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Riesz's theorem for orthogonal matrix polynomials

Abstract

We describe the image through the Stieltjes transform of the set of solutions V of a matrix moment problem. We extend Riesz's theorem to the matrix setting, proving that those matrices of measures of V for which the matrix polynomials are dense in the corresponding L2 space are precisely those whose Stieltjes transform is an extremal point (in the sense of convexity) of the image set.

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Riesz's theorem for orthogonal matrix polynomials

Author: López Rodríguez, Pedro
Publisher: Springer
Year: 1999
DOI: 10.1007/s003659900101
Source: https://idus.us.es/bitstreams/48b2c69e-060e-4324-ae79-846c624d3e84/download
Manusc ip : Ap il, 27 2001
RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS
Ped o Lopez-Rod iguez
Uni e sidad de Se illa
Abs ac . We desc ibe he image h ough he S iel jes ans o m o he se o solu ions
Vo a ma ix momen p oblem. We ex end Riesz's heo em o he ma ix se ing, p o ing
ha hose ma ices o measu es o V o which he ma ix polynomials a e dense in he
co esponding L2space a e p ecisely hose whose S iel jes ans o m is an ex emal poin (in
he sense o con exi y) o he image se .
1. In oduc ion.
Fo a posi i e Bo el measu e ºon Rwi h ¯ni e momen s o any o de sn=RR ndº( )
we deno e by V he se o posi i e Bo el measu es ¹on Rsa is ying RR nd¹( )=sn,n¸0,
ha is, he se o solu ions o he Hambu ge momen p oblem de¯ned by º.ByVnwe
deno e he se o posi i e Bo el measu es on Rsuch ha RR kd¹( )=sk,0·k·n, ha
is, he se o solu ions o he unca ed momen p oblem de¯ned by º.
We say ha he measu e ºis de e mina e i he e is no o he posi i e measu e ha ing he
same momen s as hose o º, ha is,i V= ºg, o he wise we say ha ºis inde e mina e.
This al e na i e is ela ed o he index o de¯ciency o he ope a o de¯ned on `2by he
in¯ni e Jacobi ma ix
J=0
B
B
@
b0a1
a1b1a2
a2b2a3
.........
1
C
C
A;
whe e he coe±cien s ai(6=0)andbia e he coe±cien s which appea in he h ee e m
ecu ence ela ion sa is¯ed by he o hogonal polynomials (pn)nassocia ed o º,
pn( )=an+1pn+1( )+bnpn( )+anpn¡1( );n¸0:
The index o de¯ciency o Jis 0 i he momen p oblem is de e mina e and 1 i he momen
p oblem is inde e mina e.
This wo k has been pa ially suppo ed by DGICYT e . PB-96-1321-CO2.
This wo k was pa ly ca ied ou a he Ma hema ics Ins i u e o he Uni e si y o Copenhagen unde a
g an o he spanish Minis y o Educa ion and Science o he P og ama de becas de o maci¶on de pe sonal
in es igado en el ex anje o.
1991 Ma hema ics Subjec Classi¯ca ion. 42C05, 44A60.
Typese by A
M
S-T
EX
1
2 P. LOPEZ-RODRIGUEZ
In 1922 Ne anlinna p o ed ha o a ¯xed non eal ¸ he image h ough he S iel jes
ans o m o all he measu es o Vin he poin ¸
I(V)(¸)=½ZR
d¹( )
¡¸:¹2V¾
is ei he a poin i he momen p oblem is de e mina e o a ci cle i he momen p oblem
is inde e mina e, and he occu ence o hese wo cases does no depend on he non eal
¸chosen (see [N]o [A]). The measu es ¹ o which I(¹)(¸) lies in he ci cum e ence o
his ci cle I(V)(¸) a e called N-ex emal (Ne anlinna-ex emal).
