Rates of decay in the classical Katznelson-Tzafriri theorem
aa r X i v : . [ m a t h . F A ] O c t RATES OF DECAY IN THE CLASSICALKATZNELSON-TZAFRIRI THEOREM
DAVID SEIFERT
Abstract.
Given a power-bounded operator T , the theorem of Katznelsonand Tzafriri states that k T n ( I − T ) k → n → ∞ if and only if the spectrum σ ( T ) of T intersects the unit circle T in at most the point 1. This paperinvestigates the rate at which decay takes place when σ ( T ) ∩ T = { } . Theresults obtained lead in particular to both upper and lower bounds on this rateof decay in terms of the growth of the resolvent operator R (e i θ , T ) as θ →
0. Inthe special case of polynomial resolvent growth, these bounds are then shownto be optimal for general Banach spaces but not in the Hilbert space case. Introduction
The Katznelson-Tzafriri theorem (see [25, Theorem 1]) is one of the corner-stones of the asymptotic theory of operator semigroups; for surveys, see for in-stance [6] and [12]. In its original and simplest form, the result concerns theasymptotic behaviour of k T n ( I − T ) k as n → ∞ for suitable operators T and hasapplications both in the theory of iterative methods (see [35]) and to zero-twolaws for stochastic processes (see [25] and [38]). Writing T for the unit circle { λ ∈ C : | λ | = 1 } , it can be stated as follows. Theorem 1.1.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator. Then (1.1) lim n →∞ k T n ( I − T ) k = 0 if and only if σ ( T ) ∩ T ⊂ { } . Since its discovery in 1986, the Katznelson-Tzafriri theorem has attracted aconsiderable amount of interest, and this has lead to a number of extensions andimprovements of the original result; see [12, Section 4] for an overview, and also[29], [39] and [41]. One aspect which so far has been studied only in special cases,however, is the rate at which decay takes place in (1.1); see for instance [13], [15],[35, Chapter 4], [36] and [37]. Of course, for operators T satisfying σ ( T ) ∩ T = ∅ the question is of no real interest, since in this case r ( T ) < σ ( T ) ∩ T = { } , with the aim of relating the rate of decay in (1.1) to the growthof the norm k R (e i θ , T ) k of the resolvent operator as θ →
0. Once the behaviourof the resolvent near its singularity is adequately taken into account, it turns out
Date : 6 March 2014.2010
Mathematics Subject Classification.
Primary: 47A05, 47D06; secondary: 47A10, 47A35. to be possible not only to obtain explicit bounds on the quantity k T n ( I − T ) k forsufficiently large n ≥ C -semigroups, where they can be used to study energy decay for damped waveequations; see [5], [7], [8], [11], [31] and the references therein.The remainder of the paper divides into two parts. The first, Section 2, con-tains the main general results. Most importantly, these include both a lower(Corollary 2.6) and an upper bound (Theorem 2.11) in terms of the growth of theresolvent near 1 for the quantity k T n ( I − T ) k when T is a suitable power-boundedoperator and n ≥ X ,let B ( X ) stand for the algebra of bounded linear operators on X . An operator T ∈ B ( X ) is said to be power-bounded if sup {k T n k : n ≥ } < ∞ . Denote therange and kernel of an operator T ∈ B ( X ) by Ran( T ) and Ker( T ), respectively,and write Fix( T ) := Ker( I − T ) for the set of fixed points of T , σ ( T ) for itsspectrum and r ( T ) for its spectral radius. Furthermore, given an element λ ofthe resolvent set ρ ( T ) := C \ σ ( T ), let R ( λ, T ) := ( λ − T ) − denote the resolventoperator of T . All remaining pieces of notation will be introduced as the needarises. 2. General results
Let T ∈ B ( X ) be a power-bounded operator on a complex Banach space X , andsuppose that σ ( T ) ∩ T = { } . In order to address the question of rates of decay in(1.1), it will be convenient to have in place a few non-standard pieces of notation.Thus, given an operator T as above, a decreasing function m : (0 , π ] → (0 , ∞ )such that k R (e i θ , T ) k ≤ m ( | θ | ) for all θ with 0 < | θ | ≤ π will be said to bea dominating function (for the resolvent of T ) . Likewise a decreasing function ω : Z + → (0 , ∞ ) such that k T n ( I − T ) k ≤ ω ( n ) for all n ∈ Z + will be said to bea dominating function (for T ) . The minimal dominating functions are given, for θ ∈ (0 , π ] and n ≥
0, by(2.1) m ( θ ) = sup (cid:8) k R (e i ϑ , T ) k : θ ≤ | ϑ | ≤ π (cid:9) ,ω ( n ) = sup (cid:8) k T k ( I − T ) k : k ≥ n (cid:9) , respectively. Thus, for the minimal dominating function ω of T , ω ( n ) → n → ∞ precisely when (1.1) holds. Note also that the function m defined in(2.1) is continuous. In what follows, the same will be assumed to be true of anydominating function m for the resolvent of T . In particular, any such dominatingfunction m possesses a right-inverse m − defined on the range of m . On the other ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 3 hand, given a dominating function ω for T which satisfies ω ( n ) → n → ∞ ,define the function ω ∗ : (0 , ∞ ) → Z + by(2.2) ω ∗ ( s ) := min (cid:8) n ∈ Z + : ω ( n ) ≤ s (cid:9) . Then ω ( ω ∗ ( s )) ≤ s for all s >
0, with equality for all s in the range of ω .Recall the elementary estimate(2.3) k R ( λ, T ) k ≥ λ, σ ( T )) , which holds for all λ ∈ ρ ( T ). Since 1 ∈ σ ( T ), it follows that m ( θ ) ≥ θ − for all θ ∈ (0 , π ]. Thus there is a minimal rate at which the resolvent of any operator T as above must blow up near its singularity. This may suggest that there shouldexist a corresponding minimal rate, independent of the operator T , at whichdecay takes place in (1.1). As Corollary 2.6 below will show, however, this is farfrom being the case; see also [1, Theorem 4.2]. The next result, on the otherhand, shows that instances in which the decay is faster than that of n − are ofa very special nature. It is a direct analogue of [7, Theorem 6.7]; see also [36,Remarks 2.3 and 2.4]. Theorem 2.1.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator such that σ ( T ) ∩ T = { } . Then either (2.4) lim sup n →∞ n k T n ( I − T ) k > or there exist closed T -invariant subspaces X and X of X such that X ⊂ Fix( T ) , the restriction T of T to X satisfies r ( T ) < and X = X ⊕ X . Proof . Supposing first that 1 is a limit point of σ ( T ), let λ j ∈ σ ( T ) \{ } besuch that λ j → j → ∞ and set n j := ⌊| − λ j | − ⌋ . Since r ( T n ( I − T )) ≤k T n ( I − T ) k for all n ≥
1, it follows thatlim sup n →∞ n k T n ( I − T ) k ≥ lim j →∞ n j n j + 1 (cid:18) − n j (cid:19) n j = e − , and hence (2.4) holds.If 1 is an isolated point of σ ( T ), on the other hand, then a standard spectraldecomposition argument (see for instance [2, Proposition B.9]) shows that thereexist closed T -invariant subspaces X and X of X and a bounded projection P of X onto X along X which commutes with T . In particular, X = X ⊕ X .Moreover, the restrictions T and T of T to X and X satisfy σ ( T ) = { } and σ ( T ) = σ ( T ) \{ } , respectively. Now, if (2.4) fails, thenlim inf n →∞ n k T n ( I − T ) k = 0and it follows from [24, Theorem 2.2] that T x = x for all x ∈ X , as required. (cid:3) Remark 2.2.
It is easily seen that, if X splits, then in fact X = Fix( T ) and X = Ran( I − T ). In particular, Ran( I − T ) is closed; see also [35, Theorem 4.4.2]. DAVID SEIFERT
Thus k T n ( I − T ) k decays either at least exponentially as n → ∞ or at arate no faster than n − . The case of decay at this borderline rate turns out tobe connected with a special class of operators. Recall that an operator T on acomplex Banach space X is said to be a Ritt operator if σ ( T ) ∩ T = { } andthere exists a constant C > k R ( λ, T ) k ≤ C | − λ | for | λ | >
1; various interesting results on Ritt operators may be found for instancein [3], [9], [10], [14], [16], [17], [18], [27], [32], [33] and [40]. The reason why Rittoperators are important in the present context is that a power-bounded operator T satisfying σ ( T ) ∩ T = { } is a Ritt operator if and only if k R (e i θ , T ) k = O ( | θ | − )as | θ | →
0; see for instance the proof of Lemma 3.9 below. The following resultshows that these operators are precisely those for which the rate of decay in (1.1)is no slower than n − . This characterisation was obtained independently in [30]and [34]; see also [35, Theorem 4.5.4]. Theorem 2.3.
