Source and/or detailed proof for $\exp(\lambda (A + B)) = \exp(\lambda A)\exp(\lambda B)$ for commuting elements in a Banach algebra

Question

Every time I need to revisit the proof that $\exp(\lambda (A + B)) = \exp(\lambda A)\exp(\lambda B), \lambda \in\mathbb{K}$ for commuting elements in a Banach algebra, I find myself struggling to remember the key details. I can recall certain aspects, such as why the series converges in norm (which justifies the whole equality), but I have difficulty recalling the exact justification for other details. For instance, I can't seem to remember why we can "just permute" the order of the double series $\displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(\lambda A)^k}{k!}\frac{(\lambda B)^{n-k}}{(n-k)!}$. As dull and uninteresting it may sound, I am looking for a proof which is as explicit as possible without being "too tedious". That is, I would like to see a treatment of this proof with indicator functions and the such to make life easier. Additionally, I currently can't recall any written sources that discuss this theorem. Therefore I am asking either for a reference to a book discussing the prior claim in detail (including why it makes sense to permute the two series; let the convergence theorems from measure theory be allowed if that makes it any easier) or a proof of the said claim, in detail.

Any help would be greatly appreciated. Thank you in advance!

Answer 1

Here is a fairly straightforward argument.

First of all, since $\lambda(A+B)=\lambda A+\lambda B$, keeping the $\lambda$ throughout the argument is unnecessary, as it is enough to show that $e^{A+B}=e^Ae^B$.

Second, you have (note that there is a typo in the formula in your question) $$ \sum_{n=0}^\infty\sum_{k=0}^n\Big\|\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\Big\|\leq\sum_{n=0}^\infty\sum_{k=0}^n\frac{\|A\|^k}{k!}\,\frac{\|B\|^{n-k}}{(n-k)!}=e^{\|A\|+\|B\|}<\infty, $$ so the double series converges absolutely. This guarantees that the double series converges, and it allows one to exchange the series.

After the above, the proof carries exactly as in the scalar case: if $AB=BA$, $\def\abajo{\\[0.3cm]}$ \begin{align} e^{A+B} &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{n=0}^\infty\sum_{k=0}^\infty\,1_{[0,n]}(k)\,\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{k=0}^\infty\sum_{n=0}^\infty\,1_{[0,n]}(k)\,\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{k=0}^\infty\sum_{n=0}^\infty\,1_{[k,\infty)}(n)\,\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{k=0}^\infty\sum_{n=k}^\infty\,\frac{A^k}{k!}\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{k=0}^\infty \frac{A^k}{k!}\,\sum_{n=k}^\infty\,\frac{B^{n-k}}{(n-k)!}\abajo &=\sum_{k=0}^\infty \frac{A^k}{k!}\,\sum_{m=0}^\infty\,\frac{B^{m}}{m!}\abajo &=e^A\,e^B. \end{align}

Answer 2

As with most mathematical struggles, even this one turned out to be something I can answer to, only after having explicitly asked help for it to make the whole process needlessly complicated. In any case, one way to switch the two sums is the following: Define $f(n,\lambda,A) := \frac{(\lambda A)^n}{n!}$ for $n\in\mathbb{N},\lambda\in\mathbb{K},A\in\mathcal{A}$. Then

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,\lambda,A)f(n-k,\lambda,B) = \lim_{N\to\infty}\sum_{n=0}^N\sum_{k=0}^nf(k,\lambda,A)f(n-k,\lambda,B)\leftrightarrow$$

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,\lambda,A)f(n-k,\lambda,B) = \lim_{N\to\infty}\sum_{k=0}^\infty\sum_{n=0}^N\chi_{\{k\leq n\}}f(k,\lambda,A)f(n-k,\lambda,B)\leftrightarrow$$

Then, define $S_N(k) := \sum_{n=0}^N\chi_{\{k\leq n\}}f(k,\lambda,A)f(n-k,\lambda,B)$. It follows by continuity of the norm that $\forall N,k\in\mathbb{N}:||S_N(k)||\leq \exp(\lambda (||A|| + ||B||)) < \infty$. Hence by the Lebesgue Dominated Convergence theorem with respect to the counting measure,

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,t,A)f(n-k,\lambda,B) = \sum_{k=0}^\infty\lim_{N\to\infty}\sum_{n=0}^N\chi_{\{k\leq n\}}f(k,\lambda,A)f(n-k,\lambda,B)\leftrightarrow$$

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,t,A)f(n-k,\lambda,B) = \sum_{k=0}^\infty\sum_{n=0}^\infty\chi_{\{0\leq n - k\}}f(k,\lambda,A)f(n-k,\lambda,B)$$

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,t,A)f(n-k,\lambda,B) = \sum_{k=0}^\infty\sum_{n=k}^\infty f(k,\lambda,A)f(n-k,\lambda,B)$$

Then, by taking $l = n - k$ we get

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,t,A)f(n-k,\lambda,B) = \sum_{k=0}^\infty\sum_{l=0}^\infty f(k,\lambda,A)f(l,\lambda,B)\leftrightarrow$$

$$\sum_{n=0}^\infty\sum_{k=0}^nf(k,t,A)f(n-k,\lambda,B) = \sum_{k=0}^\infty f(k,\lambda, A)\sum_{l=0}^\infty f(l,\lambda,B)$$

from which the claim follows.