Linear Algebra, Theory And Applications, 2012a

C.6. GENERAL GEOMETRIC THEORY 433
Then
( ) 1
t −M f (t) =f M (t)+O
t
where the last term means that tO ( 1
t
)
is bounded. Then
f M (t) =
m∑
c j e ib jt
j=1
It can’t be the case that all the c j are equal to 0 because then M would not be the highest
power exponent. Suppose c k ≠0. Then
∫
1 T
lim t −M f (t) e −ibkt dt =
T →∞ T 0
m∑
j=1
∫
1 T
c j e i(b j−b k )t dt = c k ≠0.
T 0
Letting r = |c k /2| , it follows ∣ ∣t −M f (t) e ∣ −ib kt >rfor arbitrarily large values of t. Thus it
is also true that |f (t)| >rfor arbitrarily large values of t.
Next consider the general case in which σ is given above. Thus
e −σt f (t) =
∑
p j (t) e bjt + g (t)
j:a j=σ
where lim t→∞ g (t) =0,g(t) being of the form ∑ s p s (t) e (a s−σ+ib s )t where a s −σ0 such that
∣
∣e −σt f (t) ∣ ∣ >r
for arbitrarily large values of t.
Next here is a Banach space which will be useful.
Lemma C.6.3 For γ>0, let
E γ = { x ∈ BC ([0, ∞), F n ):t → e γt x (t) is also in BC ([0, ∞), F n ) }
and let the norm be given by
Then E γ is a Banach space.
||x|| γ
≡ sup {∣ ∣e γt x (t) ∣ ∣ : t ∈ [0, ∞) }
Proof: Let {x k } be a Cauchy sequence in E γ . Then since BC ([0, ∞), F n )isaBanach
space, there exists y ∈ BC ([0, ∞), F n ) such that e γt x k (t) converges uniformly on [0, ∞) to
y (t). Therefore e −γt e γt x k (t) =x k (t) converges uniformly to e −γt y (t) on[0, ∞). Define
x (t) ≡ e −γt y (t) . Then y (t) =e γt x (t) and by definition,
||x k − x|| γ
→ 0.