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Two linear operators $A$ and $B$ commute when $AB=BA$, with all products taken on the stated operator domains. For finite matrices, or bounded linear operators on a Banach space, commutation impliesThe binomial theorem applies because the factors commute, and absolute convergence of the operator exponential series permits regrouping its terms. For unbounded operators, a formal equality on one common operator domain alone does not justify this exponential identity; the existence and commutation of the corresponding exponential operators require separate hypotheses.

An unbounded operator is a linear map $T:D(T)\subseteq X\to Y$ for which no finite constant $C$ bounds $\Vert Tx{\Vert }_Y\le C\Vert x{\Vert }_X$ on its domain. The domain and norms are therefore part of its specification.

On complex $L^2(0,2\pi )$, take $T=-id/dx$ with periodic $H^1$ domain. The functions $e_n(x)=e^{inx}/\sqrt{2\pi }$ have norm one, while $T{e}_n=n{e}_n$ has norm $|n|$. Hence no such constant exists. Differentiation can nevertheless be bounded when the source is equipped with an $H^1$ norm. An operator's unboundedness thus cannot be inferred from its formula without stating the spaces and domain.