A Blaschke-type condition for analytic and subharmonic functions and application to contraction operators
aa r X i v : . [ m a t h . C V ] J un To V.P. Havin on the occasion of his 75th birthday
A BLASCHKE-TYPE CONDITION FOR ANALYTICAND SUBHARMONIC FUNCTIONS ANDAPPLICATION TO CONTRACTION OPERATORS
S. FAVOROV AND L. GOLINSKII
Abstract.
Let E be a closed set on the unit circle. We find aBlaschke-type condition, optimal in a sense of the order, on theRiesz measure of a subharmonic function v in the unit disk witha certain growth at the direction of E . In particular case when E is a finite set, and v = log | f | with an analytic function f , ourresult agrees with the recent one by A. Borichev, L. Golinskii andS. Kupin. An application to contractions close to unitary operatorsin the Hilbert space is given. Introduction
In [1] the authors study zero sets of analytic functions in the unit disk D , which grow exponentially fast near a finite set E of points on theunit circle T . Here is the main result of [1]. As usual, x + = max { x, } . Theorem BGK.
Let E ⊂ T be a finite set, f ∈ A ( D ) an analyticfunction in D , | f (0) | = 1 , and (1) | f ( z ) | ≤ exp (cid:16) Dρ q ( z, E ) (cid:17) with D, q ≥ , ρ ( z, E ) = dist( z, E ) . Let Z f be the zero set of f , eachzero is counted according to its multiplicity. Then for any ε > , (2) X z ∈ Z f (1 − | z | ) ρ ( z, E ) ( q − ε ) + ≤ C ( ε, q, E ) D. Note that both (1) and (2) make sense for arbitrary (infinite) sub-sets E on T , so the question arises naturally, whether the finitenessof E , which is an essential ingredient in the proof of Theorem BGK,can be relaxed. We should also mention that in the case E = T , ρ ( z, E ) = 1 − | z | we come to the well known Blaschke-type condition Mathematics Subject Classification.
Primary: 30D50; Secondary: 31A05,47B10.
Key words and phrases. analytic function in the unit disk, subharmonic func-tion in the unit disk, Riesz measure, Blaschke-type condition, discrete spectrum,contraction operator. for analytic functions in D with radial growth. The theory of suchfunctions goes back to M.M. Djrbashian [4], see also V.I. Matsaev &E.Z. Mogulskii [16], W. Hayman & B. Korenblum [11], F.A. Shamoyan[18], A.M. Jerbashian [13].Our investigation shows that the natural setting for the above prob-lem is the class of subharmonic functions v and their Riesz measures(generalized Laplacians) µ = (1 / π ) △ v rather than analytic functionsand their zero sets. It turns out that arbitrary closed sets E can beinvolved, and the results are optimal. Note that in the case of subhar-monic functions of the form log | f | with f ∈ A ( D ), the Riesz measureis a discrete and integer-valued measure supported on Z f , and µ { z } equals the multiplicity of zero of f at z .To formulate our result we need the following quantitative character-istic of the “sparseness” of a closed set E in terms of its t -neighborhood(3) E t := { ζ ∈ T : ρ ( ζ , E ) < t } . Put(4) I ( α, E ) := Z ν E ( s ) s α +1 ds ≤ + ∞ , ν E ( t ) = | E t | , α ∈ R . Here and in what follows we denote by | A | the normalized Lebesguemeasure of a set A ⊂ T . It is not hard to see that I ( α, E ) < ∞ for any E and an arbitrary negative α . In the case when ν E ( t ) = O ( t β ), β > I ( α, E ) < ∞ for α < β . Examples of sets E and evaluationsof their characteristic I are given in Section 2.The main result of our paper is the following Theorem 1.
