Laplace transform formula for the resolvent

Convergence and bound: for $0<R<R^{\mathrm{\prime }}$ the curves $e^{-\lambda t}T(t)x$ are continuous, hence Bochner integrable on compacts, and [F1] gives $\mathrm{\parallel }{J}_{R^{\mathrm{\prime }}}x-J_{R}x\mathrm{\parallel }\le {\int }_R^{R^{\mathrm{\prime }}}M{e}^{(\omega -\lambda )t}\mathrm{\parallel }x\mathrm{\parallel }dt\le M\mathrm{\parallel }x\mathrm{\parallel }\frac{e^{(\omega -\lambda )R}}{\lambda -\omega }\to 0$ as $R\to \mathrm{\infty }$. Thus $(J_{R}x{)}_R$ is Cauchy for every $x$, so $Jx:={\lim }_{R\to \mathrm{\infty }}{J}_{R}x$ exists, and the same estimate at $R=0$ gives $\mathrm{\parallel }Jx\mathrm{\parallel }\le \frac{M}{\lambda -\omega }\mathrm{\parallel }x\mathrm{\parallel }$; the map $x\mapsto Jx$ is linear.

For $x\in D(A)$ the curve $g(t):=e^{-\lambda t}T(t)x$ is differentiable with $g^{\mathrm{\prime }}(t)=-\lambda e^{-\lambda t}T(t)x+e^{-\lambda t}T(t)Ax=-e^{-\lambda t}(\lambda I-A)T(t)x$ by [F2]. The pair curve $t\mapsto (e^{-\lambda t}T(t)x,e^{-\lambda t}T(t)Ax)$ is continuous with values in the closed graph $\mathrm{\Gamma }(A)\subseteq X\oplus X$, so its Bochner integral lies in $\mathrm{\Gamma }(A)$: approximate the pair uniformly on $[0,R]$ by step functions sampled at partition points. Each simple integral is a finite linear combination of graph vectors, hence lies in the graph; the coordinate integrals converge by the norm inequality, and closedness retains their limit. Thus consequently $J_{R}x\in D(A)$ and $A{J}_{R}x={\int }_0^R{e}^{-\lambda t}T(t)Axdt$.

Hence, for $x\in D(A)$, $(\lambda I-A)J_{R}x=\lambda J_{R}x-A{J}_{R}x={\int }_0^R{g}^{\mathrm{\prime }}(t)dt\cdot (-1)$; more explicitly $\lambda J_{R}x-A{J}_{R}x={\int }_0^R{e}^{-\lambda t}(\lambda T(t)x-T(t)Ax)dt=-{\int }_0^R{g}^{\mathrm{\prime }}(t)dt=x-e^{-\lambda R}T(R)x$ by the fundamental theorem of calculus [F2], and $\mathrm{\parallel }{e}^{-\lambda R}T(R)x\mathrm{\parallel }\le M{e}^{(\omega -\lambda )R}\mathrm{\parallel }x\mathrm{\parallel }\to 0$ as $R\to \mathrm{\infty }$.

Passing to the limit $R\to \mathrm{\infty }$ in the identity of [step 2.1]: $J_{R}x\to Jx$ and $(\lambda I-A)J_{R}x\to x$; since $A$ is closed (hence $\lambda I-A$ is closed), the pair limit gives $Jx\in D(A)$ and $(\lambda I-A)Jx=x$ for every $x\in D(A)$.

Likewise $J(\lambda I-A)x={\lim }_{R\to \mathrm{\infty }}{J}_R(\lambda I-A)x={\lim }_{R\to \mathrm{\infty }}(\lambda J_{R}x-A{J}_{R}x)=x$ for every $x\in D(A)$, by the same computation as [step 2.1]; note that $(\lambda I-A)x\in X$ is a fixed vector to which the definition of $J$ applies.

$\lambda I-A$ is injective: if $(\lambda I-A)x=0$ for $x\in D(A)$, then $x=J(\lambda I-A)x=J0=0$ by [step 3.2]. It is also surjective: for $y\in X$ choose $y_n\in D(A)$ with $y_n\to y$ (density, [F2]); then $(J{y}_n)$ converges to $Jy$ and $(\lambda I-A)J{y}_n=y_n\to y$, so closedness of $\lambda I-A$ gives $Jy\in D(A)$ and $(\lambda I-A)Jy=y$.

Therefore $\lambda I-A:D(A)\to X$ is bijective with inverse $J$, which is bounded with $\mathrm{\parallel }J\mathrm{\parallel }\le M\mathrm{/}(\lambda -\omega )$; hence $\lambda \in \rho (A)$ and $R(\lambda ,A)=J={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}T(t)xdt$ as an improper Bochner integral, with $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M\mathrm{/}(\lambda -\omega )$. As $\lambda >\omega$ was arbitrary, $(\omega ,\mathrm{\infty })\subseteq \rho (A)$.