exp(S) exp(T) = exp(S+T) for commuting operators

Question

The standard way to prove the exponential law for two bounded commuting operators $S, T$ $$ \exp(S)\exp(T) = \exp(S+T) $$ is to pass by the binomial formula and the power series of $\exp(.)$. I wonder whether one can prove it alternatively by a (Dunford-Riesz) functional calculus argument i.e. a resolvent-based proof, either directly, via $$ \exp(S)\exp(T) = \tfrac{1}{(2\pi i)^2 } \int_{|z|=r} \int_{|\lambda|=R} e^{\lambda+z} R(\lambda, S) R(z, T)\,d\lambda dz $$ for some $r> \|T\|, R > \max(r, \|S\|)$ or maybe by an operator matrix approach (for example the functional calculus of $A= \left( \begin{array}{rr} S & I \\ 0 & T \end{array}\right)$).

Answer 1

An alternative way of proving the exponential law for commuting operators is the following: Define $F(t) = \exp(tT)\exp(tS)$ for every $t\in\mathbb{R}$. Note that the commutativity of $S$ and $T$ implies that $S$ commutes with $\exp(tT)$. Using the product rule you can calculate $$\tfrac{d}{dt} F(t) = T\exp(tT)\exp(tS) + \exp(tT)S\exp(tS) = (S+T)F(t),\quad t\in\mathbb{R}.$$ Since $F(0)=I$ you obtain in particular that $F(1)=\exp(T+S)$.

Answer 2

Also: for bounded operators, $(I+T/n)^n$ converges to $e^T$ uniformly on bounded sets (however, uniform convergence on compact sets will suffice). So if $T$ commutes with $S$, $$e^Te^S=\lim_{n\to+\infty}\Big(I+\frac{T}{n}\Big)^n\Big(I+\frac{S}{n}\Big)^n=\lim_{n\to+\infty}\bigg(I+\frac{T+S+\frac{TS}{n}}{n}\bigg)^n=e^{S+T}\, .$$