# Kreiss bounded and uniformly Kreiss bounded operators

aa r X i v : . [ m a t h . F A ] D ec Kreiss bounded and uniformly Kreiss boundedoperators

A. Bonilla and V. M¨uller ∗ December 18, 2019

Abstract If T is a Kreiss bounded operator on a Banach space, then k T n k = O ( n ).Forty years ago Shields conjectured that in Hilbert spaces, k T n k = O ( √ n ). Anegative answer to this conjecture was given by Spijker, Tracogna and Welfertin 2003. We improve their result and show that this conjecture is not trueeven for uniformly Kreiss bounded operators. More precisely, for every ε > T on a Hilbert space suchthat k T n k ∼ ( n + 1) − ε for all n ∈ N . On the other hand, any Kreiss boundedoperator on Hilbert spaces satisﬁes k T n k = O ( n √ log n ).We also prove that the residual spectrum of a Kreiss bounded operator ona reﬂexive Banach space is contained in the open unit disc, extending knownresults for power bounded operators. As a consequence we obtain examples ofmean ergodic Hilbert space operators which are not Kreiss bounded.2010MSC: 47A10, 47A35Key words and phrases: Kreiss boundedness, Ces`aro mean, mean ergodic. Throughout this paper X stands for a complex Banach space, the symbol B ( X )denotes the space of all bounded linear operators acting on X , and X ∗ is the dual of X . Deﬁnition 1.1.

For an operator T ∈ B ( X ) we have three notions of Kreiss bound-edness, ordered by strength:1. T is strong Kreiss bounded if there exists C > k ( λI − T ) − k k ≤ C ( | λ | − k for all k ∈ N and | λ | > ∗ The ﬁrst author was supported by MINECO and FEDER, Project MTM2016-75963-P. Thesecond author was supported by grant no. 17-27844S of GA CR and RVO:67985840. T is uniformly Kreiss bounded if there exists C > (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X k =0 λ − k − T k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ C | λ | − n ∈ N and | λ | > T is Kreiss bounded if there exists

C > k ( λI − T ) − k ≤ C | λ | − | λ | > . Given T ∈ B ( X ) and n ≥

0, we denote the

Ces`aro mean by M n ( T ) := 1 n + 1 n X k =0 T k . We recall some deﬁnitions concerning the behavior of the sequence of Ces`aromeans ( M n ( T )). Deﬁnition 1.2.

A linear operator T on a Banach space X is called1. mean ergodic if M n ( T ) converges in the strong operator topology of X ;2. Ces`aro bounded if the sequence ( M n ( T )) n ∈ N is bounded;3. absolutely Ces`aro bounded if there exists a constant C > N + 1 N X j =0 k T j x k ≤ C k x k , for all x ∈ X and N ∈ N .An operator T is said to be power bounded if there is a C > k T n k < C for all n .The ﬁrst example of a mean ergodic operator which is not power-bounded wasgiven by Hille ([8], where k T n k ∼ n / ). An example of a mean ergodic operator T on L ( Z ) with lim sup n k T n k /n > k T n x k /n → x ∈ L ( Z )). Remark 1.1.

