A mini-max problem for self-adjoint Toeplitz matrices

0.19 MB·English

by  Dennis Courtney

A MINI-MAX PROBLEM FOR SELF-ADJOINT TOEPLITZ
MATRICES
1
1 DENNISCOURTNEY∗ ANDDONALDSARASON
0
2
Abstract. Westudyaminimumproblemandassociatedmaximumproblem
n forfinite,complex,self-adjointToeplitzmatrices. IfAissuchamatrix,ofsize
a (N+1)-by-(N+1),weidentifyAwiththeoperatoritrepresentsonPN,the
J space of complex polynomials of degrees at most N, with the usual Hilbert
8 spacestructureitinheritsasasubspaceofL2 oftheunitcircle. Theoperator
1 A is the compression to PN of the multiplication operator on L2 induced by
any function inL∞ whose Fourier coefficients of indices between −N and N
] matchthematrixentriesofA. OurminimumproblemistominimizetheL∞
A normofsuchinducers. Weshowthereisauniqueoneofminimumnorm,and
wedescribeit. Theassociatedmaximumproblemasksforthemaximumofthe
F
ratiooftheprecedingminimumtotheoperatornorm. Thatproblemremains
.
h largelyopen. Wepresentsomesuggestivenumericalevidence.
t
a
m
[
1. Introduction

2
v To begin we consider an (N +1)-by-(N +1) complex Toeplitz matrix A (N a
3 positive integer):
5 A=(a )N .
j k j,k=0
5 −
2 By L2 we shall mean the L2 space of normalized Lebesgue measure on the unit
. circle, T. The subspace of L2 consisting of the complex polynomials of degrees at
9
mostN willbe denotedby . We identify Awith the operatorit induces on ,
0 N N
P P
0 i.e., the operator on N whose matrix with respect to the monomial basis is A.
P
1 The operator A is the compression to of the multiplication operator on L2
N
v: induced by any function f in L∞ of thePcircle whose Fourier coefficients indexed
i between N and N match the matrix elements of A, in other words, for which
X −
f(n)=a for n= N, N +1,...,0,...,N 1,N. The family of such functions
n
r − − −
a f will be denoted by A. We pose the problem of minimizing f over the class
G k k∞
b . Standard reasoning shows the minimum is attained; we denote it by c . It is
A A
G
evident that A c .
A
k k≤
In the cases where A is lower triangular or upper triangular, our problem fits
into the classical interpolation problem of C. Carath´eodory and L. Fej´er, dating
back 100 years [2]. Perhaps surprisingly, the more general problem seems not to
have been considered until recently [1], [6], [7]. Although we will not do so here, it
can be treated as a special case of a general dual extremal problem studied by M.
G. Krein and A. A. Nudelman in [5] — more details later.
2000MathematicsSubjectClassification. 15B05,42A70,30E05,30H10,47A20,47A57,47B35.
Key words and phrases. Toeplitzmatrix,Toeplitzoperator, trigonometricmomentproblems,
Hardyspaces.
∗PartiallysupportedbytheUniversityofIowaDepartmentofMathematicsNSFVIGREgrant
DMS-0602242.
1
2 DENNISCOURTNEYANDDONALDSARASON
In this paper wefocus onthe casewhere Ais self-adjoint. Forthis case weshow
that the minimizing function in is unique, and we describe it: it alternates the
A
valuesc and c onafamilyofGsubarcsthatpartitionT,areeveninnumber,and
A A
−
number at most 2N.
If the matrix A is self-adjoint but not diagonal, then the minimizing function is
not a rational function and so multiplies every nonzero function in outside of
N
P
. Itfollowsthatc > A . Let denotetheclassof(N+1)-by-(N+1)self-
N A N+1
P k k T
adjoint Toeplitz matrices. We pose the problem of maximizing the ratio c / A
A
k k
over the nonzero matrices A in . As in the minimum problem, standard
N+1
T
reasoning shows that the maximum is attained; we denote it by c .
N
It is proved in [7] that c = π/2. The other cases remain open. We have
1
numericalevidence,presentedinSection7,thatc >π/2forN >1,andsuggesting
N
thatc increaseswithN. We areabletoprovethattheprecedinginequalityholds
N
for infinitely many N.
It is known, and nontrivial, that the numbers c have a common bound.
