and so for p large enough, this term on the right in the above inequality is less than
ε. Since ε is arbitrary, this shows the partial sums of 13.10 are a Cauchy sequence.
Therefore by Corollary 13.0.7 it follows that these partial sums converge. As to the last
claim,
Hence this scalar valued function equals a constant and so the constant must be
(x,y)
_{X}.
Hence for all x,y,
(Φ (− t)Φ (t)x − x,y)
_{X} = 0 for all x,y and this is so in particular for
y = Φ
(− t)
Φ
(t)
x − x which shows that Φ
(− t)
Φ
(t)
= I. ■
As a special case, suppose λ ∈ ℂ and consider
∑∞ tkλk
k!
k=0
where t ∈ ℝ. In this case, A_{k} =
tkλk-
k!
and you can think of it as being in ℒ
(ℂ,ℂ )
. Then the
following corollary is of great interest.
Corollary 13.4.4Let
∞∑ tkλk ∞∑ tkλk
f (t) ≡ k! ≡ 1 + k!
k=0 k=1
Then this function is a well defined complex valued function and furthermore, it satisfies the initialvalue problem,
y′ = λy,y(0) = 1
Furthermore, if λ = a + ib,
|f|(t) = eat.
Proof:The first part is a special case of the above theorem. Note that for f
(t)
= u
(t)
+ iv
(t)
,
both u,v are differentiable. This is because
-- --
f-+-f f −-f
u = 2 ,v = 2i .
Then from the differential equation,
(a + ib)(u +iv) = u ′+ iv′
and equating real and imaginary parts,
u′ = au− bv,v′ = av+ bu.
Then a short computation shows
( 2 2)′ ′ ′ ( 2 2)
u + v = 2uu +2vv = 2u (au− bv)+ 2v(av+ bu) = 2a u + v
( 2 2) 2
u + v (0) = |f| (0) = 1
Now in general, if
′
y = cy,y(0) = 1,
with c real it follows y
(t)
= e^{ct}. To see this,
y′ − cy = 0
and so, multiplying both sides by e^{−ct} you get
d-(ye−ct) = 0
dt
and so ye^{−ct} equals a constant which must be 1 because of the initial condition y
(0)
= 1.
Thus
( 2 2) 2at
u + v (t) = e
and taking square roots yields the desired conclusion. ■
Definition 13.4.5The function in Corollary 13.4.4given by that power series is denotedas
λt
exp(λt) or e .
The next lemma is normally discussed in advanced calculus courses but is proved here for the
convenience of the reader. It is known as the root test.
Definition 13.4.6For
{an}
any sequence of real numbers
lim sup an ≡ lim (sup{ak : k ≥ n})
n→∞ n→∞
Similarly
lim ni→nf∞ an ≡ nli→m∞(inf{ak : k ≥ n})
In case A_{n}is an increasing (decreasing) sequence which is unbounded above (below) then it isunderstood that lim_{n→∞}A_{n} = ∞(−∞) respectively. Thus either of limsup or liminf can equal
+∞ or −∞. However, the important thing about these is that unlike the limit, these alwaysexist.
It is convenient to think of these as the largest point which is the limit of some subsequence of
{an}
and the smallest point which is the limit of some subsequence of
{an}
respectively.
Thus lim_{n→∞}a_{n} exists and equals some point of
[− ∞, ∞ ]
if and only if the two are
equal.
Lemma 13.4.7Let
{ap}
be a sequence of nonnegative terms and let
r = lim sup a1p∕p.
p→∞
Then if r < 1, it follows the series,∑_{k=1}^{∞}a_{k}converges and if r > 1, then a_{p}fails to converge to0 so the series diverges. If A is an n × n matrix and
p 1∕p
r = lim spu→p∞ ||A || , (13.11)
(13.11)
then if r > 1, then∑_{k=0}^{∞}A^{k}fails to converge and if r < 1 then the series converges. Note thatthe series converges when the spectral radius is less than one and diverges if the spectral radiusis larger than one. In fact, limsup_{p→∞}
Proof:Suppose r < 1. Then there exists N such that if p > N,
1∕p
ap < R
where r < R < 1. Therefore, for all such p, a_{p}< R^{p} and so by comparison with the geometric
series, ∑R^{p}, it follows ∑_{p=1}^{∞}a_{p} converges.
Next suppose r > 1. Then letting 1 < R < r, it follows there are infinitely many values of p at
which
R < a1p∕p
which implies R^{p}< a_{p}, showing that a_{p} cannot converge to 0 and so the series cannot converge
either.
To see the last claim, if r > 1, then
||Ap ||
fails to converge to 0 and so
{∑m Ak}
k=0
_{
m=0}^{∞}
is not a Cauchy sequence. Hence ∑_{k=0}^{∞}A^{k}≡ lim_{m→∞}∑_{k=0}^{m}A^{k} cannot exist. If
r < 1, then for all n large enough,
∥An∥
^{1∕n}≤ r < 1 for some r so
∥An∥
≤ r^{n}. Hence
∑_{n}
∥An∥
converges and so by Lemma 13.4.2, it follows that ∑_{k=1}^{∞}A^{k} also converges.
■
Now denote by σ
(A )
^{p} the collection of all numbers of the form λ^{p} where λ ∈ σ
(A )
.
Lemma 13.4.8σ
(Ap )
= σ
(A)
^{p}≡
{λp : λ ∈ σ (A )}
.
Proof:In dealing with σ
p
(A )
, it suffices to deal with σ
p
(J )
where J is the Jordan form of A
because J^{p} and A^{p }are similar. Thus if λ ∈ σ
p
(A )
, then λ ∈ σ
p
(J )
and so λ = α where α is one of
the entries on the main diagonal of J^{p}. These entries are of the form λ^{p} where λ ∈ σ
(A)
. Thus
λ ∈ σ
(A)
^{p} and this shows σ
p
(A )
⊆ σ
(A)
^{p}.
Now take α ∈ σ
(A )
and consider α^{p}.
p p ( p−1 p−2 p−1)
α I − A = α I + ⋅⋅⋅+ αA +A (αI − A )
and so α^{p}I − A^{p} fails to be one to one which shows that α^{p}∈ σ