In 1923 M. Riesz p o ed ha in o de ha he se Po polynomials in L2(¹)bedense,
i is necessa y and su±cien ha he measu e ¹be N-ex emal a e e y non eal poin ¸,
and o his i is su±cien ha i should be N-ex emal in a leas one such poin (see [Ri]
o [A]).
The pu pose o his pape is o gene alize hese wo esul s o a comple ely inde e mi-
na e ma ix momen p oblem.
Gi en º=(ºi;j)1·i;j·Na posi i e de¯ni e ma ix o measu es ( o any Bo el se A he
nume ical ma ix º(A) is posi i e semide¯ni e) wi h ¯ni e ma ix momen s
Sk=ZR
kdº( )
o any o de k¸0, we deno e by V hese o posi i ede¯ni ema iceso measu esha ing
he same ma ix momen s as hose o º,andbyVn he se o posi i e ma ices o measu es
whose momen s up o deg ee na e he same as hose o º.
We say ha he posi i e de¯ni e ma ix o measu es ºis de e mina e i no o he posi i e
de¯ni e ma ix o measu es has he same momen s as hose o º, ha is, he posi i e de¯ni e
ma ix o measu es ºis uniquely de e mined by he momen s RR ndº( ), n¸0.
By (Pn)1
n=0 we deno e he sequence o o hono mal ma ix polynomials wi h espec o
º,Pno deg ee nand wi h non-singula leading coe±cien .
These polynomials (Pn)nsa is y a h ee e m ecu ence ela ion o he o m
(1.1) Pn( )=An+1Pn+1( )+BnPn( )+A¤
nPn¡1( );n¸0;
(Anand Bnbeing N£Nma ices such ha de (An)6= 0 and B¤
n=Bn), wi h ini ial
condi ion P¡1( )=µ(he e and in he es o his pape , we w i e µ o he null ma ix,
he dimension o which can be de e mined om he con ex . Fo ins ance, he e µis he
N£Nnull ma ix). I is well-known ha his ecu ence ela ion is equi alen o he
o hogonali y wi h espec o a posi i e de¯ni e ma ix o measu es: his is he ma ix
e sion o Fa a d's Theo em (see [AN], [D1]and[DL1]).
We deno e by Qn( ) he co esponding sequence o polynomials o he second kind,
Qn( )=ZR
Pn( )¡Pn(x)
¡xdº(x);n¸0;
which also sa is y he ecu ence ela ion (1.1), wi h ini ial condi ions Q0( )=µand
Q1( )=A¡1
1.
RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS 3
In he ma ix case he de e minacy o inde e minacy o he ma ix momen p oblem is
also ela ed o he index o de¯ciency o he ope a o Jde¯ned by he in¯ni e N-Jacobi
ma ix
J=0
B
B
@
B0A1
A¤
1B1A2
A¤
2B2A3
.........
1
C
C
A
on he space `2,whe eAnand Bna e he coe±cien s which appea in he h ee e m
ecu ence ela ion (1.1). In his case he index o de¯ciency can be any na u al numbe
om 0 o N, 0 in he de e mina e case and Nin he comple ely inde e mina e case. In
he la e case he wo se ies
1
X
k=0
Q¤
k(¸)Pk(´)and
1
X
k=0
P¤
k(¸)Pk(´)
con e ge uni o mly in he a iables ¸and ´on e e y bounded se o he complex plane
(see [K]).
In [B]i isp o ed ha he anko helimi ma ixR(¸)= lim
n!1 Rn(¸) exis s and is
hesame o e e ynon eal¸,whe e
(1.2) Rn(¸)=Ãn
X
k=0
P¤
k(¸)Pk(¸)!¡1
:
This esul is also men ioned by K ein in [K], who e e s o [Na] o a p oo . In his pape
we assume he ank o his ma ix is Nand consequen ly he ma ix
1
X
k=0
P¤
k(¸)Pk(¸)is
in e ible, o e e y non eal ¸, and equal o R(¸)¡1.
