Here I will consider the case where the matrix A has both positive and negative eigenvalues. First
here is a useful lemma.

Lemma D.6.1Suppose A is an n × n matrix and there exists δ > 0 such that

0 < δ < Re (λ1) ≤ ⋅⋅⋅ ≤ Re(λn)

where

{λ1,⋅⋅⋅,λn}

are the eigenvalues of A, with possibly some repeated. Then there exists aconstant, C such that for all t < 0,

|Φ (t)x| ≤ Ceδt|x |

Proof: I want an estimate on the solutions to the system

Φ′(t) = AΦ (t),Φ (0) = I.

for t < 0. Let s = −t and let Ψ

(s)

= Φ

(t)

. Then writing this in terms of Ψ,

Ψ′(s) = − A Ψ (s),Ψ (0) = I.

Now the eigenvalues of −A have real parts less than −δ because these eigenvalues are obtained
from the eigenvalues of A by multiplying by −1. Then by Theorem D.5.3 there exists a constant,
C such that for any x,

|Ψ (s)x| ≤ Ce− δs|x|.

Therefore, from the definition of Ψ,

|Φ (t) x| ≤ Ceδt|x |.■

Here is another essential lemma which is found in Coddington and Levinson [6]

Lemma D.6.2Let p_{j}

(t)

be polynomials with complex coefficients and let

∑m λjt
f (t) = pj (t)e
j=1

where m ≥ 1,λ_{j}≠λ_{k}for j≠k, and none of the p_{j}

(t)

vanish identically. Let

σ = max (Re (λ1),⋅⋅⋅,Re (λm )).

Then there exists a positive number, r and arbitrarily large positive values of t suchthat

e−σt|f (t)| > r.

In particular,

|f (t)|

is unbounded.

Proof: Suppose the largest exponent of any of the p_{j} is M and let λ_{j} = a_{j} + ib_{j}. First assume
each a_{j} = 0. This is convenient because σ = 0 in this case and the largest of the Re

(λj)

occurs in
every λ_{j}.

Then arranging the above sum as a sum of decreasing powers of t,

f (t) = tM fM (t)+ ⋅⋅⋅+ tf1(t)+ f0(t).

Then

( )
t−Mf (t) = fM (t)+ O 1
t

where the last term means that tO

(1)
t

is bounded. Then

∑m ibjt
fM (t) = cje
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 ∑m 1 ∫ T
lim -- t−M f (t)e−ibktdt = cj-- ei(bj−bk)tdt = ck ⁄= 0.
T→∞ T 0 j=1 T 0

Letting r =

|ck∕2|

, it follows

| |
|t−Mf (t)e− ibkt|

> r for arbitrarily large values of t. Thus it is also
true that

|f (t)|

> r for arbitrarily large values of t.

Next consider the general case in which σ is given above. Thus

−σt ∑ bjt
e f (t) = pj(t)e + g(t)
j:aj=σ

where lim_{t→∞}g

(t)

= 0, g

(t)

being of the form ∑_{s}p_{s}

(t)

e^{(as−σ+ibs)
t} where a_{s}− σ < 0.
Then this reduces to the case above in which σ = 0. Therefore, there exists r > 0 such
that

||−σt ||
e f (t) > r

for arbitrarily large values of t. ■

Next here is a Banach space which will be useful.

Lemma D.6.3For γ > 0,let

{ }
Eγ = x ∈ BC ([0,∞),Fn) : t → eγtx(t) is also in BC ([0,∞ ),Fn)