Link to the post: A problem regarding the "exponential" of a continous linear mapping

Using notations from this post of mine, I have trouble proving that $e^A \circ e^B = e^B \circ e^A = e^{A+B}$ where $A$ and $B$ sastifies $A \circ B = B \circ A$.

Were them be series in $\mathbb{R}$ or $\mathbb{C}$, I would have multiplied the series together to get the desired result.

Therefore, I'm wondering if I could build the "multiplication" of series in the same manner. However, I have no clue how this can be done. Or maybe I'm going a wrong direction?

Please give me a hint to a correct way. Any help is greatly appreciated. Thank you for reading.


By definition, $$e^{A+B}x=\sum_{n=0}^{\infty}\frac{(A+B)^nx}{n!}=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}C(n,k)A^kB^{n-k}x $$ Since this series is absolutely convergent, we may exchange the summation order, obtaining $$ \sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{C(n,k)A^k B^{n-k}x}{n!}=\sum_{k=0}^{\infty}\frac{A^{k}}{k!}\sum_{n=k}^{\infty}\frac{B^{n-k}x}{(n-k)!}=\sum_{k=0}^{\infty}\frac{A^k}{k!}e^Bx=e^Ae^Bx$$ Where moving $A^k$ out of the inner summation is justified because $A^k$ is continuous and the inner summation is a convergent series . Moreover, when $A$ and $B$ commute (i.e. $AB=BA$) we have $e^Ae^Bx=e^Be^Ax$ since they are both equal to $e^{A+B}x=e^{B+A}x$.


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