As in he scala case, o a ¯xed non eal ¸, we deno e by I(V)(¸) he image h ough
he S iel jes ans o m o all he ma ices o measu es o Vin he poin ¸
I(V)(¸)=½ZR
d¹( )
¡¸:¹2V¾:
Fi s ly, o a ¯xed non eal ¸, we desc ibe he se I(V)(¸). This is he se o N£N
complex ma ices !sa is ying he ma ix inequali y
(1.3) [!+C(¸)]R(¸)¡1[!+C(¸)]¤·j¸¡¸j¡2R(¸);
whe e C(¸)=B(¸; ¸)D(¸; ¸)¡1(see he P elimina ies o he de¯ni ions o B(¸; ¸)and
D(¸; ¸). A·Bmeans ha B¡Ais posi i e semide¯ni e.
The ex emal poin s (in he sense o con exi y) o he se I(V)(¸) a e he ma ices ! o
which equali y is a ained in (1.3). I ¹is a ma ix o measu es in V o which I(¹)(¸)is
4 P. LOPEZ-RODRIGUEZ
aex emalpoin o I(V)(¸), we call his ma ix o measu es N-ex emal, as in he scala
case.
Finally, we gene alize Riesz's heo em o he ma ix se ing by p o ing ha he ma ices
o measu es o V o which he se Po ma ix polynomials is dense in he co esponding
space L2(¹) a e p ecisely he N-ex emal ma ices o measu es, and ha he N-ex emali y
o a ma ix o measu es does no depend on he non eal ¸chosen.
Fo he case N= 1 one eco e s he e y well known classical o mulas exposed in [A,
Ch. 1].
2. P elimina ies.
In wha ollows, i P(¸) is a ma ix polynomial, we deno e by P¤(¸) he polynomial
ob ained om P(¸) by eplacing each o i s ma ix coe±cien s by i s he mi ian conjuga e,
so ha P(¸)¤=P¤(¸). Fo a ma ix polynomial o wo a iables P(¸; ´) he de¯ni ion is
he same, so ha we ha e P(¸; ´)¤=P¤(¸; ´). I F(¸) is a holomo phic ma ix unc ion
we de¯ne F¤(¸)=F(¸)¤.
The se o posi i e de¯ni e ma ices o measu es is endowed wi h he ague and weak
opologies. I ºa posi i e de¯ni e ma ix o measu es, he se Vo ma ices o measu es
ha ing he same momen s as ºis a compac and con ex se o hese opologies which
coincide on V(see [DL2]).
Fo ¹a posi i e de¯ni e ma ix o measu es, he space L2(¹)isde¯nedas hese o
N£Nma ix unc ions :R!MN£N(C) such ha ¿( ( )M( ) ( )¤)2L1(¿¹), whe e
M( ) is he Radon-Nikodym de i a i e o ¹wi h espec o i s ace (¿¹)( o ama ix
A=(ai;j)1·i;j·N, we deno e ¿A o i s ace, i. e. ¿A =PN
i=1 ai;i):
M=(mi;j)N
i;j=1 =µd¹i;j
d¿¹ ¶1·i;j·N
:
The space L2(¹) is endowed wi h he no m
k k2;¹ =k¿( ( )M( ) ( )¤)1
2k2;¿¹ =µZR
¿( ( )M( ) ( )¤)d¿¹( )¶1
2
and is a Hilbe space. The duali y wo ks as o he scala case (see [R]o [DL2] o mo e
de ails. Fo he de¯ni ion o he Lpspaces associa ed o ¹,1·p<1,see[DL2]).
We s ess ha since we only impose he ma ices o measu es in V2n o ha e ¯ni e
momen s up o deg ee 2n, o ¹2V2nwe can gua an ee only ha he polynomials up o
deg ee nbelong o he co esponding space L2(¹). In any case, he polynomials (Pk)k=0;:::;n
a e o hono mal wi h espec o any measu e in V2n.
We include he e he ma ix e sion o some classical o mulas o o hono mal scala
polynomials. The p oo s a e easily e i¯ed using he h ee e m ecu ence ela ion (1.1).