Let E ⊂ T be a closed set. Let v be a subharmonicfunction in D , v
6≡ −∞ , µ its Riesz measure, and for all z ∈ D andsome q > v ( z ) ≤ ρ q ( z, E ) . ( i ) If I ( q, E ) < ∞ , then (5) Z D (1 − | λ | ) dµ ( λ ) < ∞ . Moreover, if v (0) ≥ , then (6) Z D (1 − | λ | ) dµ ( λ ) ≤ q q I ( q, E ) . ( ii ) If I ( q, E ) = ∞ and I ( α, E ) < ∞ for some α < q , then (7) Z D (1 − | λ | ) ( ρ ( λ, E )) q − α dµ ( λ ) < ∞ . Moreover, if v (0) ≥ , then (8) Z D (1 − | λ | ) ( ρ ( λ, E )) q − α dµ ( λ ) < − α + q (90) q − α I ( α, E )) . BLASCHKE-TYPE CONDITION AND APPLICATION 3
The result turns out to be optimal in the following sense.
Theorem 2.
Let E ⊂ T be a closed set so that I ( α, E ) = + ∞ for some α ≥ . Then for the subharmonic function v ( z ) = ρ − q ( z, E ) with theRiesz measure µ , and q ≥ α , we have (9) Z D (1 − | λ | ) ( ρ ( λ, E )) q − α dµ ( λ ) = + ∞ . As a consequence we obtain
Theorem 3.
Let E ⊂ T be a closed set, and I ( α, E ) < ∞ for some α ∈ R . Let f ∈ A ( D ) , and (10) | f ( z ) | ≤ exp (cid:16) Dρ q ( z, E ) (cid:17) , z ∈ D . Then (11) X z ∈ Z f (1 − | z | ) ρ ( z, E ) ( q − α ) + < ∞ . Moreover, if | f (0) | ≥ , then X z ∈ Z f (1 − | z | ) ρ ( z, E ) ( q − α ) + ≤ C ( q, α, E ) D,C ( q, α, E ) does not depend on f . In the case q > α one can take C ( q, α, E ) = 36(2 − α + q (90) q − α I ( α, E )) . The classical Blaschke condition arises for E = T , α = − q < q → f ≡ Proof of the main results
There is a simple way to compute the function ν E defined in (4), interms of the complimentary arcs of E . Let T \ E = [ j γ j , | γ j | ↓ . Then ν E ( t ) = ∞ X j = N +1 | γ j | + 2 Nπ arcsin t | E | , where N = N ( t ) is taken from | γ N +1 | ≤ π arcsin t < | γ N | . An interesting case occurs when the number of complimentary arcs is infinite.
S. FAVOROV AND L. GOLINSKII
So the exponent α in (4), for which I ( α, E ) < ∞ can be easily deter-mined. For instance, for E = ( e iϕ n : ϕ n = ∞ X k = n k ) ∪ { } one can take any α <
1, for E = ( e iϕ n : ϕ n = c ∞ X k = n k γ , γ > ) ∪ { } , c − = ∞ X k =1 k γ , any α < − γ works, for E = ( e iϕ n : ϕ n = c ∞ X k = n k log k ) ∪ { } , c − = ∞ X k =2 k log k ,I ( α, E ) < ∞ for any α <
0. For the generalized Cantor set C β (for thestandard Cantor set β = 1 /
3) one can take any α < − d ( β ), d ( β ) = log 2log 2 − log(1 − β )is the Hausdorff dimension of C β (see, e.g., [6]).Let us list some elementary properties of ν E .(1) ν E is a continuous and strictly monotone increasing function on[0 , t ( E )] and ν E ( t ) = 1 for t ( E ) ≤ t < ∞ ;(2) ν E (0+) = | E | ;(3) ν E ( t ) ≥ | E | + t/π for each closed E = T and 0 < t ≤ t ( E );(4) ν E ( t ) = O ( t ), as t →