N
M. Bakonyi and D. Timotin obtain in [1] the bound c 2. The interpolation
N
≤
problem they consider is slightly different from ours, but their reasoningapplies in
ourcase. L.N.NikolskayaandYu.B.Farforovskayaobtainin[6]theboundc 3,
N
≤
andtheirreasoningapplieseveninthenon-self-adjointcase. Inbothpapers[1]and
[6], the inequality obtained is an outcome of a matrix extension problem.
In Section 2 we state what we shall need concerning the Carath´eodory–Fej´er
interpolation problem. Section 3 recasts our minimum problem in a form that
fits into the Carath´eodory–Fej´er framework. The minimizing functions for our
minimum problemareidentified in Section4 andstudied further inSection 5. The
inequality c π/2 is reprovedin Section 6 in a way that exploits the results from
1
≤
Sections4and5. TheconcludingSection8containsopenquestionsandconjectures.
The maximum part of our mini-max problem remains largely mysterious, while
strongly enticing. New ideas are needed.
We are indebted to an anonymous referee for alerting us to the relevance to our
study of references [4] and [5].
2. Carath´eodory–Fej´er Interpolation Problem
The Carath´eodory–Fej´er problem asks whether there is a holomorphic function
in the unit disk D having prescribed power series coefficients of orders 0,1,...,N
andhavingaprescribedsupremumnorm. Carath´eodoryandFej´er’ssolution,recast
in our context and in our notation, can be stated as follows:
Theorem CF. If A is lower triangular then c = A , attained by a unique
A
k k
function in , a Blaschke product of order at most N multiplied by A .
A
G k k
This is actually a mild generalization of what Carath´eodory and Fej´er prove,
because their minimization problem is over H functions, not L functions. But
∞ ∞
once one knows their result, one easily generalizes it to the one stated above. The-
orem CF forms the basis of the analysis to follow.
3. Reformulation of the Minimum Problem
Let A be a matrix in the class , and let the function f belong to . Then
N+1 A
T G
Ref is also in and Ref f . Hence c , the minimum of f over
A A
G k k∞ ≤ k k∞ k k∞
the class , equals the minimum of f over the class = f :f =f .
GA k k∞ GA∗ { ∈GA }
A MINI-MAX PROBLEM FOR SELF-ADJOINT TOEPLITZ MATRICES 3
We let P denote the orthogonal projection of L2 onto the Hardy space H2 =
+
h L2 : h(n) = 0 for n < 0 . As is customary, we identify the functions in H2
{with∈their holomorphic extens}ions in D.
Supposebf is a function in the class . Then the function h=P f a /2 is in
GA∗ + − 0
H2, and f =h+h=2Reh. We have
h(z)=a /2+a z+ +a zN +O(zN+1) (z 0).
0 1 N
··· −→
Let denote the class of functions h in H2 that satisfy the preceding condition
A
H
and for which Reh is in L . Then 2Reh is in for such an h. Accordingly, our
∞ GA∗
minimum problem can be restated as the problem of minimizing 2 Reh over
k k∞
the class .
A
H
4. The Minimum Problem
Let A be a matrix in . To eliminate a trivial case, we assume here that
N+1
T
A is not diagonal. Suppose h is a minimizing function in , i.e., a function in
A
H
satisfying 2 Reh =c . To simplify the notation we temporarily write c in
A A
H k k∞
place of c below.
A
We introduce the domain
c c
S = z C: <Rez < ,
c
∈ −2 2
n o
an open vertical strip in the plane bisected by the imaginary axis, of width c. The
domains D and S are conformally equivalent under the map h : D S given
c c c
−→
by
c 1+z
h (z)= Log ;
c
πi 1 z
(cid:18) − (cid:19)
here, Log denotes the principal branch of log. The inverse map h 1 : S D is
−c c −→
given by
πz
(h 1)(z)=itan .
−c 2c
The composite function (cid:16) (cid:17)
πh
b=h 1 h=itan
−c ◦ 2c
(cid:18) (cid:19)
is a self-map of D. We write its power series about 0 as
b(z)=b +b z+ +b zN +O(zN+1).