Let X be a complex Banach space. An operator T ∈ B ( X ) is aRitt operator if and only if it is power-bounded and k T n ( I − T ) k = O ( n − ) as n → ∞ . Thus decay in (1.1) at a rate no slower than that of n − already implies a strongcondition on the growth of the resolvent near its singularity at 1. The next resultestablishes a corresponding resolvent bound in a rather more general situation;see [7, Theorem 6.10] for an analogous result in the setting of C -semigroups. Theorem 2.4.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator. Suppose that ω is a dominating function for T such that ω ( n ) → as n → ∞ , and let ω ∗ be as defined in (2.2) . Then σ ( T ) ∩ T ⊂ { } and,for any c ∈ (0 , , (2.6) k R (e i θ , T ) k = O (cid:18) | θ | + ω ∗ ( c | θ | ) (cid:19) as | θ | → . Proof . Suppose that λ ∈ σ ( T ) ∩ T . By the spectral mapping theorem for poly-nomials, λ n (1 − λ ) ∈ σ ( T n ( I − T )) and hence | − λ | ≤ ω ( n ) for all n ≥
0. Letting n → ∞ , it follows that λ = 1, so σ ( T ) ∩ T ⊂ { } .Now let λ ∈ T \{ } . Then, for n ≥ λ n (1 − λ ) − T n ( I − T ) = (1 − λ ) λ n − n − X k =0 λ − k T k ( λ − T ) − T n ( λ − T )and hence, letting M := sup {k T n k : n ≥ } , | − λ |k R ( λ, T ) x k ≤ ω ( n ) k R ( λ, T ) x k + M (cid:0) n | − λ | (cid:1) k x k ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 5 for all x ∈ X . Fix b ∈ ( c,
1) and let n = ω ∗ ( b | − λ | ). Then k R ( λ, T ) k ≤ M − b (cid:18) | − λ | + ω ∗ ( b | − λ | ) (cid:19) and, since b | − λ | ≥ c | θ | whenever λ = e i θ for some sufficiently small θ ∈ ( − π, π ] \{ } , the result follows. (cid:3) Remark 2.5.
A similar argument shows that, given any constant
K > M , where M is as above, there exists c ∈ (0 ,
1) such that(2.7) k R (e i θ , T ) k ≤ K (cid:18) | θ | + ω ∗ ( c | θ | ) (cid:19) whenever | θ | is sufficiently small. Note also that, by (2.3), the | θ | − term in (2.6)and (2.7) cannot in general be omitted.In analogy with [7, Corollary 6.11], these observations can be used to obtain alower bound on the quantity k T n ( I − T ) k when n ≥ Corollary 2.6.
Let X be a complex Banach space, let T ∈ B ( X ) be a power-bounded operator such that σ ( T ) ∩ T = { } and let m be the minimal dominatingfunction for the resolvent of T defined in (2.1) . Suppose that (2.8) lim θ → max (cid:8) k θR (e i θ , T ) k , k θR (e − i θ , T ) k (cid:9) = ∞ . Then, given any right-inverse m − of m , there exist constants c, C > such that (2.9) k T n ( I − T ) k ≥ cm − ( Cn ) for all sufficiently large n ≥ . Proof . Let ω be as defined in (2.1). Since ω ( n ) → n → ∞ by Theo-rem 1.1, it follows from Theorem 2.4 that there exists B > m ( θ ) ≤ B (cid:0) θ − + ω ∗ ( θ/ (cid:1) for all sufficiently small θ ∈ (0 , π ], and hence(2.10) ω ∗ ( θ/ ≥ m ( θ ) (cid:18) B − θm ( θ ) (cid:19) for all such values of θ . Let C := 2 B and, for n ≥
0, let θ n := 2 ω ( n ). By (2.8), θ n m ( θ n ) > C for all sufficiently large n ≥
0, so (2.10) implies that ω ∗ ( θ n / >C − m ( θ n ) for each such n ≥
0. Since ω ∗ ( θ n / ≤ n and therefore m ( m − ( Cn )) = Cn ≥ Cω ∗ ( θ n / > m ( θ n ) , it follows that m − ( Cn ) < θ n for all sufficiently large n ≥
0. Moreover, θ n ≤ M k T n ( I − T ) k for all n ≥
0, where M := sup {k T n k : n ≥ } , which shows that(2.9) holds for c = (2 M ) − . (cid:3) Remark 2.7.
A similar argument using Remark 2.5 instead of Theorem 2.4shows that the conclusion (2.9) remains true if (2.8) is replaced by the weakercondition that
L > M , where M is as above and L := lim inf θ → max (cid:8) k θR (e i θ , T ) k , k θR (e − i θ , T ) k (cid:9) . DAVID SEIFERT
Taking T to be the identity operator shows that the conclusion can be false when L = M .Suppose that T is a power-bounded operator such that σ ( T ) ∩ T = { } and let m be the minimal dominating function for the resolvent of T defined in (2.1). If T is a Ritt operator, then it follows from Theorem 2.3 that, for any c ∈ (0 , k T n ( I − T ) k = O (cid:0) m − ( cn ) (cid:1) as n → ∞ and, in view of Corollary 2.6, this type of upper bound is in generalthe best one can hope for. The next result describes the class of functions m for which such an upper bound is satisfied in the case of a normal operator on aHilbert space; see also [7, Proposition 6.13]. Proposition 2.8.
Let X be a complex Hilbert space and let T ∈ B ( X ) be apower-bounded normal operator such that σ ( T ) ∩ T = { } . Furthermore, let m be the minimal dominating function for the resolvent of T defined in (2.1) , let m − be any right-inverse of m and let S ⊂ N .(1) Suppose there exist constants c, C > such that (2.11) k T n ( I − T ) k ≤ Cm − ( cn ) for all n ∈ S . Then, for any b ∈ (0 , c ) , there exists a constant B > suchthat (2.12) m ( θ ) m ( ϑ ) ≥ b log ϑθ − B, for all θ ∈ (0 , π ] of the form θ = m − ( cn ) with n ∈ S and for all sufficientlysmall ϑ ∈ (0 , π ] .(2) Conversely, if there exist constants b, B > such that (2.12) holds for all θ ∈ (0 , π ] of the form θ = m − ( bn ) with n ∈ S and all ϑ ∈ (0 , π ] , thenthere exists a constant C > such that (2.11) holds with c = b . Proof . Note first that, for θ ∈ (0 , π ], m ( θ ) − = min (cid:8) | λ − e i ϕ | : λ ∈ σ ( T ) , θ ≤ | ϕ | ≤ π (cid:9) , and that (2.11) is equivalent to having(2.13) n log 1 | λ | ≥ log | − λ | Cm − ( cn )for all λ ∈ σ ( T ) \{ } and all n ∈ S .Suppose this holds and let θ = m − ( cn ) for some n ∈ S . Then m ( θ ) ≥ c log | λ | log | − λ | Cθ for all λ ∈ σ ( T ) \{ } . Define the function g : (0 , → R by g ( s ) := s − s . Then g is a continuous increasing function satisfying g ( s ) → s →
1, and in fact g ( s ) = inf (cid:26) r − r : s < r < (cid:27) ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 7 for all s ∈ (0 , b ∈ (0 , c ), there exists s ∈ (0 ,
1) such that cg ( s ) > b . Now suppose that ϑ ∈ (0 , − s ), let λ ∈ σ ( T ) be such that m ( ϑ ) = | λ − e i ϕ | − for some ϕ ∈ (0 , π ] with | ϕ | ≥ ϑ , and let r := | λ | . Since m ( ϑ ) ≥ ϑ − ,it follows from the estimate 1 − r ≤ | λ − e i ϕ | that r > s . Thus, if | − λ | ≥ ϑ ,then m ( θ ) m ( ϑ ) ≥ c | λ − e i ϕ | log r log | − λ | Cθ ≥ b log ϑ Cθ , which gives (2.12) with B = b log 2 C. If | − λ | < ϑ , on the other hand, then | λ − e i ϕ | ≥ ϑ and hence m ( θ ) m ( ϑ ) ≥ ϑ θ ≥ b log (cid:18) θ bϑ (cid:19) , which gives (2.12) with B = b log 3 b . Thus, taking B = b max { log 2 C, log 3 b } , theproof the first statement is complete.Now suppose, conversely, that (2.12) holds for some constants b, B >
0, all θ ∈ (0 , π ] of the form θ = m − ( cn ) with n ∈ S and all ϑ ∈ (0 , π ]. Let λ = r e i φ ∈ σ ( T ) \{ } , and set ϑ := | φ | , so thatlog 1 r ≥ − r = | e i φ − λ | ≥ m ( ϑ ) . Hence, if ϑ ≥ | − λ | , then (2.12) gives n log 1 r ≥ b m ( m − ( bn )) m ( ϑ ) ≥ log (cid:18) ϑm − ( bn ) (cid:19) − Bb ≥ log (cid:18) | − λ | m − ( bn ) (cid:19) − Bb , thus establishing (2.13) with c = b and C = 2e B/b . On the other hand, if ϑ < | − λ | , then 1 − r ≥ | − λ | and consequently n log 1 r ≥ n (1 − r ) ≥ | − λ | b m ( m − ( bn )) ≥ | − λ | bm − ( bn ) ≥ log (cid:18) | − λ | bm − ( bn ) (cid:19) , which gives (2.13) with c = b and C = 2 b . Thus taking C = 2 max { e B/b , b } finishes the proof. (cid:3) Remark 2.9.
