Resolvent as Laplace transform of a semigroup

Remarks

Constants $M$ and $\omega$ as above exist for every $C_0$-semigroup, by the uniform boundedness principle. Let $R(\lambda) x$ denote the integral; it converges absolutely by the bound on $\norm{T(t)}$, which also gives the norm estimate. A direct computation with the semigroup law shows $\frac{1}{h}(T(h) - I) R(\lambda) x \to \lambda R(\lambda) x - x$ as $h \to 0^+$, so $R(\lambda) x \in D(A)$ and $(\lambda I - A) R(\lambda) x = x$. For $x \in D(A)$, $A$ commutes with the integral, so $R(\lambda)(\lambda I - A) x = x$. Example: for a square matrix $A$ and $T(t) = e^{t A}$, the formula reads $(\lambda I - A)^{-1} = \int_0^\infty e^{-\lambda t} e^{t A} \, dt$ when $\operatorname{Re} \lambda$ exceeds the real parts of all eigenvalues of $A$.

Going deeper

For a student

The formula is the bridge between semigroups and their generators. It gives the necessity part of the Hille–Yosida theorem, and the norm bound $M/(\operatorname{Re} \lambda - \omega)$ is exactly the resolvent estimate in that theorem. Conversely, one recovers $T(t)$ from the resolvent by a complex inversion integral or by $T(t) x = \lim_{n \to \infty} \left(\frac{n}{t} R(\frac{n}{t})\right)^n x$, an operator form of Post's inversion formula.