0 1 N
···
Therelationbetweenthecoefficientsa ,...,a ofhandb ,...,b ofbisrather
0 N 0 N
complicated. Toillustrate,weexhibitbelowtheexpressionsforb ,b ,b ,b interms
0 1 2 3
of a ,a ,a ,a .
0 1 2 3
πa
0
b =itan ,
0
4c
πia (cid:16) (cid:17)πa
b = 1 sec2 0 ,
1
2c 4c
πia (cid:16)πa (cid:17) π2ia πa πa
b = 2 sec2 0 + 1 sec2 0 tan 0 ,
2 2c 4c 22c2 4c 4c
πia (cid:16)πa (cid:17) π2a a (cid:16) πa(cid:17) (cid:16) πa(cid:17)
b = 3 sec2 0 + 1 2 sec2 0 tan 0
3 2c 4c 22c2 4c 4c
π3ia3(cid:16) (cid:17) πa (cid:16)πa (cid:17) (cid:16) πa(cid:17)
+ 1 2sec2 0 tan2 0 +sec4 0 .
3 23 c3 4c 4c 4c
· · h (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)i
4 DENNISCOURTNEYANDDONALDSARASON
However, for our purpose these complications are not an obstacle. From the rule
for composing formal power series one knows that, for 0 n N, the coefficient
≤ ≤
b depends only on c,a ,...,a , and vice versa. That is all we shall need.
n 0 n
LetB denotethe(N+1)-by-(N+1)lowertriangularToeplitzmatrixwithentries
b ,b ,...,b in the first column. By Theorem CF, there is a unique Blaschke
0 1 N
product ω of order N such that
≤
B ω(z)=b +b z+ +b zN +O(zN+1).
0 1 N
k k ···
We then have
a
((h 1) ( B ω))= 0 +a z+ +a zN +O(zN+1).
−c ◦ k k 2 1 ··· N
Butinfact B =1,fortheinequality B <1wouldimplythattwicetherealpart
k k k k
of the preceding composite function has L -norm less than c = c , contradicting
∞ A
the assumed minimizing property of h. Hence
c 1+ω
A
h=h ω = Log .
c
◦ πi 1 ω
(cid:18) − (cid:19)
Definition 4.1. For c > 0 and n a positive integer, an alternating step function
of height c and order n is a function in L that assumes alternatively the values c
∞
and c on 2n subarcs that form a partition of T.
−
For example, the function
1, 0<θ <π,
ψ(eiθ)=
( 1, π <θ <0,
− −
is an alternating step function of height 1 and order 1.
Forthefunctionh=h ωabove,wehave,onT(withArgdenotingtheprincipal
c
◦
value of arg),
2c 1+ω 2c
A A
2Reh= Arg = Arg((1+ω)(1 ω))
π 1 ω π −
(cid:18) − (cid:19)
2c 2c
A A
= Arg(ω ω)= Arg(iImω)
π − π
2c π +0 if sinω >0,
A
=
π 2 ( π if sinω <0,!
−
c if sinω >0,
A
=
( cA if sinω <0.
−
Letnbetheorderofω. TheBlaschkeproductωwrapstheunitcircleTarounditself
ntimes inthe strictly counterclockwisedirection. We thus see thatthe minimizing
function 2Reh is an alternating step function of height c and order n.
A
The following theorem summarizes what has been proved. To include the diag-
onal case in the statement, we define an alternating step function of order 0 to be
a constant function. The constant is defined to be the height of the function.
Theorem 4.2. For A a matrix in , there is a unique function in of
N+1 A
T G
minimum L -norm. It is an alternating step function of height c and order
∞ A
at most N.
A MINI-MAX PROBLEM FOR SELF-ADJOINT TOEPLITZ MATRICES 5
Theorem4.2asstatedexcludesmuchoftheinformationobtainedinthereasoning
leading up to it— information we shall use later. The theorem as stated can be
obtained by fitting it into the general dual extremum problem studied by Krein
andNudelmaninChapterIXof[5]. Thegeneralproblem,whichinvolvesaBanach
space and its dual, is formulated at the beginning of 1 of Chapter IX, after which
§
their basic theorem on the problem, Theorem 1.1, is established. Taking real L1
of the unit circle as the Banachspace in question, it follows fairly simply from this
theorem that each of our minimizers has the form Lsignν, where L is a positive
numberandν isarealtrigonometricpolynomialoforderatmostN;theminimizer
is thus an alternating step function. Establishing the uniqueness of the minimizer
takes more work but can be done on the basis of Krein-Nudelman’s Theorem 1.3
anditscorollaries. SeealsoTheorem2.1and2.2fromChapterIXof[5]. Ofcourse,
the approachweuse here,whichisspecific to oursetting,is quite distinct fromthe
approachin [5].