(2.1)
An(u; )=( ¡u)
n¡1
X
k=0
Q¤
k(u)Qk( )=Q¤
n¡1(u)AnQn( )¡Q¤
n(u)A¤
nQn¡1( ); o u; 2C;
RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS 5
(2.2)
Bn(u; )=¡I+( ¡u)
n¡1
X
k=0
Q¤
k(u)Pk( )=Q¤
n¡1(u)AnPn( )¡Q¤
n(u)A¤
nPn¡1( ); o u; 2C;
( his is G een's o mula),
(2.3)
Cn(u; )=I+( ¡u)
n¡1
X
k=0
P¤
k(u)Qk( )=P¤
n¡1(u)AnQn( )¡P¤
n(u)A¤
nQn¡1( ); o u; 2C;
(2.4)
Dn(u; )=( ¡u)
n¡1
X
k=0
P¤
k(u)Pk( )=P¤
n¡1(u)AnPn( )¡P¤
n(u)A¤
nPn¡1( ); o u; 2C;
( his a Ch is o®el-Da boux o mula). We will also use he Liou ille-Os og adsky o mula
(2.5) Qn(¸)P¤
n¡1(¸)¡Pn(¸)Q¤
n¡1(¸)=A¡1
n; o ¸2C;
and he ela ions
(2.6) Pn(¸)Q¤
n(¸)=Qn(¸)P¤
n(¸); o ¸2C;
(2.7) An(u; )D¤
n(u; )¡B
n(u; )C¤
n(u; )=I; o u; 2C;
and ¯nally
(2.8) Cn(u; )D¤
n(u; )=Dn(u; )C¤
n(u; ); o u; 2C:
We will also use ha
(2.9) Cn(¸; ¸)=¡Bn(¸; ¸)¤; o ¸2C;
and ha
(2.10) Dn(¸; ¸)=(¸¡¸)Rn¡1(¸)¡1and D¤
n(¸; ¸)=(¸¡¸)Rn¡1(¸)¡1; o ¸2C:
By A(u; ), B(u; ), C(u; )andD(u; ) we deno e he limi ma ix unc ions de¯ned om
An(u; ), Bn(u; ), Cn(u; )andDn(u; )whenn ends o in¯ni y.
3. The main heo ems.
Fo any non eal ¸,wede¯ne hese Bn(¸) obe hese o N£Ncomplex ma ices !
such ha

6 P. LOPEZ-RODRIGUEZ
(3.1) [!+Cn(¸)]Rn¡1(¸)¡1[!+Cn(¸)]¤·j¸¡¸j¡2Rn¡1(¸);
whe e Cn(¸)=Bn(¸; ¸)Dn(¸; ¸)¡1.
The calcula ions in Lemma 1 (see below) show ha Bn(¸)isalso hese o N£N
complex ma ices !sa is ying he ma ix inequali y
(3.2)
n¡1
X
k=0
(Q¤
k(¸)+!P¤
k(¸))(Qk(¸)+Pk(¸)!¤)·!¡!¤
¸¡¸:
We pu B1(¸) o he in e sec ion o all he se s Bn(¸). B1(¸) is clea ly he se o
N£Ncomplex ma ices !such ha
(3.3) [!+C(¸)]R(¸)¡1[!+C(¸)]¤·j¸¡¸j¡2R(¸);
whe e C(¸)=B(¸; ¸)D(¸; ¸)¡1.
