Exponential of a bounded operator

Remarks

The series converges because $\norm{T^n / n!} \le \norm{T}^n / n!$, and the numbers $\norm{T}^n / n!$ have the finite sum $e^{\norm{T}}$; see absolutely convergent series in a Banach space converge. This also gives $\norm{e^T} \le e^{\norm{T}}$.

Example: let $D$ be the diagonal operator $D(x_n) = (\lambda_n x_n)$ on $\ell^2$, with $(\lambda_n)$ a bounded sequence of complex numbers. Then $D^k(x_n) = (\lambda_n^k x_n)$, so summing coordinatewise gives $e^D(x_n) = (e^{\lambda_n} x_n)$.

For $X = \C^n$, the definition gives the matrix exponential. If $ST = TS$, then $e^{S + T} = e^S e^T$; in particular $e^T$ is invertible with inverse $e^{-T}$. The identity $e^{S+T} = e^S e^T$ can fail when $S$ and $T$ do not commute, already for $2 \times 2$ matrices. The map $t \mapsto e^{tT}$ from $\R$ to bounded operators is differentiable in operator norm, with derivative $T e^{tT}$.