Thereasoningthatledtothetheoremcanbereversed. Startwithanonconstant
Blaschkeproductω ofordern N. Fixapositivenumberc,andformthefunction
≤
h= c Log 1+ω . Then2Rehisanalternatingstepfunctionofheightcandorder
πi 1 ω
−
n. We show(cid:16)that(cid:17)2Reh is the minimizing function for the correspondingminimum
problem.
In view of the analysis that produced Theorem 4.2, it will suffice to show that,
among all H functions g satisfying g(k)= ω(k) for k = 0,...,N, the function ω
∞
hastheleastnorm. ByTheoremCF,thisamountstoshowingthatthecompression
B to of the operator on L2 of multiplication by ω has norm 1.
N b b
P
Let z ,...,z be the zeros of ω. Then ω has the form
1 n
n
z z
ω(z)=λ2 − j ,
1 z z
j=1(cid:18) − j (cid:19)
Y
where λ is a constant of modulus 1. Defining the polynomials r and s by
n n
r(z)=λ (z z ), s(z)=λ (1 z z),
j j
− −
j=1 j=1
Y Y
wehaveω =r/s. Thepolynomialsrandsbelongto ,andwehave r = s ,
N 2 2
P k k k k
plus Bs = r. The desired equality, B = 1, follows. (The preceding argument is
k k
of course well known.)
Thequestionwhethereverynonconstantalternatingstepfunctionisobtainedby
theprecedingprocedurenowarises. Thatquestionisaddressedinthenextsection.
5. Inner Functions and Alternating Step Functions
Theorem 5.1. Let ψ be an alternating step function of height c and order N >0.
Then there is a Blaschke product ω of order N such that
2c 1+ω
ψ = Arg
π 1 ω
(cid:18) − (cid:19)
on T.
Proof. We assume (without loss of generality) that c=1.
6 DENNISCOURTNEYANDDONALDSARASON
Step 1. Let the real numbers α ,...,α ,β ,...,β satisfy
1 N 1 N
0 α <β <α <β < <α <β <α +2π :=α .
1 1 2 2 N N 1 N+1
≤ ···
Let the arcs χ and χ+ (n=1,...,N) be defined by
−n n
χ = eiθ :α <θ <β , χ+ = eiθ :β <θ <α .
−n { n n} n { n n+1}
The function ψ that takes the value 1 on each arc χ and the value 1 on each
−n −
arc χ+ is, to within a sign, the general alternating step function of height 1 and
n
order N. We let
α=α + +α , β =β + +β .
1 N 1 N
··· ···
Step 2. We introduce the polynomials
N N
p(z)=e−iα/2 (z eiαn), q(z)=e−iβ/2 (z eiβn).
− −
n=1 n=1
Y Y
Itisassertedthatthepolynomialq iphasdegreeN. Infact,theleadingcoefficient
−
ofq ipise iβ/2 ie iα/2. Forthistovanishwemusthaveei(β α)/2 = i=e3πi/2.
− − −
How−ever 0<β −α<2π, so 0< β−α <π, and the assertion follows. −
− 2
Step 3. We show that the rational function q/p is real valued on T (except at
iintscrpeoalseisngeiαfunn,cntio=n1w,.it.h.,rNan),geanRd.that on each arc χn :=χ−n ∪χ+n ∪{eiβn} it is an
We have
N N
q(eiθ)/p(eiθ)=ei(β−α)/2 (eiθ eiβn) (eiθ eiαn)
− , −
n=1 n=1
Y Y
=ei(β−α)/2nN=1ei(αn−2βn)"eeii((θθ−−22αβnn))−ee−−ii((θθ−−22βαnn))#
Y −
N sin θ−2βn
= .
