14.3. SERIES AND SEQUENCES OF LINEAR OPERATORS 395

so it converges by Lemma 14.3.2.

Φ(t +h)−Φ(t)h

=1h

∞

∑k=0

((t +h)k− tk

)Ak

k!

=1h

∞

∑k=0

(k (t +θ kh)k−1 h

)Ak

k!=

∞

∑k=1

(t +θ kh)k−1 Ak

(k−1)!

this by the mean value theorem. Note that the series converges thanks to Lemma 14.3.2.Here θ k ∈ (0,1). Thus∥∥∥∥∥Φ(t +h)−Φ(t)

h−

∞

∑k=1

tk−1Ak

(k−1)!

∥∥∥∥∥=∥∥∥∥∥∥

∞

∑k=1

((t +θ kh)k−1− tk−1

)Ak

(k−1)!

∥∥∥∥∥∥=

∥∥∥∥∥∥∞

∑k=1

((k−1)(t + τkθ kh)k−2

θ kh)

Ak

(k−1)!

∥∥∥∥∥∥= |h|∥∥∥∥∥∥

∞

∑k=2

((t + τkθ kh)k−2

θ k

)Ak

(k−2)!

∥∥∥∥∥∥≤ |h|

∞

∑k=2

(|t|+ |h|)k−2 ∥A∥k−2

(k−2)!∥A∥2 = |h|e(|t|+|h|)∥A∥ ∥A∥2

so letting |h|< 1, this is no larger than |h|e(|t|+1)∥A∥ ∥A∥2. Hence the desired limit is valid.It is obvious that AΦ(t) = Φ(t)A. Also the formula shows that

Φ′ (t) = AΦ(t) = Φ(t)A, Φ(0) = I.

Now consider the claim about Φ(−t) . The above computation shows that

Φ′ (−t) = AΦ(−t)

and so ddt (Φ(−t)) =−Φ′ (−t) =−AΦ(−t). Now let x,y be two vectors in X . Consider

(Φ(−t)Φ(t)x,y)X

Then this equals (x,y) when t = 0. Take its derivative.((−Φ

′ (−t)Φ(t)+Φ(−t)Φ′ (t))

x,y)

X

= ((−AΦ(−t)Φ(t)+Φ(−t)AΦ(t))x,y)X

= (0,y)X = 0

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 fory = Φ(−t)Φ(t)x− x which shows that Φ(−t)Φ(t) = I. ■

In fact, one can prove a group identity of the form Φ(t + s) = Φ(t)Φ(t) for all t,s ∈R.

Corollary 14.3.4 Let Φ(t) be given as above. Then Φ(t + s) = Φ(t)Φ(s) for any s, t ∈R.