)
κ n j−1 ( 2)mwm−1
S (κ,t)= [(1 w2)e−t]j L(m+1) (2jt) P0,2j(1 2w2) P0,2j (1 2w2) − dw.
n iπ − j−m−1 n−j − − n−1−j − (w κ)m+1
ZγκXj=1 mX=0 n o −
8
In the second step, we fix j m+1 1 and consider the series of Jacobi polynomials:
≥ ≥
P0,2j(1 2w2)zn =zj P0,2j(1 2w2)zn, w γ .
n−j − n − ∈ κ
n≥j n≥0
X X
Accordingto [10](see the discussionp.2), for any fixed z D, this seriesconvergesuniformly inw onclosed
∈
subsets of the ellipse E of foci 1 and semi-axes
|z|
{± }
1 1
z .
2 z ±| |
(cid:18)| | (cid:19)
Since both semi-axes stretches as z becomes small, then given an ellipse E ,0<r <1, the series of Jacobi
r
| |
polynomials above converges absolutely in the disc z < r uniformly on the closure of the domain D
r
{| | }
enclosed by E . Thus, we fix r and choose γ such that its image under the map w 1 2w2 lies in D .
r κ r
7→ −
Doing so proves the analyticity of the series of Jacobi polynomials above in the variable w E so that the
r
∈
following equality holds by analytic continuation (see [17], p.69 for real 1 2w2):
−
(4z)j
(4.4) zj P0,2j(1 2w2)zn = , w γ ,
n − R(z,w)(1+z+R(z,w))2j ∈ κ
n≥0
X
provided
R(z,w)= (1 z)2+4w2z
−
does not vanish and is analytic in the variable z. This last condition holds true at least for small z since
p
| |
[(1 z)2+4w2z] 1 z (z +max 1 2w2 ).
| − − |≤| | | | γκ | − |
Similarly,
(4z)j
(4.5) P0,2j (1 2w2)zn =z , w γ .
n−j−1 − R(z,w)(1+z+R(z,w))2j ∈ κ
n≥j
X
Next comes the third step where we fix the curve γ and m 1, and work out the series
κ
≥
[4ze−t(1 w2)]j
L(m+1) (2jt) − , w γ .
j−m−1 (1+z+R(z,w))2j ∈ κ
j≥m+1
X
More precisely,
Lemma 4.3. For any m 1 and y <1,
≥ | |
(K (y))2 1
(4.6) 2m+1 L(m+1) (2jt)(e−ty)j = 2t − [K (y) 1]m.
j−m−1 t(K (y))2+(2 t) 2t −
2t
j≥m+1 −
X
Proof. The series in the LHS of (4.6) is aninstance of equation 1.2 from [3] where it is claimed without any
detail that it converges in D. This claim can be checked as follows: the derivation rule ([17])
1
L(m+1) (2jt)= ∂mL(1) (2jt)
j−m−1 (2j)m t j−1
together with the following integral representation ([12], p.561):
j
1 1
(4.7) L(1) (2jt)= 1+ e−2jthdh
j−1 2iπ h
ZC0(cid:18) (cid:19)
over a small closed curve C around the origin lead to
0
( 1)m 1 j
(4.8) L(m+1) (2jt)= − hm 1+ e−2jthdh.
j−m−1 2iπ h
ZC0 (cid:18) (cid:19)
When m = 0, the behavior of L(1) (2jt) as j was analyzed [12] using the saddle point method (see
j−1 → ∞
p.561-562). For general m 1, the integrand in (4.8) differs from the one in (4.7) by the factor hm which
≥
is everywhere analytic and is independent of j. According to the saddle point method, the behavior of
L(m+1) (2jt) as j is the same as that of L(1) (2jt) up to multiplication by m-powers of the saddle
j−m−1 → ∞ j−1
9
points(onesaddlepointfort 2andtwoconjugatesaddlepointswhent<2). Asamatteroffact,formulas
≥
(2.57) and (2.60) in [12] show that the series displayed in (4.6) converges for y <1.
| |
ComingintothederivationoftheRHSof (4.6),wespecializeequation(1.2)in[3]tob=0,v=m+1,x=
2(m+1)t,a=1/(m+1) in order to get:
1 (ye−t)m+1
L(m+1)(2jt+2(m+1)t)(e−ty)j+m+1 = e2(m+1)tu/(u−1).
j (1 u)2+2tu (1 u)m
j≥0 − −
X
Here, u=u (z,w) D is determined by
t
∈
e−ty =ue2tu/(1−u) y =uet(1+u)/(1−u).
⇔
Equivalently,
u+1
Z :=
1 u
−
satisfies ξ (Z)=y, (Z) 0. But since ξ is a one-to-one map from the Jordan domain
2t 2t
ℜ ≥
Γ := (Z)>0,ξ (Z) D
2t 2t
{ℜ ∈ }
onto D whose composition inverse is K ([2], Lemma 12), then Z =K (y) and in turn
2t 2t
Z 1
u= −
Z+1
is uniquely determined in the open unit disc. Substituting
u = (ye−t)e2ut/(u−1)
1 K (y)+1
2t
=
1 u 2
−
proves the lemma. (cid:3)
According to this lemma,
[4ze−t(1 w2)]j (K (y))2 1
2m+1 L(m+1) (2jt) − = 2t − [K (y) 1]m
j−m−1 (1+z+R(z,w))2j t(K (y))2+(2 t) 2t −
2t
j≥m+1 −
X
provided that
4z(1 w2)
y =y(z,w)= −
(1+z+R(z,w))2
lies in D, which holds true for z small enough.
| |
Finally, K (y) 1 becomes small enough when z does since K (0)=1. As a result
2t 2t
− | |
wm−1 1
(4.9) [1 K (y)]m =
(w κ)m+1 − 2t w(wK (y) κ)
2t
m≥0 − −
X
in some disc centered at z =0. Gathering (4.3), (4.4), (4.5) and (4.9), the proposition is proved. (cid:3)
Remark 4.4. If 0<t 2 then the map
≤
z t(K (y))2+(2 t)
2t
7→ −
doesnotvanishsinceK ,asamapfromtheclosedunitdiscintothe righthalf-plane,takesvaluesin√ 1R
2t
−
onlyonthe unitcircle. Otherwise,the rangeΓ ofK doesnotcontainthe real (t 2)/t(see[2],section
2t 2t
−
4.2.3). Thus, for any t>0 and any z in the disc evoked in the previous proposition, the map
p
1
w
7→ [t(K (y))2+(2 t)]
2t
−
isholomorphicintheinteriorofγ . Thisobservationtogetherwiththesplittingκ=(κ wK (y))+wK (y)
κ 2t 2t
−
lead to
10