1. In [15, Corollary 3.2], it is proved that an operator T is uni-formly Kreiss bounded if and only if there is a constant C such that k M n ( λT ) k ≤ C for all n ∈ N and | λ | = 1 . (1)2. In [6], it was shown that every strong Kreiss bounded operator is uniformlyKreiss bounded. The converse is not true, see [15, Section 5]. Moreover,McCarthy (see [14], [18]) proved that if T is strong Kreiss bounded then k T n k ≤ Cn / (see also [12, Theorem 2.1]) and gave also an example of a strong Kreissbounded operator which is not power bounded.2. Denote by M (2) n ( T ) := 2( n + 1)( n + 2) n X j =0 ( j + 1) M j ( T )the second Ces`aro mean. It is easy to see that M (2) n ( T ) = 2( n + 1)( n + 2) n X j =0 ( n + 1 − j ) T j . In [20], it was proved that T is Kreiss bounded if and only if there is a constant C such that k M (2) n ( λT ) k ≤ C for all n ∈ N and | λ | = 1 . (2)There exist Kreiss bounded operators which are not Ces`aro bounded, and con-versely [22].4. An operator T is called M¨obius bounded if its spectrum is contained in theclosed unit disc and ϕ ( T ) is uniformly bounded on the set of the automorphismof the unit disc. By [18], T is a M¨obius bounded operator if and only if it isKreiss bounded.5. On ﬁnite-dimensional spaces, the classes of Kreiss bounded operators and powerbounded operators coincide.6. By (1), any absolutely Ces`aro bounded operator is uniformly Kreiss bounded.Let X be the space of all bounded analytic functions f on the unit disc in thecomplex plane such that the derivative f ′ belongs to the Hardy space H , endowedwith the norm k f k := k f k ∞ + k f ′ k H . Then the multiplication operator M z acting on X is Kreiss bounded but it fails to bepower bounded. Moreover, this operator is not uniformly Kreiss bounded (see [20]).Let V be the Volterra operator acting on L p [0 , ≤ p ≤ ∞ deﬁned by( V f )( t ) = Z t f ( s ) ds ( f ∈ L p (0 , . Then I − V is uniformly Kreiss bounded. For p = 2 it is even power bounded (see[15]).It is immediate that any power bounded operator is absolutely Ces`aro bounded.In general, the converse is not true.Let 1 ≤ p < ∞ and let e n , n ∈ N be the standard basis in ℓ p ( N ). The follow-ing theorem yields examples of absolutely Ces`aro bounded operators with diﬀerentbehavior on ℓ p ( N ). 3 heorem 1.1. [3, Theorem 2.1] Let T be the weighted backward shift on ℓ p ( N ) with ≤ p < ∞ deﬁned by T e := 0 and T e k := w k e k − for k > . If w k := (cid:18) kk − (cid:19) α with < α < p , then T is absolutely Ces`aro bounded on ℓ p ( N ) and is not powerbounded. For p = 2, the adjoint of the operator in Theorem 1.1 is uniformly Kreiss bounded,mean ergodic but not absolutely Ces`aro bounded.In [9], Kornfeld and Kosek constructed for every δ ∈ (0 ,

1) a positive mean ergodicoperator T on L with k T n k ∼ n − δ . By positivity, T is absolutely Ces`aro bounded.Since T n = ( n + 1) M n ( T ) − nM n − ( T ) , (3)any Ces`aro bounded operator satisﬁes that k T n k = O ( n ).In the following picture we summarize the implications among the above deﬁni-tions. Power boundedStrong Kreiss bounded Absolutely Ces`aro boundedUniformly Kreiss boundedKreiss bounded Ces`aro bounded k T n k = O ( n )Figure 1: Implications among diﬀerent deﬁnitions related with Kreiss bounded andCes`aro bounded operators on Banach spaces. If T is a Kreiss bounded operator in a Banach space, then k T n k ≤ Cn [12, (2.4)].By Nevanlinna [16, Theorem 6], there are Kreiss bounded operators T on Banachspaces with k T n k ≥ C ′ n for some C ′ > T satisﬁes k T n k = O ( √ n ). A negative answer to this conjecture was given in [19]where it was shown that for every ε > T such that the norms of its powers k T n k grow as fast as n − ε .In this section we improve this result and construct an operator with similarproperties, which is even uniformly Kreiss bounded.In the same paper [18] Shields mentioned without proof that if T is a Kreissbounded Hilbert space operator such that the sequence of norms ( k T n k ) is increasing4nd there is a unit vector x such that k T n x || ≥ k T n k / n then k T n k = O ( √ n ) (such properties would satisfy the ﬁrst natural attempt to disprove the Shieldsconjecture). For the sake of completeness we give a proof of this result. We need toprove the following lemma. Lemma 2.1.