The result remains true, with the same proof, for any complexBanach space X and any power-bounded operator T ∈ B ( X ) satisfying k f ( T ) k = sup (cid:8) | f ( λ ) | : λ ∈ σ ( T ) (cid:9) for all functions f of the form f ( λ ) = λ n (1 − λ ) with n ≥ f ( λ ) = ( µ − λ ) − with µ ∈ ρ ( T ). This includes, in particular, the class of multiplication operatorson any of the classical function or sequence spaces. Note also that the secondof the two implications holds more generally when m is an arbitrary dominatingfunction for the resolvent of T .Thus (2.11) holds for a normal operator T if and only if the minimal dominatingfunction m ( θ ) for the resolvent of T grows in a fairly regular way as θ →
0. Thefollowing example exhibits a class of normal operators for which this is not thecase.
DAVID SEIFERT
Example 2.10.
Let X = ℓ . Given a strictly increasing sequence ( r k ) of positiveterms such that r k → k → ∞ , let λ k := r k e i /k and consider the operator T ∈ B ( X ) given by T x := ( λ k x k ). Then T is a normal contraction with σ ( T ) = { λ k : k ≥ } ∪ { } . Moreover, if r k > − k − for all k ≥
1, then 1 − r k < | e i /k − λ j | whenever j = k and hence m ( k − ) = k R (e i /k , T ) k = 11 − r k for all k ≥
1, where m is the minimal dominating function for the resolvent of T defined in (2.1). Suppose moreover that log r k !+1 ≥ r ( k +1)! for all k ≥ c >
0, let b ∈ (0 , c ) and n k := ⌈− ( b log r ( k +1)! ) − ⌉ . Then m (cid:0) (( k + 1)!) − (cid:1) = 11 − r ( k +1)! ∼ − r ( k +1)! ∼ bn k as k → ∞ , and hence m − ( cn k ) ≤ (( k + 1)!) − for all sufficiently large k ≥ | − λ k !+1 | ≥ (3 k !) − and | λ n k k !+1 | ≥ e − /b for all k ≥
1, it follows that k T n k ( I − T ) k ≥ | λ n k k !+1 (1 − λ k !+1 ) | ≥ /b k ! ≥ k e /b m − ( cn k )when k ≥ c > T such that σ ( T ) ∩ T = { } , a right-inverse m − of some dominating function m for the resolvent of T and a constant c ∈ (0 , k T n ( I − T ) k = O ( m − ( cn )) as n → ∞ .The next result shows that it is nevertheless possible to obtain an upper boundof this kind provided the function m is modified appropriately. Indeed, given anoperator T as above and a dominating function m for the resolvent of T , definethe function m log : (0 , π ] → (0 , ∞ ) by(2.14) m log ( θ ) := m ( θ ) log (cid:18) m ( θ ) θ (cid:19) , noting that this function is strictly decreasing and hence possesses a well-definedinverse m − defined on the range of m log . As Theorem 2.11 below shows, theabove upper bound on k T n ( I − T ) k for large values of n ≥ m − is replaced by m − . This raises the question by how much the asymptoticbehaviour of these two functions differs in particular instances. If m ( θ ) = C e α/θ ,for example, where C, α > m − ( s ) ∼ α log s as s → ∞ , so m − has the same asymptotic behaviour as m − in this case. On the other hand,if m ( θ ) = Cθ − α for some constants C > α ≥
1, then m − ( s ) ∼ ( log ss ) /α as s → ∞ , so m − differs from m − by a logarithmic factor. For similar examplesin the continuous-time setting, see [8, Example 1.4] and [31, Section 2].Throughout the proof of the next result, and also in various other places lateron, the letters c and C , if used without having been introduced explicitly, standfor positive constants, which will be thought of as being small and large, respec-tively, and which need not be the same at each occurrence. The result itself is a ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 9 discrete analogue of [31, Proposition 3.1], which in turn is a development of [8,Theorem 1.5]; see also [19, Chapter VI], where similar techniques are discussedin the context of combinatorial problems.
Theorem 2.11.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator such that σ ( T ) ∩ T = { } . Furthermore, let m be a dominatingfunction for the resolvent of T and let m log be as defined in (2.14) . Then, for any c ∈ (0 , , k T n ( I − T ) k = O (cid:0) m − ( cn ) (cid:1) as n → ∞ . Proof . Having fixed a dominating function m and a constant c ∈ (0 , (cid:26) r e i θ ∈ C : 0 ≤ r ≤ − cm ( | θ | ) , < | θ | ≤ π (cid:27) . Moreover, noting that σ ( T ) ⊂ Ω by a standard Neumann series argument, definethe function F n : C \ Ω → B ( X ) by F n ( λ ) := T n (2 − T ) (cid:0) I − ( λ − R ( λ, T ) (cid:1) . It then follows from the resolvent identity that(2.15) F n ( λ ) = T n (2 − T ) R ( λ, T ) (cid:0) I − R (2 , T ) (cid:1) , and hence F n (2) = T n ( I − T ). Thus, by Cauchy’s integral formula, T n ( I − T ) = 12 π i I Γ h ( λ ) λ − F n ( λ ) d λ, where Γ is any contour outside Ω around the point 2 and where h is any functionthat is holomorphic in the relevant region and satisfies h (2) = 1. In what follows,it will be convenient to take Γ = Γ in ∪ Γ out to consist of an outer contour Γ out ,which encloses both the point 2 and the set Ω, and an inner contour Γ in , whichlies in the interior of Γ out and incloses Ω but not the point λ = 2. Such a contourcan be thought of as being closed by inserting a cut from any point on Γ in to anypoint on Γ out , the contributions along which cancel out.Let ϕ be the Cayley transform defined by ϕ ( λ ) := − λ λ and, for r ∈ (0 ,
1) and
R > , let(2.16) γ r := (cid:26) λ ∈ C : (cid:12)(cid:12)(cid:12)(cid:12) λ − r − r (cid:12)(cid:12)(cid:12)(cid:12) = 2 r − r (cid:27) and Γ R := { λ ∈ C : | λ − | = R } , noting that ϕ maps γ r onto r T , the realline onto itself and the unit circle T onto the imaginary axis. Now suppose that r ∈ (0 , ) and R >
