Resolvent identity and holomorphy for a closed operator

Factorization. Fix ${\lambda }_0\in \rho (A)$ and set $T:=({\lambda }_0-\lambda )R({\lambda }_0,A)\in \mathcal{B}(X)$. For every $y\in D(A)$, writing $z:=({\lambda }_{0}I-A)y$ gives $y=R({\lambda }_0,A)z$ by [L1] and $(\lambda I-A)y=({\lambda }_{0}I-A)y+(\lambda -{\lambda }_0)y=z+(\lambda -{\lambda }_0)R({\lambda }_0,A)z=(I-T)({\lambda }_{0}I-A)y,$ so $\lambda I-A=(I-T)({\lambda }_{0}I-A)$ as maps $D(A)\to X$.

Resolvent identity. For $\lambda ,\mu \in \rho (A)$ the identity $(\mu -\lambda )R(\mu ,A)=(\mu I-A)R(\mu ,A)-(\lambda I-A)R(\mu ,A)=I-(\lambda I-A)R(\mu ,A)$ holds on $X$, because both resolvents are everywhere defined and $(\mu I-A)R(\mu ,A)=I$ by [L1]. Hence, using $R(\mu ,A)X\subseteq D(A)$ and $R(\lambda ,A)(\lambda I-A)z=z$ for $z\in D(A)$, $(\mu -\lambda )R(\lambda ,A)R(\mu ,A)=R(\lambda ,A)(I-(\lambda I-A)R(\mu ,A))=R(\lambda ,A)-R(\mu ,A)\mathrm{.}$ Exchanging $\lambda$ and $\mu$ gives $(\lambda -\mu )R(\mu ,A)R(\lambda ,A)=R(\mu ,A)-R(\lambda ,A)$; combining the two displays yields the second form $(\mu -\lambda )R(\lambda ,A)R(\mu ,A)=(\mu -\lambda )R(\mu ,A)R(\lambda ,A)$ for $\mu \mathrm{\ne }\lambda$, while for $\mu =\lambda$ both sides vanish.

Openness and the expansion. If $\mathrm{\parallel }T\mathrm{\parallel }=\mathrm{\mid }\lambda -{\lambda }_0\mathrm{\mid }\mathrm{\parallel }R({\lambda }_0,A)\mathrm{\parallel }<1$, then [L2] makes $I-T$ invertible with inverse $S:={\sum }_{n\ge 0}{T}^n$. For $x\in X$ put $y:=R({\lambda }_0,A)Sx\in D(A)$; by [step 1.1] and [L1], $(\lambda I-A)y=(I-T)({\lambda }_{0}I-A)R({\lambda }_0,A)Sx=(I-T)Sx=x,$ so $\lambda I-A$ is surjective; it is injective because $(\lambda I-A)y=0$ and [step 1.1] give $(I-T)({\lambda }_{0}I-A)y=0$, hence $({\lambda }_{0}I-A)y=0$ and $y=R({\lambda }_0,A)0=0$. Thus $\lambda \in \rho (A)$ and $R(\lambda ,A)=R({\lambda }_0,A)(I-T{)}^{-1}=R({\lambda }_0,A){\sum }_{n\ge 0}(({\lambda }_0-\lambda )R({\lambda }_0,A){)}^n={\sum }_{n\ge 0}({\lambda }_0-\lambda {)}^{n}R({\lambda }_0,A{)}^{n+1},$ the series converging in operator norm because $\parallel ({\lambda }_0-\lambda {)}^{n}R({\lambda }_0,A{)}^{n+1}\parallel \le \mathrm{\parallel }R({\lambda }_0,A)\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}^n$ and $\mathrm{\parallel }T\mathrm{\parallel }<1$.

Differentiability of the resolvent. Let ${\lambda }_0\in \rho (A)$ and let $h\in \mathbb{K}$ with $\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }<1$, where $R:=R({\lambda }_0,A)$. By [step 2.1] applied to the pair ${\lambda }_0+h,{\lambda }_0$, $R({\lambda }_0+h,A)-R({\lambda }_0,A)={\sum }_{n\ge 1}(-h{)}^n{R}^{n+1}=-h{R}^2+h^2{\sum }_{n\ge 2}(-h{)}^{n-2}{R}^{n+1},$ and the norm of the second summand is at most $\mathrm{\mid }h{\mathrm{\mid }}^2\mathrm{\parallel }R{\mathrm{\parallel }}^3(1-\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }{)}^{-1}$, so $\mathrm{\parallel }{h}^{-1}(R({\lambda }_0+h,A)-R({\lambda }_0,A))+R^2\mathrm{\parallel }\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R{\mathrm{\parallel }}^3(1-\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }{)}^{-1}\to 0$. Hence $\lambda \mapsto R(\lambda ,A)$ is differentiable at ${\lambda }_0$ with derivative $-R({\lambda }_0,A{)}^2$; when $\mathbb{K}=\mathbb{C}$ this is complex differentiability in operator norm, that is, norm-holomorphy.

Local boundedness and continuity. With $R:=R({\lambda }_0,A)$ and $\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }<1$, the expansion of [step 2.1] gives $\mathrm{\parallel }R({\lambda }_0+h,A)-R({\lambda }_0,A)\mathrm{\parallel }\le {\sum }_{n\ge 1}\mathrm{\mid }h{\mathrm{\mid }}^n\mathrm{\parallel }R{\mathrm{\parallel }}^{n+1}=\frac{\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R{\mathrm{\parallel }}^2}{1-\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }}⟶0$ and $\mathrm{\parallel }R({\lambda }_0+h,A)\mathrm{\parallel }\le \mathrm{\parallel }R\mathrm{\parallel }(1-\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }{)}^{-1}$; thus $\lambda \mapsto R(\lambda ,A)$ is continuous at every point of $\rho (A)$ and locally bounded in operator norm.

The exponential factor. Fix ${\lambda }_0\in \rho (A)$ and $z\in \mathbb{C}$, write $R=R({\lambda }_0,A)$, and let $h\mathrm{\ne }0$ with $\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R\mathrm{\parallel }<1$. Then $\frac{e^{({\lambda }_0+h)z}R({\lambda }_0+h,A)-e^{{\lambda }_{0}z}R({\lambda }_0,A)}{h}=e^{{\lambda }_{0}z}\frac{e^{hz}-1}{h}R({\lambda }_0+h,A)+e^{{\lambda }_{0}z}\frac{R({\lambda }_0+h,A)-R({\lambda }_0,A)}{h}\mathrm{.}$ As $h\to 0$ the first factor $\frac{e^{hz}-1}{h}\to z$ by [L3], the second factor $R({\lambda }_0+h,A)\to R({\lambda }_0,A)$ in operator norm by [step 3.2], and the last difference quotient tends to $-R^2$ by [step 3.1]; multiplying by the bounded scalars $e^{{\lambda }_{0}z}$ gives convergence in operator norm to $z{e}^{{\lambda }_{0}z}R({\lambda }_0,A)-e^{{\lambda }_{0}z}R({\lambda }_0,A{)}^2$.

Collecting [step 2.1] (openness and the displayed expansion, claim 1), [step 1.2] (the resolvent identity, claim 2), [step 3.1] (norm differentiability with derivative $-R^2$, and norm-holomorphy over $\mathbb{C}$, claim 3) and [step 4.1] (the exponential factor, claim 4) proves the four claims; every step used only the resolvent identities, the Neumann expansion and the scalar exponential, so no choice principle was used.