Let ( a k ) ∞ k =0 be an increasing sequence of non-negative numbers, B > ,and let ∞ X n =0 a k r k ≤ B/ (1 − r ) for ≤ r < . Then a n = O ( √ n ) .Proof. Since ∞ X n =0 a k r k ≤ B (1 − r ) , multiplying both sides by 1 − r , we see that a + ∞ X k =1 ( a k − a k − ) r k ≤ B − r . Now since { a k } k ∈ N is increasing, we have r n (cid:16) a + n X k =1 ( a k − a k − ) (cid:17) ≤ a + n X k =1 ( a k − a k − ) r k ≤ B − r . Set r = e − /n . We conclude that a n = a + n X k =1 ( a k − a k − ) ≤ B ′ n for some constant B ′ . Thus a n = O ( √ n ) . Theorem 2.1.

Let T be a Kreiss bounded operator on a Hilbert space such that {k T n k} ∞ n =0 is increasing and suppose that there exist a unit vector x and a constant A such that k T n k ≤ A k T n x k for all n . Then k T n k = O ( √ n ) .Proof. Let f ( z ) = ∞ X k =0 T k z k . Since T is Kreiss bounded we have k f ( z ) k ≤ C −| z | forall | z | <

1. If y ∈ H with k y k = 1 then ∞ X k =0 r n k T n y k = 12 π Z π k f ( re iθ ) y k dθ ≤ C (1 − r ) . Since there exists a unit vector x and a constant A such that k T n k ≤ A k T n x k forall n , we have ∞ X k =0 r n k T n k ≤ A ∞ X k =0 r n k T n x k ≤ A C (1 − r ) . Now by Lemma 2.1, we obtain k T n k = O ( √ n ).5he Shields conjecture k T n k = O ( n / ) is true for some subclasses of Kreissbounded operators:1. If T is a strong Kreiss bounded operator on a Banach space, then k T n k = O ( n / ), see [14].2. If T is an absolutely Ces`aro bounded operator on a Hilbert space, then k T n k = o ( n / ) and moreover for all ε there exist absolutely Ces`aro bounded operatorson ℓ ( N ) such that k T n k = O ( n / − ε ) [3].3. See [5] for other classes of Kreiss bounded operators where the Shields conjec-ture is true.Now we construct a uniformly Kreiss bounded operator which disproves theShields conjecture. Theorem 2.2.

Let < η < / . Then there exists a constant c > with thefollowing property: for each N ∈ N there exists an operator T N acting on a N -dimensional Hilbert space H N such that k T N − N k = N η ,T N is a weighted shift satisfying k T N k = 2 η , k M n ( T N ) k ≤ c for every n ∈ N . Proof.

Let H N be the Hilbert space with an orthonormal basis e , . . . , e N . Let w j = j η ( j = 1 , . . . , N )and w j = N η (2 N − j + 1) η ( j = N + 1 , . . . , N ) . Consider the weighted shift T N on H N deﬁned by T N e j = w j +1 w j e j +1 ( j = 1 , . . . , N − T N e N = 0.Note that w = 1, w N = w N +1 = N η and w N = N η . Then k T N − k = k T N − e k = w N = N η .Clearly k T N k = max (cid:8) w j +1 w j : 1 ≤ j ≤ N − (cid:9) = 2 η .Let n ∈ N . We have k M n ( T N ) k = sup (cid:8)(cid:12)(cid:12) h M n ( T N ) x, y i (cid:12)(cid:12) : x, y ∈ H N , k x k = k y k = 1 (cid:9) . Let x = P Nj =1 α j e j , y = P Nj =1 β j e j , k x k = P Nj =1 | α j | = 1 and k y k = P Nj =1 | β j | =1. Let x = x + x where x = P Nn =1 α j e j and x = P Nj = N +1 α j e j . Similarly, y = y + y , where y = P Nj =1 β j e j and y = P Nj = N +1 β j e j .6e have |h M n ( T N ) x, y i| ≤ A + B + C (1)where A = |h M n ( T N ) x , y i| ,B = |h M n ( T N ) x , y i| and C = |h M n ( T N ) x , y i| . To estimate

A, B and C we need two simple lemmas. Claim 1.