0, if and only if E is a finite set.Our argument relies upon some basic results from the potential the-ory in the complex plane (see [17, Chapters 3,4]). Let v be a subhar-monic function on D , v
6≡ −∞ . The Riesz measure µ = (1 / π ) △ v is known to be a positive Radon measure on D . If v has a harmonicmajorant on a subdomain Ω ⊆ D , then the following representationholds(12) v ( z ) = u ( z ) − Z Ω G Ω ( z, λ ) dµ ( λ ) , z ∈ Ω , where u is the least harmonic majorant in Ω, and G Ω ( z, λ ) is Green’sfunction for Ω, that is,(13) G Ω ( z, λ ) = log(1 / | z − λ | ) − h ( z, λ ) ,h ( z, · ) is a harmonic function in Ω such that h ( z, λ ) = log(1 / | z − λ | )for λ ∈ ∂ Ω. Note that if Ω = D and 0 ∈ Ω, then h (0 , λ ) > λ ∈ Ω. Proof of Theorem 1 . We begin with a proof of (6). It easily followsfrom I ( q, E ) < ∞ , q >
0, and (4) that now ν E (0+) = | E | = 0. Put ρ ( z ) = ρ ( z, E ). BLASCHKE-TYPE CONDITION AND APPLICATION 5
The monotone permutation theorem (also known as the “layer cakerepresentation”, see, e.g., [15, Theorem 1.13])(14) Z X f r ( x ) dσ ( x ) = r Z ∞ y r − σ ( { x : f ( x ) > y } ) dy, is of importance in our argument. Here r >
0, (
X, σ ) is a measurespace, f ≥ X . We have (see (4))(15) Z T dm ( ζ ) ρ q ( ζ ) = q Z ∞ y q − (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) ζ : ρ ( ζ ) < y (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) dy = 2 − q + qI ( q, E ) < ∞ ,dm is the normalized Lebesgue measure on T . Therefore the Poissonintegral U ( z ) = Z T − | z | | ζ − z | dm ( ζ ) ρ q ( ζ ) , z ∈ D is a well defined harmonic function.Let us prove that v admits a harmonic majorant on the whole disk D . A simple geometrical inequality shows that for all | z | ≤ < τ < ρ ( z ) ≤ ρ ( τ z ). The function v = ρ − q is obviouslysubharmonic in D since v ( z ) = sup ζ ∈ E ( | z − ζ | − q )(see [17, Theorem 2.4.7]), and continuous at any point ζ ∈ T \ E . Bythe well known property of the Poisson integrallim z → ζ U ( z ) = v ( ζ ) ≥ − q v ( τ ζ ) , ζ ∈ T \ E, < τ < . As long as ζ ∈ E , one has lim z → ζ U ( z ) = + ∞ . But v ( τ z ) is a bounded subharmonic function in D , so by the Maximum Principle U ( z ) ≥ − q v ( τ z ), z ∈ D . It remains only to tend τ → q U is a desired harmonic majorant for v .Representation (12) takes now the form v ( z ) = u ( z ) − Z D log (cid:12)(cid:12)(cid:12)(cid:12) − ¯ λzz − λ (cid:12)(cid:12)(cid:12)(cid:12) dµ ( λ ) , z ∈ D , where u ≤ U in D . For z = 0 we come to Z D (1 − | λ | ) dµ ( λ ) ≤ Z D log 1 | λ | dµ ( λ ) ≤ u (0) ≤ q U (0) = 2 q k ρ − q k L ( T ) , and (6) follows from (15).The case when I ( q, E ) = ∞ but I ( α, E ) < ∞ for some α < q ismuch more delicate. S. FAVOROV AND L. GOLINSKII
