Exercise on Banach space

Question

Let $exp:\mathcal L(E)\to \mathcal L(E)$ the exponential map, where $E$ is a Banach space. $exp(A)=e^A= \sum_{n=0}^\infty \frac{A^n}{n!}$ and $A\in\mathcal L(E)$. I proved using partial sum $s_n= \sum_{k=0}^n \frac{A^k}{k!}$, that $||exp-s_n||_{\infty,B_R}=o(1)$ for $n\to\infty$ so $exp$ is uniform limit of continuous

function so exp is continuous. Then the question is find the differential of the map exp: $Dexp(A)[H]$. I don't know how to proceed for this second point. Any ideas?

Note:($B_R=B(0,R)$) and $$Dexp(A)[H]= \ lim_{t\to0}\frac{e^{A+tH}-e^A}{t}$$

Answer 1

Borrowed from https://fr.wikipedia.org/wiki/Exponentielle_d%27une_matrice#Application_t_%E2%86%A6_etX :

$$\begin{align}D\exp(A)(H)&=\sum_{i,j\ge0}\frac{A^iHA^j}{(i+j+1)!}\\&=\sum_{i,j\ge0}\mathrm B(i+1,j+1)\frac{A^i}{i!}H\frac{A^j}{j!}\\&=\int_0^1e^{(1-t)A}H{\rm e}^{tA}\,dt\end{align}$$ where $\mathrm B$ is the beta function.

When $AH=HA,$ this boils down to $D\exp(A)(H)=H\exp(A).$