Meromorphic Functions and Linear Algebra (Fields Institute Monographs, 18)

Ir(pe i6 - aj)1 ~l Ir(pe i6 - bk)1 og Ir 2 _ a.pe i6 1 - L...J og Ir2 _ bkpei6 1' 3 k=l

(2.21)

Proof Multiply f by ITj,k ;=!~ to obtain a zerofree analytic function. Apply (2.18) to it and arrange all the terms to get (2.21). 0 We shall now apply Poisson-Jensen formula at the origin. To that end, let us assume that f is regular and nonzero at the origin. Since P(O,O) = 1 we obtain at the origin

It has been said that the value distribution theory, "Nevanlinna theory", was born when Rolf Nevanlinna arranged the terms in (2.22) in a new more symmetric form. This was done by writing, log t for t > 0 as follows logt = log+ t -log+

1

t'

where log+ t = max{log t, O} and denoting 1 m(r, f) := -2 7r

j1l" log+ If(rei'P)ld
(2.23)

-1l"

Since the zeros of f are the poles of 1/f it is natural to collect terms in (2.22) as follows: n

m(r,f) + ~log

I~I

m

=

m(r,I/f) + f;log I:jl

+ log If(O)I.

(2.24)

Here the left hand side collects information about If I being large and the right hand side about it being small. When r grows the identity remains. The terms related to poles and zeros can be written still in another natural way. First notice that if we use log+ we can write n

LIog-l;1= Llog+ -I;I' k=l k k k where bk runs thru all poles. Then, if n(r, f) denotes the number of poles (always with multiplicities) such that Ibkl ::; r, then by integrating by parts we have r lor log -dn(t, r lor --dt. n(t,f) L log+ -Ib I= f) = tot k

k

0

SECOND CHAPTER

28

This suggests to put

N(r, f):=

r n(t, f) dt.

Jo

t

However, this is under the assumptions that I was regular and nonzero at the origin. These exceptional values can be dealt with as follows. Suppose that I has a zero of order k at the origin, i.e. 00

I(z) = LCjZ j j=k

holds for Izl < Ro with Ck f 0. Then the discussion above can be carried out for the function (r/z)kl(z) leading instead of (2.22) to log ICkl = 217C'

111' log I/(rei'P)ld

f:

j=1

log lajl - i)og ~ - k logr. r k r =1

(2.25)

In case I would have a pole at the origin, this holds with a negative k. To incorporate these cases into the identity (2.23) we modify N accordingly. If I has a pole of multiplicity k at the origin then n(O, f) = k and the partial integration above now gives Llog+ ~ = n(t,f) - n(O,t) dt.

k

r

Jo

Ibkl

t

Finally setting

N(r, f):=

r n(t, f) - n(O, t) dt + n(O, f)

Jo

logr

t

(2.26)

leads to a simple identity as (2.25) contains the term klogr.

Definition 2.1 Given a meromorphic function I for Izl < R $ 00, put for r
Theorem 2.4 II I is meromorphic for Izl < R, then 1

T(r, f) = T(r, ,) + log ICkl where Ck f

(2.28)

°

is determined by the expansion

I(z)

=

Ckzk + Ck+ 1 z k+l + ...

We need a couple of simple inequalities. If Cl, C2, • •• are arbitrary complex numbers, then log+1 and

n

n

j=l

j=1

II cjl $ Llog+lcjl,

n

log+1

L cjl $ log+(n . max

. J=l

J=I, ... ,n

n

ICj!) $

L log+lcjl + logn. n=l

SECOND CHAPTER

29 n

These can be used to estimate m(r,

m(r,

n

n 1;) and m(r, j=1 E 1;): j=1

n

n

j=1

j=1

II 1;) ::; L

n

m(r, 1;),

n

m(r'LI;)::; Lm(r,1;) +logn. j=1 j=1 If I is either the product or the sum of meromorphic functions, then the order of a pole of I is at most equal to the sum of the orders of the poles of individual n

functions at the same point. This gives the required inequalities for N(r, n

and N(r,

E

j=1

n 1;), j=1

1;) so that we can conclude: T(r,

n

n

j=1

j=1

II 1;) ::; LT(r, 1;),

n

n

T(r'LI;)::; LT(r,l;) +logn. j=1 j=1 For an analytic function I we can compare T(r, f) and log+ M(r, f). Theorem 2.5

(2.29)

(2.30)

II I is analytic in Izl < Ro, then lor fJ > 1 with fJr < Ro we

have

fJ+1

T(r, f) ::; log+ M(r, f) ::; fJ _ 1 T(fJr, f).

(2.31)

Proof Here the first inequality follows from N(r, f) = 0 and

m(r, f) ::; log+ M(r, f). If M(r, f) ::; 1, then the second inequality holds automatically. Suppose M(r, f) > 1, and let Zo = re icp be a point such that I/(zo)1 = M(r, f). Applying the PoissonJensen formula (2.21) we have

1 logM(r,f)::; 271"

111" 1 -11" P(O,t-cp)logl/(fJreit)ldt,

since the terms corresponding to zeros are negative:

IfJr( re icp - aj) 1 < IfJ2r2 - o,jreiCP I. But

and (2.31) follows.

P ~ t < 1 - l/fJ2 _ fJ + 1 (fJ' ) - (1 - l/fJ)2 - fJ - 1 '

o

SECOND CHAPTER

30

The first main theorem

We can now prove R. Nevanlinna's first main theorem. Theorem 2.6 Let f be meromorphic in Izl < R :::; complex number and r < R, then

T(r, f)

=

1

T(r, f _ a)

00.

If a is an arbitrary

+ log ICk(a)1 + c(a, r)

(2.32)

where Ck (a) is the first nonzero coefficient in the expansion f(z) - a = ck(a)zk + Ck+1(a)zk+l + ... and le(a,r)l:::; log+lal Proof Applying (2.28) to

f -

+ log 2.

(2.33)

a gives

1

T(r, f - a) = T(r, f _ a) + log ICk(a)l·

(2.34)

But (2.30) implies

IT(r,J) - T(r,J - a)1 :::; log+lal

+ log 2,

as T(r, a) == log+ lal, and substituting this into (2.34) yields the claim.

D

Example 2.1. If f is a rational function of exact degree d (Le. f = p/q,d = max{deg(p),deg(q)} and p, q contain no common factors), then

T(r, J) = dlogr + 0(1),

as r

--+ 00.

Example 2.2. Consider f(z):= eZ • Then N(r,J) == 0, while 1 111"

m(r, J) = -2 11"

-11"

1 111"/2 r log+1 exp(reitp)ld

Thus T(r, eZ ) = ~r. By (2.28) we also have T(r, Then

-11" /2

e1.) =

11"

~r. Let now a

:I

0,

00.

(2.35) has a solution zo, and all the solutions are obtained by collecting Zo + 2k1l"i, where k is an integer. From this we conclude that n(r, e.=-a)' the number of roots of (2.35) smaller than r in modulus satisfies 1

1

n(r, -Z - ) = -r + 0(1) e - a 11" which gives 1

1

N(r, -Z - ) = -r + O(logr). e - a 11" Actually, a somewhat technical calculation (see e.g. [Se], p. 171-173) shows that m(r, _1_) = 0(1)

eZ

-

a

SECOND CHAPTER

31

and therefore, by Theorem 2.6, 1 1 T(r, - - ) = -r + 0(1). eZ - a 7r

For a =I 0, 00, (2.35) ha.'l solutions and T(r, f~a)' essentially tells that their 'amount'. However, for a = 0, or a = 00 (2.35) ha.'l no complex solutions, but in compensation equality holds approximatively in a half plane, i.e.

il Re z «0,

eZ ~ 0

and m(r, l/e Z ) = ~r exhibits this.

0

Cartan's identity We start with a lemma.

Lemma 2.1 If a is an arbitrary finite complex number, then we have (2.36)

Proof i) if a = 0, then (2.36) is true. ii) if nonzero and analytic in Izl ~ 1 and thus

lal >

1, then l(z) := a - z is

1 111" log la - ei'Pld
iii) if 0 <

lal < 1, then a -

-11"

Izl < 1 and thus

z has one zero and no pole in

1 111" log lal = 27r -11" log la - ei'Pld

The limit Ca.'le

lal =

o

1 holds by continuity.

Theorem 2.7 If 1 is meromorphic in 1

T(r, J) = 27r

Izl < R

~ 00

then we have for r

111" 1 -11" N(r, f _ ei9 )dO + C


where while otherwise l(z)

=

Proof Let 1(0) =I

C = log+ 11(0)1 when 1(0) C_kZ-k + ... and

00.

=I 00

Applying Poisson-Jensen formula to 1 - ei9 yields

log 11(0) - ei9 1=21 111" log 11(rei 'P) - ei9 ld

-11"

+ N(r, J) -

N(r,

1

1 _ ei9 )'

SECOND CHAPTER

32

Integrating both sides with respect to () and changing the order of integrations we obtain

By Lemma 2.1 the left hand side equals log+ 11(0)1 while the double integral equals m(r,f). Assume then that I has a pole at the origin. Then we apply Poisson-Jensen formula to (;:)k(f(z) - ei9 ). This function has the same zeros as I(z) - ei9 , but since N(r,f) = N(r,zkf) + klogr, the right hand side takes the form

m(r, f)

+ N(r, f)

1

- klogr - 21l'

111' -11' N(r, 1-1ei9 )d()

while the left hand side becomes

Ic-kl 1og-k-' r

Since these are equal, -k log r disappears, and (2.37) follows.

o

Corollary 2.1 T(r, f) is nondecreasing in r and convex in logr. Proof The first claim follows from the fact that N(r, i_Ie iB ) is nondecreasing in r for each fixed (). Differentiating (2.37) w.r.t. logr gives d 1 111' 1 -1d T(r,f) = -2 n(r'-I ogr 1l' -11' - e''9)d().

Here again n(r, i_Ie i9 ) is nondecreasing in r and thus T(r, f) is convex as a function of logr. 0

Order and type for meromorphic functions Let now

I

be meromorphic in the whole plane (Le. R = 00).

Definition 2.2 We say that

I

is meromorphic 01 order v, where

logT(r,f) . v:= r1..... 1moo sup 1ogr ' and if 0 < v < 00, then it is of type (f:=

(f

where

lim sup T(r, f) . r ..... oo

rV

If I is entire, then we can compare the orders and types as an entire and meromorphic function.

SECOND CHAPTER

33

Theorem 2.8 If f is an entire junction, then its orders as an entire and meromorphic junction equal: w = v. Furthermore, the types satisfy the following inequality 0' :::; 7" :::; C(w)O', (2.38) where

C(w) =

1rW,

for w ;::: 1/2

1rW

= - . - - , for 0

sm

1rW

< W < 1/2.

Proof By Theorem 2.5, for () > 1 and all r > 0

(}+1

T(r, I) :::; log+ M(r, I) :::; () _ 1 T((}r, I).

(2.31)

This implies immediately that w = v. We only demonstrate (2.38) for C(w) := (2w + l)e. For the sharp value of C(w), see the references given on p. 182 of [Se]. Dividing (2.31) by rW gives immediately 0' :::; 7". From the right hand side we obtain with () := 1 +

t

o

The claim follows.

Boutroux-Cartan lemma The following result has been used extensively in particular in the study of entire functions and is often called Boutroux-Cartan lemma. n

Theorem 2.9 Let p(z) = = nand h

n (z ;=1

aj) be any given monic polynomial of degree

> 0 any fixed number. Let E := {z

Eel

h n Ip(z)1 :::; (-) }. e

Then there are disks Bl. ... , B n , B;:={zllz-z;l:::;r;} such that

and

SECOND CHAPTER

34

Bound along a circle With help of Boutroux-Cartan lemma we consider bounding log+ 1!(z)1 pointwise in terms of T(r, I). Theorem 2.10 Let! be meromorphic in Izi < R and choose r such that ()r < R. Then there exists a radius p such that

<

R, () > 1

r

-
"()+1 log+I!(pe~"')1 $ () -1 m(()r, I)

()r.

+ log(

4()(()+I)e () _ 1 )n(()r, I).

Proof Suppose bb b2 , •.• , bn (n = n(()r, I)) are the poles of! satisfying Ibjl $ Put n

p(z):= II(z-b j ) j=l

and choose in Boutroux-Cartan lemma h:= 9;/r. Then there are disks B l , ... , Bn with the total sum of diameters $ 9/jlr. Thus, in the annulus ~r $ Izl $ r there must exist a circle Izi = p which does not intersect the interiors of any of these disks. Along this circle we have "

Ip(pe'''') I ~

()-1

( 4()e

n

r) .

Now the Poisson-Jensen formula gives log

· Smce

"

I!(pe t ",)

" Ln

2

-

"

1 /'11" P (()r) - bkpe'''' 1 $ -2 P( -() ,t - c.p) log 1!(()red)ldt + log I () (" b ) I· 7r -'II" r r pet'" - k k=l

..e... <1 9r - 6'

By construction, 1

n

L log Ipe k=l

i '" _

4()e bkl $ n log (() _ l)r'

This gives

+ " 1 log I!(pe'''') 1 $ 27r

/'11" () + 11log + I!(()re'"t )Idt -'II" () _

( ()r ) 2

+n log

() + 1

+ ()r2

()r

= () _ 1 m(()r, I)

4()e

+ n log (() _ l)r 4()(() + l)e

+ log(

() _ 1

)n(()r, I).

o It is useful to formulate the pointwise bound also in terms of T(r, I) directly. To that end one needs to estimate n(r, I) in terms of N(r, I). However, we first point out that in general it is not possible to bound n(r, I) by N(()r, I) for () > 1.

SECOND CHAPTER

35

In fact, if n(O, I) > 0, then for small enough r we have N(r, I) = n(O, I) logr. Likewise, it is not possible to bound log+ If(peill')1 with T(()p, I). Example 2.3. Let f(z) := z-k. Then T(r, I) = k log+ r and thus an inequality of the form (2.39) cannot be true for all functions f. We shall therefore either assume f(O) #- 00, or make the necessary modifications to (2.39). 0

Lemma 2.2 Suppose f(O)

#- 00.

For ()

> 1 and ()r < R we have

1 n(r, I) ~ log ()N(()r, I).

(2.40)

If f has a pole at the origin, then the inequality holds in the form

logr 1 n(r,l)+n(O,l)log() ~ log ()N(()r,l).

(2.41)

Proof We have N(()r, I) =

(or n(t, I) - n(O, I) dt + n(O, I) log(()r)

10

t

~[n(r, I) - n(O, fl

l r

Or

dt

t

+ n(O, I) log(()r) o

which gives (2.41).

Corollary 2.2 Let f be meromorphic in Izl < R such that f(O) #- 00. Choose () > 1, 0 < r < R such that ()r < R. Then there exist a constant C(()), only depending on (), and a radius p depending on f and satisfying that for all cp

./0

~

p

~

r, such

(2.42)

Proof We estimate ()+1 ()+1 ()+1 2 () _ 1 m(()r, I) ~ () _ 1T(()r, I) ~ () _ 1 T(() r, I) and, by Lemma 2.2 n(()r, I)

Replacing () by

()1/2

~ lo~()T(()2r, I).

Theorem 2.10 implies (2.42) for some p,

C(()) := JO + 1 JO - 1

./0 ~ p ~ r, with

+ log 4JO(JO + l)e_l_. JO - 1

log JO

o

SECOND CHAPTER

36

Representation theorems

We forinulate two theorems concerning the possibility of representing a meromorphic function f as a quotient of analytic functions fI! h such that the growth of Ii's are controlled. Definition 2.3 A function f, meromorphic in chamcteristic in Izl < R, if sup T(r, f) < 00.

Izl < R is said to be of bounded

r
Theorem 2.11 Assume f is meromorphic in Izl < R. Then f can be represented as hi h with hand h bounded and analytic in Izl < R if and only if f is of bounded chamcteristic.

o

Proof See e.g. [NR2].

The other theorem is concerned with meromorphic functions in the whole plane. Theorem 2.12 For every 0 > 1 there is a constant B(O) with the following property. If f is meromorphic for Izl < 00, then there are entire functions h, h such that f = hi h and for i = 1,2 we have for all r > 0

T(r, Ii) ::; B(O) T(Or, f). Proof This theorem is in [Mi], see [Ru] for an exposition.

o

Comment 2.1 Theorem 2.10 is taken from [Ya] (Lemma 4.2) where it is used in a discussion on an inequality of Chuang Chi-tai bounding T(r, f) in terms of T(r, /'). Comment 2.2 In addition to the Nevanlinna characteristic function T(r, f) there are other related characteristic functions in the literature. In particular when the values are considered as points in the Riemann sphere and the distances are measured accordingly, the theory gets a different, more geometric flavor.

THIRD CHAPTER Keywords: Subharmonic functions, vector valued analytic and meromorphic functions, matrix and operator valued meromorphic functions, finitely meromorphic. Analytic vector valued functions We shall next generalize the characteristic function T for operator valued meromorphic functions. The first concept is going to be denoted by Too and it is defined as such for Banach space valued functions; for operator valued functions we just use the operator norms. The discussion shall touch the properties of subharmonic functions, some of which we present below. Before that, however, let us recall what we mean by vector valued analytic and meromorphic functions. If J is defined in a domain n c C taking values in a Banach space X, then it is analytic if

(3.1) lim _l_[J(z) - J(zo)] z - Zo exists for all Zo E n. The limit of the difference quotient is in the norm topology. Furthermore, J is called meromorphic if apart from poles it is analytic and around any pole b there is a smallest positive integer m = m(b) such that z-zo

Z 1--+

(z - b)m J(z)

(3.2)

is analytic at b. It is a well known and important result that if the limits are assumed in the weak topology only, they actually exist in the norm topology as well, and so "weakly analytic are analytic".

Subharmonic functions It is an important starting point for our discussion that if J is analytic taking values in a Banach space, then the mapping

u :z

1--+

u(z)

:= log+

IIJ(z)1I

(3.3)

is subharmonic.

Definition 3.1 Let n be a domain of C. A function u from n to R U { -oo} is said to be subharmonic on n if it is upper semicontinuous and satisfies the mean inequality 1 111" u(zo + rei'P)dv; (3.4) u(.zo) :5 -2 7r

_11"

whenever the closed disc B(zo, r) is contained in n. Furthermore, it is harmonic if both u and -u are subharmonic. We recall that u is upper semicontinuous if for 37

THIRD CHAPTER

38

allzoEO

limsupu(z) $ u(zo). %-+Zo

We can now state the following result.

Theorem 3.1 Suppose I is analytic from a domain 0 to a Banach space X. Then the functions IIIII and log 11/11 are subharmonic in O. Proof Clearly IIIII is continuous when formula we have I I(zo) =-. 271"~

1

I

I<-zol=r

is analytic. From Cauchy's integral d( I(()-( - Zo

I 111" I(zo + rei'P)dcp =2 71" -11"

and so I II/(zo)1I $ 271"

111" -11" II/(zo + rei'P)lIdcp.

Thus IIIII is subharmonic. Now to conclude the same on log 11/11 the following result of P. Montel and T. Rado can be used ([AI], Theorem A.1.14): Given a positive function v in a domain the function log v is subharmonic there il and only if z t--+ leCUlv(z) is subharmonic lor all complex numbers Q. But for every Q

o

and this is subharmonic as eQZ f(z) is analytic.

We shall not use the following result of E. Vesentini, ([AI], Theorem 3.4.7) but include it without proof. We recall that in a Banach algebra the spectrum a(a) of an element a consists of complex numbers ,\ at which ,\ - a has no inverse and that the spectral radius p(a) = sup{I,\11 >. E a(a)}.

Theorem 3.2 II in addition to the assumptions 01 Theorem 3.1 the Banach space X is a Banach algebra, then also the functions Z t--+ p(f(z)) and Z t--+ log p(f(z)) are subharmonic. Since the maxi:num of two subharmonic functions is again subharmonic, and log+ u = max{logu, O} under the assumptions above, log+ 11/11 and log+ p(f) are also subharmonic. Subharmonic functions satisfy a maximum principle.

Lemma 3.1 II u is subharmonic on a domain 0 and lor some Zo E 0 holds u(zo) ~ u(z) lor all z E fl, then u is constant in O. Meromorphic vector valued functions We can now define the characteristic function for scalar case, by just replacing log+ III by log+ 11/11.

I

in the same way as in the

39

THIRD CHAPTER

Before formulating it let us see why it is well defined. We shall put, for r


Here the integrand is continuous except at poles, where it has a logarithmic singularity and thus integrable. In fact, near a pole we have log+ IIf(z)1I ::; m(b) log+

1 Iz _ bl

+ log+

lI(z -

b)m(b) f(z) II

where m(b) is as in (3.2). Let us also put m(zo) := 0 at regular points zoo Then we can define the counting function noo(r, f) :=

:E m(b)

(3.6)

Ibl~r

and its logarithmic average

loo [noo(t,f) -noo(O,J)]-dt +noo(O,J)logr. r

Noo(r,J):=

(3.7)

t

Definition 3.2 Given a meromorphic function f from Izl Banach space X, the characteristic function Too is defined by

< R ::;

00

into a

(3.8) We collect the main properties of Too into the following two theorems.

Theorem 3.3 Let f, 9 and Ii be meromorphic and X -valued for Izl < R ::; and 4J meromorphic scalar function there. Then Too(r, f) is nonnegative and nondecreasing function in r for 0 ::; r < R which is convex in the variable log r. The following inequalities hold

00

(3.9) Too(r,

k

k

1

1

:E fi) ::; :E Too(r, fi) + log k.

(3.10)

If X is also a Banach algebra, then additionally

(3.11)

Proof The inequalities (3.9), (3.10) and (3.11) are straightforward, but to conclude that Too is nondecreasing and convex in log r has to be done differently than in the scalar case, where we based it on the Cartan's identity. However, the same conclusion can be obtained from subharmonicity of log+ IIfli. We omit the details here but come back to them while discussing the characteristic function TI.D For an analytic vector valued

f we denote analogously

Moo(r, f) := sup IIf(z)lI· Izl~r

40

THIRD CHAPTER

Theorem 3.4 If f is analytic for Izl < Rand 0 < r < ()r < R, then

+ ()+1 Too(r, f) :::; log Moo(r, f) :::; () _ 1 Too (()r, f).

(3.12)

Proof As log+ IIfll is subharmonic if f is analytic, the claim follows using Poisson-Jensen formula as in the scalar case. 0

In finite dimensional matrix case all norms are equivalent but we might want to know how the characteristic function changes with changing of norm. In general, suppose that in a Banach space X we are given another norm 11.11. which satisfies

for all x EX. Let us denote the corresponding characteristic functions by Too and by Too, •. Since 11.11. is stronger or equivalent with 11.11 whenever f is meromorphic in the stronger norm it is also in the weaker one but the reverse claim does not hold. Proposition 3.1 If f is meromorphic in the norm 11.11. for Izl < R:::; it is also in the norm 11.11 and the following holds for r < R:

Too(r,f):::; Too,.(r,f)

+ log+ C.

00

then

(3.13)

Proof For moo we obtain moo(r,f)

J :::;2~ J

=2~

log+ IIfll

10g+(Cllfll.)

:::;moo,.(r, f)

+ log+ C.

For N 00 we obtain

Noo(r, f) :::; Noo,.(r, f) as the multiplicities of the poles satisfy m(b) :::; m.(b) : m(b) = lim

log+ Ilf(z)1I 10g+(I/lz - bl)

· [log+ Ilf(z)ll. < 11m 10g+(I/lz - bl)

-

=

log+ C I + --,---=------10g+(I/lz - bl)

m.(b).

o Rational functions

In Example 2.1 we pointed out that if f is rational f = p/q and the representation contains no common terms, then T(r, f) = dlogr +-0(1) where d is the degree of f, namely d = max{deg(p),deg(q)}. The growth rate O(logr) is also characteristic for rational functions. We shall now look at it in the vector valued set up.

THIRD CHAPTER

41

The terminology is somewhat unfitting but it should not cause much confusion to call functions "rational" if they are of the form f = p/q where q is a complex valued polynomial and p is an X-valued "polynomial": k

p(z) = L:XjZ j j=O

where Xj EX. Theorem 3.5 Assume f is an X -valued meromorphic function on C such that

Too(r,f) Then

=

O(logr) as r

-+ 00.

(3.14)

f is rational and as such is either analytic at 00 or has a pole there.

Proof From (3.14) we conclude that there exists an integer K and a constant C such that Noo(r, f) $ Too(r, f) $ Klog+ r + C holds for all r > 0. This implies that the number of poles of f is bounded by K. In fact, choose an arbitrary 1 < s and take r > s so that

Noo(r, f)?

