13.4. SERIES AND SEQUENCES OF LINEAR OPERATORS 339

Then the series converges for each t ∈ R. Also

Φ′ (t) ≡ limh→0

Φ (t+ h)− Φ (t)

h=

∞∑k=1

tk−1Ak

(k − 1)!= A

∞∑k=0

tkAk

k!= AΦ (t)

Also AΦ (t) = Φ (t)A and for all t,Φ (t) Φ (−t) = I so Φ (t)−1

= Φ(−t), Φ (0) = I. (It isunderstood that A0 = I in the above formula.)

Proof: First consider the claim about convergence.

∞∑k=0

∥∥∥∥ tkAk

k!

∥∥∥∥ ≤∞∑k=0

|t|k ∥A∥k

k!= e|t|∥A∥ <∞

so it converges by Lemma 13.4.2.

Φ (t+ h)− Φ (t)

h=

1

h

∞∑k=0

((t+ h)

k − tk)Ak

k!

=1

h

∞∑k=0

(k (t+ θkh)

k−1h)Ak

k!=

∞∑k=1

(t+ θkh)k−1

Ak

(k − 1)!

this by the mean value theorem. Note that the series converges thanks to Lemma 13.4.2.Here θk ∈ (0, 1). Thus∥∥∥∥∥Φ (t+ h)− Φ (t)

h−

∞∑k=1

tk−1Ak

(k − 1)!

∥∥∥∥∥ =

∥∥∥∥∥∥∞∑k=1

((t+ θkh)

k−1 − tk−1)Ak

(k − 1)!

∥∥∥∥∥∥=

∥∥∥∥∥∥∞∑k=1

((k − 1) (t+ τkθkh)

k−2θkh)Ak

(k − 1)!

∥∥∥∥∥∥ = |h|

∥∥∥∥∥∥∞∑k=2

((t+ τkθkh)

k−2θk

)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 d

dt (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