There exists a constant c such that1 n + 1 ∞ X j =1 X j ≤ j ′ ≤ j + n γ j δ j ′ j ′ η j η ≤ c for all n , γ j , δ j ′ ≥ j, j ′ ∈ N ) with P j γ j = P j ′ δ j ′ = 1. Proof.

Let H be the Hilbert space with an orthonormal basis f j ( j ∈ N ). Considerthe weighted shift V ∈ B ( H ) deﬁned by V f j = (cid:16) j +1 j (cid:17) η f j +1 . Let u = P ∞ j =1 γ j f j and v = P ∞ j =1 δ j ′ f j ′ with k u k = k v k = 1. By [3, Corollary 2.4], V is uniformly Kreissbounded. So there exists a constant c such that c ≥ k M n ( V ) k ≥ (cid:12)(cid:12) h M n ( V ) u, v i (cid:12)(cid:12) = 1 n + 1 ∞ X j =1 X j ≤ j ′ ≤ j + n γ j δ j ′ j ′ η j η . Claim 2.

There exists a constant c > M X j =1 j − η ≤ c M − η for all M . Proof.

We have M X j =1 j − η ≤ Z M t − η dt = 1 + h t − η − η i M ≤ c M − η for some constant c > M . Continuation of the proof of Theorem 2.2:

We have A = 1 n + 1 X j ≤ j ′≤ j + nj ′≤ N α j ¯ β j ′ w j ′ w j ≤ n + 1 X j ≤ j ′≤ j + nj ′≤ N | α j | · | β j ′ | j ′ η j η ≤ c (2)7y Claim 1.Similarly, B = 1 n + 1 X N

13 (2 N ) − ε and k T N k ≥ k T NN +1 k ≥ k T N +1 N +1 kk T N +1 k ≥ ( N + 1) η η > ( N + 1) − ε √ >

13 (2 N + 1) − ε . Hence k T k k ≥ ( k + 1) − ε for all k ∈ N .On the other hand, we prove a small improvement of the general estimate ofnorms of powers of Kreiss bounded operators on Hilbert spaces. The proof followsthe argument of [3, Theorem 2.3] for uniformly Kreiss bounded operators with somenecessary modiﬁcations. Theorem 2.4.

Let T be a Kreiss bounded operator in a Hilbert space. Then k T n k = O (cid:16) n √ log n (cid:17) .Proof. In [20] it was proved that T is Kreiss bounded if and only if there is a constant C ′ > k M (2) n ( λT ) k ≤ C ′ for n = 0 , , , · · · and | λ | = 1 . Thus there exists a constant

C > (cid:13)(cid:13)(cid:13) N − X j =0 ( N − j )( λT ) j (cid:13)(cid:13)(cid:13) ≤ CN for all λ, | λ | = 1 and N ≥ laim 1. Let x ∈ H , k x k = 1. Then N − X j =0 k T j x k ≤ C N for all N ≥ Proof.

Consider the normalized Lebesgue measure on the unit circle. We have N N − X j =0 k T j x k ≤ N − X j =0 (2 N − j ) k T j x k = Z | λ | =1 (cid:13)(cid:13)(cid:13) N − X j =0 (2 N − j )( λT ) j x (cid:13)(cid:13)(cid:13) dλ ≤ C N . So P N − j =0 k T j x k ≤ C N . Claim 2.

Let 0 < M < N and x ∈ H , k x k = 1, T N x = 0. Then M − X j =0 k T N x k k T N − j x k ≤ C M . Proof.

Set y = T N x . Since T ∗ is also Kreiss bounded with the same constant, wehave 16 C M k y k ≥ M − X j =0 k T ∗ j y k ≥ M − X j =0 (cid:12)(cid:12)(cid:12)D T ∗ j y, T N − j x k T N − j x k E(cid:12)(cid:12)(cid:12) = M − X j =0 (cid:12)(cid:12)(cid:12)D y, T N x k T N − j x k E(cid:12)(cid:12)(cid:12) = k y k M − X j =0 k T N x k k T N − j x k . Hence M − X j =0 k T N x k k T N − j x k ≤ C M . Claim 3.