I 8

r

dt [noo(t, f) - noo(O, f)]T

+ noo(O, r) logr

?noo(s, f) logr + [noo(O, f) - noo(s, f)] log s. This implies

noo(s,f)

(1-

logs) $ Noo(r,f) logr logr and the result follows by letting r -+ 00. Without loss of generality we can assume that f is nonzero and regular at the origin. Let now q be a polynomial which vanishes just at the poles of f and is normalized such that q(O) = 1. Then we know that T(r,q) $ Klog+ r + C1 with some C1 and that qf is entire. But then

+ 0+1 log Moo(r,qf) $0 -1 Too(Or,qf) 0+1

$ 0 _ 1 [T(Or, q)

+ Too (Or, f)]

0+1 $0 _ 1 [Klog+ r + C 1 + Klog+ r + C + 2KlogO]. Thus we obtain (3.15) by choosing 0 large enough. What remains is to conclude from this that qf is a polynomial of degree at most 2K + 1. In fact, since qf is entire it has a power series converging everywhere 00

(qf)(z) = L:ajz j o

42

THIRD CHAPTER

with some aj EX. If we now set 2K+l

p(z):=

L

ajz j ,

o

then

g(z)

:=

[(qf)(z) - p(z)]z-2(K+1)

is entire and by (3.15) Moo(r, g) = O(l/r).

o

Thus 9 = O.

Recall, see Comment 1.1 and Definition 1.1, that an operator A is called algebraic if it has a minimal polynomial.

Corollary 3.1 If A is bounded operator in a Banach space, then it is algebraic if and only if Too(r, (1 - ZA)-l) = O(logr). When is the inverse also meromorphic If f is a scalar meromorphic function (with f(O) = 1) then by the first main theorem 1

T(r, f) = T(r, 7)' Generalizing this to vector valued functions does not make sense as in general we cannot "invert" a vector. However, if we are in a Banach algebra, then a natural inversion is available. We show below a simple estimate for the inverse in case of d x d complex matrices. However, let us start with a simple example from the Banach algebra of bounded linear operators in a Hilbert space. Namely, we see that we cannot in general estimate the inverse function in terms of Too of the function. In fact, if

f : z 1-+ 1 - zA where A is a bounded operator, then Too(r, f) = log+ r + 0(1) but it does not tell whether the inverse (1 - ZA)-l is meromorphic in the whole plane or not. As A is bounded we do know that it exists and is analytic for smallizi. Example 3.1 Let H := l2' i.e. the Hilbert space consisting of complex square summable sequences x = {Xjb~l and A :=diag(O!j) where {O!jb~l is a bounded sequence of positive numbers. Now (I - ZA)-l has poles at 1/0!j and it is meromorphic if and only if there is either a finite number of poles or they accumulate at infinity. Observe that if O!j takes the same value for different j's, it corresponds to the same pole and is counted only once. Note further that moo(r, (1 - zA)-l) = 0(1). All growth in Too(r, (1 - ZA)-l) comes from counting the poles. In fact

Too(r, (1 - zA)-l) = Llog+(O!jr) + 0(1).

(3.16)

Observe that here the eigenvalues O!j are also the singular values of A. If we like to conclude from the growth of I - zA the growth of its inverse we have to see more

THIRD CHAPTER

43

growth in it. We do this later in Tl by including all the singular values into the characteristic function, and not just the largest one, the norm. Note that in the example the operator is compact exactly when lim aj = 0, but in general compactness is only a sufficient condition for the inverse to be meromorphic. The first thing, however, is to recall that if we have inverse at some point, then it is analytic in some neighborhood. To that end, let X be a Banach space and denote by B(X) the bounded linear operators in X.

Lemma 3.2 Assume that F is analytic from a domain n to B(X). If F has a bounded inverse at a point zo. then F-l exists and is analytic in some neighborhood

of zoo Proof Suppose F is invertible at zoo Define G{z) from

F{z) = F(zo)(I - (z - zo)G{z)) so that G is analytic in n. Let M := 2I1G(zo)1I and take p > 0 such that the disc B(zo, p) en and IIG(z)11 :$ M in that disc. Then the series 00

~~)z - zo)jG{z)j

o converges in B{zo, p), is analytic and represents the inverse of F{zO)-l F{z).

0

In order to state a sufficient condition for the inverse of a meromorphic function to also be meromorphic we need the concept of principal part.

Definition 3.3 If f has a pole at b with Laurent series

then the rational function -m

is the principal part of f at b.

Example 3.2 Again, consider the sequence space 12. Let

K(z) := diag(aj/{l - z)) where {aj} is a decreasing sequence of positive numbers converging to o. Then K has a simple pole at 1, elsewhere it is analytic and compact. Put F{z) = 1- K{z) so that -K is the principal part of F. Now the inverse

F(z)-l = (I - K(Z))-l = 1+ diag

(1-

;j_ aj)

has an accumulation point of poles at 1 and is therefore not meromorphic there. This can be avoided by requiring the principal parts to be of finite rank.

THIRD CHAPTER

44

Theorem 3.6 Assume that we are given a function F(z) = 1- K(z) such that it is meromorphic in a domain n into B(X) and there exists a point Zo E n at which F is analytic and invertible. A sufficient condition for the inverse F-l = I + K (I - K) -1 to exist as a meromorphic function in n is that K (z) is compact away from poles and that additionally, at poles its principal parts are of finite rank. Proof By Lemma 3.2 F-l is analytic at points where F is analytic and invertible. Follow the analytic continuation of F-l starting from Zo until you reach a point Zl where F is analytic but is not invertible. Assuming now that K(zd is compact, 1 must be an eigenvalue and the corresponding invariant subspace is finite dimensional. This simply means that not only is the singularity a pole but also the principal part is of finite rank. Since K(I - K)-l is also compact whenever we are away from singularities we see that the inverse has the same properties as F. What remains is to consider the singularities of F. So, let F have a pole at b. Without loss of generality we can assume that K(z) = A(z) + B(z) where B(z) is near b regular and such that 1- B(z) is invertible while A(z) is finite dimensional and has a pole at b. Then near the pole we have F- l = (I - A - B)-l = (I - B)-l(1 - A(I _ B)-l)-l and since (I - B)-l is analytic at b it suffices to study the behavior of (I - A(z)(1 - B(z»-l)-l as z -+ b. But notice that now C(z) := A(z)(I - B(Z»-l is finite dimensional and has a pole at b. It also follows from the structure of the principal part that Xo := C(z)X is independent of z for z near the pole. Write X = Xo E9 Xl and decompose 1- C(z) into a matrix operator I _ C= ( I

Thus

(I _ C)-l

=

~ Co ~l).

((1 - ~0)-1 -(1 - ~O)-lCl )

and the problem has been reduced to finite a dimensional one, as I - Co is invariant in Xo. The argument for finite dimensional case is given below in a separate proof of Theorem 3.7, in order to have a simple proof for the matrix case alone. D Definition 3.4 A meromorphic operator valued function F is finitely meromorphic if the principal part is of finite rank at every pole of F. We could rephrase Theorem 3.6 by saying that finitely meromorphic functions of the form Identity + Compact have inverses in the same class, provided that the inverse exists at least at one point. A simple estimate for matrices We specialize the discussion now to d x d matrices. Let Md denote the Banach algebra with norm induced from the Euclidean norm in Cd. Here F = (Ai) is meromorphic in a domain n if and only if each scalar function Ai is meromorphic in n. Theorem 3.6 takes the following simple form.

THIRD CHAPTER

is

45

n to MId. If there then F- 1 exists as a meromorphic

Theorem 3.7 Suppose F is merom orphic from a domain

a point Zo E function in n.

n such

that F{Zo)

is invertible,

Proof Since F is invertible at a point zo, det F does not vanish identically, but is a nontrivial meromorphic function. Near Zo F- 1 can be represented in the form 1 F(z)-l = det F(z) adj F(z)

where the elements of adjF(z) are sums of products of the entries of F. Thus it continues to be meromorphic in all of n. o As pointed out many times already, we cannot expect Too{r, F) to be of the same size as Too(r, F- 1 ) in general. However, we can estimate one in terms of the other if we allow dimension dependent constants to appear. Such a result is interesting for fastly growing meromorphic functions but not for slowly growing such as I - zA.

is a meromorphic MId-valued function for I. Then for r < R we have

Theorem 3.8 Suppose F 00

such that F{O)

=

Izl < R :s; (3.17)

Remark We shall later see that (3.17) actually holds with (2d - 1) replaced with d. Proof Here we use the following facts from linear algebra. If O"j{A) denote the singular values of an invertible matrix A, numbered decreasing so that O"d(A) denotes the smallest one, then

We also need the fact

IIA-11I = l/O"d{A). that I1~ O"j(A) = Idet(A)I. Then we have -1 _ I1~-1 O"j(A) < IIAll d- 1 IIA II - Idet(A) I - Idet(A)I·

But then

Too(r, F- 1 ) :s; T(r, and since F(O)

det~F)) + (d -

(3.18)

I)Too(r, F)

= I, det F(O) = 1 so that 1

T(r, det(F)) = T{r,det{F)). But Idet(A) I :s; IIAlid so that

T{r,det(F)) :s; d Too(r, F) which gives the result.

o

Comment 3.1 For those interested in knowing more on subharmonic functions we recommend [AI]. Further, there exists a value distribution theory for subharmonic functions, see [Ha-K].

46

THIRD CHAPTER

Comment 3.2 Ribaric and Vidav [Ri-V] have studied analytic properties of

F{z)-l for an operator valued function F{z). Above we used the terminology of [N08].

FOURTH CHAPTER Keywords: Canonical product Jormula, SVD, Weyl inequalities, Hom inequalities, Ky Fan norms, total logarithmic size, total conditioning. A product form for matrices

We have seen that by replacing the absolute value by the norm we obtain a characteristic function Too which shares many properties of T. In order to bring more symmetry in inversion we shall once more start from the very basic matrix function F: z 1-+ 1- zA where z E C and A is a d x d complex matrix A E Md. It is most natural to think of F as a polynomial in z of first order. But its inverse F- 1 is typically a rational function of degree d. Thus, the basic symmetry between F and F- 1 seems not to hold. There are two simple ways to try. The first one would be to reduce the problem back to scalar valued functions: take unit vectors x and y and put

J(z) := y*(I - zA)x = y* F(z)x. But now 1/J is not in any simple way related to F- 1. We shall look at scalar functions obtained in this way in the Ninth Chapter. On the other hand, if we work with the spectral norm, then z 1-+ IIF(z)11 satisfies IIF(z)1I ::; 1 + rllAIl while IIF-1(z)11 has singularities for l/z = Aj where Aj is eigenvalue of A. Thus the characteristic function T when applied to IIFII would not reveal anything essential about IIF-Ili. Example 4.1 But let us consider rank-l matrices: A = ub*. Without loss of generality we may assume that Ilull = 1 and that a = IIbll is the singular value of ub*. Introducing v by b = IIbliv we have found the nontrivial term in the singular value decomposition of ub* = oouv*. From Example 1.2 we know that

zoo

1

(1 - zoouv*)- = 1 + - - \ uv*, 1 - ZA '

(4.1)

where A = oov*u is the possibly nonzero eigenvalue. We can now simply compute Too of both F(z) := 1 - zoouv" and of F-l. As r --+ 00 we have moo(r, F)

1("

= 211" J-1r log+llF(re i9 ) lidO = log(oor) + 0(1). 47

48

FOURTH CHAPTER

On the other hand, F-l(z)

=I+

A"# 0), an.d Noo(r, F- l ) =

l~~z uv* and moo(r, F- 1 )

!,

= log I~I + 0(1) (if

dt

r

- = logr + log IAI.

l/IAI t

Hence

(4.2) Thus

Too(r,F) = Too(r, F- l ) + 0(1) holds. In fact we shall later see with help of Tl that here is actually an exact identity in this case. Our aim here is to demonstrate that there is a natural way to view F( z) = I - zA as a polynomial of degree d rather than of 1 by writing it as a product of terms considered above. Then inversion keeps the size of each individual term constant when measured by Too. Theorem 4.1 Given A E Md there are vectors al, ... , ad and Ul, ... , Ud such that ujuk = 8jk , ajuk = 0, for j > k and for all z

I - zA = (I - zUdad)'" (I - zUlai).

(4.3)

Before proving this observe that, we get by collecting the first order terms d

A= LUjaj j=1

while all the higher powers vanish. Proof Let the Schur decomposition of A be

A = Q(A+C)Q* where Q is unitary, A is diagonal and contains the eigenvalues and C is strictly upper triangular. Let the rows of A + C be denoted by bi's. Then

I - z(A + C) = (I - zedbd)'" (I - zelbi) where ej denotes the j'th coordinate vector. The higher powers of z vanish as bj+l ej = O. But then

I - zA =Q(I - z(A + C))Q* =(I - ZQed( Qbd)*) ... (I - ZQel (Qb 1 )*) which is of the required form with Uj = Qej and aj = Qbj . Since Q is unitary, the vectors Uj are orthonormal. 0 Observe that by construction bjej = ajuj = Aj gives the eigenvalues of A and in particular the following Corollary 4.1 In the decomposition of Theorem d

det(I - zA) =

II (I j=1

4.1 we have

d

zajuj) =

II (1 j=1

Taking the conjugate of (4.3) we obtain the following:

ZAj).

FOURTH CHAPTER

49

Corollary 4.2 Given A E Md, there are vectors bl , •.. ,bd and Vl,'" ,Vd such that vjvk = djk and for all z

(4.4) Substituting (4.1) into (4.3) gives still another representation for the resolvent. Corollary 4.3 In the decomposition of Theorem (I - ZA)-l = (I + 1

4.1

we have

z * ulai)··· (I + 1 z * udad)' - zal Ul - zadud

(4.5)

In principle we could now define a characteristic function for general I - zA as the infimum of sums of characteristic functions of its factors, the infimum taken over all representations of the form (4.3). This would behave "correctly" in the inversion but there are several reasons why this would be unsatisfactory. First, it would not be the natural analogue of the scalar case as it would correspond estimating polynomials on the factor level and summing up: d

d

10g+(II 11- ZAjl) ~ Elog+ll- ZAjlj j=l j=l

(4.6)

the inequality in (4.6) can be proper. Secondly, such a definition would not work for arbitrary meromorphic matrix valued functions. However, we see that at least it is possible to view 1- zA and (I - ZA)-l as equally large. Let us look at a concrete example. d-l

Example 4.2 Let C be the maximal nilpotent Jordan block: C =

L: ejej+l

j=l

so that 1- zC =(1 - zed-led) " . (I - zelei) =FdFd-l ... F l , with Fd == I.

Clearly, for large r moo(r, Fj ) = logr + 0(1), and moo(r, Fd) == O. Thus

j
d

E moo(r, Fj ) = (d -1) logr + 0(1), j=l

r

-+ 00.

As (I - zejej+l)-l = 1+ zejej+l we have moo(r,Fj) == moo(r, F j d

and thus

l )

d

L: moo(r, Fj ) == L: moo(r, Fj- l ).

Comparing with polynomials and (4.6), j=l j=l for r large the overestimation vanishes and we might want to conclude that the "answer" (d - 1) log r + 0(1) should be the right behaviour of our new characteristic function. This can be confirmed by looking at (1 - zC)-l: (1 - zC)-l = 1+ zC + ... + zd-lCd -

l

FOURTH CHAPTER

50

which implies as r

-+ 00

moo(r, (I - zC)-l) = (d - 1) logr + 0(1). In this case the norm of the inverse grows alone like the sum of the factors, while moo(r, 1- zC) = log r

+ 0(1).

The key reason for these very different growth speeds is that I - zC has d - 1 singular values of size O(r) and one very small one, namely 1 1 d-1 ad(1 - zC) = 11(1 _ zC) 111 = O((-;J ).

o We conclude that we should look at the product of all large singular values rather than just the largest one. In fact, if a1(F) :?: a2(F) :?: ... :?: ad(F) :?: 0 are the singular values of a matrix F, then we shall look at d

II max{aj(F), 1} j=l

as describing the size of F. This gives an exact inversion formula, is submultiplicative in matrix products and allows one to estimate the norm, because

Singular value decomposition Let A be a complex d x d matrix, with eigenvalues AI, A2, ... , Ad. We order the eigenvalues as IA11 :?: IA21 :?: .... The singular values ai of A are the positive square roots of the eigenvalues of A* A. We order them decreasingly: a1 :?: a2 :?: ... :?: o. Put ~ =diag(a1, a2, . .. ,ad). Theorem 4.2 (Singular Value Decomposition) There exist unitary matrices U, V such that A = U~V*. Proof Let x, y be unit vectors such that Ax = a1Y with a1 = IIAII. Since any orthonormal set can be extended to form an orthonormal basis for the whole space, there are VI E C dX (d-1), U1 E Cdx(d-1) so that the d x d-matrices [x VI] and [y U1] are both unitary. Then U* AV must be of the form U*AV =

(~1 ~)

=: AI.

Then

and IIAl112 :?: (a?

+ w;w)2 + IIBwl1 2 :?: a~ + w*w. a 1 + w*w

Since U, V are unitary, we have IIA111 = IIAII = a and therefore w = O. The 0 proof can be completed by an induction argument.

FOURTH CHAPTER

51

Theorem 4.3 (Polar Decomposition) Let P, Q be positive semidefinite such that p2 = AA * and Q2 = A * A. There exists a unitary U such that

A=PU=UQ. Proof If the SVD of A is A = VEW* , then

A

= (VEV*)(VW*) = (VW*)(WEW*) o

and we may put U := VW*, P:= VEV*, Q:= WEW*. r

E O"jUjV;

Theorem 4.4 Let A = UEV* =

j=l

be the SVD of A, where r =

rankA ~ d. Let k < rank(A) and denote k

Ak := L O"jUjvj. j=l Then O"k+l = IIA - Akll =

min

rank(B)~k

IIA - BII·

Proof We have IIA - Akll = IIU*(A - Ak)VII = Ildiag(O, ... ,0, O"k+l, O"k+2, ... )11 so that IIA - Akll = O"k+l. Now suppose rank(B) = m (~ k). Thus there exists orthonormal vectors XI. ... ,Xd-m such that ker(B) = span{xl, ... ,Xd-m}. But then span{xl, ... ,Xd-m}nspan{vl, ... ,Vk+l} -# {O} as d - m + k + 1

~

d + 1 > d. Let a be a unit vector in this intersection. But then k+l

a E ker B and writing a =

E

j=l

vjavj we obtain k+l

IIA - BI12 ~ II(A - B)a11 2 = IIAal1 2 ~ LO"]lv;aI 2 ~ O"~+l j=l

o

completing the proof.

Remark If A is not a square matrix, then it can be augmented to be a square matrix by adding a suitable number of columns or rows consisting of zeros. Basic inequalities for singular values and eigenvalues

In the following we shall denote by Mm,n the space of complex matrices, consisting of n columns of length m. The following basic but as such a simple lemma can be proved using unitary invariance and the so called interlacing property for the singular values of submatrices. Lemma 4.1 Let C E Mm,n, Vk E Mm,k, W E Mn,k be given, where k min{m,n}, and Vk, Wk have orthonormal columns. Then

(a) O"j(V':CWk) ~ O"j(C) j = 1,2, ... ,k, (b) Idet Vk*CWkl ~ O"l(C) ... O"k(C).

and

~

52

FOURTH CHAPTER

o

Proof See [Ho-J2), Lemma 3.3.1.

Let Aj = Aj{A) denote the eigenvalues of A, O'j = O'j{A) singular values and recall that we number them in the order of decreasing absolute values. Theorem 4.5 (H. Weyl, 1949) If A E Md, then k

k

j=1

j=1

III Ajl :5 II O'j

(4.7)

for k = 1,2, ... ,d,

with equality for k = d.

Proof Let A =diag (AI, ... ,Ad)' By the Schur Decomposition Theorem there exists a unitary U and a strictly upper triangular N such that

A = U{A + N)U*. Let Uk E Md,k denote the k first columns of U. Then we have

A+N=U*AU= (Uk:Uk

~)

with some matrices E, F, G. Since A + N is upper triangular, F = 0 and Uk AUk is upper triangular. Now we apply Lemma 4.1 with C := A, Vk = Wk := Uk and conclude

k

k

k

j=1

j=1

j=1

III Ajl = IdetUkAUkl :5 II O'j(Uk AUk ):5 II O'j. When k

= d the SVD gives us

Idet AI = Idet U det E det V* I = det E

o

and the equality in (4. 7) follows.

If the singular values of A and B are known, what can be said about the singular values of AB? We formulate the answer in the square matrix case. For the general case, see [Ho-J2), Theorem 3.3.4.

Theorem 4.6 (A. Horn, 1950) If A, B E Md, then for k = 1,2, ... ,d k

k

j=1

j=1

II O'j(AB) :5 II O'j {A)O'j (B)

(4.8)

with equality for k = d.

Proof Let AB = YEW· be the SVD of AB and put Vk for the k first columns of V, and W k for those of W. Then

V: ABWk = diag{O'I{AB), ... ,00k{AB)).

(4.9)

Consider BWk E Md,k' It can be written, using polar decomposition, as

BWk=Uk R where Uk E Md,k has orthonormal columns and R E Mk is positive semidefinite satisfying

53

FOURTH CHAPTER

Then Lemma 4.1 gives k

det R2 = det(W; B* BWk) ~

II Uj (B* B). j=l

But uj(B* B) = Uj(B)2 and thus det R2 ~

k

Il uj(B)

2

. From (4.9) we obtain

j=l k

II uj(AB) =ldet(Vk' ABWk) I j=l

=ldet(Vk* AUkR) I =ldet(Vk' AUk)lldet RI· But by Lemma 4.1 Idet Vk' AUk I ~

k

Il uj(A)

and since detR ~

j=l

k

Il uj(B), j=l

(4.8) follows. For k = d there must be equality as Idet(AB) I = Idet Alldet BI and d

d

j=l

j=l

Il Uj = I Il Ajl.

D

The singular values of a sum of two square matrices can be easily estimated with help of Theorem 4.4.

Lemma 4.2 If A, B E M d , then for 1 ~ j, k

Proof Let A j -

17 Bk-l

~

d, j

+k

~

d + 1 we have

be as in Theorem 4.4. Then, since

rank(A + B) ~ rank(A)

+ rank(B) = j

- 1+k - 1 = j

+k -

2,

D

Definition 4.1 For A E Md put for k = 1,2, ... ,d k

IIIAlllk := L:uj(A). j=l

These are sometimes called Ky Fan-norms. For k = 1 we have the induced operator norm IIAII (spectral norm), and with k = d we have the trace norm, also denoted IIAI11' IIAlltr.

Theorem 4.7 For k = 1, ... ,d, III . III k is a submultiplicative norm in Md, i.e., IllABlllk ~ IllAlilk IIIBlllk.

FOURTH CHAPTER

54

o

Proof See e.g. [Ho-J2], section 3.4.

The total logarithmic size of a matrix

In the value distribution theory log+lfl separates the large values of If I from those of small ones. When looking at meromorphic functions F: z 1-+ F(z) E Md we need to be able to do the same thing. Definition 4.2 For A E Md, put d

s(A)

:=

L log+ uj(A). j=1

We may call it the total logarithmic size of A.

In order to study simple properties of s(A) we need the following simple technical tool. Lemma 4.3 Let a1 ~ a2 ~ ... such that for k = 1,2, ... ,d we have

~

ad

~

k

k

j=1

j=1

0, {31

~

{32

~

...

~

{3d

~

0 be given

II aj ::; II (3j. Then d

d

Llog+(aj)::; Llog+({3j).

j=1

(4.10)

j=1

Proof If a1 ::; 1, then (4.10) holds. Otherwise, if we put ad+1 := 0, then let 1::; m ::; d be such that am ~ 1 but a m+1 < 1. Then, with k := m, d

m

m

Llog+aj = Llogaj ::; Llog{3j. j=l

j=1

j=1

But m

d

m

Llog{3j ::; Llog+{3j ::; Llog+{3j,

j=1

j=1

j=1

o

and (4.10) follows.

Theorem 4.8 For A, B E Md we have

s(AB) ::; s(A)

+ s(B).

s(A + B) ::; 2(s(2A)

+ s(2B)).

(4.11)

(4.12)

FOURTH CHAPTER

55

Proof Put OJ := O"j(AB) and (3j := O"j (A)O"j (B). Then (4.8) allows us to apply Lemma 4.3 to conclude d

s(AB) ~ I)og+(O"j(A)O"j(B)). j=l But 10g+(O"j(A)O"j(B)) ~ log+ O"j(A) + log+ O"j(B) and (4.11) follows. To obtain (4.12) notice that for any nonnegative a, b we have 1 log+ '2(a + b) ~ log+ a + log+ b.

Since O"j(A) = !O"j(2A) we have by Lemma 4.2 1 0"2j-1(A+B) ~ '2(O"j(2A)+O"j(2B)).