n=1 sin(cid:16)θ−2αn(cid:17)
Y
Thus q/p is real valued on T. (cid:0) (cid:1)
For αn < θ < αn+1, the function sin θ−2αn is positive, while the function
sin θ−2βn is negative for θ < βn and pos(cid:0)itive fo(cid:1)r θ > βn. For j 6= n and αn <
θ <(cid:16)αn+1,(cid:17)the functions sin θ−2αj and sin θ−2βj have the same sign (negative
forj <n,positiveforj >n)(cid:16). Toge(cid:17)therwitht(cid:16)he exp(cid:17)ressionaboveforq/p,thistells
us that q/p is negative on χ−n and positive on χ+n. As eiαn and eiαn+1 are poles of
q/p,wethusmusthaveq(eiθ)/p(eiθ) asθ α andq(eiθ)/p(eiθ) +
n
as θ α . Hence q/p maps the a−r→c χ−∞onto Rց. It must be monotone−o→n ea∞ch
n+1 n
ր
such arc because, being a rational function of degree N, it assumes no value with
multiplicity greater than N.
Step 4. We show that the function ω =(q ip)/(q+ip) is a Blaschke product
−
of order N satisfying
2 1+ω
(1) ψ = Arg .
π 1 ω
(cid:18) − (cid:19)
A MINI-MAX PROBLEM FOR SELF-ADJOINT TOEPLITZ MATRICES 7
In fact, since ω can be rewritten as ω = q i q +i , it follows by Step 3
p − p
that ω is unimodular on T. From the de(cid:16)finition(cid:17).of(cid:16)ω we(cid:17)see immediately that
ω(eiαn)=1 for each n and ω(eiβn)= 1 for each n.
−
We can re-expressω as ϕ q , where ϕ(z)=(z i)/(z+i), a linear-fractional
◦ p −
transformation that maps the(cid:16)up(cid:17)per half-plane to D and R to T 1 , with ϕ(z)
moving counterclockwise on T as z moves in the positive direction\o{n }R. We know
from Step 3 that on each arc χ the function q/p is increasing with range R. We
n
can conclude that argω undergoes an increment of 2π on each arc χ , hence an
n
increment of 2πN on T. By the argument principle, ω has N zeros in D, thus is a
Blaschke product of order N.
Because 1+ω = q, if q(eiθ)/p(eiθ)>0, i.e., if (according to Step 3) eiθ is in one
1 ω ip
ofthearcsχ+−,wehaveArg 1+ω(eiθ) = π+0= π. Similarly,ifeiθ isinoneof
n 1 ω(eiθ) −2 −2
−
the arcs χ then Arg 1+ω((cid:16)eiθ) = (cid:17)π +π = π. The desired equality (1) follows.
−n 1 ω(eiθ) −2 2
This concludes the pro(cid:16)of−of The(cid:17)orem 5.1. (cid:3)
Theorem 5.1 is a particular case of Theorem 9 in the paper [4] of P. Gorkin and
R.C. Rhoades. Both [4, Theorem 9] and our Theorem 5.1 assert the extistence of
certainfiniteBlaschkeproducts. Theproofsofthetwotheoremsareessentiallythe
same, the chief difference being that Gorkin-Rhoades work mainly on the real line
while we stick to the unit circle. A conformal isomorphism between the unit disk
and the upper half-plane connects the two approaches.
We havechosentoinclude the proofaboveofTheorem5.1,ratherthanreferthe
reader to [4] for a proof, for the sake of completeness, and because the aims of [4]
arerather differentfrom ours. (Since Theorem5.1is less generalthan [4, Theorem
9], its proof is also a bit simpler.)
To conclude this section we discuss the extent to which the Blaschke product ω
produced in the preceding proof fails to be unique. Note that 1+ω(0) =1. In fact
1 ω(0)
−
(cid:12) (cid:12)
1+ω(0) q(0) (cid:12) (cid:12)
= = iei(β α)/2. (cid:12) (cid:12)
−
1 ω(0) ip(0) −
−
Suppose the Blaschke product ϕ of order n satisfies (1) when substituted for ω.
Thentheargumentof 1+ϕ mustundergoajumpof πacrosseachpointeiαn anda
1 ϕ −
jump of π across each p−oint eiβn. This implies, by a standardargument, that each
pointeiαn is a simple pole of 1+ϕ and eachpointeiβn is a simple zeroof 1+ϕ. The
1 ϕ 1 ϕ
functions 1+ϕ and 1+ω are thu−s rational functions with the same zeros an−d poles,
1 ϕ 1 ω
so they are−constant−multiples of each other. It follows that one can obtain ϕ by
composing ω from the left with a conformal automorphism of D. That conformal
automorphism must fix the points 1 and 1, so it has the form z z+r , where
− 7→ 1+rz
1<r <1. We get ϕ= ω+r .