2, and let Γ out = Γ R and Γ in = C r ∪ γ + r , where γ + r denotes thepart of γ r that lies outside the unit disc D := { λ ∈ C : | λ | < } and where C r isany suitable path in D \ Ω connecting the endpoints of γ + r . Furthermore, choose for h the map h r given by h r ( λ ) := 11 + 9 r (cid:18) r ϕ ( λ ) (cid:19) , so that h r is holomorphic away from 1. Now, letting M := sup {k T n k : n ≥ } , itfollows from the series expansion of the resolvent that(2.17) k R ( λ, T ) k ≤ M | λ | − | λ | > T is power-bounded, k F n k ≤ C ( | λ | − − for all such λ , where C is independent of n ≥
0. Since h r isbounded above in modulus independently of r along Γ R , it follows that (cid:13)(cid:13)(cid:13)(cid:13)I Γ R h r ( λ ) λ − F n ( λ ) d λ (cid:13)(cid:13)(cid:13)(cid:13) ≤ CR , where C is independent of n ≥
0, and hence, by appealing to Cauchy’s theoremand allowing R → ∞ , this contribution can be neglected.Next note that, for λ ∈ r T , the function g r defined by g r ( λ ) := 1 + r λ − satisfies | g r ( λ ) | = 2 r − | Re λ | . Moreover, an elementary calculation shows that,for λ ∈ γ r ,(2.18) 1 − | λ | = 4 Re ϕ ( λ )1 + 2 Re ϕ ( λ ) + r . Since h r = g r ◦ ϕ r and ϕ ( γ r ) = r T , it follows that(2.19) | h r ( λ ) | ≤ C | Re ϕ ( λ ) | r ≤ C || λ | − | r for all λ ∈ γ r . But for each λ ∈ γ r , | λ | − ≤ Cr and | − λ | ≤ Cr so, by (2.17)and the definition of F n , k F n ( λ ) k ≤ C k I − ( λ − R ( λ, T ) k ≤ Cr | λ | − λ ∈ γ + r . Hence (cid:13)(cid:13)(cid:13)(cid:13)Z γ + r h r ( λ ) λ − F n ( λ ) d λ (cid:13)(cid:13)(cid:13)(cid:13) ≤ Cr, where C is independent of n ≥
0, and it remains to control only the contributionalong C r .Let θ r ∈ (0 , π ) denote the argument of the point at which γ r meets T in theupper half-plane and define the curve C ◦ r , for θ r ≤ | θ | ≤ π , by C ◦ r ( θ ) := (cid:18) − cm ( | θ | ) (cid:19) e i θ . Furthermore, let C ± r denote the rays given, for 1 − cm ( θ r ) − ≤ s ≤ , by C ± r ( s ) := s e ± i θ r and set C r = C ◦ r ∪ C + r ∪ C − r . Defining p n ( λ ) := n − X k =0 λ n − k − T k ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 11 for λ ∈ C \ Ω, it follows from the resolvent identity, the relation p n ( λ ) = ( λ n − T n ) R ( λ, T ) and some elementary manipulations that F n ( λ ) = 1 λ − − T ) (cid:16) ( λ − (cid:0) λ n R ( λ, T ) − p n ( λ ) (cid:1) − T n R (2 , T ) (cid:17) for all λ ∈ C \ Ω with λ = 2; see also [31, Lemma 2.2]. Hence Cauchy’s theoremgives R (2 , T ) Z C r h r ( λ ) λ − F n ( λ ) d λ = Z C r h r ( λ )( λ − λ n ( λ − R ( λ, T ) d λ − Z γ − r h r ( λ )( λ − (cid:0) ( λ − p n ( λ ) + T n R (2 , T ) (cid:1) d λ, where γ − r := γ r ∩ D . To estimate the first integral on the right-hand side, notefirst that, by a standard Neumann series argument, k R ( λ, T ) k ≤ (1 − c ) − m ( θ r )for all λ ∈ C ◦ r . Since h r is uniformly bounded independently of r along C ◦ r , itfollows that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)Z C ◦ r h r ( λ )( λ − λ n ( λ − R ( λ, T ) d λ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ Cm ( θ r ) (cid:18) − cm ( θ r ) (cid:19) n . Similarly, (cid:13)(cid:13)(cid:13)(cid:13)Z C ± r h r ( λ )( λ − λ n ( λ − R ( λ, T ) d λ (cid:13)(cid:13)(cid:13)(cid:13) ≤ C Z − cm ( θ r ) − s n d s ≤ Cn + 1 . To bound the integral along γ − r , note that k ( λ − p n ( λ ) k ≤ Cr − | λ | for all λ ∈ γ − r . Thus by (2.19) both h r ( λ ) and h r ( λ )( λ − p n ( λ ) are uniformlybounded, independently of r and n , as λ ranges over γ − r , and it follows that (cid:13)(cid:13)(cid:13)(cid:13)Z γ − r h r ( λ )( λ − (cid:0) ( λ − p n ( λ ) + T n R (2 , T ) (cid:1) d λ (cid:13)(cid:13)(cid:13)(cid:13) ≤ Cr, where C is independent of n ≥ C − r ≤ θ r ≤ Cr , combining these bounds gives k T n ( I − T ) k ≤ C (cid:18) θ r + 1 n + 1 + m ( θ r ) (cid:18) − cm ( θ r ) (cid:19) n (cid:19) , where C is independent of n ≥ r ∈ (0 , ). Now, if n ≥ r ∈ (0 , ) so as to satisfy θ r = m − ( cn ) gives exp( m ( θ r ) − cn ) =1 + θ − r m ( θ r ) and hence m ( θ r ) (cid:18) − cm ( θ r ) (cid:19) n ≤ m ( θ r ) exp (cid:18) − cnm ( θ r ) (cid:19) ≤ θ r . Since moreover ( n + 1) − ≤ Cm − ( cn ) for all n ≥
0, this completes the proof. (cid:3)
Remark 2.12.
As in [31], it is possible to obtain an analogous result when σ ( T ) ∩ T is finite by replacing I − T with a finite product of linear terms of theform e iθ − T with θ ∈ ( − π, π ]. Optimality in the case of polynomial resolvent growth
Suppose that X is a complex Banach space and that T ∈ B ( X ) is a power-bounded operator such that σ ( T ) ∩ T = { } . The purpose of this section is toinvestigate the optimality of Theorem 2.4 in the special case where the resolvent of T grows at most polynomially, which is to say it admits a dominating function ofthe form m ( θ ) = Cθ − α for some constants C > α ≥
1, where the restrictionon the parameter α is a consequence of (2.3). Corollary 2.6 and Theorem 2.11combine to give the following result, which describes the range of decay rates thatare possible in this situation. Here, given Ω ⊂ (0 , ∞ ) and functions f, g : Ω → (0 , ∞ ), the notation f ( s ) = Θ( g ( s )) as s → s → ∞ ) means that there existconstants c, C > cg ( s ) ≤ f ( s ) ≤ Cg ( s ) for all sufficiently small (orlarge) values of s ∈ Ω. Corollary 3.1.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator such that σ ( T ) ∩ T = { } . Suppose that, for some α ≥ , k R (e i θ , T ) k = Θ( | θ | − α ) as θ → . Then there exist constants c, C > such that (3.1) cn /α ≤ k T n ( I − T ) k ≤ C (cid:18) log nn (cid:19) /α for all sufficiently large n ≥ . The remainder of this section is concerned with the question whether the log-arithmic factor on the right-hand side of (3.1) is really needed. It follows fromProposition 2.8 and Remark 2.9 that it can be dropped whenever T is a suit-able multiplication operator on some function or sequence space. The followingexample exhibits a less trivial case in which the same is true. Example 3.2.
