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*}

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  • $\begingroup$ I tried something like that but, how can I justify that, indeed, $\frac{A(t+h)-A(t)}{h}\to A'(t)$? $\endgroup$ – Juan David Samboní May 1 '18 at 23:52
  • $\begingroup$ The point is that, how do you define $A'(t)$? Not of the quotient type? $\endgroup$ – user284331 May 2 '18 at 0:02
  • $\begingroup$ $A'(t)$ is such that, for all $u,v\in V$, $\langle A'(t)u,v \rangle = \frac{d}{dt}\langle A(t)u,v \rangle$, where $\frac{d}{dt}$ denotes the generalized derivative ($g$ is the generalized derivate of $f$ in $(0,T)$ if $g$ satisfies that $\int_0^T \phi'(t)f(t)dt = -\int_0^T \phi(t)g(t)dt $, for all $\phi\in C_0^{\infty}(0,T)$). $\endgroup$ – Juan David Samboní May 2 '18 at 3:04
  • $\begingroup$ Sorry then I misunderstood your question maybe, because I thought that the derivative is defined in the strong sense just like in the Hilbert space theory. Then I think the issue may be much more complicated, but I guess can you do it something like Lebesgue Dominated Convergence Theorem? $\endgroup$ – user284331 May 2 '18 at 3:15
  • $\begingroup$ Don't worry about it. Actually, in the proof it's stated that $A(t)u(t)$ is absolutely continuous. That implies that $A(t)u(t)$ is strongly differentiable, i.e. we can obtain the derivative with the quotient's limit. That's why I tried to prove it just like you wrote above. But the part that I don't ser clearly is that $\frac{A(t+h)-A(t)}{h}\to A'(t)$ $\endgroup$ – Juan David Samboní May 2 '18 at 15:39

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