Simila ly, B1(¸)isalso hese o N£Ncomplex ma ices !such ha
(3.4)
1
X
k=0
(Q¤
k(¸)+!P¤
k(¸))(Qk(¸)+Pk(¸)!¤)·!¡!¤
¸¡¸:
Looking a (3.1) and (3.3) i is immedia e ha upon a linea ma ix ans o ma ion,
any o he se s Bn(¸)o B1(¸) is in a one o one co espondence wi h he se o N£N
complex ma ices Tsa is ying TT¤·I, which is a con ex se whose ex emal poin s a e
he ma ices e i ying TT¤=I, ha is, he uni a y ma ices ( his is a well-known esul
in ope a o heo y which can be p o ed o example wi h he aid o he singula alue
decomposi ion o ma ices). This implies ha hese se s Bn(¸)andB1(¸) a e con ex se s
whose ex emal poin s (Ex Bn(¸) and Ex B1(¸)) a e hose o which equali y is a ained
in (3.1) and (3.2) o (3.3) and (3.4) espec i ely.
By using o mulas (2.1), (2.2), (2.3) and (2.4) in (3.2) i is s aigh o wa d o see ha
an equi alen condi ion o ! o be an ex emal poin o Bn(¸) is ha he ma ix
(3.5) (!P¤
n(¸)+Q¤
n(¸))A¤
n(Pn¡1(¸)!¤+Qn¡1(¸))
is he mi ian.
I is clea ha o all n¸1weha eB1(¸)µBn+1(¸)µBn(¸). I is also clea ha
!belongs o he se o in e io poin s o Bn(¸)o B1(¸)(In Bn(¸)andIn B1(¸)) i a
s ic inequali y is a ained in (3.1) and (3.2) o (3.3) and (3.4) espec i ely.
We ha e he ollowing esul s:
Theo em 1. Le Vdeno e he se o solu ions o a comple ely inde e mina e ma ix
momen p oblem de¯ned by a na ix o measu es ºand le ¸2CnR.Thenweha e
B1(¸)=I(V)(¸):
RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS 7
The key o p o e his heo em will be he inclusions
(3.6) In Bn(¸)µI(V2n¡2)(¸)µBn(¸):
The p oo s o hese inclusions p esen mo e di±cul ies han in he scala case. We will
p o e hen la e in Lemmas 1 o 7. Indeed, in he scala case he se I(V2n¡2)(¸)isgi en
by
I(V2n¡2)(¸)=Bn(¸)n½¡qn¡1(¸)
pn¡1(¸)¾:
The poin ¡qn¡1(¸)=pn¡1(¸) lies on he bo de o he ci cle Bn(¸). When amo es along
he eal axis, he quo ien
¡qn(¸)¡aqn¡1(¸)
pn(¸)¡apn¡1(¸)
desc ibes all he poin s o he ci cum e ence o he closed disk Bn(¸) excep o he limi
poin ¡qn¡1(¸)=pn¡1(¸). The well known quad a u e o mula (see [A,p. 20])gi es ha
e e y poin de¯ned by he o me quo ien o a2Rbelongs o I(V2n¡2)(¸). I is easy
o see ha ¡qn¡1(¸)=pn¡1(¸)=2I(V2n¡2)(¸), bu his is o no impo ance because aking
in o accoun ha I(V2n¡2)(¸) is a con ex se and he simple geome y o he ci cles Bn(¸)
i is immedia e o deduce ha In Bn(¸)µI(V2n¡2)(¸). This inclusion is no a all so
immedia e in he ma ix case. We p o e i in Lemmas 2 o 7 by means o new ideas.
P oo o Theo em 1
Suppose ¯ s ha !2B1(¸). B1(¸)iscon ainedinBn(¸) o all nand hus we can pu
!=!n, o all n,being!nin Bn(¸). Since he in e io se In Bn(¸)isdenseinBn(¸), we
can ¯nd ´nin In Bn(¸)such ha lim
n!1 k!n¡´nk=0. SinceIn Bn(¸)µI(V2n¡2)(¸),
he e exis s a ma ix o measu es ¾nin V2n¡2such ha ´n=I(¾n)(¸), o all n.Since he
se ¹¸µ:¿¹(R)·cgis aguely compac , whe e cis a posi i e cons an (see Lemma
3.8 in [DL2]) and ¾n(R)=S0 o n¸0, he e exis s ¾a ague accumula ion poin o
asubsequence(¾np)o (¾n). Like in he p oo o Lemma 3.10 o [DL2]weha e¾2V.