For fixed t ∈ (0 ,
1) denote by Ω = Ω t the connected component ofthe open subset { z ∈ D : ρ ( z ) > t } , which contains the origin, so Ω isa nonempty subdomain of D . PutΓ := { z ∈ D : ρ ( z ) = t } , E ct = T \ E t ,E t is defined in (3). It is easy to check that E t is a finite union ofdisjoint open arcs, so E ct is a finite union of disjoint closed arcs. Theboundary ∂ Ω is contained in the closure of E ct ∪ Γ. The function v is bounded from above in Ω, so (12) holds for v in Ω, with the leastharmonic majorant u .Let V be a harmonic function in D such that(16) V ( ζ ) = (cid:26) t − q , ζ ∈ E t , ρ − q ( ζ ) , ζ ∈ E ct .Note that V is continuous in the closed unit disk. For 0 < t < z ∈ Γ there is ζ ′ ∈ E such that | z − ζ ′ | = t . Let ω ( λ, γ z , D )be the harmonic measure of the arc γ z = { ζ ∈ T : | ζ − ζ ′ | ≤ t } . Iteasily follows from the explicit expression (see, e.g., [7, Chapter 1])that ω ( λ, γ z , D ) ≥ / | λ − ζ ′ | = t , λ ∈ D . Since γ z ⊂ E t , we get V ( z ) ≥ t − q ω ( z, γ z , D ) ≥ t − q /
6. Therefore, we havelim sup z → ζ v ( z ) ≤ V ( ζ ) , ∀ ζ ∈ ∂ Ω . By the Maximum Principle v ( z ) ≤ V ( z ) and so u ( z ) ≤ V ( z ) , z ∈ Ω . Let us prove (8). Since v (0) ≥
0, (12) for z = 0 can be written as(17) Z Ω G Ω (0 , λ ) dµ ( λ ) ≤ u (0) ≤ V (0) = 6 ν E ( t ) t q + 6 Z E ct dm ( ζ ) ρ q ( ζ ) . We proceed with another application of (14) to obtain Z E ct dm ( ζ ) ρ q ( ζ ) = q Z ∞ y q − |{ ζ ∈ T : t < ρ ( ζ ) < /y }| dy =2 − q − ν E ( t ) t q + q Z t ν E ( s ) s q +1 ds . (18)Hence, the following bound holds(19) Z Ω G Ω (0 , λ ) dµ ( λ ) ≤ · − q + 6 q Z t ν E ( s ) s q +1 ds . Our next step is to obtain a lower bound for the Green function G Ω (0 , · ) in a smaller domain Ω τ with τ = (6 π + 3) t . We assume nowthat t ∈ (0 , π +3 ), and so τ ∈ (0 , z ∈ Γ there is ζ ′ ∈ E suchthat | z − ζ ′ | = t , so t ≥ − | z | , and hencelog 1 | z | ≤ log 11 − t ≤ t . BLASCHKE-TYPE CONDITION AND APPLICATION 7