− 1+rω
A calculation produces the equality
1+ϕ 1+r 1+ω
= .
1 ϕ 1 r 1 ω
− (cid:18) − (cid:19)(cid:18) − (cid:19)
Thus 1+ϕ(0) =1 only if r =0, i.e., only if ϕ=ω. We see that ω becomes unique
1 ϕ(0)
−
if one(cid:12)(cid:12)impose(cid:12)(cid:12)s the supplementary condition 1+ω(0) =1.
(cid:12) (cid:12) 1−ω(0)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
8 DENNISCOURTNEYANDDONALDSARASON
Theorem 5.2. If ψ is an alternating step function of order N, then ψ is uniquely
determined by the Fourier coefficients ψ(0),ψ(1),...,ψ(N), in the sense that if
those coefficients are known then the coefficients ψ(N +1),ψ(N +2),... can in
principle be found. b b b
b b
Proof. Let ψ be such a function, and let c be its height. We may assume N > 0
(the case N = 0 being trivial). By Theorem 5.1, there is a Blaschke product ω of
order N such that ψ =h+h, where
c 1+ω
h= Log .
πi 1 ω
(cid:18) − (cid:19)
From this we see that, for any n, the coefficient ψ(n) can be expressed in terms of
c and the coefficients ω(0),...,ω(n). As ω is a rational function of degree N, the
coefficients ω(N +1),ω(N +2),... can be exprebssed in terms of ω(0),...,ω(N).
The desired conclusion follows. (cid:3)
b b
b b b b
Corollary 5.3. Let E be the union of N closed disjoint subarcs of T. Then E is
uniquely determined by the coefficients χ (0),...,χ (N).
E E
Proof. ItsufficestoapplyTheorem5.2tothealternatingstepfunction2χ 1. (cid:3)
E
b b −
The phenomenon described in the corollary is treated at length in [3].
6. Two Dimensions
We indicate how the equality c = π can be deduced from our characterization
1 2
of minimizing functions. The inequality states that c / A π for all nonzero
A k k ≤ 2
matrices A in . For such a matrix A we know from Theorem 4.2 that the min-
2
T
imizing function in is an alternating step function of order 1 and height c .
A A
G
We can thus try to establish the inequality by starting with an alternating step
function ψ of order 1 and a given height, computing the corresponding matrix A,
finding A , and checking the ratio of interest. And it will clearly suffice to treat
k k
the case where ψ has height 1 and has a discontinuity at the point 1 with a jump
there of 2.
Let α be a point in the interval (0,2π), and define ψ by
1, 0<θ <α,
ψ(eiθ)=
( 1, α<θ <2π.
−
Then ψ is the general alternating step function of interest. The corresponding
matrix A is given by
ψ(0) ψ( 1)
A= − .
ψ(1) ψ(0) !
b b
Calculations give
b b
ψ(0)= α−π, ψ(1)= 2e−iα/2sinα2 .
π π
Thus
b b
1 α π 2eiα/2sinα
A= − 2 .
π 2e iα/2sinα α π
(cid:18) − 2 − (cid:19)
A MINI-MAX PROBLEM FOR SELF-ADJOINT TOEPLITZ MATRICES 9
The eigenvalues of A can be found by the standard procedure. The calculations
yield
α π +2 sinα
A = | − | 2 .
k k π
(cid:12) (cid:12)
As c =1, the desired conclusion is 1 π. (cid:12)We ha(cid:12)ve
A A ≤ 2
k k
1 π
= ,
A α π +2 sinα
k k | − | 2
so we need only verify that (cid:12) (cid:12)
(cid:12) (cid:12)
α
π α+2sin 2
− 2 ≥
for0 α π. Differentiationwithrespecttoαshowsthatthe functiononthe left
≤ ≤
side is decreasing on [0,π]. At α = π it has the value 2, so the desired inequality
holds, and it reduces to an equality only for α=π.
