Holomorphic functional calculus

Remarks

The integrand is continuous, since $\lambda \mapsto (\lambda e - a)^{-1}$, the resolvent, is defined and holomorphic off $\sigma(a)$. By Cauchy's theorem for $A$-valued holomorphic functions, $f(a)$ does not depend on the choice of $\Gamma$. Example: for $f(\lambda) = \lambda^n$ and $\Gamma$ a large circle $\abs{\lambda} = R$ with $R > \norm{a}$, expand $(\lambda e - a)^{-1} = \sum_{k \ge 0} \lambda^{-k-1} a^k$; only the term $k = n$ survives the integral, so $f(a) = a^n$. Non-example: $f(\lambda) = \bar\lambda$ is not holomorphic, so the definition does not apply; for non-normal operators there is no reasonable way to define $\bar a$ by this method. The calculus is an algebra homomorphism, it is continuous for locally uniform convergence of $f$, and $\sigma(f(a)) = f(\sigma(a))$. For a normal operator on a Hilbert space it agrees with the continuous functional calculus on holomorphic functions. When $f$ is $1$ near one closed and open part of $\sigma(a)$ and $0$ near the rest, $f(a)$ is the Riesz projector of that part.