Linear Algebra, Theory And Applications, 2012a

352 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
and by the mean value theorem this equals an expression of the following form where θ k is
a number between 0 and 1.
∞∑ k (t + θ k h) k−1 λ k ∞∑ (t + θ k h) k−1 λ k
=
k!
(k − 1)!
k=1
k=1
∞∑ (t + θ k h) k λ k
= λ
k!
It only remains to verify this converges to
k=0
∞∑ t k λ k
λ
k!
k=0
= λf (t)
as h → 0.
∣ )
∞∑ (t + θ k h) k λ k ∞∑ t k λ k
∣∣∣∣∣ ∞ −
∣ k!
k! ∣ = ∑
((t + θ k h) k − t k
k!
k=0
k=0
and by the mean value theorem again and the triangle inequality
∞∑ k |(t + η
≤
k )| k−1 |h||λ| k
∞ ∣
k! ∣ ≤|h| ∑ k |(t + η k )| k−1 |λ| k
k!
k=0
where η k is between 0 and 1. Thus
k=0
k=0
λ k
∣
∞∑ k (|t| +1) k−1 |λ| k
≤|h|
k!
k=0
= |h| C (t)
It follows f ′ (t) =λf (t) . This proves the first part.
Next note that for f (t) =u (t)+iv (t) , both u, v are differentiable. This is because
Then from the differential equation,
u = f + f
2
and equating real and imaginary parts,
,v= f − f .
2i
(a + ib)(u + iv) =u ′ + iv ′
u ′ = au − bv, v ′ = av + bu.
Then a short computation shows
(
u 2 + v 2) ′ (
=2a u 2 + v 2) , ( u 2 + v 2) (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