The Dunford contour integral defines a bounded holomorphic family on the sector

Angle geometry on $K$. For $z\in K$, the upper-ray angle satisfies $\pi \mathrm{/}2<\theta +\arg z<3\pi \mathrm{/}2$, since $\theta +\arg z\ge \theta -{\delta }_K>\pi \mathrm{/}2$ and $\theta +\arg z\le \theta +{\delta }_K<\pi \mathrm{/}2+\delta +{\delta }_K<3\pi \mathrm{/}2$. The lower-ray angle satisfies $-3\pi \mathrm{/}2<-\theta +\arg z<-\pi \mathrm{/}2$, since $-\theta +\arg z\le -\theta +{\delta }_K<-\pi \mathrm{/}2$ and $-\theta +\arg z\ge -\theta -{\delta }_K>-\pi \mathrm{/}2-\delta -{\delta }_K>-3\pi \mathrm{/}2$. Thus $\cos (\pm \theta +\arg z)<0$ for both rays. The continuous function $z\mapsto -\cos (\pm \theta +\arg z)$ is positive on $K$, so $c_K:={\min }_{z\in K,{\theta }^{\mathrm{\prime }}=\pm \theta }(-\cos ({\theta }^{\mathrm{\prime }}+\arg z))>0$, and for $\lambda =s{e}^{\pm i\theta }$ one has $\mathrm{Re}(\lambda z)=s\mathrm{\mid }z\mathrm{\mid }\cos (\pm \theta +\arg z)\le -c_{K}s\mathrm{\mid }z\mathrm{\mid }$.

The integrand is holomorphic on a star-shaped sector. By [L1] the sector $S_{\epsilon }$ of [L3] lies in $\rho (A)$, and by [L2] both $\lambda \mapsto e^{\lambda z}R(\lambda ,A)$ and $\lambda \mapsto \lambda e^{\lambda z}R(\lambda ,A)$ are norm-holomorphic on it; $S_{\epsilon }$ is star-shaped with base point any positive real by [L3]. Since $\mathcal{B}(X)$ is Banach by [L5], the Banach-valued Cauchy theorem applies to these $\mathcal{B}(X)$-valued maps.

Absolute convergence and local uniformity. For $z\in K$ and $\lambda =s{e}^{\pm i\theta }$ on the rays, [step 1.1] and [L1] give $\mathrm{\parallel }{\psi }_z(\lambda )\mathrm{\parallel }\le e^{-c_{K}s\mathrm{\mid }z\mathrm{\mid }}{M}_{\epsilon }\mathrm{/}s\le e^{-c_K{m}_{K}s}{M}_{\epsilon }\mathrm{/}s$, which is integrable over $s\ge r$; on the arc $\mathrm{\mid }\lambda \mathrm{\mid }=r$ one has $\mathrm{\parallel }{\psi }_z(\lambda )\mathrm{\parallel }\le e^{M_{K}r}{M}_{\epsilon }\mathrm{/}r$, an integrable bound on a compact interval. Hence ${\int }_{\mathrm{\Gamma }}\mathrm{\parallel }{\psi }_z(\lambda )\mathrm{\parallel }\mathrm{\mid }d\lambda \mathrm{\mid }<\mathrm{\infty }$ and, by [L4], $\mathrm{\parallel }{\int }_{{\mathrm{\Gamma }}_R}{\psi }_{z}d\lambda -{\int }_{{\mathrm{\Gamma }}_{R^{\mathrm{\prime }}}}{\psi }_{z}d\lambda \mathrm{\parallel }\le 2M_{\epsilon }{\int }_R^{R^{\mathrm{\prime }}}{e}^{-c_K{m}_{K}s}ds\mathrm{/}s$ for $R<R^{\mathrm{\prime }}$, a bound independent of $z\in K$ tending to $0$. Thus the truncated integrals form a Cauchy family in $\mathcal{B}(X)$ and converge there by [L5]; this is the operator-norm limit, uniformly on $K$, and the integrals converge absolutely.

Independence of the inner radius. Fix $\theta$ and $0<r_1<r_2$ and $R>r_2$. The truncated paths ${\mathrm{\Gamma }}_R(r_1,\theta )$ and ${\mathrm{\Gamma }}_R(r_2,\theta )$ have the same initial point $R{e}^{-i\theta }$ and the same terminal point $R{e}^{i\theta }$, so their concatenation with the reversal of the second is a closed piecewise $C^1$ contour lying in the star-shaped sector $S_{\epsilon }$ of [step 1.2], where ${\psi }_z$ is holomorphic; by the Banach-valued Cauchy theorem in $\mathcal{B}(X)$, applicable by [L5], [L3] its integral vanishes, hence ${\int }_{{\mathrm{\Gamma }}_R(r_1,\theta )}{\psi }_{z}d\lambda ={\int }_{{\mathrm{\Gamma }}_R(r_2,\theta )}{\psi }_{z}d\lambda$ for every $R>r_2$; letting $R\to \mathrm{\infty }$ and using [step 2.1] gives equality of the limits.