We ha e ¾n(R)=¾(R) o n¸1, so by i ue o Theo em 3.1 o [DL2], (¾n)con e ges
weakly o ¾.Inpa icula
I(¾)(¸) = lim
p!1 I(¾np)(¸) = lim
p!1 ´np=lim
p!1 !np=!
andweha ep o ed ha B1(¸)µI(V)(¸).
Since I(V2n¡2)(¸)µBn(¸), o e e y n, he e e se inclusion is clea .
¥
Theo em 2. (Riesz's heo em o o hogonal ma ix polynomials) Le ¹be a posi i e
de¯ni e ma ix o measu es co esponding o a comple ely inde e mina e ma ix momen
p oblem. Then he ollowing condi ions a e equi alen :
(1) The e exis s ¸02CnRsuch ha I(¹)(¸0)is an ex emal poin (in he sense o
con exi y) o he se B1(¸0).
(2) Fo any ¸2CnR,I(¹)(¸)is an ex emal poin (in he sense o con exi y) o he
se B1(¸)
(3) Pis dense in L2(¹),equi alen ly(Pn( ))1
n=0 is an o hono mal basis o he Hilbe
space L2(¹).
8 P. LOPEZ-RODRIGUEZ
P oo o Theo em 2
(3) )(2) The polynomials a e dense in L2(¹) i o any unc ion in he space L2(¹)
we ha e equali y in Bessel's inequali y, which is equi alen o
(3.7)
1
X
k=0
( ;Pk)( ;Pk)¤=ZR
( )M( ) ¤( )d¿¹( ):
In pa icula , o ¸( )= I
¡¸2L
2(¹)weha e
( ¸;P
k)=ZR
I
¡¸d¹( )P¤
k( )
=ZR
d¹( )P¤
k( )¡P¤
k(¸)
( ¡¸)+ZR
d¹( )
¡¸P¤
k(¸)
=Q¤
k(¸)+I(¹)(¸)P¤
k(¸);
being I(¹)(¸)=ZR
d¹( )
¡¸ he S iel jes ans o m o ¹in he poin ¸,and
ZR
( )d¹( ) ¤( )=ZR
d¹( )
j ¡¸j2=I(¹)(¸)¡I(¹)(¸)¤
¸¡¸=ImI(¹)(¸)
Im¸;
so equali y in (3.7) is
1
X
k=0
(Q¤
k(¸)+!(¸)P¤
k(¸))(Qk(¸)+Pk(¸)!¤(¸)) = ImI(¹)(¸)
Im¸;
ha is I(¹)(¸)2Ex B1(¸).
(2) )(1) is ob ious.
(1) )(3) We suppose (1) holds and we claim n
¸0=I
( ¡¸0)n2 P, o n¸1. The
asse ion o n= 1 is he assump ion. We now p o e ha n+1
¸02 P unde he assump ion
n
¸02 P, so ha he claim is es ablished by induc ion. Fo gi en ²>0, he e exis s a
ma ix polynomial P2Psuch ha k n
¸0¡Pk2·²jIm¸0j. Di iding Pby (x¡¸0)Iwe
ge P(x)=(x¡¸0)Q(x)+A,wi hQano he polynomial o deg ee n¡1andAaN£N
complex ma ix. We ha e
k n+1
¸0¡A ¸0¡Qk2
2=
=¿ZRµI
( ¡¸0)n+1 ¡A
¡¸0
¡Q( )¶M( )µI
( ¡¸0)n+1 ¡A
¡¸0
¡Q( )¶¤
d¿¹( )
=¿ZR
1
j ¡¸0j2µI
( ¡¸0)n¡A¡Q( )( ¡¸0)¶M( )µI
( ¡¸0)n¡A¡Q( )( ¡¸0)¶¤
d¿¹( )
RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS 9
·1
jIm¸0j2kI
( ¡¸0)n¡P( )k2
2·²2;
and since A
¡¸0
belongs o he closu e o Pwe deduce ha I
( ¡¸0)n+1 also does.
