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Chapter 2. Stability
If there does not exist such a nε, then
Thus
1
N
N�
k=1
�A k dx0� ≥ ε
, for all k ≥ 0.
M
�A k dx0� 2 ≥ 1
N
N�
k=1
ε 2
M
ε2
= , for all N ≥ 0.
2 M 2
This contradicts condition (2.34). Hence, there exists a nε satisfying condition
(2.35).
Thus, for all N > nε,
�A N d x0� = �A N−nε
d A nε
d x0� < M ε
M
= ε.
This holds for all ε > 0, so limN→∞ A N d x0 = 0.
To prove the ”⇐” implication, we choose an ε > 0. Since limk→∞ A k d x0 = 0,
there exists a nε such that for all k > nε, �A k d x0� < ε.
1
N
N�
k=1
�A k dx0� 2 ds = 1
N
nε �
k=1
≤ M 2 �x0� 2 nε
N
�A k dx0� 2 ds + 1
N
N − nε
+ ε
N
N�
k=nε+1
�A k dx0� 2 ds
For N large enough, the left-hand side is smaller than 2ε. This holds for all
ε > 0, so the left-hand side goes to 0, as N goes to ∞. �
We end this section with a characterization of the boundedness of an operator
sequence.
Lemma 2.23 Let Ad be a bounded operator on the Hilbert space X. Then
the operator sequence (A n d )n≥0 is bounded if and only if the following estimates
hold for all x ∈ X:
sup (1 − r)
r∈[0,1)
sup (1 − r)
r∈[0,1)
∞�
n=0
∞�
n=0
�A n d x0�r 2n ≤ M1�x0� 2 , (2.36)
�A ∗n
d x0�r 2n ≤ M2�x0� 2 . (2.37)
Furthermore, if equations (2.36) and (2.37) hold, then the bound of (A n d )n≥0
is given by
�A n d � ≤ e � M1M2.
For the proof we refer to Guo and Zwart [21, Theorem 3.2 and Remark 3.3].
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