Independence of the angle. Fix $r$ and $\pi \mathrm{/}2<{\theta }_1<{\theta }_2<\pi \mathrm{/}2+\delta$, both admissible for the given $z$, and $R>r$. Let $A_R^{+}$ be the counterclockwise arc $R{e}^{i\alpha }$, $\alpha \in [{\theta }_1,{\theta }_2]$, and $A_R^{-}$ its reflection $\alpha \in [-{\theta }_2,-{\theta }_1]$; then ${\mathrm{\Gamma }}_R(r,{\theta }_1)+A_R^{+}-{\mathrm{\Gamma }}_R(r,{\theta }_2)+A_R^{-}$ is a closed piecewise $C^1$ contour in the star-shaped sector $S_{\epsilon }$ (for ${\theta }_2<\pi \mathrm{/}2+\delta -\epsilon$), so the Banach-valued Cauchy theorem in $\mathcal{B}(X)$ applies by [L5] and [L3] its integral vanishes; hence ${\int }_{{\mathrm{\Gamma }}_R(r,{\theta }_1)}{\psi }_z-{\int }_{{\mathrm{\Gamma }}_R(r,{\theta }_2)}{\psi }_z=-{\int }_{A_R^{+}}{\psi }_z-{\int }_{A_R^{-}}{\psi }_z$. On the arcs $\mathrm{\mid }\lambda \mathrm{\mid }=R$ with angle $\alpha$ between ${\theta }_1,{\theta }_2$ and their reflections one has $\mathrm{\mid }\alpha +\arg z\mathrm{\mid }>\pi \mathrm{/}2$ uniformly, so $\mathrm{\parallel }{\psi }_z\mathrm{\parallel }\le e^{-cR}$ for a constant $c>0$ and the right-hand side tends to $0$ as $R\to \mathrm{\infty }$; hence the two limits agree.

Norm-holomorphy and the derivative formula. Fix $K$, the angle $\theta$ of [step 1.1] and $r>0$. Put $d_K:=\mathrm{dist}(K,\mathbb{C}\setminus {\mathrm{\Sigma }}_{\delta })>0$ and choose $h_0<\min \{c_K{m}_K,d_K\}$. Then $z+h\in {\mathrm{\Sigma }}_{\delta }$ for $z\in K$, $\mathrm{\mid }h\mathrm{\mid }<h_0$, and the exponential estimate $\mathrm{\mid }{h}^{-1}(e^{\lambda h}-1)-\lambda \mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }\lambda {\mathrm{\mid }}^2{e}^{\mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }\lambda \mathrm{\mid }}$ together with [L1] gives, on the rays, $\mathrm{\parallel }{h}^{-1}({\psi }_{z+h}(\lambda )-{\psi }_z(\lambda ))-\lambda {\psi }_z(\lambda )\mathrm{\parallel }\le \mathrm{\mid }h\mathrm{\mid }{M}_{\epsilon }\mathrm{\mid }\lambda \mathrm{\mid }{e}^{-c_K\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }+h_0\mathrm{\mid }\lambda \mathrm{\mid }}\le \mathrm{\mid }h\mathrm{\mid }{M}_{\epsilon }s{e}^{-(c_K{m}_K-h_0)s}$ with $c_K{m}_K-h_0>0$, an integrable bound whose integral tends to $0$ with $h$, while on the compact arc the same estimate is bounded uniformly and contributes $O(\mathrm{\mid }h\mathrm{\mid })$; hence for each fixed $R$ the difference quotients of ${\int }_{{\mathrm{\Gamma }}_R}\psi$ tend to ${\int }_{{\mathrm{\Gamma }}_R}\lambda {\psi }_{z}d\lambda$ with an error bounded uniformly in $R$, and letting $R\to \mathrm{\infty }$ with the majorant of [step 2.1] and [L4] gives $\mathrm{\parallel }{h}^{-1}(T(z+h)-T(z))-(2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}\lambda e^{\lambda z}R(\lambda ,A)d\lambda \mathrm{\parallel }\to 0$; completeness [L5] ensures that this derivative integral and the limit lie in $\mathcal{B}(X)$, so $T$ is complex-differentiable with that derivative (uniformly on $K$).

Uniform boundedness on smaller sectors. Fix ${\delta }^{\mathrm{\prime }}<\delta$ and an angle $\theta$ with $\pi \mathrm{/}2+{\delta }^{\mathrm{\prime }}<\theta <\pi \mathrm{/}2+\delta$; for $z\in {\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$ the value $T(z)$ is, by the independence of the inner radius [step 3.1], computable with $r=1\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, so on the rays $\mathrm{Re}(\lambda z)\le -cs\mathrm{\mid }z\mathrm{\mid }$ with $c=c(\theta ,{\delta }^{\mathrm{\prime }})>0$ and the ray contribution is at most $M_{\epsilon }{\int }_1^{\mathrm{\infty }}{e}^{-c\sigma }d\sigma \mathrm{/}\sigma$, while the arc has length at most $2\theta \mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, radius $1\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ and integrand norm at most $e{M}_{\epsilon }\mathrm{\mid }z\mathrm{\mid }$, contributing at most $2\theta e{M}_{\epsilon }$; both bounds are independent of $z$, so ${\sup }_{{\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}}\mathrm{\parallel }T\mathrm{\parallel }<\mathrm{\infty }$.

Conclusion. [step 2.1] proves claim 1, [step 3.1] and [step 3.2] prove claim 2, [step 3.3] proves claim 3, and [step 4.1] proves claim 4; the argument used only the sectorial resolvent bound, the resolvent holomorphy, the contour integral over closed curves in the star-shaped sector and norm estimates, hence no choice principle beyond Dependent Choice was used.