The harmonic function h (0 , · ) from (13) does not exceed 3 t/ E ct , so(20) h (0 , λ ) ≤ t , λ ∈ Ω . We will distinguish two situations for λ .Let | λ | ≤ − t . Then by (20) G Ω (0 , λ ) = log 1 | λ | − h (0 , λ ) ≥ − | λ | − t ≥ − | λ | . Let | λ | > − t . This is where the restriction λ ∈ Ω τ is essential. Let ω ( λ, E t , D ) be the harmonic measure of the set E t . We put g ( λ ) =9 tω ( λ, E t , D ). The same argument as above implies ω ( λ, E t , D ) ≥ / λ ∈ Γ, so h (0 , λ ) ≤ g ( λ ) , λ ∈ ∂ Ω ⇒ h (0 , λ ) ≤ g ( λ ) , λ ∈ Ω , and G Ω (0 , λ ) ≥ − | λ | − g ( λ ) . We will find the upper bound for g ( λ ) in Ω τ . For λ = | λ | e iθ write g ( λ ) = 9 t Z E t − | λ | | ζ − λ | dm ( ζ ) = 9 t (1 − | λ | )2 π Z e iϕ ∈ E t dϕ (1 − | λ | ) + 4 | λ | sin ϕ − θ . Since ρ ( λ ) ≥ τ and 1 − | λ | < t , we have for some ζ ∈ Eρ ( e iθ ) = | e iθ − ζ | = | λ − ζ + e iθ − λ |≥ (6 π + 3) t − (1 − | λ | ) > (6 π + 1) t. Next, for e iϕ ∈ E t there is ζ ∈ E with | e iϕ − ζ | = ρ ( e iϕ ) ≤ t , so | θ − ϕ | ≥ (cid:12)(cid:12)(cid:12)(cid:12) sin θ − ϕ (cid:12)(cid:12)(cid:12)(cid:12) = | e iθ − ζ + ζ − e iϕ | ≥ ρ ( e iθ ) − t ≥ πt. Therefore, as | λ | > − t > /
10, we come to g ( λ ) ≤ πt (1 − | λ | )2 Z πt ≤| ϕ − θ |≤ π dϕ ( ϕ − θ ) < − | λ | )6 . Finally, the following lower bound holds for all λ ∈ Ω τ G Ω (0 , λ ) ≥ − | λ | , and we can continue (19) as(21) Z Ω τ (1 − | λ | ) dµ ( λ ) ≤ (cid:18) − q + q Z τ/ (6 π +3) ν E ( s ) s q +1 ds (cid:19) . The latter holds for each τ ∈ (0 , ρ := min { ρ, } . It is easily checked that(22) ˜ ρ ( λ ) ≤ ρ ( λ ) ≤ ρ ( λ ) ∀ λ ∈ D , S. FAVOROV AND L. GOLINSKII and { λ : ρ ( λ ) > τ } = { λ : ˜ ρ ( λ ) > τ } for 0 < τ <
1. (14) applied to themeasure dσ = (1 − | λ | ) dµ and the function ˜ ρ ≤ Z D ( ˜ ρ ( λ )) q − α dσ ( λ ) = ( q − α ) Z τ q − α − dτ Z { ρ>τ } dσ ( λ ) . Since ρ ( z ) ≤ ρ ( τ z ) for z ∈ D and 0 < τ < ρ ( τ z ) > τ /
2, as longas ρ ( z ) > τ , so the whole interval [0 , z ] belongs to the set { ρ > τ / } .The latter means that { ρ > τ } ⊂ Ω τ/ . Hence by using (22) we have Z D ( ρ ( λ )) q − α dσ ( λ ) ≤ q − α ( q − α ) Z τ q − α − dτ Z Ω τ/ (1 − | λ | ) dµ ( λ ) , and (21) leads to (8), as claimed.To prove (5), (7) in the case v (0) > −∞ , one can apply (6), (respec-tively, (8)) to the function v ( z ) = ( v ( z ) − v (0))(1+2 q | v (0) | ) − . Indeed, v ( z ) ≤ ρ − q ( z ) for all z ∈ D , and the Riesz measure of v coincides withthe Riesz measure of v up to a constant factor (which depends on v (0)).If v (0) = −∞ , take the harmonic function h in the disk {| z | < / } such that v = h for | z | = 1 /
