let the conditions (1.2) and

j(V+ )j(V’ ) < j(V+ ) + j(V’ )

17.2. Proof of Theorem 17.1.1 229

be ful¬lled. Then I ’ K is boundedly invertible in L∞ and the inverse operator

˜

satis¬es the inequality

j(V’ )j(V+ )

|(I ’ K)’1 |L∞ ¤

˜ .

j(V’ ) + j(V+ ) ’ j(V’ )j(V+ )

Lemma 17.2.2 Under condition (1.2), operator V’ de¬ned by (1.3) satis¬es

the inequality

k

M∞ (V’ )

k

|V’ |L∞ ¤ (k = 1, 2, ...). (2.1)

k!

Proof: We have

x 1

|V’ h|L∞ = ess supx∈[0,1] | K(x, s)h(s)ds| ¤ w’ (s)|h(s)|ds.

0 0

Repeating these arguments, we arrive at the relation

1 s1 sk

k

|V’ h|L∞ ¤ |h(sk )|dsk . . . ds2 ds1 .

w’ (s1 ) w’ (s2 ) . . .

0 0 0

Taking |h|L∞ = 1, we get

1 s1 sk’1

k

|V’ |L∞ ¤ w’ (s1 ) w’ (s2 ) . . . dsk . . . ds2 ds1 . (2.2)

0 0 0

It is simple to see that

1 sk’1

w’ (s1 ) . . . w’ (sk )dsk . . . ds1 =

0 0

µ

˜ z1 zk’1

µk

˜

... dzk dzk’1 . . . dz1 = ,

k!

0 0 0

where sj

zj = zk (sj ) ≡ w’ (s)ds (j = 1, ..., k)

0

and

1

µ=

˜ w’ (s)ds.

0

Thus (2.2) gives

1

w’ (s)ds)k k

( M∞ (V’ )

k 0

|V’ |L∞ ¤ = .

k! k!

As claimed. 2

17. Integral Operators in L∞

230

Similarly, the inequality

k

M∞ (V+ )

k

|V+ |L∞ ¤ (k = 1, 2, ...) (2.3)

k!

can be proved.

Relations (2.1) and (2.3) imply

|(I ’ V± )’1 |L∞ ¤ j(V± ) ¤ eM∞ (V± ) . (2.4)

The assertion of Theorem 17.1.1 follows from Lemma 17.2.1 and relations

(2.4).

17.3 The Spectral Radius

Clearly,

»I ’ K = »(I ’ »’1 K) (» = 0).

˜ ˜

Consequently, if

’1 ’1 ’1

e(M∞ (V’ )+M∞ (V+ ))|»| < e|»| M∞ (V+ )

+ e|»| M∞ (V’ )

,

˜

then due to Theorem 17.1.1, »I ’ K is boundedly invertible. We thus get

Lemma 17.3.1 Under condition (1.2), any point » = 0 of the spectrum

˜ ˜

σ(K) of operator K satis¬es the inequality

’1 ’1 ’1

e(M∞ (V’ )+M∞ (V+ ))|»| ≥ e|»| M∞ (V+ )

+ e|»| M∞ (V’ )

. (3.1)

˜ ˜

Let rs (K) be the spectral radius of K. Then (3.1) yields

˜ ˜ ˜

’1 ’1 ’1

ers (K)(M∞ (V’ )+M∞ (V+ ))

≥ ers (K)M∞ (V+ )

+ ers (K)M∞ (V’ )

. (3.2)

˜

Clearly, if V+ = 0 or ( and ) V’ = 0, then rs (K) = 0.

Theorem 17.3.2 Under condition (1.2), let V+ = 0, V’ = 0. Then the

equation

e(M∞ (V’ )+M∞ (V+ ))z = ezM∞ (V+ ) + ezM∞ (V’ ) (z ≥ 0) (3.3)

has a unique positive zero z(K). Moreover, the inequality rs (K) ¤ z ’1 (K)

˜

is valid.

Proof: Equation (3.3) is equivalent to the following one:

(eM∞ (V+ )z ’ 1)(ezM∞ (V’ ) ’ 1) = 1. (3.4)

In addition, (3.2) is equivalent to the relation

˜ ˜

’1 ’1

(ers (K)M∞ (V+ )

’ 1)(ers (K)M∞ (V’ )

’ 1) ≥ 1.

17.4. Nonnegative Invertibility 231

Hence, the result follows, since the left part of equation (3.4) monotonically

increases. 2

From (3.3) it follows that

e(M∞ (V’ )+M∞ (V+ ))z ≥ 2

and

ez(M∞ (V+ )’M∞ (V’ )) = eM∞ (V+ )z ’ 1 ≥

exp [ln 2 M∞ (V+ )(M∞ (V’ ) + M∞ (V+ ))’1 ] ’ 1.

Thus with the notation

M∞ (V+ ) ’ M∞ (V’ )

δ∞ (K) = (3.5)

ln [ exp ( M∞ (V’(V+ )ln (V+ ) ) ’ 1]

M∞ 2

)+M∞

we have

’1

z(K) ≥ δ∞ (K), (3.6)

provided

M∞ (V+ ) < M∞ (V’ ). (3.7)

Clearly, in (3.5) we can exchang the places of V’ and V+ . Now Theorem

17.3.2 yields

˜

Corollary 17.3.3 Under conditions (1.2) and (3.7), the inequality rs (K) ¤

δ∞ (K) is true.

