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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...

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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 $\mathcal L(E)$.

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  • $\begingroup$ Thanks!! Just one further question. Do you know of any simpler way to prove this? Actually, the exercise consists of two parts: verify $\exp\in C^1(\mathcal{L}(E),\mathcal{L}(E))$ and show $\partial\exp(0)=\operatorname{id}_E$. So I guess you don't have to calculate the derivative explicitly to show that it's continuously differentiable. $\endgroup$ – Colescu Jun 20 '18 at 11:06
  • $\begingroup$ A way to do it is to consider power series $\sum a_n u_(x)$ where $u_n(x)=x^n$ is the homogeneous polynomial (of the Banach algebra) of debree $n$, to compute its $p$th-derivative $u_n^{(p)}$ and to consider the power series $S_p(x) = \sum a_n u_n^{(p)}(x)$. You'll be able to prove that $S_{p-1}^\prime(x) = S_p(x)$ inside the radius of convergence which is the same for all the $S_p$. By induction, you'll be able to prove that $\exp\in C^\infty(\mathcal{L}(E),\mathcal{L}(E))$. Proving $\partial\exp(0)=\operatorname{id}_E$ is not so complicated as $0$ commutes with all the other elements. $\endgroup$ – mathcounterexamples.net Jun 20 '18 at 12:44

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