Therefore log+ 0"2j-1(A + B) ~ log+ O"j(2A) + log+ O"j(2B) holds for and since 0"2j(A + B) ~ 0"2j-1(A + B) we obtain (4.12).

o

When we deal with large dimensional matrices or with operators in trace class we often want to write them as [ + A. Corollary 4.4 s(I + A + B) ~ 2(s([ + 2A) Proof Write

[+ A + B

= (![

+ s(I + 2B)).

+ A) + (![ + B) and use

(4.12).

o

Theorem 4.9 Let A be invertible. Then

s(A) = S(A-1) + log Idet AI. Proof We have by Theorem 4.5 d

II O"j = Idet AI

j=l and thus log

d

d

j=l

j=l

II O"j = I)og O"j =

d

d

1

j=l

j=l

3

L log+ O"j - L log+;-: = log Idet AI·

But ;. 's are the singular values of A-1 and so substituting J

d

1

s(A-1) = '"'log+~ 0". j=l 3 gives the result. It is convenient to put s(A -1) = 00 if A is not invertible.

o

FOURTH CHAPTER

56

Theorem 4.10 A is unitary if and only if s(A)

+ S(A-l) = o.

Proof If A is unitary, then the SVD is A = U with E = I, and s(A) = s(A -1) = O. Reversely, if s(A) + S(A-l) = 0 holds then uj(A) = 1 for all j and the SVD is A = UEV· = UV· as E = I. But UV· is unitary and we are done. 0

Some basic properties of the total logarithmic size We start by studying how the total logarithmic size behaves in similarity transformations. Let A = SBS- 1. Then by Theorem 4.6 k

k

j=1

j=1

IT uj(A) ::; IT Uj(S)Uj(B)Uj(S-I).

(4.13)

But Uj(S-I) = I/Ud-j+l(S) and if we define Kj(S) := Uj(S)/Ud-j+l(S),

then we obtain from (4.13) k

k

j=1

j=1

IT uj(A) ::; IT Uj(B)Kj(S).

Notice that Kl (S) = IISIlIIS- 1 11 is the condition number of S. Since Kj(S) ~ Kj+1 (S) we obtain using Lemma 4.3 d

d

~)og+uj(A) ::; I)og+(uj(B)Kj(S)) j=1

j=1 d

d

::; Llog+uj(B) j=1

+ Llog+Kj(S), j=1

so that s(A) ::; s(B)

+ c(S),

(4.14)

where c(S) is defined in (4.15). Definition 4.3 For invertible S put d

c(S):= Llog+Kj(S), j=1

the total (logarithmic) conditioning of S. Theorem 4.11 Let S, Rand T be invertible matrices. Then (b)

c(SR- 1) =0 if and only if SR- 1 is unitary c(SR- 1) =c(RS- 1),

(c)

c(ST- 1) ::;C(SR-l) + c(RT- 1).

(a)

(4.15)

FOURTH CHAPTER

57

Proof (a) c(SR- 1) ~ 0 always, but if c(SR- 1) = 0 then in particular = 1 and SR- 1 is unitary. Reversely, for a unitary SR- 1, O'j (SR- 1) = 1 for all j and c(SR- 1 ) = o. (b) is trivial from the definition. (c) follows from writing ST- 1 = SR- 1RT- 1 = (SR- 1)(RT- 1) and using /'i,1 (SR- 1)

c(AB) ~ c(A)

+ c(B),

(4.16)

which holds for any invertible matrices A, B. To obtain (4.16) notice that k

k

II /'i,j(S) = II

O'j(S) . j=IO'd-j+1(S)

j=1

We have

k

k

k

j=1 d

j=1 d

j=1 d

II O'j(AB) ~ II O'j(A) II O'j(B). .n

.n

.n

U.(~B) ~ u.tA) u}B)' Thus with m = d-k+l J=m ' J=m' J=m' we obtain by multiplying both sides But in a symmetric way,

k

k

j=1

j=1

II /'i,j(AB) ~ II /'i,j(A)/'i,j(B) which implies (4.16) with help of Lemma 4.3: d

d

I)og+/'i,j(AB) ~ L)og+(/'i,j(A)/'i,j(B)) j=1 j=1 d

d

~ L)og+/'i,j(A) j=1

+ I)og+/'i,j(B). j=1

o Corollary 4.5 If A, B are invertible, then (4.16) holds. When we think of SR- 1 as the similarity transformation which takes RAR- 1 into SAS-l = (SR- 1)(RAR- 1)(RS- 1) then we may think of c(SR- 1) as the "distance" between the similarity transformations Sand R. In particular, the following shows the continuity of the total logarithmic size s in similarity transformations. Theorem 4.12 If Sand R are invertible, then for all A Is(SAS- 1) - s(RAR- 1)1 ~ c(SR- 1).

Proof From (4.14) we have s(SAS- 1) ~ s(RAR-l)

+ c(SR- 1)

and likewise

s(RAR-l) ~ s(SAS- 1) + c(RS- 1). Since c(RS-l) = c(SR- 1), the claim follows.

o

FOURTH CHAPTER

58

We conclude that s(A) behaves in a natural and controlled way under similarity transformations. Next we ask, in what ways we can possibly estimate the norm IIAII and s(A) in terms of each others. Since O'l(A) = IIAII, we have trivially IIAII ~ exp (s(A)) and

s(A) ~ dlog+ IIAII. However, if we know the function s(zA), then the norm can be obtained accurately. In fact, 1 IIAII = sup{lzl I s(zA) = o}.

(4.17)

Let us now look at the power An and the exponential e zA . The behavior of s(An) is related to the spectral radius formula: lim IIAnIl 1/ n = p(A) = max{IAI I A E O'(A)}.

n-+oo

(4.18)

Let A = diag(A1(A), A1(A), ... ).

Theorem 4.13 We have lim .!.s(An) = s(A). n

n-+oo

(4.19)

Proof The claim follows from the following generalization of (4.18): O'j(An)l/n ~ IAj(A)I, which is due to Yamamoto (1967), and generalizes to compact operators, see [Ro], Proposition 2.d.6. 0

Theorem 4.14 For z E C, Izl = r, A E Md, s(e zA ) ~ r11A111'

(4.20)

d

where IIAI11 = ~ O'j(A). j=l

Proof Since ezA = lim(1 + ~A)n, we have s(e zA ) ~ liminf n s(1 + ~A). But O'j(1 + ~A) ~ 1 + ~O'j(A) by Theorem 4.4 and thus log+ O'j(1 + ~A) ~ ~O'j(A), and the claim follows. 0 We have observed that s behaves nicely when two matrices are multiplied together, but estimates for the sum are necessarily somewhat more complicated. Consider the sum of two matrices. If A = B = I, then s(A) = s(B) = 0, but s(21) = dlog 2. By Lemma 4.2 we have

59

FOURTH CHAPTER

and since 0"2k ::; 0"2k-1 we have d

Ld/2J

j=l

j=l

L log+O"j(A+B)::;2 L

10g+(O"j(A) +O"j(B))

::;2(s(A)

+ s(B)) + rank(B) log 2.

(4.21)

Another grouping of the indices in Lemma 4.2 is also useful. Theorem 4.15 For A, BE Md we have

s(A+B)::; 2(s(A) +s(B))

+ rank(B) log 2

(4.22)

and

s(A + B) ::; s(A)

+ s(B) + rank (B) (log+ IIAII + log 2).

(4.23)

Proof Inequality (4.22) is in (4.21) while (4.23) follows from

O"j(A + B) ::; IIAII

+ O"j(B),

j::; rank(B)

and from

O"j+k-1(A + B) ::; O"j(A) + O"k(B) = O"j(A), where k > rank(B). Thus Llog+ O"j(A + B) ::; rank(B)(log+ IIAII

+ log 2) + Llog+ O"j(B) + Llog+ O"j(A). D

Finally, if we add something small into A we do have an estimate without additional terms, so that we see the natural continuity. Continuity Lemma 4.4 If A, B E Md, then

Is(A) - s(B)1 ::; IIA - Bill.

(4.24)

Proof If a and b ~ 0, then 110g+(a) -log+(b)1 ::; la - bl· So, we have

Is(A) - s(B)1 ::;

L

Ilog+ O"j(A) -log+ O"j(B)1

::; L 100j(A) - O"j(B)I· The trace norm 11.111 has the following property. Form from the singular values two diagonal matrices E(A) and E(B) respectively, arranging the diagonals in the usual decreasing order. Then

IIE(A) - E(B)111 ::; IIA - Bill see [Ho-J1J, p. 448. This completes the proof.

D

FOURTH CHAPTER

60

Direct sum, Kronecker product and Hadamard product Given two matrices A E Md1 , B E M~, operating in C d1 , C d2 respectively, their direct sum A E9 B is the linear mapping in C d1 E9 C d2 which maps as (x, y) E C d1 E9 C d2 to (Ax, By) and can be represented with a block diagonal matrix

The singular values of AE9B E Mdl+d2 are clearly O'j(A), O'k(B), j = 1,2, ... ,d1, k = 1,2, ... , d 2 • Therefore we have the following result:

s(A E9 B) = s(A) + s(B).

(4.25)

The Kronecker product of two matrices A E Mm,n and B E Mp,q is denoted by A ® B and is given by

anB A®B=

a

(

1n

B) E Mmp,nq

:

:

am 1 B

amnB

where =

(a~1

a~n) ..

A.

.

.

am1

amn

Remark In this notation the rank-l matrix xy*, with x, y E Cd becomes xy* = x®y*.

In multilinear algebra it is customary to write x ® y for the bilinear mapping. Our notation here follows the matrix analysis tradition where x is thought of as a column vector and x* as a row vector. Lemma 4.5 (A ® B)(C ® D) = AC ® BD. Proof This is of course under the assumption that the dimensions match so that the ordinary products make sense. To prove this, split into blocks and multi~

0

Corollary 4.6 If A E Md 1 and B E (A ® B)-1 = A- 1 ® B-1. Remark In general A ® B prove it:

=f.

Md2

are invertible, then so is A ® Band

B ® A so observe the order: A-I ® B- 1 • To

(A ® B)(A- l ® B- 1 ) =AA- l ® BB- l = Idl ® Id2 =(A- 1 ® B- 1)(A ® B).

= Id1d2

FOURTH CHAPTER

61

We can now compute the spectrum of A ® B for A E Mdl' B E Md2. Let Ax = AX and By = f,Ly, where x, yare eigenvectors, and A, f,L eigenvalues. We have by Lemma 4.5 (A ® B)(x ® y) = (Ax ® By) = Af,LX ® Y so that Af,L E O'(A®B), and x®y E C d1d2 is a corresponding eigenvector. If we take all products Ajf,Lk with multiplicities when needed, we have obtained all eigenvalues of O'(A ® B). This is easiest to check using the Schur decomposition. In fact let L, M be upper triangular and U, V unitary so that

A = ULU*,

B = VMV*.

Then U ® V is unitary (by Corollary 4.5) and L ® M is upper triangular, with Ajf,Lk'S on the diagonal. The conclusion follows as by Lemma 4.5

(U ® V)* (A ® B)(U ® V) = L ® M. In order to obtain the singular values of A®B observe that since (A®B)* = A*®B*, we have (A ® B)*(A ® B) = (A* A) ® (B* B) and the previous result on eigenvalues implies that the singular values of A ® B are obtained as products of singular values of A and B:

where j

= 1, ... ,dl , k = 1, ...

,d2 • We obtain the following result.

Theorem 4.16 If A E Mdl' BE

then

Md2'

s(A ® B) ~ d2 s(A)

+ dl s(B).

(4.26)

Proof Consider the product of all singular values of A®B and replace O'i(A® B) with ui(A ® B) where, Q := max{a, I}. Then s(A ® B) is the logarithm of the new product. But ui(A ® B) = (O'j(:A);;;(B)) ~ Uj(A)Uk(B) and doing this for every o'i(A ® B) yields the following product: dl

d2

IT Uj(A)d IT uk(B)d 2

j=1 The logarithm of this is d2 s(A)

1•

k=1

o

+ dl s(B).

Given two matrices A,B E Mm,n with elements (aij), (bij ) respectively, their Hadamard product A 0 BE Mm,k is defined by

(A 0 B)ij = aijbij . This is sometimes called the Schur product or the entrywise product. An important and simple result related with this product is that the Hadamard product of positive semidefinite matrices is always positive semidefinite. If A, B E Md are positive definite and the eigenvalues are ordered decreasingly then k

k

j=1

j=1

IT Aj(A)Aj(B) ~ IT Aj(A

0

B),

for k = 1,2, ... , d

FOURTH CHAPTER

62

(see [Ho-Jl], p. 316). For our purposes the reverse inequalities are of interest. To that end, for any A, BE Mm,n, already Schur showed that

0'1 (A 0 B)

~

0'1 (A)O'l (B).

Let A E Md' Put r1(A) ~ r2(A) ~ ... ~ rd(A) for row sums as follows: (rk(A))2 d

is the kth largest number among L: laij 12, i = 1, ... ,d. Let Ck (A) be defined j=l similarly for column sums. Then the following holds: for A, B E Md: k

IT O'j(A

j=l

k

0

B) ~

IT cj(A)rj(B),

k = 1,2, ... ,d

j=l

[Ho-JI], p. 355). This allows an easy upper estimate for s(A Lemma 4.3

0

B). In fact, by

d

s(A 0 B) ~ I)og+[cj(A)rj(B)]. j=l

(4.27)

Comment 4.1 In the discussion above we have used [Ho-J2] as a basic reference. Comment 4.2 Notice that s(A) is not of the form log+ IIAII in any operator norm. In that sense it is really a different "tool". Many of its properties appear here first time. Comment 4.3 The total logarithmic size generalizes for bounded operators A in Hilbert spaces. In fact, let O'j(A) :=

inf

rank(B)<j

and then set

IIA -

BII,

00

s(A)

L)og+ O'j(A). j=l For example, with compact operators K we always have s(K) < 00, and if K is in the trace class, that is, IIKll1 := L:~1 O'j(K) < 00, then also s(1 - K) < 00. Many properties of the total logarithmic size can be proved simply by approximating techniques. We refer to [N08]. See also the subsection "Extension to trace class" in the next chapter. :=

FIFTH CHAPTER Keywords: Inversion identity, trace class, finitely trace class meromorphic, Schatten class. The total logarithmic size is subharmonic We shall consider here the subharmonicity of the total logarithmic size of an analytic Md-valued function. To that end we shall first consider a problem related to the eigenvalues instead of the singular values. We follow closely [A2]. Let {>.j(AHt denote the eigenvalues of A E Md, indexed so that

If F is now an analytic function in a domain n we know from Theorem 3.2 that log+ IA1(F(z)1 is subharmonic as IA1(A)1 gives the spectral radius of A. However, the corresponding function with the other eigenvalues need not be subharmonic.

Example 5.1 Let

F(Z)=(~ ~) so that the eigenvalues are {I + z, 1 - z}. Thus IA1(F(z))1 = max{11 + zl, 11- zl} while IA2(F(z))1 = min{11+zl, 11-zl} and we see that IA2(F(z))1 is not subharmonic as it violates the mean value property at the origin with a small enough radius. Lemma 5.1 Let F be analytic from a domain

n into Md.

Then the functions

k

Uk(Z)

:=

L log IAj(F(z))1

(5.1)

1

are subharmonic for k = 1,2, ... , d. Proof Fix Zo E n and choose an eigenvalue /-to E a(F(zo)) and take a small enough radius s > 0 such that the closed disc B(/-to,s) contains no other eigenvalue of F(zo). Then we can fix a small 8 > 0 such that for Iz - zol < 8 no eigenvalue of F(z) touches the circle {}B(/-to, s), which is possible as the eigenvalues are continuous and there is only a finite number of them. When counted with multiplicities, let rno denote the multiplicity of /-to, so that d-rno eigenvalues of F(z) stay outside of B(/-to, s). If rno = 1 then /-to(z) is analytic in z, but for rno > 1 it may happen that the eigenvalue is not analytic. In such a case the eigenvalue splits into several eigenvalues, say, into /-t1 (z), ... , /-tmo (z), each of which is analytic in a small punctured neighborhood of zoo Notice that some of these eigenvalues can be multiple copies 63

64

FIFTH CHAPTER

of each others, but then they stay as copies and each one is separately analytic. In any case, if one defines a function h around Zo by setting

and for 0 <

Iz - zol < 8 by rno

h(z) := IIJ.Lj(Z) 1

then h is continuous at Zo and analytic in a punctured neighborhood. Thus the singularity is removable and h is analytic at zo, too. But then log Ihl is subharmonic at Zo and in particular 1

r

rno

mo log lJ.Lol :S 271" 1-11' ~ log lJ.Lj(zo + pei'l')ld
(5.2)

for 0 < p < 8. Consider now the function Uk in (5.1). Suppose that k happens to be equal to the sum of p first elements in u(F(ZO)), when counted with multiplicities. Apply now (5.2) p times and sum the both sides pairwise. The left hand side gives you Uk (zo). The terms on the right under the integral can all be estimated by the sum over k largest ones, this gain giving the integral over Uk. We end up with (5.3)

for all small enough p. Since the eigenvalues are continuous functions so is Uk. What remains is to look at the situation where the multiplicities do not match in the simple way as above. Namely, let l be the sum of multiplicities not exceeding k and let m be the multiplicity of Ak(F(zo)). Thus l < k < l +m.

o

We omit the extra details, and refer to [A2].

Lemma 5.2 Let A E Md be given. Then for every k :S d k

k

1

1

mrrc II IAj(UA)I = II Uj(A)

(5.4)

where the maximum is taken over all unitary matrices U E Md. Proof As the singular values are invariant under multiplication by a unitary matrix the Weyl inequalities give k

k

1

1

II IAj(UA)I :S II Uj(A). But by the polar decomposition of A we have A = (A* A)1/2UO and thus

Aj(U0 1 A)

= uj(A).

Therefore the equality in (5.4) is always obtained.

o

FIFTH CHAPTER

65

Lemma 5.3 Let F be analytic from a domain n into Md. Then the functions k

Vk(Z) := ~)ogO"j(F(z))

(5.5)

1

are subharmonic for k = 1,2, ... , d. Proof Choose k. Clearly Vk is continuous, so we need to check the mean inequality. Given a unitary U let us put k

ur (z) := I : log IAj(UF(z))l· 1

Take now a point Zo E

n and let Uo be unitary such that ufO(zo) = Vk(ZO)

which is possible by Lemma 5.2. Since ufO(z) is subharmonic we have the mean inequality for it:

(5.6) But by Lemma 5.2 we have Vk(Z) = maxu ur (z) and taking the maximum under the integral gives 1

r

.

Vk(ZO) :::; 211" 1-11: Vk(ZO + re"")dcp. and the subharmonicity follows.

D

We can now formulate the main result of this subsection. Theorem 5.1 Let F be analytic from a domain n into Md. Then its total logarithmic size s(F(z)) = I:log+ O"j(F(z))

is subharmonic in n. Proof We have, in the notation of Lemma 5.3,

s(F(z)) = max{vk(z)} k

and since the maximum of subharmonic functions is subharmonic, we are done. D

Behavior near poles So far we know that s(F(z)) is subharmonic whenever F is analytic. We shall now look at its behavior near a possible pole. Recall that the multiplicity m(b) of a pole b can be obtained by

m(b) = lim log+ IIF(z)11 . z-+b log(1/lz - bl) Here we introduce another multiplicity in the same spirit.

FIFTH CHAPTER

66

Definition 5.1 If F is a meromorphic Md-valued function in a domain define at every Zo E n

J.t(Zo) := lim sup Z-+Zo

s(F(~)) . log Iz-zol

n, we (5.7)

Clearly J.t(zo) = 0 exactly when F is analytic at zoo That it takes only integer values is not so obvious. Lemma 5.4 If F is Md-valued and has a pole at b, then J.t(b) is a positive integer, depending only on the principal part L=~ Aj(z - b)j. Proof Write

-1

L

Aj(z - b)j + G(z) j=-m with G analytic near b. Then by (4.24) in the Continuity Lemma 4.4 F(z) =:

Is(F(z)) - s(F(z) - G(z)) I ~

IIG(z)lll.

But IIGlll ~ d I Gil is bounded near b and has thus no effect on J.t(b) which shows that J.t(b) depends only on the principal part. We can now assume without restricting the generality that b = 0 and that -1

F(z) =

L

Ajzj.

j=-m Let Aj(Z) denote the eigenvalues of F(z)* F(z), ordered decreasingly. These are nonnegative and their square roots are the singular values. The characteristic polynomial can be "expanded by diagonal elements": det(AI - F(z)* F(z)) = Ad - bl(z)A d- l

+ ... + (-l)dbd(z)

where bl = L Aj, b2 = Lih AiAj etc. As all the eigenvalues are nonnegative, the functions bj are nonnegative as well. Also, if bk = 0 then bj = 0 for j = k + 1, ... , d. The coefficients bj are sums of all principal minors of order j in det(F* F). These are determinants of j x j submatrices which in turn are of the form Fj* Fj where each Fj is a d x j matrix consisting of j columns of F. Let I j denote a selection of j rows from a matrix so that Fj(Ij ) denotes a j x j submatrix of Fj . The Cauchy-Binet Theorem allows us to conclude that then

bj

=

L IdetFj(IjW

where the summation is over all j x j minors Fj(Ij ) of F. But determinants are meromorphic and therefore there exists Cj > 0 and an integer mj such that

bj(z) = cj(l + o(1))r2m; as Izl = r ---+ O. Consider now bl decreasingly we have

= L~ Aj. As the eigenvalues are numbered

which further implies Cl

-d

~

l'

1m

. f Al(Z)

III

-2- ~

z-+o r

ml

l'

1m sup

Al(Z)

- 2 - ~ Cl'

z-+o r

ml

FIFTH CHAPTER

67

For the coefficient b2 we have in the same way

),1),2::; b2::; (~),1),2 This implies

< l'lIDm . f

C2 -C1 (~) -

< lim sup

),2(Z)

z->O

r 2 (m2- m l) -

z->O

< -c2 d

),2(Z)

r 2 (m2- m d -

C1

Continuing this way we see that if ),j is not identically 0, then there exists constants aj > 0 and an integer kj such that

a~ < lim inf ),j(z) < lim sup ),j(z) < ~. z->O r 2k j

J -

-

z->o r2kj

-

a~ J

Taking the logarithm and dividing by 2 gives log aj ::; lim inf (log Uj (F(z)) z->O

1

+ k j log -r )

::; lim sup (loguj(F(z)) + k j z-+O

log~) r

::; log ~ aj

Since the eigenvalues were ordered decreasingly there is a largest J such that k j < 0 for j ::; J. Summing over j then gives J

0: ::;

lim inf ( " log+ uj(F(z)) + " k j log ~) z->O

~

~

r

j=1 J

::; lim sup (Llog+Uj(F(z)) 0:

- 'E.;=1

:=

r

j=1

Z-+O.

where

+ Lkjlog~)::;(3

'E.;=11ogaj and (3 := 'E.;=11og

;j'

Thus, in particular, J.L(O) .-

0

k j is an integer.

The proof actually gave somewhat more. Namely that limsup can be replaced by lim and that the limit process is controlled with bounds. Lemma 5.5 If F is as above, then .

1 ) p. (zo=lm

Z-+Zo

and there are constants 0: ::;

0:

s(F(z)) 1 log p-=r IZ-ZOI

(5.8)

and (3 such that

lim inf (s(I - F(z)) - J.L(zo) log I 1 I) z-+zo z - Zo

::; lim sup (s(I - F(z)) - J.L(zo) log I 1 Z-+Zo

z - Zo

I)::; (3.

Proof The inequalities are explicitly available in the previous proof and the limit in (5.8) is obtained by dividing the estimates by log(l/Iz - zol). 0 We shall need an auxiliary function.

FIFTH CHAPTER

68

Izl <

Definition 5.2 If F is meromorphic for (when z is not a pole)

u7j(z) := s(F(z)) + L

Ibl~7j

R, then for

Izl

~ 1J

< R set

p,(b) log 11J~Z - b) I. 1J - bz

At poles b define u7j(b) := limsuPz-+b u7j(z).

Lemma 5.6

u7j

Izl ~ 1J

is subharmonic in

and equals s(F) on

Izl =

1J.

Proof By Lemma 5.5 u7j is bounded near poles and as it is a sum of subharmonic and harmonic except at poles we conclude that u7j is subharmonic also at poles. 0 Introducing Tl for matrix valued functions It is now natural to count the multiplicities as follows.

Definition 5.3 If F is meromorphic for nl(r,F):=

Izl < R, then for r < R denote

L

(5.9)

p,(b).

Ibl~r

Likewise, Nl (r, F):=

Further

r

10

nl (t,

F) -

nl (0,

F) dt + nl (0, F) log r.

(5.10)

t

(5.11) and finally T1(r,F) := ml(r, F)

Lemma 5.7 If F is meromorphic for

+ Nl(r, F).