7. The Numbers c
N
Wepresentherewhatwehavelearnedaboutthemaximac ,includingnumerical
N
resultsforsmallN >1. Forf afunctioninL weletA denotethecompression
∞ f,N
to of the multiplication operator on L2 induced by f.
N
P
The inequality c π is easily seen at this point. For example, let ψ be the
N ≥ 2
alternatingstepfunctionofheight1andorderN associatedwiththeinnerfunction
ω(z)=zN as in the proof of Theorem 4.2:
1 1+zN
ψ(z)=2Re Log .
πi 1 zN
(cid:18) (cid:18) − (cid:19)(cid:19)
Referring to that proof, one sees that ψ alternates the values 1 and 1 on 2N
consecutive subarcs of T, each of length π. For z 0 we have −
N −→
1+zN
=1+2zN +O(z 2N),
1 zN | |
−
so
2 1+zN
ψ(z)= Im Log
π 1 zN
(cid:18) (cid:18) − (cid:19)(cid:19)
2
= Im(Log(1+2zN))+O(z 2N)
π | |
2
= Im(2zN)+O(z 2N)
π | |
2 2zN 2zN
= − +O(z 2N)
π 2i | |
(cid:18) (cid:19)
2
= (zN zN)+O(z 2N).
πi − | |
We see that the matrix A has only two nonzero entries, 2 in the lower left
ψ,N πi
corner and 2 in the upper right corner. In particular, A = 2, and the ratio
−πi k ψk π
kkAψψTk,∞Nhek f=ollkoAwψ1i,nNgkoebqsuearlvsatπ2io,ntepllrinodguucsesthaagtecnNer≥aliπz2a.tionof the last inequality.
Proposition 7.1. Let the function f be in L , let k be a positive integer, and let
∞
the function g on T be defined by g(eiθ)=f(eikθ). Then A = A .
f,N g,kN
k k k k
10 DENNISCOURTNEYANDDONALDSARASON
Proof. To simplify the notation, let T = A . For p in , the m-th Fourier
g,kN kN
P
coefficient of Tp is given by
(Tp)∧(m)= g(m j)p(j).
−
0 j kN
kN≤Xm≤ j kN
− ≤ − ≤ b b
If this coefficient is nonzero, there must be a j such that k divides m j, in other
−
words, such that j m (modk).
≡
For l=0,...,k 1, let S be the subspace of consisting of the polynomials
l kN
− P
p such that p(j) = 0 if j l (modk). These subspaces are mutually orthogonal,
6≡
they span , and, by the observation above, they are T-invariant. Hence T is
kN
P
the direct subm of its restrictions to the subspaces Sl.
If p is in S then there is a q in such that p(z) = q(zk). For the km-th
0 N
P
coefficient of Tp we have
N
(Tp)∧(km)= g(km kj)p(kj)
−
j=0
X
N b b
= f(m j)q(j),
−
j=0
X
showing that T S is unitarily equivalentbto A .b
0 f,N
|
For p in S with l = 0, the (l+km)-th coefficient of Tp (m = 0,...,N 1) is
l
6 −
given by
(Tp) (l+km)= g(l+km j)p(j).
∧
−
j≡lX(modk)
l j l+k(N 1)
≤ ≤ − b b
In the summation on the right, j l runs through the numbers 0,k,...,(N 1)k,
− −
and correspondingly, j runs through the numbers l,l+k,...,l+k(N 1). The
−
equality can be rewritten as
N 1
−
(Tp)∧(l+km)= g((m j)k)p(l+jk).
−
j=0
X
A polynomial p in S (l = 0) can be writbten as p(z)b= zlq(zk), where q is in
l N
6 P
and q(N)=0. This gives us
N 1
−
b (Tp) (l+km)= f(m j)q(j),
∧
−
j=0
X
showing that the restriction of T to S is ubnitarilybequivalent to the N-by-N
l
principal minor of A , hence is of norm at most that of A . The equality
f,N f,N
A = A now follows. (cid:3)
f,N g,kN
k k k k
The following corollary generalizes the inequality c π.
N ≥ 2
Corollary 7.2. For k a positive integer, c c .
kN N
≥
Proof. Itsufficestoapply Proposition7.1withf equalto analternatingstepfunc-
tion ψ of order at most N for which kkAψψk,∞Nk =cN. (cid:3)