Let X = ℓ p with 1 ≤ p ≤ ∞ and, writing S for the left-shiftoperator on X given by Sx := ( x k +1 ), define the operator T ∈ B ( X ) as T := ( I + S ) . Then T is a (non-normal) Toeplitz operator of unit norm, with σ ( T ) = (cid:26) r e i θ ∈ C : − π < θ ≤ π and 0 ≤ r ≤ θ (cid:27) . In particular, σ ( T ) ∩ T = { } . A calculation shows that, for λ ∈ ρ ( T ) and x ∈ X ,the resolvent satisfies R ( λ, T ) x = y , where, for each k ≥ ,y k = 1 λ / ∞ X n =0 ( − n +1 (cid:18) − λ / ) n +1 − λ / ) n +1 (cid:19) x k + n , the complex plane being cut along the negative real axis. Thus, for p ∈ { , ∞} , k R ( λ, T ) k = O (( | − λ / | − − ) as λ → ρ ( T ) and, by the Riesz-Thorin theorem, the same statement holds for p ∈ (1 , ∞ ). It follows, in particular,that k R (e i θ , T ) k = O ( | θ | − ) as θ →
0. Since k R (e i θ , T ) k ≥ c | θ | − for all θ ∈ ( − π, π ] by (2.3) and the geometry of σ ( T ), it follows from Corollary 3.1 that(3.1) holds with α = 2 for some constants c, C > n ≥
0. However, an explicit calculation involving Stirling’s formula shows that, for p ∈ { , ∞} , the actual rate of decay satisfies k T n ( I − T ) k ∼ πn ) − / as n → ∞ ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 13 and hence, by another application of the Riesz-Thorin theorem, k T n ( I − T ) k =Θ( n − / ) as n → ∞ also for p ∈ (1 , ∞ ). Thus the logarithmic factor in (3.1) isredundant in this case.Theorem 3.10 will show that, if the underlying space is a Hilbert space, thenthe logarithmic factor in (3.1) can in fact be dropped for any operator whoseresolvent grows at most polynomially. For general Banach spaces, however, thisis not the case, as Theorem 3.6 below establishes. The proof of this result requirestwo lemmas. The first is a variant of [2, Lemma 4.6.6], which itself is a specialform of Levinson’s log-log theorem; see for instance [26, VII D7]. Here, given aset Ω ⊂ C , ∂ Ω denotes the boundary of Ω.
Lemma 3.3.
Let X be a complex Banach space, let θ ∈ ( − π, π ] and let Ω be aneighbourhood of the point e i θ ∈ T . Furthermore, given r ∈ (0 , , let (3.2) Ω r,θ := (cid:26) λ ∈ C : (cid:12)(cid:12)(cid:12)(cid:12) λ − e i θ r − r (cid:12)(cid:12)(cid:12)(cid:12) ≤ r − r (cid:27) . Then there exists a constant
C > with the following property: If r ∈ (0 , ) is such that Ω r,θ ⊂ Ω and if F : Ω → X is a holomorphic function such that,for some constant B > , k F ( λ ) k ≤ B | − | λ || − for all λ ∈ ∂ Ω r,θ \ T , then k F ( λ ) k ≤ BCr − for all λ ∈ Ω r,θ . Proof . Assume, without loss of generality, that θ = 0 and, as in the proof ofTheorem 2.11, let ϕ denote the M¨obius transformation defined by ϕ ( λ ) := − λ λ ,so that ϕ maps the circle γ r := ∂ Ω r, onto r T for each r ∈ (0 , r replaced by 2 r , there exists a constant C ′ > r ∈ (0 , ) and such that | Re ϕ ( λ ) | ≤ C ′ || λ | − | for all λ ∈ γ r . Consider thefunction G : Ω → X defined by G ( λ ) := (cid:18) ϕ ( λ ) r (cid:19) F ( λ ) . For λ ∈ γ r , the term in brackets has modulus r − | Re ϕ ( λ ) | and hence, by theassumption on F , k G ( λ ) k ≤ BC ′ r − for all such λ . Since Ω r, ⊂ Ω r, , it followsfrom the maximum principle that k G ( λ ) k ≤ BC ′ r − for all λ ∈ Ω r, . But if λ ∈ Ω r, , then | ϕ ( λ ) | ≤ r and hence k G ( λ ) k ≥ k F ( λ ) k , which gives the resultwith C = C ′ . (cid:3) The second auxiliary result is a technical one and analogous to [11, Lemma 3.9].Given α ≥ λ ∈ C \{ } , let(3.3) K α ( λ ) := | arg λ | α π α , where the argument of a complex number is taken to lie in ( − π, π ], and definethe regions Ω α , Θ α ⊂ C by(3.4) Ω α := (cid:8) λ ∈ C \{ } : | λ | ≤ − K α ( λ ) (cid:9) ∪ { } , Θ α := (cid:8) λ ∈ C \{ } : 1 − K α ( λ ) < | λ | < (cid:9) , respectively, so that Θ α = 2 D \ Ω α . Furthermore, given a complex measure µ whose support is contained in Ω α , define the transforms C α µ , L α µ and D α µ , for λ ∈ Θ α , k ≥ n ≥
0, respectively, by(3.5) ( C α µ )( λ ) := Z Ω α d µ ( z ) λ − z , ( L α µ )( k ) := Z Ω α z k − d µ ( z ) , ( D α µ )( n ) := Z Ω α z n (1 − z ) d µ ( z ) . Lemma 3.4.
Suppose that α > and let the function K α , the regions Ω α and Θ α , and the transforms C α , L α and D α be defined as in (3.3) , (3.4) and (3.5) ,respectively. Then there exists a constant C > with the following property:Given any n ∈ N , there exists a complex measure µ whose support is containedin Ω α and which is such that(i) K α ( λ ) | ( C α µ )( λ ) | ≤ C for all λ ∈ Θ α ;(ii) | ( L α µ )( k ) | ≤ C for all k ≥ ;(iii) | ( D α µ )( n ) | α ≥ ( Cn ) − log n for some n > n . Proof . Choose θ ∈ (0 , ) and β ∈ ( α , α ) in such a way that ℓ := − βθ − α log θ is an integer satisfying ℓ > n + 2 and that θ − ( α − > αβ − + 1. Now, with B ℓ := 2 ℓ log ℓ , ζ ℓ := e π i /ℓ and λ := e i θ , define the measure µ as µ := B ℓ − ℓ ℓ / ℓ − X r =0 ζ rℓ (cid:18) ζ rℓ B ℓ λ (cid:19) δ λ + ζrℓ Bℓ , where δ λ denotes the Dirac measure concentrated at λ .Then, for any λ ∈ Θ α ,( C α µ )( λ ) = B ℓ − ℓ ℓ / ℓ − X r =0 (cid:18) B ℓ ζ rℓ B ℓ ( λ − λ ) − ζ rℓ + 1 λ ζ rℓ B ℓ ( λ − λ ) − ζ rℓ (cid:19) . However, for 1 ≤ j ≤ ℓ and λ ∈ C such that λ ℓ = 1, ℓ − X r =0 ζ jrℓ λ − ζ rℓ = ℓλ j − λ ℓ − j = 1 , C α µ )( λ ) = λλ ℓ / ℓ ( λ − λ ) ℓ − B − ℓℓ for all λ ∈ Θ α . Since B − ℓ ≤ | λ − λ | and | λ | ≤ λ ∈ Θ α , this in turnbecomes(3.7) | ( C α µ )( λ ) | ≤ Cℓ / ℓ | λ − λ | ℓ . ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 15
Let λ ∈ Θ α be given. If K α ( λ ) > θ α , then | arg λ | > πθ and an elementarygeometric argument shows that | λ − λ | ≥
12 + 12 (cid:0) − cos(( π − θ ) − K α (e i πθ ) (cid:1) ≥