17.4 Nonnegative Invertibility

We will say that h ∈ L∞ is nonnegative if h(t) is nonnegative for almost all

t ∈ [0, 1]; a linear operator A in L∞ is nonnegative if Ah is nonnegative for

each nonnegative h ∈ L∞ . Recall that I is the identity operator.

Theorem 17.4.1 Let the conditions (1.2), (1.5) and

K(t, s) ≥ 0 (0 ¤ t, s ¤ 1) (4.1)

˜

hold. Then operator I ’ K is boundedly invertible and the inverse operator

is nonnegative. Moreover,

(I ’ K)’1 ≥ I.

˜ (4.2)

˜

Proof: Relation (2.9) from Section 16.2 with A = I ’ K, W = V’ and

V = V+ implies

(I ’ K)’1 = (I ’ V+ )’1 (I ’ BK )’1 (I ’ V’ )’1

˜ (4.3)

17. Integral Operators in L∞

232

where

BK = (I ’ V+ )’1 V+ V’ (I ’ V’ )’1 .

Moreover, by (4.1) we have V± ≥ 0. So (I ’ V± )’1 ≥ 0 and BK ≥ 0.

Relations (2.4) give us the inequalities

|(I ’ V± )’1 V± |L∞ ¤ eM∞ (V± ) ’ 1.

Consequently,

|BK |L∞ ¤ (eM∞ (V+ ) ’ 1)(eM∞ (V’ ) ’ 1).

But (1.5) is equivalent to (1.7). We thus get |BK |L∞ < 1. Consequently,

∞

’1 k

(I ’ BK ) BK ≥ 0.

=

k=0

Now (4.3) implies the inequality (I ’ K)’1 ≥ 0. In addition, since I ’ K ¤ I,

˜ ˜

we have inequality (4.2). 2

17.5 Applications

17.5.1 A nonselfadjoint di¬erential operator

Consider a di¬erential operator A de¬ned by

d2 h(x) dh(x)

(Ah)(x) = ’ + m(x)h(x) (0 < x < 1, h ∈ Dom (A) )

+ g(x)

2

dx dx

(5.1)

on the domain

Dom (A) = {h ∈ L∞ , h ∈ L∞ + some boundary conditions } (5.2)

In addition,

the coe¬cients g, w ∈ L∞ and are complex, in general. (5.3)

Let an operator S be de¬ned on Dom (A) by

(Sh)(x) = ’h (x), h ∈ Dom (A).

It is assumed that S has the Green function G(t, s). So that,

1

’1

h)(x) ≡ G(x, s)h(s)ds ∈ Dom (A)

(S

0

17.5. Applications 233

for any h ∈ L∞ , and the derivative of G in x satis¬es the condition

1

sup |Gx (x, s)|ds < ∞. (5.4)

x

0

Put

1

sup |G(x, s)|ds.

b∞ (S) :=

x

0

We have

˜

A = (I ’ K)S,

where

1 1

d

˜

(Kh)(x) = ’(g(x) + m(x)) G(x, s)h(s)ds = K(x, s)h(s)ds

dx 0 0

with

K(x, s) = ’g(x)Gx (x, s) ’ m(x)G(x, s). (5.5)

According to (5.3) and (5.4), condition (1.2) holds. Take into account that

|S ’1 h|L∞ ¤ b∞ (S)|h|L∞ .

Since

A’1 = S ’1 (I ’ K)’1 ,

˜

Theorem 17.1.1 immediately implies the following result:

Proposition 17.5.1 Under (5.3)-(5.5), let condition (1.5) hold. Then op-

erator A de¬ned by (5.1), (5.2) is boundedly invertible in L∞ . In addition,

b∞ (S)eM∞ (V’ )+M∞ (V+ )

|A’1 |L∞ ¤ .

eM∞ (V+ ) + eM∞ (V’ ) ’ eM∞ (V’ )+M∞ (V+ )

17.5.2 An integro-di¬erential operator

On domain (5.2), let us consider the operator

1

d2 u(x)

(Eu)(x) = ’ K0 (x, s)u(s)ds (u ∈ Dom (A), 0 < x < 1), (5.6)

+

dx2 0

where K0 is a kernel with the property

1

|K0 (x, s)| ds < ∞.

ess supx (5.7)

0

Let S and G be the same as in the previous subsection. Then we can write

˜ ˜

E = (I ’ K)S where K is de¬ned by (1.1) with

1

K(x, s) = ’ K0 (x, x1 )G(x1 , s)dx1 (5.8)

0

17. Integral Operators in L∞

234

˜

So if I ’ K is invertible, then E is invertible as well. Clearly, under (5.4) and

(5.7), condition (1.2) holds. Since

E ’1 = S ’1 (I ’ K)’1 ,

˜

Theorems 17.1.1 and 17.4.1 yield

Proposition 17.5.2 Under (5.4), (5.7) and (5.8), let condition (1.5) hold.

Then operator E de¬ned by (5.6), (5.2) is boundedly invertible in L∞ and

b∞ (S)eM∞ (V’ )+M∞ (V+ )

’1

|E |L∞ ¤ M (V ) .

e ∞ + + eM∞ (V’ ) ’ eM∞ (V’ )+M∞ (V+ )

If, in addition, G ≥ 0 and K0 ¤ 0, then E ’1 is positive. Moreover,

1

’1 ’1

h)(x) ≥ (S

(E h)(x) = G(x, s)h(s)ds

0

for any nonnegative h ∈ L∞ .

