On a tauberian condition
for bounded linear operators
J. Malinen∗, O. Nevanlinna†, and Z. Yuan‡
Department of Mathematics
Helsinki University of Technology
P.O.Box 1100
FIN-02015 HUT, Finland
March 5, 2007
Abstract: We study the relation between the growth of sequences kT n k
and k(n + 1)(I − T )T n k for operators T ∈ L(X) satisfying variants of the
C
Ritt resolvent condition k(λ − T )−1 k ≤ |λ−1|
in various subsets of {|λ| > 1}.
Keywords: Power bounded, Ritt resolvent condition, tauberian theorem.
2000 Mathematics Subject Classification: 47A10, 40E05, 47A30, 47D03.
Abbreviated title: On a tauberian condition.
1
Introduction
Let T ∈ L(X); a bounded linear operator on a (complex) Banach space X.
It was R. K. Ritt himself who first studied the Ritt resolvent condition
(1)
k(λ − T )−1 k ≤
C
|λ − 1|
for |λ| > 1. He proved that if T satisfies (1) for |λ| > 1, then limn→∞ kT n /nk =
0, see [16]. Clearly (1) implies that σ(T ) ⊂ D ∪ {1} but, in fact, even
∗
[email protected]
[email protected]
‡
[email protected]
†
1
σ(T ) ⊂ Kδc ∩ (D ∪ {1}) for some δ > 0, where
π
+ δ};
2
see O. Nevanlinna [12, Theorem 4.5.4] and Yu. Lyubich [8].
An operator T ∈ BLO(X) is power bounded if supn≥1 kT n k < ∞.
Y. Katznelson and L. Tzafriri proved in 1986 that for a power bounded
T , we have σ(T ) ⊂ D ∪ {1} if and only if limn→∞ k(I − T )T n k = 0,
see [7]. Related to this, J. Zemánek asked in 1992 whether (1) implies
limn→∞ k(I − T )T n k = 0, too. This was answered in positive by O. Nevanlinna, and he also noted that if (1) holds in the larger set Kδ ∪ Dc for some
δ > 0, then T is power bounded, see [12, Theorem 4.5.4], [13] and [20].
It was observed in 1998 independently by B. Nagy and J. Zemánek [11],
Yu. Lyubich [8], and O. Nevanlinna that if (1) holds for all |λ| > 1, then
the same estimate holds for all λ ∈ Kδ ∪ Dc for some δ > 0 (with another
possibly larger constant Cδ in place for C). Hence, if T satisfies (1) for all
|λ| > 1, then T is power bounded. The upper bound supn≥1 kT n k ≤ (eC 2 )/2
was given by N. Borovykh, D. Drissi and M. N. Spijker, see [1]. A tighter
estimate supn≥1 kT n k ≤ C 2 was shown by O. El-Fallah and T. Ransford in
[2].
Much of these developments culminate in the following fundamental result connecting power boundedness, the Ritt resolvent condition and the
tauberian condition (3):
Kδ := {λ = 1 + reiθ : r > 0 and |θ| <
(2)
Proposition 1. The following are equivalent:
(i) T satisfies (1) for all |λ| > 1,
(ii) σ(T ) ⊂ D ∪ {1}, and there exists δ > 0 and C = C(δ) such that (1)
holds for all λ ∈ Kδ , and
(iii) T is power bounded, and it satisfies the tauberian condition
(3)
sup (n + 1)k(I − T )T n k ≤ M
n≥1
for some M < ∞.
Indeed, the equivalence (i) ⇔ (ii) has already been discussed above. The
implication (ii) ⇒ (iii) is given in [12, Theorem 4.5.4]. That (iii) implies (i)
was reported in [13, Theorem 2.1]. The proof relies on the theory of analytic
semigroups, and it follows closely [14, Theorem 5.2]1 . The equivalence (i)
⇔ (iii) is is also given in [11, p. 147].
1
However, the restrictive assumption 0 ∈ ρ(A) must be first removed from [14, Theorem 5.2] by a more careful analysis.
2
We shall show in this paper that the conditions of Proposition 1 can be
combined in a different way. Indeed, we shall prove the following tauberian
theorem and discuss some of its consequences:
Theorem 1. Assume that T ∈ L(X) satisfies tauberian condition (3), and
k(λ − 1)(λ − T )−1 k ≤ C
(4)
for all λ > 1. Then T is power bounded with the estimates
(5)
kT n k ≤ 2 + CkT k + 2M and
lim sup kT n k ≤ 2 + CkT k + (1 + 1/e) M.
n→∞
The proof of this theorem is given in Section 2 below. The main results of
this paper were given in 2002 in [19].
2
Equivalent conditions
under the tauberian condition
Let us remind the results of the classical tauberian theorem in the scalar
case. Let {an } be a complex sequence and sn = a0 + a1 + ... + an for n ≥ 0.