Izl < R,

then in the notation above, for

r~1J
(5.12) Proof This is a direct calculation, based on 1 111" log la - ei"'ldcf> = log+ -2 7r

-11"

lal·

o The following theorem summarizes the main properties of the characteristic function T1 •

Theorem 5.2 If F is an Md-valued meromorphic function in Izl < R ~ 00, then Tl (r, F) is well defined, nonnegative, nondecreasing in r < R such that it is convex as a function of log r. It satisfies Too(r,F) ~ T1(r, F).

(5.13)

If G is another such function, then T1(r,FG) ~ T1(r,F)

+ T1(r, G).

(5.14)

69

FIFTH CHAPTER

Proof Positivity of Tl is clear from the definition. It is increasing and convex in the variable log r by Lemma 5.7 as this is a general fact of mean values of subharmonic functions, see [Ha-K]' p.127. The inequality (5.14) follows from Theorem 4.8. D The analogue of Theorem 3.4 holds also for T 1 , by the same argument, since s(F) is subharmonic. However, we need to introduce the following notation: d

M1(r,F):= sup

II max{uj(F(z)),l}.

(5.15)

Izl~r j=l

Theorem 5.3 If F is an analytic Md-valued function for r

Izl < Ro

and 0 <

< (}r < Ro, then (5.16)

We are now ready to invert F.

Basic identity for inversion Theorem 5.4 Let F be a meromorphic Md-valued function for that around origin

Izl < R

such

(5.17) with

ICkl =F O.

Then (5.18)

holds for all r < R.

Proof By Theorem 4.9 we have log I det FI = s(F) - s(F-l)

(5.19)

so that the definition of ml implies immediately

Likewise, it gives

The claim now follows by summing up these identities and by using Theorem 2.4 for the meromorphic scalar function det F. D We can formulate Theorem 5.4 also in a symmetric form which does not use the determinant function. In fact F can have simultaneously a pole and yet the determinant can even vanish at such a point.

FIFTH CHAPTER

70

Definition 5.4 If F has a pole at b with multiplicity f.L(b) given by (5.7) we set

a(b,F):= lim sup (s(F(z)) - f.L(b)log(l/lz - bl)). z-+b

Note that this is well defined by Lemma 5.5. Applying (5.19) we see that log ICkl = a(O, F) - a(O, F- 1 ). This allows the following formulation.

Corollary 5.1 Let F be a meromorphic Md-valued function for we have

Izl < R.

Then

for all r < R. Any text on scalar Nevanlinna theory would continue from the first main theorem to the second one. In generalizing it to matrix valued functions one meets the need of assuming that matrices evaluated at different points would commute. We shall not go into that direction.

Extension to trace class Here we assume that H is a separable infinite dimensional Hilbert space and we look at functions of the form

F(z)

1 - K(z)

=

where 1 denotes the identity in B(H) and K(z) is compact away from poles and finitely meromorphic in n. In Theorem 3.6 we saw that such operators have inverses in the same class. We can extend T1 for these operators provided we put an extra requirement for the singular values of K. In fact, let A be a compact operator, then .lim O'j(A) = 0 J-+OO

where the singular values can be given for example as distances to finite rank operators (see Theorem 4.4). In fact, if A is bounded we can set

O'j(A) :=

inf

rank(B)<j

IIA -

BII·

Note that it follows that in infinite dimensional spaces we always have for compact operators A and

O'j(l - A) ~ 1 + O'j(A). Thus s(I - A) := L:log+ O'j(l - A) is well defined for all operators A which are in the trace class: 00

IIAI11 := LO'j(A) < 00. j=1

Since the inverse of F = 1 - K is 1 + K(l - K)-1 we see that if K has values in the trace class, so does K(l - K)-1 and we conclude that the following holds.

FIFTH CHAPTER

71

Theorem 5.5 If F = I - K is finitely merom orphic in n, K taking values in the trace class, and if there is a point Zo E n such that F(zo) is invertible, then F-l itself is such a function. Now for these functions we can define Tl exactly as in the Md-valued case, with the natural modification that all sums and products now run from 1 to 00, instead of from 1 to d. The results basically follow from the finite dimensional case by approximation techniques. In particular, the following result is useful.

Approximation Lemma 5.S If F is an analytic trace class -valued function in Iz - zol < Ro, then F can be approximated in the trace norm by finite rank polynomials uniformly in discs Iz - zol ~ TJ < Ro· We refer here to the original paper [NOB], but collect the main results as follows.

Theorem 5.6 Theorems 5.1, 5.2, 5.3 and 5.4 hold as such if we study functions of the form F = 1- K with K finitely meromorphic, taking values in the trace class instead of Md· Example 5.2 An operator valued function can be entire in the uniform norm but have a bounded domain of definition if considered as an analytic function taking values in the trace class. In fact, let Aj be a diagonal operator with positive decaying diagonal elements 0j,k such that 00

LOj,k

= 1

k=l

while

OJ,l = fro

Then

IIAjl1 = fr

while

IIAjlll =

1 and if K(z) := I:j Ajzj then

r

IIK(r)lll =

1- r

IIK(r)11 =

eT -1.

while

How to work outside the trace class When an operator valued function is still compact but is not in the trace class, then the passage from the finite dimensional case to the infinite dimensional one is more complicated. Think as a model case the function

F: z

1--+

I - zA,

where A is compact but not in the trace class. Clearly we can estimate Too(r, F) but if we try to invert F we meet difficulties in writing down an entire function, say X which vanishes at Zj = 1/ Aj and which is needed in the representation of it, see Comments in First Chapter. Namely, we have to use the Weierstrass elementary factors to make the expression to converge. An analogous thing can be done on the operator level and that would eventually yield an expression for the resolvent. This is available in the literature, see e.g. [Du-S].

FIFTH CHAPTER

72

Fortunately there is a much simpler approach that works. To that end, recall that an operator A is said to be in the Schatten class Sp if 00

IIAlip := (L:: O"j(A)P)1/P < 00. ;=1

It follows immediately that if A is in Sp and m ~ p, then Am is in the trace class S1. Now, we can build on the following identity.

m-1 (I - A)-1 = (I - A m)-1

II (I -

ei21rj/m A).

(5.20)

1

If K is now finitely meromorphic, taking values in such a Schatten class, then (5.20) can be utilized simply as follows:

m-1 Too(r, (I - K)-1) ~Too(r,

II (I -

ei21rj/mK)) + Too(r, (I - Km)-1)

1

m-1 ~Too(r,

II (I -

ei21rj/mK)) + T 1(r, (I - Km)-1)

1

where the last term can be estimated using the inversion identity. Thus we have the following.

Theorem 5.7 If K is finitely meromorphic and m is a positive integer such that K(z)m is in the trace class away from poles, then

Too(r, (I - K)-1) ~ T1 (r, 1- Km)

+ (m -

1) (Too(r, K)

+ log 2) + log(1/lckl)

where det(I - Km(z)) = Ckzk + .... Proof Notice that m-1 m-1 Too(r, (I - ei21rj/m K)) ~ (Too(r, K)

II

L

1

1

+ log 2)

= (m - 1)(Too(r, K)

+ log 2).

o Comment 5.1 Theorem 5.1 is from [A2]. Comment 5.2 T1 was introduced for matrix valued meromorphic functions in [N05] together with an inversion identity. The other main properties of T1 and extensions to operators are from [NOB].

SIXTH CHAPTER Keywords: Perturbation, finite rank, trace class, Schatten class, normal operators, bounded characteristics. Perturbation results We start here with the following question. We know the growth of F- 1 and if F is being updated by G what can we conclude on the growth of

(6.1) We shall formulate the first results in a complex Hilbert space H which we assume to be either separable or finite dimensional. The formulation requires (F + G)-l to be meromorphic and to guarantee this we assume that there is a point at which F+G is invertible. This then implies that det(1 +F-IG) does not vanish identically so that the inversion formula can be used. To that end, let Ck t- 0 be such that det(1 + F-1G) =

Ckzk

+ Ck+IZk+1 + . . .

(6.2)

Theorem 6.1 Assume that F, F- 1 and G are meromorphic B(H)-valued for Izl < R, and that rank(G) ~ q. Then (F + G)-l is meromorphic for Izl < Rand forr < R 1 Too(r, (F + G)-I) ~ (q + 1) Too(r,F- 1 ) + q Too(r,G) + q log 2 + log (6.3)

1CrJ.

Proof We write

F + G = F(1 +F-1G) and observe that since F- 1 is meromorphic, and G is both meromorphic and of finite rank, their product is both meromorphic and of finite rank and TI is defined for 1+ F-1G. But then we have with help of the inversion identity

Too(r, (F + G)-I) ~Too(r, F- 1) + Too(r, (I + F-1G)-I) ~Too(r, F- 1) + TI (r, I + F-1G) + log

1 1CrJ.

(6.4)

If a, b ~ 0 then always

10g(l + ab) ~ log+ a + log+ b + log 2. We use this in estimating log+ uj(1 + F-1G). For j > q, uj(1 + F-1G) ~ 1 (with equality if dim H = 00) while for j ~ q we have uj(1 + F-IG) ~ 1 + IIF- 1II IIGII· Thus 73

SIXTH CHAPTER

74

and

T1 (r,! + F-1G) 5: q (Too(r, F- 1 )

+ Too(r, G) + log 2).

o

The claim follows now by substituting this into (6.4).

If we think this as a prototype of the perturbation estimates there is one important point to be observed. We do not assume that F nor F-l would be finitely meromorphic. Information on F goes into the estimates only in the form of Too (r,F- 1 ) which does not control the dimensions of the principal parts at all. In this result all control of this nature is taken care by the finite rank assumption on the perturbation G. The following example aims to illustrate this. Example 6.1 Let H = l2 and take F(z) = I - zI so that F is meromorphic in the whole plane with

Too(r, F) = Too(r, F- 1 ) = T(r, 1 - z) = log+ r. We perturb F(z) now with a compact operator G(z) = z diag(oj) where OJ is a nonnegative decreasing sequence converging to zero. We have

Too(r, G) = 10g+(0Ir). Thus, only the largest term, 01 enters the estimates. Yet, the poles of (F+G)-1 = diag(l/(l-z(l-oj)) are at Zj = l/(l-oj) and thus converge to 1. Let us compute Too(r, (F + G)-I). We show below in Lemma 6.1 that for a self adjoint bounded A

moo(r, (I - zA)-I) 5: log 2. Now F

+G = I

= I -diag( OJ) so that we have moo(r, (F + G)-I) 5: log 2.

- zA with A

Suppose now that rank(G) = q so that 01 ~ ... ~ Then the point 1 is also a pole. Thus

Oq

> 0 while

OJ

= 0 for j

> q.

q

Noo(r, (F + G)-I) = log+ r + Llog+(11- ojlr). j=1

Thus

Too(r, (F + G)-I)

=

(q + 1) log+ r + 0(1).

Suppose now that all oj's are positive. Then (F + G)-1 is meromorphic for Izi but has an essential singularity at 1. We have for r < 1

<1

and we get no "warning" that (F + G) -1 is not meromorphic for larger r. Finally, observe that if we had OJ -+ 00 modelling the inverse of a compact operator, then (F + G)-1 would be meromorphic in the whole plane with 00

Too(r, (F + G)-I) = Llog+(l1- ojlr) j=1

In the example above we used the following Lemma.

+ 0(1).

SIXTH CHAPTER

75

Lemma 6.1 Let A be a bounded self adjoint operator such that (I - ZA)-l is

meromorphic for

Izl < R.

Then for

r
moo(r, (I - ZA)-l) ::; log 2.

11(1 -

Proof Since the operator is self-adjoint, we have

d(re itP ):= inf

AEu(A)

11 -

(6.5) zA)-lll =

dtz)

where

z'\l

~ l~~ 11- z'\l = Isin¢l·

Thus 1 111" 1 1 111" 1 -2 log+ d( i'P) dcp ::; -2 log -I-.-I dcp = log 2. 7r -11" re 7r -11" sm cp

o

In Example 6.1 the speed (q + 1) log+ r + 0(1) could in principle come from both F- 1 or from G as their growths are both of the form log+ r + 0(1) and we do not learn whether in (6.3) the coefficients q + 1 and q are both really needed. Example 6.2 Let now F(z) := eZ 1 while G =diag(aj) with 1 > al > ... > a q > 0 and aj = 0 for j > q. Now 1 Too(r,F- 1 ) = T(r,e- Z ) = -r, 7r

while Now

(F + G)-l = diag(

eZ

1 -

aj

)

(6.6)

and we see that growth of moo comes from the components with j > q while N 00 collects everything from the components j ::; q. We see that

Noo(r, (F + G)-I) = !I r + 0(1) 7r

as every aj creates a different sequence of poles. Since for r large enough moo(r, (F + G)

-1

1 ) = -r, 7r

we obtain

Too(r, (F + G)-I) = (q + 1) Too(r,F)

+ 0(1).

Example 6.3 We shall now expose the effect of G and see that also the term q Too(r, G) is needed in (6.3). Let F := diag(q, q - 1, q - 2, ... ,2,1,1,1, ... )

so that Too(r, F- 1 ) = 0 for all r, while G(z) := eZ 1q where 1q = diag(l, ... , 1,0, ... ) is the rank-q projection onto the q first components. Now r

Too(r,G) = 7r

while

SIXTH CHAPTER

16

On the other hand, (F + G) - I has poles at z = log j j = 1,2, ... , q. Therefore

Too(r, (F + G)-I) ~ Noo(r, (F + G)-I)

=

+ 211"in q !:.. 11"

for all n and for

+ 0(1).

This is

Too(r, (F + G)-I) ~ q Too(r, G) + 0(1), and we conclude that multiplying Too(r, G) with q is really needed. Example 6.4 We shall modify Example 6.2 a little bit. Consider the poles that are created in (6.6). If G is not of finite r~k but O'.j > 0 for all j, converging to 0, we shall still have (F + G) -1 meromorphic on the whole plane. This is due to the fact that F-l is entire. However, there are really a lot of poles being created. To see this, let 1

O'.j

p+e

=

so that G is in the trace class: IIGlll = O(!). c Each O'.j gives rise to a sequence of poles {(I + c) logj + i211"n}nEZ, If we just count the real zeros then a crude estimation gives for r large (6.7) Notice that this is vastly faster than what we had for rank-q perturbation, in which case Noo(r, (F + G)-I) = !I r + 0(1). 11"

The following theorem and its corollary show that we have

Too(r, F + G)-I) :5 Moo(r, F- I ) sup IIGlh

+ log+ Moo(r, F) + 0(1)

Izl~r

= er

O(I/c) + O(r).

Compared with (6.7) we see that the fast growth allowed by the next theorem can actually happen. Theorem 6.2 Assume that F is meromorphic, F-l analytic and G is finitely meromorphic with values in the trace class for Izl < R. Then (F + G)-I is meromorphic for r < Rand

Too(r, (F + G)-I) :5 Too(r, F- 1) + max{Moo(r,F- 1), l}(ml(r, G) + N1(r, G))

+ where

1

log f;J'

(6.8) 00

ml(r,G):= Lm(r,I+O'j(G)), j=1

and Ck is the first nonzero coefficient in the Laurent series of det(I + F-IG) at the origin.

77

SIXTH CHAPTER

Proof We start the proof in the same way as before but the estimation of T1 (r, I + F- 1G) in (6.4) is different. Let, again, a, b ~ O. Then we have 10g(1 + ab) :5 max{a, I} 10g(1 + b). Put for short, M:= max{Moo (r,F- 1 ), I}. Then this inequality gives log+ (1j(I + F- 1G) :5 Mlog(l + (1j(G)) and so

00

s(I + F- 1G) :5 MI)og(l j=l Now the claim follows from this.

for

+ (1j(G)).

o

Corollary 6.1 If, in addition to the assumptions in Theorem 6.2, G is analytic then for r < R

Izl < R,

Too(r, (F + G)-l) :5 max{Moo(r, F- 1), I} sup

IIG(z)111

Izl~r

+Too(r, F

-1) + log ~ 1

Proof We have 10g(1 + (1j(G)) :5 (1j(G) which gives the term

(6.9)

IIGIl1.

0

The previous results were all such that the information regarding F- 1 entered thru Too. However if F takes values in the trace class, then the following result is useful. Notice that due to the inversion identity we can formulate the result directly for I +F+G. Theorem 6.3 IfF and G are trace class valued finitely meromorphic for R then for r < R T1(r,I + F

Izl <

+ G):5 2T1(r,I + 2F) + 2T1(r,! + 2G).

Proof This follows immediately from Corollary 4.4.

o

Special results for resolvents In the following we specialize to resolvents. We write them in the form (I zA)-l instead of ()"I - A)-l. In Lemma 6.1 we saw that moo(r, (I - zA)-l) is bounded for all self adjoint operators and so all growth of Too comes from N oo . Recall (see (0.15)) that for scalar functions f a complex number a is called defective or deficient if

I:()

I"

fm(r,~) 0 T( r, f) > .

u a := 1m III r--+oo

In this spirit, self adjoint operators are not defective. For normal operators moo need not be bounded but a strong result (6.10) still holds, with limsup in place of liminJ We shall return to the defect relations still in the Tenth Chapter. We use the concepts order and type for operator valued functions F in the same sense as for the scalar case. For a formal definition, just replace T(r, f) by Too(r,F) in Definition 2.2.

SIXTH CHAPTER

78

Theorem 6.4 Let A be nonzero bounded normal operator in a Hilbert space, such that (I - zA)-1 is meromorphic for Izl < 00 with growth at most of finite order. Then (6.10)

Proof Let r > 0 be fixed. We shall first derive an upper bound on moo (r, (I ZA)-1) in terms of noo(r, (1 - zA)-1). Since A is normal we have

11(1 - zA)

-11 1= d(z) 1

where d(z) =

inf

>'E<7(A)

11 - zAI.

SO, if we denote n:= noo(r, (1 - zA)-1) and keep rand n fixed, then moo(r, (1 - zA)

-1

1 ) = 211"

J

+ 1 log d(rei'f') d
is maximized when the n eigenvalues are all evenly distributed on IAI = 1/r. The maximum value is clearly independent of r and is of the form O(logn) as n grows. But then we can apply Lemma 2.2 with () > 1 so that logn S log Noo((}r, (1 - zA)-1)

1

+ log log(}.

By assumption the resolvent is at most of finite order, so there exists some p such that as r --t 00 we have

< 00

But then we obtain, e.g. with () = e, moo(r, (1 - zA)-1) S O(logn) S O(log(er)P) = O(log+ r). We have two possibilities. Either Noo(r, (I - zA)-1) "" O(log+ r), which means, since Too(r, (1 - zA)-1) is logr-convex, that (6.10) holds, or Noo(r, (1 - ZA)-1) = O(log+ r), which means that Too(r, (1 - zA)-1) = O(log+ r) and we conclude by Theorem 3.5 that the number of poles is bounded and the operator is algebraic. But then for all large enough Izl = r we have d(z) S 1 and thus moo(r, (1 - ZA)-l) is bounded. Again (6.10) follows, except when also Noo(r, (I - zA)-1) is bounded and this can only happen with the 0 operator. 0 Next we look at operators in Schatten classes which we denote by Sp, see the Fifth Chapter.

Theorem 6.5 Assume A ESp. Then (1 -ZA)-l is of orderw S p and ifw = p then it is of zero type. Further, let k :2: 0 be an integer such that k < p S k + 1. Then Too(r, (1 - ZA)-1) S k + 1 IIAII~ rP + k 10g(1 + rllAII). (6.11) p

79

SIXTH CHAPTER

Proof We prove first the result in the case p $ 1. To that end it suffices to show that for any c > 0 we have

Too(r, (I - ZA)-1) $ crP + O(logr). Since A E Sp there exists an m, large enough so that 1

00

- L aj(A)P < c. p j=m+1 Then, however, we can proceed as follows:

Too(r, (1 - ZA)-1) $T1(r,I - zA) $ logM1 (r, I - zA) m

$ Llog(l + raj (A)) j=1 $O(logr) + crP •

1

00

p

j=m+1

+ -rP L

aj(A)P

Here we used the inequality 10g(1 + x) $ ~xP, valid for x > 0 and 0 < p $ 1. In the general case, let k be a positive integer such that k < p $ k + 1. Then in particular Ak+1 E S1, and in fact 00

00

Laj(Ak+1)m $ Laj(A)P, j=1 j=1 see e.g. Corollary II.4.2 in [Go-K]. We have, compare with Theorem 5.7,

Too(r, (I - ZA)-1) $ T 1(r, 1- Zk+1 Ak+1) + k 10g(1 + riIAII). Here we proceed as above and in particular use 10g(1 + r k+1aj(A k+1)) $ k + 1 rPaj(Ak+1)m p to split the sum at a proper place in order to have the growth again bounded by crP + O(logr). 0 Recall that in Example 1.5 we had a self adjoint operator A such that its eigenvalues were

Aj = -

(:jr

Thus w = 1/2, but A E Sp only for p > 1/2. Since V 2 is a rank-1 perturbation of A, the same applies to V 2 .

Powers and their resolvents The proof of Theorem 6.5 was based on

T 1(r, 1- Zk+1 Ak+1) = T 1(r, (I - Zk+1 Ak+1)-1), valid for k + 1 ::::: p. We shall next study the asymptotic behavior of kT1 (r, (I - zk Ak)-1) as k grows. Given a compact A, we denote by {Aj(A)} the sequence of its eigenvalues, indexed so that i>'1(A)1 ::::: IA2(A)1 ::::: ... and each eigenvalue repeated according

SIXTH CHAPTER

80

to the dimension of the corresponding eigenspace. If the operator has only a finite number of eigenvalues, then the sequence is continued by setting Aj{A) = 0 for the larger indeces.

Lemma 6.2 If A E SI, then 00

N l (r, (I - ZA)-I) = I)og+ IAj(A)rl.

(6.12)

j=1

Proof Choose r and take the Riesz spectral projection of A including all eigenvalues which are larger than, say, in modulus. This gives a finite rank operator A r . Then A - Ar can be approximated arbitrary well with another finite rank operator and this shows that N l (r, (I - zA)-I) only depends on A r . (Compare with the Continuity Lemma 4.4.) But since this is of finite rank, it is unitarily similar to a finite dimensional upper triangular (that is, a sum of diagonal and nilpotent) operator. But then the nilpotent part can be made arbitrarily small by another suitable similarity transformation and we conclude that Nl only depends on the eigenvalues. In fact, if S denotes a similarity transformation, and if B = SCS- l with d =dimB, then

r!1

s(B) :$ d log (lISIIIIS- l ll) + s(C) and therefore the multiplicity J.L{ Aj ~A)) is not affected by the similarity transforma&a 0 Observe that the right hand side of (6.12) makes sense for all compact operators as it is always a finite sum for any fixed r. We introduce the following notation. Given a sequence {Aj} converging to zero we set N(r,{Aj}):= I)og+ IAjrl· j

Now the following holds.

Theorem 6.6 Assume A E Sp with some p. Then lim -k1T1(r, (I - zk Ak)-I) = N(r, {Aj(A)}). k--+oo

(6.13)

Proof Recall that if A E Sp then A k E SI for k ~ p. Then for such k T l (r, (I - zk Ak)-I) and T l (r,! - zk Ak) are both well defined and equal. The proof is given by several simple lemmas, some of which have some independent interest. Lemma 6.3 If A is compact, then (6.14)

Proof of Lemma 6.3 We formulated this as Theorem 4.13 for Md. This version can be found as Proposition 2.d.6 in [8]. 0 0 The aim is to show that 1 k k kml(r,I-z A )-+N{r,{Aj(A)}).

(6.15)

SIXTH CHAPTER

81

This would imply (6.13) as Tl(r, (I - zk Ak)-I) = T l (r, I - zkA k )

and, trivially, Nl(r, 1- zk Ak) = o. We shall first reduce the claim to a finite dimensional problem. Since our basic claim is about a limit with a fixed r, we can without lack of generality set r = 1 in the following. Choose a small 0 < 0 < 1. Then take a spectral decomposition of A = Al EEl A2 as follows:

1. ( A2 := -2 'In

)"(>./ - A)-ld)".

11>\1=1-6

By the spectral radius formula we have for large enough n

~ 1- ~.

IIA2'II!.

Lemma 6.4 Assume that A E Sp and p(A) < 1. Then we have lim ml(1,I - zk Ak) = 0

k-+co

as k -

00.

Proof of Lemma 6.4 If p(A) < p < 1 then for large enough n we have IIAnl1 ~ pn. If also n ~ k where k such that Ak E Sl, then we can estimate as Izl = 1, which shows that

o

The claim follows.

Lemma 6.5 If A E SI and B is of finite mnk and they opemte in invariant subspaces H A, H B respectively with HAn H B = {O}, then s(I + (A EEl B)) ~ 8(1 + A) + s(I + B) + rank(B) (log(1 + IIAII) + log 2).

