# Proving one of the properties of a strongly continuous semigroup

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

I want to show that $T(t+\tau)=T(t)T(\tau)$.

So we have that $\displaystyle T(t+\tau)=e^{A(t+\tau)}=\sum_{n=0}^{\infty}\frac{(A(t+\tau))^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(At)^{n}+(A\tau)^{n}}{n!}$.

I don't really know where to go from here.

• You need to fix the last equation first. $(A(t+\tau))^n=A^n(t+\tau)^n=A^n\sum_{k=0}^{n}\binom{n}{k}t^k\tau^{n-k}$. Then multiply the series for $e^{At}$ and $e^{A\tau}$ and compare. Mar 27 '15 at 14:33

## 1 Answer

You made a mistake in the last step : $(A(t + \tau))^n \neq (At)^n + (A\tau)^n$. Note

$$[A(t + \tau)]^n = (At + A\tau)^n = \sum_{k = 0}^n\binom{n}{k} (At)^k(A\tau)^{n-k}$$

for all $n\in \Bbb N\cup \{0\}$, thus

\begin{align}\sum_{n = 0}^\infty \frac{[A(t + \tau)]^n}{n!} &= \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{1}{n!}\binom{n}{k}(At)^k(A\tau)^{n-k}\\ & = \sum_{k = 0}^\infty \sum_{n = k}^\infty \frac{(At)^k(A\tau)^{n-k}}{k!(n-k)!}\\ & = \sum_{k = 0}^\infty \sum_{n = 0}^\infty \frac{(At)^k(A\tau)^n}{k!n!}\\ & = \sum_{k = 0}^\infty \frac{(At)^k}{k!}\sum_{n = 0}^\infty \frac{(A\tau)^n}{n!}\\ & = T(t)T(\tau). \end{align}

N.B. You could also write $[A(t + \tau)]^n = A^n\sum_{k = 0}^n \binom{n}{k}t^k\tau^{n-k}I$, and solve the problem from there. The method I used can be used to show more generally that $e^{X + Y} = e^{X}e^Y$ whenever $X$ and $Y$ commute.