17.6 Notes

The present chapter is based on the paper (Gil™, 2001).

About well-known results on the spectrum of integral operators on L∞ ,

see, for instance, the books (Diestel et al., 1995), (K¨nig, 1986), (Kras-

o

nosel™skii et al., 1989), (Pietsch, 1987) and references therein.

References

[1] Diestel, D., Jarchow, H, Tonge, A. (1995), Absolutely Summing Opera-

tors, Cambridge University Press, Cambridge.

[2] Gil™, M.I. (2001). Invertibility and positive invertibility conditions of

integral operators in L∞ , J. of Integral Equations and Appl. 13 , 1-14.

[3] K¨nig, H. (1986). Eigenvalue Distribution of Compact Operators,

o

Birkh¨user Verlag, Basel- Boston-Stuttgart.

a

[4] Krasnosel™skii, M. A., J. Lifshits, and A. Sobolev (1989). Positive Linear

Systems. The Method of Positive Operators, Heldermann Verlag, Berlin.

[5] Pietsch, A. (1987). Eigenvalues and s-Numbers, Cambridge University

Press, Cambridge.

[6] Zabreiko, P.P., A.I. Koshelev, M. A. Krasnosel™skii, S.G. Mikhlin, L.S.

Rakovshik, B.Ya. Stetzenko (1968). Integral Equations, Nauka, Moscow.

In Russian

18. Hille - Tamarkin

Matrices

In the present chapter we investigate in¬nite matrices, whose o¬ diagonal

parts are the Hille-Tamarkin matrices. Invertibility conditions and estimates

for the norm of the inverse matrices are established. In addition, bounds for

the spectrum are suggested. In particular, estimates for the spectral radius

are derived.

18.1 Invertibility Conditions

Everywhere in this chapter

A = (ajk )∞

j,k=1

is an in¬nite matrix with the entries ajk (j, k = 1, 2, ...). Besides, V+ , V’ and

D denote the strictly upper triangular, strictly lower triangular, and diagonal

parts of A, respectively:

«

« 0 0 0 0 ...

0 a12 a13 a14 ... ¬ ... ·

a21 0 0 0

¬0 0 ... · ¬ ·

a23 a24

V+ = ¬ · , V’ = ¬ ... ·

a31 a32 0 0

¬ ·

0 0 ...

0 a34 ...

a41 a42 a43 0

. . . . ...

. . . ... .

(1.1)

and

D = diag [a11 , a22 , a33 , ...].

M.I. Gil™: LNM 1830, pp. 235“241, 2003.

c Springer-Verlag Berlin Heidelberg 2003

236 18. Hille - Tamarkin Matrices

Throughout this chapter it is assumed that V’ and V+ are the Hille-Tamarkin

matrices. That is, for some ¬nite p > 1,

∞ ∞

|ajk |q ]p/q ]1/p < ∞

[ (1.2)

j=1 k=1, k=j

with

11

+ = 1.

pq

As usually lp (1 < p < ∞) is the Banach space of number sequences equipped

with the norm

∞

|hk |p ]1/p (h = (hk ) ∈ lp ).

|h|lp = [

k=1

So under (1.2), A represents a linear operator in lp which is also denoted by

A. Clearly,

∞

p

|akk hk |p < ∞}.

Dom (A) = Dom (D) = {h = (hk ) ∈ l :

k=1

Assume that

d0 ≡ inf |akk | > 0 (1.3)

k

and introduce the notations

∞ ∞

|a’1 ajk |q ]p/q )1/p ,

+

≡(

Mp (A) [ jj

j=1 k=j+1

∞ j’1

|a’1 ajk |q ]p/q )1/p ,

’

Mp (A) =( [ jj

j=2 k=1

and

∞

(Mp (A))k

±

±

√

Jp (A) = .

p

k!

k=0

Now we are in a position to formulate the main result of the chapter.

Theorem 18.1.1 Let the conditions (1.2), (1.3) and

’ + + ’

Jp (A)Jp (A) < Jp (A) + Jp (A) (1.4)

hold. Then A is boundedly invertible in lp and the inverse operator satis¬es

the inequality

’ +

Jp (A)Jp (A)

’1

|A | ¤ (1.5)

lp + ’ ’ +

(Jp (A) + Jp (A) ’ Jp (A)Jp (A))d0

The proof of this theorem is presented in the next section.

18.2. Proof of Theorem 18.1.1 237

18.2 Proof of Theorem 18.1.1

Lemma 18.2.1 Under condition (1.2), for the strictly upper and lower tri-

angular matrices V+ and V’ , the inequalities

(vp )m

±

m

¤√

|V± |lp (m = 1, 2, ...)

p

m!

are valid, where

∞ ∞

+

|ajk |q ]p/q )1/p

vp =( [

j=1 k=j+1

and

∞ j’1

’

|ajk |q ]p/q )1/p .

vp =( [

j=2 k=1

This result follows from Lemma 3.2.1 when n ’ ∞, since

1

γn,m,p ¤ √ .

p

m!

So operators V± are quasinilpotent. The latter lemma yields

Corollary 18.2.2 Under conditions (1.2), (1.3), the inequalities

(Mp (A))m

±

’1 m

√

|(D V± ) | ¤ (m = 1, 2, ...)

lp p

m!

are valid.

Proof of Theorem 18.1.1: We have

A = V+ + V’ + D = D(D’1 V+ + D’1 V’ + I)

Clearly,

|D’1 |lp = d’1 .