Proof of Lemma 6.5 This is clear by (4.23) and (4.25).

(6.16)

o

If A = Al EEl A2 as above and rankA I = d then Lemma 6.5 gives ml(1,I _znAn) ~ml(1, 1-

+d(log(1

zn Ai) + ml(1, 1- zn A2')

(6.17)

+ IIA2'11) + log 2).

This follows because An = Af EEl A2 allows US to apply Lemma 6.5 with _zn An in place of A. By Lemma 6.4 we have limn-+ co ml(1, 1- zn A 2 ) = 0 and since IIA211- 0, then inequality (6.17) implies lim sup .!.ml(1,I - znAn ) n-+co

n

~ limsup .!.ml(1,I n-+co

n

znAi).

82

SIXTH CHAPTER

What we need still to prove is the reverse inequality liminf !ml(l,I - zn Ar) n-+oo n

~ liminf !ml(l,I n-+oo n

zn An).

(6.18)

and that the limit exists and satisfies (6.19) Consider first (6.18). Let P denote the spectral projection: Al = PA. Then for ~ d we have aj(I + A 1 ) ~ 1lPllaj(I + A)

j

while for j > d we have aj(1 + A 1 ) = 1. Thus s(I + A 1 ) ~ s(1 + A) + dlog IIPII.

Applying this to _zn An in place of A gives (6.18). In order to prove (6.19) observe first that by construction Nl (1, (I - zAd- 1 = N(I, {Aj(A)}). And recall that we have set r = 1. For Izl = 1 we have -1

+ aj(An)

~

aj(1 - zn Ar)

~

1 + aj(An)

which implies, as Al is of rank d,

By Lemma 6.3 we know that

which proves (6.19). The proof of Theorem 6.6 is now completed.

D

We shall close this topic with similar results for Too. Here it is natural to look at general bounded operators in a Banach space X. Definition 6.1 Suppose A E B(X). We denote by Poo(A) the smallest radius such that (I - zA)-l is meromorphic for Izl < 1/ Poo(A). Theorem 6.1 If A E B(X), then (1 - ZA)-l and (I - zk Ak)-l are meromorphic in the same discs: Poo(A) = Poo(Ak)-k and

(6.20) while

SIXTH CHAPTER

83

Proof Write, with,pj := 2rrj/k,

(I - zk Ak) = (I - zA)(I - ei
(I - ZA)-l = (I - zk Ak)-l(I - ei
So in particular, the order is preserved while the type might change somewhat. If the operator is quasinilpotent, so that the resolvent is entire, then we can also

look at the growth of the maximum and here the type is preserved as well. Theorem 6.8 If A E B(X) and p(A) denotes the spectral radius, then for r < p(~) we have (6.22)

and Moo(r, (I - zA)-l) :::; (1 + rllAll)k-l Moo(r, (I - zk Ak)-l).

(6.23)

Proof Here (6.23) is analogous to (6.21) while (6.22) follows from writing

(I - zk Ak)-l

1

k

k

j=l

= - ~)I

.

- e'
Bounded characteristics Returning to Example 6.1, in a slightly different notation, let A :=diag(O:j) where O:j > 1 and limO:j = 1. Then Poo(A) = 1 and for r < 1

Too(r, (I - ZA)-l) :::; log 2. Definition 6.2 If F is meromorphic for

Izl < Rand

sup Too(r, F) <

00

r
then we say that F is of bounded characteristic in Iz I < R. Here is an example how this type of information can be used. Theorem 6.9 Assume that A E B(X) is such that p(A) :::; 1. If the resolvent is of bounded characteristic and C is such that sup Too(r, (I - zA)-l) :::; C,

(6.24)

r
then, for all n

~

0,

(6.25)

SIXTH CHAPTER

84

Proof We obtain for r < 1 using (3.11) log+ Moo(r, (I - zA)-I)

~ C 1+

(6.26)

r.

l-r

The powers of A appear as Taylor coefficients of the resolvent. By the operator valued analogue of (2.10) we have

II An II· ~ r-nMoo(r, (I -

zA- 1 ).

(6.27)

The inequality now follows from (6.26) and (6.27) by setting r:= 1/(I+v'2C/n).D If we take A = I, the identity, then

Too (1, (I - zI)-I) = 'Y where

1111" log+ lei8 - I1d8 = 0.323 ... 'Y := -2 1r

-11"

If (6.24) holds with C < 'Y, then the spectral radius of the operator is necessarily smaller than 1 and we can bound the size of Moo (1, (I - ZA)-I) in terms of C.

Theorem 6.10 If C in (6.24) satisfies C the positive solution of

e=

< 'Y,

1

3(-y _ C) (1 + 10g(1 +

then peA)

<1

and if

e> 1 is

e)),

then

(6.28)

Proof Since

we conclude from

lI(ei8 I-A) -1 II

1 ~ sup 1.8 AI AEu(A) e' that u(A) is strictly inside the unit disc and in particular M := Moo(l, (I - ZA)-I) is finite. Without loss of generality we can assume that f(8) := II (e i8 I - A)-III assumes its maximum M at 8 = O. But then we have from

that

M

f(8) ~ 1 + Mle i8 Therefore C

~2~

i:

-

11·

log+ f(8)d8

1 111"/3 1 1 111"/3 ~log 1.8 11 dB - 10g(1 + 1r 0 e' 1r 0 ~'Y -

1 log (1 + I/M) a

which implies the claim.

1

MI·e'8 -

11 )d8

1

3M 10g(M + 1)

o

SIXTH CHAPTER

85

Observe that since "( = 0.323 ... < 1/3, the constant ~ > 1, for all 0 :s: c < "(. Clearly, we have C = 0 only when A = O. We close this topic by a consequence of resolvent being bounded in the unit disc, as in (6.28).

Theorem 6.11 If (6.29) then 00

B

:s: L IIAnl1 :s: 4B(1 + B).

(6.30)

i=1 Proof The idea ofthe proof is simple. Knowing the value of Moo(r, (I -zA)-1) allows us to use the estimate (6.27) with r > 1 such that B(r - 1) < 1. In fact, we obtain from 1- zA = z(I - A) - (z - 1)1 that -1 1 +B

Moo(r, (I - zA)

If we choose r := 1 +

2k we obtain

00

B =

):s: 1- (r-1 )B.

II L

00

Ai ll-1:S: L IIAili i=O i=1

00

:s: LMoo(r, (I -

zA)-1)r- i

:s: 4B(1 + B).

i=1

o What if small perturbation means small in norm One application of knowing the growth function Too (r, (I - zA) -1) is given in the following chapter: we show that there exists a sequence of monic polynomials {Pi} such that the decay of Ilpi (A) II is related to the growth of the resolvent. Above we saw that the speed of growth of Too(r, (I - ZA)-1) is robust in low rank perturbations of A. In practical computations we would however not use A itself, sayan integral operator representing the inverse of some differential operator but rather a discretization of it, say Ah. In such a case it is of interest to know what happens to the growth function, under the assumption that E := A - Ah is small in norm. The first observation is that knowing IIA - Ahll and Too(r, (I - zA)-1) alone does not imply much. In fact, Too carries no information on the dimensions of invariant subspaces related to the poles, and thus an arbitrarily small perturbation of A can split the pole into arbitrarily many poles. Thus in these terms, all we can say is that, if Too(r, (I - zA)-1) stays small for Izl :s: R o, so that we can conclude that (I - zA)-1 is actually analytic in that disc, then a simple perturbation result is possible, as a corollary of Theorem 6.9.

Corollary 6.2 If for r

:s: Ro,

then (I - ZA)-1 is analytic for r

:s:

(6.31) Ro and the following estimate holds (6.32)

SIXTH CHAPTER

86

for r ::; Ro where ~ is given in Theorem 6.10. If now

11(1 -

z(A + E))-lll

RoIIEII < ~,

then

::; ~ _ £IIEII'

(6.33)

In order to be able to estimate Too(r, (I - z(A + E))-l for larger r we must pose further restrictions on either A or on E. We shall assume that A is in the trace class.

Theorem 6.12 Assume that A E 8 1 and E E B(H). Then for

1

Too(r, (1 - z(A + E))- ) ::;

zA.

r11E11 < 1

rllEl1 rllAlh + (1 + rllAll1) 1 _ rlIEIl'

(6.34)

Proof This follows from Theorem 6.2 by choosing F(z) = 1- zE and G(z) = 0

Comment 6.1 Much of the material of this chapter is from [N08] and [N09]. About Theorem 6.5 there are early related results in the Russian literature, see e.g. [Ma] and the references given there. Comment 6.2 Theorem 6.11 is from [N09]. It would be interesting to know the exact constant(s) in (6.30), say in the form 00

L IIAil1 ::; aB + bB2. i=l

We know that this requires a

~

2, b ~ 4/9.

SEVENTH CHAPTER Keywords: Infinite products, quotient representations, spectral polynomials, Krylov solver, robust error bounds. Combining a scalar function with an operator In the following we consider functions fA which are obtained by combining a bounded operator A E B(X) with a scalar meromorphic function f as follows

fA: z For simplicity, we shall assume that at most with finite order p, that is,

1-+

f(zA) E B(X).

(7.1)

f is meromorphic in the whole plane and grows (7.2)

for all f > O. Likewise, we assume that A is almost algebraic (equivalent with assuming that the resolvent (I - zA)-l is meromorphic in the whole plane) and such that the resolvent grows at most with finite order w, i.e.

(7.3) for all f > O. For example, operators in Schatten class Sp grow at most with order p by Theorem 6.5. Now, assuming additionally that f is analytic at the origin, fA is analytic for small z. If either f is entire or A is quasinilpotent, then fA is actually entire. Otherwise singularities can occur but these are all poles. In general we can define fA as follows. We assume for simplicity that f is analytic at the origin. Thus it has an expansion 00

f(z) = I:ajz j j=O

for

Izl < Ro with some Ra :::; 00.

If p(A) denotes the spectral radius of A, then 00

fA(Z) = I:ajAjzj

(7.4)

j=O

converges for Izl < Rajp(A). Outside of this disc fA is then extended by meromorphic continuation. 87

88

SEVENTH CHAPTER

Theorem 7.1 Let 1 be a meromorphic function in the whole plane such that 1(0) i= 00, and such that it grows at most with finite order p. Let A E B(X) be an almost algebmic opemtor such that its resolvent grows at most with finite order w. Then IA in (7.4) is a well defined B(X)-valued meromorphic function in the whole plane such that it grows at most with order max{p, w}, that is,

(7.5) for all {3 > max{p, w}. Proof Let us consider first the cases in which with Theorem 2.1, that if

1A

is entire. Recall, compare

00

G(z) = L:Bjzj j=O

then G is entire of order at most w if and only if for all .lim l/(W+E) IIBj1l1/j = 3-+ 00

€

> 0

o.

(7.6)

If 1 is entire, then 1A is entire, too, and it follows from the inequality IlajAjIl1/j::; IIAlllajl1/j that the Taylor coefficients of IA cannot decay slower than those of I. Consequently the order cannot increase. Likewise, if A is quasinilpotent then limj-+oo IIAj Il1/ j = 0 and the fact that 1 is analytic near the origin guarantees that for some G we have lajl1/j ::; G. Thus now the coefficients can be estimated by

lI aj Aj ll1/ j ::; GIIAj I11/j and the decay is now dominated by the decay of the coefficients in the resolvent. Again we have an entire function with order at most that of the resolvent. Fix now {3 >max{p, w}. If 1 has only a finite number of poles, then form a polynomial n

p(z) :=

II (1- z/bj )

j=1 so that pi is entire and hence of order less than (3. Estimating the inverse of p( zA) is easy: n

Too(r,p(zA)-1) ::;L:Too(r,(I - :.A)-1) j=1 J n

=

L: O(r/lbjl).B) = O(r.B).

(7.7)

j=1

By construction pf is entire and of order less than (3. Thus

Too(r,IA) ::;Too(r,p(zA)-1) + Too(r,p(zA)/(zA)) ::;O(r.B) + O(r.B). What remains is the general case of 1 having infinitely many poles. Let us denote these by {bj }, ordered so that Ibjl ::; Ibj+11 and each pole is repeated as many times as the multiplicity requires. Without loss of generality we can assume that {3 is not

SEVENTH CHAPTER

89

an integer and that it is close enough to max{p, w} so that they have a common integer part: (7.8) m ~ max{p,w} < f3 < m+ 1. We shall form an entire function

II E(z/bj , m) j=l

where E(z, m) is a Weierstrass factor, see (2.2). We need to estimate Too(r,
I

Lemma 7.1 Assume that

is analytic lor Izl

< R. Then we have (7.9)

lor r < R/(21IAII). Proof of Lemma 7.1 The claim follows from representing I A (z) as a Cauchy integral

2~i

IA(Z) =

f

I(z>,,)(>"! - A)-ld>"

and estimating the integrand along 1>"1 = 211AII with II(>..! - A)-III ~ 1/IIAII.

0

Lemma 7.2 For m < f3 < m + 1 there exists 0(3 such that log IIE(zA, m)1I ~ 0(3 IIAII(3r(3

holds lor all A

(7.10)

and all r > O.

E B(X)

Proof of Lemma 7.2 The proof is divided into two parts, depending whether 211 A II r is smaller or larger than 1. Assume that 211Allr ~ 1. Then we can denote by F the function

F(z) = 10g(E(zA, m)) = -

~

L...J

j=m+l

1

..

-:-A3 Z3

J

which is analytic in this disc. Clearly for these values IIF(z)1I

~

E ~IIAlljrj ~ j=m+l

2l1All m +l r m+1

~ 2I1 AII(3r(3.

J

But E = eF so that IIEII ::; e llFll and thus

10gIlE(zA,m)lI::; IIF(z)1I ::;2I1AII(3r(3 which is of the form required. Assume then that 211Allr ~ 1. Here we base the estimation on Lemma 7.1 and on the fact that the claim holds in the scalar case. In fact, we have log IE(z, m)1 ::; c(3r(3

(7.11)

where c(3 = ~ for m = 0 and c(3 ::; e (log(f3 + 1) + 1) otherwise, see (5.6.13) and (5.6.16) in [NOl].

SEVENTH CHAPTER

90

We now apply Lemma 7.1 to the function E{z, m) and obtain

Since

211Allr 2: 1 we have log IIE{zA, m)11 SCi32i311Alli3ri3 + log 2

S{ci3 + log 2) 2i3l1Alli3ri3.

o

The proof of Lemma 7.2 is thus completed.

:;t

We shall now start estimating

p, we have 00

1

~ Ibj 113

(7.12)

< 00.

We write 00


=

II E{:.A,m)-l j=l

J

and so we need to estimate the inverse of E{zA, m) as well. Lemma 7.3 Assume that (7.13)

with f3 such that m < f3 < m

+ 1.

Then there exists a constant L such that

Too{r, E{zA, m)-l) S L r/3

(7.14)

holds for all r > O. Proof of Lemma 7.3 Again the proof uses two different arguments, one for small r and another for r large. Assume first that 211Allr S 1. Here we can work on the same function F as in the previous proof. So,

-F{z) = log (E{zA,m)-l) =

f: j=m+1

~Ajzj. J

This implies as before But then gives (7.15) for 211Allr S 1. Notice that for f3 < 1 the Lemma is now proved as the larger values are taken care directly by (7.13). To be able to work with the large values of r we need the following identities.

SEVENTH CHAPTER

91

Lemma 7.4 The following identities E(z,2k)-1 = E(Z2,k)-lE(-z,2k) E(z, 2k + 1)-1 = E(z2, k)-l E( -z, 2k + 1)

hold for all positive integers k.

Proof of Lemma 7.4 We have 1 E(z2,k)-1 = - - exp(-(z2 1- z2 and E(-z,2k) = (1

+ z)

Z2 exp(-z + 2

z2k k

+ ... + -)) z3

z2k

- 3 + ... + 2k).

Multiplying these pairwise gives the first identity. The other one is proved similarly. 0 We continue with the proof of Lemma 7.3. Assume that 211Allr ~ 1 and that we are given m < /3 < m+ 1. In order to estimate E(zA, m)-l we use the identities above to represent it as a long product, the first term being related to the resolvent and the others to the Weierstrass factors. We show this with an example as the necessary alternating with the identities makes the writing complicated although the matter as such is not. Given m = 10 we obtain E(zA, 10)-1 = (1 - Z16 A 16 )-1 E( _z8 A8, I)E( _z4A4, 2)E( _Z2 A2, 5)E( -zA, 10). Here the first term on the right can be estimated with help of Theorem 6.7 to give Too(r, (I - z16 A 16 )-1) ~ 16 Too(r, (I - zA)-l). Since

211Allr ~ 1 we conclude that Too(r, ( 1- zA)

(7.13) yields a constant C such that -1

f3 ) ~ Cr,

for

1 r ~ 211AII.

This takes care of the first term in the product. Consider the second one. Since 10 < /3 < 11 we use Lemma 7.2 with m = 1 < /3/8 < 2 = m+ 1. Thus log IIE( _z8 A 8, 1)1 ~ Cf318I1A811f318r¥ ~ Cf318I1AIlf3rf3. The next term requires m = 2 and with 2 < /3/4 < 3 we obtain log IIE( _z4 A4, 2)1 ~ Cf3/41IAIIf3rf3.

= 5 < /3/2 < 6 = m + 1 and therefore IIE( _z2 A2, 5)11 ~ Cf3/21IAIIf3rf3.

FUrther, with E( _z2 A2, 5) we have m

Finally IIE(-zA, 10)11 ~ Cf3IIAII f3 r f3 . Combining all the estimates we obtain for 211Allr ~ 1 Too(r,E(zA, 10)-1) ~ (C + (Cf3 18 + Cf3 /4

+ Cf3 /2 + C(3)IIAII f3 )rf3 .

(7.16)

Recall that here C contains information on the growth of the resolvent while the other constants depend only on /3. We consider Lemma 7.3 proved. 0

SEVENTH CHAPTER

92

We can now continue with the proof of Theorem 7.1. With help of Lemma 7.3 we have

Now fA Lemma 7.1

=


Furthermore, by (7.2) and Lemma 7.2

0+1

log+ M(2I1Allr,
0+1

~ 0 _ 1 [log+ M(21IAIIOr,
(}+1

~(}-1

C,8

1

(2I1AIIOr) L.J Ibj

l,8

,8

+O(r ).

Combining these estimates we have

o

and the theorem is proved.

Example 7.1 The result above is formulated for the order only. The proof actually gives more, but we see in this example that it is possible that if both f and the resolvent grow with the same order of finite type, then fA need not be of finite type. In fact, let 1

f(z) : = eZ

-

e

and

A := diag(l/ j) 1

T(r, I) = -r + 0(1) 11"

and

Too(r, (I - zA)-l) = r + O(logr) but 1

Noo(r, fA) = (- + 0(1)) r logr + 0(1). 11"

SEVENTH CHAPTER

93

Remark 7.1 In Theorem 7.1 the claim is that IA is meromorphic of order at most the maximum of p and w. The order can actually be smaller. To see this consider an entire I of order p and a quasinilpotent A such that the resolvent is entire of order w. Then IA is entire, too, and its order can be bounded from the decay of the Taylor coefficients, see (7.4) and (2.13). Thus one obtains that IA is of order wp w+p at most. Representing F as Gig At the end of the second chapter we presented two results on representing a scalar function I as hi /2. The first one, Theorem 2.11, considers functions in the unit disc with /i's bounded and the other one, Theorem 2.12, functions in the plane with Ii'S entire. We have also seen, Theorem 3.5, that a growth requirement Too(r,F) = O{logr)

implies

F=Plq with P a B(X)-valued polynomial and q a scalar polynomial. We shall here extend Theorem 2.12 to operator valued functions. Theorem 7.2 For 0 > 1 there is a constant C(O) such that if F is a B(X)valued function, meromorphic in the whole plane, then there is an entire scalar function g and an entire B(X)-valued function G such that F = Gig and T(r,g) ::; C(O) Too (Or, F)

(7.17)

and Too(r,G)::; (C(O)

+ 1) Too (Or, F).

(7.18)

Proof This is a consequence of the Theorem of Miles, Theorem 2.12, or rather its proof. In fact, the proof of Theorem 2.12 is based on a construction of a balanced sequence of zeros, {llj} which includes a given sequence {bj }. Applied to the poles of F, there is an entire scalar function g satisfying T(r,g)::; C(O) Noo(Or, F).

Then G:= gF is entire and Too(r,G)::; T(r, g)

+ Too(r, F)

and we conclude that (7.18) holds as well.

o

Theorem 7.2 guarantees a representation which is valid in the whole plane. If we are willing to work in smaller discs, then simple representations exist. In fact, if F is meromorphic in Izl < R then for every '" < R we can represent F in the form

F=Glq where G is analytic for Izl ::; '" and q scalar valued and also analytic for such that Iql = 1 for Izl = ",.

Izl ::; '" and

94

SEVENTH CHAPTER

Lemma 7.5 Assume that F is meromorphic for Izl < R, analytic at the origin and that the poles {bj } of F are repeated with according to multiplicities, and ordered so that 0< Ib1 1$lb2 1$ ... $Ibjl $ .... Take 'f/ < R and denote by n = noo ('f/, F) the number of poles satisfying Ibj I $ 'f/. If n = 0, put q"., = 1. Otherwise, define a rational function q"., (a finite Blaschke product) as follows: n Ib·1 b· - z • q".,(z) := 'f/--J J (7.19) j=1 bj 'f/2 - bjz

lI

Then q"., and G"., := q".,F are analytic in

and

Izl $ 'f/,

M('f/,q".,) = 1

(7.20)

1 ~ + 'f/ log-(O) = ~log -lb'l = Noo('f/,F).

(7.21)

j=l

q".,

J

o

Proof The claims are easy to check. We can now normalize q"., by setting X".,(z) := q".,(z)/q".,(O) so that

X".,(z) = 1 + Cl('f/)Z + C2('f/)z2

+ ...

is analytic for Iz I $ 'f/ satisfying log+ M(r, X".,) $ Noo('f/, F) for r $ 'f/.

Representations for the resolvent We shall apply Lemma 7.5 and Theorem 7.2 to the resolvent and obtain two extensions of Proposition 1.1.

for

Corollary 7.1 Assume that A E B(X) is such that (I -ZA)-1 is meromorphic Then for every 'f/ < R there exists a rational function

Izl < R $ 00.

x".,(z) which is analytic for

Izl

= 1 + al('f/)z + a2('f/)z2 + ...

$ 'f/, such that for

r$

'f/

log+ M(r,x".,) $ Noo('f/, (I - zA)-I)

and is valid for

1 1 (I - zA)- = X".,(z) A,,,.,(Z)

Izl $ 'f/.

Here 00

A,,,.,(Z) := LPj(A, 'f/)zj, j=O

is analytic for

Izl $ 'f/

with

(7.22)

95

SEVENTH CHAPTER

Further,

+ Too(r, (I - ZA)-l) ZA)-l) + Too(r, cI>A,,.,(Z))

Too(r, cI>A,,.,(Z)) ~ Noo(TJ, (I - ZA)-l) hold for

r ~

Too(r, (I - zA)-l) ~ Noo(TJ, (I TJ.

Proof The representation follows from Lemma 7.5 by multiplying with X,., and identifying the coefficients. To obtain the last estimate use also the first main theorem. 0 We shall now consider the case where R = following definition.

00.

We therefore start with the

Definition 7.1 An operator A is almost algebraic if there exists a sequence {aj} of complex numbers such that (7.23) wherepj(>') := >.j+al>.j-l+. +aj. We call the sequence {Pj} a spectral polynomial sequence. We say that A is almost algebraic of at most order w, if there is a spectral polynomial sequence such that logj lim sup 1 < w. j-+oo -J log Ilpj(A) II -

(7.24)

Example 7.2 All compact operators are almost algebraic, so are polynomially compact operators, quasinilpotent operators and algebraic ones. Operators in a Schatten class Sp are at most of order p, by Theorems 6.5 and 7.3. Theorem 7.3 A bounded operator in a Banach space is almost algebraic if and only if its resolvent is a meromorphic function in the whole plane. The operator is additionally of at most order w if and only if (7.25)

for every f > O. Proof The first part of the theorem is Theorem 5.7.2 in [NOI]. The idea of the proof is the following. If we assume that the resolvent is meromorphic in the whole plane, then there exists an entire function X such that X(z)(I - zA)-l is entire and therefore its Taylor coefficients, pj(A), satisfy (7.23). Reversely, if (7.23) holds, then we can conclude from

that and so the function

x(z) :=

L ajz j

is entire. If we now define cI>(z) := Epj(A)zj then it follows from (7.23) that cI> is entire. Since (I - ZA)-l is analytic for Izl < I/IIAII we see by multiplying the expansions termwise that for these small values of z

X(z)(I - ZA)-l = cI>(z)

SEVENTH CHAPTER

96

and by analytic continuation the equality holds everywhere. Suppose now that A is almost algebraic of order at most w. Then (7.24) and (7.26) imply that the coefficients {aj} decay rapidly enough so that X is at most of order w and the same is true for 1

(1 - ZA)-l = X(z) cp(z). The reverse direction follows easily from Theorem 7.2. In fact, given 9 > 1 there exists an entire (7.27) such that (7.28) is entire and T(r, XA) ~ C(9) Too(9r, (1 - zA)-l). (7.29) Combining (7.28) and (7.29) gives

9+1 log+ Moo(r, CPA) ~ 9 -1 (C(9)

+ 1)

Too (9 2r, (1 - zA)-l).