A. Tauber proved in 1897 that if
(i) limn→∞ (n + 1)an = 0, and
(ii) limr→1− f (r) = s, where f (r) =
P∞
0
an rn for 0 < r < 1,
then limn→∞ sn = s follows. J. E. Littlewood showed in 1910 that the
tauberian condition (i) can in fact be replaced by the weaker tauberian
condition supn n|an | < ∞ but the proof with this modification becomes
considerable harder. Good references to this and other related tauberian
theorems are [15] and [21].
If we take an = (I − T )T n , we see that the weaker tauberian condition
supn n|an | < ∞ corresponds to assumption (3), and the partial sums are
simply sn = I − T n+1 . We remark that the stronger tauberian condition (i)
above is too restrictive in operator context because a power-bounded T 6= I
with σ(T ) = {1} satisfies lim inf n→∞ (n + 1)k(I − T )T n k ≥ 1/e; see [3], [6],
and [9] for various proofs.
In this paper, we are not interested in the limit behaviour of {sn } (in
other words, the ergodicity of T ) but in the boundedness of this sequence
when (3) holds. For this reason, we may adapt the original simple approach
3
by Tauber to prove Theorem 1, without having to deal with the more complicated technique by Littlewood; see also [15, Remark 2 on p. 67]. For a
related ergodicity result, see [17, Theorem III-1 on p. 150].
Proof of Theorem 1. Define
sn :=
f (r) :=
n−1
X
j=0
∞
X
(I − T )T j = 1 − T n ,
(I − T )T j rj = I + T (r − 1)(1 − rT )−1 , and
j=0
fn (r) :=
n−1
X
(I − T )T j rj .
j=0
Then for all r ∈ (0, 1) and n ≥ 0, we have
(6)
ksn k ≤ ksn − fn (r)k + kfn (r) − f (r)k + kf (r)k.
By condition (4), the last term on the right hand side of (6) is bounded by
1 + CkT k uniformly for all r ∈ (0, 1). For the second term, we have
kfn (r) − f (r)k = k
X
(I − T )T j rj k ≤
j≥n
X M
rj
j
+
1
j≥n
M Xn+1 j
M n
=
r ≤
r (1 − r)−1
n + 1 j≥n j + 1
n+1
by (3). From now on, we choose rn := 1 − 1/n in (6). Then
(
n
→ M/e as n → ∞;
M n
M
1
rn (1 − rn )−1 =
1−
n
n+1
n+1
n
≤ M for all n ≥ 1.
So the second term on the right hand side of (6) is bounded with this choice
of r = rn .
The first term of the right side of inequality (6) (when choosing r = rn )
we have
n−1
X
sn − fn (rn ) =
(I − T )T j (1 − rnj ).
j=0
By the mean value theorem, there exists r0j ∈ [rn , 1) for any j > 0, such
that we can estimate
1 − rnj = jr0j−1 (1 − rn ) ≤ j(1 − rn ) =
4
j
.
n
This together with (3) yields
n−1
n−1
n−1
X
X
1X
j
j M
j
ksn − fn (rn )k ≤
k(I − T )T k ≤
≤M
1 = M.
n
nj +1
n j=0
j=0
j=0
So the sequence {sn }n≥0 is uniformly bounded from above, which is equivalent to the power boundedness of T . It is also clear from this argument
that the constants in (5) are as claimed.
If T ∈ L(X) satisfies the tauberian condition (3), then a number of conditions will be equivalent. The following theorem is analogous to [13, Theorem
2.1], except that now (3) is a standing assumption instead of power boundedness. We remark that condition (iv) constitutes a slight improvement to
Theorem 1.
Theorem 2. Assume that T ∈ L(X) satisfies the tauberian condition (3).
Then the following are equivalent:
(i) T is power bounded,
(ii) there exists 0 < δ ≤ 1 ≤ C < ∞ such that T satisfies the Ritt resolvent
condition (1) for all λ ∈ Kδ0 := {λ = 1 + reiθ |r > 0, |θ| < π2 + δ},
(iii) there exists CK < ∞ such that T satisfies the iterated Kreiss resolvent
condition
k(λ − T )−k k ≤
CK
for all
(|λ| − 1)k
|λ| > 1
and
k ∈ N,
(iv) for some k ∈ N there exists 0 < ηk ≤ 1 ≤ Ck < ∞ such that T satisfies
the kth order resolvent condition
k(λ − 1)k (λ − T )−k k ≤ Ck
for all
λ ∈ (1, 1 + ηk ),
(v) there exists CHY < ∞ such that A = T −I satisfies the Hille – Yoshida
resolvent condition
k(λ − 1)k (λ − T )−k k ≤ CHY
for all
λ>1
and
k ∈ N,
(vi) A = T − I generates an uniformly bounded, norm continuous, analytic
semigroup t 7→ eAt of linear operators,
Pn
1
j
(vii) the operators Mn := n+1
j=0 T are uniformly bounded, and
5
(viii) there exists CU A < ∞ such that T is uniformly Abel bounded, i.e.,
k(λ − 1)
n
X
λ−k−1 T k k < CU A
for all
n∈N
and λ > 1.
k=0
Proof. Claims (i) and (ii) are equivalent by Proposition 1 and an extension
result that can be found, e.g., in [11]. By estimating the Neumann series it
follows that (i) ⇒ (iii). It is trivial that (iii) ⇒ (iv).
