Let $X$ be a Banach space, we understand $L^1(0, T, X)$ is the space of strongly measurable functions from $[0, T]$ valued in $X$, that is integrable. Assume ${\bf u}\in L^1(0, T, X)$, we say ${\bf v}\in L^1(0, T, X)$ is the (weak time) derivative of ${\bf u}$ if for any $\varphi\in C_0^\infty(0, T)$ we have $$ \int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf v}(t)\varphi(t)dt, $$ and we will write ${\bf v}={\bf u}'$.
Now in Evans' PDE book, he discussed a situation where ${\bf u}\in L^2(0, T, H_0^1(\Omega))$ while ${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$. It seems there are two possible interpretations:
(A). The situation implies ${\bf u}\in L^1(0, T, H_0^1(\Omega))$, so we follow the DEFINITON of derivative I mentioned in the beginning and understand that ${\bf u}'\in L^1(0, T, H_0^1(\Omega))$. However in this situation is possible ${\bf u}'\notin L^2(0, T, H_0^1(\Omega))$, while from the continuous injection $H_0^1(\Omega)\to H^{-1}(\Omega)$ it happens that ${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$. In conclusion, $$ {\bf u}'\in L^1(0, T, H_0^1(\Omega))\cap L^2(0, T, H^{-1}(\Omega)). $$
(B). It just means $$ \int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf u}'(t)\varphi(t)dt $$ literally. This looks like a little miracle - the right hand side is supposed to be in $H^{-1}(\Omega)$, but the equation tells us it is indeed in $H_0^1(\Omega)$.
My question is, Assume one uses (B) as definition, is there a theorem which implies that indeed ${\bf u}'$ is also in $L^{1}(0, T, H_0^1(\Omega))$ ? Or counterexamples?
