Summability for Cauchy Product

Question

Let $\sum_{m\in M}a_m$ and $\sum_{n\in N}b_n$ be summable.

Proof that the product is summable: $\sum_{(m,n)\in M\times N}a_m b_n$

Now, let $e^{sA}:=\sum_{m\in M}\frac{s^mA^m}{m!}$ and $e^{tA}:=\sum_{n\in N}\frac{t^nA^n}{n!}$ for some bounded operator $\|A\|<\infty$.

Proof that it holds: $e^{sA}e^{tA}=e^{(s+t)A}$

Answer 1

Denote the limit by $A:=\lim_M\sum_{m\in M}a_m$ and $B:=\lim_N\sum_{n\in N}b_n$. Then: $$\|\sum_{m\in M}a_m\sum_{n\in N}b_n-AB\|\\ =\|(\sum_{m\in M}a_m-A+A)(\sum_{n\in N}b_n-B+B)-AB\| \\ =\|(\sum_{m\in M}a_m-A)(\sum_{n\in N}b_n-B)+(\sum_{m\in M}a_m-A)B+A(\sum_{n\in N}b_n-B)+AB-AB\|\\ \leq\|\sum_{m\in M}a_m-A\|\cdot\|\sum_{n\in N}b_n-B\|+\|\sum_{m\in M}a_m-A\|\cdot \|B\|+\|A\|\cdot\|\sum_{n\in N}b_n-B\|\\ <\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\quad\text{ for all } M\geq M_\epsilon\text{ and }N\geq N_\epsilon$$

Since it is summable rearrangement do not affect the sum. Thus: $$e^{sA}e^{tA}=\sum_{m=0}^\infty\frac{s^mA^m}{m!}\sum_{n=0}^\infty\frac{t^nA^n}{n!}\\ =\sum_{l=0}^\infty\sum_{k=0}^l\frac{1}{k!}\frac{1}{(l-k)!}s^kA^kt^{(l-k)}A^{(l-k)}\\ =\sum_{l=0}^\infty\frac{1}{l!}\left(\sum_{k=0}^l\binom{l}{k}s^kt^{(l-k)}\right)A^l\\ =\sum_{l=0}^\infty\frac{1}{l!}(s+t)^lA^l=e^{(s+t)A}$$

Answer 2

I think so. Assume $\sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty} |A_{m,n}|$ is finite.

A) It works if $\forall m,n A_{m,n} \ge 0$ because the sum in the triangle $\sum_{k=0}^K\sum_{l=0}^k A_{k-l,l}$ can be sandwiched between two squares $\sum_{n = 0}^{K/2} \sum_{m = 0}^{K/2} A_{m,n}$ (assume $K$ is even for ease of notation)

and $\sum_{n = 0}^{K} \sum_{m = 0}^{K} A_{m,n}$.

B) Now, use the common trick $A_{mn} = A^+_{m,n} - A^-_{m,n} $ and Dominated Convergence to show the result for arbitrary $A_{mn}$