Let N ∈ N , x ∈ H , k x k = 1 and T N x = 0. Then N − X j =0 k T j x k ≥ √ N C .

Proof.

We have N − X j =1 k T j x k ≤ (cid:16) N − X j =0 k T j x k (cid:17) / · √ N ≤ CN / . N = N − X j =0 p k T j x k p k T j x k ≤ (cid:16) N − X j =0 k T j x k (cid:17) / · (cid:16) N − X j =0 k T j x k (cid:17) / and N − X j =0 k T j x k ≥ N P N − j =0 k T j x k ≥ N CN / = √ N C .

Claim 4.

Let 0 < M < M < N , k x k = 1 and T N x = 0. Then M − X j = M k T N − j x k k T N x k ≥ ( M − M ) C M . Proof.

Let a j = k T N − j x k k T N x k . By Claim 2, M − X j = M a j ≤ M − X j =0 a j ≤ C M . We have M − M = M − X j = M √ a j √ a j ≤ (cid:16) N − X j =0 a j (cid:17) / · (cid:16) N − X j =0 a j (cid:17) / and M − X j = M a j ≥ ( M − M ) · (cid:16) N − X j =0 a j (cid:17) − ≥ ( M − M ) C M . Continuation of the proof of Theorem 2.4.

Let K ∈ N and 2 K +1 < N ≤ K +2 . Let x ∈ H , k x k = 1 and T N x = 0.For | λ | = 1 let y λ = P N − j =0 ( λT ) j x k T j x k . Then Z | λ | =1 k y λ k d λ = 2 N and Z | λ | =1 (cid:13)(cid:13)(cid:13) N − X j =0 (2 N − j )( λT ) j y λ (cid:13)(cid:13)(cid:13) d λ ≤ C N Z | λ | =1 k y λ k d λ ≤ C N .

11n the other hand, Z | λ | =1 (cid:13)(cid:13)(cid:13) N − X j =0 (2 N − j )( λT ) j y λ (cid:13)(cid:13)(cid:13) d λ = Z | λ | =1 (cid:13)(cid:13)(cid:13) N − X j =0 (2 N − j )( λT ) j N − X r =0 ( λT ) r x k T r x k (cid:13)(cid:13)(cid:13) d λ = Z | λ | =1 (cid:13)(cid:13)(cid:13) N − X j =0 ( λT ) j x min { j, N − } X r =0 N − j + r k T r x k (cid:13)(cid:13)(cid:13) d λ = N − X j =0 k T j x k (cid:16) min { j, N − } X r =0 N − j + r k T r x k (cid:17) ≥ N − X j = N − K k T j x k (cid:16) j X r =0 N k T r x k (cid:17) ≥ N N − X j = N − K k T j x k (cid:16) √ N − K C (cid:17) ≥ N C N − X j = N − K k T j x k ≥ N C k T N x k K − X k =0 N − k − X j = N − k +1 k T j x k k T N x k = N C k T N x k K − X k =0 2 k +1 X j =2 k +1 k T N − j x k k T N x k ≥ N C k T N x k K − X k =0 k C · k +2 = N C k T N x k K. Thus we have k T N x k ≤ C N K ≤ C N log N − . Hence k T N k = O ( N √ log N ).In the next diagram we show graphically the implications among various deﬁni-tions related with Kreiss boundedness on Hilbert spaces and corresponding knownestimates for the growth of k T n k .Strong Kreiss bounded ( k T n k = O ( n / )) Abs. Ces`aro bounded ( k T n k = o ( n / ))Uniformly Kreiss boundedKreiss bounded Mean ergodic k T n k = o ( n ) k T n x k = o ( n ) , ∀ x Figure 2: Implications among diﬀerent deﬁnitions related with Kreiss bounded onHilbert spaces.

Example 2.1.

