I'm trying to understand the following proposition (from the textbook "Monotone Operators in Banach Space and Nonlinear Partial Differential Equations"-Showalter, chapter III.3, proposition 3.1):
Where $V$ is a reflexive and separable Banach space; $\nu = L^2(0,T;X)$; $W^{1,2}(0,T,V)=\{f\in L^2(0,T;V) : f'\in L^2(0,T;V)\}$; the definition of a regular family of operators (in the proof they mention an "above estimate", it's the one that appears in the definition) and Proposition 1.1 are the following:
Now, I understand the proof clearly until the part it states "$A(t)u(t)(v) = A^*(t)v(u(t))$ is absolutely continuous, since the above shows the function $A^*(t)v$ is absolutely continuous". But, after that, I don't really undersatand how the proof ends. Why is $$\frac{d}{dt}((A(t)u(t)) = A'(t)u(t) + A(t)u'(t)?$$
Believe me, I'm tired of trying to prove it formally. And, how is that the result "follows easily" from that? I'd really appreciate some help with this.
\begin{align*} &\left<\dfrac{A(t+h)u(t+h)-A(t)u(t)}{h},g\right>\\ &=\left<\dfrac{(A(t+h)-A(t))u(t+h)+A(t)(u(t+h)-u(t))}{h},g\right>\\ &=\left<\dfrac{A(t+h)-A(t)}{h}u(t+h),g\right>+\left<A(t)\left[\dfrac{u(t+h)-u(t)}{h}\right],g\right>\\ &\rightarrow\left<A'(t)u(t),g\right>+\left<A(t)u'(t),g\right>\\ &=\left<A'(t)u(t)+A(t)u'(t),g\right> \end{align*}