12 (1 + θ − θ α ) . Using the fact that 1 + θ − θ α ≥ e ( θ − θ α ) for all θ ∈ (0 , ), it follows from (3.7)that | ( C α µ )( λ ) | ≤ Cℓ / e − ℓ ( θ − θ α ) = C ( − β log θ ) / θ − α + β ( θ − ( α − − . Now the choices of θ and β ensure that the exponent of θ on the right-hand side ofthis expression is strictly greater than α , and hence | ( C α µ )( λ ) | and consequently K α ( λ ) | ( C α µ )( λ ) | are uniformly bounded, independently of θ and β , for all λ ∈ Θ α satisfying K α ( λ ) > θ α . If K α ( λ ) ≤ θ α , on the other hand, then | λ − λ | ≥ (1 − θ α ) and, using the fact that 1 − θ α ≥ e − θ α for all θ ∈ (0 , ), (3.7) gives | ( C α µ )( λ ) | ≤ Cℓ / e ℓθ α = C ( − β log θ ) / θ − ( α +4 β ) . Since the choice of β ensures that α + 4 β < α , K α ( λ ) | ( C α µ )( λ ) | is uniformlybounded, again independently of θ and β , also for all λ ∈ Θ α with K α ( λ ) ≤ θ α .This establishes (i) for C = C , where C > k ≥ L α µ )( k ) = B ℓ − ℓ λ k − ℓ / ℓ − X r =0 ζ rℓ (cid:18) ζ rℓ B ℓ λ (cid:19) k . Expanding and using the fact that, for any integer s ≥ ℓ − X r =0 ζ r ( s +1) ℓ = ( ℓ if s + 1 = 0 (mod ℓ ) , L α µ )( k ) = B ℓ − ℓ λ k − ℓ / k X s =0 ℓ − X r =0 (cid:18) ks (cid:19) ω r ( s +1) ℓ (2 B ℓ λ ) s = ℓ / B ℓ − ℓ λ k − ⌊ k +1 ℓ ⌋ X r =1 (cid:0) krℓ − (cid:1) (2 B ℓ λ ) rℓ − . Next note that, for 1 ≤ r ≤ ⌊ k +1 ℓ ⌋ , (cid:0) krℓ − (cid:1) ≤ (cid:0) kℓ − (cid:1) ( ℓ − rℓ − k ( r − ℓ . Thus, for1 ≤ k < B ℓ , | ( L α µ )( k ) | ≤ ℓ / k − (cid:18) kℓ − (cid:19) ⌊ k +1 ℓ ⌋ X r =1 ( ℓ − rℓ − ≤ C ℓ / k (cid:18) kℓ − (cid:19) , where C is independent of ℓ . Now, if k ≤ ℓ −
3, then (cid:0) kℓ − (cid:1) ≤ (cid:0) k +1 ℓ − (cid:1) and, if k ≥ ℓ −
2, then (cid:0) kℓ − (cid:1) ≤ (cid:0) k − ℓ − (cid:1) , so in either case k (cid:0) kℓ − (cid:1) ≤ ℓ − (cid:0) ℓ − ℓ − (cid:1) . Hence(3.10) | ( L α µ )( k ) | ≤ C ℓ / ℓ − (cid:18) ℓ − ℓ − (cid:19) , which by Stirling’s formula is bounded above independently of ℓ . If k ≥ B ℓ , onthe other hand, then (3.8) and the fact that 1 + B − ℓ ≤ e /B ℓ give | ( L α µ )( k ) | ≤ ℓ / B ℓ − ℓ k − (cid:18) B ℓ (cid:19) k ≤ ℓ / B ℓ − ℓ e − k (log 2 − B − ℓ ) ≤ C ℓ / B ℓ − ℓ B ℓ and, by the definition of B ℓ , the right-hand side is again bounded above indepen-dently of ℓ . Thus (ii) holds with C = C for some sufficiently large C > n ≥ D α µ )( n ) = B ℓ − ℓ λ n ℓ / ℓ − X r =0 ζ rℓ (cid:18) ζ rℓ B ℓ λ (cid:19) n +1 (cid:18) − λ (cid:18) ζ rℓ B ℓ λ (cid:19)(cid:19) . Thus, if n ≥ ⌊ n +2 ℓ ⌋ = ⌊ n +3 ℓ ⌋ , proceeding as in (3.9) gives( D α µ )( n ) = ℓ / B ℓ − ℓ λ n ⌊ n +2 ℓ ⌋ X r =1 (cid:0) n +1 rℓ − (cid:1) − λ (cid:0) n +2 rℓ − (cid:1) (2 B ℓ λ ) rℓ − . Now let n := 2 ℓ −
4, so that n > n and ⌊ n +2 ℓ ⌋ = ⌊ n +3 ℓ ⌋ = 1. Then | ( D α µ )( n ) | = ℓ / ℓ − (cid:18) ℓ − ℓ − (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) − λ (cid:12)(cid:12)(cid:12)(cid:12) , and hence, by another application of Stirling’s formula, | ( D α µ )( n ) | ≥ cθ , where c is independent θ . Since the definition of ℓ implies that θ α ≥ cn − log n , itfollows that (iii) holds for C = C , where C > C = max { C , C , C } now completes the proof. (cid:3) Remark 3.5.
The estimates leading to (3.10) may also be viewed in anotherway. Indeed, given ℓ ≥
1, let Y ℓ be the random variable counting the numberof tosses of a fair coin required in order to obtain a total of exactly ℓ heads, sothat Y ℓ has the negative binomial distribution with P ( Y ℓ = k ) = k (cid:0) k − ℓ − (cid:1) for each k ≥
1. Then the aforementioned estimates amount to the observation that Y ℓ hasmode 2 ℓ −
1. This probabilistic interpretation will reappear in Remark 3.7 below.The following result, which is an analogue of [11, Theorem 4.1], shows that thelogarithmic factor in Corollary 3.1 cannot in general be omitted.
Theorem 3.6.
Given any α > , there exists a non-trivial complex Banachspace X α and a power-bounded operator T ∈ B ( X α ) such that σ ( T ) ∩ T = { } and k R (e i θ , T ) k = O ( | θ | − α ) as θ → , and for which (3.11) lim sup n →∞ k T n ( I − T ) k (cid:18) n log n (cid:19) /α > . Proof . Given any sequence x ∈ ℓ ∞ , define the function F x , for | λ | >
1, by F x ( λ ) := ∞ X k =1 x k λ k . Now, with K α , Ω α and Θ α as defined in (3.3) and (3.4), let X α denote the subspaceof ℓ ∞ consisting of sequences x for which F x extends analytically to Θ α and ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 17 satisfies sup { K α ( λ ) | F x ( λ ) | : λ ∈ Θ α } < ∞ . This space is non-trivial, containingfor instance any finitely supported sequence as well as the constant sequence(1 , , , . . . ), and, by an application of Vitali’s theorem, it is complete under thenorm k · k X α given by k x k X α := k x k ∞ + k x k α , where k x k α := sup { K α ( λ ) | F x ( λ ) | : λ ∈ Θ α } . Consider the restriction T := S | X α to X α of the left-shift operator S ∈ B ( ℓ ∞ ).Given x ∈ X α and | λ | > F T x ( λ ) = λF x ( λ ) − x , so F T x extends analyticallyto Θ α and k T x k α ≤ k x k ∞ + 2 k x k α . Thus T maps X α into itself and defines anelement of B ( X α ) with norm k T k ≤
2. More generally, having fixed some x ∈ X α and given n ≥
0, let F n := F T n x . Then F n extends analytically to Θ α and isgiven, for | λ | >
1, by(3.12) F n ( λ ) = ∞ X k =1 x n + k λ k = λ n F x ( λ ) − n X k =1 λ n − k x k . Writing A for the annulus { λ ∈ C : 1 < | λ | < } , it follows that | F n ( λ ) | ≤ k x k ∞ | −| λ || if λ ∈ A , k x k ∞ | −| λ || + | F x ( λ ) | if λ ∈ Θ α ∩ D ,and, in particular, | F n ( λ ) | ≤ | − | λ || − k x k X α for all λ ∈ Θ α \ T . Let λ ∈ Θ α begiven. If λ ∈ (1 , K α ( λ ) | F n ( λ ) | = 0. Suppose therefore that θ := arg λ satisfies 0 < | θ | ≤ π , and note that Ω r λ ,θ ⊂ Θ α , where r λ := K α ( λ ) and Ω r,θ is defined, for r ∈ (0 , | − | λ || > r λ , in which case K α ( λ ) | F n ( λ ) | ≤ C k x k α for some constant C which is independent of x ∈ X α , n ≥ λ , or | − | λ || ≤ r λ . In the latter case λ ∈ Ω r λ ,θ , so the same estimatefollows from Lemma 3.3 applied to the function F n on the disc Ω r λ ,θ . Thus k T n x k α ≤ C k x k X α for some constant C which is independent of x ∈ X α and n ≥
0. Since moreover k T n x k ∞ ≤ k x k ∞ for all n ≥
0, it follows that T ispower-bounded.Now fix x ∈ X α and let Ω := { λ ∈ C : 1 < | λ | < } . Then Ω ⊂ ρ ( T ) and( R ( λ, T ) x ) n +1 = F n ( λ ) for all λ ∈ Ω and all n ≥
