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Let $\mathcal{A}$ be a Banach algebra with $1\in\mathcal{A}$.

(a) For $a\in\mathcal{A}$ show that $\exp(a)=\sum\limits_{n=0}^\infty\frac{1}{n!}a^n$ converges in $\mathcal{A}$ and show that if $a,b\in\mathcal{A}$ satisfy $ab=ba$, then $\exp(a+b)=\exp(a)\exp(b)$.

(b) Let $a,b\in\mathcal{A}$ and $M=\max\{\|a\|, \|b\|\}$. Show that $\|\exp(a)-\exp(b)\|\le e^M\|a-b\|$ and hence that $a\to\exp(a)$ is continuous from $\mathcal{A}\to\mathcal{A}$.

I have seen some solutions on MSE which are relevant to this problem but my complex analysis is quite rusty and I'm struggling to put the pieces together, I am wondering if someone wouldn't mind walking through a "proof for dummies" so to speak. Any help is much appreciated.

The Mad Scientist
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    Suppose $a_n$ is a sequence in a Banach space so that $\sum_n |a_n|$ converges, can you see why $\sum_n a_n$ also converges? For the other part of (a) - do you know how to evaluate the product of two unconditionally convergent infinite series? ie the cauchy product – s.harp Apr 11 '21 at 20:44

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I've given an answer to part a with little machinery here (answer was for bounded linear operators, but the same exact proof follows through): Proving $\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T)$ for $T,S\in\mathcal{L}(E)$

For the next part, note that $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1}),$$ and so $$\|a^n-b^n\|\leq\|a-b\|\|a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1}\|\leq \|a-b\|nM^{n-1}.$$ Thus, $$\|\exp a-\exp b\|\leq \sum\limits_{n=1}^\infty\frac{nM^{n-1}}{n!}\|a-b\|=e^M\|a-b\|.$$

cmk
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