Derriennic [4] gave an example of a mean ergodic operator T on a realHilbert space for which n − k T n k does not converge to zero, and such that T ∗ is not12ean ergodic, only weakly mean ergodic (i.e., the Ces`aro means converge weakly).In [23, Example 3.1] it was shown that the operator T := (cid:18) B B − I B (cid:19) acting on the Hilbert spaces l ( N ) ⊕ l ( N ), where B is the backward shift in l ( N ), ismean ergodic and n − k T n k ≥

2. As a consequence of Theorem 2.4, T is an exampleof a mean ergodic operator acting on a Hilbert spaces, which is not Kreiss bounded.By [1, Remark 3.1], in Banach spaces there exists a Kreiss bounded operator suchthat lim n →∞ k M n +1 ( T ) − M n ( T ) k 6 = 0 . However, in Hilbert spaces this is not possible.

Theorem 2.5. If T is a Kreiss bounded operator on a Hilbert space then lim n →∞ k M n +1 ( T ) − M n ( T ) k = 0 . Proof.

We have n + 2 n + 1 M n +1 ( T ) − M n ( T ) = 1 n + 1 T n +1 . So M n +1 ( T ) − M n ( T ) = 1 n + 1 T n +1 − n + 1 M n +1 ( T ) . Now if T is a Kreiss bounded operator on a Hilbert space, then we have k T n k = o ( n )by Theorem 2.4 and k M n ( T ) k = O (log( n + 2)) (see, [20, Theorem 6.1, 6.2]). Thuslim n →∞ k M n +1 ( T ) − M n ( T ) k = 0 . The following characterization of ergodic operators was proved in [11].

Theorem 3.1. [11, Theorem 2.1.3, page 73] An operator T in a Banach space X ismean ergodic if and only if it is Ces`aro bounded, k T n x k = o ( n ) for all x ∈ X and X = N ( I − T ) ⊕ R ( I − T )Recall that the residual spectrum σ R ( T ) of an operator T ∈ B ( X ) is the set ofall λ ∈ C such that T − λ is injective and R ( T − λ ) = X .13 orollary 3.1. If T is mean ergodic in a Banach space, then / ∈ σ R ( T ) Corollary 3.2. [3, Corollary 2.5], [3, Corollary 2.7] Let T be a bounded operator ona Banach space X . If1. either T is a uniformly Kreiss bounded operator on a Hilbert space,2. or T is an absolutely Ces`aro bounded operator on a reﬂexive Banach space,then λT is mean ergodic for all λ, | λ | = 1 . Consequently, σ R ( T ) ⊂ D . The next result generalizes the above observation as well as the results of [13],[6] for power bounded operators.

Theorem 3.2. If T is a Kreiss bounded operator on a reﬂexive Banach space then σ R ( T ) ⊂ D .Proof. If T is a Kreiss bounded operator on a Banach space X , then k M (2) n ( λT ) k areuniformly bounded and k M n ( λT ) k = O ( log ( n + 2)) for all λ ∈ T (see, [20, Theorem6.1, 6.2]).Now, since X is a reﬂexive Banach space, M (2) n ( λT ) converge strogly in X = N ( I − λT ) ⊕ R ( I − λT ) (see, [21, Theorem 2.1]).So for all λ ∈ T , we have X = N ( λI − T ) ⊕ R ( λI − T ). Hence if λ / ∈ σ p ( T ), then X = R ( λI − T ) and λ / ∈ σ R ( T ).Thus σ R ( T ) ⊂ D .The condition on the residual spectrum is optimal. The forward shift in l ( N ) isa power bounded operator with residual spectrum equal to the open unit disc. Example 3.1.

There exists a power bounded operator T on c ( N ) such that 1 ∈ σ R ( T ). Proof.

The operator T : c ( N ) → c ( N ) deﬁned by T ( a , a , a , · · · ) = ( a , a , a , a , · · · )is power-bounded and 1 ∈ σ R ( T ). Example 3.2.

There exists a Kreiss bounded operator T on a non-reﬂexive Banachspace, which is not power bounded and 1 ∈ σ R ( T ). Proof.