0, where F n is as above. Inparticular, (3.12) remains true for each n ≥ R ( λ, T ) x ) n +1 , so the argument in the previous paragraph shows that(3.13) K α ( λ ) k R ( λ, T ) x k ∞ ≤ C k x k X α , where C is independent of both x ∈ X α and λ ∈ Ω. The aim now is to showthat K α ( λ ) k R ( λ, T ) x k α ≤ C k x k α for all λ ∈ Ω, from which it will follow thatthe norm of the resolvent of T grows at most polynomially. Since the estimateholds trivially when λ is real, assume that λ ∈ Ω satisfies 0 < | arg λ | ≤ π and let F λ := F R ( λ,T ) x . Then, for | µ | ≥ , F λ ( µ ) = ∞ X n =1 ∞ X k =1 x k + n − λ k µ n = ∞ X n =1 λ n − µ n F x ( λ ) − n − X k =1 x k λ k ! = − F x ( λ ) − F x ( µ ) λ − µ , so F λ extends analytically to Θ α , taking the values(3.14) F λ ( µ ) = ( − F x ( λ ) − F x ( µ ) λ − µ if µ ∈ Θ α \{ λ } , − F ′ x ( λ ) if µ = λ .Now let µ ∈ Θ α and set M α ( λ, µ ) := max { K α ( λ ) , K α ( µ ) } , which is positive bythe assumption on arg λ . If | λ − µ | > M α ( λ, µ ), then by (3.14)(3.15) K α ( λ ) K α ( µ ) | F λ ( µ ) | ≤ K α ( λ ) + K α ( µ ) M α ( λ, µ ) k x k α ≤ C k x k α , where C is independent of λ and µ . Now suppose that | λ − µ | ≤ M α ( λ, µ ). ByCauchy’s formula,(3.16) F λ ( µ ) = 12 π i I Γ F x ( λ ) − F x ( z )( λ − z )( z − µ ) d z, where Γ is any contour in Θ α whose interior contains the point µ and is itselfcontained in Θ α . If | λ − µ | ≤ K α ( λ ) , choose Γ to be the circle with centre λ and radius K α ( λ ), so that Γ ⊂ Θ α by the definitions of Θ α and Ω. Elementaryestimates show that, for any z ∈ Θ α satisfying | λ − z | ≤ K α ( λ ), c | arg λ | ≤| arg z | ≤ C | arg λ | and hence cK α ( λ ) ≤ K α ( z ) ≤ CK α ( λ ), where c and C areindependent of λ and z . This applies in particular to all z ∈ Γ and also to z = µ .Since moreover | µ − z | ≥ K α ( λ ) for all z ∈ Γ, it follows from (3.16) that(3.17) K α ( λ ) K α ( µ ) | F λ ( µ ) | ≤ C k x k α K α ( λ ) I Γ (cid:18) K α ( µ ) K α ( λ ) + K α ( µ ) K α ( z ) (cid:19) | d z | ≤ C k x k α , where C depends neither on λ nor on µ . A similar argument applies when K α ( λ ) < | λ − µ | ≤ K α ( µ ), this time taking Γ to be the circle with centre λ and radius K α ( µ ). Then | µ − z | ≤ K α ( µ ) for all z ∈ Γ, so Γ ⊂ Θ α as before.Moreover, K α ( µ ) ≤ CK α ( z ) for all z ∈ Γ, where C is independent of λ and µ ,and K α ( λ ) < K α ( µ ), giving(3.18) K α ( λ ) K α ( µ ) | F λ ( µ ) | ≤ C k x k α K α ( µ ) I Γ (cid:18) K α ( λ ) K α ( z ) (cid:19) | d z | ≤ C k x k α . Combining (3.15), (3.17) and (3.18) shows that K α ( λ ) k R ( λ, T ) x k α ≤ C k x k α forall λ ∈ Ω. Together with (3.13), this gives K α ( λ ) k R ( λ, T ) x k X α ≤ C k x k X α , where C is independent of x ∈ X α and λ ∈ Ω, and hence sup { K α ( λ ) k R ( λ, T ) k : λ ∈ Ω } < ∞ . In particular, it follows from by (2.3) that σ ( T ) ∩ T ⊂ { } , and asimple approximation argument shows that k R (e i θ , T ) k = O ( | θ | − α ) as θ → , , , . . . ) is a fixed point of T , 1 ∈ σ ( T ).Finally, let the transforms C α , L α and D α be as defined in (3.5) and note that,given any complex measure µ whose support is contained in Ω α and for whichsup {| ( L α µ )( k ) | : k ≥ } < ∞ , there exists an associated sequence x µ ∈ ℓ ∞ whoseentries are given, for each k ≥
1, by x µk := ( L α µ )( k ). By Fubini’s theorem, F x µ ( λ ) = ∞ X k =1 Z Ω α z k − λ k d µ ( z ) = Z Ω α d µ ( z ) λ − z = ( C α µ )( λ ) ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 19 whenever | λ | >
1, so x µ ∈ X α provided sup { K α ( λ ) | ( C α µ )( λ ) | : λ ∈ Θ α } < ∞ .Note also that, for each n ≥
0, ( D α µ )( n ) = x µn +1 − x µn +2 , which coincides with thefirst entry of T n ( I − T ) x µ . Now, by Lemma 3.4, it is possible to find a sequence( n j ) of integers, with n j → ∞ as j → ∞ , and associated measures µ j such that { x µ j : j ≥ } is a bounded subset of X α and moreover | ( D α µ j )( n j ) | α ≥ cn − j log n j for each j ≥
1. By rescaling if necessary, there is no loss of generality in assumingthat k x µ j k X α ≤ j ≥
1, so that k T n j ( I − T ) k ≥ k T n j ( I − T ) x µ j k ∞ ≥ | ( D α µ j )( n j ) | . Hence (3.11) holds and the proof is complete. (cid:3)
Remark 3.7.
It is possible to replace Lemma 3.4, which here gives rise to thesequences x µ ∈ X α used to establish to (3.11), by a simpler, more ad-hoc con-struction. Indeed, using the notation introduced in the proof of that result, let x k := ℓ / λ k − ℓ ℓ − (cid:18) kℓ − (cid:19) for each k ≥
1, so that x k equals the first term of the final sum in (3.9) whichdefines x µk in the above proof. In the notation of Remark 3.5, this becomes x k = 4 ℓ / (2 λ ) k − ℓ P ( Y ℓ = k + 1) , so the formula for the probability generating function of Y ℓ (see for instance [21,Section 4.2]) gives F x ( λ ) = λλ ℓ / ℓ ( λ − λ ) ℓ whenever | λ | >
1, which should be compared with the right-hand sides of (3.6)and (3.7). Since the estimates for x µ established in Lemma 3.4 apply equally to x , it follows that (3.11) may also be obtained using sequences of this simpler formin the final paragraph of the above proof. Remark 3.8.
It is unclear whether Theorem 3.6 can be extended, for instanceby modifying the construction in Lemma 3.4, to the range 1 < α ≤
2. Notehowever that, by Theorem 2.3, the case α = 1 is necessarily excluded. See [15,Theorem 1.2] for a result relating specifically to the case α = 2.Theorem 3.10 below shows that the situation is different when X is a Hilbertspace. It relies on the following preparatory result, which is analogous to [11,Lemma 2.3] (see also [5, Lemma 1.1] and [28, Lemma 3.2]) and holds for generalBanach spaces. Recall that, if T ∈ B ( X ) is a power-bounded operator with M := sup {k T n k : n ≥ } and if λ ∈ C satisfies Re λ <
0, then it follows from(2.17) that k R ( λ, I − T ) k ≤ M | Re λ | − , and hence that the operator I − T issectorial. Thus, given any s >
0, the fractional power ( I − T ) s is defined as( I − T ) s := 12 π i I Γ λ s R ( λ, I − T ) d λ, where the complex plane is cut along the negative real axis and where Γ is anysuitable contour that contains the point 1 and otherwise encloses σ ( T ) without touching it. Fractional powers coincide with the usual ones whenever s ∈ N , andmoreover ( I − T ) s + t = ( I − T ) s ( I − T ) t for all s, t >
0; see for instance [23] fordetails.
Lemma 3.9.