Now, i 2L
2(¹) is o hogonal o Pand we conside he S iel jes ans o m
I( ¹)(z)=ZR
( )
¡zd¹( );z2CnR
using ha n
¸0; n
¸02 P o n¸1, we see ha
I( ¹)(n)(z)=µ; o z=¸0; ¸0;and n¸0;
bu hen we ha e an analy ic unc ion I( ¹)inadomainDsuch ha I( ¹)(n)(z0)=µ
o n¸0andace ainz02D.SoI( ¹)isequal oµin D.Weconclude ha I( ¹)
is iden ically ze o in each o he wo hal planes CnR,and hen =µ,¹a.e., hence he
polynomials a e dense in L2(¹).
¥
We also ha e he ollowing Theo em:
Theo em 3. I ¹2V2n¡2is such ha I(¹)(¸)2Ex Bn(¸), henPn¡1=L2(¹)
P oo o Theo em 3
The hypo hesis means ha equali y is a ained in (3.2), ha is, he unc ion ¸( )= I
¡¸
can be app oxima ed by ma ix polynomials up o deg ee n¡1. Now he p oo ¯nishes
exac ly in he same way as he p oo o Theo em 2.
¥
4. P oo s o he inclusions.
In his las sec ion we s udy in de ail he se Bn(¸) and o he ela ed se s, wi h he
pu pose o p o ing he inclusions (3.6). We ema k ha hese inclusions a e alid wi hou
supposing he ma ix momen p oblem o be comple ely inde e mina e. We ¯ s p o e he
second inclusion o (3.6)
Lemma 1.
I(V2n¡2)(¸)µBn(¸)
P oo
Le 's suppose ¹is a measu e in V2n¡2. We know ha he ¯ s no hono mal ma ix
polynomials P0;:::;P
n¡1 o m an o hono mal sys em in he space L2(¹). A e he abo e
calcula ions, om Bessel's inequali y o he unc ion ¸( )= I
¡¸,wededuce ha
n¡1
X
k=0
(Q¤
k(¸)+I(¹)(¸)P¤
k(¸))(Qk(¸)+Pk(¸)I(¹)(¸)¤)·ImI(¹)(¸)
Im¸:
16 P. LOPEZ-RODRIGUEZ
Since ibelongs o Ke (!P¤
n¡1(¸)+Q¤
n¡1(¸)), and using ha Tis he mi ian, we ha e ha
o any m+1·i·Nand o any 1 ·j·m,
i(!P¤
n(¸)+Q¤
n(¸))A¤
nu¤
j= i(!P¤
n(¸)+Q¤
n(¸))A¤
n(Pn¡1(¸)!¤+Qn¡1(¸) ¤
j
= i(!P¤
n¡1(¸)+Q¤
n¡1(¸))An(Pn(¸)!¤+Qn(¸)) ¤
j=µ:
This means ha i(!P¤
n(¸)+Q¤
n(¸))A¤
nbelongs o Im(!P¤
n¡1(¸)+Q¤
n¡1(¸))?, and con-
sequen ly (4.9) holds i we p o e ha o any ec o uin Im(!P¤
n¡1(¸)+Q¤
n¡1(¸))?we
ha e
lim
p!1 u(AnPn(¸)Pn¡1(¸)¡1¡Hp)¡1P¤
n¡1(¸)¡1=µ:
Since P¤
n¡1(¸)isanin e iblema ixand um+1;:::;u
Ngis a basis o Im(!P¤
n¡1(¸)+
Q¤
n¡1(¸))?i is enough o p o e ha o m+1·i·Nwe ha e
lim
p!1 ui[AnPn(¸)Pn¡1(¸)¡1¡Hp]¡1=µ:
Obse e ha
ui[AnPn(¸)Pn¡1(¸)¡1¡Hp]¡1
=ui[AnPn(¸)Pn¡1(¸)¡1¡C¤MpC]¡1
=uiC¤[CAnPn(¸)Pn¡1(¸)¡1C¤¡Mp]¡1C
=ei[CAnPn(¸)Pn¡1(¸)¡1C¤¡Mp]¡1C;
whe e ei=(0;:::;1;:::;0), being he 1 in he posi ion i.