2, and put v ( z ) = (cid:26) max( v ( z ) , h ( z )) for | z | < / ,v ( z ) for | z | ≥ / . Clearly, v ( z ) ≤ v ( z ) ≤ ρ − q ( z, E ). Moreover, v is subharmonic in D (see, e.g., [17, Theorem 2.4.5]), and the restriction of its Riesz measure µ on the set { z ∈ D : | z | > / } is the same as one for µ . Therefore,the integrals in (5), (7) taken for µ and µ differ by a bounded term.Since v (0) = ∞ , the result follows. The proof is complete. Proof of Theorem 2 . As it was mentioned above, the function v = ρ − q is subharmonic in D . We invoke again the harmonic function V (16). It is clear that V = v on E ct , and V ≤ t − q = v on Γ. Put u forthe least harmonic majorant of v on Ω = Ω t , which is defined above.We have V ( ζ ) ≤ v ( ζ ) ≤ u ( ζ ) , ζ ∈ ∂ Ω ⇒ V ( z ) ≤ u ( z ) , z ∈ Ω . Since h (0 , · ) ≥ v (0) = 1, then by (12) for z = 0(23) V (0) ≤ u (0) ≤ Z Ω log 1 | λ | dµ ( λ ) + 1 . For V (0) we see (cf. (17)–(18)) that(24) V (0) = 2 − q + q Z t ν E ( s ) s q +1 ds . Thereby, in the case q = α we have Z D log 1 | λ | dµ ( λ ) = + ∞ . BLASCHKE-TYPE CONDITION AND APPLICATION 9
Let now q > α . We apply (14) with dσ = log | λ | − dµ , keeping inmind (23)–(24) Z D (cid:0) ρ ( λ, E ) (cid:1) q − α dσ ( λ ) = ( q − α ) Z y q − α − σ ( { λ : ρ > y } ) dy ≥ ( q − α ) Z y q − α − dy Z Ω y log 1 | λ | dµ ( λ ) ≥ ( q − α ) Z y q − α − (cid:26) − q − q Z y ν E ( s ) s q +1 ds (cid:27) dy = q Z ν E ( s ) s α +1 ds + 2 − q − , and hence Z D log 1 | λ | (cid:0) ρ ( λ, E ) (cid:1) q − α dµ ( λ ) = + ∞ . To go over to (9) we note that in view of (12) and (13) the function ρ − q in the disk { z : | z | < / } differs from R | λ | < / log(1 / | λ − z | ) dµ ( λ )by a harmonic function. Hence, the latter integral is finite at z = 0,and (9) follows. The proof is complete.The following example shows that the result in Theorem 3 is optimalas well. Example . Let z n = 1 − / ( n + 1), n = 1 , , . . . . Consider an infinitecanonical product (see, e.g., [12], p. 132) f ( z ) := ∞ Y n =1 P (cid:18) − z n − z n z (cid:19) , P ( z ) = (1 − z ) e z . It follows from the Taylor expansion of P that | − P ( z ) | < | z | for | z | ≤
2. Since | − z n | < | − z n z | for z ∈ D , we get(25) log | f ( z ) | ≤ ∞ X n =1 (cid:12)(cid:12)(cid:12)(cid:12) − P (cid:18) − z n − z n z (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) < ∞ X n =1 (cid:12)(cid:12)(cid:12)(cid:12) − z n − z n z (cid:12)(cid:12)(cid:12)(cid:12) . Furthermore, for z ∈ D (cid:12)(cid:12)(cid:12)(cid:12) − z n − z n z (cid:12)(cid:12)(cid:12)(cid:12) ≤ | n (1 − z ) + 1 | ≤ √ n | − z | + 1Hence for n ( z ) = [ | − z | − ] + 1 ∈ N the right hand side of (25) doesnot exceed32 n ( z ) + 32 | − z | ∞ X n = n ( z )+1 n ≤ | − z | + 32 n ( z ) | − z | ≤ | − z | . So f satisfies (10) with E = { } , q = 1, and the choice α = 1 − ε isoptimal. The growth of certain canonical products in the unit disk was studiedin [8, 9].3.