(7.30)

Together with (7.25) this gives log+ Moo(r, CPA) = O(rW+E) and therefore the Taylor coefficients pj(A) of CPA decay with at least the speed corresponding to the order w + E. D The proof above contains the following representation result for the resolvent. Corollary 7.2 For 9 > 1 there exists a constant C(9) such that the following holds. Given an almost algebraic operator A there exists for each 9 > 1 an entire function XA(Z), and corresponding spectral polynomials {Pj}, such that

(1 - ZA)-l =

~()CPA(Z) XA z

holds for all z with CPA (z) = E~o Pj (A) zj. Furthermore Too(r, (1 - zA)-l) ~ C(9)Noo(9r, (1 - zA)-l)

+ Too(r, CPA)

and Too(r, CPA) ~ C(9)Noo(9r, (1 - ZA)-l)

+ Too(r, (1 -

ZA)-l)

hold for all r.

Decay of spectral polynomials Our aim is to demonstrate that the error in good Krylov space iteration methods can be bounded using Too(r, (1 - ZA)-l). Since this growth function is robust in low rank perturbations of A, so are these error bounds. We shall formulate the results for almost algebraic operators. Analogous results can be formulated for resolvents which are meromorphic in a disc Izl < R. Here we give bounds for monic polynomials. Then in the next section we show how these

SEVENTH CHAPTER

97

polynomials can be normalized to give results for Krylov solvers for the equations of the form x=Ax+b. The idea of the bounds is simple. Knowing the growth of Too(r, (I - ZA)-l) we have control over the decay of the Taylor coefficients of CI> A.

Corollary 7.3 Assume that

Too(r, (I - ZA)-l) :5 c rW + 1 for all r > O. Given 0

(7.31)

> 1 there is a spectral polynomial sequence {Pj} such that 02aecw)j/w IIpj(A) I :5 ea ( - jfor j = 1,2, ...

where

0+1

a = a(O,w) :5 0 -1 (C(O)

(7.32)

+ 1)

and C(O) is such that (7.29) holds.

Proof The inequality follows from

IIpj(A)1I :5 Moo(r, CI>A)r- j by using (7.30) and substituting 02wrw := j / ac 2': 1.

o

If A is not almost algebraic, then there still can exist a sequence of polynomials

{pj} such that IIpj(A)1I1/i _ 0,

(7.33)

but then the polynomials are of the general form .

pj(A) = AJ

. 1

+ aj,lAJ - + ... + aj,j.

Operators for which (7.33) holds with some sequence of monic polynomials are called quasialgebraic and it is a property of spectrum only, namely, an operator is quasialgebraic if and only if the capacity of its spectrum vanishes [H]. In particular, if the capacity of he spectrum is positive then the resolvent cannot be meromorphic in the whole plane and there exists largest R such that it is meromorphic for Izl < R. Example 7.3 A simple operator with spectrum equal to a small disc is provided by B := p8 where 8 is the unilateral shift in l2 satisfying 11811 = 1 and 0"(8) = {z I Izl :5 I}. Then the polynomials {Aj} form the sequence of monic polynomials which minimize the operator norm of IIpj(B)1I at every j so that

IIBili

=,oi.

Clearly here R = 1/p. Suppose now that our operator is a sum of almost algebraic operator and a small bounded operator. If we assume additionally that the almost algebraic operator is in some Schatten class then we can formulate bounds for the sum, which say that the bound first allows superlinear decay but then the bound saturates on linear decay dominated by the small bounded perturbation. We formulate here such a result directly with help of Theorem 6.12 in which we assumed that the operator is a perturbation of a trace class operator.

SEVENTH CHAPTER

98

Corollary 7.4 Given C 1 > 1 we set

C2 := 2(

va;:- + 1)/( va;:- - 1)2.

If A E 8 1 and E E B(H) are given, then there are monic polynomials {pj} depending on A, E and C 1 , satisfying for j ~ C211AlldllEII

IIpj(A + E) II

~ eC2 (C1C2~AIl1er

(7.34)

and for j ~ C211AlldllEII

Ilpj(A + E) II ~ eC2 (HIIAlh/Il E IJ)(CIIIEll)j .

(7.35)

Proof Here the polynomials Pj are not obtained from just one function X but rather from a sequence of such functions, X7)' with help of Lemma 7.5 as in the proof of Corollary 7.1. By Theorem 6.12 we have for rllEIl < 1 1 rllEIl Too(r, (I - z(A + E))- ) ~ rllAll1 + (1 + r1lA1l1) 1 _ rllEIi'

(6.34)

Let C1 > 1 be given and choose () > 1 such that ()2 = C1 • Then for "I ~ 1/(()IIEII) we may assume X7) given so that (7.22) holds. But then for r ~ "I we have

Too(r'X7)(z)(I - z(A + E))-l) ~ 2Too (TJ, (I - z(A + E))-l)

(7.36)

and further ()+1 log+ Moo (TJ/(), X7)(z)(I - z(A + E))-l) ~ () _ 1 2 Too ("I, (I - z(A + E))-l)

2(() + 1) ~ (() _ 1)2 (1 + ()TJIIAlld Choosing "I = 1/()IIEII we obtain

IIpj(A + E) II

~ exp (~~()-+1~~ (1 + IIAlldIlEII)) (()2I1EII)j

which holds for all j. For short, put c := 2(() + 1)/(() - 1)2. Then with j < ciiAlldllEIl we have TJj := j/(c()IIAI11) ~ 1/()IIEII and we obtain

IIpj(A + E) II

~e

C

(c()2 11:11 Ie

r.

o Robust bounds for Krylov solvers Krylov subspace methods is a class of iterative methods for solving linear systemsof equations. Among them conjugate gradient method is widely used for positive definite problems, while GMRES and QMR are examples of methods suitable for general nonsingular problems. A typical step of such an iterative method involves applying a matrix to a vector and doing linear algebra operations in the low

99

SEVENTH CHAPTER

dimensional subspace created. The methods are often used with preconditioning. For example, suppose we have a nonsingular problem in the form

Bx=c. If we additionally have an approximate inverse for B, Le. we have an M such that M B = I - A with I - A invertible and A "small" , then we can write the equation equivalently in the form (7.37) x=Ax+b where b = M c. Often the preconditioner is not given explicitly but requires running a short subroutine. One special property of good Krylov methods is the following. If A is small except possibly in a low dimensional subspace, then the methods converge rapidly. Traditionally the convergence analysis in the case of conjugate gradient method has been based on approximation theory on the spectrum - a technique which cannot be used for highly nonnormal problems. Our analysis covers both cases simultaneously. In fact, a low rank perturbation of A may change the operator from self adjoint to highly nonnormal and in such a case one would have to change the method e.g. from conjugate gradient method to GMRES, but the error bounds remain essentially unchanged. Practical computations in the low dimensional subspaces created assume inner product structure. Our bounds, however, are based on spectral polynomials: they are upper bounds for the best polynomials and they can be formulated in general Banach spaces. We outline now our setting. Given a bounded operator A and a vector b we may create the sequence {Ajb}~o. If 1 ¢ a(A), we can ask for approximations to the solution of (7.37) from the subspaces

Kk(A,b):= span{Ajb}J:J. There are a lot of different methods for different kind of problems which associate an approximation Xk E Kk(A, b) for (7.37). These typically aim to satisfy

Ilxk -

AXk -

bll :::; Ily -

Ay -

bll

(7.38)

for all y E Kk(A, b). We shall give a bound assuming that (7.38) holds exactly.

Lemma 7.6 If x satisfies (7.37), then

IIx - yll :::; 11(1 - A)-llilly -

Ay -

bll.

(7.39)

Proof The claim follows from

(x - Ax - b) - (y - Ay - b) = (I - A)(x - y) = -(y - Ay - b).

o Note that any vector y in Kk(A,b) can be written in the form y = qk-I(A)b with some polynomial qk-l and that all vectors of this form are in Kk(A, b). It then follows from (7.38) that if Pk-l is any polynomial of degree k - 1 and we set Yk := Pk-I(A)b then necessarily

IIx -

xkll

:::;11(1 - A)-IIlIIYk - AYk - bll :::;11(1 - A)-IIiIII - AIIII(I - A)-lb - Pk-I(A)bll :::;11(1 - A)-III III - AIIII(1 - A)-l - Pk-I(A)lllIbll

SEVENTH CHAPTER

100

We conclude that if we can give an estimate for Ek := inf 11(1 - A)-l - p(A) II

where the infimum (actually minimum) is over all polynomials p of degree less than k, then (7.40) IIx - xkll ~ Ek ll(1 - A)-III 111 - Alillbil where Xk satisfies (7.38).

Theorem 7.4 Assume A is almost algebmic and {ai} is a sequence such that for all j = 1,2, ... IIpi(A)1I ~ Co

+

( c ew)i/W

(7.41)

holds where Pi (A) = Ai + alAi - l + ... + ai' Then X(z) = 1 + alZ + a2z2 entire. Assume also that 1 - A is nonsingular and that X(l) -:f O. Then

+ ...

is

(7.42)

Proof We have 1

00

.

(1 - zA)-l = - () Lpi(A)zJ X z i=O and so Ek

~11(1 1

k-l A)-l -

xt1)

~Pi(A)1I

00

~ IX(l)IL IIPi(A) II J=k

o

which implies the estimate (7.42).

We call this error bound "robust" as it has the following property: we have shown above that there are spectral sequences whose decay is bounded by the growth of Too(r, (1 - zA)-l). Then we have shown that this growth is insensitive in low rank updatings. Thus the only part in the error bound which is obtained by combining (7.40) and (7.42) which is not robust is in the term 11(1 - A)-lll/lx(l)l. Of course it may happen that some low rank updating brings a problem nearly singular, and then this would be large.

A bound for spectral projectors In the first chapter we discussed shortly Riesz projections: 1 . [ (AI - A)-IdA p = -2 1n

lr

(7.43)

where r surrounds an eigenvalue of A. Here we consider the following situation. We ask whether it is possible to give a bound for such a projection in terms of the growth function of the resolvent. To that end, let A be a bounded operator in

101

SEVENTH CHAPTER

a Banach space X and assume that the resolvent (1 - zA)-1 is meromorphic for Izl < R ~ 00. Choose any radius r < Rand () > 1 such that ()r < R. Then we take

(7.44) where p =

1/s satisfies

r

v'o ~ s ~ r

(7.45)

and is such that I (>.1 - A) -111 can be controlled along r in terms of its characteristic function Too (()r, (1 - zA)-I).

Theorem 7.5 Given () > 1 there is

C(()) < v'o + 1 + 10 4ev'o( v'o + 1) - v'o-1 g v'o-1

(7.46)

such that the following holds. Let A be a bounded linear operator in a Banach space such that the resolvent (1 - ZA)-1 is meromorphic for Izl < R ~ 00. Then for any r such that ()r < R there exists an s satisfying (7.45) so that for cp E (-7r, 7r]

(7.47) Proof Observe that the claim is essentially the same as in Corollary 2.2. So is the proof, too. However, we have here an operator valued function and therefore u:

z ~ log (

Ip(z)III(1 -

zA)-III)

is only subharmonic; here again we denote by P the monic polynomial vanishing at Zj = l/bj for poles bj with Ibjl 2: 1/()r. Thus we cannot use Poisson-Jensen formula as in the proof of Theorem 2.10 but we get the exactly same inequality by arguing as follows. Since u is subharmonic it stays below the harmonic function h S h(se'°t ) := 21 111" P( -() , t - cp)u(()re'°t )dt

_11"

7r

r

and so we obtain ° 1 111" P( -() S log 11(1 - se"P A)- 1 II ~, t - cp) log 11(1 - ()re'°t A)- 1 Iidt

27r

_11"

~l

(()r)2 - bk seicp og ()r(seicp - bk)

+~

r

k=1

The rest is then identical to that of the scalar case.

o

This is related to projections onto invariant subspaces as follows. Assume p is such that (7.48) a(A) nrp = 0 and denote 1 (7.49) Pp = ~ (>.1 - A) -1 d)". 7rZ

1 rp

Then Pp projects onto the invariant subspace corresponding to the part of spectrum which is smaller than p in modulus.

SEVENTH CHAPTER

102

Corollary 7.5 Given 0 > I there is C(O) satisfying (7.46) such that the following holds. Let A be a bounded linear operator in a Banach space such that the resolvent (I - zA)-l is meromorphic for Izl < R ~ 00. Then for any r > 0 such that Or < R there exists p such that I r

VB

~p~-,

r

(7.48) holds and

log IIPpl1 ~ C(O) Trx;;(Or, (I - ZA)-l). Furthermore, the number of eigenvalues outside r p is bounded by

(7.50)

I

noo(r, (1 - ZA)-l) ~ logO Too (Or, (I - ZA)-l). Proof This is clear by Theorem 7.5.

D

Comment 7.1 Almost algebraic operators were discussed in [NOI]. Bounds for Krylov solvers, based directly on Too(r, (I - ZA)-l), were discussed in [Hy-N], without help of theorem of Miles.

EIGHTH CHAPTER Keywords: Approximate polynomial degree, approximate rational degree. Approximate polynomial degree of an analytic function If p is a polynomial of degree = d, then

log+ M(r,p)

= (1 + o(l))d log r

as r

--+ 00

and reversely, if log+ M(r, 1)/ logr is bounded as r --+ 00 then I is a polynomial. Suppose we look at a given analytic I in a tiny neighborhood Izl :$ r. Then obviously, just one evaluation of I, e.g. at origin is sufficient to approximately represent I. In a larger disc one needs more evaluations. Likewise, we may want to know, how the work increases with increasing accuracy. This is achieved simply by looking at 1/c in place of I. We shall make this precise by introducing the following notation and terminology.

Definition 8.1 Let

I

be analytic for

Izl < Ro :$ 00.

Put for r < Ro

do(r, 1) := min{deg pip is a polynomial and such that M(r, 1- p) :$ I}.

We shall call do the approximate polynomial degree. We can relate do to M as follows.

Theorem 8.1 Suppose ()r

I

is analytic lor

Izl < Ro

:$

00.

Then lor () > 1 and

< Ro we have 1 + (M(()r, 1)) do(r,1) < log () log () _ 1

+ 1,

(8.1)

and

log+ M(()r, 1) :$log+ M(r, 1) + do (()r, 1) log() + log 2.

(8.2)

Proof Let d be an integer such that 1 + M 1 + M log()log () -1 :$ d < log()log () -1

+ 1,

(8.3)

where for short M = M(()r, 1). Then ()d =

M

M

exp(dlog()) ;::: exp(1og+ () _ 1) = max{() _ l' I}. d

Let us now put p(z) := the coefficients satisfy

E

akzk where ak's are the Taylor coefficients of

k=O

103

I.

Since

104

EIGHTH CHAPTER

see (2.10), we obtain with the help of (8.3) 00

2:

M(r,f-p)~

JakJr k

k=d+l 00 O-d ~ M ' " O-k = M L...J 0-1

k=d+l

1 0-1 ~ M 0 -1 min{ 1\{' 1}

~ min{l, 0 ~ 1 } which implies the first claim. In order to prove (8.2) observe that if M(Or, f -p) with deg p = d, then

M(Or, f)

~

1

M(Or,p) + 1 ~ Od M(r,p) + 1 ~

~ Od(M(r, f)

+ 1) + 1

which implies the second claim. Here we used the inequality

M(Or,p) ~ OdM(r,p) which is a special case of Bernstein's lemma and can in this form be concluded as follows. The function g(z) := z-dp(z) is analytic and bounded for JzJ ~ r. By the maximum principle we have

(Or)-dM(Or,p)

=

sup Jg(z)J ~ sup Jg(z)J Izl~(lr

Izl~r

= r-dM(r,p).

o Our main interest in formulating Theorem 8.1 is the fact that we shall later be able to formulate an analogue of it for meromorphic functions, approximated by rational functions. However, in that case we cannot in general code the growth in terms of the Taylor coefficients. The following examples illustrate the inequalities (8.1) and (8.2). Example 8.1 If p is a polynomial of degree d, then clearly do(r,p) ~ d for all r with equality for all r large enough. Consider first (8.1) with a fixed r:

do(

)

_1_1

+ (M(Or,f)) 0_ 1

r, p < log 0 og

Letting here 0 -+

00

+

1 = _1_1

log 0 og

+ ((Or)d(1+0(1))) 0- 1

gives

do(r,p) < d + 1 or, as both are integers,

do(r,p)

~ d.

On the other hand, from (8.2) we obtain, for r > 1

do(r,p) logr ~ log+ M(r,p) - C with C = log+ M(l,p)

+ log 2.

0

+

1

105

EIGHTH CHAPTER

While log+ M(r, f) is bounded for an entire f only when f is constant, do(r, f) is bounded for polynomials and thus for very slowly growing functions do(r, f) is essentially slower than log+ M(r, f). It is then not without interest that for entire functions of positive order, do(r, f) grows with the same speed as log+ M(r, f) (without any logr term) and that also the type can be correctly recovered from do(r,f).

Example 8.2 In order to see that do(r, f) codes both the order and type faithfully, it essentially suffices to consider the function

We show the following:

do(r, F) = (1 + o(l))'Tewrw,

as r

-+ 00. d

Let do be fixed and put for short d := do(r, F). If P(z) =

E Cjzj

is the corre-

j=O

sponding approximating polynomial, then the Parseval's identity gives us d

1 ~ M(r,F - p)2 ~

00

L laj - cjl2r2j + L j=O

lakl2r2k

k=d+1

~ lad+11 2 r 2 (d+1).

This gives us immediately

do(r, F)

~

'Tewrw - 1,

for all r > O.

To bound do(r, F) from above, we use inequality (8.1). By Theorem 2.2 we have for 1/2 ~ c > 0 and r > 0 13 log+ M(r, F) :5 (1 + e)'Trw + log+( -w). c Inequality (8.1) now implies with ():= exp(l/w) that there exists Ce such that

o

holds for all r > O.

Theorem 8.2 If f is entire of order w, then log do(r, f) . w = 11m sup 1 . r-+oo

ogr

(8.4)

If f is of finite positive order w, and of type 'T, then 1 l' do(r, f) 'T = 1m sup . ew r-+oo rW

(8.5)

EIGHTH CHAPTER

106

8.2.

Proof We leave this as an exercise: try to modify the discussion in Example D Some properties of the approximate polynomial degree

If we want to approximate f within tolerance e, i.e. that M(r, f - p) ~ e, then the minimum degree possible is given by do(r, fie). From (8.1) we obtain with () > 1, r fixed such that ()r < Ro,

1 1 1 do(r, - I) ~ - 1 ()log+ - + C e og e

(8.6)

where C = C(r, (), I) is independent of c. Observe that, apart from C, the right hand side of (8.6) depends on f only through (). We can relate these notions to the standard setting in approximation theory. To that end, let Ed(r, 1):= inf M(r, f - p). deg(p)~d

It is well known that if f is analytic in a slightly larger disc, then Ed(r, I) decays fast with increasing d, see e.g. [Wa], p. 75. Here is a simple version with explicit constants. Theorem 8.3 Assume f is analytic in Izl < such ()r < Ro. Then for d = 0,1,2, ... we have E ( f) d

r,

< M(()r, I) -

() -1

Ro ()-d

~ 00.

Choose r <

Ro, () > 1, (8.7)

.

Proof Let () := M~9r~j)' so that by Theorem 8.1 for e E (0,1] we have () 1 1 do(r, - I) < - 1() log - + 1. e og e Denote

cd

(8.8)

:= ()-d. Then by (8.8)

and D

In the following we formulate some simple inequalities for do(r, I). Expressions of the form do(r, c(r)1) are to be understood as follows: we look at functions z 1-+ c(r)f(z) with fixed r for Izl ~ r and consider c(r) as a constant in the approximation process. Theorem 8.4 Let f and g be analytic for Then

do(r, f + g) do(r, fg) where cf

:=

Izl < Ro

~ 00

~

max{do(r, 21), do(r,2g)}, ~ do(r, cgl) + do(r, cfg),

max{3M(r, I), va}, cg := max{3M(r, g), va}.

and ()r <

Ro, () > 1. (8.9) (8.10)

EIGHTH CHAPTER

107

Proof If M(r,21 -p) ~ 1, M(r,2g-q) ~ 1, then M(r, I +g- !(P+q» ~ 1, while 1 deg 2(P+q) ~ max{deg(p), deg(q)}. To prove (8.10), suppose M(r, 1- p)

~

1/cg and M(r, 9 - q) ~ 1/cf. Then

M(r,/g - pq) M(r,/g - Iq) + M(r, Iq - pq) ~ M(r, f)M(r,g - q) + M(r,g)M(r, 1- p) + M(r, 1- p)M(r,g - q) 111 ~"3 +"3 +"3 = 1, ~

while deg(pq)

~

deg(p) + deg(q).

0

Theorem 8.5 Let I be analytic lor Izl < Then do(r,J')

Ro

~ 00

and () > 1 such that (}r < Ro.

~ max{do((}r, (() ~ l)rf) -

1, O}

(8.11)

and do(r,f) ~ do(r,rl')

+ 1.

(8.12)

Proof Differentiating the Cauchy integral

r

I(z) = ~

I(() d(

2n i1z-
gives the Cauchy inequality 1

II'(z)1 ~ P1<1~1~+p I/(()I, and further, with

Izl =

(8.13)

r, p = (() - l)r, M(r, I')

If now p satisfies M((}r, 1- p)

M(r, I'

-

~

~

(()

~ l)r M((}r, f).

(() - l)r, then

pi)

~

(()

~ l)r M((}r, 1- p) ~ 1.

Since deg(p') = max{O, deg(p) - I} we obtain (8.11). To prove (8.12), let p be such that deg(p) = do(r, r I') and M(r, I' - p) Then with P(z) := 1(0) + Joz p we obtain (with Izl ~ r) I/(z) - P(z)1 = while deg(p) = do(r, r 1') + 1.

~

1/r.

'foz (I' - p)1 ~ rM(r, I' - p) ~ 1, o

We can combine the Cauchy inequality (8.13) and it Bernstein's inequality for polynomials r M(r,p/) ~ deg(p)M(r,p) (8.14) to obtain an inequality for rM(r, 1') where the information regarding I is over the same disc.

EIGHTH CHAPTER

108

Theorem 8.6 Let I be analytic lor r <

Ro ~ 00.

+ 1)

(M(r,f)

rM(r,f') ~ (do(r,rf')

Then we have lor r <

+ 1) + 1.

Ro

(8.15)

Proof Let p be such that M(r,/' - p)

~!r

and deg(p) = do(r, r I'). Put further P(z) = 1(0) +

J; p. Then, by (8.14),

rM(r, I') ~ rM(r, f' - p) + rM(r,p) = rM(r,f' - P') + rM(r,P') ~ 1 + (deg(p) + l)M(r, P) ~ 1 + (deg(p) + l)(M(r, f) + 1) where the last inequality follows from

M(r, P)

~ M(r, f) + max I

r(f' - p)l.

Izl::5r 10

o Example 8.3 Consider I(z) := eTZk • We have rM(r, 1') = rkrk M(r, f), while (8.15) gives rM(r, 1') ~ (1 +o(l))rekrk M(r, f), as r --+ 00, by Theorem 8.2. Thus, (8.15) has here an extra multiplicative constant e. 0 Approximate rational degree of a meromorphic function Parallel to Definition 8.1 we give the following definition for meromorphic functions.

Definition 8.2 Let

I

be meromorphic for

Izl < R ~ 00.

d(r, f) := min{deg q I q is a rational function such that

r
We shall call d the approximate rational degree. Now we can bound the growth of T in terms of d as follows.

Theorem 8.7 Suppose such that (Jr < R we have T( (Jr, f)

I is ~

meromorphic lor

T(r, f)

Izl < R

~ 00.

+ d( (Jr, f) log (J + 2 log 2.

Then lor (J > 1 (8.16)

Proof We use Cartan's identity, see Theorem 2.7. Differentiating (2.37) with respect to log r gives d 1 -Id T(r, f) = -2 ogr 7r

j1r n(r, -1I'f) )d(J. -1r - e'

(8.17)

Let q be a rational function satisfying deg(q)=d((Jr, f) and such that M((Jr, I -q) 1. Clearly 1 n(r'--'f)) ~ deg(q). q- e'

~

EIGHTH CHAPTER

109

Applying (8.17) to q gives d

-1d T(r, q) :5 deg(q). ogr

Integrating this gives

+ deg(q) log().

