Is matrix exponentiation $\exp:\mathcal{L}(E)\to\mathcal{L}(E)$ continuously Fréchet differentiable?

Question

This is exercise VII.3.8 in Amann & Escher, Analysis II.

Let $E$ be a Banach space. Denote by $\mathcal{L}(E)$ the space of all bounded linear maps from $E$ to itself. Then the exponential map $\exp:\mathcal{L}(E)\to\mathcal{L}(E)$ is continuously Fréchet differentiable.

For $A,B\in\mathcal{L}(E)$, if $AB\neq BA$ then the expression $e^{A+B}-e^A$ is difficult to manipulate.

I read the wikipedia page, where it says for any matrices $X,Y$ $$\|e^{X+Y}-e^X\|\leq\|Y\|e^{\|X\|}e^{\|Y\|},$$ so that $\exp$ is continuous when $E=\mathbb{C}^n$. But this does not solve the problem.

I don't know how to proceed. Any help will be apprecated! By the way, I'm not sure about whether the claim is correct, because I've already seen a few wrong exercises in this book...

Answer 1

Reference

If you denote $\mathrm{ad} X \in \mathcal L(E)$ by $$\mathrm{ad} X(H) = XH-HX$$

Then you can prove that $$D \mathrm{exp}(X).H= \mathrm{exp}(X)\sum\limits_{k=0}^\infty \frac{(-\mathrm{ad } X)^k}{(k+1)!}.H$$

See Différentielle de l'exponentielle de matrice for a reference... in French. The proof is for $E$ finite dimensional space, but can be adapted for the Banach algebra $\mathcal L(E)$.