Let X denote the Banach space of analytic functions f in the open unitdisc and continuous on the boundary, such that f ′ belongs to the Hardy space H ,equipped with the norm k f k := k f k ∞ + k f ′ k If M z denotes the multiplication operator and T = ( I + M z ), then T is a Kreissbounded operator, see [16, Example 4].Moreover, N ( I − T ) = { } and R ( I − T ) is not dense because every function inthis closure necessarily veriﬁes that f (1) = 0. Thus 1 ∈ σ R ( T ).14 roposition 3.1. There exists a Ces`aro bounded operator T on a Hilbert space whichis mean ergodic, N ( T + I ) = { } and N ( T ∗ + I ) = { } .Proof. Let H be the Hilbert space with an orthonormal basis e j ( j = 0 , , . . . ). Let ε j = 2 − j , c j = 1 − ε j ( j ≥ T = − ε ε ε · · · c · · · c · · · c · · ·· · · Clearly

T e = − e , so N ( T + I ) = { } . We have ( T + I ) e j = (1 − c j ) e j − ε j e = ε j e j − ε j e . So ( T + I )( − ε − j e j ) = e − ε j e j → e . Thus e ∈ R ( T + I ) and it is easyto see that R ( T + I ) = H . Hence N ( I + T ∗ ) = { } .For n ≥ T ) n ) j,j = ( − c j ) n ( j ≥ , (( T ) n ) ,j = ( − n ε j (1 + c j + c j + c n − j ) = ( − n ε j − c nj − c j ( j ≥ T ) n ) i,j = 0 otherwise.So T n = ( − n ε (1 + c + · · · + c n − ) ε (1 + · · · + c n − ) · · · c n · · · c n · · ·· · · · · · We show that n − k T n k →

0. Let δ >

0. Find j such that ε j < δ and n satisfying c n δj < δ and 2 n − < δ .Let n ≥ n . Clearly | ( T n ) j,j | ≤ j ≥ j ≥ j then | ( T n ) ,j | = ε j (1 + c j + · · · + c n − j ) ≤ ε j n = 2 − j n < δn j − j . If 1 ≤ j < j then | ( T n ) ,j | = ε j (1 + c j + · · · + c n − j ) = ε j X ≤ i

There exists a mean ergodic operator T on a Hilbert space such that σ R ( T ) ∩ ∂ D = ∅ . Example 3.3.

By Theorem 3.2, the operator of Corollary 3.3 is another example ofa mean ergodic operator on a Hilbert space, which is not Kreiss bounded.

As consequence of results of this paper we gave two examples in Hilbert spaces ofmean ergodic operators which are not Kreiss bounded. By [3, Corollary 2.5], alluniformly Kreiss bounded operators on Hilbert spaces are mean ergodic. However,the following problem is open:

Question 4.1.

Does there exist a Kreiss bounded operator on a Hilbert space whichis not mean ergodic?Observe that by Theorem 2.4, the above problem is equivalent to the questionwhether there exists a Kreiss bounded Hilbert space operator which is not Ces`arobounded.If the answer of the above question is positive then such an operator is notuniformly Kreiss bounded. If the answer is negative then it is natural to ask

Question 4.2.

Does there exist a Kreiss bounded Hilbert space operator which isnot uniformly Kreiss bounded? 16nother open question is whether it is possible to generalize the Jacobs-de Leeuw-Glicksberg theorem for uniformly Kreiss bounded operators.

Question 4.3.