0

From Lemma 18.2.2, it follows that

j(D’1 V± ) ¤ Jp (A)

±

Now Corollary 16.2.2 and condition (1.4) yield the invertibility of the operator

D’1 V+ + D’1 V’ + I

and the estimate (1.5). 2

238 18. Hille - Tamarkin Matrices

18.3 Localization of the Spectrum

Let σ(A) be the spectrum of A. For a » ∈ C, assume that

ρ(D, ») ≡ inf |» ’ amm | > 0,

m

and put

∞ ∞

+

|(ajj ’ »)’1 ajk |q ]p/q )1/p ,

Mp (A, ») =( [

j=1 k=j+1

∞ j’1

’

|(ajj ’ »)’1 ajk |q ]p/q )1/p ,

Mp (A, ») =( [

j=2 k=1

and

∞

(Mp (A, »))k

±

±

√

Jp (A, ») = .

p

k!

k=0

Clearly

± ± ± ±

Mp (A, 0) = Mp (A), Jp (A, 0) = Jp (A).

Lemma 18.3.1 Under condition (1.2), for any µ ∈ σ(A) we have either

µ = ajj for some natural j, or

’ + + ’

Jp (A, µ)Jp (A, µ) ≥ Jp (A, µ) + Jp (A, µ). (3.1)

Proof: Assume that

’ + + ’

Jp (A, µ)Jp (A, µ) < Jp (A, µ) + Jp (A, µ)

for some µ ∈ σ(A). Then due to Theorem 18.1.1, A ’ µI is invertible. This

contradiction proves the required result. 2

±

Recall that vp are de¬ned in Section 18.2 and denote,

∞

(vp )k

±

±

√ (z > 0)

Fp (z) = (3.2)

z k p k!

k=0

Lemma 18.3.2 Under condition (1.2), for any µ ∈ σ(A), either there is an

integer m, such that, µ = amm , or

’ + ’ +

Fp (ρ(D, µ))Fp (ρ(D, µ)) ≥ Fp (ρ(D, µ)) + Fp (ρ(D, µ)). (3.3)

Proof: Let µ = akk for all natural k. Then

Mp (A, µ) ¤ ρ’1 (D, µ)vp .

± ±

Hence,

± ±

Jp (A, µ) ¤ Fp (ρ(D, µ)). (3.4)

18.3. Localization of the Spectrum 239

In addition, (3.1) is equivalent to the relation

’ +

(Jp (A, µ) ’ 1)(Jp (A, µ) ’ 1) ≥ 1.

Now (3.4) implies (3.3). 2

Theorem 18.3.3 Under condition (1.2), let V+ = 0, V’ = 0. Then the

equation

’ + ’ +

Fp (z)Fp (z) = Fp (z) + Fp (z) (3.5)

has a unique positive root ζ(A). Moreover, ρ(D, µ) ¤ ζ(A) for any µ ∈ σ(A).

In other words, σ(A) lies in the closure of the union of the discs

{» ∈ C : |» ’ akk | ¤ ζ(A)} (k = 1, 2, ...).

Proof: Equation (3.5) is equivalent to the following one:

’ +

(Fp (z) ’ 1)(Fp (z) ’ 1) = 1. (3.6)

The left part of this equation monotonically decreases as z > 0 increases;

so it has a unique positive root ζ(A). In addition, (3.3) is equivalent to the

relation

’ +

(Fp (ρ(D, µ)) ’ 1)(Fp (ρ(D, µ)) ’ 1) ≥ 1. (3.7)

Hence the result follows. 2

Rewrite (3.5) as

∞ ∞

(vp )k

’

(vp )j

+

√ √ = 1.

z j p j!

z k p k! j=1

k=1

Or

∞ k’1

(vp )k’j (vp )j

+ ’

k

Bk z = 1 with Bk = (k = 2, 3, ...).

j!(k ’ j)!

p

j=1

k=2

Due to the Lemma 8.3.1, with the notation

δp (A) ≡ 2 sup j

Bj ,

j=2,3,...

we get ζ(A) ¤ δp (A). Now Theorem 18.3.3 yields

Corollary 18.3.4 Under condition (1.2), let V+ = 0, V’ = 0. Then for any

µ ∈ σ(A), the inequality ρ(µ, D) ¤ δ(A) is true.

In other words, σ(A) lies in the closure of the union of the sets

{» ∈ C : |» ’ akk | ¤ δp (A)} (k = 1, 2, ...).

240 18. Hille - Tamarkin Matrices

Note that Theorem 18.3.3 is exact: if A is triangular: either V’ = 0, or

V+ = 0, then we due to that lemma σ(A) is the closure of the set

{akk , k = 1, 2, ...}.

Moreover, Theorem 18.3.3 and Corollary 18.3.4 imply

rs (A) ¤ sup |akk | + ζ(A) ¤ sup |akk | + δp (A), (3.8)

k=1,2,... k=1,2,...

provided D is bounded. Furthermore, let the condition

∞

|ajk | < ∞

sup (3.9)

j=1,2,...

k=1

hold. Then the well-known estimate

∞

rs (A) ¤ |ajk |

sup (3.10)

j=1,2,...

k=1

is valid, see (Krasnosel™skii et al. 1989, Theorem 16.2). Under condition

(3.9), relations (3.8) improve (3.10), provided

∞

|ajk |

ζ(A) < sup

j=1,2,...

k=1,k=j

or

∞

|ajk |.