T«()r,q):5 T(r,q)

o

But T«()r, J - q) = 0 and (8.16) follows.

Example 8.4 We see from Theorem 8.7 that rational approximation of entire functions is not much more accurate than what we obtain with polynomials. For example with J(z) = e% we have T(r, e%) = ~r. Hence we obtain from (8.16) 1 «() - 1) d«() %) :5 r, e og

- -l-()-r ~

and further

d(r, e%) 2':

1 + 0(1)

2 log 2 +1 () og

r as r

-+ 00.

~

(8.18)

(8.19)

Notice, that for polynomials we obtain from Theorem 8.2

do(r, e%)

= (e + o(l))r

so they are both of first order and of finite type. Next we ask for the reverse inequality, the analogue of (8.1). By replacing log + M by T we might hope to have

d(r, f) :5 C 1 «())T«()r, f) + C2 «()).

(8.20)

We shall actually have exactly this if J does not have a pole at the origin. Without this assumption the claim is not true. Example 8.5 Clearly, for all

lal

d(r, (z - a)-n) = n for all r 2':

lal.

On the other hand,

T(r, z-n) = n log+ r for all r > 0 while

_

1

T(r, (z - a) n) = n log Tal for 0 < r :5 1 -Ial. Thus, in general we cannot bound d in terms of T alone. Since trivially d(r, f) 2': n(r, f) we may try to bound d in terms of both T and n. Theorem 8.8 For () > 1 let ()i > 1 be such that ()1 ()2()3 = (). Suppose J is meromorphic Jor Izl < R:5 00. Then Jor ()r < R we have

d(r, f) < C1 «())T«()r, f)

+ 2n«()r, f) + C2 «()).

where

C1 «()) = _1_()2 + 1 [2 + 10g()1 ()2 - 1

and

+ 1]

()3 ()3 -

1

1 ( + 1 ()2 + 1 ) C2 «()) = - 1 () log ( - ( )1) + -()1 10g2 + 1. Ogl

1-

2-

(8.21)

EIGHTH CHAPTER

110

Proof Let () > 1 be given. Choose (}i > 1 such that (}l (}2(}3 = () and assume that r is such that 1] := (}r < R. To start, let q be a finite Blaschke product such that it is analytic in Izl < 1], vanishes at the poles bj of f in that disc so that g:= qf

is analytic there and Iql = 1 along Izl = 1] and of minimal degree. Thus deg(q) n(1], f) and T(1], q) = O. By Theorem 2.4 we have 1

1

T(1], -) = log - I1 = N(1], f)· q Ck

~

(8.22)

In fact, if f is regular at origin, then this is part of Lemma 7.5, since then

q(O) =

IIn .J..b j=l 1]

so that 1 ~ + 1] log Iq(O)1 = {:rlog Ibjl = N(1],f).

f

if, on the other hand, Ie q = ~q. So, writing

has a pole at origin of degree k, then q is of the form

q(O)

q () z =

-kz 1]

k

+ Ck+l Z k+l + ...

and using (2.28) we again get (8.22). Put p := (}1(}2r and consider the NevanlinnaPick interpolation problem: Find a w, analytic in Izl

~ p

such that

w(bj ) = g(bj ) for j = 1, ... , n(p, f) and such that M(p, w) is minimal (with natural modifications if some poles are multiple). It is well known that the solution is unique and that the solution is a rational function of degree at most n(p, f). Furthermore

M(p,w)

~

M(p, g)

(8.23)

since 9 itself is a feasible function. By construction

w

1 = -(g - w) q q

f - is analytic for Izl ~ p =

(}1(}2r

log+ M((}l r ,f _~) q

~

~ ~

and we can estimate it pointwise as follows:

+ 1 T(p,/ _

(}2 (}2-

1

+ 11

(}f)2 2-

(}2 + 1 01 2 -

w) q

(T(p, ~ ) + T(p, 9 - w)) q

( N(1], f)

+ T(p, g) + T(p, w) + log 2) . (8.24)

111

EIGHTH CHAPTER

Since T(T/, q) = 0 we have

T(p,g) S T(T/, g) S T(T/,j), while by (8.23) we obtain

T(p, w) S log+ M(p, w) S log+ M(p, g) S :: ~~ T(T/, j). Substituting these into (8.24) gives log+ M((}1 r ,I - w) S (}(}2 + 1 ((2+ (}(}3 + 1) T(T/,j) + log 2 ). q 2- 1 3- 1

(8.25)

We have now an analytic function I - w / q which we shall still approximate with a polynomial p. By Theorem 8.1 we have wI + wI + 1 do(r, 1- -) < - 1 () log M((}1 r,/ - -) + - 1 () log ( - ( )1) + 1. q og 1 q og 1 1-

Thus we have approximated

I by a rational function

(8.26)

w / q + P and we have

w d(r, j) S degq + degw + degp S 2n(T/, j) + do(r, 1- -). q

o

This completes the proof.

Corollary 8.1 For () > 1 let (}i > 1 be such that (}1(}2(}3(}4 = (). Assume I is meromorphic lor Izl < R S 00 and 1(0) =1= 00 • Then lor (}r < R inequality (8.20) holds with

and C2((})=-1 1() (log+(-() 11)+(}(}2+111og2)+1. og1 12-

Proof In (8.21) we can now estimate

o Example 8.6 Let us apply these bounds for rational functions. First, if T(r, q) satisfies a lower bound T(r,q) ~ d log+ r - C (8.27) then we have from Theorem 8.7

d logr S T(l, q) + d(r, q) logr + 21og2 + C which gives immediately liminf d(r, q) r ...... oo

~

d.

(8.28)

Reversely, suppose that T(r, q) satisfies an upper bound

T(r,q) S d log+ r + C.

(8.29)

EIGHTH CHAPTER

112

Of course, we can conclude from this that q is a rational function of degree at most d, but we look what the bound in Theorem 8.8 gives. First, the bound contains the term 2n(()r,q). For r ~ 1 we have, see Lemma 2.2, 1

n(r,q)::; 10g()N(()r,q) and so we obtain, by using (8.29) and letting ()

-+ 00,

n(r, q) ::; d. We then show that inf {C1 (())T(()r,q)+C2 (())} ::;3d+1.

8>1

(8.30)

Thus, combined we have

d(r, q) ::; 5d + 1. To obtain (8.30) choose t::

(8.31)

> 0 and take ()2 and ()3 large enough so that C1 (())

::;

3 + t:: .

log ()1

Then (8.29) gives limsup C1 (())T(()r,q)::; (3 + t::)d. 8 1 -+00 But limsupC2 (()) 81 -+00

::;

1

and (8.30) follows.

Example 8.7 We saw earlier that do codes both the order and type of entire functions accurately. For meromorphic functions the approximate rational degree codes the order accurately but leaves a gap for the type. In fact, suppose T(r, J) grows with a positive order w and with a positive type a. Then we obtain from Theorem 8.7

. d(r, J) 11m sup - W -

(8.32) ~ aew, r-+oo r analogously to (8.5). To get an upper bound for limsuPr-+oo d(r, J)/r w we use Theorem 8.8. We take ()1 = e, ()2 = ()3 = 2 so that () = 4e. Thus

C1 (())T(()r, J) ::; 15 (4e)WarW Further, since n(()r, J) ::; T(()er, J) for r lim sup r-+oo

d(r~J) r

+ o(rW).

> 1, see Lemma 2.2, we conclude

::; (15 + 2eW)(4e)W a.

Spijker's lemma Polynomials satisfy Bernstein's inequality

rM(r,p') ::; deg(p)M(r,p) which we used to get Theorem 8.6

rM(r,f')::; (do(r,r!')

+ I)(M(r,J) + 1) + 1.

(8.33)

EIGHTH CHAPTER

For meromorphic by M. Spijker.

113

I an analogous result can be obtained from the following lemma

Lemma 8.1 II w is a rational/unction, then (8.34)

o

Proof The original is in [Sp].

Theorem 8.9 II I is meromorphic in Izl < Rand 0> 1 is such that Or < R, then

rl11"· .) -11" 1!,(re''P)ldcp :5 d(Or, f) ( s~p I/(re''P) I + 1 + 0 _1 1·

211"

(8.35)

Proof Suppose deg(w) = d(Or,f) and M(Or,J - w) :5 1. Then using (8.34) we obtain r 111" Iw'(rei'P)ldcp -r 111" 1/'(rei'P)ldcp:5211" -11" 211" -11" + ~ 111" 1!,(rei'P) - w'(rei'P)ldcp 211" -11"

:5d(Or, f) sup Iw(rei'P) I + rM(r,!, - Wi)

'P

:5d(Or, f)(s~p I/(rei'P) I + 0 ~ 1 M(Or, 1- w)). To have the last inequality we used the Cauchy inequality

rM(r, I'

- Wi) :5 0 ~ 1 M(Or, 1- w).

o Remark 8.1 Observe that if I is rational then for € > 0 we can apply ( 8.35) to ~I and recover (8.34) as d(Or, ~f) :5 deg(f). This scaling technique can also be used in the following result.

Theorem 8.10 Let

I be meromorphic in Izl < R such that it is analytic in

Izl < Ro, and has there the expansion 00

I(z) =

I>k Zk . k=O

Then lor r < Ro we have (8.36)

EIGHTH CHAPTER

114

Proof Let w be rational, of degree d(r, f) such that M(r, f - w) :::; 1. Then 1. ( 1. { Ck = -2 Ck-1w(()d( + -2 Ck-1(f(() - w(())d( 7rZ J1(I=r 7rZ J1(I=r where by partial integration and Spijker's Lemma

111

12 . 7rZ

while

Ck-1w(()d(1 :::; -k deg(w)r- kM(r, w)

1(I=r

I~ (

Ck-1(J(() - w(())d(1 :::; r-k. J1(I=r The claim then follows as M(r, w) :::; M(r, f) + 1. 27rZ

o

Theorem 8.12 below contains a different variant of this mechanism.

Power bounded operators and bounds for the Laurent coefficients Let A be a bounded operator in a Banach space. It is a very basic task to give conditions on the resolvent that guarantee power boundedness

IIAnl1 :::; K

for n = 0,1,2, ...

(8.37)

(see Prologue). A necessary condition is obtained easily from (8.37). In fact, for Izl < 1 we obtain

This is often called the Kreiss resolvent condition and we may write it here as follows:

K

M(r, (1 - ZA)-l) :::; - - for r < 1. (8.38) 1-r This does allow a linear growth IIAnll = O(n). We shall assume that the resolvent is additionally meromorphic in some neighborhood of the unit disc. Together with this the Kreiss condition is sufficient. We make this quantitative by assuming for some () > 1 and L = L(()) < 00 (8.39)

Theorem 8.11 For each () > 1 there are constants C i (()), i = 1,2,3 such that if the resolvent is meromorphic for Izl :::; () and the conditions (8.38) and (8.39) hold, then for n = 1,2, ... (8.40)

Proof This is a special case of the following result on Laurent coefficients of meromorphic operator valued functions. 0

EIGHTH CHAPTER

115

Theorem 8.12 For each () > 1 there are constants Ci (()), i = 1, 2, 3 such that the following holds. Assume that F is a B(X)-valued junction, meromorphic for Izl < () and analytic for 0 < Izl < 1, satisfying the quantitative estimates limsup (1 -lzI)IIF(z)1I ~ K 1%1-->1-

(8.41)

sup Tco(r,F) ~ L.

(8.42)

and r<(J

Then for all j = -m, -m + 1, ... (8.43) where Aj denote the Laurent coefficients of F: co

L

F(z) =

Ajzj for 0 < Izl < 1.

j=-m

Proof Let us put G := zm F so that G is analytic for Izl < 1. If x E X and y" E X" are of unit length, then we set g(z) :=< G(z)x, y" > and we have Ig(z)1 ~ IIG(z)1I and

T(r,g)

~

Tco(r, G).

But for Izl = r < 1 we have by maximum principle that

M(r,g) ~ rmsupIIF(rei'l')1I 'I'

which together with (8.41) gives limsup (1- r) M(r,g)

~

K.

r-+1-

Finally, by using maximum principle once more,

M(r,g)

~

K

-1- for r < 1.

(8.44)

-r

Likewise we obtain from (8.42) for r < ()

T(r,g) ~ T(r, zm) since by definition of

+ Tco(r, F)

~ mlog+ r

+ L ~ 2L

Nco

mlog+ r ~ Noo(r, F) ~ Too(r, F). Denote by Ck the Taylor coefficients of g:

and apply (8.36) to 9 with r

< 1:

ICklrk~ d(~g)( l~r +1)

+1.

(8.45)

EIGHTH CHAPTER

116

For k

> 1 we can choose r

= I-11k and obtain, with d:= d(l,g), ICkl ::; 4dK + 2d + 4.

(8.46)

By Corollary 8.1 we have d(l,g) ::; C l (9)2L + C 2 (9)

>1 ICkl ::; 8Cl (9)KL + 4C2 (9)K + 4Cl (9)L + 2C2 (9) + 4

and substituting this into (8.46) gives, for k

where the right hand side is of the form used in (8.43). Notice that for k = 0, 1 we have simply leol ::; K and ICll ::; 4K. To finish the proof, recall that the coefficients Cj depend on the vectors x, y*:

and thus the claim follows, as

IIAk-mll =

sup ICkl. Z,tI*

o For the resolvent the additional condition (8.42) can often be estimated. Here is an example.

Corollary 8.2 Assume that A is a bounded operator in a Hilbert space such that for some k O"k(A):= inf IIA - Ell < 1.

rankB
If A satisfies the Kreiss condition (8.38), then A is power bounded and an estimate is obtained from Theorem 8.11, since Too(9, (I - zA)

-1

1+kO"l(A)9 )::; 1 _ O"k(A)9

for 90"k(A)

< 1.

(8.47)

Proof This follows immediately from Theorem 6.12. Notice that we need not to know 0"1 (A) = IIAII in (8.46) since the Kreiss condition implies

IIAII ::; 4K. Combined we obtain a bound for the powers in terms of K, O"k(A) and k.

Comment 8.1 Part of the material in this chapter has appeared in [N03].

0

NINTH CHAPTER Keywords: Locally almost algebraic, locally algebraic. In this chapter we study the following question. We assume given a Banach space X, a meromorphic function F with values in B(X), and we ask whether there are vectors x E X and y* E X* such that the associated scalar function

Ix,y.:

z ~ y*(F(z)x)

(9.1)

grows like the operator valued function F. The answer to this question is positive and as corollaries we obtain results on the "local" operator theory. Growth of associated scalar functions Our main result is that the growth of F as a meromorphic function can approximatively be seen already with just one function Ix,y •. The nature of this claim can be seen e.g. from the following implication. Recall that a bounded linear operator A is said to be algebraic if it has a minimal polynomial and it is said to be locally algebraic if for every x there exists a monic polynomial p such that p(A)x = o. A theorem by Kaplansky says that locally algebraic operators are always algebraic, see [Al],[NOl]. Applying our result to the resolvent of an operator which is not algebraic implies a growth which is faster than that of rational functions and then there must exist Ix,y. which also grows faster than rational functions implying that A is not locally algebraic. We shall come back to this application below. Let now F be a meromorphic B(X)-valued function for Izl < R. In the following it is natural to assume that the vectors x and y* have unit length. Then we have in particular that (9.2) I/x,y·1 ~·IIFII and thus for r < R (9.3) Our aim is to show that there are unit vectors x and y* for which this inequality can be approximatively reversed. In order to achieve this we have to allow I to be estimated in a slightly larger disc than F. Notice also that if 1(0) = 1, then by the first main theorem we have 1 T(r, f) = T(r,

7)

and we conclude in particular that when Ix,y. shows the growth of F; the function 1/lx,y. is in general not related to the growth of F- 1 .

NINTH CHAPTER

118

Theorem 9.1 Suppose F is a B(X)-valued meromorphic function in Izl < R ~ 00. Given an increasing sequence 0 < r1 < r2 < ... < r j < ... < R and a positive constant p < 1/2, there exist unit vectors x E X and y* E X*, depending on F, {rj} and p, such that for fx,y. in (9.1) we have for every n = 1,2, ... T(rn+1,fx,y·)~

rn+1 - rn ( Too (r ,F)-4nlog-+21og(1-2p). 1 ) n rn+1 +rn P

(9.4)

We can formulate a simple corollary for the case where F is meromorphic in the whole plane. Corollary 9.1 If F is meromorphic in the whole plane, choose ro > 0, 0> 1 and p < 1/2. Then there are unit vectors x, y* such that for r > ro we have 0+1 2 Too(r, F) ~ 0 _ 1 T{O r, fx,y·)

+ c(r)

(9.5)

where

log.! Or 1 2 =O{logr) c(r) =4 1 OPlog(-)+21og1 og ro - p Proof of Corollary 9.1 Set rn = onro, choose n such that rn-1 rn+1 ~ 02r and use the fact that T and Too are nondecreasing in r.

~

r

~

rn < 0

Proof of Theorem 9.1 We shall now construct the vectors x and y* promised in Theorem 9.1. We start with some auxiliary results. 0 Lemma 9.1 Given {Aj}f C B{X) and p, 0 < p < 1/2, there exists x E X, 1 such that for j = 1,2, ...

IIxll =

(9.6)

Proof Denote s := p/(l- 2p) and let {Xj} be a sequence of unit vectors such that

(9.7) Then set 00

OjXj (9.8) j=l where c > 0 shall be chosen to give x unit length and OJ'S are determined as follows. Put Uo := 0 and n (9.9) Un := LOjXj. j=l If (9.1O) IIAnXnl1 ~ pnllAnll then set On = 0, otherwise there exists On satisfying IOnl ~ (2s + l)pn /28 such that

x

:= c L

(9.11) In fact, if (9.12)

NINTH CHAPTER

119

then by (9.7)

IIAnun-l - 28 + 1 pnAnxnl1 > 2 282+ 1 pnllAnxnll- pnllAnl1 28 8 Now we can estimate Anx: n

~

(2pn _ pn)IIAnll.

00

IIAnxll ~cllAn Lajxjll- cllAnll L lajl j=l j=n+l

~c pn(1_ 28 + 1_P_)II A nll

28 1- p 1 =c (1- 2(1- p))IIAnli. Since E~llajl ~

2(1:'P)

(9.13)

and therefore c ~ 2(1- p) we obtain 1

IIAnxll ~ 2(1 - p)(1- 2(1 _ p) )pnllAnll = (1 - 2p)pn IIAnll·

o Lemma 9.2 Given {an}f 1, such that for all n = 1,2, ...

C

X and p E (0,1/2) there exists y* E X*, lIy* II =

(9.14) Proof The proof is essentially identical to the former one.

o

Combining the previous results we can formulate still one result.

Ilxll

Lemma 9.3 Given {Aj}f C B(X) and p, 0 < p < 1/2, there exist x E X, = 1 and y* E X*, lIy*1I = 1, such that for j = 1,2, ...

ly*(Ajx)1 ~ (1 - 2p)2p2j IlAjll.

(9.15)

We continue now with the proof of Theorem 9.1. We are given a meromorphic F for Izl < R, an increasing sequence 0 < rl < r2 < ... < R and a fixed number satisfying 0 < p < 1/2. We shall start by defining a family of auxiliary rational functions. Assume F has a pole at b of multiplicity m, that is there exists a nontrivial bounded operator D satisfying

F(z) = (z - b)-mD + O(lz - W-m). Let

Ibl

~ t

(9.16)

and put

z-b . t - bz/t Then Rt(z, b)m F(z) is analytic near b. Collecting all poles bj in that disc, we form Rt(z, b) :=

I

Xt{z) :=

II Rt(z,bj)m

j

j=l

so that XtF is then analytic for Izl ~ t. (If there are no poles in such a disc, just put Xt(z) = 1.) We shall need the following result.

120

NINTH CHAPTER

Lemma 9.4 For r ::; t we have 1

T(r, Xt) = T(r, -) - Noo(t, F). Xt Proof of Lemma 9.4 By construction N(r, ;.) = Noo(r, F) for r ::; t. The claim then follows from the first main theorem. Next we shall define a sequence of operators. To that end let {D j } denote the operators, associated with poles {b j } as in (9.16). If there are only a finitely many poles, just put Dj = 0 for large j. Another set of operators is determined as follows. Since Xrn+l F is analytic for Izl ::; rn+b let Zn be a point along Izl = rn which maximizes IIXrn+l (z)F(z)lI. We put (9.17) Now for j = 1,2, ... we set A 2j +1 := Dj and A 2j := Bj . By Lemma 9.3 there are unit vectors x,y* such that (9.15) holds. Denoting J(z) := y*(F(z)x) we obtain around bj IJ(z)1 = ly*(Djx)(z - bj)-mj + O(lz - bjI1-mj)1 which, together with IJ(z)1 ::; IIF(z)1I and

ly*(Djx)1 ~ (1- 2p) 2p4i+ 2I1Djll implies that J has a pole at bj , also of multiplicity mj and so N(r, f) = Noo(r, F) for all r < R. Using the even numbered sequence we obtain respectively that

IXr n+l(Zn)f(zn)1 ~ (1- 2p)2p4nMoo(rn,Xrn+lF)

(9.18)

where

Moo(r, G) = sup IIG(z) II·

Izl:S;r We already noted that N(r, f) = Noo(r, F), so all we need to do is to bound m(r, f) from below. The tool we need here allows one to estimate the maximum in terms of T on a slightly larger disc. In fact, if 9> 1 and G is analytic for Izl ::; 9r, then

+ 9+1 Too(r, G) ::; log Moo(r, G) ::; 9 _ 1 Too (9r, G).

(9.19)

Using (9.19) and the fact that IXrn+l (z)1 = 1 along Izl = rn+1 we estimate

m(rn+1,f) = m(rn+b Xrn+J) ~

rn+l - rn + ( ) log M rn, Xrn+J . rn+l +rn

(9.20)

Let us put for short en := log((I- 2p)2p4n). Then noticing that

Too(r, F) ::; Too(r, Xrn+lF)

1

+ T(r, - - ) Xr n +l

and using Lemma 9.4 we obtain from (9.18) log+ M(rn' Xrn+J) ~ log+ Moo (rn' Xrn+l F) ~Too(rmXrn+1F)

+ en

+ en 1

- T(rm - - ) + en Xrn+l =Too(rn' F) - Noo(rn+bF) + en. ~Too(rmF)

(9.21)

NINTH CHAPTER

121

Combining everything we finally have T(rn+b f) =m(rn+l' f)

+ N(rn+l' f)

~ r n+l - rn (Too (rn, F) - Noo(rn+b F) rn+l +rn rn+1-rn (Too (rn, F) ~ rn+l +rn

+ Cn) + Noo(rn+b F)

+ Cn ) o

which is what we had to prove. Locally algebraic and locally almost algebraic operators

Recall that a bounded operator A is said to be almost algebraic if there exists a sequence of complex numbers {aj} such that (9.22) where (9.23) Definition 9.1 A bounded operator A E B(X) is called locally almost algebraic if for every x there exists a sequence {aj}, depending possibly on x, such that if {pj} is as in (9.23), then .lim IIpj(A)xll l / j = 3-+00

o.

(9.24)

Corollary 9.2 A bounded linear operator is almost algebraic if and only if it is locally almost algebraic. Proof By Theorem 5.7.2 in [NOI) a bounded linear operator is almost algebraic if and only if (I - zA)-l is meromorphic in the whole plane. The proof of that statement can be carried thru in the same way for the vector valued function (I - zA)-lX. We leave it to the reader. 0 Recall that an operator A is said to be quasialgebraic if there is a sequence of monic polynomials {Pj}, not assuming (9.23) but still satisfying degpj = j and (9.22). Likewise, it is locally quasialgebraic if for every x such polynomials exist so that (9.24) holds. An operator is quasialgebraic if and only if its spectrum is of zero logarithmic capacity, [H), and further, it is locally quasialgebraic if and only if it is quasialgebraic, [Mii). Notice that there is no requirement on the behavior of the resolvent near the spectrum and so the singularities can be essential. Let us recall that we called an operator valued function F rational if there are polynomials p(z) = zn + b1zn - 1 + .. . bn and P(z) = Aozm + ... + Am where bj's are complex numbers and Ak'S are bounded linear operators, such that F can be represented in the form F( ) = P(z) (9.25) z p(z) . Corollary 9.3 If fx,'Y. is rational for all x,y·, then F is rational.

NINTH CHAPTER

122

Proof A scalar function

f

is rational if and only if

T(r, f) = O(logr) as r

-+ 00.

Likewise, F is rational by Theorem 3.5 if and only if

Too(r,F) = O(logr) as r

-+ 00.

o

The claim follows now immediately from Corollary 9.1.

