# Uniformly sectorial family: Differentiability of Resolvent

Suppose that the operator family $$(A(t))_{t \in \mathbb{R}}$$ with dense domain of definition $$D(A(t)) = D \subseteq X$$ is uniformly sectorial with angle $$\phi < \pi/2$$.

Additionally assume that their resolvents $$R(\lambda,A(t)) = (\lambda - A(t))^{-1}$$ commute, i.e. for $$t,s \in \mathbb{R}$$ and $$\lambda, \mu$$ in the sector $$\Sigma_\phi$$ we have that $$R(\lambda, A(t))R(\mu, A(s)) = R(\mu,A(s)) R(\lambda,A(t))$$.

If for all $$x \in D$$ the mapping $$t \mapsto A(t)x$$ is differentiable, can we also (strongly) differentiate the mapping $$t \mapsto R(\lambda, A(t))x$$? Maybe further assumptions are needed?

What would the derivative look like? ( a formal calculation suggests that $$\frac{d}{dt} R(\lambda, A(t)) = A'(t) R(\lambda, A(t))^2$$ )

Edit: I think this could work:

We use the second resolvent identity for $$h > 0$$ and all $$t \in \mathbb{R}, \lambda \in \Sigma_\phi$$: \begin{align*} R(\lambda,A(t + h)) - R(\lambda,A(t)) &= R(\lambda,A(t + h)) \; (A(t + h) - A(t)) \; R(\lambda,A(t)) \end{align*} In particular, we can see that $$R(\lambda,A(t + h)) \to R(\lambda,A(t))$$ strongly as $$h \to 0$$ by the uniform sectoriality and the strong continuity of the family $$(A(t))$$.

Furthermore, we have \begin{align*} \frac{1}{h}(R(\lambda,A(t + h)) - R(\lambda,A(t))) &= \; R(\lambda,A(t + h)) R(\lambda,A(t)) \; \frac{1}{h} (A(t + h) - A(t)) \end{align*} where we use the commuting resolvents to rearrange the terms.

Taking the limit the uniform boundedness of the resolvents implies \begin{align*} \frac{d}{dt} R(\lambda,A(t)) &= R(\lambda,A(t)) R(\lambda,A(t)) A'(t)\\ &= R(\lambda,A(t))^2 A'(t)\\ &= - \frac{d}{d\lambda} R(\lambda,A(t)) A'(t) \end{align*} where I used that $$t \mapsto A(t)$$ is strongly continuous and commutes with its resolvent on $$D$$.

Did I miss something?