We argue next that if condition (iv) holds with some
k ∈ N, then it holds
P∞
−k−1
j −n−j−1
with k = 1, too. Using the identity (λ − T )
= j=0 ( n+j
n )T λ
for |λ| > 1 we get from (3) the estimate
C
1
1
−k−1
k(I − T )(λ − T )
k≤
−
for λ > 1,
k (λ − 1)k (λ)k
see [13, Theorem 2.1]. From this it follows for all k ≥ 1 that
k(λ − 1)k (λ − T )−k − (λ − 1)k+1 (λ − T )−k−1 k ≤
C
for all λ > 1.
λ
Having shown this, it follows directly from Theorem 1 that (iv) ⇒ (i),
bacause (4) is only used near point 1 in the proof of Theorem 1.
Assume (i). The semigroup generated by T − I is norm continuous
P
j j
since T ∈ L(X), and it is uniformly bounded since ketT k ≤ j≥0 kT j!kt ≤
supj≥0 kT j k · et for all t ≥ 0. Moreover, it is not difficult to see that (3)
implies kAetA k ≤ M t−1 (1 − e−t ) where A := T − I. This implies that etA
is analytic, by a slight generalization of [14, Theorem 5.2]. We conclude
that (i) ⇒ (vi). We have (vi) ⇒ (v) the classical theorem of E. Hille and
K. Yoshida on strongly continuous semigroups; see, e.g., [5]. The implication
(v) ⇒ (iv) is trivial. We have now shown that all the conditions (i) – (vi)
are equivalent.
It is trivial that (i) ⇒ (vii). The implication (vii) ⇒ (viii) is given in [4].
That (viii) ⇒ (iv) with k = 1 is trivial, and the proof is now complete.
A number of remarks are now in order. That (iii) vith k = 1 implies
(i) was first proved by using a Cauchy integration argument, see [19]. It is
shown in [10, Theorem 3.1], conditions (vii) and (viii) are equivalent even
without assuming (3). Likewise, condition (vii) implies always (iv) (with
k = 1) by [13, Theorem 4.2].
If T satisfies the following weaker form of tauberian condition
√
sup n + 1k(I − T )T n k < ∞,
n≥0
6
then (i) ⇔ (v) follows from [15, Theorem III.5 on p. 68]. However, then the
bounded semigroup generated by I − T need not be analytic as can be seen
by applying [13, Theorem 2.3] on a contraction T with σ(T ) = D.
Recall from the ergoficity theory the identity (n + 1)(Mn − Mn−1 ) =
T n − Mn−1 where Mn is defined as in claim (vii). Thus claims (i) and
(vii) are equivalent if lim supn→∞ (n + 1)kMn − Mn−1 k < ∞. Note that (3)
implies lim supn→∞ (n + 1)kMn − Mn−1 k < ∞. All this was pointed out in
[18, Proposition 6.1].
If the Banach space X is reflexive, condition (i) implies that the operator sequence Mn in claim (vii) converges in strong operator topology to a
bounded projection without assuming (3). Then (vii) holds by the Banach–
Steinhaus theorem.
We remark that the tauberian condition (3) implies kT n k = O(ln n),
and by [6, Theorem 3.3], the growth can really be there for an operator in
a Banach space. Condition (3) “almost” implies condition (iv) of Theorem
2, too. Indeed, as (1 − r)(I − rT )−1 = I − r(I − T )(I − rT )−1 for all |r| < 1,
we obtain the estimate
k(1 − r)(I − rT )−1 k ≤ 1 + M
X |r|j+1
j≥0
j+1
= 1 + M ln
1
1 − |r|
for all 0 ≤ |r| < 1. Setting r = 1/λ for λ > 1 gives now
k(λ − 1)(λ − T )−1 k ≤ 1 + M ln
λ
.
λ−1
Hence k(λ − T )−1 k = O ((λ − 1) ln (λ − 1)) as λ → 1+. Again, the logarithmic term can really be present on the right hand side, as can be seen by
studying more carefully the example given in [6, Theorem 3.3].
Finally, the tauberian condition (3) “almost” implies condition (vi) of
Theorem 2. Indeed, as kAetA k ≤ M t−1 (1 − e−t ) where A := T − I, and the
function t 7→ t−1 (1 − e−t ) is decreasing for t ≥ 0, it follows that
Z t
tA
ke k ≤ 1 +
kAetA k dt ≤ 1 + M + M (1 − e−1 ) ln t for all t ≥ 1.
0
Acknowledgment
The authors are grateful to Professor Jaroslav Zemanek and an anonymous
reviewer for improving Theorem 2.
7
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