We can now obtain Kaplansky's theorem from this.

Corollary 9.4 A bounded operator is algebraic if and only if it is locally algebraic. Proof Ifp(A)x

= 0 andp(A) = An+a1An-l+ .. '+an , write for j = 0, ... , n-1 .

. 1

pj(A) := AJ + alAJ- + ... + aj. Then denoting 1 n-1 . R(A):= (A) LPn-l-j(A)AJ p j=O we conclude by multiplying R(A)X by p(A)(AI - A) that R(A)X Thus y*(R(A)X) is rational and so is

= (AI -

A)-lx.

fx,y. (z) := y*( ~R(l/Z)X). But by Corollary 9.3 F(z) = (I - zA)-l is then rational and therefore operator A is algebraic. In fact, if the resolvent (AI - A)-l is rational, say of the form

(AI _ A)-1 = Q(A, A) q(A) then the Cauchy formula implies immediately that q(A) = 0:

q(A)

= 2~i

J

q(A)(A - A)-ldA

= 2~i

J

Q(A, A)dA = O.

Thus locally algebraic operators are algebraic. The other direction is trivial.

0

We can still focus on the almost algebraic operators and ask what happens to the order and type. Recall that in combining a scalar function with an operator

(f, A)

1-+

fA,

where fA(z) = f(zA), we noticed that the order did not exceed the maximum of the two but the type could increase from finite to infinite. Here not only the order is recovered but finiteness of the type as well.

Corollary 9.5 If for a meromorphic function F all the scalar functions fx,y. are of finite order, then F is of finite order. If all scalar functions are of order w and of finite type, then F is of order wand of finite type and if all types of scalar functions are of type ~ c, then F is of type T ~ (1 + 2w)ec. Notice that this is to be understood in the way that a function is of order w and type T if it is at most of order w and if the order is equal to w then the type is at most the given bound.

NINTH CHAPTER

123

Proof This follows immediately from Corollary 1. The type is concluded as follows. We have from (9.5)

r < 0+ 1 02Wr(x y*) -0-1 ' which yields the bound by substituting 0 := 1 + l/w.

D

Comment 9.1 It is natural to ask what is the best conclusion which can be made on F, knowing that all scalar functions ix,y. are of bounded characteristic. The technique above seems not to be able to answer this.

TENTH CHAPTER Keywords: Eigenvalues, exceptional values, deficiencies, defective eigenvalues, defective operators. We could call this a defective chapter. As explained in the Prologue, one of the starting points of this work was the apparent parallelism between Picard exceptional values and an operator, namely V 2 , being quasinilpotent. In the former case 1/(/ -a) is entire, in the latter the resolvent is entire. We shall therefore at the end have a look at the deficiency concepts, and connect these to familiar concepts in linear algebra. Exceptional values We recall first the following well known result of E. Picard. Theorem 10.1 Meromorphic function attains in a neighborhood of an essential singularity all values infinitely often, except possibly two. Values which are not attained, i.e. f(z) =I- a, are called Picard exceptional values. Such an omitted value is deficient with maximal deficiency, deficiencies equal 1. Here is the formal definition. Definition 10.1 If f is a meromorphic (scalar) function in C then the deficiency of the value a E C is m(r, f~a) T(r, f)

(10.1)

. f m(r, f) T(r,f).

(10.2)

8(a) := li~~f

and likewise

r()

I·

u 00 := ~~

Values with positive deficiencies are called Nevanlinna exceptional values or deficient values. Example 10.1 Exponential function gives here an illustrative example: it has no zero nor pole, thus 8(0) = 1, 8(00) = 1. Furthermore, 8(a) = 0 for a =I- O. Notice that then ~ 8(a) = 8(0) + 8(00) = 2. aECU{oo}

The next result, a corollary of the second main theorem by R. Nevanlinna, generalizes Theorem 10.1 as follows. 125

126

TENTH CHAPTER

Theorem 10.2 If f is meromorphic in the whole plane, then the cardinality of deficient values is at most countable and

L

(10.3)

8(a):S 2.

aEICU{oo}

Since for entire functions 8(00) = 1, entire functions satisfy

L 8(a) :s 1. aEIC

Since the deficient values are at most countable we can ask how fast or slowly the deficiencies decay. We copy here one result of such nature, by Eremenko 1986. Theorem 10.3 Let {aj} be an arbitrary sequence of distinct complex numbers and {8j } a sequence of real numbers such that

0< 8j < 1

and

L(8j )l <

00.

Then there exists a meromorphic function f of finite order such that for j = 1,2, ...

while other values a "# aj are not deficient: 8(a) = O.

Many of the results related to defects are difficult. Their starting point is the second main theorem which we have not discussed here. Simple asymptotics for resolvents of matrices In this section we look at resolvents of matrices. When we estimate the growth of the resolvent it does not make much difference whether we write the resolvent as

z~(~I-A)-l

(10.4)

z

or

z

~

(I - ZA)-l

(10.5) as Too grows essentially in the same way with both of them. However, the portions carried by moo vary and the corresponding deficiencies are also different. Using also T1 in place of Too we get altogether four different sets of deficiencies. We need to know the asymptotics of these functions as r ~ 00. In order to present the results we return to the formula for the resolvents, given in the first chapter. We write it here in terms of the variable z. Let us write the minimal polynomial q as follows I

q(A) := An + ... + alA n- 1 = An- l

II (A - Aj) j=l

(10.6)

TENTH CHAPTER

127

with al =F 0 (and if l < n then we put aj := 0 for j = l + 1, ... , n). In (10.6) we only emphasize the possible multiplicity of 0 as the possible multiplicity of other eigenvalues does not effect the formula. In fact, we have n-l (I - zA)-l = 1 "qj(A)zj, (10.7) 1+···+alzl }=o ~ where qO(A) = 1 and qj+l(A) = Aqj{A) + aj+1. Note, that deg(A) = n and that A is nonsingular if and only if lim (I - zA)-l = 0 as z ~ 00.

Theorem 10.4 Let A E Md be given and n = deg(A), qn-l as above. If A is nonsingular, then as r ~ 00

+ log lanl + 0(1),

(10.8)

+ log Ilqn-l(A)1I + 0(1).

(10.9)

Too(r, (I - ZA)-l) = nlog+ r while if A is singular, then

Too(r, (I - zA)-l)

= (n -

1) log+ r

Proof If A is nonsingular, then limr -+ oo moo{r, (I - zA)-l) = 0 and (10.8) follows from n

Noo(r, (I - ZA)-l) = Llog+ IrAjl j=l as for all large enough r we have n

Llog+ IrAjl = nlogr + log lanl. j=l

If A is singular, consider first the case l = n - 1. Then for large r n-l Noo(r, (I - zA)-l) = log+ IrAjl = (n - 1) logr + log lan-II j=l

L

while lim moo(r, (I - zA)-l) = log+ IIqn-l{A)II. r-+oo lan-II If we can conclude that IIqn-l(A)1I ~ lan-II then the claim follows by summing moo and N oo . But in the nullspace of A the matrix qn-l(A) operates like multiplying by an-l and the inequality follows. If finally l ~ n - 2, then II (I - ZA)-lll = r n- 1 -

1(

IIqn-l (A) II

lad

+ 0(1))

so that moo(r, (I - ZA)-l) = (n -1 -l) logr + log

IIqnl~,jA)1I + 0(1).

(10.10)

Thus the claim follows as now for large r

Noo(r, (I - ZA)-l) = llogr + log lad.

o

128

TENTH CHAPTER

Theorem 10.5 Let A E Md be given and n = deg(A), qn-l as above. If A is nonsingular, then (10.11) where while if A is singular, then 1 Too(r, (-I - A)-I)

z

= nlog+ r + log IIqn-l(A) II + 0(1).

(10.12)

Proof Multiply (10.7) by z and proceed in the same way as in the previous proof. 0 Notice that (10.11) and (10.12) could be given by the same formula if we would denote an = 0 in case of A singular. Theorem 10.6 Let A E Md be given and let k:= rank(A). Then k

T1(r, (I - zA)-I) = klog+ r + LloglTj(A) + 0(1) j=1

(10.13)

and

(10.14) Proof Both claims follow immediately from the inversion identity for Tl'

0

Example 10.2 1 T(r, -1-) = log+ r

--a z

+ log+ lal + 0(1)

and for a =f:. 0 1 T(r, -1- )

-az

= log+ r + log lal + 0(1).

Eigenvalues and exceptional values In this section we demonstrate that choosing our perturbations suitably the deficiency concepts in value distribution theory correspond to familiar concepts in linear algebra. All results are simple consequences of the explicit formulas from the previous section. We begin with the following notion. Definition 10.2 Given A E Md let q be its minimal polynomial, given as follows: k

II (A -

Aj yl<j j=1 in which all Aj'S are distinct and aj's are positive integers such that q(A) =

(10.15)

k

Laj = n = deg(A). j=1

(10.16)

TENTH CHAPTER

129

We call an eigenvalue Aj defective if Qj

~

2.

In this language A is similar to a diagonal matrix if and only if all eigenvalues are nondefective. We shall see that defective eigenvalues can be seen as deficient values of the resolvent. To that end we need to specify how we define them. In fact, in the scalar theory one divides by T(r, f) but one could equally well divide by T(r, 1/ f), due to the first main theorem. In the matrix valued case we get natural dimension independent results if we use the following definitions. Definition 10.3 Given A E Md we put for A E C

and

Further, . moo(r, 1 - zA) 800 (00):= r-oo hm T.00 ((1 r, - z A)-I) and

._. moo(r, ~1 - A) COO (00).- r-oo hm Too r, z1 - A - 1)' Using ml and Tl in place of moo and Too we obtain analogously 81 and Cl'

(e

)

We start with 800 , Theorem 10.7 Let 0 #- A E Md. Then for A E C

800 (A) > 0 if and only if A is a defective eigenvalue.

(10.17)

Further,

(10.18) AEC

with equality if and only if A is a nilpotent matrix of degree 2, i.e. A2 = O. Proof Let q be the minimal polynomial of A and given in the form (10.15). If a is not an eigenvalue it is clear that 800 (a) = O. Let therefore a := Aj. Denote now the minimal polynomial of A - Ajl by q[jl. It follows that it is of the form

(10.19) and we see that moo behaves like a constant if Qj = 1, that is, the eigenvalue is nondefective, while otherwise it grows like (Qj - 1) logr, compare with (10.10). If A is nonsingular, we then get

and if A is singular we obtain

130

TENTH CHAPTER

Likewise, 800 (00) is either l/n or 1/(n - 1). Summing over all eigenvalues gives immediately 00 (.\) :::; 1

L8 C

and in particular (10.18) holds. To get equality in (10.18) we need to have n = 2 and A singular so that 800 (00) = 1. But since

Laj =n=2 and at least one eigenvalue must be defective, there cannot be other eigenvalues than O. The proof is now complete. 0 Example 10.3 The only examples of resolvents being deficient in such a way that the sum equals 2 are those of the form (1 - zA)-1 = 1 + zA

as A2

= O. Notice that these matrices also satisfy 1 + zA = ezA .

Next we look at the exceptional values of (~1 - A) -1 . Theorem 10.8 Let A E Md' Then for.\ E C

coo(.\) > 0 if and only if .\ is an eigenvalue of A.

(10.20)

Further, Coo (00) = 0 and (10.21)

Lcoo(.\) = 1. >'EC

Proof We have at eigenvalues, using the previous notation, 1 moo(r, (-1 - (A - .\j))-l) = aj logr + 0(1) z which, together with L: aj = nand (10.12), implies (10.20). Away from the eigenvalues, and near 00, moo stays bounded and that proves the rest. 0

Theorem 10.9 Let 0 f:. A E Md. Then 81(00) = 1 and for.\ E C 1 "i oo (.\) :::; 8

1 (.\) :::;

Furthermore

8(00) + L8

d 800 (.\),

1 (.\):::;

1

2

(10.22)

(10.23)

C

with equality if and only if A is nilpotent. Proof We leave the proof for the reader.

o

Theorem 10.10 Let A E Md' Then for .\ E C Cl (.\)

Further,

Cl ( 00)

> 0 if and only if .\ is an eigenvalue of A.

(10.24)

= 0 and

(10.25)

TENTH CHAPTER

131

Proof Compare with the proof of Theorem 10.8.

D

Deficient operators Above we perturbed the matrix A by a scalar matrix >"1 and noticed that the sum of defects stayed below 2 in the same way as for scalar functions. However, this perturbation is not possible with operators, since the perturbed resolvent is no longer meromorphic in the whole plane (except if A is algebraic). We can, however, perturb the function F(z):= 1- zA (10.26) with a constant aI and then look at Z f-+

(F(z) - aI)-l.

After all, this is what one does in the scalar case. Here a new phenomenon shows up. Namely, either there are no deficient values a or the set of deficient values is unc01:1I1table. Example 10.4 Let

A=(~ ~), then

(F(z) - aI)-l = (

l~a

(l~a)2 ) I-a

and we see that lim moo(r, (F(z) - aI)-I) = 1 r-+oo Too(r,F(z)-I) ,

for a E C, a

::f. 1.

Notice further that if we consider F = I - K with K of the form

K=

(~ ~)

with f entire, the same situation occurs and all a ::f. 1 are again deficient. It is the structure, which is defective, not individual values. We only discuss the case K = zA in the following. Definition 10.4 Given an almost algebraic operator A, put for a E C, a ::f. 1 ( ) '-1' 'nf moo(r, (F(z) - aI)-l) "Yoo a .- 1m1 ,.,., ( F( )-1) r-+oo .Loo r, Z

where F(z) = 1- zA. If "Yoo(a) > 0 for some a, then we call A defective, otherwise it is nondefective. Theorem 10.11 Let 0 ::f. A E Md. Then for a ::f. 1

"Yoo(a) = "Y00(0).

132

TENTH CHAPTER

Proof Write F(z) - aI = (1- a)(I - l':aA), so that

moo(r, (F(z) - aI)-I) = moo ( 11: ai' F- 1 ) + 0(1).

(10.27)

From the proof of Theorem 10.4 we know that with some integer k we have moo(r, F- 1 ) = k log+ r

+ 0(1)

which substituted into (10.27) gives moo(r, (F(z) - aI)-I) = moo(r,F(z)-I) + 0(1). Hence 'Yoo(a) is independent of a.

D

Corollary 10.1 A E Md is defective, in the sense of Definition 10.4, if and only if 0 is a defective eigenvalue. Proof This follows from 'Y00(0) = 600 (0).

D

Example 10.5 The operator V 2 is quasinilpotent and thus (I - zV2)-1 is entire. Clearly all quasinilpotent operators are defective as 'Y00(0) = 1. Recall that V 2 is a rank-l perturbation of a self-adjoint operator A (see Example 1.5). Theorem 10.12 If A is almost algebraic and defective, then the set of values a for which 'Yoo (a) > 0 contains a circle. Proof If a::j:. 1 is such a value, then set p:= 11 - al. Writing as in (10.27) we see that all b's on the circle p = 11 - bl satisfy 'Yoo(b) = 'Yoo(a) > O. D

We end this with a natural statement on diagonalizable operators. Theorem 10.13 Let A be an almost algebraic operator of at most finite order in a Hilbert space. If it is similar to a normal operator, then it is nondefective. Proof Clearly defectiveness is preserved under similarity transformations. Assume thus that A is normal. Now the statement is essentially that of Theorem 6.4, except that we need to check all a ::j:. 1. Comparing with the proof of Theorem 6.4 we see that moo(r, F- 1 ) = o (log Too (er, F- 1 )). Using again (10.27) yields the result. D Comment 10.1 Some recent developments in the value distribution theory, in particular for holomorphic curves and quasiregular maps, are summarized in [Er].

EPILOGUE Keyplaces: Toronto, Karjalohja. Lecturing and typing in Toronto During October 2001 I gave ten lectures at Fields Institute in Toronto. Each lecture formed the basis of a chapter in this book. My former book [N01] was intended to be an easy-to-read text book on waveform relaxation - but it transformed into a difficult-to-read research monograph on convergence theory for iterative methods in an abstract setting. Likewise, this book was intended to be an easy-to-read text book on matrix valued meromorphic functions - but it transformed into an extended version of [N08] instead, where we alternate between matrix- and operator-valued functions. To be exact, one chapter contains material from two lectures, and the tenth chapter is an exceptional chapter, or simply defective one, written later as a partial and simple minded answer to a natural question: the word defective appears both in the value distribution theory and in linear algebra, so, are they related? Fishing and finishing in Karjalohja A year later I am finishing this monograph at our summer home in Karjalohja. My grandfather worked winters in an insurance company in Helsinki but spent his summers here doing mathematics. During the winters he and Rolf Nevanlinna had their offices close by in Helsinki - at times they even shared an office. Rolf's summer home was across the lake - the ride on a motor boat, at six knots, took twenty minutes. OlIi Lehto has written (in Finnish) a biography on Rolf Nevanlinna which appeared in the fall 2001 when I got back from Toronto. There is an interesting section on the birth of Nevanlinna theory with a discussion on the mutual relations between the two brothers. My grandfather kept diary all his adult life. Unfortunately, the diaries from years around 1925 are missing, but Lehto's book includes diary quotations from later years. I decided to include the Prologue in this book for several reasons. One is this: if a Nevanlinna writes about Nevanlinna theory three quarters of a century after its birth, some explanation is wanted, and I had already published a version of the Prologue in Finnish. During the seven or so years on this project many people have been of great help. I want to thank them all, but especially my hosts and the personnel at the Fields Institute and Bob, Carl, Jarmo, Marja, Marko, Nikolai, Olli-Pekka, Saara, Ulla, Timo and Xiaoushu. 133

134

EPILOGUE

It would be only natural to dedicate this book to the brothers Frithiof and Rolf. However, I dedicate this to my father. He lost his elder brother in the war and was wounded himself. I grew up in the independent Finland. When we go fishing we use a rowboat, and the lake remains quiet.

BIBLIOGRAPHY [AI] [A2] [Bo] [Dr] [Du-S] [Er]

[F] [Go-K]

[H] [Ha-K] [Ho-J1] [Ho-J2] [Hu-N] [Hy] [Hy-N] [Ko] [Ma] [Mi] [Mii] [NF] [N01] [N02] [N03] [N04] [N05]

[N06] [N07] [N08]

Aupetit, B. [1991] A Primer on Spectral Theory, Springer-Verlag, New York. Aupetit, B. [1997] On log-subharmonicity of singular values of matrices, Studia Mathematica 122(2), 195-200. Boas Jr., R. P. [1054] Entire Functions, Academic Press. Drasin, D. [1987] ProoLoJ..aS!!!}jecture of F. Nevanlinna concerning junctions which have deficiency sum ~~Mat:h:~-Q4. Dunford, N an«SchwiiJ:tz', 'J:;.'t'z.[19p3j zan"/lr Operators, Part II: Spectral Theory, Interscience. ." ~. . Eremenko,~. [2002) Va(~~.·R¥.~butio_n'~nd ~otential Theory, Proceedings of ICM 2002, Vol II, Hig~ lM!Jcation ,PresSi.:i>p",~681-:69Q; Faddeeva, V. ~959tComp~~ti"na!.~ods of Linear Algebra, Dover. Gohberg, 1. and KrlJioB;.;~LU.~.l1J,..1ntt"oduction to the Theory of Linear Nonselfadjoint Operators, AMS Translations of Mathematical Monographs, Vol. 18. Halmos, P. R. [1971] Capacity in Banach algebras, Indiana Univ. Math. 20, 255-863. Hayman, W. K. and Kennedy, P. B. [1976] Subharmonic Functions, Vol. I., Academic Press. Horn, R. A. and Johnson, C. R. [1985] Matrix Analysis, Cambridge Univ. Press. Horn, R. A. and Johnson, C. R. [1991] Topics in Matrix Analysis, Cambridge Univ. Press. Huhtanen, M. and Nevanlinna, O. [2000] Minimal decompositions and iterative methods, Numer. Math. 86(2), 257-282. Hyvonen, S. [1997/98] Case studies on growth properties of meromorphic resolvents, Insitute Mittag-Leffler Report No. 18. Hyvonen, S. and Nevanlinna, O. [2000] Robust bounds for Krylov methods, BIT 40(2), 267-290. Konig, H. [1986] Eigenvalue Distribution of Compact Operators, Birkhiiuser. Matsaev, V. 1. [1964] Doklady Akademii Nauk SSSR 154/5, 1034-1037. Miles, J. [1972] Quotient representations of meromorphic junctions, J. Analyse Math. 25, 371-388. Miiller, V. [1987] On quasialgebraic operators in Banach spaces, Operator Theory 17, 291-300. Nevanlinna, F. [1930] Uber eine Klasse meromorpher Funktionen, Den syvende skandinaviske matematikerkongress i Oslo 19-22 August 1929, A. W. Broggers, Oslo. Nevanlinna, O. [1993] Convergence of Iterations for Linear Equations, Birkhiiuser. Nevanlinna, O. [1991] Tiede 2000, vol. 3, s. 51. Nevanlinna, O. [1996] Meromorphic resolvents and power bounded operators, BIT 36(3), 531-54l. Nevanlinna, O. [1996] Convergence of Krylov methods for sums of two operators, BIT 36(4), 775-785. Nevanlinna, O. [1996] A characteristic junction for matrix valued meromorphic junctions, XVIth Rolf Nevanlinna Colloquium, Eds. Laine/Martio, Walter de Gruyter & Co, Berlin, pp. 171-179. Nevanlinna, O. [1998] Juhlien jalkeen, Arkhimedes 3, 12-19. Nevanlinna, O. [1997] On the growth of the resolvent operators for power bounded operators, Banach Center Publications 38,247-264. Nevanlinna, O. [2000] Growth of operator valued meromorphic junctions, Ann. Acad. Sci. Fenn. Math. 25, 3-30. 135

136

[N09] [NRl] [NR2] [Ri-V]

[Ru] [Se] [Sp]

[Wa] [Ya]

BIBLIOGRAPHY

Nevanlinna, O. [2001] Resolvent conditions and powers of operators, Studia Mathematica 145(2), 113-134. Nevanlinna, R. [1925] Zur Theone der meromorphen Funktionen, Acta Math. 46, 1-99. Nevanlinna, R. [1970] Analytic Functions, Springer-Verlag. Ribaric, M. and Vidav, I. [1969] Analytic properties of the inverse A(z)-l of an analytic linear operator valued function A(z), Arch. Rational Mech. Anal. 32, 298-310. Rubel, L. A. [1996] Entire and Meromorphic Functions, Springer-Verlag, Universitext. Segal, S. [1996] Nine Introductions in Complex Analysis, North-Holland Publ. Co.. Spijker, M. N. [1991] On a conjecture by LeVeque and Trefethen related to the Kreiss matrix theorem, BIT 31, 551-555. Walsh, J. L. [1935] Interpolation and approximation by rational functions in the complex domain, AMS Colloquium Publications, vol. XX. Yang, L. Value Distribution Theory" Springer-Verlag, Berlin, and Science Press, Beijing.

Titles in This Series 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4

3

2 1

Olavi Nevanlinna, Meromorphic functions and linear algebra, 2003 Vitaly I. Vol08hin, Coloring mixed hypergraphs: theory, algorithms and applications, 2002 Neal Madras, Lectures on Monte Carlo Methods, 2002 Bradd Hart and Matthew Valeriote, Editors, Lectures on algebraic model theory, 2002 Frank den Hollander, Large deviations, 2000 B. V. Rajarama Bhat, George A. Elliott, and Peter A. Fillmore, Editors, Lectures in operator theory, 2000 Salma Kuhlmann, Ordered exponential fields, 2000 Tibor Krisztin, Hans-Otto Walther, and .Jianhong Wu, Shape, smoothness and invariant stratification of an attracting set for delayed monotone positive feedback, 1999 .Jiff Patera, Editor, Quasicrystals and discrete geometry, 1998 Paul Sellck, Introduction to homotopy theory, 1997 Terry A. Loring, Lifting solutions to perturbing problems in C·-algebras, 1997 S. O. Kochman, Bordism, stable homotopy and Adams spectral sequences, 1996 Kenneth R. Davidson, C*-Algebras by example, 1996 A. Weiss, Multiplicative Galois module structure, 1996 Gt§rard Besson, .Joachim Lohkamp, Pierre Pansu, and Peter Petersen Mir08lav Lovric, Maung Min-Oo, and McKenzie Y.-K. Wang, Editors, Riemannian geometry, 1996 Albrecht Bottcher, Aad DlJksma and Heinz Langer, Michael A. Dritschel and .James Rovnyak, and M. A. Kaashoek Peter Lancaster, Editor, Lectures on operator theory and its applications, 1996 Victor P. Snaith, Galois module structure, 1994 Stephen Wiggins, Global dynamics, phase space transport, orbits homoclinic to resonances, and applications, 1993