 The formula shows ("z"1−"T")−1 and ("z"2−"T")−1 commute, which hints at the fact that the image of the functional calculus will be a commutative algebra.Letting "z"2 → "z"1 shows the resolvent map is (complex-) differentiable at each "z"1 ∈ ρ("T"); so the integral in the expression of functional calculus converges in "L"("X").Stronger statement than differentiability can be made regarding the resolvent map.