Let X be a reﬂexive Banach space and T ∈ B ( X ) a uniformly Kreissbounded operator such that n − T n x → x ∈ X . Is it true that X can bedecomposed as X = _ | λ | =1 N ( T − λ ) ⊕ \ | λ | =1 R ( T − λ ) ?Clearly for Hilbert space operators the condition n − T n x → T on a Banach space X satisﬁes the condition k n − T n k → References [1] L. Abadias and A. Bonilla, Growth orders and ergodicity for absolutely Ces`arobounded operators, Linear Algebra and its Applications, (2019), 253-267[2] A. Aleman and L. Suciu, On ergodic operator means in Banach spaces, IntegralEquations Operator Theory, (2016), 259-287.[3] T. Berm´udez, A. Bonilla, V. M¨uller and A. Peris, Cesaro bounded operators inBanach spaces, J. d’Analyse Math., to appear.[4] Y. Derriennic, On the mean ergodic theorem for Ces`aro bounded operators,Colloq. Math., (2000), part 2, 443–455.[5] S. Ghara, The orbit of a bounded operator under the M¨obius group modulosimilarity equivalence, ArXiv1811.05428v1[6] A. Gomilko and J. Zem´anek, On the uniform Kreiss resolvent condition, (Rus-sian) Funktsional. Anal. i Prilozhen., (2008), no. 3, 81–84; translation inFunct. Anal. Appl. (2008), no. 3, 230-233.[7] M. Haase and Y. Tomilov, Domain characterizations of certain functions ofpower-bounded operators, Studia Math, 196 (2010), no. 3, 265–288.[8] E. Hille, Remarks on ergodic theorems, Trans. Amer. Math. Soc., , (1945).246–269.[9] I. Kornfeld and W. Kosek, Positive L operators associated with nonsingularmappings and an example of E. Hille, Colloq. Math., (2003), no. 1, 63–77.[10] W. Kosek, Example of a mean ergodic L operator with the linear rate of growth,Colloq. Math. (2011), no. 1, 15–22.[11] U. Krengel, Ergodic theorems. De Gruyter Studies in Mathematics, 6. Walterde Gruyter and Co., Berlin, 1985. 1712] C. Lubich and O. Nevanlinna, On resolvent conditions and stability estimates,BIT, (1991), no. 2, 293–313.[13] A. Mello and C.S. Kubrusly, Residual spectrum of power bounded operators.Funct. Anal. Approx. Comput. 10 (2018), no. 3, 21–26.[14] C.A. McCarthy, A strong resolvent condition does not imply power-boundedness, Chalmers Institute of Technology and the University of G¨oteborg,Preprint No 15 (1971).[15] A. Montes-Rodr´ıguez, J. S´anchez- ´Alvarez and J. Zem´anek, Uniform Abel-Kreissboundedness and the extremal behavior of the Volterra operator, Proc. LondonMath. Soc., (2005),761-788.[16] O. Nevanlinna, Resolvent conditions and powers of operators, Studia Math., (2001), no. 2, 113–134.[17] A. L. Shields, Weighted shift operators and analytic function theory, Topics inOperator Theory,Math. Surveys Monographs, vol. 13, Amer. Math. Soc, Provi-dence, RI, 1974, pp. 49-128.[18] A. L. Shields, On M¨obius Bounded operators, Acta Sci. Math., (Szeged), (1978), 371-374.[19] M. N. Spijker, S. Tracogna and B. D. Welfert, About the sharpness of thestability estimates in the Kreiss matrix theorem, Math. Comp., (242)(2003),697-713.[20] J. C. Strikwerda and B. A. Wade, A survey of the Kreiss matrix theorem forpower bounded families of matrices and its extensions, Linear operators (War-saw, 1994), 339-360, Banach Center Publ., , Polish Acad. Sci., Warsaw, 1997.[21] L. Suciu, Saturation for Ces`aro means of higher order. Oper. Matrices (2013),no. 3, 557–572.[22] L. Suciu and J. Zem´anek, Growth conditions on Ces`aro means of higher order,Acta Sci. Math (Szeged), (2013), 545-581.[23] Y. Tomilov and J. Zem´anek, A new way of constructing examples in operatorergodic theory, Math. Proc. Cambridge Philos. Soc., (2004), no. 1, 209-225. A. Bonilla

Departamento de An´alisis Matem´atico, Universidad de la Laguna, 38271, La Laguna(Tenerife), Spain.e-mail: abonilla@ull.esV. M¨ullerMathematical Institute, Czech Academy of Sciences, Zitn´a 25, 115 67 Prague 1, CzechRepublic. e-mail : muller@math.cas.cz: muller@math.cas.cz