δp (A) < sup

j=1,2,...

k=1,k=j

In conclusion, note that Theorem 18.1.1 is exact: if A is upper or lower

triangular, then A is invertible, provided D is invertible.

18.4 Notes

The present chapter is based on the paper (Gil™, 2002).

About other results on the spectrum of Hille-Tamarkin matrices see, for

instance, the books (Diestel et al., 1995), (K¨nig, 1986), (Pietsch, 1987), and

o

references therein.

Note that Hille-Tamarkin matrices arise, in particular, in recent investi-

gations of discrete Volterra equations, see (Kolmanovskii et al, 2000), (Gil™

and Medina, 2002), (Medina and Gil™, 2003).

18.4. Notes 241

References

[1] Diestel, D., Jarchow, H, Tonge, A. (1995), Absolutely Summing Opera-

tors, Cambridge University Press, Cambridge.

[2] Gil™, M.I. (2002), Invertibility and spectrum of Hille-Tamarkin matrices,

Mathematische Nachrichten, 244, 1-11

[3] Gil™, M.I. and Medina, R. (2002). Boundedness of solutions of matrix

nonlinear Volterra di¬erence equations. Discrete Dynamics in Nature

and Society, 7, No 1, 19-22

[4] Kolmanovskii, V.B., A.D. Myshkis and J.P. Richard (2000). Estimate

of solutions for some Volterra di¬erence equations, Nonlinear Analysis,

TMA, 40, 345-363.

[5] K¨nig, H. (1986). Eigenvalue Distribution of Compact Operators,

o

Birkh¨user Verlag, Basel- Boston-Stuttgart.

a

[6] Krasnosel™skii, M. A., J. Lifshits, and A. Sobolev (1989). Positive Linear

Systems. The Method of Positive Operators, Heldermann Verlag, Berlin.

[7] Medina, R. and Gil™, M.I. (2003). Multidimensional Volterra di¬erence

equations. In the book: New Progress in Di¬erence Equations, Eds. S.

Elaydi, G. Ladas and B. Aulbach, Taylor and Francis, London and New

York, p. 499-504

[8] Pietsch, A. (1987). Eigenvalues and s-Numbers, Cambridge University

Press, Cambridge.

19. Zeros of Entire

Functions

The present chapter is devoted to applications of our abstract results to the

theory of ¬nite order entire functions. We consider the following problem: if

the Taylor coe¬cients of two entire functions are close, how close are their

zeros? In addition, we establish bounds for sums of the absolute values of

the zeros in the terms of the coe¬cients of its Taylor series. These bounds

supplement the Hadamard theorem.

19.1 Perturbations of Zeros

Consider the entire function

∞

ck »k (» ∈ C; c0 = 1)

f (») =

k=0

with complex, in general, coe¬cients ck , k = 1, 2, ... . Put

Mf (r) := max |f (z)| (r > 0).

|z|=r

Recall that

ln ln Mf (r)

ρ (f ) := limr’∞

ln r

is the order of f . Moreover, the relation

n ln n

ρ (f ) = limn’∞

ln (1/|cn |)

is true, cf. (Levin, 1996, p. 6).

M.I. Gil™: LNM 1830, pp. 243“252, 2003.

c Springer-Verlag Berlin Heidelberg 2003

244 19. Zeros of Entire Functions

Everywhere in the present chapter it is assumed that the set

{zk (f )}∞

k=1

of all the zeros of f taken with their multiplicities is in¬nite.

’1

Note that if f has a ¬nite number m of the zeros, we can put zk (f ) = 0

’1

for k = m, m + 1, ... and aplly our arguments below. Here and below zk (f )

means zk1 ) .

(f

Rewrite function f in the form

∞

ak »k

f (») = (a0 = 1) (1.1a)

(k!)γ

k=0

with a positive γ, and consider the function

∞

bk » k

h(») = (b0 = 1). (1.1b)

(k!)γ

k=0

Assume that

∞ ∞

2

|bk |2 < ∞.

|ak | < ∞, (1.2)

k=0 k=0

Relations (1.1) and (1.2), imply that functions f and h have orders no more

than 1/γ.

De¬nition 19.1.1 The quantity

’1 ’1

zvf (h) = max min |zk (f ) ’ zj (h)|

j k

will be called the variation of zeros of function h with respect to function f .

For a natural p > 1/2γ, put

∞

|ak |2 ]1/2 + 2 [ζ(2γp) ’ 1]1/2p ,

wp (f ) := 2 [ (1.3)

k=1

where ζ is the Riemann Zeta function, and

p’1 2p

wk (f ) 1 wp (f )

ψ(f, y) := exp [ + ] (y > 0). (1.4)

y k+1 2y 2p

2

k=0

Finally, denote

∞

|ak ’ bk |2 ]1/2 .

q := [

k=1

19.1. Perturbations of Zeros 245

Theorem 19.1.2 Let conditions (1.1) and (1.2) be ful¬lled. In addition, let

r(q, f ) be the unique positive (simple) root of the equation

qψ(f, y) = 1.

