22.7. DIFFERENTIAL EQUATIONS IN BANACH SPACE 601

Then S is called a continuous semigroup and A is said to generate S.

Then we have the following corollary of Theorem 22.7.3. First note the following. Fort ≥ 0 and h≥ 0, if x ∈ D(A) , the semigroup identity implies

limh→0

S (t +h)x−S (t)xh

= limh→0

S (t)S (h)x− x

h= S (t) lim

h→0

S (h)x− xh

≡ S (t)Ax

As shown above, L (X ,X) is a Banach space with the operator norm whenever X is aBanach space.

Corollary 22.7.5 Let X be a Banach space and let A ∈L (X ,X) . Let S (t) be the solu-tion in L (X ,X) to

S′ (t) = AS (t) , S (0) = I, t ≥ 0 (22.29)

Then t → S (t) is a continuous semigroup whose generator is A. In this case that A isactually defined on all of X, not just on a dense subset. Furthermore, in this case whereA ∈ L (X ,X), S (t)A = AS (t) . If T (t) is any semigroup having A as a generator, thenT (t) = S (t). Also you can express S (t) as a power series, S (t) = ∑

∞n=0

(At)n!

n.

Proof: The solution to the initial value problem 22.29 exists on [−b,b] for all b soit exists on all of R thanks to the uniqueness on every finite interval. First consider thesemigroup property. Let Ψ(t)≡ S (t + s) ,Φ(t)≡ S (t)S (s) . Then

Ψ′ (t) = S′ (t + s) = AS (t + s) = AΨ(t) , Ψ(0) = S (s)

Φ′ (t) = S′ (t)S (s) = AS (t)S (s) = AΦ(t) , Φ(0) = S (s)

By uniqueness, Φ(t) = Ψ(t) for all t ≥ 0. Thus S (t)S (s) = S (t + s) = S (s)S (t) . Nowfrom this, for t > 0

S (t)A = S (t) limh→0

S (h)− Ih

= limh→0

S (t)S (h)− I

h= lim

h→0

S (h)− Ih

S (t) = AS (t) .

As to A being the generator of S (t) , letting x ∈ X , then from the differential equationsolved,

limh→0+

S (h)x− xh

= limh→0+

1h

∫ h

0AS (t)xdt = AS (0)x = Ax.

If T (t) is a semigroup generated by A then for t > 0,

T ′ (t)≡ limh→0

T (t +h)−T (t)h

= limh→0

T (h)− Ih

T (t) = AT (t)

and T (0) = I. However, uniqueness applies because T and S both satisfy the same initialvalue problem and this yields T (t) = S (t).

To show the power series equals S (t) it suffices to show it satisfies the initial valueproblem. Using the mean value theorem,

∞

∑n=0

An ((t +h)n− tn)

n!=

∞

∑n=1

An (t +θ n (h))n−1

(n−1)!