Discrete spectrum of contraction operators
The relation between the spectral theory of non-selfadjoint operatorsand the theory of analytic functions via perturbation determinants iswell established due to the works of I. Gohberg, M. Krein, V. Matsaevand others. The variants of this approach were used recently in [2, 3].The results from Section 2, in particular Theorem 3, can be appliedto the operator theory in the same fashion as it is done in [1, 2, 3].Let U be a unitary operator in the Hilbert space H , σ ( U ) = E its spectrum. The resolvent R U ( z ) = ( U − z ) − is an analytic in D operator-function, and (see, e.g., [14, Section V.3.8]) k R U ( z ) k = ρ − ( z, σ ( U )) z ∈ D . A bounded, linear operator T in H is called a contraction if k T k ≤ k · k is the standard operator norm. Assume that T − U is a compactoperator. If E = T , then by a general result from the perturbationtheory σ ( T ) = E ∪ σ d ( T ), where the discrete component of the spectrum σ d ( T ) ⊂ D \ E is at most countable set with all possible accumulationpoints in E , and each z k ∈ σ d ( T ) is an eigenvalue of finite algebraicmultiplicity.Our goal here is to obtain some quantitative bound for the rate ofconvergence of z k to E in the case, when T − U ∈ S q , q >
0, theSchatten-von Neumann classes of compact operators, i.e., the sequenceof eigenvalues of the operator | T − U | belongs to ℓ q . The norm in S q isdenoted by k · k q , and the standard operator norm k · k = k · k ∞ . Thefamily {S q } is known to be nested: S q ⊂ S q for 0 < q < q ≤ ∞ , and k A k q ≤ k A k q . If A ∈ S q , q ≥
1, and S is a bounded linear operatorthen AS ∈ S q and k AS k q ≤ k A k q k S k . An extensive information onthe subject is Gohberg–Krein [10], Simon [19], and Dunford–Schwartz[5, Chapter XI.9].The key analytic tool is the so called renormalized determinantdet n ( I + A ), defined for A ∈ S n and n = 1 , , . . . . In particular, weneed the following continuity property (cf. [19, Theorem 9.2, (c)])(26) | det n ( I + A ) − det n ( I + B ) | ≤ k A − B k n exp ( C n (1 + k A k nn + k B k nn )) , a constant C n depends only on n . If T − U ∈ S n , then the perturbationdeterminant u n ( z ) := det n ( T − z )( U − z ) − = det n (cid:0) I + ( T − U )( U − z ) − (cid:1) is a well defined and analytic function in D , and its zero set in D is Z u n = σ d ( T ) ∩ D , with the order of each zero equal the algebraicmultiplicity of the eigenvalue. BLASCHKE-TYPE CONDITION AND APPLICATION 11
Theorem 4.
Let U be a unitary operator in H , σ ( U ) = E = T , and I ( α, E ) < ∞ in (4) for some α ∈ R . Let a contraction T be a S q -perturbation of U : T − U ∈ S q , q > . Then (27) X z ∈ σ d ( T ) (1 − | z | ) ρ ( z, E ) ( q − α ) + < ∞ . Moreover, there is a constant δ = δ q , < δ q < , which depends onlyon q , such that for k T − U k q < δ q the inequality holds (28) X z ∈ σ d ( T ) (1 − | z | ) ρ ( z, E ) ( q − α ) + ≤ C ( q, E ) max( k T − U k q , k T − U k qq ) . Remark . Obviously, the eigenvalues of T with | z | = 1 do not enterthe left hand side, so we claim nothing about this part of the discretespectrum. However, this part is missing as long as T is a completelynon-unitary contraction. Proof . Take a positive integer n from q ≤ n < q + 1, so T − U ∈ S n .By [5, Lemma XI.9.22,(d)] | u n ( z ) | ≤ exp (cid:0) c q k ( T − U )( U − z ) − k qq (cid:1) ≤ exp (cid:0) c q k T − U k qq k ( U − z ) − k q (cid:1) ≤ exp (cid:0) c q k T − U k qq ρ − q ( z, E ) (cid:1) . An application of the first statement of Theorem 3 proves (27).Throughout the rest of the proof c q stands for different positive con-stants which depend only on q . Since k T − U k <
1, then T is invertible,and u n (0) = det n (cid:0) I + ( T − U ) U − (cid:1) = 0 . Put f ( z ) = u n ( z ) /u n (0), | f (0) | = 1, so thatlog | f ( z ) | ≤ c q k T − U k qq ρ q ( z, E ) + log 1 | u n (0) | . Bound (26) with A = ( T − U ) U − , B = 0 gives in view of the mono-tonicity of norms | u n (0) − | ≤ k T − U k q exp (cid:0) C ′ q (1 + k T − U k nq ) (cid:1) < , as soon as δ q is small enough. Hencelog 1 | u n (0) | ≤ log 4 | u n (0) − | ≤ c q k T − U k q . An application of the second statement of Theorem 3 gives (28).
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