Then zvf (h) ¤ r(q, f ). That is, for any zero z(h) of h there is a zero z(f )

of f , such that

|z(h) ’ z(f )| ¤ r(q, f )|z(h)z(f )|. (1.5)

The proof of Theorem 19.1.2 is presented in the next section. Substitute in

(1.5) the equality y = xwp (f ) and apply Lemma 8.3.2. Then we have

r(q, f ) ¤ δ(q, f ), (1.6)

where

if wp (f ) ¤ epq,

epq

δ(q, f ) := .

wp (f ) [ln (wp (f )/qp)]’1/2p if wp (f ) > epq

Theorem 19.1.2 and inequality (1.6) yield

Corollary 19.1.3 Let conditions (1.1) and (1.2) be ful¬lled. Then zvf (h) ¤

δ(q, f ). That is, for any zero z(h) of h, there is a zero z(f ) of f , such that

|z(h) ’ z(f )| ¤ δ(q, f )|z(h)z(f )|. (1.7)

Relations (1.5) and (1.7) imply the inequalities

|z(f )| ’ |z(h)| ¤ r(q, f )|z(h)||z(f )| ¤ δ(q, f )|z(h)||z(f )|.

Hence,

|z(h)| ≥ (r(q, f )|z(f )| + 1)’1 |z(f )| ≥ (δ(q, f )|z(f )| + 1)’1 |z(f )|.

This inequality yields the following result

Corollary 19.1.4 Under conditions (1.1) and (1.2), for a positive number

R0 , let f have no zeros in the disc {z ∈ C : |z| ¤ R0 }. Then h has no zeros

in the disc {z ∈ C : |z| ¤ R1 } with

R0 R0

R1 = or R1 = .

δ(q, f )R0 + 1 r(q, f )R0 + 1

Let us assume that under (1.1), there is a constant d0 ∈ (0, 1), such that

|ak | < 1/d0

k

limk’∞ (1.8)

and

|bk | < 1/d0

k

limk’∞

246 19. Zeros of Entire Functions

and consider the functions

∞

ak (d0 »)k

˜

f (») = (1.9)

(k!)γ

k=0

and

∞

bk (d0 »)k

˜

h(») = .

(k!)γ

k=0

˜ ˜ ˜ ˜

That is, f (») ≡ f (d0 ») and h(») ≡ h(d0 »). So functions f (») and h(»)

satisfy conditions (1.2). Moreover,

∞

d2k |ak |2 ]1/2 + 2[ζ(2γp) ’ 1]1/2p .

˜

wp (f ) = 2[ 0

k=1

˜ ˜

Thus, we can apply Theorem 19.1.2 and its corollaries to functions f (»), h(»)

and take into account that

˜ ˜

d0 zk (f ) = zk (f ), d0 zk (h) = zk (h). (1.10)

19.2 Proof of Theorem 19.1.2

For a ¬nite integer n, consider the polynomials

n n

ak »n’k bk »n’k

F (») = and Q(») = (a0 = b0 = 1). (2.1)

(k!)γ (k!)γ

k=0 k=0

Put

n n n

2 1/2 2γp 1/2p

|ak ’bk |2 ]1/2 .

|ak | ]

wp (F ) := 2 [ +2 [ 1/k ] and q(F, Q) := [

k=1 k=2 k=1

In addition, {zk (F )}n n

k=1 and {zk (Q)}k=1 are the sets of all the zeros of F

and Q, respectively taken with their multiplicities. De¬ne ψ(F, y) according

to (1.4).

Lemma 19.2.1 For any zero z(Q) of Q, there is a zero z(F ) of F , such that

|z(F ) ’ z(Q)| ¤ r(Q, F ),

where r(Q, F ) be the unique positive (simple) root of the equation

q(F, Q)ψ(F, y) = 1. (2.2)

19.2. Proofs of Theorem 19.1.2 247

Proof: In a Euclidean space Cn with the Euclidean norm . , introduce

operators An and Bn by virtue of the n — n-matrices

«

’a1 ’a2 ... ’an’1 ’an

¬ 1/2γ 0·

0 ... 0

¬ ·

An = ¬ 0 0·

1/3γ ... 0

¬ ·

. .

. ... .

... 1/nγ

0 0 0

«

and

’b1 ’b2 ... ’bn’1 ’bn

¬ ·

1/2γ 0 ... 0 0

¬ ·

Bn = ¬ ·.

1/3γ

0 ... 0 0

¬ ·

. . ... . .

... 1/nγ

0 0 0

It is simple to see that

F (») = det (»I ’ An )

and Q(») = det (»I ’ Bn ), where I is the unit matrix. So

»k (An ) = zk (F ), »k (Bn ) = zk (Q) (k = 1, 2, ..., n), (2.3)

where »k (.), k = 1, ..., n are the eigenvalues with their multiplicities. Clearly,

An ’ Bn = q(F, Q).

Due to Theorem 8.5.4, for any »j (Bn ) there is an »i (An ), such that

|»j (Bn ) ’ »i (An )| ¤ yp (An , Bn ), (2.4)

where yp (An , Bn ) is the unique positive (simple) root of the equation

p’1

(2N2p (An ))k (2N2p (An ))2p

q(F, Q) exp [(1 + )/2] = 1,

y k+1 y 2p

k=0

where N2p (A) := [T race(AA— )p ]1/2p is the Neumann-Schatten norm and the

asterisk means the adjointness. But An = M + C, where

«

’a1 ’a2 ... ’an’1 ’an

¬0 0·

0 ... 0

M =¬ ·

. .

. ... .

0 0 ... 0 0

«

and

0 0 ... 0 0

¬ ·

1/2γ 0 ... 0 0

¬ ·

C=¬ ·.

1/3γ

0 ... 0 0

¬ ·

. . ... . .

... 1/nγ

0 0 0

248 19. Zeros of Entire Functions

Therefore, with

n

|ak |2 ,

c=

k=1

«

we have

c 0 ... 0 0

¬0 0 ... 0 0 ·

MM— = ¬ ·

. . ... . .