Let X be a complex Banach space and let T ∈ B ( X ) be a power-bounded operator such that σ ( T ) ∩ T = { } . Furthermore, let α ≥ and supposethat k R (e i θ , T ) k = O ( | θ | − α ) as θ → . Then sup {k ( I − T ) α R ( λ, T ) k : | λ | > } < ∞ . Proof . By (2.17) it suffices to prove that sup {k ( I − T ) α R ( λ, T ) k : λ ∈ A } < ∞ , where A := { λ ∈ C : 1 < | λ | < } . A first step towards this result is to establishthat, under the above assumptions, sup {k (1 − λ ) α R ( λ, T ) k : λ ∈ A } < ∞ . Thus,given r ∈ (0 , r := (cid:26) λ ∈ C : 1 ≤ | λ | ≤ (cid:12)(cid:12)(cid:12)(cid:12) r − r − λ (cid:12)(cid:12)(cid:12)(cid:12) ≥ r − r (cid:27) and let the map H r : Ω r → B ( X ) be defined by H r ( λ ) := (cid:18) r ϕ ( λ ) (cid:19) (1 − λ ) α R ( λ, T ) , so that H r ( λ ) = h r ( λ )(1 − λ ) α R ( λ, T ) , where h r := g r ◦ ϕ with g r and ϕ asin the proof of Theorem 2.11. Then sup {| h r ( λ ) | : r ∈ (0 , ) , λ ∈ Ω r } < ∞ and, by the argument leading to equation (2.19), | h r ( λ ) | ≤ Cr − ( | λ | −
1) for all λ ∈ ∂ Ω r ∩ γ r , where γ r is as defined in (2.16). Note also that | − λ | ≤ Cr forall λ ∈ ∂ Ω r ∩ γ r . Thus, by (2.17) and the assumption on the resolvent, k H r ( λ ) k is uniformly bounded, independently of r , for all λ ∈ ∂ Ω r and hence, by themaximum principle, sup {k H r ( λ ) k : r ∈ (0 , ) , λ ∈ Ω r } < ∞ . Since, given any λ ∈ A , there exists r ∈ (0 , ) such that λ ∈ Ω r and | h r ( λ ) | ≥ , the claim follows.Now let n ∈ N and β ∈ [0 ,
1) be such that α = n + β , and note that, for any k ≥ λ ∈ ρ ( T ), k ( I − T ) k R ( λ, T ) k ≤ | − λ | k k R ( λ, T ) k + k − X j =0 (cid:18) kj (cid:19) | − λ | j k λ − T k k − j − . Setting k = n − k = n , this shows, respectively, that k ( I − T ) n − R ( λ, T ) k ≤ C | − λ | β and k ( I − T ) n R ( λ, T ) k ≤ C | − λ | β for all λ ∈ A . In particular, if β = 0, the proof is complete. If β = 0, on the otherhand, the moment inequality (see for instance [22, Corollary 7.2]) gives k ( I − T ) α − R ( λ, T ) k ≤ C k ( I − T ) n − R ( λ, T ) k − β k ( I − T ) n R ( λ, T ) k β , and hence k ( I − T ) α − R ( λ, T ) k ≤ C | − λ | − for all λ ∈ A . Since k R ( λ, T )( I − T ) α k ≤ | − λ |k R ( λ, T )( I − T ) α − k + k ( I − T ) α − k for all λ ∈ ρ ( T ), the result follows. (cid:3) ATES OF DECAY IN THE KATZNELSON-TZAFRIRI THEOREM 21
The final result shows that the phenomenon described in Theorem 3.6 cannotarise on Hilbert space. For analogous results in the continuous-time setting see[11, Theorem 2.4] and [7, Theorem 7.6]; compare also with [37, Theorem 9].
Theorem 3.10.
Let X be a complex Hilbert space and let T ∈ B ( X ) be apower-bounded operator such that σ ( T ) ∩ T = { } . Furthermore, let α ≥ .Then k R (e i θ , T ) k = O ( | θ | − α ) as θ → if and only if k T n ( I − T ) k = O ( n − /α ) as n → ∞ . Proof . Suppose that k R (e i θ , T ) k = O ( | θ | − α ) as θ →
0, so that, by Lemma 3.9,sup {k ( I − T ) α R ( λ, T ) k : | λ | > } < ∞ . For n ≥ | λ | >
1, let F n ( λ ) := λR ( λ, T ) n X k =0 λ − k T k . Then a simple calculation using the series expansion for the resolvent shows that F n ( λ ) = ∞ X k =0 (min { k, n } + 1) λ − k T k , and hence, by Parseval’s identity, ∞ X k =0 (min { k, n } + 1) k T k x k r k = 12 π Z π (cid:13)(cid:13) F n (cid:0) r e i θ (cid:1) x (cid:13)(cid:13) d θ for all n ≥ x ∈ X and r >
1. Replacing x with ( I − T ) α x and letting B := sup {k ( I − T ) α R ( λ, T ) k : | λ | > } , it follows from the definition of F n that n X k =0 ( k + 1) r k k T k ( I − T ) α x k ≤ B r π Z π (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =0 r − k e − i kθ T k x (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d θ, or indeed n X k =0 ( k + 1) r n k T k ( I − T ) α x k ≤ B r n X k =0 k T k x k r k , by another application of Parseval’s identity. Letting r → n X k =0 ( k + 1) k T k ( I − T ) α x k ≤ M B ( n + 1) k x k , where M := sup {k T n k : n ≥ } . Now, for y ∈ X and n ≥ (cid:0) ( n + 2) T n ( I − T ) α x, y (cid:1) = 2 n + 1 n X k =0 (cid:0) ( k + 1) T k ( I − T ) α x, ( T ∗ ) n − k y (cid:1) , where T ∗ denotes the adjoint of T . By (3.19) and Cauchy’s inequality, the right-hand side is bounded above in modulus by 2 M B k x kk y k , and hence(3.20) k T n ( I − T ) α k ≤ M Bn + 2for all n ≥
0. Thus the proof is complete in the case α = 1. If α >
1, on theother hand, the moment inequality gives k T n ( I − T ) k ≤ C k T n k ( α − /α k T n ( I − T ) α k /α for all n ≥
0, and the result now follows from (3.20) and the fact that T ispower-bounded.The converse implication is a consequence of Theorem 2.4. (cid:3) Remark 3.11.
The above proof follows the method used in [7]. An alternativeapproach, analogous to that of [11], is to consider the operator Q ∈ B ( X × X )given by Q ( x, y ) := ( T x + T ( I − T ) α y, T y ). Then, for n ≥ Q n is representedby the matrix Q n = T n nT ( I − T ) α T n ! and, in particular, Q is power-bounded if and only if sup {k nT n ( I − T ) α k : n ≥ } < ∞ . Since the latter is equivalent, by the moment inequality, to having k T n ( I − T ) k = O ( n − /α ) as n → ∞ , the main implication of Theorem 3.10can be deduced from results in [20], which characterise power-boundedness of anoperator on a Hilbert space in terms of a certain integrability condition on itsresolvent. Remark 3.12.
As in [11, Theorem 2.4], the equivalent statements in Theo-rem 3.10 are also equivalent to the condition that, for every x ∈ X , k T n ( I − T ) x k = o ( n − /α ) as n → ∞ , which in turn is equivalent, by another applicationof the moment inequality, to having nT n ( I − T ) α → n → ∞ . One implication follows from the general observation that,given any complex Banach space X and a power-bounded mean ergodic opera-tor T ∈ B ( X ) satisfying σ ( T ) ∩ T ⊂ { } , the powers T n converge strongly, as n → ∞ , to the projection P onto Fix( T ) along the closure of Ran( I − T ); see [4,Theorem 4.1]. Indeed, if k T n ( I − T ) k = O ( n − /α ) as n → ∞ , then the opera-tor Q ∈ B ( X × X ) defined in Remark 3.11 is power-bounded, and furthermore σ ( Q ) = σ ( T ) and Fix( Q ) = Fix( T ) × Fix( T ). Hence applying this observationto Q shows that, for any x, y ∈ X , T n x + nT n ( I − T ) α y → P x as n → ∞ .Since T n x → P x as n → ∞ by the same observation applied to T , it followsthat nT n ( I − T ) α → n → ∞ . The converseimplication is a simple consequence of the Uniform Boundedness Theorem. Acknowledgements
The author is grateful to Professor C.J.K. Batty for his guidance and his carefulreading of an earlier version of this work, to Professor Y. Tomilov for sharing anobservation that led to Remark 3.7 and for a number of helpful discussions duringhis visit to Oxford in autumn 2012, and finally to the EPSRC for its financialsupport.
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