# Finding the infinitesimal generation of a strongly continuous semigroup

Let $X$ be a Hilbert space, $A\in\mathcal{L}(X)$ and $\displaystyle T(t)=e^{At}=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}$.

I have already shown that $T(t)$ defines a $C_{0}$ semigroup. But now I need to find the infinitesimal generator of $T(t)$.

I would appreciate any hints and/or tips on how to tackle this problem.

When you have to differentiate a series, it often pays to look at the series of derivatives, integrate, and justify changing the order of integration and summation. It's a trick that helps you avoid interchanging differentiation and summation, which seems always to be more difficult to justify. For example, you know the derivative should be $$F(t)=\sum_{n=0}^{\infty}A\frac{(tA)^{n}}{n!}.$$ This series converges uniformly in $t$ on any interval $t \in [a,b]$. The resulting function $F : [a,b]\rightarrow\mathcal{L}(X)$ is continuous and, therefore, the following is justified \begin{align} \int_{0}^{t}F(t')dt' & = \sum_{n=0}^{\infty}\int_{0}^{t}A\frac{(t'A)^{n}}{n!}dt' \\ & = \sum_{n=0}^{\infty}\frac{t^{n+1}A^{n+1}}{(n+1)!} \\ & =\sum_{n=0}^{\infty}\frac{(tA)^{n}}{n!}-I \end{align} Because $F$ is continuous, the derivative of the left side exists in the operator topology, which means that the derivative of the series exists in the operator topology, and $$F(t) = \frac{d}{dt}\int_{0}^{t}F(t')dt' = \frac{d}{dt}\sum_{n=0}^{\infty}\frac{(tA)^{n}}{n!}.$$ So the generator is $A$ because $$\frac{d}{dt}\sum_{n=0}^{\infty}\frac{(tA)^{n}}{n!}=A\sum_{n=0}^{\infty}\frac{(tA)^{n}}{n!}.$$

If $B$ denotes the infinitesimal generator of $T$ then we know that:

If $f \in \mathcal{D}(B)$ and $t \geq 0$, then $T(t)f \in \mathcal{D}(B)$ and

$$B T(t) f = \frac{d}{dt}T(t)f .$$

(c.f. Ethier and Kurtz page $9$).

Here $\mathcal{D}(B)$ is the domain of the operator $B$.

So you just need to check the domain is the whole space, and justify the interchange of the order of summation and $\frac{d}{dt}$.