0 0 ... 0 0

«

and

0 0 ... 0 0

¬ ·

1/22γ

0 ... 0 0

¬ ·

CC — = ¬ ·.

0 0 ... 0 0

¬ ·

. . ... . .

0 1/n2γ

0 0 ...

Hence,

n

√

1/k 2γp ]1/2p .

N2p (An ) ¤ N2p (M ) + N2p (C) = c+[

k=2

Consequently yp (An , Bn ) ¤ r(Q, F ). Therefore (2.3) and (2.4) imply (2.2),

as claimed. 2

Proof of Theorem 19.1.2: Consider the polynomials

n n

ak »k bk » k

fn (») = and hn (») = . (2.5)

(k!)γ (k!)γ

k=0 k=0

Clearly, »n fn (1/») = F (») and hn (1/»)»n = Q(»). So

zk (F ) = 1/zk (fn ); zk (Q) = 1/zk (hn ). (2.6)

Take into account that the roots continuously depend on coe¬cients, we have

the required result, letting in the previous lemma n ’ ∞. 2

19.3 Bounds for Sums of Zeros

Again consider an entire function f of the form (1.1a) and assume that the

condition

∞

|ak |2 ]1/2 < ∞

θf := [ (3.1)

k=1

holds. Let the zeros of f be numerated in the increasing way:

|zk (f )| ¤ |zk+1 (f )| (k = 1, 2, ...). (3.2)

19.3. Bounds for Sums of Zeros 249

Theorem 19.3.1 Let f be an entire function of the form (1.1a). Then under

conditions (3.1) and (3.2), the inequalities

j j

’1

(k + 1)’γ (j = 1, 2, ...)

|zk (f )| ¤ θf +

k=1 k=1

are valid.

The proof of this theorem is presented in this section below. Note that

under condition (1.8) we can omit condition (3.1) due to (1.9) and (1.10).

To prove Theorem 19.3.1, again consider the polynomial F (») de¬ned in

(2.1) with the zeros ordered in the following way:

|zk (F )| ≥ |zk+1 (F )| (k = 1, ..., n ’ 1).

Set

n

|ak |2 ]1/2 .

θ(F ) := [

k=1

Lemma 19.3.2 The zeros of F satisfy the inequalities

j j

(k + 1)’γ (j = 1, ..., n ’ 1)

|zk (F )| ¤ θ(F ) +

k=1 k=1

and

n n’1

(k + 1)’γ .

|zk (F )| ¤ θ(F ) +

k=1 k=1

Proof: Take into account that according to (2.3),

j j

|»k (An )| ¤ sk (An ) (j = 1, ..., n), (3.3)

k=1 k=1

where sk (An ), k = 1, 2, ... are the singular numbers of An ordered in the

decreasing way (Marcus and Minc, 1964, Section II.4.2). But An = M + C,

where M and C are introduced in Section 19.2. We can write

s1 (M ) = θ(F ), sk (M ) = 0 (k = 2, ..., n).

In addition,

sk (C) = 1/(k + 1)γ (k = 1, ..., n ’ 1), sn (C) = 0.

Take into account that

j j j j

sk (M + C) ¤

sk (An ) = sk (M ) + sk (C),

k=1 k=1 k=1 k=1

250 19. Zeros of Entire Functions

cf. (Gohberg and Krein, 1969, Lemma II.4.2). So

j j

(k + 1)’γ (j = 1, ..., n ’ 1)

sk (An ) ¤ θ(F ) +

k=1 k=1

and

n n’1

(k + 1)’γ .

sk (An ) ¤ θ(F ) +

k=1 k=1

Now (2.3) and (3.3) yield the required result. 2

Proof of Theorem 19.3.1: Again consider the polynomial fn (z) de-

¬ned as in (2.5). Now Lemma 19.3.2 and (2.6) yield the inequalities

j j

’1

(k + 1)’γ (j = 1, ..., n ’ 1).

|zk (fn )| ¤ θf + (3.4)

k=1 k=1

But the zeros of entire functions continuously depend on its coe¬cients. So

for any j = 1, 2, ...,

j j

’1

|zk (f )|’1

|zk (fn )| ’

k=1 k=1

as n ’ ∞. Now (3.4) implies the required result. 2

19.4 Applications of Theorem 19.3.1

Put

„1 = θf + 2’γ and „k = (k + 1)’γ (k = 2, 3, ...).

Corollary 19.4.1 Let φ(t) (0 ¤ t < ∞) be a convex scalar-valued function,

such that φ(0) = 0. Then under conditions (1.1a), (3.1) and (3.2), the

inequalities

j j

’1

)¤

φ(|zk (f )| φ(„k ) (j = 1, 2, ...)

k=1 k=1

are valid. In particular, for any r ≥ 2,

j j j

’r r ’γ r

(k + 1)’rγ (j = 2, 3, ...). (4.1)

|zk (f )| ¤ „k = (θf + 2 )+

k=1 k=1 k=2

19.4. Applications of Theorem 19.3.1 251

Indeed, this result is due to the well-known Lemma II.3.4 (Gohberg and

Krein, 1969) and Theorem 19.3.1.