The ma ix CAnPn(¸)Pn¡1(¸)¡1C¤¡Mpis o he o m
0
B
B
B
B
B
B
B
B
@
®1;1::: ®
1;m ®1;m+1 ::: ®
1;N
.
.
.....
.
..
.
.....
.
.
®m;1::: ®
m;m ®m;m+1 ::: ®
m;N
®m+1;1::: ®
m+1;m ®m+1;m+1 ¡p::: ®
m+1;N
.
.
.....
.
..
.
.....
.
.
®N;1::: ®
N;m ®N;m+1 ::: ®
N;N ¡p
1
C
C
C
C
C
C
C
C
A
:
The de e minan o his ma ix is a polynomial in he a iable po deg ee N¡m,whe eas
he p incipal mino s Ai;j o io jbigge han ma e polynomials in he a iable po
deg ee N¡m¡1. Fo his eason, any en y Ei;j o [CAnPn(¸)Pn¡1(¸)¡1C¤¡Mp]¡1
wi h io jbigge han m ends o 0 when p ends o in¯ni y, and consequen ly
lim
p!1 ei[CAnPn(¸)Pn¡1(¸)¡1C¤¡Mp]¡1=µ; o m+1·i·N
which p o es he esul .
¥
We ¯nish by p o ing ha he con ex hull o he se ¡n(¸)con ains hese In Bn(¸)o
in e io poin s o Bn(¸).

RIESZ'S THEOREM FOR ORTHOGONAL MATRIX POLYNOMIALS 17
Lemma 7.
In Bn(¸)µco(¡n(¸))
P oo
Since ¡n(¸)isdenseinEx Bn(¸), ha is we ha e ¡n(¸)=Ex Bn(¸), we deduce ha
co ¡n(¸)=co¡n(¸)=co(Ex Bn(¸)) = Bn(¸)
he las equali y by i ue o K ein-Millman's heo em. Now, applying well-known a gu-
men s o con exi y we ha e ha
In (co(¡n(¸)) = co ¡n(¸)=Bn(¸):
As we ha e p e iously men ioned, he e exis s an in e ible linea ope a o Lde¯ned on he
N£Ncomplex ma ices ans o ming Bn(¸) bijec i ely on o he se B= T:TT¤·Ig.
I is immedia e ha he image se L(¡n(¸)) is dense in Ex B= T:TT¤=Ig,and ha
In (co(L(¡n(¸))) = B. To p o e ha In Bn(¸)µco(¡n(¸)) i is enough o p o e ha
In Bµco(L(¡n(¸))).
Fo his, i he e exis s x2In BnIn (co(L(¡n(¸))), we can sepa a e xand In (co(L(¡n(¸)))
wi h a linea ope a o ¤ such ha ¤(x)=1and¤(z)·1 o any zin In (co(L(¡n(¸))).
Since In (co(L(¡n(¸))) = Bwe ha e ha ¤(z)·1 o any zin Band hus k¤k·1, bu
his is in con adic ion wi h ¤(x)=1becausexin an in e io poin o B.
¥
Acknowledgemen s
The au ho exp esses his g a i ude o P o esso An onio J. Du ¶an o p oposing he
p oblem and o help ul sugges ions o he ¯nal d a , and o P o esso Luis R. Piazza o
ui ul discussions abou ques ions o con exi y.
Re e ences
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Ped o L¶
opez Rod ¶
³guez, Depa amen o de An¶
alisis Ma em¶
a ico, Uni e sidad de Se illa,
Apdo. 1160. 41080-Se illa, Spain. E-mail: plo[email p o ec ed]