Exponential of commuting operators with holomorphic calculus

Question

Let assume, for simplicity, a finite dimensional Banach space, and two commuting linear operators $A$ and $B$. By defining the exponential of some operator $T$ as $$ \exp(T):=\frac{1}{2\pi {\rm i}}\int_\Gamma \exp(z) (z\mathbb{I}-T)^{-1} dz, $$ where $\Gamma$ is a positively oriented contour enclosing the all points in the spectrum of $T$, I'm trying to prove the well-known identity $$ \exp(A+B)=\exp(A)\exp(B). $$ Any idea?

Answer 1

The holomorphic functional calculus is an extension of the polynomial and continuous functional calculi. This is enough to check that $$ \frac{1}{2\pi {\rm i}}\int_\Gamma \exp(z) (z\mathbb{I}-T)^{-1} dz= \sum_{k=0}^\infty \frac{T^k}{k!} $$ where the convergence is in the operator norm and $T^0=\Bbb{I}$.

Now the proof goes essentially the same way as the one you know for the exponential function expressed as a series:

Since $AB=BA$ we have that binomial theorem is valid for computing $(A+B)^j$. Thus, \begin{align} \exp(A)\exp(B) & =\left( \sum_{k=0}^\infty \frac{A^k}{k!}\right)\left( \sum_{n=0}^\infty \frac{B^n}{n!}\right)\\ & =\sum_{j=0}^\infty \left( \sum_{l=0}^j \frac{A^l}{l!}\frac{B^{j-l}}{(j-l)!}\right)\\ & =\sum_{j=0}^\infty \frac{1}{j!}\left( \sum_{l=0}^j { j \choose l}A^lB^{j-l}\right)\\ & =\sum_{j=0}^\infty \frac{(A+B)^j}{j!} \\ &= \exp(A+B) \end{align}