Furthermore, assume that

rγ > 1, r ≥ 2. (4.2)

Then the series

∞ ∞

r ’γ r

(k + 1)’rγ = (θf + 2’γ )r + ζ(γr) ’ 1 ’ 2’rγ

„k = (θf + 2 )+

k=1 k=2

converges. Here ζ(.) is the Riemann Zeta function, again. Now relation (4.1)

yields

Corollary 19.4.2 Under the conditions (1.1a), (3.1) and (4.2), the inequal-

ity

∞

|zk (f )|’r ¤ (θf + 2’γ )r + ζ(γr) ’ 1 ’ 2’γr (4.3)

k=1

is valid. In particular, if γ > 1, then due to (3.4)

∞

|zk (f )|’1 ¤ θf + ζ(γ) ’ 1. (4.4)

k=1

Consider now a positive scalar-valued function ¦(t1 , t2 , ..., tj ) with an integer

j, de¬ned on the domain

0 ¤ tj ¤ tj’1 ¤ t2 ¤ t1 < ∞

and satisfying

‚¦ ‚¦ ‚¦

> > ... > > 0 for t1 > t2 > ... > tj . (4.5)

‚t1 ‚t2 ‚tj

Corollary 19.4.3 Under conditions (1.1a), (3.1), (3.2) and (4.5),

¦(|z1 (f )|’1 , |z2 (f )|’1 , ..., |zj (f )|’1 ) ¤ ¦(„1 , „2 , ..., „j ).

Indeed, this result is due to Theorem 19.3.1 and the well-known Lemma II.3.5

(Gohberg and Krein, 1969).

In particular, let {dk }∞ be a decreasing sequence of non-negative num-

k=1

bers. Take

j

¦(t1 , t2 , ..., tj ) = d k tk .

k=1

Then Corollary 19.4.3 yields

j j j

’1

dk (k + 1)’γ

dk |zk (f )| ¤ „k d k = d 1 θ f +

k=1 k=1 k=1

(j = 2, 3, ...).

252 19. Zeros of Entire Functions

19.5 Notes

The variation of the zeros of general analytic functions under perturbations

was investigated, in particular, by P. Rosenbloom (1969). He established

the perturbation result that provides the existence of a zero of a perturbed

function in a given domain. In the present chapter a new approach to the

problem is proposed.

The material in the present chapter is taken from the papers (Gil™, 2000a,

2000b, 2000c and 2001). Corollary 19.4.2 supplements the classical Hadamard

theorem (Levin, 1996, p. 18), since it not only asserts the convergence of the

series of the zeros, but also gives us the estimate for the sums of the zeros.

References

[1] Gil™, M.I. (2000a). Inequalities for imaginary parts of zeros of entire

functions. Results in Mathematics, 37, 331-334

[2] Gil™, M.I. (2000b). Perturbations of zeros of a class of entire functions,

Complex Variables, 42, 97-106

[3] Gil™, M.I. (2000c). Approximations of zeros of entire functions by zeros

of polynomials. J. of Approximation Theory, 106, 66-76

[4] Gil™, M.I. (2001). Inequalities for zeros of entire functions, Journal of

Inequalities, 6 463-471.

[5] Gohberg, I. C. and Krein, M. G. (1969). Introduction to the Theory of

Linear Nonselfadjoint Operators, Trans. of Math. Monographs, v. 18,

Amer. Math. Soc., R.I.

[6] Levin, B. Ya. (1996). Lectures on Entire Functions, Trans. of Math.

Monographs, v. 150. Amer. Math. Soc., R. I.

[7] Marcus, M. and Minc, H. (1964). A Survey of Matrix Theory and Matrix

Inequalities, Allyn and Bacon, Boston.

[8] Rosenbloom, P.C. (1969). Perturbation of zeros of analytic functions. I.

Journal of Approximation Theory, 2, 111-126.

List of Main Symbols

A operator norm of an operator A

(., .) scalar product

|A| matrix whose elements are absolute values of A

’1

A inverse to A

—

A conjugate to A

AI = (A ’ A— )/2i

AR = (A + A— )/2

C1 Trace class

C2 Hilbert-Schmidt ideal

C∞ the set of all compact operator

Cp Neumann-Schatten ideal

n

C complex Euclidean space

det (A) determinant of A

det2 (A) generalized determinant of A 93

Dom (A) domain of A

g(A) 11, 83

gI (A) 106

H separable Hilbert space

I = IH identity operator (in a space H)

m.r.i. -maximal orthogonal resolution of identity 98

ni(A) nilpotency index of A 102

Np (A) Neumann-Schatten norm of A

N (A) = N2 (A) Hilbert-Schmidt (Frobenius) norm of A

n

R real Euclidean space

R» (A) resolvent of A

rsvA (B) relative spectral variation of B with respect to A 167

RPTO 164

rs (A) spectral radius of A

rl (A) lower spectral radius of A

sj (A) s-number (singular number) of A

svA (B) spectral variation of B with respect to A

T r A = T race A trace of A

w(», A) 165

±(A) = sup Re σ(A)

˜

βp , βp 108

γn,k 12

»k (A) eigenvalue of A

σ(A) spectrum of A

(p)

θk = √ 1 where [x] is the integer part of x

[k/p]!

ρ(A, ») distance between a point lambda and the spectrum of A

Index

Hille-Tamarkin integral

Carleman inequality 33, 93

operator 215

Hille-Tamarkin matrix 235

diagonal part of

compact operator 82

Lidskij™s theorem 79

matrix 8

noncompact operator 99

matrix-valued function 4

maximal resolution

estimate for norm of function of

of identity (m.r.i) 98

Hilbert-Schmidt operator 91